[hide]
$24*60=1440$ minutes in a day. Divide that by 14 for each game and round down because it has to be an integral amount to get $\boxed{102}$ minutes.[/hide]

oh btw this was meant to be a mental math question, so yeah whatever...

[quote="Klebian"][quote="Treething"]Treething has a collection of fourteen violent video games. If he spends an equal integral number of minutes per day playing each game, what is the maximum number of minutes he can spend on one violent video game in a day?[/quote]
[hide]
So to do mental math, it's $\frac{24*60}{14} = \frac{24*30}{7}$
That's 24 * $4\frac27$
= 96 + $\frac{48}{7}$
= 96 + 6 (not quite 7) = 102[/hide][/quote]

[hide="Hint"]
Try finding the total minutes in a day first, it'll suddenly appear magically quicker...
[/hide]

[hide]
There are 24 hours, 24*60=1440.
1428 is the closest multiple of 14 (it's 14*102).
So, the answer is 102 minutes.[/hide]

This is a little off topic, but speaking of hours, WORLDCUP'S IN 3 HRS.

[quote="mathgeniuse^ln(x)"]As of now, Germany vs. Costa Rica.  Half-time, Germany over Costa Rica 2-1.

Oliver Khan, is not goalkeeper for germany.  that is just scary. :o[/quote]
Germany won 4-2!
There was a beautiful shot made at the end....40 yds away....that guy pummeled the ball into the goal...goalie got owned :D

soccer isn't that violent, go discuss it somewhere else =p

Well, as of now, there are so many yellow cards being given because of fights between Poland and Ecuador.

Surprisingly, Ecuador wins over Poland 2-0.

soccer is violent...it's just that most people use a ball instead of an iron mace.

No point to triple post especially if they're all spam

Treething: [hide]Fine, I long divide 1440 by 14 and take away the remainder to get 102[/hide]

Yeah, England beat Paraguay 1-0, on an own goal. :o

Lets see what happens between Sweden and Trinidad and Tobago.

I know I'm not a mod or anything, but can you please start a topic in G&FF? This topic is waaaaaaaaaaaaaaaaaaaaaay off topic

Yeah, England beat Paraguay 1-0, on an own goal[/quote]

 :rotfl:  :rotfl:

[quote="mathgeniuse^ln(x)"]Fine, Treething I will do this mentally, and I get an answer of 102.

$102$.

I am sorry Ignite168, I just like soccer.

Sweden ties with Trinidad and Tobago 0-0.  A red card was issued to one of Trinidad's best players.

Ivory Coast loses to Argentina 1-2.[/quote]

And you'll get more people discussing it with you if you discuss it in the topic created for it in G&FF

Actually, I'd go to the round table. Bigger, more active topic in the round table.

[quote="mathgeniuse^ln(x)"]What does G&FF stand for?

I am not that computer like as you.[/quote]
Games and Fun Factory :P  :D


factorial
inequalities
number theory
prime factorization

i think this should be moved to the high school getting started forum. looks too hard for basics.

[hide]The only solution I got was $(0,1)$. [/hide]

[quote="easyas3.14159..."][quote="nxxx"][color=red]Find all $x,y\in\mathbb{Z}$ such that \[ x^2+1=y^3  \][/color][/quote]

I don't think many people here know what $\in\mathbb{Z}$ means.[/quote]
$\mathbb{Z}$ just means the set of integers I think. So $x,y\in\mathbb{Z}$ basically means find all the $(x,y)$ pairs that are integers (and for the question, satisfying $x^2+1=y^3$).

[quote="derezin"]I think it's an easy problem. We can factor it as $(x-\sqrt{-1})(x+\sqrt{-1})=y^3$. The numbers $(x-\sqrt{-1})$ and $(x+\sqrt{-1})$ are coprime. Therefore, using unique factorisation in $\mathbb{Z}[\sqrt{-1}]$, $(x+\sqrt{-1})$ is a cube. Now it isn't very difficult so you can solve it.[/quote]

Not sure, you're using prime factorization with nonreal numbers....

In a ring $R$ an element $p$ is said to be prime if from $p|ab$ (where $a$, $b$ are from $R$) results $p|a$ or $p|b$.
$\mathbb Z[i]$ is known to be factorial so you can use unique prime decomposition...
Of course there are solutions without using $\mathbb Z[i]$.

Thanks for moving it!  Sorry if I seemd overly upset, I just thought the math was not at the appropriate level for a middle school forum.

oh, I'm sorry.. I didn't want to seem arrogant.. I was just saying to Jose that what derezin wrote it's correct...
to be honest I saw that I'm in this subforum only after I posted  :blush:

...sorry...just made an educated guess and hoped I was right...

Actually, your guess was right. You have found the only solution. :)

But, the hardest part is to prove it's indeed the only one.
For this, you have several methods, such as the one by [b]derezin[/b] and explained by [b]Kuba[/b], but you can also do it by Fermat's infinite descent, or congruences, etc.

We can assume $x \ge 0$.  We have $1 = y^3 - x^2$.  Obviously $y > x$.  Then $1 = y^3 - x^2 \ge (x+1)^3 - x^2 = x^3 + 2x^2 + 3x + 1 \ge 1$, where the first inequality is from $y>x$, the second from $x\ge 0$.  It follows $y = x+1$, and $x = 0$.  Hence $(0,1)$ the only solution.

[quote="Singular"]Obviously $y > x$.[/quote]

This is just wrong -- when $y^3$ is "close" to $x^2$, $y$ will most often be less than $x$.

[hide]Ummm...I factored the expression into $x^2=(y-1)(y^2+y+1)$. This is only true if $y-1$ is a perfect square and $y^2+y+1$ is a perfect square, $p^2$. So $y^2+y+1=p^2$ and $y(y+1)=(p+1)(p-1)$. I think this is never true unless both sides equal zero, so $p=-1$ or $p=1$. Substituting $p$ for $y$, $-1$ doesn't work and $1$ does, so...the solution is...$\boxed{(0,1)}$...[/hide]

[quote="ch1n353ch3s54a1l"]I factored the expression into $x^2=(y-1)(y^2+y+1)$. This is only true if $y-1$ is a perfect square and $y^2+y+1$ is a perfect square[/quote]

This doesn't have to be true.  If $y \equiv 1 \mod 3$, then both $y - 1$ and $y^2 + y + 1$ will be divisible by 3, so each could be 3 times a perfect square, as well.

[quote="Arne"]What is "Bedzio" in fact?[/quote]

I agree. I wonder what the thread means.

$(0,1)$ is indeed obvious fact after you factored as $x^2 = (y-1)(y^2+y+1)$. I have an idea but I'm not sure how to finish it. If $y > 1,$ we can divide the cases into two: odd and even. If $y$ is odd, then $y-1$ is even and $y^2+y+1 = y(y+1) + 1$ is odd. This means that the product has "extra" 2 so it can't be a perfect square. Now, the problem is when $y$ is even because this gives both $y-1$ and $y^2+y+1$ odd. Maybe someone can finish the cases for the odd to show that there is only one solution.

$gcd(y^2+y+1,y-1)=gcd(2y+1,y-1)=gcd(3,y-1)$

then i think that we can do case work looking at y in mod 3

Well more interesant, is to solve is equation ,in$Z$  we want to solve $y^3=x^2+2$....as in the first one it it's easy to find out, that we need here to use complex numbers....in this one, all we need is a $Z[\sqrt{2}]$...pretty cool way of solving equations in integers..:)

BTW. I'd rather ask "WHO is Bedzio??" ;)

Thanks to derezin, we can write that $x+\sqrt{-1}=(a+bi)^3$ for some $a,b\in\mathbb{Z}$. Comparing imaginary parts we obtain that $3a^2b-b^3=1$. Therefore, $b=-1, a=0$ so we are done.

As unique factorization in $\mathbb{Z}[i]$ was previsouly mentioned, I will post a solution for interested people.

We have that $\gcd (x+i,x-i)=1$.  Since we have that $x^2+1=y^3$, we know that $x+i$ must be a cube. As a result,
$x+i=(a+ib)^3=a^3+3a^2bi-3ab^2-ib^3$

Collecting real and imaginary components, we have
$x=a^3-3ab^2$ and $1=3a^2b-b^3=b(3a^2-b^2)$
If $b=1$, then $1=3a^2-1\implies a^2=\frac{2}{3}\notin\mathbb{Z}$.
If $b=-1$, then $-1=3a^2-1\implies a=0\implies x=0\implies y=1$

Therefore, the only solution is $(x,y)=(0,1)$.

EDIT: I didn't see cvix's post on the second page.

Masoud Zargar

[quote="boxedexe"]As unique factorization in $\mathbb{Z}[i]$ was previsouly mentioned, I will post a solution for interested people.

We have that $\gcd (x+i,x-i)=1$.  Since we have that $x^{2}+1=y^{3}$, we know that $x+i$ must be a cube. As a result,
$x+i=(a+ib)^{3}=a^{3}+3a^{2}bi-3ab^{2}-ib^{3}$

Collecting real and imaginary components, we have
$x=a^{3}-3ab^{2}$ and $1=3a^{2}b-b^{3}=b(3a^{2}-b^{2})$
If $b=1$, then $1=3a^{2}-1\implies a^{2}=\frac{2}{3}\notin\mathbb{Z}$.
If $b=-1$, then $-1=3a^{2}-1\implies a=0\implies x=0\implies y=1$

Therefore, the only solution is $(x,y)=(0,1)$.

EDIT: I didn't see cvix's post on the second page.

Masoud Zargar[/quote]

GREAT :!:  :!:  :!:


quadratics
modular arithmetic
function
abstract algebra
Euler
Galois Theory
superior algebra

Let $ n$ be an odd integer, and let $ \xi \equal{} e^{\frac {2i\pi}{n}}$.
Prove that $ \left|\sum_{k \equal{} 1}^{n}{\xi}^{k^2}\right| \equal{} \sqrt {n}$.

[Moderator edit: Fixed the problem (the original question forgot to require the oddity of $ n$).]

For your problem, alekk.
\[\left|\sum_{k=1}^{n}{\xi}^{k^2}\right|^2=\left(\sum_{k=1}^{n}{\xi}^{k^2}\right)\left(\sum_{k=1}^{n}{\xi}^{-l^2}\right)=\] \[=n+\sum_{k\neq l}\xi^{k^2-l^2}=n+\sum_{k-l\neq 0}\sum_{k+l}\left(\xi^{k-l}\right)^{k+l}.\]

For fixed $k-l$ we have $S=\sum_{k+l}\left(\xi^{k-l}\right)^{k+l}=\sum_{j=1}^{n}\left(\xi^{k-l}\right)^{2j+t}$, where $t=1$ for odd $k-l$ and $t=0$ for even $k-l$, so $S=S\cdot \xi^{2(k-l)}$. It means that $S=0$ for $\xi^{2(k-l)}\neq 1$.

[b]Case 1.[/b] $n$ is odd. Then $\xi^{2(k-l)}\neq 1$, thus $S=0$ and $ \left|\sum_{k=1}^{n}{\xi}^{k^2}\right|=\sqrt{n}$.

[b]Case 2.[/b] $n=4n'$. Then $\xi^{2(k-l)}=1$ iff $k-l=2n'$ (for all other cases $S=0$), but for $k-l=2n'$ we have $S=\sum_{j=1}^{n}\left(\xi^{2n'}\right)^{2j}=\sum_{j=1}^{n}1=n$, thus $ \left|\sum_{k=1}^{n}{\xi}^{k^2}\right|=\sqrt{2n}$.

[b]Case 3.[/b] $n=2(2n'+1)$. Then $\xi^{2(k-l)}=1$ iff $k-l=2n'+1$, but then $S=\sum_{j=1}^{n}\left(\xi^{2n'+1}\right)^{2j+1}=\sum_{j=1}^{n}-1=-n$, thus $ \left|\sum_{k=1}^{n}{\xi}^{k^2}\right|=0$.

[Corrected the awful mistake. Thanks to Peter VDD]

[quote="Myth"]Let $p$ be a prime number, $p=4k+1$, $\xi=e^{\frac{2\pi i}{p}}$. Prove that
a) $\sum_{k=0}^{p-1} Im(\xi^{k^2})=0$;
b) $\left|\sum_{k=0}^{p-1} Re(\xi^{k^2})\right|=\sqrt{p}$.[/quote]

Part (a) follows from the fact that if $t$ is a quadratic residue $\pmod p$, then so is $-t$, so we can decompose the sum $\sum_{k=1}^{p-1}\xi^{k^2}$ in sums of the form $\xi^t+\xi^{-t}$, which are real, and part (b) follows from the problem posted by alekk and part (a).

yes but how do you solve the original pb ?

Hmmm? What do you mean? Did you understand my solution?

Ok, let's continue. Compute the exact value of the sum alekk spoke about. THis shouldn't be that easy.

Is it just a question? Or you know the answer?

hmm... but... if I say n=2

then I have ksi = exp(i*pi) =-1 ...

so then the sum is simply (-1)^1 + (-1)^4 = 0

The modulus of which isn't quite sqrt(2) :?

where am I wrong?

My answers:

=> n=0 mod 4 : sum=sqrt(n)*(1+i)
=> n=1 mod 4 : sum=sqrt(n)
=> n=2 mod 4 : sum=0
=> n=3 mod 4 : sum=sqrt(n)*i

all almost trivial, but the original question only appears to be true for odd n here :?

I have found a gap in my solution. Thanks Peter for noticing case $n=2$ :)

and that makes the originial question true only for odd m I suppose? (since 2 can't be a factor)

Still, then I do not see why the sum without abs. val is that much more difficult? I had solved the original one calculating these and seeing that they were always sqrt(n) * something, something being what I wrote there, and for odd n the modulus of something is 1.

I invite you to prove what you observed and we will see if it is that easy.

the only proof i know deduces it from two functional equations for some zeta- and L-functions... let's see if i can elementarize that proof...

Well... my solution is based on some other "well-known" summations for that matter :P (which I cannot prove)

[ but I thought in contests you could use those? Or was that only for the olympiads and no longer in college contests? ]

[accidentally clicked on Submit rather than Preview]

ok, just for those interested in the proof:

define the L-series $L(s,\chi)=\sum_{n\geq 1} \frac{\chi(n)}{n^s}$ where $\chi: \mathbb{N}\to \mathbb{C}$ is a dirichlet character (meaning that $\chi(mn)=\chi(m)\chi(n)$ for all $m,n$ and $\chi(n+f)=\chi(n)$ for some $f$; the minimal such $f$ is called the conductor of $\chi$). the first thing to prove is that (for the analytic continuation of $L(s,\chi)$)

\[\left(\frac f{\pi}\right)^{s/2}\Gamma\left(\frac{s+\delta}2\right)L(s,\chi)=W_{\chi}\left(\frac f{\pi}\right)^{(1-s)/2}\Gamma\left(\frac{1-s+\delta}{2}\right)L(1-s,\overline{\chi})\]

where we set $\delta=0$ if $\chi(-1)=1$, $\delta=1$ if $\chi(-1)=-1$, $W_{\chi}=\frac{\tau(\chi)}{\sqrt{f}i^{\delta}}$ where $\tau(\chi)=\sum_{a=1}^{f} \chi(a) e^{2\pi i a/f}$, $\overline{\chi}(n)=\overline{\chi(n)}$.

the next thing is to define the zeta function of a number field $K$.  we define $\zeta_K(s)=\sum_{\mathfrak{A}\neq\{0\}} \frac 1{N\mathfrak{A}}$ where the sum is taken over all ideals $\mathfrak{A}$ of the ring of algebraic integers $\mathcal{O}_K$ of $K$ and $N$ denotes the norm from $K$ to $\mathbb{Q}$.
$\zeta_K(s)$ satisfies a functional equation(or more exactly, its analytic continuation):

\[A^s\Gamma\left(\frac s2\right)^{r_1}\Gamma(s)^{r_2}\zeta_K(s)=A^{1-s}\Gamma\left(\frac {1-s}2\right)^{r_1}\Gamma(1-s)^{r_2}\zeta_K(1-s);\]

here $r_1$ is the number of real embeddings of $K$, $2r_2$ the number of complex embeddings, $A=2^{-r_2}\pi^{-N/2}\sqrt{|d(K)|}$ where $N=\deg{K/\mathbb{Q}}$, $d(K)$ is the discriminant of the number field $K$.

on the other hand, in the case that $K$ is a galois number field over $\mathbb{Q}$ with $Gal(K/\mathbb{Q})$ abelian it is connected to the L-series by a simple identity. to write down this identity, the first step is to define the associated class of dirichlet characters. we know that we can embed $K$ into some cyclotomic number field $\mathbb{Q}(\zeta_n)$(here $\zeta_n$ denotes a primitive $n$-th root of unity) by the kronecker-weber theorem. there the galois group is of the form $(\mathbb{Z}/n\mathbb{Z})^{*}$ for some $n$; a multiplicative map $(\mathbb{Z}/n\mathbb{Z})^{*}\to\mathbb{C}$ then may be naturally extended to a dirichlet character; those whose kernel contain $Gal(\mathbb{Q}(\zeta_n)/K)$ form a group which only depends on the number field $K$; we will denote it by $X$. now we get from euler-products for both sides that

\[\zeta_K(s)=\prod_{\chi\in X} L(s,\chi).\]

now one can compare the two functional equations and get some nice results. e.g., it follows that $\prod_{\chi\in X} \tau(\chi)=\sqrt{|d(K)|}$ if $K$ has only real embeddings, $\prod_{\chi\in X} \tau(\chi)=i^{\deg(K/\mathbb{Q})/2}\sqrt{|d(K)|}$ if $K$ has only complex embeddings(since $K$ is galois, we always have one of these cases). now, take the appropriate quadratic number fields to conclude.

both functional equations follow from some quite straightforward analysis; the identity between the zeta- and the L-functions is not hard to prove. however, the proofs are quite long and boring, so i will omit them :D (one can look them up somewhere).

[quote="ZetaX"]Can you please tell me what summations that are ¿[/quote]Well, seeing the huge complicity of Peter Scholze's proof, I suppose the sums I have used will need the original problem to prove them somehow. (some of them are more ugly than this problem)
In the actual paper it said that the proof was not written there "due to the need for higher mathematics".

So when I said "easy", I actually meant "easy starting from some given sums that are a hell to prove on their own". :)
(and which could be said to "trivialize" the problem compared to the expected solution)

 [ but what can one expect a first year student to know about the complexity of the above things, I can't even read ]

[quote="Peter VDD"]
 [ but what can one expect a first year student to know about the complexity of the above things, I can't even read ][/quote]

i didn't even start studying - in fact, i still have more than two boring years of school in front of me :(

Waiting for the marvelous solution Peter VDD found, I must say that in the ENS exam they gave 2 different solutions, one analytical using Fourier series and another one, purely algebraic. Maybe I will have the time this week-end to post the steps. Peter, your solution is very nice, but extremely complicated.

[quote="harazi"] the ENS exam [/quote]

wich year ?

[quote="harazi"]Waiting for the marvelous solution Peter VDD found...[/quote]as I said, it'll probably not be a solution... You know, if a theorem isn't proven in a paper, it's hard to know wether it has a one-line solution with higher mathematics, or it is a real work on its own.

@ Zetax: well I don't know which ones exactly are based on what, I just have a long "resource list" here, but this is one that resembles quite good the original prob (and therefor may be "forbidden" to use)

$\sum^{mn}_{k=1} e^{\frac{\pi i k^2}{n}}$
is -1 for all odd n and odd m
is 0 for all odd n and even m
is $nim^2$ for even n

To Moubi: I think 2001. Anyway, it is recent. It is test of 6 heures.


you are given a suffled (randomized) regular deck of cards (ie. 52  normal cards and 2 jokers). Let $ P_k$ be the probability that, counting from the top of the deck, the first joker is in the $ k$th position. For which $ k$ is $ P_k$ maximum?

[hide]
k = 1.

We can split the deck into two kinds of cards:

52 non-joker cards.
2 joker cards.

Now, we can consider the 52 other non-joker cards identical. Now, the number of ways the 1st joker can be in the ith place is 2(54-i). (i, i+1), (i, i+2),...(i, 54), (i+1,i), (i+2,i)...,(54,i).

Therefore, P_k is minimal for k = 1.[/hide]

can u latexify i cant understand ur post

Yes, Pythag is right. Another question: what is the average position that the first card will be in?

Because it's dealing with probability you can let the other cards be considered identical. You can also let the jokers be identical. If the first joker is in the kth place, the other joker can only be in 54-k positions (think of 54 spaces between and outside the cards). So k=1.

For average position, I got $ \frac{\displaystyle\sum_{k\equal{}1}^{53}k(54\minus{}k)}{\dbinom{54}{2}}$. I evaluated the top using hockeystick and got 18.333...


calculus
derivative
algebra
polynomial
ratio
geometric series
geometric sequence

find the following sum:

$\frac{1}{4}+\frac{3}{8}+\frac{6}{16}+\cdots+\frac{\binom{n}{2}}{2^n}+\cdots$

[hide="Solution"]
$\sum\binom{n}{2}\left(\frac{1}{2}\right)^{n}=\frac{1}{2}\sum n\left(n-1\right)\left(\frac{1}{2}\right)^{n}$

$\sum n\left(n-1\right)x^{n}=\frac{2x^{2}}{\left(1-x\right)^{3}}$

So the sum is

$\left(\frac{1}{2^{2}}\right)\left(8\right)=\boxed{2}$[/hide]

[quote="amcavoy"]

$\sum n\left(n-1\right)x^{n}=\frac{2x^{2}}{\left(1-x\right)^{3}}$
[/quote]

How do you know this?  Is it just an identity?

I cheated and used calculus to derive it :).

$\sum x^{n}=\frac{1}{1-x}$

then take the derivative of both sides twice and multiply by $x^{2}$ to get the desired result.

And if you don't know calculus, you can just use my simple algebra solution (completely learned from AoPS vol. 2!) which comes in handy quite a bit.

[quote="Hokkage"]Let $S$ be the sum, so
$S= \frac14 + \frac38 + \frac{6}{16} + \frac{10}{32} + \cdots$
$\frac{S}{2} = \frac18 + \frac{3}{16} + \frac{6}{32} + \cdots$
$S-\frac{S}{2} = \frac{S}{2} = \frac14 + \frac28 + \frac{3}{16} + \frac{4}{32} + \cdots$
$\frac{S}{4} = \frac18 + \frac{2}{16} + \frac{3}{32} + \cdots$
$\frac{S}{2}-\frac{S}{4}=\frac{S}{4}= \frac14 + \frac18 + \frac{1}{16} + \frac{1}{32} + \cdots$
And then by geometric series, $\frac{S}{4} = \frac{\frac{1}{4}}{1-\frac{1}{2}} = \frac{1}{2} \Rightarrow S=2$[/quote]

Any spaecific reason you did that or is there any ay you can identify what to do in such questions?It would really help if someone tells this.

and yeah hokkage,hide your sollutions :)

Sorry for not hiding, I'll edit after I post this.
As for the method, it pretty much involves taking successive differences until you get degree 0. For example, in the given problem, the polynomial is degree 2. So if we take the difference two times, we're going to get a constant, at which point it becomes a geometric series which is easy to evaluate.

Hokkage's method required recursion in series, which is often very handy in solving many different series problems that may not be as clean as an infinite geometric series. Often setting the sum equal to some variable $S$ helps a lot.

Sum of nth degree in numerator and geometric progression in denominator will require n iterations of divisions/multiplications of the common ratio in the denominator.


rotation
geometry
geometric transformation

$Question 1$
There are three parallel st lines at varying distances. Draw an equilateral triangle with one vertex on each of the lines.

$Question 2$ Four points are given in the plane. Draw one straight line through each of the points such that the pairwise intersection of the lines form a square.

[hide="#1"]Let the lines be $p,q,r$ with $q$ being the middle one. Pick an arbitrary point $A$ on $q$ and rotate $p$ around $A$ for $60$ degrees. Let the image of $p$ be $p'$. If $p'$ intersects $r$ at $B$, then $AB$ is the side of the desired equilateral triangle. Since $p$ can be rotated either clockwise or counterclockwise, there are two symmetric solutions (up to the translation along the lines).[/hide]


Propose a reasonable synthesis of vinyl ether.

Base-catalyzed nucleophilic addition of alcohols to alkynes?

Or maybe a process [b]resembling[/b] Alkoxymercuration/Demercuration of Alkenes?

I don't think so. The problem is much more simple than it seems.

Shouldn't the first one do the job? I even looked it up.

Then show the equations of your synthesis, together with the conditions. Just saying "Base-catalyzed nucleophilic addition of alcohols to alkynes" is nothing.

$ \ce{CH3} \equiv \ce{CH} \plus{} \ce{ROH} \rightarrow \ce{CH3} \minus{} \ce{C(OR)} \equal{} \ce{CH2}$

Where we can catalyze the process using $ OH^{\minus{}}$ and heat at $ 150^{\circ}C$.

The purpose is to prepare vinyl ether, then how would you use your proposal?

[quote="BanishedTraitor"]$ \ce{CH3} \minus{} \ce{C(OR)} \equal{} \ce{CH2}$[/quote]

Isn't that a Vinyl Ether :maybe:

One more time, the purpose is to prepare vinyl ether, $ CH_2\equal{}CH\minus{}O\minus{}CH\equal{}CH_2$, not [i]a[/i] general vinyl ether.

Add $ \ce{H2SO4} , \Delta$

$ \ce{2CH2\equal{}CHOH} \plus{} \ce{H\plus{} \minus{}> CH2\equal{}CH\minus{}O\minus{}CH\equal{}CH2} \plus{} \ce{H2O}$

Reasonable?

[quote="Grandiloquence73"]Add $ \ce{H2SO4} , \Delta$

$ \ce{2CH2 \equal{} CHOH} \plus{} \ce{H \plus{} \minus{} > CH2 \equal{} CH \minus{} O \minus{} CH \equal{} CH2} \plus{} \ce{H2O}$

Reasonable?[/quote]

Don't forget about tautomerization (keto-enol).

Yes, that's right: vinyl alcohol does not exist.

how about this?
ethylene oxide + ethylene glycol -> HO-CH2-CH2-O-CH2-CH2-OH
foll by socl2 + alk. koh->vinyl ether

Very good, that seems to be a plausible method.


What horizontal force $ F$ (applied to the axle) is needed to push a wheel of radius $ R$ over a block of height $ h$?

$ \tau_{1}\equal{}\minus{}mg\cdot D$ where $ D\equal{}\sqrt{h(2R\minus{}h)}$
$ \tau_{2}\equal{}F(R\minus{}h)\Rightarrow\sum\tau\equal{}0\equal{}F(R\minus{}h)\minus{}mg\sqrt{h(2R\minus{}h)}$
Equate the two and divide, and you get $ \boxed{F\equal{}\frac{mg\sqrt{h(2R\minus{}h)}}{R\minus{}h}}$ where $ m$ is the mass of the axle.  I took this solution from my mechanics homework so it might be somewhat unclear.  Draw a diagram and my solution should make sense.  Otherwise, just post and I can try to clarify.

[quote="JRav"]$ \tau_{1} \equal{} \minus{} mg\cdot D$ where $ D \equal{} \sqrt {h(2R \minus{} h)}$
$ \tau_{2} \equal{} F(R \minus{} h)\Rightarrow\sum\tau \equal{} 0 \equal{} F(R \minus{} h) \minus{} mg\sqrt {h(2R \minus{} h)}$
Equate the two and divide, and you get $ \boxed{F \equal{} \frac {mg\sqrt {h(2R \minus{} h)}}{R \minus{} h}}$ where $ m$ is the mass of the axle.  I took this solution from my mechanics homework so it might be somewhat unclear.  Draw a diagram and my solution should make sense.  Otherwise, just post and I can try to clarify.[/quote]

Thanks for your help! But what are $ r_1$ and $ r_2$ again?  :blush:

What are you referring to?  I didn't mention any $ r_{1}$ or $ r_{2}$.

he means t_1 and t_2 which I believe are the two torques// torque is a rotational force sort of like when you push a door and it turns rather than move forward.

^^You are correct.  Tau represents the torques which must be considered, since they cause the rotation.


linear algebra
matrix
algebra unsolved
algebra

If $A=\left[\begin{matrix}a_{11}&a_{12}&\dots &a_{1n}\\a_{21}&a_{22}&\dots &a_{2n}\\ \vdots &\vdots & &\vdots\\a_{n1}&a_{n2}&\dots &a_{nn}\end{matrix}\right]$, where $a_{ij}\in\mathbb{Z}$, then we have to prove

$\det(A-{1\over 2}I)\neq 0$

But that's almost obvious, since if we keep expanding the determinant by the first column, finally we'll obtain (among the others) the term $(a_{11}-{1\over 2})(a_{22}-{1\over 2})\dots(a_{nn}-{1\over 2})$, and that one is the only one whose expansion contains $\frac{(-1)^n}{2^n}$, which can't be possibly canceled out by any other term in the determinant expansion.

Sorry,Farenhajt.I haven't learnt about matrix yet.Can you change your solution into a synthetic solution.Thanks a lot!

Have you learned about systems of linear equations? Determinants? Cramer's rule? It's basically that, only written in the matrix form.


Why not

\[ \sum {\frac{1}{a} \plus{} \frac{3}{{2a \plus{} b}}}  \ge \sum {2\sqrt 3 \frac{1}{{\sqrt {a(2a \plus{} b)} }}}\]


inequalities limit calculus algebra polynomial

[quote="veryfriendly"]Find the minimum value of $xy+x+y+\frac{1}{xy}+\frac{1}{x}+\frac{1}{y}$
 $x, y > 0$ real.[/quote]

[hide]AM-GM inequality can be used
\[xy+x+y+\frac{1}{xy}+\frac{1}{x}+\frac{1}{y}\ge 6 \cdot \sqrt[6]{1}= 6. \]
Therefore the minimum possible value is $6$ when $x=y=1$.[/hide]

well you are going to have to try something different, your am-gm has hte equality case: x=y=1

EDIT: sorry the equality case does not come out nicely

[quote="vishalarul"][quote="Altheman"]what if the problem was: find the minimum value of

xy+x^2+y+1/x+1/y+1/[xy]

?[/quote]

[hide="Err..."]I applied the AM-GM and got:

$xy+x^{2}+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\geq 6\sqrt[6]{x}$

I'm not sure, but if you make $x$ "sufficiently small", there is a value of $y$ (hopefully real) that satisfies the requirement.

So I don't think there is an answer.[/hide][/quote]

This was precisely Altheman's point.  AM-GM works on the original problem because the equality case produced a bound that was constant (the constant RHS).  Here, because the RHS is not a constant, you are not guaranteed that the equality case produces the minimum.

could someone please post a solution to altheman's question?

I doubt my solution can work sufficiently, but...

[hide="Solution"]

From vishalarul's AM-GM use I used limits to deduce that $\lim_{x\to0}6\sqrt[6]{x}= 0$, so the answer would be 0?

[/hide]

As I have already stated, vishalarul's use of AM-GM is not valid.  If you are taking $x \to 0$ then the $\frac{1}{x}$ term tends to infinity.

[hide="In fact..."] By AM-GM we have

$y+\frac{1}{y}\ge 2$
$xy+\frac{1}{xy}\ge 2$
$x^{2}+\frac{1}{2x}+\frac{1}{2x}\ge 3 \sqrt[3]{ \frac{1}{4}}$

So the minimum is [b]at least[/b] $4+3 \sqrt[3]{ \frac{1}{4}}$; however, this minimum is not attainable because the equality case gives the contradiction $x = 1 = \sqrt[3]{ \frac{1}{2}}$. [/hide]

I can't think of an elementary solution to Altheman's problem.

Oh. I knew something was wrong...

Do you know the solution anyway, even if it isn't elementary? Just out of curiosity...

Of course, I take $x,y$ to be positive or else there is no minimum.
Hold $x$ constant, so we must minimize an expression of the form $ky+y+\frac{1}{y}+\frac{1}{ky}=\left(1+\frac{1}{k}\right)\left(ky+\frac{1}{y}\right)$ which is equivalent to minimizing $ky+\frac{1}{y}$. Since $k$ is arbitrary, we must be careful with Am-Gm:\[ky+\frac{1}{y}\geq 2\sqrt{k}\] nicely has equality at $y=\frac{1}{\sqrt{k}}$.
Okay so going back to the original gives (for $x$ constant):
\[xy+x^{2}+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\geq x^{2}+\left(1+\frac{1}{x}\right)2\sqrt{x}\](this is equivalent to finding the minimum point on an arbitrary curve in the family of curves $f(y)=xy+x^{2}+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$)
Now we have an expression in $x$ to minimize:
\[=x^{2}+2\sqrt{x}+\frac{2}{\sqrt{x}}=\frac{x^{2}\sqrt{x}+2x+2}{\sqrt{x}}\]This is the reason all the AM-Gm attempts earlier have been stumped; normally, we would minimize the last two terms easily with equality, but the $x^{2}$ in the front is the issue.
It is equivalent to minimizing \[\frac{a^{5}+2a^{2}+2}{a}\]
Since this problem was conceived on the spot, it doesn't have a nice elementary solution but using standard calculus one finds that the minimum occurs when $a$ is the single real root of the quintic $2a^{5}+a=1$

yes, sorry, i realized and posted that the solution did not come out, i had made a mistake in my algebra

So suppose instead of $xy+x^{2}+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$ you propose \[xy+p(x)+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\]
then the minimum would be nice whenever the polynomial $f(x)+\left(1+\frac{1}{x}\right)2\sqrt{x}$ is a constant times a perfect square... for instance a better choice would be something along the lines of $f(x)=x\sqrt{x}$ so the problem would read:

[code]what if the problem was: find the minimum value of

xy+x*sqrt(x)+y+1/x+1/y+1/[xy]

?
[/code]
Wait oops that one isn't so nice either...
any ideas of a nicer choice :P


[hide]The area of the hexagon is $\frac{3\cdot10^{2}\sqrt{3}}{2},$ or $150\sqrt{3}.$

In order for the circle to be entirely contained within this hexagon, it's center must be inside another hexagon (inside the big one), with side length 8. The area of this smaller hexagon is $\frac{3\cdot8^{2}\sqrt{3}}{2},$ or $96\sqrt{3}.$

The probability is $\frac{96\sqrt{3}}{150\sqrt{3}},$ or $\frac{16}{25}.$[/hide]

[quote="t0rajir0u"]A point is chosen at random in a regular hexagon with sides of length $10$.  A circle of radius $1$ is drawn around this point.  What is the probability the circle is entirely contained within the hexagon?[/quote]

[hide]We see that in order for the circle to be completely inside the hexagon, the center of the circle must be contained inside a hexagon of length 8.  The ratio of side lengths of the small and large haxahedrons is $\frac{4}{5}$.  Therefore, the ratio or probability that the point is inside the small hexahedron is $\boxed{\frac{16}{25}}$.[/hide]

[quote="mathgeniuse^ln(x)"][hide]Let us consider only one of the six equilateral triangles of the hexagon.

Each of the sides has a length of 10.

So the altitude to the base of this triangle is $5\sqrt{3}$.

So, now we have to consider all points in the equilateral triangle that are greater than 1 unit away from the edge.

This is just going to be an equilateral triangle with altitude of $5\sqrt{3}-1$.

So the side length of the equilateral triangle is $10-\frac{2}{\sqrt{3}}$.

The area of ratio of this equilateral triangle to the bigger equilateral triangle is $(\frac{10-\frac{2}{\sqrt{3}}{10})^{2}=\frac{304-40\sqrt{3}}{300}$.

This is the probability it occurs like the way the problem stated.  Now if we draw the same figure for all six triangles in the equilateral triangle, we still get the same probability of

$\frac{304-40\sqrt{3}}{300}$[/hide][/quote]

Correct.  i_like_pie and jli, draw a diagram and actually draw some circles before you reach a hasty conclusion. ;)

[quote="mathgeniuse^ln(x)"]Yeah, I got it right...

I think they might have been confused because maybe they thought it was a square and not a hexagon.[/quote]
I knew it was a hexagon... :roll:

I wasn't too sure about the side length of the smaller hexagon, though.


Prove, that $5$ is the only natural number, which is the product of all prime numbers less than this number.

No, he's right when you assume that he means what the headline says: "[u]Sum[/u] of prime numbers"

the solution for the correct problem.

let $p_1, p_2, ...$ the primes in increasing order.

let $S(i) = p_1+p_2+..+ p_i$.

$LEMA$: $S(i) > 2p_i$, for $i \ge 4$.

$PROOF$: induction.

with simple computations we find that for $i = 4,5$ the result is valid.
suppose that for $n = k$ the proposition is valid.
we have $S(k) > 2p_k$.

by Bertrand, we have $2p_k > p_{k+1}$, so, $S(k+1) = S(k) + p_{k+1} > 2p_k + p_{k+1} > 2p_{k+1}$,
and the LEMA is proved.

__

suppose that for one integer $m$ we have $m = S(j)$, with $j>3$. $p_j$ is the greatest prime less than $m$.

so, we have $S(j) > 2p_j$.
this implies that, by Bertrand, between $p_j$ and $S(j)$ exists one prime, contradiction!


probability
conditional probability
probability and stats

it depends on the probability that B shoots the first... otherwise it's stupid

Unfortunately, it is my midterm and I can't understand it. Stupid me...

Is there an explanation?

hungkhtn: Crux said that you didn't give enough information. You can't answer this question without knowing the person starting to shoot. It's most likely to be A, but that's never said.

Crux: You just gave me a good laugh, "shoots the first" means shooting A   :lol:  :D  :wink:

ZetaX. Yes, I understand Crux's sayings. I meant it is however a problem of my midterm, and I can't make it. It took me almost time to think.

Thank you, Crux and ZetaX. At least, I was not too stupid to not be able to solve this problem.

By the way, if we assume that A starts and then they are alternately shooting, we end up with the answer being $ \frac{7}{12}$.

Besides, the problem doesn't make clear whether the deer is distinct from A and B.

  darij

I think the problem is: A and B [b]both[/b] shoot at a deer and kill it. After examining the corpse, they see that only one bullet hit the deer. What is the probability that this bullet was shot by $ B$? If so, this is a simple exercise on conditional probabilities. If I didn't make any stupid mistake in fraction arithmetic, the answer is $ 7/9$.

I interpreted the question the same as Fedja, and have the same answer.  As a general principle, though, wouldn't it make more sense to ask the [i]author of the exam[/i] to explain its meaning to you?

I am sorry, but even though, the problem still seems to be very obscure to me...  :(

OK, let $ P_\varnothing$, $ P_A$, $ P_B$, and $ P_{AB}$ be the probabilities that nobody hit the deer, only A hit the deer, only B hit the deer, and both A and B hit the deer respectively.

Now answer the following questions:
a) what are the numerical values of all these probabilities?
b) how to express the probability we are interested in in terms of them?

But with this explanation, the definition of $ P_{AB}$ seems to be unreasonable.

Ok, I admit that you do not need to know $ P_{AB}$ and $ P_\varnothing$ to find the answer of the problem but still I want you to tell me all the numbers. Note that my $ P$s are the probabilities before the body was examined (or even before it became clear that the deer was killed). They are not conditioned on the event that the deer was killed by exactly one bullet yet. This conditioning is the second step in the solution.

[quote="hungkhtn"]A,B are two hunters, hunting one deer, which is shoot by just one bullet. The probability that A shoots on target is .3, and for B is .6. What is the probability that B killed the deer?[/quote]

The probability that B kills the deer is the same as the conditional probability that B’s shot hits the deer given that A’s shot does not, i.e.

	 \[ P\left(B|\neg A\right)
\]

	\[ P\left(B|\neg A\right) \equal{} \frac{P\left(B\cap\neg A\right)}{P\left(\neg A\right)}
\]

	\[ P\left(B\right) \equal{} P\left(B\cap\left(A\cup\neg A\right)\right)\equal{}P\left(B\cap A\right) \plus{}P\left(B\cap\neg A\right)
\]

	\[ P\left(B\cap\neg A\right) \equal{} P\left(B\right)\minus{}P\left(B\cap A\right)
\]

Assuming A and B are independent, $ P\left(B\cap A\right) \equal{}P\left(B\right)\cdot P\left(A\right)$.

	 \[ P\left(B\cap \neg A\right) \equal{} P\left(B\right)\minus{}P\left(B\right)\cdot P\left(A\right)\equal{}P\left(B\right)\left(1\minus{}P\left(A\right)\right)\equal{}P\left(B\right)P\left(\neg A\right)
\]

We find that B and not-A are independent as well. Hence,

	\[ P\left(B|\neg A\right) \equal{} \frac{P\left(B\cap \neg A\right)}{P\left(\neg A\right)} \equal{}\frac{P\left(B\right)P\left(\neg A\right)}{P\left(\neg A\right)} \equal{} P\left(B\right)
\]

	\[ \boxed{\therefore P\left(B|\neg A\right) \equal{} P\left(B\right) \equal{} 0.6}
\]

[quote="fedja"][quote="Dr. No"]
The probability that B kills the deer is the same as the conditional probability that B’s shot hits the deer given that A’s shot does not, i.e.
\[ P\left(B|\neg A\right)
\]
[/quote]
Nope. You are conditioning correctly but on the wrong event. All we know is that the deer was killed by one bullet, which means conditioning on $ (B\cap\neg A)\cup(A\cap\neg B)$ ;)[/quote]

Okay,
\[ P\left(B|\left(B\cap\neg A\right)\cup\left(A\cap\neg B\right)\right) \equal{} P\left(B\cap\neg A\right)\div P\left[\left(B\cap\neg A\right)\cup\left(A\cap\neg B\right)\right] \equal{} \frac {\left(.6\right)\left(.7\right)}{\left(.6\right)\left(.7\right) \plus{} \left(.3\right)\left(.4\right)} \equal{} 0.777...
\]

[quote="hungkhtn"]Thank you very much for your careful advise. The solution Dr No posted seems to be very logical, but I still think that the problem is posed irrational.

As at the first step, we have to assume that $ P(AB) \equal{} 0$, but $ P(AB)$ is not $ P(A)P(B)$? So it is impossible that
\[ P\left(B\cap\neg A\right) \equal{} P(B)P(\neg A).
\]
[/quote]

In this case, however, $ A\cap B \neq \emptyset$. $ \,A$ is the event that the first shooter hits the target. $ B$ is the event that the second shooter hits the target. Therefore, $ A\cap B$ is not necessarily empty; it is merely that in the case we are  considering only one of the two shooters hits the target. That is why fedja pointed out that one needed to condition upon $ \left(B\cap\neg A\right)\cup\left(A\cap\neg B\right)$ -- this is literally the event that $ B$ hits and $ A$ misses or $ A$ hits and $ B$ misses. In essence, the event that one shooter hits the target while the other misses irrespective of which one hits or which one misses. I had not thought to condition $ B$ in this way, but it is the correct approach.

[quote="fedja"]You seem to confuse two things: the conditional probability $ P(AB|\text{the deer is killed by 1 bullet})$, which is $ 0$ and the unconditional probability $ P(AB)$ which is $ 0.18$. Look: suppose that your friend has a die, throws it and asks you what is the probability that he got 1. It is 1/6, isn't it? Suppose now that he does exactly the same but also tells you that the number he got is even. What is the probability now? It is $ 0$. The point is that having extra information changes your probability space. You do not need numbers to understand that. If I choose a living creature at random giving every creature equal chance and ask you what is your estimate of the chance that it is less than 5 inches tall, you'll say almost 100% (most living creatures are bacteria). Now if I tell you that the creature I chose just tore my new trousers with its teeth, your answer will, probably, change a bit. The reason is that with this extra information you still have a uniform distribution but on a narrower sample space. If the deer story still confuses you, imagine that A and B go hunting every day. Sometimes they kill a deer, sometimes not. So, there will be some unkilled deers, some deers killed by 1 bullet and some deers killed by 2 bullets. The question (equivalent to the problem) is the following: [i]out of all deers killed by one bullet, what portion was killed by $ B$?"[/i][/quote]

You are very kind of me, fedja! Thank you very very much.

I will study Probability more seriously!


function
algebra
polynomial
algebra unsolved

Hello,

I have a question:
Is there a possibility to transform the following function into a form where one can immediately see, that there a no positive solutions?
$ f(k)\equal{}k^9\plus{}k^8\minus{}7k^7\minus{}5k^6\plus{}21k^5\plus{}11k^4\minus{}27k^3\minus{}5k^2\plus{}20k\plus{}4\equal{}0$ with $ k\geq 0$

This is the best I've got:

$ 4 \plus{} k (20 \plus{} k (\minus{}5 \plus{} k (\minus{}27 \plus{} k (11 \plus{} k (21 \plus{} k (\minus{}5 \plus{} k (\minus{}7 \plus{} k \plus{} k^2)))))))\equal{}0$ but I have no idea if it helps, though.

Thanks, but I think, your transformation cannot fulfill the purpose to show, that there are no positive solutions. Unfortunately, it seems to be not fully suitable.

I know... but I posted it anyway. Few other things I've noted: only one solution is real (and negative), other are complex. the real one is somewhere between 0 and 1...

And here's a use for my factorization: it's enough to prove that for positive $ k$ that giant factorization is positive, or a big negative number. But still, it doesn't seem quite doable.

[quote="Atan"]Hello,

I have a question:
Is there a possibility to transform the following function into a form where one can immediately see, that there a no positive solutions?
$ f(k) \equal{} k^9 \plus{} k^8 \minus{} 7k^7 \minus{} 5k^6 \plus{} 21k^5 \plus{} 11k^4 \minus{} 27k^3 \minus{} 5k^2 \plus{} 20k \plus{} 4 \equal{} 0$ with $ k\geq 0$[/quote]
yes,actually if $ P$ is a real polynomial without positive root, then there exist
$ A,a_1,...,a_k,b_1,...,b_m,c_1,...c_m\in\mathbb{R}$ with $ a_i\geq0,c_i > 0$ such that:
$ P(x) \equal{} A(x \plus{} a_1)(x \plus{} a_2)...(x \plus{} a_k)((x \plus{} b_1)^2 \plus{} c_1)...((x \plus{} b_m)^2 \plus{} c_m)$ that one can immediately see that there is no positive root...
but maybe $ a_i$ or $ b_i$ or $ c_i$ can not expressible as you want...

$ f(k) \equal{} k^9 \plus{} k^8 \minus{} 7k^7 \minus{} 5k^6 \plus{} 21k^5 \plus{} 11k^4 \minus{} 27k^3 \minus{} 5k^2 \plus{} 20k \plus{} 4$
$ f(k) \equal{} k \cdot g(k) \plus{} 4$

$ g(k) \equal{} k^8 \plus{} k^7 \minus{} 7k^6 \minus{} 5k^5 \plus{} 21k^4 \plus{} 11k^3 \minus{} 27k^2 \minus{} 5k \plus{} 20$

$ g(k) \equal{} k^8 \plus{} k^7 \plus{}k^4 (\minus{}7k^2\minus{}5k\plus{}21) \plus{}11k^3 \plus{}(\minus{}27k^2 \minus{} 5k \plus{} 20)$

if $ k <\equal{} 3/4$  then $ g(k) > 0$ because each term is positive

use this fact:
if $ P$ is a  polynomial with real coefficinet.such that:
$ x\geq0\Rightarrow{P(x)>0}$
then,there exist natural $ m$ such that all coefficient  $ (1\plus{}x)^mP(x)$ are nonnegative.

therefore we can transform $ P$ to $ \frac{(1\plus{}x)^mP(x)}{(1\plus{}x)^m}$

I rather mean a transformation like that:
If we would have another equation like that one:
$ x^9\plus{}x^8\minus{}7x^7\minus{}5x^6\plus{}21x^5\plus{}11x^4\minus{}27x^3\minus{}5x^2\plus{}20x\plus{}10\equal{}0$
Then we could also write it as:
$ x^5(x^2\minus{}\frac{7}{2})^2\plus{}x^4(x^2\minus{}\frac{5}{2})^2\plus{}\frac{35}{4}x(x^2\minus{}\frac{54}{35})^2\plus{}\frac{19}{4}(x^2\minus{}1)^2\plus{}\frac{9}{2}(x\minus{}\frac{29}{315})^2\plus{}\frac {229843}{44100} \equal{}0$
And we can immediately see, that there a no positive solutions.

Is there a similar possibility for my equation, there is only one single factor different, instead of 10 it is 4 at the last position. So I hope, we can find a possibility.

Perhaps we can reduce the degree of our equation with the following idea:
f(0)=4>0
If we can schow, that $ f'(k) \geq 0$ then the purpose is also fulfilled, because the function ascends monotoniously with our starting value f(0)=4.

Perhaps, it is easier, to show, that:
$ f'(k)\equal{}9k^8\plus{}8k^7\minus{}49k^6\minus{}30k^5\plus{}105k^4\plus{}44k^3\minus{}81k^2\minus{}10k\plus{}20>0$ for all $ k\geq0$

Isn't there any possibility that can help me to solve the given problem? No idea?


parameterization
real analysis
real analysis unsolved

how can i parametrize this ellypse $ \{ \begin{array}{ll}
                			(x+1)^2+y^2=4 \\
                			z=x-y+3 \\
                			  \end{array}
                		 $ ?

$ x\equal{}2 cos(t) \minus{}2   ,\ y \equal{} 2 sin(t) ,\ z \equal{} 2cos(t)\minus{}2 \minus{} 2 sin(t) \plus{}3$   :cool:


let A be square matrix have form :
 $A = \left(\begin{array}{ccc}0&1&0\\ 0&0&1\\ 0&0&0\\ \end{array}\right)$
existence or nonexistence  a matrix B such that :$B^{2}=A$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=131267]here[/url] or one of many older threads. Please look at least through the last month or so before posting.


:roll: This is not true, for counter-example, $x=\frac{2}{3},\, y=\frac{1}{2},\, z=3$.

You didn't wrote it anywhere  but this is a style of inequalities you kill .So if it was true I will dedicated for you . But it is wrong as it seems


[color=green][b]What can run but never walks, has a mouth but never speaks, has a head but never weeps, and has a bed but never sleeps? :?: ?[/b][/color]

that is a brilliant answer  :clap2:  :clap2:  :clap2:

[quote="chess64"][hide="turkey"]A river?[/hide][/quote]

yup :10:  :D

oh well..it was probably just put in there in reference to the source, but hey, it is a riddle, and it had to keep the rhyming scheme going so, yea, u know? it works....because the answere is not differed becucase of that....

[url=also there]http://www.braingle.com/brainteasers/teaser.php?op=2;id=535;comm=0[/url]
\I cheated :)

[quote="chess64"][quote="SplashD"][quote="chess64"]well that was easy (i guessed that after reading the mouth part) but how do rivers have heads? :D[/quote]

I think their head is the point where they start.[/quote]

i though that was the source :D[/quote]
"Head" or headwaters is a another name for the source of the river.


Simplify this summation sequence:

$1^5+2^5+3^5+4^5+$...$N^5$

This solution assumes that you know the expansions of
 $\displaystyle\sum^n_{k=1} n^4$
 and the expansions for the first $n$ cubes, squares and numbers.
 now,
 $(k+1)^6 - k^6 = 6k^5 + 15k^4 + 20k^3 + 15k^2 + 6k + 1$
 put in a row $k = 1, 2, ... , n$ and add all of them to get the
 required expression.

[quote="shobber"]First we need to know the formulas of $\sum_{i=1}^n i$ and $\sum_{i=1}^n i^2$ and $\sum_{i=1}^n i^3$ and $\sum_{i=1}^n i^4$. Then use $m^6-(m-1)^6= \cdots$, take sum and togather with the sums above, to get a formula for $i^5$.[/quote]

Mind if i ask, what's the closed form for: $\sum_{i=1}^n i^4$?
I kinda forgot it....

[quote="violetcraze"]Mind if i ask, what's the closed form for: $\sum_{i=1}^n i^4$?
I kinda forgot it....[/quote]

$\displaystyle\sum_{i=1}^n i^4=\frac{1}{30}(6n^5+15n^4+10n^3-n)=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$

conceive

Alternately, the summation is a sixth-degree polynomial in $n$, so it can be Lagrange Interpolated.  Messy without a calculator...


$\displaystyle \frac {kR}{r}\leq  \sum_{cyclc}{\frac{a^{2}}{(p-b)(p-c)}}\leq\frac{6R}{r}$

Find the greatest possible value of $k$ if $a,b,c$ are sides of a triangle and $R,r$ is the circumradius/inradius and $p$ is the semiperimeter, respectively.

I am not sure(maybe I am wrong in some calculations) but maybe k=4?  :)

By the triangle inequality three times, we get:

$(a+b-c)(b+c-a)(a+c-b)\geq 0$

which implies

$\sum_{sym} a^2(p-a)\geq abc$

Dividing through by $(p-a)(p-b)(p-c)$, we get:

$\sum_{sym} \frac {a^2}{(p-b)(p-c)}\geq \frac {abcp}{A^2}$

Subtituting $R=\frac {abc}{4A}$, we get

$\sum_{sym} \frac {a^2}{(p-b)(p-c)}\geq \frac {4Rp}{A}$

substituting $A=pr$, we get $K=4$

inequality is not attainable, but 4 cannot be replaced by any greater constant.

Okay I've proved that maximum of k such that inequality is right is 4.But I think [b]blahblahblah  [/b] your proof is not sufficient for proving that k is maximum, am I wrong?  :?  ;)

I think you are wrong, if we have $a>4$, then we can reduce it to

$(a+b-c)(b+c-a)(a+c-b)<0$, which contradicts the triangle inequality.

1)Maybe you mean k>4 and not a>4?
2) [quote]I think you are wrong, if we have , then we can reduce it to

, which contradicts the triangle inequality.[/quote]
That's what we need if you show how you bring it to contradiction then your solution will be completed (that's what I mean't in my previous post).I hope you understand me. :)[/quote]

Ah.  I felt that to be trivial, hence I didn't show it :lol:

In my solution I denoted $b+c-a=x,a+c-b=y,a+b-c=z$,note that $x,y,z>0$.From this after grouping we get $(x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y)(4-k)+(24-2k)xyz\geq{0}$ and obviously if $k>4$ it is not right for all x,y,z. :)
I am waiting for you,because my solution is not so nice.And I want to know a nice solution to a nice problem ;)

I'm surprised that you can't finish it off, Armo  :D

Okay, say that $k<4$

Then this implies, after the usual simplification

$\sum_{sym} a^2(p-a) \geq \frac{k}{4}abc$

But by the triangle inequality, as I already showed,

$\sum_{sym} a^2(p-a)\geq abc$, with equality only for the degenerate triangle.  Thus we must have $k>4$ (if k=4, then the triangle must be degenerate, which implies that one of the terms in the denominator will be zero)

I think for $k>4$ you get that $(a+b-c)(b+c-a)(c+a-b)\geq{\frac{k-4}{2}\*abc}$ not
$(a+b-c)(b+c-a)(c+a-b)<0$ .What you think?


algebra
function
domain
complex analysis
complex analysis unsolved

Is there any canonical domain in the complex plane, which if conformally isomorphic to the $ \mathbb{C}$ without compact interval and a single point ?

If you include $ \infty$, then the domain can be mapped onto $ \mathbb D\setminus \{0\}$. You can map the exterior of interval onto the unit disk $ \mathbb D$ (even explicitly if you want). Then send the removed point into $ 0$ by Mobius transformation.

If the original domain does not include $ \infty$, then there are uncountably many conformal types of such domains. Each of them is equivalent to $ \mathbb D\setminus \{0,t\}$ for some $ 0 < t < 1$.


Can anybody help me with this vector problem?

 [url=http://img124.imagevenue.com/img.php?image=11703_clip_image002_122_818lo.jpg][img]http://img124.imagevenue.com/loc818/th_11703_clip_image002_122_818lo.jpg[/img][/url]

I'd really appreciate any help
Thanks,
steve

First note that $ \vec{OA}= \vec{OB}+\vec{BA}= \vec{OC}+\vec{CA}$.

Since $ \vec{BA}= \frac{3}{10}\vec{BC}= \frac{3}{10}( \vec{OC}-\vec{OB})$ and $ \vec{CA}= \frac{7}{10}\vec{CB}= \frac{7}{10}(\vec{OB}-\vec{OC})$.

Combining the two we get:

$ \vec{OA}= \vec{OB}+\frac{3}{10}( \vec{OC}-\vec{OB})$

$ \vec{OA}= \vec{OC}+\frac{7}{10}(\vec{OB}-\vec{OC})$.

Plus the two together we have and do a bit of algebraic manipulation and we have:

$ \vec{OA}= \frac{7}{10}\vec{OB}+\frac{3}{10}\vec{OC}$.

Answer

$ BA: AC = 3: 7$

$ \Leftrightarrow 7\vec{BA}= 3\vec{AC}$

$ \Leftrightarrow 7(\mathbf{a-b}) = 3(\mathbf{c-a})$

$ \Leftrightarrow 10\mathbf{a}= 7\mathbf{b}+3\mathbf{c}$

$ \Leftrightarrow \vec{OA}= \frac{7}{10}{\vec{OB}}+\frac{3}{10}{\vec{OC}}$

@brocool: you can also try and solve the general problem.

@sludgethrower: brilliant, better than mine at least.


Adele can mow her entire yard in three hours. Her sister, Brooke,
can mow 3/4 of the same yard in two hours. Using two identical
mowers, what part of the yard can they mow in one hour working
together? Express your answer as a common fraction.

if her sister can do 3/4 in 2 hours, she can mow the whole lawn in 2/.75=8/3 hours. that means in 1 hour she does 1/(8/3)=3/8 of it an adele does 1/3 of it. 1/3+3/8=17/24


Why are all the posts from 1 January till now deleted?

Probably because we tried to make a Fun Factory, which basically disintegrated to spam.

"we" tried to make a fun factory?  *COUGH TINYTIM AND MULTIS COUGH*

Oh, and 5000th post now...


conics hyperbola ellipse parabola rotation

What conic section is created by the equation -4=3x^2-18xy+27y^2-5x+7y?
A)cardiod
B)circle
C)ellipse
D)hyperbola
E)parabola

Also, can someone give me a list of the conic sections standard formulas in expanded form? It would really helpful because the questions are always in expanded form.

Thanks a lot!

Here are the equations:
Circle:$ (x \minus{} h)^2 \plus{} (y \minus{} k)^2 \equal{} r^2$
Hyperbola:$ \frac {(x \minus{} h)^2}{a^2} \minus{} \frac {(y \minus{} k)^2}{b^2} \equal{} 1$ or $ \frac {(y \minus{} k)^2}{a^2} \minus{} \frac {(x \minus{} h)^2}{b^2} \equal{} 1$ (a is under positive term)
Ellipse: $ \frac {(x \minus{} h)^2}{a^2} \plus{} \frac {(y \minus{} k)^2}{b^2} \equal{} 1$ or $ \frac {(x \minus{} h)^2}{b^2} \plus{} \frac {(y \minus{} k)^2}{a^2} \equal{} 1$ ($ a > b$)
Parabola: $ y \minus{} k \equal{} \frac {1}{4a}(x \minus{} h)^2$ or $ x \minus{} h \equal{} \frac {1}{4a}(y \minus{} k)^2$
Where $ (h,k)$ is always center of conic section

As for $ \minus{}4\equal{}3x^2\minus{}18xy\plus{}27y^2\minus{}5x\plus{}7y$
I would say its rotated ellipse.

[quote="iamhe"]can someone give me a list of the conic sections standard formulas in expanded form? It would really helpful because the questions are always in expanded form.[/quote]

That probably won't help you at all, especially with an $ xy$ term there.  See [url=http://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_conic_section]the discriminant of a conic section[/url].


function
calculus
derivative
inequalities unsolved
inequalities

If t is positive real number find maximum of function:

t*t*(3t-2)*(2-t)

For which t is this maximum attained?
(here * is multiplication sign)

Do you mean $f(t) = t^{2}(3t-2)(2-t)$?

[hide="If so"]

$f(t)=-3t^{4}+8t^{3}-4t^{2}$

So $f'(t) =-12t^{3}+24t^{2}-8t$ and $f''(t)=-36t^{2}+48 t-8$

Looking at the first derivative:

$f'(t)= t(-12t^{2}+24t-8)$

Solving for the cubic we have $t=0, 1 \pm \frac{1}{\sqrt{3}}$ then we can plug then into the original equation and check.

So maximum occurs at $t= 1+\frac{1}{\sqrt{3}}$[/hide]

You are so right, I didn't check my working  :blush:  .  Thanks, it is edited now.

Thank you very much for this help.
I am not so familiar with these f'(t) and finding extreme values of functions but I understand everything in this solution.


topology
advanced fields
advanced fields unsolved

How do I show that $S^{n}$ is not contractible.
For $S^{1}$ it is trivial how do I do it for higher n.

[quote="ZetaX"]In short: a contractible space must have trivial homology groups.
But $\tilde{H}_{n}(S^{n}) = \mathbb{Z}$.[/quote]
Do we have to use homology.Can we not do it otherwise?

You can use homotopy instead  :)  $\pi_{n}(S^{n})=\mathbb Z$.

How do you know that $S^{1}$ is not contractible?

If it were contractible, then $\text{id}_{\mathbb{S}^{1}}$ would be nullhomotopic. On the other hand, $\text{deg}(\text{id}_{\mathbb{S}^{1}})=1$ proving it cannot be nullhomotopic.

Degree is even stronger than homology ;)
And I think mlok's question was for the game, it would be strange if he couldn't show for himself.

It due to the fact that pie-1(S-1) is Z,but pie-1(S-n) is trivial for n>1


ensan if p is prime i have the solution .but p isn't prime it's so hard to have a solution

for p is prime you have 4*x^4-p*y^4=-1=>4*x^4+1=p*y^4=>(2*x^2+1)^2=py^4+(2*x)^2;
put 2*x^2+1=A;y^2=B;2*x=C;
then we have A^2=p*B^2+C^2=>(A^2-C^2)=p*C^2=>(A-C)*(A+C)=p*B^2;
let (A-C,A+C)=d then A-C=d*m;A+C=d*n;
hence m*n*p is perfect square then we have m=p*u^2;n=v^2 or m=u^2;n=p*v^2;
continue we have the solution :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:


do y'all talk about mathematical linguistics here in the sense of formal theories of language and so forth or does "linguistics" (in the sense of this forum title) more commonly refer to something distinct?

read gabe carroll's harvard thesis sometime

i think linguistics just means anything to do with languages in this case


geometry
circumcircle
geometric transformation
geometry proposed

$ AD$, $ BE$ and $ CF$ are altitudes of acute $ \triangle{ABC}$. $ P$ and $ Q$ are on segments $ DF$ and $ EF$ respectively.

Show that $ \measuredangle{PAQ}\equal{}\measuredangle{DAC}$ if and only if $ AP$ bisects $ \angle{QPF}$.

[quote="shobber"]$ AD$, $ BE$ and $ CF$ are altitudes of acute $ \triangle{ABC}$. $ P$ and $ Q$ are on segments $ DF$ and $ EF$ respectively.

Show that $ \measuredangle{PAQ} \equal{} \measuredangle{DAC}$ if and only if $ AP$ bisects $ \angle{QPF}$.[/quote]

Let $ Q$ be the foot of perpendicular from $ A$ to $ FE$, $ P$ is a point on $ DF$ such that $ \angle PAQ \equal{}\angle DAC$.
Suppose that $ AP$ is a bisector of $ FPQ$. Then $ A$ is the excenter of triangle $ FPQ$ and $ A$ is the excenter of triangle $ FDE$ $ \Longrightarrow$
$ \Longrightarrow$ $ QP$ and $ QE$ are two tangent lines from $ Q$ to circle with center $ A$ and radius $ AQ$???

The problem is correct.
[hide="Solution"]
Let the orthocenter of $ \triangle ABC$ be $ H$, $ AQ$ meet $ FC$ at $ X$, and $ AP$ meet $ FC$ at $ Y$.  $ \angle PAQ \equal{} \angle DAC$ implies that $ \angle PAD \equal{} \angle QAC$.  Since $ \angle AFC \equal{} 90 \equal{} \angle ADC$, $ AFDC$ is cyclic, and, hence, $ \angle ADP \equal{} \angle ACX$.  This means that $ \triangle APD\sim \triangle AXC$, so a spiral similarity maps $ \triangle APX \sim \triangle ADC$, so $ \angle APX \equal{} 90$.  Now, $ \angle HFA \equal{} 90 \equal{} \angle HEA$, so $ AFHE$ is cyclic, so $ \angle AHY \equal{} \angle AEQ$.  Thus, $ \triangle AHY\sim \triangle AEQ$, so a spiral similarity maps $ \triangle AEH\sim \triangle AQY$.  Therefore, $ \angle AQY \equal{} \angle AEH \equal{} 90 \equal{} \angle APX$, so $ YQXP$ is cyclic and it follows that $ \angle APQ \equal{} \angle FXA \equal{} \angle FPA$. [/hide]

EDIT:
[b]Nak[/b]: If $ Q$ is the foot of the perpendicular from $ A$ to $ FE$, then since $ \angle AFQ \equal{} \angle ACB$, we would have that $ \angle FAQ \equal{} 90 \minus{} \angle ACD \equal{} \angle DAC$, so $ F \equal{} P$.

It is sufficient to show that if $\angle PAD = \angle QAC$, then $\angle FPA = \angle APQ$. Let $R$ be the intersection of $AC$ and $FC$. $\angle ACR = \angle ADF$ and $\angle RAC = \angle PAD$, so triangles $APD$ and $ARC$ are spirally similar, whence $APR$ and $ADC$ are similar.

Let $PR$ intersect $AB$ at $S$, and let $FR$ intersect $AP$ at $X$. $FECB$ is cyclic, since $\angle BFC = \angle BEC = 90^{\circ}$; $FXPS$ is cyclic, since $\angle SFX = \angle XPS = 90^{\circ}$; $FQRS$ is cyclic, since $\angle AFQ = \angle AFE = \angle ACB = \angle ARP$; $AFPR$ is cyclic, since $\angle APR = \angle AFR = 90^{\circ}$; and $XQRP$ is cyclic, since $AQ \cdot AR = AF \cdot AS = AX \cdot AP$. Hence, $\angle ARX =  \angle APQ = \angle FPA$, as desired.

I have an other solution to this nice problem :D
----------------
First $Q$ and $C$ must be on the same side of line $AP$
----------------
If $\widehat{PAQ}=\widehat{DAC}$
Let $O$ be circumcenter of triangle $AFQ$.
We have $\widehat{OAQ}=90^{0}-\frac{\widehat{AOQ}}{2}=90^{0}-\widehat{AFQ}=90^{0}-\widehat{ACB}=\widehat{DAC}=PAQ$ so $A,O,P$ are collinear
Hence $\widehat{PFQ}=180^{0}-\widehat{PFB}-\widehat{QFA}=180^{0}-2\widehat{ACB}=2\widehat{DAC}=2\widehat{PAQ}=\widehat{POQ}$
So $PFOQ$ is cyclic but $OF=OQ$ thus $\widehat{OPF}=\widehat{OPQ}$ or $AP$ bisects $\widehat{QPF}$.
----------------
If $AP$ bisects $\widehat{QPF}$ then let $Q'$ on segment $EF$ such that $Q'$ and $C$ are on the same side of line $AP$ and $\widehat{PAQ'}=\widehat{DAC}$ and it's easy to prove that $Q\equiv Q'$.:)


function
topology
real analysis
real analysis unsolved

Let $(X,d)$ be an arbitrary metric space, and let $A \large \subset X$ be compact. Let $f : X$ → $R^{+}$ be an arbitrary (not neccesarily continuous) nonnegative function, such that
                 $Inf$ { $f(x) | d(x,A)$ ≥ $a$ } > $0$
for each $a$ > $0$. Prove : each sequence { $x_{n}$ } in $X$  for which $f(x_{n})$ → $0$ contains a subsequence converging to some point of $A$.

should be read $\inf\{f(x) : d(x,A) \geq a\}$ I think. The condition says that if $f(x_{n}) \to 0$, then $d(x_{n}, A) \to 0$. Now choose $a_{n}\in A$ such that $d(x_{n},A)=d(x_{n},a_{n})$ (possible since $A$ is compact). Because $A$ is compact, you can suppose (extract a subsequence) that $a_{n}$ converges towards a point $x \in A$. The it is clear that $a_{n}\to x \in A$ because $d(x_{n},a_{n}) \to 0$.


Heres my code:

\documentclass[11pt]{article}
\usepackage[pdftex]{graphicx}
\begin{document}
Hello, world!
\includegraphics{faces.png}
\includegraphics{image1.png}

\end{document}

The problem is that image1 shows up but faces doesn't.
Here are the files

I don't get the same effect here with the code you've written. It looks as if the images are on top of each other. Is there any other code you are using? What happens if you add a new line using \\ between the images? What if you put the images in a figure environment \begin{figure}...\end{figure}?

I tried changing my code and now it says

==> Fatal error occurred, the output PDF file is not finished!

I know this usually happens when the PDF file is open while you're building it but I don't have any PDF files open.

Could you tell us the errors TeX is giving you? They will be in the log file. Do you get the same problems with other images?

I'm only having trouble with this image.  In other files I've used images successfully.

Here's what's in the LOG file:

This is pdfeTeX, Version 3.141592-1.30.6-2.2 (MiKTeX 2.5) (preloaded format=pdflatex 2006.11.4)  1 JAN 2007 17:14
entering extended mode
**C:/Users/Ofir/LaTeX/Hello/HelloWorld.tex
(C:\Users\Ofir\LaTeX\Hello\HelloWorld.tex
LaTeX2e <2005/12/01>
Babel <v3.8g> and hyphenation patterns for english, dumylang, nohyphenation, ge
rman, ngerman, french, loaded.
("C:\Program Files\MiKTeX 2.5\tex\latex\base\article.cls"
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
("C:\Program Files\MiKTeX 2.5\tex\latex\base\size11.clo"
File: size11.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
\c@part=\count79
\c@section=\count80
\c@subsection=\count81
\c@subsubsection=\count82
\c@paragraph=\count83
\c@subparagraph=\count84
\c@figure=\count85
\c@table=\count86
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
)
("C:\Program Files\MiKTeX 2.5\tex\latex\graphics\graphicx.sty"
Package: graphicx 1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)

("C:\Program Files\MiKTeX 2.5\tex\latex\graphics\keyval.sty"
Package: keyval 1999/03/16 v1.13 key=value parser (DPC)
\KV@toks@=\toks14
)
("C:\Program Files\MiKTeX 2.5\tex\latex\graphics\graphics.sty"
Package: graphics 2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR)

("C:\Program Files\MiKTeX 2.5\tex\latex\graphics\trig.sty"
Package: trig 1999/03/16 v1.09 sin cos tan (DPC)
)
("C:\Program Files\MiKTeX 2.5\tex\latex\00miktex\graphics.cfg"
File: graphics.cfg 2005/12/29 v1.2 MiKTeX 'graphics' configuration
)
Package graphics Info: Driver file: pdftex.def on input line 90.

("C:\Program Files\MiKTeX 2.5\tex\latex\graphics\pdftex.def"
File: pdftex.def 2002/06/19 v0.03k graphics/color for pdftex
\Gread@gobject=\count87
))
\Gin@req@height=\dimen103
\Gin@req@width=\dimen104
) (HelloWorld.aux)
LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 3.
LaTeX Font Info:    ... okay on input line 3.
LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 3.
LaTeX Font Info:    ... okay on input line 3.
LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 3.
LaTeX Font Info:    ... okay on input line 3.
LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 3.
LaTeX Font Info:    ... okay on input line 3.
LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 3.
LaTeX Font Info:    ... okay on input line 3.
LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 3.
LaTeX Font Info:    ... okay on input line 3.
 ("C:\Program Files\MiKTeX 2.5\tex\context\base\supp-pdf.tex"
("C:\Program Files\MiKTeX 2.5\tex\context\base\supp-mis.tex"
loading : Context Support Macros / Miscellaneous (2004.10.26)
\protectiondepth=\count88
\scratchcounter=\count89
\scratchtoks=\toks15
\scratchdimen=\dimen105
\scratchskip=\skip43
\scratchmuskip=\muskip10
\scratchbox=\box26
\scratchread=\read1
\scratchwrite=\write3
\zeropoint=\dimen106
\onepoint=\dimen107
\onebasepoint=\dimen108
\minusone=\count90
\thousandpoint=\dimen109
\onerealpoint=\dimen110
\emptytoks=\toks16
\nextbox=\box27
\nextdepth=\dimen111
\everyline=\toks17
\!!counta=\count91
\!!countb=\count92
\recursecounter=\count93
)
loading : Context Support Macros / PDF (2004.03.26)
\nofMPsegments=\count94
\nofMParguments=\count95
\MPscratchCnt=\count96
\MPscratchDim=\dimen112
\MPnumerator=\count97
\everyMPtoPDFconversion=\toks18
)
Error:  (file faces.png): libpng: internal error
 ==> Fatal error occurred, the output PDF file is not finished!

It looks like faces.png is damaged. Try recreating it. For example, re-download the version you posted here or else open up the file in a different program and re-save again.

When I downloaded it from here it only let me download it as a bitmap so I made it into a pdf file using Virtual PDF Printer and the picture still did not show.  I tried converting image1 into a pdf file and it also did not show as a pdf file.  Is this the problem?  Should I use a different program for converting files to pdf?  Because I saw that the download for it is not shown anymore in the LaTeX section of AoPS.  Is there a bug in this program?

I don't use Virtual PDF Printer so I'll leave others to comment on it.

But you don't have to convert to pdf. When I tested your code I right-clicked on the image and saved the picture in png format. This process may uncorrupt the file. I then compiled the code and everything worked fine. What happens if you do the same?


Okay, so I had to row reduce the matrix below to upper-triangular form in order to find the determinant.

[0, 0, 2, 0, 0]
[2, 0, 58, 96, -1]
[2, 1, 58, 96, -1]
[0, 0, 2, -1, 0]
[1, 0, 58, 96, 2]

Row-reducing it to upper-triangular form, I wound up with:
[1, 0, 29, 48, -1/2]
[0, 1, 0, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 0, 1, 0]
[0, 0, 0, 0, 3/2]

I know that to calculate the determinant of an n by n upper-triangular matrix, all you do is multiply the entries along the diagonal.  However, when I found the determinant of the first matrix using cofactors, I got -10.  Multiplying the entries along the diagonal of the row-reduced upper triangular form, I got 3/2.  Am I doing something wrong here?  I've done out the row-reduction several times, and I keep getting a determinant of 3/2.

If you're trying to find a determinant, you have to keep track of scalings and transpositions. It looks like your first two steps were to swap the second row to the top and divide it by 2; both of those change the determinant.


[hide]
This is clearly true for points A_1 and B_1. Let g: R->P be such that g(x) = the point P on line A_1B_1 such that the DIRECTED distance A_1P = x. (Let A_1B_1 be positive). Define f:P->R = distance from point to AB - distance from AC - distance from BC. Note that f(g(x)) is linear and f(g(A_1)) = 0 and f(g(B_1)) = 0. Therefore, f(g(x)) = 0 for all x, which is what we wanted.[/hide]

your solution seems elegant but beyond my comprehension.....  :blush:

kindly someone post a simpler geometrical proof.....

P.S Kindly use latex... it makes solutions look good......

[quote="anurags92"]In $ \bigtriangleup$ $ ABC$ bisectors $ AA_{1}$ and $ BB_{1}$ are drawn. Prove that the distance from any point $ M$ of $ A_{1}B_{1}$ to line $ AB$ is equal to the sum of distances from $ M$ to $ AC$ and $ BC$.[/quote]
$ MN \perp AB;MV \perp BC;MU \perp AC;MN\equal{}z;MV\equal{}x;MU\equal{}y$
We need prove that $ z\equal{}x\plus{}y$
Suppose that $ MB_1\equal{}p;MA_1\equal{}q$;$ B_1S;A_1K$ are perpendicular to $ AB$;$ A_1L;B_1T$ are perpendiculars to $ AC$ and $ BC$
Suppose that $ SB_1\equal{}B_1T\equal{}b;A_1K\equal{}A_1L\equal{}a$,we have
From $ B_1S//MN//A_1K$ we have $ z\equal{}\frac{ap\plus{}bq}{p\plus{}q}$
In triangle $ B_1A_1T$ we have $ MV //B_1T$ so
$ \frac{x}{b}\equal{}\frac{q}{p\plus{}q}$=>$ x\equal{}\frac{bp}{p\plus{}q}$
Similar we have $ y\equal{}\frac{ap}{p\plus{}q}$
So $ z\equal{}x\plus{}y$
Done!

kudos...... to you.......

great solution...


This problem I believe is from the HMMT, but I'm not sure...

"Somehwere in teh universe, n students are taking a 10-question math competition. Their collective performance is called laughable if, cofr some pair of questions, there exist 57 studnets such that either all of themj answered both questions correctly or none fo them answerered both questions correctly. Compute the smallest n such that the performance must be laughable, no matter how the studnets performedon the competition."

I understand how one can prove that the answer is 253. Then comes the part where you have to show that 252 is insufficient. And that's the part I don't get. Can someone explain that to me?

I would greatly appreciate your help.

Note that there are $ 252$ distinct $ 5$-question subsets of the $ 10$ questions. Therefore, if there are $ 252$ students, then each student can answer a distinct $ 5$-question subset correctly. Note that, for a given pair of questions, there are only $ 56$ subsets that contain any given pair, and by symmetry, since no pair is correctly answered, no pair is incorrectly answered.

Don't you also have to pove 253 is sufficient?

[quote="theax"]I understand how one can prove that the answer is 253. Then comes the part where you have to show that 252 is insufficient. And that's the part I don't get. Can someone explain that to me?[/quote]

I felt lazy. :P


function
calculus
derivative
calculus computations

find all $ f: \mathbb{R} \to \mathbb{R}$ functions for which:

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

Where $ f^{\minus{}1}$ is the inverse function of $ f$. that is $ f(f^{\minus{}1}(x))\equal{}x$

Did you mean [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37916]this[/url]? As asked, the question hardly makes sense because the derivative then should take both positive and negative values, so $ f$ cannot be 1 to 1, so what inverse are we talking about?

If you meant the problem I think you meant, then
a) It has been discussed already
b) It is too hard for Computations and Tutorials, so it should be posted in Unsolved and Proposed.


Find the units digit (base 10) in the sum $ \sum_{k\equal{}1}^{99}{k!}$

its $ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}...\plus{}99!$.

However, since there are at least one 2 and 5 in their factorization, 5!, 6!, ...'s units digits are all 0.

so we only need to find the units digit of $ 1!\plus{}2!\plus{}3!\plus{}4!$, which is $ 1\plus{}2\plus{}6\plus{}4\equal{}13$

answer : 3

This, by no means, should be in intermediate, in fact i think there was an MC problem like that...

Hm maybe the notation itself qualifies this to be in Intermediate...  :idea:


Well ,My friends and I ,Came accros , this problem which ididnt managed to find a solution for it .

[color=blue]Problem [/color]

Let $AB$ be a given segment  .just with compass find the midpoint of $AB$.

[quote="lomos_lupin"]Well ,I found this problem whichI didnt managed to find a solution for it .

[color=blue]Problem [/color]

Let $AB$ be a given segment  .just with compass find the midpoint of $AB$.[/quote]
i really wanted to post this problem as soon as i could but you did it sooner. :D

[color=blue]Hint.[/color]At first try to  construct a segment with length 2 times bigger the first one.Then...

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53829]Midpoint of a segment with only circles[/url] in the high school section.


I obtained this problem from kiran's notes.

Let x_1, x_2, ..., x_n (n>=2) be positive integers satisfying 1/(x_1 + 1998) + 1/(x_2 + 1998) + ... + 1/(x_n + 1998) = 1998.

Prove that [(x_1 * x_2 *.... * x_n)^(1/n) / (n-1)]  :ge: 1998.

Q1: from the given, x_1 can be nonpositive?
Q2: how do i solve it? heh.

i believe i'm supposed to apply some inequality results involving convex functions (but there's a warning associated with this quesotin asking to beware of nonconvexity). i just learnt the theorems today, and it's pretty deep material, would be glad if someone can point me to more introductory material. thanks in advance for help.

what a coincidence..I'm learning about smoothing right now too!  As well as convexity, etc...Well, anyways, smoothing is where you take two of the variables and "smooth" one of them to a desired value and keep a sum or product constant while changing the expression such that it becomes smaller to a minimum or larger to a maximum.  Ex: AM-GM
We want to show that
(x_1+x_2+...x_n)/n>=nthroot(x_1*x_2*...*x_n).
We would smooth the left side until it reaches its minimum (the right side) by doing this:
suppose we have x_i<a<x_j, where a is the arithmetic mean.  We smooth x_i to a and keep the sum constant so that x_j'=x_i+x_j-a.  So we have the products x_i*x_j and x_i'*x_j'=a(x_i+x_j-a).  Now we want to show that a(x_i+x_j-a) is smaller than x_i*x_j.  Notice that a(x_i+x_j-a)=x_i*x_j+(x_j-a)(a-x_i)>x_i*x_j since x_j-a and a-x_i are positive.  So we do this for all x_i's until we reach n^2/n=n=nthroot(n^n).
Hope this helps a bit.

-interesting_move

I came up with this:

let a be the arithmetic mean of x_1, x_2, ..., x_n and replace each x_i by a so that we have 1/(x_1 + 1998) + 1/(x_2 + 1998) + ... + 1/(x_n + 1998) = 1998 <= n/(a+1998), equality when all x_i same.

n/(a+1998) >= 1998 => a+1998 <= 1998n => a <= 1998(n-1)
by the am-gm inequality, we have (x_1 * x_2 *...*x_n)^(1/n) <= a <= 1998(n-1) and the desired result follows.

1) is the solution i outlined above correct?
2) can someone present me a formal smoothing argument for the part about replacing each x_i by a?

Isn't your first inequality backward?  For example, 1/1999 + 1/2001 is greater than 2/2000, not less than.

Also, isn't there something wrong in your step going from n / (a + 1998) :ge: 1998 to a + 1998 :le: 1998 n?  Shouldn't that last inequality be a + 1998 :le: n / 1998?

Also, isn't your last inequality backward?  That is, you seem to be trying to show the opposite inequality to what your original post states.

Finally, is the original inequality even true?  What if you choose all the xi = 1, where n = 1998 * 1999?  Doesn't that choice violate the proposed inequality?

argh messed up pretty much back there. thanks for correction.

the question was wrong! Here's the correction.

Let x_1, x_2, ... , x_n (n :ge: 2) be positive integers satisfying
 :Sigma: 1/(x_i + 1998) = 1/1998 from i=1 to n.

Prove that (x_1 * x_2 * .. * x_n)^(1/n) / (n-1)  :ge: 1998

my solution:
Take reciprocal on both sides, we have lhs = 1998. and if all x_i are replaced by their arithmetic mean, a/(n-1)  :ge:  1998. and i'm stuck. anyone? also, i need help formulating an argument that if all x_i are replaced by their arithmetic mean, the inequality results (as above). possibly through smoothing?

thanks in advance.

Well, the part about replacing each xi by their arithmetic mean works because of the AM-HM (arithmetic-harmonic mean) inequality.  That will show that the arithmetic mean of the xi is at least 1998(n - 1).  But that won't help you with the original problem, because the geometric mean is at most the arithmetic mean, not at least.


Geom 3

The symmedian corresponding to A intersects the circumcircle of ABC at A1. The middle of AA1 is M. Prove that circle AMC is tangent to AB and circle AMB is tangent to AC.

Geom 1, 2 and 3 were taken from the book ?Pagini inedite? or, in English, ?Intriguing pages? (or something like that :) ), which contains articles and letters and stuff written by the great Romanian mathematician (and poet!) Dan Barbilian.

The symmedian through A is the isogonal conjugate of the median through A.

If we apply inversion with center A then simedian AA1 will become median again! That is all what we need!


$ 2007^2$ unit squares are arranged forming a $ 2007\times 2007$ table. Arnold and Bernold play the following game: each move by Arnold consists of taking four unit squares that forms a $ 2\times 2$ square; each move by Bernold consists of taking a single unit square. They play anternatively, Arnold being the first. When Arnold is not able to perform his move, Bernold takes all the remaining unit squares. The person with more unit squares in the end is the winner.

Is it possible to Bernold to win the game, no matter how Arnold play?

This is my solution of the problem:

A [i]snake[/i] is a subboard that consists of the first $ 2k$ rows plus the fields
$ (2k\plus{}1,i)$ and $ (2k\plus{}2,i)$ for all $ i\equal{}1,2,\ldots,l$ for some natural numbers $ k,l$.

A $ 2\times 2$ square claimed by player A (Arnold) during one of his moves we call an [i]A-block[/i].
We will assign to each A-block a [i]charge[/i], an integer value from the set $ \{0,1,2\}$.
Each unit of charge is associated with one square that has not been claimed yet
and that player A can not claim anymore.

Player B (Bernold) maintains a snake S during his play with the following properties:
(1) No fields on the snake can be claimed by player A in future moves (the fields
are not necessarily claimed by player B yet). All fields that player B has claimed
yet are on the snake.
(2) The fields on S that are not claimed by player A fall into one of two categories:
Category RB (reserved by B): fields that are already claimed by player B or that
will be claimed at the end of the game when player A cannot move anymore
Category CB (charged by B): fields that are not yet claimed by player B and that
will turn into category RB fields later
(3) After every move of player B the number of RB fields is at least as large as the
total number of fields that player A has claimed already.
(4) If the head of the snake is on rows $ 2k\plus{}1$ and $ 2k\plus{}2$, then each A-block
on rows $ 2k$ and $ 2k\plus{}1$ carries a charge equal to the number of squares from
this block that are outside of S. Each A-block on rows $ 2k\plus{}2$ and $ 2k\plus{}3$ carries
a charge equal to the number of squares it has in common with S. The total charge
equals the number of category CB squares.

If we can show how player B can make the snake grow in every move such that all these properties
are maintained up to the last move of player A, then player B will win the game. This is because in his
last move A gains an advance of at most 4 points. In the next move of player B the snake grows to a
size of $ 2006\times 2007$, and there is only one row left. The squares on this row are either free to
be claimed by player B or they have been claimed by player A before, in the latter case the corresponding
A-blocks intersect with the snake in two fields, hence they carry a charge of 2, this corresponds to two CB
fields that turn into RB fields now. Hence player B finally claims all RB fields he has not claimed yet and
wins with an advance of at least 2007-4=2003 points.

This is how player B makes the snake grow in every move (if you reach the end of the board jump
around to the next pair of rows):

(start)
Distinguish the type of the squares $ X: \equal{}(2k\plus{}1,l\plus{}1)$ and $ Y: \equal{}(2k\plus{}2,l\plus{}1)$ ahead of the snake:

(i) X and Y are both claimed by player A.
If X and Y belong to the same A-block, set $ \Delta: \equal{}1$ and goto (grow), otherwise transfer one
unit of charge from the A-block that X belongs to to the A-block that Y belongs to, set $ \Delta: \equal{}1$
and goto (grow).

(ii) X is free and Y is claimed by player A.
Put X into the set CB and charge the A-block that Y belongs to with one unit of charge.
Set $ \Delta: \equal{}1$ and goto (grow).

(iii) Y is free.
Let $ X': \equal{}(2k\plus{}1,l\plus{}2)$ and $ Y': \equal{}(2k\plus{}2,l\plus{}2)$.
Distinguish the type of the square Y':

(iii a) Y' might be claimed by future moves of player A (in particular, Y' is free).
If we have already claimed any field in this move goto (end).
Otherwise claim Y' and put Y and Y' into the set RB. If X is free, put it into the set RB, otherwise
discharge the A-block that X belongs to by one unit of charge, and transfer the associated square
from CB to the set RB. Do the same with X'. Hence RB increases by exactly four elements.
Set $ \Delta: \equal{}2$ and goto (grow).

(iii b) Y' can not be claimed by future moves of player A (either because it has been
claimed by player A already, or it is not part of a free 2x2 square).
Put Y into the set RB. If X is free, put it into the set RB, otherwise discharge the A-block
that X belongs to by one unit of charge, and transfer the associated square from CB
to the set RB. Set $ \Delta: \equal{}1$ and goto (grow).

(grow) Let the snake grow by $ \Delta$ and goto (start).

(end) End of the move of player B.

It can easily be checked that each of the properties (1)-(4) is maintained during the
algorithm. Hence player B can win the game, regardless of A's strategy.
q.e.d.

Note, that our strategy works on all boards with one odd side length.

Torsten

[quote="singapore1728"]Now, name each of the square on the board by its centroid, and put the centroid of the lowest left corner at (0,0). No matter how A plays, B first takes a sqaure at (1,1) (if that corner is occupied, choose one of the other 3), as it is the same as occupying 4 square at one go since A cannot take (0,0), (0,1), (1,0). After that, if unobstructed, take squares at (1,3), (1,5)... and always try to take the square that unable A to take the other 3 'associated with it', just like the example previously shown. For the most optimal case for A, the game will finish at the (2007^2-1)/4 step, ended by B. But then when B take all the remaining squares, which A cannot take, B would have one more square than A. So B always win.[/quote]

i think this is wrong. if the game finishes at the (2007^2-1)/4 step, that means A will take 4 * (2007^2-1)/4 unit squares, which is equal to 2007^2-1, so A wins then.

Label the squares from (0,0) to (2006,2006).  Color all squares (a,b) black if a and b are both odd, white otherwise. There are 1003*1003 black squares. All of the 2x2 blocks will occupy 1 black square exactly, no matter where it is placed. B's optimal strategy is to place a tile on any unoccupied black square.  One black square will get occupied every turn, and once there are no more black squares A cannot place any more tiles.  A will have only 4*(1003*1003+1)/2 tiles occupied, which is less than half of 2007*2007.

Here is a different solution than the above.

We claim that it is indeed possible for Bernold to win the game. At any point in time, call a point $P$ [b] admissible [/b] if we are able to place a $2 \times 2$ square centered there. When Arnold places a $2 \times 2$ square centered at $P$, we will say that he has [b] played on [/b] $P$. Then, notice that there are $2006^2$ admissible points in the beginning, and they form a $2006 \times 2006$ lattice grid. Observe that Arnold wins if and only if he plays at least $\frac{1003 \cdot 1004}{2} + 1$ times. We will show that Bernold can prevent this. First, partition the grid into $\frac{2006^2 - 2^2}{8}$ $2 \times 4$ subgrids (of either orientation) and one $2 \times 2$ subgrid (this is easy to do). Then, we will describe a strategy which allows Bernold to ensure that Arnold ends up playing on at most one point in each of the subgrids. Indeed, his strategy will be as follows. Whenever Bernold plays in a subgrid, he makes at least an entire half of the subgrid (if it's a $2 \times 4$ grid) no longer admissible. Then, Bernold will place his $1 \times 1$ square in the unit square bounded by the remaining $2 \times 2$ subgrid that is still admissible to make it un-admissible, as long as this space is vacant. Otherwise, the entire subgrid is already un-admissible and Bernold will just play wherever is empty. In this way, Bernold can restrain Arnold to at most $\frac{2006^2 - 2^2}{8} + 1 < \frac{1003 \cdot 1004}{2} + 1$ moves, as desired.

$\square$

Let's put a mark in center cell of each 3*3 square. in every move Arnold is reducing one of the marked cells so Bernold shall do the same. when all marked cells are gone Arnold has (1003^2+1)/2 * 4 cells and it's obviously less than 2007^2/2 so Bernold will have more cells and win


geometry
circumcircle
symmetry
geometric transformation
reflection
perpendicular bisector
geometry unsolved

Let $ M,N$ be two distinct points on the circumcircle $ (O)$ of $ \triangle ABC$. $ A_1;B_1;C_1$ are three images of $ M$ through the symmetry with pole $ BC;CA;AB$ respectively. $ NA_1$ intersects $ BC$ at $ A_2$. $ B_2;C_2$ are defined similarly. Prove that $ A_2;B_2;C_2$ are collinear.

Change of notation, to compare with previously posted problems:

Let $ P,Q$ be two distinct points on the circumcircle $ (O)$ of $ \triangle ABC$. $ Q_1,Q_2,Q_3$ are three reflections of $ Q$ in $ BC, CA, AB,$ respectively. $ PQ_1$ intersects $ BC$ at $ S_1$. Points $ S_2, S_3$ are defined similarly. Prove that $ S_1, S_2, S_3$ are collinear.

$ (O_1), (O_2), (O_3)$ are circumcircles of $ \triangle PQS_1, \triangle PQS_2, PQS_3$ coaxal with circumcircle $ (O)$ of $ \triangle ABC.$ Perpendiculars to $ BC, CA, AB$ at $ S_1, S_2, S_3$ cut $ (O_1), (O_2), (O_3)$ again at $ K, L, M$ on the center line $ o \equiv OO_1O_2O_3.$ Let $ AK, BL, CM$ cut $ (O_1), (O_2), (O_3)$ again at $ U, V, W.$ According to  [url]http://www.mathlinks.ro/viewtopic.php?t=186477[/url], points $ S_1, S_2, U, V$ are collinear. Similarly, $ S_2, S_3, V, W$ are collinear and $ S_3, S_1, W, U$ are collinear $ \Longrightarrow$ all six points $ S_1, S_2, S_3, U, V, W$ are collinear.

Alternately, according to [url]http://www.mathlinks.ro/viewtopic.php?t=186883[/url], the orthopole $ R$ of the perpendicular bisector $ o$ of $ PQ$ lies on $ S_1S_2.$ (The proof uses result of the previous link.) Similarly, $ R$ lies on $ S_2S_3, S_3S_1$ $ \Longrightarrow$ $ S_1, S_2, S_3, R$ are collinear.

Dear Mathlinkers,
The line A2B2C2 or S1S2S3 is the Seimiya's line of ABC.
For more information see my website : http://perso.orange.fr/jl.ayme  vol. 3  the Seimiya's line.
Sincerely
Jean-Louis


inequalities
vector
triangle inequality
inequalities proposed

[b][i]The positive reals a,b,c such that a+b+c=1 are given.[/i][/b]

[color=blue][b][u]Show that inequality holds:[/u][/b][/color]

  $\sqrt{a^2+b^2+(1-c)^2}+\sqrt{b^2+c^2+(1-a)^2}+\sqrt{c^2+a^2+(1-b)^2}> 2$

[i]This inequality was created by me when we were having an inequality marathon's with my friend.[/i]

Davron

[i][u]Nice solution now try to prove this one:[/u][/i]

$\sqrt{a^2+b^2+(1-c)^2}+\sqrt{b^2+c^2+(1-a)^2}+\sqrt{c^2+a^2+(1-b)^2}\geq 1+\sqrt3$

Davron

For $a=b=c=\frac{1}{3}$ this inequality

[quote]Nice solution now try to prove this one:

$\sqrt{a^2+b^2+(1-c)^2}+\sqrt{b^2+c^2+(1-a)^2}+\sqrt{c^2+a^2+(1-b)^2}\geq 1+\sqrt3$

Davron
[/quote]

does not hold. :)

[quote="Davron"][color=red][i][u]NO. I don't mean that. Try to prove the second case which is greater or equal. Or try to show that it's wrong. [/u][/i][/color]

Davron Latipov[/quote]

  That is what I want to say. It has only minimize at $x+y+z=1$ and three variable are equal. :)

[quote="Davron"][i] Yes i know that. Try to prove the last inequality i gave. You had proven the first one which is greater
than two. [/i][/quote]

Try to prove what ???   :?   Please explain  :)

[quote="kunny"]You can use triangle inequality for vector.

I have been posted answer for like this problems at least two times. I remember the answer is $\geq \frac{3\sqrt{3}}{2}.$[/quote]

  Kunny, that is what I try to prove too. But I don't even understand what he (Davron) wants to do either. :(
  My answer is $\sqrt{6}$ ( See my post)

Indeed $\sqrt{6}$ . You can also prove this by  Minkowsky. ;)

$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3}{\sqrt2}?(*)$

[color=red][i]Anyway: if you mean this inequality i have proof not using the Vectors but using Cauchy Swarz Inequality.[/i][/color]

$[1] (a^2+(1-b)^2)(2)\geq(a+1-b)^2 , [2] (b^2+(1-c)^2)(2)\geq(b+1-c)^2, [3] (c^2+(1-a)^2)(2)\geq(c+1-a)^2$

[i][b]Taking 1,2,3 under the squareroots and summing up we get the desired inequality (*).[/b][/i]

[i][b]And the solution idea of the inequality i sent is similar to this one.
Yes i am agree with all of you about \sqrt6. Now please try for the case [/b][/i]$abc=1$ .

Davron Latipov


When I was checking my percentage on the report page, I noticed that some of the percentages changed significantly.
When I calculated $ \frac{\text{correct}}{\text{correct\plus{}incorrect\plus{}gave up}}$ the value I got was 3 percent off my "official" percentage.
Also, in some of the subtopics, the percent was [i]really[/i] off.  (like more than 50 percent)
Last, when I clicked on the subtopic, some of the problems I did were listed twice.
 :help:

P:S: Mr. Levans, what is your avatar?  :huh:

Same here. Suddenly my percentage dropped $ 3\%$ from what it should be.

Also, my Give Up count just reset to 0, from 4...


Find the value of log 24 24

Sorry, the first 24 is suppose to be smaller, I'm just learning how to use latex.

Logarithms are simply defined as
$y = k^{x}$
The way this log is set up, we want to solve for x where
$24 = 24^{x}$
Take a logarithm with base 24,
and see that x MUST be equal to one.

This is most likely not a contest problem, so please don't post homework problems on here.

just solve 24^x=24, so x obviously must be 1.

This should be in aneasier forum, but it's too hard for getting started....


linear algebra
matrix
trigonometry
linear algebra unsolved

So I'm reviewing for the GRE and this was in a practice test. I got it right, but I wanted to share since I think it is much too beautiful for a standardized right/wrong situation. It stuck out as out of place since the problems around it tend to miss the point with horrid solutions.

For $0<t<\pi$, the matrix $\left(\begin{matrix}\cos t &-\sin t \\ \sin t & \cos t\\ \end{matrix}\right)$ has distinct complex eigenvalues $\lambda_{1}$ and $\lambda_{2}$. For what value of $t$, $0<t<\pi$, is $\lambda_{1}+\lambda_{2}=1$ ?

Enjoy, I'll post my solution tomorrow if no one has done it my way by that time. (I won't put the multiple choices, but it does work out quite nicely).

The sum of the two eigenvalues is the trace of the matrix, and the trace of the matrix is $2\cos t.$ So solving $2\cos t=1$ gives $t=\frac{\pi}3$ in the stated interval.

Although in this case the eigenvalues themselves are easy enough to find, you don't need to.


okay, so this year will be my first year taking the AMC 10 and AMC 12. right now I'm in 8th grade. for all you veterans out there, do you have any tips to give to a noob? any particular kinds of questiosn I can look out for? any tricks?(i know that's vague) i can do a lot of practice tests but i want to get an insider's point of view on all this.
Thanks!

The best thing to do is practice.  Read the AoPS books and try the problems.  For all of the problems (even the ones you get), check the solutions so you can see new ways to solve problems.  As you see more and more ways, you'll pick up on tricks and formulas and methods that you might not have known before.

hey mihail! how's ur break? lol. um, well,  my sister's friend went to USAMO, and i asked him about tips for the AMC stuff. He said to finish AoPS 1 and 2, and be sure to know the basics of math like algebra 1 and 2 and geometry. Yea...and to practice "day and night" with previous tests. lol.

uh, thanks for the help, but it isn't a good idea to revive topics that are like 3 weeks old, when i already know the advice. :)  thanks anyways


[quote="Parkdoosung"]$a,b,c$ are real numbers and $a^{2}+b^{2}+c^{2}=1$.
Prove that $a^{5}+b^{5}+c^{5}-5abc(a^{2}-bc)\leq 1$.[/quote]
I proved it by $(a^{5}+b^{5}+c^{5}-5abc(a^{2}-bc))^{2}\leq (a^{2}+b^{2}+c^{2})^{5}$ but it's ugly.
What your proof, Parkdoosung?. Thank you.

Can you post your proof? Also, I think a simple solution requires cases.

Use determinant.
\[a^{5}+b^{5}+c^{5}-5abc(b^{2}-ca) \]

\[=\left|\begin{array}{ccccc}a & b & c & 0 & 0 \\ 0 & a & b & c & 0 \\ 0 & 0 & a & b & c \\ c & 0 & 0 & a & b \\ b & c & 0 & 0 & a \end{array}\right| \]

\[\leq \parallel (a,b,c)\parallel_{2}^{5}= (a^{2}+b^{2}+c^{2})^{5/2}\]
This inequality is Hadamard's inequality. See [url]http://planetmath.org/encyclopedia/ProofOfHadamardsInequality.html[/url].

how did you observe that the expression is a determinant?  :o

The problem is most likely well known in some circles...

[quote="Parkdoosung"]$ a,b,c$ are real numbers and $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$.
Prove that $ a^{5} \plus{} b^{5} \plus{} c^{5} \minus{} 5abc(a^{2} \minus{} bc)\leq 1$.[/quote]
The hint is to prove the following for all $ a ,b,c \ge 0$
\[ (a^2\plus{}b^2\plus{}c^2)^{3/2} \ge a^3\plus{}b^3\plus{}c^3 \plus{} \left( 3\sqrt{3}\minus{}3\right) abc\]
:)
then consider two cases
Case 1. $ a^2 \le 2(b^2\plus{}c^2)$.
Case 2. $ a^2 \ge 2(b^2\plus{}c^2)$.
:)


calculus
integration
trigonometry
calculus computations

Evaluate $ \sum_{k\equal{}1}^n \int_0^{\pi} (\sin x\minus{}\cos kx)^2dx.$

The value of $ \int_0^{\pi} (\sin x\minus{}\cos kx)^2\ dx$ seems to be incorrect.

[quote="stephencheng"][quote="kunny"]Evaluate $ \sum_{k \equal{} 1}^n \int_0^{\pi} (\sin x \minus{} \cos kx)^2dx.$[/quote]

$ (\sin x \minus{} \cos kx)^2 \equal{} 2 \minus{} \cos 2x \plus{} \cos 2kx \minus{} 2[ \sin(k \plus{} 1)x \minus{} \sin(k \minus{} 1)x]$

So $ \int (\sin x \minus{} \cos kx)^2dx$
$ \equal{} 2 x \minus{} \frac {1}{2} \sin 2x \plus{} \frac {1}{2k} \sin 2kx \plus{} 2 [\frac {1}{k \plus{} 1} \cos(k \plus{} 1)x \minus{} \frac {1}{k \minus{} 1} \cos(k \minus{} 1)x]$

So $ \int_0^{\pi} (\sin x \minus{} \cos kx)^2dx \equal{} 2 \pi$ when $ k$ is odd (the case that $ k \equal{} 1$ is easy to check).
$ \equal{} 2 \pi \plus{} 4(\frac {1}{k \minus{} 1} \minus{} \frac {1}{k \plus{} 1})$ when $ k$ is even.

So $ \sum_{k \equal{} 1}^n \int_0^{\pi} (\sin x \minus{} \cos kx)^2dx \equal{} 2n \pi \plus{} 4 \minus{} \frac {4}{n \plus{} 1}$ when $ n$ is even.
$ \equal{} 2n \pi \plus{} 4 \minus{} \frac {4}{n}$ when $ n$ is odd.[/quote]
hello stephencheng
u need to divide by 2 in $ 2 x \minus{} \frac {1}{2} \sin 2x \plus{} \frac {1}{2k} \sin 2kx \plus{} 2 [\frac {1}{k \plus{} 1} \cos(k \plus{} 1)x \minus{} \frac {1}{k \minus{} 1} \cos(k \minus{} 1)x]$
$ \int (sin(x) \minus{} cos(kx))^2 \equal{} \int \frac {1 \minus{} cos2x}{2} \plus{} \frac {cos2kx \plus{} 1}{2} \minus{} sin(k \plus{} 1)x \plus{} sin(k \minus{} 1)x dx$
$ \equal{} x \minus{} \frac {cos(k \minus{} 1)x}{k \minus{} 1} \plus{} \frac {cos(k \plus{} 1)x}{k \plus{} 1} \minus{} \frac {1}{4} \sin 2x \plus{} \frac {1}{4k} \sin 2kx \plus{} C$
now i think answer will come out from ur method only .
thank u

Simplicity is beauty...

$ \sum_{k\equal{}1}^{n}\int_{0}^{\pi }\left( \sin x\minus{}\cos kx\right) ^{2}dx$

$ \equal{}\pi \plus{}\sum_{k\equal{}2}^{n}\left[ \pi \plus{}2\left( \frac{(\minus{}1)^{k}\plus{}1}{k^{2}\minus{}1}\right) \right]$

$ \equal{}n\pi \plus{}2\sum_{k\equal{}2}^{n}\frac{(\minus{}1)^{k}\plus{}1}{k^{2}\minus{}1}$

$ \equal{}\allowbreak n\pi \plus{}2\minus{}\frac{2n\plus{}1\plus{}\left( \minus{}1\right) ^{n\plus{}1}}{\left( n\plus{}1\right) n}$


You step into the elevator on the twenty-eighth floor of an office building and ride it down five floors to the print shop. From there, you take the elevator up nine floors to the purchasing department. On what floor number is the purchaing department, if floors are numbered with consecutive integers?


ABC is a triangle and the points M, N, P are on [AB] , [BC], [CA] . If the all 4 small triangles obtained have the same surface, prove that M,N,P are middle points.

(nice) :)

Let $x=\frac{AM}{AB}$, $y=\frac{AP}{AC}$, $z=\frac{BN}{BC}$. Then we have: $xy=\frac{1}{4}$, $(1-x)z=\frac{1}{4}$ and $(1-y)(1-z)=\frac{1}{4}$.

From the first equation we get $x=\frac{1}{4y}$, put it into the second we get: $z=\frac{y}{4y-1}$, plug this into the third we get: $(2y-1)^2=0$, therefore $y=\frac{1}{2}$. Hence $x=y=z=\frac{1}{2}$, thus M, N, P are all midpoints.

Here`s another aproach:$\frac{S[ABC]}{S[AMP]}=\frac14=\frac{AB\cdot AC\cdot sinA}{AM\cdot AP\cdot sinA} \Rightarrow  AM\cdot AP=\frac{AB\cdot AC}4$.By the same thinking we get $BM\cdot BN=\frac{BA\cdot BC}4$ and $CN\cdot CP=\frac{CB\cdot CA}4$ and by multiplying the 3 relations we get that $AM\cdot BM \cdot BN\cdot NC \cdot CP\cdot PA=\frac{AB^2BC^2AC^2}{64}(*)$.But by AM-GM we have $AM\cdot BM\leq\frac{(AM+BM)^2}4 \iff AM\cdot BM\leq \frac{AB^2}4$ and the analogues.By multiplying this ineq with the two similar ones get that $AM\cdot BM \cdot BN\cdot NC \cdot CP\cdot PA \leq \frac{AB^2BC^2AC^2}{64}$ Therefore,judging by (*), we must have equality in all the inequalities therefore $AM=BM,BN=CN,CP=PA$ thus the problem is solved.


How do you make a header in LaTeX like those in books i.e.,

[u](Author)             (Title)                         (Page x of y)[/u]

\*text.....
...
.....
......
*\

thanks for ur help...

Try the fancyhdr package (I think it's in the AoPS $\LaTeX$ guide.)


algorithm
inequalities
combinatorial geometry

I am new and so , don't know if this is the right forum to post my question. Anyway, I want to know what the greedy algorithm is and when it is used.

[url]http://www.artofproblemsolving.com/Wiki/index.php/Rearrangement_Inequality[/url]
Check out the AoPS Wiki.
Welcome!

Greedy algorithms is a technique of one branch of computing scince named Desing and Analys of Algorithms to solve some type of problems, generally optimus problems. The main idea is to divide the problems in stages  and  each one we  find one solution  of the partial problem who depended of the previou solution and so even the final solution.

There is one algebraic concepts named [b]matroids[/b] who play one important roll in the gredy algorithms.

Problems who can be solves by this techniques are for example

1- Convex Hull
2- Continuos knapsack  Problems

and so much