[quote="GameBot"]The sequence shown was formed by writing the first letter of the alphabet followed by writing the first two letters of the alphabet and continuing the pattern by writing one more letter of the alphabet each time. Continuing this pattern, what letter is the 280th letter in this sequence?
$ \begin{tabular}{c} A, A, B, A, B, C, A, B, C, D, A, B, C, D, E\ldots \end{tabular}$[/quote]

Summing triangular numbers got me D/.

i got D too
23*24/2=276
280-276=4
4th letter of the alphabet is D

[quote="GeorgiaTechMan"]hmmmm, i got e.  I dont understand why it is d.[/quote]
never mind.  I messed up.


approximation
number theory
decimal representation
rational number
irrational number
IMO Shortlist
IMO Longlist

Which fractions $ \dfrac{p}{q},$ where $p,q$ are positive integers $< 100$, is closest to $\sqrt{2} ?$ Find all digits after the point in decimal representation of that fraction which coincide with digits in decimal representation of $\sqrt{2}$ (without using any table).

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Consider the Pell's equation $x^2-2y^2=1$ and also the Pell's Negative equation $x^2-2y^2=-1$.
Consider the union of this two equations which has solutions
${x_n=\frac{(1+\sqrt 2)^n+(1-\sqrt 2)^n}{2}},y_n=\frac{(1+\sqrt 2)^n-(1-\sqrt 2)^n}{2\sqrt 2}$
Lemma. If $|\frac uv-\sqrt 2|<0.015$ then
$\frac{|u^2-2v^2|}{2.85 q^2}<|\frac uv-\sqrt2 |<|\frac{|u^2-2v^2|}{2.b1 q^2}$.
Proof: We have ${|\frac uv-\sqrt 2}=\frac{|\frac{p^2}{q^2}-2|}{\frac pq+\sqrt 2}$ and from now it's easy.
Now let $p,q$ realize the minimal value of $|\frac pq-\sqrt 2|$.
We claim $p=x_n,q=y_n$ or $p=2y_n,q=x_n$ for some $n$.
Suppose not. Then $|p^2-2q^2|\neq 1,2 $. Playing with residues we show that $|p^2-2q^2|=3,4,5,6$ has no solutions  for $(p,q)=1$.
That means $|p^2-2q^2|\geq 7$.
Now let $x_n<p<x_{n+1}$ then we easily show $y_n<q<y_{n+1]}$.
Since $\sqrt2-\frac 75<0.15$ using the lemma we get
$|\frac pq-\sqrt 2|>\frac{7}{2.85 q^2}>\frac 7{2.85 y_{n+1}^2}$.
But one easily shows that $y_{n+1}<2.5 y_n$ and using this we show that
$|\frac pq-\sqrt 2|>\frac 1{ 2.81 y_n^2}>|\frac {x_n}{y_n}-\sqrt 2|$.
Hence $p,q$ are not minimal. Contradiction.
So $(p,q)$ are of form $(x_n,y_n)$ or $(2y_n,x_n)$ and we easily investigate this cases.

[hide=a more elementary solution]According to [url=https://en.wikipedia.org/wiki/Continued_fraction#Best_rational_approximations]the Wikipedia page about continued fractions[/url], the closest fraction with the condition given should be a semiconvergent which its last term has a value at least half of the corresponding term in the original continued fraction. Because the continued fraction of $\sqrt2$ is $[1;2,2,2,\ldots]$, the last semiconvergent with former property having a numerator less than $100$ is:
$$[1;2,2,2,2,2]$$
So the best rational approximation with numerator less than $100$ is:
$$\sqrt2\approx1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}=\boxed{\frac{99}{70}}$$
Using a calculator we have:
$$1.41421<\sqrt2<1.41422<1.41428<\frac{99}{70}<1.41429$$
[/hide]


Prove for any integer $n>1$ that $(n-1)^2|n^{n-1}-1$

It is equivalent to proving $n-1 \mid n^{n-2}+n^{n-3}+...+n+1$. Now $n \equiv 1 \pmod{n-1}$ so all the terms are $n^k \equiv 1^k \equiv 1$, and there are $n-1$ terms. Then their sum is $S \equiv n-1 \equiv 0$.


1) A book is lying open on a desk. The product of the page numbers showing is 930. What are the two page numbers?

2) What pair of whole numbers with a sum of 45 have the greatest possible product?

[hide]x*(x+1) = 930 ... I tried to guess what x is... and figured that since its product is 930, it has be around 30 something... and I was right, it's 30 and 31... right?[/hide]

[hide]hm... the larger both number is the bigger the product, so it's gonna be 22 and 23[/hide]

1
[hide]x(x+1) = 930 so x^2 + x - 930 = 0 that factors out to (x+31)(x-30) so the page numbers must be 30 and 31.[/hide]
2
[hide]The closer two numbers are together, the larger the product will be. So the answer is 22*23 =  506[/hide]

[quote="mathprincess09"]hey tarquin i like ur method that u used for no.1
factoring! i could have never thought of that.
i did the way Lucy did it! :D[/quote]
Unless you used the formula/completing the square etc., factorising is the same as trial and error.


Show that if $(a,b)=1$ and n is an odd positive integer,then $(a+b,\frac{a^n+b^n}{a+b})$ divides n.


[hide]
We have $(x^3+4x+14)^3+4(x^3+4x+14)+14=x$  (*)
The root equation to this recursion is $x=x^3+4x+14$
The solutions to this equation are obviously solutions to (*).  We can easily solve this cuibc.  We have
$x^3+3x+14=0$
$(x+2)(x^2-2x+7)$
From this we find the three roots: $\boxed{-2, 1 \pm i\sqrt{6}}$
[/hide]


[quote="carlosbr"]Buenos dias Sres.

Estos son los 2 examenes de la IMO 2006.

REVISENLO ........[/quote]

Solucion oficiales de la Prueba:

http://www.mathlinks.ro/Forum/viewtopic.php?t=101108

Rivisenla  :wink:

Américo Arone


articles
continued fraction
irrational number

1)Let $ {f_1}=\frac{3}{2}$.
And $ {f_n}={1}+\frac{1}{f{}_n{}_-{}_1}$
express $ {f_n}$ in terms of $ {n}$. And find the sum of the first $ {n}$  terms.
2)how do you write the continued fraction of an irrational number? :?:

hello, one can prove that
$ f(n)\equal{}1\plus{}\frac{\text{Fibonacci(n)}\plus{}\text{LucasL(n)}}{\text{Fibonacci(n\plus{}1)}\plus{}\text{LucasL(n\plus{}1)}}$.
Sonnhard.

Alternately, we can write $ \boxed{f_n \equal{} \frac {F_{n \plus{} 3}}{F_{n \plus{} 2}}}$, where $ F$ is the Fibonacci sequence, indexed such that $ F_0 \equal{} 0$ and $ F_1 \equal{} 1$

To prove this, note that it works for the base case, and then show that it satisfies the rule:
$ \frac {F_{n \plus{} 3}}{F_{n \plus{} 2}} \equal{} \frac {F_{n \plus{} 2} \plus{} F_{n\plus{}1}}{F_{n \plus{} 2}} \equal{} 1 \plus{} \frac{1}{F_{n\plus{}2} / F_{n\plus{}1}}$

You can read about constructing a continued fraction representation [url=http://en.wikipedia.org/wiki/Continued_fraction#Calculating_continued_fraction_representations]in this article[/url]. Life becomes easier if the quotients of the continued fraction (the series $ a_n$) eventually enters a cycle.

[hide="@Dr. Sonnhard"]You are incredibly unhelpful on everyone of your posts.  Where are your solutions? [/hide]


Let $ x,y,z>0$,prove that
$ \frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y}\plus{}3\geq 3(\frac{y\plus{}z}{z\plus{}2x\plus{}y}\plus{}\frac{z\plus{}x}{x\plus{}2y\plus{}z}\plus{}\frac{x\plus{}y}{y\plus{}2z\plus{}x}).$

BQ

[quote="xzlbq"]Let $ x,y,z > 0$,prove that
$ \frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y} \plus{} 3\geq 3(\frac {y \plus{} z}{z \plus{} 2x \plus{} y} \plus{} \frac {z \plus{} x}{x \plus{} 2y \plus{} z} \plus{} \frac {x \plus{} y}{y \plus{} 2z \plus{} x}).$

BQ[/quote]
we only need prove that
$ \sum \frac {1}{x \plus{} y} \plus{} \frac {9}{x \plus{} y \plus{} z} \ge 6\sum \frac {1}{x \plus{} 2y \plus{} z}$ :)


inequalities
geometry
inradius
circumcircle
trigonometry
geometry proposed

In a triangle $ ABC$ denote $ 2s \equal{} a \plus{} b \plus{} c$ , the length $ r$ of  inradius , the length $ R$ of circumradius and the length

$ h_a$ of $ A$-altitude. Prove that there is the inequality $ \boxed {R \minus{} 2r\ \ge\ \sqrt {s(s \minus{} a)} \minus{} h_a \plus{} \frac {r}{4a(s \minus{} a)}\cdot (b \plus{} c \minus{} 2a)^2}$ .

[u]Consequence[/u].  $ \boxed {R \minus{} 2r\ \ge\ \sqrt {s(s \minus{} a)} \minus{} h_a}$ ([u][b]L.Panaitopol's [/b]inequality[/u]).

[color=darkblue]I will use the following trigonometrical identities : $\boxed{\ \begin{array}{cccc}
1\blacktriangleright & R-2r=R\left(1-4\sin\frac A2\cos\frac{B-C}2+4\sin^2\frac A2\right) \\
 \\
2\blacktriangleright & \sqrt{s(s-a)}=2R\cos\frac A2\cdot\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2} \\
 \\
3\blacktriangleright & h_a=2R\left(\cos^2\frac{B-C}2-\sin^2\frac A2\right) \\
 \\
4\blacktriangleright & \frac{r}{4a(s-a)}\cdot(b+c-2a)^2=R\left(\cos\frac{B-C}2-2\sin\frac A2 \right)^2\ \end{array} }$ .

Therefore, the proposed inequality is equivalent to :
$\begin{array}{cccc}
1-4\sin\frac A2\cos\frac {B-C}2+4\sin^2\frac A2\ \ge\ 2\cos\frac A2\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}-2\left(\cos^2\frac {B-C}2-\sin^2\frac A2\right)+\left(\cos\frac {B-C}2-2\sin\frac A2\right)^2 \\
 \\
\iff\ 1+\cos^2\frac {B-C}2-2\sin^2\frac A2\ \ge\ 2\cos\frac A2\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}\  \\
 \\
\iff\ \cos^2\frac A2+\left(\cos^2\frac {B-C}2-\sin^2\frac A2\right)\ \ge\ 2\cos\frac A2\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}\ \end{array}$
Now, the last inequality follows directly from the well-known [u][b]AM-GM[/b] inequality[/u] . The expression under the square root is positive according to identity [b]3[/b] .[/color]

[u]Thank you[/u], [b]C. Mateescu[/b]. Another proof ?


just state whether you would kiss, hug, kick, or punch the person above you.

hmm..this is gonna hurt....

EDIT: i should add more popular choices, you can now kiss, hug, kick, punch, give the person a piece of bread, or jump on the person.

I would...
[hide]KILL (^_^)!!![/hide]

(LL=CK if I can't)

punch punch punch. We must not have violence. punch punch punch.

here ra have a piece of bread.... :D  :D

lol we need to get ubemaya to post here after me

You scare me...  I'm afraid to do something to you.

[quote="tinytim"][b]TELETUBBIE HUG[/b][/quote]

It ends with a y you failure :wink:

HEY! I TURNED MY OTHER CHEEK!

Should I follow the bible? Hmm....

hug

kiss.  :blush:

but if I could I would enlist you in club Voldemort and persuade you to use Aveda Kedavra on your chinchilla. Then you would Imperio your mom to buy you snacks and a ds. you eat too many snacks and get fat.  :D

Don't know what a chinchilla is, I don't know how to Imperio my mom, she would never buy me a ds, and I hate snacks. Instead I eat the rest of the cake in the fridge.

punch

Shakehand in signal of apologize (no matter that it isn't an option)


modular arithmetic
number theory solved
number theory

Hello all,
can you help me to solve this problem:
If x is primitive root mod(p^2) ,then x is primitive root mod(p).

thank you

Suppose $x$ is not a primitive root $\pmod{p}$. Then for some $n < p-1$ we have $x^n \equiv 1 \pmod{p}$ so $x^n = pk+1$ for some $k$. That gives us $x^{np} = (pk+1)^p \equiv 1 \pmod{p^2}$. But $np < p(p-1) < p^2-1$ so $x$ wouldn't be a primitive root $\pmod{p^2}$. Contradiction.


How do I make the symbol { appear over some latex?  Also, how can i add text to the top of it?  Sorry if this is hard to understand.

Do you mean $ \overbrace{x\plus{}y\plus{}z}^{12}$ ? It should be obvious how to get the same thing below the expression.

Yes, that's exactly what I wanted!  Thanks!


The interior angles of a polygon form an arithmatic sequence with the smallest angle 97 degrees and the largest 143 degrees. How many sides does the polygon have?

Source: 2003 Iowa GPML sprint round

[hide]
If there are N sides (angles) the total of the measures of the angles is 180(N-2).
Also for this polygon, since 97 and 143 are the start and end of an arithmetic seq, and their average is 120, then the average of all the angles is 120.
Multiply it by the number of sides and you have the total measure of all angles.
Then 120N = 180(N-2)
120N = 180N - 360
360 = 60N
N = 6

So it's a hexagon.

Check:
97 + 106.2 + 115.4 + 124.6 + 133.8 + 143 = 720, total angles in a hexagon.
QED. (Whatever that means)
[/hide]

[quote="Hexaditidom"]
QED. (Whatever that means)
[/quote]

it means quod erat demonstrandum ("which was to be demonstrated") in latin, but nowadays, people just say "quite elegantly done" as QED. It is a way to show that you have finished your proof.

How do you know that the average of all the angles is 120 degrees?

I edited in an explanation of that, thanks for pointing it out


function
integration
calculus
calculus computations

Prove that : there is only one function $ f\in C^1([0,1], \mathbb{R})$such that: $ \forall x \in [0,1]$ we have $ f'(x) \equal{} f(x \minus{} x^2)$

Certainly false. What is true is that, given any $ f(0)$, you can construct exactly one such function. Please, pay attention if not to spelling, then, at least, to the way you state the problems!

yes i foget the condition $ f(0)\equal{}1$ so now what is you solution?

$ (Tf)(x)\equal{}a\plus{}\int_0^x f(t\minus{}t^2)\,dt$ is almost a contraction in $ C[0,1]$ with the usual $ \max$ norm. If you change the norm just a tiny bit giving it some more weight near $ 0$, it'll become a true contraction. Alternatively, solve the equation in $ C[0,\tfrac 14]$ (where $ T$ is a contraction already) and note that the right hand side extends to $ [0,1]$ using only the values on $ [0,\tfrac 14]$.

i see what you mean but unfortunatly $ U$ isn't a a contraction !!!

What do you mean? :? $ Sf(x)\equal{}\int_0^x f(t\minus{}t^2)\,dt$ is a linear operator on $ C[0,1]$ and If $ |f|\le 1$ everywhere on $ [0,1]$ then $ |(Sf)(x)|\le x\le 1$ on $ [0,1]$. It is true that $ \|S\|\equal{}1$, so you cannot just refer to the fixed point theorem right away but , as I said, there are 2 (actually now I see 3) ways to proceed:

1) Integrate again and show that $ S^2$ is a contraction.
2) Change the norm to $ \max w|f|$ where $ w$ is positive continuous on $ [0,1]$ and $ w\equal{}1$ on $ [0,1/4]$ and $ w<1$ on $ (1/4,1]$. $ S$ becomes a true contraction in $ C[0,1]$ with respect to this norm.
3) Consider $ C[0,1/4]$. Then $ \|S\|\equal{}1/4$, so $ S$ is a contraction there. Now note that the values of $ f$ on $ [0,1/4]$ determine $ f$ on $ [0,1]$ uniquely.


inequalities
search
inequalities proposed

give $a,b,c \in R^{+}$ then prove that
              $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \leq \frac{5}{4}\sqrt{a+b+c}$

Please Never post the problems which is too well-known and posted many times before. Try to search it.  :lol:

Here maybe ;) : http://www.mathlinks.ro/Forum/viewtopic.php?t=1459


What is the product of the first eight terms of the arithmetic sequence $ 12,10,8,\ldots$?

$ 12, 10, 8, 6, 4, 2, 0,$ etc.

One of the first eight terms is $ 0$. So the product is $ \boxed{0}$.


Mike leaves home every day at $7: 00$ each morning to get to school.  If he travels at $60 mph$, he arrives 5 minutes early.  If he travels at $40 mph$, he arrives 3 minutes late.  The speed at which he should travel to get to school on time is $\frac{m}{n}mph$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n$.

What number would this be on the AIME?

Decent problem, probably #1 or #2, though.

[quote="miyomiyo"]Did you get the solution?[/quote]

just a fyi, you might want to check problems with somebody, rather than posting them on the forum

[quote="miyomiyo"]Did you get the solution?[/quote]

Yes.

[hide] If $d$ is the distance, we have

$\frac{d}{60}+8 = \frac{d}{40}\Rightarrow d = 960$.

So the "right" amount of time would be $\frac{960}{60}+5 = 21$ minutes, so he would have to go at

$\frac{960}{21}= \frac{320}{7}$

miles per hour. I just noticed that I totally ignored units throughout the problem, but the good thing is that it does not change the answer.

I think you can also find the answer with a weighted harmonic mean:

$\frac{1}{\frac{3}{8}\cdot \frac{1}{60}+\frac{5}{8}\cdot \frac{1}{40}}= \frac{320}{7}$.
[/hide]

[hide="I did..."]
$60$ mph = $5$ minutes early.
$40$ mph = $3$ minutes late.
Harmonic mean($40$, $60$) = $1$ minute early.
Harmonic mean($40$, $60$) = $48$.
$48$ mph = $1$ minute early.
Harmonic mean($48$, $60$) = $1$ minute late.
Harmonic mean($48$, $60$) = $\frac{160}{3}$.
$\frac{160}{3}$ mph = $1$ minute late.
Harmonic mean($48$, $\frac{160}{3}$) = on time.
Harmonic mean($48$, $\frac{160}{3}$) = $\frac{960}{19}$
So the answer is $960+19 = 979$.
[/hide]

Hmmm...  I got a different answer.

[quote="miyomiyo"]Harmonic mean($48$, $60$) = $1$ minute late.
Harmonic mean($48$, $60$) = $\frac{160}{3}$.[/quote]

Whew, took me a long time to figure out why your answer was different. The problem is in the lines quoted. You should be taking the harmonic mean of $48$ and $40$, not of $48$ and $60$. Then your answer will agree with mine.


Given n > 1, find the maximum value of sin2x1 + sin2x2 + ... + sin2xn, where xi are non-negative and have sum π. (those sin2 stand for the square of the sin)

[quote="Pascual2005"]Given $n > 1$, find the maximum value of $sin^2x_1 + sin^2x_2 + ... + sin^2x_n$, where $x_i$ are non-negative and have sum π.[/quote]

Is it : $n \cdot sin^2\left(\frac 1n\right)$? :)


Hello everyone.

I bought a keyboard a few years ago that has buttons on the top of the F keys looks something like...
http://www.compusa.com/products/product_info.asp?ref=froogle&pfp=froogle&product_code=50206022&cm_ven=froogle&cm_cat=&cm_pla=&cm_ite=feed

Basically, I'm wondering if I can customize the buttons so that when I push the calculator button it doesn't open Microsoft's crappy calculator, but Powertoy... Anyone know how?

Doesn't it have a manual? probably the answer is either there or in Windows Help (sorry, I don't use windows that much!)

Best,

Maybe if you go into Control Panel>Keyboard it will allow you to configure the keys.  I have a laptop with special function keys that you can customize like this.


linear algebra
matrix
search
algebra
polynomial
induction
vector

already posted ... it's a nilpotent matrix  :)

But where can I find the topic? ("Search" turned up nothing)

[quote="perfect_radio"]Let $A \in \mathcal{M}_n \left( \mathbb C \right)$. Suppose that $\textrm{Tr}A=\textrm{Tr}A^2=\ldots=\textrm{Tr}A^n=0$. Find $A^n$.

sorry if it was already posted  :blush:[/quote]

Sorry, what do you mean by finding $A^n$? You mean pointing out it's nilpotent or finding the exact matrix?

[quote="perfect_radio"]alekk was refering to $A$ (not to $A^n$) when he said nilpotent. this implies the answer $A^n=0$ (which is sugested by considering small cases)[/quote]

oh, sorry, i read the question wrongly.. :oops:

I hope this is right.
As the characteristic polynomial splits (in $\mathbb{C}$), we consider A',  the Jordan canonical form of A. Suppose the distinct eigenvalues of A' are $e_1,...,e_k$, where $k \le n$. Then as we can note that if the multiplicity of $e_i$ is $m_i$, then we have
$m_1e_1 + ... +m_k e_k = 0$
$m_1e_1^2 + ... + m_k e_k^2 = 0$
...
$m_1e_1^n + ... + m_k e_k^n = 0$
as each jordan block is upper triangular.
Consider the first k equations, then we see that the matrix \[ \left( {\begin{array}{*{20}c} {e_1 } & {..} & {e_k } \\ {..} & {..} & {..} \\ {e_1^k } & {..} & {e_k^k } \\ \end{array}} \right)  \]
is singular. However, as this is a vandermonde-type determinant, we conclude that at least one of the eigenvalues is 0 (otherwise two of them are equal, which contradicts the assumption). If so, WLOG assume $e_1 = 0$, we put it back into the equation, and consider the matrix \[ \left( {\begin{array}{*{20}c} {e_2 } & {..} & {e_{k-1} } \\ {..} & {..} & {..} \\ {e_2^{k-1} } & {..} & {e_{k-1}^{k-1} } \\ \end{array}} \right)  \]
this time. By the same argument we must arrive at a contradiction. ($e_1 = e_2 = 0$ or there exists two equal eigenvalues. )
thus there is only one eigenvalue for the matrix A', then it becomes apparent that this eigenvalue must be 0. Thus we conclude that the characteristic polynomial of A' is $t^n$. As A and A' have the same characteristic polynomial, we conclude that A is nilpotent.

(I hope I am right because I am highly unfamiliar with linear algebra stuff..)

It is easy to show that if $\lambda_1,\lambda_2,\ldots,\lambda_n$ are the eigenvalues of $A$ then $\lambda_1^t,\ldots$ are the eigenvalues of $A^t$. In what cases is it true that $\lambda_i$ and $\lambda_i^t$ have the same multiplicity (hope it's the right name)? If always true, how do you prove it? If false, any simple counter-example?

[quote="Soarer"](I hope I am right because I am highly unfamiliar with linear algebra stuff..)[/quote]
Well, you are more familiar than me. I just started studying eigenvalues the other day. And I don't understand much from what is said on Mathworld about Jordan form
 :blush:

The proof using Jordan form and Vandermonde seems fine.

Jordan form does provide a proof of the result on taking eigenvalues to the $t^{th}$ power with multiplicity (since taking a matrix to the $t^{th}$ power essentially takes each Jordan block to the $t^{th}$ power).  An alternative method is to use a (slightly weaker) result known as Schur's theorem:   For any matrix $A$ there is a (unitary) matrix $U$ and an upper triangular matrix $T$ such that $A=U T U^{-1}$.  Assuming that theorem is true, than we have $A^t = U T^t U^{-1}$, so $A^t$ and $T^t$ share the same eigenvalues with the same multiplicities.  The $\lambda_i^t$ appear with the right multiplicity on the diagonal of $T^t$, so we're done.

If you haven't seen Schur's theorem before, the proof goes by induction on the size of the matrix.  The inductive step consists of first letting $V$ be a unitary matrix whose first column is any eigenvector of $A$ (constructable by Gram-Schmidt, for example).

$V^{-1} A V$ has as its first column ${(\lambda}, 0, \dots 0)$.  Now let $U_0$ be a matrix whose lower right $n-1 \times n-1$ minor is chosen to upper triangularize the lower right $n-1 \times n-1$ minor of $V^{-1} A V$, with the upper left entry being 1 and the remaining entries in the first row/column equal to 0.   Direct computation gives that $U_0^{-1} (V^{-1} A V) U_0 = (V U_0)^{-1} A (V U_0)$ is upper triangular.

Some other results that follow pretty quickly from Schur's theorem are the spectral theorem for Hermitian matrices (every Hermitian matrix has a full set of orthonormal eigenvectors w/real eigenvalues) and the Cayley Hamilton theorem.

To kevinatcausa, thanks for your proof via Schur's theorem, I still haven't understood linear algebra, so I always can't use things I have learnt.. :( By the way, may I know how can I use Schur's theorem to prove Cayley-Hamilton?

To perfect_radio, Basically jordan form is a matrix of this form
\[ \left( {\begin{array}{*{20}c} {A_1 } & O & O \\ O & {..} & O \\ O & O & {A_n } \\ \end{array}} \right)  \]
with each $A_i$ (called a jordan block) is something like this
\[ \left( {\begin{array}{*{20}c} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \\ \end{array}} \right)  \]
(eigenvalue on the diagonal, 1s on a "sub-diagonal")
This form always exists for any linear transformation whose characteristic polynomial splits, that means, there exists a basis (called Jordan canonical basis), whose matrix representation is a Jordan canonical form.

there is a proof of cayley-hamilton via schur's theorem in "matrix analysis and applied linear algebra" by carl d. meyer. (unfortunately, i'm too lazy to write it... i'm struggling with the proof of schur's theorem right now) i have a pdf with the book, and if you can't find it, i'll send it to you. there are many interesting things in it

I think to use Schur's theorem we need the fact that the characteristic polynomial of the linear transformation splits. However, I don't think Cayley Hamilton needs this condition. Does the proof via Schur's theorem prove the special case that the char. poly. splits?

[quote="perfect_radio"]what do you mean by "splitting a polynomial"? does it refer to factorization? if so, in the proof eigenvalues are used[/quote]

a n-degree polynomial splits means that it can be factored into n linear factors

my point is the same. Without any assumption of the characteristic polynomial (or the underlying field is algebraically closed), you cannot use schur's theorem. Otherwise this should be easy as the vector space can be decomposed into the direct sum of characteristic subspaces. (generalized eigenspace)

aren't you using strong theorems for an easy thing?
i mean...
let $A$ be an $n\times n$ matrix on a field $F$, and let $\Lambda(A) = \{\lambda_1, \dots, \lambda_n\}\in \bar{F}$ be its spectrum ($\bar{F}$ is algebraic closure of $F$).
by induction, eigenvalues of $A^k$ are ${\lambda_1}^k, \dots, {\lambda_n}^k$: just take an eigenvector $v_i$ relative to $\lambda_i$ (in ${\bar{F}^n}$), and notice that $A^kv_i = A^{k-1}(Av_i) = \lambda_i A^{k-1}v_i$, so we're done...
then, $\textbar{tr }B = \sum \Lambda(B)$ (well known fact).
so you have that $\sum_k \lambda_k^i = 0$ for all $i = 1,\dots,n$, so you have $n$ symmetric equations in $n$ variables. since $(0, \dots, 0)$ is one solution, and the others are its permutation, you have the only solution, so matrix is nihilpotent, and its index of nihilpotence has to be at most $n$.

[quote="ma_go"]so you have that $\sum_k \lambda_k^i = 0$ for all $i = 1,\dots,n$, so you have $n$ symmetric equations in $n$ variables.[/quote]

are you sure that we have $n$ variables? we can have $\lambda_i = \lambda_j$ for $i \neq j$. in that case we consider only different eigenvalues. to solve the new system we must know that the multiplicity stays the same. can you prove it without schur?

(sorry if i said smth stupid above)

you're right...
maybe jordan form is the best way to solve this kind of problems...

You're right that we need the fundamental theorem of algebra to hold to guarantee Schur's theorem (we need it in order to be sure we can pick an eigenvector for $A$ to use in our $V$).  However, if whatever field we're working over isn't algebraically closed we can just prove Cayley-Hamilton for the closure of that field, which automatically implies it for the original field.

[quote="alekk"]beware, because the result is not true in arbitrary fields. Take for example $K=\frac{Z}{7Z}$ and the matrix $M \in M_8(K)$ defined by $M=Id-E_{8,8}$.[/quote]

which result?

how do we prove that the vectors we obtain through Gram-Schmidt are orthogonal?

Consider the Gram-Schmidt process is running so that $v_1,...,v_k$ has undergone the process and they are normalized and $v_{k+1}$ is having the process, now we show for all $1 \le i \le k$, $<v_{k+1},v_i> = 0$, here I suppose linearity in first component and antilinearity in second component.

After gram-schmidt without normalization, $v_{k+1}$ becomes $v = v_{k+1} - <v_{k+1},v_1>v_1 - ... -<v_{k+1},v_k>v_k$

$<v,v_i> = <v_{k+1},v_i> - \sum_{m= 1}^k <v_{k+1},v_m><v_m,v_i>$
which equals $<v_{k+1},v_i> - <v_{k+1},v_i><v_i,v_i> = 0$

[quote="perfect_radio"]thanks, Soarer.

it seems that I was making a very stupid calculation error  :blush:

ps: what do you mean by antilinear in the second component?[/quote]

$<u,v+w> = <u,v> + <u,w>$
$<u,cv> = c^{-}<u,v>$ (I mean conjugate of c, but i don't know how to type it)

$\langle u,cv\rangle=\bar{c}\langle u,v\rangle$

\langle and \rangle look better than < and >, and that's \bar{c}.

i also use \overline{c} \left< u,v \right>

$\overline{c} \left< u,v \right>$


Can someone explain to me step-by-step how to solve this question. Eg. How to combine the statements to get the final answer etc. etc.

If Dr.Peter climbs the volcano, then Dr. Karston will collect soil samples. If Dr. Leary packs his truk, then Dr. Karston will collect soil samples.

The answer the book provides- If Dr. Karston does not collect soil samples, then Dr. Peters did not climb the volcano and Dr. Leary did not pack his trunk.

Another one-

If Alex watches an action movie this week, then he will watch a comedy next week. If Alex watches an action movie this week, then he will watch a drama next week.

[quote="Bugi"][quote="vip747"]idk if this helps but

If Dr. Karston will collect soil samples:
Dr.Peter climbs the volcano and Dr. Leary packs his trunk

The answer the book provides- If Dr. Karston does not collect soil samples, then Dr. Peters did not climb the volcano and Dr. Leary did not pack his trunk.

If Alex watches an action movie this week:
He will watch a comedy next week and he will watch a drama next week.

 :ninja:[/quote]

I can't agree with you in first question.
If Dr.Karston collects soil samples, then Dr.Peter climbs the volcano [b]or[/b] Dr.Leary packs his trunk [b]or[/b] they both do their jobs.[/quote]

Dr. Peter climbs the volcano = P
Dr. Kartson will collect soil samples = K
Dr. Leary packs his truck = L

$ P\implies K$ and $ L \implies K$

This doesn't mean $ K\implies P or L$

if $ P\implies K$ is true, $ \sim K \implies \sim P$ is also true.
similary, $ \sim K \implies \sim L$ is also true.

therefore, $ \sim K \implies \sim P and \sim L$

answer to #1 : if Dr. Kartson doesn't collect soil samples, Dr. Peter didn't climb the volcano and Dr. Leary didn't pack his truck.

[quote="Textangle"]If Alan does not watch a comedy and a drama next week, then he did not watch an action movie this week.[/quote]

A = Alan watches action movie this week
C = Alan watches comedy movie next week
D = Alan watches Drama movie next week

$ A \implies C$ and $ A \implies D$

your saying $ \sim C and \sim D \implies \sim A$, which is also right.

but simpler; you can just say $ A \implies C or D$.

answer to #2 : Alan watches comedy or drama movie next week if alan watches action movie this week


AMC AIME geometry rectangle calculus integration algebra

Source: 1997 AIME #15

The sides of rectangle ABCD have lengths 10 and 11.  An equilateral triangle is drawn so that no point of the triangle lies outside of ABCD.  The maximum possible area of such a triangle can be written in the form p*:sqrt:(q) - r, where p, q, and r are positive integers, and q is not divisible by the square of any prime number.  Find p + q + r.

This is a tough one.  A simple suggestion:  [hide]make sure what you think is the maximal case really is the maximal case.[/hide]

A more helpful suggestion: [hide]angles are better than sides.[/hide]

Perhaps an even better hint:

[hide]The maximum area of such a triangle is on a slant...[/hide]

[color=cyan]That hint wasn't better than my second hint.  :-[/color][color=red]p[/color][color=cyan]

Sorry, that was silly.  Someone who didn't do this problem over the summer say something . . .

Anyone want to try to generalize this, by the way?[/color]

im not sure about p,q, r, but is the area [hide] about 52.78 ? [/hide]  i dont know the exact answer (had to use calculator to solve) and therefore cant get p, q, r.  Probablly wrong anyways, it was too messy.

But i dont understand your hint Joel?

Don't use calculators on AIME problems please.

cats, your answer is almost certainly correct -- now you have to figure out how to do it without a calculator.

As for my hint:  [hide]one approach would consider lengths as extra variables (in addition to the side of the triangle).  It doesn't work.  Considering an angle instead will work, however.[/hide]

This is the first time i've seen this problem, I will attempt it now.

wow, it was lots of work... and im glad i figured out a way:

[hide]

221:rt3:-330

sum = 554

[/hide]

im lazy!!!!

[hide]

ok ok we have rect ABCD, and triangle AEF, where E lies on BC and F lies on CD.. just trying to give a visual picture. This will maximize area, lets solve for sides.

letting:

AB = 10

BE = x

EC = 11-x

CF = 10-y

FD = y

AD = 11

we can use pythagorian theorem 3 times to find the sides of our triangle... they are: AF=:sqrt:(121+y^2), AE=:sqrt:(100+x^2), EF=:sqrt:((11-x)^2+(10-y)^2)...

we want to make this equilateral so AF=AE=EF, we can solve algebraically for X and Y.... (this would require more typing than i'll do)

we get x = -10:rt3:+22    y=-11:rt3:+20

so all sides are 2:sqrt:(221-110:rt3:)

well, u guys know how to find the height, its :rt3::sqrt:(221-110:rt3:)

1/2bh = :rt3::sqrt:(221-110:rt3:):sqrt:(221-110:rt3:)/2

which equals = 221:rt3:-330

sum = 554

[/hide]

Very nice.  You just solved the (hypothetically, theoretically, choose one) hardest problem on the 1997 AIME.  Oh, by the way, you are [hide]correct[/hide].

I'm still wondering why no one's asked about my nomenclature.

Someone did on another question, and you said there was no connection. Is this one different? You probably solved this problem when you were supposed to be studying bugs, didn't you?

is the aime scored like the amc? 6 points per answer right and 2 per blank????

Or do you get like 10 points per answer right and nothing for blank or wrong.

[url=http://www.unl.edu/amc/e-exams/e7-aime/factoidsaime.html]Look here[/url]

...kind of weird all answers are always integral and it's between 0 and 999...I mean, we'll know if it's right or wrong just by the domain, but it probably doesn't matter because of it's difficulty...

I just thought it was random spam titles :) , I mean I'd understand if it was named "Bacteriological Challenge" and it was about exponential functions, but other than that I have no clue.

[quote="mathfanatic"]Yes, there is still no connection between the word in the title and the problem.  However, there is a pattern...you will either bang your head on the table or slap me when I tell you it...

And Tare, no, we won't know if something is necessarily [i]right[/i], but we will know if an answer is wrong (that is, if it's non-integral).  Also, the AIME can have non-integral answers, like 7pi*3*sqrt(2), but it would tell you to find the sum of 7+3+2 by saying "find p+q+r when the answer is p*pi*q*sqrt(r).  The answer you put into the fill-in bubbles may be integral, but the actual answer to the question doesn't have to be, which forces you to do the problem elegantly instead of trying to guess it.[/quote]

I KNOW THE PATTERN!!!!!

sometimes on the Aime, it helps to know what form your answer is in. b/c they will tell you the answer is of the form p:pi::sqrt:(q)/r, or something like that, and that helps with the problem

Mmm, I see. I'll prefer the "banging head on desk" punishment...odd it was also in my today's school newspaper article called "Stress Reduction Kit" :lol:

I know this may be sort of a lot of typing to ask, but can someone please do the algebraic part for me? Im having a lot of trouble with this and I keep just circling around. Quite aggravating too. Thanx in advance


logarithms
algebra
polynomial
quadratics
function
domain
geometric sequence

Ok, so, not counting stuff i couldnt solve, from careless errors, i missed a one pointer and a 3 pointer (sadly it was mass points and i was just being dumb...it was like the only 3 pt problem I knew immediately how to solve). I suppose it could have been worse, though, so I guess I'll be content and put average :|

edit: on a slightly off topic question, can anyone solve the last 2 on national level mandelbrot?

here is what I remember of it:

find the smallest $ n$ such that $ 201^n \minus{} 152^n$ is divisible by $ 2009$

find the only irrational $ x$ such that $ \log \frac{x\plus{}2}{x \plus{} 1}, \log \frac{x \plus{} 4}{x \plus{} 3}, \log \frac{x \plus{} 8}{x \plus{} 7}$ are in geometric progression in that order

I admit that my memory is pretty horrible and I probably did not remember those questions right

meh. can't solve those (in regional). i got 11-11-9-9-14...which i guess, isn't TOO terrible for an eighth grader...right? last round failed.

[quote="hurdler"]
find the smallest $ n$ such that $ 201^n \minus{} 152^n$ is divisible by $ 2009$[/quote]

$ 2009\equal{}7^2*41$, so $ 2009$ divides iff 41 and 49 both do. Modulo 49, we get $ 5^n\equiv5^n$, so 49 always divides it. Modulo 41, we get $ (\minus{}4)^n\equiv(\minus{}12)^n$, or dividing by $ (\minus{}4)^n$, which we can do since it's relatively prime to 41, $ 3^n\equiv1$. Writing out (or just knowing) small powers of 3, $ 3^4\equal{}81\equiv\minus{}1$, so $ 3^8\equiv1$. You can easily verify by brute force that nothing smaller than 8 works, or just check the $ n$ which divide 40. I thought it was fairly straightforward compared to a usual #7 national.

I ended up with a 41 (11,9,13,8) over the last 4 contests on national (missed the first one.) I'm quite happy with that, as I made second tier having missed a competition, but I sorta messed up the last one.

I don't remember seeing your second question, however I remember there being the only irrational x such that $ \log(x\plus{}1),\log(3x\plus{}4),\log(7x\plus{}8)$ are in arithmetic progression in that order. An outline to this problem is algebra everything out to get a polynomial (cubic, or quartic I forget) then factor out the trivial roots (0,1 or something) and get a quadratic, which you solve, and get one answer which isn't in the domain of the original, and another which works.

ok so like i seem to failing the algebra:

the arguments are in geo progression so like we have \[ \frac{x \plus{} 1}{3x \plus{} 4} \equal{} \frac{3x \plus{} 4}{7x \plus{}  8} \Longleftrightarrow 9x^2 \plus{} 24x \plus{} 16 \equal{} 7x^2 \plus{} 15x \plus{} 8 \Longleftrightarrow 2x^2 \plus{} 9x \plus{} 8 \equal{} 0 \Longleftrightarrow x \equal{} \frac{\minus{}9 \pm \sqrt{17}}{4}\] but both of those are less than one and not in the domain?  Or did I set it up wrong, since I did not get a cubic or quartic....

[quote="hurdler"]ok so like i seem to failing the algebra:

the arguments are in geo progression so like we have
\[ \frac {x \plus{} 1}{3x \plus{} 4} \equal{} \frac {3x \plus{} 4}{7x \plus{} 8} \Longleftrightarrow 9x^2 \plus{} 24x \plus{} 16 \equal{} 7x^2 \plus{} 15x \plus{} 8 \Longleftrightarrow 2x^2 \plus{} 9x \plus{} 8 \equal{} 0 \Longleftrightarrow x \equal{} \frac { \minus{} 9 \pm \sqrt {17}}{4}
\]
but both of those are less than one and not in the domain?  Or did I set it up wrong, since I did not get a cubic or quartic....[/quote]

Wasn't it like something completely different.

like $ log_{2}{x\plus{}2}$, $ log_{4}{3x\plus{}4}$, $ log_{8}{7x\plus{}8}$

oh yea thats rights, it should be x + 2 and with those bases ( i only glanced at it and tried and failed and did not want to algebra it out...i need to work on my algebra skils i guess). Thanks

was the answer $ \frac{\minus{}8 \plus{} 2\sqrt2}{7}$?

Yes it was.

I ended up with 67 on National, 14-11-14-14-14

I did regional mandelbrot, since this was the first time my school did it.

i ended up with 14-9(phail)-12-14-11=60.

6th in my region. So, average, i guess

yay mandelbrot winner!!!

are top 6 all declared mandelbrot winners?


geometry
3D geometry
modular arithmetic
algebra
factorization
difference of cubes
special factorizations

[quote="ch1n353ch3s54a1l"][quote="kohjhsd"]:arrow: An integer $7^{(3^{527})}-1$ is divisible by integer $3^{m}$. Find the largest such $m$.[/quote]
[hide="Doubtful"]
$7^{(3^{527})}\equiv 1\mod 3^{m}$

$7^{2\cdot 3^{n-1}}\equiv 1 \mod 3^{m}$

I doubt this is right, but I'm thinking:

$3^{527}>2\cdot 3^{n-1}$, so $3^{528-n}>2$ and $\max n=\boxed{527}$?
[/hide][/quote]

The answer is actually
[hide]528[/hide]
which is really close to what you got.

I don't have the official solution though :(

[hide="sketch"]
Use difference of cubes to split $7^{3^{527}}-1$ into 528 factors, the first being 6, the rest being of the form $x^{2}+x+1$ where $x$ is 7 raised to a power of 3 (so $x\equiv 1\pmod{3}$ in all cases). So all 528 factors are divisible by 3, but none divisible by 9, and you get $\boxed{528}$.
[/hide]

I really like ch1n353ch3s54a1l's solution. There's gotta be some way to modify it and make it work.


inequalities
geometry
geometric transformation
inequalities proposed

If $ x,y$ are real numbers, prove the inequality

\[ (x^7 \plus{} y^7)^3 \le (x^3 \plus{} y^3)^7 \]

The generalisation also holds(and are pretty easy to prove):
$ \left ( \sum_{i\equal{}0}^n x_i \right )^a \ge \sum_{i\equal{}0}^n x_i^a$
for $ a > 1, x_i > 0$.
With equality iff at most one of $ x_i$ is positive.

In your example we have $ n\equal{}2$, $ x_1 \equal{} x^3, x_2 \equal{} y^3, a \equal{} \frac{7}{3}$..

Just a note on notation:
We don't have $ 7[14,0] \succ 7[13,1]$. What we do have is $ (14,0) \succ (13,1)$ which implies $ 7[14,0] \ge 7[13,1]$ :)


Suppose that $f : \mathbb R^+ \to \mathbb R^+$ is a decreasing function such that
\[f(x+y)+f(f(x)+f(y))=f(f(x+f(y))+f(y+f(x)), \quad \forall x,y \in \mathbb R^+.\]
Prove that $f(x) = f^{-1}(x).$

Substitute $y$ for $x$ we have:
$f(2x)+f(2f(x))=f(2(f(x+f(x)))$ (1)
replacinf $x$ by $f(x)$ again, we have:
$f(2f(x)+f(2f(f(x)))=f(2f(f(x)+f(f(x))))$ (2)
Let (2) minus (1), we get:
$f(2f(f(x)))-f(2x)=f(2f(f(x)+f(f(x))))-f(2f(x+f(x)))$
If $f(f(x))>x$ by decreasing property of function, then $f(2f(f(x)))<f(2x)$
$\Rightarrow f(2f(f(x)))-f(2x)<0$
$\Rightarrow f(2f(f(x)+f(f(x))))-f(2f(x+f(x)))<0$
$\Rightarrow 2f(f(x)+f(f(x)))-2f(f+f(x)) >0$
$\Rightarrow f(x)+f(f(x))-x-f(x)<0$
$\Rightarrow f(f(x))<0$, that is contradictory with considered case.

if $f(f(x))<x$ has proved similarly

Done :huh:

[quote="feliz"]Did you mean $f(f(x+f(y))+f(y+f(x)))$ in the right side?[/quote]
That's right, the problem as stated in the first post does not even have matching brackets  on the RHS :wink:
(There are $5$ opening brackets and only $4$ closing brackets.)


A tree is located 50 feet east of a fance that runs north to south. Kelly tells her brother Billy that their dog buried Billy's hat a distance of 25 feet from the fence and also 20 feet from the tree. How many locations for the hat are possible?

[hide]
The locus of points 25 feet away from the fence form a line halfway between the fence and the tree.
The locus of points 20 feet away from the tree form a circle with radius 20, centered at the tree.
This radius fails to extend out 25 feet, so there is no intersection.
Thus, both conditions are not met simultaneously, and thus zero positions exist.
[/hide]


geometry
geometric transformation
reflection
function
vector

Let $K,L,M,N$ four given points . Construct a quadrilateral ABCD such that
$K,L,M,N$ are the midpoints of $AB,BC,CD,DA$ respectively .

well denote $O_{S}(A)=B$ as [b]$B$ is the reflection of $A$ wrt point $S$[/b]
then for this problem we have that:
$O_{K}\circ O_{L}\circ O_{M}\circ O_{N}(A)=A$ but trivialy the only constant point of reflection is the center itself so we get that:
$O_{K}\circ O_{L}\circ O_{M}\circ O_{N}(A)=O_{A}(A)=A$
so its sufficient to find $f(X)=O_{K}\circ O_{L}\circ O_{M}\circ O_{N}(X)$ (its a reflection with center $A$)...

there are many ways to find it,two of them are said below:
[hide]
$1)$we can give the function $f$ an arbitary point $X$ to get $f(X)=X'$,then as said above we know that $O_{A}(X)=X'$ so $A$ is the midpoint of segment $XX'$ so we can find $A$ and then the other vertices...
$2)$we can find $O_{M}\circ O_{N}$ and then find $O_{K}\circ O_{L}$ and etc.[/hide]

Well, you can still do [b]vectors[/b] pretty easily with constructions. So we have that $A+B=2K$, $B+C=2L$, $C+D=2M$, $D+A=2N$.

Take $A=A'$ arbitrarily. Then reflect that over $K$ to get $B'=2K-A'$, reflect that over $L$ to get $C'=2L-2K+A'$, reflect that over $M$ to get $D'=2M-2L+2K-A'$, reflect that over $N$ to get $A''=2N-2M+2L-2K+A'$. However, $K+M=L+N=A+B+C+D$, so $A''=A'$. Hence we can take any point for $A$ and follow that construction.

[It is necessary that $K+M=L+N$ for the construction to be possible. In other words, $KLMN$ is a parallelogram.]

Nice .What about constructing an 7-gonon from its midpoints ? :wink:


Two grade seven students were allowed to enter a chess tournament otherwise composed of grade eight students .Each contestant played once with each other and received one point for a win , one half point for a tie and zero for a loss. The two grade seven students together gained a total of  eight points and  each grade eight student scored the same number of points as his classmates. How many students from grade eight  participated in the chess tournament? Is the solution unique?

$ n\equal{}$ the number of students from grade eight
$ p\equal{}$ the score of each of the $ n$ students from grade eight (notice that $ 2p$ is a natural number)
There are $ \frac{n(n\plus{}1)}{2}$ matches with a total of $ 1$ per match.
Then $ 8\plus{}p n\equal{}\frac{n(n\plus{}1)}{2}$ $ 2p n\equal{}n(n\plus{}1)\minus{}16$ then $ n|16$
On the other hand from the total of points from the matches between those $ n$ students
$ \frac{n(n\minus{}1)}{2}\leq  p n$ $ n(n\minus{}1)\leq  2p n$
$ n^{2}\minus{}n\leq  n^{2}\plus{}n\minus{}16$ then $ 16\leq  n$
so $ n\equal{}16$ and $ p\equal{}8$

mszew, there are at least 2 errors in your solution.  The first is simply that a total of $ \binom{n\plus{}2}{2}\equal{}\frac{(n\plus{}2)(n\plus{}1)}{2}$ matches were played, not $ \binom{n\plus{}1}{2}$.  Thus we have first that $ 16\plus{}2pn \equal{} n^{2}\plus{}3n\plus{}2$ or $ n^{2}\plus{}(3\minus{}2p)n \equal{} 14$, so $ n\mid 14$.  We can also write from this equation that $ 2pn \equal{} n^{2}\plus{}3n\minus{}14$.
For the other side, we still have $ \binom{n}{2}\leq pn$, so $ n^{2}\minus{}n\leq 2pn$.  Then $ n^{2}\minus{}n\leq n^{2}\plus{}3n\minus{}14$ and so $ 7\leq 2n$.  This leaves the possibilities $ n \equal{} 7, n \equal{} 14$, with associated values $ p \equal{} 4$ and $ p \equal{} 8$.  The first is clearly possible (everyone ties everyone else); I assume the second is possible, as well, but I haven't constructed an example.

Each grade-7 student loses 7 grade-8 students and ties with everyone else
Each grade-8 student wins 1 grade-7 student and ties with everyone else


How do I install Adobe Premiere on my SuSE linux 10.1? :maybe:

Contact Adobe for product support. Locked.


The interior angles of a quadrilateral form an arithmetic sequence. If the measure of the largest angle is $ 129^{\circ}$, what is the measure of the second largest angle, in degrees?

largest angle + smallest angle should be 180.

so the measure of smallest angle is 51.

x+3y=129, x=51.
so 3y=78, and y=26.

so the measure of second largest angle is 129-26=103.

answer ; 103


\[ {\rm{Prove: }} \& \frac{{\cot \theta  + {\rm{cosec}}\theta  - 1}}{{\cot \theta  - {\rm{cosec}}\theta  + 1}} = \frac{{1 + \cos \theta }}{{\sin \theta }}
\]
tan}}^2 \theta  + 3{\rm{ tan}}^4 \theta
\]

$ \Leftrightarrow sin\theta(cot\theta\plus{}cosec\theta\minus{}1)\equal{}(1\plus{}cos\theta)(cot\theta\minus{}cosec\theta\plus{}1)$
$ \Leftrightarrow sin\theta(cos\theta\plus{}1\minus{}sin\theta)\equal{}(1\plus{}cos\theta)(cos\theta\minus{}1\plus{}sin\theta)$
$ \Leftrightarrow sin\theta(cos\theta\plus{}1)\plus{}(cos^2\theta\minus{}1)\equal{}(1\plus{}cos\theta)(cos\theta\minus{}1\plus{}sin\theta)$
It is OK!


calculus
derivative
limit
real analysis
real analysis unsolved

f:[a,b]->R is continuous on [a,b] and differentiable on (a,b). Show $ \lim_{x \rightarrow\ a}f'(x)\equal{}A$, then f'(a) exists and equals A.

The hint we got was to use the definition of f'(a) and the Mean Value theorem, but I"m still having no luck. Can someone please help?

By definition,
$ f'(a)\equal{}\lim_{h\rightarrow 0}\frac{f(a\plus{}h)\plus{}f(a)}{h}$.
For $ h$ sufficiently small, $ f$ will be continuous on $ [a,a\plus{}h]$ and differentiable on $ (a,a\plus{}h)$.  By the MVT, there is a number $ \alpha_{h}\in (a,a\plus{}h)$ such that
$ \frac{f(a\plus{}h)\minus{}f(a)}{h}\equal{}f'(\alpha_{h})$
Now $ \alpha_{h}\rightarrow a$ as $ h\rightarrow 0$.  Then it follows that
$ f'(a)\equal{}\lim_{h\rightarrow 0}\frac{f(a\plus{}h)\minus{}f(a)}{h}\equal{}\lim_{h\rightarrow 0}f'(\alpha_{h})\equal{}\lim_{x\rightarrow a}f'(x)$.


hello, it is $ \binom{10}{8}\equal{}\binom{10}{2}\equal{}\frac{10\cdot9}{1\cdot2}\equal{}45$.
Sonnhard.

[hide=sol]plugging in the values of the formula $\binom{n}{m}=\binom{\text{number of values for  }n-m}{(n-m)!}$ will give us $\frac{10*9}{2*1}=45$[/hide]


algebra
polynomial
real analysis
real analysis unsolved

Prove that the sequence x_n=sin n +sin (n^2) doesn't have a limit. Can anyone help me, please?
  Source: My thoughts

We have to look at n and n 2 "modulo 2 \pi ". We can show that there are infinitely many n's for which n and n 2 (mod 2 \pi ) >  \pi , so sin(n)+sin(n 2 ) <0, infinitely many n's for which n and n 2 (mod 2 \pi ) <  \pi , so sin(n)+sin(n 2 ) >0, and from these 2 we find that the limit should be 0 if it exists. But we can also find infinitely many n s.t. n (mod 2 \pi ) is really close to  \pi /2 and n 2 (mod 2 \pi ) <  \pi , so sin(n)+sin(n 2 ) >1- \varepsilon for  \varepsilon >0 as small as we want, so the limit couldn't be 0.

I could post the details (don't really have that much time), so tell me if that's necessary.

I also found that the limit should be 0, but I don't understand why can we find infinitely many n with (n mod 2pi) close to pi/2 and (n^2 mod 2pi)<pi

I'm not really sure, now that you mention it :D Let's see.

We have n as close as we want to 2k \pi + \pi /2 (because 2 \pi is irrational, and there's Jacobi's thm which states this). This means that n 2 is close to 4k 2 * \pi  2 +2k \pi  2 + \pi  2 /4 = (4k 2 +2k+1/4) \pi  2 Let's try to see if this can be made to be < \pi (mod 2 \pi ).

P.S. Not even my argument for showing that the limit should be 0 work! At least, I'm not sure! :D Really sorry for posting rubbish here! Back to the drawing board! :)

The fact that the limit is 0 can be deduced more easy:
  Let n_k a sequence such that sin n_k thends to -1 when k tends to infinity. Since sin n_k+sin (n_k)^2<=1+sin n_k, the limit is at most 0. Similarly, the limit is at least 0, so it is 0. But I have no idea of how to continue.

We have (4k 2 +2k+1/4) \pi  2 =
(2k 2 +k+1/8)* \pi *2 \pi . In order to achieve what we want we should have {(2k 2 +k+1/8)* \pi } as small as possible.

Can't we prove that this set is dense in (0,1)? Actually, all we need is to prove that that fractional part above is <1/2 for infinitely many k, but it would really help to prove that it's also dense in (0,1).

Maybe you have a different approach, Harazi?

Unfortunately, I don't have. But I think there is a theorem of Weyl about the density of {p(n)} with p polynomial, but I'm not sure.

Ok, I'm back with a clearer mind! :D Let's take L=lim{n->00}sin(n)+sin(n 2 ).

(i) If the limit exists, then it must be 0.

Proof:

Assume that there are infinitely many n s.t. n (mod 2 \pi ) and n 2 (mod 2 \pi ) < \pi and infinitely many n s.t. n (mod 2 \pi ) and n 2 (mod 2 \pi ) > \pi. This means that there are infinitely many positive terms of the sequence, and infinitely many negative terms, so the limit must be 0.

Assume that one of the 2 conditions is not satisfied. Then we can assume WLOG that for n>=n_0 whenever n (mod 2 \pi )< \pi we have n 2 (mod 2 \pi )> \pi . We can find infinitely many n>=n_0 s.t. n(mod 2 \pi ) is in ( \pi - \varepsilon , \pi ), with  \varepsilon as small as we want, and for these n we have n 2 (mod 2 \pi )> \pi , so sin(n) is >0 but really close to 0, and sin(n 2 ) is <0, so we can find infinitely many terms of the sequence which are <d, with d as small as we want, d>0, so the limit must be <=0. Let's assume L<0.

We can also find infinitely many n>=n_0 s.t. n(mod 2 \pi ) is really close to  \pi /2, so for these n we have sin(n) really close to 1, let's say we have infinitely many n>=n_0 s.t. 1>sin(n)>1+L/2, so sin(n)+sin(n 2 )>1+L/2 -1=L/2 for all sufficiently large n, so the limit can't be L<0, and since we know L<=0, we get L=0.

So, in both cases, L must be 0. I hope that's clear now. I'll get back with the rest. I'm off browsing the rest of the forum, now. :D

I should continue (I had forgotten :) ).

So we know sin(n)+sin(n 2 0->0 as n->00. This means that for n(mod 2 \pi ) really close to 0 and n sufficiently large, n 2 (mod 2  \pi ) should also be very close to 0 or  \pi because sin(n) is close to 0 and for n large enough so must be sin(n 2 ). This means that for n large enough and very close to 2k \pi with integer k, n 2 must also be very close to t \pi , with an integer t. It's easy to see that this can't happen.

I know this isn't a solution, because it not the least bit rigorous, but it can be translated in " \varepsilon language" :) At least I hope so..

[quote="harazi"]I think there is a theorem of Weyl about the density of {p(n)} with p polynomial, but I'm not sure.[/quote]

There is indeed a thm of Weyl which states that for a polynomial P with at least an irrational coefficient the sequence {P(n)} is equally distributed in (0,1). I saw this in some papers from some preparation camp I went to 2 years ago, but I don't have a proof, and my intuition tells me it's troublesome :D


probability
expected value
algebra unsolved
algebra

This is equivalent to the fact that a binomially distributed random variable taking values $\{0, 1, \ldots, n\}$, with expected value $np$, has variance $np(1-p)$.    :lol:

Using (mostly) your notation,
\begin{eqnarray*} \sum k{n\choose k}x^k(1-x)^{n-k}&=&nx\sum {n-1\choose k-1}x^{k-1}(1-x)^{n-k}\\ &=&nx \end{eqnarray*}
\begin{eqnarray*} \sum k^2{n\choose k}x^k(1-x)^{n-k}&=&nx\sum k{n-1\choose k-1}x^{k-1}(1-x)^{n-k}\\ &=&nx\sum (k-1){n-1\choose k-1}x^{k-1}(1-x)^{n-k}+nx\sum {n-1\choose k-1}x^{k-1}(1-x)^{n-k}\\ &=&n(n-1)x^2\sum {n-2\choose k-2}x^{k-2}(1-x)^{n-k}+nx\\ &=&n(n-1)x^2+nx \end{eqnarray*}
\begin{eqnarray*} \sum (k-nx)^2{n\choose k}x^k(1-x)^{n-k}&=&\sum k^2{n\choose k}x^k(1-x)^{n-k}-2nx\sum k{n\choose k}x^k(1-x)^{n-k}+(nx)^2\sum {n\choose k}x^k(1-x)^{n-k}\\ &=&n(n-1)x^2+nx-2nx(nx)+(nx)^2\\ &=&nx(1-x) \end{eqnarray*}

If you're interested in a similar problem, try proving that the variance of a random variable with Poisson distribution (one which takes on values $\{0, 1, 2, 3,\ldots\}$ with probability $P(X=k)=\frac{\lambda ^ke^{-\lambda}}{k!}$, for some $\lambda$) is $\lambda$.


geometry
geometric transformation
reflection

Be the acutangulo triangle ABC, in the cevian BD is designed such that AB = DC; m (CAB) = 48 º;  m (ACB) = 18 º. Find m(DBC)

[quote="Nawahd Werdna"][hide="Answer:"]
$ 12^{\circ}$[/hide]

My solution is very ugly, and requires a calculator... so I won't post it. :D[/quote]

THANKS

Solution by my son:
Take $A'$ on $[BC]$ so that $A'B=AB$. It implies $\angle A'BC=30^\circ$.
Take $A"$ the reflection of $A'$ in $BC\implies\triangle A'BA"$ equilateral.
Take $D'$ in $[A'C]$ so that $A"D'=A'A"$ $\implies \angle A'D'A"=\angle CA'A"=72^\circ$ $\implies \angle CA"D'=\angle A"CD'=36^\circ\iff CD'=A"D'=...=AB$. Consequently $D'\equiv D$ and, by easy angle chasing we get $\angle DBC=12^\circ$.

Best regards,
sunken rock

Oops! It seems $A'\in [AC]$!.
I do apologize for the mistake.

Best regards,
sunken rock

See details in the book here: https://www.facebook.com/photo.php?fbid=2777142695699952&set=pcb.2777143602366528&type=3&theater

Best regards,
sunken rock


I have use the folliwng command for the header
\thispagestyle{myheadings}
\markright{\textbf{\textit{MCS 4}} \hspace{95mm} \textit{\textbf{Lisbon, October 6-10 2005}}}
But his leads to page number in the first page.
How can I avoid that.
Pratap

to create better headers/footers, I use

\usepackage{fancyhdr}

\pagestyle{fancy}

\lhead{\slshape \rightmark}
\chead{}
\rhead{\thepage}
\lfoot{}
\cfoot{}
\rfoot{}

If you don't want the page number, just take off the "\thepage".  Naturally, "lhead" stands for left header, "chead" for center header, and so on.  You can also put pictures in the page style.

ledurt


More practice, the better!

Seriously, I bombed the A and B tests; I am interested doing the C test.
I hope the C test is as hard as or harder than the A and B (so it would be a good practice)

Since AIME is tomorrow (July 31), and I can't do the C test before that, so how about next week(starting August 1)

sure why not.  i'll do it this afternoon after 3

send it to ndkvjbn@hotmail.com

or send it to xXnosepuppyXx over aim

thanks

[quote="beta"]I wrote a Mock AMC, anyone interested in doing it? If you are, please tell me when you can do it also![/quote]

Well, the mock AIME is tomarrow... I might be able to do it today at... oh... 6 PM EST?

I dunno what mock amc is, but if it's a competition like IMO with beautiful questions, I'm your man :D

[quote="beta"]this mock AMC will be after the mock AIME, I just finished writing it today[/quote]

Oh well.  Strangely, when I do AMC 12's at home, I ALWAYS get a score of 104.5.  But when I do a Mock AMC, I get 80-90.  Maybe it's the pressure of the competition.  The more practice the better!

[quote="beta"]How about next Tuesday, August 3rd at 4:00 to 5:15pm EST?[/quote]

You got it man.

I'd like to join in maybe I can do better than my 49.5 on the last Mock AMC.
4:00-5:15 EST would be fine with me.

[quote="beta"]How about next Tuesday, August 3rd at 4:00 to 5:15pm EST?[/quote]

I want to do this too.

But I'll need the questions sent to me just like a half an hour before.  Is that possible?

Yes, I will PM you the questions 30 minutes earlier and you send the answers to me 30 minutes earlier.

It'a a combination. It's AMC12 level, about the same level of difficulty as yours, last few questions are harder though.

I guess I will try this. Could you allow me to take this a little earlier in the morning or noon - around 12:00? I have camp in the afternoon.

count me in...i'll fail miserably, but that's ok...

omg why do all the mock amc's have to be when i'm at math camp? is there a mock amc D? probably not...

Have you looked at the Mock Contests thread?  A lot of people are trying to make two AMC's a month and stuff.

Somehow, I felt this one was easier than the previous 2. I just took it.

The average of people who had taken it so far is 116.1. I am going to say that this is easier than the other two.

I'm in...but is it over already?

-interesting_move

No, the test will be posted in 25 minutes [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14764]here[/url]


This is the second part of the preview ...

-------------
Let $x,y,z >0$ real numbers, such that $xyz=1$.
Prove that:
\[ 1 \leq \frac{1}{1+ 2x} + \frac{1}{1+ 2y} + \frac{1}{1+ 2z}. \]

Please ... put the references about it here...

Carlos Bravo :)
Lima-PERU

This one is easy too, well at least the same trick works out. Now, put $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$ and the inequality becomes the following one: $\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{2a+c}\geq 1$ and from here apply Cauchy-Schwarz and it's over. ;)

Let $X=\frac{1}{x}$, $Y=\frac{1}{y}$ and $Z=\frac{1}{z}$. Then, $XYZ=\frac{1}{x}\cdot\frac{1}{y}\cdot\frac{1}{z}=\frac{1}{xyz}=\frac11=1$. Hence, applying the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=55741 to the numbers X, Y, Z, we get $\frac{1}{X+2}+\frac{1}{Y+2}+\frac{1}{Z+2}\leq 1$.

Now, $\frac{1}{1+2x}=\frac{\left(\frac{1}{x}\right)}{\frac{1}{x}+2}=\frac{X}{X+2}=1-\frac{2}{X+2}=1-2\frac{1}{X+2}$. Similarly, $\frac{1}{1+2y}=1-2\frac{1}{Y+2}$ and $\frac{1}{1+2z}=1-2\frac{1}{Z+2}$. Hence,

$\frac{1}{1+2x} + \frac{1}{1+2y} + \frac{1}{1+2z} = \left(1-2\frac{1}{X+2}\right)+\left(1-2\frac{1}{Y+2}\right)+\left(1-2\frac{1}{Z+2}\right)$
$=3-2\underbrace{\left(\frac{1}{X+2}+\frac{1}{Y+2}+\frac{1}{Z+2}\right)}_{\leq 1}\geq 3-2\cdot 1=1$,

and we are done.

  darij

we can also develop all and arrive a simple inequality wich is:

ab+bc+ac>=3 :D

Great solution's cezar lupu
and Darij .... :)

I found it in one training paper (Peru - 2000)

Carlos Bravo :)
Lima-Peru


Euler
number theory unsolved
number theory

Determine all pairs $ (n,p)$ where $ n\in\mathbb{N}$ and $ p$ is a prime such that the following condition is satisfied:
\[ n^{p\minus{}1}|(p\minus{}1)^{n}\plus{}1\,.\]

[hide]

We will prove that $ n\equal{}1$, $ n\equal{}2$, and $ n\equal{}3$ are only the eligible answers. Suppose contrary that there are other solutions. We choose a solution for which $ n$ is minimum. Keep in mind that $ n>3$.

Let $ d \equal{}\gcd(n,\phi(n))$, where $ \phi$ is the Euler totient function. With some work, which I don't show here, we have $ n$ divides $ (p\minus{}1)^{d}\plus{}1$. Therefore, $ k: \equal{}n/d$ is odd and thus,
\[ x^{k}\plus{}1 \equal{} (x\plus{}1)(x^{k\minus{}1}\minus{}x^{k\minus{}2}\plus{}...\minus{}x\plus{}1)\,.\]
For $ x \equal{} (p\minus{}1)^{d}\plus{}1$, we get that $ n$ divides $ x\plus{}1$ and $ n^{p\minus{}1}$ divides $ x^{k}\plus{}1$. Observe that
\[ x^{k\minus{}1}\minus{}x^{k\minus{}2}\plus{}...\minus{}x\plus{}1\equiv k \equal{}\frac{n}{d}(\mathrm{mod}\,n)\,.\]
Thus, $ x^{k}\plus{}1 \equal{} (x\plus{}1)\left(ny\plus{}\frac{n}{d}\right) \equal{}\frac{n}{d}(x\plus{}1)(dy\plus{}1)$ for some $ y$. Note that
\[ n^{p\minus{}1}|(x^{k}\plus{}1)\Rightarrow dn^{p\minus{}1}|n(x\plus{}1)(dy\plus{}1)\Rightarrow dn^{p\minus{}2}|(x\plus{}1)(dy\plus{}1)\,.\]
Because $ d$ divides $ n$, hence, $ d^{p\minus{}1}|(x\plus{}1)(dy\plus{}1)$. Thus, $ d^{p\minus{}1}$ divides $ x\plus{}1 \equal{} (p\minus{}1)^{d}\plus{}1$, as $ \gcd(d,dy\plus{}1)\equal{}1$. Consequently, $ d$ is a solution to the condition which is less than $ n$. Therefore, $ d\equal{}1$, $ d\equal{}2$, or $ d\equal{}3$.

Case 1: $ d\equal{}1$.
$ n$ divides $ (p\minus{}1)^{d}\plus{}1\Rightarrow n|p\Rightarrow n\equal{}1$ or $ n\equal{}p$. Since $ n>3$, then $ n\equal{}p$. Therefore, $ p^{p\minus{}1}$ divides $ (p\minus{}1)^{p}\plus{}1$. Note that $ p>3$, therefore, the largest power of $ p$ that divides $ (p\minus{}1)^{p}\plus{}1$ is $ p^{2}$. This implies $ p^{p\minus{}1}|p^{2}$ or $ p\minus{}1\leq 2$. This is then again a contradiction.

Case 2: $ d\equal{}2$.
$ d$ divides $ (p\minus{}1)^{d}\plus{}1\Rightarrow 2$ divides $ (p\minus{}1)^{2}\plus{}1 \equal{} p^{2}\minus{}2p\plus{}2$.  Thus, $ p\equal{}2\Rightarrow n$ divides $ 2$, a contradiction again.

Case 3: $ d\equal{}3$.
$ d$ divides $ (p\minus{}1)^{d}\plus{}1\Rightarrow 3$ divides $ p^{3}\minus{}3p^{2}\plus{}3p\Rightarrow p\equal{}3$.
Since $ n$ divides $ (p\minus{}1)^{d}\plus{}1$, then $ n$ divides $ 2^{3}\plus{}1 \equal{} 9$. Because $ n>3$, thus, $ n\equal{}9$.
However, $ 9^{2}$ does not divides $ 2^{9}\plus{}1$!

Therefore, the only solutions are
1) $ n\equal{}1$ and $ p$ is arbitrary,
2) $ n\equal{}p\equal{} 2$ or $ 3$. [/hide]

Different solution .

-$ n = 1\Rightarrow p+1-1\vdots 1$ then $ (n,p) = (1,p)$ satisfy the condition.

-$ n\ge 2$.

[u][b]1[/b][/u] $ p = 2$ satisfy.

[u][b]2[/b][/u] $ p\ge 3$

We have $ (p-1)^{n}+1$ is a odd number, then $ n$ also is a odd number $ \Rightarrow n < 2p$.

Let $ q$ is smallest prime divides by $ n$.
$ \Rightarrow q|(p-1)^{n}+1\Rightarrow (p-1)^{n}\equiv-1(\tex{mod q})$ and $ \gcd(q,p-1) = 1$.

Because : $ \gcd(n,q-1) = 1$ so there exists $ u,v\in\mathbb{Z}$ so that $ un+v(q-1) = 1$
$ \Rightarrow p-1\equiv (p-1)^{un}.(p-1)^{v(q-1)}\equiv-1(\tex{mod q})$( because $ u$ is a odd number)
Hence $ q|p\Rightarrow q = p\Rightarrow p|n\Rightarrow n = p$

So $ p^{p-1}|(p-1)^{p}+1$
But $ (p-1)^{p}+1 = p^{2}(pQ+1)$ with $ Q$ is a positive integer.Then $ p\le 3\Rightarrow p = 3\Rightarrow n = p = 3$


What is the greatest number of Sundays that can occur in the first $ 49$ days of a year?

Think of the sequence 1, 8, 15, ... , 43. We find that the greatest number is $ \boxed{6}$.

The sequence 1,8, ...43 is correct. If the first day were Sunday the 50th day would be, too. Subtract one from each number of the set. The sequence is now 0, 7, ...42. From 7 to 42 there are $\frac{42}{7} = 6$ multiples of 7. Add one to the final answer of $\boxed{7}.$

49 days is exactly 7 weeks. In 7 weeks there are 7 Sundays.


Let a_n mean "a with the subset of n".

Given the sequence 1, 2, 4, 7, 11, 16, 22....

Create a general equation for finding the nth term of the sequence.

By "a with the subset of n" you mean "a with an n subscript," right? I see another participant has given an answer, in spoiler. Thanks for contributing a problem to the forum.

h_s_potter2002 may have made a typo--I believe the correct answer is [hide]1 + n(n-1)/2[/hide]

Hint: [hide]subtract 1 from every term--do you recognize the resulting sequence?[/hide]


algebra
polynomial
algorithm
graph theory

A bipartite graph with classes $ A$ and $ B$, $ |A|\equal{}|B|\equal{}n$, is given. We are asked to find (if it exists) $ \varnothing\neq A_1\subset A$ (strict subset) such that all the edges from $ A_1$ lead to at most $ |A_1|$ different elements of $ B$.

What is the complexity of this? Is it polynomial?

Yes. One way is to start with the constructive proof of the Hall matching theorem and either find a subset $ A_1$ such that the edges from it lead to fewer than $ |A_1|$ elements of $ B$ or find a matching between $ A$ and $ B$ after which the problem reduces to finding a subset in an oriented graph with no outbound edges. I suspect that this algorithm is far from optimal but it is already polynomial.


geometry
geometric transformation
\/closed

Is there a part of this forum where i can disscus old and new ToT problems ?

Not sure what you mean by ToT. But if you refer to Tournament of Towns you can post them into the [url=http://www.mathlinks.ro/index.php?f=217]Olympiad Section.[/url] And please put in the source field (subject description) when you are posting Tournament of Towns, Spring/Fall Year, Junior/Senior O-Level/A-Level, Problem X e.g. Tournament of Towns, Fall 2001,  Senior A-Level, Problem 1 as an example. The reason that I want to add those problems to the [url=http://www.mathlinks.ro/resources.php]resources section[/url] later. Tournament of Towns problems are some of the most exciting problem sets around. Thanks!

[quote="Wichking"]Yes I meant Tournament of Towns :lol: .

Another question : Should I really post Junior problems(I mostly solve those, Senior level is a little a bit to hard for me) in Olyimpiad section?[/quote]

Please post all TOT problems in the olympiad section. To improve you should solve problems slightly above your level so senior is fine. Thus post senior problems and we leave junior ones for somebody else.

[quote="Wichking"]I probably didn't wrote that very well (my english is really bad,so i didn't know how too expres it corectly). Junior towns still represent quite a challange to me, and thare aren't many senior problems that i can solve. However I have quite a few senior problems and if you want i can post them. :)[/quote]

Both levels are appreciated! For a few years back you may want to compare your translation with [url=http://www.math.toronto.edu/oz/turgor/archives.php]this archive.[/url] Writing the problems in the community has the advantage to learn math and English at the same time! How cool is that?  :D

[quote="Wichking"]I started posting some of them. Hope you're not just using me as a type machine. :P[/quote]

Don't worry, you are a good typing machine!  :D But seriously, it depends on the community whether people reply to your threads or not. Thus some of your postings will get replies others don't.  :)

[quote="Johan Gunardi"]where do you get those problems?[/quote]

You can find the problems for 1980-2007 on the Russian [url=http://www.turgor.ru/problems.php]Tugor site.[/url] If somebody would help posting those problems we could discuss them on Mathlinks. For 2001 to 2008 you can find them in the English [url=http://www.math.toronto.edu/oz/turgor/archives.php]University Toronto archive.[/url] :)


hi guys

i was wondering,

what is ur favortie game (not vedio game)?

mine is chess

go is a popular board game in asia(not all of asia but w/e)

mine would be chess too

go did sound familiar

anyways, i liev in the us

There are even manga and anime about Go [Hikaru no Go] (But I don't know is there any about chess  :P )

Anyway, I like chess.  :D

[quote="BoesFX"]There are even manga and anime about Go [Hikaru no Go][/quote]

One of my favorite mangas  :lol:

I also like games similar to monopoly

wall ball or hall ball, one of those two games....(you bluntly said games, so I chose ball games)

Well, i think that Go is well knows around the world right now, i live in Brazil and So says the legend, the guys from the Institute of advanced Studies in Princeton plays go all day long!

favortie game...has got to be (if it's a board game) risk, (card game) magic the gathering, because i learned it at camp, and it's not as bad as it sounds. otherwise, hold'em

hold em is pretty fun

this should be a more narrowed topic

*sigh*

Set.  Definitely Set.

But barring that, probably Hold 'Em.

i might have liked bridge if people taught me how to during the trip back from arml....the people thought I was so bad at spades, they kicked me out of the game and didn't teach me bridge :lol:

p.s. hi silas!! I was on your team at arml  :)

Why don't you like ball games?

P.S. I could've gone to ARML, but I was 10 then, so I couldn't.

P.S.S. I was on the first place team in div. B at NYSML

ball games aren't that great....

p.s. they should have arml 3 times a year

by ball games, do u mean baseball or ball games like volleyball, basketball or ball games like tip, wallball, etc

thats a lot of confusion

weird ones. Like Catcher the flag, Hall ball, wall ball, cat ball, back toss, etc. ones that you never imagined.


calculus
derivative
calculus computations

Hello,

I am confused in as how to do this question.... any help, pointers etc.  would be great.

[img]http://img87.imageshack.us/img87/637/picture4nd7.png[/img]

Thanks

sarah :D


I just recieved a letter from the Canada/USA mathcamp. Can anyone tell if this is a good program or not?

The merits of the various programs have been discussed on other threads, but I'll plug Mathcamp - I went last summer and loved it.  You have a TON of freedom, and there are classes for all different interests.  You can go to whatever class(es) you want, and you're not required to be anywhere at any time (except for the weekly assemblies).  There are a lot of other activities, such as relay competitions and proof competitions, and the various night-time activities that are both counseler and camper-planned.  There's no "bed time" or "lights out," although you must sign in between 8 and 10 PM every night, and you're supposed to be out of the lounges by midnight (you can hang out in someone's room, though).  The classes vary in difficulty, and are rated by a star system - one star is the easiest, four stars is the hardest.  All of the people at MC are really nice and cool, and it's interesting to meet people from all different places/backgrounds.

Osiris and zabelman also went last summer (and probably some other people that I haven't noticed around here), and I think are considering returning.

Very funny. I just told my mom about the camp today and she wanted me to go, so she pushed me to solve the problems. I solved 5 problems spending the afternoon, but then she said forget it since it's so expensive($3,195).

But I am going next year!

They have good financial aid.  Apply, see what they'll give you.

Yeah, but I am too lazy to write up all the proofs in 3 days :)

You probably don't need to do all of them.  Do math this summer!  :)

I went to Canada/USA mathcamp last summer and I'm going back. It's really awesome. I highly suggest you turn in an application (with the proofs written!), even if you don't go, for three reasons:

1) You may later change your mind to go even if you have decided now that you dont want to, and if you dont send in an application you're sure not to go. :(

2) From what I've seen on this board, you'll easily be accepted.

3) The camp is awesome!

And as JBL mentioned, they do give financial aid, so dont think of the cost as too much of a deterant. And also, I think they accept late applications, although they do not give them equal consideration. (Im not sure about this, though, so dont take my word on it!)

APPLY! Take my word for it.

yeah it never hurts to try.  Too bad USAMO's coming up :-(.

Wow! This illustrates the difference in channels of information between homeschoolers (who get habituated to looking for their own information about academic programs) and people in the school system (who too often aren't told ANYTHING about even the most worthwhile programs). I knew about most of the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Programs.php]better summer programs[/url] before my oldest son was old enough to do any math contest. (It helped that I was on a homeschooling email list in which Alison's mom posted messages--that helped IMMENSELY.)

Anyway, yes, don't delay, apply as soon as possible to whatever program interests you, and see what happens. And don't let list price deter you (your parents) either. For families that are genuinely in the middle class (as is my family), there is usually enough financial aid to make it possible to stretch the budget to attend. (My son will be off to two programs this summer, [url=http://www.dcu.ie/ctyi/summer/academic/c_desc04.htm#Modern%20Mathematics]one[/url] with a full scholarship, and [url=http://www.jhu.edu/gifted/summer/catalogs/osmath.html]the other[/url] with a partial financial aid award.)

The only reason not to apply is that you really, really, really don't want to have any chance to meet other young people who like math as much as you do over the summer and you really, really, really, don't want to meet any math teachers who might be better than the teachers in your local school. Applying takes some time and effort (and some deadlines have already passed), but the cost of applying is a lot less than the opportunity cost of not applying.

I live in Utah. So obviously since I'm poor, I'm going to the one in Utah. It called Summer Mathematics Program for High School Students. Plus during teh year they offer free Math Circles, and Contests (=prizes). Its only morning and afternoon. But its cheap $500 since I live here. And you get Math 3900 credit at the University of Utah.

Yeah, I just decided to apply a couple days ago. I started the problems last night, and as of right now, I have a total of...3 solved and latexed.  :mrgreen:

Grr...I stop practicing for a few weeks and my mind starts rusting immediately. Bah, back to work.

[quote="tetrahedr0n"]Applied :) Anyone have any idea how much we have to wait after April 29 until we find out? I told my parents about this camp, but they don't think that I will make it, and I'm hoping to prove them wrong this weekend.  :mrgreen:[/quote]

You shouldn't have to wait more than a week or two after the application deadline, especially if you sent it in earlier (though I guess that doesn't make much difference since it's not rolling) :)

[quote]1.3. How many problems from the qualifying quiz do I need to answer?

Most students who are admitted have given at least partially correct solutions to at least six problems on the quiz. This does not mean that once you're done with six problems you should stop working! Not only is there a danger that some of your solutions may be incorrect, but, more importantly, the qualifying quiz is your chance to show us that you are really serious in your desire to explore new mathematics, and are willing to think hard about challenging topics. In general, it is not only your score that matters: if you love math and if this shows through in your application, then you have a good chance of being admitted regardless of your score on the qualifying quiz.[/quote]

From the Mathcamp FAQ.  I am applying too and I did 8, to be on the safe side.

[quote="JSRosen3"]Funny, I just received the same thing today. I wish they'd sent it before now, though--I'd never heard of this camp before, and I'd definitely consider going, but their application (which requires solving mulitple questions w/ proofs) is due in 4 days.[/quote]

Hi.  I'm one of the organizers of Mathcamp.

Yeah, we're a little annoyed at the postal system right now.  The brochure you just received was mailed in March!  Bulk mail is supposed to take two or three weeks, but this time most of it seems to have taken closer to a month.

So you're far from alone in not having gotten your brochure until this week.  If you (or anyone reading this) need a few extra days to get your application done, that should be fine.   We will start working on admissions on the weekend, but we won't get it all done in just a day or two, so if your application is submitted by the end of the weekend you'll be OK.

Maybe I should have read this board first thing when I came home.  I just realized today that the 29th was Thursday, and not Friday.  I was assuming that the 29th was Friday because I know that I have a math contest coming up, which is supposedly scheduled for 'Friday, April 29th'.

Now I am mailing it through UPS overnight delivery so that it arrives tomorrow.

I wanted to apply but knowing that my math skills aren't quite as good as I only was able to solve 2 or 3 questions with maybe 3 thoughts that might have lead into the answer....

Therefore, I decided not to apply since I might not have gotten in anyways.

I don't know if I should have applied and tried or not....what do you guys think?

You will never know unless you send in an application.  What can you lose by trying?  Keep in mind that although the 29th deadline is passed, they still accept applications after that.   And if you don't decide to apply this year, I think you should still work on the quiz, because the questions are pretty fun.

On an unrelated note, did anyone else have this problem?  I sent in my quiz solutions before my online application.  After sending in my online application, it said to write your name and phone # on your quiz.  But I did not know to do that, so I just have my name on the title page for identification.  I hope that this won't be a problem for matching the online + mailed-in materials.

I had the same "time" problem, I received those brochures april 20th, so I had 9 days, but May 1st I was supposed to compete in the State Math Contest along with some other kids from my school.  I had the busiest week in letsee, my life?  It kinda fries your brain when you have to finish all your homework in class (and make sure that the teachers don't see you), run home after school at 4:00 to do math problems till 10 p.m., then go to sleep, then wake up at 6:30 to go to school.  But anyway, I sure hope I got in, I managed to do 7 problems.  If I didn't...that would stink.


$x,y$are positive real number with $x+y=1$ and $x <= y$
prove or disprove
$x^x <= y^y$


the problem is:

$ (-1)^{2*\frac{1}{2}}=((-1)^{2})^\frac{1}{2}$

This is because putting in a half power, which only considers the positive solution, as it is taken as the square root function:

$ ((-1)^{2})^\frac{1}{2}=\sqrt{(-1)^{2}}=|-1|$

Which is like saying:

$ -1=|-1|$

Which is untrue.

Oh, $ -1=1$ is untrue already  :lol:
[hide="Answer"]
$ a^{1}=(a^{2})^{\frac{1}{2}}$ only true when $ a>0$
[/hide]

oh it's nothing extraordinary, when you progress from the 3rd term to the fourth, you change the order of operations....it [i]should[/i] be:

$ -1^{2*\frac{1}{2}}= i^{2}=-1$

EDIT: lol a 2 as base instead of -1
EDIT#2: I think this is the difference between real and imaginary. this makes the answer imaginary, the other, real.

[quote="igiul"]oh it's nothing extraordinary, when you progress from the 3rd term to the fourth, you change the order of operations....[/quote]

That's not the real issue.  The real issue is that $ a^{mn}= (a^{m})^{n}$ doesn't hold generally.

This is probably really obvious, but...

you can do the same thing like this: 1 = 1. (-1)^2 = 1^2. sqrt(-1*-1) = sqrt(1*1). -1 = 1.

More invalid proofs (useless, but fun) can be found here: [url]http://en.wikipedia.org/wiki/Invalid_proof[/url]

Another stupid proof that -1=1
1=sqrt(1)=sqrt((-1)(-1))=sqrt(-1)sqrt(-1)=(i)(i)=-1
The wrong step is sqrt((-1)(-1)) is not equal to sqrt(-1)sqrt(-1)
[quote="chickendude"]my favorite one of these tricks is

$ \sum_{i=1}^{x}x = x^{2}$

$ \frac{d}{dx}\sum_{1}^{x}x = \frac{d}{dx}x^{2}$

$ \sum_{i=1}^{x}1 = 2x$

$ x = 2x$

$ 1 = 2$

although I guess it shouldn't be on this forum[/quote]
It is wrong because x is not differentiable as x is not continuous as x is positive integers.


If $ x\plus{}y\equal{}\frac{7}{12}$ and $ x\minus{}y\equal{}\frac{1}{12}$, what is the value of $ x^2\minus{}y^2$? Express your answer as a common fraction.

$ x^2\minus{}y^2\equal{}(x\plus{}y)(x\minus{}y)$

answer ; $ \frac{7}{144}$


[quote="nonIMO"]It's even more lazy than some IMO sites  :( When will be the problems and solutions presented on this webpage? In a year? In five years? Never?  :([/quote]I don't think it's important as long as we have them here ;) that's why mathlinks exists :P


inequalities
geometry
perimeter
circumcircle
inradius
inequalities proposed

See also here: http://mathvn.org/forum/viewthread.php?thread_id=1149

[quote="earldbest"][quote="mathVNpro"] $ pRr$ method
[/quote]

what's this?  :lol:[/quote]
You will write everything in terms of $ p,R,r$ where $ p$ is the half perimeter, $ R$ the circumradius, and $ r$ the inradius of some triangle.
In this example you could do it the following way: Let $ a \equal{} p\minus{}x,b\equal{}p\minus{}y,c\equal{}p\minus{}z$ and substitute it into the equation. Then write everything in terms of $ x\plus{}y\plus{}z,xy\plus{}yz\plus{}zx,xyz$ (very similar to the $ p,q,r$ method). Then you use the identities $ x\plus{}y\plus{}z \equal{} 2p, xy\plus{}yz\plus{}zx \equal{} p^2\plus{}r^2\plus{}4Rr, xyz\equal{}4pRr$.
Then you can use various inequalities. For instance blundons inequality: $ p^2 \ge 16Rr\minus{}5r^2$.

[quote="mathVNpro"]Let $ a,b,c$ be $ 3$ non-negative numbers. Prove that:
$ (a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)(ab \plus{} bc \plus{} ca)^2\ge 8a^2b^2c^2(a^2 \plus{} b^2 \plus{} c^2)^2$

Some friends in Vietnam have several very nice solutions about this problem. I found a way but accidentially to be one of the solutions that have been offered. But all the way to prove this inequality basically depend on using Schur, Holder, mix-variable method, $ pRr$ method (solved by [b]HophinhanLHP[/b]- which is also a mathlinker).
Surprisingly, it happens that there is $ 1$- line solution to this problem (According to [b]Mr. Namdung[/b]). I am eager to see that if anyone can solve this problem just in $ 1$ line!
Enjoy[/quote]
Not one line, but close:

Written in terms of $ u,v^2,w^3$ it is concave in $ w^3$. Hence we only have to check when $ b \equal{} c \equal{} 1$: $ (a \minus{} 1)^2(4a^3 \plus{} a^2 \plus{} 6a \plus{} 1)$ and when $ a \equal{} 0$: $ b^2(b^2 \plus{} c^2)c^2 \ge 0$. Equality iff $ a \equal{} b \equal{} c$ or $ a \equal{} b \equal{} 0,c$.

And it took me 5 minutes to solve it, because I used computer ;)  :lol:


inequalities
logarithms
induction
algebra
functional equation
inequalities theorems

This one can be transformed to $\frac{t-1}{\ln t}\leq \frac{t+1}{2}$, where $t=x/y>1$. So we are to show \[f(t)=(t+1)\ln t-2(t-1)\geq 0\]
for $t\geq 1$. Indeed, $f(1)=0$ and $f'(t)=\ln t+1+\frac{1}{t}-2$. We see $f'(1)=0$ and $f''(t)=1/t-1/t^2>0$ for $t>1$. So $f'(t)$ increases, thus $f'(t)>f'(1)=0$. It follows $f(t)$ increases too and $f(t)>f(1)=0$.

Here is another way (inspired by a functional equation posted by vess) to find stronger inequalities: consider the recurrence $ a_1=x, b_1=y, a_{n+1}^2=(a_n^2+a_nb_n)/2, b_{n+1}^2=(b_n^2+a_nb_n)/2$. These two sequences converge to $ (x-y)/(lnx-lny)$ Moreover, from the reccursive definition we find $ a_n+b_n$ decreasing. Thus $ a_1+b_1$ is greater than 2 times the common limit, which is your inequality.

Harazi, your proof seems attractive but I don't understand it :blush:! Could you develop this one? ;) In the same way, why introduce this inductions sequences there?!  :?  :what?:  :wacko:
I excuse myself if my questions seem to you absurd :blush:.

Prove for $x,y \in R ,0<y<x$:$\frac{x-y}{lnx-lny}>xy$.
This inequality is so easy.

Do you mean $\sqrt{xy}$?  ;)
Actually, originally it was a double inequality. I remember it was on one of the rounds of ARO long time ago.

...And for the method suggested by Harazi?  :blush:  ;)


limit
logarithms
trigonometry
function
calculus
calculus computations

$ \lim_{x\to1} \frac {ln(x)}{sin(cos(x \minus{} 1 \plus{} \frac {\pi}{2}))}$

[hide="Possible argument"]When x goes to 1, (x - 1 + pi/2) goes to pi/2 and so cos(x - 1 + pi/2) goes to zero. Therefore, sin[cos(x - 1 + pi/2)] is approximately cos(x - 1 + pi/2). But cos(x - 1 + pi/2) = -sin(x-1), and so our limit becomes

$ L \equal{} \lim_{x \to 1} \frac{\ln x}{\sin\left[\cos\left(x\minus{}1\plus{}\frac{\pi}{2}\right)\right]} \equal{} \minus{} \lim_{x \to 1} \frac{\ln x}{\sin(x\minus{}1)} \equal{} \minus{} \lim_{x \to 1} \frac{\ln[1\plus{}(x\minus{}1)]}{\sin(x\minus{}1)}$

$ \equal{}> L \equal{} \minus{} \lim_{x \to 1} \frac{x\minus{}1}{x\minus{}1} \equal{} \minus{}1$.[/hide]

quick way...

$ \lim_{x\to1} \;\frac {\ln(x)}{\sin(\cos(x \minus{} 1 \plus{} \frac {\pi}{2}))}\;\equal{}\;\lim_{x\to0}\; \frac {\ln(1\plus{}x)}{\sin(\cos(x \plus{} \frac {\pi}{2}))}\;\equal{}\; \minus{}\;\lim_{x\to0} \;\frac {\ln(1\plus{}x)}{\sin(\sin\,x)}$

$ \;\equal{}\; \minus{}\;\lim_{x\to0}\; \frac {x\minus{}\mathcal{O}(x^2)}{\sin(x\minus{}\mathcal{O}(x^3))}\;\equal{}\; \minus{}\;\lim_{x\to0}\; \frac {x\minus{}\mathcal{O}(x^2)}{x\minus{}\mathcal{O}(x^3)}\;\equal{}\; \minus{}\;\lim_{x\to0}\; \frac {1\minus{}\mathcal{O}(x)}{1\minus{}\mathcal{O}(x^2)}$

$ \;\equal{}\; \minus{}1$

[quote="TTsphn"][quote="FOURRIER"]$ \lim_{x\to1} \frac {ln(x)}{sin(cos(x - 1 + \frac {\pi}{2}))}$[/quote]
Let ${{ \{sin}{\{cos}(x - 1 + \frac {\pi}{2}) = - \{sin}(\{sin(x - 1))$
Use l'Hospital rule we have:
${ \lim_{x\to\1}f(x) = \lim_{x\to\1} - \frac {\frac {1}{x}}{\{cos}(x - 1)\{cos (sin (x - 1))} = - 1$[/quote]

Ok its my solution too.

Since $ \lim_{x\rightarrow 1} \frac {\ln x}{x \minus{} 1} \equal{} \lim_{x\rightarrow 1} \frac {\ln x \minus{} \ln 1}{x \minus{} 1} \equal{} \frac {d}{dx}\ln x|_{x \equal{} 1} \equal{} 1$,

$ \therefore \lim_{x\rightarrow 1} \frac {\ln x}{\sin(\cos(x \minus{} 1 \plus{} \frac {\pi}{2}))} \equal{} \minus{} \lim_{x\rightarrow 1}\frac {\ln x}{\sin (\sin (x \minus{} 1))}$

$ \equal{} \minus{} \lim_{x\rightarrow 1} \frac {\sin (x \minus{} 1)}{\sin (\sin (x \minus{} 1))}\cdot \frac {\frac {\ln x}{x \minus{} 1}}{\frac {\sin (x \minus{} 1)}{x \minus{} 1}}$

$ \equal{} \minus{} \frac {1}{1}\cdot \frac {1}{1} \equal{} \minus{} 1.$

[color=darkblue][b]Sorry, this limit is no hardly, is very easily, near evidently ![/b]

Denote $ f(x)\equiv\frac {\ln x}{\sin \left[\cos \left(x - 1 + \frac {\pi}{2}\right)\right]}\ .$

Observe that $ \sin \left[\cos \left(x - 1 + \frac {\pi}{2}\right)\right] = - \sin [\sin (x - 1)]\ .$

Thus, $ f(x) = - \frac {\ln x}{x - 1}\cdot\frac {x - 1}{\sin (x - 1)}\cdot\frac {\sin (x - 1)}{\sin [\sin (x - 1)]}$ $ \implies$ $ \lim_{x\to 1}f(x) = - 1\ .$

[b][u]Remark.[/u][/b] I used only the remarkable limits $ \lim_{x\to 1}\frac {\ln x}{x - 1} = \lim_{x\to 0}\frac {\sin x}{x} = 1\ .$

[b][u]A similar exercise.[/u][/b] $ \lim_{x\to\frac {\pi}{2}}\frac {\ln [\sin (4m + 1)x]}{\ln [\sin (4n + 1)x]}$, where $ \{m,n\}\subset \mathcal Z\ .$

(easily, but no evidently !).[/color]

[quote="Virgil Nicula"][color=darkblue][b]Sorry, this limit is no hardly, is very easily, near evidently ![/b][/color][/quote]
Ok  :D

[quote="kunny"]
$ \equal{} \minus{} \lim_{x\rightarrow 1} \frac {\sin (x \minus{} 1)}{\sin (\sin (x \minus{} 1))}\cdot \frac {\frac {\ln x}{x \minus{} 1}}{\frac {\sin (x \minus{} 1)}{x \minus{} 1}}$
[/quote]

This remark can be generalized to the [u]L'Hospital's Rule[/u] wich states that if $ f$ and $ g$ are two functions differentiable at $ x_0$ and $ f(x_0) \equal{} g(x_0) \equal{} 0$

Then: $ \boxed{\lim_{x\to x_0} \frac {f(x)}{g(x)} \equal{} \frac {f'(x_0)}{g'(x_0)}}$

 :)


what is the maximum area of a quadrilateral with sides 1,4,7 and 8

[hide="One-liner so dont look if you want to try"]$ 1^2\plus{}8^2\equal{}4^2\plus{}7^2$[/hide]

The maximum area ocurred when the quadrilateral is cyclic.So the area is 18


logarithms
calculus
calculus computations

[hide="Quick."]Just write it out.

$\ln\left(\frac32\right)+\ln\left(\frac23\right)+\ln\left(\frac54\right)+ \ln\left(\frac45\right)+\cdots=0.$

Of course, this is conditional convergence, not absolute convergence.[/hide]


geometry
geometric transformation
reflection
ratio
angle bisector
perpendicular bisector
geometry proposed

On the plane, given an angle $ xOy$. $ M$ be a mobile point on ray $ Ox$ and $ N$ a mobile point on ray $ Oy$. Let $ d$ be the external angle bisector of angle $ xOy$ and $ I$ be the intersection of $ d$ with the perpendicular bisector of $ MN$. Let $ P$, $ Q$ be two points lie on $ d$ such that $ IP \equal{} IQ \equal{} IM \equal{} IN$, and let $ K$ the intersection of $ MQ$ and $ NP$.

$ 1.$ Prove that $ K$ always lie on a fixed line.

$ 2.$ Let $ d_1$ line perpendicular to $ IM$ at $ M$ and $ d_2$ line perpendicular to $ IN$ at $ N$. Assume that there exist the intersections $ E$, $ F$ of $ d_1$, $ d_2$ from $ d$. Prove that $ EN$, $ FM$ and $ OK$ are concurrent.

a) Let $ MO$ meet the circle centered at I of radius IN (call it $ \omega$) at $ N'$. Then $ N'$ is the reflection of $ N$ into $ d$.
However if $ O'$ is the altitude from $ K$ in $ \triangle{KPQ}$ then $ MO'\cap{\omega}\equal{}N'$ too. Thus $ O\equal{}O'$, i.e., K lies on the bisector of $ \angle{xOy}$.

b) We have that $ EP\equal{}\frac{EQ}{tan(\angle{KPQ})}\equal{}\frac{PQ}{tan(\angle{KPQ})\minus{}1}$. Thus if $ EM\cap{KO}\equal{}X$, then we can find, by Menelaos the ratio $ \frac{KX}{XO}$ in terms of the elements of $ \triangle{KPQ}$.
An analogous reasoning for $ FQ$, etc. yields the result.

Here is my solution for this problem
[b]Solution[/b]
1) Let $M'$ be reflection of $M$ through $d$
It's easy to see that: $M'$ lies on ($I$, $IM$)
Since: $d$ is external bisector of $\widehat{xOy}$ then we have:
($ON$ ; $OM'$) $\equiv$ ($ON$ ; $OP$) + ($OP$ ; $OM'$) $\equiv$ ($ON$ ; $OP$) + ($OM$ ; $OP$) $\equiv$ ($ON$ ; $OP$) + ($OQ$ ; $ON$) $\equiv$ ($OQ$ ; $OP$) $\equiv$ 0 (mod $\pi$)
So: $O$, $N$, $M'$ are collinear
Hence: $NP$ is bisector of $\widehat{ONM}$
Similarly: $MQ$ is bisector of $\widehat{OMN}$
So: $K$ is incenter of $\triangle$ $OMN$ \\
Then: $OK$ is internal bisector of $\widehat{xOy}$ \\
It means that: $K$ lies on a fixed line
2) Let $G$ $\equiv$ $EM$ $\cap$ $FN$
It's easy to see that: $O$, $M$, $G$, $N$ lie on a circle and $GM$ = $GN$
We have: ($MP$ ; $MK$) $\equiv$ ($OP$ ; $OK$) $\equiv$ ($OK$ ; $OQ$) $\equiv$ ($NK$ ; $NQ$) $\equiv$ $\dfrac{\pi}{2}$ (mod $\pi$)
Hence: ($O$, $P$, $M$, $K$) and ($O$, $Q$, $N$, $K$) lie on a circle
So: ($ON$ ; $OM$) $\equiv$ ($ON$ ; $OK$) + ($OK$ ; $OM$) $\equiv$ ($QN$ ; $QK$) + ($PK$ ; $PM$) $\equiv$  ($IN$ ; $IM$) (mod $\pi$)
It leads to: $O$, $M$, $N$, $I$ lie on a circle or $O$, $M$, $G$, $N$, $I$ lie on a circle
Combine with: $GM$ = $GN$ then: $OG$ is internal bisector of $\widehat{xOy}$
Hence: $O$, $K$, $G$ are collinear
It means that: $EM$, $FN$, $OK$ concurrent


vector
function
algebra
polynomial
linear algebra

Here's another one that seems so easy that I'm beginning to question my solution:

Prove that the real vector space consisting of all continuous real valued functions on the interval $ [0,1]$ is infinite dimensional.

Isn't the interval part unnecessary as it is the same as proving $ P(F)$ is infinite?

This problem is indeed easy, and polynomials with real coefficients indeed form an infinite dimensional family of functions. However, in other contexts one has to be more careful. The set of all polynomials with coefficients in $ \mathbb Z_2$ is infinite dimensional, but the set of all functions $ f\colon \mathbb Z_2\to\mathbb Z_2$ is finite dimensional. That is because two different polynomials may define the same function.

Thus, it is important that you have an [b]infinite[/b] set of which the polynomials are evaluated, such as an interval.

So continuity is superfluous.

mlok say more about how two different polynomials over certain fields can define the same function.

Over the field with two elements, $ x$ and $ x^2$ define the same function - and it's the same function as $ x^3,x^4,\dots.$

The set of $ \mathbb{Z}_2$-valued functions over $ \mathbb{Z}_2$ is a two-dimensional vector space with only four elements. The three nonzero elements (only two at a time of which can be independent) could be written as $ 1,x$ and $ 1\plus{}x,$ among many possible notations.

[quote="§outh§tar"]So continuity is superfluous.[/quote]
I would not call it "superfluous" because removing the continuity condition would makes the statement weaker, not stronger.

I see what you mean. I meant to say it holds with or without that requirement


group theory
abstract algebra
geometry
geometric transformation
reflection
superior algebra
superior algebra unsolved

I am having some fundamental misconceptions about (the application of) Sylow's theorems.

* How can there be distinct p-Sylows of order p (=prime)? More general: how can there be distinct groups of prime order? Any group of order p is isomorphic to $ \mathbb{Z}_p$ (the cyclic group). Obviously my interpretation of "isomorphic" as being "the same" is fundamentally incorrect. I even read in Dummit & Foote "[i]any two Sylow p-subgroups of a group (for the same prime p) are isomorphic[/i]."
Perhaps "distinct" here means "distinct as subsets"?  A group G can have two distinct subsets (which means there are elemetns of G in one that are not in the other), both isomorphic to the same group. Now I'm thinking about it, it's actually not so strange...

* Often it is used that any two distinct p-Sylows (for the same p) of order p (i.e. not of some power of p) intersect trivially. I don't see this yet.
For example [section 4.5, example, p.143, Dummit & Foote]: G is a group of order 30, assume $ n_5$ (numer of 5-Sylows) equals 6. "By Lagrange's Theorem, distinct Sylow 5-subgroups intersect in the identity."
Later on they say "This (...) works particularly well when the Sylow p-subgroups have order p (as in this example), since the intersection of two distinct Sylow p-subgroups must be the identity. If the order of the Sylow p-subgroup is $ p^\alpha$ with $ \alpha\geq 2$, greater care is required (...) since in this case distinct Sylow p-subgroups may have many more elements in common, i.e., the intersection may be nontrivial."

\\edit: I think I figured it out: if $ P < G$ and $ Q < G$, then not only $ P\cap Q < G$ (I knew this), but also $ P\cap Q < P$ and $ P\cap Q < Q$, since they all have the same group structure (inherited from $ G$), and by definition of intersection it certainly is a subset of $ P$ and of $ Q$.
Hence by Lagrange, we have $ |P\cap Q|$ divides $ |P|$ and $ |Q|$, which means it divides p, hence it is either one or p. But $ |P\cap Q|$ must be smaller than p, since P and Q are distinct. Conclusion: $ |P\cap Q| \equal{} 1$. Correct?  :)

Another (similar, but short) reasoning I picked up, was "every group of order p is cyclic, hence distinct p-Sylows of order p intersect trivially". Could someone shed some light on this argument?

[hide][PS: I studied Group Theory in a very inefficient and weird way. Now I have all these 'gaps' in my knowledge, often missing basic arguments. I would love to relearn Group Theory all over again, this time in the right way with a good book (e.g. Dummit and Foote) instead of Armstrong's Groups and Symmetry. But first I'm trying to pass my upcoming exam.][/hide]

[quote="BlueVelvet"]Perhaps "distinct" here means "distinct as subsets"?  A group G can have two distinct subsets (which means there are elemetns of G in one that are not in the other), both isomorphic to the same group. [/quote]
Exactly.  Sylow subgroups are isomorphic to each other, but the isomorphism isn't the identity map:  it is conjugation.  For example, the reflections in $ D_n$ are Sylow $ 2$-subgroups when $ n$ is odd and are isomorphic but not identical.

[quote="BlueVelvet"]Often it is used that any two distinct p-Sylows (for the same p) of order p (i.e. not of some power of p) intersect trivially. I don't see this yet.[/quote]
Every nonzero element of a cyclic group of prime order generates the whole group.

Thank you, that makes sense!

Is my other reasoning also correct? ($ |P\cap Q|$ must divide both $ |P|$ and $ |Q|$, so must divide $ p \equal{} |P| \equal{} |Q|$, cannot be p since $ P$ and $ Q$ are distinct, hence must be 1.)


vector
function
linear algebra
matrix
calculus
calculus computations

Let $ A, B \in \mathbb R^{3}$, with angle $ 0 < \theta < \pi$ between them, and suppose the position function $ \mathbf{r}(t) : \mathbb R \rightarrow \mathbb R^{3}$ satisfies the differential equation
$ \mathbf{r}'(t) = A \times \mathbf{r}(t)$, with $ \mathbf{r}(0) = B$.  Find an expression for the speed $ v = \| r'(t)\|$ for every $ t$.

Hint: $ L(\bold{x}) = A\times \bold{x}$ is a linear transformation, so there is a $ 3\times3$ matrix $ C$ such that $ A\times \bold{x}= C \bold{x}$. Then the differential equation becomes $ \bold{x}' = C \bold{x}$.