1.    Let $ a,b,c,d,e,f\in\mathbb{R}$ such that $ (a,b,c)$ majorizes $ (d,e,f)$. Prove that $ ab\plus{}bc\plus{}ca\leq de\plus{}ef\plus{}fd$

2.    This time, let $ a,b,c,d,e,f\in\mathbb{R^\plus{}}$ with $ (a,b,c)$ majorizing $ (d,e,f)$. Prove that $ \frac{(ab\plus{}bc\plus{}ca)^2}{abc}\geq\frac{(de\plus{}ef\plus{}fd)^2}{def}$

3.    Let $ x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n\in\mathbb{R^\plus{}}$ be two sequences such that $ (x_1,x_2,\dots,x_n)$ majorizes $ (y_1,y_2,\dots,y_n)$. Prove that $ \sum_{1\leq i<j<k\leq n}x_i x_j x_k\leq\sum_{1\leq i<j<k\leq n}y_i y_j y_k$

4.    We have the same sequences as in (3). For all $ x_1,x_2,\dots,x_n\in\mathbb{R}$, let $ \sigma_i(x_1,x_2,\dots,x_n)$ be the coefficient of $ x^i$ in $ \prod_{i\equal{}1}^n (1\plus{}x_i x)$. Prove that $ \sigma_i(x_1,x_2,\dots,x_n)\leq\sigma_i(y_1,y_2,\dots,y_n)$

[quote="bakerbakura"]

4.    Let $ x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n\in\mathbb{R^ + }$ be two sequences such that $ (x_1,x_2,\dots,x_n)$ majorizes $ (y_1,y_2,\dots,y_n)$. For all $ x_1,x_2,\dots,x_n\in\mathbb{R}$, let $ \sigma_i(x_1,x_2,\dots,x_n)$ be the coefficient of $ x^i$ in $ \prod_{i = 1}^n (1 + x_i x)$. Prove that $ \sigma_i(x_1,x_2,\dots,x_n)\leq\sigma_i(y_1,y_2,\dots,y_n)$[/quote]
Nice idea!  :)

My proof is not very nice, hopefully you have a better one. It is sufficient to prove $ \sigma_i(a)\le\sigma_i(b)$ for an “elementary majorization” $ a\succ b$, i.e. for $ a = (a_1,\dots,a_n)$ with $ a_1\ge\cdots\ge a_n$ and $ b = (a_1, \dots,a_j - t, \dots,a_k + t ,\dots,a_n)$ for a certain $ 1\le j < k\le n$ and $ t\le a_j - a_k$. (This is what Darij calls a ‘compression’ [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118784]here[/url]). In this case we find
\[ \sigma_i(b) - \sigma_i(a) = t\cdot(a_j - a_k - t)\cdot\sum_{\stackrel{\nu_1 < ... < \nu_{i - 2}}{\nu_\ell\ne j,k}}a_{\nu_1}\cdots a_{\nu_{i - 2}}\ge0\quad qed.\]
Now, I have looked into this for other symmetric functions. (I prefer writing $ m$ instead of $ i$.) For a partition $ \pi\vdash m$, denote by $ \sigma_\pi$ the corresponding elementary symmetric function. So the above is $ \sigma_{(1_m)}$, based on the partition $ (1_m) = (1,...,1)\vdash m$.

On the opposite side of the 'partition scale', if we take $ \pi = (m)$, then $ \sigma_{(m)}$ (not to be confused with $ \sigma_{i}$ above) is the $ m$-th power sum, so we have by Karamata trivially  the inverse: $ \sigma_{(m)}(b)\le \sigma_{(m)}(a)$.
For $ \pi = (m - 1,1)$, $ \sigma_\pi(b)\le\sigma_\pi(a)$ is already false, just choose $ a = (2,0,...,0)$ and $ (b = (1,1,0,...,0)$.

Back to the other side of the 'partition scale', we can look at the next 'bigger' one, $ \pi = (21_{m})\vdash m + 2$.
[b]I claim that we have
\[ \sigma_{(21_{m})}(b)\ge\sigma_{(21_{m})}(a)\qquad(*)\]
for all sequences $ a\succ b$ with $ n$ parts iff $ \boxed{m + 2\le n\le\frac {m^2 + m + 2}2}$. E.g., for $ (2,1)$, $ (*)$ is false, and for $ (2,1,1)$, $ (*)$ only holds if $ n = 4$.[/b]

[hide="Proof."]Taking again $ b$ as a compression of $ a$ as above, we find
\[ \sigma_{(21_{m})}(b) - \sigma_{(21_{m})}(a) = t\cdot(a_j - a_k - t)\cdot\left((a_j + a_k)x + y - 2z\right)\]
where

$ x = \sum_{\stackrel{\nu_1 < ... < \nu_{m - 1}}{\nu_\ell\ne j,k}}a_{\nu_1}\cdots a_{\nu_{m - 1}} = \sigma_{1_{m - 1}}(\hat a)$,

$ y = \sum_{\stackrel{\nu_1 < ... < \nu_{m - 1}}{\nu_\ell\ne j,k}}a_{\nu_1}\cdots a_{\nu_{m - 1}}(a_{\nu_1} + \cdots + a_{\nu_{m - 1}}) = \sigma_{21_{m - 2}}(\hat a)$

$ z = \sum_{\stackrel{\nu_1 < ... < \nu_{m}}{\nu_\ell\ne j,k}}a_{\nu_1}\cdots a_{\nu_{m}} = \sigma_{1_{m}}(\hat a)$,

writing $ \hat a$ for the sequence where $ a_j$ and $ a_k$ are omitted.

$ a_j$ and $ a_k$ can be arbitrarily small, so we need $ y\ge 2z$ for $ (*)$ to hold. By AM-GM, $ y\ge 2z$ holds as soon as the number of terms on the RHS doesn't exceed the one of the LHS, i.e.
\[ (m - 1){n - 2\choose m - 1}\ge2{n - 2\choose m},\]
equivalently $ m - 1\ge2\frac {n - m - 1}m,$ i.e. ${ n\le\frac {m^2 + m + 2}2}$. The other condition $ m + 2\le n$ is obvious. This ends the proof. [/hide]

I would guess that a similar pattern occurs for partitions $ \pi = (31_m)$ etc. Doable the same way, but too many calculations.  :maybe:

Actually my proof is the same as yours  :blush:

Unfortunately I don't know the notation you have used for partitions. Could you perhaps explain it in a different way? Thanks

I just omit commas sometimes  :blush: and use indices for multiple parts, e.g. $ (21_3)\equal{}(2,1,1,1)\vdash 5$. The sign "$ \vdash$" means "is a partition of".


geometry
rectangle
combinatorics proposed
combinatorics

Some 46 squares are randomly chosen from a $9 \times 9$ chess board and colored in [color=red]red[/color]. Show that there exists a $2\times 2$ block of 4 squares of which at least three are colored in [color=red]red[/color].

To solve this, it helps to notice that 46 is the smallest integer that is greater than half of 81 (number of squares on the board). If we colored it so that every other row was all red, then we would have 5 rows times 9 per row=45. Anywhere we add the last one, there would be a 2x2 with three red squares. QED

Another related problem is on a 9x9, what is the minimum number of squares colored red such that there exists a 2x2 that is all red?

we can divide the board into 4 symmetrical parts ie. 4x5 rectangles with one square at the middle left over.
that square can have atmost 1 colour. ie. we have to fill atleast 45 squares in the 4 symmetrical parts.
by php we have that atleast one part having 12 coloured squares
it can be shown that such a rectangle can have atmost 12 such squares.
so we get that atleast one such rectangle has exactly 12 coloured squares.
it can be shown that if such a rectangle having 12 coloured squares exists then one of the rectangles adjacent to it can have atmost 10 coloured squares.
so we get that the other 2 parts(that are adjacent to each other) must have atleast 23 coloured squares.
by php we again get that one of the rectangles has 12 coloured squares and then its adjacent square has atmost 10
thereby making a total<=12+10+12+10+1=45
thus 46 can never be selected not satisfying the question.

[quote=nthd]Assume that have at most 2 red  squares in any$2\times 2$-squares.
Cut the first row.[b]Devide the remain chessboard to18  $2\times 2$-squares[/b]
$\rightarrow$there are at most$18\times 2 +9=45$red squares :) contraction![/quote]
Can someone explain?

[quote=nthd]Assume that have at most 2 red  squares in any$2\times 2$-squares.
Cut the first row.Devide the remain chessboard to18  $2\times 2$-squares
$\rightarrow$there are at most$18\times 2 +9=45$red squares :) contraction![/quote]

This is wrong as after removing the the first row $8\times 9$ grid is remaining which can't be covered by $2\times 2$ squares.

If we colour the 5 alternate rows red. then 45 red squares with each square have 2 red blocks at most. hence the last block wherever placed will yield a 2x2 square with 3 red blocks


function
integration
complex numbers
complex analysis
complex analysis unsolved

let u be a non constant harmonic function on the whole complex plane. show that the set of zeros of u is unbounded.

That should be a real-valued harmonic function. Complex-valued harmonic functions make sense, and the problem is false for them.

The proof comes in two parts.
Part 1: Suppose $ u(z_0)\equal{}0$ for some $ z_0$. Then the set of zeros of $ u$ is unbounded.
Proof: WLOG, $ z_0\equal{}0$; if it isn't, we can set $ v(z)\equal{}u(z\plus{}z_0)$. By the averaging principle, $ \frac1{2\pi r}\int_{|z|\equal{}r}u(z)\,|dz|\equal{}u(0)\equal{}0$. There are two possibilities- either $ u$ is identically zero on the circle (impossible by the nonconstant condition) or $ u$ changes sign on the circle. Either way, $ u$ has a zero on the circle of radius $ r$ for each $ r>0$, and the set of zeroes is unbounded.

Part 2: There exists some $ z_0$ such that $ u(z_0)\equal{}0$.
Proof: Suppose not. WLOG, $ u$ is positive everywhere. We can construct an analytic function $ f$ on the whole complex plane that has $ u$ as its real part: let $ g(x,y)\equal{}\frac{\partial u}{\partial x}(x,y)\minus{}i\frac{\partial u}{\partial y}(x,y)$, treating $ u$ as a function of two real variables. This $ g$ is analytic (replacing $ (x,y)$ with $ x\plus{}iy$), and we can find an antiderivative $ f$ which has $ f(0)\equal{}u(0)$.
Now, the image of $ f$ is contained in the right half-plane. The function $ h(z)\equal{}\frac{f(z)\minus{}1}{f(z)\plus{}1}$ is bounded (by the unit circle) and entire, so it's constant. This implies that $ u$ is constant, a contradiction.


What do you think you should answer if you were asked:

"Tell me about yourself."

This is one question I hate because it's tricky and I don't know where I have to start and how long I have to go on. What do you think guys?

If this is a question for a college interview, you can tell the interviewer that you are curious (which is true) and that you like challenges (also true). What would you tell a new friend who asked you the same question? In general, you just have to be relaxed and YOURSELF to make a good impression in a college interview--usually the interview is NOT designed to trick you.

I don't think this is meant to be a "tricky" question. The interviewer is not trying to elicit a monolgue from you of fixed length and content, without telling you the parameters. And they are not looking for some deep introspective self-analysis. It's just meant to get the conversational ball rolling. Hopefully you'll say something that will allow the conversation to veer off naturally into a more specific topic. View it as an opportunity to clue the interviewer about things you would enjoy talking about. If you have an unusual hobby, for instance, this would be a good chance to bring it up which might lead to a much livelier conversation than if s/he had no clue about your interests.

Thanks!

I admit the question isn't meant to be tricky but it was just hard to answer (starting point).  :blush:

[quote="tokenadult"]<...> that you like challenges (also true) <...>[/quote]

It's such an American cliche. Everybody writes that he/she loves challenges in a college application, everybody says so on a job interview.
In fact, I beleive that people like the satisfaction that comes after overcoming a new challenge but not the challenge itself.
Most people would prefer to get paid for doing nothing. It's just at some point they become wiser and understand that facing challenges
is the only way to evolve and be successful. Then they face challenges deliberately.

completely untrue. if that was the case, i would relish the times at work when i am doing nothing. instead, i run around like a chicken trying to find something to do because i am BORED TO DEATH.
also completely untrue is that i don't like challenges. if i didn't like challenges, i would've stayed at my old school, where my lowest grade was a 96 (in PE, no less). i would have had plenty of opportunities there. but i didnt, i went to a (more challenging) school.

I think the statement that Silverfalcon likes challenges is true based on what he does here on the AoPS forums. I think Texas137's advice is even better, which is to simply describe examples of daily activities, which will give the message about liking challenges loud and clear with specific details.


[quote="nonogomaez"]BD bisects ?ABC in triangle ABC, D on AC, m?ABC =80 ,
AD=BD+BC
find measure of an angle C[/quote]
Sorry that I could not understand any head or tail of this question,.....
What EXACTLY do you mean?
Do you mean:
$ BD$ bisects $ \angle ABC$ in $ \Delta ABC$. D is a point on $ AC$, such that $ AD\equal{}BD\plus{}BC$ . If $ m\angle ABC\equal{}80$, find the ,measure of $ \angle C$....

I can get it using law of sines, but I'd much rather see a geometric approach. Does anyone have a solution?

This problem boil down to the question of following picture.
$ BD\plus{}BC\equal{}C'D$.

$ \angle ACB\equal{}80^\circ$.

I wish a solution does not make a mistake.


LaTeX
number theory unsolved
number theory

What about meaningfull titles¿ $\LaTeX$¿ Edit your post!
Some letters like "γ" are not standard ones, please also don't use them.

I try to learn Latex.(γ_1=x_1, .......,& γ_ν=x_n)

I can't understand......if  you want explanation of the problem please ask me....because  the problem is nice........

You should edit the [b]title[/b] of the thread. At the moment the title is "interesting.........", and this is meaningless. Titles are supposed to say something about the problem and should give the reader some idea what the topic is about. Imagine everyone naming his topic "interesting........."...

Yah it's my false.I wait anybody for a solution...... :wink:

What about finally changing the title ;)

actually it is very interesting.can anyone give a solution?

[quote="Αρχιμήδης 6"]Find all possitive integers γ_1,γ_2, ..... γ_ν & κ  of the equation:  (γ_1^2+γ_2^2+........γ_ν^2)/γ_1γ_2......γ_ν=κ  for examble for ν=2 we have the equation γ_1^2+γ_2^2=κγ_1γ_2..........Solve the equation for ν>2.[/quote]

This is called Hurwitz equation, and it doesn't seem like its solutions can be given in an explicit way in this generality...

[b]EDIT:[/b] See http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/hurwitz-equation .

  darij


An engineer is designing a spring to be placed at the bottom of an elevator shaft.  If the elevator cable should break when the elevator is at a height h above the top of the spring, calculate the value that the spring stiffness constant k should have so that passengers undergo an acceleration of no more than 5.0 g when brought to rest.  Let M be the total mass of the elevator and passengers.

When I try to solve this problem, this is my approach:
The potential energy and the total energy of the system when the elevator is at height h is Mgh.
When the elevator is just about to reach the spring, it has a KE of 1/2 Mv^2, which is equal to Mgh.  Therefore, v^2 is 2gh.
When the elevator compresses the spring by x amount of distance, the spring elastic energy is 1/2kx^2 which is equal to Mgh.
x = square root of (2Mgh/k)

since the final velocity squared = the intial velocity squared + 2a(x) and Vf =  0, I plug in the known values to solve for k.
My equation comes out to be
0=2gh + 10g(square root of 2MgH/k)  (2a = 10g)
i solve for k and get 50 * Mg/h

the answer is 12 * Mg/h

I can't figure out where I messed up.  Any ideas?
Also, I realize that 1/2kx^2 should not be equal to MgH.
rather, 1/2Kx^2 + MgX should = MgH, but this would make the problem ridiculously complicated...

You're taking the maximum acceleration in the whole compression process; but, after the point where the elastic force equilibrates the weight, the acceleration (in the upward sense) increases until the maximum deformation.

So the string actually compresses farther than x, and then goes back to x?
That makes sense, but how would I find how far the string actually initially compresses?

[quote="Immanuel Bonfils"]No. Your value of x is the maximum deformation (when the elevator stops falling and begins backing up). What I'm saying (rather posting...) is that the maximum acceleration, that the problem demands to be 5g, is achieved at that very moment.
Btw, I'm getting 18 instead of 12...[/quote]

"h above the top of the spring"

if any of you guys are getting the right answer can you post the solution?
Kevin?

Okay, I'll post the solution.
It appears that while the elevator is on the spring, its acceleration (or more accurately, its deceleration) is not constant, which is why my solution is invalid.

Let x be the vertical distance the spring "shrinks."

Fs - Fw = ma (up is positive)
kx - mg = 5mg (a = 5g)
kx = 6mg
x = (6mg) / k

we also know that mg(h + x) = 1/2 k x^2 by conservation of energy
substituting we get
mg ( h + 6mg / k) = 1/2 k (6mg / k)^2
for simplicity i will now say M = m^2 and G = g^2
mgh + 6MG/k = 18MG/k
kmgh = 12MG
k = 12mg/h

to do something to squared (or to any power) in latex:

do a^n, where n is whatever your power (don't forget the dollar signs)

although you probably know this already...

for fractions

use (dollar) \frac { numerator}{denominator} (dollar)

I would type in latex, but it just takes so much more time (for me at least)


[size=150]a,b,c,x,y,z be positive numbers.

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{a} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{x} \ge 3 \plus{} \frac {{x \plus{} a}}{{y \plus{} b}} \plus{} \frac {{y \plus{} b}}{{z \plus{} c}} \plus{} \frac {{z \plus{} c}}{{x \plus{} a}}$[/size]

[size=150]So nice,Should everyone not be proof?[/size]

[size=150]BQ[/size]


trigonometry
analytic geometry
graphing lines
slope
symmetry

Main Street and Central Avenue intersect, making an angle measuring34 degrees Angela lives at the intersection of the two roads, and Caitlin liveson Central Avenue 10 miles from the intersection. If Leticia lives 7 miles from Caitlin, which conclusion is valid?

(1) Leticia cannot live on Main Street.
(2) Leticia can live at only one location on Main Street.
(3) Leticia can live at one of two locations on Main Street.
(4) Leticia can live at one of three locations on Main Street.

Which direction does Caitline live?  Where its $34^{\circ}$ or $146^{\circ}$?

The only methods I'm seeing right now involve trig....

Find $\tan 34^{\circ}$ with a calculator to find the slope of Main Street and equation of Main Street

Then find the equation of a circle with radius $7$ centered at $(10,0)$ or $(-10,0)$ and find the points of intersection of the equations fo the lines and circles.

Put Central avenue as the x axis, and main street as the line $y = x\tan 34^{\circ}$.  By symmetry, Caitlin can live at (10,0) and we dont need to do the work for the other cases.  Now consider the locus of points 7 mi. from Caitlin ; its a circle with the equation $(x-10)^2 + (y^2) = 7^2$.  Obviously two points of intersection are $(17,0), (3,0)$.   Otherwise, it intersects the other line, and we solve $x^2 + 100 - 20x + x^2\tan^2 34^{\circ} = 49$.  It gives us 2 values of x, which correspond to y values.  Are these two x values real?  Yes, because the discriminant is $400 - 4(1 + \tan^2 34^{\circ})(51)$ which is positive.

Good job singular.  I thank you for your notes.
I am not posting homework.  Got it???


If $ g(x)  and h(x)$ are polynomials with real coefficients and $ f(x)\equal{}g(x^3) \plus{} xh(x^3)$ is divisible by $ (x^2\plus{}x\plus{}1)$, show that $ g(x) and  h(x)$ are both divisible by $ (x\minus{}1)$.

$ f(x)$ is divisible by $ x^2 \plus{} x \plus{} 1$ is equivalent to $ k(x) \equal{} g(1) \plus{} x f(1) \vdots x^2 \plus{} x \plus{} 1$. If $ f(1) \not\equal{}0$ then $ 2 \leq \deg k(x) \equal{} 1$ is a contradiction. Hence, $ f(1) \equal{} g(1) \equal{} 0$. That is to say $ f(x), g(x)$ are divisible by $ x\minus{}1$.

An other one : Let $ x_1\neq x_2$  the roots of $ x^2\plus{}x\plus{}1$ , so $ f(x_1)\equal{}f(x_2)\equal{}0$. By using  $ x^3 \equal{} x^3\minus{}1\plus{}1 \equal{} (x\minus{}1)(x^2\plus{}x\plus{}1) \plus{}1$ we obtain :
$ 0 \equal{} g(1)\plus{}x_1h(1)$ , $ 0 \equal{} g(1)\plus{}x_2h(1)$ so $ g(1)\equal{}h(1)\equal{}0$ and here you are .. :cool:

Well I substituted \omega and \omega^2 for $ x$ and u will get the result.


inequalities
number theory
number theory unsolved

Let $ S$ be the set of all positive integers $ k$ such that the number of positive divisors of $ k$ is greater than the number of positive divisors of $ d$ for any positive integer $ d < k$.

(a) Find the minimal $ n$ such that $ n!$ is not in $ S$

(b) Prove that $ n!$ is not is $ S$, for $ n>100$

(c) For any positive integer $ n$, let $ S_n\subseteq S$ be the set such that $ n$ doesn't divide any of the elements in $ S_n$. Prove that $ S_n$ is finite

Lemma
if $ a \equal{} \prod_{1\leq k \leq n}p_{k}^{a_{k}}$ then numbers of divisors of a is equal to $ \prod_{1\leq k \leq n} (a_{k} \plus{} 1)$
proof:   via induction. First step is obvious, so suppose that it goes for every $ k\leq n$ and $ n \equal{} \prod_{1\leq k \leq t}p_{k}^{a_{k}}$  for  $ a_{i} > 0$ Then we know that number of divisors of $ n' \equal{} \prod_{1\leq k \leq t \minus{} 1}p_{k}^{a_{k}}$ is equal to $ \prod_{1\leq k \leq t \minus{} 1} (a_{k} \plus{} 1)$ and for every divisor $ d$ of $ n'$ we can find $ a_t \plus{} 1$ divisors of $ n$ and they will be $ d,dp_t,...,dp_{t}^{a_t}$.

a)
 $ 8! \equal{} 2^7*3^2 *5 *7$ So number of divisors is equal to $ (7 \plus{} 1)(2 \plus{} 1)(1 \plus{} 1)(1 \plus{} 1) \equal{} 96$
$ 7!*6 < 8!$ and $ 7!*6 \equal{} 2^5*3^3*5*7$ so numbers of divisors is equal to $ (5 \plus{} 1)(3 \plus{} 1)(1 \plus{} 1)(1 \plus{} 1) \equal{} 96$
You can check that for $ n < 8$   $ n! \in S_n$.  This is not that kind of number theory which I like..

b) Much stronger statement goes : for $ n > 7$  $ n!$ is not in $ S$
$ \frac {9!}{6}*5$ have more divisors than $ 9!$
$ \frac {10!}{8}*7$ have more divisors than $ 10!$
$ \frac {11!}{8}*7$ have more divisors than $ 11!$
$ \frac {12!}{8}*7$ have more divisors than $ 12!$
$ \frac {13!}{8}*7$ have more divisors than $ 13!$
$ \frac {14!}{16}*11$ have more divisors than $ 14!$
$ \frac {15!}{16}*11$ have more divisors than $ 15!$
$ \frac {16!}{16}*11$ have more divisors than $ 16!$

$ a_{1} \equal{} \sum_{1\leq k} [\frac {n}{2^k}] \geq \frac {3n}{4}$  for every $ n \geq 16$

Now let $ a_{t}$ be smallest $ a$ equal to $ 1$.
Let $ u$ be the smallest value for which $ 2^u\geq p_{t}$

$ a_{t} < n$ so  $ u < log(2n)$

$ n! > \frac {n!}{2^u}*p_{t}$ and
number of divisors of $ n!$ is equal to $ N \equal{} (a_{1} \plus{} 1)(a_2 \plus{} 1)...(a_t \plus{} 1)..(a_k \plus{} 1)$
number of divisors of $ \frac {n!}{2^u}*p_{t}$ is equal to $ n \equal{} (a_1 \plus{} 1 \minus{} u)(a_2 \plus{} 1)..(a_t \plus{} 2)..(a_k \plus{} 1)$  so  $ \frac {n}{N} \equal{} \frac {3(a_1 \plus{} 1 \minus{} u)}{2(a_1 \plus{} 1)}$
we need to proof that
$ 3(a_1 \plus{} 1 \minus{} u) > 2(a_1 \plus{} 1)$
$ 3a_1 \plus{} 3 \minus{} 3u > 2a_1 \plus{} 2$
$ a_1 > 3u \minus{} 1$
$ \frac {3n}{4} > 3*log(2n) \minus{} 1$
$ \frac {n}{4} > log(2n) \minus{} 1$
for $ n \equal{} 16$ we have equality, so for every $ n > 16$ this inequality goes.

c)
lemma
if $ n \equal{} \prod_{1\leq k \leq n}p_{k}^{a_{k}}$ and for some $ i < j$ there is $ a_i < a_j$ then $ n$ is not in $ S$.  Proof:
$ p_{1}^{a_1}*...*p_{i}^{a_{j}}*...*p_{j}^{a_{i}}*...$ is smaller then $ p_{1}^{a_1}*...*p_{i}^{a_{i}}*...*p_{j}^{a_{j}}*...$ and have the same number of divisors.

Let $ p_n$ be smallest prime divisor of $ n$
Suppose that $ S_n$ is infinite.  Then we have 2 cases:

1.
there exist $ k\in S_n$ which is divisible by $ p_k > p_n$. Then $ \frac {k}{p_k}*p_n$ at least so much divisors as $ k$ and is smaller, so we have contradiction.

2.
all $ k\in S_k$ are divisible onle per finite number of primes less than $ p_n$. But there are infinitely many elements of $ S_n$ so there must be element which is divisible by $ 2^{10000*[log(2p_k)]}$
so $ k' \equal{} \frac {k}{2^{[log(p_k)] \plus{} 1}}*p_k$ have more divisors then $ k$.


Esto debe ser conocido, pero quisiera ver varias soluciones.
Considere una cuadricula de n por n casillas. Sean A el vertice inferior izquierdo y B el vertice superior derecho. De cuantas maneras se puede llegar, mediante la cuadricula, de A hacia B, si solo esta permitido moverse hacia arriba o hacia la derecha, y en ningun momento se debe estar mas arriba de la diagonal AB.

bueno en este caso hare para un tablero de $m$ por $n$ (creoque es mas general):
es cierto que el recorrido seria recorrer $n+m$ bordes de cuadricula del tablero (por la forma del recorrido que sigue) entonces cada borde de cuadricula lo representaremos como un segmento horizontal $l , l , l ....,l$ ($m+n$ segmentos) los cuales para el recorrido se juntaran extremo con extremo para formar el recorrido partiendo desde A, ahora sin perdida de generalidad en el tablero de $m$ por $n$ sean el lado de longitud $m$ vertical y $n$ horizontal , entonces al juntar los segmentos  nesecitaremos $n$ horizontales (para el recorrido) , por ello la manera de hacer esto sera $\binom{m+n}{n}$ ahora para el caso $m=n$ tenemos que esto es $\binom{2n}{n}$

Creo que no leiste la ultima linea:

"....en ningun momento se debe estar mas arriba de la diagonal AB."

[quote]entonces como hay recorridos simetricos ( gracias al tablero ) solo tendras que dividir entre dos el resultado  :D [/quote]

[color=blue]No entiendo a q te refieres con recorridos simetricos, pero definitivamente la solucion no es dividir entre dos, basta con que mires el caso $n=2$ el total de caminos q cumplen es $2$ y no $\frac{\binom{4}{2}}{2}=3$[/color]

La respuesta es los catalanes.
Es un problema muy conocido... si nadie  los soluciona lo posteo mas tarde...
La respuesta es $C_{n}=\frac{\binom{2n}{n}}{n+1}$

Ya que nadie puso nada y me da pereza escribir voy a poner un link.
[url]http://en.wikipedia.org/wiki/Catalan_number[/url]
Suerte.


inequalities
number theory proposed
number theory

Solve in $ \mathbb{Z}$ this inequation :

$ a^2\plus{}b^2\plus{}2bc\plus{}2c^2\plus{}a\plus{}b\le 2ac$

Rearrange the inequality like this:

$ (a\minus{}c)^2\plus{}(b\plus{}c)^2\plus{}(a\plus{}b) \le 0$

Let $ p\equal{}a\minus{}c$, $ q\equal{}b\plus{}c$ it becomes $ p^2\plus{}q^2\plus{}(p\plus{}q) \le 0$

As $ p,q$ are integers, $ p^2 \ge \minus{}p$ and $ q^2 \ge \minus{}q$ and equality of both holds iff $ (p,q)\equal{}(0,0),(0,\minus{}1),(\minus{}1,0),(\minus{}1,\minus{}1)$

So $ (a,b,c)\equal{}(k,\minus{}k,k),(k,\minus{}k\minus{}1,k),(k\minus{}1,\minus{}k,k),(k\minus{}1,\minus{}k\minus{}1,k)$ for any integers $ k$ are the solutions.


We colour every square of the $ 2009$ x $ 2009$ board with one of $ n$ colours (we do not have to use every colour). A colour is called connected if either there is only one square of that colour or any two squares of the colour can be reached from one another by a sequence of moves of a chess queen without intermediate stops at squares having another colour (a chess quen moves horizontally, vertically or diagonally). Find the maximum $ n$, such that for every colouring of the board at least on colour present at the board is connected.

[quote="FelixD"]We colour every square of the $ 2009$ x $ 2009$ board with one of $ n$ colours (we do not have to use every colour). A colour is called connected if either there is only one square of that colour or any two squares of the colour can be reached from one another by a sequence of moves of a chess queen without[b] [color=red]intermediate stops[/color] [/b]at squares having another colour (a chess quen moves horizontally, vertically or diagonally). Find the maximum $ n$, such that for every colouring of the board at least on colour present at the board is connected.[/quote]

I am quite not sure about the part 'intermediate stops' of a sequence of queen's moves.

Could anyone (like those who has solved it) explain a bit more about this, please?  :)

[quote="epsilonist"]
I am quite not sure about the part 'intermediate stops' of a sequence of queen's moves.

Could anyone (like those who has solved it) explain a bit more about this, please?  :)[/quote]
Well, a color is called connected if a chess queen can move to any square of that color without stopping on squares of different color.
So, for example, A is connected on the following board whereas B isn't (because the B in the first line is "separated" from the others):

A A B A
B A A A
A B A B
B B A B

A counterexample for 4 colours is as following

ABCDABCDABCD...
CDABCDABCDAB...
ABCDABCDABCD...
CDABCDABCDAB...
ABCDABCDABCD...
..................

It seems 3 colours should work, but don't know how to prove.

Counterexample for 3 colors: Color square (1,1), the top left corner, with A. Color (2008, 2008) with B. Color (1, 2008) and (2008, 1) with C. Of the remaining squares, color all of them on the top row and left column B, all squares on the main diagonal C, and color the rest A. Then all three colors are disconnected, as the first four squares I noted form connected components for each color.

To show 2 works: take a connected component of As. If this component spans all columns, it can reach every square on the board, meaning A is connected. Otherwise, we can find two adjacent columns, one of which has A's from this component (call this column $x$) and one of which has no A's (call this column $y$). In each row, take the square from column $x$ if it has a B instead of an A and otherwise take the B on column $y$. This gives us a B in every row, and these Bs form a connected component, so they can reach every square. Thus the Bs are connected.

Counterexample for 2 colors:
aaaaaaa......
bbbbbbb......
aaaaaaa......
bbbbbbb......
.
.
.
.
.

is it right?

I have found a solution in https://skmo.sk/dokumenty.php.
But this is written in Slovak, can someone help to translate it into English?
The first page:


algebra
polynomial
superior algebra
superior algebra unsolved

Let $k$ be an algebraically closed field( let us even make it $\mathbb{C}$  :maybe: )

I want to find the singular points of $f=x^{2}y+x y^{2}-x^{4}-y^{4}=0$

so that means I want to find $(x,y)$ satisfying that equation, and also:

$g=2 x y+y^{2}-4 x^{3}=0$
$h=2 x y+x^{2}-4 y^{3}=0$

Normally, just $(0,0)$ should be the answer, but for some reason I can't figure out how to manipulate the system to show uniqueness. :(

PS: do systems of equations belong here?
Edit: I added names $f,g,h$ to facilitate possible answers.

[quote="owk"]My CAS tells me that $(x^{3},x^{2}-y^{2},x^{2}+2xy)$ is a Groebner basis for the ideal $(f,g,h)$. owk[/quote]
What is a CAS?
Groebner basis is not familiar to me, nor should it be (as far as I know :maybe: ) and yet I am expected to be able to solve this problem. :maybe:

PS: owk, you really are good at giving really short answers. :wink:

it seems like $0=3f-(xg+yh) = x^{4}+y^{4}$, which gives 4 cases to discuss, right?

CAS = computer algebra system

A Groebner basis is a (in some specific sense) particularly simple set of generators of an ideal in a polynomial ring.

So, the essence of my answer was: use a computer, this is their purpose. You should not have been assigned to do this by hand in the first place.

If you insist, use ma_go's hint $x^{4}+y^{4}=0$ to conclude that $xy(x+y)=0$ as well (from $f=0$), and these two equations imply $x=y=0$ unless $2=0$ (for characteristic 2, use $g=h=0$). owk


A round pizza is $ \frac13$ of an inch thick and has a diameter of 12 inches. It is cut into 12 congruent pieces. What is the number of cubic inches in the volume of one piece? Express your answer in terms of $ \pi$.

The whole cylinder has volume

pi(r^2)*h
pi(6^2)*(1/3)
12pi

Each slice is 1/12 of the whole, for a volume of pi cubic inches


probability
calculus
integration
function
probability and stats

let P and Q an equivalent porbabilties ( for every A in F, P(A)=0 <=> Q(A)=0 ) , we can find a z postif random variable so that dQ=zdP,  my question is , is it true that in the integral of fdQ where f mesurable , we can replace dQ by zdP ?  i tried to prove it with the definition of the integral but i'm not sure of my proof ! the original problem that i try to solve is : for all X integrable r.v  Eq(X|B)=Ep(XZ|B)/ Ep(Z|B) , so i tried to do it for X an charactiristic function to get it by limite but i need to be sure that we can replace dQ  by zdP in integral ...can any one clarify  ?

dQ=ZdP is just a notation, in fact , Z is so that for all A , Q(A)=Ep(Z over A)


AMC USA(J)MO USAMO geometry inequalities function induction

Time to bring this forum back to life.

The Canadian Mathematics Olympiad 2008 will be written on Wednesday, March 26.

Good luck to everyone writing it! :)

Good luck everyone!!!

By the way, when can we start to discuss problems???

Good luck to everyone writing the contest~

:blush: it;s tomorrow ..!!
=____________=@!!!
no luckk!!

ahhhhhh!!!

Please do not discuss the problems of the contest on the forum until Thursday evening, so like 7pm Eastern just in case someone ends up writing it on Thursday.

good luck everyone!

Alex, are you sure Thursday is enough time? Sometimes people wait more than that I think...

Hmm ok... make it Friday then.
It's just last year we waited till Friday but the yr before people started discussing on the same day!
[b]OK then no discussion till Friday evening![/b]

http://www.math.toronto.edu/barbeau/paper2008.pdf

I guess you can discuss after all.

What score on the CMO is usually required to get invited to (unofficially) write the USAMO?

Although the problems are posted, notice this sentence:
[quote]provided you have not consulted any solutions to these problems from another source.[/quote]
This is designed for those who had a bad day on COMC and thus could send in their solutions for this year's CMO. This is not a declaration that it is okay to talk about it yet, since these people need to do the problems themselves, and if we post the answers, we're robbing them of their chance at the CMO.  :)

I do have a formula, and that's what they ask for, but the proof's kinda iffy, so I'll leave it to other people to decide when it should be good to post answers and post their better solutions. :lol:

Foundation

So even if we wrote CMO, do we still send in solutions???

i have a dumb question...
for the teachers who r marking CMO.. do they check every single page of the solution booklet..
or they just get a look at every pink sheet... and if  there's nothing on there then they wont bother to flip it over ..... =____=

we were suposed to write on the pink sheets? uh.... o dang

Uhoh for me too. I only wrote on the white sheets.

[quote="Foundation"]Anyone have very rigourous solutions for 2 and 4?[/quote]

Two it ends up being the cauchy functional equation, and then you can easily get the two solutions I believe, but I think that there are more than two solutions.

For #4, I thought about first using primes, and then using the fact that they could be extended to composites, and I got two functions that work, but I don't think that those are the only two solutions, and there could be more.

For #5, I was wondering, if anyone could prove that their recursion was correct.

just wondering,
for #2 i put "It is trivial that f(n)=n"
for #4 i put "By fermat, it is clear that f(n)=n"

how many points would i get?

[quote="Phelpedo"]#2 - that isn't even correct so definitely 0.

#4 - probably 0, [i]possibly[/i] 1.[/quote]

ok...i got owned :stretcher:

Well i think for question 2 you first show that f(0) = 0, and then show if f(a) = b then f(b) = a. After that if f(a) = b then f(na) = nb. suppose now if f(a) = b, then there exist integers m,n, since a and b are rational, such that ma = nb. So then mb = na since f(ma) = f(nb), thus a^2 = b^2: a is +/- b. Then we just show that if f(a) = -a, then f(b) != b (i.e. it's either all f(a) = -a or all f(a) = a).

For question 4 you set n = p, then f(p)^p = p (mod f(p)), but f(p) = 0 (mod f(p)), so p = kf(p), so either f(p) = 1 or  f(p) = p. If f(p) = 1 for one p it can be shown that f(p) = 1 for all p by shown a contradication if f(q) = q for some prime q. But if f(p) = 1 for all prime then f(n) for non-primes can be anything. Otherwise if f(p) = p, then we can show f(n) = n but supposing f(n) = k != n. Then k is finite, thus there must be a prime larger than k. Now use this prime, then (k)^p = n (mod p), but k,n <p, and k^p = k (mod p) and n != k (mod p) contradiction.

For question 5 I kind of used induction by columns, kind of hard to sketch. But it turns out R(n) = f(n) - 1. Where f(n) = 2f(n-1) + f(n-2), f(1) = 2, f(2) = 5. And then you can take the roots of x^2 - 2x - 1, call them a, b and show that some ma + nb = 2 and ma^2 + nb^2 = 5. Since x^2 - 2x - 1 = 0 for a and b, the x^n+1 - 2x^n - x^n-1 = 0 for a and b as well, since this is linear, then ma^n + nb^n is your solution. I ended up with (1/2 + sqrt(2)/4)(1+sqrt(2))^n + (1/2-sqrt(2)/4)(1-sqrt(2)) - 1. If you want to expand this to integer form, it  becomes nC1 + 2(nC2) + 2(nC3) + 4(nC4) + 4(nC5)... or something like that. I'm pretty sure it's right but am open to suggestions. The inductive step was showing a bijection between a possible movement in a column (the last column) to a possible movement across the (last column and the added column), and you get 2 possible movements for all possible movements on the last column plus a little extra, which is all possible movements on the column right before it, hence f(n) = 2f(n-1) + f(n-2).

[quote="mathgeniuse^ln(x)"][quote="Phelpedo"]#2 - that isn't even correct so definitely 0.

#4 - probably 0, [i]possibly[/i] 1.[/quote]

Well, the thing is that for #2 you forgot f(n)=-n as well, so yeah maybe 0

But for #4 if you actually did some work to show that f(n)=n is a solution, and showed that others don't work, then you may get 1

Buf for #2 definitely 0.

Oh well.

For #5 though, all it said was find an explicit formula, so I don't know...[/quote]

what if ... my answer for #2 is correct but the solution doesnt make sense..
do i get a 0 too?~
and also, when the results will be posted?

I knew #5 had to do with recursion.

I found the wrong recursion unfortunately.

I guess when they asked for an explicit formula, those complex radicals were what they meant by explicit.

Sigh

The official solutions are now posted at:

http://www.math.toronto.edu/barbeau/sols2008.pdf  :D

Does anyone know when the results are out? Are they released to the math teachers?

Thanks.
 :maybe:  :maybe:

I heard that they sent emails to the top 8, as in who were in the top 8.

But I don't know for sure.

don't revive old posts, even in our uber in-active forum.

wow, we typed it at exactly the same time

it wasn't taht old

and we still have no idea on the CMO, sot htat is why we are wondering.


MATHCOUNTS
geometry
AMC
AIME
search
AMC 10
area of a triangle

[quote="egghead91"]Yeah, I'm a total noob in MATHCOUNTS, so I feel like a major achiever when I figure out how to do a problem, let alone get it right. :blush: This is pathetic, because I admit, because my district is deficient in true math ppl. I know 83 digits of pi, but nerdiness is relative... who can relate? PLEASE! SOMEONE HAS TO BE IN THE SAME BOAT (cuz i know little about boats, either :oops:) !!!!![/quote]
A mathematician once said "If you don't feel stupid you aren't doing math". This statement is plain and true.

yeah i feel stupid when talking about geometry and trying to understand with other people who are stronger than me

[quote="egghead91"]Yeah, I'm a total noob in MATHCOUNTS, so I feel like a major achiever when I figure out how to do a problem, let alone get it right. :blush: This is pathetic, because I admit, because my district is deficient in true math ppl. I know 83 digits of pi, but nerdiness is relative... who can relate? PLEASE! SOMEONE HAS TO BE IN THE SAME BOAT (cuz i know little about boats, either :oops:) !!!!![/quote]

It's ok. After awhile you'll get smarter. :D

Experience will help you. That means the more problems you solve and the more you learn from mistakes, the better you will do. Just try your best and never give up!

The truth:
*Advising other people is easier than actually doing it your self*

[quote="13375P34K43V312"]Don't worry, feeling stupid is a good thing. I first joined AoPS in April, I found the but MC forum to my liking, but found HSB (then called Getting Started) way too hard. I felt stupid, and in a way, felt jealous because so many people were so much smarter than me. I worked hard, because I don't like it when people are smarter than me  :D , and now find HSB very easy. Now to intermediate...[/quote]

Haha I joined at the end of September while searching for results of the Fall Startup Event. I thought I was really really stupid, and I kept posting many geometry problems in HSB and still do. However, I am finding that I am improving at a rapid pace and understand many concepts I learned on AIM chatting with AoPsers and on AoPS. in geometry, that is. Now , after next week's geometry tournament, I will start preparing for AMC 10 and AIME and devote 4 hours each day since I have like 8 hours free time at home in Middle school.  :P

I still find myself really dumb(actually more like the word inexperienced and younger)  compared to lots of people on here. But they are helpful, and I know as long as I work hard that I am improving.

[quote="indianamath"]I used to be like that too.  You just need to learn a little more about math and you will fit in.[/quote]
I disagree. Anyone can fit in, regardless of present ability. If nobody asked questions, a bunch of people on this forum would have nothing to answer!

yeah. it's great to have people from many different levels of math   constantly learning and answering questions. Unless you are Einstein, of course. or just abnormally good

A lot of people on this forum, when people post a problem that seems easy for them or for the forum it's in, they give a short solution that in the way they give it, implies "duh, how could you be so stupid to miss that?" I think that's what he means by fitting in.

when they give such a short solution, I just ask them on AIM or private message. Or, if they don't have aim , or if i need the solution fast, I can just communicate with friends and AoPSERS through messaging to get a more clearer view of the solution and how you get to it. step by step  :P

thanks everybody.... but I did ask if anyone else felt like a noob.
I'll keep this in mind.

[quote="13375P34K43V312"]Don't worry, feeling stupid is a good thing. I first joined AoPS in April, I found the but MC forum to my liking, but found HSB (then called Getting Started) way too hard. I felt stupid, and in a way, felt jealous because so many people were so much smarter than me. I worked hard, because I don't like it when people are smarter than me  :D , and now find HSB very easy. Now to intermediate...[/quote]

Of course, that was when HSB was flooded by olympiad problems, prompting rruscyk and mods to step in and reduce the difficulty to what it is now.

I was saying your EXACT words to myself quite recently, I even had pi memorized to over 90 digits, lol. I still am pretty noobish but have come a really long way in a short amount of time. My advice is to try as many problems as you can. Even if you don't get anywhere with it, check out the solution and if you see a formula or theorem used to solve the problem that you didn't know, search for other problems that use it or just search on AOPS or google for lessons that go along with it. Then put it on a notecard with a question on one side (eg. Heron's Formula?) and then on the other side write out the formula $A={(s(s-a)(s-b)(s-c))}^{1/2}$ where, A is area of triangle, s is semipermeter and a,b,c are sides of the triangle. And then you will shortly have a nice group of notecards to go through whenever. However, do not think that knowing every trick to every problem is what being math is all about. You will quickly learn that these problems (especially as they increase in difficulty) become more like puzzles rather than memorized methods for each type of problem. Whenever I solve most of the problems that I work on, I have no idea exactly how I'm going to solve it, I just write down what I know, play with it, and end up stumbling upon the solution. Once you start to do that, you are beginning to understand the art of problem solving.


[b]Let [/b]  $ A,B \in M_{2008} \mathbb{(R)}, AB=BA$[b] and polynonical P(x) has degree is 2009 [/b]($ P(x ) \in R(x)$ )
[b]such that [/b]:  [b]x.P(x)=(x-2009).P(x+1)[/b] ,$ \forall x \in R$ ,
[b]Put [/b]:$ S={x \in R : P(x)=0}$
[b]Prove that :[/b]
 [b] If [/b] $ det(A+x.B)=2008 ,\forall x \in S$
 [b]Then[/b] $ B^{2008}=0$
"...Học Toán để tìm vẻ đẹp trong toán ..."

We see easily that $ P(1)\equal{}\cdots\equal{}P(2009)\equal{}0$; then $ S\subset\{1,\cdots,2009\}$. Let $ Q(x)\equal{}det(A\plus{}xB)\minus{}2008$; $ Q$ is a polynomial of degree $ 2008$ which has $ 2009$ roots; then $ Q\equal{}0$. $ AB\equal{}BA$ implies that we may assume that $ A,B$ are upper triangular complex matrices. $ det(A\plus{}xB)\equal{}(a_{11}\plus{}xb_{11}).\cdots.(a_{nn}\plus{}xb_{nn})\equal{}2008$. This polynomial has no root then the $ b_{ii}$ are zero and we are done.

One typo loup blanc, it is
$ \{1,\cdots,2009\} \subset S$.
and not [quote="loup blanc"]$ S\subset\{1,\cdots,2009\}$[/quote]but the rest is correct.

Thanks JC_math.
Now I have a look on your problem about $ H(A)$.

You can prove that :det(A)=2008 from Prove that $ (A^{\minus{}1}B)^{2008}\equal{}0$(clay-haminton) and use AB=BA


Recently, Frank took a one-hundred question aptitude test where each correct answer scored $ 5$ points, each incorrect answer scored $ \minus{}2$ points, and each question not answered scored zero points. Frank answered $ 80$ questions and scored $ 232$ points. How many questions did he answer correctly?

Solve a system of linear equations:

$ x\plus{}y\equal{}80$
$ 5x\minus{}2y\equal{}232$

Answer happens to be:

$ x\equal{}56$


A rocket satellite travels in a circular orbit of radius $ r_0$. The rocket motor suddenly increases the rocket velocity by 8 % in its direction of motion. What is the apogee distance of the new orbit?

I think you can use COAM, so set Ro * v= (R')(v*1.08).
I'm sure sb else will have a better explanation though.

conservation of angular momenta you get
Vo Ro = V R
i.e velocity in the orbit initially be Vo and dist from centre init be Ro and velocity and radius be V and R at the apogee.
now apply energy conservation
-GMm/Ro +{m[Vo+8%Vo]^2}/2=-GMm/R +{mv^2}/2
now solve both eqns to get reqd dist choose from the smaller and larger the larger value

Oops, the 2 $ v$'s in my variables should be different.
Let me latex:

$ (1) R_o v_o \equal{} R_a v_a \Longleftrightarrow v_a \equal{} \frac {R_o v_o}{R_a}$
Where $ R_o$ and $ v_o$ are the inital conditions and $ R_a$ and $ v_a$ are the apogee radius and velocity.  Equation (1) results from conservation of angular momentum, because with central force motion, not torque can be exerted relative to the center.

From conservation of energy
$ (2) \frac 12 m{v_o}^2 \minus{} \frac {GMm}{R_o} \equal{} \frac 12 m {v_a}^2 \minus{} \frac {GMm}{R_a}$

Substitute in $ v_a$ in (1), so that you have 1 equation with only 1 unknown, $ R_a$. Obtain $ R_a$ as a function of $ R_o$ and $ v_o$ and solve for the maximum $ R_a$

BTW, you will get a very nasty quadradic.  As unearthly said, choose the larger values.  Once you have $ R_a$ as a function of $ R_o$ and $ v_o$, set
$ v_o \equal{} 1.08 \sqrt {{GM}/{R_o}}$


I found an image of all of them:

[img]http://www.math.rutgers.edu/~erowland/images/platonicsolids.gif[/img]

But how do I get it smaller and onto the wiki?

May I ask where exactly you found that image?

They look like they were from Wolfram MathWorld.

Look at his image tag.

[code]

[img]http://www.math.rutgers.edu/~erowland/images/platonicsolids.gif[/img]

[/code]

So he got it from the rutgers site.

So apparently it was created by Eric Rowland: http://www.math.rutgers.edu/~erowland/. I'm afraid that we can't use his work without his permission; are you up to emailing him, 1=2?

whoops. I'm too timid(:P) to email him, someone else do it and I will take the image down.

No need - azjps made some nice ones. I'd prefer that all images be our own anyways til the the admins decide more specific copyrights for the content - which may be never.

Of course, the bottom two are "unparseable or potentially dangerous"

Yeah, the last one hasn't been written yet and the second-to-last needs fixing. I don't think that really needed pointing out; either Azjps or one of the resident Asy wizards can fix it.


i think ill get the volume2. it's good right?
and is the solution manual worth it, or is it nothing special and i can do most of them myself (assuming that i know what's going on and partially gifted)
thanks

Umm, not sure this the right place for you to post this question but I would say it is definitely worth it.  And get the solution manual, I mean you are not going to figure them all out (yeah, even if you are gifted) and alot of times they will have a different(aka much nicer) solution in the solution manual than the one you came up with, so it can help you out.

I dont think you can do most of them by yourself, especially if you just did Book 1. Book 2 takes u to a different level and there will always be problems that you cannot do. So consider buying  it together. Cheapter together, too :D (If you regret not buying it, then you would have to buy it separately at a higher price)


If $ a_1,a_2,...,a_n$ are positive numbers , prove the inequality:

$ \frac{a_1\plus{}a_2\plus{}...\plus{}a_n}{n}\le e^{\frac{a_1\plus{}a_2\plus{}...a_n}{n}\minus{}1}$

Or $ e^x \ge 1\plus{}x$ because $ e^x$ is convex, and $ 1\plus{}x$ is the tangent line to $ e^x$ at the point $ (0,1)$.

[quote="Mij"]Substitute in the Maclaurin series and you're done.[/quote]

What is Maclaurin series ? Can u explain me ,cuz i didn't understand !!

Go to http://www.google.lt write "Maclaurin series" and you'll find information you need :cleaning:

yeah...
isn't it \[ e^{x\minus{}1}\ge x?\]
not \[ e^x\minus{}1\ge x?\]

[quote="RoFlLoLcOpT"]yeah...
isn't it
\[ e^{x \minus{} 1}\ge x?
\]
not
\[ e^x \minus{} 1\ge x?
\]
[/quote]

You are right, but $ e^x\geq x\plus{}1$ for any real numbers $ x$, so you can change $ x$ into $ x\minus{}1$, or

$ e^{x\minus{}1}\geq (x\minus{}1)\plus{}1$.


How many times does the digit 6 appear in the list of all integers from 1 to 100?

The number $ 6$ appears once in every interval of $ 10$ from $ 0$ to $ 100$, with the exception of the $ 60$'s where it appears $ 10$ times more. Thus our answer is $ 10\plus{}10\equal{}\boxed{20}$

actually, you counted 66 twice
therefore, the answer would be $ 10 \plus{} 10 \minus{} 1 \equal{} 19$
our answer is 19

edit: oh, srry
i should read the question...
the answer is 20

[quote="vallon22"]actually, you counted 66 twice
therefore, the answer would be $ 10 \plus{} 10 \minus{} 1 \equal{} 19$
our answer is 19[/quote]

It asked how many times the digit 6 appears, not the number of numbers with 6 in it, so the answer is 20.


Find all ${x}$ such that ${x^{2}}+2$ is prime.

I know that ${x}$  must be a multiple of ${3}$. This is easy to see applying Fermat.
The problem is that there are some multiples of ${3}$ for which ${x^{2}}+2$ is not prime.
For example, if ${x=27}$ then ${27^{2}}+2=731=(17)(43)$

Does anyone know how to find conditions for ${x}$?

[quote="ZetaX"]This problem is more or less open ;)

You could look $\mod$ other numbers to get more conditions.[/quote]

This is problem 7.2.8 in "The Art and Craft of Problem Solving" by Paul Zeitz. So  I thought the problem to be doable using standard techniques...

i don't know if this is obvious: $x$ is odd.

edit: yea, it is obvious, you must have seen that.

[quote="Ilthigore"]Are you sure it isn't "find all [i]prime[/i] $x$ such that...".

This would closely resemble a (very easy) problem I've seen before, and would actually be soluble within a finite time.[/quote]

As you said if x were prime this problem would be very easy. Since x must be a multiple of three then x=3.

Could someone give me a link to one of the forums where this problem or a similar problem is discussed or solved.

Thxs...

There is http://www.mathlinks.ro/Forum/viewtopic.php?t=90688 , and I have the strong impression that someone does this on purpose :D


geometry calculus integration trigonometry

hey i have attached an image to the post.
I have to find the area of the shaded region
kindly guide me about how to find it ?

This is calculus. Calculus problems should not go in this forum.


Let $a,b,c > 0$ so that $abc=1/3$. Prove that $3(a+b)(b+c)(c+a)+1\geq 3(a+b+c) \sqrt{a+b+c}$.

The given inequality is equivalent to:
$(a+b)(b+c)(a+c)+abc\geq (a+b+c)\sqrt{a+b+c}$ $\leftrightarrow$
$(a+b+c)(ab+bc+ac)\geq(a+b+c)\sqrt{a+b+c}$ $\leftrightarrow$
$ab+bc+ac\geq \sqrt{a+b+c}$ $\leftrightarrow$ $(ab+bc+ca)^{2}\geq 3abc(a+b+c)$ which is easy by AM-GM

I know these problems are easy with substitutions, but can anyone find a "super beginner" solution which doesn't use anything but powers less than 2? Like a "trick" solution?


How do I get the little numbers over and under the $\int$  sign, $\sum$ sign, etc?

Limits are seen as subscripts and superscripts so you use _ and ^. It doesn't matter which of _ and ^ come first
\int_0^{\infty} gives $\int_{0}^{\infty}$ and \sum_{r=1}^n gives $\sum_{r=1}^{n}$

The only problem is that you sometimes get $\textstyle\sum_{r=1}^{n}$ (only in your document, not here). There are 2 solutions: either use \displaystyle (which the forum uses) or use \limits as in  \sum\limits_{r=1}^n
[quote="qwang"]What is the code for the radicand of a  square root?[/quote]
Not sure what you mean. The radicand of $\sqrt{3}$ is 3


geometry
geometric transformation
rotation

A Senate committee has 5 different Democrats and 5 different Republicans.  In how many ways can they sit around a circular table without restrictions?

Since it says without restrctions, does that mean that rotations are not the same?  If that is the case, I believe eli is correct.  But if rotations are the same, then we have $ (10\minus{}1)!\equal{}9!$, which means that the answer would be $ \boxed {362880}$ instead.

I missed this the first time, as the previous problem stated that all members of the same party had to sit together and I carried over that restriction.:( But it is really a very easy problem- 10! divided by 10.

Are each group distinguishable? If not then that’s a totally diff answer.

[quote=AlcumusRunner]Was there a reason for it to say 5 democrats and 5 republicans? I don't see how that affects the problem.[/quote]
It was part of one of the book problems, it led into another problem with restrictions.
[hide=sol]Fix one person at the top to account for rotations, then there are simply $\boxed{9!}$ ways to sit the remaining people.[/hide]

I think a better approach (than the solution given by them) would be to place a person on the table and use him/her as a reference point. then it's just regular permutation of the remaining 9 which is 9!
the important bit here is that you can place a person in 1 way and NOT in 10 ways, for it doesn't really matter whoever you place first.
Think about it.
Circular permutations can have surprising connotations.

i trolled this problem on my first try because i didn't read that everyone was distinguishable and thought democrats were indistinguishable and so were republicans

typoed this first try, wrote 3632880 not 362880 oof

[hide=sol]you have 10 ways to seat the first person, then 9 ways, then 8, then 7, and so on. This gives you 10! = $3628800$ ways. (PoV: that one guy who made a typo, wrote 362880 and got it right lol)

Now, due to rotations, every 10 seatings are the same, so divide by 10 to get $9! = \boxed{362880}$. [/hide]


For what real values of $ k$, other than $ k \equal{} 0$, does the equation $ x^2 \plus{} kx \plus{} k^2 \equal{} 0$ have real roots?

$ \textbf{(A)}\ {k < 0}\qquad
\textbf{(B)}\ {k > 0} \qquad
\textbf{(C)}\ {k \ge 1} \qquad
\textbf{(D)}\ \text{all values of }{k}\qquad
\textbf{(E)}\ \text{no values of }{k}$

hello, solving the given equation for $ x$ we get
${ x_{1,2}=-\frac{k}{2}\pm\sqrt{\frac{k^2}{4}-k^2}}=-\frac{k}{2}\pm\sqrt{-\frac{3k^2}{4}}$
so we get no values of $ k$ as the right answer, $ \boxed{E}$.
Sonnhard.


I just need help with this problem
(1,3) (4,6) (5,9) it is a triangle
I need to find the area of it.
Please reply :)
Reventon

Check out the [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]shoelace theorem[/url], or read [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=232614]this[/url].

Thanks for you help. I finally understand :)
Reventon


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that there is no simple group such that $|G|=4n+2|, n\geq 1$.

I've posted a few days ago a much stronger result. For this one, one can also prove that it has a subgroup of index 2, it has been done several times on the forum.

Can you give the link, with your stronger result, please? :)


Segment $ AB$ has midpoint $ C$, and segment $ BC$ has midpoint $ D$. Semi-circles are constructed with diameters $ \bar AB$ and $ \bar BC$ to form the entire region shown. Segment $ CP$ splits the region into two sections of equal area. What is the degree measure of angle $ ACP$? Express your answer as a decimal to the nearest tenth.

[asy]draw((0,0)--10dir(180),linewidth(2));
draw((0,0)--10dir(67.5),linewidth(2));
draw((0,0)--10dir(0),dashed);

draw(10dir(180)..10dir(90)..10dir(0),linewidth(2));

draw((5,0)+5dir(180)..(5,0)+5dir(-90)..(5,0)+5dir(0),linewidth(2));

dot((0,0));
dot((5,0));

label("A",10dir(180),W);
label("B",10dir(0),E);
label("C",(0,0),SW);
label("D",5dir(0),NE);
label("P",10dir(67.5),NE);[/asy]

First, I substituted x for the radius of semicircle D. The area of semicircle D would be 4x^2 pi and the area of semicircle C would be 16x^2 pi. The area of the region would be 20x^2 pi. Segment CP splits the region into two sections of equal area. The area of section ACP would be 10x^2 pi. I then divided 16x^2 pi (the area of semicircle D) by 10x^2 pi (the section ACP). The section ACP would be 10/16 or 5/8 of semicircle D. Angle ACP would be 5/8 of 180 degrees (the degree measure of semicircle D. 5/8 of 180 is 112.5 degrees.


Four segments measure 16, 19, 43 and 50 centimeters. What is the probability that a triangle can be formed if three of these segments are chosen at random? Express your answer as a common fraction.

:arrow: first: you have a total of 4 numbers
 :arrow: second: you want to choose 3 of them is see which one would be possible to make a triangle

 :arrow: So now, we will start solving. Since you want to choose 3 out of four, we use the combination law. 4 choose 3, which would be (4*3*2)/(3*2*1). The answer is 4. So now we know that 4 is the denominator of this fraction.

 :arrow: The number of ways you can make a triangle with 16, 19, 43, 50 are:
 :arrow: 1. 16,19,43
 :arrow: 2. 16,19,50
 :arrow: 3. 19, 43, 50
 :arrow: 4. 16, 43, 50
 :!: You can eliminate 1 and 2 by using the triangle side theorem from geometry. 3 and 4 work just fine, so there are 2 ways to make a triangle.
 :idea: The answer is 2/4, which simplifies to 1/2.


Let $ S$ be the set of all positive integers $ n$ such that $ n\minus{}1,n,n\plus{}1$ can be expressed as the sum of 2 perfect squares.Prove that if $ n\in{S}$,then $ n^2 \in {S}$.


I recently took this test, and had a little trouble with this problem.

#17: A right triangle ABC with hypotenus AB has a side AC=15. Altitude CH divides AB intosegments AH and HB, with HB=16. The area of ABC is?

[hide]
Anyway they used the facct that a leg of a right triangle is the geometric mean of the hypotenuse and the projection of the leg on the hypotenuse. I suppose the real question is hat is the proof of this fact, m thinking that you can do it through similiar triangles and also through vectors. However Im not really satisified that the problem can be solved in only that way, i tried to sketch a solution but get stuck Help would be awesome.
[/hide]

Oh by the way does anyone know if i can get copy of geometers sketchpad from somewhere on the cheap...maybe can just bum one off my college.

Well, by similar triangles, you can say $\frac{15}{16+x}=\frac{x}{15}\Rightarrow x^2+16x-225=0=(x-9)(x+25)\Rightarrow x=9$, so it's a 15-20-25 triangle, with area $150$. Or, you can just see intuitively that it has to be a 15-20-25 without working it out, like I did.

[quote="jamiemorgan88"]
Oh by the way does anyone know if i can get copy of geometers sketchpad from somewhere on the cheap...maybe can just bum one off my college.[/quote]

that's not endorsed here, as geometers sketchpad is commercial software. However, there is a student discount, that brings the software down to $\$$40 for students; heres a link:

http://www.keypress.com/catalog/products/software/Prod_GSP_Student.html


calculus
integration
geometry
ratio
geometric transformation
articles
vector

also, what about Hitler? what do you think that would have happened if Einstein would have worked for Hitler? we would be saying Heil! instead of Hi! now ? :)

I do believe that Einstein was a genius, but I doubt that Einstein working for Hitler would've changed history in a decisive way. You're probably thinking about the 20'th century's nemesis: the atom bomb, but lots of scientists contributed to that and the attitude of repulsion towards the pain and suffering which can be unleashed by such a device is a common characteristic of all of those scientists (including Einstein!).

I think this is probably because a really smart man, one which may be considered a genius, does indeed have moral principles, the lack of empathy usually being the sign of lack of intelligence.

Hitler was a demagog from Austria. But I have to admit that he has a little talent. The military headquarters sometimes did wrong judgement but Hitler gave a correct judgement. Normandy battle, for instance, the Germany military headquarters thought that the allied forced would land in Calais only Hitler thought it was Normandy!

i'm no fan of Hitler, but I recently found out (maybe the german members can suport/contradict) that there are some expressways in Germany built in 1940-s which are still used now ...

In the question of whether Einstein was a genius, I would have to say that if the word has any meaning at all it must apply to him. He was an original thinker of the highest order, and possibly one of the greatest of his profession since at least Isaac Newton.

As for Hitler, that is more arguable. Hitler was a good military strategist (which I'm assuming is the context in which we are discussing his genius). However he was probably not the best military strategist of even his own generation. Some of his military strategies were impressive, but other times they were quite irrational.

Einstein however would probably never have worked for Hitler, under any circumstances. First of all, he had Jewish ancestry. Second of all Einstein was a very outspoken pacifist, and argued against the development of the atomic bomb even here in America. He played no part in it's development other than his original theoretical development of Special Relativity. Most people don't know about Einstein's life as a political activist because it is overshadowed by his scientific accomplishments. But his political views, and even philosophical views are quite poignant. I highly recommend people to research and read about his life beyond the realm of physics.

Well, a man who starts a "revolution" in physics and changes our view of the world by his work HAS to be a genius...

On the other hand, even if Hitler did have some attributes which could qualify him as a very intelligent man, his fanatism and beliefs about the German people prevented him from doing anything really constructive, and I think that one of the main qualities of a genius is the ability to employ his intelligence in order to create something truly original.

I don't believe Einstein was a genius.  What he was was a good thinker/problem solver.  I read his biography last summer and I do suggest it for anybody who wants a good read.  Einstein was nothing special in undergrad or graduate school . . . he attended a university similar to but not as good as Cambridge University . . . and he had marks similar to his classmates.  After not getting a professorship he finally was able to land a job at a patent office where he looked at material that was made by inventors/companies who wanted their innovations patented.  Just like any other physicist he diligently worked in his fields contemplating connections and solving problems in his own time.  What he did in his own time enabled him to develop the foundations for quantum physics with his intuition and his learned ability.  When he was able to legitimize his findings and when they were finally proven by other scientists, then he all the sudden got massive attention from the public.  To this day, people believe he was a "genius" for developing the foundations for QED, however, some one in the future could of reciprocated his action if Einstein was never born . . . we just don't know that.

As I stated in an earlier discussion about "Genius, Fate, and condition"  . . . I don't think that when people recognize relationships between variables that they are automatically genuises.  They are just normal people that have worked hard (not always) and probably enjoy what they do.  Do you consider the cavemen that built the foundations for arithmetic geniuses?  No?  I don't think so either.

What makes Einstein so special in my mind is that while everybody else stepped left and took one path, he soley stepped right and took the other path.  We all know that people in the 19th century and early 20th century believed everything in physics was already done (completed).  Physics was a science that was complete . . . everything had been thought of.  As we all know, Einstein experimented with his ideas and defied the majority of scientists who had believed in soley classical physics.  Therefore, Einstein should be regarded as a unique innovator who took the path not travelled before not necessarily as a genius (born-to-be (which is crap)).

If I discovered a new theory that was revolutionary while working my physics text and doing my math text . . . would I be a genius also? (rhetorical question)

[quote="integral0"]If I discovered a new theory that was revolutionary while working my physics text and doing my math text . . . would I be a genius also? (rhetorical question)[/quote] yes you would :D:D

Well, I think our ideas differ because of our incompatible thoughts on what a genius is. I, for one, think that originality is as important a quality of the genius as superior intelligence, and I believe Einstein had both. Grades you get during school don't always say something. I can't think of any other cases right now, but I'm sure not all famous scientists, who are generally considered geniuses, did well in school.

in my eyes, genius is a "pretty" title given to people for status quo only . . . its merely a word added along with someones name

I usually don't post to 'trivial' questions such as this but I just couldn't let this one slip by (forgive me integral0, I post along side you on other boards and usually don't disagree with you but...)
 Is Einstein a genius??????    Stephen Hawking isn't worthy to carry his pocket-protector! In my (and more importantly many modern scientists) mind, only Newton was his match -which is ironic since some of Einsteins theories proved Newton wrong. Many argue that Einstein saw farther because he was propped up by Newton who, as Newton himself said, was sitting on the shoulders of  the giants who came before him. He not only saw farther however, his vision extended into areas never touched by humankind before him.
 Einsteins theory of Relativity demonstrated that measurements of time and distance vary as things travel relative to anything else (Sorry Newton!- Space and time are not absolute!)

 He made contributions in the quantum theory of light,  Brownian motion, not to mention his proof that energy and matter are linked via the most famous relationship in physics: E=mc^2, and many other areas. Furthermore, he came upon these earth shattering ideas all while putting in 40 hrs a week at a patent office at 25 years old! Many of Einsteins theories (ideas on gravity relativity, even string theory etc), I must admit, are beyond my ability to understand. It doesnt take a lot of web skills to find papers outlining Einsteins genius however.
 He also is attacked for not recognizing, as other geniuses such as Heisenberg etc did, the importance of quantum physics but, to be honest, he is responsible for much of the groundwork in that area,  making the works of the future quantum physicists possible. The following quote is funny:
[quote]
                 Just like any other physicist he diligently worked in his fields contemplating connections and solving problems in his own time. What he did in his own time enabled him to develop the [b]foundations for quantum physics[/b] (sic) with his intuition and his.[/quote]

 Wow, I should be so lucky!

  I once had an instructor who counted Paul Erdos among his friends. He considered him to be one of the greatest minds of all time. Yet Erdos never considered himself a genius, just driven- and in love with his discipline.  Most people in the know would consider Einstein to be among the rarest types of people - driven AND with unequalled intellect.

 Here is a quote from the pbx website on him

[quote]Between 1905 to 1925, Einstein transformed humankind's understanding of nature on every scale, from the smallest to that of the cosmos as a whole. Now, nearly a century after he began to make his mark, we are still exploring Einstein's universe. The problems he could not solve remain the ones that define the cutting edge, the most tantalizing and compelling.

You can't touch that. Who's smarter? No one since Newton comes close.[/quote]

 'Nuff said.

I definitely COULD NOT have put it better!:D

Einstein is, believe it or not, one of those "giants" who will be mentioned whenever you speak about physics, and not just in our time, but for centuries to come.

I've heard about Newton being the only one to have ever matched Einstein, but I don't think it's fair to say stuff like :"Einstein saw farhter than Newton", since the minds of the 2 can never be compared. Newton discovered as much as he could, given the time period he was living in. The resources available to a scientist have almost the same importance as his intellect.

20'th cetnury's physics was governed by 2 huge theories: the Theory of Relativity and Quantum Physics. Einstein is, of course, the creator of the first one and Max Planck is considered to be the father of the second one. From the tiny bits I can understand, I can see that both have tremendous importance in our lives. I think Einstein did recognize the importance of quantum physics, but he was puzzled by it: the theory he had spent a lifetime developing was somewhat incompatible with the other major theory in 20'th century's physics. I think it was quantum physics he was referring to when he said (very famous quote): "God does not play dice" (that's how you translate it in Romanian, I don't know if "playng dice" makes sense in English :D).

yes, it makes sense.
the original words are : "God does not play dice with the Universe"

To quote Stephen Hawking:

"God does not only play dice,he doesnt know how far he throws them"-Stephen Hawking - A Brief History Of Time (the movie)

My statement still stands.   Listen to this allegory . . . when businesses fail, they usually pinpoint a person(s) to put the blame on.  For the non-American's here, I'm sorry if you haven't heard of the Enron scandal in the USA.

Similiar to this "pinpointing" nature that human beings like to do, Einstein undoubtedly didn't invent all "his" ideas soley through his own intuition.  Thomas Edison is considered a genius because of the many patents he put out toward his inventions that he gave the world.  However, "we" fail to mention that there were a lot of people helping him "invent" his "inventions" in the background.  Edison was more of a figurehead/manager toward his "business".  Einstein is similar to this concept.  Many other physicists/mathematicians/chemists (etc . . .) were working on similar material at the time of his great strides in physics (and some were not).  Don't forget Richard P. Feynman who was a major component in the understanding of Einstein's and other's works.

Its very ludicrous to even compare Einstein with Newton.  First of all, they were at different time periods!!!  I believe Pythagoras and his pupils were ten times smarter than Einstein and Newton ever were.  Do you want to know my reasoning?  Consider this, is it harder to start a problem or continue the problem in the middle?  Most students would say its harder to start a problem, for example a math word problem, then it is to have been already given variables, constants, and equations.  Now I'm not saying I can pick up on Einstein's work . . . for I am not one of the great minds of my day.

All of our viewpoints have meaning but I don't think we should be caught up over the logistics; therefore, these are my last words on this topic.

i think einstein was a genius because he proved many things.  I think this should be a poll.

in my opinion < NEWTON is Definitely the greatest scientist ever , look at his merits...

He Gave theories in numerous subjects, amazing versatility( Astro , MECHANICS , Pure maths , Optiks)

2nd Important factor time- Newton was a couple of centuries behind Einstein , you cant expect him to say everything correct

Though i  do respect Einstein Quite a lot , he  did extremely detailed study about only 1 topic though

However Newton was un mistakably in my opinion the greatest scientist and the greatest brain ever

P.s. My signature...... (was there befoere this topic was there) shows i am a true Newton admirer

UP WITH NEWTON!!!!

As per Hitler Though i despised his views , i quite admire the guy , it takes guts to completely sway a nation so tremendously , the little man had plenty of pluck!

U cant admire Hitler, that is not possible even though had propaganda abilities. Einstein had a lot of different works on different topics, I am not an expert and therefore will not go into details.

talking about inventing new ideas in the ancient times...how can we not mention my favorite mathematician: Archimedes

Eccenttric, dedicated, revered in his time, and definitely ahead of his times (by almost two milennia)

That, my friends, definitely fits the genius category in my humble opinion.

i read an article once that Archimedes was making some minor discoveries in calculus before he died. sadly, his papers on calculus were used for paper in religious texts. they have been found about 100 years ago and now they are trying to decipher them. wow, where we would be right now in the world of math and science if those papers hadn't been written over........

[quote="Drunken_Math"]I believe that Einstein wasn't technically a genius...

My mom always says that "If you're smart, and you don't try, you'll never make it. If you're not too smart, and you try really hard, you're gonna succeed".

Like my uncle, he graduated at the top of his class with top honors from the best university in China. (he's a genius) He now is broke and lives off of the small amounts of money he makes from the stock market... *shakes head*

Maybe Einstein wasn't too smart, but he worked really hard and became sucessful?[/quote]

:huuh:     :what?:

  On the topic of hitler, I was once reading this magazine that had this quiz inside of it.  Anyways, the point of the quiz was to see if you could identify Hitler, Roosevelt and another great American president.  Anyways, here were the clues:

   a)  This person was a heavy alchoholic.  He drank two bottle of whisky a day.  He smoked a chain each afternoon.  He was charged under accusations of spousal abuse, and cheated on his wife several times.  He was nearly kicked out of power, when he was caught stealing.

b)  This person was a vegetarian.  He was a very light drinker, he limited himself to wine on special occasions.  He did not smoke.  He never raised a hand on his wife.

  There was someother things that I can't remember at the moment.  Anyways, heres the question!

    [b]Which ones Roosevelt and which ones Hitler?[/b]

And the answer is...................[hide="?"] [hide="Are you sure you want to know?"] [hide="Brace yourself!"]  a) Roosevelt,    b) Hitler  [/hide] [/hide] [/hide]

   Thats the truth folks!  I did not make this up.  Obviously the auntehnticity of the magazine can be questioned, but (this was very long ago, I can no longer remember its name) was fairly respected and had good writers.

[quote="h78reg"]yea, i remember a quiz like this, except it was with roosevelt, churchill, and hitler.[/quote]

  Yup, thats who the other guy was, Churchill.  Do you remember where you found/took the quiz?

Back to the the topic of geniuses. I am sure that history of mankind is blooming with geniuses, but I only want to concentrate on Newton and Einstein (because I know not much about the others).

If you ask me, Newton was more of a genius than Einstein. The guy invented Calculus for crying out loud! 400 years in the future there is still no other mathematical branch more powerful than calculus!!

Not only that, he also discovered the laws of gravity and motion! (Who could have thought that the motion of the moon and the falling of the objects were infact goverened by the same phenomenon!) He also created the concepts of vectors to explain those phenomenons of gravity and motion using calculus (which he himself created).

Newton was truly way ahead of his predecessors. :clap:

P.S: I think that Newton's genius was discovered (i.e. he became interested in mathathematics) at the age of sixteen or something when he picked up Euclid's [u]Elements[/u] and was amazed by the beauty of mathematics.

F.Y.I: Did you know that Newton predicted that the world is going to end in 2064 (i.e. apocalypse)??

I agree, Newton was definitely smarter than Einstein.  Einstein made a bigger impact on the world today, as he had more technology and other scientists to build off of.

I think that Einstein was a genius - about as intelligent as most well-known and influential historical mathematicians (Newton, Liebniz, Bernoulli(s), Descartes, etc.), but not close to Euler or Gauss (who I see as the most intelligent mathematicians). Einstein was certainly very smart, as even a moderate understanding of such complicated physics is not possible for an average thinker (and Einstein was able to understand it enough to develop it). However, nothing that I know suggests that he was exceptional among exceptionals.

Einstein WAS a genius. Since he isn't alive, he ISN'T a genius now. He was considered a genius in his time, because there weren't any books to help develop those ideas(we do now)

But if I was to pick the smartest, I would pick Gabrielle Cramer. She invented Cramer's rule. It's faster than elimination, and that's fast!

Undoubtedly Einstein was a genius, but he didn't contribute much in terms of science to the Manhattan Project. He did write the letter to FDR requesting research on an atomic bomb, but that's about where his affiliation ends. Other scientists carried the project from there...meaning that if he had worked under Hitler, the Nazi would be unlikely to listen to a Jew.

Come on you all!!!!!!!! A lot of people consider Ciprian Manolescu as a genius (I think he IS) and he does not do nearly the 1% Einstein did!!!!!!!!!!

If you are not a genius and you try and practise a lot, you can do big things but not huge thins like e=mc^2..... I can'y really understand people who say that Einstein wasn't a genius.... who was a genius then????


complex numbers
algebra unsolved
algebra

Does there exist non-zero complex numbers $a, b, c$ and natural number $h$ such that if integers $k, l, m$ satisfy $|k| + |l| + |m| \geq 1996$, then $|ka + lb + mc| > \frac {1}{h}$ is true?

At 2007 Western China MO  was problem like this.
 Dirichle's method can help for proving this fact:

$ X_{1},X_{2},X_{3},n$ - real numbers
 there exist $ b_{i}\le n,a_{i}$, such that $ |a_{i}\minus{}b_{i}X_{i}|<\frac{1}{n}$

Something seems messed up in this problem.

There do not exist such complex numbers. In fact, there do not even exist $a,b\in\mathbb{C}\setminus\{0\}$ satisfying $|ka+lb|>x$ for some positive real constant $x$ whenever $|k|+|l|\ge 1996$. (This is stronger because we forced ourselves to choose $m=0$.) Here, we may assume $b=1$ because we may divide by $|b|$. Now, we just need to show that $|ka+l|$ can be arbitrarily small for arbitrarily large $|k|,|l|$, which follows from [url=https://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem]Dirichlet's Approximation Theorem[/url].


geometry
analytic geometry
graphing lines
slope
circumcircle
ratio
geometric transformation

I have a nice problem, but now i can't find the solution.
PROBLEM:
Given 3 same center circles O, with radius recpectively R, 2R, 3R. Let B, A, C are three point contructed on (O,2R), (O, R) and (O,3R), respectively, such as triangle ABC is right isoceles at B. Compute area triangle ABC with R

Put the coordinate origin O at the commn center of the 3 circles $K_1(O, R),\ K_2(O, 2R),\ K_3(O, 3R)$. Select the vertex A as the intersection of the circle $K_1$ with the positive x-axis, A = (R, 0). Let C' be an arbitrary point on the circle $K_3$ and let the triangle $\triangle AB'C'$ be isosceles with the right angle $\angle AB'C' = 90^\circ$, labeled clockwise. Let the point C' move on the circle $K_3$. Since the vertex A is fixed, the vertex C' moves on a circle and the triangle $\triangle AB'C'$ remains similar for all positions of the vertex C', the locus of vertex B' is also a circle. Let P be the center and r the radius of this circle $K(P, r)$. The circle K intersects the circle $K_2$ at 2 points $B_1, B_2$, the vertices of the isosceles right triangles $\triangle AB_1C_1, \triangle AB_2C_2$ satisfying the problem conditions. We know the equation of the circle $K_2: \ x^2 + y^2 = 4R^2$. We have to find the equation of the circle K. To do this, we choose 3 positions of the point C' and find the coordinates of the corresponding points B'.

$(1)\ \ C_1' = (3R, 0)\ \Longrightarrow B_1' = (2R, R)$

$(2)\ \ C_2' = (-3R, 0)\ \Longrightarrow B_2' = (-R, -2R)$

$(3)\ \ C_3' = (0, 3R)$

The equation of the line $AC_3'$ is

$\frac{x}{R} + \frac{y}{3R} = 1,\ \ y = -3(x + R)$

The midpoint M of the segment $AC_3'$ has the coordinates M = (R/2, 3R/2). The perpendicular bisector of the segment $AC_3'$ passes through M and has the slope 1/3. Hence, the equation of this perpendicular bisector is

$y - \frac{3R}{2} = \frac 1 3 (x - \frac R 2),\ \ y = \frac 1 3\ (x + 4R)$

The point $B_3'$ lies on this perpendicular bisector at a distance $\frac{AC_3'}{2}$ from M:

$AC_3' = \sqrt{R^2 + (3R)^2} = R \sqrt{10},\ \ MB_3' = \frac{AC_3'}{2} = \frac{R \sqrt{10}}{2}$

Thus the coordinates of the point $B_3'$ are the solutions of the perpendicular bisector equation and the equation

$(x - \frac R 2)^2 + (y - \frac{3R}{2})^2 = \frac{10R^2}{4}$

Substituting x = 3y - 4R,

$(3y - \frac{9R}{2})^2 + (y - \frac{3R}{2})^2 = \frac{10R^2}{4}$

$10 (y - \frac{3R}{2})^2 =\frac{10R^2}{4},\ \ y - \frac{3R}{2} = \pm \frac R 2$

Since the triangle $\triangle AB_3'C_3'$ is labeled clockwise and the slope of the line $MB_3'$ is positive, the y-coordinate of $B_3'$ has to be less than the y-coordinate of M, i.e., we have to choose the minus sign:

$y - \frac{3R}{2} = - \frac R 2,\ \ y = R,\ \ x = 3y - 4R = 3R - 4R = -R$

Consequently, the coordinates of the point $B_3'$ are (R, -R). Now we have to find the equation of the circumcircle $K(P, r)$ of the triangle $\triangle B_1'B_2'B_3'$. From the coordinates of its vertices, $B_1'B_3' = B_2'B_3'$ and $\angle B_1'B_3'B_2' = 90^\circ$. Hence, the center P of this circumcircle is the midpoint of the segment $B_1'B_2'$ and its radius is $r = \frac{B_1'B_2'}{2}$:

$P = (R/2,\ -R/2),\ \ B_1'B_2' = \sqrt{(3R)^2 + (3R)^2} = 3R \sqrt 2,\ \ r = \frac{3R \sqrt 2}{2}$

The equation of the circle $K(P, r)$ is

$(x - \frac R 2)^2 + (y + \frac R 2) = r^2 = \frac{9R^2}{2}$

Coordinates of the intersections $B_1, B_2$ of the circles $K, K_2$ are solutions of the equations of these 2 circles:

$K: \ \ \ x^2 - xR + \frac{R^2}{4} + y^2 + yR + \frac{R^2}{4} = \frac{9R^2}{2}$

$K_2: \ \ x^2 + y^2 = 4R^2$

Subtracting the 2nd equation from the 1st one,

$-xR + yR = 0,\ \ y = x$

and substituting this to the equation of either circle ($K_2$ is simpler),

$2x^2 = 2y^2 = 4R^2,\ \ x = y = \pm R \sqrt 2$

$B_1 = (R\sqrt 2,\ R\sqrt 2),\ B_2 = (-R\sqrt2,\ -R\sqrt 2)$

Areas of the triangles $\triangle AB_1C_1, \triangle AB_2C_2$ are then

$|\triangle AB_1C_1| = \frac{AB_1^2}{2} = \frac{(R - R \sqrt 2)^2 + 2R^2}{2} = \frac{5 - 2 \sqrt 2}{2}\ R^2$

$|\triangle AB_2C_2| = \frac{AB_2^2}{2} = \frac{(R + R \sqrt 2)^2 + 2R^2}{2} = \frac{5 + 2 \sqrt 2}{2}\ R^2$

thanks Yetti alots. Do you think the problem has another solutions. I wish this prblem has a nice solution, only used elementary geometry. Do you agree with me?

There is always more than one way to skin a cat. The problem can be stated in the following way:

Given an isosceles right triangle $ \triangle ABC$ with the right angle $ \angle ABC$, the base CA = b and the shoulder sides a = BC = AB, find a point P, which has the ratio of distances from the vertices A, B, C equal to PA : PB : PC = 1 : 2 : 3.

The point P is the common center of the 3 concentric circles with radii R, 2R, 3R. Given the triangle side a, we can calculate the radius R and the express the triangle side a or the triangle area $ |\triangle ABC$| in terms of R. Let (A), (B), (C) be 3 circles centered at the triangle vertices A, B, C and with radii r, 2r, 3r (r is arbitrary).

(1) The locus of points with the constant ratio of distances $ \frac {PB}{PC} \equal{} \frac 2 3$ from the points B, C is a circle $ (O_1)$ centered on the line BC and passing through the internal and external homothety centers $ G_1, H_1$ of the circles (B), (C).

(2) The locus of points with the constant ratio of distances $ \frac {PC}{PA} \equal{} \frac 3 1$ from the points C, A is a circle $ (O_2)$ centered on the line CA and passing through the internal and external homothety centers $ G_2, H_2$ of the circles (C), (A).

(3) The locus of points with the constant ratio of distances $ \frac {PA}{PB} \equal{} \frac {1}{2}$ from the points A, B is a circle $ (O_3)$ centered on the line AB and passing through the internal and external homothety centers $ G_3, H_3$ of the circles (A), (B).

The point P is the intersection of any 2 circles $ (O_1), (O_2), (O_3)$, because by transitivity of equvivalence, the 3rd circle also passes through this intersection. In general, the 2 selected circles intersect at 2 points P, P' and the remaining circle also passes through these 2 points. Hence, the 3 circles $ (O_1), (O_2), (O_3)$ belong to the same pencil. It is clear that we need only 2 of these circles, say $ (O_3), (O_1)$, to find the intersection points P, P'. We choose the circles $ (O_1), (O_3)$ because the angle $ \angle ABC$ is right, so that we can use the 2 lines $ BA \perp BC$ as coordinate axes. If the problem was to [b]construct[/b] the isosceles right triangle with vertices A, B, C on 3 concentric circles with radii R, 2R, 3R, given the segment length R, we would not have to use a coordinate system. But since the problem calls for [b]calculating the area[/b], a coordinate system makes it easier to set up an equation for the triangle area in terms of the given radius R. Let B be the coordinate origin, A on the positive x-axis, C on the positive y-axis.

Let $ g_3 \equal{} BG_3, h_3 \equal{} BH_3$ be the x-coordinates of the homothety centers $ G_3, H_3$ on the line BA (the x-axis). Since $ \frac {AG_3}{BG_3} \equal{} \frac 1 2$ and $ BG_3 \plus{} AG_3 \equal{} a$, we get $ g_3 \equal{} \frac {2a}{3}$. Since $ \frac {AH_3}{BH_3} \equal{} \frac 1 2$ and $ BH_3 \minus{} AH_3 \equal{} a$, we get $ h_3 \equal{} 2a$. The center and radius of the circle $ (O_3)$ are then

$ o_3 \equal{} BO_3 \equal{} \frac {BG_3 \plus{} BH_3}{2} \equal{} \frac {4a}{3},\ \ r_3 \equal{} \frac {|G_3H_3|}{2} \equal{} \frac {|BH_3 \minus{} BG_3|}{2} \equal{} \frac {2a}{3}$

Thus the equation of the circle $ (O_3)$ is

$ (x \minus{} o_3)^2 \plus{} y^2 \equal{} r_3^2,\ \ \left(x \minus{} \frac {4a}{3}\right)^2 \plus{} y^2 \equal{} \frac {4a^2}{9}$

$ x^2 \minus{} \frac {8ax}{3} \plus{} y^2 \plus{} \frac {4a^2}{3} \equal{} 0$

Likewise, let $ g_1 \equal{} BG_1, h_1 \equal{} BH_1$ be the y-coordinates of the homothety centers $ G_1, H_1$ on the line BC (the y-axis). Since $ \frac {CG_1}{BG_1} \equal{} \frac 3 2$ and $ BG_1 \plus{} CG_1 \equal{} a$, we get $ g_1 \equal{} \frac {2a}{5}$. Since $ \frac {CH_1}{BH_1} \equal{} \frac 3 2$ and $ BH_1 \minus{} CH_1 \equal{} a$, we get $ h_1 \equal{} \minus{} 2a$. The center and radius of the circle $ (O_1)$ are then

$ o_1 \equal{} BO_1 \equal{} \frac {BG_1 \plus{} BH_1}{2} \equal{} \minus{} \frac {4a}{5},\ \ r_1 \equal{} \frac {|G_1H_1|}{2} \equal{} \frac {|BH_1 \minus{} BG_1|}{2} \equal{} \frac {6a}{5}$

Thus the equation of the circle $ (O_1)$ is

$ x^2 \plus{} (y \minus{} o_1)^2 \equal{} r_1^2,\ \ x^2 \plus{} \left(y \plus{} \frac {4a}{5}\right)^2 \equal{} \frac {36a^2}{25}$

$ x^2 \plus{} y^2 \plus{} \frac {8ay}{5} \minus{} \frac {4a^2}{5} \equal{} 0$

The coordinates of the points P, P' are solutions of the equations of the circles $ (O_3), (O_1)$. But what we actually need is the distance $ p \equal{} BP \equal{} \sqrt {x^2 \plus{} y^2}$ of the point P from the origin B, because this distance is equal to p = 2R. Hence,

$ (O_3): \ \ p^2 \minus{} \frac {8ax}{3} \plus{} \frac {4a^2}{3} \equal{} 0$

$ (O_1): \ \ p^2 \plus{} \frac {8ay}{5} \minus{} \frac {4a^2}{5} \equal{} 0$

Calculating x, y form the 1st and 2nd equations, squaring the results and adding the squares, we obtain

$ x \equal{} \frac {4a^2 \plus{} 3p^2}{8a},\ \ y \equal{} \frac {4a^2 \minus{} 5p^2}{8a}$

$ p^2 \equal{} x^2 \plus{} y^2 \equal{} \frac {16a^4 \plus{} 24a^2p^2 \plus{} 9p^4 \plus{} 16a^2 \minus{} 40a^2p^2 \plus{} 25p^4}{64a^2} \equal{}$

$ \equal{} \frac {32a^4 \minus{} 16a^2p^2 \plus{} 34p^4}{64a^2} \equal{} \frac {16a^4 \minus{} 8a^2p^2 \plus{} 17p^4}{32a^2}$

$ 32a^2p^2 \equal{} 16a^4 \minus{} 8a^2p^2 \plus{} 17p^4$

$ 16a^4 \minus{} 40a^2p^2 \plus{} 17p^4 \equal{} 0$

Substituting $ p^2 \equal{} 4R^2$ and $ \sigma \equal{} |\triangle ABC| \equal{} \frac {a^2}{2},$ we arrive to a quadratic equation for the triangle area $ \sigma$ in terms of the radius R:

$ 4\sigma^2 \minus{} 20\sigma R^2 \plus{} 17R^4 \equal{} 0$

$ \sigma \equal{} \frac {20R^2 \pm \sqrt {400R^4 \minus{} 16 \cdot 17 R^4}}{8} \equal{} \frac {5 \pm \sqrt {25 \minus{} 17}}{2}\ R^2 \equal{} \frac {5 \pm 2 \sqrt 2}{2}\ R^2$

which is the same result as before. Take your pick of the solution methods, I like the 1st one better. Compare to the similar solution of a more complicated [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33328]Problem in Geometry[/url].

Yetti

First I think the problem can solved by applied rotation but haved some problems. I look point C is a projective of the point A by rotation center B and rotation angle 90. but i can't find some relation between circle (O, R) and (O,3R). Anyway Yetti gives me a way to solve this problem. Thank you very much. :blush:

[code]Yetti writed "If the problem was to construct the isosceles right triangle with vertices A, B, C on 3 concentric circles with radii R, 2R, 3R, given the segment length R, we would not have to use a coordinate system"[/code]

How do you construct the triangle?

This construct-problem is a particular case of a remarkable geometrical locus.[hide="Here is the enunciation of the geometrical locus"]
Let the fixed $\triangle ABC$, the fixed point $F$ and the fixed circle $w=C(O,r)$ (a fixed line $d$). For a mobile point $M\in w$ ($M\in d$) we define the point $L$ so that $\triangle LFM\sim \triangle ABC$. Then the geometrical locus of the point $L$ is the reunion of the two circles which are symmetrically w.r.t. the line $FO$ (the reunion of the two lines which are symmetrically w.r.t. the line $FM_0$, where $M_0\in d$, $FM_0\perp d$).

In point of fact, here are two problems with the geometrical locus ! [/hide]

Send to yetti. I inform to you that i haved find another solution for this problem. A nice solution, that only used fimilat triangle knowledges. because that solution need a picture so that tomorow I will post it at this topic. Please waiting, ok

Consider the case 1: O be inside triangle ABC.
Contruct point A' on the circle (O,R) so that triangle AOA' right isoceles at O( the point A' in order side with O by AC). We have:
$\hat{CAO}+\hat{OAB} = 45^0, \hat{CAO}+\hat{CAA'}=45^0$, We infer:
$\hat{OAB}=\hat{CAA'}$. Consider two triangleS AOB and AA'C that:
$\frac{OA}{AA'}=\frac{\sqrt{2}}{2}$,  $\frac{AB}{AC'}=\frac{\sqrt{2}}{2}$ and $\hat{OAB} = \hat{CAA'}$. So that two triangles are congruent $\Rightarrow A'C = \sqrt{2}.OB=2\sqrt{2}.R$
Consider triagle OCA':
$OC^2=9R^2,OA'^2=R^2,A'C^2=8R^2 \Rightarrow OC^2=OA'^2+A'C^2$.
So that $\hat{CA'A}=135^0$ $\rightarrow\hat{BOA}=135^0$.
Accoding to Cosin theorem: $AB^2=5R^2+2\sqrt{2}R^2$
$\Rightarrow S_{ABC} = \frac{(5+2\sqrt{2})}{2}.R^2$


vector
algebra
polynomial
algorithm
linear algebra
linear algebra unsolved

While prepping for upcoming finals, I stumbled upon an old problem from two summers ago that is simply driving me crazy.
The problem is at the end of these old notes, but I'll type it out as well,
http://math.berkeley.edu/~ecarter/Summer08/110/notes/lec19.pdf

Let $ T$ be a linear operator on a finite-dimensional vector space $ V$, and let $ x$ be a nonzero vector in $ V$. The polynomial $ p(t)$ is called a $ T$-annihilator of $ x$ if $ p(t)$ is a monic polynomial of least degree for which $ p(T)(x) \equal{} 0$
Prove the following results.
a) The vector $ x$ has a unique $ T$-annihilator.
b) The $ T$-annihilator of $ x$ divides any polynomial $ g(t)$ for which $ g(T) \equal{} T_0$
c) If $ p(t)$ is the $ T$-annihilator of $ x$ and $ W$ is the $ T$-cyclic subspace generated by $ x$, then $ p(t)$ is the minimal polynomial of $ T_W$, and $ dim(W)$ equals the degree of $ p(t)$.
d) The degree of the $ T$-annihilator of $ x$ is $ 1$ if and only if $ x$ is an eigenvector of $ T$.

So far I've only managed a solution for d), which I'll write down for the record. Any help/hints/solutions for a-c would be mighty helpful. Thanks.

d) $ \Rightarrow$| Suppose $ deg(p(t)) \equal{} 1$, that is, $ p(t) \equal{} t \plus{} a_0$. So $ p(T)(x) \equal{} T(x) \plus{} a_0x \equal{} 0$, which implies $ T(x) \equal{} \minus{}{a_0}x$.
$ \Leftarrow$| Suppose $ x$ is an eigenvector of $ T$, so $ T(x) \equal{} {\lambda}x \rightarrow (T \minus{} {\lambda}I)x \equal{} 0$. It follows that $ p(t) \equal{} t \minus{} \lambda$.

First, welcome to this forum. Second, what exactly is $ T_0$ in b)? Third, in your proof of d), shouldn't $ a_1$ be 1, since p(t) is monic?

For a), let $ p(t)$ and $ g(t)$ be two T-annihilators of vector $ x$. Then $ p(T)(x) \equal{} g(T)(x) \equal{} 0$. Let now $ q(t)$ and $ r(t)$ be the quotient and remainder on division of $ p(t)$ by $ g(t)$: $ p(t) \equal{} q(t)g(t) \plus{} r(t)$, where $ r(t) \equiv 0$ or the degree of $ r(t)$ is less than the degree of $ g(t)$. Since, by definition, $ p(t)$ and $ g(t)$ have the same degree and are both monic, we must have $ q(t) \equal{} 1$ and so $ p(t) \equal{} g(t) \plus{} r(t)$. If $ r(t) \not\equiv 0$, then $ p(T)(x) \equal{} 0 \equal{}> g(T)(x) \plus{} r(T)(x) \equal{} 0 \equal{}> r(T)(x) \equal{} 0$, and $ r(t)$ would be a T-annihilator of x with degree less than $ p(t)$ and $ g(t)$, contradicting the hypothesis that both $ p(t)$ and $ g(t)$ have the minimum degree. So $ r(t) \equiv 0$, $ p(t) \equal{} g(t)$, and the T-annihilator of x is unique.

Thanks for the welcome Carcul, as well as the feedback and help. I knew this would be a good forum to visit if I got stuck.

As for b), $ T_0$ is the zero-operator on $ V$, sorry about the lack of explanation. Thanks also for explaining a). I had seen a similar type of proof for the uniqueness of minimal polynomials, but hadn't been able to piece it together.

Now to edit d)...

From Carcul's method, I think I may have made some progress on b),

Let $ g(t)$ be any polynomial for which $ g(T)\equal{}T_0$, and let $ p(t)$ be a $ T$-annihilator for some vector $ x$. By the division algorithm for polynomials, there exist polynomials $ q(t)$ and $ r(t)$ with $ deg(r(t)) < deg(p(t))$ such that $ g(t)\equal{}q(t)p(t)\plus{}r(t)$.

It follows that $ g(T)(x)\equal{}q(T)(x)p(T)(x)\plus{}r(T)(x)$
$ \Rightarrow 0\equal{}q(T)(x) \cdot 0 \plus{} r(T)(x)$
$ \Rightarrow r(T)(x)\equal{}0$

But since $ p(t)$ is the $ T$-annihilator for $ x$, it follows that $ r(t) \equiv 0$, and hence $ g(t)\equal{}q(t)p(t)$.

Anyone for c), while it's still up in the air?


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

In the questions below: $G$ is a ﬁnite group; $H \leq G$ a subgroup of $G; |G : H |$ the index of $H$ in $G; |X |$ the number of elements of $X \subseteq G; Z (G)$ the center of $G; G'$ the commutator subgroup of $G; N_{G}(H )$ the normalizer of $H$ in $G; C_{G}(H )$ the centralizer of $H$ in $G$; and $S_{n}$ the $n$-th symmetric group.

Show that every element of $S_{n}$ is a product of $2$-cycles.

[quote="N.T.TUAN"]
Show that every element of $S_{n}$ is a product of $2$-cycles.[/quote]

Given $A=\{1,2,..,n\}$ and its symettric group $S_{n}$, for two elements $a,b \in A$ and $\sigma \in S_{n}$ define the relation $a R_{\sigma}b$ iff $a = \sigma^{n}(b)$ for some integer $n$. It is trivial to check that $R_{\sigma}$ is an equivalence relation on $A$. The point is that each premuation $\sigma$ defines a partition on $A$. Let $P_{1},P_{2},...,P_{k}$ be the partitions for $A_{\sigma}$ then $\sigma = P_{1}P_{2}...P_{K}$. (Here we treat $P_{i}$ as cycles and their product as standard function composition.)

If we drop each $1$-cycle in $P_{1},P_{2},...,P_{k}$ we see that (assuming it is not the identity premutation) by reordering that $\sigma = P_{1}P_{2}... P_{j}$. Now if each $P_{i}$ is a $2$-cycle we are done. If not we can employ the following trick:
$(a_{1},a_{2},...,a_{m}) = (a_{1},a_{m})(a_{1},a_{m-1})...(a_{1},a_{3})(a_{1},a_{2})$
Hence we see that each $P_{i}$ is expressiable as a product of $2$-cycles. Hence their total product is also a product of $2$-cylces. So he have succesfully shown that each non-trivial premuation (non identity) is a product of $2$-cycles. Checking the idenity we see that $1=(1,2)(1,2)$. Q.E.D.

[quote="N.T.TUAN"]Show that every element of $S_{n}$ is a product of $2$-cycles.[/quote]
This is equivalent to the (rather obvious) statement that any array of $n$ numbers can be sorted by swapping pairs of elements.


Let $ ABC$ be a triangle, and point $ P$ such that the projections $ Q,R,S$ of $ P$ onto $ AB,BC,CA$ are concurrent. Is it necessary that $ P$ is on the circumcircle of $ ABC$?

You mean collinear?
if yes then it is true http://mathworld.wolfram.com/SimsonLine.html

The quadrilateral $ PQBR$ is inscriptible $ \implies$ $ \angle BPR\equal{}\angle BQR$ and the quadrilateral $ PSAQ$ inscriptible $ \implies$ $ \angle AQS\equal{}\angle APS$.
We know that $ \angle BQR\equal{}\angle AQS$ therefore $ \angle APS\equal{}\angle BPR$ $ \implies$ $ \angle PBR\equal{}\angle PAS$ $ \implies$ $ APBC$ inscriptible therefore $ P$ is on the circumcircle of the triangle $ ABC$