What is $ \frac{23074356_8}{2}$when expressed in hexadecimal?

[quote="erdogankerem123"]What is $ \frac {23074356_8}{2}$when expressed in hexadecimal?[/quote]
[hide]
$ \frac {23074356_8}{2}$
$ 11436167_8$
Every 4 digits make up 3 digits in hexidecimal. $ 1143$ is $ 263$, and $ 6167$ is $ C77$
so in Hex, it shuld be $ 263C77$
[/hide]

IRT ilikepie's post: I like your solution better :D

I've never been really good at bases; what's a fast way of dividing/multiplying in another base?

The same way you divide and multiply in base $ 10$, only the multiplication tables are different.

This problem is special, though; dividing by $ 2$ in binary is as easy as dividing by $ 10$ in base $ 10$, so you don't really have to worry about general division.


[quote="ridgers"]Can you explain me why the function f(f(y))=y is a onto function???[/quote]

Because, $ \forall a\in\mathbb R$, $ \exists b\mathbb R$ (just take $ b\equal{}f(a)$) such that $ f(b)\equal{}a$

[quote="ridgers"]Thanks, and sorry for my ignorance.[/quote]

No prob.

You can also conclude that $ f(x)$ is a bijection. (it's onto as we saw and it's also injective since $ f(x_1)\equal{}f(x_2)$ $ \implies$ $ f(f(x_1))\equal{}f(f(x_2))$ and so $ x_1\equal{}x_2$)

[quote="ridgers"]Can you explain me why the function f(f(y))=y is a onto function???[/quote]
How about $ f(f(x))\equal{}ax$, where $ a$ is some constant value?? Is it on-to or bijection??

Yeah i see that it is also injektiv and thaks a lot.

[quote="mathVNpro"][quote="ridgers"]Can you explain me why the function f(f(y))=y is a onto function???[/quote]
How about $ f(f(x)) \equal{} ax$, where $ a$ is some constant value?? Is it on-to or bijection??[/quote]

Yeah this is also a one to one and injective function so it is a bijection, to prove it use the same way as for the function I asked.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14770[/img]

this is my 93º post (5 were double?)

 :(

The length of the hypotenuse is $ AB \equal{} \frac {2r^2}{h \minus{} 2r}$ and is done.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14789[/img]
\[ \frac {{AB}}{r} \equal{} \frac {{2r}}{{h \minus{} 2r}}
\]
[b]AB r h  datum[/b]

[b]and arc capable 90º of AB[/b]

 :thumbup:


Three coins are selected from a collection consisting of 2 pennies, 2 nickels, 2 dimes, 2 quarters, and 2 half-dollars. How many different sums are possible?

10*9/2=45 as there aren't 4 coins a,b,c and d such that a+b=c+d

Will anyone answer this problem?
This has remained unanswered.

[quote="Binomial-theorem"][hide="Solution"]
We start off with seeing if we can have a+b+c=d+e+f

Penny:
1+b+c=d+e+f
Because the RHS is going to be $\equiv (1,6)\pmod{10}$ and no other cases that don't have a penny have this, no other cases work.

5+b+c=d+e+f

None of the values of (d,e,f) can equal (a,b,c), therefore we are going to have:
b=5 (c+10=d+e+f$\implies$ c=5 gives us (d,e,f)=5, c=10 gives us (10,5,5) and so forth)
No combination can be chosen without repeating.

If we continue to do this, we will realize that all cases will be the same as the two above.
Therefore we have:
$\binom{(2+2+2+2+2)}{3}=\binom{10}{3}=\frac{10*9*8}{3*2}=\frac{10*3*8}{2}=$
$4*3*10=\boxed{120}$ [/hide][/quote]

I have no idea what you did, but I am sure that you are incorrect. $\binom{10}{3}$ is overcounting a lot...

Edit: Oh, after rereading your solution many times, I think I finally understand what you're saying. Well, your statement is still incorrect. Take for example the number $60$. You can acquire $60$ through $50+5+5$ or through $25+25+10$. $2$ different ways to acquire the same number.

[hide="I'm pretty sure the only way to do this is by bashing"]
So, to start off, we can have $0$, $1$, or $2$ pennies. Now, let's bash...
Assume we have $0$ pennies. With $3$ coins, we can get $20$ cents, $25$ cents, $35$ possible, $40$ possible, $45$ possible, $55$ possible, $60$ possible, $65$ possible, $70$ possible, $80$ possible, $85$ possible, $100$ possible, $105$ possible, $110$ possible, $125$ possible, and that's it. So, $15$ cases here.

Next, $1$ penny. (We're not going to count the extra $1$ cent here as it doesn't matter) So, we can have $10$ cents, $15$ cents, $20$ cents, $30$ cents, $35$ cents, $50$ cents, $55$ cents, $60$ cents, $75$ cents, and $100$ cents here. So that's $10$ here.

Finally, $2$ pennies. This is pretty simple, only $4$ options here (what kind of other coin you want).

Summing these we get $15+10+4=\boxed{29}$.

I probably missed a case, but .....
[/hide]

My way:

If you pick $3$ different numbers, you have $\binom{5}{3}=10$ different combinations.

Now, you pick $2$ of the same number and $1$ different number. There are $5\cdot 4=20$ ways to do this.

You still need to check if there are any duplicates. This can be done with logic. There can't be any duplicates with case $1$. However, with case $2$, you can notice that $2$ nickles equal $1$ dime and $2$ quarters equal $1$ half dollar, so $60$ can be acquired twice, and is also the only one that can be acquired multiple times (this can be a bit tricky).

Therefore the answer is $10+20-1=29$.


calculus
derivative
trigonometry
calculus computations

Find the general solution of differential equation:
\[\frac{d^5y}{dx^5}+\frac{d^2y}{dx^2}=x\]
and the solution of the system:
\[\{\begin{array}{l}\displaystyle\frac{d^2y}{d^2x}-5\frac{dy}{dx}+13y+\frac{dz}{dx}+20z=0\\
\\
\displaystyle\frac{dy}{dx}+2y+\frac{d^2z}{dx^2}+3\frac{dz}{dx}+3z=0
\end{array}\]

On the second problem: I don't particularly feel like solving it right now, but the first thing I thought of was to rewrite it as a system of four first order (linear, constant coefficient, homogeneous) differential equations, and then find the eigenvalues and eigenvectors of a $4\times4$ matrix.

The ways Kent and Kunny suggested are, in my opinion, good, but too long. I prefer
using the beautiful trick of reapeted differentation that, in the case of our diff. equation, gives us:
\[\displaystyle\frac{d^7y}{dx^7}+\frac{d^4y}{dx^4}=0\]
Follow the caratteristic equation $m^7+m^4=0$ whose solutions are $m = 0 \,(4\, \txt{times}),\quad m = -1,\quad \bigg|m+\frac 12\bigg|=\frac {i\sqrt{3}}2$.
Then
\[y=A+Bx+\frac C2x^2+\frac D6x^3+Ee^{-x}+e^{-\frac x2}\bigg(F\cos\frac{\sqrt{3}}2x+G\sin\frac{\sqrt{3}}2x\bigg)\quad d (1)\]
Replacing the derivate of y in the original equation with the help of (1) we get $\text{C\&=0\&}\,\,\text{and}\,\,\text{D\&=1\&}.$
Almost the same for the system.


6. Prove that for any real number  $x,y,z$

$3(x^2-x+1)(y^2-y+1)(z^2-z+1) \geq (xyz)^2+xyz+1$


geometry
perimeter
vector
trigonometry
integration
geometry solved

Prove (or disprove :) ) that the perimeter of a convex n-gon is smaller than the perimeter of any disc that contains this n-gon

Let $\ell$ be a variable line, and consider one of the vectors $\vec{v}$ that makes up one of the sides of the polygon. What is the length of the projection of $\vec{v}$ onto $\ell$? It is $|v\cos\theta|$, where $v=|\vec{v}|$ and $\theta$ is the angle that $\vec{v}$ makes with the line. As the line spins all the way around, we can find that the [i]average[/i] length of this projection is $\frac{1}{2\pi}\int_{0}^{2\pi}v|\cos\theta|\,d\theta=\frac{2v}{\pi}$. Thus, the average length of the projection of the polygon on this variable line is $\frac{2p}{\pi}$, with p=perimeter. (Notice that the projection is a line segment, and this value is twice the average length of this segment.) Next, the length of the projection of the disk containing the polygon is greater than the length of the projection of the polygon for each orientation of the line, so the average length of the projection of the circle ($\frac{2c}{\pi}$, c=circumference) must be larger than the average length of the polygon's projection ($\frac{2p}{\pi}$), so we're done.

Or, without calculus.

Let the polygon be $A_1A_2\ldots A_n$. Draw line $A_1A_2$, and let it intersect the disk at points $X_1$ and $Y_1$. Replacing arc XY with segment XY decreases the perimeter of the disk, changing it into a convex figure that still contains the polygon. Now, let line $A_2A_3$ intersect this outer shape at new points $X_2$ and $Y_2$, and replace the portion of hte perimeter from $X_2$ to $Y_2$ with the segment between these two points, thus not increasing the perimeter again. Now continue all the way around the polygon. The perimeter of the outer disk-like figure will decrease (or at least not increase) at each step, and indeed [i]will[/i] decrease at the first step, and it will end up being exactly the polygon. So the perimeter was larger to start with.


Find the greatest real number $ K$ such that for any triple of positive real numbers $ a$, $ b$, $ c$ such that $ Kabc > a^{3}\plus{}b^{3}\plus{}c^{3}$, there exists a triangle with sides of lengths $ a$, $ b$ and $ c$.

The problem is not so hard. Here is a solution:

We will show that the maximal K for which any three positive reals a, b, c satisfying $ Kabc > a^{3}\plus{}b^{3}\plus{}c^{3}$ must be the sidelengths of a triangle is K = 5.

In fact, for K > 5, the three positive reals 2, 1, 1 satisfy $ K\cdot 2\cdot 1\cdot 1 > 2^{3}\plus{}1^{3}\plus{}1^{3}$, without being the sidelengths of a triangle. So the required maximal K must be $ \leq 5$. In order to show that it equals 5, we have to prove the following assertion:

[b](*)[/b] If three positive reals a, b, c satisfy $ 5abc > a^{3}\plus{}b^{3}\plus{}c^{3}$, then they are the sidelengths of a triangle.

In fact, assume the contrary: Assume that there are three positive reals a, b, c satisfying $ 5abc > a^{3}\plus{}b^{3}\plus{}c^{3}$, but they are not the sidelengths of a triangle. Since everything is symmetric, we can WLOG assume that $ a\geq b\geq c$. Then, c + a > b (since $ a\geq b$) and a + b > c (since $ a\geq c$), so that we must have $ b\plus{}c\leq a$ (else, the positive reals a, b, c would be the sidelengths of a triangle). Hence, c + a - b > 0, a + b - c > 0 and $ b\plus{}c\minus{}a\leq 0$, so that

$ \left(b\plus{}c\minus{}a\right)\left(c\plus{}a\minus{}b\right)\left(a\plus{}b\minus{}c\right)\leq 0$.

Thus,

$ 5abc\minus{}\left(b\plus{}c\minus{}a\right)\left(c\plus{}a\minus{}b\right)\left(a\plus{}b\minus{}c\right)\geq 5abc$.

On the other hand, $ 5abc > a^{3}\plus{}b^{3}\plus{}c^{3}$. Thus,

$ 5abc\minus{}\left(b\plus{}c\minus{}a\right)\left(c\plus{}a\minus{}b\right)\left(a\plus{}b\minus{}c\right) > a^{3}\plus{}b^{3}\plus{}c^{3}$
$ \Longleftrightarrow\ \ \ \ \ 10abc\plus{}\left(a^{3}\plus{}b^{3}\plus{}c^{3}\right)\minus{}\left(a\plus{}b\plus{}c\right)\left(bc\plus{}ca\plus{}ab\right) > a^{3}\plus{}b^{3}\plus{}c^{3}$
$ \Longleftrightarrow\ \ \ \ \ 10abc >\left(a\plus{}b\plus{}c\right)\left(bc\plus{}ca\plus{}ab\right)$.

Division by abc yields

$ 10 >\left(a\plus{}b\plus{}c\right)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)$.

Now, since $ 10 \equal{} 3^{2}\plus{}1$, the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14091]IMO 2004 problem 4[/url] yields that a, b, c are the sidelengths of a triangle, contradicting our assumption. Thus, our assumption was wrong, and the assertion [b](*)[/b] must be true. Qed.

[b]EDIT:[/b] Obviously, the above yields: Let $ K$ be a real. Then, the following two assertions are equivalent:
- For any triple of positive real numbers $ a$, $ b$, $ c$ satisfying $ Kabc > a^{3}\plus{}b^{3}\plus{}c^{3}$, there exists a triangle with sides of lengths $ a$, $ b$ and $ c$.
- We have $ K\leq 5$.

  Darij

Firstly, we try to prove :
If $ 5abc>a^3+b^3+c^3 $ then $a,b,c$ must be three sides of a triangle.
Assume that $ a\geq\ b\geq\ c $ and $ c=1 $
The inequality becomes $ f(a)=a^3-5ba+b^3+1 <0 (*)$
If there exist a triple (a,b,c) which cannot form three sides of a triangle and satisfies (*), there must exist $ a\geq\ b+1 $ and  $ f(a)<0 $
We have : $ f'(a)=3a^2-5b \geq\ 3(b+1)^2-5b\geq\ 0 $
So $ f(a)\geq\ f(b+1)=2(b+1)(b-1)^2\geq\ 0 $
So for all $ a\geq\ b+1, f(a) \geq\ 0 $ so a must be smaller than b+1 and then a,b,1 are three sides of a triangle.

If K>5, take (a,b,c)=(2,1,1). This triple satisfies $ Kabc>a^3+b^3+c^3 $ but cannot form a triangle. So K is the greatest.

For the smallest, I do not believe it exits.


Forty-One Rooks are placed on a 10 X 10 chessboard. Prove that there must exist 5 rooks, none of which attack each other.

[quote="0714446459923"][hide]
Becuase of the pgeonhole principle, at least 5 rows are occoupied by a rook. Take a rook from each of these rows. SInce they are on different rows, they can't possibly attack each other.[/hide][/quote]
What if the chosen rooks are on the same column?

You're solution doesn't make much sense... or maybe I'm just dumb.

[hide="Cool Solution"]
Divide the chessboard up into 10 groups in the following way:

Make one of the two long diagonals group 1.
Make the diagonal of 9 right above it and the corner below it group 2.
Make the diagonal of 8 above that and the diagonal of two above the corner group 3.
...

All of the squares in each group are in different rows or columns.
By the pigeonhole principle, 5 of the rooks must all be one of these groups, so those 5 don't attack each other.[/hide]


What do you mean? Round 1? See appropriate topic.

Are the problems of round 1 5th edition discussed somewhere on the forum ?  :?
If yes can anyone post the links here ??  :lol:

[quote="Myth"]Hmm... Where is that post by mathpk? :?[/quote]
I had deleted it because it's wrong  :blush: .
[url=http://www.mathlinks.ro/Forum/topic-34789.html#]Solutions from Mecrazywong[/url]

I've ment the solution to exercise no1 round one

[quote="zhaobin"]yeah.my solution is only $x=y$
I don't know if it is correct. :)[/quote]

I agree.

[quote="rockymarcianocosmin99"]Yes , but we have (2005,0) and (0,2005) as solutions  :P[/quote]

why?!

[quote="rockymarcianocosmin99"]Well didn't we have \[x^3-y^3=2005(x^2-y^2)\]??? :?[/quote]

oh, i remember the question wrongly, sorry.

But 0 is not positive integer,isn't it?
Maybe I don't understand something. :?

The only solutions are positive integers $x=y$.

Thanks for siuhochung and mecrazywong.My brain has been developed.

O , so positive integers do not contain 0?Excuse me I'm not so good at english! :blush:

[quote="Xixas"]For Q2 you do not show that every prime of the form $6k+1$ satisfies the conditions, siuhochung.[/quote]

i could not solve question 2 actually.

for problem1 round 1, I have a little different proof from Suihochung
It is to prove that
If $x^2+xy+y^2\vdots 401(or 5)$ then $x\vdots 401(or 5)$ and $y\vdots 401(or 5)$

[quote="mathpk"]I think answer are $5,7$ and $13$.[/quote]

you mean question 2? I think 31 is also correct

my solution is 5,7,13,31,too.
and here is my solution except 3.2,
hope you like it.

Here is my solutions (only to exercise one)

zhaobin, I am wondering how it is possible!!!

For first one i almost did same as u. For 2, i was stucked at one point and u overcome that point. And for 3, i also solved a) and failed b) and u too!!! Hmmm..  :D  :D  :D

Anyhow, i hope we will keep up our matching in round2 huh? :P

here are my lengthy solutions
i hope that they are correct

Edit: They were here... If you really want to see them, PM me.


Becca is standing in a hole that is 50 inches deep.  She will dig the hole 54 inches deeper so that the top of her head will be the same distance below ground level as it is now above ground level.  What is Becca's height, in inches?

so i just looked at the problem and visualized two points.  The first is where her head is above ground.  the second is below ground.  The difference is 54 inches.  the midpoint is the ground.
Becca is 50+27 inches or 77 inches.

Now her head pops up 27+ground
dig down 54 feet and 27-ground

54/2=27, that's what i got after thinking about it.

uhh I got a different answer
x-50=104-x
x=77

I get 77 too, and that's after I just did something just like this one (except a little harder).

It can't be 27 or else Becca's head wouldn't be out of the hole in the first place, and that's a little short too.

What? Have you never seen a female at 6'5"?
At least this time her head is out of the hole.

Here's a problem like this one:

A 182 cm tall man is digging a hole in the ground. He stops for a moment and says: "I am done with one quarter of the hole. When I finish the job the top of my head is going to be three times as far under the ground as far it is above the ground now." How deep is the hole going to be?

[color=indigo]I get 77 also for the first problem. for the second problem:
182-x = 3x-182
4x = 364
x = 83
so the hole will be 249 inches deep.[/color]

[hide="different answer"]The man is 3/4 above the ground. so you find 3/4 of his height, multiply by three, then add the man's height to the product.
$3/4*182*3+182=591.5cm$
[/hide]

alright im new to this forum (btw i found this site thru google and it makes me feel like im retarded.. but.. ) .. Becca's height is 77" i think. 27 inches isnt her height, its how much taller she is than the hole (and, later, below the hole). I hope that clarifies things a bit..

Height of becca= 77"

as to the second problem - i have no idea.. im working on it tho

I agree with 236factorial. Though I did it using one unknown instead of two.

1 unknown is also possible, but I just like 2 variables  :)

[quote="236factorial"]That is incorrect, although you are somewhat close.
you cannot use only one variable, because a depths of the holes are different.

[hide="consider"]
let x be the depth of the hole at one-fourths completion,
let y be the depth of the hole when it's finished,

3(182-x) = y-182
4x = y

simplifying,
-3x - y = -728
4x - y = 0

-7x = -728
x = 104.

4x = y = 416 cm.
[/hide][/quote]

yeah i got 416

[quote="236factorial"]That is incorrect, although you are somewhat close.
you cannot use only one variable, because a depths of the holes are different.
[/quote]

I apologize upon saying this statement, because it is possible to use just one variable to solve this problem (actually easier too). I thought of such an equation during a ride home on my school bus (or was it when I was in bed?)
I always think about 2 variables, so I'm very bad at making equations with 1 variable in word problems.

hi 236factorial:

 You need not apologize. Actually either way is acceptable in solving the problem.
It's always a good idea, though, to know more than one way to solve a problem.


geometry
algebra
polynomial
inequalities
rectangle
number theory

Though it might be stupid to start a pre exam discussion approximately 1month before the exam ,students seeing the competitive and padips (studious) responses  might start working vigorously in this month :D :lol: ...
Ok how all you have all prepared and pour in your views regarding this year's RMO ..wish you all the best for your serious preparation . :lol:

tell me .... how have u prepared for d olympiad?? :huh:

do tell, bcoz i wanna c how other people r preparing :help:

 :gathering:

Annnnnnnnd...
i have not prepared much...  :read:
u c that's why i wanna know (its a first time for me!)

Its a first time for me also ....anyway I have not prepared anything specially for RMO ,daily if I have some mood  I will do some math..... :lol:

coz i'm in Xth and giving RMO 1st time,i had to prepare some stuff..like P and C from NCERT XI..but no exotic stuff...only basics...but i guess..RMO Q's test ur IQ and thot process..rather than knowledge...(seeing previous papers)..but its anything but easy!!

Im in Xth too....
i've just started on some basics of topics like P&C n others...
i still have a lot to do to get the hang of problems of RMO level...
hehe
 :play_ball:

Even I am in 10th......I have not prepared anything and my geometry is horrible to the core -(so for me RMO sure 2 sums out

Finally ppl seem to have realised dat rmo threads need to be revived. :)
Dunno abt u ppl...but I'm kind of enjoyin life now coz I've been given 1 month off frm skool studies for rmo :P
Srinath will definetely solv atleast 1 in geometry and the 2 which I'm xpctin frm number theory this yr.
But what abt polynomials and functional eqs?Any1 done them yet?I'm skippin them :D And I'm sort of worried abt series based inequalities in particular and the entire inequalities in general. :(
And as far as the trend goes.....these rmo ppl set xttra hard paprs in the prime no years usally and moreover last yr's paprs was relatively easy...so I think this year they may make a papr a bit xceptionally tuff. :(  :(

[quote="Pocha"]
Srinath will definetely solv atleast 1 in geometry and the 2 which I'm xpctin frm number theory this yr.
[/quote]
What made you say that ...oh good nice joke ... :rotfl:
Anyway I hardly touch my school textbooks except before the day of exam...

i dont know why pocha u take th pain of guessing....let it come as it does...lets see what RMo has :p..its game ON!!.i cant be more prepared than i'm right now :p...lving aside mugging up formulae :p

@salil :P
Ahh! Guessing doesn't pain much if u're used to it....patterns tell a lot.It simply prepares u to face the worst and to give more attention to the topics frm where more questions r likely to come.U see,dat's the 1 and only way to prepare(if u wish to prepare intelligently i.e. :P ),coz afterall,it is an exam.
@Srinath
U're marvellous when it comes to number theory and of the 2 questions which cum frm geometry,1 is usally quite solvable to evn gadhas like me.So there is no reason for u to take tension..and as far as my case goes...my conditin wont improve evn if I start to prepare now coz it's already bad enaf :(

[quote="Pocha"].
@Srinath
U're marvellous when it comes to number theory [/quote]
Really why do you say so ?....I am not marvelous and all ...I just like a hell lot of NT ...I dont think you can guarantee that I will surely solve NT.....as far as geometry is concerned (No comments  :rotfl:  :D )
Anyway dont feel bad if you dont clear RMO as mathematics is not just clearing olympiads (applies to both of us  :lol: )

I am M.Ramchandran
i am giving RMO this year
[quote="Pocha"]Finally ppl seem to have realised dat rmo threads need to be revived. :)
Dunno abt u ppl...but I'm kind of enjoyin life now coz I've been given 1 month off frm skool studies for rmo :P [/quote]
wth??? in my skl u get screwed for taking ONE day off and here you are with a month?? gosh i am jealous. :P

number theory.... :(
theres a very low chance that ill solve its problem....

Anyway how much do you have to score to qualify for INMO??  (roughly) :?:

hey Pocha and M.Ramchandran, have u given rmo before?

Hello to all and wish you all the best for the RMO.
zzzzzzz, cutoffs very from state to state and depends on the relative performance of everyone.

i am aadil from 10th.im giving rmo for the first time.
[quote]Hello to all and wish you all the best for the RMO.
zzzzzzz, cutoffs very from state to state and depends on the relative performance of everyone[/quote]
what were the cutoffs last year?

So everyone exited abt RMO............
And I hvent got my admit still :(

@Srinath
ye u're rit indeed.But u see,I may be able to forgive my own failure but all the prob lies in my skool.U see,I am the only 1 who has taken an interest in math olympiad in the entire histry of my skool and my math interest has bcome a major worrry to my teachers,parents and "well wishers".They wud rather prefer me to be a lamb who studies in a conventional fashion and will score high in the boards.I know I wont score very remarkably in my boards now( coz it's too late ) and my skool will remember me as a black sheep,atleast my RMO sucess(if it hapens tho,chances r near 0%) will make me remembered by my skool folks as "A sucessful Black sheep" :P
@sleep well zzzzzzz
I officially appeared for RMO last year with almost nil knowledge of Euclidean Geometry and no knowledge of beyond 10th level math.So u can guess dat my rmo experience last year was equivalent to 3 hrs of bench warming :rotfl:
@aadil
some 1 said dat last year's cutoffs were around 3-4 ques since last's year's was relatively easy......but the usal cutoff is 2-3 ques.
@Ramachandran
I am havin time off frm studies and not skool!I'm havin to go to skool for atleast 4 days a week with great reluctance. :(
well..and I'm havin to pay heavily for it as well....parents shoutin after me all day and my teachers r worrying a hell lot.But 1 thing worth mentioning...they've lost hope on me! :P (esp as far as icse is concerned)

I've been preparing various concepts regarding RMO in past few months..
Can I know where I could get Past RMO Papers?? :lol:

to savy2020
Hi ... pleez tell me how u have been preparing various concepts regarding RMO?  :huh:
(like books u have been referring to .... etc.) :?:

TO ALL
Any advice regarding preparation of RMO????

@savy you can get the RMO papers   here [url]http://olympiads.hbcse.tifr.res.in/subjects/mathematics/previous-question-papers-and-solutions[/url]
Before 2000 can be got here [url]http://www.mathlinks.ro/resources.php?c=78&cid=48[/url]

@zzzzz
my advice is that dont be mechanical and dont something for the sake of exams ....Also this is not an IIT JEE nor School Exam to try millions and millions of problems  monotonously ....if you like math you have mathlinks- sit on a problem-try your level best to solve it -if it is interesting once you solve it dont jump immediately to another problem-the solved problem explore its boundaries and look for generalizations and try to learn something from a problem-this can be one way to build your own strength-take up RMO cooly without any expectations -thats all ....
This is what I came to know from a lot of brilliant people.
No cross questions whether I follow it or not  :rotfl:

My response to u,zzzz is that dont prepare for RMO....except tofr some very basic formulae,and some very trival formulae,i dont think any SPECIFIC prep is REQ..ya solving previous 20 yrs pprs is highly reccomended..

This is my first time too and my preparation is somewhere between 0 and .... 0   :P

hmm... thanx

I generally do sumthin like wat u were saying srinath.r ....  :)

ok .....I dont know ...I am simply sitting ...also having NSEA(jr) and NESEJS......and ZIO and RMO and
(argh-school pre halfyearly :arrow:  :wallbash_red:  :furious: )

actually,i think RMO is meant to be a brain-churner,not an exam where practice pays off a lot..and @ tukai,i dont think the RMO guys will tell u the pattern.if u think that observing pattern and predicting is a good thing .. :P  :rotfl: best of luck....kyunkii i dont think u'll have much succes...not that i dont wish u success...:) I DO wish u success..if u deserve it,u 'll get it  :lol:  :)  :lol: ...and i seriously dont call that intelligent studies..u are exopecting 2 from NT and 2 from geometry and 1 frm geometry to b easy..if u r so infexible in ur plan..in ur study pattern...u'll suffer in case that there is 1 tough Q frm geo only, 2 from INEQ or something like that..i wont like to think the pattern if forzen and wont change

yes correct.....so everyone dont predict the difficulty of the questions....prepare yourself for the worst and highly difficult paper................... :D

@Salil
Practice makes perfect....and pattern does tell a lot.And btw...I never said dat I'm absolutely sure,did I? :wink: Preparin for 2 frm NT = Preparin for the worst
And I think these IMO ppl r quite oldies and there has probably not been a very distinct change in IMO pattern as far as geometry is concerned.So..since this ist the 1st step towards IMO.....geometry may be xpcted to remain unchanged.I cant assure the others...infact I cant assure anything.....coz I have'nt really prepared myself on the basis of these predictions :P
And The best suggestion in store......this year's paper's gonna be a beauty(in terms of torturing my brain)! :lol:

my name's not salil!!!!!!!!!!!!!!!!!!!!n btw its a fool's saying,"practice makes u perfect"..it wud never..coz u REQ more brain THAN practice for RMo(which i'm sure u HAVE!!!!!!!!!)

@the guy proud of his name :rotfl:
nay...what makes u say dat I've got brains?Had I got brains..my  condition wudn't be so miserable..esp in combinatorics :(  :( and geometry.....

abbe yaar...ur condition is well and good in combinatorics..yaad hai??u tld me hw to calculate the No. of rectangles in a square of n sides...it was amazing application of combinatorics...

Good Luck Everyone for the exam. Please post the questions once the exam is over , Ill post ours too (Im in Maharashtra , we've got a different paper). Make sure you rock the exam and manage your time well. And yeah be cool  :P

Hi everyone....
I study in the 10th standard...

Just aiming to get through the paper...

My thinking is that the RMO paper has more calculations than other papers(INMO,IMO...), but still requires knowledge of somewhat depth of all the topics....so a person who doesn't prepare nicely for it will certainly have a hard time...

Also, there's a lot of chance that somebody might fail just because of his proof-writing skills...so that's another area in which marks are generally cut...

Anyways...keep preparing...
[hide="click here"]
[hide][hide]
[b]BOL!![/b] :rotfl:  :rotfl:  :rotfl:

[/hide][/hide][/hide][/hide][/hide]

@Maharashtra...do you have a different paper ?....I thought for entire india it is common....

well the states r allowed to set their own paprs or use the 1 set by NBHm or whatevr.
And when there is sumthing wonderful like BP in a state...dat's the least u can xpct :D But their paprs r a great deal tuffer than the 1 we get...I went thru the 2006 papr...t'was really heart breaking :( There was a horrifying chessboard prob and seeing it,I gave up all hopes of cracking dat papr :(  :(

yea man...this chessboard i know not,evn till today :blush:  makes me feel like a 8th-standard guy.. :oops: anyways...i'm caught b/w CTPCM and NCV right now...i guess i'll leave RMO to my uppers tory and work towards NSEA :p

Ok...the exam's tomorrow...So
[color=red][size=200]BEST OF LUCK!!![/size][/color]

[size=134][b][color=blue]VERY VERY GOOD LUCK TO ALL[/color][/b][/size]

All the best to everyone who is writing RMO this year...
I am very happy that 3 hrs of undivided attention can be given ..would be a remarkable 3 hrs (if i get something or my brain ticks .... :D)....

@goutham Who are you by the way ?
Gouthamphilomath....hmmmmm....wild guesses...but keeping to myself.

Its already posted by keyree10...check the respective olympiad sections.

[color=#0000FF][Moderator edit: See [url=http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=48&year=2009]here.[/url]][/color]


Let $ a,b,c$ be nonnegative real numbers. Prove that

$ 1\plus{}\frac {3abc}{a^2b\plus{}b^2c\plus{}c^2a} \ge \frac {2(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}$.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ 1 \plus{} \frac {3abc}{a^2b \plus{} b^2c \plus{} c^2a} \ge \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2}$.[/quote]
We can prove it as follow:
Rewriting the inequality as
\[ \frac{3abc}{a^2b\plus{}b^2c\plus{}c^2a} \ge \frac{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2}{a^2\plus{}b^2\plus{}c^2}\]
If $ 2(ab\plus{}bc\plus{}ca) \le a^2\plus{}b^2\plus{}c^2$, it is trivial.
If $ 2(ab\plus{}bc\plus{}ca) \ge a^2\plus{}b^2\plus{}c^2$, applying Schur's Inequality:
\[ 3abc \ge \frac{(a\plus{}b\plus{}c)[2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2]}{3}\]
It suffices to show that
\[ \frac{(a\plus{}b\plus{}c)[2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2]}{3(a^2b\plus{}b^2c\plus{}c^2a)} \ge \frac{2(ab\plus{}bc\plus{}ca)\minus{}a^2\minus{}b^2\minus{}c^2}{a^2\plus{}b^2\plus{}c^2}\]
\[ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2) \ge 3(a^2b\plus{}b^2c\plus{}c^2a)\]
\[ b(a\minus{}b)^2\plus{}c(b\minus{}c)^2\plus{}a(c\minus{}a)^2 \ge 0\]
That's all. :)

After expand, it's becomes:
$ \sum\ a^4b \plus{} \sum\ a^2b^3 \plus{} abc\sum\ a^2 \geq\ 2\sum\ a^3b^2 \plus{} abc\sum\ ab$
Trues with Am-Gm because:
$ \sum\ a^4b \plus{} \sum\ a^2b^3 \geq\ 2\sum\ a^3b^2$
and $ abc\sum\ a^2 \geq\ abc\sum\ ab$
We can see it's weaker than can_hang's result.
(but also njce because althought is an cyclic ineq, equality occurs when $ a\equal{}b\equal{}c$ or $ abc\equal{}0$)
 :)

This is stronger:

Let $ a,b,c$ be nonnegative real numbers. Prove that

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

After enpand, I realize that it's also can_hang's problem  :oops:

[quote="nguoivn"]After expanding, I realized that it's also can_hang's problem  :oops:[/quote]
You are right, Nguoivn.  :oops:

But this is a little more interesting, I think.

Find the greatest value of $ p$ such that

$ \frac {a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac {3pabc}{a^2b\plus{}b^2c\plus{}c^2a} \ge 1\plus{}p$

for all positive real numbers $ a\le b\le c$.

Because inequality is homogeneous we assume that $ a+b+c=3 $
According to a well-known inequalities resulting that:
$ a^2b+b^2c+c^2a\le 4-abc $
and
$ LHS\ge 1+\frac{3abc}{4-abc}\Leftrightarrow LHS\ge\frac{4+2abc}{4-abc} $
It is sufficient that:
$ \frac{4+2abc}{4-abc}\ge\frac{2(ab+bc+ca)}{a^2+b^2+c^2} $
With $ t=\frac{ab+bc+ca}{a^2+b^2+c^2} $ result $ ab+bc+ca=\frac{9t}{2t+1},0<t\le 1 $
we have shown that:
$ abc\ge\frac{4t-2}{1+t},0<t\le 1$
Obviously for  for $ t\le\frac{1}{2} $ , the inequality is true, it remains  demonstrate it for $ t\ge\frac{1}{2} $
We apply Schur's inequality third order:
$ (a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca)\Leftrightarrow abc\ge\frac{6t-3}{2t+1} $
Because $ \frac{6t-3}{2t+1}\ge\frac{4t-2}{t+1}\Leftrightarrow (t-1)(2t-1)\le 0 $ is true,
the proof is ended!
_________
Sandu Marin


Find all pairs of prime numbers (p,q) satisfying the condition that $pq$ divides $2^p + 2^q$  .

Try [url=http://www.mathlinks.ro/Forum/topic-57586.html]this[/url]  ;)


You know how theres the "zombie" website?

How many brains did you eat?
I got 490 brains....if you dont know what i'm talking about then click on my sig.

You think we're stupid, don't you.

[hide]Did you check your spelling and capitalization?[/hide]

that website also has a zombie game that is pretty fun

how did you get so many, i only have 85 :(

[quote="zeb"][quote="math92"]click mr. green
[hide=" :D "]
hi, click the ninja
[url=http://kevan.org/brain.cgi?math_92]:ninja:[/url]
[/hide][/quote]

Haha, my browser tells me where links lead :P[/quote]
Of course, every single browser I'ved used does the same. Down at the bottom to the left. Or just a small mouse over popup.

Let me guess: half of them from people who were browsing this thread?

And when I first saw your sig, I clicked it, but I haven't fallen for anything like that ever since.


Hi,
Can anyone here tell me how can a Mac user uses the LaTex since the MiKTex clearly does not support Macintosh, please?
Thank you! :(

Have a look at [url=http://www.uoregon.edu/~koch/texshop/]TeXShop[/url] which is a TeX editor but it also tells you about downloading TeX Live, see [url=http://www.uoregon.edu/~koch/texshop/obtaining.html]Obtaining TeXShop and TeX Live[/url].


what is the age range in this site, i mean how old is everyone ? what is the average age range i should say. coz everyone in here is like maths genius.i dont know if i am stupid or you lot are very clever.i dont know how to make a poll about this, if anyone reads this and knows how to make a poll, please do so. or is it too personal thing? i hope not.

I think there are some polls in the Games and Fun Factory forum. Try searching for "age poll" within that forum, or something like that.

Most of the people who post in the College Playground are undergraduates or graduate students, so probably 18-25. Most of the people who post in the Olympiad forum are high school students, so maybe 15-19. In the other forums there are some much younger people.

Since most people only read certain parts of the forum, you will have a hard time accurately polling the users of any particular subforum.

But anyway, don't be daunted by the people seeming so smart. The people who post the most are the ones who are most advanced for their age, usually. And a lot of the people on this forum have been doing this kind of math for quite some time. It's not really your age that matters but how long you've been doing it. Any time is a good time to start loving math!!! ;)

Thanks for the well explained answer Xevarion


trigonometry
inequalities unsolved
inequalities

Let $A \rightarrow 0$, $B \rightarrow 0$, and $C \rightarrow \pi$.  Then $3 \tan^2 A + 4 \tan^2 B + 5 \tan^2 C \rightarrow 0$.

[quote="Lovasz"]Let $\triangle ABC$. Find minimum value of:
\[ T = 3\tan^2A+4\tan^2B+5\tan^2C  \][/quote]

Sorry, it should be:

[quote]Let $\triangle ABC$. Find minimum value of:
\[ T = 3\tan^2 \frac{A}{2}+4\tan^2\frac{B}{2}+5\tan^2\frac{C}{2}  \][/quote]

I think this is a "true" problem.


Hi there,

I am trying to figure out whether the following are constructable with ONLY a compass and straigtedge (not a ruler).

a) a segment of length sin((2pi)/7)

I dont think this is constructable becuase the regular 7 gon is not constructable with only a straightedge and compass. cos((2pi)/7) is not constructable, so does this imply that sin((2pi)/7) is also not constructable?

b) an angle that measures 96

I think this is constructable, but I am somewhat confused here as to how to do it. 360-96 = 264 but I am not sure how to create that angle.

c) an angle that measures (7pi)/4

This should be constructable since we can construct an angle pi , then we can multiply this angle 7 times becuase it is a semicircle, then we can bisect it twice, which is like dividing by 4. pi divided by 4 is 45 degrees which is 90 degrees bisected which we can definately construct.

d) a segment of length cuberoot[2*sqrt(2)]

We cannot create this due to the highest degree of X cubed is 3 which is not a power of 2.

Does all this seem correct?

[quote="astnfan"]a) a segment of length sin((2pi)/7)

I dont think this is constructable becuase the regular 7 gon is not constructable with only a straightedge and compass. cos((2pi)/7) is not constructable, so does this imply that sin((2pi)/7) is also not constructable?[/quote]

Right! If, for an angle x, the value cos x is not constructable, then sin x is not constructable either; in fact, if sin x was constructible, then, since $\cos x=\pm\sqrt{1-\sin^2 x}$ and since square roots are constructable, we could also construct cos x.

[quote="astnfan"]b) an angle that measures 96

I think this is constructable, but I am somewhat confused here as to how to do it. 360-96 = 264 but I am not sure how to create that angle.[/quote]

Yes, the angle 96° is constructable. Here are two possible ways to construct it:

- If you know how to construct a regular 15-gon, then just construct one; then, the angle under which a side of this 15-gon is seen from its center equals $\frac{360^{\circ}}{15}=24^{\circ}$. Now, bisecting this angle two times, we arrive at the angle $\frac{24^{\circ}}{2\cdot 2}=6^{\circ}$, and adding 90°, we get 6° + 90° = 96°.

- If you only know how to construct a regular 5-gon (this is a classical construction problem; see http://www.cut-the-knot.org/pythagoras/pentagon.shtml ), then construct one; then, the angle under which a side of this 5-gon is seen from its center equals $\frac{360^{\circ}}{5}=72^{\circ}$. On the other hand, I hope you know how to construct an angle of 60°. Thus, we can also construct the difference of these two angles, 72° - 60° = 12°. Bisecting this 12° angle, we arrive at the angle $\frac{12^{\circ}}{2}=6^{\circ}$, and adding 90°, we obtain 6° + 90° = 96°.

[quote="astnfan"]c) an angle that measures (7pi)/4

This should be constructable since we can construct an angle pi , then we can multiply this angle 7 times becuase it is a semicircle, then we can bisect it twice, which is like dividing by 4. pi divided by 4 is 45 degrees which is 90 degrees bisected which we can definately construct.[/quote]

Correct.

[quote="astnfan"]d) a segment of length cuberoot[2*sqrt(2)]

We cannot create this due to the highest degree of X cubed is 3 which is not a power of 2.[/quote]

We can construct it. In fact, $\sqrt[3]{2\cdot\sqrt{2}}=\sqrt[3]{\sqrt8}=\sqrt{\sqrt[3]{8}}=\sqrt{2}$, and this is surely constructable.

  Darij

thanks a lot for the help, the algebra on part d tricked me, I looked at it, and didnt try to actually work it out, I assumed that since cuberoot of 2 cant be constructed, any cuberoot cannot be constructed.

How about this one:

A triangle ABC with A= 108 degrees, b = 2, c = sqrt(3+sqrt(2))/sqrt(5)

I think it can be constructed, 108 degrees is 96 degrees plus 12 degrees. 2 is constructable, and c is constructable. Does this imply that we can use Side Angle Side to make ABC?

[quote="astnfan"]A triangle ABC with A= 108 degrees, b = 2, c = sqrt(3+sqrt(2))/sqrt(5)

I think it can be constructed, 108 degrees is 96 degrees plus 12 degrees. 2 is constructable, and c is constructable. Does this imply that we can use Side Angle Side to make ABC?[/quote]

Yes, it does. If we can construct the angle A and the sides b and c, then we can construct triangle ABC.

  Darij


function
integration
real analysis
real analysis unsolved

Let $f,g$ two integrable functions on $[0,1]$ such that

                                        \[ \int_{0}^{1} f =\int_{0}^{1} g =1.  \]

 Show that there is some interval $I$ included in $[0,1]$ such that \[ \int_{I} f =\int_{I} g=\frac{1}{2}.  \] :)

That's manifestly false : for ex . $f(x)= 0$ .  What's the correct statement ?!
 :cool:

i guess he meant to say that $\int_0^1 f = \int_0^1 g = 1$, right cezar?
(at least, i have a solution for this case, which have actually something to do with the topic :D)

ps. really a cute problem!

The correct initial condition is $\int^1_0f=\int^1_0g=1$.

The problem has been posted before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31268]here[/url], but there was no answer there. I wonder why I never tried applying a strong topological result, such as the Borsuk-Ulam Theorem befpore :).

I'll use the theorem in the following form:

$(*)$ Given a continuous map $\varphi$ from a closed disk $\mathcal D$ to the plane such that $f(v_1)=-f(v_2)$ for every pair of antipodal points $v_2,v_2$, there is a $v\in\mathcal D$ such that $\varphi(v)=0$.

Let $\mathcal T$ be the triangle $\{(a,b)\in[0,1]^2\ |\ x\le y\}$. On $\mathcal T$ we define $\varphi(a,b)=\left(\frac 12-\int_a^bf,\frac 12-\int_a^bg\right)$. Since $\varphi$ is constant on $\ell=\{(x,y)\in\mathcal T\ |\ x=y\}$, it's clear that $\varphi$ induces a continuous map $\tilde\varphi$ from $\mathcal D=\mathcal T/\ell$ to the plane, and that $\mathcal D$ can be regarded as a disk in which the points $(0,y),(y,1)$ from $\mathcal T$ are antipodal. Now we simply apply $(*)$ to $\mathcal D$ and $\tilde\varphi$.


[quote="Ligouras"]Prove that

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq3\geq\frac {a^2 \plus{} b^2 \minus{} c^2}{ab} \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{bc} \plus{} \frac {c^2 \plus{} a^2 \minus{} b^2}{ca}$[/quote]

The Letf hand side is obvious due to Am-Gm inequality

Th other side

we have By schur inequality

$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \ge ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(a\plus{}c)\equal{}c(a^2\plus{}b^2)\plus{}b(a^2\plus{}c^2)\plus{}a(b^2\plus{}c^2)$
$ \Longleftrightarrow 3abc \ge \sum_{cyc}a(b^2\plus{}c^2\minus{}a^2)$
$ \Longleftrightarrow 3 \ge \sum _{cyc}\frac{b^2\plus{}c^2\minus{}a^2}{bc}$


Prove that in the following product
$ P = (1-x+x^{2}-x^{3}+...-x^{99}+x^{100})(1+x+x^{2}+x^{3}+...+x^{99}+x^{100})$
after multiplying and collecting terms, there does not appear a term in $ x$ of odd degree.

[hide]For any polynomial $ Q(x)$, then $ P(x)=Q(x)Q(-x)$ is always an even polynomial, that is $ P(-x)=P(x)$ ($ Q(-x)Q(x)=Q(x)Q(-x)$).[/hide]


Let $ a,b,c > 0$ such that $ a^4 \plus{} b^4 \plus{} c^4\equal{}3$. Prove that
 $ \sum_{cyc} \frac {a^3 \minus{} b}{b \plus{} c} \geq 0$

[quote="2424"][quote="2424"]Let $ a,b,c > 0$ such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$. Prove that
 $ \sum_{cyc} \frac {a^3 \minus{} b}{b \plus{} c} \geq 0$[/quote]
It's easy by CS!
no one interested in this ineq?[/quote]
It's easy
Ineq
$ <\equal{}>\sum (a^3\minus{}b)(a\plus{}b)(a\plus{}c) \ge 0$
$ <\equal{}>\sum a^5\plus{}\sum a^4(b\plus{}c)\plus{}abc(a^2\plus{}b^2\plus{}c^2)\minus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}(a^2b\plus{}b^2c\plus{}c^2a) \ge 0$
$ <\equal{}>(a^4\plus{}b^4\plus{}c^4)(a\plus{}b\plus{}c)\plus{}abc(a^2\plus{}b^2\plus{}c^2)\minus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}(a^2b\plus{}b^2c\plus{}c^2a) \ge 0$)1)
We have:
$ a^3\plus{}b^3\plus{}c^3 \ge a^2b\plus{}b^2c\plus{}c^2a$
We need prove that:
$ LHS(1) \ge (a^4\plus{}b^4\plus{}c^4)(a\plus{}b\plus{}c)\plus{}abc(a^2\plus{}b^2\plus{}c^2)\minus{}[a^3\plus{}b^3\plus{}c^3\plus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)] \ge 0$
$ <\equal{}>(a\plus{}b\plus{}c\minus{}abc)(a^4\plus{}b^4\plus{}c^4\minus{}a^2\minus{}b^2\minus{}c^2) \ge 0$
It's always true because $ a^4\plus{}b^4\plus{}c^4\equal{}3$


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra solved

For which values of the real a does there exist an endomorphism f of the set of real polynomials (seen as vectorial space) with the property that $ f(f(P))=P'+aP$ for all polynomials P?

Positive $a$, right?

By composing $f$ with that expression, we get $[f(P)]'=f(P'),\ \forall P$. From here we find that the degree of $f(X^n)$ increases by $1$ when $n$ increases by $1$. From here we find that $f$ must preserve degrees (i.e. it takes polynomails of degree $n$ to polynomials of degree $n$). In particular, $a\ne 0$, so we need not worry about that anymore.

Let ${\bf 1}$ be the constant polynomial taking the value $1$. We have $f({\bf 1})=b\cdot {\bf 1}$, so $(f\circ f)({\bf 1})=a\cdot{\bf 1}=b^2\cdot{\bf 1}$, so $a$ must be positive.

On the other hand, if $a$ is positive, and $b$ is such that $b^2=a$, then it looks like (I haven't checked it properly) we can construct $f(X^n)$ inductively. I haven't figured out whether there are nice expressions for $f(X^n)$.

If we consider $g=f_{|R_{n}[X]}$ and $G$ the matrix associate in the canonical
basis $(1,x,...,x^n)$

$G^2(x^k)=ax^k+kx^{k-1}$

$G^2=a(I_{n+1}+1/aN)$ where $N$ is a nilpotent matrix $N^n=0$
$a\neq 0$
$det(G^2) \geq 0$ so $a^n \geq 0$ for any n even or odd

$a>0$


I found this very nice method of calculating difficult multiplications ... in an easy, fast way... :lol:

You really have to see it!
watch: http://www.metacafe.com/watch/308408/Easy%20Multiplying.%20Don

I see it has just been removed ... strange.

This is essentially the same as they teach you in elementary school, I guess you could say it reduces mistakes, but it takes longer. Nice job realizing this though.

Hey, weren't you taught long multiplication in the elementary school? This is precisely it, only taken from right to left instead from left to right... Day-before-yesterday's news ;)

awesome   :P

although it is basically the same as regular multiplication. and i have yet to look at the others.


geometry
perimeter
LaTeX
geometry proposed

What is the condition that the positive numbers $ S$ and $ p$ have to satisfy, such that there is a  right -angled triangle whose area is $ S$ and perimeter is $ p$.?

Denote $ a,b$ the cathetus and $ c$ the hipotenus $ a + b + c = p$, $ ab = 2S$ $ a^2 + b^2 = c^2$

It is easy to obtain  $ c = \frac {p^2 - 4S}{2p}$.(1)

$ p^2 - 4S \ge 0$

Combine $ a + b = p - c = \frac {p^2 - 4S}{2p}$ and $ ab = 2S$, we have the following  cuadratic equation

$ x^2 - \frac {p^2 - 4S}{2p}x + 2S = 0$, then the discriminat has to be non negative

${ \Delta = (p^2 + 4S)^2 - 32Sp^2 = (p^2 - (12 - 8\sqrt {2})S)(p^2 - (12 + 8\sqrt {2})S}$

If $ S^2 - 4p \ge 0$ also $ p^2 - (12 - 8\sqrt {2})S \ge 0$

Hence if $ S,p$ are non-negative numbers and satisfy $ p^2 - (12 - 8\sqrt {2})S \ge 0$ , using the equation 1 we get $ c$ and then
$ a,b = \frac {p^2 + 4S \pm \sqrt {p^4 + 24p^2S + 16S^2}}{4p}$

This is my first latex  :lol: , sorry if there is a typo here.


let $ A_1,A_2,...,A_n$ are n constant point in plane and $ K_1,...,K_n$ are n  real numbers.find the Locus of points like M that $ K_1|A_1M|^2\plus{}...\plus{}K_n|A_nM|^2$ be Constant if:
1)$ K_1\plus{}...\plus{}K_n$ =nonzero
2)$ K_1\plus{}...\plus{}K_n\equal{}0$


Let $ a,b,c\geq\ 0$ prove that
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)\geq\ a^4(b \plus{} c) \plus{} b^4(a \plus{} c) \plus{} c^4(a \plus{} b)$
It is very easy

well using schur one time and it's finished...


A toy car is released off a ramp of 130 cm length and 60 cm height. Assume no friction or air resistance. Determine the instantaneous speed of the car after 30 cm.

Please show how you got it, and how it is possible to find the answer to any distance that the car rolls to.

[hide="Solution"]
I assume you mean after it goes 30 cm down the length of the ramp? And by "length" we're talking hypotenuse?
From triangle similarity, the car is at a height of $ \frac {600}{13}$ cm = $ \frac{6}{13}$ m after going $ 30$ cm down the hypotenuse.
From the conservation of mechanical energy, $ m_\text{car}\cdot g\cdot (.6) \equal{} m_\text{car}\cdot g\cdot \frac {6}{13} \plus{} \frac {m_\text{car}\cdot v^2}{2}$.

$ v \equal{} \sqrt {2g\left(.6 \minus{} \frac {6}{13}\right)} \equal{} \boxed{1.65\ \frac {\text{m}}{\text{s}}}$.
[/hide]

Edit: Fixed the units.

Is it possible that the speed is 1.664 m/s?

I was thinking the following:
First we calculate the angle of the slope,(b).
sin(b)=60/130=6/13
the acceleration of the car: a=g*sin(b)

X(t)=(1/2)*a*t^2

t=sqrt(2x/a)

V=a*t=a*sqrt(2x/a)

V=sqrt(2ax)

V=sqrt[2*(60/13)*(0.3)]=1.664m/sec

Yeah, I think that's right. We had the same answer, but mine was in cm/s and I accidentally wrote m/s  :D

New question; what would the equation for a graph of distance over time be?

Knowing that $ x\equal{}v_0t\plus{}\frac{1}{2}at^2$, the distance is proportional to the square of the time if initial velocity is 0, so the graph would be a parabola.


geometry
3D geometry
vector
trigonometry

A lamp is situated at point $A$ and shines inside the cube. A (massive) square is hung on the midpoints of the 4 vertical faces. What's the area of its shadow?

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=285[/img]

Let a = AB be the lenght of the cube edge. In a central projection from the point A, the segment PQ projects into the segment PG in the plane (BCGF), the segment QM projects into the segment FH in the plane (EFGH), the segment MN projects into the segment HN in the plane (CDHG). Let S be the midpoint of the segment MN. The segment NS projects into a segment NT in the plane (CDHG) and the segment SP into a segment TP in the plane (BCGF). The point T lies on the intersection line CG of the planes (BCGF) and (CDHG). To calculate the height of the point T above the plane (ABCD), we consider the equation of the line passing through the points $A = (0, 0, 0),\ S = \left(\frac 3 4 a, \frac 3 4 a, \frac 1 2 a \right)$ and its intersection with the plane (BCGF) or (CDHG). The unit vector in the direction $\vec{AS}$ is $\vec n = \dfrac{\vec{AS}}{|AS|}$.

$|AS| = a \sqrt{\dfrac{9}{16} + \dfrac{9}{16} + \frac 1 4} = a \sqrt{\dfrac{22}{16}} = a \dfrac{\sqrt{22}}{4}$

$\vec n = (n_1, n_2, n_3) = \left(\dfrac{3}{\sqrt{22}}, \dfrac{3}{\sqrt{22}}, \dfrac{2}{\sqrt{22}}\right)$

$\vec{AT} = \dfrac{a}{n_1} \vec n = \left(a, a, \frac 2 3 a \right)$

As a result, the point T is at the height $CT = \frac 2 3 a$. The area of the shadow is the sum of the areas of the right angle triangle $\triangle FGH$ and the congruent quadrilaterals PTGF, NTGH.

$S(\triangle FGH) = \frac 1 2 \cdot GF \cdot GH = \dfrac{a^2}{2}$

$S(PTGF) = S(\triangle CGF) - S(\triangle CPT) = \frac 1 2 \cdot GC \cdot GF - \frac 1 2 CP \cdot CT \cdot \sin 45^\circ =$

$= \dfrac{a^2}{2} - \frac 1 2 \cdot \dfrac{a \sqrt 2}{2} \cdot \dfrac{2a}{3} \cdot \dfrac{\sqrt 2}{2} = \dfrac{a^2}{2} -\dfrac{a^2}{6} = \dfrac{a^2}{3}$

$S(\text{shadow}) = S(\triangle FGH) + S(PTGF) + S(NTGH) = S(\triangle FGH) + 2S(PTGF) =$

$= \dfrac{a^2}{2} + 2 \cdot \dfrac{a^2}{3} = \frac 7 6 a^2$

A lamp is situated at point $A$ and shines inside the cube.
A (massive) square is hung on the midpoints of the $4$ vertical faces.
What's the area of its shadow?
[img]https://1.bp.blogspot.com/-D67kYgxTSWw/XWuvNu1hEeI/AAAAAAAAKo4/N_RYf9T-U6E5_fNsOW2t0K4b6KIa-XHfACK4BGAYYCw/s400/2002%2Bflanders%2Bp4.png[/img]


trigonometry
function
geometry proposed
geometry

$R$ is the radius of the big circle and $BM=CM$. Show that: $R=r_1+r_2+r_3$

[quote="Ph-An"][u]A [b]big[/b] problem with a big circle ![/u] I note:
$w=C(O,R),\ w_k=C(O_k,r_k),\ k\in \overline {1,3},\ w'=C(O',r_2), M\in BC\cap w\cap w'$, where the line $BC$ separates the points $O_2,O'$ and don't separate the points $O_2,O_1;\ U\in BC\cap w_1,\ V\in BC\cap w_3;\ x=m(\widehat {AOM})$. Thus,

$r_2=R\sin^2\frac x2$ and $UV=2\sqrt{r_2} (\sqrt r_1 +\sqrt r_3 ),\ UV=UM+MV\Longrightarrow$

$UV=2\sin \frac x2(\sqrt{R^2\cos^2\frac x2-Rr_1}+\sqrt{R^2\cos^2\frac x2 -Rr_3})\Longrightarrow$

${UV=2\cdot \sqrt{\frac{r_2}{R}}(\sqrt{R^2 (1-\frac{r_2}{R})-Rr_1}}+\sqrt{R^2( 1-\frac{r_2}{R})-Rr_3})\Longrightarrow$

$\sqrt r_1 +\sqrt r_3=\sqrt{R-(r_1+r_2)}+\sqrt{R-(r_3+r_2)}$.

The function $f(t)=\sqrt{t-(r_1+r_2)}+\sqrt{t-(r_3+r_2)}-(\sqrt r_1 +\sqrt r_3)$

is a injection and $f(R)=f(r_1+r_2+r_3)=0\Longrightarrow R=r_1+r_2+r_3.$[/quote]
And with a big solution!!! :lol:


In an analysis book i found this proof for the uniqueness of a limit of a function.

"It is easy to show that when a Limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that

lim f(z) =l and lim f(z)=m as z----> a (z goes to a).

Then,for any positive number ε,there are positive numbers r,δ such that

           |f(z)-l |< ε   whenever 0< |z-a |<r

                       and

             |f(z)-m |<ε whenever 0< |z-a |<δ

     So if   0< |z-a |<θ ,where θ denotes the smaller of the two Nos r and δ,we find that

 |l-m |= |(f(z)-m)-(f(z)-l) |=< |f(z)-l | + |f(z)-m | < ε +ε=2ε

          But   |l-m | is a nonnegative constant, and ε can be chosen arbitrarily small.
Hence

l-m =0 , or l=m."

is that proof correct??

[quote="t0rajir0u"]Why not?  The key here is the triangle inequality step.[/quote]

Isn't there a particular check up according to which we can say for sure whether a proof is right or wrong?

To your "Why not"  presumably correct, i could produce my "Why not" wrong proof??

[quote="t0rajir0u"]Each step in the proof is a valid application of definitions and the rules of logic.  Which step seems strange to you?[/quote]

..............................Excellent!!!!!!!!!........................................................

Write,if you can,the above proof in a manner you just mentioned ,i.e in steps justifying each step.

THEN we will find out if the proof is RIGHT OR WRONG.

[quote="t0rajir0u"]The proof is already written in such a manner.  I don't understand why you won't elaborate on what you don't trust about this proof; the burden is on you to clarify your questions and not on me to clarify an already pretty clear proof.  The use of English does not invalidate the mathematical validity of an argument.[/quote]

Let me then elaborate :

What the Author of that proof really proves here is:

If limf(z) = l and limf(z) = m as z goes to a ,then given ε>0 and also GIVEN

0<lz-al < θ ,where θ denotes the smaller of the two Nos r and δ,

THEN l=m.

...............and NOT..............................

If limf(z) = l and limf(z) = m as z goes to a, THEN l=m

Further more the proof is not written in a stepwise manner where each step is justified by a theorem,or axiom ,or definition ,or a law of logic.

IN the above proof those thing are NOT mentioned explicitly ,hence they cannot be checked for  their correctness,or not.

Many proofs in analysis ,if scrutinized in the above way can be found incorrect.

And the question is; MUST university students keep on learning proofs violating the laws of logic ??

Shouldn't that state of affairs change ?

Also ,if you wish, i can give you a complete stepwise proof of the above theorem justifying my claims about the proof

[quote="τρικλινο"]What the Author of that proof really proves here is:

If limf(z) = l and limf(z) = m as z goes to a ,then given ε>0 and also GIVEN

0<lz-al < θ ,where θ denotes the smaller of the two Nos r and δ,

THEN l=m. [/quote]
I don't think you really understand quantifiers.  The definition of a limit contains three quantifiers:  two are universal and one is existential.

"[b]For every[/b] $ \epsilon > 0$, [b]there exists[/b] $ \delta > 0$ such that [b]for every[/b] $ z$ such that $ |z \minus{} a| < \delta$, it is true that $ |f(z) \minus{} l| < \epsilon$."

This is all that is contained in the notion of a limit.  I think your confusion is that the proof appears to be using the existential quantifier in a universal way.  That's not what's happening.  All we need is the existence of some small neighborhood about $ a$ where our claims are true; we don't need our claim to be true on every neighborhood about $ a$.  The reason local reasoning can imply our result is because of the universal quantifier that's already there.  What this proof does is straightforward:  it uses some perfectly valid logical implications of the above two statement applied first to $ l$ and then to $ m$ to show the following statement:

"[b]For every[/b] $ \epsilon > 0$, [b]there exists[/b] $ \theta > 0$ such that [b]for every[/b] $ z$ such that $ |z \minus{} a| < \theta$, it is true that $ |l \minus{} m| < \epsilon$."

One then needs the following additional statement:

"[b]For every[/b] $ \theta > 0$, [b]there exists[/b] $ z$ such that $ |z \minus{} a| < \theta$."

One then arrives at the following statement:

"[b]For every[/b] $ \epsilon > 0$, [b]there exists[/b] $ \theta > 0$ such that $ |l \minus{} m| < \epsilon$."

Note that this statement is no longer a statement about $ \theta$.  It is therefore equivalent to the following statement:

"[b]For every[/b] $ \epsilon > 0$, it is true that $ |l \minus{} m| < \epsilon$."

This implies $ |l \minus{} m| \equal{} 0$.

If that was gibberish, then let me put it this way:  [url=http://en.wikipedia.org/wiki/First-order_logic#Quantifier_axioms]learn first-order logic.[/url]

As a side note, don't blame the book for your own lack of understanding.  It's arrogant to presume that you know better than the author of a textbook.  A humble student questions himself before his teachers.

YOU blame me of not understanding quantifiers,but yet yourself cannot write QUANTIFIED FORMULAS in a purely formalized way.

CAN you write the limit of a function at a point a, using formal language and not metalanguage?

You advise me to learn 1st order logic. Can you write ,lets say, the formal proof in 1st order logic of 0x=0 ,for all x in real Nos??

DO YOU know that if one opens a supposition in a,proof and then does not close the supposition one violates the law of logic called conditional proof? A thing that the Author did in his proof??

Do you know what the law of conditional proof is and how and where is used in analysis?

In the above proof can you EXPLICITLY mentioned the laws of logic involved??

At the end i suppose i have to write   a correct proof  of the above ,but before i do so  lets fight it out

I am waiting for your answer ,thanks

I am sorry i forgot to add that in the book of SALAS/ HILLE , A COMPLETELY different proof is offered,not 100% CORRECT.

The page is 69 and the title of the book is :CALCULUS of  one and several Variables.

[quote="τρικλινο"]In an analysis book i found this proof for the uniqueness of a limit of a function.

"It is easy to show that when a Limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that

lim f(z) =l and lim f(z)=m as z----> a (z goes to a).

Then,for any positive number ε,there are positive numbers r,δ such that

           |f(z)-l |< ε   whenever 0< |z-a |<r

                       and

             |f(z)-m |<ε whenever 0< |z-a |<δ

     So if   0< |z-a |<θ ,where θ denotes the smaller of the two Nos r and δ,we find that

 |l-m |= |(f(z)-m)-(f(z)-l) |=< |f(z)-l | + |f(z)-m | < ε +ε=2ε

          But   |l-m | is a nonnegative constant, and ε can be chosen arbitrarily small.
Hence

l-m =0 , or l=m."

yes perfectly right

is that proof correct??[/quote]

[quote="ouboujlal"]hey
 if a function f has two limits m and n in certain point a then
 for any ε>0 there is r>0    lx-al<r  ---> lf(x)-ml< ε
and for any ε>0   there is q>0   lx-al<q  --> lf(x)-nl<ε
 if  n<>m
      for ε=(1/2)*(lm-nl)
         we have for  lx-al< k       (k=min{r;q})       we have      lf(x)-ml<ε  and
lf(x)-nl<ε
so
  lm-nl =< lf(x)-ml + lf(x)-nl < 2ε=  lm-nl
 then we have  lm-nl<lm-nl
  Impossible then n=m .
done![/quote]

Your proof is wrong,a litle different from the previous proof ,but on the same tracks


Using the cost structure shown below, a customer must use more than $ x$ minutes per month for Tellfon to be the better deal. What is the value of $ x$?

\begin{tabular}{|c|c|c|} \hline \textbf{Company} & \textbf{Monthly Charge} & \textbf{Cost per Minute of Use} \\
\hline Tellfon &$ \$6$&$ \$0.05$\\
\hline Celltell &$ \$0$&$ \$0.10$\\
\hline \end{tabular}

So the chart says that Tellfon's monthly charge is 6 dollars, plus 5 cents for each minute. Celltell's monthly charge is 0 dollars, but each minute costs 10 cents. We can set up this inequality.

$ 6 \plus{} 0.05x < 0.10x$

$ 6 < 0.05x$

$ 120 < x$

Thus our answer is $ \boxed{120}$.


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

[hide="Similar inequality"]Let $ a,b,c\geq0$ s.t. $ ab\plus{}bc\plus{}ca\equal{}3$.Prove that:
$ \frac{1}{a^2\plus{}1}\plus{}\frac{1}{b^2\plus{}1}\plus{}\frac{1}{c^2\plus{}1}\geq \frac{3}{2}$.[/hide]

Inequality is equivalent with:

 $ \sum {\frac {1}{2}\cdot\frac {2 \plus{} a^2 \minus{} a^2}{a^2 \plus{} 2}}\leq1$ or $ \sum {\frac {a^2}{a^2 \plus{} 2}}\geq1$

 wich is obvious by CBS.

Very easy indeed.
Try to solve the "similar inequality"(post #1).

[quote="Marius Mainea"]Inrquality is equivalent with: $ \sum {\frac {1}{2}\cdot\frac {2 \plus{} a^2 \minus{} a^2}{a^2 \plus{} 2}}\leq1$ or $ \sum {\frac {a^2}{a^2 \plus{} 2}}\geq1$ wich is obvious by CBS.[/quote]what is CBS?


I was looking through some of my old school papers, and found this problem:

[quote]In September of 1995, M&Ms added a new color, blue to its mixture.  A survey of consumers helped choose the color.  Or the phone-in-votes, about 5,400,000 selected blue, 3,200,000 purple, 1,000,000 pink, and 400,000 wqanted no change.  About how many votes were cast?

1. Which method of computation do you think is most appropiate? Justify your choice.

2. Solve the problem using the four-step plan[/quote]

Anyone can see that all you need to do is add the numbers together.  So why do they include explaining out of all the choices (I don't even see any other way than adding them up) which one is the best, and why, and then going through a four-step plan?

And if anyone wants the four-step plan here it is:

[quote]1.  Explore    Determine what information is given in the problems and what you need to find.  Do you have all the information you need to solve the problem? Is there too much information?

2.  Plan    Make an estimate of what you think the answer should be.  Then select a strategy for solving the problem. There may be any number of ideas or strategies that you can use.

3. Solve  Solve the problem by carrying out your plan.  If your plan doesn't work, try another, and maybe even another.

4.  Examine    Examine your answer carefully. See if its fits the facts given in the problem.  Compare it to your estimate.  If the answer is not reasonable, make a new plan.  You may also want to check your answer by solving the problem again in a different way.[/quote]

the 4-step plan, or other plans similar to it, is found frequently in middle or elementary school math textbooks. i don't know about high school though.......


ratio
geometric sequence
geometric series

What is the sum of the first n terms of a geometric sequence if the first term is 250, the nth term is 75 and the common ratio is 4/5?

There is something wrong with the question. I got 6.395507946... as the no. of terms in the sequence. This is not possible. Are you sure this is the question?   :?

The ratio of 75 to 250 is $\displaystyle\frac{3}{10}$. $0.8^n\not=0.3$ when n is an integer so yes, this problem is unsolvable.

this is what the solution says:

let An=the nth term of a Geometric Sequence and r=the common ratio

A1*r^(n-1)=An

A1*r^n=(An)r

Then you substitute that into the formula for the partial sum of a geometric series

(250-(75)(4/5))/(1-(4/5))=950

That's all correct, except for the statement that the nth term is 75.


superior algebra
superior algebra unsolved

if $ E/F$ be an extension.then proof these:
1)if $ E/F$ be algebraic and F be perfect then E is perfect.
2)if $ E/F$ be finite and E be perfect then F is perfect too.

1) is trivial.
2) in characteristic p, use: perfect = frobenius is surjective


function
number theory unsolved
number theory

Given a function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x^{2}+y^{2})=f(x^{2})+f(y^{2})$ for all $x,y  \epsilon  \mathbb{N}$.
Prove or disprove: $f(x^{2})=x^{2} f(1)$ for all $x  \epsilon  \mathbb{N}$

[edit: I originally mis-typed what I wanted to be proved; it was originally written $f(x)=x f(1)$ which is not necessarily true]

Maybe the equation that edummit had in mind is $f(x^2+y^2) =f(x)^2 + f(y)^2$? In that case, it is easy to show that $f(x) = x$.

[quote="Ravi B"]I don't think it's true.  Let's define $f(x) = x$ for all $x \neq 3$, and define $f(3) = 17$.  That should be a counterexample.  The idea is that 3 is neither a square nor a sum of two squares.

Hmm, that looks too easy.  Have I missed something?[/quote]
No, at least not in the way I had originally mistyped it... sorry....  :oops:
[quote="zscool"]verify that f(n) = n for small n, then use the identity:

(5x-2y)^2 + y^2 = (3x-2y)^2 + (4x-y)^2

which should give you something to induct on.[/quote]
The hard part is finding a base case to start with... I'm having problems showing that (for example) $f(4)=4 f(1)$.


Suppose $ a$, $ b$ and $ c$ are numbers satisfying $ \frac{a}{b} \equal{} \frac{3}{8}$ and $ \frac{b}{c} \equal{} \frac{12}{21}$. What is the value of $ \frac{a}{c}$? Express your answer as a common fraction.

hello, we get by multiplying
$ \frac{a}{b}\cdot\frac{b}{c}\equal{}\frac{a}{c}\equal{}\frac{3}{8}\cdot\frac{12}{27}\equal{}\frac{3}{14}$
Sonnhard.


Hi. I have these three easy (not for me) problems. How do I show that

1. [tex]ac \equiv bd (mod 6)[/tex]
2. [tex]a^5 + \equiv b^5 (mod 6)[/tex]
3. [tex]a + c \equiv b + d (mod 6)[/tex]

if
[tex]a \equiv b (mod 6)[/tex] and [tex]c \equiv d (mod 6)[/tex]

This is the first time I started with mod. And I never do this kind of problems (I never du "problems", just basic school math) so I don't have the logic. I don't want the answers, I just want a push (or two) in the right direction. Thank you. :)

Well, knowing that a=b (mod 6) just means that c | a-b, so why don't you let a=6m+b, c=6n+d for some integers m and n, and plug a,c into the three identities? (cancel out the multiples of 6....and eureka!)  :D


superior algebra
superior algebra solved

a)What is [b]ring ordered[/b]?
b)What is [b]division ring[/b]?

a) to be (totally) ordered means that you can compare elements by $<$ and $>$ as used (see http://mathworld.wolfram.com/TotallyOrderedSet.html ).
b) a division ring is ring such that for any nonzero element there is an inverse (so a field with not necessary commutative multiplication)

[quote="ZetaX"]a) to be (totally) ordered means that you can compare elements by $<$ and $>$ as used (see http://mathworld.wolfram.com/TotallyOrderedSet.html ).
[/quote]

Thank you very much.But what is ring ordered? :)


linear algebra
matrix
Functional Analysis
real analysis
real analysis unsolved

[I didn't know whether this (functional analysis) should be in the analysis section or here.]

On $ \ell^2 \equal{} \ell^2(\mathbb{C})$, use the compelete orthogonal system $ (e_n)_n$ defined by $ e_n \equal{} (\delta_{kn})_{k\in\mathbb{N}}$ to define the bounded operator $ T: \ell^2\to\ell^2$ as follows:
$ Te_{2k \minus{} 1} \equal{} \frac {1}{k}(e_{2k \minus{} 1} \minus{} ie_{2k})$
$ Te_{2k} \equal{} \frac {1}{k}(ie_{2k \minus{} 1} \plus{} e_{2k}).$
(i) Give the matrix $ A$ that represents $ T$ with respect to $ e_{2k \minus{} 1}$ end $ e_{2k}$ on the space spanned by these two vectors.
(ii) Show $ T$ is compact and self-adjoint.
(iii) Calculate the spectral representation $ T \equal{} \sum\limits_{\lambda\in\sigma(T)}\lambda\pi_\lambda$.

(i) Well, this is just the definition: $ A \equal{} \frac {1}{k}\left(\begin{array}{cc}1 & i \\
\minus{} i & 1\end{array}\right).$
(ii) $ A$ obviously satisfies $ a_{ji} \equal{} \overline{a_{ij}}$ ($ i,j \equal{} 1,2$), so $ A$ is self-adjoint. Is this enough to conclude that $ T$ is self-adjoint?
Also, $ A$ has finite rank, so is compact. But I guess this doens't prove $ T$ to be compact. My idea was to approximate $ T$ with finite rank operators $ T_n: \ell^2\to\ell^2$ defined by $ T_ne_k \equal{} Te_k$ if $ k\leq n$ and $ T_ne_k \equal{} 0$ if $ k > n$.
(iii) This boils down to diagonalizing $ A$, which gives $ \left(\begin{array}{cc}0 & 0 \\
0 & \frac {2}{k}\end{array}\right)$, and then calculating the projections explicitly. This is pretty straightforward, my final answer is $ T(a_j)_j \equal{} \sum\limits_{k\in\mathbb{N}}\frac {2}{k}\pi_{\frac {2}{k}}(a_j)_j \equal{} \sum\limits_{k\in\mathbb{N}}\frac {a_{2k \minus{} 1} \plus{} ia_{2k}}{k}(e_{2k \minus{} 1} \minus{} ie_{2k}).$

Could anyone comment on my answers/approach? Thanks in advance.

The solution is basically correct. There are only few things that could be improved (in my opinion). From (i) you can deduce that T possesses an orthonormal basis consisting of eigenvectors with corresponding eigenvalues -- zeros and $ \frac2k$. For such a diagonal operator self-ajointness is equivalent to the fact that all the eigenvalues are real and compactness -- to the fact that the sequence of eigenvalues tends to zero. This solves (ii). Your solution of (iii) is OK.

Your comment has helped me improved my solution; thanks again!


Dear Mathlinkers ,i am looking for a proof that there exist infinitly primes on the form $ an\plus{}b$.
could anyone post it .

you should have searched just a little harder:

http://planetmath.org/?op=getobj&from=lec&id=11


At the figure below, A,B,C and D are collinear and AB = BC = CD = 1. Besides, the arches AB,AC,BD and CD are semi circunferences.
The area of the region highlighted is ??

answer : $ \frac{7pi}{12}$+ $ \sqrt { \frac {3}{4}}$ ; how to get in these number?
Thanks

[url=http://img10.imageshack.us/i/circlez.jpg/][img]http://img10.imageshack.us/img10/9789/circlez.th.jpg[/img][/url]

edit: i got a better picuture :D

[url=http://img33.imageshack.us/i/pictureaqe.jpg/][img]http://img33.imageshack.us/img33/8634/pictureaqe.th.jpg[/img][/url]

[quote="Likon"]At the figure below, A,B,C and D are collinear and AB = BC = CD = 1. Besides, the arches AB,AC,BD and CD are semi circunferences.
The area of the region highlighted is ??

answer : $ \frac {7\pi}{12}$+ $ \sqrt {\frac {3}{4}}$ ; how to get in these number?
Thanks

[url=http://img10.imageshack.us/i/circlez.jpg/][img]http://img10.imageshack.us/img10/9789/circlez.th.jpg[/img][/url][/quote]

are you sure of your answer?

total area of semicircles $ AB,CD \equal{} \frac {\pi}{4}$

everything else $ \equal{} \frac {\pi}{3} \plus{} \frac {\sqrt {3}}{2}$ (because the white area between $ B$ and $ C$ is composed of two arcs that have an equilateral triangle as intersection)

add it up, and we're done  :D

[quote="earldbest"][quote="Likon"]Yes im sure about the answer[/quote]

oh yeah, you're right :blush:[/quote]

That was ironic? , ;\
Actually, the answer is that informed.


Find all functions f: Z -> Z such that f(f(n)) = n and f(f(n+2)+2) = n for every integer n and f(0) = 1.

I see that it is easy.
I have similar result: f(n)=-n+1 all n in Z.

This problem was solved by a user named 'I think...' What a shame we don't know who that was  :D

[quote="I think..."]we have: f(1)=0;f(0)=1;
f(f(1)+2)=-1 ->f(2)=-1.
Similiarly,we have:f(3)=-2....f(n)=-n+1 ,n>=0.
From: f(f(n))=n with n>=0
->f(-n+1)=n,n>=0.
-->f(n)=-n+1 all n in Z.[/quote]

i will solve this for my pupils.. they will like..  :D

First, suppose $ f(a) \equal{} f(b)$. Then $ a \equal{} f(f(a)) \equal{} f(f(b)) \equal{} b$, so $ f$ is injective. Then we can say $ f(f(n)) \equal{} f(f(n\plus{}2)\plus{}2)$, implying that $ f(n) \equal{} f(n\plus{}2) \plus{} 2$ since $ f$ is injective. Plugging $ n \equal{} 0$ into the first given equation gives $ f(1) \equal{} 0$. From $ f(n) \equal{} f(n\plus{}2) \plus{} 2$, we can induct both ways from the base cases $ f(0)$ and $ f(1)$ to show that $ f(n) \equal{} 1 \minus{} n$ for all integers $ n$.


[quote="mr. math"]How would you do this problem?

In square ABCD, a right triangle (triangle AFB) with hypotenuse segment AB is drawn, with its right angle inside the square. If  line segment AF = 6 and line segment BF = 8, then what is the distance from the intersections of the square's diagnol to point F.

Please include an explanation.    :chief:  :yoda:  :harhar:[/quote]
[hide]
I used coords. D: (0,0), so the intersection of diagonals: (5,5) because square has length 10. I drew a parallel line crossing point F parallel to the side of the square AB/CD. I labeled the side between A and the point where this parallel line crosses AD as x and the point where the parallel line crosses AD to F as y. Therefore the point where the parallel line crosses BC to point F is 10-y and the same point to B is also x. so $x^2+y^2=36, (10-y)^2+x^2=64$, the coords of F are $(3.6,4.8)$ and the distance between this point and (5,5) is $\boxed{\sqrt{2}}$[/hide]


Solve this:

http://presents.elliian.com/riddle/10031.htm

Level 35

There are 5 worlds in MapleStory (set in a specific order). In each world there is a GM of a different job/class. Each of the 5 GMs drinks a certain type of beverage, eat a certain kind of food, and has a different favorite pet. None of them have the same pet, eat the same food or drink the same beverage.

THE QUESTION IS :

Who loves Dragons?

CLUES:

The Korean works in Broa.

The American loves Lions.

The Chinese drinks tea.

Windia is on the left of Khaini.

The GM of Windia drinks milk.

The GM who eats Kimchee likes Gryphons.

The Scanian GM eats Pho.

The GM in the world in the middle drinks soda.

The GM in the first world, is Vietnamese.

The GM who eats rice, is next to the one likes the Phoenix.

The GM that loves the tigers, is next to the one that eats Pho.

The GM that eats Pizza drinks Coffee.

The Japanese one eats Noodles.

The Vietnamese GM is next to Bera.

The GM who eats rice is next to the GM that drinks water.

This is the final level. Pass this and you win the internets

If we solve this, we WIN THE "INTERNETS". I've already solved a bunch of them, and this is the LAST LEVEL! Calling all AoPSers for help.

I guess it is, Shobber.

http://www.reed.edu/~mcphailb/puzzles/einstein.html

Both riddles have 15 clues, the format is the same. So we just have to figure out who is the German in this riddle, right?  :lol:

Wait... are all the worlds lined up in a line? :huh:

This reminds me of the "who owns the zebra" problem--there are also 15 clues. :huuh:

Just a guess, but comparing the two puzzles, I'd say the Japnese GM.  :lol:


There is an 8x8 chess board. What is the maximum number of 1X1 squares that a straight line contained in the board will go through?

[hide]Along the diagonal... using the inclusion-exclusion principal $A + B - GCF$ is the maximum number of squares that the line can go through so the answer should be $8 + 8 - 8 = 8$.[/hide]

If you move the line containing a diagonal to the right of left just a bit, the line will go thru several more squares instead of just going thru the vertices of the squares on the diagonal.  This way, the line goes thru 14 squares.

I did it like deej21, but I got [b]15[/b].  If the line is a little bit to the side of the main diagonal, it goes through the 8 squares on the main diagonal as well as the 7 squares on a secondary diagonal, thus 15 total.

just think of it this way: there are 7 vertical and 7 horizontal lines going through the interior of the chessboard. To have the line going through the most amount of 1x1 squares, we need that line to go through every one of those 14 interior lines. Going through 14 interior lines mean going through $14 + 1 = 15$ squares at the most.

By the way, this was one of the problems at the Eotvos Competition. Good job for handling it very well, the monster.

Well, the Fianchetto Diagonal has 8 squares on it, and the diagonal next to it has 7, so I guess you could have a line going in between them that hits a total of 15 squares.


Prove that 1+1/3+1/5+......+1/(2n-1) can never be an integer if n>1

that's 1 + 1/2 + ... + 1/(2n) - 2(1/2 + 1/4 + ... + 1/(2n)) = 1 + 1/2 + ... + 1/(2n) - (1/1 + 1/2 + ... + 1/n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n) and now the easiest way to kill this problem is to say by bertrand's conjecture, there exists some prime p in that range (so the denominator is divisible by p while the numerator isn't). If you don't want to use hard theorems, consider the largest power of 2 in that range (from n+1 to 2n); obviously, no multiples of it are in that range or it would contradict maximality, so the denominator will be divisible by this largest power of 2 while the numerator will not be (and will actually be odd, if I am not mistaken).

I had another idea consider $3^r$ which $3^{r+1}>2n-1$. we know that that just $3^r$ itself is a numberlike $k$ between $1,...,2n-1$ that $3^r|k$ . ;)
It is like proving that $\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ is not integer but there we consider powers of $2$ and here we consider power of $3$. :lol:

I'm having trouble trying to follow the first part of  your argument. Maybe you could clarify for me....

The sum I'm talking about only involves odd denominators,
that is, for a given N you would get 1+1/3+1/5+....1/(2N-1).

How does 1+1/3+1/5+....1/(2N-1) relate to your SUMS:

[quote="MysticTerminator"]that's 1 + 1/2 + ... + 1/(2n) - 2(1/2 + 1/4 + ... + 1/(2n)) = 1 + 1/2 + ... + 1/(2n) - (1/1 + 1/2 + ... + 1/n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n) [/quote]


i am not sure what you are asking to solve for? is it (a,b,n,m) or (a,b) or (n,m)?

I suppose it is to find $(a,b,m,n)$. Now, my guess,
[hide="Guess"]$(0,0,0,0)$ is the only one in both of them[/hide]

Good as your guess may be, proof is required! :lol:  Try mods

[quote="seamusoboyle"]1: Find all solutions to $a^2+b^2=3(n^2+m^2)$
2: Find all solutions to $a^2+b^2=7(n^2+m^2)$[/quote]

A general statement: the equation $a^2+b^2=p(m^2+n^2)$ has only the trivial solution in integers, where $p$ is a prime number of the from $4k+3$.

Myth, does $p$ have to be in the form $4k+3$ because then we can't have a square congruent to $-1 \mod p$? How can this be proved?

I tried considering $x^2 \equiv -1 \mod p$, then raising that an odd power to get $x^{2(2k+1)} \equiv x^{4k+2} \equiv x^{p-1} \equiv -1 \mod p$, which contradicts FLT. Is this correct?

[quote]Myth, does p have to be in the form 4k+3 because then we can't have a square congruent to -1 \mod p? How can this be proved? [/quote]

Or simply:

Notice:  if $n^2+m^2$ is odd it can NOT be $3 \bmod 4$  (and thus had to be 1 mod 4)

Proof:

$(2k+1)^2 = 4k^2+4k+1 \equiv  1 \bmod 4$   and  $ (2k)^2  \equiv 0 \bmod 4 $
Thus $n^2+m^2 $ has to be   $0,1,2 \bmod 4 $   ( [b]and can not be  $3 \bmod 4$ [/b])

Now if  $n^2+m^2 = 2x$ is even and $x>0$  , Then   $n+m$ and $n-m$  both would be even too and so   get a new relationship:
$n_1 = (n+m)/2 $ and $m_1 = (n-m)/2$  and $n_1^2 + m_1^2 = x$

(And similarly reduce $a$ and $b$ too by having $(a+b)/2$ and $(a-b)/2$  as new variables)

Keep doing this till  till RHS is odd,  if if $n_k^2 + m_k^2 $ is 1 mod 4 ;

  LHS would be  3 mod 4 .. Contadiction... (Unless original $x$ was $0$)
(

Way to take apart my problem! Ill post a different one...

Thanks, Myth. :)

I just don't feel that my proof is sufficient, I mean I just proved that it's impossible for -1, but I can't find proper reasoning why this should apply to, say, $-2^2$. I had some weak reasoning when I began my proof, but I just wondered if you could explain it better.

If we have $a^2+b^2=0\pmod{p}$ and $p=4k+3$, then $(ab^{-1})^2=-1\pmod{p}$, where $b^{-1}$ is such residue that $b^{-1}\cdot b=1$ (I assume $a,b,\neq 0\pmod{p}$).

[quote="seamusoboyle"]Also handy in competitions is the fact that if

$a^2+b^2=0 (\mod 3)$

then $a=0 (\mod) 3$  and  $b=0 (\mod 3)$... the proof is obvious...[/quote]

BTW that is correct, not only for $3$ but  for any number of the form $4k+3$ as Myth proves above ..
And that gives  yet another interesting proof in terms of  Fermat's "Infinite Descent" for theoriginal problem
[hide]Eg if $a^2+b^2 = 3(m^2+n^2)  \equiv 0 \pmod 3$  we get $a=3a_1 , b=3b_1$
and get  $m^2+n^2 = 3(a_1^2+b_1^2)$
which will get even smaller ($m=3m_1 , n=3n_1$)
$(a_1^2+b_1^2) = 3(m_1^2+n_1^2)$
And so on .. Therefore:
 apart from all zeros there is no other solution ...[/hide]


geometry
circumcircle
trigonometry
geometry proposed

Let $ AH$ be altitude of triangle $ ABC$, $ d$ is a line which is through $ A$, $ d$ cut circumcircle of $ ABC$ second point $ D$, $ M,N$ are projections of $ B,C$ on $ AD$, resp, $ P$ is projection of $ D$ on $ BC$.

a) Prove that $ M,N,H,P$ are concyclic.

b) Let $ O$ be circumcenter of $ ABC$, $ DP\cap AO \equal{} O'$, $ Y,Z$ are projections of $ O'$ on $ AB,AC$. Prove that $ ZYHP$ is isoceles trapezoid.

Note that, this is generalization of the problem [url=http://www.mathlinks.ro/viewtopic.php?t=302748]concyclic ..[/url]

a, $ \angle MHP\equal{}\angle BAM\equal{}\angle DCP\equal{}\angle MNP$
b, Let $ O'P\cap YZ\equal{}\{G\}$
We have $ GZ\equal{}O'Z.\sin \angle GO'Z\equal{}AO'.\sin \angle O'AC.\sin C$
$ YN\equal{}AY.\sin \angle YAN\equal{}AO'.\sin \angle AZY. \sin \angle CAO\equal{}AO'.\sin \angle CAO.\sin C$
Therefore $ GZ\equal{}YN$ and we are done.


limit
function
real analysis
real analysis unsolved

Suppose that $ f: [0,1]\rightarrow \mathbb{R}$ satisfies $ lim_{x\rightarrow} f(x) \in \mathbb{R}$, for all c

$ \forall$ $ c \in (0,1)$,
and $ f(0\plus{})\equal{}lim_{x\rightarrow 0^{\plus{}}}f(x)$ ,  $ f(1\minus{})\equal{}lim_{x\rightarrow 1^{\minus{}}}f(x)$ in $ \mathbb{R}$.

Can $ f$ be Riemann integrable on $ [0,1]$?
Prove or disprove your answer.

[quote="This is an amended version of what lelouch"]Suppose that $ f: [0,1]\rightarrow \mathbb{R}$ satisfies the conditions that $ \lim_{x\rightarrow c} f(x)$ exists, for all $ c \in (0,1)$,
and $ f(0 \plus{} ) \equal{} \lim_{x\rightarrow 0^{ \plus{} }}f(x)$ ,  $ f(1 \minus{} ) \equal{} \lim_{x\rightarrow 1^{ \minus{} }}f(x)$ also exist.

Can $ f$ be Riemann integrable on $ [0,1]$?
Prove or disprove your answer.[/quote]
I hope I have interpreted the question correctly. Here's an outline of a proof that $ f$ is Riemann integrable.

Let $ \varepsilon > 0$. The aim is to find a partition of [0,1] for which the upper and lower Riemann sums of $ f$ differ by less than $ \varepsilon$.

Let $ g(x) \equal{} \lim_{x\rightarrow c} f(x)\ (c\in[0,1])$ (the limits being one-sided at the endpoints). The set $ E$ of points at which $ |g(x) \minus{} f(x)|\geqslant\varepsilon$ is finite (otherwise $ E$ would have a limit point $ c$, at which $ \lim_{x\rightarrow c} f(x)$ would not exist). Define $ f_0$ to be the function equal to $ f$ on $ [0,1]\setminus E$, and equal to  $ g$ on $ E$. It is then true that for every $ c$ in [0,1] there is a neighbourhood of $ c$ in which the value of $ f_0$ varies by less than $ \varepsilon$.

Now let $ S$ be the set of points $ c$ in [0,1] for which there is a partition of [0,$ c$] such that $ f_0$ varies by less than $ \varepsilon$ on each interval of the partition. Let $ b \equal{} \sup S$. It follows from the statement at the end of the previous paragraph firstly that $ b\in S$, and secondly that $ b \equal{} 1$.

Thus there is a partition of [0,1] for which the upper and lower sums of $ f_0$ differ by less than $ \varepsilon$. To complete the proof, you just need to refine this partition by adding small intervals around the points in $ E$ to get a partition of $ f$ with that property.

[The motivating example that any proof of this result has to cope with is the function $ f(x) \equal{} 0$ if $ x$ is irrational, $ f(x) \equal{} 1/q$ if $ x \equal{} p/q$ is rational (in its reduced form). This function is discontinuous everywhere, but the limit $ \lim_{x\rightarrow c} f(x)$ exists everywhere, and is identically 0.]

[quote="Opalg"]Define $ f_0$ to be the function equal to $ f$ on $ [0,1]\setminus E$, and equal to  $ g$ on $ E$. It is then true that for every $ c$ in [0,1] there is a neighbourhood of $ c$ in which the value of $ f_0$ varies by less than $ \varepsilon$.[/quote]

Can you justify this?

[quote="Opalg"]The motivating example that any proof of this result has to cope with is the function $ f(x) \equal{} 0$ if $ x$ is irrational, $ f(x) \equal{} 1/q$ if $ x \equal{} p/q$ is rational (in its reduced form). This function is discontinuous everywhere, but the limit $ \lim_{x\rightarrow c} f(x)$ exists everywhere, and is identically 0.][/quote]

This function is continuous at the irrationals. Isn't it?

[quote="limsup"][quote="Opalg"]Define $ f_0$ to be the function equal to $ f$ on $ [0,1]\setminus E$, and equal to  $ g$ on $ E$. It is then true that for every $ c$ in [0,1] there is a neighbourhood of $ c$ in which the value of $ f_0$ varies by less than $ \varepsilon$.[/quote]

Can you justify this?[/quote]
Since $ \lim_{x\to c}f_0(x) \equal{} g(c)$, there is a neighbourhood of $ c$ within which $ |f_0(x) \minus{} g(c)| < \min\{\varepsilon/2,\,\varepsilon \minus{} |g(c) \minus{} f_0(c)|\}$. The result then follows from the triangle inequality.

[quote="limsup"][quote="Opalg"]The motivating example that any proof of this result has to cope with is the function $ f(x) \equal{} 0$ if $ x$ is irrational, $ f(x) \equal{} 1/q$ if $ x \equal{} p/q$ is rational (in its reduced form). This function is discontinuous everywhere, but the limit $ \lim_{x\rightarrow c} f(x)$ exists everywhere, and is identically 0.][/quote]

This function is continuous at the irrationals. Isn't it?[/quote]
Yes, of course. Thoughtless slip on my part.


Source: Furman University Wylie Math Tournament Jr. Exam 1997 #26

Find the value of

:sqrt:(1+:sqrt:(1+:sqrt:(1+:sqrt:(1+...))))

(without symbols, this is sqrt(1+sqrt(1+sqrt(1+sqrt(1+...)))))

I feel the urge to do a lot of math tonight...

Spoiler:

[hide] let sqrt (1+sqrt(1+sqrt(1... = x

x^2= 1+x

x^2-x-1=0

use the quadratic equation to factor, and take the positive root, which I believe is the golden ratio...[/hide]

That should be it unless I managed to do the arithmetic wrong.

Well I think the golden ratio is [hide]:rt5:+1:2[/hide], but otherwise your correct. (The answer should come out to [hide](:rt5:+1)/2[/hide].)

I've seen that used as a representation of Phi.

Right.  [hide]Phi, Golden Ratio, (1+:rt5:)/2...[/hide]all right.

yeah, golden ratio, phi, (sqrt[5]+1)/2, all the same thing

Wait. a:b is a/b? Isn't it technically different?


vector
algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ T$ be an operator on a vector space $ V$, let $ p$ be a polynomial over $ \mathbb{C}$. Prove that $ a \in \mathbb{C}$ is an eigenvalue of $ p(T)$ if and only if $ a\equal{}p(\lambda)$ for an eigenvalue $ \lambda$ of $ T$. Give a counterexample if we replace $ \mathbb{C}$ with $ \mathbb{R}$.

I'm a little unclear about exactly what you mean, but would you consider this a counterexample?

Let $ T\equal{}\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}$ and let $ p(x)\equal{}x^2\plus{}1,$ which is the characteristic polynomial of $ T.$ We have $ p(T)\equal{}T^2\plus{}I\equal{}0,$ which has eigenvalue $ 0,$ but $ T$ itself has no real eigenvalues.

About the proof- if $ \lambda$ is an eigenvalue of $ T$ and $ p(x) = a_kx^k + ...a_0$, $ Tv = \lambda v$ then, clearly,
$ P(T)v = a_kT^kv + ...a_ov = p(\lambda )v$.
Let's suppose that for some $ a\in\mathbb {C}$ $ P(T)v = av$. We can rewrite the last equality: $ (T - a_1I)\cdot...\cdot (T - a_k\I)v = o$ where $ a_i$ are the roots of equation $ P(x) = a$. Now if $ (T - a_k)v = 0$ we are done.If this is not the case consider maximal $ l$ for wich $ (T - a_lI)\cdot...\cdot (T - a_kI)v = 0$. Then $ a_l$-eigenvalae of $ T$.

This is an exercise from Axler.

[quote="Kent Merryfield"]I'm a little unclear about exactly what you mean, but would you consider this a counterexample?

Let $ T \equal{} \begin{bmatrix}0 & \minus{} 1 \\
1 & 0\end{bmatrix}$ and let $ p(x) \equal{} x^2 \plus{} 1,$ which is the characteristic polynomial of $ T.$ We have $ p(T) \equal{} T^2 \plus{} I \equal{} 0,$ which has eigenvalue $ 0,$ but $ T$ itself has no real eigenvalues.[/quote]

I wasn't considering this a counterexample because $ p(\pm i) \equal{} 0$ is an eigenvalue of $ p(T)$ and $ \pm i$ are eigenvalues of $ T$. However, perhaps this is what Axler had in mind.


inequalities
function
logarithms
inequalities unsolved

Let $a,b,c \geq 0$. Show that
\[(a+b)(b+c)(c+a) \geq 2 \sqrt{2}\left(abc(a+b+c)\right)^{3/4}\]

the stronger inequality is $(a+b)^4(b+c)^4(c+a)^4 \ge \frac{2^{12}}{27} a^3b^3c^3(a+b+c)^3$
Take log on both sides we see that it is exactly popoviciu's inequality on concave function $f(x)=\ln x$ and variables $a,b,c$


vector
geometry
analytic geometry
trigonometry
complex numbers
geometry theorems

We will show some nice properties of equilateral triangle from a fundamental identity of vector, and we will extent them to regular polygon.

[b]Main problem:[/b][color=blue]Let $ O$ be center of equilateral triangle $ ABC$ and $ M$ is an arbitrary point in plane. Let $ M_{a},M_{b},M_{c}$ be projections of $ M$ on the lines $ BC,CA,AB$, respectively. Prove that
\[ \overrightarrow{MM_{a}}+\overrightarrow{MM_{b}}+\overrightarrow{MM_{c}}=\frac3{2}\overrightarrow{MO}\][/color]

[b]First proof:[/b]We will use the concept of algebraic area of triangles,we denote algebraic area of triangle $ MBC,MCA,MAB,ABC$ be $ [S_{a}],[S_{b}],[S_{c}],[S]$ respectively and called $ AA',BB',CC'$ be altitude of triangle $ ABC$. We have
\[ \frac{[S_{a}]}{[S]}=\frac{\overline{M'M_{a}}}{\overline{AA'}}\]
therefore
\[ \overrightarrow{MM_{a}}=\frac{\overline{MM_{a}}}{\overline{AA'}}\overrightarrow{AA'}=\frac{3[S_{a}]}{2[S]}\cdot\overrightarrow{AO}\]
Similarly we have
\[ \overrightarrow{MM_{b}}=\frac{3[S_{b}]}{2[S]}\cdot\overrightarrow{BO},\qquad\overrightarrow{MM_{c}}=\frac{3[S_{c}]}{2[S]}\cdot\overrightarrow{CO}\]
Thus
\[ \sum\overrightarrow{MM_{a}}=\sum\frac{3[S_{a}]}{2[S]}\cdot\overrightarrow{AO}=\frac{3}{2[S]}(\sum[S_{a}](\overrightarrow{AM}+\overrightarrow{MO}))=\]
\[ =\frac{3}{2[S]}(\sum[S_{a}](\overrightarrow{AM})+\frac{3}{2[S]}(\sum[S_{a}])(\overrightarrow{MO})=\frac{3}{2}\overrightarrow{MO}\]
Here we used two fundamental formulas of algebraic area
\[ [S_{a}]\overrightarrow{MA}+[S_{b}]\overrightarrow{MB}+[S_{c}]\overrightarrow{MC}=\overrightarrow{0},[S_{a}]+[S_{b}]+[S_{c}]=[S]\]

[b]Second proof:[/b]We will give a solution by complex number, choose a coordinate system such that $ A(a),B(b),C(c)$ are three root of equation $ x^{3}=1$. We well-know the formula of projection $ S(s)$ from a point $ M(m)$ to the line connecting $ A(a),B(b)$ is
\[ s=\frac{1}{2}(a+b+m-ab\overline{m})\]
Now using it when $ M_{a},M_{b},M_{c}$ are projection of $ M(m)$ to the lines $ AB,BC,CA$
\[ m_{a}=\frac{1}{2}(b+c+m-bc\overline{m}),m_{b}=\frac{1}{2}(c+a+m-ca\overline{m}),m_{c}=\frac{1}{2}(a+b+m-ab\overline{m})\]
Thus
\[ \sum\overrightarrow{MM_{a}}=\sum(m-m_{a})=\sum m_{a}-3m=a+b+c-\overline{m}{ab+bc+ca}-\frac{3}{2}m\]
Easily seen $ a+b+c=0$ and $ ab+bc+ca=0$ because $ a,b,c$ are three root of $ x^{3}=1$ thus
\[ \overrightarrow{MM_{a}}+\overrightarrow{MM_{b}}+\overrightarrow{MM_{c}}=-\frac{3}{2}m=\frac3{2}\overrightarrow{MO}\]
We are done.

With the solution of complex numbers, by the similar way we can prove the general problem for the regular polygon:

[b]Problem 1: [/b][color=blue]Let $ O$ be center of reagular polygon $ A_{1}A_{2}...A_{n}$ and $ M$ is an arbitrary point in plane. Let $ M_{1},M_{2},...,M_{n}$ be projections of $ M$ on the lines $ A_{i}A_{i+1},(i=\overline{1,n},n+1\equiv 1)$, respectively. Prove that
\[ \overrightarrow{MM_{1}}+\overrightarrow{MM_{2}}+...+\overrightarrow{MM_{n}}=\frac{n}{2}\overrightarrow{MO}\][/color]

[b]Problem 2:[/b] [color=blue]Given equilateral triangle $ ABC$ and $ M,N$ are two arbitrary point point on plane. Let $ M_{a},N_{a}$ be projections of $ M,N$ on the line $ BC$, similarly we have $ M_{b},N_{b},M_{c},N_{c}$ we will have:
\[ \overrightarrow{M_{a}N_{a}}+\overrightarrow{M_{b}N_{b}}+\overrightarrow{M_{c}N_{c}}=\frac{3}{2}\overrightarrow{MN}\][/color]

[b]Proof:[/b] By main problem we have
\[ \overrightarrow{MM_{a}}+\overrightarrow{MM_{b}}+\overrightarrow{MM_{c}}=\frac3{2}\overrightarrow{MO},\overrightarrow{NN_{a}}+\overrightarrow{NN_{b}}+\overrightarrow{NN_{c}}=\frac3{2}\overrightarrow{NO}\]
And
\[ \sum\overrightarrow{M_{a}N_{a}}=\sum(\overrightarrow{M_{a}M}+\overrightarrow{MN}+\overrightarrow{NN_{a}})=\sum\overrightarrow{M_{a}M}+3\overrightarrow{MN}+\sum\overrightarrow{NN_{a}}=\]
\[ =-\frac{3}{2}\overrightarrow{MO}+3\overrightarrow{MN}+\frac{3}{2}\overrightarrow{NO}=\frac{3}{2}\overrightarrow{MN}\]
We are done.

We will have the similar problem and solution for regular polygon.

[b]Problem 3: [/b][color=blue]Given regular polygon $ A_{1}A_{2}...A_{n}$ and $ M,N$ are two arbitrary point point on plane. Let $ M_{i},N_{i}$ be projections of $ M,N$ on the line $ A_{i}A_{i+1},(i=\overline{1,n},n+1\equiv 1)$ we will have:
\[ \overrightarrow{M_{1}N_{1}}+\overrightarrow{M_{2}N_{2}}+...+\overrightarrow{M_{n}N_{n}}=\frac{n}{2}\overrightarrow{MN}\][/color]

[b]Problem 4: [/b][color=blue]Let $ O$ be center of equilateral triangle $ ABC$ with cricumradius $ R$ and $ M$ is an arbitrary point in plane. Let $ M_{a},M_{b},M_{c}$ be projections of $ M$ on the lines $ BC,CA,AB$, respectively. Prove that
\[ MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}=\frac3{2}MO^{2}+\frac{3}{4}R^{2}\][/color]

[b]Proof:[/b] Let $ A',B',C'$ be midpoint of $ BC,CA,AB$ we have
\[ \quad MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}\]
\[ =\sum(\overrightarrow{MO}+\overrightarrow{OA'}+\overrightarrow{A'M_{a}})^{2}\]
\[ =\sum(MO^{2}+OA'^{2}+A'M_{a}^{2}+2\overrightarrow{MO}\cdot\overrightarrow{OA'}+2\overrightarrow{OA'}\cdot\overrightarrow{A'M_{a}}+\overrightarrow{A'M_{a}}\cdot\overrightarrow{MO}\]
\[ =3MO^{2}+\frac{3}{4}R^{2}+\sum A'M_{a}^{2}+2\overrightarrow{MO}\cdot(\sum\overrightarrow{OA'})+2\sum\overrightarrow{OA}\cdot\overrightarrow{A'M_{a}}+2\sum\overrightarrow{A'M_{a}}\cdot\overrightarrow{MO}\ (*)\]

Here we see $ \sum\overrightarrow{OA'}=\overrightarrow{0}$ and $ A',M_{a}$ are projection of $ O,M$ on line $ BC$ therefore $ \overrightarrow{A'M_{a}}\cdot\overrightarrow{MO}=-A'M_{a}^{2}$ and $ \overrightarrow{OA'}\perp\overrightarrow{A'M}\Rightarrow \overrightarrow{OA'}\cdot\overrightarrow{A'M}=0$ and now do similar for vertex $ B,C$ we get from $ (*)$
$ MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}=3MO^{2}+\frac{3}{4}R^{2}+(\sum\overrightarrow{A'M_{a}})\cdot\overrightarrow{MO}=3MO^{2}+\frac{3}{4}R^{2}-\frac{3}{2}MO^{2}=\frac{3}{2}MO^{2}+\frac{3}{4}R^{2}$
Here note that $ A',B',C'$ are projection of $ O$ to lines $ BC,CA,AB$, respectively too, therefore by problem 2 we get $ \sum\overrightarrow{A'M_{a}}=-\frac{3}{2}\overrightarrow{MO}$.We are done.

We will show two three corollaries from this problem
[b]Corllolary 1:[/b] When $ M$ lie on the circle center and $ O$ radius $ r$ then
\[ MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}=\frac{3}{2}r^{2}+\frac{3}{4}R^{2}=constant\]

[b]Corllolary 2: [/b]The locus of point $ M$ such that $ MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}=k\ge\frac{3}{4}R^{2}$ is the circle center $ O$ with radius $ \frac{2}{3}\sqrt{k-\frac{3}{4}R^{2}}$

[b]Corllolary 3:[/b] $ MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}\ge \frac{3}{4}R^{2}$ for all $ M$ on plane equality hold iff $ M\equiv O$.

By this ideal from problem 2 we will get similar problem and corollaries for regular polygon with similar solution like equilateral triangle.

[b]Problem 5:[/b] [color=blue]Let $ O$ be center of ragular polygon $ A_{1}A_{2}...A_{n}$ with cricumradius $ R$ and $ M$ is an arbitrary point in plane. Let $ M_{i}$ be projections of $ M$ on the lines $ A_{i}A_{i+1}(i=\overline{1,n},n+1\equiv 1)$. Prove that
\[ MM_{1}^{2}+MM_{2}^{2}+...+MM_{n}^{2}=\frac{n}{2}MO^{2}+n\cos^{2}\frac{\pi}{n}R^{2}\][/color]

[b]Corllolary 1:[/b] When $ M$ lie on the circle center and $ O$ radius $ r$ then
\[ MM_{1}^{2}+MM_{2}^{2}+...+MM_{n}^{2}=\frac{n}{2}r^{2}+n\cos^{2}\frac{\pi}{n}R^{2}=constant\]

[b]Corllolary 2:[/b]The locus of point $ M$ such that $ MM_{1}^{2}+MM_{2}^{2}+...+MM_{n}^{2}=k\ge n\cos^{2}\frac{\pi}{n}R^{2}$ is the circle center $ O$ with radius $ \frac{2}{n}\sqrt{k-n\cos^{2}\frac{\pi}{n}R^{2}}$
\bigskip

[b]Corllolary 3:[/b] $ MM_{1}^{2}+MM_{2}^{2}+...+MM_{n}^{2}\ge n\cos^{2}\frac{\pi}{n}R^{2}$ for all $ M$ on plane equality hold iff $ M\equiv O$.
[b]
Problem 6:[/b] [color=blue]Given equilateral triangle $ ABC$ and $ M,N$ are two arbitrary point point on plane. Let $ M_{a},N_{a}$ be projections of $ M,N$ on the line $ BC$, similarly we have $ M_{b},N_{b},M_{c},N_{c}$ we will have:
\[ M_{a}N_{a}^{2}+M_{b}N_{b}^{2}+M_{c}N_{c}^{2}=\frac{3}{2}MN^{2}\][/color]

[b]Proof:[/b] Because $ M_{a},N_{a}$ are projection of $ M,N$ to the line $ BC$ therefore $ \overrightarrow{M_{a}N_{a}}\cdot\overrightarrow{MN}=M_{a}N_{a}^{2}$
Thus
\[ \sum M_{a}N_{a}^{2}=\sum\overrightarrow{MN}\cdot\overrightarrow{M_{a}N_{a}}=\overrightarrow{MN}(\sum\overrightarrow{M_{a}N_{a}})=MN\cdot\frac{3}{2}\overrightarrow{MN}=\frac{3}{2}MN^{2}\]
Here by the problem 1 $ \overrightarrow{M_{a}N_{a}}+\overrightarrow{M_{b}N_{b}}+\overrightarrow{M_{c}N_{c}}=\frac{3}{2}\overrightarrow{MN}$

The proof is quite simple but we can get nice result from this problem, now we take $ N\equiv O$ and let $ A',B',C'$ be the projections of $ O$ to line $ BC,CA,AB$ by that above problem we have $ \sum M_{a}A'^{2}=\frac{3}{2}MO^{2}$ now by Pythagore theorem we have $ M_{a}A'^{2}=OM_{a}^{2}-OA'^{2}$ and similarly for vertex $ B,C$ we will have
\[ \sum(OM_{a}^{2}-OA'^{2})=\frac{3}{2}MO^{2}\Leftrightarrow OM_{a}^{2}+OM_{b}^{2}+OM_{c}^{2}=\frac{3}{2}MO^{2}+3(\frac{R}{2})^{2}=\frac{3}{2}MO^{2}+\frac{3}{4}R^{2}\]

Now combine this problem with problem 3 we will get the corollary

[b]Corollary:[/b] $ MM_{a}^{2}+MM_{b}^{2}+MM_{c}^{2}=OM_{a}^{2}+OM_{b}^{2}+OM_{c}^{2}$

We will get similar problem and corollary for regular polygon

[b]Problem 7:[/b][color=blue]Given regular polygon $ A_{1}A_{2}...A_{n}$ and $ M,N$ are two arbitrary point point on plane. Let $ M_{i},N_{i}$ be projections of $ M,N$ on the line $ A_{i}A_{i+1},(i=\overline{1,n},n+1\equiv 1)$ we will have:
\[ M_{1}N_{1}^{2}+M_{2}N_{2}^{2}+...+M_{n}N_{n}^{2}=\frac{n}{2}MN^{2}\][/color]

[b]Corollary:[/b] $ MM_{1}^{2}+MM_{2}^{2}+...+MM_{n}^{2}=OM_{1}^{2}+OM_{2}^{2}+...+OM_{n}^{2}$