algebra
polynomial
quadratics
analytic geometry
graphing lines
slope
function

21.  Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $(x-19)$, the remainder is 99; and when $P(x)$ is divided by $(x-99)$, the remainder is 19.  What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$?

NOTE:  There is multiple choice, but I'll omit those  :D

[hide]
since the divisor is quadratic, the remainder will be linear in the form ax + b

$P(19) = 99 = a(19)+b$
$P(99) = 19 = a(99)+b$
solving:
$a =-1$ and $b = 118$

yielding a remainder of
$118-x$
[/hide]

Damn, beaten to it.
There's an extremely similar example in AoPS II, from Mu Alpha Theta.

[hide]
\begin{align*}P & =f\cdot(x-19)+99 \\ P & =g\cdot(x-99)+19 \\ P & =h\cdot(x-19)(x-99)+k_{1}x+k_{2}\\ & \text{For given polynomial }P\text{ and some polynomials }f,g,h\end{align*}\text{To determine }k_{1},k_{2}\text{ notice the system of equations: }\par\begin{align*}P(19) &=99=19k_{1}+k_{2}\\ P(99) &=19=99k_{1}+k_{2}\\ &\implies 19-99=(99-19)k_{1}\implies k_{1}=-1\\ &\implies 99+19=118=k_{2}\end{align*}
[/hide]

[quote="GrantCBHS"]Here's a quicker way that should be a nice little reminder if your Algebra II teacher mentioned it in class (this exact problem with different numbers was on a MAO Test that I took a few months ago for practice; I couldn't solve it, but my team member Ari helped me remember the information that I needed for this problem):

[hide] We know the following:
[b]P(19) = 99[/b] and
[b]P(99) = 19[/b] because dividing by (x-19) and (x-99) is the same thing as plugging 19 and 99 into your polynomial/function.

Surprisingly, just those two pieces can generate us a polynomial.
From P(19) = 99; we can deduce that the point (19,99) is on our polynomial, and from P(99) = 19 we can derive the point (99,19).
[b]
(99-19)/(19-99) = -1[/b] [u]This is our slope[/u], now we can write the equation of our function:
[b]y-19 =  -1(x-99)[/b] ->
[b]y-19 = -x+99[/b] ->
[b]y = -x+118[/b] is you polynomial; when -x+188 is divided by that quadratic, the remainder will be itself ([b]-x+188[/b]).

[/hide][/quote]

Just a little something to clarify - you are merely constructing the [b]simplest[/b] polynomial which satisfies the conditions, which happens to be the remainder (well I guess it always is the remainder but...)

That's neat - since the question implies that the remainder is always the same, you [b]assume[/b] that in your solution, by constructing a very simple polynomial. I like that method :P but it won't work on a proof contest lol


While shopping at the local mall, Ann walked slowly (at constant speed) down a descending escalator, taking ten steps to reach the last step. While on the lower level, she heard that free concert tickets were being given away on the next level up. She ran up the same (descending) escalator with a speed five times greater than the speed with which she had walked down, taking 25 steps to reach the top.

Find the number of steps showing on the escalator at any one moment.

No one tried to solve this so i'll just post the solution.. was is that hard? It's similar to an AIME problem.

Let $ n$ be the number of steps showing on the escalator at any moment. In the time Ann took to walk down 10 steps, the escalator must have descended through a distance equivalent to $ n \minus{} 10$ steps. Hence the escalator descended $ \frac {n\minus{}10}{10}$ steps in the time Ann took to walk down one step. When running up the escalator, Ann took 25 steps, which, since she ran five times faster than she walked, took her the same time as walking $ \frac {25}{5} \equal{} 5$ steps. In that time, the escalator moved [i]downward[/i] through $ 5 \times \frac {n\minus{}10}{10} \equal{} \frac {n\minus{}10}{2}$ steps. Hence she actually ascended through a distance equivalent to $ 25 \minus{} \frac {1}{2}(n \minus{} 10)$ steps. As Ann made a complete ascent of the escalator she had risen a distance equivalent to $ n$ steps. So $ 25 \minus{} \frac{1}{2}(n \minus{} 10) \equal{} n$, giving $ 50 \minus{} n \plus{} 10 \equal{} 2n$, $ 60 \equal{} 3n$, and so $ n \equal{} 20$.


Find all possible pairs $(a,b)$ of integers satisfying $a^{2}-3ab-a+b=0$.

I've got
[hide] no solutions... (i used the discriminant)
$\Delta=\sqrt{9b^{2}+2b+1}$
but since $(3b)^{2}<9b^{2}+2b+1<(3b+1)^{2}$, $\sqrt{9b^{2}+2b+1}$ cannot be an integer. so $a$ is not an integer.
[/hide]

you can't say it cant be a perfect square. if we consider negitive numbers or 0 the inequallity written by you fails. (0,0) is one of the solution.

I do not have a full sketch but here is an idea.

[hide="Read"]

Writing this as $a^{2}+b = 3ab+a$ and simplifying:

$a+\frac{b}{a}= 3b+1$

This shows that $a|b$ so that's something to consider. Now, $a=b=0$ is an exception because when we simplified to get above expression, we divide by 0, assuming that it will work. So even though it doesn't work in our new equation, $(0,0)$ fits in original one. [/hide]

you are changing the given equation. ok i will clearify by an example.

let us assume that the equation given is $2b = b$
not it can be said that $b=0$ but according to your mathod $b$
cancells from both the sides. and remains $2=1$. which is absolutely wrong.

so firstly do not change the form of the equation without checking solution for $0$.

[quote="kimby_102"]Find all possible pairs $(a,b)$ of integers satisfying $a^{2}-3ab-a+b=0$.[/quote][hide]The given equation is a quadratic in $a$ with discriminant \[(-3b-1)^{2}-4(1)(b)=9b^{2}+2b+1.\] Consider the equation in three different cases, defined by the sign of $b$.

[b]Case 1: $b>0$.[/b]
If $b>0$, then \[9b^{2}<9b^{2}+2b+1<9b^{2}+6b+1,\]  \[(3b)^{2}<9b^{2}+2b+1<(3b+1)^{2}.\] Therefore, the discriminant will be positive but not a perfect square, because $b$ must be an integer. Hence, if $b>0$, then $a$ is forced to be irrational, and there are no integer solutions.

[b]Case 2: $b=0$[/b]
If $b=0$, then the original equation implies \[0=a^{2}-a=a(a-1).\] Thus, $a=0$ or $a=1$.

[b]Case 3: $b<0$[/b]
If $b$ is a negative integer then \[9b^{2}>9b^{2}+2b+1>9b^{2}+6b+1,\]  \[(3b)^{2}>9b^{2}+2b+1>(3b+1)^{2}.\] Again, the discrimant is not a perfect square, and $a$ has to be irrational. There are no solutions to this case, too.

The ordered pair solutions $(a,b)$ are $(1,0)$ and $(0,0)$.[/hide]

in case 3 you have considered that $b<0$. if we take $b =-1$ then also it do not satisfy your inequality.

oops i have commited a mistake. your solution is fine.

[hide="NT  :showoff:  "]We want to solve $a^{2}-3ab-a+b = 0$ for integers $a, b$. We know that $0 = a^{2}-3ab-a+b \equiv b\mod{a}$, so $b=ak$ for an arbitrary [u]integer[/u] $k$. The equation becomes $0 = a^{2}-3ab-a+b = a(a-3ak-1+k)$. Therefore, we have $a=0$ or $a-3ak-1+k = 0$.
If $a=0$, then $b=ak=0$, so $(a,b) = (0,0)$ is a solution.
If $a-3ak-1+k=0$, then $a = \frac{k-1}{3k-1}$. But $a$ is an integer, so $3k-1 | k-1$. Therefore, we must have that $|3k-1| \leq |k-1|$. Square both sides to get that $9k^{2}-6k+1 \leq k^{2}-2k+1$, so $8k^{2}-4k \leq 0$, so $k(2k-1) \leq 0$, so $k\in\left[0,\frac{1}{2}\right]$. But $k$ was an integer, so we must have $k=0$, which means $a=\frac{k-1}{3k-1}= 1$ and $b=ak=0$. So $(a,b)=(1,0)$ is another solution. And we're done. [/hide]

the solution given by Kurt Gödel is really exellent :)  :)  :)

Also the question reduces to finding out the factors of 2 :)!  ..using   the Standard technique;   ..Rewriting the above ...we want to  find  integer solutions for ...
$(3a-1)(3a-9b-2)=2$
so  $3a-1 = \pm 1$  and  $(3a-9b-2)=\pm 2$
 or $3a-1 = \pm 2$ and  $(3a-9b-2)=\pm 1$ ... rest is easy.
So only  possible values of $(a,b)$ are  $(0,0)$ or $(1,0)$ etc.:)


Ring Theory
superior algebra
superior algebra unsolved

Let $ k$ be a field and $ f(x) \in k[x]$. Show that $ k[x]/f(x)$ is a field $ \iff f(x)$ is irreducible.

This is a standard result, but...

The following are equivalent:

1. $ k[x]/f(x)$ is a field
2. $ (f)$ is a maximal ideal.
3. $ f$ is irreducible.

(1) <=> (2) is standard: ideals in $ k[x]/(f)$ correspond to ideals in $ k[x]$ containing $ f$.
(2) <=> (3): $ k[x]$ is a PID, so if $ (f)$ is not maximal, there is some $ g$ such that $ (g) %Error. "contains" is a bad command.
(f)$ as a strict inclusion. Therefore there is a non-unit $ h$ such that $ gh = f$; so $ f$ is reducible. The converse follows similarly.


[url]http://www.divisibilitybyseven.mat.br/[/url]

EDIT:

fractions:

[url]http://www.mathpath.org/concepts/Num/frac.htm[/url]

[quote="pkerichang"][quote="srulikbd"]there is a lot of information on the net for this subject, how ever for this rule method I just couldn't find proof:
When divided into groups of three digits from right to left, with alternating signs before each group, the result is divisible by 7.  2911272: "272" - "911" + "2" = -637 is divisible by 7.
it's pretty the same as for 13.
sorry if it's spam...and thanks!
ok for it won't be just spam-question: can you determine if a number $\frac{a}{b}$ is a periodic fraction, what about other bases other than 10.[/quote]

Hmm...it's similar to the way I do it, but I guess for mine the number decrease faster.
[hide="proof for divisibility"]
take the number $10^{n}a_{n}+\dots+10a_{1}+a_{0}$, and mod it by 7 you get the pattern (starting from 1 to $10^{n}$) $1, 3, 2, 6, 4, 5, 1, \dots$.  Note that some numbers is greater than $3$ so instead of writing $k$, we write $7-k$ (since it's still the same mod 7).  So the new pattern is $1, 3, 2,-1,-3,-2, \dots$.  Therefore, the original number is divisible by 7 if and only if $a_{0}+3a_{1}+2a_{2}-(a_{3}+3a_{4}+2a_{5})+\dots$ is divisible by $7$.  and you can rewrite this again in base ten form and get your rule.  But I use this one because it decreases faster. (For example, 2911272 becomes 7 immediately, instead of -637).
[/hide][/quote]
thanks but this method I understood, but I think it isn't what I mentioned...


[hide="A direct proof"] $ \sum_{k\equal{}1}^{n} k \cdot k! \equal{} \sum_{k\equal{}1}^{n} (k\plus{}1)! \minus{} k! \equal{} (2! \minus{} 1!) \plus{} (3! \minus{} 2!) \plus{} ... \plus{} ( (n\plus{}1)! \minus{} n! ) \equal{} (n\plus{}1)! \minus{} 0!$ [/hide]
[hide="Discussion"] The value in this result is that it can be used as a lemma in proving that every number has a unique expression in "factorial base," that is,

$ n \equal{} d_1 \plus{} d_2 \cdot 2! \plus{} d_3 \cdot 3! \plus{} d_4 \cdot 4! \plus{} ...$

where $ 0 \le d_k \le k$. [/hide]
[hide="Generalization"] $ \sum_{k\equal{}0}^{n} b_1 b_2 ... b_k (b_{k\plus{}1} \minus{} 1) \equal{} b_1 b_2 ... b_{k\plus{}1} \minus{} 1$

The factorial case follows when $ b_k \equal{} k$ and the base-$ b$ case follows when $ b_k \equal{} b$.  This can be used to prove that every number has a unique expression in a generalized base. [/hide]

[quote="Brut3Forc3"]Prove that:
$ 1(1!) \plus{} 2(2!) \plus{} 3(3!) \plus{} ... \plus{} n(n!) \equal{} (n \plus{} 1)! \minus{} 1$[/quote]
[hide="induction"]
Clearly this is true for n=1.
Add $ (n\plus{}1)(n\plus{}1)!$ to both sides.
$ (n\plus{}1)!\plus{}(n\plus{}1)(n\plus{}1)!\minus{}1$
$ (n\plus{}1\plus{}1)(n\plus{}1)!\minus{}1$
$ (n\plus{}2)!\minus{}1$
Q.E.D.[/hide]


I'm sure that sometimes you all get nervous before performing.  While waiting your turn, what are something you think about or do that calm you down?  I'm curious.

I basically think about anything cept the person playing before me, messing up, or the piece i'm performing itself... usually before a performance i fiddle with a deck of cards or something... to take my mind off of negative thoughts

Always have your towel- to dry those sweaty hands. My issues have always been the physical ones of how my body reacts, not the psychological ones of how to mentally prepare.

It can be a little distracting sometimes to be shaking uncontrollably while I play, but there's not really anything I can do about that without losing concentration on my piece.

I'm not sure it would help everybody, but eating a high-carbohydrate snack before performing helps me calm down a bit.

I usually just breathe in and out and do something before it's my turn. If you haven't warmed up.. That's the time to do so, finger it a few times (if its a woodwind) or unplug it if its a guitar or something.

I typically have my Rubik's cube handy, or stare at my music, or fold small pieces of paper into birds, or stare at a wall as long as I can.  Of course, none of these ever work (I always get nervous and play too fast)  :?

On stage, when I was acting, I have no jitters at all.

But on a piano recital, I started shaking all over and when it was my turn, I walked forcefully to the bench and my mind went blank.

The amount of time you spend practicing and who's been teaching you the piece really affect the performance.


http://www.math.umd.edu/highschool/mathcomp/2005winner.html

Congratulations to John, Brian, and Matt for their uber-math skills!

Did anyone else participate? I'm happy with my honorable mention.

oof! results are back =)
HM

haha, john is awesome

[quote="chess64"]Nice job. I didn't know you went to Montgomery Blair... lucky! :)

I didn't participate, maybe I can get our math club to go next year.[/quote]

It really seems like this contest has been dominated by Blair kids, do they study more, or are they just better ?
I think it's a combination.

[quote=mathleticguyyy][quote=daveed]oof![/quote]

DID THIS DUDE PREDICT A MEME[/quote]

ur mom predicted a meme


inequalities
geometry
circumcircle
trigonometry
geometry solved

Suppose $ABC$ is an acute triangle with circumradius $R$. Let altitudes $AD$, $BE$, and $CF$ have lengths $h_1$, $h_2$, and $h_3$, respectively. Let $t_1$, $t_2$, and $t_3$ be the lengths of the tangents from $A$, $B$, and $C$, respectively to the circumcircles of triangle $DEF$. Show that

$\sum^3_{i=1} \left( \frac{t_i}{\sqrt{h_i}}\right)^2 \leq \frac{3}{2}R$

The circumcircle of $\triangle{DEF}$ is the Euler's circle of $\triangle{ABC}$.
So $\frac{t_1^2}{h_1}=\frac{\frac{AH}{2}\cdot{h_1}}{h_1}=R\cos{A}$. So the inequality reduces to:
$\cos{A}+\cos{B}+\cos{C}\leq\frac{3}{2}$ which follows by Jensen.


CT had a difference from #1 to #4 of 2 points.
The difference between the lowest and highest scores at our state was 31.

Michigan had a difference of 5 pts to going to Nationals a 7th Place got 38 going to nationals who won countdown, and first place (me) got a score of 43 going to nationals.
I don't know the difference between lowest and highest

Oh, CT has unofficial countdown, so that's the difference between the top 4 written.

In MO:

The first place guy got 37.
The second place guy (meeee!!!) got 36.

Fourth and fifth place tied.

The eleventh place guy was three points away from the fourth place guy.

[quote="ragnarok23"]FL: I'm pretty sure first place got a 37. 4th place was around 35.[/quote] Actually first was a 38. Also, 4th got EXACTLY 35 (me). Fifth and 6th got 35's as well though. After countdown though I got :winner_third:

heard from xmorg something that 1st was 44 and 4th 31.

That's in AR though, not some high level state.  *Wonders how other states have no one above 43*

Maine:
1st- 39 - 1st after countdown
2nd- 36- 3rd after countdown
3rd- 35/36- 4th after countdown
4th- 35?- 2nd after countdown

Wow. 2 target errors drop you from 1st with 43 to 23rd with 39.  :o

Wow, that's really crazy! Every little thing matters in California state I guess. You'd think with the 22nd place person getting 40 pnts, the 1st place would be a lot higher. Is that California north and south combined?

[quote="Klebian"]If you read my response, you'd see that I included about 5 other states, not just California.[/quote]
[hide=":ninja:"]touché[/hide]

[quote="Zmastr"]I agree with Klebian.  34 is actually not a great score for 10th.  I heard that in California 1st got 42, 2nd-8th got 41, and 9th-30th got 40.[/quote]

Erm,, Michigan's tenth place was a 36, 7th place a 38.

Just wondering, who got a 44 on the state round?

I got 34 at State because I did absolutely no preparation...I got seventh and moved up to second in Countdown. And I got a 46 at Chapter. >_>

[quote="easyas3.14159..."]He's in colorado.[/quote]

How did you know that???

It's good to see that I'm not the only Colorado AOPSer (sort of) now!

is there anyone else from oklahoma?...

i don't really know my state's range, but i think the #1 got a 34. i was #7 with a 28.

[quote="pgpatel"]Look at his profile. His location is denver.
If he got second does that mean that you beat him XxHarryxX in cd. :D[/quote]

I did win in CD, though just barely. 3-2.  My winning state is a good thing, especially considering I'm from quite a small town compared to Denver. :)

kk, so here is MN....well, MN got a range of lyk 2 pts...

1st=40 (me)
2nd=40 (tiebreaker)
3rd=39
4th=38 (?)
5th=38 (tiebreaker)

Countdown was top 16, but did not count...was just for fun.

I <3 MIDWESTERN STATES!!!


geometry
circumcircle
inradius
trigonometry
inequalities proposed
inequalities

Given are triangle $ABC$ with circumradius $R$ and inradius $r$ prove that
$\frac{2R}{r}\ge\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge 6$
$(\frac{R}{r})^{3}\ge (\frac{b}{c}+\frac{c}{b})(\frac{c}{a}+\frac{a}{c})(\frac{a}{b}+\frac{b}{a})\ge 8$
 :)

[quote="gemath"]Given are triangle $ABC$ with circumradius $R$ and inradius $r$ prove that
$\frac{2R}{r}\ge\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge 6$
 :)[/quote]

I think that there is $\frac{3R}{r}$...and for proving the both you use the following ineq: $\frac{R}{r}\geq \frac{b}{c}+\frac{c}{b}$ and the analogues.

A proof for this one...see here, if u understand a little romanian :)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=122568[/url]

[quote="evarist"][color=purple]The 2nd problems is easy.My solution is 8 lines.How about you?[/color][/quote]
I'm fine thanks  :D , I think in the solution the importance isn't the number of line.
ok here we use $\frac{R}{r}\ge\frac{b}{c}+\frac{c}{b}$ but follow  pohoatza's link we need to prove
$\frac{s}{a}\ge \sin B+\sin C$ how can we do?

forget that link  :D , here`s another proof of : $\frac{R}{r}\geq \frac{b}{c}+\frac{c}{b}$

we start from : $l_{a}=\frac{4pbc(p-a)}{(b+c)^{2}}$ and using $\frac{4bc}{(b+c)^{2}}\leq 1 \Rightarrow l_{a}^{2}\leq p(p-a)$

but $h_{a}\leq l_{a}$ and the analogues so: $h_{b}^{2}+h_{c}^{2}\leq l_{b}^{2}+l_{c}^{2}\leq ap$, but $h_{b}=\frac{2S}{b}\Rightarrow \frac{4S^{2}}{b^{2}}+\frac{4S^{2}}{c^{2}}\leq ap (\star)$
and we multiply $(\star)$ by : $\frac{bc}{4S}\Rightarrow \frac{b}{c}+\frac{c}{b}\leq \frac{abc}{4S}\cdot \frac{p}{S}=\frac{R}{r}$

Can you help me?
Prove that: $\boxed{\frac{R}{2r}\ge\frac{m_{a}}{h_{a}}\ge\frac{1}{2}\left(\frac{b}{c}+\frac{c}{b}\right)}$


limit
integration
trigonometry
calculus
function
calculus computations

$ a_n\equal{}\frac{1.3.5.\ldots.(2n\minus{}1)}{2.4.6.\ldots.(2n)}$ for all $ n\in \mathbb{Z^\plus{}}$

Find $ \lim_{n\to\infty}a_n$.

I found $ a_n\equal{}\frac{2}{\pi}\int_0^\frac{\pi}{2}\sin^{2n}\theta \ d\theta$.  But I don't know where to go from here.

You posted the problem twice:
http://www.mathlinks.ro/viewtopic.php?t=249701

But this being the calculus forum, we might as well give the calculus-analysis answer, starting with cosinator's observation.

We note that $ \lim_{n\to\infty}\sin^{2n}\theta \equal{} 0$ everywhere except $ \theta \equal{} 1,$ and that furthermore, the sequence of functions $ \sin^{2n}\theta$ is decreasing: $ \sin^{2n \plus{} 2}\theta\le\sin^{2n}\theta.$

If we could only justify interchanging the limit and the integral, we'd be done. If we knew either the Dominated Convergence Theorem or the Monotone Convergence Theorem, we'd be done - but this is the "computations and tutorials" forum, so we probably shouldn't assume that knowledge.

We still have plenty of ways to argue this. For one thing, we have that $ \sin^{2n}\theta$ tends uniformly to zero on any interval of the form $ \left[0,\frac {\pi}2 \minus{} \delta\right].$ Use that.

=========

Or, there's yet another calculus argument, an even simpler one:

(This is essentially what kirill posted on the other thread.)

$ a_n \equal{} \prod_{k \equal{} 1}^n\frac {2k \minus{} 1}{2k} \equal{} \prod_{k \equal{} 1}^n\left(1 \minus{} \frac1{2k}\right)$

Hence, $ \lim_{n\to\infty}a_n \equal{} \prod_{k \equal{} 1}^{\infty}\left(1 \minus{} \frac1{2k}\right)$ and we have the following reasonably well-known result:

If $ 1 > b_k > 0,$ then $ \prod_{k \equal{} 1}^{\infty}(1 \minus{} b_k)$ converges to a positive number if and only if $ \sum_{k \equal{} 1}^{\infty}b_k$ converges.

$ \prod_{k \equal{} 1}^{\infty}(1 \minus{} b_k)$ diverges to zero iff $ \sum_{k \equal{} 1}^{\infty}b_k \equal{} \infty.$

The proof involves taking the logarithm.


Let $x,y,z$ be positive integer numbers. Find minimum of:
$A=\frac{1}{x+y+z+1}-\frac{1}{(1+x)(1+y)(1+z)}$

If we change condition as $x,y,z$ are positive numbers? Any one can solve the problem?

[hide="My thoughts on the problem"]
$x+y+z+1 < (x+1)(y+1)(z+1)= xyz+xy+xz+yz+x+y+z+1$
Hence $\frac{1}{x+y+z+1}> \frac{1}{(x+1)(y+1)(z+1)}$.
Hence $A >0$

But as $x,y,z \rightarrow 0$, $A \rightarrow 0$. ($A$ cannot reach $0$ with $x,y,z >0$, but it can get arbitarily close, and so $0$ is the infimum)

Alternatively, we may think of $x,y,z \rightarrow \infty^{+}$. Since $A < \frac{1}{x+y+z+1}$, we could get $A \rightarrow 0$ this way as well, which would give the solution for integers as well.

It might be much more interesting to try and find the [i]maximum[/i] value of the expression: I would conjecture that this value if $\frac{1}{8}$, obtained at $x=y=z=1$. Could anyone prove (or refute) this?[/hide]


[quote="kheg_k"]for some reason i got $-134$[/quote]

Can you show your work?

[quote="kheg_k"]for some reason i got $-134$[/quote]

I agree.

[quote="goldendomer"][quote="Mike33"]-c^2-5(c^0+d)^2 if c = -3 and d=4[/quote]

[hide]-c^2-5(c^0+d)^2=
[color=red]3^2[/color]-5(-3^0+4)^2=
3^2-5(-1+4)^2=
9-5(3)^2=
9-45=[size=150]-36[/size]

Is that right??? :? [/hide][/quote]

no that isn't. you did everything right, but the first part.(highlighted above)
you do the square first. $-(-3)^2=-(9)$


If I roll 5 standard 6 sided dice and multiply the number on the face of each dice, what is the probability that the result is a composite number?

The only way it couldn't be a composite number is if we get any of the following:

(1,1,1,1,2) 5 ways
(1,1,1,1,3) 5 ways
(1,1,1,1,5) 5 ways
(1,1,1,1,1) 1 way

This gives a total of 16 combinations we DON'T want out of a total of $ 6^5 \equal{} 7776$ ways. So there are 7760 favorable outcomes, giving us a probability of 485/486.

Make sure that you count 1 along with the primes. Although 1 is not prime, it is not composite, either, and that is what we are really counting.

You just necroed a 7 year post.  C'mon.  You know better!

There's nothing wrong with necroing a problem thread...

[quote=cutiemouse888]same oof i got this wrong cuz i didn't count 1 as composite[/quote]

1 is [b]not[/b] an composite

I got it wrong cause I accidentally typed 385/486  :mad:


Well, keep in mind that generally contest math is A LOT more advanced than school math. Most, nay, ALL of the Intermediate classes cover things that cannot even be FOUND in the standard curriculum.

[quote="mathswimmer"]Is intro to algebra the equivalent of algebra 1 and intermediate algebra the equivalent of algebra 2? Algebra one is so easy at school but I looked at the "Is this class for you" problems for intro to algebra and they were a lot harder...[/quote]

Intro Algebra covers the fundamentals that most students will see in Algebra 1 and at least half of Algebra 2, as well as extending these fundamentals to much harder problems.  Intermediate Algebra is Algebra 2 + non-trig precalculus + a lot of topics the standard curriculum doesn't really touch at all.


probability
pigeonhole principle
expected value
absolute value
calculus
calculus computations

We have n balls. Suppose that balls are tossed into b bins. Each toss is independent, and each ball is equally likely to end up in any bin. What is the expected number of ball tosses before at least one of the bins contains two balls?

=============
My answer is: Expected number of ball tosses before at least one of the bins contains two balls
            =1–[(n-i)/b] = (b-n+i)/b

is that correct?
thank so much
 :D

What Fire has written down is not an expected value at all, but rather a conditional probability.

Let $X$ be the number of the toss on which we first get two balls in one bin. Let $f(k)=P(X>k).$

Consider this:

$P(X>k|X>k-1)=\frac{b-k+1}b$

The explanation? If $X>k-1$ then we have already performed $k-1$ tosses with no doubling up. Thus we had $k-1$ bins with one ball and $b-k+1$ empty bins. In order to get $X>k$ we would need to toss the ball into one of the empty bins - with the probability given above.

Thus for $k\ge 2,\ f(k)=\frac{(b-1)(b-2)\cdots(b-k+1)}{b^{k-1}}$

Note that $f(1)=1.$

Then, $P(X=k)=f(k-1)-f(k)= \frac{(b-1)(b-2)\cdots(b-k+2)k}{b^{k-1}}$

This is for $2\le k\le b+1.$ The experiment must conclude on or before the $b+1$ st toss. (The classic pigeonhole principle.)

Then we have one expression for the absolute value:

$E(X)=\sum_{k=2}^{b+1}\frac{(b-1)(b-2)\cdots(b-k+2)k^{2}}{b^{k-1}}.$

I haven't tried to compute this sum yet. There are several things that could be tried, including generating functions.

thanks Kent, i think your solution is correct, all i have to do is compute your sum and get a nice answer  :D

My solution:
Suppose 	n = number of balls
			b = number of bins

	The ith stage is the stage when the first bin contains 2 balls.
	So in ith stage there are (n-i) balls remain

	We have the expected number of balls (n) in a given bin (b) = n/b
	So expected number of ball tosses after at least one of the bins contains two balls = (n-i)/b
	So expected number of ball tosses before at least one of the bins contains two balls
			=1–[(n-i)/b] = (b-n+i)/b

Fire, there at least aspects of that that give it no chance of being correct.

1. The statement of the problem had two variables: $b,$ the number of bins, and $n,$ number of balls. But $n$ is problematic. If $n\ge b+1,$ then we are certain to double up in some bin before we run out of balls - and it doesn't matter how many balls we have left. But if $n\le b,$ then it is possible that we never do double up in any bin. In that case, there is a positive probability that we will never double up in any bin - in which case, the expected value asked for appears meaningless. I would then assume that $n\ge b+1$ - and then note that it doesn't matter what $n$ is. The answer - the expected number of tosses - is a number that depends only on $b.$ Your answer must be an expression that depends only on $b.$ And yet, you have given an expression that contains another variable, $i.$ Since $i$ is without meaning in the statement of the problem, that answer has no meaning.

2. It takes a minimum of two tosses to double up in a bin. The expected number of tosses must be greater than or equal to 2. Your expession is meaningless, but if it had any meaning, it would appear to be less than 1. That can't be right.

--

Now it is possible that you never meant to say "expected value" in the first place. A question that does have an answer if $n\le b$ is what is the probability that we will double up in some bin if we toss in all $n$ balls? And that probability is given in my previous post:

$P(X>n)=\frac{(b-1)(b-2)\cdots(b-n+1)}{b^{n-1}}$

is the probability that we don't double up in any bin. The probability that we do double up is 1 minus that number.


At each stage, the square at the lower left is divided into 4 congruent square regions, 2 of which are shaded. The area of the entire square (including shaded and unshaded parts) is 256 square units. How many square units are in the shaded area at the fifth stage? Express your answer as a decimal.

[asy]path A, B, C,D,E; unitsize(1.25cm);
C = (0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(1,0)--(1,2)--(2,2)--(2,1)--(0,1);
B = (0,1)--(1,1)--(1,2)--(0,2)--cycle;
A = shift((1,-1))*B;
for(int i = 0; i < 3; ++i){
for(int k = i; k < 3; ++k){
D = shift((3*k,0))*(scale(.5^i)*A);
E = shift((3*k,0))*(scale(.5^i)*B);
fill(D,grey); fill(E,grey);
}
}
draw(C); draw(shift((3,0))*C); draw(shift((6,0))*C);
label("Stage 1", (0,0)--(2,0),S);
label("Stage 2", (3,0)--(5,0),S);
label("Stage 3", (6,0)--(8,0),S);[/asy]

Each stage adds $ \frac{1}{4}$ of what the previous stage added.  The first stage added $ 128$ to the shaded region, so the shaded area of the fifth stage is \[ 128\plus{}\frac{1}{4}\cdot 128\plus{}\left(\frac{1}{4}\right)^2\cdot 128\plus{} \left(\frac{1}{4}\right)^3 \cdot 128 \plus{} \left(\frac{1}{4}\right)^4 \cdot 128\equal{}\boxed{170.5}\]

Or you could count the white region which is 1/4+1/16+1/64+1/256+1/1024+1/1024=171/512. Therefore the shaded region is 341/512. This means that the shaded region is 341/512 of the square. So the answer is 341/512*256=170.5 as mentioned.


[quote="anuonhu"]If n is the natural number.
then find the value of 1/1+1/2+...+1/n=?[/quote]

There is no closed expression for $ S_n\equal{}\sum_{k\equal{}1}^n\frac{1}{n}$

You can have an approx value with the inequality : $ \ln(n\plus{}1)<S_n<1\plus{}\ln(n)$

[quote="pco"][quote="anuonhu"]If n is the natural number.
then find the value of 1/1+1/2+...+1/n=?[/quote]

There is no closed expression for $ \boxed{S_n \equal{} \sum_{k \equal{} 1}^n\frac {1}{n}}$

You can have an approx value with the inequality : $ \ln(n \plus{} 1) < S_n < 1 \plus{} \ln(n)$[/quote]

Hi patrick :) !!!

Sorry  but i think that you want to write $ Sn\equal{}\sum_{k\equal{}1}^n \frac{1}{k}$

and thanks

I think pco's $ S_n$ is correct. However, I believe that the expression should be $ \ln(n\plus{}1) < S_{n} \leq 1\plus{}\ln(n)$

No, it's actually incorrect.

$ \sum_{k\equal{}1}^{n}\frac{1}{n}\equal{}\frac{n}{n}\equal{}1$.

However, $ \sum_{k\equal{}1}^{n}\frac{1}{k}\equal{}H_n$, the $ n^{\mathrm{th}}$ [url=http://en.wikipedia.org/wiki/Harmonic_number]harmonic number[/url].

Oh, I was looking at $ Sn$ as opposed to $ S_n$.

Maybe some moderator could transfer this thread, since mathema's remarks, in some "typo_error_discussion" forum.


Let $n$ be an integer, and assume that there exist integers $x$ and $y$ such that $3n=x^{2}+2y^{2}$. Show that there exist integers $u$ and $v$ such that $n=u^{2}+2v^{2}$.

even though i have already solved this problem i would like to see if there are any other methods of solving it.
[hide="my solution"]
use the fact that $(a+b)^{2}+2(a-b)^{2}=3\left((a-b)^{2}+2b^{2}\right)$. (i attained this after a little experimentation.)
now all we have to do is let $u=a+b$ and $v=b$.
this gives use that $(u+2v)^{2}+2(u-v)^{2}=3(u^{2}+2v^{2})$.
now once again we make a substitution, this time we let $x=u+2v$ and $y=|u-v|$. after this we are done because we have,
$x^{2}+2y^{2}=3(u^{2}+2v^{2})=3n$
[/hide]

Your solution is incomplete in one detail. The given are $x,y$, hence you must prove that $u,v$ obtained by your formulas are indeed integer.

Some of his latex isn't rendering on my browser, the part before "I attained with with a little experimentation." I'm using firefox 1.5, does anyone know why this is?

I can verify it exists by quoting him...

Also, how did you come up with the idea of using that identity?

I still don't understand how you came up with that first identity, it seems very involved. Though the substitutions from there are very obvious...

well i got the idea from knowing that if $n=x^{2}+y^{2}$ and $m=a^{2}+b^{2}$ then $nm=(ax+by)^{2}+(bx-ay)^{2}$.

also in my solution i forgot to mention that I made $a$ and $b$ integers.
so then $u$ and $v$ are integers.

Thats just multiplicativity $N(xy)=N(x)N(y)$ of norms of field extensions  :D

The norm of $\mathbb{Q}[\sqrt{-2}]$ is given by $N: (a+b \sqrt{-2}) \mapsto (a^{2}+2b^{2})$, in general that in $\mathbb{Q}[\sqrt d]$ is $N: (a+b \sqrt d) \mapsto (a^{2}+db^{2})$.
To check $N(xy)=N(x)N(y)$ is just a routine check here, and it gives your identity with $y=1+1 \sqrt{-2}$.

For a general finite extension $L|\mathbb{Q}$, one considers the different mappings $\sigma: L \to \mathbb{C}$ with $\sigma(1)=1, \sigma(a+b)=\sigma(a)+\sigma(b), \sigma(ab)=\sigma(a)\sigma(b)$ (called field embeddings) and defines $N(x)=\prod_{\sigma}\sigma(x)$.
To check that it is multiplicative by this is is easy if the set of those $\sigma$ is shown to be finite, but to say more about them is part of superior algebra (just believe that it works ;)).
This generalizes to arbitrary extensions, too.

@maokid7: this doesn't work that way, you set $x,y$ to some values, but these values are given. In fact, you can't be sure that you get integers that way.

i dont quite understand what you are saying.

[b]maokid7[/b], please review your solution once more. Youre final conclusion is:

$n=u^{2}+2v^{2}\implies 3n=x^{2}+2y^{2}$

and what's asked for is the contrary statement.

What you've been looking for is

$3n=x^{2}+2y^{2}\implies n=\left({x+2y\over 3}\right)^{2}+2\left({x-y\over 3}\right)^{2}$, which is easily provable. Now you only have to show that the two fractions on the RHS are integer.


inequalities calculus derivative function trigonometry

Let $x$ be a real numbers satisfying  $1 \leq x \leq 2$.Find min of:
 $P = \sqrt{4x-x^{3}}+\sqrt{x+x^{3}}$

This is sorta cheap-ish and probably not the nice, inequalities solution you wanted; however, if I saw this on a contest and it was an open-answer contest, this is how I would go about solving it.

[hide="Cheap"]
$x+x^{3}$ is strictly increasing on this interval, so it's squareroot must be, also.

$4x-x^{3}=-x(x-2)(x+2)$ clearly cannot have a point of inflection in this interval. Since it is zero at $x=0$ and $x=2$ and positive in between, $4x-x^{3}$ is either:

i. first increasing then decreasing
   [b]or[/b]
ii. strictly decreasing on this interval.

However, because finite differences (or derivative, for that matter) of cubics are parabolic, and this cubic is centered at the origin, we know it must be the second (the maximum will occur more than halfway between the roots). Hence, we know that $4x-x^{3}$ is first increasing and then decreasing on the interval (possibility i).

Thus, its squareroot is also first increasing then decreasing.

Then, we know that $P$ is either:
i. strictly increasing
or
ii. increasing then decreasing.

Either way, we only have to try the two endpoints:
$P(1)=\sqrt{2}+\sqrt{3}$ and $P(2)=\sqrt{10}$

Squaring both and manipulating makes it clear that $P(1)<P(2)$, so the minimum value is in fact $P(1)=\sqrt{3}+\sqrt{2}$
[/hide]

I don't think I understand your argument.  You are saying that $\sqrt{4x-x^{3}}$ is decreasing on the interval, and that $\sqrt{x+x^{3}}$ is increasing, which I agree with.  But what does that tell you about the sum?

No, it is first [i]increasing[/i] and [i]then[/i] decreasing.

(sorry, I've spaced out the solution so that it's more easy to follow visually)

So basically, $P'(1)=\text{positive}+\text{positive}>0$ and $P'(2)=\text{positive}+\text{negative}=\text{?}$, and there can be no points of inflection in between.

So $P(x)$ either first increases and then decreases, in which case the minimum is at one of the two endpoints, or, $P(x)$ is strictly increasing, in which case the minimum is necessarily at $P(1)$.

Either way, checking $P(1)$ and $P(2)$ will give the minimum value.

Sorry, I said it wrong: $\sqrt{4x-x^{3}}$ is first increasing and then decreasing, while $\sqrt{x+x^{3}}$ is increasing.  (I was a little misled because your first post referred to possibility ii, when I think you meant possibility i.)

But, again, I don't see what they tells you about the sum.  How do you know that $P$ has no inflection points?  Why must $P$ either first increase and then decrease or be increasing?  Why can't it oscillate up and down many times?

:oops: My bad, sorry!

The squareroot function is strinctly increasing over non-negative reals, so it suffices to analyze the way what is under the squareroot is changing.

It's quite clear that $x^{3}+x$ is strictly increasing, and $4x-x^{3}$ can be analyzed by a combination of factoring it and noticing it that it's point-symmetric about the origin.

In any case, I'm curious to see a more... well, "appropriate" solution to this problem  :maybe: I'm assuming that it was written to be solved with one of, or a combination of, well-known inequalities

Okay, we agree on when the individual summands go up or down.  But that by itself tells us little about the sum.  As an example, the function $2x+\sin x$ is increasing, while $-2x$ is decreasing.  But their sum oscillates back and forth.  So your argument is incomplete.

That's true...  oops :(

Thanks

Ok, so I guess I'd have to mention concavity too. Which is very doable, but it'll be making a big deal out of something that probably has a very straightforward solution.

Let $x$ be a real numbers satisfying  $1 \leq x \leq 2$.Prove that
 $$\sqrt{3}+\sqrt{2}\leq \sqrt{4x-x^{3}}+\sqrt{x+x^{3}}\leq 3\sqrt[3]{3}$$
Let $x$ be a real numbers satisfying $\frac{1}{2} \leq x \leq 2$.Prove that
$$\frac{\sqrt{5(2+\sqrt{3})}}{2}\leq \sqrt{4x-x^{3}}+\sqrt{x+x^{3}}\leq 3\sqrt[3]{3}$$


geometry parallelogram rhombus rectangle

Let E be the point of intersection of the diagonals of convex quadrilateral ABCD, and let P,Q,R,and S be the centers of the circles circumscribing triangles ABE, BCE, CDE, and ADE, repectively.  Then?

A. PQRS is a parallelogram
B. PQRS is a parallelogram if and only if ABCD is a rhombus
C. PQRS is a parllelogram if and only if ABCD is a rectangle
D. PQRS is a parallelogram if and only if ABCD is a parallelogram
E. None of the above is true

Dialate everything out of point E by a factor of 2...the sides of O_1O_2O_3O_4 are perpendicular to AE, BE, CE, DE, and pass through A,B,C,D. It is easy to finish the problem now with angle chasing.

Hmmm, I may be missing something...

Let the midpoints of AE,BE,CE,DE be A',B',C',D'. Clearly PA' and SA' are both perpendicular to AC so PA'S are collinear, similarly B' is on PQ, C' is on QR, D' is on RS. Thus PQ and RS are perpendicular to BD are thus parallel, similarly PS and QR are parallel and it's a parallelogram.


search
combinatorics proposed
combinatorics

Given a set of n elements, find the largest number of subsets such that no subset is contained in any other

The answer is  $ \binom{2k}{k}$ for $ n \equal{} 2k$ and  $ \binom{2k \plus{} 1}{k}$ for $ n \equal{} 2k \plus{} 1$

A proof of this can be found on:

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=743837099&t=110415


inequalities
inequalities proposed
algebra

[quote="vo thanh van"][b]BJ2. [/b] Positive real numbers $ a,b,c$ satisfy inequality $ \frac {3}{2}\geq a \plus{} b \plus{} c$. Find the smallest possible value for
                       $ S \equal{} abc \plus{} \frac {1}{abc}$[/quote]
we only check minimum of $ S_{a,a,c}$ and $ S_{a,a,a}$($ ABC$ method)
But i think it can be killed by AMGM

By AM-GM we have:
$ \frac{3}{2} \geq a\plus{}b\plus{}c \geq 3\sqrt[3]{abc}$=>$ abc \leq \frac{1}{8}$
We have:
$ S\equal{}abc\plus{}\frac{1}{abc}\equal{}(abc\plus{}\frac{1}{64abc})\plus{}\frac{63}{64abc}$
$ \geq \frac{1}{4}\plus{}\frac{63}{8}\equal{}\frac{65}{8}$
Equality occurs when $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$

$ abc\plus{}\frac{1}{abc}\equal{}\frac{(a^2b^2c^2\minus{}\frac{abc}{4}\plus{}\frac{1}{64})\plus{}\frac{abc}{4}\plus{}\frac{63}{64}}{abc}$ $ \equal{}$

$ \frac{(abc\minus{}\frac{1}{8})^2}{abc}\plus{}\frac{1}{4}\plus{}\frac{63}{64abc}$

$ \ge0\plus{}\frac{1}{4}\plus{}\frac{63}{8}\equal{}\frac{65}{8}$.

We use that $ \frac{1}{abc}\ge 8$.Equality holds for $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$

[quote=sqing]Let $ a,b,c>0$ and$a^2+b^2+c\leq\frac {3}{2}$. Prove that
$$abc + \frac {1}{abc}\geq\frac{1105}{288}$$
Equality holds when $a=b=\frac {\sqrt 6}{4},c=\frac {3}{4}.$[/quote]

[quote=sqing][quote=vo thanh van][b]BJ2. [/b] Positive real numbers $ a,b,c$ satisfy inequality $ \frac {3}{2}\geq a \plus{} b \plus{} c$. Find the smallest possible value for
                       $ S \equal{} abc \plus{} \frac {1}{abc}$[/quote]

Moldavian Junior Balkan First TST？
[b]Does anyone know what year this is the question？[/b]
Thanks.[/quote]

2008

[quote=Rofler]By AM-GM we have $ \frac {1}{8}\ge abc$. Therefore, abc must lie between 0 and 1/8 inclusive.

Now, I claim $ 1/8 \plus{} 8 \le x \plus{} 1/x$ where x is between 0 and 1/8.

Proof:

$ 1/8 \plus{} 8 \le \frac {x^2 \plus{} 1}{x}$

$ 65x \le 8x^2 \plus{} 8$

$ 0 \le 8x^2 \minus{} 65x \plus{} 8$

$ 0 \le x^2 \minus{} (8 \plus{} 1/8) \plus{} 1$

$ 0 \le (x \minus{} 1/8)(x \minus{} 8)$

$ 0 \le (1/8 \minus{} x)(8 \minus{} x)$

Which is true. Therefore,

$ 8 \plus{} 1/8 \le abc \plus{} 1/abc$. You can get equality by letting a=b=c=1/2.[/quote]

Nice sol! I did the same thing.

Let $ a,b,c>0$ and$a^4+b^4+c\leq\frac {3}{2}$. Prove that
$$abc + \frac {1}{abc}\geq\frac{5}{2}$$Equality holds when $a=b=\frac{1}{\sqrt 2},c=1.$
Let $ a,b>0$ and  $ab+ \frac {1}{ab}\leq 3$. Prove that
$$a+b\geq \sqrt{5}-1$$
$$a^2+b^2\geq 3-\sqrt{5}$$Equality holds when $a=b=\frac{\sqrt  5-1}{2}.$

[quote=sqing]Let $ a,b,c>0$ and$a^4+b^4+c\leq\frac {3}{2}$. Prove that
$$abc + \frac {1}{abc}\geq\frac{5}{2}$$Equality holds when $a=b=\frac{1}{\sqrt 2},c=1.$
[/quote]


Find a polynomial with nonzero integral coefficients, such that a root of the polynomial is $\sqrt{2}+\sqrt[3]{3}$.

Show work

Thank You.

Could you explain to me how you would be able to tackle a question like this?

Find a polynomial with nonzero integral coefficients, such that two roots of the polynomial are $\sqrt{6}+\sqrt{5}$, and $\sqrt{3}+\sqrt{2}$.

Would you just do the same method and then multiply the two polynomials?

Thanks.

[quote="mathgeniuse^ln(x)"]Would you just do the same method and then multiply the two polynomials?[/quote]

Yes, that's what I would do too.

[quote="mathmanman"]You may want to have a look at that thread :
http://www.mathlinks.ro/Forum/viewtopic.php?t=89383[/quote]

Thank You Very Much.

I will learn from this.

Isn't it necessary that all coefficients are nonzero? :maybe:

Well if I didn't say that then someone would just say 0, as the polynomial and that is a polynomial.

So I had to do that.

What you could say is just that the polynomial is nonzero, not that its coefficients are nonzero.  In particular, the polynomial Farenhajt gave has a zero coefficient after the leading coefficient.  We could fix this by multiplying by $x+1$, say, but that's pretty artificial and distorts the obvious intent of the question.  (This entire issue is pretty minor, of course.)


geometry
power of a point
radical axis
geometry unsolved

Let be given a triangle $ ABC$ and a point $ M$ on its side $ BC$. Circle $ (O_1)$ passes through $ A$, $ B$, $ M$ and circle $ (O_2)$ passes through $ A$, $ C$, $ M$. Points $ E$ and $ F$ are on $ (O_1)$, $ (O_2)$, respectively such that $ ME\parallel AC$ and $ MF\parallel AB$. Let $ N$ be an arbitrary point on $ AM$. Denote by $ P$ the intersection of $ BN$ with $ (O_1)$ and $ Q$ the intersection of $ CN$ with $ (O_2)$. Prove that lines $ AM$, $ EP$ and $ FQ$ are concurrent.

[img]http://img136.imageshack.us/img136/9064/sketch1wg6.png[/img]
$ 1)$ It's easy to see that $ E,A,F$ are collinear
$ 2)$ $ BN.NP \equal{} MN.NA \equal{} NC.NQ.$ Hence $ BCPQ$ is cyclic
$ 3)$ $ \angle QAM \equal{} \angle QCB \equal{} \angle QPN.$ Hence $ AQNP$ is cyclic.
$ 4)$ $ \angle FQP \equal{} \angle AQP \minus{} \angle AQF \equal{} \angle BNM \minus{} \angle BEM \equal{}$ $ \angle BNM \minus{} \angle BAM \equal{} \angle ABP \equal{} \angle FEP.$ Hence $ PQEF$ is inscribed in circle $ O_3$.
$ 5)$ $ FQ$ - radical axis of $ O_2$ and $ O_3$, $ EP$ - radical axis of $ O_1$ and $ O_3$, $ AM$ - radical axis of $ O_1$ and $ O_2$. Hence $ AM\cap EP\cap FQ$=$ S$


geometry
rhombus
projective geometry
geometric transformation
homothety
geometry proposed

Points $ X, Y, Z$ are marked on the sides $ AB, BC, CD$ of the rhombus $ ABCD$, respectively, so that $ XY \| AZ.$ Prove that $ XZ, AY$ and $ BD$ are concurrent.

direct application of desargues theorem..

i solve it nice but easy.let $ M,N$ and $ P,Q$ be the projections of $ B,D$ on $ XZ,YA$ if $ O$ is the intersect of $ XZ,AY$ we must prove (M,N is on XZ) $ (MO)(QO)\equal{}(NO)(PO)$ and its realy is with similitude or Tales

Let's call $ T\equal{} AZ \cap BC$ and $ AB \equal{} a$; from $ \triangle ABT \sim  \triangle ZDA$ we get $ BT \cdot DZ \equal{} a^2$ ( 1 ). From $ XY \parallel{} AT$ we get $ \frac{BX}{AB} \equal{} \frac{BY}{BT}$ ( 2 ). Substituting (1) in (2) we get $ \frac{BX}{DZ}\equal{}\frac{BY}{a}$ ( 3 ). From (3), with $ a\equal{}AD$, we get that $ AY$ and $ XZ$ intersect $ BD$ at the same point, done.

Best regards,
sunken rock

We see that two triangles $ ADZ,YBX$ are homothetic,so $ BD,XZ,AY$ concur at the one point which is the center of homothety maps $ ADZ \to YBX$.
 :)


inequalities
trigonometry
geometry proposed
geometry

If $ a,b,c$ are the sides of $ \triangle{ABC}$ and $ r_a,r_b,r_c$ are the radii of excircles, then

$ \left( \cos{B}\cos{C} \plus{} \cos{C}\cos{A} \plus{} \cos{A}\cos{B}\right)^2\leq\frac {\sqrt {3}}{8}\left(\frac {r_a}{a} \plus{} \frac {r_b}{b} \plus{} \frac {r_c}{c}\right),$

with equality if and only if $ \triangle{ABC}$ is equilateral.


I don't know if it's my computer, but I went ot an Alcumus problem the words were overlapped.  I couldn't do the problem because of that.  I don't know what to do.  I don't want to just give up.

Can anybody help me???

try refreshing the screen
if that doesn't work, close and restart your browser
if that doesn't work, refresh your whole computer

if that doesn't work, post here again and tell us

Please let us know what browser, browser version, and operating system you are using.

[quote="geotrig"]I don't know what browser, browser version, or operating system it is.
I don't know what that means but my computer at home shows the problem just fine.[/quote]

How do you not know?

The browser is either "Internet Explorer", "Mozilla FireFox", "Opera", "Safari", and "Google Chrome".

If you go to Help->About, it should tell you what version and type.

As for operating system, do you have a computer that is totally white and has an Apple logo on it?  IF so, you have  Mac.  If it is not, it is most probably Windows.  To tell what operating system in windows it is, go to Control Panel--> System.  that should say your OS and everything else.

OR

Send a screen shot of your browser with the task bar included (or your other thingy if you have a mac)

If someone doesn't know what browser they are using, then I doubt they can send you a screenshot  :wink:

For browser, all you really need to do is take a quick look at the icon in the top left corner of the internet thing.

If it's a picture of the Earth with a fox sort of hanging on to it, it's Mozilla.
If it's a picture of an E with a kind of horizontal line, it's IE.
If it's a picture of a blueish dot (I'm not totally sure about this), it's Safari.
I think Google Chrome has a much more colorful logo with arrows, I might need to check.

For operating system, if you see a blue bar across the bottom of the screen that has a "start" button on the left of it, then it's Windows.
If you have a bar going across the top of the screen, that has a logo on it that kind of reminds you of Macromedia Flash's logo, it's Linux.
If you have a little rectangular see-through menu-like bar on either the bottom or the sides or the top of the screen that have icons in them that enlarge a bit when you move your cursor over, it's Mac.

BTW - It's not called "refreshing" your computer.

If you are using Internet Explorer, which you probably are, please go to the Help menu and select About Internet Explorer. Under there you will find version information. Please provide that information.

If you are using AOL, there is probably a similar thing under Help. Although, I don't use AOL so can't say for certain.


If a body falls under the action of gravity (and air resistance is neglected), the distance traveled is given by the formula $s=16t^2$, where s is the distance in feet and t is the time in seconds.  How far does a falling body drop during the fifth second?

[color=white]
To show how fast it's falling...

f(x) = 16t^2
f'(x) = 32t
t = 5
32(5)
160 feet per second

[/color]


Euler
function
number theory unsolved
number theory

Let [tex]n[/tex] be a positive integer with [tex]n \geqslant 3[/tex]. Show that [list][tex]n^{n^{n^n}} - n^{n^n}[/tex][/list] is divisible by [tex]1989[/tex].

I have seen this before on the forum a very long time ago.

I was quite sure it has been discussed before... anyway providing me a link would be very nice  :roll:

Actuall that topic is old and didn't present a solution. Actually people were kind of doubting whether the problem was true or not. Anyway, grobber has presented a solution in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1989&t=300 :)

I think your solution is too long. I have a solution ,I think this solution is interesting and general.
     Consider $a_1=n^n,a_2=n^{n^n},......$ .We had some results
1, With all $m$ this consequen $a_n$ is stop follow mod $m$
2,Suppose $k_m$ is stop constant ( not sure is minimum).
   Let $p_m$ is k s.t $a_{k}\geq m$
  Then $k_m\geq max\{p_m,k_{f(m)}\}$ with $f(m)$ is EULER function
We can prove thatb by induction.
 Then ,$f(1989)=1320,f(1320)=420,f(420)=96,f(96)=64,f(64)=32,f(32)=16,f(16)=8,f(8)=4,f(4)=2,f(2)=1$
Then we have with $ k_2=1$$k_4=max\{2,1\}=2  ,k_8=2,k_{16}=2,k_{32}=3,,k_{64}=3,k_{96}=3,k_{420}=3,k_{1320}=3,k_{1989}=3$  then we have done(note $n\geq3$


inequalities
function
search
limit
modular arithmetic
algorithm
inequalities theorems

[size=150][b]Strong(er) Mixing Variables theorem[/b][/size]
[size=117][b](or SMV-theorem, from Hungkhtn's book)[/b][/size]

Hi every one,

The method that will be showed as follow is very effective for 4-variable inequalities, but very elementary. It's comed from the uncomfortableness of the old mixing variables, that is

[u][b]Theorem 1 (Mixing variables theorem).[/b][/u]
If $F: I\subset R^{n}\to R$ is a sysmetric, continuous function and for all $(x_{1},x_{2},...,x_{n})$ we have \[F(x_{1},x_{2},...,x_{n}) \ge F\left(\frac{x_{1}+x_{2}}{2},\frac{x_{1}+x_{2}}{2},x_{3},...,x_{n}\right)\] Then \[F(x_{1},x_{2},...,x_{n}) \ge F(x,x,...,x)\] where $x=\frac{x_{1}+x_{2}+...+x_{n}}{n}$.

Why I tell that this theorem is uncomfortable? It's because, if someone asks you that, "explain me this theorem, I can't proved it", what will you say? Oh, dont tell me that, by "superior mathematics" it's obvious. I know that fact, but I dont fell any satisfied.

Now SMV theorem resolved all this problem, one side it can refer mixing variables theorem, further more it's almost useful for 4-variables inequalities.

The following lemma is the main principal of SMV-theorem. It's not very hard, but very nice. I dont show the solution right now, it's for your try in some days.

[u][b]Lemma (the main principal).[/b][/u]
Let $(x_{1},x_{2},...,x_{n})$ be a sequence of real numbers. Carry out the following transformation, (hereinafter, we call it $\Delta$ transformation)
(i). Choose the maximum and minimum of numbers from this sequence, denote $x_{i}$ and $x_{j}$ (with $x_{i}\le x_{j}$)
(ii). Replace $x_{i}$ and $x_{j}$ to $\frac{x_{i}+x_{j}}{2}$ (but their orders don't change).
If we perform this transformation infinitely, then all number $x_{k}(k=1,2,...,n)$ tend to the same limit \[x=\frac{x_{1}+x_{2}+...+x_{n}}{n}.\] Now try to prove this lemma, because the solution is far apart from inequality, but very nice and special.

This lemma shows SMV-theorem immediately,

[u][b]Theorem 2 (SMV theorem).[/b][/u]
If $F: I\subset R^{n}\to R$ is a sysmetric, continuous function and for all $(x_{1},x_{2},...,x_{n})$ we have \[F(x_{1},x_{2},...,x_{n}) \ge F(y_{1},y_{2},...,y_{n})\] where $(y_{1},y_{2},...,y_{n})$ is performed from $(x_{1},x_{2},...,x_{n})$ by $\Delta$ transformation, then \[F(x_{1},x_{2},...,x_{n}) \ge F(x,x,...,x)\] where $x=\frac{x_{1}+x_{2}+...+x_{n}}{n}$.

After proving the sequence lemma, I will show that why SMV-theorem is almost effective for 4-variables inequalties. But firstly, prove it!

[quote="Rzeszut"]I proved some time ago a similar lemma:

[i]Let $(x_{0},\ldots,x_{n-1})$ be a sequence of real numbers with indices modulo $n$ and let $m$ be such positive integer that $\gcd(m,n)=1$. Then after applying transformation $f$ defined as follows: $f(x_{0},\ldots,x_{n-1})= \left(\frac{x_{0}+x_{m}}{2},\ldots,\frac{x_{n-1}+x_{n-1+m}}{2}\right)$ infinitely many times all terms of sequence tend to the same limit $\frac1n\cdot\sum_{0\leqslant i\leqslant n-1}x_{i}$. [/i]

It works also for transformation $f(x_{0},\ldots,x_{n-1})= \left(px_{0}+qx_{m},\ldots,px_{n-1}+qx_{n-1+m}\right)$ for $p+q=1$. Is my result useful here?[/quote]

It's very nice, Rzeszut. I dont know it before, but it's really amazing to me.

Please send some problems(inequalities), where we can solved with SMV theorem!
:wink:

I want to say, all 4-variables symetric inequality you show in mathlinks.

For instance, a quick search shows (Other are so much, but I only seach key words "hungkhtn"). Sorry, I haven't so much time,

VASC's problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=90766

My problem, Part (a)
http://www.mathlinks.ro/Forum/viewtopic.php?t=112730

My problem, very hard
http://www.mathlinks.ro/Forum/viewtopic.php?t=64926

My problem, especially two other improved parts
http://www.mathlinks.ro/Forum/viewtopic.php?t=84043

My problem, dedicated to Mircea
http://www.mathlinks.ro/Forum/viewtopic.php?t=81539

My problem, improred Nesbitt inequality,
http://www.mathlinks.ro/Forum/viewtopic.php?t=76277

VASC's inequality, 5-variables
http://www.mathlinks.ro/Forum/viewtopic.php?t=68701

Also, the last one is very nice and unsolved in Mathlinks now.
http://www.mathlinks.ro/Forum/viewtopic.php?t=32525

Turkevici's inequality follow immediately, also for Suranji's inequality (of 4-variables) and VASC-inequality (of 4-variables).

But I'm so tired to check more now... Find any 4-variables symetric inequality and then compare...

[quote="hungkhtn"][size=117][b](or SMV-theorem, from Hungkhtn's book)[/b][/size][/quote]
Thank you.  :D

Not at all, Xevaxion. I want to post it before, but I have no time. This theorem can be improved to a new one, called n-SMV theorem. It's only changed a bit, but really more effective.

I have no time. :(

[Post 998] A proof of SMV-theorem, shared from my book

Following result only uses elementary mathematics, understanding the theorem's usage and its meaning is more important to you than remembering its detailed proof.

[quote="Hungkhtn's book"]
[b]Lemma. (General mixing variables lemma).[/b]
[i]Suppose that $(a_{1},a_{2},...,a_{n})$ is an arbitrary real sequence. Carry out the following transformation consecutively
(i). Choosing $i,j \in \{1,2,...,n\}$ to be two indeces satisfying \[a_{i}=\min(a_{1},a_{2},...,a_{n}),\ a_{j}=\max(a_{1},a_{2},...,a_{n}).\] (ii). Replacing $a_{i}$ and $a_{j}$ by $\frac{a_{i}+a_{j}}{2}$ (but their orders don't change).
After doing infinitely above transformations, each number  $a_{i}$ comes to the same limit[/i] \[a=\frac{a_{1}+a_{2}+...+a_{n}}{n}.\] Henceforward, the above transformation is called the $\Delta$ transformation.
Symbol the first sequence as $(a^{1}_{1},a^{1}_{2},...,a^{1}_{n})$. After one transformation, we have a new sequence symboling as $(a^{2}_{1},a^{2}_{2},..,a^{2}_{n})$.
Similarly, from the sequence $(a^{k}_{1},a^{k}_{2},...,a^{k}_{n})$ we have a new one symboling as $(a^{k+1}_{1},a^{k+1}_{2},...,a^{k+1}_{n})$.
Thus, for every integers $i=1,2,...,n$, we need to prove \[\lim_{k\rightarrow \infty}a^{k}_{i}=a,\ a=\frac{a_{1}+a_{2}+...+a_{n}}{n}.\] Let $m_{k}=\min(a^{k}_{1},a^{k}_{2},...,a^{k}_{n})$ and $M_{k}=\max(a^{k}_{1},a^{k}_{2},...,a^{k}_{n})$.

Clearly, the transformations $\Delta$ don't make the value of $M_{k}$ increase or the value of $m_{k}$ decrease. Because both $m_{k}$ and $M_{k}$ are bound,  there exist \[m=\lim_{k \rightarrow \infty}m_{k},\quad M=\lim_{k \rightarrow \infty}M_{k}.\] We must prove that $m=M$. By contradiction, suppose that $M > m$. Denote $d_{k}=M_{k}-m_{k}$.
We have a simple notice

Comment. Suppose that after carrying out some transformations $\Delta$, the sequence $(a^{1}_{1},a^{1}_{2},...,a^{1}_{n})$ turns into the new one $(a^{k}_{1},a^{k}_{2},...,a^{k}_{n})$ satisfying that $m_{k}=\frac{M_{1}+m_{1}}{2}$, then we will have $m_{2}=\frac{M_{1}+m_{1}}{2}$.

Indeed, WLOG we may assume that $M_{1}=a^{1}_{1}\ge a^{1}_{2}\ge ... \ge a^{1}_{n}=m_{1}$. To be briefer, replace $a_{i}$ by $a^{1}_{i}$.
If $m_{k}=\frac{a_{1}+a_{n}}{2}$ and $k$ is the smallest index satisfying that then $a^{2}_{i}\ge m_{k}\forall i=\oval{1,n}$.
It comes from $\{m_{k}\}$ is a non-decreasing sequences. Note that $\frac{a_{1}+a_{n}}{2}$ is a term of the sequence $(a^{2}_{1},a^{2}_{2},..,a^{2}_{n})$, so we have done.

By the above property, we have more important result. Denote \[S=\{k : \exists l>k | m_{k}+M_{k}=2m_{l}\}\Ra S=\{k | m_{k}+M_{k}=2m_{k+1}\},\]  \[P=\{k : \exists l>k | m_{k}+M_{k}=2M_{l}\}\Ra P=\{k | m_{k}+M_{k}=2M_{k+1}\}.\] If $S$ or $P$ has infinite elements, assume $|S|=\infty$ then, for each $k \in S$ \[d_{k+1}=M_{k+1}-m_{k+1}=M_{k+1}-\frac{m_{k}+M_{k}}{2}\le \frac{M_{k}-m_{k}}{2}=\frac{d_{k}}{2},\] because $(d_{r})_{r=1}^{+\infty}$ is a decreasing sequence. Thus, if $|S|=\infty$ then $\lim_{r \rightarrow \infty}d_{r}=0$ and hence $M=m$, the conclusion follows.

Otherwise, we must have $|S|, |P| <+\infty$. Hence, we can suppose that $|S| = |P|=0$ without affecting the result of the problem.
Then, for every $k>1$ the number $\frac{a_{1}+a_{n}}{2}$ can't be the smallest or greatest number in the sequence $(a^{k}_{1},a^{k}_{2},...,a^{k}_{n})$. So we can consider the confined problem with $n-1$ numbers when we rejected exactly {\it one} number $(a_{1}+a_{n})/2$.
By a simple induction method, we have a desired result. \cq

From the above lemma, we have a direct result

[b]Theorem (Stronger mixing variables - S.M.V theorem).[/b]
[i]If $f: \rr^{n}\rightarrow \rr$ is a continuous, symmetric, under-limitary function satisfying \[f(a_{1},a_{2},...,a_{n}) \ge f(b_{1},b_{2},...,b_{n}),\] in which $(b_{1},b_{2},...,b_{n})$ is a sequence obtained from the sequense $(a_{1},a_{2},...,a_{n})$ by the transformation $\Delta$,
then we always have \[f(a_{1},a_{2},...,a_{n}) \ge f(a,a,...,a)\] with $a=\frac{a_{1}+a_{2}+...+a_{n}}{n}.$[/i]

By this theorem, when using the mixing variables method, we only need to choose the smallest and greatest numbers to perform. By using elementary knowledge, the old mixing variables theorem is proved and improved to have a stronger result. So it can be applied freely.

Moreover, the transformation $\Delta$ can be different. For example, we can change it to $\sqrt{ab},\sqrt{\frac{a^{2}+b^{2}}{2}}$ or any arbitrary average's forms. Depending on the supposition of problem, we can choose a suitable way of mixing variables.[/quote]

Complete!

To Zhaobin, for the below post of you.

For the sequence $(0,1,3.1,3.9)$ then the new one is \[(1.95,1,3.1,1.95)\] (we change both $0$ and $3.9$ to $1.95$).

$m_{k}$ is certainly non-decreasing, not non-increasing.

I'm agree that maybe SMV is not belong to me, because by superior mathematics, it's quite obvious. But when I created it, I dont know who invent it before and furthermore, I think my proof is the first solution using elementaty mathematics.

Hungkhtn,Thanks for sharing your method.But I am sorry to say that I don't think the SMV method is yours.I think others may use it in proving inequalities,too.But we may think it is obvious,Then we didn't consider to give a strick proof for the SMV-Theorem as you named.Now I also want to point out the mistake of your proof for the lemma.
[quote="hungkhtn"]I have no time. :(

[Post 998] A proof of SMV-theorem, shared from my book

Following result only uses elementary mathematics, understanding the theorem's usage and its meaning is more important to you than remembering its detailed proof.

[quote="Hungkhtn's book"]
Comment. Suppose that after carrying out some transformations $\Delta$, the sequence $(a^{1}_{1},a^{1}_{2},...,a^{1}_{n})$ turns into the new one $(a^{k}_{1},a^{k}_{2},...,a^{k}_{n})$ satisfying that $m_{k}=\frac{M_{1}+m_{1}}{2}$, then we will have $m_{2}=\frac{M_{1}+m_{1}}{2}$.

Indeed, WLOG we may assume that $M_{1}=a^{1}_{1}\ge a^{1}_{2}\ge ... \ge a^{1}_{n}=m_{1}$. To be briefer, replace $a_{i}$ by $a^{1}_{i}$.
If $m_{k}=\frac{a_{1}+a_{n}}{2}$ and $k$ is the smallest index satisfying that then $a^{2}_{i}\ge m_{k}\forall i=\oval{1,n}$.
It comes from $\{m_{k}\}$ is a non-decreasing sequences. Note that $\frac{a_{1}+a_{n}}{2}$ is a term of the sequence $(a^{2}_{1},a^{2}_{2},..,a^{2}_{n})$, so we have done.

.[/quote]

Complete![/quote]
Sorry,I think this proposition is wrong! :D
You can try the sequcence $(0,1,3.1,3.9)$(Which mean $x_{1}=0,x_{2}=1,x_{3}=3.1,x_{4}=3.9$)
Then $m_{2}=1\neq \frac{0+3.9}2$,$m_{3}=\frac{0+3.9}2$
So you are wrong.
Where you mistake in the proof,I think is you can't get $a^{2}_{i}\ge m_{k}\forall i=\oval{1,n}$.Because  $\{m_{k}\}$ is a $non-decreasing$ sequences.(I think when you do you mistake  $\{m_{k}\}$ is a $non-increasing$ :D )

Anyway,I have a more directly proof for the lemma,If somebody interested in,I will post it :lol:

[quote="zhaobin"]

Anyway,I have a more directly proof for the lemma,If somebody interested in,I will post it :lol:[/quote]

Zhaobin, I am so curious and please let me see the proof. Thanks! :D

From my understanding the proof given by khung is just showing that $m=M$, but doesn't show that the limit is $\frac{a_{1}+\cdots+a_{n}}{n}$.

Correct me if I am wrong.

Thanks Zhaobin, it is clear now. Nice proof! :D

But I still have a lot of problem with hungkhtn's proof

Here another point that I dont't understand
Let say the set $S$ in hungkhtn's proof is turn out to be
$\{i: i\equiv 0\pmod 3\}$

So then we have $d_{4}<\frac{d_{3}}{2}$ and $d_{7}<\frac{d_{6}}{2}$, but there is no way to relate $d_{4}$ and $d_{7}$. So why $d_{k}\to 0$?

I'll try to generalize a bit that formula (I'm not very good at analysis, so correct me if I'm wrong somewhere).

I think it is true that for any symmetric and continuous function $f: I\subset \mathbb{R}_{+}^{n}\to \mathbb{R}$ and for any $p\in\mathbb{R}$:

if
\[f(x_{1},x_{2},x_{3},...,x_{n})\geq f(\sqrt[p]{\frac{x_{1}^{p}+x_{2}^{p}}{2}},\sqrt[p]{\frac{x_{1}^{p}+x_{2}^{p}}{2}},x_{3},...,x_{n}) \]
then
\[f(x_{1},x_{2},x_{3},...,x_{n})\geq f(\sqrt[p]{\frac{x_{1}^{p}+x_{2}^{p}+x_{3}^{p}+...+x_{n}^{p}}{n}},...,\sqrt[p]{\frac{x_{1}^{p}+x_{2}^{p}+x_{3}^{p}+...+x_{n}^{p}}{n}}) \]
The set of values that the smallest element may assume has an upper bound, that is to say $\sqrt[p]{\frac{x_{1}^{p}+x_{2}^{p}+x_{3}^{p}+...+x_{n}^{p}}{n}}=M$, so it has also a minimum upper bound.

Let's call it $b$

Let's suppose $b<M$ and we will reach an absurd.

Let $x_{1}$ be the biggest element and $x_{2}$ the smallest.

We can suppose $\sqrt[p]{2b^{p}-x_{2}^{p}}<M$

Now we should have:
\[\sqrt[p]{\frac{x_{1}^{p}+x_{2}^{p}}{2}}\leq b \]

\[x_{1}^{p}\leq 2b^{p}-x_{2}^{p}\]

\[x_{1}\leq \sqrt[p]{2b^{p}-x_{2}^{p}}<M \]
Absurd.

If there is more than a minimum element, we just repeat the algorithm for the others.

[quote="barra"]I also have the same generalization, which in fact indicated by hungkhtn.
BTW why we can assume $2b^{p}<M^{p}+x_{1}^{p}$?

Are you also trying to say that the same argument can be apply to show that the infimum of maximal elements is equal to $M$, so the limit of the sequence is $(M,M,\cdots,M)$?[/quote]

I assumed a different thing, and I can because $b-x$ can be as little as you want, so $\sqrt[p]{2b^{p}-x_{2}^{p}}$ can be as near as you want to $b$ and $b<M$

I read only zhaobin proof and I didn't see the generalization of hungkhtn.  :oops:

I know that the p-mean of the element (that is to say M) does not change, so if the smallest elemet equals the mean, then.... so I do not need to prove also the biggest element.

Anyway I forgot to say that the function $f(I)$ has to be continuous, otherwise, although $|x_{1}-M|+...+|x_{n}-M|$ goes as near as you want to $0$, you can't be sure that $f(x_{1},...,x_{n})$ goes as near as you want to $f(M,...,M)$

I think you can prove that, if the main condition is verified, then the function is continuous. I'll try to prove it later, but I'm not sure that it is always true.

Thank you for your explanation.
[quote="piever"]
I know that the p-mean of the element (that is to say M) does not change, so if the smallest elemet equals the mean, then.... so I do not need to prove also the biggest element.
[/quote]

What you proved is the supremum of the minimum elements equal to $M$. In order to use your argument need to mention that the set of all sequences with the same $p$-sequence is closed.

The generalization of the lemma:

[b]Lemma(Improved general mixing variable lemma) [/b]
Let $(a_{1},a_{2},...,a_{n})$ be a sequence of real numbers and $\epsilon_{1},\epsilon_{2}$ are two constants such that $\epsilon_{1},\epsilon_{2}\in(0,1)$. Carry out the following transformations
1. Chosing $i,j\in\{1,2,...,n\}$ to be two different indeces satisfying
\[a_{i}=\max(a_{1},a_{2},...,a_{n}), a_{j}=\min(a_{1},a_{2},...,a_{n}).\]
2. Replacing $a_{i}, a_{j}$ by a certain number $\alpha$ (but their order don't change) for which
\[\alpha\in [a_{i},a_{j}], \left|\frac{a_{i}-\alpha}{a_{i}-a_{j}}\right|<\epsilon_{1}<1, \left|\frac{a_{i}-\alpha}{a_{i}-a_{j}}\right|<\epsilon_{2}<1.\]
After doing infinitely above transformations, all numbers $a_{i}$ tend to the same limit.

To handle this general one, I am afraid that Zhaobin's way is unable to apply.

[quote="barra"]I also have the same generalization, which in fact indicated by hungkhtn.
BTW why we can assume $2b^{p}<M^{p}+x_{1}^{p}$?

Are you also trying to say that the same argument can be apply to show that the infimum of maximal elements is equal to $M$, so the limit of the sequence is $(M,M,\cdots,M)$?[/quote]

Since there is some ideas here, I want to assert that my initial solution to this theorem is true. Zhaobin's is true, too. But I wonder if someone can solve the general lemma in my previous part. I think it is really hard.

I am probably misunderstanding something:

What if $ f(a,b,c,d) \equal{} abcd$ then wouldn't we have $ f(a,b,c,d) \geq f(x,x,x,x)$ where $ x\equal{}\frac{a\plus{}b\plus{}c\plus{}d}{4}$?

Hi to, everybody!

At the past time, the teacher in the university (i'm a student at university XXX, at the faculty Telecomunication) to mathematics students at PhD that always there are an extensionary case to empty the includings from the line of the inequality very similiar to what you said. But i don't understand what he wanted to say. Can you please explain me if you have some time?

[quote="BanishedTraitor"]I am probably misunderstanding something:

What if $ f(a,b,c,d) \equal{} abcd$ then wouldn't we have $ f(a,b,c,d) \geq f(x,x,x,x)$ where $ x \equal{} \frac {a \plus{} b \plus{} c \plus{} d}{4}$?[/quote]

Not. It should be $ f(a,b,c,d) \leq f(x,x,x,x)$, since $ f(a,b,c,d) \leq f(a,b,\frac{c\plus{}d}{2},\frac{c\plus{}d}{2}).$ It is a direct corollary.

[quote="BanishedTraitor"]Yeah, I know that but I was wondering how I am using SMV theorem wrong. In fact: $ F: I \subset R^{n} \rightarrow R$ what does it mean?[/quote]

How did you use SMV?

Just wondering in

http://reflections.awesomemath.org/2006_6/2006_6_mixing.pdf

Why is the $ \Delta$ transformation only performed on the last $ 3$ variables and not on the whole sequence?


Any recommendations on books to read to get good at combinatorics and just discrete math in general?

this is a ok book http://www.amazon.com/exec/obidos/tg/detail/-/0072930330/qid=1124588874/sr=8-1/ref=pd_bbs_1/103-5321865-5748669?v=glance&s=books&n=507846

they use it, or did use it, at MIT but i think if ur looking for hard contest problems the counting problems are probablly too easy.


Three persons (Billy, Chris and John) in a pizza restaurant produce all the pizzas there. Billy produces $60%
$ from the total of pizzas the restaurant produces, Chris $30%
$ and John $10%
$. But, as they cannot be perfect, some of the pizzas lack some products (we will call it "not perfect"). Billy produces $2%
$ from his pizzas not perfect, Chris $3%
$ and John $4%
$. A customer takes a pizza:

1) What is the probability that the pizza is not perfect?

2) We take a pizza and it lacks some products. What is the probability that the pizza is made by John?


Jennifer earned $ \$180$ working a few days in her Dad's shop. If her daily wage had been $ 2$ less, she would have taken one more day to earn the same amount. How many days were the ``few days'' she worked?

here is another way:

dw=180
(d+1)(w-2)=180
dw=dw+w-2d-2
w-2d=2
look at factors and you find w=20 and d=9


Can anyone explain to me why there is a difference in the heats of formation of cis and trans-1,4-dimethylcyclohexane? The heat of formation of the cis isomer is -42.2 kcal/mol and that for the trans isomer is -44.0 kcal/mol. Is it because the most stable of the trans isomer is diequatorial and lower in energy than either conformation of the cis isomer, which has one methyl axial and one equatorial?

Also, how can I calculate the heat of combustion of these stereoisomers?, given that I know the heat of formation of both? I am really confused. Is the standard state of the elements zero, then the heat of combustion and formation are both negative?

Thanks.

Yes, the trans isomer is more stable due to the eq,eq conformation (not possible for the cis).


Let $0 < a, b, c < 1$ with $ab + bc + ca = 1$.  Prove that
\[\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2} \geq \frac {3 \sqrt{3}}{2}.\]

Determine when equality holds.

as it turns out the function $ \displaystyle f(x) = \frac x{1-x^2}$ is convex on $(0,1)$ because \[ f''(x) = \frac { 4a(1-a^2)}{(1-a^2)^4} >0, \] so we can apply Jensen to obtain that

\[ \sum \frac a{1-a^2} \geq 3 \frac { \frac s3}{ 1 - \frac {s^2}9 } \geq \frac {3\sqrt 3}2 \Leftrightarrow \]
\[ \frac {s^2}9 + \frac 2{3\sqrt 3} s \geq 1 \quad (1)  \] where $s=a+b+c$. But $(a+b+c)^2 \geq 3 (ab+bc+ca)=3$, so $s\geq 3$, so obviously (1) takes place. The equality can only follow when $a=b=c= \displaystyle \frac 1{\sqrt 3}$.

Another approach is following: $\frac{x}{1-x^2}\geq\frac{3\sqrt{3}}{2}\cdot x^2$.

When did the two of you become beginners? ;) :P

And yet another simple approach is to use the trigonometric substitution $a = \tan A, b = \tan B, c = \tan C, \mbox{ where } A + B + C = \pi/2$.

[quote="Myth"]I am not guilty, Valentin did it first.[/quote] sorry, never gonna happen again  :blush:  :blush:  :blush: :P

Could you bring the Singapore Selection Tests 2004?
thank you very much...
:):):)
Moldova rulezzzz :)

_________________________
"Our days are never coming back.... (System of a down -> Highway Song)"

I posted all 6 questions from the Singapore TST 2004, so they're lying in various threads around the forum.  You can just do a search to find them.

Maybe you can gather up all the links like I did with the Romanian TST 2004 and open up a thread on the National Olympiads forum :P :)

By the way, Singapore TST is it just for beginners ??!!
 :D

Obviously it's not just for beginners, it just so happens that the 1st 2 questions were rather easy (as far as I see, so that everybody's marks would look a bit better).  There were harder questions such as [url=http://www.mathlinks.ro/viewtopic.php?p=18493]this[/url] and [url=http://www.mathlinks.ro/viewtopic.php?p=18394]that[/url].

It's a very old thread

But I'm going to post a different (but easy to obtain) solution

\[ \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq\frac{3\sqrt{3}}{2}. \] $\iff \sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge 0$

Let $a\ge b\ge c$
$\implies 2a+\sqrt 3 a^2 - \sqrt 3 \ge 2b+\sqrt 3 b^2 - \sqrt 3 \ge 2c+\sqrt 3 c^2 - \sqrt 3$
And $1-a^2 \le 1-b^2 \le 1-c^2$

Thus by Tchebychef's inequality
$3\sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge (\sum_{cyclic}2a+\sqrt 3 a^2 - \sqrt 3)(\sum_{cyclic} \frac{1}{1-a^2})$

Since $ab+bc+ca=1 \implies a+b+c \ge \sqrt 3$ and $a^2+b^2+c^2 \ge 1$
Combining the above two results
$(\sum_{cyclic}2a+\sqrt 3 a^2 - \sqrt 3)(\sum_{cyclic} \frac{1}{1-a^2}) \ge 0$
$\implies \sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge 0$

QED

Easy. Just a Cauchy-Schwartz and it is equivalent to a+b+c>=sqrt{3}, which is trivial.

Similar, generalisated: sum a/(b^2+c^2)>=9/2sqrt(a^2+b^2+c^2).

Let $f(x)=\frac{x}{1-x^2}\text{ furthermore notice that }f''(x)=\frac{4x(1-x^2)}{(1-x^2)^4}>0\text{ for } x\in(0,1)$ thus the function is convex on this interval.
Furthermore $\sum_{cyc}\frac{a}{1-a^2}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f\left(\frac{\sum_{cyc}a}{3}\right)=\frac{\sum_{cyc}a}{1-\frac{\left(\sum_{cyc}a\right)^2}{9}}$
Therefore the inequality boils down to $\frac{\sum_{cyc}a}{1-\frac{\left(\sum_{cyc}a\right)^2}{9}}\ge\frac{3\sqrt{3}}{2}\Longrightarrow2\sum_{cyc}a\ge3\sqrt{3}-\frac{\sqrt{3}\left(\sum_{cyc}a\right)^2}{3}\Longleftrightarrow\left(\sum_{cyc}a\right)\left(6+\sqrt{3}\sum_{cyc}a\right)\ge9\sqrt{3}$, which is clearly true as $\left(\sum_{cyc}a\right)^2\ge3\sum_{cyc}ab\Longrightarrow \sum_{cyc}a\ge\sqrt{3}$ $\blacksquare$.


logarithms
limit
calculus
derivative
function
real analysis
real analysis unsolved

Prove without l'hopital: $\lim_{x \to 0}x\ln x=0$

Let $f(x)=x\ln x.$ Then $f'(x)=1+\ln x.$ From examining where the derivative is positive and negative, we see that $f$ is decreasing on the interval $(0,e^{-1}].$ We also have that $f(x)<0$ on that interval. A monotone bounded function must have a limit, so $\lim_{x\to0^+}f(x)$ exists and is $\le0.$

Let's try this again. This time, let $g(x)=\sqrt{x}\ln x.$ Then $g'(x)=\frac1{\sqrt{x}}\left(1+\frac12\ln x\right).$ We have that $g(x)$ is negative and decreasing on $(0,e^{-2}].$ Again, by the existence of limits for monotone bounded functions, $\lim_{x\to0^+}g(x)=L\le0.$

But then $f(x)=g(x)\sqrt{x}$ and $\lim_{x\to0^+}f(x)=L\cdot\lim_{x\to0^+} \sqrt{x}=0.$

$\lim_{x\to 0} x\ln x=0\Longleftrightarrow (x: =\frac 1x)\ \lim_{x\to\infty} \frac{ln x}{x} =0\Longleftrightarrow$

$\left(\forall\right)x_n\rightarrow \infty ,\ \frac{\ln x_n}{x_n}\rightarrow 0$ ([u]the Heine's[/u] [u]definition[/u]).

I suppose (w.l.o.g.) that $x_n>1,\ n\in N^*$.

I note $p_n=[x_n]\le x_n<p_n+1,\ n\in N^*.$

Then $\frac{\ln p_n}{p_n+1}<\frac{\ln x_n}{x_n}<\frac{\ln (p_n+1)}{p_n}\Longleftrightarrow$

$\frac{p_n}{p_n+1}\cdot \frac{\ln p_n}{p_n}<\frac{\ln x_n}{x_n}<\frac{p_n+1}{p_n}\cdot \frac{\ln (p_n+1)}{p_n+1}$

(the sequence $(p_n)_n$ is a subsequence of the sequence $(n)_n$ and $p_n\rightarrow \infty ).$

You can easily prove that $y_n =\frac{\ln n}{n}\rightarrow 0$.

Therefore, the sequence $y_n$ is decresing (?), $y_{2n}=\frac{\ln\ (2n)}{2n}=\frac{\ln 2}{2n}+\frac 12\cdot y_n$

and for $y_n\rightarrow y$ results $y=\frac y2\Longrightarrow y=0.$


Show that if three digit numbers abc, bca, cab are divisible by n, then so is a^3+b^3+c^3-3abc

Let's be clear:  $3abc$ means $3 \cdot a \cdot b \cdot c$?

[hide="Solution"]$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)( (a-b)^{2}+(b-c)^{2}+(c-a)^{2})$

Now, we are given that $n | (100a+10b+c)$, etc.  This tells us $n | 111(a+b+c)$.

[b]Case:[/b]  $gcd(n, 111) = 111$.  Then $a = b = c$ and our given expression is $0$.

[b]Case:[/b]  $gcd(n, 111) = 37$.  Then

$100a+10b+c \equiv 10(b-a)+(c-a) \bmod 37$
$100b+10c+a \equiv 10(c-b)+(a-b) \bmod 37$
$100c+10a+b \equiv 10(a-c)+(b-c) \bmod 37$

Since these are all $\equiv 0$, we can rewrite this as

$10(b-a) \equiv (a-c) \bmod 37$
$10(c-b) \equiv (b-a) \bmod 37$
$10(a-c) \equiv (c-b) \bmod 37$

Squaring and adding,

$99( (a-b)^{2}+(b-c)^{2}+(c-a)^{2}) \equiv 0 \bmod 37$

Hence our original expression is divisible by $37$.  The "rest of" $n$ divides $(a+b+c)$.

[b]Case:[/b]  $gcd(n, 111) = 3$.  Of course, $a+b+c \equiv 0 \bmod 3$.  The "rest of" $n$ again divides $(a+b+c)$.

[b]Case:[/b]  $gcd(n, 111) = 1$.  Then $n | (a+b+c)$.

Edit:

[b]Minor addendum:[/b]  We need to address the case when $n$ is even, since $n | (a+b+c)$ but not necessarily the entire expression because of the $\frac{1}{2}$.  Of course, if $n$ is even then $a, b, c$ must be even, so

$4 | (a-b)^{2}+(b-c)^{2}+(c-a)^{2}$

So we're fine. [/hide]

[quote="t0rajir0u"]
Edit:

[b]Minor addendum:[/b]  We need to address the case when $n$ is even, since $n | (a+b+c)$ but not necessarily the entire expression because of the $\frac{1}{2}$.  Of course, if $n$ is even then $a, b, c$ must be even, so

$4 | (a-b)^{2}+(b-c)^{2}+(c-a)^{2}$

So we're fine. [/hide][/quote]

You can also write $\frac{1}{2}((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$ as $a^{2}+b^{2}+c^{2}-ab-bc-ca$.

Oh yeah.  I forgot about that.   :rotfl:


Try these ;-)
1. $ a, b, c$ are in range $ (0; 1), a \plus{} b \plus{} c \equal{} 1, p \geq 1$. Prove that:
$ \frac {1 \plus{} a^{p}}{1 \minus{} a^{p}} \plus{} \frac {1 \plus{} b^{p}}{1 \minus{} b^{p}} \plus{} \frac {1 \plus{} c^{p}}{1 \minus{} c^{p}} \geq 3.\frac {3^{p} \plus{} 1}{3^{p} \minus{} 1}$

2. $ a, b, c > 0$ and $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$
Find minimal value of the expression:
$ M \equal{} \frac {a^{6}}{(b^{2} \plus{} c^{2})^{2}} \plus{} \frac {b^{6}}{(c^{2} \plus{} a^{2})^{2}} \plus{} \frac {c^{6}}{(a^{2} \plus{} b^{2})^{2}}$

These are not mine - creator is Janeta Germanova - Gulubovo, Bulgaria

the first one follows from jensen's on
$ f(x)\equal{}\frac{1\plus{}x^{p}}{1\minus{}x^{p}}$

Is it possible without Jensen? (I don't know Jensen...or give me some good link to Jensen)

Thank you! I'll see this...

After sawing I think:
http://mathworld.wolfram.com/JensensInequality.html is easier for understanding for novices in Jensen as me...


[quote="easternlatincup"]Let $n\in N$.

Find the maximum product $n_{1}n_{2}...n_{k}$ such that $n_{1}+n_{2}+...+n_{k}=n$ and $n_{i}\geq 1$, $k\geq 1$[/quote]
$n_{1}n_{2}...n_{k}\le (\frac{1}{k}\sum_{i}n_{i})^{k}$ maximum, when $n_{i}=\frac{n}{k}$ and $max n_{1}n_{2}...n_{k}=(\frac{n}{k})^{k}.$
If $n_{i}$ must be natural, then let $m=[\frac{n}{k}], r=n-k*m$, then maxim is $m^{k-r}(m+1)^{r}$.

I'm sorry. I didn't state the problem carefully.

$n_{i}$ must be natural, and $k$ is variable.  :)

I'm pretty sure it was posted before. The answer is to use $2$'s and $3$'s only, with at most two $2$'s.

[quote="harazi"]Rust, when will you read the statement of the problems carefully?  :)  Here $k$ is also variable.[/quote]
autor don't say about $n_{i}$ - natural, and about $k$ fixed or variable.
In this case if $n_{i}\ge 4$ we can change $n_{i}$ to $2$ and $n_{i}-2$, because $2(n_{i}-2)>n_{i}$. If $n_{i}=1$ we can change $n_{i},n_{j}$ to $n_{l}=n_{i}+n_{j}$ because $(n_{j}+1)>1*n_{j}$. Therefore maximum value when all $n_{i}=2 \ or \ n_{i}=3$. Let $n=2m+3r, k=m+r$ then $P=\prod_{i}n_{i}=2^{m}3^{r}.$ We have $3*3>2*2*2$, therefore $m\le 2$. It give
if $3|n$ $m=0,r=\frac{n}{3}, P=3^{r}$.
if $n=1(mod \ 3)$, then $n=1, P=1$, if n>1 $m=2,r=[\frac{n}{3}]-1,P=4*3^{r}.$
If $n=2(mod \ 3)$ then $m=1,r=[\frac{n}{3}],P=2*3^{r}.$


Sound travels faster in water than in air. which aspects of v, lambda , and f changed when part of the sound travelled through the water? If the speed in water is faster by a factor of four, by how much did the wavelength and the frequency change?

Frequency remains same.
frequency*wavelength= speed of light

Now when the ratio of speed of light is given. we can find the ratio of wavelength

Prior answer is almost correct.  However the frequency times the wavelength equals the velocity, which is the SPEED OF SOUND, not light.

If the speed increases by a factor of 4 the wavelength increases by a factor of 4.

If the frequency changed as well as the wavelength, our complex ear structure would not accurately reproduce the frequency for us to hear what is going on around us.

Remember when submerged in a tub or pool that the sounds are muffled (energy loss at the interface, but the sounds are not shifted in frequency, either blue or red to borrow from light again.


function
geometry
geometric transformation
reflection
rotation

Show why the graphs of $ y\equal{}x^{\frac{1}{2}}$  and $ y\equal{}x^2$ ($ x\geq 0$) are reflections over the line $ y\equal{}x$.

Does it count to say that $ x \equal{} \sqrt {y}$ and $ x \equal{} y^2$ when flipped about the line $ y\equal{}x$ and so basically those are equal when x>0 ?

First, we can just consider all points with $ x\ge 0$.

Now, reflecting a point over $ y \equal{} x$ is a mapping from $ R^2\rightarrow R^2$ with $ f(x,y) \equal{} (y,x)$. Then the image of the graph of $ y \equal{} \sqrt {x}$ would be the graph of $ x \equal{} \sqrt {y}$ or $ y \equal{} x^2$ for $ x \ge 0$. This is a 'reflection' but is really a bijective mapping :P

I don't know if this is too overly complicated or not... Really, mathemagician1729 simplified it... =D

I just showed that they are inverse functions (when $ x \geq 0$) and then applied the property that the graphs of $ y\equal{}f(x)$ and $ y\equal{}f'(x)$ are symmetrical about the line $ y\equal{}x$.

[quote="someperson01"]

Now, reflecting a point over $ y \equal{} x$ is a mapping from $ R^2\rightarrow R^2$ with $ f(x,y) \equal{} (y,x)$. [/quote]

Is it possible to reflect over the line $ y\equal{}2x$

Why not? The relevant formula is $ f((x,y)) \equal{} \left( \frac{\minus{}3x\plus{}4y}{5} , \; \frac{4x\plus{}3y}{5} \right)$, which can be derived from a change of basis followed by a reflection, or a rotation by an angle of $ \minus{} \arctan 2$ followed by a reflection.

Suppose you wanted to reflect $ y \equal{} \sqrt{x}$ over that line. We can parametrize this as the set of points $ (t^2,t)$ for $ t \geq 0$.

$ f((t^2,t)) \equal{} \left( \frac{\minus{}3t^2\plus{}4t}{5} , \; \frac{4t^2\plus{}3t}{5} \right)$

So for the new curve:
$ x \equal{} \frac{\minus{}3t^2\plus{}4t}{5}$
$ y \equal{} \frac{4t^2\plus{}3t}{5}$

In this particular case, the result is actually a function, so it is still meaningful to solve for $ y$ explicitly, which we may do by eliminating $ t$ from the equation above. The result is $ y \equal{} \frac{10 \minus{} 12x \minus{} 5 \sqrt{4 \minus{} 15 x}}{9}$ for $ x \leq 0$.

Wow thats a lot! :)

One question; what is a change of basis?

Also when i tried rotating the function and reflecting i got

$ f(x,y) \equal{} \left(\frac { \minus{} 2x \plus{} y}{\sqrt {5}},\frac {x \plus{} 2y}{\sqrt {5}}\right)$... :?:


Let $n$ be a natural number. Solve in whole numbers the equation \[x^{n}+y^{n}=(x-y)^{n+1}.\]

hmmmm.... somewhat similar to the method for solving JMO 2017 #2


analytic geometry
geometry
probability
expected value

I am given one trillion purely random integers between 0 and 99 and two "counter"s [i]c[/i] and [i]d[/i] which are so far set to 0.  I split the trillion integers into five hundred billion pairs of two.  For each pair, I call the first integer [i]a[/i] and the other [i]b[/i].  If the sum of the squares of [i]a [/i] and [i]b[/i] is greater than 10000, I add one to [i]d[/i].  Otherwize, I add one to [i]c[/i].  After I do this with all five hundred billion pairs, what is 4[i]c[/i] divided by [i]c[/i]+[i]d[/i]?

We know that $c+d$ will always equal 500 billion, but won't the value of $c$, and so $4c$, depend on the distribution of the numbers? :? The value of $c$ will be different when all the numbers are $99$ than when all the numbers are $0$. Or maybe I'm misunderstanding the question  :blush:

oops... edited

stupid mistake... that can't easily be fixed

miyomiyo:  You should be asking for the [b]expected value[/b].

calculuslover:  We're looking for the [b]sum[/b] of $a^2$ and $b^2$, which can be larger than $99^2$.

[hide="Hint 1"] Interpret on a coordinate grid. [/hide]
[hide="Hint 2"] Monte Carlo method ;) [/hide]

[quote="t0rajir0u"]Something of a giveaway.

http://math.fullerton.edu/mathews/n2003/MonteCarloPiMod.html[/quote]

Cool!!! Thanks.

So would anyone like to post the solution?  The link makes it rather obvious now ;)

[hide]
For every pair $(a, b)$ such that $a^2+b^2 < 10000$, we add $1$ to $c$. We have to keep in mind that art of the circle represented by the equation $a^2+b^2 = 10000$ lies outside the square represented by the possible values of $(a, b)$ since $a$ and $b$ can only equal $99$ at the greatest.

Setting $x=99$ to find the y-coordinate of the point of intersection between the circle and the square, $y = \sqrt{199}$. The area of the region of the square that lies outside the circle is then equal to $(99-\sqrt{199})^2 - \frac{(99-\sqrt{199})^2\pi}{4} = \displaystyle \left(1-\frac{\pi}{4} \right) \displaystyle \left(99 - \sqrt{199}\right)^2$

The expected value of $c$ is then equal to $\displaystyle \left(99^2 - \displaystyle \left(1-\frac{\pi}{4} \right) \displaystyle \left(99 - \sqrt{199}\right)^2 \right) \times 5 \times 10^{11}$

We know that $c+d = 5 \times 10^{11}$ since there are that many pairs, so the value of $\frac{4c}{c+d}$ is equal to:

$\frac{\displaystyle \left(99^2 - \displaystyle \left(1-\frac{\pi}{4} \right) \displaystyle \left(99 - \sqrt{199}\right)^2 \right) \times 5 \times 10^{11} \times 4}{5 \times 10^{11} \times 99^2} = \displaystyle \frac{\displaystyle \left(99^2 - \displaystyle \left(1-\frac{\pi}{4} \right) \displaystyle \left(99 - \sqrt{199}\right)^2 \right) \times 4}{99^2}$

That doesn't look like a very nice answer, did I mess up somewhere :?
[/hide]


[quote="ashegh"]suppose L is a line.prove that:

the locus of points like M ,inwich the distance betwin M and L is fixed ,is two paralell line with L.[/quote]
this questions are easy but very useful. :D

[quote="ashegh"]suppose L is a line.prove that:

the locus of points like M ,inwich the distance betwin M and L is fixed ,is two paralell line with L.[/quote]

It's easy to prove that every three distinct points like M, at the same side of L, are collinear. Then the locus of points M will be a straight line. Any point on this straight line will be at a fixed non-zero distance of L, i.e it don't have common point with L, then this line and L are pararel lines. The another straight line is symmetrical.


function
search
complex analysis
complex analysis unsolved

Let's $ f$ be an entire function bounded on every line of complex plan. Is $ f$ constant?

That is to say :

$ \forall z_0 \in \mathbb{C}$ , $ \forall a \in \mathbb{C}$ , $ \exists C_{(z_0,a)} > 0$ such that
$ \forall x \in \mathbb{R}$ : $ |f(z_0 \plus{} a x)| < C_{(z_0,a)}$.

Is f constant?

Just a remark: D.J. Neumann constructed an entire nonconstant function which is bounded on every line passing through the origin, so refining Liouville's theorem in such a way is not obvious at all. However, I see that you ask for more, that is for any line, but I believe that the answer remains the same.

Yes, harazi, it is the same. If you look at my hint [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=96420331&t=133261]here[/url], you'll see that it is enough to take $ f(z^2)$. The construction of an entire function bounded outside that strip is not hard at all, so I leave it to you or somebody else :)

Thank you again for your help fedja  :) , i knew for the entire function bounded on every line through the origin,  but i did not succeed in constructing a counter-example for every line (even not through the origin).

In any case, Popolux, seing the level of the problems you post, I invite you  take a look at the problems left unsolved on the forum, you may find interesting challenges. And some people wait for solutions for more than  1 year.  :wink:


inequalities
n-variable inequality
Holder
Convexity

Let $n\ge2$ be a natural number, and $a_i$ be positive numbers, where $i=1,2,\cdots,n.$ Show  that
\[\left(a_1^3+1\right)\left(a_2^3+1\right)\cdots\left(a_n^3+1\right) \geq \left(a_1^2a_2+1\right)\left(a_2^2a_3+1\right)\cdots\left(a_n^2a_1+1\right)\]

We have $\left(a_1a_2^2+1\right)^2\le\left(a_2^3+1\right)\left(a_2a_1^2+1\right),$ and so on. Multiply these and find that \[\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc} \left(a_1a_2^2+1\right)\geq\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2.\]Similarly, we get \[\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc}\left(a_2a_1^2+1\right)\ge\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2.\]And from now on everything is simple.

A  liitle  more  help  would  be  fine!  thanks

So, we have established that $\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc} \left(a_1a_2^2+1\right)\geq\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2,$ and $\prod_{cyc} \left(a_1^3+1\right)\cdot\prod_{cyc}\left(a_2a_1^2+1\right)\ge\left[\prod_{cyc}\left(a_2a_1^2+1\right)\right]^2.$ If $\prod_{i=1}^n\left(a_ia_{i+1}^2+1\right)\geq\prod_{i=1}^n \left(a_{i+1}a_i^2+1\right),$ then we are done with the second relation. Otherwise, the first relation saves us.

Just posting the problem as it appeared in the contest, for the resources section.

Hello Sorry to ask but Can you tell me what is holder inequality here I ain't gettin how you are going about this.
By what formula or derivation you are gettin can you explain me the case of holder one pl's .

Holder inequality here it's the following.

For $a_i>0$, $b_i>0$, $\alpha>0$ and $\beta>0$ we have:
$$(a_1+a_2)^{\alpha}(b_1+b_2)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}.$$
In our case $\alpha=2$ and $\beta=1$.


Can you please give a solution? Also, if you learn to use Latex, your answers (and especially solutions) will be easier to read.

[quote="solafidefarms"]Jos. A Bank is holding a Labor Day Sale. They said in their ad to take an additional 50% off the suits, already 50% off regular price. If they are now 123.95, what was the original price? Express your answer as a common fraction.[/quote]

[hide]

  Say the original price is designated by $x$

$ \frac12 \cdot \frac12 \cdot x = \frac{x}{4}=123.95$

$x=4 \cdot 123.95 = 495.8 = \boxed{\frac{2479}{5}}$ [/hide]

[hide]the answer is $\frac{2479}{5}$ you can figure out what x is with $\frac{1}{2}*\frac{1}{2}*x=123.95=\frac{x}{4}$ we multiply by 4 to get x so we have 495.8, and that equals my answer above :idea: [/hide]


Um....

If it just 364 / (30-3+1) = 11.375???

I don't understand what you are saying....

[quote="nutz_for2.718281828"]Craze E. Gye just got home from Japan. He thinks that it's Halloween a week from now, so he starts preparing. He starts preparing at Midnight at October 3rd. Every time someone walks by his house, he or she goes into hysterics. After a week, he learns that it's only October 10th, he gets so ad that he's gonna get real busy for the rest of the month preparing. If Craze stops preparing at midnight of Halloween, and a total of 364 people passed, how many people on average went into hysterics?

P.S. At midnight of a day, the day specified starts.[/quote]

Crazy E. Gye=brother of Fun guy (read it aloud if you don't get it)

So then 364 people have gone crazy within 28 days?(sorry I'm terrible with these...)
[hide="If so..."] then we know 364/28=average so 13 (which is "unlucky)[/hide]

I hope I put the right amount of days...your hint is a good one though... :D


If your school has a high AMC 12 team score (top 3 scores), please  post it here. I'd like to see how our school did relative to the schools of AoPS students.

lol no thats 282. And anyways a team score is top 3 in a school

Our team score is 423
(Me-144, Nathan-141, David-138)

[quote="kyyuanmathcount"]Me and Neil Gurram add to 272.  138+144? is that right?[/quote]
:rotfl:
Your two people beat my school's top three
:rotfl:
Considering that the 3rd out of our top three was a 60, and I was the sole AIME qualifier, it isn't that shocking...
But still
:rotfl:

James S. Rickards High School
150
150
132

= 432

Of course, TJ beat us again. I heard they had a 150, 144, 144  :|

[quote="bpms"]My School PWNED with a 270.[/quote]

I thought you took AMC 10.

It is the 12 that determines team score.

and yes kyyuan and I toook the test for the same school.

Our team score was 240ish. 40+, 60, and 132.

[quote="paladin8"]
Haha yeah. Are you from around here? The two 150's were me and t0rajir0u, though. Jonathan has a couple more years to achieve this :P.[/quote]

I would assume so. :P  I am indeed from around here, otherwise it would be a bit scary.  Bellevue definitely is a powerhouse, our school relies on individuals.  However, at Blaine and MIC Master's the individual thing worked out quite nicely.

[quote="paladin8"][quote="Aegor"][quote="paladin8"]
Bellevue High :P. We're a public school; the three scores you see are all AoPSers, though. Me, t0rajir0u, and ch1n353ch3s54a1l.[/quote]

Your math team is easily the most active in the Bellevue School District.  Which one of you got the 126?  If JH aced the AMC12, even more kudos to him as a freshman.[/quote]

Haha yeah. Are you from around here? The two 150's were me and t0rajir0u, though. Jonathan has a couple more years to achieve this :P.[/quote]

That's pretty impressive. I think you guys are tied for 2nd right now. So far I've heard 432 (Rickards - 150, 150, 132 and Lynbrook - 150, 144, 138) and you guys and TJ (A) at 426.

my school is 99

since im the only one who take amc

my school didnt even know
amc EXIST
untill i told them so

so yeah it kinda sucks and kinda cool
cuz i get this nice office room all by myself to take

Our school got raped!

114+112.5+109.5 = 336

Is it sad that I looked for a calculator to add those?

It's probably pointless to ask, but does anyone know of B scores for TJ/Exeter and A/B scores for Blair?

our school was horrible (and so was i) at the 12B; we got a 121.5+121.5+109.5(me :( )=352.5

i think my schools team score on the amc 10a was 391.5.

i wonder what honey creek's team score is.

[quote="yhh3002"]how do you take the AMC average?

take the top four and average them?[/quote]

It is composite, top 3 and sum it.

our school got a 372 on the AMC 12-B or... somewhere around there


Να βρεθούν όλες οι συνεχείς συναρτήσεις $ f,g: \mathbb{R}\to\mathbb{R}$ που είναι τέτοιες ώστε

                                 $ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)g(y)$      για κάθε $ x,y\in \mathbb{R}$.

[quote="silouan"]Λοιπόν η βασική ιδέα είναι να οδηγούμαστε κάθε φορά στην συναρτησιακή του D'Alembert αλλά έχει αρκετή δουλειά με τις διάφορες περιπτώσεις και θέλει πολλή χαμαλοδουλειά  :P
Εν κατακλείδι οι λύσεις είναι $ f(x) \equal{} k\cos mx \plus{} l\sin mx$ , $ g(x) \equal{} \cos mx$
ή μία aνάλογη με το υπερβολικό συνημίοτονοή $ f(x) \equal{} 0$ και g mia οποιαδήποτε  ή  $ f(x) \equal{} k \plus{} lx$ , $ g(x) \equal{} 1$ .[/quote]

ontos exei polli douleia!kai prepei na einai idiaitera prosektikos kapoios!


probability geometry inequalities rectangle parallelogram

David arrives at the train station sometime between 3:00 and 5:00 and stays for 15 minutes, waiting for a train. He knows a train is expected sometime from 2:00 to 5:00 and will wait 5 minutes for passengers. What is the probability that David is able to catch the train?

[hide="A general hint"]Use geometric probability. Draw a graph that has two axes, one for the train and one for David. Shade in certain regions that coincide with possible times of arrival for each and the time period they would be at the train station, and find an intersection of the two regions. Find the area of that intersection, and divide it by the total area of the union of the two shaded regions, giving you the desired probability. (Sorry if I worded this too vaguely to be understood; I'm not really that good at explaining things)[/hide]

ummm, could someone post a solution for this problem, because i was never good at these problems... :blush:

[hide="Solution"]
We will use geometric probability to find the solution (as gauss1181 suggested). Let the x-axis be the time in hours after noon that David arrives at the train station, and let the y-axis be the time in hours after noon that the train arrives at the station. We have two cases to consider:

[b]Case 1:[/b] The train arrives after David.
In this case, David waits for 15 minutes (1/4 of an hour). If David catches the train, then $ 0 \leq y \minus{} x \leq \frac{1}{4}$. Thus, $ x \leq y \leq x \plus{} \frac{1}{4}$.

[b]Case 2:[/b] The train arrives before David.
The train waits for only 5 minutes (1/12 of an hour). If David to catches the train, we have $ 0 \leq x \minus{} y \leq \frac{1}{12}$. That tells us that $ x \minus{} \frac{1}{12} \leq y \leq x$.

In Case 1, $ x \leq y \leq x \plus{} \frac{1}{4}$, and in Case 2, $ x \minus{} \frac{1}{12} \leq y \leq x$. The union of these two double inequalities is $ x \minus{} \frac{1}{12} \leq y \leq x \plus{} \frac{1}{4}$, which gives us the two inequalities:
$ y \geq x \minus{} \frac{1}{12}$

$ y \leq x \plus{} \frac{1}{4}$

The total region to be considered is the region such that $ 3 \leq x \leq 5$ and $ 2 \leq y \leq 5$, since David arrives between 3 and 5, and the train arrives between 2 and 5. If we plot this region (green) and the inequalities previously obtained (blue), we get:
[asy]import graph;
size(150);

xaxis("$x$", -1, 6, Ticks(0, 0, 0, 1, false, false, 3, 0));
yaxis("$y$", -1, 6, Ticks(0, 0, 1, 0, false, false, 3, 0));

fill((3, 2)--(5, 2)--(5, 5)--(3, 5)--cycle, rgb(0.7, 1.0, 0.7));
fill((-1, -1)--(-11/12, -1)--(6, 71/12)--(6, 6)--(5.75, 6)--(-1, -0.75)--cycle, rgb(0.4, 0.7, 1.0));
draw((3, 2)--(5, 2)--(5, 5)--(3, 5)--cycle, rgb(0.0, 0.6, 0.0)+linewidth(1));[/asy]

Here is a enlarged view of the green rectangle:
[asy]import graph;
unitsize(72);
filldraw((3, 2)--(5, 2)--(5, 5)--(3, 5)--cycle, rgb(0.7, 1.0, 0.7), rgb(0.0, 0.6, 0.0)+linewidth(2));
filldraw((3, 35/12)--(5, 59/12)--(5, 5)--(4.75, 5)--(3, 3.25)--cycle, rgb(0.4, 0.7, 1.0), rgb(0, 0, 1)+linewidth(2));
draw((4.75, 5)--(4.75, 14/3), rgb(0, 0, 1)+linewidth(1));
draw((5, 59/12)--(4.75, 59/12), rgb(0, 0, 1)+linewidth(1));
label("$\frac{1}{4}$", (3, 37/12), W);
label("$\frac{1}{4}$", (39/8, 5), N);
label("$\frac{1}{12}$", (5, 119/24), E);
label("$\frac{7}{4}$", (31/8, 5), N);[/asy]

The blue region within the rectangle has been divided into a parallelogram, rectangle, and isosceles right triangle, in order to facilitate the computation of its area. The area of the parallelogram is
$ bh \equal{} \frac{1}{4} \cdot \frac{7}{4} \equal{} \frac{7}{16}$.

The area of the rectangle is
$ bh \equal{} \frac{1}{4} \cdot \frac{1}{12} \equal{} \frac{1}{48}$.

The area of the isoceles right triangle is
$ \frac{1}{2}bh \equal{} \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} \equal{} \frac{1}{32}$.

Thus, the total area of the blue region within the rectangle is
$ \frac{7}{16} \plus{} \frac{1}{48} \plus{} \frac{1}{32} \equal{} \frac{47}{96}$.

The desired probability is the area of the blue region within the rectangle divided by the area of the rectangle. This is equal to
$ \frac{47}{96} \div 6 \equal{} \boxed{\frac{47}{576}}$.
[/hide]