[hide]$\frac{1}{\cos 6^\circ}+\frac{1}{\sin 24^\circ}+\frac{1}{\sin 48^\circ}$

$=\frac{1}{\cos 6^\circ}+\frac{\sin 24^\circ +\sin 48^\circ}{\sin 24^\circ\sin 48^\circ}$

$=\frac{1}{\cos 6^\circ}+\frac{2\sin 36^\circ \cos 12^\circ}{2\sin 12^\circ\cos 12^\circ\sin 48^\circ}$

$=\frac{1}{\cos 6^\circ}+\frac{\sin 36^\circ}{\sin 12^\circ\sin 48^\circ}$

$=\frac{\sin 12^\circ\sin 48^\circ +\cos 6^\circ\sin 36^\circ}{\cos 6^\circ\sin 12^\circ\sin 48^\circ}$

$=\frac{\sin 12^\circ\sin 48^\circ+ \cos 6^\circ\cos 54^\circ}{\cos 6^\circ\sin 12^\circ\sin 48^\circ}$

$=\frac{\frac{1}{2}(\cos 36^\circ -\cos 60^\circ)+\frac{1}{2}(\cos 48^\circ +\cos 60^\circ)}{\cos 6^\circ\sin 12^\circ\cos 42^\circ}$

$=\frac{\frac{1}{2}(\cos 36^\circ +\cos 48^\circ)}{\cos 6^\circ\sin 12^\circ\cos 42^\circ}$

$=\frac{\cos 42^\circ\cos 6^\circ}{\cos 6^\circ\sin 12^\circ\cos 42^\circ}$

$=\frac{1}{\sin 12^\circ}$
[/hide]

 :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that a group of order 33 has an element of order 3(and I want to use Lagranges theorem in the proof)

The proof of Cauchy's Theorem is way too short and way too beautiful to dismiss the theorem in such cases :flex:

Anyway: $o(x) | 33$, so $o(x) \in \left\{ 1, 3, 11, 33 \right\}$. Suppose that $o(x) \neq 3, \, \forall x$. That also excludes the posibility of $o(x) = 33$. So all non-trivial elements have order $11$. Let $A,B$ be two subgroups generated by two elements $\neq e$. We have that $A \cap B \leq A$, so $\left| A \cap B \right| \Big| |A| = 11$. Hence, either $A \cap B = \left\{ e \right\}$ or $A=B$. The contradiction is easy to derive now; we can partition our group into $\left\{ e \right\}, H_{1}, \ldots, H_{k}$, where $|H_{i}| = 10$; that's just not possible, since we can't write $33$ as $1+10+\ldots+10$ :)

If the group is abelian we're done. So pick $x\in G$ such that $x\notin Z(G)$. Denote the conjugates by $x^{G}$, then $|x^{G}|=[G: C_{G}(x)]$. So since $x$ is not in the center $x^{G}$ has more than one element, so $|C_{G}(x)|<|G|$. Suppose $3\nmid |C_{G}(x)|$ for all $x$ not in the center.

Then since $3$ is prime we have $3 | [G: C_{G}(x)]$ (since $|G|=[G: C_{G}(x)]|C_{G}(x)|$). Now from $|G|=|Z(G)|+\sum_{i}[G: C_{G}(x_{i})]$. Now since each term in the sum is divisible by $3$, and $|G|$ is divisible by $3$, $|Z(G)|$ must be divisible by $3$. But $Z(G)$ is an abelian subgroup of $G$ and so since an element of order $3$ is in $Z(G)$, an element of order $3$ must be in $G$.

I think I made this far far too complicated, and possibly there is an error.

perfect_radio: thanks, that was nice. Can't believe I couldnt come up with that :-)


This is all about the discriminant in quadratic equations. Specifically the discriminant is  :sqrt:(b:^2:-4ac) when you have the equation ax:^2:+bx+c=0. You might or might not recognize this is part of the quadratic formula ( x=(-b:pm::sqrt:(b:^2:-4ac))/2a I suggest you look into this, figure out why it's in the quadratic formula and how you use this discriminant, not just for this problem but for all problem.

8.) If the quadratic equation 2x:^2:+ax+3=0, where a is an integer, has no real solutions, then a can take on how many different values?

[hide]

You want the discriminant to be <0 which means there would be no real roots.

a2-4*2*3<0

a2-24<0

a has to be an integer -->

a can equal 0, +-1, +-2, +-3, +-4

so there are 9 total values

[/hide]


geometry
circumcircle
geometric transformation
homothety
geometry unsolved

A triangle $ ABC$ has $ \angle BAC > \angle BCA$. A line $ AP$ is drawn so that $ \angle PAC \equal{} \angle BCA$ where $ P$ is inside the triangle. A point $ Q$ out side the triangle is constructed so that $ PQ$ is parallel to $ AB$ and $ BQ$ is parallel to $ AC$. $ R$ is the point on $ BC$ (separated from $ Q$ by the line $ AP$) such that $ \angle PRQ \equal{} \angle BCA$.

Prove that the circumcircle of $ ABC$ touches the circumcircle of $ PQR$.

Extend $ AP$ meet $ BG$ at $ G$.It's easy to see that $ G$ on $ (ABC)$.Consider a homothety $ (G;\dfrac {GQ}{GB})$: $ (ABC) \rightarrow (PQR)$ => $ O,I,G$ collinear ($ I,O$ is circumcenter of $ (PQR),(ABC)$ respectively)

Observation: There is no need for $R$ to be on $BC$! $ABGC$ is isosceles trapezoid.

Best regards,
sunken rock


named the most darned statistics or logic you've ever found on a website or in a book.

[quote]90% of the stadistics are clearly invented[/quote]I don't remember where I read it first, but it is easy to find in several websites and books. And other about humor, that is from a group of comediants called [b]Les Luthiers[/b]: [quote]From each ten people who watch TV, five ... are half of them.[/quote]

Did you know that 50% of all people who live in Los Angeles also live in California? We haven't been able to poll the other 50% yet. :lol:

[quote="djimenez"][quote]90% of the stadistics are clearly invented[/quote]I don't remember where I read it first, but it is easy to find in several websites and books. And other about humor, that is from a group of comediants called [b]Les Luthiers[/b]: [/quote]

are u making that one up?? :P

[quote="sheepwarrior"][quote="djimenez"][quote]90% of the stadistics are clearly invented[/quote]I don't remember where I read it first, but it is easy to find in several websites and books. And other about humor, that is from a group of comediants called [b]Les Luthiers[/b]: [/quote]are u making that one up?? :P[/quote]I am not making that up, but I doubt it is real. It is a very famous joke, because it is a stadistic saying something bad about stadistic, and Irony!


I did states Mar. 7 and made nationals (yay!) But I want to compare scores...  When do I get the scores back????????????????????????????????

you should have already gotten your score... my coach got it on the day of state

My coach had to bug the chapter coordinator and we just got them 4 weeks later.

Chapters vary enormously in how (or even whether) they distribute scores to coaches.

For states, your coach should be given an envelope or something with your score.  Your nats coach should also have your score.  Or you can be like me and peek at it right after the countdown round. :D


integration
calculus
calculus computations

Find $ \sum_{(n,k) \in \mathbb{N}^2} \left( \prod_{j\equal{}0}^k \frac{1}{n\plus{}j} \right)$

It has been established elsewhere on this site that $ \sum_{n\equal{}1}^{\infty}\left(\prod_{j\equal{}0}^k\frac1{n\plus{}j}\right)\equal{}\frac1{k\cdot k!}.$

Hence, the sum in this problem is $ \sum_{k\equal{}1}^{\infty}\frac{1}{k\cdot k!}.$

This can be rewritten as $ \int_0^1\frac{e^x\minus{}1}{x}\,dx$ and does not appear at first glance to be representable in any simpler terms.


A lemma
$ \lim_{n\to\infty}\frac{a_{n\plus{}1}}{a_{n}}\equal{} a$ then $ \lim_{n\to\infty}a_{n}^{\frac{1}{n}}\equal{} a$
Consider sequence
$ a_{n}\equal{}\frac{(2n\plus{}1)!}{n^{n}.n!}$
$ \lim_{n\to\infty}\frac{a_{n\plus{}1}}{a_{n}}\equal{}\lim_{n\to\infty}\frac{(2n\plus{}3)!n!n^{n}}{(2n\plus{}1)!(n\plus{}1)^{n\plus{}1}(n\plus{}1)!}\equal{}\frac{4}{e}$
So $ \lim_{n\to\infty}\equal{}\frac{4}{e}$


How many ways are there to put 5 balls in 3 boxes if the balls are distinguishable but the boxes are not?

Since the boxes are indistinguishable until we put the balls in them, we can call the box with ball 1 "A". Then there are two cases:
[hide="case 1"]
Ball 2 is in a different box. Call it "B". Then we can tell apart all 3 boxes, so there are 3^3=27 ways to put the remaining 3 balls in the boxes.[/hide]
[hide="case 2"]
Ball 2 is in box A. Then we have a similar situation as before:
[hide="case 2a"]
Ball 3 is in a different box. Call it "C". There are 3^2=9 ways.[/hide]
[hide="case 2b"]
Ball 3 is in box A. Again:
[hide="case 2bi"]
Ball 4 is in a different box. Call it "C". There are 3^1=3 ways[/hide]
[hide="case 2bii"]
Ball 4 is in box A. Then ball 5 can go into either A or a different one, for 2 total ways.[/hide]
So we have 3+2=5 ways in this case.[/hide]
So we have 9+5=14 ways in this case.[/hide]
So we have 27+14=41 ways in total.

I did it with casework.

Our cases are $(5,0,0),(4,1,0),(3,2,0),(3,1,1),(2,2,1)$ since the boxes are indistinguishable, and we can arrange these in $1,5,\binom52,\binom52,\frac{\binom52\cdot\binom32}{2}$ ways respectively, which altogether sum to $\boxed{41}$.

I don't get why you divide by two.... help?

Put in $\text{\LaTeX}$ for easier reading and put in hidden text.

[hide=Click this text to see afroromanian's solution.]Say the boxes are labeled $1$, $2$, $3$ and the balls $a$, $b$, $c$, $d$, $e$.  Pretend for a second that the boxes are distinguishable. There are $3^5 = 243$ ways to assign each of the $5$ balls to each of the $3$ boxes ($1^{st}$ ball can go in any of $3$ ways, $2^{nd}$ in any of $3$ ways, and so on). Now out of these $243$ cases, there are only $3$ ways to put all $5$ balls in one of the three boxes and leaving $2$ boxes empty. That leaves $240$ ways to put balls in boxes such that at most one box is empty. Now just notice that whatever way you put the balls in the $3$ boxes, you have $6$ permutations if the boxes are distinguishable. But since they are not, you have overcounted $6$ times, so you need to divide by $6$. Hence $\frac{240}{6} = 40$. And don't forget the $3$x overcount on the $\frac{5}{\frac{0}{0}}$ split (if the boxes are indistinguishable, there is only one way to throw all $5$ in a box). So that gives $41$ in total.[/hide]

Is there a way to do this without casework-bashing?

[quote=monkeyl]You divide two because options can be the same but in another order.

For example, if I say I wanted Vanilla Ice Cream with Cherry Toppings, that would be the same as saying I want Cherry Topping on a Vanilla Ice Cream.[/quote]

Except for the fact that no ice cream shop would ever give you Cherry [i]Toppings[/i] before Vanilla Ice Cream.
@Above Casework is pretty much the fastest way. I believe droid347 basically pointed out the fastest way.

About the alternate solution, wouldn't $(1,1,3)$ and $(1,2,2)$ also be counted 3 times? For those who haven't yet seen the alternate solution from alcumus: [hide]There are $3^5 = 243$ arrangements to put 5 distinguishable balls in 3 distinguishable boxes. Among these 243 arrangements, there is one case in our problem that is counted three times: if all 5 balls are placed in one box and the other two boxes both contain nothing. This leaves 240 other arrangements.

For every other case, the contents of each box is different, and so these cases are each counted $3! = 6$ times. Therefore there must be 40 of these cases, and we have $\boxed{41}$ cases total.[/hide]

[quote=The Solution]$(3,2,0)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes.

$(3,1,1)$: There are $\binom{5}{3} = 10$ choices for the 3 balls in one of the boxes, and we simply split the last two among the other indistinguishable boxes.

$(2,2,1)$: There are $\binom{5}{2} = 10$ options for one of the boxes with two balls, then $\binom{3}{2} = 3$ options for the second box with two balls, and one option remaining for the third. However since the boxes with two balls are indistinguishable, we are counting each pair of balls twice, and must divide by two. So there are $\dfrac{10 \times 3}{2} = 15$ arrangements of balls as $(2,2,1)$.[/quote]

Can someone explain to me what exactly happened here?

[quote=fishy15]the alcumus solution was copied from stack exchange
[url]http://math.stackexchange.com/questions/340890/how-many-ways-are-there-to-distribute-5-balls-into-3-boxes-under-additional-con[/url][/quote]

It could have the other way around.

I tried using reverse Balls and Urns, like what the Distinguishability article shows, but I get a different answer. I think it is because I do not understand what is meant by "with(out) duplicates". What does that mean?

[quote=fishy15]the alcumus solution was copied from stack exchange
[url]http://math.stackexchange.com/questions/340890/how-many-ways-are-there-to-distribute-5-balls-into-3-boxes-under-additional-con[/url][/quote]

 I think its the other way around...

[quote=fishy15]the alcumus solution was copied from stack exchange
[url]http://math.stackexchange.com/questions/340890/how-many-ways-are-there-to-distribute-5-balls-into-3-boxes-under-additional-con[/url][/quote]

No, people search up these alcumus questions on websites like that, causing this illusion.

[quote=fishy15][quote=Kagebaka]hi fishy!!!!!

also this problem hurt my soul :coolspeak:[/quote]

what the moo how did you find me[/quote]

what the moo why did it take you so long to respond

I have bumped this so that others can see the problem. I have asked a question somewhere else that relates to this.

[hide=Solution]First, we count the number of possible quantities of balls among the boxes, then we will count the amount of ways to get those quantities. There are 5 different quantities of balls: $(5,0,0), (4,1,0), (3,1,1), (3,2,0), (2,1,2).$ We know count the number of was to reach these quantities with casework:

[i]Case 1: $(5,0,0)$.[/i]

If we have $(5,0,0)$, then there are only $\binom{5}{5}=1$ ways to reach this.

[i]Case 2: $(4,1,0)$.[/i]

If we have $(4,1,0)$, then there are only $\binom{5}{4}=5$ ways to reach this.

[i]Case 3: $(3,1,1)$.[/i]

If we have $(3,1,1)$, then there are only $\binom{5}{3}=10$ ways to reach this.

[i]Case 4: $(3,2,0)$.[/i]

If we have $(3,2,0)$, then there are only $\binom{5}{3}=10$ ways to reach this.

[i]Case 5: $(2,1,2)$.[/i]

If we have $(2,1,2)$, then there are only $\frac{\binom{5}{1}\cdot\binom{4}{2}}{2!}=15$ ways to reach this.

Adding the cases, we get a final answer of $\boxed{41}. \blacksquare$[/hide]

oops :( i got all the partitions this time around but rearranged them all wrong

[quote=TheReaper10]wow you said that one year later[/quote]

lol and now I'm back at it again but I got it right this time ;)

[quote=JustKeepRunning][quote=afroromanian][quote=BillytheChamp]Can someone please explain the alternate solution?[/quote]

Say the boxes are labeled 1,2,3 and the balls a,b,c,d,e.  Pretend for a second that the boxes are distinguishable. There are 3^5 = 243 ways to assign each of the 5 balls to each of the 3 boxes (1st ball can go in any of 3 ways, 2nd in any of 3 ways, and so on). Now out of these 243 cases, there are only 3 ways to put all 5 balls in one of the three boxes and leaving 2 boxes empty. That leaves 240 ways to put balls in boxes such that at most one box is empty. Now just notice that whatever way you put the balls in the 3 boxes, you have 6 permutations if the boxes are distinguishable. But since they are not, you have overcounted 6x, so you need to divide by 6. Hence 240/6 = 40. And don't forget the 3x overcount on the 5/0/0 split (if the boxes are indistinguishable, there is only one way to throw all 5 in a box). So that gives 41 in total.

So basically the alternate solution is using the overcounting and correcting strategy.[/quote]

If the cases are $(3,1,1)$ and $(2,2,1)$, then there are $3$ ways to arrange them. So there are some cases that are multiplied by $3$ and not $6.$ I am confused...[/quote]

It might seem so at first glance, but remember the balls are distinguishable so e.g the 1's in (3,1,1) above are actually different. => Everything is overcounted 6x except for the case where two of the boxes are empty.

Is this the problem which had that slick fast solution & is level 25? If so, could someone please post the fast solution? The AoPS page for it should have two solutions.

For (2,2,1), I did 5 choose 2 and it gets 10 choices, without considering putting it in a box. Then I did 3 choose 1 to get the next ball, again, without considering the box selection. It gives me 10x3 ways to simply select 2 balls followed by 1 ball. There's no over counting in this case. I don't understand why it needs to be divided by two.

What the heck did I just get 500 power from this problem I promise a problem ago it was at 260 or 300 and now it's at 861...


Find all polynomials $ P(x)$ with nonnegative coefficients such that $ P(x)P(\frac{1}{x})\leq (P(1))^2$ holds for all $ x\in R^\plus{}$

Let deg P=n and $ P(x)\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_nx^n, a_i\ge 0$, then $ P(x)P(\frac 1x )\equal{}a_0a_nx^n\plus{}(a_0a_{n\minus{}1}\plus{}a_1a_n)x^{n\minus{}1}\plus{}..$
Therefore, from $ P(x)P(\frac 1x )\le P(1)^2$ we get $ a_0\equal{}a_1\equal{}...a_{n\minus{}1}\equal{}0$ or $ P(x)\equal{}a_nx^n$ and $ P(x)P(\frac 1x )\equal{}a_n^2\equal{}P(1)^2.$


[quote="lotrgreengrapes7926"][quote="i_like_pie"]Is there a rational $x$ such that $\sin{x}$ is also rational?[/quote]
[hide="in Radians"] $(0,0)$ :D [/hide]
[hide="in Degrees"] $x=180\pm30k, k\in\mathbb{Z}$ [/hide][/quote]

Correction, sin(60*) is not rational  :wink:

The actual answer is:

In degrees:

x= 0, 30+60k


[color=darkblue]Prove that :

$1.\blacktriangleright$ $\boxed{\math S\equiv\left\|\begin{array}{c}0\le x\le a\ \ ;\ \ 0\le y\le b\\\\ x^{2}+b^{2}=2(ax+by)=y^{2}+a^{2}\end{array}\ }$ $\Longrightarrow$ $\boxed{\ ab+xy=2(bx+ay)\ }\ .$

$2.\blacktriangleright$ The system $\mathbb S$ is compatibly if and only  if $\frac{\sqrt 3}{2}\le\frac{a}{b}\le\frac{2}{\sqrt 3}$ and in this case the system $\mathbb S$ has only one solution.[/color]

[color=red]I corrected after the second Farenhajt's reply.[/color]

Part 1:

Let $\vec u=\langle a,y\rangle,\vec v=\langle x,b\rangle$. From $a^{2}+y^{2}=x^{2}+b^{2}$ it follows $|\vec u|=|\vec v|=r$. Since $\vec u\cdot \vec v=ax+by={x^{2}+b^{2}\over 2}={r^{2}\over 2}$ and $\vec u\cdot\vec v=r^{2}\cos(\vec u,\vec v)$, we get $(\vec u,\vec v)={\pi\over 3}$ (the angle is positive since $0<x<a$ and $b>y>0$).

Therefore $x+bi=(a+yi)\left({1\over 2}+i{\sqrt{3}\over 2}\right)$, giving $x=\frac{a-y\sqrt{3}}{2},b=\frac{y+a\sqrt{3}}{2}$. Plugging that into $ab+xy=2(bx+ay)$ we get a correct identity.

[color=darkblue]Very nice your proof, [b]Farenhajt[/b] !

See "[b][u]inspiration source[/u][/b] " : http://www.mathlinks.ro/Forum/viewtopic.php?t=118078 [/color]

For Part 2 I must say I think you're wrong - but it's possible that [i]I'm[/i] wrong.

See the diagram. In Part 1 we've proven that we have an equilateral triangle inscribed in a rectangle in the way presented on the picture. (One joint vertex, the other two on non-adjacent sides of the rectangle.)

Obviously, for such an inscribing to be possible, we must have $0\leqslant\varphi\leqslant{\pi\over 6}$, hence

$a=r\cos\varphi$
$b=r\cos({\pi\over 6}-\varphi)=r({\sqrt{3}\over 2}\cos\varphi+{1\over 2}\sin\varphi)$

yielding

${b\over a}={\sqrt{3}\over 2}+{1\over 2}\tan\varphi$

For $0\leqslant\varphi\leqslant{\pi\over 6}$ we have $0\leqslant\tan\varphi\leqslant{1\over\sqrt{3}}$, giving ${\sqrt{3}\over 2}\leqslant{b\over a}\leqslant{2\over\sqrt{3}}$ (taking the reciprocals doesn't change anything)

[b]APPENDIX[/b]: Now I see the source of your mistake. From my final expressions for $x$ and $b$ we get $y=2b-a\sqrt{3}, x=2a-b\sqrt{3}$. Now $y\leqslant b\iff 2b-a\sqrt{3}\leqslant b\iff b\leqslant a\sqrt{3}$. However, if you put $b=a\sqrt{3}$ into the expression for $x$, you get $x=-a$, which is out of range.

[color=darkblue]You are right, [b]Farenhajt[/b] ! I corrected. I think the source of the my mistake is ... the [u]late night[/u]
(when I established this interesting compatible case). [u]Thank you very much for the yours rectifying ![/u][/color]


[color=darkred]Let $ p$ be a prime divisor of $ n\ge 2$. Prove that there exists a set of natural numbers $ A \equal{} \{a_1,a_2,...,a_n\}$ such that product of any two numbers from $ A$ is divisible by the sum of any $ p$ numbers from $ A$.[/color]

I don't understand this problem very well . Are there any conditions about the set A ? If not, I think this problem is trivial !
Take $ d$ be the product of all sum has p terms from the set $ \{1,..,n\}$ and $ \{a_1,..,a_n\} \equal{} \{d,2d,..,nd\}$

I also think there must be some other condition on the set $ A$, as otherwise we may take $ a_1\equal{}a_2\equal{}...\equal{}a_n\equal{}p$!

[quote="Ahwingsecretagent"]I also think there must be some other condition on the set $ A$, as otherwise we may take $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n \equal{} p$![/quote]
A set can't contain equal element. The problem didn't contain some other condition. TTsphn's solution is the same as the official one.

Well then it is far too easy for a number 4 with a one line solution which is easy to find.

According to TTsphn's solution, the conditions that $ p $ is prime, and moreover is a divisor of $ n $ are useless, isn't it?


Well...i was looking at the Sophie German Identity when i find this problem:

Is $ 4^{545} \plus{} 545^4$ a prime number ?

I realize that since 545 is odd then $ 4^{545}$ finish in 4.....and $ 545^4$ in 5, so it would be a number finishing in 9 and i don't know how can i relate the German Sophie Identity with this problem. Could someone provide me some help?

[hide="solution"] The Sophie Germain Identity states that $ a^4+4b^4=(a^2+2b^2)^2-4a^2b^2=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$.

In this case, $ 4^{545}+545^4=545^4+4^{545}=545^4+4\cdot{4^{544}}$. Now the above factorization finishes off the problem.[/hide]

[hide="Note"]You still need to prove that $ a^{2}\minus{}2ab\plus{}2b^{2}$ is greater than 1, or else the number could still be prime.  Sometimes this step is not that easy, so don't leave it out.[/hide]

xpmath.....i thought that ,but then if the identity starts with $ a^4 + 4b^4$

lets say that a=545......but in the expression  $ 4\cdot{4^{544}}$ the degree of 4 is 544 and not 4, and i think we can follow the identity only if the degree is 4.

@matt276eagles: Yeah, I realize that you have to show that that is greater than 1, but in this case it's not that hard to just plug in and bash, so I just left it out due to my laziness  :P . Of course if I was writing a real solution, I would show that.

@Chalao: Let $ a \equal{} 545$, $ 4^{136} \equal{} b$. Then $ 545^4 \plus{} 4\cdot{4^{544}} \equal{} a^4 \plus{} 4b^4$. So, as long as the degree is a multiple of $ 4$, you can write it as $ b^4$. Sorry for not being clear  :oops: .

:oops:  I didn't noticed xpmath. You are right; then it follows :

($ 545^2 \plus{}2*4^{272}\plus{}2 * 545^2 *4^{136}$)   ($ 545^2 \plus{}2*4^{272}\minus{}2 * 545^2 *4^{136}$)

So it is mostly proved that the sum isn't a prime number.

Thank you xpmath  :P

matt276eagles i will try to prove it...

[quote="matt276eagles"]You still need to prove that $ a^{2} \minus{} 2ab \plus{} 2b^{2}$ is greater than 1 ...[/quote]

[hide="does this help?"]
$ a^2\minus{}2ab\plus{}2b^2 \equal{} a^2\minus{}2ab\plus{}b^2\plus{}b^2 \equal{} (a\minus{}b)^2\plus{}b^2 > 0$
[/hide]

$ a^{2}\minus{}2ab\plus{}2b^{2}$ is greater than 1

let $ 545^2$ =m    and     $ 4^{272}$  =n   then    $ m^2 \minus{}2mn \plus{} 2n^2$ =   $ (n) (2n\minus{}2m) \plus{} m^2$

If we demonstrate that n-m is positive then the whole expression would be greater than 1:

 ($ 4^{272}$)  - ($ 545^2$) =  $ (1024)^2 * 4^ {262} \minus{}545^2$

Since  $ 1024^2$ is greater than $ 545^2$  it follows that n>m and $ a^{2}\minus{}2ab\plus{}2b^{2}$ is greater than 1

[quote="Chalao"]:oops:  I didn't noticed xpmath. You are right; then it follows :

($ 545^2 \plus{} 2*4^{272} \plus{} 2 * 545^2 *4^{136}$)   ($ 545^2 \plus{} 2*4^{272} \minus{} 2 * 545^2 *4^{136}$)

So it is mostly proved that the sum isn't a prime number.

Thank you xpmath  :P

matt276eagles i will try to prove it...[/quote]

Greater than one: In this case, and many others, we can see the number ends in 7, and with Nora's >0 , it's done

[hide]For those who do not know the Germaine-sophie...
Use the fact that for all primes >7, p^4 is congruent to 1 modulo 240....
So just expand the exp bionomially, followed by congruencies modulo 240....[/hide]


Determine all positive integral solutions to $ x$ for which $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ yields a perfect square.

This seems really easy, but I cant seem to find the proper factors. I factored it a bit to

$ (x \plus{} 1)^2(x^2 \plus{} 1) \minus{} x(x^2 \plus{} x \plus{} 1)$ and then I dont know what to do.

I also need help solving these general types of problems, so if your solution could be more general rather than specific I would really appreciate it  :lol:

Factoring isn't really the way.  Bound it between two polynomials which are perfect squares.

[hide="maybe?"]The upper bound could be $ x^{2}\plus{}x\plus{}1$ and the lower one is $ x^{2}\plus{}1$ because $ x^3>x^{2}$ for a positive integer?[/hide]

Im a complete beginner. What exactly is bounding and how do you do it?

[quote="t0rajir0u"]That doesn't quite work.  You want the bounds to be different by $ 1$; that's how you can conclude that there are no squares between them.[/quote]
O, sorry, I've done these problems before but I forgot that this time.

@thunder The general idea is that you find two polynomials (with integer coefficients,) with one of them 1 greater than the other (ex: $ x^{2}\plus{}3$ and $ x^{2}\plus{}4$,) They have to be selected such that the smaller polynomial, when squared, is smaller than your target, and the bigger polynomial is bigger. Because there are no perfect squares between two consecutive perfect squares, we can then conclude that the polynomial can't be a perfect square.

You can't prove it is never a perfect square because x=3 makes the polynomial have a value of 121, obviously a perfect square...

The bounds only work for all but finitely many $ x$.

Here is a complete solution .
I knew this problem before and the mathlinks user Agr_94_math ,had said his beautiful solution to me .
Here it goes
For convenience sake ,let $ x \equal{} n$ and $ y \equal{} m$ ,so we have to solve for $ m \in \mathbb{N}$ for which
$ m^{4} \plus{} m^{3} \plus{} m^{2} \plus{} m \plus{} 1$ is a perfect square .
Consider the expansion of $ (m^{2} \plus{} \frac {m}{2})^{2} \equal{} m^{4} \plus{} m^{3} \plus{} \frac {m^{2}}{4} \equal{} m^{4} \plus{} m^{3} \plus{} m^{2} \plus{} m \plus{} 1 \minus{} (\frac {3m^{2}}{4} \plus{} m \plus{} 1)$
Since the $ \text{L.H.S}$ is positive and $ m$ is natural ,we have
$ m^{4} \plus{} m^{3} \plus{} m^{2} \plus{} m \plus{} 1 \equal{} n^{2} > (m^{2} \plus{} \frac {m}{2})^{2}$ ,since $ m$ is natural ,
$ n > m^{2} \plus{} \frac {m}{2}$
If $ m$ is even ,
$ n \ge m^{2} \plus{} \frac {m}{2} \plus{} 1$,
Squaring both sides we get
$ n^{2} \ge n^{2} \plus{} \frac {5m^{2}}{4}$
$ \implies 0 \ge \frac {5m^{2}}{4}$ ,so no solutions in naturals .
If $ m$ is odd ,
we have  $ n \ge m^{2} \plus{} \frac {m}{2} \plus{} \frac {1}{2}$ ,
again repeating the same step we dont get any solutions in naturals .

I'll finish your solution.

In the even case, 5m^2/4=0 has no solutions in naturals, but $ m^2 /4\minus{}m/2 \minus{}3/4$ has one solution, namely m=3, and hence n=11.

That is the only solution.

Thank you Srinath for posting the solution.


It's fun!!!

[url]http://gprime.net/game.php/pandagolf[/url]

I'm at level 54.


geometry
geometric transformation
reflection
inequalities
logarithms
inequalities proposed

I am not able to find this problem in that PDF file.
Hint:
[hide]Substitiute $ a\equal{}\dfrac{x}{y}$ and similarly for $ b,c$ and simplify your new expression by applying AM GM here and there properly, you get the desired result.[/hide]

[quote="Aravind Srinivas L"]I am not able to find this problem in that PDF file.
Hint:
[hide]Substitiute $ a \equal{} \dfrac{x}{y}$ and similarly for $ b,c$ and simplify your new expression by applying AM GM here and there properly, you get the desired result.[/hide][/quote]
O103 ;)

[quote="thaithuan_GC"]Let positive real numbers $ a,b,c$ satisfying $ abc \equal{} 1$. Prove that $ \sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\ge\sqrt [4]{4(a \plus{} b \plus{} c \plus{} 1)}$[/quote]
Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3.$
Hence, $ \sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\ge\sqrt [4]{4(a \plus{} b \plus{} c \plus{} 1)}\Leftrightarrow$
$\Leftrightarrow v^2\geq\frac{1}{3}\left(\left(4(3u+w)\right)^{\frac{3}{4}}w^{\frac{5}{4}}-2w^2-3uw\right).$
But $v^2$ gets a minimal value when two numbers from $\{a,b,c\}$ are equal.
Since the last inequality is homogeneous, it remains to prove him for $b=c=1.$
Let $a=x^3.$ Hence, $v^2\geq\frac{1}{3}\left(\left(4(3u+w)\right)^{\frac{3}{4}}w^{\frac{5}{4}}-2w^2-3uw\right)\Leftrightarrow$
$\Leftrightarrow2x^3+1+2x^2+(x^3+2)x\geq\left(4(x^3+2+x)\right)^{\frac{3}{4}}x^{\frac{5}{4}}\Leftrightarrow$
$\Leftrightarrow(x^2+1)(x+1)^2\geq\left(4(x^3+x+2)\right)^{\frac{3}{4}}x^{\frac{5}{4}}\Leftrightarrow  f(x)\geq0,$
where $f(x)=\ln(x^2+1)+2\ln(x+1)-\frac{5}{4}\ln x-\frac{3}{4}\ln(x^3+x+2)-\ln\sqrt8.$
Hence, $f'(x)=\frac{2x}{x^2+1}+\frac{2}{x+1}-\frac{5}{4x}-\frac{3(3x^2+1)}{4(x^3+x+2)}=$
$=\frac{(x-1)(x^5-2x^4-x^3+3x^2+6x+5)}{2x(x+1)(x^2+1)(x^3+x+2)}.$
Since $x^5-2x^4-x^3+3x^2+6x+5>0$, we obtain $a_{min}=x_{min}=1$
and $f(x)\geq f(1)=0.$ Done!

[quote="shaaam"][quote="thaithuan_GC"]Let positive real numbers $ a,b,c$ satisfying $ abc \equal{} 1$. Prove that $ \sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\ge\sqrt [4]{4(a \plus{} b \plus{} c \plus{} 1)}$[/quote]

Dear Mathlinkers, this is the problem O103 from the Mathematical Reflections journal...
see:
[url]http://reflections.awesomemath.org/reflections/2008_6/MR_6_2008_problems.pdf[/url]

I think this topic should be locked.[/quote]
There is nothing wrong in discussing problems from older issues, I would assume. Because the issue was published in 2008(!), I am sure that it is OK to discuss it after a year.
The solutions(official) are available [url=http://reflections.awesomemath.org/2008_6/MR_6_2008_solutions.pdf]here[/url]. (Pg. 34)
:)

[quote=thaithuan_GC]Let positive real numbers $ a,b,c$ satisfying $ abc \equal{} 1$. Prove that $ \sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\ge\sqrt [4]{4(a \plus{} b \plus{} c \plus{} 1)}$[/quote]

[img]http://s8.sinaimg.cn/middle/006ptkjAzy76vJfjUmbb7&690[/img]


function
induction
real analysis
real analysis solved

f(x-1) >= f(x)+f(-1) , f(x) >= f(x-1)+f(1) , so f(x)=f(x-1)+1, for all x.
So by induction, it is easy to see that f(m+y)=m + f(y) , for all m integer and y real . Put y=0 , m=1 ===> f(0)=0
Let y be in ]0,1[ . We have f(y) =< 0 , f(1-y) =< 0 and  f(y)+f(1-y)+1 >= f(1)=1 ===> f(y)+f(1-y) >=0 , so f(y)=f(1-y)=0 .
Note [x] = integer part , {x} = fractionar part . Finally f(x)=f([x]+{x}) = [x] + f({x}) = [x]

[quote="lephuocdaint1987"]find all functions f:R->R s.t.:
       1/f(x) + f(y) +1  \geq  f(x+y) \geq f(x)+f(y).
       2/f(x)  \leq  f(0) with all x in [0,1).
       3/f(1)=1 and f(-1)=-1.[/quote]
Aww, I got confused I thought he meant [b]1/f(x)[/b] when it should have been [b] 1) f(x) [/b] (similarly the others) making me conclude some ridiculous results. BTW f(1)=3 caused me a lot of headaches :?  :(
Still great pb and sln both of you.


function
algebra
domain
symmetry
real analysis
real analysis unsolved

[quote="Peter VDD"]Find all functions $\mathbb{R}^+\rightarrow\mathbb{R}^+$ for which we have for all $x,y\in \mathbb{R}^+$ that $f(x)f(yf(x))=f(x+y)$.[/quote]
Answer:$f(x)=\frac{1}{1-xt}$ :D

[quote="Peter VDD"][quote="Tiks"]Answer:$f(x)=\frac{1}{1-xt}$ :D[/quote]that's not even a defined function for $t>0$. ;)[/quote]
Oh,I am sorry I havn't seen the condition $R^{+}$ :blush: .

even for R it was false. Simply because it's not a function on the given domain, the denumerator gets zero.

The correct answer is $f(x)=\frac1{ax+1}$, with $a\in\mathbb{R^+}\cup\{\infty\}$. Perhaps it was what you meant, but that's hard to figure with no proof. ;)

Actually for the second approach I forgot to mention that if $\inf(S)=0$, we use the fact that $f(x)=f(x+t)$ to conclude that $f(x)=1$ for all $x$. I was just looking at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=85071&start=0]2005 A2[/url], so I remembered this...

Sorry. The second approach is clearly incorrect when $\inf(S)=0$. I was being stupid. :blush:

[quote=math154]Let $S$ be the set of $x\in\mathbb{R^+}$ such that $f(x)=1$. Also let $P(x,y)\implies f(x)f(yf(x))=f(x+y)$.

First, if $f(x)>1$ then
\[P\left(x,\frac{x}{f(x)-1}\right)\implies f(x)=1,\]a contradiction. So $f(x)\le1$ for all $x$.

[b]Case 1:[/b] $|S|=0$. Then $f$ is injective, or else there exist $a<b$ such that $f(a)=f(b)$, whence
\[P(a,b-a)\implies f((b-a)f(a))=1.\]By symmetry,
\[P\left(x,\frac{y}{f(x)}\right)\implies f(x)f(y)=f\left(x+\frac{y}{f(x)}\right)=f\left(y+\frac{x}{f(y)}\right),\]so because $f$ is injective,
\[x+\frac{y}{f(x)}=y+\frac{x}{f(y)}\implies \frac{xf(x)}{1-f(x)}=\frac{yf(y)}{1-f(y)}=c>0.\]So
\[f(x)=\frac{c}{c+x},\]which indeed satisfies the original equation.

[b]Case 2:[/b] $|S|>0$.

[b]First Approach:[/b] Since $f(x),f(yf(x))\le1$, $f$ is a non-increasing function. If we take $t$ such that $f(t)=1$, then $P(t,x)\implies f(x)=f(x+t)$ for all $x>0$. Since $f$ is non-increasing, $f(x)=1$ for all $x$.

[b]Second Approach:[/b] We can take $t$ arbitrarily close to $\inf(S)$ with $f(t)=1$. Then $P(t,x)\implies f(x)=f(x+t)$, and
\[P\left(x,\frac{t}{f(x)}\right)\implies f(x)=f\left(x+\frac{t}{f(x)}\right).\]Thus for all integers $j,k$ with $j+kf(x)>0$, we have
\[f(x)=f\left(x+\frac{jt}{f(x)}+kt\right),\]whence
\[P\left(x,\frac{jt}{f(x)}+kt\right)\implies f((j+kf(x))t)=1.\]If $f(x)$ is irrational, then we can make $(j+kf(x))t$ an arbitrarily small positive number, contradicting the fact that $t$ can be arbitrarily close to $\inf(S)$. So let $f(x)=p/q$ with $\gcd(p,q)=1$. By Bézout's identity, we can find $j,k$ with $jq+kp=1$ so that
\[(j+kf(x))t=\frac{t}{q}.\]If $q\ge2$, then we can find $t$ such that $t/q<\inf(S)$, so $q=1$ and $f(x)\in\mathbb{Z^+}$ for all $x\in\mathbb{R^+}$. The rest is simple.[/quote]

For case $|S| \geq 1$ we can take one $s \in S$, then setting $x + y = s$ we get $f(xf(y))f(y) = 1$. With $f \leq 1$ it implies $f(y) = 1$ and so $f(x) = 1$.
So $\forall x \leq s$ we have $f(x) = 1$. Then setting $y = s$ we get $f(x) = f(x + s)$, so with result from last substitution we get $f(x) = 1 \forall x$

Let $P(x,y)$ denote the assertion $f(x+y)=f(x)f(yf(x)).$ Suppose $f(x)>1$ for some $x$, then by $P(x,\tfrac{x}{f(x)-1})$ we get contradiction. So $f(x)\leqslant 1$ for all $x.$

Note that this means $f$ is decreasing. So if $f(x)=1$ for some $x$ then $f(x+y)=f(y)$ which means $f$ is constant. Thus $f\equiv 1$ is the only solution.

We consider $f(x)<1$ for all $x$, then $f$ is strictly decreasing. Comparing $P(x,1/f(x))$ and $P(1,x/f(1))$ and using injectivity we get $f(x)=1/(kx+1)$ for all $x$ and some constant $k.$ This works.

For $f:\mathbb R\to \mathbb R$ this is trivial.

Clearly $f\equiv 0$ is a solution. So assume for some $x_0,$ we have $f(x_0)\neq 0.$ Then if $f(x_0)\neq 1$ by $P(x_0,\tfrac{x_0}{f(x_0)-1})$ gives contradiction. So $f\equiv 1;$ which works.

[quote=GianDR][quote=ZETA_in_olympiad]Clearly $f\equiv 0$ is a solution.[/quote]

But the codomain of $f$ is $\mathbb{R}^+$, not $\mathbb{R}_0^+$.[/quote]

Uhh, for the problem in #1, I gave solution in #10. And the solution in #11 is for the function $f$ mapping [b]reals $\to$ reals[/b], which I mentioned [quote=ZETA_in_olympiad]For $f:\mathbb R\to \mathbb R$ this is trivial. [/quote]


inequalities solved
inequalities
4-variable inequality
Leningrad olympiad
2000

It is clear that $\frac12<\frac{a}d<2$, from where we find that $a^2+d^2<\frac52ad$. Similarly, $b^2+c^2<\frac52bc$. Adding to this the fact that $(ab+cd)^2 \ge 4abcd$, the problem is solved.


Solve the following equation:

$3.4^{x}+(3x-10)2^{x}+3-x=0.$


calculus
derivative
function
parameterization
integration
limit
calculus computations

im having trouble finding the derivative of this equation:

f(x) = $ \sqrt{2x\minus{}3}$

Do you know the [url=http://en.wikipedia.org/wiki/Chain_rule]chain rule[/url]?
- If so, what would be a useful way of writing f as a composition of two simple functions?
- If not, familiarize yourself with it; it is one of the most important tools in differential calculus, and you'll revisit it when you learn changes of variables, parametrization, and integration by substitution.

This is the wrong section. There's a calculus forum.

Anyway, hint: Write the square root as a 1/2 power. Let $ u\equal{}2x\plus{}3$ and use $ dy/dx\equal{}dy/du\cdot du/dx$

(grn_turtle, it was $ 2x \minus{} 3$, not $ 2x \plus{} 3$)

I think it would be something like [hide="this"]
$ \frac {1}{2} (2x \minus{} 3)^{ \minus{} \frac {1}{2}} \cdot (2) \equal{} \frac {1}{\sqrt {2x \minus{} 3}}$

First, we take the derivative of $ u$ with respect to $ y$ or just the derivative of $ \sqrt {u} \equal{} u^{\frac {1}{2}}$, which is (after replacing $ 2x \minus{} 3 \equal{} u$) $ \frac {1}{2} (2x \minus{} 3)^{ \minus{} \frac {1}{2}}$. Then, we multiply that by derivative $ u \equal{} 2x \plus{} 3$, which is simply $ 2$. Simplifying that gives $ \frac {1}{\sqrt {2x \minus{} 3}}$.[/hide].

It's kind of interesting how the derivative of this equals its reciprocal.

Implicit differentiation is also a viable option.  However, this is even further along the calculus path than the chain rule.

[quote="PowerOfPi"]It's kind of interesting how the derivative of this equals its reciprocal.[/quote]  $ y' \equal{} \frac{1}{y}$ implies $ y' y \equal{} 1$ and so $ (y^2)' \equal{} 2y' y \equal{} 2$.  Thus $ y^2 \equal{} 2x \plus{} C$ ;)

You can also use the definition of the derivative:

$ f'(x) \equal{} \lim_{h \to 0} \frac{f(x\plus{}h) \minus{} f(x)}{h}$.

Letting $ f(x) \equal{} \sqrt{2x\minus{}3}$,

$ \frac{d}{dx}\left[\sqrt{2x\minus{}3}\right] \equal{} \lim_{h \to 0} \frac{\sqrt{2(x\plus{}h) \minus{} 3} \minus{} \sqrt{2x \minus{} 3}}{h}$

$ \equal{} \lim_{h \to 0} \frac{(\sqrt{2(x\plus{}h) \minus{} 3} \minus{} \sqrt{2x \minus{} 3})(\sqrt{2(x\plus{}h) \minus{} 3} \plus{} \sqrt{2x \minus{} 3})}{h(\sqrt{2(x\plus{}h) \minus{} 3} \plus{} \sqrt{2x \minus{} 3})}$

$ \equal{} \lim_{h \to 0} \frac{2(x\plus{}h) \minus{} 3 \minus{} (2x \minus{} 3)}{h(\sqrt{2(x\plus{}h) \minus{} 3} \plus{} \sqrt{2x \minus{} 3})}$

$ \equal{} \lim_{h \to 0} \frac{2}{\sqrt{2(x\plus{}h) \minus{} 3} \plus{} \sqrt{2x \minus{} 3}}$

$ \equal{} \frac{1}{\sqrt{2x\minus{}3}}$.

$ y \equal{} \sqrt {2x \minus{} 3}\Longrightarrow x \equal{} \frac {y^2 \plus{} 3}{2}$.

$ \therefore \frac {dy}{dx} \equal{} \frac {1}{\frac {dx}{dy}} \equal{} \frac {1}{y} \equal{} \frac {1}{\sqrt {2x \minus{} 3}}$.

Comment:

Learn by heart:
\[ \boxed{\boxed{\frac {d}{dx}\sqrt {f(x)} \equal{} \frac {f'(x)}{2\sqrt {f(x)}}}}\]

[quote="PowerOfPi"](grn_turtle, it was $ 2x \minus{} 3$, not $ 2x \plus{} 3$)[/quote]

Right. Doesn't matter though.

My way was to do this:

$ u\equal{}2x\plus{}3$

$ \frac{du}{dx}\equal{}2$

$ y\equal{}\sqrt u\equal{}u^{\frac 12}$

The power rule states that if $ y\equal{}u^a$, $ \frac {dy}{dx}\equal{}au^{a\minus{}1}$, so we have $ \frac{dy}{du}\equal{}\frac 12 u^{\minus{}\frac 12}$

And $ \frac{dy}{du}\cdot\frac{du}{dx}\equal{}\frac{dy}{dx}\equal{}u^{\minus{}\frac 12}\equal{}\frac{1}{\sqrt u}$. Now substitute back in.

This is, of course, basically the chain rule in a more intuitive form.

In my opinion, the least "labor intensive" method is as follows:

Call $ p \equal{} \sqrt {2x \minus{} 3}$.
Then $ p^2 \equal{} 2x \minus{} 3$. Taking derivates w.r.t. $ x$,
\[ 2p \ dp/dx \equal{} 2\]
Thus, $ dp/dx \equal{} \frac {1}{p}$
Substituting,
\[ dp/dx \equal{} \frac {1}{\sqrt{2x \minus{} 3}}\]


analytic geometry
geometry
rectangle
puzzles

You are asked to mend a hole in a floor exactly 20 cm wide and 120 cm long. All you have available is a board 30 cm wide and 80 cm long.

How can you accomplish the task by just making one cut to divide the board into two pieces?

The hole must be entirely covered, not two parts of the board may overlap and the board may not overlap the edge of the hole.

[hide]let the 30x80 board be placed on the coordinate plane so that it is a rectangle with vertices at $(0,0), (30,0), (0,80),$ and $(30,80)$

the desired cut should go from $(10,0)$ to $(10,40)$ to $(20,40)$ to $(20,80)$.[/hide]

Wait, your answer is the solution given in the book so it must be correct.  :blush:


What Indian movies or songs do you like or have you seen lately?

I like Main Hoon Na, and Bunty aur Bubli.

I dont really like Hindi Movies that much but I have seen Bunty Aur Babli which was unique from the rest.

[quote="star99"]it might be one of the few movies not copied from hollywood.  but anyways i like bollywood better than hollywood.  did you know theres a kollywood?[/quote]
Really I didnt know. :lol:

I watch a lot of Indian Movies..actually my parents do, I just watch with them, I dont really know any singers, and I really dont like the songs much either, but I really liked Lagaan, and Im also interested in seeing Mangal Pande-The Rising, I heard it was a really good movie.

mangal pandey was really good.  in music, I like Salaam Namaste - it's like the new fad.  I swear.  Anyway, it's really good, but it just came out.

I can believe it.  I want to watch it too, cuz anyways preity and saif are some of my favorites.  Bunty aur Bubli and Main hoon na music is really good too.  Any girls (or guys) out there learn Indian dance?  (classical, folk, or filmi)  I learn folk, and me and my friends do a little filmi for parties.

ok instead of indian music, what western music do you guys listen to?

i like rock- Guns N Roses , metallica , etc

eminem stop writing those unsacred names everywhere....
I can't believe star99,u feel bunty aur babli was unique!!It was almost like catch me if u can .You want to know my choices for western music eminem?

Have you guys seen "Kaante" and "Munnabhai"?The're one of the few hindi movies i can tolerate.I've seen almost all hindi movies though :huh:

What's the worst hindi movie anyone's watched ever?

[quote="GoBraves"][quote="shadysaysurspammed"]eminem stop writing those unsacred names everywhere....
[/quote]

I know, seriously. Stop trying to tell everyone you listen to music, nobody cares.[/quote]

 :rotfl:  :rotfl: Way to "go"braves...i'm gonna make sure eminem reads that....no one does care :rotfl:  :rotfl:
Sorry for this spam...

No one said which is the worst movie they've watched...since 1 hour :D

Indian movies are so pathetic. They are so unoriginal and so predictable, (always a love story with a hot girl, and someone tries to stop them from marrying each other, most of the time it's their families. And these people fall in love on first sight, which is really stupid. Now Indian songs aren't horrible, I think that the old indian songs from like ten years ago were decent, but now adays, Indian entertainers are trying to copy american pop like the backstreet boys and brittany spears, which is just about the worst music ever. :mad:  :mad:  :mad:

I agree about the part of english in the hindi songs.  the B&B rap was just wrong.

the worst movie i have watched is khushi (with kareena kapoor--yuck)

like I said before...nobody cares about you and your music

This topic is about Indian movies and songs. Its almost like your trying to show off(i said almost because I dont know if this was your intention, but it sure seems like it)

eminem, no one said you should be banned.  mr. high-and-mighty, can't you tell when someone's just kidding?

anyways, does anyone have favorite old-is-gold movies/songs/actors?  I do, mainly cuz of my parents (and my dad sings randomly a lot).  my friends and I play abc antakshari too.  the old songs are much easier to pick out on the piano as well.

well,okay , guess  i have been toooo high and mighty , i apologise , ne way i didnt mean to show off that is ,that wasnt my intention as go braves said , but  maybe  shady and go braves could have said it more politely , if they found me irritating?

as for the slang , well its a bore to write "are not" cant u simply write "aint?" whats so wrong or stupid in that ? i cant understand whats to show off in this, all i wrote is a smaller word than a longer one

dont think i have been writing what i wrote for "Showing off" . if its that offensive , i'll delete it

ok, no hard feelings from me

[quote="Iversonfan2005"]eminem, I know more about rap and well, Eminem, than you do, so please stop trying to show off, especially with your usage of "ain't" and other slang on this forum ;)[/quote]

Anyone who roasts eminem is my friend :D

Anyway eminem,99.9% on this site are civilized people and would like to see nice posts and not feel that they are on a steert side forum...ain't "ain't" too nice to use 'round here....

Iversonfan,You say u know a lot about rap,eh?I have this topic called "YOur songs" in the games and fun factory...So if you feel like "Showing off" ( :D ) your talent,please do bcos i haven't found an equal yet,You know,so right now i feel like the king of rappers.

Old hindi songs had really deep lyrics and feelings though they were boring to listen to...but now a days songs like "Ishq kammeena" are coming out...what the heck!!!

[quote="shadysaysurspammed"]...but now a days songs like "Ishq kammeena" are coming out...what the heck!!![/quote]

agreed

[quote="Iversonfan2005"]Nah, I don't really write any raps, and if I do, they'd be too profane for this forum, I simply am a fan and only listen to rap for the last 7 years.[/quote]

7 years! thats cool, i myself $heard$ of rap last year  :D

who's your favourite rapper? and who's the guy on your avatar?

[quote="riddler"]I think the guy is 2pac right?[/quote]

It's $\text{50 Cent}$ who is part of $\text{G-Unit}$...but he's not my favorite rapper ever

Here's my top 5 favorite rappers of all time:

1) 2Pac
2) Notorious B.I.G.
3) Eminem
4) 50 Cent
5) Jay-Z

[quote="star99"]again, this is about [b]indian[/b] entertainment (and don't say that this is entertainment for some indians).  if you want to discuss rap, do it in the games and fun factory.[/quote]

Well, riddler and shady asked me a question, I had to answer it ;)

u didnt have to answer it.  end of disscussion.

who likes remixes?  i personally hate them.  they make it all 'modern' like the B&B rap - really messed up

any one learn indian vocal, music, or dance?

i used to do vocal, but now i do indian folk dance

Haha, I like know ntohing about Indian entertainment.

I used to learn carnatic Music.
   Ashwath

hey eminem, here's another music fan.
i like linkin park,evanesence etc.mostly rock.
anyways ive got an equal craze for hindi music as well. bands like jal , strings r good.(at least i find them good)
(good to have some music on a maths-link!!!:-)

any one watch rang de basanti?  i [u]loved[/u] the movie...except for the end.  why not let them all live and instead say that india is changing, ...

IQBAL was great.

I wonder why no one mentioned it.

BLACK was also a good movie.

black was great too.  i didnt watch iqbal.

[quote="star99"]black was great too..[/quote]

I discovered a new way to feel sleepy incase u have a sleeping disorder------> by watching that movie :rotfl:  :rotfl:  :rotfl:


Prove that if $ 2n\plus{}1$ and $ 3n\plus{}1$ are perfect squares, then $ 40|n$.

[quote="Nora.91"]Prove that if $ 2n \plus{} 1$ and $ 3n \plus{} 1$ are perfect squares, then $ 40|n$.[/quote]

[hide="brutal force......."]
$ 2n\plus{}1\equal{}1 \pmod{8} \Rightarrow 4|n \Rightarrow 3n\plus{}1\equal{}1 \pmod{8} \Rightarrow 8|n$

As squares $ \equal{}\minus{}1,0,1 \pmod{5}$,

$ 2n\plus{}1\equal{}\minus{}1 \pmod{5} \Rightarrow n\equal{}\minus{}1 \pmod{5} \Rightarrow 3n\plus{}1\equal{}\minus{}2 \pmod{5}$, contradiction.

$ 2n\plus{}1\equal{}0 \pmod {5} \Rightarrow n\equal{}2 \pmod{5} \Rightarrow 3n\plus{}1\equal{}2 \pmod{5}$, contradiction.

$ 2n\plus{}1\equal{}1 \pmod{5} \Rightarrow 5|n$

$ \begin{cases}8|n \\ 5|n \end{cases} \Leftrightarrow 40|n$

[/hide]

My solution:
From $ 2n \plus{} 1 \equal{} a^2$, $ a$ is odd.
So $ 2n \plus{} 1 \equal{} 8k \plus{} 1 \Leftrightarrow n \equal{} 4k$.
So $ 12k \plus{} 1 \equal{} b^2$, $ b$ is odd.So: $ 12k \plus{} 1 \equal{} 8m \plus{} 1 \Leftrightarrow 3k \equal{} 2m$,so $ 2|k$
Thus $ k \equal{} 2l \rightarrow n \equal{} 8l$ $ \boxed{1}$

$ 2n \plus{} 1 \equal{} a^2$
$ 3n \plus{} 1 \equal{} b^2$
So $ a^2 \plus{} b^2 \equiv 2(mod 5)$
But $ a^2 \equiv 0,1,4(mod 5)$.So $ a^2 \equiv b^2 \equiv 1(mod5)$
But $ n \equal{} b^2 \minus{} a^2 \equiv 0(mod 5)$ $ \boxed{2}$
From $ \boxed{1},\boxed{2}$,we have $ 40|n$.
I hope it is right.

great solutions . thanx NORA.91 for the question.


given that $7ab73$ is a number of 5 digits such that a,b are unknown . Determine a,b such that the number shall be divisible by 99

the number is divisible by 9, so [b]a + b = 1[/b] or [b]a + b = 10[/b]. if we check the first posibility (numbers [b]70173[/b] and [b]71073[/b]), they are not divisible by 11. => [b]a + b = 10[/b]

now let's take [b]x[/b], such that [b]99 * x = 7ab73[/b]. we know that x is somewhere between 700 and 800 (if x was greater that 800, [b]a + b[/b] would be greater than 10)

[b]99 * x = 7ab73[/b] => [b]100 * x - x = 7ab73[/b] =>
[b]???00 - x = 7ab73[/b], so the last two digits of x should be 27.
so [b]x = 727[/b].

check: 99 * 727 = 71973.

[b]a = 1
b = 9
[/b]

if you remember the divisibility test for 11, then this is easy (it is if you sum the digits, alternating pos and neg, and that is divisible by 11, then the whole is divisible by 11) like 121 would be 1-2+1 = 0 which is divisible by 11 so 121 is divisible by 11, the proof of this is pretty easy, just look at a number in $\mod 11$, and note $-1 \equiv 10 \mod{11}$
i.e. $121 \mod 11$
$1*10^2+2*10^1 +1 \mod{11} \equiv 1(-1)^2 + 2(-1) +1$


calculus
derivative
vector
advanced fields
advanced fields unsolved

What is the chain complex for $ \mathbb{R}^n\setminus\{x_1,\ldots,x_m\})$?

I guess I'm not sure what is meant by "the" chain complex and "for". Are you referring to the chain complex $ \Omega^*(\mathbb{R}^n\setminus \{x_1, ..., x_m\})$ with $ d$ ? Then it would just be, say $ U\equal{}\mathbb{R}^n\setminus \{x_1, ..., x_m\}$, $ \Omega^0(U)\to \Omega^1(U)\to \cdots$ where each map is just exterior differentiation $ d$ and the $ p$-th vector space is just the vector space of $ p$ forms on $ U$.

Or maybe you actually are referring to something completely different and I'm just babbling on about nothing you care about?


A set of natural numbers 1,2,…….,n are considered at the outset. Every second integer in the said set is deleted in the first operation; every second integer in the remaining set is deleted in the second operation and the process is continued till the pth operation.
(i)  If integers less than 10000  are considered(  i.e. n < 10000), and the operation is continued 7 times (i.e. p = 7); what will be :
(a) The 59th term in ascending order ; (b) The last term  in ascending order
in the remaining sequence?
Also, determine the perfect squares which belong to this sequence.

(ii) If integers less than 100,000 are considered instead ( i.e. n < 100,000) and the operation is continued 9 times(i.e. p = 9), what will be the 48th term in ascending order  in the remaining sequence.
Determine the perfect squares that belong to the sequence.

The set of natural numbers at the start can be represented as Tn = n. (n < 10000)

After the 1st operation we would be left with numbers like 1, 3, 5, 7, .....
(i)
T1n = 2n - 1.
T2n = 4n - 3.
T3n = 8n - 7.
T4n = 16n - 15.
T5n = 32n - 31.
T6n = 64n - 63.
And, after the 7th operation we would be left with 1, 129, 257, 385, 513, ....
T7n = 128n - 127.

(a) The 59th term in ascending order = 128*59 - 127 = 7425.

(b) For the last term in ascending order in the remaining sequence, 128*n - 127 < 10000 ==> n < 79.1171875. Therefore, n = 79 and the required last number is 128*79 - 127 = 9985.

The sequence is t(n) = 128*n - 127. The perfect squares belonging to this sequence would satisfy 128*n - 127 = m^2 where m is a positive integer.
This implies n = (m^2 + 127)/128 ==> m^2 - 1 should be divisible by 128.

We will find perfect squares in the sequence when (m+1)*(m-1) is divisible by 2^7.
This holds good for m = 1. Say, it holds good for odd values of m = (2*k + 1).

==> (2*k + 2)*(2k) is divisible by 2^7.
==> k*(k+1) is divisible by 2^5 = 32.

This would hold good only when k = 32*x or 32*x - 1.

We have the following table for the values of k, m, n.
[b]k	m	n[/b]
0	1	1
31	63	32
32	65	34
63	127	127
64	129	131
95	191	286
96	193	292
127	255	509
128	257	517

Since, n <= 79, we will have 3 perfect squares for n = 1, 32 and 34.
The numbers being 1, 3969 and 4225 which are perfect squares of 1, 63 and 65, respectively.

(ii) n < 100,000, p = 9
T(n) = (2^p)*n - (2^p - 1) = 512*n - 511

48th term in ascending order in the remaining sequence = 512*48 - 511 = 24065.

For the last term in ascending order in the remaining sequence, 512*n - 511 < 100000 ==> n < 196.3105469 . Therefore, n <= 196.

[b]k	m	n[/b]
0	1	1
127	255	509
128	257	517
255	511	2041
256	513	2057
383	767	4597
384	769	4621

The only perfect square is 1 as n <= 196.


Show that there are infinitely many squares that are the sum of a square and a prime.

Let $n^2=m^2+p$ where $p$ is prime
$(n-m)(n+m)=p$ so p to be prime $n-m=1$ or $n=m+1$
if $n=m+1$ to assume that there a finite number of solutions is to assume that there are finite number of primes in the form $p=2m+1$

For any odd prime n,
1+3+5+7+...+n=a square
1+3+5+7+...+n-2=a square
if you add n to the second square, then it forms the first square


Hi there.

I'm currently refreshing math from my commercial school and i've now started from the very bottom. When i ended the commercial school my math knowledge wasn't as good as when i ended the elementary school (in Denmark, the elementary school consists of 8-10 levels (years)).

So i'm starting off with the first year from the commercial school and got to this equation:

Reduce:

(a+b / a-b) - (a-b / a+b)

I really got stuck at this one and asked everybody on messenger on how to do this one. No one could tell me anything but one single contact who could make his computer come up with the following result (might remember it uncorrectly, so plz don't take a big look at this one if you are having problems reaching that result):

-4ab / a^2 + b^2

It was something like this, but not sure it is how to told me.

Now, can anyone show me the equation step by step (not just the true result). Thx a lot in advance to anyone that can contribute in solving the equation that i've spent so many hours on figuring out how to do.

$\frac{a + b}{a - b} - \frac{a - b}{a + b}$

In order to simplify this expression, you need to find a common denominator.  So:

$\frac{a + b}{a - b} \cdot \frac{a + b}{a + b} - \frac{a - b}{a + b} \cdot \frac{a - b}{a - b}$

$= \frac{(a +b)^2 - (a - b)^2}{a^2 - b^2}$

$= \frac{4ab}{a^2 - b^2}$

This looks great, and very close to how i remembered it (actually more like to be this one, and furthermore you are probably right). But how can you allow to make the a+b / a-b be timed with a+b / a+b and on the other side of the minus in the middle you use a-b / a-b? I've learnt you may only do the same on each side, but guess this is only if you have the equalsmark. The book i use isn't that detailed, but obviously from your suggestion (which looks very right indeed) you are allowing yourself to do the same above and beneath the dividend line. This one surprices me since i thought you would have to do the same on the other side aswell.

But thx a lot for your reply, which furthermore was fast indeed :)

Well, Sporally notice that

$\frac{a-b}{a-b}$ and $\frac{a+b}{a+b}$ both $= 1$

Because a number divided by itself can be seen to always equal 1 (0 is the exception)

That is why we can multiply them to the two expressions in our original equation, because we know that when we multiply a number by 1, the number isn't changed!

Perhaps this should be in middle school? But as a new member, I guess it's okay  :)

Right, as Windslicer said:

$\frac{a + b}{a - b}  = \frac{a + b}{a - b}\cdot\frac{a+b}{a+b}$
and
$\frac{a - b}{a + b} = \frac{a - b}{a + b}\cdot\frac{a-b}{a-b}$,
so
$\frac{a + b}{a - b} - \frac{a - b}{a + b} = \frac{a + b}{a - b}\cdot\frac{a+b}{a+b} - \frac{a - b}{a + b}\cdot\frac{a-b}{a-b}$,
and we can carry on from there.  The basic point here is that you haven't in fact done anything to change the value of your expression -- you've just multiplied things by 1, which doesn't change them, in order to get them into the form you need.  (In general, to combine fractions, you need to get them all with the same denominator, which is why we carried out this process.)

What do you mean by "commercial school?"  Are you going to school and studying for a particular career?  (In America, we usually call that "vocational school.")

Well, this one is a low difficulity, but it was first year study in the commercial school or business school. In DK, that's a quite high thing. Though this book i'm studying is quite low in difficulity it will speed up in the rest of the book for the first year and end on the third year with stuff like integral calculation and stuff like that, which i consider quite high (of course not the highest, but quite high).

About this equation i knew about the fact that you are allowed to do as much as you want with 1 (times 1 or divided by 1). I just didn't see it that way yesterday when i looked at the equation that Dakyru wrote, but i noticed it was two different things he timed it up with, but of course it will always be 1 once you put in numbers.

I can...

When you have in the numerator $(a+b)^2-(a-b)^2$, you know that $(a+b)^2=a^2+2ab+b^2$ and that $(a-b)^2=a^2-2ab+b^2$, however, you have to distribute the $-$ sign to the second part, thus everything cancels out except for $2ab+2ab=4ab$. In the denominator, $(a+b)(a-b)$ is the difference of two squares, therefore it would be $a^2-b^2$, or you could just foil it out to make sure.

[quote="Sporally"][u]@ Dakyru[/u]

Can you tell me more specifically what you do between the last two lines you have? I can follow you this far, but going from (a+b)^2-(a-b)^2 to 4ab is out of my area of comprehension. Can you make it more detailed with an extra step or two? Or maybe someone else can.

BTW: How do you make those real equations instead of something like (a+b)^2-(a-b)^2?[/quote]

Here's more of a visual to add to what Iversonfan2005 said:

$\frac{(a + b)^2 - (a - b)^2}{(a + b)(a - b)}$

$= \frac{(a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)}{(a^2 - ab + ba - b^2)}$

$= \frac{a^2 - a^2 + 2ab + 2ab + b^2 - b^2}{a^2 - b^2}$

$= \frac{4ab}{a^2 - b^2}$

I'm not exactly sure what you mean by "making them real equations".  Firstly, it's best to make sure you have your terminology right.  These are [i]expressions[/i].  Equations require something set equal to something else with the use of an $=$ sign.  You can't create an equation when you are just given an expression to simplify.  I hope that answers your question.   :)

Well, let's see if i can rewrite it better. When i write an expression, i write something like this:

(a+b)^2/(a-b)^2

And my question is: How do you write it with the fraction lines and stuff like that, where you will instead have the (a+b) above the fraction line, and furthermore you have the ^numbers raised like it should be. Hope you understand what i mean (english isn't my mother language you know ;) ).

Oh, you're talking about formatting.  When you want to type math, start with a $.  The list of symbols and commands to use are listed [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php]here[/url].  For example, if you want to create a fraction, you would use \frac{numerator}{denominator}, replacing my text with whatever you want to put in the numerator and denominator.  After you're done with the math, close it with another dollar sign.

Ah great. Doesn't look easier, but surely is better when you have wrote it with that format. Let me just give it a try:

$\frac{(a+b)^2}{(a-b)^3}$

Ah great, works! Thx for the tip. It will give a better overview whenever i get back to this forum with another question (which surely will come sooner or later when i run into a new problem).


calculus
integration
inequalities
real analysis
real analysis solved

Evaluate the follwing definite integral.

\[\int^1_0\frac{x^4(1-x)^4}{1+x^2} dx\]

The calculation is a little tedious but straightforward. We mutliply out the numerator and do the long division:

$\frac{x^4(1-x)^4}{x^2+1}= x^6-4x^5+5x^4-4x^2+4-\frac4{x^2+1}$.

Integrate that from 0 to 1 to get $\frac17-\frac46+\frac55-\frac43+4-\pi = \frac{22}7-\pi.$

Now go back to the original integral. Given the identity

$\int_0^1x^m(1-x)^ndx=\frac{m!n!}{(m+n+1)!}$, we have that

$\int_0^1x^4(1-x)^4dx = \frac{4!4!}{9!}=\frac1{630}.$ Thus

$\frac1{1260}<\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx<\frac1{630}.$

(This because $\frac12<\frac1{1+x^2}<1.$)

The inequality we have can be rearranged to the following:

$\frac{22}7-\frac1{630}<\pi<\frac{22}7-\frac1{1260}.$


inequalities
LaTeX
inequalities proposed

Please type this solution by LATEX, i can't read it. :mad:

me neither.

[quote="Sung-yoon Kim"]For positive reals $ a,b,c$, prove that
\[ \frac{4(a\plus{}b\plus{}c)(bc\plus{}ca\plus{}ab)}{(b\plus{}c)^{2}\plus{}(c\plus{}a)^{2}\plus{}(a\plus{}b)^{2}}\leq{((2a\plus{}b)(2b\plus{}c)(2c\plus{}a))}^{\frac{1}{3}}.\]
[/quote]

anyhow, my solution is really simpler.

[quote="Sung-yoon Kim"]For positive reals $ a,b,c$, prove that
\[ \frac{4(a\plus{}b\plus{}c)(bc\plus{}ca\plus{}ab)}{(b\plus{}c)^{2}\plus{}(c\plus{}a)^{2}\plus{}(a\plus{}b)^{2}}\leq{((2a\plus{}b)(2b\plus{}c)(2c\plus{}a))}^{\frac{1}{3}}.\]
[/quote]
We aim at proving a slightly stronger inequality
\[ \frac{a\plus{}2b}{a\plus{}2c}\plus{}\frac{b\plus{}2c}{b\plus{}2a}\plus{}\frac{c\plus{}2a}{c\plus{}2b}\le\frac{3(a^{2}\plus{}b^{2}\plus{}c^{2})}{ab\plus{}bc\plus{}ca}\]
Adding 3 to both sides, the above becomes
\[ 2(a\plus{}b\plus{}c)\left(\frac{1}{a\plus{}2c}\plus{}\frac{1}{b\plus{}2a}\plus{}\frac{1}{c\plus{}2b}\right)\le\frac{3(a^{2}\plus{}b^{2}\plus{}c^{2})}{ab\plus{}bc\plus{}ca}\plus{}3\]
which together with the AM-GM inequality imply the original posted inequality.
Since $ (a\plus{}b\plus{}c)^{2}\equal{}(a\plus{}2b)(a\plus{}2c)\plus{}(b\minus{}c)^{2}$ it follows that
\[ \frac{a\plus{}b\plus{}c}{a\plus{}2c}\equal{}\frac{a\plus{}2b}{a\plus{}b\plus{}c}\plus{}\frac{(b\minus{}c)^{2}}{(a\plus{}2c)(a\plus{}b\plus{}c)}\]
Summing up, we get
\[ \sum_{cycl}\frac{a\plus{}b\plus{}c}{a\plus{}2c}\equal{}3\plus{}\sum_{cycl}\frac{(b\minus{}c)^{2}}{(a\plus{}2c)(a\plus{}b\plus{}c)}\]
Thus the inequality to prove is equivalent to
\[ 6\plus{}2\sum_{cycl}\frac{(b\minus{}c)^{2}}{(a\plus{}2c)(a\plus{}b\plus{}c)}\le\frac{3(a^{2}\plus{}b^{2}\plus{}c^{2})}{ab\plus{}bc\plus{}ca}\plus{}3\]
or $ x(b\minus{}c)^{2}\plus{}y(c\minus{}a)^{2}\plus{}z(a\minus{}b)^{2}\ge 0$ where
\[ x\equal{}\frac{3(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)}\minus{}\frac{1}{a\plus{}2c}\]
Without loss of generality, assume $ (b\minus{}a)(b\minus{}c)\le 0$. Then $ (a\minus{}b)(a\minus{}c)\ge 0$ and $ (c\minus{}a)(c\minus{}b)\ge 0$. Using $ (a\plus{}2b)(a\plus{}2c)\equal{}3(ab\plus{}bc\plus{}ca)\plus{}(a\minus{}b)(a\minus{}c)$ we have
$ \frac{3(ab\plus{}bc\plus{}ca)}{a\plus{}2c}\equal{}a\plus{}2b\minus{}\frac{(a\minus{}b)(a\minus{}c)}{a\plus{}2c}\le a\plus{}2b$
$ \le 2(a\plus{}b\plus{}c)\le\frac{9(a\plus{}b\plus{}c)}{4}$
that is, $ x\ge 0$. Similarly, $ z\ge 0$.
[b]Case 1.[/b] $ a\ge b\ge c$. Then $ y\ge 0$ since
\[ \frac{1}{b\plus{}2a}\le\frac{1}{a\plus{}b\plus{}c}\le\frac{a\plus{}b\plus{}c}{3(ab\plus{}bc\plus{}ca)}\le\frac{3(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)}\]
and we are done.
[b]Case 2.[/b] $ c\ge b\ge a$. If $ y\ge 0$, we are done. Suppose $ y\le 0$. Then
\[ x\plus{}2y\equal{}\frac{9(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)}\minus{}\frac{1}{a\plus{}2c}\minus{}\frac{2}{b\plus{}2a}\ge 0\]
since
\[ \frac{1}{a\plus{}2c}\plus{}\frac{2}{b\plus{}2a}\equal{}\frac{4a\plus{}b\plus{}4c}{(a\plus{}2c)(b\plus{}2a)}\le\frac{2(a\plus{}b\plus{}c)}{ab\plus{}bc\plus{}ca}\le\frac{9(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)}\]
and
\[ z\plus{}2y\equal{}\frac{9(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)}\minus{}\frac{1}{c\plus{}2b}\minus{}\frac{2}{b\plus{}2a}\ge 0\]
since
\[ \frac{1}{c\plus{}2b}\plus{}\frac{2}{b\plus{}2a}\equal{}\frac{2a\plus{}5b\plus{}2c}{(c\plus{}2b)(b\plus{}2a)}\le\frac{2(a\plus{}b\plus{}c)}{ab\plus{}bc\plus{}ca}\le\frac{9(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)}\]
The proof is complete.

Nice solution, Ductrung. :)

Sung-yoon Kim, is your solution like Ductrung? :?:

Thank you very much.

good solution.

[quote="Sung-yoon Kim"]For positive reals $ a,b,c$, prove that
\[ \frac{4(a\plus{}b\plus{}c)(bc\plus{}ca\plus{}ab)}{(b\plus{}c)^{2}\plus{}(c\plus{}a)^{2}\plus{}(a\plus{}b)^{2}}\leq{((2a\plus{}b)(2b\plus{}c)(2c\plus{}a))}^{\frac{1}{3}}.\]
[/quote]

$ ((a\plus{}b\plus{}c)(bc\plus{}ca\plus{}ab))^{3}\equal{}((a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}abc)^{3}\leq ((a\plus{}b)^{3}\plus{}a^{3})((b\plus{}c)^{3}\plus{}b^{3})((c\plus{}a)^{3}\plus{}c^{3})$
$ \equal{}(2a\plus{}b)(2b\plus{}c)(2c\plus{}a)(a^{2}\plus{}ab\plus{}b^{2})(b^{2}\plus{}bc\plus{}c^{2})(c^{2}\plus{}ca\plus{}a^{2})$, so we need to show that $ 8(a^{2}\plus{}ab\plus{}b^{2})(b^{2}\plus{}bc\plus{}c^{2})(c^{2}\plus{}ca\plus{}a^{2})\leq (a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}bc\plus{}ca\plus{}ab)^{3}$. This can be checked by the direct calculation, as in the solution of http://www.mathlinks.ro/Forum/viewtopic.php?t=162989.


superior algebra
superior algebra solved

[quote="Mitzah"]How many simetrizable elements are there in the ring $Z_{9}$ when we look at multiplication?[/quote]

Hi Mitzah,

May be I just forgot or I never knew, but what do you mean by a simetrizable element in a ring?

Best,

I think simetrizable here means invertible.

In $\mathbb{Z}n$ there are $\varphi (n)$ multiplicative invertible elements.

$\varphi (n)$ is the number of positive integers smaller than $n$ coprime to it.

This is because if $gcd(m,n)=1$ then there exist $a,b\in\mathbb{Z}$ such as $am+bn=1$ so $am\equiv 1(\mod n)$.

We will have to wait for Mitzah to clarify that point, because I don't really think it should be invertible (otherwise the computation is quite easy  :lol: )

Best,

How difficult could the computations be if we work with only 9 elements?

May be I didn't explain myself. I mean that if you say simetrizable=invertible, then, with 9 elements, everything you have to do is to check which elements are coprimes with 9 in the set {1,2,3,4,5,6,7,8}, and that is inmediate, the only not coprime with 9 are 3 and 6, the others are coprimes  :lol: .

That is why I suppose he didn't ment to say invertible.

Best,

what I ment to say was that whatever his definition for simetrizable is, to find the simetrizable elements can't be a dificult task since we only work with 9 elements.

I have seen in romanian publications the word "simetrizable" used for invertible.

[quote="amfulger"]I have seen in romanian publications the word "simetrizable" used for invertible.[/quote]

OK, I didn't know that. I have listen the term as a synonim of "diagonalizable", and of course, you can have invertible matrices non diagonalizable (at least not over the original field) and diagonalizable matrices non invertible  :D .

But every day one learns something new. Best,


Let a be an element in some group G and say that a has finite order n. Let q = a^k for some k.

Prove that the order of q divides k.

I dunno if its me or what, but this has been driving me batty.

[quote="Ourobolus"]this has been driving me batty.[/quote]

Perhaps this is because the statement is nonsense?

The order of $a^k$ is $\frac n{\gcd(n,k)}$.

Let $d=(k,n)$

$(q^k)^\frac{n}{d} = q^{k\frac{n}{d}} = (q^n)^\frac{k}{d} = e$.

Hence $s=o(q^k)$ is a divisor of $n/d$.

$q^s = a^{ks} = e \Rightarrow n|ks$ because $o(a)=n$.
thus $\frac{n}{d}|\frac{k}{d}s$ and $(\frac{n}{d},\frac{k}{d})=1$, we must have $\frac{n}{d}|s$.

Thus $s=\frac{n}{d}$.


I recently purchased the book First Steps for Math Olympians and was reading the number theory section with no problem till i came to an example problem and solution that seemed odd to me. bare with me i don't know how to type in math symbols so for reference the three parallel line structure will be replaced with an = sign.

First the book notes that [i]-11 mod 3= 4, since -11 -4= -15[/i], see that seems to me like a kind of a stretch, if that's true could any number be the remainder as long as it is a multiple of 3?

Next: the problem is 3^2006 mod 11
the book has since [i]3^5= 243= 2x121+1= 2x11^2+1 thus 3^5 mod 11=1 and 3^2006 mod 11= 3^2005x 3 mod 11= (3^5 mod 11)^401 x 3 mod 11= 1^401 x 3= 3
so therefore 3^2006 mod 11= 3[/i]
sorry but i don't understand how [i]3 mod 11= 3[/i] as in the last line of the equation, i understand that [i](3^5 mod 11)^401= 1[/i] but i dont see how multiplying that by [i]3 mod 11 [/i]can give an answer of 3. :huh:
Please help shed some light on this problem for me
 :lol:

The $ \equiv$ sign is a more subtle notion than remainder.  We say that $ a \equiv b \bmod m$ (read: $ a$ is congruent to $ b$ modulo $ m$) iff $ m | a \minus{} b$.  In other words, $ a, b$ leave the same remainder upon division by $ m$.  so it's true that $ \minus{}11 \equiv 4 \bmod 3$ because $ 3 | \minus{}11 \minus{} 4$.  It's also true that $ \minus{}11 \equiv 7 \equiv 10 \equiv 1000 \bmod 3$.

It turns out that this definition has a lot of useful properties:

- It is transitive, symmetric, and reflective,
- It respects addition, and
- It respects multiplication.

So we can perform arithmetic with the $ \equiv$ sign just as we would perform arithmetic with the $ \equal{}$ sign.  For example, since

$ 3^5 \equiv 1 \bmod 11$

we know that

$ 3^6 \equal{} 3 \cdot 3^5 \equiv 3 \cdot 1 \equiv 3 \bmod 11$
$ 3^7 \equal{} 3 \cdot 3^6 \equiv 3 \cdot 3 \equiv 9 \bmod 11$

and so forth.  With numbers this small, you can verify that this works for yourself and perhaps even venture a proof as to why this works.  :)

yeah, that makes sense as does the explanation in wiki, i don't like the way the author explained number theory plus he seems to have his mods written differently he has them written as a mod n = b instead of a= b mod n and i prefer the latter. Thanks for the information and quasi proof, it helped :lol:

i think i understand it now, however i have just one question, when you put
that 3^6= 3^5 x 3= 3 x 1 = 3 mod 11
how did you do that, i mean i get that you substituted 1 for 3^5 becase 3^5=1 mod 11 but when you multiply that by three, is three the remainder for 3= x mod 11? (x represents the remainder in the last line)
edit*- actually i think i know what you did, correct me if i'm wrong but  in the equation 3^6= 3^5 x 3= 3 x 1 = 3 mod 11, when you do 3^5 x 3 is that the same as ( 3^5= 1 mod 11) x 3?  or is it 3^5=1 mod 11 times 3^1= x mod 11?

Essentially.  The underlying lemma, written out explicitly, is this:

[b]Lemma:[/b]  If $ a \equiv b \bmod m$ then $ ac \equiv bc \bmod m$.

Proof:  If $ m | a \minus{} b$ then $ m | c(a \minus{} b)$.

cool, thanks :lol: i think however that i'm going to get the AOPS Intro to Number because the book i use is written a different way than i prefer, i've seen the aops and like that.


Hi every One. Can you tell me the way easy to find this sum:

$ S \equal{} \frac {1}{1} \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{10}$

Thank you!

Note:

I know that $ S\equal{}\frac{7381}{2520}$ but i don't like with my solution. I think it can be done by a nice way but I don't remember it.

$ H_n$, the partial sums of the harmonic series, doesn't really have a nice closed form.


inequalities
geometry
geometric transformation
reflection
cyclic quadrilateral
angle bisector
geometry proposed

$ \triangle ABC$, $ \angle A, \angle B, \angle C < 90^\circ$, $ M$ is a point in it.

$ F,G$ is the orthogonal projection of M on $ BC$ and $ AC$. Prove that:

$ AB-FG\geq \frac{MF\cdot AG+MG\cdot BF}{CM}$

[hide]Substitute $ AG\equal{}AC\minus{}GC$ and $ BF\equal{}BC\minus{}FC$ to the right hand side to get

$ \frac{MF\cdot AC\plus{}MG\cdot BC}{CM}\minus{}\frac{MF\cdot GC\plus{}MG\cdot FC}{CM}$

From Ptolemy's Theorem on cyclic quadrilateral $ MGCF$ we get $ \frac{MF\cdot GC\plus{}MG\cdot FC}{CM}\equal{}FG$ so that our inequality is equivalent to

$ AB\ge \frac{MF\cdot AC\plus{}MG\cdot BC}{CM}$

Reflect line $ CM$ across the angle bisector from $ C$ to line $ CM^\prime$ and let $ CM^\prime$ intersect $ AB$ at $ C^\prime$.  Let the projection of $ A$ and $ B$ to line $ CM^\prime$ be $ X$ and $ Y$, respectively.  Then, since $ \angle ACX\equal{}\angle MCF$ we have $ ACX\sim MCF$ so we have $ \frac{MF}{CM}\equal{}\frac{AX}{AC}\Rightarrow \frac{MF\cdot AC}{CM}\equal{}AX$.  Similarly, we find $ \frac{MG\cdot BC}{CM}\equal{}BY$.  Thus, our inequality is equivalent to

$ AB\ge AX\plus{}BY$

But, this is obvious since $ AC^\prime\ge AX$ and $ BC^\prime\ge BY$ since $ AX,BY$ are legs of the triangles with respective hypotenuses $ AC^\prime,BC^\prime$.

[/hide]


For those of you who have taken the Stats exam, do they provide you with the standard normal probabilites table? and t-distribution critical values? and $\chi^2$ critical values?

I don't really remember, but I know that you can do that on your TI-83+...

I haven't used my TI-83 plus for some 3 years now... oh well, I guess I could figure it out.

I think they do, but I'm not sure if they're available for the multiple choice as well. Based on previous multiple choice sections, I think you need them to answer a few.

i know this topic is more questions about stats than whether to take it or not,  but do u think its a good idea for me to take AP Stats and Pre-Calculus at the same time? or should i only do 1? i'm a pretty quick learner but im also doing APUSH and heard its very timeconsuming

Hey,
I'm pretty sure that they give you the tables and everything for both the multiple choice and the free-response, as that's what I've been told...I also recently took a mock, and they were given to me for both sections then.   Also, on another note, I'm currently taking both AP Stat. and Pre/Calculus, and I'm doing okay, but I don't know what'll happen if you take AP US as well....

Don't worry, stats really won't require any advanced math.  If you're not looking to know stats very well, the class should not be very challenging.

I heard it was easy, but I don't know, since I'm self-studying it... and I started about 2 weeks ago :?. Hope I can pull off a 5 on the exam.

You should be able to.  I believe getting 70% or higher of the points virtually guarantees a 5.  So that's not much you have to aim for :).

They do give you the reference tables, even the formulas for the Z, T, etc. test statistics.  There's also a statistics program thingy on the TI-89, which you can use.

Sheepwarrior, I'm sure it'll be fine if you take Stats, Pre-calc, and APUSH.  I know lots of kids who do that.


geometry
incenter
geometric transformation
reflection
geometry proposed
U sviti matematyky

Construct the triangle $ABC$ if known are the vertice $A,$ the incenter $I$ and the intersection point of the medians $M.$

See all the problems from Vol. 12(2006), #1-3 at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=465391#p465391]Ukrainian journal "In the World of Mathematics" problems contest[/url]

Since the vertex A and the centroid M are known, the midpoint X of BC is easily construced as $MX =-\frac{MA}{2}.$ Let D be the (unknown) tangency point of the incircle (I) with BC. Then XI is the Newton line of degenerate tangential quadrilateral ABDC with the diagonals AD, BC and incircle (I), so that XI cuts AD at its midpoint Y. Hence one locus of D is a line $k \parallel XI$ passing through the reflection of A in XI. On the other hand, $\angle IDX = 90^\circ$ is right, so that another locus of D is a circle $\mathcal C$ with diameter XI. This circle can intersect the line k in two points D, D', touch it in one point D, or miss the line completely. Since the midpoint X of BC and the tangency point D of the incircle (I) with BC must lie on the opposite sides of the bisector AI of $\angle A$, only the intersections of $D, D' \equiv k \cap \mathcal C$ on the opposite side of AI than X are acceptable. Thus the problem can have two, one, or no solution. Once the point D is constructed, the line $BC \equiv XD$ is known, the incircle (I) with radius r = ID is also known, and AB, AC are the incircle tangents from A intersecting XD at B, C.


Let $k$ a natural number and $ a_1,...,a_k,b_1,...,b_k$ rational numbers such that for any irational numbers $ x_i>1$ there exist positive integers $n_1,...,n_k, m_1,...,m_k$ such that $ a_1 f(x_1^{n_1})+...+a_kf(x_k^{n_k})=b_1 f(x_1^{m_1})+...+f(x_k^{m_k})$ where $ f(x)$ is the integer part of x. Prove that $a_i=b_i$.

Harazi, here is what i did for this very nice problem:

1) for every big prime P, there exist an irrational $x$, such that $\left\lfloor x^k\right\rfloor=1$(mod p)

2) For every big prime P, there exist an irrational $y$, such that $\left\lfloor y^k\right\rfloor=2$ (mod p)

Plug in $x$ and $y$, for a big enough prime in the equation to find out $a_i=b_i$.

If you want i can post the proof of 1) and 2).

BTW, whats your solution?

My idea is to use irrational numbers of form $x=\sqrt{m^2-1}+m$. We know that $x^n$ is almost integer, very very close to integer.

I really liked this problem a lot Harazi, why do you call it ugly? :P :)

Really, it is not ugly, but just technical. :)

I'm sorry guys for my stupidity, but I can't finish solution even with these hints.
Myth, Pascual and Harazi, could you please explain your solutions in details?

I said ugly in appearance! I really liked it a lot and I agree with Myth, it is quite technical. Iura, I am sorry, but I do ot have the time to post the solution now. Myth, could you please do that? Thanks a lot.

Can someone explain to me how we use the "almost integer" idea to solve the problem?

Thanks.


geometry
3D geometry
tetrahedron
rectangle
ratio
trigonometry

a.) It is well known in Euclidean geometry that the sum of the angles of a triangle is constant. Prove, however, that the sum of the dihedral angles of a tetrahedron is not constant.

b.) Let $ABC$ be the right-angled isoceles triangle whose equal sides have length $1$. $P$ is a point on the hypotenuse, and the feet of the perpendiculars from $P$ to the other sides are $Q$ and $R$. Consider the areas of triangles $APQ$ and $PBR$, and the area of the rectangle $QCRP$. Prove that regardless of how $P$ is chosen, the largest of these three areas is at least $\frac{2}{9}$.

c.) If the three equal cevians of a triangle divide the sides in the same ration and the same sense, must the triangle be equilateral?

For Question b) let AQ = x. Then suppose [QCRP]<2/9. Therefore $x-x^{2}<\frac{2}{9}$ so $0<(3x-1)(3x-2)$. Therefore x<1/3 or x>2/3. Since these cases are effectively symmetric (with substitution of A for B), consider x>2/3. Therefore $[APQ]=\frac{x^{2}}{2}>\frac{2}{9}$
So, one of the three areas is greater than 2/9 with equality if x = 2/3.

Dihedral angle = the angle formed between two faces.

Consider the tetrahedron formed by two right angled isosceles triangles and an equilateral triangle. Three of the dihedral angles are 90, the other three are given by $\arccos\frac{1}{\sqrt{3}}$ You can then work out the cosine of the sum of all these angles, and I think it is $\frac{\sqrt{2}}{3\sqrt{3}}$. Do a similar thing for a regular, platonic tetrahedron, and the cosine of the sum is $\frac{329}{729}$. Therefore, clearly they are different! However, there is a possibility of some arithmetical foibles somewhere...

I got a solution, slightly similar to Inquisitors for number 1:

[hide]
Consider the case when the tetrahedron's altitude is close to zero. The three dihedral angles in the base will all approach $0$, while the three in the top will approach $\pi$ radians. Thus sum of this is $3 \pi$. Now consider the case in which the altitude of the triangle is thousands of times higher than the maximum width of the base. In this case the dihedral angles in the base will approach $\frac{\pi}{2}$ while the angles at the vertex will approach $0$. Thus sum of this is $\frac{3 \pi}{2}$, half of the sum when the altitude is close to zero. Thus the dihedral angles of a tetrahedron are not constant.
[/hide]

I haven't solved number 3 yet, but I'm quite certain that it must be equilateral.

Sorry, but looking at Q3, what is an "equal cevian"?

Equal cevians are cevians that are equal in length.

[hide="2"]The maximum area of the quadrilateral is 1/4. When the quadrilateral has area 2/9, an area has to have at least that or more, so that's settled.

But what about the times when the quadrilateral does NOT have an area of 2/9 or more? The biggest triangle will have an area of at least 2/9, so, no matter how p is chosen, there will be a maximum area of at least 2/9.[/hide]


function
integration
inequalities
calculus
geometry
trapezoid
real analysis

Let $ f$ be function such that $ f''\geq 0$ and $ a<b$.Prove that
         $ \int_{a}^{b}f(x)dx\leq (b-a)f(\frac{a+b}{2})$.

Let $ f(x) = x^{2}$

$ \Rightarrow f''(x) = 2$

Hence $ f(x) = x^{2}$ certainly satisfies $ f''(x) \geq 0$

Let $ a =-1$ and $ b = 1$

$ \Rightarrow \int_{a}^{b}f(x)\,dx = \int_{1}^{-1}x^{2}\,dx = \left[\frac{x^{3}}{3}\right]_{-1}^{1}= \frac{2}{3}$

But $ (b-a)f\left(\frac{a+b}{2}\right) = 2f(0) = 0$

$ \frac{2}{3}> 0$ thus the conjecture is false

To the original poster - is this what you meant to ask? Counterexamples abound. Perhaps you meant $ \frac{f(a)+f(b)}{2}$ in place of $ f(\frac{a+b}{2})$. In this case the inequality basically states that the mean of the values of a convex function at two points is greater than the average value of the function in between those two points - draw a sketch.

Actually, we can make it true by simply reversing the inequality sign: $ \frac1{b-a}\int_{a}^{b}f(x)\,dx\ge f\left(\frac{a+b}2\right)$.

Yes ,I think $ f''\leq 0$,too.And now i'll post solution:
Sence $ f''(x)\leq 0$ therefore $ f(x)\leq f(x_{0})+(x-x_{0})f'(x_{0})$.
Hence I have
$ \int_{\frac{a+b}{2}}^{b}f(x)dx\leq \int_{\frac{a+b}{2}}^{b}f'(\frac{a+b}{2})(x-\frac{a+b}{2})dx+\int_{\frac{a+b}{2}}^{b}f(\frac{a+b}{2})dx$(1)
$ \int_{a}^{\frac{a+b}{2}}f(x)dx\leq \int_{a}^{\frac{a+b}{2}}f'(\frac{a+b}{2})(x-\frac{a+b}{2})dx+\int_{a}^{\frac{a+b}{2}}f(\frac{a+b}{2})dx$(2)
From (1),(2) thus
$ \int_{a}^{b}f(x)dx\leq (b-a)f(\frac{a+b}{2})$ :lol:
 Probl is proved.

Context: we are approximating an integral by a midpoint rule partial sum. If $ f''>0,$ then the midpoint rule underestimates the actual integral and if $ f''<0$ the midpoint rule overestimates the actual integral. If we go ahead and also talk about the trapezoid rule, then we find that the trapezoid rule errs in the opposite sense to the midpoint rule.

For instance, if $ f''(x)\ge 0$ on $ [a,b],$ then

$ (b-a)f\left(\frac{a+b}2\right)\le\int_{a}^{b}f(x)\,dx\le (b-a)\left(\frac{f(a)+f(b)}2\right)$

If $ f''\le0,$ then reverse the inequalities.

Now let's be a little more analytical about this - numerical analysis, to be specific. Suppose that $ f''(x)\in[c,d]$ for all $ x$ in the interval. Can you give sharp estimates of the size of

$ \int_{a}^{b}f(x)\,dx-(b-a)\left(\frac{f(a)+f(b)}2\right)$ and

$ \int_{a}^{b}f(x)\,dx-(b-a)f\left(\frac{a+b}2\right)$ ?

[The answer is printed in many mainstream calculus textbooks, if you know where to look - but the derivation usually isn't in those books.]


floor function
number theory unsolved
number theory

If n is a positive integer and p is a prime, prove that n! divides p^n(p^n - 1)(p^(n-1) - 1).....(p^2 - 1)(p - 1).

we have to show that for any prime $q \leq n$, the power of $q$ dividing $p^n (p^n - 1)(p^{n-1}-1)\cdots (p-1)$ is at least the power of $q$ dividing $n!$. in general the power of $q$ dividing $n!$ is

$\sum_{i=1}^{\infty} \lfloor \frac{n}{q^i} \rfloor \leq \sum_{i=1}^{\infty} \frac{n}{q^i} = \frac{n}{q-1}$ (**)

case 1. $q = p$. in this case, notice that $\frac{n}{q-1} \leq n$ and we're done.

case 2. $q \neq p$. in this case $q | p^{a} - 1$ if $q-1 | a$. the number of $a$ from 1 through n for which $q-1 | a$ is of course $\lfloor \frac{n}{q-1} \rfloor$. so the power of $q$ dividing $p^n (p^n - 1)(p^{n-1}-1)\cdots (p-1)$ is at least $\lfloor \frac{n}{q-1} \rfloor$. since the power of $q$ dividing $n!$ is, by (**), at most $\lfloor \frac{n}{q-1} \rfloor$, we're done with this case too.


Anne and Bea are playing a game with a 10x5 square-sheet of chocolate. They are alternating in breaking one piece off of the sheet of chocolate along the grid-lines. They agreed that whoever breaks off a 1x1 piece first wins the whole chocolate. Anne starts and they both play their best. Does Anne have a chance to win the chocolate? Would Anne have a chance to win if the rule was that the first person breaking off a 1x1 piece loses the whole chocolate?

the goal is to get them a 2x2 piece so that they have to give you a 2x1 piece. This can be done by getting any square piece. Thus, if Anne breaks off a 5x5 piece leaving a 5x5 piece, Bea loses. If the person who breaks off a 1x1 piece loses then Bea controls who wins (i think...)

I see what you're trying to say, because the square piece does make a person have a chance of winning. There might be other ways though

The easiest way to set up this problem (for me) is to have two piles of 10 and 5 of lets say chocolate stones.  You can take as many stones as you want (except 0or all of them), from only one pile.  To win (in the first variation of the problem)  you have to take a stone, while there only remains 1 stone in a certain pile.  For example if there are 5 stones in one pile and 1 stone in the other (a 5 by 1 rectangle)  if you are forced to make a move from the "5 stone pile" (taking a 1 by 1 square)  but since you made a move while there "1 stone" in a certain pile you are victorius.  The winning strategy for this scenario:  You must leave your opponent with 2 stones in one pile and 2 stones in the other.  He is forced to take at least 1 stone from either pile meaning you can make your move and win.  But to preven your opponent from leaving you with 2 stones in both piles, you must leave him 3 stones in both piles.  Leaving 1 extra in each pile prevents your opponent from trapping you because he can only make a move in one of the two piles.  [b]So for Anne to win she must leave Bea with a 5 by 5 piece of chocolate and then mimic the number of pieces she takes out of one direction, to the other until there is only 1 square in either the length or the width, allowing her to take a 1 by 1 square and win [/b].

The second variation:

Before I begin this I have to ask what happens if someone is left with exactly a 1 by 1 square what happens?  I will exclude this from an possible scenarios because of its ambiguous nature.

  To win in this scenario you must leave ur opponent with a 2 stones in one pile and 2 stone in another pile so that he is forced to take a 1 by 1 square.  But as we learned earlier we dont want this to happen to us, so leave leave him with a 3 by 3 rectangle adding two to each pile.  Why?  Because this way he cannot leave us with a 2 by 1 rectangle.  Leaving him with a 3 by 2 recatangle woud allow him to take 2 from the 3 pile leaving 1 us with a 1 by 2 rectangle, and ultimately our defeat.  With a 3 by 3 rectabgle he is unable to put us in a 2 by 1 position, but we are.  But as we explained earlier we dont want him to put us in a 3 by 3 rectangle so we put him in a 4 by 4 rectangle and so on.[b]  Thus, the winning strategy for Anne is to leave him in a 5 by 5 rectangle, 4 by 4 or a 3 by 3 but to reduce to 2 by 1 as soon as she can.[/b]

Concise answer. It makes a lot of sense.


So how exactly are the possible combinations determined?

[quote="pinkmuskrat"]We must investigate the series. Key- A=teamA wins, B=teamB wins.
If B won  a game(the 2nd), then A must have won the series 3-1, or 3-2. The possibilites are shown below. Since each result has the same likliness, the 3-2 series is weighted.
ABAA
BBAAA
ABBAA
ABABA

$ 1/4$[/quote]

that is actually not the correct answer

consider the following:
is the probability that team A wins the series 3-1 equal to that of them winning 3-2?

[quote="jamesshin97"][quote="az1337"]copied from Solution Page

Problem:
Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

Solution:
There are four possible outcomes,$ \text{ABAA, \quad ABABA, \quad ABBAA, \quad and \quad BBAAA, }$ but they are not equally likely. This is because, in general, the probability of any specific four-game series is $(1/2)^4 = 1/16$, whereas the probability of any specific five-game series is $(1/2)^5 = 1/32$. Thus the first listed outcome is twice as likely as each of the other three. Let $p$ be the probability of the occurrence $ABBAA$. Then the probability of $ABABA$ is also $p$, as is the probability of $BBAAA$, whereas the probability of $ABAA$ is $2p$. So $2p + p + p + p= 1$, $\quad \text{and} \quad p = \frac{1}{5}$. The only outcome in which team B wins the first game is $BBAAA$, so the probability of this occurring is $\frac{1}{5}$.

OR

To consider equally-likely cases, suppose that all five games are played, even if team A has won the series before the fifth game. Then the possible ways that team A can win the series, given that team B wins the second game, are $\text{BBAAA, \quad ABBAA, \quad ABABA, \quad ABAAB, \quad and \quad ABAAA. }$ In only the first case does team B win the first game, so the probability of this occurring is$ \boxed{\frac{1}{5}} $.[/quote]

In the problem it says the FIRST team to win 3 games. It makes no sense to have ABAAA because the A team wouldve already won after the 4th game. ABAAB doesnt make sense either because the A team wouldve won after the 4th game
I'm really confused..[/quote]
@jamesshin97, just think of the series ABAAB as ABAA which would be different from all the others.  And think of the series ABAAA as just ABAA because it really does not make a difference.
OR
Just think of it as if the teams had to play the entire series even if a team already won 3 games.

@JSGandora ok. It wouldnt make sense logically but I guess you can look at it that way. thanks!


Source: Baltic Way 1996

1. Let a be the angle between two lines containing the diagonals of a regular 1996-gon.  Let b be another such angle.  Prove that a/b is a rational number.

Let two consecutive vertices be A and B. Then, since the 1996-gon can be inscribed in a circle, for any other vertex O, the measure of AOB will be constant, since it subtends the same arc AB. Let this measure be :theta:.  If we have two nonconsecutive vertices P and Q, POQ will be constant and an integral multiple of :theta: Now, if we have an intersection between diagonals AC and BD, the included angle will be (AB-CD)/2 by a well-known theorem. This will be :theta: times a rational number.  So, the ratio between the two angles will be :theta: times a rational number divided by :theta: times another rational number, which will be rational.

sigh...isn't anyone going to congratulate me on my beautiful solution? oh well... jk


i've downloaded the package mhchem and when i try to use it, it doesn't work. Do I have to configure TexNicCenter or something? Here was my trial document:

\documentclass[11pt]{article}
\begin{document}
\usepackage{rsphrase}

\ce{H2O}

\end{document}

\usepackage must come [i]before[/i] \begin{document}. Also, for chemical formulae you need \usepackage{mhchem}. The rsphrase package is for risk and safety labels for chemicals.

So this will work:[code]\documentclass[11pt]{article}
\usepackage{mhchem}
\begin{document}

\ce{H2O}

\end{document}[/code]
Incidentally, if any chemists are reading this, mhchem is a great package for anyone who wants to typeset chemistry. For example, \ce{SO4^2- + Ba^2+ -> BaSO4 v} produces [img]http://sixthform.info/steve/wordpress/latexrender/pictures/70b25aea524c7b2d208ec46a31dfafae.gif[/img]
(only on your computer, not this forum)

is there any program where i can be able to type like electron pairs and other bonds connecting elements above and below?

You can put things on top of one another in LaTeX by using an array or possibly by using \stackrel. Try an array first. The author of mhchem may also be able to advise.

Have a look at [url=http://portal.mytum.de/onlinehelp/renderformat/latex/chemtex]Typesetting Chemical Equations using LaTeX[/url] which may help.

If not, it seems [url=http://imt.chem.kit.ac.jp/fujita/fujitas3/xymtex/indexe.html]XyMTeX[/url] does draw lots of chemistry diagrams (have a look at the [url=http://imt.chem.kit.ac.jp/fujita/fujitas3/xymtex/xym403/xympdf/xymtx402403.pdf]manual[/url]) but not being a chemist I can't be sure it's what you want.


Functional Analysis
advanced fields
advanced fields unsolved

Compute the homology groups of $ R^4-S^1$ where:
$ S^1={(x,y,0,0)\in \mathbb R \mid x^2+y^2=1 \}}$

uhm.. our space $ X$ is $ S^4\setminus\{p\}\setminus S^1$, where $ p$ is just a point, which in turn can see as $ \mathbb{R}^4\setminus(\ell\cup\{p\})$ where $ \ell$ is now a line disjoint from the point $ p$.

if you take an hyperplane separating them (which exists by hahn-banach  :D ) and shrinking it, you see that $ X$ is homotopically equivalent to $ S^3\wedge S^2$.
the conclusion follows from mayer-vietoris sequence: choosing obvious open neighborourhoods of $ S^2$ and $ S^3$, with contractible intersection, every of the maps $ H_i(S^2)\oplus H_i(S^3)\to H_i(X)$ (for $ i\ge 1$) is an isomorphism, while $ X$ is connected, so $ H_1(X)$ and all homologies in dimensions al least $ 4$ are trivial, and $ H_0(X)\simeq H_2(X)\simeq H_3(X) \simeq \mathbb{Z}$.

probably there's some more technical solution dealing with mayer-vietoris and alexander duality, but i've not been trying.


[size=150][color=darkblue][u][b]Problem 3[/b][/u]:[/color][/size]

Let $ a,b,c>0$. Prove that $ \frac{1}{5a}\plus{}\frac{1}{3b}\plus{}\frac{7}{15c} \ge \frac{15}{3a\plus{}5b\plus{}7c}$

[quote="thanhnam2902"][size=150][color=darkblue][u][b]Problem 3[/b][/u]:[/color][/size]

Let $ a,b,c > 0$. Prove that $ \frac {1}{5a} \plus{} \frac {1}{3b} \plus{} \frac {7}{15c} \ge \frac {15}{3a \plus{} 5b \plus{} 7c}$[/quote]

It's very easy  :)
By caychy-swarchz we have:
$ LHS \geq \frac{(\sqrt{\frac{3}{5}}\plus{}\sqrt{\frac{5}{3}}\plus{}\sqrt{\frac{49}{15}})^2}{3a\plus{}5b\plus{}7c}\equal{}\frac{15}{3a\plus{}5b\plus{}7c}$


trigonometry
algebra
polynomial
complex numbers

Come on guys? You had 1.5 years time and still nothing? :P

[quote="MysticTerminator"][quote="Silverfalcon"]This is a year old thread..  :?

I thought reviving old threads were bad thing..  :? or is this different in advanced forum?[/quote]

personally I think nothing is a bad thing unless it's actually fragrantly offensive or is really obviously simple spam. how can it be bad to talk about math ideas?[/quote]

Agree. I just posted that because when I wasn't moderator and accidentally revived an old thread, people said not to do it...  :) so that kind of became habit for me.

This looks a challenging problem that propably has even a simple solution.
I think it comes from GM......

[quote="Silverfalcon"]This is a year old thread..  :?

I thought reviving old threads were bad thing..  :? or is this different in advanced forum?[/quote]

Why would it be a bad thing? If people have given several different solutions to some problem, then it's better for the problem to rest in peace, but this is an unsolved problem which hasn't been discussed by anybody yet - for me, that's the same as posting a completely new problem? And, what's the use of some old unanswered posts on page 48 of the pre-olympiad forum? I think these problems really want to be solved!

As always, we can crush the problem with complex numbers...

$ z^{13}\equal{}\minus{}1$, etc then square and subtract everything till you get a polynomial that should be zero then divide...

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Let $ \omega \equal{}\cos\frac{2\pi}{13}\plus{}i\sin\frac{2\pi}{13}$. First, define $ A(x),B(x)\in\mathbb{Z}_{13}[x]_{x^{13}\minus{}1}$ (polynomials modulo $ x^{13}\minus{}1$):
\[ A(x)\equal{}x\plus{}x^{3}\plus{}x^{4}\plus{}x^{9}\plus{}x^{10}\plus{}x^{12}\]
\[ B\equal{}x^{2}\plus{}x^{5}\plus{}x^{6}\plus{}x^{7}\plus{}x^{8}\plus{}x^{11}\]
We have $ A(\omega)\plus{}B(\omega)\equal{}\minus{}1$ and $ A(\omega)B(\omega)\equal{}\minus{}3$. No, the second equation isn't immediately obvious. You could multiply this out... A trickier way is to notice that
\[ A(\omega )B(\omega )\equal{}A(\omega^{2})B(\omega^{2})\equal{}\cdots \equal{}A(\omega^{12})B(\omega^{12})\]
\[ \equal{}\frac{A(\omega )B(\omega )\plus{}A(\omega^{2})B(\omega^{2})\plus{}\cdots\plus{}A(\omega^{12})B(\omega^{12})}{12}\equal{}\frac{13C\minus{}A(1)B(1)}{12}\]
where $ C$ is the constant term in $ A(x)B(x)$. But then $ C\equal{}0$ since the additive inverse of a square is another square modulo 13. Since $ A(1)\equal{}B(1)\equal{}6$, the result follows.

Now we have $ A(\omega)$ and $ B(\omega)$ are the roots of $ t^{2}\plus{}t\minus{}3$ so that
\[ A(\omega),B(\omega )\equal{}\frac{\minus{}1\pm\sqrt{13}}{2}\]
Now note that
\[ A(\omega)\equal{}2\cos\frac{2\pi}{13}\plus{}2\cos\frac{6\pi }{13}\plus{}2\cos\frac{8\pi}{13}\]
\[ \equal{}2\left(\cos\frac{2\pi}{13}\minus{}\cos\frac{5\pi}{13}\plus{}\cos\frac{6\pi}{13}\right)\]
which is positive, so $ A(\omega )\equal{}\frac{\minus{}1\plus{}\sqrt{13}}{2}$ and $ B(\omega )\equal{}\frac{\minus{}1\minus{}\sqrt{13}}{2}$.

What does this have to do with the given problem? The trigonometric expression is equal to
\[ \sin\frac{4\pi}{13}\plus{}\sin\frac{10\pi}{13}\plus{}\sin\frac{12\pi}{13}\]
the imaginary part of $ \omega^{2}\plus{}\omega^{5}\plus{}\omega^{6}$. Now let
\[ P(x)\equal{}x^{2}\plus{}x^{5}\plus{}x^{6},\ Q(x)\equal{}x^{7}\plus{}x^{8}\plus{}x^{12}\]
where polynomials are again considered modulo $ x^{13}\minus{}1$. Then $ P(x)\plus{}Q(x)\equal{}B(x)$, so that $ P(\omega)\plus{}Q(\omega)\equal{}\frac{\minus{}1\minus{}\sqrt{13}}{2}$. It remains to find $ P(\omega)Q(\omega)$. This time, multiplying out is short enough:
\[ (\omega^{2}\plus{}\omega^{5}\plus{}\omega^{6})(\omega^{7}\plus{}\omega^{8}\plus{}\omega^{11})\]
\[ \equal{}\omega^{9}\plus{}\omega^{10}\plus{}1\plus{}\omega^{12}\plus{}1\plus{}\omega^{3}\plus{}1\plus{}\omega\plus{}\omega^{4}\]
\[ \equal{}3\plus{}A(\omega)\equal{}\frac{5\plus{}\sqrt{13}}{2}\]
It now follows that $ P(\omega)$ and $ Q(\omega)$ are the roots of
\[ t^{2}\plus{}\frac{1\plus{}\sqrt{13}}{2}\cdot t\plus{}\frac{5\plus{}\sqrt{13}}{2}\]
with discriminant
\[ \left(\frac{1\plus{}\sqrt{13}}{2}\right)^{2}\minus{}4\cdot\frac{5\plus{}\sqrt{13}}{2}\equal{}\frac{\minus{}13\minus{}3\sqrt{13}}{2}\]
so that
\[ P(\omega)\equal{}\minus{}\frac{1\plus{}\sqrt{13}}{4}\pm i\cdot\frac{1}{2}\sqrt{\frac{13\plus{}3\sqrt{13}}{2}}\]
Of course, the given trig expression is positive, so it must equal
\[ \frac{1}{2}\sqrt{\frac{13\plus{}3\sqrt{13}}{2}}\]

Whew.
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