inequalities
probability
induction
function
probability and stats

Hello. Our lecturer stated the Bonferroni Inequalites as follows:

Let $ A_1, A_2, A_3 \ldots$ be events, and $ \cal P$ be a probability distribution. Let $ \cal A$ be an odd/even down-set consisting of subsets of $ \{1, 2, 3, \ldots, n\}$. Let $ A_X = \bigcap_{x \in X} A_x$. Then:

${ {\cal P} \left(A_1 \cup \ldots \cup A_n \right) \leq / \geq \sum_{X \in {\cal{A}}, X \neq \phi} (-1)^{|X| + 1} {\cal{P}} (A_X)}$

Is this correct, as when he comes to prove it, he only considers down-sets with only one maximilised set? (i.e. the power set of a subset of $ \{1, 2, 3, \ldots, n\}$). If it is correct, can you provide a proof for me...it probably can be done by induction, but I am too lazy to construct a proof myself.

P.S. our lecturer is the esteemed field medalist Timothy Gowers.

I was under the impression he retracted his claim of the theorem as you have stated it; Imre Leader appeared to think there was something wrong with it...

Oh ok...thanks...I never turn up for lectures...

i gave a very handwavy proof of what are usually called the bonferroni inequalities [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169137]here[/url].

(the idea is to prove the corresponding pointwise inequality for the indicator functions of the $ A_i$s and use that $ 1_{A_i\cap A_j}\equal{}1_{A_i}1_{A_j}$ + a bit of elementary combinatorics)


I want a pure geometric solution for the following question

1)If $ BC$ is the greatest side of $ \triangle ABC$ and $ D, E$ are points on $ BC, CA$ respectively prove that $ BC \ge DE$

2)If $ BC$ is the greatest side of $ \triangle ABC$ and $ E, F$ are points on $ AB, CA$ respectively prove that $ BC \ge EF$

3)Prove that no straight line can be drawn within a triangle which is greater than the greatest side.

I want separate solutions but its Ok if u just give one solution for the first two
I'll work it out

That's a simple problem.
In problem a-) (using the same strategy again in the other items) see when $ \angle DEA$ is obtuse or not, spliting in some cases. Like this you will have an easy but non-beautiful solution.

I dont understand
Its trivial that when $ \angle EDC \ge 90$ then $ ED < BC$ but what if it is acute??  :?:
Also I understand that when the $ \angle EDC$ tends to $ 0$ then it become equal....
Please help me out....... :maybe:  :(

$ \angle DEA$ is obtuse $ \Rightarrow DA \geq DE$, and $ DA \le AB$ or $ AC$, and both are less than or equal to $ BC$.
Now $ \angle DEA$ is acute $ \Rightarrow DE \le DC \le BC$, and it's done.

Thanx erudito for all ur help

can someone help me on these questions
http://www.mathlinks.ro/viewtopic.php?t=248523
http://www.mathlinks.ro/viewtopic.php?t=247475
http://www.mathlinks.ro/viewtopic.php?t=247471

please i need all the help u can give me...........


calculus
derivative
function
inequalities
complex analysis
complex analysis unsolved

Let $ H\equal{}\{ z: \Im (z) \ge 0 \}$ denote the upper half plane. Suppose $ F: H \to H$ is analytic and $ a \in H$. Prove that $ |F'(a)| \le \frac{ \Im F(a)}{ \Im a}$.

I only can take a quick guess because I don't have time to check it and even if it's wrong you can solve it as an exercise. $ H$ and the unit disk are the same, I mean there's a suitable Mobius fractional function which transforms one of them into other and in unit disk we have a very useful inequality, known as the Pick's lemma which states, $ \frac {|f'(z)|}{1 \minus{} |f(z)|^2} \leq \frac {1}{1 \minus{} |z|^2}$ and the equality holds when $ f$ is a Mobius function and I think, it should derive from this lemma. :wink:

Take a conformal map $ D\to H$ that sends $ 0$ to $ a$.
Follow it with $ F$.
Then map back onto the disk, sending $ F(a)$ to $ 0$. Now the ordinary Schwarz lemma (rather than Schwarz-Pick) can be applied.

Of course, Schwarz-Pick is just Schwarz augmented by a conformal mapping.


$ a^{3}\plus{}b^{3}\plus{}c^{3} \geq a^{2}b\plus{}b^{2}c\plus{}c^{2}a$
for nonnegative a, b, c.

Consider the sequences:

$ a,b,c$, and $ a^2,b^2,c^2$
$ b,c,a$ and $ a^2$, $ b^2$, $ c^2$

If either $ a\ge b\ge c$ or $ c\ge b\ge a$, the top sequences are like sorted, and the bottom array isn't, so we are done by the rearrangement inequality.

Well for positive $ a,b,c$, $ a,b,c$ and $ a^2,b^2,c^2$ are always similarly sorted, and are bigger than any other scalar product of them...

In general, I think

$ a^k\plus{}b^k\plus{}c^k \ge \sum_{cyc}{a^{k_1}b^{k_2}c^{k_3}}$ for $ k\equal{}k_1\plus{}k_2\plus{}k_3$

holds for positive $ a,b,c,k$.

You can do rearrangement on three sequences too so that proves your generalization (as well as an n variable one).

[quote="not_trig"]Well for positive $ a,b,c$, $ a,b,c$ and $ a^2,b^2,c^2$ are always similarly sorted, and are bigger than any other scalar product of them...

In general, I think

$ a^k \plus{} b^k \plus{} c^k \ge \sum_{cyc}{a^{k_1}b^{k_2}c^{k_3}}$ for $ k \equal{} k_1 \plus{} k_2 \plus{} k_3$

holds for positive $ a,b,c,k$.[/quote]

If $ k_1,k_2,k_3$ are nonnegative, of course. With AM-GM, we get $ k_1a^k\plus{}k_2b^k\plus{}k_3b^k\geq ka^{k_1}b^{k_2}c^{k_3}$ etc.

If one is negative, of course not (consider $ a\equal{}b\equal{}1,c\equal{}\epsilon$ for $ \epsilon$ really close to 0).


[quote="mychrom"]Let $ ABCD$ be a quadrilateral and let $ P$ be the intersection of $ AB$ and $ CD$. For any $ R\in [AD]$ let $ f(R) \in [CD]$ be the intersection between $ PR$ and $ CD$. What is the locus of the midpoints of segments $ Rf(R)$?[/quote]

I think that $ f(R)\in [BC].$ It's an interesting problem. If you heard about that problem from a book the locus can be a circle (if $ ABCD$ is a trapezoid then the radius of the circle is $ r\equal{}\infty$). You may try to prove that points $ M,K_{1},K,N$, from my figure are concyclic using complex numbers. The locus we're looking for is marked with [color=red]red[/color].

[url=http://imageshack.us][img]http://img178.imageshack.us/img178/3999/maximbogdan2xh9.png[/img][/url]


trigonometry
integration
algebra
linear equation
calculus
calculus computations

(1) Solve $ t^2\frac {dx}{dt} \minus{} 2tx \equal{} 3$.

(2) Show that $ \frac {dy}{dt} \plus{} P(t)y \equal{} Q(t)y^n$ will become a linear equation by proper transformation.

(3) In (2), find the solution for $ P(t) \equal{} \frac {2}{t},\ Q(t) \equal{} t^2\sin t,\ n \equal{} 2$.

For the first one:
$ \displaystyle t^2\frac {dx}{dt} \minus{} 2tx \equal{} 3$
$ \displaystyle \frac {dx}{dt} \equal{} \frac {3}{t^2} \plus{} 2\frac {x}{t}$
Let $ v \equal{} x/t$ then $ \displaystyle \frac {dx}{dt} \equal{} v \plus{} t\frac {dv}{dt}$
Thus, $ \displaystyle v \plus{} t\frac {dv}{dt} \equal{} \frac {3}{t^2} \plus{} 2v\Rightarrow t\frac {dv}{dt} \equal{} \frac {3}{t^2} \plus{} v\Rightarrow \frac {dx}{dt} \minus{} \frac {v}{t} \equal{} \frac {3}{t^3}...(1)$
The integrating factor is $ \displaystyle e^{\int \minus{} \frac {1}{t}dt} \equal{} \frac {1}{t}$
Take the differential of $ (1)$, and multiply both sides of the resulting equation by $ 1/t$.
Then integrate to get $ \displaystyle \int d(v/t) \equal{} \int\frac {3}{t^4}dt\Rightarrow v \equal{} \minus{}t^{\minus{}2}\plus{}Ct$
Therefore, $ \displaystyle x \equal{} \minus{} \frac {1}{t} \plus{} Ct^2$.

2) $ \dfrac{\mathrm{d}y}{\mathrm{d}t} \plus{} P(t)y\equal{}Q(t)y^n$
Divide throughout by $ y^n$ to get
$ \dfrac{1}{y^n}\cdot\dfrac{\mathrm{d}y}{\mathrm{d}t} \plus{} \dfrac{1}{y^{n\minus{}1}}P(t)\equal{}Q(t)$
Set $ \dfrac{1}{y^{n\minus{}1}}\equal{}u\ \Rightarrow \ \dfrac{1}{y^n}\,\dfrac{\mathrm{d}y}{\mathrm{d}t}\equal{}\dfrac{1}{1\minus{}n}\cdot\dfrac{\mathrm{d}u}{\mathrm{d}t}$
Thus the given equation
$ \dfrac{1}{1\minus{}n}\cdot \dfrac{\mathrm{d}u}{\mathrm{d}t}\plus{}P(t)u \equal{} Q(t)$
which is the same as the linear equation
$ \dfrac{\mathrm{d}u}{\mathrm{d}t}\plus{}(1\minus{}n)P(t)u \equal{} (1\minus{}n)Q(t)$
P.S: This is the Bernoulli's eqn.

3) Setting $ P(t) \equal{} \dfrac{2}{t}$, $ Q(t)\equal{}t^2\sin t$ and $ n\equal{}2$, we get, after setting $ u\equal{}\dfrac{1}{y}$,
$ \dfrac{\mathrm{d}u}{\mathrm{d}t} \minus{}\dfrac{2u}{t}\equal{}\minus{}t^2\sin t$
$ \dfrac{1}{t^2}\cdot\dfrac{\mathrm{d}u}{\mathrm{d}t} \minus{}\dfrac{2u}{t^3}\equal{}\minus{}\sin t$ (multiplying throughout by $ \dfrac{1}{t^2}$)
$ \Rightarrow\ \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{u}{t^2}\right) \equal{} \minus{}\sin t$
$ \Rightarrow \dfrac{u}{t^2} \equal{} \cos t \plus{}C$
$ \Rightarrow\ y \equal{} \dfrac{1}{t^2\cos t \plus{} Ct^2}$


Any trivia buffs out there? IM me at P3nguinP0wered on AIM and we can play!!

go to:

http://www.funtrivia.com/private/main.cfm?tid=960

AoPS has it s own trivia thing


integration
calculus
function
inequalities
real analysis
real analysis unsolved

$\int_{\pi^{e}}^{e^{\pi}}\frac{dt}{\sqrt{\zeta^{3}(x^{2}(t))-1}}\int_{\pi^{e}}^{e^{\pi}}\sqrt{\zeta^{3}(x^{2}(t))-1}dt\to min$ if
$2\leq x(t) \leq 4$
find also for wich $x(t)$ hold  equality.

By Caushy-Schwarz, the product of the integrals is never less than $(b-a)^{2}$ where $a=\pi^{e}$ and $b=e^\pi$ are the limits of integration. Any $x$ that makes the integrands constant functions gives an equality, so the answer to the second part is any constant function $x(t)=c\in[2,4]$. Again, I don't quite understand what's the point in piling up $\zeta$'s, squares and square roots. There is no real difficulty in the question, just a lot of irrelevant notation. I could also ask: what's the minimum of
\[\left(x\frac{e^{e^{x-239}}+x^{2}\log(1+x^{32})}{\sin(cos(e^{x}))+2}\right)^{2002}\]
but what's the point? The problem basically reduces to the fact that every square is non-negative and $0$ is attained. Let me make myself even more clear: if you continue to post the problems, which, stripped of all fancy notation and other irrelevant things, are trivial, nobody will read your posts. A good problem has no added artificial complexities in it, the intrinsic difficulty of the question should be enough to make it interesting. I loved your question about $\sum_{k}(-1)^{[kx]}$ but what you posted lately is not really what I would call attractive and, judging from other people's lack of response, almost no one else seems to get interested. :(

Sorry, I don't posted right  my problem. You must find $max$ .
I mean:

$\int_{\pi^{e}}^{e^{\pi}}\frac{dt}{\sqrt{\zeta^{3}(x^{2}(t))-1}}\int_{\pi^{e}}^{e^{\pi}}\sqrt{\zeta^{3}(x^{2}(t))-1}dt\to max$ if
$2\leq x(t) \leq 4$
find also for wich $x(t)$ hold  equality.
you only need to use http://www.mathlinks.ro/Forum/viewtopic.php?t=135955
I understand that my problems is not so interesting because it's based on combinations Holder's inequality and etc... sorry

.

Thanks for your comment fedja.


if I have f(x)=x^4 +x +3
They asked me find the f(-x), what is the result of it, and why? if you please giving me the formulas to find f(-x).

Here you are. Simply to find f(-x) you have to change x into -x and simplify the equation.
If f(x)=x^4 +x +3 than f(-x)=(-x)^4+(-x)+3=x^4-x+3. Good luck!

[quote="ashegh"]i confused...

is the answer X^4-X+3, or not...

because the problem isnt hard,and i think,the problem  that Camytran said,is another thing...and need to be

 discussed... :)[/quote]

Yes, you're rite, the problem is not easy to solve like that, because the answer teacher gives me is differerent, and i dunno y it's like that.  Can u give me the other solution?

[quote="ashegh"]

is the answer X^4-X+3, or not...

[/quote]

Yes, it is! :)

pupOK's method is the good one...

[quote="camytran"][quote="ashegh"]ok ok ,i will do my best...

but could u plz tell me what did u teacher say about it exactly?

and could u plz say again,what the problem want me to do?

because i want to be sure about it ;) .[/quote]

Hey, if u dun mind, can u give me your y!m nick? so i can tell you clearer and faster.[/quote]
I know that you are new, but next time try not to use the phrases: "u" instead of "you" and "dun" instead of "don't" becuase it just makes reading it a whole lot easier.  :)

yea sometimes people mis read things like u might be a variable and you would use it to say you so people would get confused

[quote="camytran"][quote="ashegh"]ok ok ,i will do my best...

but could u plz tell me what did u teacher say about it exactly?

and could u plz say again,what the problem want me to do?

because i want to be sure about it ;) .[/quote]

Hey, if u dun mind, can u give me your y!m nick? so i can tell you clearer and faster.[/quote]

you can look at his profile or the bottom of his post. Look for "YIM".

I don't think there's any other solution! The mean I proposed is the easiest. And what answer were you given?

the problem is too easy,and dont need to be discussed more...


geometry
geometric transformation
reflection
algebra proposed
algebra

This is from Mathematical Reflections, see [url]http://reflections.awesomemath.org/2006_2/2006_2_solutions.pdf[/url]

Hmm... has it appeared somewhere before?


Hi,

How would you show that if $ \mathcal{F}\subset H(G)$ is normal then $ \mathcal{F'}\equal{}\{f': f\in\mathcal{F}\}$ is also normal?

I'm assuming that I need to use Montel's theorem, but I don't know how to.

Thanks for the help.

You have locally uniform estimates for $ f$, and you want locally uniform estimates for $ f'$. The Cauchy integral formula is the key here.

We can be more direct than Montel's theorem here: if $ f_n$ converges uniformly on compact sets to a limit $ f$, $ f_n'$ converges uniformly on compact sets to $ f'$. That's enough convergent sequences right there.


Let $ a,b,c\in Z_\plus{}$ such that $ a^3\plus{}b^3\plus{}c^3\equal{}d^3$. Does there exist $ n>3$ and $ n\in Z_\plus{}$ such that
$ a^n\plus{}b^n\plus{}c^n\equal{}d^n$

$ d^{3\plus{}n} \equal{} a^3d^n \plus{} b^3d^n \plus{} c^3d^n > a^{3\plus{}n} \plus{} b^{3\plus{}n} \plus{} c{3\plus{}n}$ because $ d>a$, $ d>b$, $ d>c$.


This problem is from Olimpiads in 2004 from Romania!
  Find all real numbers x,y,z sug as   :
             x+y+z=3 and x^8+y^8+z^8=3

C'mon guys, this is so easy!
What about Cauchy (or, even better, the power mean inequality)? :D

By Power Mean, ((x^8+y^8+z^8)/3)^(1/8) >= (x+y+z)/3, but here we have equality. Therefore, x=y=z=1.

:D  :D  :D Hehe Arne getting impatient on the beginners. Very unlike him...


All integer numbers [tex]c[/tex],give equation:[tex]3y^2+4cy^3+2xy+48=0[/tex].Find all c in Z such that:the amount of solution [tex](x,y)[/tex] satisfying A,B conditions  is maximum:
A.\x\ is perfect square
B.there is no integer p such that:[tex]p^2[/tex] divides [tex]y[/tex]


I am homeschooling a middle school student who is trying to increase his speed at solving math problems. He knows the concepts and can answer all the problems he faces on tests such as the AMC8, SATs, and MathCounts, but he's too slow and methodical to finish within the allotted time. This is keeping him from scoring beyond a 700 on the SAT or making it to countdown in MathCounts because he doesn't finish all the problems to get the top scores.

Does anyone have suggestions to help him? He's almost finished with the Intro to Geometry book and has completed all the other Intro books. Again, he doesn't need help with concepts and basic math, just techniques to improve his speed.

Thanks!

Don't second guess what you're doing.  If you carry out a computation, assume it's right unless you have reason not to.  If you spend too much time re-checking what you're doing before you've had a chance to complete other problems, you're going to waste time.  Don't bother checking yourself until you've answered as many problems as you can.  Maybe some people have other strategies but this helps me go faster.

I think that if you just do really really a lot of practice, your speed will increase. Because when you do lots of problems, you gain experience, and your intuition gets better, as well as you get more used to doing routine calculations (or parts of proofs that consistently occur, e.g. induction, standard combinatorial sums). As a result, when you you see a problem, you will be thinking faster and it is more likely you will be thinking along the right lines, as a result you will come up with the solution faster.
As for techniques, again, if you do a lot of problems, your will learn more techniques as well as figure out which techniques are useful to remember and you will master them better. For checking, I find it difficult to check the first few problems after going through the whole test, as I sometimes forget what I did and need to resolve it. Maybe an efficient idea is checking your answers/solutions after doing a certain number of problems, so that the solution to them is still fresh in your head (e.g. after doing 5 problems, check them, and then move on, on the AMC. Depending on your preferences, this number could be less or greater. You can figure out what works best for you again by trying different styles on past contests).
However, this was just my experience, I am not sure how efficient this is. Because the amount of practice you need for certain things to get really good at them, is sometimes quite large, as a result when you compare your speed at the moment to what it was a week ago, there will probably not be much change. But if it is compared to a year ago, and if you have consistently practiced, then yeah, it will probably show a pretty significant increase.


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \sqrt {\frac {8a^2 \plus{} bc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {8b^2 \plus{} ca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {8c^2 \plus{} ab}{a^2 \plus{} b^2}} \ge \frac {9}{\sqrt {2}}
\]
Equality holds for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or any cyclic permutations. :)

PS: This is a problem I posted in the topic "Some new results". I like it very much. I hope we will discuss more about it. Thank you very much! :)[/quote]
[hide="hint"]\[ bc \ge \frac{2b^2c^2}{b^2\plus{}c^2}\] :)[/hide]

Did you use Holder in your solution and p,q,r method?

Thank you very much.

P.S. Really a beautiful inequality :wink:


function
complex analysis
complex analysis unsolved

Hi,
suppose $ L\subset\mathbb C$ is a lattice.
$ f, g$ are two elliptic functions with $ L$ as periodic lattice.

How to show that
i) if $ f$ and $ g$ have the same poles and the same principal part,
then $ f\equal{}g\plus{}c$ for a constant $ c$.

and

ii) if $ f$ and $ g$ have got the same poles and zeros (with the same multiplicities),
then $ f\equal{}cg$ for a constant $ c$.

In case i), what properties does $ f\minus{}g$ satisfy? What do you know about elliptic functions with these properties?

Also do the same for $ \frac{f}{g}$.

[quote="calc rulz"]In case i), what properties does $ f \minus{} g$ satisfy? What do you know about elliptic functions with these properties?

Also do the same for $ \frac {f}{g}$.[/quote]

Hi,
i get it.  :lol: thanks very much.


calculus
integration
real analysis
function
real analysis theorems

I'm hoping someone can help me address this query: How many different ways are there of introducing Measure Theory? I am aware of the following 3 methods, and wonder if these are it, or there are more.

1. Paul Halmos' approach using measures on rings of sets
2. Walter Rudin's approach using linear functionals
3. Daniell's approach

I hope to hear soon. Thanks!

Most of books in China use (1) and (maybe therefore) I feel that it is easier to understand than (2).

What is (3)?

Does anyone know some good books with theory and also with exercices for Measure Theory? ;)

I like Rudin's Real and Complex Analysis.

In general, any book with "Real Analysis" in the title is likely to be useful. The details will very, since there are a lot of these textbooks, but all should have measure theory.

I'm partial to Wheeden and Zygmund, [i]Measure and Integral[/i]. The approach there is to work out from the beginning the details of Lebesgue measure in $\mathbb{R}^n.$ This differs on the one hand from approaches that start with just $\mathbb{R}$ and on the other hand from approaches that start with abstract measure spaces.

The integral of a function $f: E\mapsto [0,\infty]$ where $E\subset\mathbb{R}^n$ is then defined as the $(n+1)$-dimensional measure of the set $\{(x,y): x\in E,0\le y \le f(x)\}.$ One advantage of this is that it produces Fubini's theorem earlier and more naturally.


Given the equation $ x^2\plus{}y^2\plus{}z^2\plus{}t^2\minus{}nxyzt\minus{}n\equal{}0$ where $ x,y,z,t$ are positive integers.Prove that this equation is solvable when $ n\equal{}2008$ and isn't solvable when $ n\equal{}2007$

For $ n \equal{} 2008$, try $ x \equal{} y \equal{} 6$ and $ z \equal{} 44$.


Let $ a,b,c\geq0$ s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \dfrac{1}{2ab^2\plus{}1}\plus{}\dfrac{1}{2bc^2\plus{}1}\plus{}\dfrac{1}{2ca^2\plus{}1}\geq1$.

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \dfrac{1}{2ab^2 \plus{} 1} \plus{} \dfrac{1}{2bc^2 \plus{} 1} \plus{} \dfrac{1}{2ca^2 \plus{} 1}\geq1$.[/quote]
$ <\equal{}>(ab^2\plus{}bc^2\plus{}ca^2)\plus{}1 \ge 4 (abc)^3$ ; $ t \equal{} abc \le 1$;
$ ab^2\plus{}bc^2\plus{}ca^2 \ge 3t$  ; $ 3t\plus{}1 \ge 4t^3$  $ (0 \le t \le 1)$;

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \dfrac{1}{2ab^2 \plus{} 1} \plus{} \dfrac{1}{2bc^2 \plus{} 1} \plus{} \dfrac{1}{2ca^2 \plus{} 1}\geq1$.[/quote]

$ \frac{1}{2ab^{2}\plus{}1}\equal{}1\minus{}\frac{2ab^{2}}{2ab^{2}\plus{}1}\equal{}1\minus{}\frac{2ab^{2}}{ab^{2}\plus{}ab^{2}\plus{}1}\geq 1\minus{}\frac{2ab^{2}}{3\sqrt[3]{a^{2}b^{4}}}\equal{}1\minus{}\frac{2\sqrt[3]{a^{2}b}}{3}\geq 1\minus{}\frac{2(a\plus{}a\plus{}b)}{9}\equal{}1\minus{}\frac{2(2a\plus{}b)}{9}$

$ \dfrac{1}{2ab^2 \plus{} 1} \plus{} \dfrac{1}{2bc^2 \plus{} 1} \plus{} \dfrac{1}{2ca^2 \plus{} 1}\geq 1\minus{}\frac{2(2a\plus{}b)}{9}\plus{}1\minus{}\frac{2(2b\plus{}c)}{9}\plus{}1\minus{}\frac{2(2c\plus{}a)}{9}\equal{}3\minus{}\frac{2}{9}(3a\plus{}3b\plus{}3c)\equal{}3\minus{}2\equal{}1$

$ a\equal{}b\equal{}c\equal{}1$


There are 25 stones in  a heap. The heap is divided into two parts, then one of the parts is divided in two again,etc,until we have 25 separate stones.After each division of one of the heaps into two smaller heaps we write the product of the numbers of stones in these two heaps  .Prove that at the end the sum of all the numbers  is 300.

[hide="Easy"]Use Second Principle of Finite Induction, knowing the result for number of stones n=1,2,3 and use $ f(n)\equal{}\frac {n(n\minus{}1)}{2}$[/hide]


MATHCOUNTS
articles
probability
percent
geometry
3D geometry
dodecahedron

16 hours after the scheduled end of the Countdown Round, and still no results posted here or on the MATHCOUNTS Web Site.  I find an online news article that reports:

Illinois Eighth - Grader and Illinois Team Awarded Mathematics Champions at 2004 MATHCOUNTS National Competition -- Friday May 7, 10:06 pm ET[quote]Gregory Gauthier was victorious in the intense, one-on-one oral Countdown Round where the top 12 students competed for the National Championship title. Jeffrey Chen of College Station, Tex., was awarded the second-place individual title with Haitao Mao of Falls Church, Va. and Sam Elder of Ft. Collins, Colo., advancing to the Semi-finals.[/quote]

I hope those "in the know" will provide further details.  Thanks!

I have heard just a bit more through the grapevine - consider none of these results 'official'.

I heard that Greg (IL) won both the written round and the countdown round.

Illinois won the Team competition with all 4 students probably ranked 15th or better.  The Individual (combined) score for 15th was around 36.

Any other news out there?

California was second-place team. New York was sixth place team and Danny Zhu from NY was seventh place individual.

I got this site from Alison:

http://biz.yahoo.com/prnews/040507/lnf007_1.html

Didn't he do well in last year's competition? I think I've heard his name somewhere before...

ya, he was in the top 10 last year too (figures that he'd win this year). didn't know that he was a 7th grader last year though.

yeah, I'd thought Shaunak was the only sevie last yr...

maybe he stayed back a grade to win MC :P

lol...i've thought about doing that myself...lol...

yeah, greg won everything. everything. written round, final standings, masters, team, everything.

rcv - Sam did Colorado proud.  He finished 12th on the written and mowed through the countdown round until running into Greg in the semis.

Greg put on one of the most impressive shows I've ever seen at Nationals.

Several AoPSers made the top 12, and I was very happy to meet many of you there (though I wish I'd had more time to talk to you all!)

According to the [url=http://mathcounts.org/Competitions/2003StateWinners.html]2003 state winners list[/url], Greg Gauthier was a 7th grader last year.  He did not stay back an extra year.  He was just a three-year competitor, I think.

I seem to remember he placed in 6th grade also.  Or no.  He was just on the 2nd place Illinois team with Albert Ni.

I vaguely remember that the ESPN broadcast last year mistakenly said that Greg was an 8th grader that year.  But I could be the one that is mistaken.

5th place written here.  Got booted first round of countdown.  I also would've been 2nd written if only I had realized that the question asked for a percent, not a common fraction.  Oh well, MC is done for me!

Wasn't #2 about the dodecahedron net and the shortest distance between a and b? The answer to that was 3.

[quote="captcha000"]yea i missed it i got 4....[/quote]
lol... so did i.
can't believe that was the only target i missed

i talked to so many people who got 4 for that one including two of my teammates.

[quote="white_horse_king88"]How do you do that one. -1 :le: x :le: 3
and -2 :le: y :le: 4?

I keep getting 14/35.

For x, there is -1,0,1,2,3 and for y there's -2,-1,0,1,2,3,4. So that is 5*7=35 pairs of xy. How can you get 12 as the denominator?[/quote]
that's exactly what i did. i thought it meant integers so i got screwed there

I got 98th, but Nebraska got 12th because we had the 19th place kid (alc) and we got a 10 on team.

10 on team is vital for a good team ranking...

Results are posted.  [url]http://mathcounts.org/Competitions/NatlCompRankings04.pdf[/url]

[quote="Syntax Error"]Well, the thinking was more of a "there isn't much of a difference between 1st and 10th place written" because usually they aren't separated by too many questions.

Personally, I favor the new countdown system.[/quote]

I haven't been on this forum for a whole week. WHat's the new countdown system?

[quote="Chinaboy"]Well... did well on sprint, but totally bombed the target... so i got 24 + 6 = 30... (51st)

OK got 18th... 10 higher than last year...[/quote]

Huh? You got 18th, 28th, and 51st somehow... and it seems that two are in the same year...

Edit: Ahhh... OK = Oklahoma... makes sense now ;-).

???

OK = Oklahoma

hopes that clears everything up


actually the information is more than enough to solve it!

i forgot how to do surface area, and i didn't know what you needed

[quote="BOOYA"][hide]yes, it is more information than you need. Since the question states that it is a cube, then you wouldn't need both the width and the height, just one is enough. 10 x 10 x 6 = 600 sq units[/hide][/quote]

Yes, that's true.

...by the way, the surface area of a cube of a side $s$ is $6s^2$ as you have seen.

[quote="math92"]Find the surface area of a cube with a length of 10, and a height of 10.  (is that enough information to solve it?)[/quote]

Area of one side of the cube is $10^2$ or $100$

So the entire surface area would be $100*6$ or $600$

[hide]The Surface Area of the cube is composed of $6$ squares with Side Lengths $10$. Thus, the Surface Area is $6\cdot 10^2$ or $600$[/hide]


calculus
integration
function
limit
inequalities
real analysis
real analysis unsolved

Hello, let $ f_n$ a sequence of nonnegative measurable function on a measurable set E, and let the summable function $ f$ be  the pointwise limit of the sequence. Suppose that
$ \lim_{n \rightarrow \infty} \int_E f_n dx \equal{} \int_E fdx$
Prove that for every measurable subset $ F \subseteq E$ it is

$ \lim_{n \rightarrow \infty} \int_F f_n dx \equal{} \int_F f dx$

My question:
Is it obvious that $ \lim_{n \rightarrow \infty} \int_F f_n dx$ exists? BTW I just could prove that

$ \int_F fdx \leq \liminf_{n \rightarrow \infty} \int_F f_n dx$ using Fatou's Lemma, but how do I prove the opposite inequality?

Existence of the limit is a part of what you have to prove. Indeed, by Fatou's lemma $ \int_F fdx \leq \liminf_{n \rightarrow \infty} \int_F f_n dx$. What's more, the same lemma tells you that $ \int_{E\setminus F} fdx \leq \liminf_{n \rightarrow \infty} \int_{E\setminus F} f_n dx$. Think about what this last inequality means for the integrals over $ F$.

thanks for the hint :)

it is for all n:
$ \int_F f_n dx \equal{} \int_E f_n dx \minus{} \int_{E \backslash F} f_n dx$
which gives

$ \limsup \int_F f_ndx \leq \int_E f dx \plus{} \limsup \left ( \minus{} \int_{E \backslash F} f_n dx \right ) \leq \int_E f dx \minus{} \liminf \int_{E \backslash F} f_n dx \leq \int_E f dx \minus{} \int_{E \backslash F} f dx \equal{} \int_F f dx$

Hence:
$ \int_F f dx \leq \liminf \int_F f_n dx \leq \limsup \int_F f_n dx \leq \int_F fdx$
so equalities hold and we finish.

right?

Right. One of the inequalities in the chain is actually an identity, since $ \limsup(\minus{}\dots)\equal{}\minus{}\liminf(\dots)$

is it still true if we don't require the $ f_n$s to be nonnegative?
I cannot adapt any counterexample of Fatou's Lemma to disprove this :)

Let $ E\equal{}[\minus{}1,1]$ and $ f_n(x)\equal{}\begin{cases}n,&0<x<\frac1n\\\minus{}n,&\minus{}\frac1n<x<0\\0,&\text{otherwise}\end{cases}.$ Note that $ f_n\to0$ [i]a.e.[/i]

Let $ F\equal{}[0,1].$ Do you see where this example is going?


$ \sum_{i \equal{} 0}^{n \minus{} 1}{i \plus{} 2\choose2}(n \minus{} i)$

Hint: try using a combinatorial argument...

hello, your sum is $ \sum_{i\equal{}0}^n{i\plus{}2\choose 2}(n\minus{}i)\equal{}\frac{n(n\plus{}1)(n\plus{}2)(n\plus{}3)}{24}$.
Sonnhard.

That's the answer I obtained; could you post your solution?

Try using the Hockey Stick Theorem twice.  (This can be readily converted into a combinatorial argument if necessary.)

Is there a proof of the Hockey Stick Theorem?  I tried http://www.math.uconn.edu/~troby/Hidden/4MattH/PTW/Identities.html
but I don't think it's there.

It's right there: [url=http://www.math.uconn.edu/~troby/Hidden/4MattH/PTW/proofs/pf3400001.html]click me[/url]

It is the repetitive application of the fundamental identity $ {n \choose k} \plus{} {n \choose k \plus{} 1} \equal{} {n \plus{} 1 \choose k \plus{} 1}$

It is straightforward to prove the Hockey Stick Theorem by a direct combinatorial argument by extending the combinatorial argument for Pascal's identity:  given a subset of a set with $ n\plus{}1$ elements of size $ k\plus{}1$, put them in order from left to right and consider the leftmost element.  If the leftmost element is in the $ (n\minus{}i)^{th}$ position, you have to pick $ k$ more elements from the remaining $ i$ elements to the right of that one; that is,

$ {n\plus{}1 \choose k\plus{}1} \equal{} \sum_{i\equal{}0}^{n\minus{}1} {i \choose k}$

as desired.  This is also the limit as $ p$ goes to $ 1$ of one of the identities proven [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1222350#1222350]here[/url] by the same argument.


Anyone have a nice proof for IMO 1993/1?

Let f(x) = x^n + 5x^(n-1) + 3, where n > 1 is an integer. Prove that f(x) cannot be expressed as the product of two non-constant polynomials with integer coefficients.

The ``Eisenstein'' hint is quite tantalizing. I'm trying to play around with prime factors of n after replacing x by x+1 and expanding things out by the binomial theorem. Do you know what the Eisenstein Irreducibility Criterion is?

You can't apply Eisenstein directly here, but the proof of Eisenstein might work.   That is, work by contradiction, assume a factorization, etc.  Or, since the problem has the letter n in it, induction might work, too.

It doesn't. I don't believe this problem is well-suited for induction.

[quote="Nukular"][quote]
 Or, since the problem has the letter n in it, induction might work, too.[/quote]

Why does that aphorism of proof always hold true?[/quote]

It doesn't really, and Simon is right that it probably doesn't in this case.  But remarkably much of the time, if there is a positive integer variable labelled n in a problem, you can induct on n.

[quote="TripleM"]Even more true is that if the year number occurs in it, you will be able to induct in some way.. either that or its to do with its factors.[/quote]

True enough -- 2004 is the first year in ages I haven't felt obligated to  memorize the factors of :)

[quote="ComplexZeta"]I'm guessing that's because you haven't taken any math contests in 2004 yet. Are there any contests for undergraduates other than the Putnam?[/quote]

Yep.  And not as far as I know.  The Kalva site lists two others, one in Vietnam and one organized by a university in London with the University of Warsaw.  So, that doesn't make me overly optimistic.

There are quite a few mathematics contests for undergraduates in Europe. But there aren't as many here in the States. There are some other than the Putnam, but most of them are somewhat local. For example, there is the University of Illinois at Urbana-Champaign (UIUC) Mathematics Contest http://www.math.uiuc.edu/~hildebr/putnam/ - which is significantly easier than the Putnam, but still has some nice questions.

No one ever answered this, but I just cheated and went and looked at Kalva's solution, and I was basically correct -- assume a factorization and it's easy to show that it doesn't work, much like Eisenstein.


I asked somebody for their old avatar, which was given to me.  I uploaded the file, but it said it must be .jpg, .gnf etc. I even went to the site and hovered over the picture and put what came up in. It didn't work.  I know this avatar fits, as it was algebra2000 old one.

[url]http://www.gifs.net/gif/index.php3?n=image.php3&image_id=8015&image_name=Blue_flame[/url]

Next time use the correct url for the image (in this case http://www.gifs.net/Animation11/Everything_Else/Fire/Blue_flame.gif ) or download it to your computer and using the browse window to find it. Fixed right now.


inequalities
induction
inequalities unsolved

Prove:
$ 0 \leq x_j \leq 1 \ \Rightarrow \ \sum_{j\equal{}1}^n \frac{1}{1\plus{}x_j} \leq \frac{n}{1\plus{}\sqrt[n]{x_1...x_n}}$

[quote="Sylwek"]Prove:
$ 0 \leq x_j \leq 1 \ \Rightarrow \ \sum_{j \equal{} 1}^n \frac {1}{1 \plus{} x_j} \leq \frac {n}{1 \plus{} \sqrt [n]{x_1...x_n}}$[/quote]
 we can prove it by Jensen ineq
First prove this ineq is true with $ n\equal{}2$
Then use hensen ineq we have it is true for all $ n$

this is similar to the one from shortlist but the condition of the one from shortlist was $ x_i\geq 1$.
anyway I think it can be proved in a similar solution:

let $ x_i\equal{}e^{a_i}$,from $ 0\leq x_i\leq 1$ we get that $ a_i\leq 0$,now we have to show that:

$ \sum_{i\equal{}1}^n\frac 1{e^{a_i}\plus{}1}\leq\frac n{e^{\frac 1n(a_1\plus{}a_2\plus{}\ldots \plus{}a_n)}\plus{}1}$

now let $ f(x)\equal{}\frac 1{e^x\plus{}1}$ then we have:

$ f''(x)\equal{}(e^x\plus{}1)^{\minus{}3}e^x(e^x\minus{}1)$

so for $ x\leq 0$ we have $ f''(x)\leq 0$ so $ f$ is concave on $ [\minus{}\infty,0]$ hence:

$ \frac 1n\left(f(a_1)\plus{}f(a_2)\plus{}\ldots \plus{}f(a_n)\right)\leq f\left(\frac{a_1\plus{}a_2\plus{}\ldots \plus{}a_n}n\right)$

$ \Rightarrow\sum_{i\equal{}1}^n\frac 1{e^{a_i}\plus{}1}\leq\frac n{e^{\frac 1n(a_1\plus{}a_2\plus{}\ldots \plus{}a_n)}\plus{}1}$

I remember it in shortlist about 7 years ago!
  Also  try Induction!

no navid,the one you remember is different,it has different conditions and the sign is reversed...


Let $ a,b,c$ be real numbers .prove that

\[ (\sum_{cyc}a^6)\plus{}6(\sum_{cyc}a^4b^2\plus{}a^4c^2)\plus{}3(\sum_{cyc}a^4bc)\plus{}9a^2b^2c^2\ge 3(\sum_{cyc}a^5b\plus{}a^5c)\plus{}7(\sum_{cyc}b^3c^3)\plus{}3(\sum_{cyc}a^3b^2c\plus{}a^3c^2b)\]

I would like finally to know that someone discovered a general theorem for this kind of annoying problems.

Can Schur and Muirhead kill it? It seems very bad.

The problem is equivalent to
\[ (a \minus{} b)^6 \plus{} (b \minus{} c)^6 \plus{} (c \minus{} a)^6 \minus{} 3(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 0
\]
$ AM\minus{}GM$


function
real analysis
real analysis unsolved

Let $ f: \mathbb{R} \minus{}> \mathbb{R}$ be a surjective function such that for every sequence $ (x_n)_n$ of real numbers , if $ (f(x_n))_n$ converges , then $ x_n$ itself should also converge.Prove that $ f$ is continuous.

your some definition is not so  clear for me. is there difference between?:
 sequance $ \{x_{n}\}_{n \equal{} 0}^{ \plus{} \infty}$ and  your $ (x_{n})_{n}$?
if not , then why you don't  formulate problem as follows:
[quote="Svejk"]Let $ f: \mathbb{R} \minus{} > \mathbb{R}$ be a surjective function such that for every sequence $ x_n$ of real numbers , if $ f(x_n)$ converges , then $ x_n$ itself should also converge.Prove that $ f$ is continuous.[/quote]

His $ (x_n)_n$ means the same as your $ \{x_n\}_{n \equal{} 0}^\infty$.  The way he wrote it is more proper (although not notably clearer, and more notationally burdensome) because "$ x_n$" is just a number (the $ n$th member of a given sequence) not the sequence itself.


induction
pigeonhole principle
number theory
number theory solved

Prove that for every n \in N there is a n-digits number in in base n including only the digits 1 and 2, and it is divisible by 2^n !

a.) n = 10
b.) 4k+2 (k \in N)

my abbreviations:

mathematical requirements: mr
heuristical requirements: hr

comment:

mr: number theory, residues, induction
hr: pigeonhole, systematic attempts, restructuring

The number should be in base n?

Here's a soln for a) (n being 10):
We can prove that for each k there exists a number in base 10 of k digits (the digits are 1 or 2) which is divisible by 2^k. We do this by induction: let's assume we built such a number with k-1 digits, call this number x_(k-1). We know 2^(k-1)|x_(k-1). If 2^k|x_(k-1) then we add 2 at the beginning pf x_(k-1). If 2^k does not divide x_(k-1) we add 1 at its left side, and we got x_k.

I think sth similar can be done for the other case (I mean b).

You are right, grobber.
(4k+2)^s=2^s*(2k+1)^s is divisible by 2^s and is not divisible by 2^(s+1). The same property is used for n=10.

Yeah, so the soln is EXACTLY the same! :D


I need examplez of corner points :S!!

i knw tht |x| givez a corner point graph .. :S!!

but i cant find other examplez!!

ill be thankful to anyone who can help ..!!

Ah, you must mean those fun shapes you can make with absolute value!
Try $ |x|\plus{}|y|\equal{}4$


ratio
trigonometry
geometry unsolved
geometry

Let ABC be an isosoles triangle at angle A , and  $angleA=36^o$.
prove that the ratio $\frac{BA}{BC}$ is irrational.

$AB=AC=1;\ D\in AB,\ BC=CD=DA=x\ (BD=1-x)$

$\triangle ABC\sim \triangle CBD\Longrightarrow\frac{BA}{BC}=\frac 1x=\frac{x}{1-x}\Longrightarrow x=\frac{-1+\sqrt 5}{2}\Longrightarrow$

$\frac{BA}{BC}=\frac 1x=\frac{1+\sqrt 5}{2}\not\in Q.$

[b]Remark.[/b] $E\in (AC),\ EA=EC\Longrightarrow \frac{AE}{AD}=\frac{1}{2x}=\overline {\underline {\left| \cos 36^{\circ}=\frac{1+\sqrt 5}{4}\right| }}.$

[u]Otherwise:[/u] $t=36^{\circ}\ (5t=180^{\circ})\Longrightarrow \sin 2t=\sin 3t\Longrightarrow$

$2\sin t\cos t=\sin t(3-4\sin^2t)\Longrightarrow 2\cos t=3-4(1-\cos^2 t)\Longrightarrow$

$4\cos^2 t-2\cos t-1=0\Longrightarrow \underline {\overline {\left| \cos 36^{\circ}=\frac{1+\sqrt 5}{4}\right| }}.$


In a computer game, you try to bounce as high and as far as you can.  You don't know how high the part that you see (the screen) is, but when a bounce is so high, it goes off the screen, and the distance from where it is to the top of the part you see (the screen) is measured.  How high is the part you see (the screen) if on the first bounce you are 80 feet above the top, next bounce 34 feet above and next bounce 12 feet above, if the next bounce is always the same percentage as the previous one?

not too hard, just long

[hide]I set up a proportion. let d be the height of the screen,

$\frac{80+d}{34+d}=\frac{34+d}{12+d}$

When we cross multiply, we get
$1156+68d+d^2=960+92d+d^2$
$24d=196$
$d = 8\frac{1}{6}$[/hide]

You made an arithmetic error. ;)

[hide="Answer"]Say that $k$ is the percentage of the previous bounce that the next bounce is, with $0<k<1$, and $x$ is the height of the screen. Therefore, we have $k(80+x)=34+x \Rightarrow k=\frac{34+x}{80+x}$, and $k(34+x)=12+x \Rightarrow k=\frac{12+x}{34+x}$, so $\frac{34+x}{80+x}=\frac{12+x}{34+x}$. Cross-multiply to get $(80+x)(12+x)=(34+x)(34+x) \Rightarrow 1156+68x+x^2=960+96x+x^2 \Rightarrow 28x=196$, and $x=7$.[/hide]

However, this answer also does not substitute correctly back into the original equation... :huh:

Okay, so which answer looks correct?
Check mine  :D

I think 236factorial's solution was correct. The error that happened was that it should be 960 + [b]92[/b]x + x^2 instead of 960 + 96x + x^2. Also, if you plug 49/6 into the equation, it works.

Oh, whoops. :P

That's odd... I checked the other solution with the ratio, and it didn't work, but now that I check it again, it does...

Correct! 236! is right, but Jesusfreak made a mistake, why did he do that??!

Everyone knows that $80+12=96$! Right...? :roll:

maybe a new way of adding is a+b=ab/10 :rotfl:


calculus MATHCOUNTS function derivative integration search trigonometry

hi

i need a graphing calculator and i was wondering which one i should get

thanks

TI-89's are nice - they can operate symbolically and do 3-D graphing stuff.  Very powerful.  I don't think they're allowed on some tests because of how much they can do, can someone verify this?  :maybe:

I personally have a TI-83+ Silver, but I really want an 89.

TI-89s are basically the calculator of choice.  They are the most powerful calculator that is allowed to be used on the SATs, the Calculus AP tests, and on certian rounds of MathCounts and other competitions.  The TI-89 can do almost anything Mathematica or Maple can, although it is slower and not as pretty, of course.

It really depends on your tastes, however.  While the TI-89 is powerful, it's not as user-friendly as calculators like the TI-32, so it will take a little while to adjust to the layout and function of the TI-89.

would it be allowed on AMC 8?

if the TI89 is top of the TI series, what is the top of the GRAPHING  calculator series?

EDIT: is the TI89 even allowed on sat?

ti89 is allowed on sat and AP but not IB.

Some teachers might not allow ti89 (it does derivatives and integration for you!)

search for the other topics that have been made in this forum.  i got a TI-84+SE and that is really great, but takes a long time to get used to.  the other topic has a lot more info, pros, cons, and tips.

The TI-89s are nice for basically everything. The only exceptions are the ACT and AP Statistics (at least at my school :lol:, our book is TI-83 enchanced ). It has some pretty well-programed programs that you can download online.

Getting a TI-84 (or 83 plus) isn't bad either. You can get programs for practically anything online (integration, differentiation, trig equation solving, 3D graphing). Plus the 84 is cheaper.

The TI-89 seems to be more common in calculus classes now. I really have no preference for any of these two calculators. Just pick the one that you like the most.

The TI-84 Plus Silver Edition is great as well.  I just got one a few weeks ago after losing my last one at the beach :(      Yes, that's right, I do math at the beach ;)

[quote="math92"]ti-83+ is a nice calculator and has a lot of basic functions
ti-84 has more memory
and ti 89 titanium has a million more functions.  I have a ti 89 but i have had a 83+. I like the 89 better. :)[/quote]
And how much money does ti89 cost ? :huh:

[quote="hmmm"][quote="Eros88"]And how much money does ti89 cost ? :huh:[/quote] The TI-89 is about $\$$70-80, or about $\$$120-130 if you want the titanium version (which I'm pretty sure has no more power or features than the regular 89, it's just prettier).  But you can get them way cheaper used as well.[/quote]
Well ,100 bucks sounds like reasonable price for that little monster.
Thanks for the information.

[quote="1/(ln x)"]The TI-89s are nice for basically everything. The only exceptions are the ACT and AP Statistics (at least at my school :lol:, our book is TI-83 enchanced ).[/quote]

Yeah, our book was made for 83's as well but I was a rebel and used my 89 anyway :P

Compared to the regular 89, the titanium has 3 times the user memory, a USB port and some extra programs pre-loaded onto it.  Not much, but if you're planning on using your calculator extensively (i.e. doing a bunch of programming or other stuff on it) you wont really notice the difference between it and the regular 89.

There isn't much more you can add to the 89 before it becomes a handheld computer ;)

[quote="akalra1"]but the question is still there... which is the best calculator?

scrolling through your posts, there are so many opinions, I am confused[/quote]

As far as the TI series goes, the TI-89 and TI-92 are at the top, hoever the TI-92 is not allowed on tests because of the layout of the keyboard.  Also, the TI-89 does just about everything for you: Analyticaly solves equations, factors, does partial fractions, calculates derivatives and integrals, and a bunch more.  The most importand difference between the TI-89 and the previous models is that it can perform symbolic manipulations.

I recently got my first graphing calculator.  :D

It is a TI-83-plus.

[quote="lingomaniac88"][quote="hmmm"][quote="Eros88"]And how much money does ti89 cost ? :huh:[/quote] The TI-89 is about $\$$70-80, or about $\$$120-130 if you want the titanium version (which I'm pretty sure has no more power or features than the regular 89, it's just prettier).  But you can get them way cheaper used as well.[/quote]
Actually, I've seen an ad for it, and it cost $\$150$.

I start school in about a week and a half, so I'll probably ask my math teacher if they're allowed.  If so, I'll probably get it.

I was just wondering about something else.  I have familiarized myself with TI-Basic on my TI-84+.  Will I get to understand it quickly on the TI-89?[/quote]

the 84 basic is easy compared to the 89 as there are a lot more features on the 89 but you need much more memory and commands to write the same program.

[quote="SplashD"][quote="moogra"]geta ti-89 but they are expensive. I got ati-84+ silver[/quote]

only gamers get ti-89s as they have basically no other really useful purposes.[/quote]

you, sir, are an idiot on many different levels

1. you're from montgomery

2. TI-89s can do a lot of incredibly useful stuff aside from phoenix 89.

and 3. gamer usually means computer/console games, definitely not calculator games

ti-84 silver compared to ti-89 titanium?

hands down ti-89 is better, but you have a chance of getting a free calculator next year so i wouldn't be worried, i know i'm satisfied with the ti-84 silver i got

[quote="SplashD"]a ti-89 (normal) compared to the TI-89 Titanium[/quote]

ohhhh... next time don't quote somebody talking about ti-84s then lol

yeah i guess you're right except ti-89s are faster and look a lot sexier (for a calculator, whatever that's supposed to mean) and generally are better, but if you alreay have a ti-89 there's not really any point in getting a titanium unless it's free

k guys i got the ugly one (by treething)

i got the... ti84+se

i am still kinda confused on whether i shoulda get it or not

i mean like really, spalshd and treething fighting it out

[quote="bubka"]We used fx-84-TL for high school.[/quote]

india always uses casio...

[quote="Ars"]Get the Ti-89 titanium it is the best calculator allowed on the SAT's and is worth its money because of its speed. Also you can get it a lot cheaper on Ebay.[/quote]

of course there is a chance that the title for a "unused" product is probably "used"

and it mite be broken

it isnt safe


P is a point in the plane of an equilateral triangle ABC

Prove that PA,PB, and PC are the sides of a triangle.

When is the triangle degenerate?

Is there any simple way to do this not using [hide]Ptolemy's Inequality[/hide]?

[quote="darij grinberg"]Moderator's Note -

Solution censored to give less experienced a chance.[/quote]

Why censor any solutions. Inexperienced solutions really benefit from the creative ideas provided by excellent solvers as Darij. You might put them in spoiler if you really want it. A given solution should not prevent any other people from solving it again. Otherwise you could not publish any problem-solving books with solutions anymore as students can look up the solution. But only few do it without having struggled hard with all the problems. Otherwise you do not learn it.  :)

It's very intimidating for beginning solvers to approach a problem and find that someone has already posted a solution to it.  Because of the peculiar format of this sight, it would be almost impossible to avoid Darij's solution, especially since he didn't put it in spoiler and solves the questiosn so quickly.  Beginning solvers will benefit far more from having more experienced people give advice on solutions-in-progress than they will from reading someone else's solutions.  People at Darij's level (and yours, and even mine) should keep their hands off problems aimed at younger or less-experienced students.  That's why AOPS had lower-level forums -- so that lower-level solvers could have a place where more advanced people wouldn't butt in.  There is no sense at all in having an intermediate forum if people are going to come down from the advanced forum to solve all of the questions there.

Hello - just to clear up a thing: I did put the solution into a spoiler. The only things not spoiled were the  and  signs, but I think it would require some notable intellect to recover a solution from these signs only. There are actually some ways without using Ptolemy ([hide]for instance, you can project P on the sides of triangle ABC and show that the projections form a triangle whose sides are in the ratios PA : PB : PC[/hide]), but I don't wanna post a complete solution since it will most likely be deleted here.

  Darij

I totally agree with orl, solutions should not be censored anyhow. Besides, if you had to do the effort of editting his post, you could as well have spoilered the things that weren't spoilered enough to your opinion, instead of removing the whole post :roll:

i agree with mod.the soln should be censored.but it can be shown to JBL who posted.By the way i didnt understand the question.if P is a point inside the triangle ABC then how can PA ,PB and PC form the sides of the same triangle.please explain  :maybe:

[quote="sriram_mask"]i agree with mod.the soln should be censored.but it can be shown to JBL who posted.By the way i didnt understand the question.if P is a point inside the triangle ABC then how can PA ,PB and PC form the sides of[b] the same triangle[/b].please explain  :maybe:[/quote]

Note the words that you have used that weren't in the question :) It never says it has to be the sides of the same triangle, it just says they are the sides of a triangle - ie, the three lengths can be the lengths of a triangle (any two add up to more than the third).

I hope we have to consider the following cases

Case 1:PA and PB form sides of One triangle.
Case 2:PA and PC form sides of another triangle.
Case 3:PB and PC form sides of one triangle.
Finally
Case 4:PA PB PC form sides of diff triangles.

Sriram[/quote]


this is my first picture and it was a test to see if it worked o and its my backround image for my comp

ah i didnt check what this was i just uploaded it

this one looks alot like a sucking doughnut

dunno what it is
so don't want to try it
so can't judge its yumminess
sorry guys =[

SCRUMPTIOUS!
its like a bomb in my tummy...

eh...

yummiest  :lol:
zesty flavor

kirby loves blackholes.

kirby has more suction power than blackholes ;)

thats true. tonights meal: black hole. it is well done and was cooked by chef kirby!

It's a vortex. according to the file name.


projective geometry
geometry
similar triangles
geometry solved

im very fond of this problem because i solved it in less than an hour and it was supposed to be a very hard problem; so hard my instructor even told that it will take at least a week to solve it¡¡¡¡¡ he underestimeted my potential :D

here it is:

Let ABC  be a triangle, we draw parallel lines passing through B and C. Let any line through A intersect the parralel lines in M  and  N  with closer to AB  than  N. Draw the parallel to AB  passing through N . The point of intersection of this parallel with  BC  is P. Prove that MP is parallel to  AC. ENJOY¡¡¡ :rotfl:  :lol:

Applying the Pappos-Pascal theorem to the collinear point triples (B; P; C) and (N; A; M), we see that the points $PM\cap CA$, $CN\cap BM$ and $BA\cap PN$ are collinear; but the points $CN\cap BM$ and $BA\cap PN$ lie on the line at infinity (since CN || BM and BA || PN), and thus the point $PM\cap CA$ must also lie on the line at infinity, i. e. we have PM || CA; in other words, MP || AC.

This was very hard.  :rotfl:  :rotfl:  :rotfl:

  Darij

obviously this looks easy to you  :?  , but youre a master at geometry plus your solution uses too much 'technology' as always  ;)  try finding a solution only with similar triangles it´s not as easy as you think  :lol:  :D

Yes, this is actually much harder. ;)

Let the lines MN and BC intersect each other at a point Q. Then, BM || CN yields $\frac{QM}{QN}=\frac{QB}{QC}$, and NP || AB yields $\frac{QN}{QA}=\frac{QP}{QB}$, so that $\frac{QM}{QA}=\frac{QM}{QN}\cdot\frac{QN}{QA}=\frac{QB}{QC}\cdot\frac{QP}{QB}=\frac{QP}{QC}$; thus, MP || AC, and the problem is solved. Actually, this solution worked only for the case when the lines MN and BC have a common point, i. e. they are not parallel; but the case when they are parallel is even easier (use parallelograms).

  Darij

oops i didnt know it was that easy¡¡¡ i had to draw two parallels to get my solution but youre solution works just as well¡¡ :blush:  very sorry; i guess since my instructor it was so hard it got into my head  :P  sorry darij  :lol:


Here's a small part of an IMO problem this year I really liked:

Prove:

$2^{p-2}+3^{p-2}+6^{p-2}-1$ is divisible by p, where p is a prime $\ge5$

This is in the IMO 2005 forum. Anyways it is a easy application of fermat's little.

[hide]$A^{p-1} \equiv 1 \pmod {p}$
Divide by $A$ because it's relatively prime.
$A^{p-2} \equiv 1/A \pmod {p}$

 er nevermind :blush: [/hide]

Basically that's right.  [hide]Suppose $3^{p-2} \equiv x \mod p, 2^{p-2} \equiv y \mod p, 6^{p-2} \equiv z \mod p$.  Then $3x = ap+1, 2y = bp+1, 6z = cp+1$.  Then $x+y+z \equiv 1 + \frac{2ap + 3bp + cp}{6}\equiv 1 \mod p$.[/hide]

Or  [hide]6 times the expression = $3\cdot 2^{p-1} + 2\cdot 3^{p-1} + 6^{p-1} -6 \equiv 3+2+1-6 \equiv  0 \bmod p $  ...:) .[/hide]


Define the sequence of positive integers a_n recursively by a_1=7, a_n=7^a_(n-1)) for all n => 2.  Determine the last two digits of a_2007.

Basically what is $ 7^{7^{7..}}$ mod 100 where its a tower of 2007 7s.

[hide="hint"]
the last two digits of $ 7^n$ repeat in cycles of 4
[/hide]


[quote="Scrambled"]Protoss are out of the OSL!

Predictions for the MSL
EffOrt 2-1 Upmagic
Flash 2-1 Kwanro
Bisu 1-2 Fantasy
Iris 2-1 Hwasin
Jaedong 2-0 Kal
Canata 2-0 Piano\
Calm 2-0 ForGG
Pure 2-1 ZerO

Ro8
Jaedong 3-1 Pure
Flash 3-2 Canata
Fantasy 3-1 Iris
EffOrt 3-1 Calm

Ro4
Jaedong vs EffOrt 3-1
Flash vs Fantasy 3-2

Finals!
Jaedong vs BackHo 0-3[/quote]

I'm assuming the backho part isn't serious

EffOrt 2-0 Upmagic
Flash 2-1 Kwanro
Bisu 2-1 Fantasy
Iris 2-1 Hwasin
Jaedong 2-0 Kal
Canata 2-0 Piano
Calm 2-0 ForGG
Pure 1-2 ZerO

Jaedong 3-0 ZerO
Flash 3-1 Canata
Fantasy 2-3 Iris
EffOrt 3-1 Calm

Jaedong vs. EffOrt 3-2
Flash vs. Iris 3-1

Jaedong vs. EffOrt 0-3.

Dude what happened in Ro8 first day:

Canata > Jaedong
Calm > EffOrt
Iris > Bisu
Kwanro > Zero

Uh at least Ro8 is Bo5... hmm hopefully Jaedong/EffOrt/Kwanro will win next 3 games (And iris will beat beat bisu 3-2 hopefully)

[quote="junggi"]effort 2-0 upmagic
flash 1-2 kwanro
bisu 1-2 fantasy
iris 2-0 hwasin
jaedong 2-0 kal
canata 2-0 piano
calm 2-0 forgg
pure 0-2 zero

jaedong 3-1 canata
effort 3-2 calm
zero 1-3 kwanro
fantasy 1-3 iris

jaedong 2-3 effort
kwanro 3-1 iris

effort 3-2 kwanro[/quote]

whoa dang you were pretty much correct in the winners except for replace effort with calm :O :O :O

in other news MSL and OSL group selection this week~

sup guys

why is flash so good?

also why is stork sometimes silly? :(

hmm hello guys

now that pretty much every dual TL/AoPS member has failed FPL (even pythag! lol CJ fanboy team) let's do some starleague predictions

MSL Ro8:

Jaedong > Stats
Hwasin < Kal
Light < Kwanro (... darn this is a difficult one)
Flash > Best

then
Jaedong > Kal
Flash > Kwanro

Flash > Jaedong YEAH

OSL Ro4:

Flash > Calm
Movie > Shine (please? beat shine? please?)

Flash > all (ok wait just kidding maybe I want shine to win so he can fail epically vs flash)

dude i'm still in top 2000

MSL Ro8:

Jaedong 3-1 Stats
Hwasin 1-3 Kal
Light 3-0 Kwanro
Flash 0-3 Best

then
Jaedong 3-0 Kal
Best 3-0 Light

Jaedong 3-0 Best

OSL Ro4:

Flash 3-1 Calm
Movie 3-0 Shine (El nino and hbr are not zerg favored maps at all, obviously)

FINALS:

Movie 3-2 Flash

EDIT: And hamster is raping everyone in fpl dude

OSL finals today! and I predicted everything right so far, too. Clearly my predictions will win.


[quote="FelixD"]There is just a term missing (or they confused $ \le$ and $ \equal{}$). In fact, it should be $ \frac {k^k}{(k \minus{} 1)^{k \minus{} 1}} \left( \frac {1}{k \minus{} 1} \minus{} \frac {1}{k} \plus{} \dots \minus{} \frac {1}{n} \right) \equal{} \left( 1 \plus{} \frac {1}{k \minus{} 1} \right)^{k \minus{} 1} \minus{} \frac {k^k}{n(k \minus{} 1)^{k \minus{} 1}} \le e$.

Where did you find this solution? In a book or somewhere else? (and if it appeared in a book, which book?^^)[/quote]

Dear Felix
this problem in:《 Secrets in Inequalities (volume 1)》------Pham Kim Hung
but your explain ,I cann't understand!
I think like this:
$ \frac {k^k}{(k \minus{} 1)^{k \minus{} 1}} \left( \frac {1}{k \minus{} 1} \minus{} \frac {1}{k} \plus{} \dots \minus{} \frac {1}{n} \right) \equal{} \left( 1 \plus{} \frac {1}{k \minus{} 1} \right)^{k} \minus{} \frac {k^k}{n(k \minus{} 1)^{k \minus{} 1}} \\
\\
..................$

but next I  cann't do!
help?
thanks

[quote="sxgfly"][color=red][size=150]  see  that   (*),  How to obtain  it ?
or this solution has some mistakes, please tell me?
thanks !
 [/size][/color][/quote]

The solution is false!!!!!!!!!

[quote="FelixD"]Yes, my post above was stupid^^ The solution is indeed false, but the problem holds still true.  :wink:[/quote]

"but the problem holds still true"

Can you give me a solution ?

best regards

how to prove this inequality ,please tell me?


Prove that if n is not a multiple of 3, then the angle Pi/n can be trisected with ruler and compasses.

constructions are fun  :D

i posted this one a while back. the general idea is to [hide]construct 60 and 120 degree angles to the horizontal.[/hide]

Yes, you just draw very many of the angle to make a full angle, and then draw a 60 degree angle, and it must trisect.  This was up here before.


For $a,b,c>0$ prove that

$\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}} \leq 3$

Very nice!!!!

Just a reference: "Old and New Inequalities", problem 113.

  Darij

[quote="manlio"]For $ a,b,c > 0$ prove that

$ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}} \leq 3.$[/quote]Using the arithmetic mean - geometric mean inequality,

$ \sum_{cyc}{\sqrt {\frac {2a}{a \plus{} b}}}\leq\sum_{cyc}{\left[\frac {3a(b \plus{} c)}{4(bc \plus{} ca \plus{} ab)} \plus{} \frac {2(bc \plus{} ca \plus{} ab)}{3(a \plus{} b)(b \plus{} c)}\right]}$

$ \equal{} \frac {17}{6} \plus{} \frac {4abc}{3(b \plus{} c)(c \plus{} a)(a \plus{} b)}\leq 3.$

See also : http://www.mathlinks.ro/viewtopic.php?t=56184


The value of cosθ is
a pic of the figure can be seen at the link below

http://img.villagephotos.com/p/2006-12/1234083/6sb14.gif

any ideas?

Just remember the definition of cosine on the unit circle. Also remember the 3-4-5 right triangle!

ohh yeahh. i forgot about the 3 4 5. hey thanks a lot


function
calculus
integration
limit
real analysis
real analysis theorems

Suppose that $ f: \mathbb{R}\rightarrow\mathbb{R}$ is a function such that $ \int^{\infty}_af(x)dx$ is convergent. Does $ \lim_{x\rightarrow \infty}f(x)$ necessarily exist? If so, does it equal zero? If not, what conditions are required for these propositions to be true? (If possible, please provide counter-examples and proofs).

[quote="Kent Merryfield"]Counterexample: let $ g(x) \equal{} \max(1 \minus{} |x|,0)$ be a simple "tent" function. Note that $ \int_{\mathbb{R}}g(x)\,dx \equal{} 1.$

Now let $ f(x) \equal{} \sum_{k \equal{} 2}^{\infty}kg(k^3(x \minus{} k)).$

The graph of $ f$ consists of isolated spikes centered on the integers, with patches of zero in between the spikes. $ f$ is continuous on $ [0,\infty)$ but not only does $ \lim_{x\to\infty}f(x)$ not equal zero, we even have $ \limsup_{x\to\infty}f(x) \equal{} \infty.$

But $ \int_0^{\infty}f(x)\,dx \equal{} \sum_{k \equal{} 2}^{\infty}\int_{\mathbb{R}}kg(k^3x)\,dx \equal{} \sum_{k \equal{} 2}^{\infty}\frac1{k^2} < \infty.$

===============================

For an additional condition worth investigating: what if $ f$ is uniformly continuous on $ [0,\infty)?$ (The $ f$ in my counterexample is most assuredly not uniformly continuous.)[/quote]

Weren't you trying to look at $ \lim_{x \to 0} f(x)$?

Ah, I didn't notice that - probably because it's not a very interesting question. Of course you can have functions that behave like, say, $ \frac1{\sqrt{x}}$ near zero and are integrable. I don't know what else you can say about that.

There's no good reason to quote the entire post immediately preceding yours -- all it does is create unnecessary clutter.  The existence of the limit as $ x \to a$ is irrelevant to the question of the convergence of the given improper integral, so Kent assumed (almost certainly correctly) that the original poster had meant to ask about $ \lim_{x \to \infty} f(x)$.

Edit: too slow.

Kent how did you come up with the "tent" function as an example? Are there any simpler examples? It seems like it came out of thin air. Also what happened to the $ x\minus{}k$ term? Why did it go to $ x$?

Thanks

It's actually a rather natural example. Try working out a construction yourself, starting by listing out all the properties that such function needs to have. You'll probably get something similar to Professor Merryfield's example.


[hide]
$1+\sum_{cyc}x+\sum_{cyc}xy+xyz\geq 3-2\sum_{cyc}x^{2}+\sum_{cyc}x^{4}$
$2\sum_{cyc}x^{2}+\sum_{cyc}xy-1+xyz\geq \sum_{cyc}x^{4}$
$2\sum_{cyc}x^{2}+\sum_{cyc}xy-(\sum_{cyc}x^{2}+2\sum_{cyc}xy)+xyz\geq \sum_{cyc}x^{4}$
$\sum_{cyc}x^{2}+xyz\geq \sum_{cyc}xy+\sum_{cyc}x^{4}$
$\sum_{cyc}x^{2}(\sum_{cyc}x^{2}+2\sum_{cyc}xy)+\sum_{cyc}x^{2}yz\geq \sum_{cyc}xy(\sum_{cyc}x^{2}+2\sum_{cyc}xy)+\sum_{cyc}x^{4}$
$\sum_{sym}x^{3}y\geq 2\sum_{cyc}x^{2}yz$
$\sum_{cyc}xy(x-y)^{2}+\sum_{cyc}x^{2}(y-z)^{2}\geq 0$
equality cases.
if two of them are not $0$ lets say $x,y$ then from $x^{2}(y-z)^{2}=0$ we have y=z and from
$y^{2}(z-x)^{2}$ we have $z=x$. if any two of them are $0$ equality holds so
if $x=0,y=0,z=1$ and cyclic permutations or $x=y=z=\frac{1}{3}$ equality holds...
[/hide]

APMC(Austrian Polish Mathematical Competition) 2000. I'm hoping that this year it will be MEMO(Middle European Mathematical Competition so my students can participate) :)
$a,b,c\geq 0$ $a+b+c=1$
prove that
$2\leq \sum_{cyc}(1-a^{2})^{2}\leq (1+a)(1+b)(1+c)$
i think my friend pulled out some sort of generalization on the left inequality for $n$ variables.

[quote="kim_ina_88"]x,y,z>0 x+y+z=1
                                           prove that
\[(1+x)(1+y)(1+z) \ge (1-x^{2})^{2}+(1-y^{2})^{2}+(1-z^{2})^{2}\]
[/quote]

Here is my solution, I think it's o.k., is it?

[hide="solution"]
\[\sum_{cyc}xy+xyz \geq 1+\sum_{cyc}x^{4}-2 \sum_{cyc}x^{2}\,\,\,\,\,(*) \]
- now we have:
$\sum_{cyc}x^{2}= 1-2 \sum_{cyc}xy$,
and:
$\sum_{cyc}x^{4}= 1-4 \sum_{cyc}xy+2 \sum_{cyc}x^{2}y^{2}+8 \sum_{cyc}x^{2}yz$
\[(*) \Leftrightarrow \sum_{cyc}xy+xyz \geq 2 \sum_{cyc}x^{2}y^{2}+8 \sum_{cyc}x^{2}yz\,\,\,\,\,(**) \]
- now choose $a,b,c \in R^{+}$ so that $x= \frac{a}{a+b+c}$, $y= \frac{b}{a+b+c}$ and $z= \frac{c}{a+b+c}$
\[(**) \Leftrightarrow \frac{abc}{(a+b+c)^{3}}+\frac{ \sum_{cyc}ab}{(a+b+c)^{2}}\geq \frac{2 \sum _{cyc}a^{2}b^{2}+8 \sum _{cyc}a^{2}bc }{(a+b+c)^{4}}\]
Expanding gives the equivalent inequality:
\[\sum_{sym}a^{3}b \geq 2 \sum_{cyc}a^{2}bc = \sum_{sym}a^{2}bc \]
Last inequality holds by Muirhead $(3,1,0) \succ (2,1,1)$, with equality iff $a=b=c$ i.e. $x=y=z= \frac{1}{3}$.
Q.E.D.

Sorry for my bad English  :)
[/hide]


Find the sum of the reciprocals of the roots of $ x^2\minus{}13x\plus{}4\equal{}0$.

$ \frac {1}{\alpha} \plus{} \frac {1}{\beta} \equal{} \frac {\alpha \plus{} \beta}{\alpha \beta} \equal{} \boxed{\frac {13}{4}}$.

This is because of [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_Formulas]Vieta's Formulas[/url], the sum of the roots is equal to $ \minus{}\frac{b}{a}$ and the product of the roots is equal to $ \frac{c}{a}$, in the quadratic equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$.


During our AoPS class at school, we were doing the exercises of section 26.5 of Volume 1. For most of us, the answer we got for every question was the reciprocal of the answer given in the book. Our teacher couldn't explain to us what we were doing wrong (or rather, she couldn't explain it in a way that convinced us). We did read the warning on the top of page 320, but this seemed to justify the answers most of us got, rather than those the book gave.
could you, perchance, give us a basic explaination of odds and how to compute them??
many, many thanks!

I'll try to explain this using an example.

I have a bag with 10 marbles: 3 red and 7 green.

If I draw one marble from the bag at random there is a 3/10 chance that it will be red.

The "odds" that I draw a red marble are 3:7 because there are 3 chances of affirming the proposition and 7 chances of negating it.

On the other hand, I could also say the "odds against" me drawing a red marble are 7:3 because the propositions have been reversed.

"odd against" are not as common in problems, but common in gambling.

Springboarding from this example, answer each of the following using the same bag of 10 marbles described above.

1.(a) What is the probability of drawing a green marble with one draw?
(b) What are the odds that you draw a green marble out of the bag with one draw?
(c) What are the odds against drawing a green marble out of the bag with one draw?

2. (a) What is the probability of drawing at least one green marble given two draws?
(b) What are the odds that you draw at least one green marble when drawing two marbles from the bag?

3. (a) What is the probabilty of drawing a marble of each color when drawing two marbles from the bag?
(b) What are the odds that you draw a marble of each color when drawing two marbles from the bag?

In the book, however, it says on page 319
"the odds of an event is the ratio of the probability it does not happen to the probability that it does."
According to this definition, the odds of drawing a red marble would be
[u](7/10)[/u]
(3/10)
=7:3

whereas the odds against would be 3:7.

The original edition of AoPS Volume 1 was not exactly correct on odds.  Newer volumes have been edited to include the formal definition of odds which agrees with the way I defined it above.

What it really comes down to is that many gamblers butcher the language of "odds" through slang.  This makes it difficult use as a precise mathematical terms.

I wouldn't lose sleep over this.  My exlpanation is correct, but you can usually gauge from the problem statement or the answer choices which you are expected to use.

Ultimately, there isn't much reason to test the definition of odds on math team tests and you won't see it at all (without an accompanying definition or context) on any serious contest.


trigonometry
Pythagorean Theorem
geometry

Hi, I forgot how to do this, if you could point me in the right and simplest direction, it would be appreciated.

The question:
 Find the six trigonometeric values of theta=inverse cos (3/7). Give exact answers.

What I know:
 The six values I need to find are:Cos theta, sin theta, tan theta, cosecant theta, cotangent theta, and secant theta.

 Also, inverse cos (3/7) equals 64.6 degrees or 1.13 radians.

Draw a right triangle with a hypotenuse of seven and an adjacent side of three (after all, $ \cos \theta \equal{}\frac{3}{7}$).

You can use Pythagorean Theorem to find the third side, and then you have all the lengths needed to find the ratios.


Find the constant numbers $a,b,c$ and $d$ which satisfy the following two equations.

$\lim_{x\rightarrow 1} \frac{ax^3+bx^2+cx+d}{x^2+x-2}=1.$

$\lim_{x\rightarrow -2} \frac{ax^3+bx^2+cx+d}{x^2+x-2}=4.$

[hide]
Note that $x^2+x-2 = (x+2)(x-1)$.

-----BETTER SOLUTION BELOW-----

Let $f(x) = ax^3+bx^2+cx+d$. The first limit tells us that $f(x)$ is divisible by $x-1$, so

$f(x) = (x-1)(ax^2+(a+b)x+(a+b+c))$

and that

$\lim_{x \to 1} \frac{ax^2+(a+b)x+(a+b+c)}{x+2} = 1$.

We can simply plug $x = 1$ to get

$3a+2b+c = 3$.

Similarly, $f(x)$ is divisible by $x+2$, so

$f(x) = (x+2)(ax^2+(b-2a)x+(c+4a-2b))$

and do the same as above and plug in $x = -2$ to get

$12a-4b+c = -12$.

Equating the constant coefficients in the above factorizations, we get

$d = -(a+b+c) = 2(c+4a-2b) \Rightarrow 3a-b+c = 0$.

We now have a system:

$3a+2b+c = 3$
$12a-4b+c = -12$
$3a-b+c = 0$

which we can solve to get $a = -1, b = 1, c = 4, d = -4$ so

$f(x) = -x^3+x^2+4x-4$.

-----BETTER SOLUTION (with less details)-----

EDIT: Maybe it's easier if we write $f(x) = (x+2)(x-1)(ex+f)$. Then we get

$e+f = 1$

$2e-f = -4$

so $e = -1, f = 2$. Then $f(x) = (x+2)(x-1)(2-x)$ as found above. [/hide]


Hint:

[hide]

Write each term in a different row and add like this:

1 +

1 + 3 +

1 + 3 + 5 +

...[/hide]

Me:  "GAH! THREE VARIABLES"

Yeah, I know what you mean.


Let $ f$ be a real-valued function such that for any real $ x$,

$ (a)$ $ f(10\plus{}x) \equal{} f(10\minus{}x)$

$ (b)$ $ f(20\plus{}x) \equal{}\minus{}f(20\minus{}x)$.

Prove that $ f$ is odd ($ f(\minus{}x)\equal{} \minus{}f(x)$)and periodic (there exists $ T > 0$ such that $ f(x \plus{} T) \equal{} f(x)$)

from a we get :
*) $ f(x)\equal{}f(20\minus{}x)$
so $ f(x)\equal{}\minus{}f(20\plus{}x)\equal{}\minus{}(\minus{}f(20\plus{}20\plus{}x))\equal{}f(40\plus{}x)$
and $ f(x)\equal{}f(20\minus{}x)\equal{}\minus{}f(20\plus{}x)\equal{}\minus{}f(\minus{}x)$

I found some special cases
f(30)=f(20+10)=-f(20-10)=-f(10)=f(10)=-f(50)=-f(30) and f(0)=-f(0)
and other nums would be like this
I think it is not hard
maybe I'm wrong


function
logarithms
limit
integration
floor function
calculus
calculus computations

Gamma

\[ \sum\limits_{k \equal{} 2}^\infty  {\frac{{\left( { \minus{} 1} \right)^k \left[ {\log _2 k} \right]}}
{k} \equal{} \gamma }\]

prove


solve the following inequality:

$ \sqrt[3]{x^{3}\minus{}x^{2}\plus{}x\plus{}1}\ge x\minus{}1$

thanx.

Cubing yields $ 2x^2\minus{}2x\plus{}2\ge 0$, which is true.

yes,but I think we have to solve the equation $ x^{3}\minus{}x^{2}\plus{}x\plus{}1\equal{}0$ firstly .


calculus
function
inequalities proposed
inequalities

Is this what you want me to prove?
$\sum_{cyc}\frac{a^{2}+b^{2}}{a+b}\leq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$

A bit more generally, for real $k>1$ we have
$\sum_{cyc}\frac{a^{k}+b^{k}}{a+b}\leq \frac{3(a^{k}+b^{k}+c^{k})}{a+b+c}$
$\Leftrightarrow \sum_{cyc}c\frac{a^{k}+b^{k}}{a+b}\le \sum_{cyc}c^{k}$
$\Leftrightarrow \sum_{cyc}\frac{ca(a^{k-1}-c^{k-1})+cb(b^{k-1}-c^{k-1})}{a+b}\le 0$
$\Leftrightarrow \sum_{cyc}\frac{c(a^{k-1}-c^{k-1})(a-c)}{(b+c)(a+b)}\ge 0$ :)

We can see that the sign flips if $k<1$.

Now what about a further generalisation to $n$ variables? i.e. for $k\ge1$,
$\sum_{{n\choose2}}\frac{a^{k}+b^{k}}{a+b}\leq{n\choose2}\frac{\sum a^{k}}{\sum a}$
Already for $n=4$, this seems more tricky than expected.
Is it possible that we even have for integers $1\le\ell<m\le n$ \[\sum_{{n\choose \ell}}{n\choose \ell}^{-1}\frac{a_{1}^{k}+\cdots+a_\ell^{k}}{a_{1}+\cdots+a_\ell}\leq \sum_{{n\choose m}}{n\choose m}^{-1}\frac{a_{1}^{k}+\cdots+a_{m}^{k}}{a_{1}+\cdots+a_{m}}\] :?:

And what about if the sums are only [i][b]cyclical [/b][/i]instead of summing over all $\ell$-tuples and $m$-tuples :roll: :D

[quote="spanferkel"] \[\sum_{{n\choose \ell}}{n\choose \ell}^{-1}\frac{a_{1}^{k}+\cdots+a_\ell^{k}}{a_{1}+\cdots+a_\ell}\leq \sum_{{n\choose m}}{n\choose m}^{-1}\frac{a_{1}^{k}+\cdots+a_{m}^{k}}{a_{1}+\cdots+a_{m}}\] :?:  [/quote]
:yup: It is sufficient to prove this for $\ell+1=m=n$, as the other cases can be deduced from that by iterating and adding up symmetrically. Let's assume wlog $\sum_{i=1}^{n}a_{i}=1$ and put $\sum_{i=1}^{n}a_{i}^{k}=: A\le1$. So we just have to show $\sum_{i=1}^{n}\frac{A-a_{i}^{k}}{1-a_{i}}\le nA$. But if I'm right, this follows directly from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=208384#208384]Vasc's n-1EV Principle[/url] with some calculus :blush: to check the condition.

Now the [b]cyclical [/b]one remains, certainly tougher. (Of course the above would be a consequence of the cyclical one as well.) So to reformulate it:
For $a_{1},...,a_{n}>0$, real $k>1$ and integers $1\le\ell<m\le n$, prove \[\sum_{i=1}^{n}\frac{a_{i+1}^{k}+\cdots+a_{i+\ell}^{k}}{a_{i+1}+\cdots+a_{i+\ell}}\leq \sum_{i=1}^{n}\frac{a_{i+1}^{k}+\cdots+a_{i+m}^{k}}{a_{i+1}+\cdots+a_{i+m}},\] the indices being cyclical $\mod n$. Good luck...

Darij has pointed out that the symmetrical version can be easily proved the same way as the $n=3$ case above. :blush: No need for n-1EV here.

On the contrary, the [i]cyclical [/i]version doesn’t hold in general.  :( But it may be possible that something similar to the Shapiro ineq happens. Let’s conjecture that the following holds :

[b]There exists a universal constant $C>0$ such that [/b]
\[C\cdot\sum_{i=1}^{n}\frac{a_{i+1}^{k}+\cdots+a_{i+\ell}^{k}}{a_{i+1}+\cdots+a_{i+\ell}}< \sum_{i=1}^{n}\frac{a_{i+1}^{k}+\cdots+a_{i+m}^{k}}{a_{i+1}+\cdots+a_{i+m}}. \]
If my computer is right, $C=\frac89$ is best possible (i.e. asymptotically best as $k\to\infty$) for the cases both $(\ell,m,n)=(2,3,4)$ and $(\ell,m,n)=(2,3,5)$. But for a proof? :roll:

By the way, is it true that $\frac{RHS}{LHS}$ as a function of $k$ is always monotonous for $k>0$ ?


What is the greatest prime factor of the positive difference between 2,000 and 1,400?


[quote="IOO"]Does there exist one to one mapping between a unite circle and $ [0, 1]$?[/quote]

yes but then this doesn't belong in real analysis

Note that there is a natural bijection from the unit circle to the set $ [0,1)$ namely $ f(x) \equal{} e^{ix}$.  So, you are really asking whether $ [0,1)$ and $ [0,1]$ have the same cardinality.

If you believe in the Continuum Hypothesis, which says that the cardinality of the reals is the smallest uncountable cardinality, then its clear that $ |[0,1)| \equal{} |[0,1]|$.

If you don't, see the second proof offered here

http://planetmath.org/encyclopedia/ClosedOpen.html

which constructs a bijection from $ [0,1]\cap \mathbb{Q}$ to $ (0,1) \cap \mathbb{Q}$ and then extends that to a bijection from $ [0,1]$ to $ (0,1)$ by using the identity map on irrationals. The argument can be easily modified when the range is $ [0,1)$.

Don't involve the continuum hypothesis. We have the Cantor/Schroder/Bernstein theorem, which says that if you can find an injection from $ A$ into $ B$ and an injection from $ B$ into $ A$, there is a bijection between the two sets. The proof includes a construction, but it can get very complicated.


calculus
integration
calculus computations

Evaluate $ \int_0^1 \frac{(x^2\minus{}3x\plus{}1)e^x\minus{}(x\minus{}1)e^{2x}\minus{}x}{(x\plus{}e^x)^3}\ dx$.

well, ummm... i don't know if this was supposed to be a trick question   , but    it appears to me, provided i am not wrong,   [u]after[/u] multiplying both num and den by $ x\plus{}e^x$, the num could be written as

$ (x\plus{}e^x)^2 \cdot(x\plus{}xe^x)\, '\minus{} (x\plus{}xe^x) \cdot ((x\plus{}e^x)^2)\, '$

when then leads to me to believe i can write it as $ \int_0^1 \left(\dfrac { x\plus{}xe^x}{(x\plus{}e^x)^2}\right)\,' \equal{}\frac{1}{1\plus{}e}$

or maybe i am wrong...    :(

it may be checked using integration software...but this integral definitely not seems to be easy.

[quote="misan"]well, ummm... i don't know if this was supposed to be a trick question   , but    it appears to me, provided i am not wrong,   [u]after[/u] multiplying both num and den by $ x \plus{} e^x$, the num could be written as

$ (x \plus{} e^x)^2 \cdot(x \plus{} xe^x)\, ' \minus{} (x \plus{} xe^x) \cdot ((x \plus{} e^x)^2)\, '$

when then leads to me to believe i can write it as $ \int_0^1 \left(\dfrac { x \plus{} xe^x}{(x \plus{} e^x)^2}\right)\,' \equal{} \frac {1}{1 \plus{} e}$

or maybe i am wrong...    :([/quote]

That's the correct answer. :lol:


What is the number of square units in the area of the region determined by the following system?
\[ |x|\plus{}|y|\leq 4\]
\[ y \leq 0\]

The first region represents a square region with vertices $ (0,4)$, $ (4,0)$, $ (0,\minus{}4)$, and $ (\minus{}4,0)$. The upper triangle contains the coordinates with the positive $ y$-values required by the 2nd equation. The triangle resultant in the union of the two regions has an area of

$ \frac{1}{2}(4\minus{}(\minus{}4))(4\minus{}0)$

$ \equal{}\frac{1}{2}(8)(4)$

$ \equal{}4(4)$

$ \equal{}\fbox{16}$.


Let $x,y<0$ so that $7(x+y)=3(x^3-y^3)$. Prove that $x^2+y^2> \frac{7}{3}$.

cheers! :D :D :cool:

This one is not too difficult.Suppose that x^2+y^2<=7/3.So
        7(x+y)(x^2+y^2)=3(x^3-y^3)(x^2+y^2)>=7(x^3-y^3)
<=> x^3+y^3+x^2y+xy^2>=x^3-y^3.
<=> y^3+x^2y+xy^2=-y^3>0,this is unreasonable.
So x^2+y^2 must be larger than 7/3  :)

$x^2+y^2> \frac{7}{3}\Leftrightarrow x^2+y^2>  \frac{x^3-y^3}{x+y}\Leftrightarrow 2y^3+xy^2+x^2y<0$
$\Leftrightarrow 2y^2+xy+x^2>0\Leftrightarrow 2(y+\frac{x}{4})^2+\frac{7}{8}x^2>0$.


Six points on a circle are given. 4 of the chords joining pairs of the six points are selected at random. What is the probability that the four chords are the sides of a convex quadrilateral?

[hide]There are $\binom62$ $=15$ possible chords and $\binom{15}{4}=1365$ ways to choose 4 of them.  Of those, $\binom{6}{4}=15$ will form a convex quadrilateral.  Thus we have a probability of $\frac{15}{1365}=\frac{1}{91}$.
[/hide]


Find all polynomials P(x)  \in R[x] with property: if F(x) \in Z then F(x+1) in Z (Z is the set of intergers)

Suppose deg F > 1. In such case for sufficient large negative n segment [F(n),F(n+1)] contains arbitrary many integer points which correspond arguments n<x1<...<xm<n+1 ==> values F(x1-n),...,F(xm-n) must be integer. But F have definite number of integer values on [0,1].

So F(x)=kx+b ==> k is integer.


[hide="[color=blue"]I know the solution you are hinting at, but I like this one better[/color] :lol: ]
Let $f(x)=x^{2}x_{1}+(x+1)^{2}x_{2}+...(x+6)^{2}x_{7}$
Notice that $deg(f)=2$.
We are given: $f(1)=1, f(2)=12, f(3)=123$
Now we have everything we need, since 3 points determine a quadratic. In a quadratic, the second differences are a constant (trivial to see, even more trivial with super basic calc).
Knowing this fact, we easily determine $\boxed{f(4)=434}$[/hide]
edit: as of this minute, farenhajt has exactly 500 more posts than me!!

[quote="cincodemayo5590"][quote="me@home"][hide="[color=blue"]I know the solution you are hinting at, but I like this one better[/color] :lol: ]
Let $f(x)=x^{2}x_{1}+(x+1)^{2}x_{2}+...(x+6)^{2}x_{7}$
Notice that $deg(f)=2$.
We are given: $f(1)=1, f(2)=12, f(3)=123$
Now we have everything we need, since 3 points determine a quadratic. In a quadratic, the second differences are a constant (trivial to see, even more trivial with super basic calc).
Knowing this fact, we easily determine $\boxed{f(4)=434}$[/hide][/quote]

Nice solution! Though, what are "second differences"?[/quote]

Let $f(n)$ be some sequence.

Then $f_{1}(n) = f(n+1)-f(n)$ is the [b]first difference[/b], and $f_{2}(n) = f_{1}(n+1)-f_{1}(n)$ is the [b]second difference[/b].

It's a nice lemma that if $f(x)$ is a polynomial of degree $n$ then $f_{k}(x)$ is a polynomial of degree $n-k$.

The difference between f(1) and f(2) is a first difference, the difference between the first difference of f(1) and f(2) and the first difference of f(2) and f(3) is the second difference defined by f(1), f(2), f(3).

Ok, since everybody is elaborating here, let me do the same :)

[hide]$ai^{2}+b(i+1)^{2}+c(i+2)^{2}\equiv (i+3)^{2}$

$(a+b+c)i^{2}+(2b+4c)i+(b+4c)\equiv i^{2}+6i+9$

$a+b+c=1, 2b+4c=6, b+4c=9\implies a=1, b=-3, c=3$

$\sum_{i=1}^{7}(i+3)^{2}x_{i}=1\cdot 1-3\cdot 12+3\cdot 123=334$[/hide]

[b]me@home[/b] has an extra hundred in his answer ;)

oops about the extra hundred, no more mental math for me!? :)


[quote="zhaobin"]You just want to prove:
$y,z>0,yz \ge 1$
then:
$\frac{1}{\sqrt{1+8y}}+\frac{1}{\sqrt{1+8z}}<1$
I think it is not hard :)[/quote]
Try $y=\frac{1}{z}=117.$ :wink:

[quote="arqady"]
Try $y=\frac{1}{z}=117.$ :wink:[/quote]
You are right, :blush:
It's not as easy as I think.

zhaobin, i took the problem from a chinese mathematical magazine.
So I dont think it's easy..

This solution is by Harazi.

In fact, we will prove a stronger inequality:

Suppose $x, y, z$ are positive reals such that $xyz=1$. Then $\frac{1}{\sqrt{1+2x}}< 2$.

Your inequality is weaker because if you let $x = \frac{bc}{a^{2}},....$, then $xyz = 1$ and $LHS = \sum \frac{1}{\sqrt{1+8x}}< \sum \frac{1}{\sqrt{1+2x}}< 2$.

Let $X = \frac{1}{\sqrt{1+2x}}, ....$. Then $0 < X, Y, Z < 1$. Consider the contrary that $X+Y+Z \geq 2$. Hence, $x = \frac{1-X^{2}}{2X^{2}}, ...$. We will now show that $xyz = \prod{\frac{1-X^{2}}{2X^{2}}}< 1$ which is contradictory with the given condition $xyz = 1$.

Since $1+X < 2$, it follows that we only need to prove $\prod \frac{1-X}{X}<2$. Since $0 < X, Y, Z<1$ and $X+Y+Z \geq 2$, it is easy to see that there exist positive $a, b, c$ such that $a+b+c \geq 1$ and $X = b+c, ...$. Replacing this to the inequality, we obtain it is sufficient to prove $(a+b)^{2}(b+c)^{2}(c+a)^{2}> abc (*)$.

Notice that

$[(a+b)(b+c)(c+a)]^{2}\geq [\frac{8}{9}(a+b+c)(ab+bc+ca)]^{2}$

$\geq \frac{64}{27}abc(a+b+c)^{3}> \frac{64}{27}abc > abc$.

Thus $(*)$ is proved, and we are done.

Nice solution.And I think there is a small type in the proof,but is not important. :) I think we can also understand it.

[quote="treegoner"]

Suppose $x, y, z$ are positive reals such that $xyz=1$. Then $\frac{1}{\sqrt{1+2x}}< 2$.

[/quote]

Very nice solution harazi .
I have a question..Is $2$ the best constant in the above inequality ??

very good solution! thank to treegoner and harazi..  :)

[quote="RDeepMath91"][quote="silouan"][quote="treegoner"]

Suppose $x, y, z$ are positive reals such that $xyz=1$. Then $\frac{1}{\sqrt{1+2x}}< 2$.

[/quote]

Very nice solution harazi .
I have a question..Is $2$ the best onstant in the above inequality ??[/quote]

I think that must be 1 and it's much harder... (IMO problem, isn't it ??)[/quote]

Silouan meant if 2 is the best constant at the right side of the inequality.

And also, actually i think the inequality of imo 2001 is much easier than this one. :D

[quote="silouan"][quote="anthony19891019"]
Silouan meant if 2 is the best constant at the right side of the inequality.
[/quote]

Exactly I meant the number $2$ in the RHS of the inequality .
By Harazi's solution I think it can be improved  :maybe:[/quote]

Try $a=b \rightarrow 0$.

[quote="Lovasz"]
If I'm right, the last inequality is true, \[1527abc+128\cdot{a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}\over abc}\ge 16(a^{3}+b^{3}+c^{3}).\] [/quote]
I think you are wrong:$a>0,$ $b>0$ and $c\rightarrow+\infty$

I think, the  harazi's  reasoning gives
that following inequality holds for all $\alpha\geq\frac{3}{2},$ $xyz=1$ and $\{x,y,z\}\subset(0,+\infty)$
$\frac{1}{\sqrt{1+\alpha x}}+\frac{1}{\sqrt{1+\alpha y}}+\frac{1}{\sqrt{1+\alpha z}}\leq2.$
But Vasc approves, that for $\{x,y,z\}\subset(0,+\infty)$ such that $xyz=1$ holds:
$\frac{1}{\sqrt{1+1.25x}}+\frac{1}{\sqrt{1+1.25 y}}+\frac{1}{\sqrt{1+1.25z}}\leq2.$

[quote="treegoner"]This solution is by Harazi.

In fact, we will prove a stronger inequality:

Suppose $ x, y, z$ are positive reals, then

$ \frac {x}{\sqrt { x^{2} + 2yz }} + \frac {y}{\sqrt {y^{2} + 2zx}} + \frac {z}{\sqrt {z^{2} + 2xy}} < 2.$[/quote]$ \sum{\frac {x}{\sqrt {x^2 + 2yz}} < \sum{\frac {x^2(y + z) + x\left(y^2 + z^2\right) }{x^2(y + z) + y^2(z + x) + z^2(x + y)} = 2.}}$

[hide="Proof."]$ \left(x^2 + 2yz\right)\left[x(y + z) + y^2 + z^2\right]^2 - \left[x^2(y + z) + y^2(z + x) + z^2(x + y)\right]^2$

$ = yz\Big[2x(y + z)\left(y^2 + z^2\right) + 2y^4 - y^3z + 2y^2z^2 - yz^3 + 2z^4\Big]$

$ = yz\left[2x(y + z)\left(y^2 + z^2\right) + (y - z)^2\left(y^2 + yz + z^2\right) + \left(y^2 + z^2\right)^2\right] > 0.$
[/hide]

[quote="arqady"]Vasc approves, that following inequality holds

$ \frac {x}{\sqrt { 4x^{2} + 5yz }} + \frac {y}{\sqrt {4y^{2} + 5zx}} + \frac {z}{\sqrt {4z^{2} + 5xy}} \leq 1$

for $ \{x,y,z\}\subset(0, + \infty)$.[/quote]Using the Cauchy inequality,

$ \left(\sum{\frac {x}{\sqrt { 4x^{2} + 5yz }}}\right)^2\leq\sum{x(y + z)}\sum{\frac {x}{(y + z)\left(4x^2 + 5yz\right)}},$

we only need to prove

$ \sum{yz}\sum{\frac {x}{(y + z)\left(4x^2 + 5yz\right)}}\leq\frac {1}{2}.$

[hide="Proof."]$ \frac {1}{2} - \sum{yz}\sum{\frac {x}{(y + z)\left(4x^2 + 5yz\right)}}\equiv\frac {xyzF(x,y,z)}{2(y + z)(z + x)(x + y)\left(4x^2 + 5yz\right)\left(4y^2 + 5zx\right)\left(4z^2 + 5xy\right)}$.

$ F(x,y,z) = F(x,x + s,x + s + t)$

$ = 396\left(s^2 + st + t^2\right)x^4 + (2s + t)\left(403s^2 + 403st + 778t^2\right)x^3$

$ + \left(532s^4 + 1064s^3t + 2679s^2t^2 + 2147st^3 + 490t^4\right)x^2$

$ + 5(2s + t)\left(13s^4 + 26s^3t + 121s^2t^2 + 108st^3 + 20t^4\right)x$

$ + 2s(s + t)\left(4s^4 + 8s^3t + 109s^2t^2 + 105st^3 + 25t^4\right)\geq 0$,

which is clearly true for $ x\leq y\leq z$.[/hide]

[quote="treegoner"]
Since $ 1 \plus{} X < 2$, it follows that we only need to prove $ \prod \frac {1 \minus{} X}{X} < 2$. [/quote]

   " $ \prod \frac {1 \minus{} X}{X} < 2$. " is wrong .
    $ \prod \frac {1 \minus{} X}{X^2} < 1$. is  correct.

[quote="treegoner"]This solution is by Harazi.

In fact, we will prove a stronger inequality:

Suppose $ x, y, z$ are positive reals such that $ xyz \equal{} 1$. Then $ \frac {1}{\sqrt {1 \plus{} 2x}} < 2$.

Your inequality is weaker because if you let $ x \equal{} \frac {bc}{a^{2}},....$, then $ xyz \equal{} 1$ and $ LHS \equal{} \sum \frac {1}{\sqrt {1 \plus{} 8x}} < \sum \frac {1}{\sqrt {1 \plus{} 2x}} < 2$.

Let $ X \equal{} \frac {1}{\sqrt {1 \plus{} 2x}}, ....$. Then $ 0 < X, Y, Z < 1$. Consider the contrary that $ X \plus{} Y \plus{} Z \geq 2$. Hence, $ x \equal{} \frac {1 \minus{} X^{2}}{2X^{2}}, ...$. We will now show that $ xyz \equal{} \prod{\frac {1 \minus{} X^{2}}{2X^{2}}} < 1$ which is contradictory with the given condition $ xyz \equal{} 1$.

Since $ 1 \plus{} X < 2$, it follows that we only need to prove $ \prod \frac {1 \minus{} X}{X} < 2$. Since $ 0 < X, Y, Z < 1$ and $ X \plus{} Y \plus{} Z \geq 2$, it is easy to see that there exist positive $ a, b, c$ such that $ a \plus{} b \plus{} c \geq 1$ and $ X \equal{} b \plus{} c, ...$. Replacing this to the inequality, we obtain it is sufficient to prove $ (a \plus{} b)^{2}(b \plus{} c)^{2}(c \plus{} a)^{2} > abc (*)$.

Notice that

$ [(a \plus{} b)(b \plus{} c)(c \plus{} a)]^{2}\geq [\frac {8}{9}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)]^{2}$

$ \geq \frac {64}{27}abc(a \plus{} b \plus{} c)^{3} > \frac {64}{27}abc > abc$.

Thus $ (*)$ is proved, and we are done.[/quote]
[img]http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=15513[/img]

[quote="treegoner"]Suppose $ x, y, z$ are positive reals such that $ xyz \equal{} 1$. Then $ \sum{\frac {1}{\sqrt {1 \plus{} 2x}} < 2}$. [/quote]Notate $ \frac {1}{\sqrt {1 \plus{} 2x}} \equal{} 1 \minus{} u,\frac {1}{\sqrt {1 \plus{} 2y}} \equal{} 1 \minus{} v,\frac {1}{\sqrt {1 \plus{} 2z}} \equal{} 1 \minus{} w,$ where $ 0 < u,v, w < 1$.

Hence, $ x \equal{} \frac {u(2 \minus{} u)}{2(1 \minus{} u)^2},$ etc. $ xyz \equal{} 1\Longrightarrow uvw(2 \minus{} u)(2 \minus{} v)(2 \minus{} w) \equal{} 8(1 \minus{} u)^2(1 \minus{} v)^2(1 \minus{} w)^2.$

We need to prove that $ (1 \minus{} u) \plus{} (1 \minus{} v) \plus{} (1 \minus{} w) < 2\Longleftrightarrow u \plus{} v \plus{} w > 1.$

In fact, $ 8(1 \minus{} u)^2(1 \minus{} v)^2(1 \minus{} w)^2\geq uvw(3 \minus{} u)(3 \minus{} v)(3 \minus{} w)>uvw(2 \minus{} u)(2 \minus{} v)(2 \minus{} w)$

holds for all positive numbers $ u,v,w$ with $ u \plus{} v \plus{} w\leq 1$. See : http://www.mathlinks.ro/Forum/viewtopic.php?t=229647

[quote="treegoner"]This solution is by Harazi.

In fact, we will prove a stronger inequality:

Suppose $ x, y, z$ are positive reals such that $ xyz \equal{} 8$. Then $ \sum\frac {1}{\sqrt {1 \plus{} x}} < 2$. [/quote][b]Lemma[/b](Ming-Bin Jiang$ ^{[1]}$) Suppose $ a, b$ are positive reals such that $ ab \leq2,$ then

$ \frac {1}{\sqrt {1 \plus{} a^2}} \plus{} \frac {1}{\sqrt {1 \plus{} b^2}} \leq \frac {2}{\sqrt {1 \plus{} ab}},$

with equality if and only if $ a \equal{} b.$

[b]New Proof[/b] Squaring both sides,

$ \frac {2}{\sqrt {(1 \plus{} a^2)(1 \plus{} b^2)}}\leq \frac {4}{1 \plus{} ab} \minus{} \frac {1}{1 \plus{} a^2} \minus{} \frac {1}{1 \plus{} b^2}$

$ \left[ \equal{} \frac {2 \minus{} 2ab \plus{} (a^2 \plus{} b^2)(3 \minus{} ab) \plus{} 4a^2b^2}{(1 \plus{} ab)(1 \plus{} a^2)(1 \plus{} b^2)}\geq\frac {2 \minus{} 2ab \plus{} 2ab(3 \minus{} ab) \plus{} 4a^2b^2}{(1 \plus{} ab)(1 \plus{} a^2)(1 \plus{} b^2)} \equal{} \frac {2(1 \plus{} ab)}{(1 \plus{} a^2)(1 \plus{} b^2)} > 0.\right]$

But this is true since

$ \left(\frac {4}{1 \plus{} ab} \minus{} \frac {1}{1 \plus{} a^2} \minus{} \frac {1}{1 \plus{} b^2}\right)^2 \minus{} \frac {4}{(1 \plus{} a^2)(1 \plus{} b^2)}$

$ \equal{} \frac {(a \minus{} b)^2[8 \minus{} 6ab \plus{} 12a^2b^2 \minus{} 6a^3b^3 \plus{} (a^2 \plus{} b^2)(3 \minus{} ab)^2]}{(1 \plus{} ab)^2(1 \plus{} a^2)(1 \plus{} b^2)}$

$ \geq\frac {(a \minus{} b)^2[8 \minus{} 6ab \plus{} 12a^2b^2 \minus{} 6a^3b^3 \plus{} 2ab(3 \minus{} ab)^2]}{(1 \plus{} ab)^2(1 \plus{} a^2)(1 \plus{} b^2)} \equal{} \frac {4(a \minus{} b)^2(2 \minus{} ab)}{(1 \plus{} a^2)(1 \plus{} b^2)}\geq0.$

[b]Proof of Harazi's inequality[/b] WLOG, we may assume $ z \equal{} \max\{x,y,z\}.$

From $ xyz \equal{} 8$ we have $ xy\leq4.$ Using the lemma,

$ \frac {1}{\sqrt {1 \plus{} x}} \plus{} \frac {1}{\sqrt {1 \plus{} y}} \plus{} \frac {1}{\sqrt {1 \plus{} z}}\leq\frac {2}{\sqrt {1 \plus{} \sqrt {xy}}} \plus{} \cfrac{1}{\sqrt {1 \plus{} \cfrac{8}{xy}}}.$

This comes down to

$ \frac {2}{\sqrt {1 \plus{} c}} \plus{} \frac {c}{\sqrt {c^2 \plus{} 8}} < 2,$

where $ c \equal{} \sqrt {xy}.$ Square it,

$ \frac {4c}{\sqrt {(c \plus{} 1)(c^2 \plus{} 8)}} < 4 \minus{} \frac {4}{c \plus{} 1} \minus{} \frac {c^2}{c^2 \plus{} 8} \equal{} \frac {c(3c^2 \minus{} c \plus{} 32)}{(c \plus{} 1)(c^2 \plus{} 8)},$

which is clearly true for

$ (3c^2 \minus{} c \plus{} 32)^2 \minus{} 16(c \plus{} 1)(c^2 \plus{} 8) \equal{} 9c^4 \minus{} 22c^3 \plus{} 177c^2 \minus{} 192c \plus{} 896$

$ \equal{} (2c^2 \minus{} 5c \plus{} 27)(4c^2 \minus{} c \plus{} 33) \plus{} c^4 \minus{} 2c^2 \plus{} 5 > 0.$

[b]Remark[/b] Harazi's inequality be used as the last problem of Jiangxi College Entrance Examination 2008.

[b]Reference[/b]

[1] Ming-Bin Jiang, Source and Generalization of a Problem of China Western Mathematical Olympiad (Chinese), Studies in Middle School Math (Nanchang), 2006, No. 2,  48-50.

[2] Marcin Mazur, Problem 10944, Amer. Math. Monthly, Vol. 109 (2002), No. 5, 475.


Suppose that every positive integer can be written as the product of two positive integers belonging to a set $A$. Then for all $n$ there exists a positive integer that can be written in at least $n$ ways as a product of two elements of $A$.

I think, that A is subset of N.
Prime p $p=xy\Longrightarrow x=1,y=p \ or \ x=p,y=1$ therefore $1\in A$ and $p\in A \ \forall p\in P.$
Let $B_{n}=\{k\le n |p^{k}\in A$. It is easy to show, that $B_{n}$ had at least [(n+1)/2] members. Let $C_{n}=\{x,y|x\le y, x,y\in B_{n}\}$. C had $m(m+1)/2>\frac{n^{2}}{8}$ (m number of members $B_{n}$) members. Because $x+y\le 2n$ from Diriclet's princip exist M<=2n, suth that we have at least n/16 representation $M=x+y,x\ley, x,y\in B_{n}$. It mean
exist M<=16n, suth that $p^{M}=x*y,x\le y, x,y\in A$ had at least n representation.


inequalities
geometry
incenter
inequalities proposed

The inequality question 6 in 1983 can be transformed to the form

AI^2/b^2 + BI^2/c^2 +CI^2/a^2 >=1  where I denotes the incentre of triangle ABC.
Is there a purely geometric proof of this result?

Eum... i think not...

would surprise me if you could solve an inequality purely geometric...

there are solutions doing the greater part geometric, but you still need to do the trick with inequalities I guess