geometry
probability
AMC
AIME
number theory
graph theory

that's not necessarily a bad thing
if you can ever do almost all the problems in a book easily, find a harder book
you're meant to struggle with AoPS
thats what makes it fun

try using each of the independent books. like intro to number theory, geometry, counting and probility, etc. These books explain the solutions a bit deeper.

yeah, it's hard to concentrate when you get 50% of the problems.

I agree with indianamath.
If you don't have a problem solving background, you should probably do more in-depth books.  AoPS 1 is very fast-paced.

Geometry and algebra are really good, number theory and probability should have been explored more, I think.

I'd recommend getting Intro to Number Theory and Intro to Counting/Probability; if you go through them step by step and really try solving all the problems, you'll be very well prepared for even the AIME.

yeah, it does take getting used to do AoPS. I finished 1/4 of the book, became increasingly frustrated with 75% of the problems, then I took some time off trying to calm down, then try again with more patience. It definitely works for me. Good luck!  :D

[quote="cdymdcool"]I can't even work out 50% of the problems on AoPS's Volume one!
 :([/quote]
if you want to prepare for the amcs and aime you dont need to do certian sections in volume one or two (like matrices, graph theory, paramectric equations..)

[quote="simulacrum"]intro to NT and intro to C and P are good enough for the AIME as well? Don't you mean the early problems on the AIME?[/quote]
foundation for them

I don't think it's necessary to go buy all the books in the Intro series.  Sure, that will help, but you can get a lot out of Volume 1.  The fact that you can't solve half of the problems doesn't mean the book is too hard for you, it just means there's a lot you can learn.  Take your time on the problems, and don't expect to find a solution immediately.  The most rewarding problems are the ones that stump you for a long, long time, but then you finally figure them out.  Try to get something out of every problem - this means work on a problem until you solve it or are completely convinced that you're not going to solve it, and then learn the actual solution so that you know it well enough to explain to someone else.  Good luck!

I don't know.

Right after MOP, I started looking through AoPS Volume 2.

It's crazy, in chapters 1 and 2 I had trouble grasping the concepts behind half the questions, never mind actually getting the answers. Just keep bashing at it, and read the solutions if you're really stuck (and it will happen). You'll get there eventually.

I have AoPS volume one. It [i]is[/i] challenging. I suggest that you read each and every explanation they give. That always helps.

yeah..
i almost finished volume one..
and i get most of it..
the first time i struggled a LOT!!
but now i get it... especially now that i'm doing the book at an academy thing..
i tried looking at vol. 2, but it was crazy.
i couldn't do anything..

[quote=diophantient][quote="paladin8"]:surrender: :stretcher: :help:

Since when could you only have 3 smilies in a message?[/quote]
:huh:
:?
:|
:) No one ever said that[/quote]

:) :( :D :| :P

Yes, I agree. I am currently doing Volume I which does get really challenging after a certain point, but one strategy I always use is if I didn't understand how to do the problem, I will look through the solution, understand it, then do the problem again without looking at the solution.

If you do the problem right after looking at the solution you'll still remember the solution so there's no real benefit. If you do it like a few days after when you don't remember the solution, if you're able to solve it, that means progress!

[quote=CantonMathGuy][quote=whiteawesomesun][quote=Calculus123]How do you keep bashing, and not lose motivation?

I can barely do 30 percent of Volume 2 problems...[/quote]

I feel like I'm experiencing a similar thing. I just got Volume 2, and some questions, especially the proof section of the combinations chapter is really driving me crazy. I am not used to this content yet, as I have only done AMC 10 (12- at most)-level math before. How can we make a successful transition into harder math?[/quote]

Are you talking about proofs? Because proofs aren't really different from problems you've been doing right now.

Pick your favorite AMC 10/12 problem and write a solution for it. Bam. You now have a proof.[/quote]

No, what I mean is that, I am not used to the difficulty level of Problems>AMC.

I just have a hard time attempting these problems due to many things, one of which is a lack of patience, experience, and how to start the problem (also content knowledge). I just want some tips regarding how I can proceed from where I am now.

Don't worry the marked problems are usually beyond the level of AMC 10. Although I don't want to disappoint you, it's a little problematic if you can't solve the unmarked problems and maybe at least half of the marked ones.

[quote=whiteawesomesun]What do you mean "marked" problems?[/quote]

A diagonal line with a symbol at the end next to a problem means it's difficult.

Okay, but I am asking how can we overcome this frustration?

[quote=whiteawesomesun]Well, will doing AIME problems necessarily give me more stamina? I'm not actually sure where the roots of my problem are, it's perhaps just my personality :([/quote]

Yes.

Another thing you can do to increase stamina is to do long bashes, but I wouldn't suggest that for practice purposes (other than on the contest itself)

[quote=CantonMathGuy]Hm... Your issue sounds like you don't spend enough time on problems: if you can't solve it in ~5 min, you go to the solution (lack of concentration - you also mentioned lack of patience)

To remedy this, I would suggest trying the early problems on AIMEs: this will help you develop the ability to concentrate on a single problem longer.[/quote]

Yeah this becomes very important. Being a college freshman, one of my (math) classes generally assigns a problem set with 6-7 problems a week, and it's normal to spend 2 hours on a problem (not just for me, but for the majority of students), so I know building this stamina is a good idea for student life in general.

As far as the AIME goes, often times they require the same skills//computations as easier problems but are more involved or require some sort of clever trick. This, I've learned, comes with practice. Good luck.

We all start somewhere, and it's important to know that you aren't alone. Do you think people who go to Nationals for Mathcounts, qualify for the USAJMO, IMO, and more competitions are born that way? We all either started where you are, or are where you are now. So don't push yourself down. You can't get better without challenging yourself, and AoPS gives you both the community(and family) to get better and improve :).

wait who are people talking to, since the OP was made in 2007.

recent posters who bumped the thread to ask questions

[quote=shiningsunnyday]@whiteawesomesun

Rome was not built in a day.[/quote]

Ok SSD.

[quote=Konigsberg]Lol it's really hard... I'm an IMO medallist and can only get 70-90% of vol 2, so don't expect to get everything...[/quote]

This was the bump....


The length of a rectangular prism is four times the height, and the width is twice the height. If a space diagonal is 42 cm, find the volume of the prism. Express your answer to the nearest cubic centimeter.


\[ {\rm{Solve for }}x: 4\left( {x^2  \plus{} \frac{1}{{x^2 }}} \right) \plus{} 8\left( {x \minus{} \frac{1}{x}} \right) \equal{} 29
\]

[quote="sanbittu"]
\[ {\rm{Solve for }}x: 4\left( {x^2 \plus{} \frac {1}{{x^2 }}} \right) \plus{} 8\left( {x \minus{} \frac {1}{x}} \right) \equal{} 29
\]
[/quote]
[hide="Solution"]
Allow $ \frac {1}{x} \minus{} x \equal{} y$.  Notice that $ y^2 \plus{} 2 \equal{} \frac {1}{x^2} \plus{} x^2$.  Hence, we obtain that
\[ 4(y^2 \plus{} 2) \plus{} 8y \equal{} 29\implies 4y^2 \plus{} 8y \minus{} 21 \equal{} 0 \equal{} (2y \plus{} 7)(2y \minus{} 3) \equal{} 0\implies y \equal{} \frac { \minus{} 7}{2}, \frac {3}{2}
\]
Solving for both, we notice that \[ x \minus{} \frac {1}{x} \equal{} \frac {3}{2}\implies 2x^2\minus{} 2 \minus{} 3x \equal{} 0\implies (2x\plus{}1)(x\minus{}2)\equal{}0\implies x\equal{}\frac{\minus{}1}{2}, 2\] or \[ x \minus{} \frac {1}{x} \equal{} \frac { \minus{} 7}{2}\implies 2x^2 \minus{} 2 \plus{} 7x \equal{} 0\implies x \equal{} \frac { \minus{} 7\pm \sqrt {65}}{4}\]  Therefore, our solutions for $ x$ are $ \boxed{x \equal{} \frac { \minus{} 7\pm \sqrt {65}}{4}, \frac{\minus{}1}{2}, 2}$. [/hide]

EDIT: Thanks, [b]kunny[/b]!  My solution has been edited.  Merry Christmas  :)

Maybe you wanted to set $ y\equal{}x\minus{}\frac{1}{x}$

[hide="Different Way"]
$ x^2 \plus{} \frac {1}{x^2} \equal{} \frac {x^4 \plus{} 1}{x^2}, x \minus{} \frac {1}{x} \equal{} \frac {x^3 \minus{} x}{x^2}$

$ \frac {4x^4 \plus{} 4}{x^2} \plus{} \frac {8x^2 \minus{} 8x}{x^2} \equal{} 29 \implies 4x^4 \plus{} 8x^3 \minus{} 29x^2 \minus{} 8x \plus{} 4 \equal{} 0$

Rational Roots Test deduces this to $ (x \minus{} 2)(2x \plus{} 1)(2x^2 \plus{} 7x \minus{} 2)$
Therefore $ x \equal{} \boxed{2, \minus{} \frac {1}{2}, \frac {\sqrt{65}\minus{}7}{4}, \minus{}\frac{7\plus{}\sqrt{65}}{4}}$
[/hide]
Merry Christmas! :)


Alan earns 4 dollars more in 4 hours than Joy earns in 3 hours. Joy earns 0.50 cents more in 4 hours than Alan earns in 5 hours. How much does Alan earn per hour?

When a positive interger x is divided by a positive integer y, the quotient is u and the remainder is v, where u and v and integers. What is the remainder when x+2uy is divided by y?

Number one:
[hide="Hint"]
For this type of problem the hardest part is turning the English into math.

If we let $r_A$ be how much Alan earns in one hour and $r_J$ be how much Joy earns in one hour, then we can make some equations.

In four hours Alan earns $4r_A$ dollars.  In three hours Joy earns $3r_J$ dollars.

See if you can finish up the rest from there...you will need to make two equations and then solve.
[/hide]

Number two:
[hide="Hint"]
We are given that: $\frac xy=uy+v$ and we want to evaluate the remainder of $\frac{x+2uy}{y}$.  What can you do with the second one?
[/hide]

Well, for the first one, I still have no idea of what to do...

For the second one, can you break the second fraction into the following:

$\frac{x}{y}+ \frac{2uy}{y}$ = $uy+v+2u$

Which means that the remainder is v?

[hide]1. Let a be Alan's hourly wage, and j be Joy's hourly wage. From the given information, we can set up to equations:

$4a=3j+4$ (1)
$4j=5a+0.5$ (2)

Adding these two together, we get
$5a+4j=3j+5a+4.5$ or
$j=a+4.5$

Substituting this into (1), we get $4a+18=5a+0.5$, or $a=17.5$. Thus, Alan's hourly wage is 17.50 dollars.

2. From the given information, we can write that $x\equiv v(mod y)$, and $x+2uy\equiv v+2uy(mod y)$. Applying a modular arithmetic identity, this is the same as $x(mod y)+2uy(mod y).$ Obviously, y is divisible by y, as is u, so $2uy\equiv 0(mod y)$, making the remained of x + 2uy = v.[/hide]

Are you posting these as challenges or as problems you need help with?

Hint:

[hide]Let A be the dollars Alan earns per hour, and let J be the dollars Joy earns per hour.

Make equations out of the statements and solve.[/hide]

for number one[hide]let x be alan's wage and y be joy's wage.
4x=3y+4
and
4y=5x+.5
lets figure out what y is in terms of x....
y=(5/4)x+(1/8)
substitute
4x=3((5/4)x+(1/8))+4
4x=(15/4)x+(35/4)
(1/4)x=(35/4)
x=35
y=(5/4)(35)+(1/8)
y=43.875
check
4(35)=3(43.875)+4
140=135.625
either math is screwy or i did something wrong

probably the latter
some correct me i looked at it a few times and i don't see it

[/hide]

o yes that would do it.  thx i always make a stupid mistake :D


nice, but what's ascii and the point of this?  :?

ascii: Its an abbreviation which I can't remember right now. Anyway its basically a code using the keys on your keyboard

point: Because im bored. <.<

sorry but it is supposed to be an animal of some sort...? and you might want to disable smiles.

ASCII stands for American Standard Code for Information Interchange.
Why do they call those pictures ASCII? I always thought ASCII codes are the corresponding numbers to each character especially in computer programming.

yeah i dunno, im guessing b/c ascii is used to display the graphics and text, they associated the actual code with the stuff its programmed to display?

Put it in "code" tags, or try changing the font to fixedsys.

Anyways, ASCII is pretty cool.

Here is a site I found.
http://www.network-science.de/ascii/

I'm not sure exactly why it's called ascii art, but my guess is that it has to do with the processors involved.  Normally ascii art is displayed in a font which has a uniform size for all the letters, a property which is typical of more primitive word processors.  These processors usually store content in ascii mode, as opposed to byte mode or whatever the other one is called.


Prove that:$a^{2}+b^{2}+c^{2}+3 \geq a+b+c+ab+bc+ca$
where $a;b;c$ are positive numbers satisfying $abc=1$

If $a=\min\{a,b,c\}$. Denote $f(a,b,c)=a^{2}+b^{2}+c^{2}+3-a-b-c-ab-bc-ca$. We have $f(a,b,c)-f(a,\sqrt{bc},\sqrt{bc})=(\sqrt{b}-\sqrt{c})^{2}(b+c+2\sqrt{bc}-1-a)\geq 0$. Now, We only need prove $f(a,t,t)\geq 0$ with $at^{2}=1$. But it is easy!

Let a=1/bc
LHS-RHS, we have
$\frac{1}{b^{2}c^{2}}+b^{2}+c^{2}+3-\frac{1}{bc}-b-c-\frac{1}{b}-\frac{1}{c}-bc\geq 0$
$\iff(\frac{1}{bc}-1)^{2}+(b-c)^{2}+(1-b)(1-c)+(1-\frac{1}{b})(1-\frac{1}{c})+1\geq 0$
$\iff(\frac{1}{bc}-1)^{2}+(b-c)^{2}+\frac{(bc+1)(1-b)(1-c)}{bc}+1\geq 0$
since the it's symmetric, there are only two situations we need to consider:
a>=1,b<=1,c<=1 or a<=1,b>=1,c>=1;

PS. using the same method, it's easy to prove a somewhat similar inequality
$a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$ when abc=1

EDIT: typo and misusing "stronger"

[quote="andy_tok"]

PS. using the same method, it's easy to prove a stronger inequality
$a^{2}+b^{2}+c^{2}+3>2(ab+bc+ac)$ when abc=1[/quote]
What about if $a=b=c=1$, I think It is $a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$ . Am I wrong?  :maybe:

You're right.  :blush: I just edited my post.
Thanks

[quote="andy_tok"]

 it's easy to prove a [b]stronger inequality[/b]
$a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$ when abc=1
[/quote]
i.e. $ab+bc+ca\geq a+b+c$ ? Can you post prove?

[quote="andy_tok"]PS. using the same method, it's easy to prove a stronger inequality
$a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$ when abc=1

[/quote]
$\sum{a^{3}}+3abc\geq \sum{a^{2}b}$ (Schur)
Let $a=x^{3}, b=y^{3}, c=z^{3}$ We have $xyz=1$
So  $\sum{x^{6}}+3(xyz)^{2}\geq \sum{x^{4}y^{2}}$
Because $(4,2,0)\succ\(3,3,0)$
So we have $\sum{x^{4}y^{2}}\geq 2\sum{(xy)^{3}}$ (Muirhead)
That is $a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$

Sorry, that inequality won't hold. i think I misused the word "stronger" here.
What I was referring to is that $a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$
and $a^{2}+b^{2}+c^{2}+3\geq 2(a+b+c)$(AM-GM) will automatically give the inequality in the original post. Sorry for misleading

[quote]
$\sum{a^{3}}+3abc\geq \sum{a^{2}b}$ (Schur)
Let $a=x^{3}, b=y^{3}, c=z^{3}$ We have $xyz=1$
So  $\sum{x^{6}}+3(xyz)^{2}\geq \sum{x^{4}y^{2}}$
Because $(4,2,0)\succ\(3,3,0)$
So we have $\sum{x^{4}y^{2}}\geq 2\sum{(xy)^{3}}$ (Muirhead)
That is $a^{2}+b^{2}+c^{2}+3\geq 2(ab+bc+ac)$[/quote]
ya this solution is well-known, here is another one which may be familiar to lots of people too.

Let a=1/bc, use pretty much the same method
LHS-RHS, we have
$(\frac{1}{bc}-1)^{2}+(b-c)^{2}+2(1-\frac{1}{b})(1-\frac{1}{c})\geq 0$

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=128441 .

  darij


If two quadrilaterals $ ABCD$ and $ PQRS$ have angles $ A, B, C, D$ equal to angles $ P, Q, R, S$ respectively and $ AB\equal{}PQ, CD\equal{}RS$ and if $ AD$ is not parallel to $ BC$, prove that the quadrilaterals are congruent

Hint: Use this easy lemma - let $ \angle AOB \equal{} \alpha$ be fixed, and a line $ l$ variable meet $ AO$ at $ P$ and $ BO$ at $ Q$, wich must have the property that always $ \angle OPQ$ is the same. Than there is just one line $ l$ for wich $ PQ \equal{} x$ for some $ x \ge 0$.

$ \mbox{Let }\{E\}\equal{}AD\cap BC; \ \{T\}\equal{}PS\cap RQ, \mbox{ then: } \\
\triangle{EAB}\sim \triangle {TPQ}\mbox{ and } \frac{|AB|}{|PQ|}\equal{}1\Rightarrow \triangle{EAB}\equiv \triangle{TPQ}\Rightarrow \\
\Rightarrow |EA|\equal{}|TP| \mbox{ and }|EB|\equal{}TQ|. \ \ (1)$
$ \mbox{Analog } \triangle{ECD}\equiv\triangle{TRS}\Rightarrow |EC|\equal{}|TR|\mbox{ and }|RD|\equal{}|TS|. \ \ (2)\\
(1) \mbox{ and } (2)\Rightarrow |AD|\equal{}|PS| \mbox{ and }|BC|\equal{}|QR|.$

[quote="Rijul saini"]If two quadrilaterals $ ABCD$ and $ PQRS$ have angles $ A, B, C, D$ equal to angles $ P, Q, R, S$ respectively and $ AB \equal{} PQ, CD \equal{} RS$ and if $ AD$ is not parallel to $ BC$, prove that the quadrilaterals are congruent[/quote]

___________
this is my answer   :)


Auggie spent all of his money in 5 stores. In each store, he spent $ \$1$ more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?

[hide="Answer"]$ \$62$ dollars[/hide]

[quote="alanchou"][hide="working backwards again"]
0 -> 0
1 -> 2
2 -> 6
3 -> 14
4 -> 30
5 -> $ \boxed{62}$
[/hide][/quote]
Gee.. I feel stupid.. okay. I got it. Thanks.


calculus
integration
LaTeX
logarithms
calculus computations

Hello, $ \text{\LaTeX}$ please!

Also, is that the inverse hyperbolic tangent?

So, is it $ \int \frac{x^\frac37 }{\tanh x} \, dx$ or $ \int x^\frac37 \tanh^{-1} x \, dx$?

the second part x^(3/7) multiplied by tanh^-1(x)

For x < 1, $ \tanh^{\minus{}1}x \equal{} \frac12 \ln \left|\frac{1\plus{}x}{1\minus{}x}\right|$, and so your antiderivative is

$ P \equal{} \frac12 \int x^{3/7} \ln \left|\frac{1\plus{}x}{1\minus{}x}\right|\,dx \equal{} \frac12 \cdot \frac{7}{10} x^{10/7} \ln \left|\frac{1\plus{}x}{1\minus{}x}\right| \minus{} \frac12 \cdot \frac{7}{10} \int x^{10/7} \cdot \frac{2}{1\minus{}x^2}\,dx$,

and you can now proceed from here.


What is $ \text{Periplanar Geometry}$ and how does it apply to $ E2$ reactions (during which anti-periplanar geometry is favoured).

The following picture might be relevant:

Anti Periplanar geometry is one of the essential requirements between the leaving groups.consider dehydrohalogentaion of $ \text{2\minus{}Bromo Butane}$
say the reagent used is $ \text{MeO}^\minus{}/\text{MeOH}/\Delta$([b]@Carcul[/b] is this good enough fr $ \text{E}_2$)  :?: ) ensuring $ \text{E}_2$ now if the $ \text{H}_{3}$ and $ \text{Br}$ are antiperiplanar(same plane) then just the moment the $ \text{H}$ is being pulled away since the two $ \text{p}$ orbitals of the $ \text{C}_2$ and $ \text{C}_3$can overlap best leading to perfect $ \pi$ bond.

[quote="Pardesi"]@Carcul is this good enough fr[/quote]

For a secondary halide, it is enough.

[quote="Pardesi"]are antiperiplanar(same plane)[/quote]

Both anti periplanar and periplanar are in the same plane, but the direction of the orbitals is opposite. See the picture I posted above.

So what are the evidences that support the mechanism of $ E2$ reactions?

In fact I don't think I quite get how the need for $ \text{antiperiplanar geometry}$ supports the proposed mechanism.

[quote="BanishedTraitor"]So what are the evidences that support the mechanism of E2 reactions?[/quote]

The distinction between antiperiplanar and periplanar is made on basis of the alkene formed.


function
geometry
geometric transformation
vector
real analysis
real analysis solved

Prove that for all n there exists a continuous function from $R^2$ into the same set such that $ f^n$ is unbounded, but $f^{n+1}$ is bounded. Nice problem...

I haven't checked this thoroughly, but it looks like it works.

Consider the regions $\mathcal R_1=\{(x,y)||x|\le 1\},\ \mathcal R_2=\{(x,y)|1<|x|\le 2\},\ldots,\mathcal R_n=\{(x,y)|n-1<|x|\le n\},\ \mathcal R_{n+1}=\{(x,y)||x|>n\}$.

Now pu $f(x,y)=(0,0),\ \forall (x,y)\in\mathcal R_1$. For the next regions, consider this: assume a ray from $O=(0,0)$ through $(x,y)\in\mathcal R_2$ cuts the lines $x=1,x=2$ in $A,B$ respectively. Then define $f(x,y)$ to be the translation of $(x,y)$ by the vector $\overrightarrow{BA}$. Do the same for the other regions to make $f(\mathcal R_m)\subseteq \mathcal R_{m-1},\ \forall m\in\overline{1,n}$. As for the last region, $\mathcal R_{n+1}$, there are, of course, lots of ways of defining $f$ on it s.t. it's compatible with the previous definitions.

I hope it's Ok... :?


geometry
trapezoid
rectangle
geometry proposed

Let $A_0B_0C_0$ and $A_1B_1C_1$ be any two acute-angled triangles. Consider all triangle $ABC$ that are similar to $\triangle A_1B_1C_1$ (so that vertices $A_1, B_1, C_1$ correspond to vertices $A,B,C,$ respectively) and circumscribed about triangle $A_0B_0C_0$ (where $A_0$ lieson $BC$, $B_0$ lies on $CA,$ and $C_0$ lies on $AB$). Of all such possible triangles, determine the one with the largest area and construct it.

Since the angles $\angle B_0AC_0=\angle B_1A_1C_1,$ $\angle C_0BA_0=\angle C_1B_1A_1$ and $\angle A_0CB_0=\angle A_1C_1B_1$ are constant, then the circles $(X) \equiv \odot(B_0C_0A),$ $(Y) \equiv \odot(C_0A_0B),$ $(Z) \equiv \odot(A_0B_0C)$ are fixed. Their construction is straightforward.

As $\triangle ABC$ are all similar, then the area $[ABC]$ will attain its maximum when $BC$ attains its maximal length. If $N,L$ are the projections of $Y,Z$ on $BC,$ then from the right trapezoid $YZLN,$ we get $YZ \ge NL=(NA_0+LA_0)=\tfrac{1}{2}BC$ $\Longrightarrow$ $BC \le 2 \cdot YZ$ with equality iff $YZLN$ is rectangle, i.e. $YZ \parallel NL.$ Thus, the desired triangle $\triangle ABC$ is that having its sides parallel to $\triangle XYZ.$


I`am Polish and I can make some mistakes so coorect me or something...

I am writing some kind of scientific essay which topic is: "Zero-knowledge proof"
(For example:
In Mexico you don`t have to carry driving license with yourself. When you drive too fast, police stop you and you can lie him that you are eg. your neighbour. ) How can we check if I am not lying (from mathematical point of view and without any knowledge about me)

I`ve found some informations, but it`s not enough to write good essay, so I would like you to give me materials. For me its difficult to search in "English" math..

Edit: I`ve found pages like wikipedia, but i hope you have materials wchih are not so complicated

I think the basic idea behind it is two request 2 pieces of information from the prover and randomly decrypt one of them, such that it is easy to forge either one of the pieces of information but computationally hard to forge both simultaneously


trigonometry
trig identities
Law of Cosines
geometry unsolved
geometry

Hi!
Every body know that, Pitago theorem in triangle is very important in Geometry. [b]But, how many way we can prove this theorem? I want help from you.[/b]
[u][b]Note[/b][/u]
The Pitago theorem in triangle: Let $ ABC$ is triangle. We have $ AB\perp AC$ $ \Longleftrightarrow$ $ AB^2\plus{}AC^2\equal{}BC^2$.

Thank. But, I want you give me the full solution.

Hi thanhnam2902!
(sin'A')²=AB²/BC²,
(cos'A')²=AC²/BC²,
by adding we have :AB²/BC² + AC²/BC² =1,
multiplying by BC² we have:AB²+AC²+BC². :wink: .

Thank rachid. Who can give me other solution?

Rachid, I think you made mistake. You forgot, that to prove that $ cos^2 \alpha \plus{} sin^2\alpha \equal{} 1$ you need Pitago theorem. So you proof is false.

Could you prove it as special case of Law of Cosines?

[quote]Rachid, I think you made mistake. You forgot, that to prove that $ \cos^2 \alpha \plus{} \sin^2 \alpha \equal{} 1$ you need Pitago theorem. So you proof is false.[/quote]

You can show that with product to sum formulas. (Unless you need Pytagora to prove those formulas too?)

$ \cos^2 \alpha \plus{} \sin^2 \alpha \equal{} \cos \alpha * \cos\alpha \plus{} \sin \alpha * \sin \alpha \equal{} \frac {1}{2}(\cos 0 \plus{} \cos 2\alpha) \plus{} \frac {1}{2}(\cos 0 \minus{} \cos 2\alpha) \equal{} \frac {1}{2}(2\cos 0) \equal{} 1$

Dear Mathlinkers,
accordling to the geometric litterature, there is around 500 proofs of the Pythagore's theorem.
Sincerely
Jean-Louis

[quote="turcas_c"]See [url=http://terrytao.wordpress.com/2007/09/14/pythagoras-theorem]here[/url] an ingenious proof ( by Dr. Terry Tao)  :)

http://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/[/quote]
What's so ingenious about it? We have [b]exactly[/b] this proof in our 10th grade school syllabus for [b]years[/b]. And this is merely seven months old. I am sorry if I have misunderstood it. But it appeared to me the same.
[quote="NapoleonXIV"]$ \cos^2 \alpha \plus{} \sin^2 \alpha \equal{} \cos \alpha * \cos\alpha \plus{} \sin \alpha * \sin \alpha \equal{} \frac {1}{2}(\cos 0 \plus{} \cos 2\alpha) \plus{} \frac {1}{2}(\cos 0 \minus{} \cos 2\alpha) \equal{} \frac {1}{2}(2\cos 0) \equal{} 1$[/quote]
I think all the trigonometrical basic formulas are interlinked and at some point uses the pythagoras's(or Pitago's, as you say) theorem and in some cases other geometric proofs. So, to establish such a proof, one would need to find a bypass around the theorem and prove every theorem he uses, for example $ \cos A \cos B \equal{} \frac {1}{2} (\cos(A \minus{} B) \plus{} \cos(A \plus{} B))$ used here.

there is another proof using inversion you know. :)

[quote="Akashnil"]
[quote="NapoleonXIV"]$ \cos^2 \alpha \plus{} \sin^2 \alpha \equal{} \cos \alpha * \cos\alpha \plus{} \sin \alpha * \sin \alpha \equal{} \frac {1}{2}(\cos 0 \plus{} \cos 2\alpha) \plus{} \frac {1}{2}(\cos 0 \minus{} \cos 2\alpha) \equal{} \frac {1}{2}(2\cos 0) \equal{} 1$[/quote]
I think all the trigonometrical basic formulas are interlinked and at some point uses the pythagoras's(or Pitago's, as you say) theorem and in some cases other geometric proofs. So, to establish such a proof, one would need to find a bypass around the theorem and prove every theorem he uses, for example $ \cos A \cos B \equal{} \frac {1}{2} (\cos(A \minus{} B) \plus{} \cos(A \plus{} B))$ used here.[/quote]

Ive seen nice geometric proof for $ \cos (\alpha \plus{} \beta)$, which should be enough to derive formula for $ \cos \alpha \cos \beta$, but haven't seen proof for $ \sin(\alpha \plus{} \beta)$
.

[quote="dogcat"]See [url=http://www.cut-the-knot.org/pythagoras/index.shtml]this[/url] for example :wink: [/quote][quote="t0rajir0u"][url=http://www.cut-the-knot.org/pythagoras/index.shtml]A collection of proofs[/url].[/quote]Dogs and cats will be angry with you t0rajir0u. :P  :D

Sorry, I missed that  :oops:

[url=http://mathworld.wolfram.com/PythagoreanTheorem.html]Mathworld[/url] has a fair collection of proofs as well as a discussion.

Thank for link: http://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/


are the solutions on the cemc website for euclid, examples of solutions we should write to gain full marks on the question?

or are they just rough solutions, to explain the answers?

i mean some of them are kind of short for what i would expect to be worthy of full marks

i mean when they prove one thing, they say shortcuts like this:

Using similar reasoning and properties of parallel lines we get right angles at Q, R and S.
Thus PQRS is a rectangle.

(2000, question 8 a)
[url]https://cemc.math.uwaterloo.ca/english/contests/past_solution/2000/euclid.pdf[/url]

is that even a reasonably complete solution?

i'm sorry if it's an obvious or easy question but i just wanted to understand what is needed in a complete solution

Yeah, they are good solutions that are good enough to write up on the contest.

A good solution has to have all its steps explained and is supposed to be clear and concise.

And that shortcut is valid -- as one can use analogy for solutions.

so we can say things like

"using the same reasoning..."

to prove a bunch of stuff you've already proven?

isn't there some subjectivity here...
i mean is it that obvious to the marker that you can apply the same reasoning to  proving other things?

[quote="SM4RT"]if its something that is well known i dont think you have to prove it..like you can use properties that are well known with out proving..if thats what you are asking??[/quote]

no... actually it's more like

from the solution page
[quote]
2000 Euclid Solutions 12
For any parallelogram, any two
consecutive angles add to
180°, ∴2x+2y=180
or, x+y=90 .
Therefore in Δ PAF, ∠APF=90°.

Using similar reasoning and properties of parallel lines we get right angles at Q, R and S.
Thus PQRS is a rectangle.[/quote]

this part:

[quote]
Using similar reasoning and properties of parallel lines we get right angles at Q, R and S.[/quote]

it just says use similar reasoning and properties of parallel lines  we can get what we wanted.

skipping over a bunch of steps, is that reasonable?
i mean it's like proving half of the proof and say: using the above "half proof", you can prove the rest[/quote]

[quote="princevegetam"]

it just says use similar reasoning and properties of parallel lines  we can get what we wanted.

skipping over a bunch of steps, is that reasonable?
i mean it's like proving half of the proof and say: using the above "half proof", you can prove the rest[/quote]
yes...i think that is allowed...i use the word "similarly" a lot in my proofs...frankly speaking...the markers will have a hard time reading ur proof if u repeat the same thing over and over again...and its more likely that u'll make a mistake in the repetition of that similar proof... :D happy usamo!!!

ya [b][u]he/she[/u][/b] got a point i think u can use similarly when you know that the other parts require the same procedure..

[quote="SM4RT"]ya he;s got a point[/quote]
how do u kno im a he???  :D  :P

hmmm u got a point again i'll change it  :rotfl:

i see what you mean

that it's senseless to repeat the same thing again

and alot of times it is obvious that the same logic/reasoning can be reapplied to solve a different part of the problem

but when is this "allowed", to just bluntly say similarly, using same reasoning, so and so is proven..

maybe i'm just really inexperienced...

[quote="bryant124"]actually i am a he :rotfl:[/quote]

lol i know


Find $ f:R \rightarrow R $
    $ (x + y) [f(x) - f(y)] = f(x^2) - f(y^2) $

A really easy one :

Clearly, if $f(x)$ is a solution then, for each constant $k$,  $f(x)+k$ is a solution. Thus, we may assume that $f(0)=0$.
For $y=0$ we deduce that, for all $x$ : $f(x^2)=xf(x)$.
Back to the general case, for all $x,y$ we then have :
$xf(x)-xf(y)+yf(x)-yf(y) = f(x^2) - f(y^2) = xf(x) - yf(y)$
so that $xf(y)=yf(x)$.
Thus, for $y=1$ it leads to $f(x) = xf(1)$, which means that $f$ is linear. Fropm our initial hypothesis, we deduce that if $f$ is a solution then $f(x)=ax+b$ for some constants $a,b$.

Conversely, it is easy to verify that all these functions are indeed solutions.

Pierre.


geometry
3D geometry
tetrahedron
number theory
relatively prime

Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i+1}$, replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $P_{3}$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

sorry too revive this thread after so many years, but the volume of the next tetrahedra is 1/8^n+1, not 4.

Very, very, very long bump, but what is the easiest way to do this (AoPS wiki down R.I.P.)


given that $xyz=1$ find the minimum value of $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$

[hide="here is how i did it"]
ok so i got an answer but im not sure if it is totally correct.
i had to use Chebyshev's Inequality. its sorta like Cauchy-Schwarz

it states $\displaystyle\frac{1}{n}\sum{a_ib_i}\ge\left(\frac{1}{n}\sum{a_i}\right)\left(\frac{1}{n}\sum{b_i}\right)$

ok now that that has been stated ill start.

$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$
we make the substitution $x=\frac{1}{xy}$
 and we now have,
$\frac{y+z}{yz}+\frac{x+y}{xy}+\frac{x+z}{xz}$
next we simplify into one term by getting common denominator

$\frac{(y+z)(xy)(xz)+(x+y)(yz)(xz)+(x+z)(xy)(yz)}{(xyz)^2}$
or
$(x+y)(z)+(x+z)(y)+(y+z)(x)$ this was done by using the fact that xyz=1.
this is where we use Chebyshev's
so we have
$\frac{(y+z)(x)+(x+z)(y)+(x+y)(z)}{3}\ge\left(\frac{x+y+z}{3}\right)\left(\frac{2x+2y+2z}{3}\right)$
so from here we want to find the minimum value of the RHS
this is done with AM-GM
so we have $\frac{(x+y)(z)+(x+z)(y)+(y+z)(x)}{3}\ge\left(\frac{3}{3}\right)\left(\frac{6}{3}\right)=2$
and from here we multiply both sieds by 3 and find that the minimum value is $6$.
[/hide]
im not 100% sure if this is correct, could someone please verify or give the correct answer.

yeah i just noticed that problem with my proof. could you please explain how to do this problem.

Add $1$ to each of the terms, then AM-HM:
\[ \sum \frac{x+y}{x+y+z} \cdot \sum \frac{x+y+z}{x+y} \geq 9  \]
\[ 2 \cdot \sum \frac{x+y+z}{x+y} \geq 9  \]
\[ \sum \frac{x+y+z}{x+y} \geq \frac{9}{2}  \]
\[ \sum \frac{z}{x+y} \geq \frac{3}{2}  \]

Or you could use AM-GM since the inequality is equivalent to:
\[ \frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b} -3 \geq \frac{3}{2}  \]
\[ \Leftrightarrow (a+b+c) \left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) -3 \geq \frac{3/2}  \]
\[ \Leftrightarrow \frac{1}{2} \left((a+b)+(b+c)+(c+a) \right) \left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) -3 \geq \frac{3}{2}  \]
Let $a+b=x$, $b+c=y$, and $a+c=z$ and multiply by 2 to get
\[ (x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) -6 \geq 3  \]
\[ \Leftrightarrow \frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{z}-6 \geq 0  \]
which holds by AM-GM.

Personally I like the latter since it's shorter  :lol:

$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\geq\frac{(x+y+z)^2}{2(xy+yz+zx)}$

So it remains to prove that $(x+y+z)^2\geq 3(xy+yz+zx)$  which is trivial :)

OR you can use the rearrangement inequality twice and add them together.

sorry but i do not see how your proofs solve the problem.
i am sorta new at inequalities.

They all prove that the expression you posted is $\geq \frac{3}{2}$. Hence, the minimum value it can attain is $\frac{3}{2}$.

i know that. but what does $\frac{x+y}{x+y+z}$ have to do w/ this problem. that is what i dont get

He is just transforming a well known inequality into the expression you want and establishing the minimum through the inequality.

I think the Cauchy (Jensen's?) way is the nicest, followed by rearrangement, but here is a good (effective, not aesthetic) way to solve inequalities.

$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq \frac 32$

is equivalent to $2[x(z+x)(x+y)+y(x+y)(y+z)+z(y+z)(z+x)]\geq 3(x+y)(y+z)(z+x)$

Multiplying out, we get

$2(x^3+y^3+z^3)+2[x^2(y+z)+y^2(z+x)+z^2(x+y)]+6xyz\geq 3[x^2(y+z)+y^2(z+x)+z^2(x+y)]+6xyz$

or

$2(x^3+y^3+z^3)\geq x^2(y+z)+y^2(z+x)+z^2(x+y)$

This last inequality is not hard - see if you can handle it yourself!

Dpopov,
in your proof could you show me how you proved that
$\frac{a+b+c}{b+c}+\frac{a+b+c}{b+c}+\frac{a+b+c}{b+c}-3\ge\frac{3}{2}$

i like your proof its just that i dont get this first step.

What you wrote is $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}$ (with x,y,z) which is equivalent to the following (I'm already putting it in inequality so I can proceed with the proof; your expression is just the LHS):

\[ \frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{a+c}+\frac{a+c}{a+c}+\frac{c}{a+b}+\frac{a+b}{a+b}-3 \geq \frac{3}{2}  \]
\[ \frac{a+b+c}{b+c}+\frac{a+b+c}{b+c}+\frac{a+b+c}{b+c} -3 \geq \frac{3}{2}  \]
\[ \Leftrightarrow (a+b+c) \left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) -3 \geq \frac{3}{2}  \]
\[ \Leftrightarrow \frac{1}{2} \left((a+b)+(b+c)+(c+a) \right) \left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) -3 \geq \frac{3}{2}  \]
Let $a+b=x$, $b+c=y$, and $a+c=z$ and multiply by 2 to get
\[ (x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) -6 \geq 3  \]
\[ \Leftrightarrow \frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{z}-6 \geq 0  \]

From the AM-GM inequality we know that $\frac{m}{n}+\frac{n}{m} \geq 2$ for $m,n>0$ so the above is clearly true.

I hope that cleared everything up; I had a minor error in my first line since I put everything under one denominator and it shouldn't have been like that but now it's fixed.


in triangle ABC, D is on AB so that AD :  DB = 1:2, and G is on CD so that CG:GD = 3:2.  if ray BG intersects AC at F, find BG:GF

On problems like this, if it comes up on a competition such as the AIME, just assume the triangle is equilateral. In this problem, a sidelength of 6 or 12 looks like a convenient length. It's not a proof that it's invariant, but you'll get the answer.

The problem is so easy with the use of mass points.

Well, using just normal Euclidean geometry, the problem is still rather easy. USing COmplexZeta's tip, assuming the triangle is equilateral, you can easily apply Stuart's Theorem to find the actual lengths in question. It's actaully not as bad as you think...

i dont understand why i can assume it is equilateral?  That would be necessary in order for me to use stewart's theorem.

And i did use mass points, without assuming anything about the lengths, and it did not work (unless if i messed up)

The point is that the problem tells you that it will be the same if it's an equilateral triangle or some other kind of triangle. Therefore, it may be useful to assume that it's a special case and see what you can figure out to get the answer. Remember that you don't have to write a proof. Only the answer matters.

alright, i'll do that.  But do you know of a way to do it without assuming its equilateral?  Thanks.


function
calculus
integration
search
real analysis
real analysis theorems

let $ I\equal{}[a,b]$, let $ f: I\rightarrow\mathbb{R}$ be continuous, and let $ f(x)\geq 0$   $ \forall x\in I$. Prove that if $ L(f)\equal{}0$, then $ f(x)\equal{}0$  $ \forall x\in I$

I gave my favorite hint [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1795952106&t=61448]here.[/url]


Maybe this topic should be at the "Inequalities---Theorems and Formulas" forum, so that other people can go there and look them up for reference

Chebyshev:

Assuming rearrangment and assuming WLOG that the sequences are increasing, it's clear that:

$n\sum a_ib_i\geq \sum^{n-1}_{j=0} \sum a_ib_{i+j}=\left(\sum a_i\right)\left(\sum b_i\right)$ which is Chebyshev.  I used rearrangement $n$ times to show that each of the middle inequalities hold (for $j=0$, $j=1$, etc).  I've been taking indices mod $n$; for example, $b_{i+n}=b_i$ and so on.

Holder: We have that $\frac {1}{p}+\frac{1}{q}=1$.  We can normalize $\sum a^p$ and $\sum b^q$ so that $\sum a^p=\sum b^q=1$.

Now, by the convexity of $\ln x$ (Jensen's inequality), we have that $a_ib_i\leq \frac {a_i^p}{p}+\frac{b_i^q}{q}$.  Applying this to the inequality at hand and summing, we get:

$\sum a_ib_i\leq \frac {1}{p}\cdot\sum a_i^p+\frac {1}{q}\cdot\sum b_i^q=\frac {1}{p}+\frac{1}{q}=1=\left(\sum a^p\right)^{\frac{1}{p}}\left(\sum b_q\right)^{\frac{1}{q}}$

Schur: I posted a proof earlier today, it's somewhere in the intermediate forum in the inequality about logs.

Minkowski: Try to solve it from Holder.

I apologize if I've been too brief; tell me which parts are unclear and I'll try to tidy them up for you.

[quote="xxreddevilzxx"]I need to know how to prove some of these theorems!! (If you can then great!!)

1. Tchebychef's Inequalty
2. Holder's Inequality
3. Schur's Inequality
4. Minkowski's Inequality

thanks...[/quote]

Can you also post the math forms of the inequalities? not only the words...
thanks!

You can find the math forms of some of them [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=7942]here[/url]


solve the systeme in R.
[img]http://www.gnux.be/latex/data/183e1f1aa6e71541b130d2f6021f37bf.png[/img]
[img]http://www.gnux.be/latex/data/37ceb98098f9ed078a21e5d0bf62c4e9.png[/img]
[img]http://www.gnux.be/latex/data/0c433767f479d5016a6cb9668bc0c010.png[/img]

[hide="solution partielle."]nous ajustons la première expression:
$ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2xz\plus{}2yz\equal{}1$
et nous soustrayons la deuxième expression:
$ 2xy\plus{}2xz\plus{}2yz\equal{}\minus{}13\Rightarrow xy\plus{}xz\plus{}yz\equal{}\minus{}\dfrac{13}{2}$
Maintenant nous cubons la première expression:
$ (x\plus{}y\plus{}z)^3\equal{}x^3\plus{}y^3\plus{}z^3\plus{}3x^2y\plus{}3x^2z\plus{}3y^2x\plus{}3y^2z\plus{}3z^2y\plus{}3z^2x\plus{}6xyz\equal{}1$
Maintenant nous multiplions les deux premières expressions ensemble:
$ x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}x^2z\plus{}y^2x\plus{}y^2z\plus{}z^2y\plus{}z^2x\equal{}14$
Nous soustrayons la troisième équation d'elle :
$ x^2y\plus{}x^2z\plus{}y^2x\plus{}y^2z\plus{}z^2y\plus{}z^2x\equal{}\minus{}22$
Et nous substituons dedans $ (x\plus{}y\plus{}z)^3$:
$ 36\minus{}66\plus{}6xyz\equal{}1\Rightarrow xyz\equal{}\dfrac{31}{6}$
Nous avons maintenant la somme, la somme du produit de deux à la fois, et du produit trois à la fois des trois variables. Nous employons les formules de Vieta pour obtenir un polynôme avec x, y, et z comme racines.
$ q(a)\equal{}a^3\minus{}a^2\minus{}\dfrac{13}{2}a\minus{}\dfrac{31}{6}$
Nous pouvons nous multiplier par une constante et les racines ne changent pas.
$ p(a)\equal{}6a^3\minus{}6a^2\minus{}39a\minus{}31$
Maintenant nous devons trouver les racines de cela polynômes.[/hide]

Pardonnez mon mauvais français, j'a utilisé un traducteur. :(

yes very well, it is the same solution what I made. but I am going to give you an easier solution to calculate $ xyz$.
[img]http://www.gnux.be/latex/data/6a677483b86933aba83a84c5066d251e.png[/img]


linear algebra
matrix
linear algebra unsolved

Let $ p$ be a prime. I ask for a formula giving the number of solutions of $ A^2\plus{}A\plus{}I_2\equal{}0$ with $ A\in\mathcal{M}_2(\mathbb{Z}/p\mathbb{Z})$  (When I read $ p$ I know immediately the sought value).

[color=red]Result: 9 for $ p \equal{} 3$, 0 otherwise.
[/color]
EDIT: Needs a tremendous amendment since the case $ p\neq 3$ is not complete . ... hasty mistake in identifying all the diagonal matrices  :blush:  Please see Kuba's complete counting procedure.

[color=green]Corrected (partial  - and poor - result): 9 for $ p \equal{} 3$[/color]

[hide="Proof:"]$ A$ has finite order 3, such that it is diagonalizable if $ p\neq 3$. [color=red]It turns out that $ A \equal{} \pm I_2$ in this case[/color]. Plugging into the equation gives a contradiction.
If $ p \equal{} 3$ then 1 is the only eigenvalue. Then $ A$ is equal to $ I_2$ or a conjugate of $ I_2 \plus{} N$ in $ GL_2(\mathbb Z/3\mathbb Z)$ with $ N \equal{} \left(\begin{array}{cc}0 & 1 \\
0 & 0\end{array}\right)$, by Jordan. Counting $ GL_2(\mathbb Z/3\mathbb Z)$ and the isotropy group of $ N$ w.r.t. conjugation gives 8 different non-diagonalizable matrices. There is exactly 1 diagonalizable matrix, namely $ I_2$, thus 9 altogether in this case.[/hide] :(

If $ p\equal{}3$ then $ x^2\plus{}x\plus{}1\equal{}(x\minus{}1)^2$ and the minimal poly of $ A$ is either $ (x\minus{}1)^2$ or $ x\minus{}1$. In the later case the only solution is $ I_2$ and in the former one the solutions are the conjugates of $ U: \equal{}\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ under the action of $ GL_2(p)$. This number is (easy computation) $ |GL_2(p)|/|Stab(U)| \equal{} (p^2\minus{}1)(p^2\minus{}p)/((p\minus{}1)p) \equal{} p^2\minus{}1$.
Summing up we get a total of $ p^2\equal{}9$ solutions.

If $ \minus{}3$ is a square modulo $ p$ denote by $ u,v$ the roots of $ z^2\plus{}z\plus{}1$.
The matrix $ A$ can be diagonalized and we have three cases:
1. $ A$ is $ uI_2$
2. $ A$ is $ vI_2$
3. $ A$ is $ diag(u,v)$.
Now $ GL_2(p)$ has a natural action by conjugation on the solutions we are looking for, but since the matrix can be diagonalized the number of solutions will be the sum of the cardinals of the orbits $ GL_2(p)\cdot diag(u,v), GL_2(p)\cdot uI_2,GL_2(p)\cdot vI_2$ (I used here the fact that these orbits are disjoint since $ u\neq v$).
Denote by $ U\equal{}diag(u,v)$
The stabilizer of $ U$ in $ GL_2(p)$ has order $ (p\minus{}1)^2$ (simple computations and use that $ u\neq v$) and then the orbit has cardinal $ \frac{(p^2\minus{}1)(p^2\minus{}p)}{(p\minus{}1)^2}\equal{}p(p\plus{}1)$. So the number of solutions in this case is $ 1\plus{}1\plus{}p(p\plus{}1)\equal{}p^2\plus{}p\plus{}2$.

If $ \minus{}3$ is not a square modulo $ p$ then the Jordan canonical form of $ A$ is $ U: \equal{}\begin{pmatrix}0 & \minus{}1\\1 & \minus{}1\end{pmatrix}$.
Now we have to compute the cardinal of the stabilizer of $ U$ (w.r.t $ GL_2(p)$):
$ \begin{pmatrix} a & b\\ c& d\end{pmatrix}\begin{pmatrix}0 & \minus{}1\\1 & \minus{}1\end{pmatrix} \equal{} \begin{pmatrix}0 & \minus{}1\\1 & \minus{}1\end{pmatrix}\begin{pmatrix} a & b\\ c& d\end{pmatrix}$
 gives
$ b\equal{}\minus{}c$
$ a\plus{}b\equal{}d$
We also have the restriction $ ad\minus{}bc\neq 0\Leftrightarrow a^2\plus{}ab\plus{}b^2\neq 0\Leftrightarrow a\neq 0$ or $ b\neq 0$ (here we use again that $ \minus{}3$ is not a square modulo $ p$).
Hence we have $ p^2\minus{}1$ posibilities for $ \begin{pmatrix}a & b\\c & d\end{pmatrix}$, i.e. the stabilizer has cardinal $ p^2\minus{}1$.
Therefore the number of solutions of $ A^2\plus{}A\plus{}I_2\equal{}0$ in this case is $ p^2\minus{}p$.

In conclusion, the number of solutions is:
$ p^2$ if $ p\equal{}3$.
$ p^2\plus{}p\plus{}2$ if $ \minus{}3$ is a square modulo $ 3$.
$ p^2\minus{}p$ if $ \minus{}3$ is not a square modulo $ 3$.

Hi Kuba,
1)  $ p^2\minus{}p$ if  $ \minus{}3$ is not a square mod $ p$ is true.
     $ p^2\plus{}p\plus{}2$ if $ \minus{}3$ is a square mod $ p$ is true.
2) But If I read $ p$ I don't know immediately if $ \minus{}3$ is a square mod $ p$.

$ \minus{}3$ is a square modulo $ p$ precisely when $ p\equal{}1, 5$ modulo $ 12$ (if I solved correctly the congruences).
If this is not what you meant by 2) then please be more explicit.

Hi Kuba,
 this is what I meant by 2) but the result is not $ 1,5$ mod $ 12$.

Yes, it's not 1,5...
well, we need $ 1\equal{}\left(\frac{\minus{}3}p\right)\equal{}(\minus{}1)^{(p\minus{}1)/2}(\minus{}1)^{(p\minus{}1)/2}\left(\frac{p}3\right) \equal{} 1$ iff $ p\equal{}1$ modulo 3.
So $ \minus{}3$ is a square modulo $ p$ iff $ p\equal{}1$ modulo 3.


A charged particle$ (q,m)$ is released from origin with velocity $ V_o \hat{i}$ in a uniform magnetic field $ B \equal{} \frac{B_0}{2} \hat{i} \plus{} \frac{\sqrt{3}B_0}{2}\hat{j}$

1) Find the pitch of the helical path described by the particle.
2)Find the time after which z component of velocity is $ \frac{\sqrt{3}V_0}{2}$
3)Maximum z - coordinate of the particle is?

I have  my answers for 1 and 3 but they dont match.

The projection of the particle's trajectory in the plane $ \sqrt {3}y\plus{}x\equal{}0$ is a circle and its time period is $ \frac {2\pi m}{qB_0}$. Its velocity along the direction of the field is $ v_0 \cos \frac {\pi}{3}$. It moves in circle with velocity $ v_0 \sin \frac {\pi}{3}$. Radius of the circle is $ \frac {mv_0 \sin \frac {\pi}{3}}{qB_0}$

(1) Pitch of helix = $ v_0 \cos \frac {\pi}{3} \cdot \frac {2\pi m}{qB_0}$
(2) It is when it completes quarter of circle. $ \frac T4$
(3) the diameter of the circular path.

[hide="My Solution:"]

Circular motion will be caused due to the component of velocity perpendicular to the field.

$ R \equal{} \frac {v_0 \pi m}{B_0q} \\
T \equal{} \frac {2\pi m}{B_0 q} \\
Pitch\equal{} 2R$
The rest depends on these formulae.
[/hide]


If A and B are matrix with with size $ 3$ by $ 3$, that statisfies : $ A^3\equal{}B^2\equal{}0$ and $ rank(A)\equal{}rank(B)$, then prove or disprove that $ A$ and $ B$ are similar. Thanks in advance.

$ B$ is either the zero matrix (rank $ 0$) or has Jordan normal form consisting of a single $ 2 \times 2$ block (rank $ 1$).  $ A$ is either the zero matrix, has the above form, or has Jordan normal form consisting of a single $ 3 \times 3$ block (rank $ 2$, which is not a possible rank of $ B$).  Hence $ A$ and $ B$ have the same Jordan normal form.


modular arithmetic
number theory
least common multiple

Find the smallest positive integer greater than one which yields a remainder of one when divided by any single digit positive integer greater than one.

[hide]The smallest number divisible by all single-digit numbers + 1 = 2521.[/hide]

[hide]
$ x\equiv 1\pmod{2}$
$ x\equiv 1\pmod{3}$
$ x\equiv 1\pmod{4}$
$ x\equiv 1\pmod{5}$
$ x\equiv 1\pmod{6}$
$ x\equiv 1\pmod{7}$
$ x\equiv 1\pmod{8}$
$ x\equiv 1\pmod{9}$
Now the LCM of 2, 3, 4, 5, 6, 7, 8, 9 is $ 2520$ or $ 2^{3}\cdot3^{2}\cdot5\cdot7$. Since you know that x is congruent to 1 in mods 2, 3, 4, 5, 6, 7, 8, and 9, you can simply add 1 to $ 2520$, giving you $ \boxed{2521}$.
[/hide]
Here's another approach, but I'm not sure if it would work consistently.
[hide] So we know that $ x\plus{}2519$ would be congruent to 0 in mods 2, 3, 4, 5, 6, 7, 8, and 9. However, $ x\plus{}2519\neq2520$ because $ x\neq1$. We can however multiply 2520 by 2 (we want the smallest positive integer so we simply multiply by 2, but not something like 1.5, because that wouldn't give us something that's congruent to 1 in mods 2, 3, 4, 5, 6, 7, 8, and 9), giving us $ 5040$. From this we get the equation $ x\plus{}2519 \equal{} 5040$ and $ x \equal{}\boxed{2521}$.
[/hide]


The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $ 20$ cents. If she had one more quarter, the average value would be $ 21$ cents. How many dimes does she have in her purse?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\  1 \qquad \textbf{(C)}\  2 \qquad \textbf{(D)}\  3\qquad \textbf{(E)}\ 4$

[hide]let s = total sum
let d = number of coins
then $s/d = 20$
and $2+25/d+1 = 21$
Rearranging, we get:
$s = 20d$
$s = 21d - 4$
so using linear combinations, etc. we get d = 4 and s = 80
so,
$\frac{80+25}{4+1} = 21$
so we have to have three quarters for the 4, which gets us 75 cents, and an extra dime for the 5 cents :) so the answer is $%Error. "block" is a bad command.
{B}$
But that's wrong because she has to have pennies and dimes... :|
[/hide]

Apprentice: dimes are 10 cents. You'd add another nickel to get 5 cents more.

[hide]Let n be the number of coins she has how.

$20n+25=21(n+1)$.
$n=4$.

So she has 4 coins totally 80 cents before the quarter was added.

She must have three quarters (two quarters would require 3 dimes, which is over 4 coins, 4 quarters is over 80 cents). Then the remaining coin would be a nickel. Thus she has no dimes. [/hide]


$ABCD$ is a convex quadrilateral. We draw its diagnals to divide the quadrilateral to four triabgles. $P$ is the intersection of diagnals. $I_{1},I_{2},I_{3},I_{4}$ are excenters of $PAD,PAB,PBC,PCD$(excenters corresponding vertex $P$). Prove that $I_{1},I_{2},I_{3},I_{4}$ lie on a circle iff $ABCD$ is a tangential quadrilateral.

Similar to [url=https://artofproblemsolving.com/community/c6h1113224p5083627]Indian TST 2015 Day 1 P1[/url].


Pretty much what I did too...

[hide]$\log 2^{512}<m \Rightarrow 512\log 2<m \Rightarrow 512(.3010)<m \Rightarrow 154.112<m \Rightarrow \boxed{155=m}$[/hide]

I thought the idea was to not use a calculator...

Just use the knowledge that we were given in the original problem: $\log 2\approx.3010$

what this is essentially getting at is the number of digits that $2^{512}$ contains :)

[quote="calculuslover800"][quote="José"]What about generalizing? Find $m$: $10^{m-1}<2^k<10^m$, for some positive integrer $k$.[/quote]

In general, we are trying to solve $2^k = 10^x$ for $x$.  Simply taking the log of both sides:

$k*\log{2} = x$

Now we need to take the ceiling of $x$ to find $m$ so that $m-1<x<m$.

So, in general, the solution is $[k*\log{2}]$ (the brackets are the ceiling function).  Or, more generally, replace 2 with any base.

I think this type of problem has been posted a thousand times before, but oh well...[/quote]

I see.


[quote="moldlee"]A group of friends bought a crystal bowl, but when they had to pay

for it 4  of them were not present, therefore, everybody there had to

give in an extra $36. (This way they paid for it fully.) Then 3 of the missing friends arrived and after dividing the cost evenly again among those who are currently present, each of those who paid for it the first time got$30  back.

How many people are there in the group, and how much did

the bowl cost?[/quote]
The answer is undefined because your LaTeX is messed up. :P

[quote="lotrgreengrapes7926"][quote="moldlee"]A group of friends bought a crystal bowl, but when they had to pay for it 4 of them were not present, therefore, everybody there had to give in an extra $\$$36. (This way they paid for it fully.) Then 3 of the missing friends arrived and after dividing the cost evenly again among those who are currently present, each of those who paid for it the first time got $\$$30 back. How many people are there in the group, and how much did the bowl cost?[/quote]
[hide] Let's call the cost of the bowl $d$ and the number of people $x$. So,
\begin{eqnarray*}\frac{d}{x-4}&=&36+\frac{d}{x}\\ \frac{d}{x-4}&=&\frac{36x+d}{x}\\ dx&=&36x^{2}+dx-144x-4d\\9x^{2}-36x&=&d\end{eqnarray*}

and

$\begin{eqnarray*}\frac{d}{x-4}&=&30+\frac{d}{x-1}\\ \frac{9x^{2}-36x}{x-4}&=&30+\frac{9x^{2}-36}{x-1}\\9x^{2}-9x&=&30x-30+9x^{2}-36\\39x&=&66\\ x&=&\frac{22}{13}$

 :maybe: What did I do wrong?[/hide][/quote]

They got $\$30$ back; they didn't pay an additional $\$30$. In the second equation a $6$ should replace the $30$. You should end up with $x = 10$.


I am finding the AMC/AIME geometry pretty difficult. I am planning on getting challenging problems in geometry. How much does that book help?

I'm also pretty concerned about AIME/AMC Geometry.  Right now, I can get like 4s-6s on past AIME's but I mainly skip geometry questions and the rather difficult ones. What book(s) should I read for it aside from AoPS 1 & 2?  How much would it improve my score?  2-3 points? What practice book(s) should I get too?

I'm also planning on taking Intro to Geometry in the spring... if I feel I still need it then.

7's and 8's would be nice occasionally haha.

[quote="WarpedKlown1335"]Two words, [u]Geometry Revisited[/u].[/quote]

For the AMC/AIME? I haven't worked through that book but I have given it a glance.  I seem to think that much of it would be irrelevant in terms of the AMC and the AIME.

I suggest going through the geometry sections in AoPS Vol 1/2.  After that just start hitting the problems.  That is really all you need to solve the problems that I've come across on those tests (and a ton of ingenuity/hard work/etc.).

Geometry Revisited will lay the foundation for solid geometry skills, so I would recommend going through that book regardless of what you're attempting. If you can really get good at all the Geo Revisited stuff, then getting used to some of the more basic tricks and techniques required for AMC/AIME problems will be no problem.


limit
logarithms
calculus
calculus computations

$ \lim_{x \to \infty}x \left ( e-\left ({ \frac{x+2}{x+1}}\right )^{x}\right )$

[quote="pureblack2003"]$\lim_{x \to \infty}x \left ( e-\left ({ \frac{x+2}{x+1}}\right )^{x}\right )$[/quote]
answer is $e$
by taylor is simpl:
$\lim_{x \to \infty}x \left ( e-\left ({ \frac{x+2}{x+1}}\right )^{x}\right )$~$x(e-e^{\frac{x}{x+1}})$~$-xe(e^{\frac{-1}{x+1}}-1)$~$\frac{xe}{x+1}\to e$

[quote="misan"]the answer is $0$.

we know $\lim_{x\to\infty}\;\left( 1+\frac{a}{x}\right)^{x}\;=\;e^{a}$

and so,
$\lim_{x\to\infty}x\left( e-\left(\frac{x+2}{x+1}\right)^{x}\right) \;=\;\lim_{x\to\infty}x\left( e-\left(\frac{1+\frac{2}{x}}{1+\frac{1}{x}}\right)^{x}\right) \;=\;\\ \\ \\ \hspace* \;\;\;\quad \lim_{x\to\infty}x\left( e-\frac{e^{2}}{e}\right)\;=\boxed{0}$[/quote]

$\infty \cdot 0$ is not necesarily $0$,and I don't see how your proof can hold! :ninja: .

misan your proof is wrong. I already have solved it.

Extremal, I do not think you are right, but I'm not sure how to solve, either  :oops: -- I suspect that you've made essentially the same error as misan by taking part of the limit before other parts.

We may replace $x$ to $n$ to consider this as a limit of a sequence. Then

$n \left( e-\left( 1+\frac{1}{n+1}\right)^{n}\right)$
$= n \left( \left( \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots \right)-\left( \frac{1}{0!}+\frac{n}{1!(n+1)}+\frac{n(n-1)}{2!(n+1)^{2}}+\frac{n(n-1)(n-2)}{3!(n+1)^{3}}+\cdots \right) \right)$
$= n \left( \frac{1}{1!}-\frac{n}{1!(n+1)}\right)+n \left( \frac{1}{2!}-\frac{n(n-1)}{2!(n+1)^{2}}\right)+n \left( \frac{1}{3!}-\frac{n(n-1)(n-2)}{3!(n+1)^{3}}\right)$ $+\cdots$

Now let's focus on each corresponding summand:

$n \left( \frac{1}{k!}-\frac{n(n-1)\cdots(n-k+1)}{k!(n+1)^{k}}\right)$

By simple observation, we know that this expression can be reduced to

$\frac{1}{k!}\frac{k(k+1)}{2}\frac{n^{k}+\cdots}{(n+1)^{k}}$

Taking $n \to \infty$, each summand converges to $\frac{1}{k!}\frac{k(k+1)}{2}$. So we have

$\lim_{n\to\infty}n \left( e-\left( 1+\frac{1}{n+1}\right)^{n}\right) = \sum_{k=0}^{\infty}\frac{1}{k!}\frac{k(k+1)}{2}$

It is easy to calculate the right side. Let's consider $\frac{1}{2}xe^{x}= \sum_{k=0}\frac{1}{k!}\frac{x^{k+1}}{2}$. Differentiating both side twice, we have

$\frac{2+x}{2}e^{x}= \sum_{k=0}\frac{1}{k!}\frac{k(k+1)x^{k-1}}{2}$

Plugging $x = 1$ gives the answer $\frac{3e}{2}$.

[i](Edited)[/i] Oh, my first attempt also gave me $e$... and I still don't know where is wrong.

[quote="JBL"]Proof by Mathematica sucks.  Proof by explaining what the other person did wrong is much more interesting.

[quote="Extremal"]I don't agree with you.
$(\frac{x+2}{x+1})^{x}=e^{x\ln{\frac{x+2}{x+1}}}=e^{x(\frac{1}{x+1}+o{\frac{1}{x}})}$
because By taylor
$\ln\frac{x+2}{x+1}=\ln{1+\frac{1}{x+1}}=1+\frac{1}{x+1}+o(\frac{1}{x})$[/quote]
Okay.  But then you should get $x(e-e^{\frac{x}{x+1}+o(1)})$ on the next line, and I believe that's where you lost the constant factor that should be there.[/quote]
 :blush:
yes I Have forgot something:
$x\ln(\frac{x+2}{x+1})=x(\frac{1}{x+1}-\frac{1}{2(x+1)^{2}}+O(\frac{1}{x^{3}})$
therefore
$-ex(e^{(\frac{-1}{x+1}-\frac{x}{2(x+1)^{2}}+O(\frac{1}{x^{3}})}-1)=-xe((\frac{-1}{x+1}-\frac{x}{2(x+1)^{2}}+O(\frac{1}{x^{3}}))\to \frac{3e}{2}$

[quote="JBL"]Proof by Mathematica sucks.  Proof by explaining what the other person did wrong is much more interesting.[/quote]

You are right. I was too rude, forgetting mathematical mind.  :blush: I replaced it by something else.


[color=darkblue] Let $ M$ be the set of positive integers which consist of $ 2006$ digits, one of which is $ 7$, but others are $ 1$. Prove that there are at least 666 composite numbers in $ M$.
[/color]

Any such number $ M$ can be uniquely written as $ M_k\equal{}\sum_{i\equal{}0}^{2005}10^i\plus{}6\cdot 10^k$ with $ k\equal{}0,1,\ldots,2005$, This implies $ 9M_k\equal{}10^{2006}\minus{}1\plus{}54\cdot10^k$.

For $ k\equiv 4\bmod{6}$ we calculate $ 9M_k\equiv 3^{2006}\minus{}1\plus{}5\cdot 3^k\equiv 2\cdot 8^{334}\minus{}1\plus{}5\cdot 3^4\equiv 0\bmod{7}$, i.e. $ 7|M_k$ and $ M_k>7$ is composite.

For $ k\equiv 2\bmod{6}$ we compute $ 9M_k\equiv (\minus{}3)^{2006}\minus{}1\plus{}2\cdot (\minus{}3)^k\equiv 9\cdot 27^{668}\minus{}1\plus{}2\cdot 3^2\equiv  0\bmod{13}$, i.e. $ 13|M_k$ and $ M_k>13$ is composite.

Putting alltogether, the numbers $ M_{s\plus{}6t}$ with $ s\equal{}2,4$ and $ t\equal{}0,1,\ldots,333$ are different composite numbers, hence, [b]we have at least $ 668$ composite numbers[/b].

From this code
a = (10^2006 - 1)/9;
Table[{k, PrimeQ[a + 6 10^k]}, {k, 0, 2005}];
Count[%, {_, False}]
Cases[%%, {_, True}]

amazingly all the numbers are composite but the one for $ k \equal{} 42$
then there are $ 2005$ such cases...


We have to prove that $ \sum\frac {x^2}{y^2} \ge \sum\frac {x}{y}$ holds true. By Cauchy, $ \sum \frac {x^2}{y^2} \ge \frac {1}{3} \left( \sum \frac {x}{y} \right)^2$. Hence it's left to prove $ \sum \frac {x}{y} \ge 3$, which is obviously true. (all sums are cyclic)

[quote="jagdish"][b](1) If x,y ,z are (+ve ) real no. then prove that
   (x^2)/(y^2) +(y^2)/(z^2) +(z^2)/(x^2)  >= (x/y)+ (y/z)+(z/x)[/b][/quote]

WLOG, assume $ x\equal{}\max\{x,y,z\}$

\[ \frac{{x^2 }}{{y^2 }} \plus{} \frac{{y^2 }}{{z^2 }} \plus{} \frac{{z^2 }}{{x^2 }} \minus{} \left( {\frac{x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x}} \right)\]
\[ \equal{} \frac{{\left( {y^2  \plus{} yz \plus{} z^2 } \right)\left( {x\left( {y \plus{} z} \right) \minus{} yz} \right)\left( {y \minus{} z} \right)^2  \plus{} \left( {y^4  \minus{} y^3 z \plus{} x^2 z^2  \plus{} z^3 x \plus{} z^4 } \right)\left( {x \minus{} y} \right)\left( {x \minus{} z} \right)}}{{x^2 y^2 z^2 }}\]
\[ \ge 0\]


probability
vector
linear algebra
matrix
expected value
linear algebra unsolved

Let $ (a,b)$ denotes the vector inner product of two Vectors $ a,b \in \mathbb{R}^n,n \in \mathbb{N}^{*}$
Take $ a_1,a_2,...a_k \in S \equal{} \{0,1\}^{n},k\in \mathbb{N}^{*}$
 Put :
  $ A \equal{} \begin{bmatrix}(a_1,a_1) & (a_1,a_2) & ... & (a_1,a_k) \\
(a_2,a_1) & (a_2,a_2) & ... & (a_2,a_k) \\
... & ... & .... & ... \\
(a_k,a_1) & (a_k,a_2) & ... & (a_k,a_k)\end{bmatrix}$
Find expected value of the $ \det(A)$ for random $ k$x$ k$ matrix $ A$ whose vectors $ a_k \in S \equal{} \{0,1\}^{n}$ are independently sistributed random variables taking values $ \{0,1\}^{n}$ each with probability $ \frac {1}{2}$
[hide="Answer of me"] $ k \equal{} n$, $ T(\det(A)) \equal{} \frac {(n \plus{} 1)!}{2^{2n}}$
 $ k \neq n$ (I Known this result) May be hard and general $ S \equal{} \{i_1,i_2,..i_m\}^n$ very hard [/hide]
 If you have a solution please post it  :)

If $ k > n$ then it is zero, since $ A \equal{} B^TB$ where $ B$ has columns $ a_1,\dots,a_k$. $ B$ is an $ n \times k$ matrix, so if $ k > n$ it has a nontrivial nullspace; then so does $ A \equal{} B^TB$, so $ \det (A) \equal{} 0$.


Find the coefficient of $x^{8}$ in the expansion $(\frac{1}{2}+2x+x^{10})^{10}$

[hide]
The only way to get an $x^{8}$ term is to multiply 8 $2x$'s and 2 $\frac{1}{2}$'s.  The coefficient for each arrangement of these will be $\frac{2^{8}}{2^{2}}=2^{6}=64$.  There are $\binom{10}{2}=45$ ways to choose the two $\frac{1}{2}$'s, so the total coefficient is $64*45=2880$.
[/hide]

[u]Another question:[/u]

Find the coefficient of $x^{6}$ in the expansion of $\left( x+\frac{1}{x}+1 \right)^{12}$


At Brokaw Hospital, six babies were born to six different women on Monday through Thursday of a particular week.  Assuming that each baby was equally likely to be born on any of the four days, what is the probability that there was at least one baby born on each day?

Please help :)

no offense, but this doesn't seem olympiad level...

[hide="Solution"]
Let A be the property "none born on Monday". Let B be the property "none born on Tuesday". Similarly for C and D.

Then the number of possibilities with atleast one of the properties A, B, C or D can be found using PIE you get:

\[ X = \binom{4}{1} 3^6 - \binom{4}{2} 2^6 + \binom{4}{1} 1^6\]

Since the total number of possibilities is $ 4^6$, the final answer is:

\[ 1 - \frac{X}{4^6} = \frac{195}{512} \approx 0.381\]
[/hide]

Could someone tell me what's wrong with this solution?

Using balls and urns method, the probability would be $ \frac{(4\plus{}2\minus{}1)!/(2!3!)}{(4\plus{}6\minus{}1)!/(6!3!)} \equal{} 5/42$.

mathwizarddude: You've counted the total number of ways for the combinations to occur, but this won't give you the right answer. For example, having 6 babies being born all on Monday is not as likely as an even 1-1-2-2 distribution, but your solution treats both cases as having equal weight. (It's like saying 2 through 12 are all equally likely results of rolling 2 dice.)

Well, I think the problem is that I treated all the babies as non-distinct.  Of course, for the 1-1-2-2 distribution, there are more than one possible way for this to happen if we treat all the babies as distinct.  Am I correct?


Let $ A\equal{}\left(\begin{array}{cc}
1 & 2 \\
0 & 1 \end{array}
\right)$. Solve in $ M_2(\mathbb{C})$ the equation $ Z^n\equal{}A$.

$ Z\equal{}\left(\begin{array}{cc}
\omega & \frac{2\omega}{n} \\
0 & \omega \end{array}
\right)$

where $ \omega$ is the n-th root of 1.  (i.e. $ \omega^n\equal{}1$)


algebra polynomial function calculus derivative inequalities search

1/ Suppose that $f(x,y,z)$ $\in$ $Z[x,y,z]$ . Prove that exits g(x,y,z) and h(x,y,z) $\in$ Z[x,y,z] such that $f(x,y,z).g(x,y,z)=h(x^2,y^2,z^2)$
2/ Find the polynomial $f(x,y)$ $\in$ $Z[x,y]$, deg f as little as possible such that:
	f(x,y)+f(y,x)=0
	f(x,x+y)+f(y,x+y)=0
$\forall$ $x,y$ $\in$ $R$
3/ Find all the polynomial $P(x,y)$ $\in$ $R[x,y]$ such that:
	$x^{100}+y^{100}\leq P(x,y) \leq101(x^{100}+y^{100})$
	$(x-y)P(x,y)=(x-1)P(x,1)+(1-y)P(1,y)$
$\forall$ $x,y$ $\in$ $R$
I didnt solve these problem so someone help me!!!!!!!!!!!!

1. Take $g(x,y,z)=f(x,y,-z)\cdot f(x,-y,z)\cdot f(-x,y,z)\cdot f(x,-y,-z)\cdot f(-x,y,-z)\cdot f(-x,-y,z)\cdot f(-x,-y,-z)$.
Then $F(x,y,z)=f(x,y,z)\cdot g(x,y,z)$ has property $F(\pm x,\pm y,\pm z)=F(x,y,z)$, i.e. F contains even degrees only.

For the nasty problem 3. If you substitute $ Q(x,y)=(x-y)P(x,y)$ you will find $ Q(x,y)=Q(x,1)+Q(1,y)$. Thus, Q has the form $ a_0+...+a_k\cdot x^k+b_0+...+b_n\cdot y^n$. From the condition $ Q(x,x)=0$ it follows that $ a_i=-b_i$ and $n=k$. Thus, your polynomial P has the form $ P(x,y)=c_0+c_1 (x+y)+...+c_{n-1}\cdot (x^{n-1}+...+y^{n-1})$. We also have that $ 2x^{100}\leq P(x,x)\leq 202\cdot x^{100}$. But it is easy to see that a polynomial which verifies $ 2x^{100}\leq f(x)\leq 202x^{100}$ must be of the form $ kx^{100}$. Thinking with big x you will arrive that $deg f=100$ and then thinking with x close to 0 you will find that all coefficients of $x^0, x^1,..., x^{99}$ are 0. This shows that in fact P has the form $ k(x^{100}+x^{99}y+...+xy^{99}+y^{100})$. By substituting $y=tx$ all we need is to find the extreme values of the function $\frac{t^{100}+1}{1+t+...+t^{100}}$ and from here it will follow k and thus P. I will let you compute the derivative and solve the 200-th degree equation!

I dont think the problem is so nasty as you said harazi!
After we have the form of P(x,y) I think we should use the follow inequality to prove
$c^{100}+1\leq2(c^{100}+c^{99}+...+c+1)\leq101(c^{100}+1)$
So we have
$P(x,y)\leq k(x^{100}+y^{100})\frac{101}{2}$
and $P(x,y)\geq k(x^{100}+y^{100}).\frac{1}{2}$
The both equal hold only for k=2
=> $P(x,y)=2(x^{100}+x^{99}y+...+y^{100})$ :)

And where is the beauty of this problem??(here nasty=ugly!) Anyway, , the same value of k can be obtained using the study of the function from my solution, so

I know harazi but....
OK, thanks Myth and harazi for helping me. Now what do you think about the problem 2 ???
I found it : f(x,y)=xy(x-y)(x+y)(x+2y)(2x+y) but I still can't prove that
 :(

Why do you think  f(x,y)=xy(x-y)(x+y)(x+2y)(2x+y) works? Maybe wrong signs?
I found that $f(x,y)=xy(x+y)P(x,y)$. I think it is quite easy to prove that $P(x,y)\neq ax+by+c$, so $\dep P\geq 2$ and you need to find appropriate $P$ of degree 2.

How you found the factor $(x+y)$, Myth
I found only the factorisation $f(x,y)=xy(x-y)(x-2y)(2x-y)P(x,y)$.
Both together we can get $f(x,y)=xy(x+y)(x-y)(x-2y)(2x-y)P(x,y)$   (ugly !  :? ) and we can search for $P(x,y)$. But the more I think about it, the more I believe that the only solution is $f(x,y)=0$.

I think, I did that conlusion because $f(x,x)=0$.

Oh, my solution works for $P(x,y)=1$... :blush:
But for the factor $(x+y)$ wouldnt you need $f(x,-x)=0$

Ops... Indeed, I obtained factor $x-y$, but $x+y$ also can be easily obtained:
take $y=-2x$, then $f(x,-x)+f(-2x,-x)=0$, but $0=f(x,x+x)+f(x,x+x)=2f(x,2x)$, so $f(x,-x)=0$.


geometry
circumcircle
parameterization
trigonometry
ratio
function
analytic geometry

A circle is drawn in the center of an equilateral triangle $ \triangle ABC$ (that is, they are concentric).  Show that for any point $ P$ on the circle, $ PA^2 \plus{} PB^2 \plus{} PC^2$ is a constant.  Does the result still hold if the circle is not entirely contained in the triangle?

[hide]Let $ r$ be the radius of the circle and $ R$ be the circumradius of the triangle.

We can write $ PA^2$, $ PB^2$, $ PC^2$ in terms of a single parameter $ \theta$ as follows using the Law of Cosines:
$ PA^2 \equal{} R^2 \plus{} r^2 \minus{} 2Rr \cos \theta$
$ PB^2 \equal{} R^2 \plus{} r^2 \minus{} 2Rr \cos (\theta \plus{} 60^{\circ})$
$ PC^2 \equal{} R^2 \plus{} r^2 \minus{} 2Rr \cos (\theta \plus{} 120 ^{\circ})$

Adding;
$ PA^2 \plus{} PB^2 \plus{} PC^2 \equal{} 3(R^2 \plus{} r^2) \minus{} 2Rr\left(\cos \theta \plus{} \cos (\theta \plus{} 60^{\circ}) \plus{} \cos (\theta \plus{} 120 ^{\circ}) \right)$

It is necessary and sufficient to prove that $ \cos \theta \plus{} \cos (\theta \plus{} 60^{\circ}) \plus{} \cos (\theta \plus{} 120 ^{\circ})$ is a constant.

In fact, this expression is equal to zero. Either expand using the cosine angle-addition formula or equate real parts of the LHS and RHS in $ e^{i \theta} (1 \plus{} \omega \plus{} \omega^2) \equal{} 0$, where $ \omega^3 \equal{} 1, \; \omega \neq 1$.

Since the ratio $ r/R$ does not affect the application of the Law of Cosines above, it's perfectly fine for the circle to spill out of the triangle.[/hide]

[b]Follow-up:[/b]  Is the same true of $ PA \cdot PB \cdot PC$?  (I would be especially interested in a geometrical method that did not explicitly use complex numbers.)

This result is obviously true for a regular $ n$-gon. Assume that this n-gon is inscribed in a unit circle.

Let $ r\in \mathbb{R}^{\plus{}}$. $ p$ is such that $ |p|\equal{}r$.

Let $ z,z^2,...,z^{n}$ be the $ n$th roots of unity that represent the verticies $ V_1$, $ V_2$,..., $ V_n$ of the regular $ n$-gon. Now \[ \sum PV_i^2\equal{}\sum (p\minus{}z^i)\overline{(p\minus{}z^{i})}\equal{}\sum p\cdot\overline{p}\minus{}p\overline{z^i}\minus{}\overline{p}z^i\plus{}z^i\overline {z^i}\]
Since we are summing over all $ n$th roots of unity, the conjugation of that sum is still the same, so
\[ \equal{}n r^2\minus{}(p\plus{}\overline{p})\sum z^i\plus{}n\equal{}n(r^2\plus{}1)\]

[quote="Altheman"]This result will follow for a regular $ n$-gon. Assume that this n-gon is inscribed in a unit circle.

Let $ r\in \mathbb{R}^{ \plus{} }$. $ p$ is such that $ |p| \equal{} r$.

Let $ z,z^2,...,z^{n}$ be the $ n$th roots of unity that represent the verticies $ V_1$, $ V_2$,..., $ V_n$ of the regular $ n$-gon. Now
\[ \sum PV_i^2 \equal{} \sum (p \minus{} z^i)\overline{(p \minus{} z^{i})} \equal{} \sum p\cdot\overline{p} \minus{} p\overline{z^i} \minus{} \overline{p}z^i \plus{} z^i\overline {z^i}
\]
Since we are summing over all $ n$th roots of unity, the conjugation of that sum is still the same, so
\[ \equal{} n r^2 \minus{} (p \plus{} \overline{p})\sum z^i \plus{} n \equal{} n(r^2 \plus{} 1)
\]
[/quote]
No, there is no "obviously true" about it!

I made a sensible substitution in its place.
When you type mathematics, and certainly to explain it to
an audience, leave out your biases of subjective words
and phrases.  Use objective ones.

If you use the complex numbers argument, you'll realize that none of the argument depends on the choice of roots of unity, so in that sense it's "obvious" from an algebraic perspective, but again, what I'm interested in is whether this result can be explained using geometric intuition.

t0rajir0u,

I worked on the PA*PB*PC supplemental problem,
but I did not get a constant at the end of it.

In the xy-plane, I let the circle be x^2 + y^2 = 1, and I had
the vertices of the equilateral triangle, A, B, C,  be:
A:  (-3 - sqrt3)/3, -1)
B:  (3 + sqrt3)/3, -1)
C:  (0, sqrt3)

Without loss of generality, I used (x, sqrt(1 - x^2)) for point P.
I worked out expressions for the length of line segments
PA, PB, PC, and multiplied them using the "Expand" feature
of the http://www.quickmath.com website.

In more ways than one, the variable terms were not cancelling
out to zero for me as I condensed them.

For Constant = (PA)^2 + (PB)^2 + (PC)^2,

when I use A = (-3 - sqrt3)/3, -1), B = (3 + sqrt3)/3, -1),
C = (0, sqrt3), and vary between two different P points
of (1,0) and (0,1), I got two different sums (constants):

P = (1,0), Constant = (32 + 4sqrt3)/3

P = (0,1), Constant = (44 - 2sqrt3)/3

Why the difference?

I have the length of line segment AB equal to the length
of line segment BC, when I applied the distance formula.
Also, I worked out that the circle is internally tangent to the
triangle at (-1/2, (sqrt3)/2), (1/2, (sqrt3)/2), and (0, -1).

You are correct that I may not have chosen my words carefully. I apologize. But I still say that my generalization did not need a huge leap of faith.

If you look at TZF's proof carefully, it almost directly follows that it is true for any n-gon.

It relied on the fact that $ e^{i\theta}(1\plus{}\omega\plus{}\omega^2)\equal{}0$

Now by analogy, it is reasonable to see what is actually going on here... In particular,

$ e^{i\theta}(1\plus{}z\plus{}z^2\plus{}...\plus{}z^{n\minus{}1})\equal{}0$.

When we expand the necessary sum, we get this, and we're done. This intuition gets formalized into a proof which I presented.

[quote="TZF"]
It is necessary and sufficient to prove that $ \cos \theta \plus{} \cos (\theta \plus{} 60^{\circ}) \plus{} \cos (\theta \plus{} 120 ^{\circ})$ is a constant.
[/quote]

Is this really constant? I don't know a whole lot about complex numbers so i couldn't follow that bit of the proof, but expanding the equations gives;

$ \cos(x) \minus{} \sqrt {3}\sin(x)$

I think it should be: $ \cos(x) \plus{} \cos(120 \plus{} x) \plus{} \cos(120 \minus{} x)$

[quote="ocha"][quote="TZF"]
It is necessary and sufficient to prove that $ \cos \theta \plus{} \cos (\theta \plus{} 60^{\circ}) \plus{} \cos (\theta \plus{} 120 ^{\circ})$ is a constant.
[/quote]

Is this really constant? I don't know a whole lot about complex numbers so i couldn't follow that bit of the proof, but expanding the equations gives;

$ \cos(x) \minus{} \sqrt {3}\sin(x)$[/quote]
The correct expansion is in fact $ 0$.  See a non-complex numbers approach to the corresponding sine problem [url=http://en.wikipedia.org/wiki/Three-phase#No_neutral_current]here[/url].

The follow-up is incorrect; my apologies.  (The answer is not constant, but it is a straightforward function of the location of $ P$ as a complex variable.)

[quote="Arrange your tan"]when I use A = (-3 - sqrt3)/3, -1), B = (3 + sqrt3)/3, -1),
C = (0, sqrt3)[/quote]
Unless I'm sorely mistaken, these are not the coordinates of an equilateral triangle.  (And as far as Ravi B's generalization, their centroid is not the origin.)

[quote="t0rajir0u"][quote="ocha"][quote="TZF"]
It is necessary and sufficient to prove that $ \cos \theta \plus{} \cos (\theta \plus{} 60^{\circ}) \plus{} \cos (\theta \plus{} 120 ^{\circ})$ is a constant.
[/quote]

Is this really constant? I don't know a whole lot about complex numbers so i couldn't follow that bit of the proof, but expanding the equations gives;

$ \cos(x) \minus{} \sqrt {3}\sin(x)$[/quote]
The correct expansion is in fact $ 0$.  See a non-complex numbers approach to the corresponding sine problem [url=http://en.wikipedia.org/wiki/Three-phase#No_neutral_current]here[/url].
[/quote]

The three-phase-power artice only backed up my theory that the corrct expression is $ \cos(x) \plus{} \cos(120 \plus{} x) \plus{} \cos(120 \minus{} x)$

 [geogebra]220d08f40dd19aa2e0a3d988024f607110e90024[/geogebra]

This is how i interpret TZF's answer

Ah, my apologies again; what we really want is $ \cos x \plus{} \cos (x \plus{} 120) \plus{} \cos (x \plus{} 240)$ (in the interests of symmetry), which is the real-variable expression equal to the real part of what TZF gave.

[quote="t0rajir0u"]

[quote="Arrange your tan"]when I use A = (-3 - sqrt3)/3, -1), B = (3 + sqrt3)/3, -1),
C = (0, sqrt3)[/quote]
Unless I'm sorely mistaken, these are not the coordinates of an equilateral triangle.  (And as far as Ravi B's generalization, their centroid is not the origin.)[/quote]
You are "sorely mistaken."  They are the coordinates of an equilateral
triangle.  However, this triangle does not have (0,0) as that center.
For the circle inscribed, its center should be (0, (sqrt3 - 2)/3).

My previous work doesn't count because I used the wrong center.

[quote="Ravi B"]I think we can generalize beyond regular polygons.  Let $ A_1$, $ A_2$, ..., $ A_n$ be arbitrary points.  Let $ G$ be the centroid of these points.  Let $ \omega$ be a circle with center $ G$.  Then for all points $ P$ on $ \omega$, the sum $ \sum {P A_i}^2$ is a constant.[/quote]
How would we prove this general case?

I did it using vectors.  We can shift so that the centroid $ G$ is at the origin.  Then we have
\[ PA_i^2 \equal{} \left\Vert P \minus{} A_i \right\Vert^2 \equal{} \left\Vert P \right\Vert^2 \minus{} 2 P \cdot A_i \plus{} \left\Vert A_i \right\Vert^2.
\]
Now sum over all $ i$.


[quote="morheli"]i have a question here, can we juz normalise it to a^2+b^2+c^2=1?[/quote]
Yes,we can.

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \sum \frac {a^3}{a^2 \plus{} b^2} \ge \frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}
\]
[/quote]

$ f(a,b,c)\equal{}\sum \frac {a^3}{a^2 \plus{} b^2} \minus{} \frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}$

$ f(ka,kb,kc) \equal{} \frac{k^3a^3}{k^2(a^2\plus{}b^2)} \plus{} \frac{k^3b^3}{k^2(b^2\plus{}c^2)} \plus{} \frac{k^3c^3}{k^2(c^2\plus{}a^2)} \minus{} \frac{\sqrt { 3 ( k^2 ( a^2\plus{}b^2\plus{}c^2) ) } }{2}$

$ f(ka,kb,kc) \equal{} k ( f(a,b,c) )$

Since $ a,b,c > 0$ as long as $ k>0$ we may normalise.

Very nice inequality, Can_hang2007! :lol:

It is proved if we can show that
$ \sum S_a(b\minus{}c)^2\ge 0$,
where $ S_a\equal{}\frac {2c(a\plus{}b\plus{}c)}{b^2\plus{}c^2}\minus{}1$.

[quote="Vasc"]Very nice inequality, Can_hang2007! :lol:

It is proved if we can show that
$ \sum S_a(b \minus{} c)^2\ge 0$,
where $ S_a \equal{} \frac {2c(a \plus{} b \plus{} c)}{b^2 \plus{} c^2} \minus{} 1$.[/quote]
Thank you very much, VasC.
But there is a nice solution only with AM-GM. :)

[quote="zaizai-hoang"]No, I don't think so  :wink: the inequality is cyclic not symmetric  :wink:[/quote]
you is wrong! :roll:

[quote="shfdfzhjj"][quote="Vasc"]Very nice inequality, Can_hang2007! :lol:

It is proved if we can show that
$ \sum S_a(b \minus{} c)^2\ge 0$,
where $ S_a \equal{} \frac {2c(a \plus{} b \plus{} c)}{b^2 \plus{} c^2} \minus{} 1$.[/quote]

Can you tell me how to prove this?
can_hang2007:I want to see your proof.[/quote]
Here is my solution. :)

Using the Can_hang method, we can prove it by multiplying the inequalities:

$ \prod (2ab\plus{}bc\plus{}ca)\ge 2(\sum ab)^3$,

$ 9\prod (a^2\plus{}b^2) \ge 8(\sum a^2)(\sum a^2b^2)$,

$ 3\sum a^2 \ge (\sum a)^2$,

$ 9\prod (a\plus{}b) \ge 8(\sum a)(\sum ab)$,

$ \prod (a^2\plus{}b^2) \ge \prod (a^2\minus{}b^2)$,

$ 16(\sum ab)^4(\sum a^2b^2) \ge 9(\sum a^2b^2\plus{}abc\sum a)^3$.

The last inequality is equivalent to
$ 16(1\plus{}2y)^2 \ge 9(1\plus{}y)^3$,
which is trivial for $ 0\le y\le 1$.

Can, did you try this directly with the $ p,q,r$ method?

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \sum \frac {a^3}{a^2 \plus{} b^2} \ge \frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}
\]
[/quote]

The tronger and harder problem:  Let $ a,b,c$ be positive real numbers. Prove that

$ \sum \frac {a^4}{a^3 \plus{} b^3} \ge \frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}$

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}+\frac{a+b+c}{6}\ge2\sqrt{\frac{a^2+b^2+c^2}{3}}$$

[quote=arqady]The following inequalities are also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
1. \[\sum_{cyc}\frac{a^3}{a^2+b^2}\ge\frac{\sqrt{4(a^2+b^2+c^2)-ab-ac-bc}}{2}\]
2. \[\sum_{cyc}\frac{a^5}{a^4+b^4}\ge\frac{\sqrt{3(a^2+b^2+c^2)}}{2}\][/quote]

Dear arqady!
Can we prove it by Cauchy-Schwarz

[quote=zaizai-hoang]No, I don't think so  :wink: the inequality is cyclic not symmetric  :wink:[/quote]

Masson

Or maybe by Holder  :maybe:  I have read some following inequality in a documentary of VQBC.


List all multiples of 72 which have more than 12 but less than 24 factors.

[hide]Well. . So, you realize that 72 has 4*3 = 12 factors already. If you have 2^3 * 3^2 * c, you have 4*3*2 = 24, which is too many factors, so you have to have the multiple in the form 2^x*3^y, where (x+1)(y+1) <24, and possible solutions are 72 * (2,3,4,6,8,9) which is 144, 288, 432, 576, 216, and 648.[/hide]

[hide]144,216,288,432,648

These are 2,3,4,6, and 9 times 72.

First we factor 72 into 2^3*3^2. 4*3=12 factors. Adding another prime would give us at least 24, so the multiple of 72 can only be divisible by 2 and 3. 12, which is 2^2*3, gives us 6*4=24 factors, so 2,3,4,6,and 9 work.[/hide]

How would you find out how many factors a number has?

I ll give you some examples and we ll see if you can't figure it out.

\[360=2^3\cdot3^2\cdot5^1\]

360 has 24 factors

\[17=17^1\]

17 has 2 factors

\[625=5^4\]

625 has 5 factors

\[290=2\cdot5\cdot29\]

290 has 8 factors

See if you can figure out what I am doing. Then try to give a good reason why that method works.


[color=darkred]Let $ABC$ be a right-angled triangle $(AB\perp AC,\ AB<AC)$. Denote the projection $D$ of the vertex $A$ on the hypotenuse $BC$. Define the point $M\in (DC)$ for which $\frac{MD}{MC}=\frac{h_a}{a}$ and define the projection $N$ of the point $B$ on the line $AM$. Prove that the quadrilateral $BINC$ is cyclically.[/color]

Erect a normal to the line BC at the vertex C and let P be a point on this normal, on the opposite side of the line BC than the vertex A, such that PC = BC = a. The line AP meets the line BC at a point M such the $\frac{MD}{MC} = \frac{AD}{PC} = \frac{h_a}{a}$. Let (O) be the circumcircle of the right triangle $\triangle ABC$ and let (K) be the circumcircle of the triangle $\triangle BIC$. The angle $\angle BIC = \frac{\angle A}{2} + 90^\circ = 135^\circ$ spans the minor arc BC of the circle (K), hence its center K is the midpoint of the arc BC opposite to the vertex A of the circumcircle (O). The angle spanning the major arc BC of the circle (K) is equal to $180^\circ - \angle BIC = 45^\circ$. Since the triangle $\triangle BCP$ is right and isosceles, $\angle BPC = 45^\circ$, so that the point $P \in (K)$ and BP is a diameter of the circle (K). The circle (K) meets the line $AP \equiv AM$ again at a point N and the angle $\angle BNP \equiv \angle BNM = 90^\circ$ on the diameter BP is right, so that N is the given projection of vertex B to the line AM.


algebra
polynomial
linear algebra
linear algebra unsolved

Let $ A \in M_{n}(C)$ .Prove or disprove that always the exist a polynomical $ P \in C[X]$ Such that : $ P(e^A)\equal{}A$
 :?:  Thanks !

This is a rather trivial question as stated since you can just take $ P \equal{} A$ (or, more generally, you can tweak the constant term to account for the error in all of the other terms).  And you clearly can't take $ P$ nonconstant and independent of $ A$ since you can take $ A \equal{} tI$ and differentiate.

Let $ A,B \in M_{n}(C)$ and  $ e^Ae^B\equal{}e^Be^A$ then $ AB\equal{}BA$ ??? :?: [b]Question of Ticuren[/b]

Hi t0rajir0u, I dont understand your post.
in fact we have the following result:
$ A$ is a polynomial in $ e^A$ iff
for all $ \lambda,\mu\in{spectrum(A)}$, $ \dfrac{\lambda\minus{}\mu}{2i\pi}$ is not a non zero integer.

Oops, my apologies; I forgot that the constant term is a multiple of the identity  :oops:


Find all natural numbers $N$ which satisfy the following properties :
(i)  $N$ has exactly $6$ distinct positive factors $1,d_1,d_2,d_3,d_4,N$
(ii) $1+N=5(d_1+d_2+d_3+d_4)$


Two guys are playing chess, only that each one moves two times. Prove that there exists a strategy for the player with white pieces, for him to win, with no matter what black moves.

This is a classic use of a game theory technique:

Suppose that this is false, and black had a winning strategy (I am assuming someone has a winning strategy).  Then white could move his knight out front, and then back into place, thereby essentially "passing" his move to black.  But now white is the second player, and so he now has the winning strategy.  This is a contradiction, since both players cannot have a winning strategy.

Note that this doesn't not work when there is only one move, because it is uncertain that a pawn in the middle of the board, for instance, is always an advantage.

To make that approach work, you would have to prove that the game is not a draw with best play.

This is probably a stupid comment, but do they have to make it before move 50 ?


Let $ f \in W^{1,p}(\mathbb{R}^n)$ and $ f$ compact support.
Let $ f_m \in C_c^{\infty}$ be such that  $ f_m \rightarrow f$in $ W^{1,p}$.

Suppose $ \|f_m \minus{} f_n\|_{C^{0,\gamma}} \leq C \|f_m \minus{} f_n\|_{W^{1,p}}$
Let $ \bar{f}\in C^{0,\gamma}$ be such that $ f_m \rightarrow \bar{f}$ in $ C^{0,\gamma}$.

Then $ f \equal{} \bar{f}$ $ a.e.$??? WHY ?? please help me.

I am studying a PDE book by Evans . This question is from 269 page.. This is easy but I don't know..

Convergence in $ W^{1,p}$ implies convergence in $ L^p$. Convergence in $ C^{0,\gamma}$ implies uniform convergence. Now, assume that $ m\{|f\minus{}\bar f|>\delta\}\equal{}\mu>0$ for some $ \delta>0$. Since $ f_n$ converge to $ \bar f$ uniformly, we have $ \|f_n\minus{}\bar f\|_{L^\infty}<\frac\delta 2$ for large $ n$ and, thereby, $ \|f_n\minus{}f\|_{L^p}\ge \mu^{1/p}\frac\delta 2$, which contradicts the convergence of $ f_n$ to $ f$ in $ L^p$.

$ f_n \to f$ in $ L^p$ implies $ f_{n_k} \to f$ pointwise a.e. for some subsequence. Since $ f_{n_k} \to \overline{f}$ uniformly, $ f \equal{} \overline{f}\, a.e.$


trigonometry
inequalities
geometry
trig identities
Law of Sines

Let ABCD be a convex cyclic quadrilateral. Prove that
|AB-CD| + |AD-BC| >= 2|AC-BD|

A question to TripleM :

I think I've shown (using law of sines and a bit of trigonometry) that :
|AB-CD| >= |AC-BD| and
|AD-BC| >= |AC-BD|
which appears to be stronger than the problem and so it makes me feel I'm wrong somewhere. Do you confirm ?

I haven't worked on it much, but I think that approach would be perfectly valid, and it isn't necessarily a stronger condition since if the equality cases hold in both of your two inequalities, then the equality case will hold in the actual problem. So it looks good to me.

I knew I had seen this before...
This problem appeared on the 1999 USAMO.  Just thought I'd mention it.

Pointless post, I know. But it's interesting information.

i/3:

Yeah, thats fine, it is true for each part separately. Maybe the problem was stated like that to make you think it was harder or something.. no idea.

It's a stronger condition because the original statement admits, say, |AB-CD| < |AC-BD|, |AD-BC| >= 2|AC-BD| - |AB-CD|.