I'm in California, but i think all the states have the same questions and answers. I want to check the answer. Thank you very much.

I think the answers were:

[hide]
1) 1
2) (1,0)
3) 3
4) 288/5
5) 2:12
6) 11,-7,13,-9, -17, 21

I got all but number 6- I ran out of time and got all the solutions but 21.  :([/hide][/hide]

The official answer key says the answer to question nr. 3 is - I'm not sure what they mean by that.

Perfect squares are congruent to 1 or 0 mod 4. So you only look at the last two digits. I picked 3 because I forgot that 93 is conguent to 1 mod 4, so I got lucky. To make sure, I think you need to check mod 8 or 16 or something like that.

The answers to this contest should not be discussed until the next day, and maybe even later.  As was pointed out, this contest is given throughout the country.  In the West Coast, it is still only about 2:40 pm.

On Contest 2, folks started discussing the contest too early, which I believed forced yifman to disqualify himself, as he had seen the questions.

without saying what the answer should be, I will say that there was an error in the answer key for number 3. TxML is aware, and when coaches go to the website to enter scores, this is addressed in a way intended to make everyone happy.

[quote]The answers to this contest should not be discussed until the next day, and maybe even later.[/quote]

Sorry about that.  I edited my post and hid the answers above.


ratio
geometry
number theory
relatively prime

Let $ABCDE$ be a convex pentagon with $AB\parallel CE$, $BC\parallel AD$, $AC\parallel DE$, $\angle ABC=120^\circ$, $AB=3$, $BC=5$, and $DE=15$. Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

[hide]Let the intersection of AD and CE by F.

Since AB is parallel to FC and BC is parallel to AF, AFBC is a parallelogram.

Therefore triangle ABC is congruent to triangle AFC.

By vertical angles angle AFC is equal to EFD.  And since AC is parallel to ED, angle CAD is equal to angle ADE and triangles FED and AFC are similar by AAA.

Since we only need to ratio of the two areas we can compare the corresponding bases and heights of the two triangles.

We know that ED=15 so the base of EFD is 15/7 of the base of ABC.

The heigh of EBD is equal to the sum of the heights of ABC, AFC, and EFD because all three triangles have their attitudes facing the same way.

The answer is (15/7)(2+15/7)=435/49

484[/hide]

Let $EC\cap AD=F$ and $[ABC]=\alpha$. Now, because $ABCF$ is a parallelogram, $[AFC]=\alpha$ as well. Remark that $\triangle EFD\sim\triangle ABC$ as $AB\parallel CE$ and $BC\parallel AD$. Note that $EACD$ is a trapezoiod so that means $[EFD]=\frac{225}{49}\alpha$. Also note that that $CF=3$ and $AF=5$ trivially. Now, by similar triangles, we find that $DF=\frac{75}{7}$. Hence, $[AFE]=[DFC]=\frac{225}{49}\alpha\cdot\frac{5}{\frac{75}{7}}=\frac{15}{7}\alpha$. Now, note that $[ABCDE]=\alpha+\alpha+\frac{15}{7}\alpha+\frac{15}{7}\alpha+\frac{225}{49}\alpha$. Note that $[ABE]=[BCD]=\alpha$ because $ABCE$ and $ABCD$ are trapezoids. Hence, $[EBD]=[ABCDE]-[ABE]-[BCD]=\frac{435}{49}\alpha$. Finally, $\frac{[ABC]}{[EBD]}=\frac{49}{435}$ so our answer is $49+435=484$ as desired.

The $120^\circ$ is literally quite useless except for extracting $\overline{AC}$. :P

Denote $O = \overline{CE} \cap \overline{AD}$. The ratio of the bases of $\triangle ABC$ and $\triangle EBD$ is obviously $\frac 7{15}$. Note that by simple angle chasing, $\triangle EOD$ is similar to $\triangle ABC$. Henceforth, one can compute $EO = \frac{45}7$ and $DO = \frac{45}7$.

Henceforth, we find $AD = \frac{110}7$, and $$\frac{h_1}{h_2} = \frac{BC}{AD} = \frac 5{\frac{110}7} = \frac 7{29},$$ so the desired answer is $\frac{49}{435}$. Extract $\boxed{484}$.

[b]Walk-Through:[/b]
[list]
[*] Let $AD$ and $CE$ meet at $X$. It's clear that $ABCX$ is a parallelogram.
[*] LoC on $ABC$ yields $AC = 7$.
[*] Because $CAB \cong ACX \sim DEX$, we know $$\text{dist}(B, DE) = \text{dist}(B, AC) + \text{dist}(X, AC) + \text{dist}(X, DE)$$ $$= 2 \cdot \text{dist}(B, AC) + \frac{15}{7} \cdot \text{dist}(X, AC) = \frac{29}{7} \cdot \text{dist}(B, AC).$$
[*] Extract $$\frac{[ABC]}{[BDE]} = \frac{7}{15} \cdot \frac{7}{29} = \frac{49}{435} \implies \boxed{484.}$$
[/list]


geometry
geometric transformation
rotation
reflection

If two figures can be made to look the same by rotations or reflections, they are not considered unique. How many unique two-dimensional figures can be formed by fitting four congruent squares together so that adjacent squares share an entire edge? (Squares may not overlap.)

[quote="GameBot"]If two figures can be made to look the same by rotations [size=150][b]or reflections[/b][/size], they are not considered unique.[/quote]

So no. Those two pieces would be unique in Tetris because you cannot reflect pieces there.

Just to add, the answer counting reflections as the same is $5$.

If you count reflections as different, the answer isn't $6$...it's $7$.  There's both L and backwards-L in Tetris.


Do $x,y$ belong to the reals? Or positives? Or just integers? Or complex? etc.

I suppose $x,y$ are integers. Then the solutions are $(x,y)\in\left\{(0,k),(k,0),(k,-k)\right\}$ if $n$ is odd and $(x,y)\in\left\{(0,k),(k,0)\right\}$ if $n$ is even, where $k$ is an integer.

The problem was Problem 4 of the Final Round of the Belgian Mathematical Olympiade (OMB) in Wallony, MIDI category (i.e. for students aged 14-16 approximately) The problem was not really difficult, but rather much of a case-analysis. The solution by the winner of the contest is available here: [url=http://dev.ulb.ac.be/urem/IMG/pdf/OMB10.pdf]Solution by the winner of the OMB[/url] (scroll to the bottom)

[hide]
If $n=1$, everything trivially works. If $x+y=0$, then $x^{n}+y^{n}=0$. When $n$ even, only $(0,0)$ works (otherwise rhs positive), but when $n$ odd, all $x,y$ such that $x+y=0$ works (factor $x+y$ out of rhs).

Otherwise, let $a=\frac{x}{x+y}$ and $b=\frac{y}{x+y}$. We then have $a+b=1$ and $a^{n}+b^{n}=1$. In particular,
\[a(1-a^{n-1})+b(1-b^{n-1})=0\]
Now factoring $1-a^{n-1}$ into $(1-a)(1+a+a^{2}+\cdots+a^{n-2})=b(1+a+a^{2}+\cdots+a^{n-2})$, we have
\[ab\left( \sum_{k=0}^{n-2}(a^{k}+b^{k})\right) =0\]
On the last factor, each term in the summation is strictly positive! For even $k$, it's the trivial inequality, while for odd $k$, we have $a+b=1>0\Rightarrow a^{k}+b^{k}>a^{k}+(-a)^{k}=0$. Therefore either $a=0$ or $b=0$, so that either $x=0$ or $y=0$.

It follows that $n=1$, or $n$ odd and $x+y=0$, or $x=0$, or $y=0$.
[/hide]

[quote="Christine"]Solve the equation

(x + y)^n = x^n + y^n

for all n in N.[/quote]
$ (x+y)^{n}\geq x^{n}+y^{n}$


geometry
superior algebra
superior algebra theorems

I am once more on an automorphism of C trip

http://www.mathlinks.ro/Forum/viewtopic.php?t=56352

Here I started a topic on that subject too
However a new development came up

What I thought was, if $\theta$ is an involutive automorphism of $\mathbb{C}$ and $\tau$ too, there is an automorphism $\sigma$ such that
$\sigma^{-1}\theta\sigma= \tau$

Now my professor told me that, although included in a syllabus of geometry last year, it is incorrect,

However, this caused a problem in class, here it is :

$\sqrt{2}$, what is the image of that under involutive automorphism
There are only two options, the same or the opposite

But does the last case occur too?

Yes, I think the image can be either $\sqrt{2}$ or $-\sqrt{2}$. I'll write the proof later.

I was wrong  :blush: . The only possible value is $\sqrt{2}$. Assume the contrary. Let $\phi$ be an involutive automorphism, s.t. $\phi\left(\sqrt{2}\right) = - \sqrt{2}$. What is the value of $\phi\left(\sqrt{2-\sqrt 2}\right)$ ? Clearly, it is either $\sqrt{2+\sqrt 2}$ or $-\sqrt{2+\sqrt 2}$. Observe, that for every $z$, $\phi\left(z\phi(z)\right) = \phi(z)\phi\left(\phi(z)\right) = z \phi(z)$. So $z\phi(z)$ is a fixed point of $\phi$. We have
\[ \phi\left(\sqrt{2-\sqrt 2}\right) \cdot \sqrt{2-\sqrt 2} = \pm \sqrt{2+\sqrt 2} \cdot \sqrt{2-\sqrt 2} = \pm \sqrt{2} \]
is a fixed point of $\phi$. But this means that $\phi (\sqrt{2}) = \sqrt{2}$, which contradicts to our assumption.

More generally, those $\alpha\in\mathbb R$ which are not moved by any involutive automorphism of $\mathbb C$ are precisely those algebraic reals all of whose conjugates (over $\mathbb Q$) are real. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=115877]this topic[/url].

It's not hard to see the connection between the two problems (using the results of Artin and Schreier mentioned in that other thread): a real number $\alpha$ is moved by at least one involutive automorphism of $\mathbb C$ iff there is a proper subfield $K$ of $\mathbb C$ such that $K(\alpha)=\mathbb C$.

[quote="fredbel6"]
What I thought was, if $\theta$ is an involutive automorphism of $\mathbb{C}$ and $\tau$ too, there is an automorphism $\sigma$ such that
$\sigma^{-1}\theta\sigma= \tau$
[/quote]

This is indeed not true. It would imply that the fixed field of an involutive automorphism of $\mathbb C$ is isomorphic to $\mathbb R$. There are, in fact, involutive automorphisms of $\mathbb C$ whose fixed fields cannot even be embedded in $\mathbb R$.

Let $K$ be a hyperreal field such as the one whose construction is outlined in [url=http://en.wikipedia.org/wiki/Hyperreal_number#The_ultrapower_construction]this[/url] Wikipedia article. What's important here is that it is a non-Archimedean real closed field of continuum cardinality. Being non-Archimedean, it cannot be embedded in $\mathbb R$.

The algebraic closure of $K$ has continuum cardinality, so it's an algebraically closed field of characteristic zero whose transcendence degree over its prime field has continuum cardinality. All such fields are easily seen to be isomorphic to $\mathbb C$, so we can embed $K$ in $\mathbb C$ such that $\mathbb C=K(i)$. Now fix $K$ pointwise and send $i$ to $-i$. This is the desired automorphism of $\mathbb C$.


It's just a challenge to know who is the most creative and logic here.

If someone decipher my code, he could make his our code.
If someone dont decipher, he cannot make his our code.

Try
[quote]
741593 852 37241 31742
32154978 396 34271 27413
[/quote]
We have some hints:

1 - "One die alone, just together we have life"
2 - "Lock at your hands"

3 - "DAQ" - Just somes will understand, but this hint is not necessary

Let's see who can decipher it, just more one hint "Think Easy, not Hard"

It has something to do with [hide="a"]calculator; 741, 852, and 963 appear repeatedly.[/hide]

the order is not important

so 741 = 471 = 147 = 174 = etc...

All permutations are equal

hmm...

yes. make now your code

and please, give less hints. because i gave too many :)

uahuuah

I am stunned. I have been planning to start this on a large scale for the past week!! Don't worry, I will anyway, and pretty soon. :)

Meanwhile, a kindrgarten one with an interesting history:
RBTAZ

Hint?
[i]You need a hint for this??[/i]

Very well, how about
[hide="..."]Et tu brute?  :rotfl:  :rotfl: [/hide]

Rirelbar xabjf guvf pbqr.

Hint - I encoded that in my head.

Edit: well, he didn't make a code, so, HUMPT

no one get the way of the game.

Only people that decode a code can make a code.

humpft. a trash topic.

need to be locked. humpt


LaTeX
trigonometry
calculus
calculus computations

Find the equations for the tangent plane and the normal line at the point (0, 1, 2) of the surface with
equation cos(pi*x) - (x^2)y + e^(xz) + yz = 4.

$ \frac{\partial f}{\partial x}(x\minus{}x_0)\plus{}\frac{\partial f}{\partial y}(y\minus{}y_0)\plus{}\frac{\partial f}{\partial z}(z\minus{}z_0)\equal{}0$

[quote="mathbeauty"]Find the equations for the tangent plane and the normal line at the point (0, 1, 2) of the surface with
equation cos(pi*x) - (x^2)y + e^{xz} + yz = 4.[/quote]
You could make this look better using the $ LaTeX$ code-set.

Just put $ \$$ signs around the equation....

It'll appear like this:

$ \cos ( \pi x) \minus{} (x^2)y \plus{} e^{xz} \plus{} yz \equal{} 4$

[quote="Rajiv"]You could make this look better using the $ LaTeX$ code-set.[/quote]
You can actually make that look better by using the $ \text{\LaTeX}$ command, which is[code]\LaTeX[/code]


linear algebra
matrix
superior algebra
superior algebra solved

X,Y \in Mn(R) such that for any t \in R, exp(tX).exp(tY)=exp(tY).exp(tX)
Prove that XY=YX

exp(t.X) is the exponential of the matrix t.X

If A is a matrix, exp(A) =  \sum _(k \geq 0) A^k/k!  :)

we have exp(tA)exp(tB)=exp(tB)exp(tA) => (exp(tA)-I)(exp(tB)-I)=(exp(tB)-I)(exp(tA)-I) => ((exp(tA)-I)/t)((exp(tB)-I)/t)=((exp(tB)-I)/t)((exp(tA)-I)/t). [b](1)[/b]

It is easy to see that lim x->0(exp(tA)-I)/t=A.

Go to limit in (1) and get AB=BA.

Notice that I replaced X,Y by A,B. Don't know why.

Hehe amfulger, A,B...
It is like denoting x,y,z for unknowns
or a,b,c for integerss
or n for naturals
or t for time
or F, K for fields
or G for groups
or ... and so on ... endless and mindless, sameness, substitution.
 :D  :D  :D  :blush:  :blush:  :blush:  Man I love the redface.


am I allowed?

and congratulations to you Treething for being the new moderator :lol:

[quote]shinwoo - make sure you don't have 'Disable Smilies in this post' marked in the options when you post[/quote]

Alright sir

Hm....

Am I alright to be part of this forum? ( I'm a middleschooler )

[quote="rrusczyk"]We've decided to make this forum and The Basics 'middle schoolers and teachers/parents only', in the expectation that this will keep the conversation at the appropriate level.[/quote]
[quote="rrusczyk"]
I would also add that middle schoolers who have been to State or Nationals in MATHCOUNTS, or think they have a very good chance of making it to State or Nationals this year, should leave the problems marked with the asterisk symbol in this forum for less experienced students. If you are one of these students, focus on those problems in The Basics, MATHCOUNTS, or Getting Started, instead. [/quote]

And a certain moderator who made it to Nationals :P ?

Yes, shinwoo, you may post here :) .

Would it be okay if I were a moderater. I know I don't post nearly as many problems as I would like to or should, but I would still like to be a moderater.

[quote="Treething"][color=darkred]EDIT: Don't answer them at all. This is an order from the mighty Mr. Ruscyk.[/color][/quote]

Once again, you spelled your name wrong. As I reminded you at the end of a math chat
[color=darkblue]
[size=200]Rusczyk[/size][/color]

[quote="A+MATH"]Hey, thats a tough name (no offence Mr. Rusczyk) even tokenadult has spelled it wrong before.[/quote]

In one math chat, Treething pretended to be Mr.Rusczyk. At that time, he obviously wasn't Rusczyk because he had just said he was Treething. (You can't misspell Rusczyk as Treething.)

That sounds like something Treething would do.  :lol:  ;)

Guys, chill out. It took me months to spell correctly Richard's surname :P :blush: so don't make a big issue out of that :)
As for being moderators, trust me it's just editing / deleting bad posts. Nothing really big.

PS If you really want to be one in one of these forums, you'ld better try to gather more people to populate them, so that there would be need for more moderators ;) Up untill now, only a handfull of kids (including you) use them, so there's really no need for more than 2 moderators.

Then can I be the moderator in The Basics section with Treething even though Treething is like 100 times smarter than I am? :blush:

It seems that Valentin only listed the BASICS so I'll tell more responsibilities.

A) You should be posting regularly (if possible) a challenge question.
B) You should try to help other people out.
C) Your major posts should not answering the other people's question but posting the challenges so that people can know what you're doing and learn it.

And just wondering, do you want to be a moderator? :lol:

wait we're allowed to ask about moderating? I always thought that was rude, cause Mr. Rusczyk would hand-pick the mods by there contribution, not if they asked first. but o well.... I guess I would like to be a moderator....But I guess I'm too late since ppl already asked....
 :blush:

[quote]As I've said above we already have sufficient moderators per number of active users in these forums (middle-school). Untill your numbers grow, we probably won't be putting any more moderators.

Please stop making requests for this in public.[/quote]

sorry.. :blush:

[quote="dinoman"]...wow I feel stupid.....sorry......
 :blush:[/quote]No, no don't feel stupid or anything. It's just that we can't obviously make everyone a moderator, but we would gladly do all of you, if there were like an extra-hundred of middle schoolers posting in these forums :)

[quote="byrdgyrl"]i'll try and get some more people from my school on here so you can all be moderators... but i'm not promising anything.  people can really be stubborn at times.[/quote]

You should try to get them on even if no one wants to be moderators.


linear algebra
algebra
system of equations
IMO Shortlist

Let $n$ be a positive integer that is not a perfect cube. Define real numbers $a,b,c$ by

\[a=\root3\of n\kern1.5pt,\qquad b={1\over a-[a]}\kern1pt,\qquad c={1\over b-[b]}\kern1.5pt,\]

where $[x]$ denotes the integer part of $x$. Prove that there are infinitely many such integers $n$ with the property that there exist integers $r,s,t$, not all zero, such that $ra+sb+tc=0$.

Rename
$\{a\}=\alpha$
$[a]=m$
$[b]=p$

Now we rewrite the equality in:

$r ( \alpha + m) + s \frac 1{\alpha} + t \frac{\alpha}{1-p\alpha} =0$

Mutiplying per $\alpha(1-p\alpha)$ we get:

$tp\alpha^3+(tmp-t-r)\alpha^2 + (ps-tm)\alpha - s=0$

Now if $(m+0.5)^3<n<(m+1)^3$ we have that $p=1$ in fact we have $\alpha>0.5$ and so $1<b<2$.

Now we take $t=1$, $r=-1-2m$ and $s=3m^2+m$ and we have:

$\alpha^3+(m-1+1+2m)\alpha^2 + (3m^2+m-m)\alpha - 3m^2-m=0$
$\alpha^3+3m\alpha^2+3m^2\alpha-3m^2-m=0$
$(\alpha+m)^3-3m^2-m-m^3=0$
$n=m^3+3m^2+m$

and so because we have $(m+0.5)^3<n<(m+1)^3$ if $m>0$ we can say that for $n=m^3+3m^2+m$ with $m \in N$ we can find integers $(r,s,t)=(-1-2m,3m^2+m,1)$ such that $ra+sb+tc=0$

is it ok?

btw... nice problem!  :D

[quote="Simo_the_Wolf"]Mutiplying per $\alpha(1-p\alpha)$ we get:

$tp\alpha^3+(tmp-t-r)\alpha^2 + (ps-tm)\alpha - s=0$
is it ok?

[/quote]
I think you've made a typo.Am I right? :?

Since $ c $ is the "nastiest" of the variables we want to artificially make it nice.  A good method to do so would be to force $ [b] = 1 $.   Let $ \{\} $ denote the fractional part of a real number.  Since $ b = \frac{1}{\{a\}} $ we want $ \{a\} > \frac{1}{2} $ so for now assume this to be the case.

Now, let $  \alpha = \{a\} $ and $ m = [a] $.  We have that $ a = m + \alpha $ and $  b = \frac{1}{\alpha} $ and $ c = \frac{\alpha}{1 - [b]\alpha} = \frac{\alpha}{1 - \alpha} $.  So we want $ r(m + \alpha) + \frac{s}{\alpha} + \frac{t\alpha}{1 - \alpha} = 0 $.  Multiplying both sides by $ \alpha(1 - \alpha) $ we get that $ -r\alpha^3 + (r - rm + t)\alpha^2 + (mr - s)\alpha + s = 0 $.

Since we want to figure out what $ n $ should be in terms of $ m $ we should choose $ r, s, t $ so that we can find a $ (m + \alpha)^3 $ in our equation.  So we want to solve the simultaneous set of equations in terms of $ m $:

\[ -r = -1 \] \[ r - rm + t = -3m \] \[ mr - s = -3m^2 \]

Therefore we find that $ r = 1, s = 3m^2 + m, t = -2m - 1. $  Plugging these in our equation becomes $ m^3 + 3m^2 + m - (m + \alpha)^3 = 0 \Longrightarrow n = m^3 + 3m^2 + m $.  Now we need to check that $ \alpha > \frac{1}{2} $ like we assumed but this is clear since $ m^3 + 3m^2 + m > \left(m + \frac{1}{2}\right)^3 $ for all positive integers $ m $.

Therefore for every $ n $ of the form $ m^3 + 3m^2 + m $ for some positive integer $ m $, letting $ r = 1, s = 3m^2 + m, t = -2m - 1 $ yields a desired triple $ (a, b, c) $.  Therefore we have found an infinite number of solutions and so we are done.

My solution is almost the same, but motivated in another way. We see that it suffices to show that $xa + yb + zc=0$, for some $(x,y,z) \in \mathbb{Q}- \{(0,0,0)\}$, or equivalently \[\frac{xa}{bc} + \frac{y}{c} + \frac{z}{b} = 0  \qquad (1)\].The integer part bothers us, so we let $n=k^3+i$ for $k$ and $0 < i < 3k^2+3k+1 $ to be constructed. We have easily $[n]=k$, and \[ b = \frac{1}{a-[a]} = \frac{1}{\sqrt[3]{n} - k} = \frac{\sqrt[3]{n^2}+k \sqrt[3]{n} + k^2}{i} \]. Now let $M = [b]$, to be easier to deal with for now. We have $(1)$ becomes after expanding, rearranging and multiplying by $i$;  \[ \left[ y-iMkx \right]\sqrt[3]{n^2} + \left[ x(i+iMk)+yk+z \right] \sqrt[3]{n} + \left[ (k^2-iMk)y - kz \right] = 0\qquad (2) \].The following is a known fact;

 So in order to have a solution for problem, It is necessary to find solutions in $x,y,z$ in rational number set, for the equations;
\[\begin{cases}
\begin{array}{lcl}
 -iMkx + y  & = & 0 \\
 x(i+iMk)+yk+z & = & 0 \\
 (k^2-iMk)y \quad  - kz & = & 0 \\
\end{array}

\end{cases} \]

From linear algebra we know in order to get a non-trivial solution (in fact infinitely many of them) for the system of equations above, we must have \[ \begin{vmatrix}
i+iMk  \quad & k & 1 \\
 -iMk \quad & 1 & 0 \\
 0 \quad  &  k^2-iMk& -k
\end{vmatrix} = 0 \]

After evaluating the above determinant, we get equivalent equation after dividing by $-i$ ; \[k + 3Mk^2 - iM^2 =0 \], It is now plausible to think of $i=3k+k$ and $M = 1$. In fact that can be achieved for infinitely many $n$'s, where we note that if we set $n=k^3 + 3k^2 + k$, we have \[ \frac{\sqrt[3]{n^2}+k \sqrt[3]{n} + k^2}{i} = \frac{\sqrt[3]{(k^3+3k^2+k)^2}+k \sqrt[3]{k^3+3k^2+k} + k^2}{3k^2+k} \rightarrow 1  \] as $k \rightarrow \infty $, where the limit clearly approaches 1 from above. So for large enough $k$, we have $M=1$, so we get the following family of infinitely many non-cubic integers, for which the condition is satisfied which is \[ \{k^3 + 3k^2 + k : k \geq N \} \], where $N$ is sufficiently large integer, It is clear one can let $N=1$.

[quote=bobthesmartypants]One of the failed paths that I walked down produced this problem:

Let $n=m^3+k$. Show that if $k\mid 3m^2$, then $a, b, c$ are [b]never[/b] linearly dependent.[/quote]

I hit the same problem during my solution. I just eventually set $\lfloor b \rfloor = \frac{3m^2+r}{k}$ and bashed from there, which gave the interesting condition that $9m^4+4mk$ is a square (which gives $n=m^3+3m^2+m$, but is kind of interesting in its own right).


let $ n$ be a positive integer. prove that $ (x^2 \plus{} x)^{2^n} \plus{} 1$ is irreduciable :arrow:


A triangle has sides length  x+4, 2x+2, and 3x-1.

If two of the angles of the triangle are equal, what are the possible values of x?

hint:

[hide]if two of the angles are the same, that means two of the sides are the same.

Now, consider all possibilities, then eliminate any degenerate triangles.

The answer should be there then.[/hide]

Wait, I think the answer is [hide]2, 2.5 and 3:

From the supposition "If two of the angles of the triangle are equal" we get that at least two sides of the triangle must be equal by the Isosceles Angle Theorem or the equivalent (I'll explain what it is if anyone's wondering).

So either x+4=2x+2, 2x+2=3x-1, or x+4=3x-1. The answers are 2, 3, and 2.5, in that order.[/hide]

P.S. [hide]I don't get how you could get x such that it would be a degenerate triangle (meaning line, for those of you who never heard of "degenerate" as in a "degenerate hyperbola").

If you suppose that one of the sides equal 0 (either x+4=0, 2x+2=0, or 3x-1=0) you have to have -4, -1, or 1/3 as x, which I didn't get.[/hide]

oops, yeah, i thought  x+4=3x-1 yielded x=2/5. my bad. i really should have used paper, oh well

Agreed, because I think you flipped the 2 and the 5.
Although I do think this can be done without any, because on the real thing I don't think you'd have enough time for it...oh well.

pfft, im prone to make silly mistakes unless im actually concentrating. basically, when im not on aim


geometry
topology
limit
real analysis
real analysis unsolved

This is a problem from Falconer's book The Geometry of Fractal Sets (or a minor variant; I don't have the book in front of me).

Exhibit a set $E \subset \mathbb{R}^{2}$ such that $0 < \mathcal{H}^{1}(E) < \infty$, $E$ is totally disconnected, and $E$ is regular. Is it possible for such an $E$ to be compact?

[hide="definitions"]
$\mathcal{H}^{1}$ is the 1-dimensional Hausdorff measure. If you aren't familiar with this already, I don't advise attempting this problem. But you can read about it on Wikipedia or in most measure theory textbooks. I recommend Mattila or Falconer. Or Eli Stein's book for a brief intro.

A set is totally disconnected if for any two points $x, y \in E$, the statements "$x = y$" and "$x$ and $y$ are in the same connected component" are equivalent. That is, there is no nontrivial continuous curve that is a subset of $E$.

A set is regular if for almost every point $x \in E$, the density of $E$ at $x$ exists and equals 1. That is,

$\lim_{r \to 0}\frac{\mathcal{H}^{1}(E \cap B_{r}(x))}{2r}= 1$

for almost every $x \in E$.
[/hide]

You can totally disconnect a line segment by removing a countable dense subset from it. The real question is whether a compact example of this kind exists.

Isn't a Cantor subset of the line of positive 1-dimensional Lebesgue measure an example? :?

Isn't it every Lebesgue measurable set regular? Isn't this the content of the density theorem? What am I misunderstanding?

@Kalle: This is a set in $\mathbb{R}^{2}$. Its Lebesgue measure is 0. The Lebesgue density theorem doesn't help us. The density theorems about Hausdorff measures are weaker.

However, now I think fedja is right. At first I discarded that example because I thought that one was an irregular set. But actually that was stupid of me; we know that any set of positive $\mathcal{H}^{1}$ measure which is contained in a countable union of rectifiable curves is in fact a regular set.  :blush:

A set in $\mathbb R$ is also a set in $\mathbb R^{2}$. As Kalle pointed out, Lebesgue density theorem says that any measurable subset of $\mathbb R$ with positive measure is regular.

I was interpreting Kalle's question as regarding the original question (where we didn't know $E$ was a subset of a line), not a reiteration of fedja's "this is actually easy isn't it?"


factorial
number theory
prime factorization

Prove that from the product $ 1!2!...120!$, we can remove on of the $ k!$ factors such that the remaining product is a perfect square.

Anybody have a solution? This problem was pretty. As a comment: what is the significance of the choice of $ 120$?

It is $ 5!$? It has to do something with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=278806]this[/url] thread that I posted.

No, it's not 5.
A solution's already posted in a thread a few lines down, actually.

[hide="hint"]
$ 1!2!...120!\equal{}1^{120}\times 2^{119}\times 3^{118}\times\cdots \times 120^1$
Try to make each number in the prime factorization have an even exponent, so it will be a perfect square.[/hide]

By "Is it $ 5!$?" I meant 120, not the answer.

Lets say we replace $ 120$ with $ n$. For this problem to be true is it necessary that $ n$ itself is a factorial of some integer?

If you rewrite $ 1! \cdot 2! \cdot 3! \cdot  \ldots \cdot 120!$ as $ 2  (1!)^2 \cdot 4  (3!)^2 \cdot 6 (5!)^2 \cdot \ldots \cdot 120 (119!)^2$, it is apparent that dividing this product by $ 60!$ will produce the perfect square $ 2^{60} \cdot (1! \cdot 3! \cdot 5! \cdot 7! \cdot \ldots \cdot 119!)^2$. For this particular method to work with arbitrary $ n$ instead of 120, 4 must divide $ n$; however, this does not prove that values not divisible by 4 will not work.

so we have to divide by (n/2)! to get a perfect square always.


MATHCOUNTS
number theory
greatest common divisor

Here are two problems which I don't get:
1. I don't understand this one.
[hide="Click here to see the problem"]A newsletter consists of 1 ledger size sheet (11" by 17") and 3 letter size sheets (8 1/2" by 11").  If 275 copies of this news letter are needed, and each sheet of paper os printed on both sides, how many standard 8 1/2" by 11" single copies were made?[/hide]
[hide="Click here to see my solution"]  So three pages times two and you get 6.  6*275=1650 [/hide]
[hide="Real answer"]2750[/hide]
2.I think they have this answer wrong...
[hide="Click here to see the problem"]John and his older sister Laura split a paper route, alternating days that they throw the paper.  John takes 2 hours to complete the route on bike while Laura can finish in only 80 minutes by car.  How long would it take to complete the route if they worked together?[/hide]
[hide="Click here to see my solution"]  John: 1/120, Laura: 1/80.  1/120+1/80=1/48, so 48 minutes [/hide]
[hide="Real answer"]40[/hide]

[hide]
For #1, I think they counted the 11 by 17 as equivalent to two 8 1/2 by 11 sheets.
[/hide]

WHAT ABOUT QUESTION 2?????  <subscript>no one answer :(</subscript>

[quote="wiseidiot"]You're correct on the second question. Are these mathcounts problems?[/quote]

These are not MATHCOUNTS problems, but I believe that you would get these questions in a easy chapter sprint.  This is actually from a contest from KCATM (who I believe has a bad reputation for solving problems wrongly, that problem was evidence).

For the 2nd question, the makers probably found the gcd of 80 and 120 for some reason.


I'm just starting out with C++ and I need some help.
I want to find the decimal part when a number is divided by eight using bitwise operators. (don't ask :P ) For some reason, my program always gives me the decimal part zero  :( . I declared an integer variable "number" whose value is input by the user and then:

double first = 0.0, second = 0.0, third = 0.0;
third = (number & 1) * (1/8);
second = ((number & 2) >> 1) * (1/4);
first =  ((number & 4) >> 2) * (1/2);
double sum = 0.0;
sum = first + second + third;

"sum" is supposed to give the decimal part. Because shifting binary code three places to the right is like dividing by eight, I tried to find the digits that fell off the end by taking the original number and masking each of the last three digits (I think I messed up somewhere there), then shifting to the right if needed to produce eigther a value of one or zero. I multiplied by the necessary fractional value and added the results. I think I messed up with the bitwise AND operator. Can someone help :?

'1', '2', '4', and '8' are integers in your code. So (1/2) will do the integer division, and give 0. So will (1/4) and (1/8). Try using 1.0 instead (or leave out the 1 and use 2.0, 4.0, 8.0, etc).

By the way, you don't even need to do each of the three digits separately. You could just do (number&7)/8.0. Or if you really want to do it only using bitwise operations, if you defined number to be a double instead (which you can still do even if you are reading an integer), then you get the nicest way: (number&7)>>3.

Thanks so much :)
Now it works (Yay)! It's also kind of a cool way prove divisibility by eight...


let denote m(a) , m(b) , m(c) -medians of the sides BC, AC, AB respectivetily. then prove the inequality
(m(a))/a + (m(b))/b+(m(c))/c>=(1/2)(3)^(3/2)

[quote="kimnimalar"]let denote $ m_a$ , $ m_b$ , $ m_c$ -medians of the sides BC, AC, AB respectivetily. then prove the inequality
$ \frac {m_a}{a} \plus{} \frac {m_b}{b} \plus{} \frac {m_c}{c}\geq\frac {3\sqrt3}{2}.$[/quote]
See my last post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56265


function
algebra
rational function
algebra proposed

A quick calculation shows

$ x_0 \equal{} a$
$ x_1 \equal{} b$
$ x_2 \equal{} \frac {1 \plus{} b}{a}$
$ x_3 \equal{} \frac {1 \plus{} a \plus{} b}{ab}$
$ x_4 \equal{} \frac {1 \plus{} a}{b}$
$ x_5 \equal{} a$
$ x_6 \equal{} b$

and so forth.  Hence $ x_{1998} \equal{} x_3 \equal{} \boxed{\frac{1 \plus{} x_0 \plus{} x_1}{x_0 x_1}}$.

This particular problem is fairly trivial, but suppose we try to generalize it:  what other second-order non-linear recursions $ x_{n \plus{} 2} \equal{} f(x_n, x_{n \plus{} 1})$ given by $ f$ a rational function are periodic for arbitrary $ x_0, x_1$?


function
logarithms
number theory
number theory proposed

Be $ n\in \mathbb{N^*}\setminus\{2,6\}.$
Prove $ \varphi(n)\ge\sqrt{n}$ where $ \varphi$ is the Euler's indicator.

Because $ \varphi (n)$ and $ \sqrt{n}$ is multiple function then only need to prove for $ n\equal{}p^a$
Consider $ n\equal{}2^a b$ where $ b\equiv 1 (\mod 2 )$
Then consider 2 case :
1. $ a\equal{}1$
2.$ a>1$

[quote="TTsphn"]Because $ \varphi (n)$ and $ \sqrt {n}$ is multiple function then only need to prove for $ n \equal{} p^a$
Consider $ n \equal{} 2^a b$ where $ b\equiv 1 (\mod 2 )$
Then consider 2 case :
1. $ a \equal{} 1$
2.$ a > 1$[/quote]
Is this a proof? Or u just increasing ur posts number? :rotfl:

Still, his first sentence yields a proof quite quickly.

Indeed, we have :
$ \varphi(p)^2 \equal{} (p \minus{} 1)^2 \geq p,$
and $ \varphi(p^a)^2 \equal{} (p^{a \minus{} 1}(p \minus{} 1))^2 \geq p^{2a \minus{} 2} \geq p^a.$
$ (p$ is a prime and $ a \geq 2$ is an integer$ )$
Hence, the fact that $ \varphi$ and $ n \mapsto \sqrt {n}$ are both multiplicative (and positive) functions allows us to conclude that :
\[ \forall n \geq 7, \varphi(n) \geq \sqrt {n}.
\]
The remaining cases can be checked by hand.

Now, as mentioned [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]there, in the 8th post,[/url] more is true, namely :
\[ \varphi(n) \gg \frac n{\log \log n}.
\]
On a side note, this is a(n analytic) number theory problem, so I'll move this topic to where it belongs.


greatest common divisor
arithmetic sequence

What is the greatest common divisor of all sets of numbers where each member of the set is the sum of three consecutive positive integers?

There are three methods to solve this problem:

[b]First Method[/b]: Notice that in any set of three consecutive positive integers, inevitably one number is a multiple of $ 3$ and the other two numbers add up to $ 3$. That means that the sum will be a multiple of $ \fbox{3}$!

[b]Second Method[/b]: Let's try to crank out some numbers and look for a pattern. $ 1\plus{}2\plus{}3\equal{}6$, $ 2\plus{}3\plus{}4\equal{}9$, $ 3\plus{}4\plus{}5\equal{}12$, and we see a pattern of an arithmetic sequence with a common difference of $ 3$. Notice that all the numbers are also multiples of $ 3$. Therefore, $ \fbox{3}$ is the greatest common divisor.

[b]Third Method[/b]: Use algebra. Three consecutive positive integers can be written as $ x\minus{}1$, $ x$, and $ x\plus{}1$. If we add the three terms together, we find:
$ (x\minus{}1)\plus{}x\plus{}(x\plus{}1)$
$ \equal{}3x\minus{}1\plus{}1$
$ \equal{}3x$
Now, notice that since $ x$ is a positive [i]integer[/i], $ 3x$ must be a multiple of $ \fbox{3}$.

This is pretty picky, but in the first method, the other two numbers [b]do not[/b] half to add up to $ 3$.  Their sum must be DIVISIBLE by $ 3$.


calculus quadratics conics parabola function rotation symmetry

I need help with finding an equation of a graph. I was wondering if anyone was able to give me an equation to a graph that has a [b]turning point at (0,0)[/b] (or a gradient of zero at (0,0)) and a [b]turning point at (60,10)[/b]. Plus it is not aloud to pass through a circle which has a center at points (24,9) with a radius of 3 units. It would be great if you could show your basic working out so I can see how you came to your answer.

There is more than one answer and I was hoping for as many as people could come up with.

If anyone could help it was be very much appreciated.

Thanks

this is a bit cheap but
[hide]

$y=$
$x \in [-.1,.1] : x^{2}$
$x \in [59.9,60.1] :-(x-60)^{2}+10$

how do you format that properly?
[/hide]

[hide="To Chickendude"][code]y= \left \{ \begin{array}{ll} \; x^2& \text{for}\ \ \; -0.1 \leq x \leq 0.1 \\\ \; -(x-60)^2+10& \text{for}\ \ 59.9 \leq x \leq 60.01 \end{array} \right. [/code]

makes

$y= \left \{ \begin{array}{ll}\; x^{2}& \text{for}\ \ \;-0.1 \leq x \leq 0.1 \\\ \;-(x-60)^{2}+10& \text{for}\ \ 59.9 \leq x \leq 60.01 \end{array}\right.$[/hide]

Easy, no calculus solution
[hide="Adjusted quadratic"]

We'll do this. We'll take a parabola with vertex at $(0,0)$ that goes through $(30,5)$, and consider the interval $(-\infty,30]$. Then, we'll use the absolute value function to rotate 180º it around this point, and create another branch for the interval $[30, \infty)$.

First, we start with $y=Ax^{2}$. Plugging in $(30,5)$, we get $5=900A \implies A = \frac{1}{180}$. Note that $x^{2}=((30-x)-30)^{2}=(30-x)^{2}-60(30-x)+900$. Now, to create symmetry about $x=30$, we need to first perform $(30-x) \to |30-x|=|x-30|$

$y=\frac{1}{180}[|x-30|^{2}-60|x-30|+900]=\frac{1}{180}|x-30|^{2}-\frac{1}{3}|x-30|+5$

But we're not done yet. We need to flip the right side up-side down. First, we have to move it down $5$, to get $y=\frac{1}{180}|x-30|^{2}-\frac{1}{3}|x-30|$. Now, we need to create a filter that's negative for $x>30$ and positive for $x<30$. This can be done with $\frac{|x-30|}{30-x}$.

Lastly, we multiply by the filter and add back our 5 to get:

$y=\frac{1}{180}\frac{|x-30|}{30-x}|x-30|^{2}-\frac{1}{3}\frac{|x-30|}{30-x}|x-30|+5$

Or, if you prefer,

$\boxed{y=\frac{|x-30|^{3}}{5400-180x}-\frac{|x-30|^{2}}{90-3x}+5}$

Notice that it's not defined at $x=30$. You can go in and define it with a piecewise, or use a limit, or just leave it as is.
[/hide]

Different, calculus solution
[hide="Cubic Solution"]Anyway, consider a cubic centered at $(30,5)$. It's equation will be of the form $y=a(x-30)^{3}+b(x-30)+5$. Note that this will easily steer clear of $(24,9)$

The two points $(0,0)$ and $(60,10)$ are now symmetric with respect to this cubic's center point. Let's temporarily transpose this cubic to the origin, as well as the two points, for convenience: $Y=AX^{3}+BX$. Our two points are $\pm (30,5)$. By symmetry, we only need to force an extremum at $(30,5)$, and the other follows automatically.

$Y'=3AX^{2}+B$. Plugging in $30$ needs to give $0$. $2700A+B=0$.

Also, plugging in $30$ needs to make $Y=5$. $5=27000A+30B$
$\frac{1}{2}=2700A+3B=2B \implies B=\frac{1}{4}$
$\implies A =-\frac{1}{4(2700)}=-\frac{1}{10800}$

Hence, $y=-\frac{1}{10800}(x-30)^{3}+\frac{1}{4}(x-30)+5$ satisfies your problem.[/hide]


limit
integration
calculus
calculus computations

Evaluate

\[\lim_{n\to\infty} \sum_{k=1}^n \frac{[\sqrt{2n^2-k^2}\ ]}{n^2}\]

$[x]$ is the greatest integer $\leq x$.

[quote="kunny"]Evaluate

\[\lim_{n\to\infty} \sum_{k=1}^n \frac{[\sqrt{2n^2-k^2} ]}{n^2}\]

$[x]$ is the greatest integer $\leq x$.[/quote]

\[\int_0^1\sqrt{2-x^2}dx=\lim_{n\to\infty}\sum_{k=1}^n\frac1n\sqrt{2-\frac{k^2}{n^2}}=\lim_{n\to\infty}\{\sum_{k=1}^n\frac1n\sqrt{2-\frac{k^2}{n^2}}-\frac1n\}=\lim_{n\to\infty}\sum_{k=1}^n \frac{\sqrt{2n^2-k^2}-1}{n^2}\leq\lim_{n\to\infty} \sum_{k=1}^n \frac{[\sqrt{2n^2-k^2} ]}{n^2}\leq\lim_{n\to\infty}\sum_{k=1}^n\frac1n\sqrt{2-\frac{k^2}{n^2}}=\int_0^1\sqrt{2-x^2}dx\]


a quick check that sweeps 7, 11, AND 13 is to divide by 1001 (the product of 7, 11, and 13)
the remainder is 429 which infact is divisible by 11 and 13 so 22451 is too!
then after dividing by that you are left with 157, a prime

i dont know if there is a faster way

Is there anything special about the number?
[hide]$ 22451\equal{}22500\minus{}49\equal{}150^2\minus{}7^2\equal{}(150\plus{}7)(150\minus{}7)$[/hide]

Small enough to brute force with stevenmeow's method, in my opinion.

[quote="xxxyyyy"]Which is the largest factor of $ 22451$?

Can you think of a simple method to ascertain without actually trying division by odd primes ?[/quote]

Generally, if there is a number $ x$, the largest factor of $ x$ is $ x$  :lol:

applying this brilliant formula, we see the largest factor of $ 22451$ is $ 22451$

Mewto is right :)

I would have checked the divisibility rule for 11 before trying any primes above 11 and before checking 7. I believe recognizing it as 49 less than a perfect square is the easiest way for this problem *(if you meant to ask for the largerst PRIME factor)


[quote=can_hang2007][quote="ductrung"]For positive real numbers $ a,b$ and $ c$, prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\sum_{cycl}\frac {1}{a^2 \plus{} 2b^2 \plus{} 2c^2}\ge\sum_{cycl}\frac {b^2 \plus{} bc \plus{} c^2}{a^2 \plus{} (b \plus{} c)^2}\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-39-p-h-duc.html[/quote]

A link failure（链接失效）

[quote=ductrung]For positive real numbers $a,b$ and $c$, prove that
\[ (a^2+b^2+c^2)\sum_{cycl}\frac{1}{a^2+2b^2+2c^2}\ge\sum_{cycl}\frac{b^2+bc+c^2}{a^2+(b+c)^2} \][/quote]

[img]http://s9.sinaimg.cn/middle/006ptkjAzy772rNqIJOa8&690[/img]


A long uniform rod of length $l$ and mass $m$ is free to rotate in a horizontal plane about a vertical axis through its center . A spring with force constant $k$ is connected horizontally between the end of thr rod and a fixed wall .What is the period of the small oscillation that result when the rod is pushed slightly to one side and released ?

I got the answer $2\pi\sqrt{\frac{ml^2}{3k}}$ but the answer given is $2\pi\sqrt{\frac{m}{3k}}$ . Can anyone explain why ?

Hm, I don't quite understand the problem statement, a picture would be helpful... but your result has obviously incorrect units (it's in meter x second but the period should be in seconds).

Let the spring is extended by $x$ units, and the rod rotates about its rotational axis (here the centre of the rod) with angular frequency $\omega$.

1. By restoring torque method

The restoring torque, $\tau = -kx{\frac l{2}}$
$i.e, I\alpha = -kx{\frac l{2}}$
where I=moment of inertia of the rod about the rotational axis
& $\alpha$= angular acceration of the rod

$i.e, \frac 1{12}ml^2 {\alpha} = -kx{\frac l{2}}$
$i.e, \alpha = -(\frac{6kx}{ml})$
Let us consider that the extention $x$ in the spring subtends an angle $\theta$ about the rotational axis. Then,
$\theta = \frac x{(\frac l{2})}$
so, $\alpha = -(\frac{6k(\frac l{2}\theta)}{ml})$
$i.e, \alpha = -(\frac{3k}{m})\theta$
so, $\omega = \sqrt{\frac{3k}{m}}$
Thus time period, $T = 2\pi \sqrt{\frac m{3k}}$

2. By energy method

Total energy $E$ of the system remains constant.
$E = \frac 1{2}kx^2 + \frac 1{2}I\omega^2$
Let $v$ be the speed of the end point of the rod where the spring is attached. Then, $v = \frac l{2}\omega$
So, $E = \frac 1{2}kx^2 + \frac 1{2}(\frac 1{12}{ml^2})(\frac{2v}{l})^2$
Thus, $\frac{d(E)}{dt} = 0$
substituting $\frac{d(x)}{dt} = v$ & $\frac{d(v)}{dt} = a$
we get, $kxv - \frac{mva}{3} = 0$
$i.e, a = - (\frac{3k}{m})x$
so, $\omega = \sqrt{\frac{3k}{m}}$
Thus time period, $T = 2\pi \sqrt{\frac m{3k}}$

For diagram please refer 53P, Oscillations, Fundamentals of Physics.

[quote="shyong"][quote="anjani_kumarsingh"][quote="shyong"]I dont understand how $\theta=x/(l/2)$ in your first solution .Can you explain ?

Also why dE/dt = 0 in secind solution ?

Thanks  :)[/quote]
You must be knowing that $\theta = \frac{arc}{radius}$. Using this relation i have written $\theta=x/(l/2)$.

By Principle Of Conservation Of Energy total Energy of the system remains conserved. So, E=constant.
Hence $\frac{d(E)}{dt} = 0$.
Bye.  :D[/quote]

Well ya but i cant see that arc = x  :?  Isnt x is the extension of the spring ? Then how come it become the arc ?[/quote]

The most basic thing for shm is that for the equations $a = -\omega^2 x$ and $a = -\omega^2 \theta$.
$x$ and $\theta$ are very small.
i.e, is how we approximate $\sin{\theta}$ as ${\theta}$ in solving many problems of SHM.
So, Here the extension of the spring is also very small so it van be assumed to be arc length of a very small sector subtending an angle $\theta$ at the centre with radius $\frac l{2}$.
Bye... :lol:

[quote="shyong"]Actually , my doubt is , the extension spring , x is downward direction but the arc of oscilation is left right direction . And thats why I cant relate x to arc  :?[/quote]

Can you explain what is your confusion. I didn't understand it from your last post.
Bye... :lol:

[quote="shyong"]ok , I dont actually know how the diagram looks like  :?  . Maybe that makes me abit confused since the diagram from the book isnt very clear[/quote]

The diagram given the book very clearly explains the situation.
Now see the diagram and follow what I did.
What I did is pushed the end of the rod attached to the spring avery small distance $x$(so that SHM is established). Then you can visualise that the rod rotates about the centre and the end point of the rod describes a path which is an arc of a circle of radius $\frac l{2}$ centred on the centre of the rod.
It is very simple thing. I know you can understand it. If not I will help you again and again till you understand completely.
Thaks for your prompt reply. Bye... :lol:

[color=red]Wish you a MERRY CHRISTMAS And A HAPPY NEW NEAR[/color]

[quote="shyong"][quote="anjani_kumarsingh"][quote="shyong"]Does this mean that , there are two things rotate : one is the rod , another is the spring which attached to the rod .
So when the rod rotate small distance x , the spring also rotate x ?[/quote]

Actually only the rod is rotating and the spring suffers extension and compression. But the extension or compression is so small that we can assume it to be an arc with centre at the centre of the rod.
If you do not get it. Then next time I would send you the diagram depicting the situation and also what I did.
Bye... :lol:[/quote]

I would be very appreciate if you can send me a diagram  :lol:

Bye ...[/quote]

The diagram is in attachment.
Bye.... :lol:

Hm, me neither :(, and I've been really looking forward to finally see the picture...

Ensure that you have Winzip and MS paint.

[quote="varun"]Anjani there is any problem in your file. Please check it.
I also tried to download it.
TIP: Try to upload without compressing.[/quote]
Varun the file is a bitmap file and $AOPS$ doesn't support it.

You can always rename it - for example from file.bmp to file.txt or whatever file type is accepted, upload it and tell everybody that in fact it is a bitmap and they'll have to rename it after they download it - however, the file will be probably way too big if it is a bmp. Hm, is there no way to convert it to jpg or zip it correctly so that it's readable? ...

Lets see if it works.
If it doesn't then I will put the original file.
:lol:

[quote="varun"]Anjani there is any problem in your file. Please check it.
I also tried to download it.
TIP: Try to upload without compressing.[/quote]

Varun, does your mathsgroup on MSN exists. Then we can use to post img.
If exists then do send the link.

The powerpoint presentation works, thanks :).

do we disregard the fact that the true torque isnt -kx (l/2) since the spring exerts a horizontal force on a rod that is angled away from the horizontal by angle theta? i guess if you consider that the true torque would be -kx (l/2 * cos (theta))<-- by torque arm approach, and use the small angle approximation cos (theta) = 1


Suppose that $n\geq 2$ be an integer number and $M={\{1,2,...,n}\}$.We define for every $k\in {\{1,2,...,n-1}\} , x_k=\frac {1}{n+1}\sum_{A\subset M}(minA+maxA)$(where $\mid A\mid=k$) .Show that $x_1,x_2,...,x_{n-1}\in \mathbb{Z}$ and at least one of them isn't divisible by $4$.

For every $A \subset M$, $A'=n+1-A=\{n+1-j,j\in A\}$ is also in the sum
Then we can partitionate the set $\{A\subset M, |A|=k\}$ in 3 sets $A_1, A_2, A_3$ such that
$\forall A\in A_1, n+1-A=A$
$\forall A\in A_2, \exists ! A'\in A_3, A'=n+1-A$

then on $A_1$ we have $min A + max A=n+1$
on $A_2$ and $A_3$ we have
$min A + max A + min A' + max A' = min A + max A + min (n+1-A) + max (n+1-A) =  min A + max A + n+1 - min A + n+1 - max A = 2(n+1)$
Then $x_k$ is an integer.

$x_1=1$ isn't divisible by $4$

Nice solution...i have done it before in a  more counting way but yours is quite good but the last thing you said isn`t true $k_1$ is not $1$

$\sum(minA+maxA)$ is $2(1+2+...+n)$ so  $k_1=n$ and it can be multiple of 4

i`m going to type my solution

Here is my solution:

 For some k, let`s count how many times $x$ appears as a $\min A$ or as $\max A$ for all the posibles subsets $A$. For $x$ being $\min A$ there are $\binom{n-x}{k-1}$,    and if $x$ is the $\max A$ then there are $\binom{x-1}{k-1}$, so summing we get:

 $\displaystyle \sum_{x=1}^{n} x\binom{n-x}{k-1} + \sum_{x=1}^{n} x\binom{x-1}{k-1}$.

 see that both sums are equal and we can join two by two, and see that we can arrange the numbers $x\binom{n-x}{k-1}$ and $(n+1-x)\binom{(n+1-x)-1}{k-1}$, and the whole sum is:

 $\displaystyle x_k=\sum_{x=1}^{n} \binom{n-x}{k-1} =\binom{n}{k}$

 (and that is because the representation of the sum in the Pascal`s triangle is a diagonal so you can sum with Pascal`s theorem to get that.) so to see that there`s a $x_k$ not multiple by 4, let`s suppose no, that all of them are multiples of 4, and see that their sum is multiple of 4 too and:

 $\displaystyle \sum_{k=1}^{n-1} \binom{n}{k}=2^n-\binom{n}{0}-\binom{n}{n}$ that is $2 \mod 4$ for $n>1$ absurd.


Which skin do you think is better? Math links or AoPS?

Mathlinks, because I think it's clearer.

I was an AoPS skinner, but then I came to the blue side :D

[quote="solafidefarms"]Mathlinks people are more partisan about Mathlinks being better than AoPSers are. I'm an AoPSkinner[/quote]

Perhaps this is like the whole Firefox thing...I am a Firefoxer, but I'm also an AoPSkinner.   :roll:

huh? i just like looking at ml but have problems doing classes with it

i think mathlinks is better but im too used to AoPS

I switched to aops, And I like it better.

Mathlinks needs a link to AoPS classes and the Wiki.

i have no idea wat mathlink is, so could someone tell me?

but im sure mathlinks doesnt compare to AoPS  :lol:

link to mathlinks: [url]http://www.mathlinks.ro/Forum/[/url]

i totally like AOPS better. its way more clear, and mathlinks is so monotonous it makes my eyes hurt...[/code]

i like mathlinks too. so i vote for mathlinks :roll:

Everyone likes mathlinks so i shall be different and choose AoPS...i am a loyal AoPSer

I like AoPS. The layout for Mathlinks [kind of] confuse me. I think AoPS is clearer (at least to me).

[quote="pingala"]AoPS :rotfl:  :roll:[/quote]

My thoughts too.


Ok... because the whole task is to prove:

$ 365 ~|~9^{n}+1$

$ 9^{n}$ with arbitrary $ n$ is always odd, so $ 9^{n}+1$ is even. So for odd $ n$ we can say:

$ 9^{n}+1 = (9+1)(9^{n-1}-9^{n-2}+-...+1) \Rightarrow 10~|~(9^{n-1}-9^{n-2}+-...+1)$

$ \Rightarrow 10(9^{n}+1)~|~9^{n}+1$

That's a contradiction, because that would mean $ a|b$ with $ a>b$, so there are no solutions for $ n$??

Ok, I think that is completely wrong, because $ 10~|~(9^{n-1}-9^{n-2}+-...+1)$ is not an equation.^^

[quote="Ikarus"]$ 9^{n}+1 = (9+1)(9^{n-1}-9^{n-2}+-...+1) \Rightarrow 10~|~(9^{n-1}-9^{n-2}+-...+1)$^[/quote]

I'm not sure what you did here.  All the first part implies is that $ 10 | 9^{n}+1$ when $ n$ is odd.

[quote="Ikarus"]$ 365 ~|~9^{n}+1$[/quote]

Well, this is clearly not true for all $ n$ (say, $ n = 1$).  Do you mean to find what conditions on $ n$ under which this is true?

[hide="Solution"] $ 365 = 5 \cdot 73$

$ 5$ divides $ 9^{n}+1$ iff $ n$ is odd.

We know that $ 3^{72}\equiv 1 \bmod 73 \Leftrightarrow 73 | 9^{36}-1 = (9^{18}-1)(9^{18}+1)$

At this point we should try to find the order of $ 9 \bmod 73$.  Well, it turns out that

$ 9^{3}\equiv 81 \cdot 9 \equiv 8 \cdot 9 \equiv 72 \equiv-1 \bmod 73 \implies$
$ 73 | 9^{3}+1$

So $ 365 | 9^{n}+1 \Leftrightarrow n \equiv 1 \bmod 2, 3 | n$. [/hide]


function
limit
logarithms
real analysis
real analysis unsolved

Show that $ (1 \plus{} \frac {1}{x})^x$ is an increasing function of $ x$. Show also that its limit as $ x\rightarrow \infty$ is $ e$.

The limit is in the form of $ \infty^\infty$. Are you sure that it has a limit $ e$?

I think you mean $ (1\plus{}\frac{1}{x})^{x}$

hello, we have $ \lim_{x \to \minus{}\infty}\left(1\plus{}\frac{1}{x}\right)^x\equal{}\lim_{x \to \infty}\left(1\plus{}\frac{1}{x}\right)^x\equal{}e$.
Sonnhard.

$ \ln (1 \plus{} x) \approx x \minus{} \frac {x^2}{2}$ from taylor ..
so we can show $ (1 \plus{} \frac {1}{x})^x$ like $ e^{x\ln(1 \plus{} \frac {1}{x})}$ and from taylor :
$ (1 \plus{} \frac {1}{x})^x \equal{} e^{x(\frac {1}{x} \minus{} \frac {1}{2x^2})} \equal{} e^{1 \minus{} \frac {1}{2x}}$
 $ \rightarrow \displaystyle\lim_{x\to\infty}(1 \plus{} \frac {1}{x})^x \equal{} \displaystyle\lim_{x\to\infty}(e^{1 \minus{} \frac {1}{2x}}) \equal{} e$

Bounded monotonic therefore has limit; same applies to its natural log.  Thus
L'Hop applied to the indeterminate form 0/0 gives:

$ \lim_{x\to\infty}\left(1\plus{}\frac{1}{x}\right)^x
\equal{}e^{\lim_{x\to\infty}\ln\left(1\plus{}\frac{1}{x}\right)^x}
\equal{}e^{\lim_{x\to\infty}\frac{\ln\left(1\plus{}\frac{1}{x}\right)}{\frac{1}{x}}}
\equal{}e^{\lim_{x\to\infty}\frac{\frac{\frac{\minus{}1}{x^2}}{\left(1\plus{}\frac{1}{x}\right)}}{\frac{\minus{}1}{x^2}}}
\equal{}e^{\lim_{x\to\infty}\frac{1}{\left(1\plus{}\frac{1}{x}\right)}}
\equal{}e^1$


induction
algebra
binomial theorem
combinatorics theorems
combinatorics

I am sorry if it should be posted smwhere else and if it was posted before.
How we prove that
\[ (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^k \equal{} \sum_{i_1 \plus{} ... \plus{} i_n \equal{} k} \frac {k!}{i_1!\cdot i_2! \cdot ... \cdot i_n!}a_1 ^{i_1} \cdot ... \cdot a_n^{i_n}
\]
I tried to use induction but I can't see the identity like $ \binom{n \plus{} 1}{k} \equal{} \binom{n}{k} \plus{} \binom{n}{k \minus{} 1}$ for the cofficients. :blush:

The identity you are looking for is $ \frac {k!}{i_1!\cdot i_2! \cdot ... \cdot i_n!} \equal{} \sum_{j \equal{} 1}^n\frac {\left(k \minus{} 1\right)!}{i_1!\cdot i_2!\cdot ...\cdot i_{j \minus{} 1}!\cdot\left(i_j \minus{} 1\right)!\cdot i_{j \plus{} 1}!\cdot ...\cdot i_n!}$.

  darij

Thanx a lot!
Now I think I can prove that $ D^m([x_1,...x_n]) \equal{} \sum_{i_1 \plus{} ...i_n \equal{} m} \frac{m!}{i_1! ... i_n!}[D^{i_1}(x_n),..., D^{i_n}(x_n)]$ for any linear transformation D from an algebra A with n-ary multiplication [-,-,...,-] into it which satisfies the identity for m=1

The easier way to induct is over $ k$, actually -- write $ a_1 \plus{} \ldots \plus{} a_{k \minus{} 1} \plus{} a_k \equal{} (a_1 \plus{} \ldots \plus{} a_{k \minus{} 1}) \plus{} a_k$ and apply the binomial theorem, then the multinomial theorem for $ k \minus{} 1$.

Is it needed to apply induction? For example, paritite $ k$ into $ n$ parts.


calculus
trigonometry
geometry
trig identities
Law of Sines

Given $ a,b\ge0$ such that $ a^2\plus{}b^2\plus{}ab\equal{}12$, find the maximum and minimum values of $ (\sqrt3\plus{}1)a/2\plus{}b$.

hello, with the help of calculus i have got
$ (\sqrt{3}\plus{}1)\frac{a}{2}\plus{}b\geq2\sqrt{3}$ and the minimum will be obtained for
$ a\equal{}\frac{4\sqrt{3}\minus{}2(\minus{}1\plus{}\sqrt{3}\plus{}\frac{1}{6}\sqrt{144\plus{}72\sqrt{3}})}{1\plus{}\sqrt{3}}$
$ b\equal{}\minus{}1\plus{}\sqrt{3}\plus{}\frac{1}{6}\sqrt{144\plus{}72\sqrt{3}}$
Sonnhard.

Very cool problem! Is this original?
[hide="Hint"]
Extremal cases aside the expression $ a^{2} \plus{} b^{2} \plus{} ab \equal{} 12$ should remind you of the law of cosines. From here we can use the law of sines and some trig.
[hide="Answer"]
The maximum is $ 2\sqrt {6}$ it's attained when the angles of your triangle have measures $ 15^\circ, 45^\circ,$  and $ 120^\circ$. This corresponds to $ a\equal{} 2\sqrt {2}$ and $ b\equal{}\sqrt {6}\minus{}\sqrt {2}$
The minimum is $ 2\sqrt{3}$ and is attained when the angles of your triangle are $ 60^\circ$ and $ 120^\circ$. This corresponds to $ a\equal{}0$ and $ b\equal{}2\sqrt {3}$
[/hide]
[/hide]

[hide="Hint"]
 Let $ x \equal{} \left(\frac{a}{2} \plus{} b \right)$ and $ y \equal{} \frac{\sqrt 3 a}{2}$

Then we are asked to maximize $ x\plus{}y$ given $ x^2\plus{}y^2 \equal{} 12$ This can be done by Cauchy-Schwarz or a Trig substitution.

[/hide]


[quote="dingzhou"]how many ppl attend mathzoom anyway?[/quote]

We have about 30 students.


Nobody want to help????????!!!!!!! :maybe:

[quote="mohammad reza"]$ n \equal{} 3$[/quote]
could you write your solution????

[quote="behdad.math.math"]Find minimum of $ n \in N$ such that the equation have the solution in natural numbers.

$ y^2 \equal{} x ^{1992} \minus{} x^{1990} \plus{} (2n \minus{} 1)x^3 \minus{} x$[/quote]
$ n \equal{} 3$, $ x \equal{} 1$, $ y \equal{} 2$ is obviously a solution. Hence there are two cases left: $ n \equal{} 1$ and $ n \equal{} 2$.

If $ n \equal{} 1$, the equation is equivalent to $ y^2 \equal{} x(x^{1989} \plus{} 1)(x^2 \minus{} 1)$. Hence $ x$ is a perfect square itself. Moreover, $ x^{1989} \plus{} 1 \equal{} (x \plus{} 1)\left( \sum_{i \equal{} 0}^{1988}( \minus{} 1)^{i}x^{i}\right)$. Hence the equation ist equivalent to $ y^2 \equal{} x(x \plus{} 1)^2(x \minus{} 1)\left( \sum_{i \equal{} 0}^{1988}( \minus{} 1)^{i}x^{i}\right)$. Suppose a prime $ p$ divides $ x \minus{} 1$. Hence $ x \equiv 1 \mod p$. Then $ \sum_{i \equal{} 0}^{1988}( \minus{} 1)^{i}x^{i} \equiv 1 \mod p$, hence they share no common divisor. Therefore $ x \minus{} 1$ has to be a square itself, $ x \minus{} 1 \equal{} z^2$. But as shown before, $ x$ is a square, a contradiction, unless $ x \equal{} 1$, which yields $ y \equal{} 0 \not \in \mathbb{N}$.

If $ n \equal{} 2$, the equation is equivalent to $ y^2 \equal{} x(x^{1991} \minus{} x^{1989} \plus{} 3x^2 \minus{} 1)$. Hence $ x$ is a square, which leads to $ x \equiv 0 \vee 1 \mod 4$. If $ x$ is q square, $ S: \equal{} x^{1991} \minus{} x^{1989} \plus{} 3x^2 \minus{} 1$ is a square too. But for $ x \equiv 0 \mod 4$, we get $ S \equiv 3 \mod 4$, contradiction. For $ x\equiv 1$, we get $ S \equiv 2$, contradiction again.

Hence the minimum is $ n \equal{} 3$.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

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

[2] $\int \frac{e^{3x}}{\sqrt{e^x+1}}dx$

[3] $\int \sin 2x\cos 3xdx$

[4] $\int x\ln (x+1)dx$

[5] $\int \frac{x}{(x+3)^2}dx$

[quote="amirhtlusa"]2
let $u=e^x+1$
answer:
$2e^3\sqrt{e^x+1}$[/quote]
but $e^{3x}\neq e^3\cdot e^x$...

[quote="skylarth"]4:

parts first then $u = 1 + x$

messy but:

$ \frac 12[ln|x+1|(x^2-1) - \frac {(x+1)^2}{2} + 2(x+1)]$

hope thats right..[/quote]

Thank you for your reply, skylarth.

In this case since there is a guarantee for the condition of anti-logarithm, you can ommit the sign of absolute value.

[quote="litoxe2718281828"]3.
$=\int \left(\frac{\sin 5x}{2}-\frac{\sin x}{2}\right)dx=-\frac{\cos 5x}{10}+\frac{\cos x}{2}$[/quote]

That's right :)

[quote="amirhtlusa"]1
expand, then devide then integrate:
$x+\frac{2}{x}-\frac{1}{3x^3}$[/quote]

O.K. :)

[quote="amirhtlusa"]2
let $u=e^x+1$
answer:
$2e^3\sqrt{e^x+1}$[/quote]

 :(

[quote="amirhtlusa"]oops. sorry, i made a  mistake on 2
the answer for 2 is:
$\frac{2}{5}\sqrt{(e^x+1)^5}-\frac{4}{3}\sqrt{(e^x+1)^3}+2\sqrt{(e^x+1)}$[/quote]

No problem. This time you got it. :)

[quote="litoxe2718281828"][quote="kunny"][quote="litoxe2718281828"]5.
$u=x+3$ and integrate.
$-\frac{1}{x+3}+\frac{3}{2(x+3)^2}$[/quote]

Your answer is incorrect, litoxe2718281828.

By the way, 2718281828.... is $e$,isn't it?[/quote]

Yes, I thought the denominator was cubed. It should be:

$\log|x+3|+\frac{3}{x+3}$

$e$ is such a nice number to type, isn't it?  :lol:[/quote]

That's right. :)


calculus
integration
trigonometry
real analysis
real analysis unsolved

this integral looks scary, my friend said he found it in some text book in library.  can someone here solve it??

\[ \int\frac{x^2\plus{}1}{x^5\plus{}x^4\plus{}x^3\plus{}x^2\plus{}1}\cdot\frac{\arctan\sqrt{e^x\plus{}1}}{\cosh{x}}\,{\rm d}x\]

i will try this when i have time, but i am not sure where to start , need some insight

he lost the answer from the book, but he said it is something like natural log as a result, but he doesnt know how to do it

[quote="kelvin911"]this integral looks scary, my friend said he found it in some text book in library.  can someone here solve it??
\[ \int\frac {x^2 \plus{} 1}{x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} 1}\cdot\frac {\arctan\sqrt {e^x \plus{} 1}}{\cosh{x}}\,{\rm d}x
\]
i will try this when i have time, but i am not sure where to start , need some insight

he lost the answer from the book, but he said it is something like natural log as a result, but he doesnt know how to do it[/quote]

i can solve your integral, but only if it says
\[ \int\frac {x^2 \plus{} 1}{x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1}\cdot\frac {\tan^{ \minus{} 1}\,\sqrt {e^x \plus{} 1}}{\cosh{x}}\,{\rm d}x
\]
(i have included a $ ( \plus{} x)$ in the denominator)

then what is it?  my friend gave this integral to me, i am not sure if he got it wrong or not

Quoting the previous post in full [b]never[/b] makes sense, you are replying to it by default! I removed all unnecessary quotes leaving only the one by misan which is also rather poorly justified :).


abstract algebra
group theory
Ring Theory
complex numbers
absolute value
superior algebra
superior algebra theorems

A nice, not too difficult (but certainly not trivial!) result whose generalizations play an important role in the so-called 'isomorphism problem of group rings' (which is: given a ring $R$, decide whether the group ring isomorphism $RG \cong RG'$ implies the group isomorphism $G \cong G'$, where $G,G'$ are finite groups; for example, when $G,G'$ are abelian and $R = \mathbb{Z}$, then the answer to this question is affirmative, but in general it is not). Anyway, here is the problem:

[b]Higman's Theorem: [/b] Let $G$ be a finite abelian group. Let $\alpha \in \mathbb{Z}G$ be a non-zero element in the group ring $\mathbb{Z}G$ such that $\alpha^k = 1$ for some $k>0$. Prove that $\alpha = \pm g$ for some $g \in G$.

This is really delightful, and not too difficult -- give it a try  ;) . Enjoy!

--Vesselin

P.S. A generalization of this result: if $G$ is a finite group (not necessarily abelian) and $\alpha = \sum_{g \in G} a_gg$ is a torsion element of $(\mathbb{Z}G)^*$, then either $\alpha \in \pm G$ or $a_e = 0$. But this is already quite difficult to prove, I suppose ...  :)

I agree that, using characters, it is straightforward to show the following: if $G$ is any finite group (not necessarily abelian), and if $T$ is a subgroup of finite order of the multiplicative group $(\mathbb{Z}G)^*$, then $T$ is contained in $\pm G$. This, of course, automatically implies that, in the abelian case, every torsion element of $(\mathbb{Z}G)^*$ is contained in $\pm G$.

The result I quoted in the P.S. belongs to Saksonov (or maybe dates earlier to Higman, I'm not sure; I think his thesis is dated to the 1940s). Of course, the proof uses character techniques; but is there a 'straightforward' proof? Perhaps I'm just overlooking something?

[quote="cimabue"]
in the abelian case, one can prove that if $\alpha=\sum_{g\in G}a_gg\in\mathbb C[G]$ has the property that  $\alpha^k=1$, then $\sum_{g\in G}|a_g|^2=1$, and the result follows.
[/quote]

Since I'm new to the field, I'll post the proof just to see if I get it right :).

Let $X$ be the set of irreducible characters of $G$ (which must have order $1$, and are simply homomorphisms from $G$ to the multiplicative group of complex numbers with absolute value $1$). Now, since $(\sum_{g\in G}\alpha_gg)^k=1$, we have $(\sum_{g\in G}\alpha_g\chi(g))^k=1,\ \forall\chi\in X$. In particular, this means that for all $\chi\in X$ the complex number $\sum_{g\in G}\alpha_g\chi(g)$ has absolute value $1$, so we can write $(\sum_{g\in G}\alpha_g\chi(g))(\sum_{g\in G}\bar\alpha_g\chi(g^{-1}))=1,\ \forall\chi\in X$. We sum after $\chi\in X$ and divide by $|X|=|G|$ to get precisely what we want: $\sum_{g\in G}|\alpha_g|^2=1$ (by using the orthogonality relations $\sum_{\chi\in X}\chi(g)\chi(h^{-1})=\{\begin{array}{c}|X|=|G|,\ g=h\\0,\ g\ne h\end{array}$).

Is it Ok?

In the non-abelian case, the following is entirely analogous:

If $T$ is a finite subgroup of the multiplicative group $(\mathbb{C}G)^*$ [b]and[/b] $T$ contains $G$, then every element $\alpha = \sum_g a_gg \in T$ satisfies $\sum_g |a_g|^2 = 1$; consequently, if $T$ is a finite subgroup of $(\mathbb{Z}G)^*$ containing $G$, then $T \subset \pm G$.

[I forgot to mention the condition that $T$ contains $G$ in my previous post, but this condition is absolutely essential. ]

Of course, when $G$ is abelian, the initial assertion is just a special case, for if $\alpha \in \mathbb{Z}G$ has finite order, then $\langle G, \alpha \rangle$ is still a finite group (but this is not necessarily so when $G$ is not commutative). What I wanted to stress is that the abelian case is substantially easier not because representation-theoretic techniques are more convenient in that case, but for a deeper reason. In fact, it is not true in general that every torsion element of $(\mathbb{Z}G)^*$ is contained in $\pm G$! Saksonov's result, in a sense, is optimal:

[b]Saksonov's Theorem: [/b] If $\alpha = \sum_g a_gg$ and $\alpha^k = 1$ for some $k$, then either $a_e = 0$ or $\alpha \in \pm G$.

--Vesselin

Here is a sketch of a proof of the following

[b]Theorem. [/b] Let $G$ be a finite group. Let $T \subset (\mathbb{C}G)^*$ be a finite subgroup of the multiplicative group of $\mathbb{C}G$, such that $T$ contains $G$. If $\alpha = \sum_g a_gg \in T$ is arbitrary, then $\sum_g |a_g|^2 = 1$.

[i]Sketch of proof. [/i] Write $\alpha^{-1} = \sum_g b_gg \in T$. Then, $\sum_g a_gb_{g^{-1}} = 1$; therefore, it will be sufficient to show that $b_{g^{-1}} = \overline{a_g}$.

Let $(\pi_i,V_i)$, $1 \leq i \leq N$, be an enumeration of the irreducible representations of $G$ over $\mathbb{C}$. Recall that $\mathbb{C}G$ is canonically isomorphic with $\bigoplus_{i=1}^N \mathrm{End}(V_i)$ via $\oplus_{i=1}^N \pi_i$. We have a Fourier inversion formula: if $\alpha = \sum_g a_gg \in \mathbb{C}G$, then
\[ a_g = \frac{1}{|G|} \sum_{i=1}^{N} \dim{V_i} \, \mathrm{tr} \big( \pi_i(g^{-1}\alpha)\big). \]
Now let $g \in G$ be arbitrary. Because $T$ is a finite group containing $G$, we have that $g^{-1}\alpha^{-1} \in T$, and therefore $g^{-1}\alpha^{-1}$ has finite order, showing that (for arbitrary $i$) the eigenvalues of $\pi_i(g^{-1}\alpha^{-1})$ are roots of unity and therefore are on the unit circle. But taking complex conjugates on the unit circle is the same as taking inverses! Therefore, as the trace is the sum of eigenvalues on the unit circle,
\[ \overline{\mathrm{tr}(\pi_i(g^{-1}\alpha^{-1}))} = \mathrm{tr}(\pi_i (\alpha g)) = \mathrm{tr} (\pi_i(g \alpha)). \]
By the Fourier inversion formula above, it follows that $b_{g^{-1}} = \overline{a_g}$, as claimed. Hence, $\sum_g |a_g|^2 = \sum_g a_g\overline{a_g} = \sum_g a_gb_{g^{-1}} = 1$, completing the proof.

Note that we essentially used that $G$ is contained in $T$, as otherwise we would not be able to conclude that every $g^{-1}\alpha^{-1}$ has finite order.

[b]Corollary: [/b] If $T \subset (\mathrm{Z}G)^*$ is a finite group containing $G$, then $T \subset \pm G$.

Well, $\pm G = \{\pm g \mid g \in G\}$ is a finite subgroup of $(\mathbb{Z}G)^*$ strictly larger than $G$. I suppose you intended to ask for an example of a (non-abelian) group $G$ and an $\alpha \in (\mathbb{Z}G)^*$ such that $\alpha^k = 1$ for some $k$ but $\alpha \notin \pm G$? Well, I have to admit that I haven't thought of this, but I alleged it in my previous post because I think I have seen such examples somewhere -- or at least I have seen a remark that Saksonov's theorem is optimal. Nevertheless, I will try to find such examples --- I don't think it should be too difficult, we shall see  :).

--Vesselin


[u]$Question 1$[/u]A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel which returns him to his cell after three days travel, the second that returns him to his cell after four days of travel and the third immediately leads to freedom.

Assuming that the prisoner will select doors $1, 2, 3$ with the
respective probabilities $\frac{1}{2}$, $\frac{7}{20}$ and $\frac{3}{20}$; determine the expected
number of days till he reaches freedom.

[u]$Question 2$[/u]
An urn contains $8$ blue balls and $7$ black balls and a series of drawings of one ball at a time is made, the ball removed being returned immediately after the taking of the next following ball.

Determine the probability that the 10th ball drawn is blue, when:

(A) Nothing is known about the preceeding drawings.

(B) The 4th ball drawn is known to be blue.

[quote="NightFlarer"]For question 1 are we assuming that once he picks door one, he will never choose it again because he knows it's wrong?[/quote]

I think we're assuming he's stupid and we're practicing infinite geometric series. :D

[hide="Solution to (1)"]Expected value is the sum of all the probabilities multiplied by their respective # of days:
$E=(\frac{1}{2})(3 days)+(\frac{7}{20})(4 days)+(\frac{3}{20})(0 days) = 2.9 days$
[/hide]

[hide="Solution to (2)"](a) Since we don't care about the preceding drawings, the probability is $\frac{8}{15}$.

(b) We want the 4th ball to be blue. Since we don't care about the 1st, 2nd, or 3rd drawings, the probability of that happening is $\frac{8}{15}$. Now we want the 10th ball to be blue. Since the probabilities are all independent, we can just multiply the probability of the 4th ball being blue, with the 10th ball to be blue:
$(\frac{8}{15})(\frac{8}{15}) = \frac{64}{225}$[/hide]

[quote="vishalarul"][hide="Solution to (2)"](a) Since we don't care about the preceding drawings, the probability is $\frac{8}{15}$.

(b) We want the 4th ball to be blue. Since we don't care about the 1st, 2nd, or 3rd drawings, the probability of that happening is $\frac{8}{15}$. Now we want the 10th ball to be blue. Since the probabilities are all independent, we can just multiply the probability of the 4th ball being blue, with the 10th ball to be blue:
$(\frac{8}{15})(\frac{8}{15}) = \frac{64}{225}$[/hide][/quote] But ball 1 is put back after ball 2 is drawn, so the probability of choosing a blue ball on the second drawing isn't $\frac{8}{15}$.

[quote="vishalarul"]Oh, I realize my mistake.  :blush: So, how would you do the problem? :maybe:[/quote]
Possibly recursion. :maybe:
[hide] Never mind. The probabilities work out so that the probability of getting blue on the 2nd draw is $\frac{8}{15}\frac{7}{14}+\frac{7}{15}\frac{8}{14}=\frac{8}{15}$. So, the probability is constant. [/hide]

Hm, so I did a ton of work for #1, most likely making it more complicated than it is, but I got $\frac{34}{3}$.  If that's right, I'll show my solution... don't really want to show it not knowing if it's right, since it's kinda long.

edit:

For 2B, you have to use recursion.  Let $U_{n}$ and $A_{n}$ represent probabilities of blue and black on nth ball.  $U_{4}=1$.
Then we have the property
$U_{n+1}= \frac{7}{14}\cdot U_{n}+\frac{8}{14}\cdot A_{n}$
$= \frac{1}{2}\cdot U_{n}+\frac{4}{7}\cdot (1-U_{n})$
$= \frac{4}{7}-\frac 1{14}\cdot U_{n}$
Then $U_{5}= \frac{1}{2}, U_{6}= \frac{15}{28}, U_{7}= \frac{209}{392}$, and then after the 9th term, the fractions get too big for my calculator to calculate.  Needless to say, $U_{10}\approx 0.5333333953$

That's correct for 2B, deej.  The exact answer is $\frac{573679}{1075648}= \frac{8}{15}+\frac1{16134720}$.

If the prisoner doesn't learn each time, it should take him just less than 20 days (expected).  If he does learn each time, it's not clear what we're supposed to do with those probabilities should he end up coming back to the cell (as is highly likely).

How did you get $\frac{34}{3}$ for #1, deej21? Is there some fault in my method? :maybe:

[quote="vishalarul"] Is there some fault in my method? :maybe:[/quote]

Yes: you didn't read the question carefully.

Yes, JBL is right, you assumed he learned from his mistakes, which I don't think is the case.

I don't know if my answer is right, but here's how I started my solution... Sum of $n$ from 1 to infinity, where $n$ is the number of doors picked wrong before picking the right one.  For each $n$, we can consider all cases $a^{n-k}b^{k}$, where $0 \le k \le n$, and $a$ and $b$ represent the probabilities of opening doors 1 and 2.  For the expected value, we multiply each case by the number of days it will take.  For each a, it takes 3 days, for each b, 4 days.  The number of days will be $3(n-k)+4(k) = (3n+k)$.  Finally, everything needs to be multiplied by $\frac{3}{20}$ for picking the final door.
Then the value should be
$\frac{3}{20}\sum_{n=1}^{\infty}\sum_{k=0}^{n}(3n+k)a^{n-k}b^{k}$, which after a lot of simplification after subbing in a and b, gets
$\frac{3}{2}\sum_{n=1}^{\infty}\frac{n+\frac{7}{9}}{2^{n}}$
$= \frac{25}6$
Well this is obviously wrong, but I redid it.  The first time I did it, I had a $n$ multiplied in between the two sigmas, which I think is wrong because I already multiply by the number of days, and I think that's what I was using it for... anyway it's definitely not 25/6, but that work seems kinda right.
Oh, maybe I made a mistake in the simplification.  That part I didn't redo.

[hide="Number 1"]I think the probability of the first question would be equivalent to a situation where there are 20 doors in which 10 lead to door 1 area, 7 to door 2 area, and 3 to door 3 area. There for over an infinite period of time, the number of days is equal to the number of times the man enters each door.

(Number of entries to Door 1)(3) + (Number of entries to Door 2)(4) + (Number of entries to Door 3)(0) = total days.

To find the average number of days, we divide by the total number of days, which should reduce the coefficients of the days in the above equation to their respective probabilities. Thus, the average and expected number of days is:

$\frac{1}{2}(3)+\frac{7}{20}(4)+\frac{3}{20}(0) = \frac{29}{10}$ days.[/hide]

Breez, that was the exact same mistake visalarul made:  re-read the question.  The 3-day and 4-day doors don't send him to freedom, they return him to his cell.  (Contra deej21, this is not quite the same thing as his learning from his mistakes because it neglects the possibility that he could go through one bad door, return to his cell and then go through the other bad door.)

Deej, I haven't tried to figure out where you went wrong, but here's the simple solution to number 1, (assuming he doesn't learn):

[hide]Let $x$ be the expected number of days.  Then we have immediately that $x = \frac12(x+3)+\frac7{20}(x+4)+\frac3{20}(0)$.  It's all smooth riding after that.[/hide]


I'm using "Patty" to send you this message, as I cannot log in as myself because I cannot remember my username.  I know I must have registered with my e-mail address because I tried to register (again) using my e-mail address, and it told me that was already assigned to another username.  Suggestions?  Thank you so much.  I'll check back here using Patty.

I've sent you email at the above address.


inequalities
function
geometry
3D geometry
calculus
floor function
ceiling function

Let be given $ a,b,c \in [1,2]$ . Prove that
	$ (a+b+c)(\frac1a+\frac1b+\frac1c) \leq 10$

Or you can see, that
$
(a+b+c)(\frac{1}{a}+\frac{1}{a}+\frac{1}{a})-10
$
is a convex function of three variables, so it attains its maximum in vertices of cube (two opposite points are (1,1,1) and (2,2,2)). So there are just 8 points where it can attain maximum... Its in (1,1,2) and cyclic so its maximum is 0. And its solved :)

[quote="prowler"]You can easily to see that a,b,c can be the sides of a triangle
    a+b > 2 > c.
So you can look in the solution to the problem 4 given on IMO 2004.
Here is link:
     http://www.mathlinks.ro/Forum/viewtopic.php?t=14091[/quote]

What you use is not IMO 2004 problem 4, but rather the converse of IMO 2004 problem 4. Alas, I must say that this converse is not generally true. For instance, if a, b, c are the sides of a triangle such that a = b is very large and c is very small, the inequality $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 10$ is wrong. And indeed, demanding $a, b, c \in \left[1;\;2\right]$ is more than demanding a, b, c to be the sides of a triangle.

Ondrob's solution is completely correct except that the function should be $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) - 10$ rather than $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{a}\right) - 10$, of course.  :)

  Darij

Hi,Ondrob,
Thanks a lot but can you prove it without using calculus in many variables function?
you make me feel that your proof look like "using nuclear weapon to kill a little dog"!  :D  :D  :D

The proof does not really use calculus. It just uses some easy facts about convex functions. See Lemma 2 in my posting at http://www.mathlinks.ro/Forum/viewtopic.php?t=15431 . The proof is completely elementary.

  Darij

Very nice solution! I have nothing to say.

Why should we prove particular case?  ;)
Prove general inequality!

Let $M\geq a_k\geq m>0$. Then
\[(a_1+a_2+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})\leq n^2 + (n-1)\frac{(M-m)^2}{mM}.\]

The proof with convex functions works again. We only have to consider cases: $a_1=a_2=...=a_k=m$ and $a_{k+1}+...+a_n=M$ for all $k$ in $[m,M]$. we get: \[(km+(n-k)M)(\frac{k}{m}+\frac{n-k}{M})=k^2+(n-k)^2+k(n-k)(\frac{m}{M}+\frac{M}{m})\]
\[=(k+n-k)^2+k(n-k)(\frac{m}{M}+\frac{M}{m}-2)\].
Now we have to maximize \[k(n-k)(\frac{m}{M}+\frac{M}{m}-2)=k(n-k)\frac{(M-m)^2}{mM}.\] So the maximum of the left side is \[n^2+\frac{n^2}{4}\frac{(M-m)^2}{mM}.\] for even $n$ and \[n^2+\left\lfloor\frac{n}{2}\right\rfloor\left\lceil\frac{n}{2}\right\rceil\frac{(M-m)^2}{mM}\] for odd $n$.

Myth, please correct me if I'm wrong but your expression on the right side seems to be wrong.

Misha

I think you mean
\[n^2+\left[\frac{n^2}{4}\right]\cdot\frac{(M-n)^2}{mM}.\]
;)

It seems I miscalculated when I was computing constants.

Can you prove that mentioned function is convex? My solution doesn't need that.

It is easy to see that the left side is convex: it has the form $\frac{c_1}{a_i}+c_2a_i$. The addition of the linear term doesn't take influence on convexity and the other term is clearly convex: so the left side is convex in all variables.

Misha

[quote="Myth"]I think you mean
\[n^2+\left[\frac{n^2}{4}\right]\cdot\frac{(M-n)^2}{mM}.\]
;)
[/quote]

And I think you mean \[n^2+\left[\frac{n^2}{4}\right]\cdot\frac{(M-m)^2}{mM}.\] ;)

Misha

I think the general ineq can be proved by the comment :

   $  (a_k - M) (a_k - m) \leq 0  <----->  (a_k)^2  - (M+m)a_k + mM \leq 0  =>  a_k - (M+m) + \frac{mM}{a_k} \leq 0 $
   Let k= 1,2,..,n
  After that , using AM_GM for $ \sum_{k=1}^n(a_k)   $ and $ Mm \sum_{k=1}^n\frac{1}{a_k}  $ we will have Myth's ineq .[/quote]

[quote="Myth"]@Misha! I don't believe you that $(a_1+a_2+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})$ is a convex function on $n$ variables. In my opinion, your reasoning is insufficient, because we vary all variables at the same time, so your coefficients $c_1$ and $c_2$ vary.[/quote]

The function needn't be convex in [i]all[/i] n variables, it must be convex in [i]every[/i] of the n variables. See Lemma 2 in my posting at http://www.mathlinks.ro/Forum/viewtopic.php?t=15431 .

  Darij

[quote="nttu"]After that, using AM_GM for $ \sum_{k=1}^n(a_k)   $ and $ Mm \sum_{k=1}^n\frac{1}{a_k}  $ we will have Myth's ineq.[/quote]
I hate when answer for all questions is "using AM-GM we obtain what we want".
What is AM-GM in your opinion, nttu?
Your approach leads us to estimate $\frac{(M+m)^2}{4mM}\cdot n^2$. In the same time we are proving more sharp estimate.

[quote="darij grinberg"]The function needn't be convex in [i]all[/i] n variables, it must be convex in [i]every[/i] of the n variables. See Lemma 2 in my posting at http://www.mathlinks.ro/Forum/viewtopic.php?t=15431.[/quote]
I [i]know[/i] that fact [i]very well[/i].
I said that it is possible that mentioned function isn't a convex function.

[quote="Myth"]I said that it is possible that mentioned function isn't a convex function.[/quote]

Then show me an example.  ;)

  darij

[quote="darij grinberg"][quote="Myth"]I said that it is possible that mentioned function isn't a convex function.[/quote]

Then show me an example.  ;)[/quote]
Pridi i posmotri!

Nu, tak gde kontrprimer?  :D

Mne vsjo eshjo kazhetsja chto funkzija eta ($\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)$) vypukla v kazhdoj iz peremennyh (kak summa postojannoj, linejnoj i $\frac{k}{x}$-funkzii). Mozhet Vy imeete vvidu druguju funkziju?

PS. This thread is becoming more and more chaotic...  :D

  darij

Meanwhile a few little remarks to make it all less confusing:

[b]1.[/b] Myth's inequality in the form he posted it,

$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\leq n^2 + \left(n-1\right)\frac{\left(M-m\right)^2}{mM}$,

is indeed wrong. Take n = 4, $a_1=a_2=1$ and $a_3=a_4=2$ as a counterexample.

[b]2.[/b] The correct form is indeed Misha's

$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\leq n^2 + K\left(n\right)\frac{\left(M-m\right)^2}{mM}$,

where $K\left(n\right)=\frac{n^2}{4}$ for even n, and $K\left(n\right)=\left\lfloor\frac{n}{2}\right\rfloor\left\lceil\frac{n}{2}\right\rceil=\frac{n^2-1}{4}$ for odd n. Both times, the inequality is sharp.

By the way, an equivalent form of this inequality is

$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\leq n^2 + K\left(n\right)\left(\sqrt{\frac{M}{m}}-\sqrt{\frac{m}{M}}\right)^2$,

since $\frac{\left(M-m\right)^2}{mM}=\left(\frac{M-m}{\sqrt{mM}}\right)^2=\left(\frac{M}{\sqrt{mM}}-\frac{m}{\sqrt{mM}}\right)^2=\left(\sqrt{\frac{M}{m}}-\sqrt{\frac{m}{M}}\right)^2$.

[b]3.[/b] Since, for any natural n, we have either $K\left(n\right)=\frac{n^2}{4}$ or $K\left(n\right)=\frac{n^2-1}{4}$, we always have $K\left(n\right)\leq\frac{n^2}{4}$, and thus we can weaken our inequality as follows:

$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\leq n^2 + K\left(n\right)\frac{\left(M-m\right)^2}{mM}$
$\leq n^2+\frac{n^2}{4}\cdot\frac{\left(M-m\right)^2}{mM}=n^2\left(1+\frac{\left(M-m\right)^2}{4mM}\right)=n^2\frac{\left(M+m\right)^2}{4mM}$.

[b]4.[/b] Misha's proof is, in my opinion, correct, except of a few typos in the beginning. Here is a correction of the text:

[quote="Misha123, corrected"]The proof with convex functions works again. We only have to consider cases: $a_1=a_2=...=a_k=m$ and $a_{k+1}=...=a_n=M$ for all $k$ in $[1,n]$. we get: [...][/quote]

  Darij

I have [i]already[/i] said that I miscalculated in constants (I verified my calculations and found silly mistake). And everyone who read this topic has already understood that Misha's form is correct.

You said that [i]it seems[/i] that you have miscalculated the constant. And Nttu proved a bound which is weaker than Misha's one (at least, for odd n). So I thought it would be better to recall Misha's bound again. Anyway, please don't take this as an offence.

However, I still think there is a misunderstanding: Why, in your opinion, isn't the function $\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)$ convex in every variable?

  Darij

Yeah, Misha's ineq is stronger than mine  :maybe:

I think with darkmaster's problem and general ineq , we can also prove by the comments :
  $ a, b , c \in [1,2]
=>  1)  \frac{a}{b} + \frac{b}{a} \leq \frac{5}{2}
       2) \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \leq \frac{7}{2} $
   Prove 1) :
  We have : $ 1\leq a , 0<b\leq 2 => b\leq 2x => \frac{1}{2} \leq \frac{a}{b}  (*) $
    and $ \frac{a}{b}\leq 2 $
  $ => (\frac{a}{b} - \frac{1}{2} ) (\frac{a}{b} -2) \leq 0  => \frac{a}{b} + \frac{b}{a} \leq \frac{5}{2} $
  Using (1) to prove (2)


1. For the following random variables, find the stated probabilities:
X ~ Geom(0.4) and Y ~ Geom(0.6) with X and Y being independent; P(X<Y)

2. During a practise session a rugby player kicks at goal at distances ranging from 25m to 50m from the goal posts. Independently for each kick from a distance of y metres his probability of succeeding is 0.025(60 - y). The coach uniformly at random picks a distance between 25m and 50m for the player to take a kick from. What is the probability that the player is successful with his kick?

Any help would be appreciated!

[hide="number 2"]
I'm not one hundred percent sure this is correct
For any kick the total probability is $ 1$ so the probability that he makes the kick is

$ \displaystyle\int^{50}_{25} 0.025(60 \minus{} y)\,dy \div \displaystyle\int^{50}_{25} 1\,dy \equal{} \dfrac{9}{16}$[/hide]

That's interesting, I obtained the same answer using a different method. Another thing we can observe which I thought is interesting, is this:

If we take the mean distance (that is, $ \frac{25\plus{}50}{2} \equal{} 37.5$) and plug this into our formula $ 0.025(60\minus{}y)$ we get the same answer, namely $ \frac{9}{16}$


inequalities
trigonometry
geometry proposed
geometry

[color=purple] Let a triangle ABC and an arbitrary point M in plane. Prove that :
 $ \frac{MA}{\sin\frac{A}{2}}\plus{}\frac{MB}{\sin\frac{B}{2}}\plus{}\frac{MC}{\sin\frac{C}{2}}\ge\frac{a\plus{}b\plus{}c}{\sqrt{3}}$
       A conjecture of gemath : $ MA\sin\frac{A}{2}\plus{}MB\sin\frac{B}{2}\plus{}MC\sin\frac{C}{2}\ge\frac{6r^2}{R}$
  Hope everybody interested in  :D  :) [/color]

Use $ 2sin{\frac{A}{2}} \geq cos{MAB} \plus{} cos{MAC}$ to prove the second problem.

[color=purple] Thank you treegoner , can you show me the minimum of the following expression :
 $ \frac {MA}{\cos\frac {A}{2}} \plus{} \frac {MB}{\cos\frac {B}{2}} \plus{} \frac {MC}{\cos\frac {C}{2}}$
    And are you interested in the 1st problem :D
[hide] Treegoner, You are Nguyen Lu Khoa aren't you  :D [/hide] [/color]

[quote="treegoner"]Use $ 2sin{\frac {A}{2}} \geq cos{MAB} \plus{} cos{MAC}$ to prove the second problem.[/quote]
You are wrong treegoner, it must be $ 2\cos{\frac {A}{2}} \geq\cos{MAB} \plus{}\cos{MAC}$


AMC
AIME
calculus
AMC 10
USA(J)MO
USAMO
MATHCOUNTS

Alright I am beginning a new game. This is the last one. I promise. I will only let up to 24 people play. If there are less, than we will just play with less. Each round of the competition will consist of 5 parts.

Part 1: The Challenge - Each person will compete against the others in a challenge.
Part 2: The Announcement - I will announce the people that are in the Bad Circle and the people that go in the Good Circle. The half of the players that do the best in the challenge will go to the Good Circle and the rest will go to the Bad Circle.
Part 3: The Choice - Each person will be paired up with another person, and they will be put up against a special challenge. The winner of each pair moves on, and the loser stays. Any person that does not get a partner will compete against one of the people that stays. The winner of this will move on. In the special challenge, it is not guaranteed that one person will win. Under some circumstances both people will win, and other different circumstances neither will win.
Part 4: Bottom Four - Part 3 will keep on going until there are four people left in the losers circle. These four will each compete  in a challenge called "Push". In this challenge, the four of them are standing on an elevator that will take them to the winners circle, but the weight capacity is two people. So any person may post who they would like to push off the elevator. Then another person chooses the second person that they push. The two people that push people off have to be different. If this challenge doesn't finish in 4 days, then all four, except anybody that pushed somebody off, will go to the losers circle. The person that pushed somebody will go to the winners circle.
Part 5: Elimination - The 2-4 people that are left will try to convince the people in the winners circle to keep them in. The people in the winners circle will decide the 1-3 that will come to the winners circle, and the remaining person will be eliminated.

When there are fifteen people left, then it will be a little bit different.

Part 0: Teaming Up - This part only happens once in the whole game. The fifteen people will divide themselves into 3 teams of five people.
Part 1: Exchange - One team will take one of the other team's players, and then that team will take one of theirs.
Part 2: Challenge - They compete in a challenge.
Part 3: Announcement - The team that does the worst will go to the losers circle.
Part 4: The Choice - The team will pick two people to go to part 5.
Part 5: Elimination - Just like the above. The person that does not get eliminated picks the team that they go on.

When there are eight people left, then it will change again.

Part 0: Teaming Up - This part only happens once in the whole game. The eight people will divide themselves into four teams.
Part 1: Challenge - They will play a challenge.
Part 2: Announcement - I will announce the bottom two teams.
Part 3: The Choice - Each team will do this challenge. The losers of this challenge will be eliminated.
    *If nobody looses, then we will do it again. If nobody looses the second time, then it will happen a third time. If nobody looses this time, then they will all be eliminated.
    *If nobody wins. They all get eliminated.
    *If one person from each pair wins, then these two will team up.
    *If both people on one team lose, but one person in the other team wins, then this person will autimatically proceed without having to do anything, until there is another person that doesn't have a partner later in the competition.

When we get to the top four, then there will be no challenge, and they will autimatically go to part 3.

When we get to the top two, then they will each give their statements as to why they should win, and then I will make a poll in the FF, that will last 14 days, and whoever has the most votes at the end will be the winner.

Alright sign-ups begin now. It will not last as long as you think.  :D
    *

Players;

AIME15

Almost there.  :lol:

EDIT; Wait there is another part that I forgot to add. The players get to pick the challenges. You can't do the same challenge twice in a row.

You'll understand it once we begin.

Players;
AIME15
eggylv999
crazychinesemaniac
erabel
mewto55555
Bugi

I hope we get more people than this.

@eggylv999 - it depends on what challenges you guys decide on.

Hmm, why isn't anybody signing up. If you want to play then please sign up, and if you know somebody that might want to play, then tell them about this game.

If nobody signs up for a while, then pretty soon.

No, that is not true.

AoPS City

Umm... I'm sure there is at least one more.

OK. Let's not get off this thread, but really stop making games which quickly fail.

It isn't my fault this game failed. I didn't make a commitment by signing up, then pick a challenge I would not want to do, and then quit the game for no reason.

OK. You are right, I am wrong, sorry now continue with less spam.

Fantasy, why don't we just then ignore those games?

[quote="Wickedestjr"][quote="FantasyLover"][quote="Wickedestjr"]Alright I am beginning a new game. This is the last one. I promise. [/quote]

[quote="Wickedestjr"]Well, sign-ups for a new competition begin now. [/quote]

enough said[/quote]

I don't know if you understand what has happened, but you are the main reason why everybody quit. You made a commitment to a game, and then you helped decide the challenge, and then you quit. Do you see how annoying that seems. This is why games die, because people like you prevent others from playing (you guys made me start this game early), and then you guys quit so nobody gets to play.[/quote]

how is that my fault

its ur fault making the game boring enough for everyone to stop playing

@Bugi

exactly
this is the first long-term game i joined, and the only reason i did was this was supposed to be 'challenging'

You were the very first person to quit, and that is why everyone else decided to quit. I didn't make the challenges. It is not my fault if this game is boring to you guys. If you read the thread (and the rules), then you would know what was going to happen in this game.

even if i hadnt quit, somebody was going to quit eventually

and seriously

who reads instructions and directiosn

Well, there is  probably at least one person. (I only read the first half of directions before stopping.)

Mods, please lock this thread. It is getting out of control.

I'd let you, but unfortunately you would be the only competitor. Unless anybody else would like to play.


ratio
symmetry
geometry
congruent triangles

Starting with two points A and B, some circles and points are constructed as shown in
the figure:[list][*]the circle with centre A through B
[*]the circle with centre B through A
[*]the circle with centre C through A
[*]the circle with centre D through B
[*]the circle with centre E through A
[*]the circle with centre F through A
[*]the circle with centre G through A[/list]
[i][size=75](I think the wording is not very rigorous, you should assume intersections from the drawing)[/size][/i]

Show that $M$ is the midpoint of $AB$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=291[/img]

The triangle $\triangle ABC$ is equilateral, hence $\angle ABC = 60^\circ$. Similarly, the triangles $\triangle BCD, \triangle BDE$ are equilateral, hence, $\angle CBD = \angle DBE = 60^\circ$. Therefore, the angle

$\angle ABE = \angle ABC + \angle CBD = \angle DBE = 60^\circ + 60^\circ + 60^\circ = 180^\circ$

is straight and the points A, B, E are collinear. In addition, AB = BC = BD = BE and B is the midpoint of the segment AE. The congruent triangles $\triangle AEF \cong \triangle AEG$ are isosceles with EA = EF = EG (radii of the circle (E)) and the ratio of the shoulder side to the base is $\dfrac{EA}{AC} = \dfrac{EA}{AB} = 2$. The congruent triangles $\triangle AFM \cong \triangle AGM$ are also isosceles with FA = FM = GA = GM (radii of the congruent circles (F), (G)). In addition, they are similar to the isosceles triangles $\triangle AEF \cong \triangle AEG$, having the base angles at the vertex A common. Thus the ratio of the shoulder side to the base is also equal to $\dfrac{AF}{AM} = \dfrac{AG}{AM} = 2$. But since AF = AG = AB (radii of the circle (A)), we get $\dfrac{AB}{AM} = 2$. The fact that the point M lies on the segment AB follows from symmetry of the congruent circles (F), (G) with respect to the line AB.

[quote="joml88"][quote="jensen"]i think this problem is very cool.i really enjoyed it and it's solution(i mean yetti's solution for it.) :D[/quote]

Please cut down on the essentially useless comments.  It's nice to be polite but reviving a thread just to say that a solution is nice is, in fact, impolite and improper board etiquette.

Thank you.[/quote]
what do you exactly mean joml88.i was polite but i think you are not. :D

He means that you should stop SPAMing!
a) bringing an old thread up again for some reason like yours is SPAM
b) answers of the form 'nice problem', 'nice solution' or similar ones are SPAM too

and responses to spam are by definition also spam. (and even more annoying)

please, next post should be about the problem again ;)

I cannot see the image.
How can I get it?

It was posted more than two years ago. The image URL is most likely a dead link by now.

[quote="Peter"]Starting with two points A and B, some circles and points are constructed as shown in
the figure:[list][*]the circle with centre A through B
[*]the circle with centre B through A
[*]the circle with centre C through A
[*]the circle with centre D through B
[*]the circle with centre E through A
[*]the circle with centre F through A
[*]the circle with centre G through A[/list]
[i][size=75](I think the wording is not very rigorous, you should assume intersections from the drawing)[/size][/i]

Show that $ M$ is the midpoint of $ AB$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=291[/img][/quote]

[color=red][i] A little history:[/i][/color]

 This problem is essentially the [b]Mascheroni  construction [/b]with compasses only of the midpoint of a segment $ AB$.
 I had read this solution many years ago in Kostovski's book ( ed. MIR). How can we find the midpoint of $ AB$ using only a ruler(not graded) and a given circle with known center ? I suggest this challenge for you.

 Babis

Starting with two points A and B, some circles and points are constructed as shown in
the figure:[list][*]the circle with centre A through B
[*]the circle with centre B through A
[*]the circle with centre C through A
[*]the circle with centre D through B
[*]the circle with centre E through A
[*]the circle with centre F through A
[*]the circle with centre G through A[/list]
Show that $M$ is the midpoint of $AB$.
[img]https://cdn.artofproblemsolving.com/attachments/d/4/2352ab21cc19549f0381e88ddde9dce4299c2e.png[/img]


modular arithmetic
algebra
binomial theorem
number theory unsolved
number theory

for odd prime, p, and $a\in\mathbb{Z},n\in\mathbb{Z^{+}}$; show:

$a\equiv 1\pmod{p^{n}}\Leftrightarrow a^{p}\equiv1\pmod{p^{n+1}}$

We can say $a=k*p^{n}+1$, therefore $a^{p}=(k*p^{n}+1)^{p}$ and this is equivalent to $a^{p}=\binom{p}{0}*k*p^{n}+\binom{p}{1}*k*p^{n}+\cdots+1$. All but the last member are divisible by $p^{n+1}$ because they are $\binom{p}{i}*k*p^{n}$ and we know that $p|\binom{p}{i}$. So we can say that $a=s*p^{n+1}+1$ which is equivalent to the solution.

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$\Rightarrow$ case: trivial from $a=1+m\cdot p^{n}$ and binomial theorem.

$\Leftarrow$ case: We have
\[(a-1)(1+a+a^{2}+\cdots+a^{p-1})\equiv 0\pmod{p^{n+1}}\]
If we can show the second factor is not divisible by $p^{2}$, then we'll be done.

To do this, note that $a^{p}\equiv 1\pmod{p^{n+1}}$ implies $a\equiv a^{p}\equiv 1\pmod{p}$. Now write $a=1+mp$. Binomial Theorem gives us
\[a^{k}=(1+mp)^{k}\equiv 1+kmp\pmod{p^{2}}\]
so that
\[1+a+a^{2}+\cdots+a^{p-1}\equiv \underbrace{1+1+1+\cdots+1}_{p}+p(1+2+\cdots+(p-1))\]
\[\equiv p+p^{2}\cdot \frac{p-1}{2}\equiv p\pmod{p^{2}}\]
(because $p$ is odd, $\frac{p-1}{2}$ is integer). So the second factor is not divisible by $p^{2}$.
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Sorry, I didn't see the other direction. Use scorpious solution, it's great.