What's the length of the longest diagonal of a regular hexagon with side length 1?

Is it [hide]sqrt(3)[/hide]?  Oh and since its regular all the diagonals are the same length.

Duck-Billed Platypus wrote:Is it [hide]7/3[/hide]?

No, but how did you come up with that?

[hide]I made a mistake... Revised version

There are two kinds of diagonals--ones that connect points one vertex away, and ones that connect points two vertices away. For example, in hexagon ABCDEF, the first type of diagonal would be AC, and the second type would be AD. A diagonal of the first type makes a triangle with two sides length 1. The measure of an interior angle is 120, so the other two angles are each 30. This triangle makes two smaller ones (30-60-90), with the leg opposite the 90 degree angle having length 1. The length of the side opposite the 30 degree angle is 1/2, and the length of the side opposite the 60 degree angle is  :rt3: /2, so diagonals of the first kind have length 2 :rt3:/3.

The second kind of diagonal divides the hexagon into two trapezoids. You can divide the trapezoid into two triangles and a rectangle with sides length 1 and  :rt3: /3. The two triangles are 30-60-90, so the side opposite the 30 degree angle is 1/3, so the length of the second kind of diagonal is 1 and 2/3, or 5/3.

Comparing 2 :rt3: /3 and 5/3, the /3 part can taken out by multiplying both sides by 3, then divide both sides by 2 to get  :rt3:  and 5/2, or 2.5. Squaring both sides gives 3 and 6.25, so the right side is larger. So the answer is 5/3...

I think I did something wrong, though...[/hide]

[hide]I'm going with 2 also.  Call the hexagon ABCDEF.   The longest is obviously AD or some equivelent one.  If you break up the hexagon into 6 equilateral triangles, then it's immediately apparant that the length is 2.[/hide]

My bad!  Yes, [hide]2[/hide] is the correct answer.  Its all based on finding [hide]30-60-90 triangles[/hide].

[hide]or you could just note that a hexagon can be split into 6 congruent equilateral triangles with their sides equal to the side length of the hexagon. [/hide]

[quote="Duck-Billed Platypus"]The length of the side opposite the 30 degree angle is 1/2, and the length of the side opposite the 60 degree angle is  :rt3: /2, so diagonals of the first kind have length 2 :rt3:/3.[/quote]

This is wrong, and that's what has messed you up.

Argh, my bad. I did it again and found my mistake. Thanks for pointing that out. And I didn't even think about the equilateral triangle thing...

[quote="tokenadult"]I just have to comment that Duck-Billed Platypus wins my award for most useful online signature.[/quote]
Gee, I'm flattered. Thanks.
(I'm not being cynical, if that's what you're thinking...)

gauss202 wrote:nr1337 wrote:Wouldn't it be [hide]2, since the longest diagonal is twice the length of a side[/hide]?

How do you know that?

[hide]

The hexagon can be broken up into 6 equilateral triangles whose length is the length of a side of a hexagon.  Since there are two rows of triangles, the length of the diagonal will be twice the length of a side.[/hide]

This was an easy question imo.

[hide]A hexagon is made up of 6 equilateral triangles.  Since the diameter is always the longest diagonal, we need to find the length of it.  Break the hexagon up into 6 equilateral triangles and you will see, since each side has 1, and the diameter is comprised of 2 of these sides, the diameter's length is 2.[/hide]

NightFlarer wrote:This was an easy question imo.[/hide]

This is getting started.  They need to be at this level.

I only made a 92 on the AMC10 in 9th grade, so I'm a beginner too and I should be in the get started forum.
Actually, I find many of the questions in the get started forum difficult, but this one I happened to find easy lol


thanks, how did you know how to do it so quickly?

[hide]
We can find the probability that none of them is in their proper seat, then subtract from one.
This probability will be the derangements (arrangements of a set where each element is in a different position from its starting position) divided by the total arrangements, which will be 30!.
The derangements of a set of 30 is defined as:
$30! \sum_{k=0}^{30}\frac{ (-1)^{k}}{ k! }$
Then the final probability becomes:
$1-\sum_{k=0}^{30}\frac{ (-1)^{k}}{ k! }$

Might be a way to simplify that... or there might be an easier way to solve the problem  :blush:
[/hide]


What is the maximum product of two integers whose sum is 18?

The maximum product correlates with the smallest difference. For an even number sum, this is easy since then the two integers are just identical. So if we call one of the $ x$, the other is also $ x$. Now, we set an equation to solve for $ x$: $ 2x\equal{}18$, and therefore $ x\equal{}9$. The product would be $ 9(9)\equal{}9^2\equal{}\fbox{81}$.

Or, by AM-GM,

$ \frac{18}{2} \geq \sqrt{ab}$
$ 81 \geq ab$

What do you mean by AM-GM? (What does it stand for?)

Arithmetic Mean - Geometric Mean. The statement is that

$ \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt{a{}_1a{}_2%Error. "hdotsa" is a bad command.
_n}$, for positive $ a_k$.

Equality holds when all the numbers are equal.


We use $ 1\plus{}i \equal{} \sqrt{2} \text{cis} \frac{\pi}{4}$.

[quote="darkdieuguerre"]I haven't seen this terminology before: what is an "acgumen"?[/quote]
I think "argument"  :)

if $ \pi$ is the argument, then the number is just a negative real.  note $ (1\plus{}i)^4 \equal{} \minus{}4$ so we want our answer to have an exponent divisible by 4.  So our answer is $ \boxed{D}$


geometry
geometric transformation
percent
probability

[b]A translation of a translated story.[/b]
[i]Let me try my best to describe the story.[/i]
Solomon was described as [b]the man of wisdom [/b]in the holly Bible. There was a beautiful  Queen, whose name is unknown to me, wanted to verify the wisdom of Solomon, and of course she was keen on Solomon.

The Queen came to Solomon, ordered her servant to poured 20 white pebbles into a jug, and another 20 black pebbles into another jug, with the pebbles of identical size, and then requested Solomon to blind-fold his eyes and pick one pebble from any one of the jugs.

"I will marry you if you can get a black pebble," promised the Queen.

Solomon closed his eyes for a while, then opened his eyes and replied to the Queen," My dear, would you allow me to re-arrange the pebbles in the jugs?"

The Queen pondered," The number of pebbles is half in white and half in black, it seems immaterial to change the distribution of the pebbles. Anyway he has fifty percent of chance to get the black pebble."
"Sure, as you like." The Queen said.

Then Solomon lifted one of the jugs and poured some of the pebbles into another jug. Surprisingly he has increased the chance from 1/2 to a much larger  value!
[i]
Question:
What was the maximum value of the probability that Solomon has attained? And of course, why?[/i]

He was looking at the pebbles when he rearranged them, right?

The Queen allowed Solomon to re-arrange the pebbles freely and of course Solomon didn't close his eyes when he poured some of the pebbles from one jug to the other.  :)

[hide]
Put 39 into one jug, and 1 black one in another.
He measures the weight of the jug, and can tell which one has the black one.
So 100% probable to win.
[/hide]

Yeah, or else this would be in the brainteasers and puzzles section. Anyway, I'll have to think about the answer

[quote="luimichael"]Solomon didn't use any dirty trick.

He simply made the probability of getting a black pebble as large as possible according to the rule of game.

The final probability wasn't 100% but Solomon did win the love of the Queen.

No dirty trick!  :)[/quote]

He could have poured the black marbles on top of the white ones...

That's not dirty...

but he doesn't know which jar is black and which is white. he could feel to see which jug absorbed more heat, because black absorbs more

Hint
[hide]The probability can't exceed 3/4.[/hide]

it depends. if the jug he pours out of is the black jug, his probability can increase to 29/39(average of 1/1 and 19/39). But if the jug he pours out of is the white jug, he's in for a nasty surprise.

[quote="luimichael"]If Solomon did pour out the white pebbles and left th jat jug empty, then his probability could have become $ \frac 1 2 \times 0 \plus{} {\frac 1 2} \times \frac {20}{20} \equal{} \frac 1 2$.
The Queen would not have adored Solomon any more...... :blush:

In fact, Solomon simply poured some black pebbles into the other jub in this way:
[hide]
He poured 19 black pebbles into the other jug and left one pebble in the jug.
The probability become $ \frac 1 2 \times \frac {19}{39} \plus{} \frac 1 2 \times \frac 1 1 \approx 0.7436$
[/hide][/quote]
thats still $ \dfrac{29}{39}$ xP


Prove that $ \lim_{n\to\infty}\frac1n\equal{}0.$

Let $ \epsilon >0$ be given.
Observe that
               $ |\frac{1}{n} - 0|=\frac{1}{n} <\epsilon$
               will hold if ${ N(\epsilon}) = \frac{1}{\epsilon}$ for all ${ n >N(\epsilon})$.


Suppose that $ S, T \in L(V)$ are such that $ ST \equal{} TS$. Prove that null$ (T \minus{} \lambda I)$ is invariant under $ S$ for every $ \lambda \in F$.

Choose $ \lambda.$ Suppose $ v\in\text{nul}(T\minus{}\lambda I).$ Then $ Tv\equal{}\lambda v.$

$ T(Sv)\equal{}TSv\equal{}STv\equal{}S(Tv)\equal{}\lambda (Sv).$ Hence $ Sv\in\text{nul}(T\minus{}\lambda I).$

That is precisely what we were asked to prove.

By the same proof, $ \text{nul}(S\minus{}\lambda I)$ is invariant under $ T.$ The summary is that if $ S$ and $ T$ commute, then they have the same eigenvectors.


geometry
circumcircle
geometric transformation
reflection
incenter
Ross Mathematics Program
Euler

Let $ ABC$ be a triangle. Let $ X$, $ Y$, $ Z$ be the midpoints of the arcs $ BC$, $ CA$, $ AB$ (not containing the points $ A$, $ B$, $ C$) on the circumcircle of triangle $ ABC$, respectively. Let $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ be the reflections of the points $ X$, $ Y$, $ Z$ in the sidelines $ BC$, $ CA$, $ AB$, respectively. Denote by $ F$ the circumcenter of triangle $ X^{\prime}Y^{\prime}Z^{\prime}$. Let $ I$ be the incenter of triangle $ ABC$, and $ N$ the center of the nine-point circle of triangle $ ABC$. Prove that the point $ N$ is the midpoint of the segment $ IF$.

[i]Remark.[/i] This can be solved by complex numbers.

[quote="sam-n"]Let $ ABC$ be a triangle. Let $ X$, $ Y$, $ Z$ be the midpoints of the arcs $ BC$, $ CA$, $ AB$ (not containing the points $ A$, $ B$, $ C$) on the circumcircle of triangle $ ABC$, respectively. Let $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ be the reflections of the points $ X$, $ Y$, $ Z$ in the sidelines $ BC$, $ CA$, $ AB$, respectively. Denote by $ F$ the circumcenter of triangle $ X^{\prime}Y^{\prime}Z^{\prime}$. Let $ I$ be the incenter of triangle $ ABC$, and $ N$ the center of the nine-point circle of triangle $ ABC$. Prove that the point $ N$ is the midpoint of the segment $ IF$.[/quote]

In

  Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], Washington 1995,

Chapter 6, it is proven that the points X', Y', Z' lie on the so-called [b]Fuhrmann circle[/b] of triangle ABC. This is the circle with diameter HM, where H is the orthocenter and M is the Nagel point of triangle ABC. Hence, the center of this Fuhrmann circle is the midpoint of the segment HM. In other words, the circumcenter F of triangle X'Y'Z' is the midpoint of the segment HM. Now, if we consider the centroid G and the circumcenter O of triangle ABC, then we know that the points H, G, O, N lie on one line (the Euler line of triangle ABC), that HG / GO = 2, and that the point N is the midpoint of the segment HO. We also know that the points M, G, I lie on one line (the 2nd Euler line alias Nagel line of triangle ABC), and that MG / GI = 2.

Since HG / GO = 2 = MG / GI, we have HM || OI, or, in other words, HF || OI.

Now let L be the reflection of the point H in the point O. Then, the point O is the midpoint of the segment HL. On the other hand, the point F is the midpoint of the segment HM. Hence, the line FO is a midparallel in triangle MHL; it follows that FO || ML.

Using directed segments, we have LO = OH (since L is the reflection of H in O). Also, we have 2 OG = GH (since HG / GO = 2). Thus,

  LG = LO + OG = OH + OG = (OG + GH) + OG
     = 2 OG + GH = GH + GH = 2 GH.

Thus, LG / GH = 2 = MG / GI, and it follows that ML || IH. Together with FO || ML, this yields FO || IH. And this, combined with HF || OI, shows that the quadrilateral HFOI is a parallelogram. In a parallelogram, the diagonals bisect each other. Thus, the diagonals HO and IF of our parallelogram HFOI bisect each other, i. e. the midpoint of the segment HO coincides with the midpoint of the segment IF. But the midpoint of the segment HO is the point N; thus, N is the midpoint of the segment IF, qed..

  Darij


Ring Theory
superior algebra
superior algebra unsolved

suppose the ring R is abelian.if I be a primary ideal show $ %Error. "display" is a bad command.
\sqrt {I}$={x$ %Error. "display" is a bad command.
\in$ R $ %Error. "display" is a bad command.
\mid$ exist  m $ %Error. "display" is a bad command.
\in$ $ %Error. "display" is a bad command.
\mathbb{N}$ $ %Error. "display" is a bad command.
\ x^{m}$ $ %Error. "display" is a bad command.
\in$ I} is a prime ideal in ring  R.

This is really only working with definitions.


counting
derangement
combinatorics open
combinatorics

I just caome up with such a permutation problem below:

Given an ordering of $n$ numbers $(i_{1},\i_{2},\ldots i_{n})$. how many number of ways are there to reshuffle the $i$'s in such as way that $r$ out of those $n$ numbers must be in the same line, i.e. we require the fixed piece $(i_{j_{1}},i_{j_{2}},\ldots i_{j_{r}})$ to be a part of the ordered tuple $(i_{1},i_{2},\ldots i_{n})$ above  :(

Is this different than the question, "How many permutations of an $n$-set are there with exactly $r$ fixed points?"?  Because then the answer is just $\binom{n}{r}D_{n-r}$, where $D_{k}$ is the $k$-th derangement number.


invariant
induction
function
geometry
geometric transformation
group theory
combinatorics proposed

In a word formed with the letters $a,b$ we can change some blocks: $aba$ in $b$ and back, $bba$ in $a$ and backwards. If the initial word is $aaa\ldots ab$ where $a$ appears 2003 times can we reach the word $baaa\ldots a$, where $a$ appears 2003 times.

I think the answer is "no", but I'm really not sure about the solution.

If the first word can't be turned into the second word by a series of transformations which include the initial transformations, then it can't be done with the given transformations either.

I considered the transformations: $bb\leftrightarrow e,\ aaa\leftrightarrow e,\ aba\leftrightarrow b$, where $e$ is th identity, i.e. the null word. As you can already see, I considered $a,b$ as elements of a group with identity $e$ s.t. $a^3=b^2=e,\ bab=a^2=a^{-1}$. I consider two words equal if they can be obtained from one another by the three given transformations, i.e. if their corresponding elements in the group are equal.

We need to show that the two given words aren't equal. We consider the smallest group generated by $a,b$. This is $S_3$. With this new, extended set of transformations the two given words are equal iff $aab=baa$, because $a^{2001}=e$, so it can be eliminated by the transformation $a^3\rightarrow e$. Take $a=\left (\begin{array}{ccc}1&2&3\\ 2&3&1\end{array}\right ),\ b=\left (\begin{array}{ccc}1&2&3\\ 2&1&3\end{array}\right )$. It's easy to see that $aba=b$ and $a^2b\neq ba^2$.

I think this solves the problem, but I'm still not sure about the method. I keep thinking that maybe I'm missing something (I'm not sure about the equivalence: two words can be obtained from one another $\Leftrightarrow$ the corresponding elements of the group are equal). What do you think?

Yes, your solution is correct. The original proof is very simple: just note that the number of $a$'s on odd positions remains invariant modulo 2 under the given transformations, whereas this is not the case in the given two words.

My own solution goes as follows. We will prove that, for any $n \geq 1$, it is
not possible to obtain $ba^n$ from $a^nb$. Consider the group $G$
with presentation
\[
G = \langle x,y \, | \, y^2 = (xy)^2 = 1 \rangle.
\]
In this group, we have the relations
\[
y^2x = x, xyx = y.
\]
Assume now that $ba^n$ can be reached from $a^nb$ after finitely
many transformations of the form $aba \leftrightarrow b$ or $a
\leftrightarrow bba$. Then, $yx^n = x^ny$, because the relations
$xyx = y$ and $x = yyx$ are true in $G$. Thus,
\[
yx^n = x^ny.
\]
By performing a right multiplication by $x$ to this identity, we
have
\[
yx^{n+1} = x^nyx = x^{n-1}(xyx) = x^{n-1}y.
\]
If we multiply from the right by $x$ again, we get
\[
yx^{n+2} = x^{n-1}yx = x^{n-2}(xyx) = x^{n-2}y.
\]
By induction, we have $yx^{n+k} = x^{n-k}y$ for all $k \geq 0$,
so, for $k=n$, we get
\[
yx^{2n} = y \, \Leftrightarrow \, x^{2n}=1.
\]
This shows that the order of $x$ is finite.

On the other hand, we will show that the presentation (1) directly
implies that $|x|= \infty$, which will finish the proof by
obtaining a contradiction. Let us make a change of basis by
letting $g=y$ and $h=yx^{-1}$. Then, $G$ also has the presentation
\[
G = \langle g,h \, | \, g^2 = h^2 = 1  \rangle.
\]
A word of the  form $ghgh \cdots gh$ cannot be reduced to a word
of smaller length by using the generator relations $g^2=h^2=1$.
Therefore, for each $k \in \NN$, $x^{-k} = (gh)^k \neq 1$, which
means that the order of $x$ is infinite. This obtains the desired
contradiction, and the proof is complete.

In fact, this solution (together with grobber's and the original as well) shows, more generally, that, for any word $v$, we cannot reach $va^nb$ from $vba^n$.

i guess my group theory isn't strong - how do you change the basis? I don't understand, what conditions and steps are needed? Also I don't understand the first proof oops.

Please, write for whom your question is addressed. :?

well, just anyone who passes by and tell me about bases.... as this thread is very old :?

change of basis. why is $h^2=e$ (from the second group proof) and$bab=b$ (from the first group proof)? unless I am just missing something oops.

I did not read the second proof, and regarding the first proof, nobody said $bab=b$. :?

For the second proof, we have $h^2=(yx^{-1})^2 = (y^{-1}x^{-1})^2=(xyxy)^{-1}=1$, where the second equality follows from $y^2=1$ and the last from $(xy)^2 = 1$.

And in the first proof, if what you're asking is how you get from $aba=b$ to $bab=a^2$, just notice that $aba=b\iff (ab)^2=1\iff aba=b^{-1}=b$.

[quote="Myth"]Classical russian solution.
Let $a(x)=x+1$ and $b(x)=-x$. Then $a\circ b\circ a=b$ and $b\circ b\circ a=a$; moreover $a\circ...\circ a\circ b(x)=-x+2003$ and $b\circ a\circ...\circ a(x)=-x-2003$.
That's all.[/quote]

How did you find that pair of functions a,b?  (Unless this one was "well known in Russia").

[b]Added in editing:[/b] actually it may not be so hard.  You just need to look for a representation of aba = b  and   bba = a,  there is obviously a finite dimensional example (by setting various other words to 0) and if dimension is 2 or 3 you should get something like the above.  Is that how you did it?

I was searching for geometric interpretation, but the problem is so straightforward, that geometrical transformations I found are simply translation and simmetry, hence I wrote them in a functional way.
I remember I was solving similar problem many years ago, and my supervisor told me about this possible approach.

Saw a proof along these lines a while ago, so I thought I might post a write up here.  A truly fantastic problem, although I do believe that this should have been more user friendly to those who don't know any group theory yet.

We claim the answer is [b]no[/b].  To do this though, we will need to impose even more conditions than what was given in the problem statement.  This is perfectly fine though, since if we can prove the claim impossible using more restrictions, we can certainly disprove the original claim as well.  To wit, we are proving something stronger that will be significantly easier to prove so don't get mad about what we're about to do :)

To begin, we add inverses.  We see that
\begin{align*}
bba=a&\implies b^2=e \\
b=aba &\implies  e=abab^{-1}.\\
\end{align*}
So we can construct the group presentation $G=\langle a,b~|~abab^{-1},b^2\rangle=\langle a,b~|~abab,b^2\rangle$ since $b^2=e\implies b=b^{-1}$.  Finally, we add the relation $a^n=e$ to $G$ to get
$$G=\langle a,b~|~abab^{-1},b^2,a^n\rangle.$$
Thus it suffices to observe whether or not the equation
$$\underbrace{a\ldots ab}_{2003~\text{many times}}=\underbrace{ba\ldots a}_{2003~\text{many times}}\implies a^{2003}b=ba^{2003}$$
is true under this presentation.  Since this presentation behaves like the Dihedral Group, we can pass $a$ through $b$ to get
$$a^{2003}b=ba^{2003}\implies ba^{-2003}=ba^{2003}\implies b=ba^{4006}\implies e=a^{4006}.$$
To prove that this is impossible, consider $D_3$ (the motivation here is that $n$ isn't a divisor of $2003$).  The Dihedral Group $D_3$ has generators $a$ and $b$ and satisfy all $4$ conditions given in the original problem statement. Hence, if any two words are equivalent under the relations given in the problem statement, they must create the same words in $D_3$. This means we get
$$a^{2003}b=a^2b\in D_3~\text{and}~ba^{2003}=ba^2=ba\in D_3$$
but we see that $a^2b\ne ba$ in $D_3$ which means that the condition proposed is impossible $\square$


What is the 20th digit of the sum of the decimal equivalents for the fractions $ \frac{1}{7}$ and $ \frac{1}{3}$?

The sum $ \frac{1}{7}\plus{}\frac{1}{3}\equal{}\frac{10}{21}$ has a decimal equivalent of $ 0.\overline{476190}$.  The 20th digit is also the 2nd digit of this string which is $ \boxed{7}$.


group theory
superior algebra
superior algebra unsolved

Let (G,*) be group and $a,b \in G$ such that $a^{2}=e$ and
$a*b^{4}*a=b^{7}$
(i) Prove that $b^{33}=e$
(ii) Find the order of the element b.

By $a^{2}=e$, we have that $a=a^{-1}$ and hence $ab^{4}=ab^{7}$ and $b^{3}=e$ hence the order of $b$ is 3 and of course $b^{(3\times 11)}=e$.

On #2: any of 1,3,11,33 is possible. Examples, all in $G=D_{1}1\times \mathbb{Z}_{3}$ ($D_{1}1$ is the dihedral group with 22 elements).
Let the generators of $G$ be $x,y,z$ with $x^{2}=y^{11}=z^{3}=e$.

$a=x,b=e$ (order 1)
$a=x,b=z$ (order 3)
$a=x,b=y$ (order 11)
$a=x,b=yz$ (order 33)

For part 1, I offer this hint: Look at the cyclic group generated by $b$, and the action of $a$ by conjugation on it.

For $(i)$ i think we can do like this:

$ab^{4}a=b^{7}\Rightarrow b^{4}=ab^{7}a \Rightarrow b^{11}=ab^{11}a$.
Then $b^{15}=ab^{18}a \Rightarrow b^{30}=ab^{36}a \Rightarrow ab^{36}ab^{3}=b^{33}=ab^{33}a \Rightarrow b^{3}ab^{3}=a \Rightarrow (b^{3}a)^{2}=e(*)$
So $abb^{3}ab^{3}a=ab \Rightarrow (ab^{4}a)b^{3}a=ab \Rightarrow b^{10}a=ab \Rightarrow b^{10}=aba \Rightarrow b^{30}=ab^{3}a \Rightarrow b^{33}=ab^{3}ab^{3}=e$.

small solution

$ab^{4}a=b^{7}\Rightarrow b^{4}=ab^{4}b^{3}a$
but $a^{2}=e$ so if we stick one e in the middle there we get
$ab^{4}a \cdot ab^{3}a=b^{7}ab^{3}a$
but also derived from first relation we have $ab^{4}=b^{7}a$
so now if we put it in the last relation we get
$b^{7}a^{2}=b^{7}ab^{3}a$ so now we have after multiplying with inverses:
$a^{2}=ab^{3}a$ so $ab^{3}a=e$,now combining first relation with this we have
$ab^{4}aab^{3}a=b^{7}$ so $ab^{7}a=b^{7}$.
coupling this with what we have in the first line we find
$b^{4}=b^{7}$ so $b^{3}=e$
ofcourse now $b^{33}=e$

ottem have in mind that what you wrote there about ord(b)=3 works only because 3 is prime

See my examples: the order of $b$ is not necessarily three, and can be any of $1,3,11,33$.

jmerry if $b^{n}=e$ that does not make n the order of b.
actually the order of b is a divisor of n or n itself.
so how do you think that if $b^{3}=e$ than there could be some
k>3 so that $b^{k}=e$ ?!?!? wouldn't it contradict the definition?

My definition of order includes being minimal, and my examples contradict the statement that $b^{3}=e$; in my third and fourth examples, $b^{3}\neq e$.

When I wrote $x^{2}=y^{11}=z^{3}=e$, I meant that those exponents were minimal. The other relations defining that group are that $xz=zx,yz=zy$, and $xyx=y^{-1}$.


superior algebra
superior algebra theorems

I need your help. I need some info (e-material) on the "Ideal class group" topic (perhaps something aside from what you can find with google in the first few pages, i.e. [url=http://en.wikipedia.org/wiki/Ideal_class_group]wiki[/url] & [url=http://planetmath.org/encyclopedia/IdealClassesFormAnAbelianGroup.html]planet math[/url] definitions).

Thanks.


Solve the following equation for real and finite negative integer
values of $ x$, $ y$ and $ z$.

$ (\frac{(x^4 \minus{} 1)}{z^6})^3 \plus{} (\frac{(z^2 \minus{} 4)}{y})^2
\plus{} (y^2 \minus{} 9)^4 \equal{} 0$

Since x,y,z are integers (if i understood the problem correctly) x^4>=1 and all of the summands are positive or zero (since they are raised to an even power).But since they add up to  zero they are all zero, so $ (x^4\minus{}1)/z^6\equal{}0\equal{}>x^4\equal{}1\equal{}>x\equal{}\minus{}1$ and similarly we get that $ z^2\equal{}4\equal{}>z\equal{}\minus{}2$ and $ y^2\equal{}9\equal{}>y\equal{}\minus{}3$ since all of them are negative
So $ (x,y,z)\equal{}(\minus{}1,\minus{}2,\minus{}3)$


if α and β are the roots of ax^2+ bx+c=0, form the equation whose roots are α^2+ β^2 and α^(-2)+ β^(-2)

Please help

I still dont know what to do. the answere needs to be with a, b, c not β,α

[hide="Solution"] Since $x^{2}+\frac{b}{a}x+\frac{c}{a}= (x-\alpha)(x-\beta)$ we have

$\alpha+\beta =-\frac{b}{a}$
$\alpha \beta = \frac{c}{a}$

Then

$\alpha^{2}+\beta^{2}= (\alpha+\beta^{2})-2 \alpha \beta = \frac{b^{2}}{a^{2}}-2 \frac{c}{a}= \frac{b^{2}-2ac}{a^{2}}$

$\alpha^{-2}+\beta^{-2}= \frac{\alpha^{2}+\beta^{2}}{ \alpha^{2}\beta^{2}}= \frac{ \frac{b^{2}-2ac}{a^{2}}}{ \left( \frac{c}{a}\right)^{2}}= \frac{ b^{2}-2ac}{c^{2}}$

So if we want a polynomial with those roots, it needs to be

$x^{2}-px+q$

Where

$p = \frac{b^{2}-2ac}{a^{2}}+\frac{b^{2}-2ac}{c^{2}}$
$q = \frac{b^{2}-2ac}{a^{2}}\cdot \frac{b^{2}-2ac}{c^{2}}$  [/hide]


algebra
polynomial
number theory unsolved
number theory

For each prime number $ p$, let $ S$ be the set of squares modulo $ p$ (i.e. the numbers $ k$ such that there is $ x$ such that $ x^2\equal{}k(\mod p)$). Let $ T$ be the set of nonsquares modulo $ p$. (Here we consider everything $ \mod p$, so the numbers that can be in $ S,T$ are only $ 0,1,2,...,p\minus{}1$.

(a) Evaluate $ \prod_{x\in S}(x\plus{}1)$ for each $ p\equal{}3\mod 4$
(b) Evaluate $ \prod_{x\in T}(x\plus{}1)$ for each $ p\equal{}1\mod 4$

a)  The polynomial with roots $ x \plus{} 1, x \in S$ is $ (x \minus{} 1)^{ \frac{p\minus{}1}{2} } \minus{} 1 \bmod p$, which has constant term $ \minus{}2$, so this product is $ 2$.

b)  The polynomial with roots $ x \plus{} 1, x \in T$ is $ (x \minus{} 1)^{ \frac{p\minus{}1}{2} } \plus{} 1 \bmod p$, which has constant term $ 2$, so this product is $ 2$.

[quote="nguyentrang"]For each prime number $ p$, let $ S$ be the set of squares modulo $ p$ (i.e. the numbers $ k$ such that there is $ x$ such that $ x^2 = k(\mod p)$). Let $ T$ be the set of nonsquares modulo $ p$. (Here we consider everything $ \mod p$, so the numbers that can be in $ S,T$ are only $ 0,1,2,...,p - 1$.

(a) Evaluate $ \prod_{x\in S}(x + 1)$ for each $ p = 3\mod 4$
(b) Evaluate $ \prod_{x\in T}(x + 1)$ for each $ p = 1\mod 4$[/quote]
$ \prod_{x\in A}(1+x)=1+\sigma_1(x\in A)+\sigma_2(x\in A)+...+\prod_{x\in A}x$.
In these case $ \sigma_i(x\in A)=0, \ \ 1\le i<\frac{p-1}{2}.$
Therefore $ \prod_{x\in A}(1+x)=1+M_A, \ M_A=\prod_{x\in A}x$.
If $ x\in A$, then $ x^{-1}\in A$. $ x=x^{-1}\to x=\pm 1$.
Therefore for a) and b) $ 1+M_A=2$.


calculus
integration
function
linear algebra
calculus computations

My textbook has now tried to explain to me what integration with substitution is. But unfortunately i don't feel like i get the point of it. I find it very complicated. If you could show me how to make the first steep in this integral i can compare it to the examples in the textbook and see if i can find any logic in it.

$\int_1^2(\frac{1}{x\sqrt{x}})dx$

Shouldn't i use substitution here or what do you think? I rewrote it first of as it might be easier in this way:

$\int_1^2(\frac{1}{x^{1.5}})dx$

And i also have problems with partiel integration. Here is where i get to:

$\int_0^1(4^x\cdot 3^x)dx=[4^x\cdot \frac{1}{ln(3)}\cdot 3^x]-...$

Integration by parts is one of the most important tools in analysis. You pick two functions $u(x)$ and $v(x)$ so that $u \frac{dv}{dx}$ equals the integrand, and write your integral as

$I = \int udv$

(with bounds if you want, it hardly matters).

Note that $\frac{d}{dx}uv = u\frac{dv}{dx} + \frac{du}{dx}v$, or in shorthand $d(uv) = udv + vdu$. So

$I + \int vdu = \int d(uv)$

from which we obtain

$I = [uv] - \int vdu$

I'm absorbing the constant of integration into the integrals we haven't evaluated yet. If you have bounds on the original integration, instead evaluate $[uv]$ at those bounds (and subtract the lower from the upper, as usual). It should be clear why you have to do that from the derivation of the formula.

In your example, you'd actually end up with an expression like
$I = C_1 - C_2I$

(where $C_1$ and $C_2$ are real numbers.) This is a perfectly legitimate way to solve the problem, but it seems simpler to me to note that $3^x4^x = 12^x$ and then you're just integrating one thing. The sort of situation where it is best to integrate by parts is when you have a power of $x$ times something you know you can integrate; then you can reduce the power of $x$ until you have something you can deal with. Try, for example, the integral of $x^5e^{x^3}$. (edit: of course, there are many more complicated and interesting situations where you'd use it, but there's no way I could list them all and the power of $x$ is a good simple example.)

But again, integration by parts is really important. I didn't realize this when I first learned it, but now that I've studied some higher level analysis I am amazed by how often you get to use it. My linear algebra professor this semester told us that if we ever have a problem and we don't know how to approach it, "First, draw a diagram. Then, integrate by parts."

You're not being clever enough -- both of these problems are easy integrals, disguised a little to make your life more difficult.

[quote="Sporally"]I rewrote it first of as it might be easier in this way:

$\int_1^2(\frac{1}{x^{1.5}})dx$[/quote]
It would be even easier if you re-wrote it $\int_1^2 x^{-\frac32}dx$.

[quote="Sporally"]And i also have problems with partiel integration. Here is where i get to:

$\int_0^1(4^x\cdot 3^x)dx=[4^x\cdot \frac{1}{ln(3)}\cdot 3^x]-...$[/quote]

As Xevarion points out, there is no reason for you to use integration by parts here (although, as he also said, it will work provided you do it correctly -- you might want to write out explicitly what your $u$ and $v$ are in order to make it easier to apply).

Of course, why didn't i see that. I know $\frac{1}{x}=x^{-1}$, and so i should know how to rewrite it into a simple function with a $x$ and a power. But ok, honorstly didn't know that $3^x\cdot 4^x=12^x$, but now i know :)


let a,b,c>0 show that: a+b+c=3
prove that: $ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2} \geq a^2\plus{}b^2\plus{}c^2$


invariant
geometry
geometric transformation
function
real analysis
real analysis unsolved

Can it be proven that there doesn't exists measure $ \mu : \mathcal{P}(\mathbb{R})\rightarrow \overline{\mathbb{R}}$ , invariant to translation, that is $ \mu (A+c)=\mu (A) , A\in \mathcal{P}(\mathbb{R}),c\in \mathbb{R}$, without(!!) using axiom of choice.

You forgot to include the condition that the measure of some interval be finite. For the way you stated it, counting measure measures all sets and is translation invariant.

Thank you for the answer , but i realized that I wasn't clear with my question.

So, to each set $ A\subseteq \mathbb{R},A \rightarrow \mu (A)$, where $ \mu (A)\in \mathbb{R}$, such that function $ \mu : A\rightarrow \mu(A)$ , has this properties:
$ \forall A\in\mathbb{R},\mu (A)\geq 0, \mu(\emptyset)=0, \{A_{n}\}_{n\in\mathbb{N}}\subseteq \mathbb{R}, A_{i}\cap A_{j}=\emptyset,i\neq j , \mu (\cup_{n\in\mathbb{N}}A_{n})=\sum_{n\in\mathbb{N}}\mu(A_{n})$
for $ A\subseteq \mathbb{R},c\in\mathbb{R},\mu(A+c)=\mu(A),\mu(-A)=\mu(A)$, and
$ a\leq b,a,b\in\mathbb{R},\mu([a,b])=b-a$.
Then we have showed that there doesn't exists, function ${ \mu : \mathcal{P}(\mathbb{R}})\rightarrow \overline{\mathbb{R}}$, that satisfies these properties and in proof we used axiom of choice. But so far, I haven't seen any proof  that didn't use axiom of choice....

As far as I remember, if you remove AC, there is a model of ZF in which all subsets of $ \mathbb R$ are Lebesgue measurable, so some form of AC is necessary to prove the existence of non-measurable sets. Unfortunately, I cannot give a precise reference to this result but I'm 99% sure my memory doesn't betray me here.

[quote="fedja"]As far as I remember, if you remove AC, there is a model of ZF in which all subsets of $ \mathbb R$ are Lebesgue measurable[/quote]

Yes.  See http://en.wikipedia.org/wiki/Axiom_of_choice#Stronger_forms_of_.C2.ACAC .

Thank you! That was exactly what I was looking for.But I have another question, or statement to prove :

Prove that there doesn't exists a finite(!) aditive function $ \mu : \mathcal{P}(\mathbb{R}^{3})\rightarrow \overline{\mathbb{R}}$, that satisfies properties in the last post.

Edit: Again, I wasn't very clear ; it satisfies all of the properties and $ \mu([a_{1},b_{1}]\times[a_{2},b_{2}]\times[a_{3},b_{3}])=\prod_{i=1}^{3}(b_{i}-a_{i})$


Stie cineva exemplu de corp cu 25 de elemente? :o  :ninja:

Da, la nivel elementar, fie $A=\mathbb{Z}_5\times \mathbb{Z}_5$, cu legile:
\[(a,b)+(c,d)=(a+c,b+d)\]
\[(a,b)\cdot (c,d)=(ad+bc-ac,bd-ac)\].
Are 25 de elemente si e corp (poti verifica ca nu are divizori ai lui zero, si fiind inel finit...).
La nivel mai putin elementar, exemplul este, de fapt, obtinut prin factorizarea inelului $\mathbb{Z}_5 [X]$ la idealul generat de polinomul ireductibil $f=X^2+X+1$, dar asta e deja mai complicat de explicat. Incearca sa postezi (in engleza) in College playground-Superior algebra daca vrei detalii.

d-l  B.Enescu,puteti,va rog, sa dati mai multe detalii referitor la treaba cu idealul,ca m-ati facut chiar curios.

E complicat de dat, pe scurt, detalii. Totusi: multimea $\mathbb{Z}_5[X]$ e un inel, cu proprietati "asemanatoare" inelului $\mathbb{Z}$. Daca $f$ este un polinom ireductibil in $\mathbb{Z}_5[X]$ , el joaca un rol similar celui jucat de un numar prim $p$ in $\mathbb{Z}$. Astfel, factorizarea inelului  $\mathbb{Z}[X]$ la $f$ este similara obtinerii corpului $\mathbb{Z}_p$ prin factorizarea lui $\mathbb{Z}$ la idealul generat de numarul prim $p$ (adica $p\mathbb{Z}$).
Pentru detalii, citeste un curs de algebra.


Create a function whose graph has a vertical aymptote at x=0 and a slant aymptote at y=x-1

$y = x - 1 + \frac{1}{x}$

Is this just a homework problem of yours?  You should probably be working them out yourself.  And this is almost definitely in the wrong forum.


ARML
geometry
email
analytic geometry
AMC
AIME
USA(J)MO

Every ARML team is run a little differently than every other ARML team. Each team represents a particular geographic area. So how do you find out who runs the team for your area, and what do you have to do to join a team? This information is available from the ARML organization (http://www.arml.com) if you ask, but I though it would be useful to create a topic here with that information. If you know, and especially if you are one of the organizers, post that information here.

If you live in Southern California (including Santa Barbara and the Inland Empire but not including San Diego County) and would like to join an ARML team, here's what to do:

1. Sent a private message through AoPS to me, Kent Merryfield. Tell me who you are and where you're from. There is a hidden "Southern California ARML" forum in the National and Local Communities section. I will give you access to that. You could also use my email address, kmerry@csulb.edu.

2. Take the AMC exams next February. If your school doesn't offer them, see if you can get your school to particpate; if that doesn't work, we'll try to find a place for you.

3. Attend practices. We have practices in Long Beach throughout the school year, approximately once a month, usually on Saturday afternoons. The next practice will be held in September and October. I will announce them on our website [url=http://www.csulb.edu/depts/math/arml/index.htm]here[/url]. That website includes directions to the location on the CSULB campus.

3a. If you live in or near Santa Barbara, attend the practices run by Skona Brittain. Send me a message, and I will put you in contact with her.

3b. There is also a Los Angeles Math Circle, run by Ely Dahan, that meets on the UCLA campus. He and I stay in contact. I can give you his contact information, if you ask.

4. if your school sends a team to [url=http://www.csulb.edu/depts/math/mathday/index.htm]CSULB Math Day at the Beach[/url], do everything you can to convince your teacher/sponsor that you belong on your school's team. If your school doesn't send a team, see if you can talk someone into sending one; ask me for advice about that.

New York City Math Team: Mr. David Linker of the City College of New York is in charge of the team, and I am currently the captain of the team

Taking AMC/NYCIML is recommended.

feel free to PM me for more information :)

If you live in Iowa... just ask the guy in charge of ARML.
Usually we can't get enough people to even form three teams... :(

[quote="The Original Pi Guy"]... just ask the guy in charge of ARML.[/quote]
Can you make the identification a little more precise than that? I'd like to make this topic useful, which would include offering some help to someone in Iowa who [i]doesn't[/i] know who you mean when you say that.

[quote="The Original Pi Guy"]Usually we can't get enough people to even form three teams... :([/quote]

don't worry. Washington State do not even have a 2nd team  :(

By the way, for next year, if you live in Washington State and you want to be in ARML, you should talk to the Hesterbergs. Although I do not know if they will coordinate again next year, it would still be a good thing to let them know. Also I advise you to take AIME and AMC, as those are probably how we picked the team this year.

For Indiana, its John Rickert from Rose Hulman

Indeed.  As a coach working under Linker's supervision, he usually chooses the members based on AMC and the IML, but he can also look at other contests as well.  AIME is also a good indicator as well.

If any of the seniors are interested in coming back as a coach, like I did, talk to him too.  I'm sure that he will be able to accept all of the help he can get.

If you're in Minnesota, do the AMC and the state high school math league.  And do really well.

If you are in Michigan, do well on the MMPC. Otherwise, talk to prof. Ruth Favro of LTU. Join the Michigan forum for some details.

[quote="chess64"]jack_po.[/quote]
he's a very nice person, [hide="except"]when I lose Burger King slips for lunch :)[hide]haha I'm only kidding, Jack[/hide][/hide]

First off check with your school to see if they know anything, or at least if they could find out. Secondly check with mathematical friends in your area. If they know nothing, try looking on the stats list for ARML 05 to see if there was a team from your nearby area. If you get information there I figgure you can find contact information with a quick google serch. If there isnt a nearby team and you really want to go, try to get a team togeather, and if not peraphs see if you can get on a friends team or something  :)

Sorry about before. The person in charge of ARML for Iowa is James Kirpes. He teaches at West High School in Iowa City.

[size=150][b]South Carolina[/b][/size]

Usually we have a first round of all state in the fall.  Those who qualify go to the second round test at Irmo.  Then, the coaches determine who made the all state team.  Usually about 60 people are named to all state.  Everyone on the all state team are invited to go to ARML.  Usually we get around 30 to go (33 this year).

The process might change a little this year but I think it should stay roughly the same.  We are planning on holding some practices during the year so contact Patrick Rybarczyk by emailing either prybarczyk@rcsd1.org or ronskimomo@hotmail.com as he is the head coordinator for all of this.

If you can't get a hold of him give me a PM and I might be able to help.  Here is this year's all state website: http://155.225.48.46/statemat/stuf2005/teampage.htm .

I don't think it has to be a region from what I've heard.

[quote="renatzu"][quote="neelnal"]Can individual schools register? Does it have to be a region?[/quote]I believe AAST has its own ARML team.[/quote]
On the whole, I would strongly discourage the idea of single school teams. There are some unique situations, such as Philips Exeter or AAST, but in general I think that single school teams are a bad idea.

I say that as an organizer who helped manage that transition from a region (Southern California) having two or more single-school or near-single-school teams to having one region-wide team. Single school teams are by their nature exculsionary, even if they try not to be. What are the opportunities for students from other schools? (Note: I learned about ARML in the first place in 2000 when jmerry participated as the one "outsider" on the Peninsula South Bay team - so it is possible for such a team to include someone not from that school - but there are still problems.)

High school students with strong interests and abilities in mathematics may well find themselves isolated in their home schools. An ARML team can help break that isolation, by giving such students a community that they might not otherwise have. They should learn from and be inspired by students they might otherwise never have met.

Even those at an exceptional school that is already a community should open that community. It's all right if Vestavia Hills provides the core of the Alabama team or East Chapel Hill the core of the North Carolina team - but those teams should still welcome others - not just welcome them, but actively recruit them. And if the team is called the Alabama team or the North Carolina team, then maybe 10 years from now there will be some different school, one we're not even thinking about at the moment, which will be where the top people come from.

So it is legal to come as a school though right?

My school has a pretty strong math group with a couple USAMOers and mostly close to USAMOers.

[quote="neelnal"]So it is legal to come as a school though right?[/quote]
It is the American [b]Regions[/b] Math League, i.e., teams are supposed to draw on students from a large area.  If your school lies in a region that already has an ARML team, then no, it is not legal.  Even if you do not live in an ARML region (which you probably do, anyhow), it would be probably a very ill-advised idea to do this.  A "pretty strong math team" is probably not enough—you would need one of the best teams in the country.  Besides, every American state that had two or more USAMO qualifiers from the same school last year who were not seniors at the time sent at least one team to ARML last year, so I'm pretty sure that one way or another, you already live in an ARML region.

Boy Soprano II has pointed out the competition side of it. Let me give you an example to gauge the level of competition. Last year our A team (Southern California) included 5 USAMO qualifiers, including two MOPpers and one USAMO winner, three others who certainly qualify as "almost USAMO" (indices > 205), and I think everyone else on the A team had an AIME of at least 6. We finished 14th overall.

In fact, when there were single-school teams from Southern California, those teams only rarely finished in the top 50 teams. It's just too narrow a base.

But even so, it's not the level of competitiveness that I would object to - its the exclusiveness.

PM me, tell me where you're from, and I'll tell you what the situation for established teams is in your territory. And I'll tell you that the ARML officials would both enforce existing territories (that is, not letting you establish a team in a region that already has a team) and strongly discourage any new single-school teams.

If you want to win, regional teams are almost a necessity.  If you just want to go to ARML and have fun, it doesn't matter.  Last year I was on what was certainly the smallest-based ARML team ever: a team drawn from a homeschool group with only 11 high schoolers, so that we needed to take middle schoolers just to fill out a fifteen-person team.  We finished 65th in Division B, but we had fun and it worked out fine.

Are there any FL ARML teams? My stats may are not too amazing but here they are:
8th grade
Top 30 at nationals mathcounts
Those are my only official stats. For AIMEs I can like usually do four of the first five, 2 out of the next five and brute force 1 out of the last five. So 7 :(. I haven't been able to find an official place to do the AMCs, but this year I hopefully will be able to.

[quote="neelnal"]Wow, are you really on in 8th grade bpms?

By the solutions you post in the pre-olympiad and intermediate sections, I thought you were older...[/quote]
I don't post solutions in pre olympiad, and kinda rarely in intermediate.  :huh:
Anyways, yes I"m an 8th grader, I just got started on math competitions early (the first one I went to was county mathcounts in 2nd grade). It got me hooked early.

Well, I can't speak for other states, but in Georgia eighth graders don't get on the team very often.  Unfortunately, MathCounts isn't at a challenging enough level to give an indication of strong ARML performance.  In Georgia, you would have to do well on the AIME or at high school tournaments (in the varsity level, if there is more than one division).

I've run the Southern California team for five years; there's been a B team for four years and a C team for three years. So far, the C team has not been highly selective - it's been more a matter of willingness to participate; and the C team hasn't always been filled. That gives us a total of $15\times5=75$ members of the A teams, $15\times4=60$ members of the B team, and $38$ members of the C team. Those 173 team members (many fewer distinct people) have included seven 8th graders and four 7th graders. Three of the 8th graders were on the B team; the rest of the middle schoolers were on the C team. All three of the 8th grade B teamers have become USAMO qualifiers (one already was) - and one of them moved to North Carolina, so he's not on our team any more.

We've had 23 ninth graders: 5 on the A team (including two of those who had been on the B team as 8th graders), 7 on the B team and 11 on the C team.

Hey if you're in Florida and you want to go to ARML, email the organizer Mr. Lin at lin.chenyu212@gmail.com.

The information in this thread should probably also be in the [url=http://www.artofproblemsolving.com/Wiki/index.php/ARML_teams]wiki[/url]. A few teams have pages with the appropriate information, but most don't.


To get to a letter, the number of ways to do so is the sum of the letters above it.  Thus, we can use Pascal's Triangle to find that the sum is $ \binom{10}{0}\plus{}\binom{10}{1}\plus{}\binom{10}{2}\plus{}\ldots\plus{}\binom{10}{10}\equal{}2^{10}\equal{}\boxed{1024}$.

Or, if you notice that there are two paths for each letter, and then that the path is of length 10, you have $2^10 = \boxed{1024}.$


Prove that there are infinitely many square-free numbers of the form $ n^2\plus{}1$, where $ n$ is a positive integer. (square-free numbers are not divisible by any perfect square other than 1)

I gave this problem some thought and could get nowhere. I would love a solution if you have one!


everybody help me out with this how do you calculate the time constant for charging and discharging of capacitor in a given circuit is the time constant same for bot charging and discharging or different explain with an example and diagram !!!! :D

i can't show a diagram, but if u r still willing to see my explanation, then
[hide]
in any circuit,consider any arbitrary point of time, and label charges on each capacitor, using conservation of charges and sometimes, potential difference b/w plates. then apply kirchoffs laws over loops, to get the desired no. of equations (when taking i=dq/dt, take care of which q u have taken, and i is corresponding to which q) upon integrating and solving, u get an expression for charge, which is generally of the form
$ q\equal{}a\left[1 \minus{} e^\frac{t}{b} \right ]$
in case of charging.
for discharging, similar procedure is followed, except for the limits in the integration are changed.

[/hide]

doubt clarified topic and case closed thanks to the physics genius madness!!!! :D


any1 got question on possibility. I am weak in this area. Question level around AMC 12.
What topics are covered under AMC 12 huh??

The AMC 12 does cover basic probability and counting, including conditional probability and probability using geometry.

What's your question?

I don't have any question.. Could anyone just give me some sample AMC 12 question to try??


\begin{align*}\int_0^1 {x^x \,dx} & = \int_0^1 {e^{x\ln x} \,dx}\quad a = e^{\ln a},\,a > 0 \\
& = \int_0^1 {\sum\limits_{k = 0}^\infty {\frac {{(x\ln x)^k }}{{k!}}} \,dx}\quad e^x = \sum_{k = 0}^\infty\frac {x^k}{k!} \\
& = \sum\limits_{k = 0}^\infty {\frac {1}{{k!}}\int_0^1 {x^k \ln ^k x\,dx} }\quad(\text{interchange the sum by the integral})\end{align*}

How can I align the things what I'm sayin' in every line?

Use & to align formulae, rather than \quad
\begin{align*}\int_0^1 {x^x \,dx} & = \int_0^1 {e^{x\ln x} \,dx} && a = e^{\ln a},\,a > 0 \\
& = \int_0^1 {\sum\limits_{k = 0}^\infty {\frac {{(x\ln x)^k }}{{k!}}} \,dx} && e^x = \sum_{k = 0}^\infty\frac {x^k}{k!} \\
& = \sum\limits_{k = 0}^\infty {\frac {1}{{k!}}\int_0^1 {x^k \ln ^k x\,dx} }&& (\text{interchange the sum by the integral})
\end{align*}

Perfect! Thanks Steve.

One more question: using the eqnarray environment, does not work?

It works but usage of \begin{align} \end{align} is recommanded.


Did anyone ever wonder where some of the old people went?  Tare.... Arne... not to mention some others.

No longer likes the forums anymore????? Busy with school??

i think one of them was having troubles getting along with some ppl on here and was being tormented so he/she left.


How hard were AMC 10/12 from 2000 and 2001? What was the AIME qualify score? I got a 121.5 on 2001 :blush:

[quote="AIME15USAMO"]Did you get a 121.5 on the AMC 10 or the AMC 12?[/quote]

Sorry for not being clear. The AMC 10:blush:

Can you believe it.

Grr, those mistakes.... :mad:

[quote="Ihatepie"]That's not a score to be ashamed of. On AOPS, lots of people treat it as a fail, but that's much better than I would guess 99.9% of the people in the world could do.

Anyways, the 2001 AMC 10/12 are known for being hard.[/quote]

Thanks. Does anyone know what the AIME Qualifying scores were that year?

[url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2001-12a/01-1012stats.html]Here you go.[/url]

What you want is buried in the respective tables somewhere.  The 12 score was an 84, and the 10 score was 116.  Which doesn't make sense, because it's impossible to get a 116.

[quote="dysfunctionalequations"]This was before the point value of an answer that was not filled in was decreased from $ 2.5$ to $ 1.5$.[/quote]

Yeah, I get confused about that sometimes when checking practice test answers :P


search
number theory proposed
number theory

Find all positive integers $ x$ and $ y$ such that $ y$ divides $ x^{2}\plus{}1$ and $ x^{2}$ divides $ y^{3}\plus{}1$

well...
63 is not the sum of two squares so the equation has no solution  :wink:

this fact is well known but for a complete treatment see for example:
http://www.numbertheory.org/pdfs/negative_pell.pdf

Thanks for reply [b]Albanian Eagle[/b] I made mistake This topic should be in Pre Olympiad forum I have edited it now and changed question

 For other readers, initial question asked is there any solution of equation $ x^{2} \minus{} 63y^{2} \equal{} \minus{} 1$ and [b]Albanian Eagle[/b] gave answer to that one.

It's not a good idea to change the problem of the topic. :)  but in any case
this problem has been posted
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2095105677&t=151582
and reposted
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2095105677&t=194248
and probably reposted again.
it's not hard to search. (i just typed Mediterranean  :wink: )

Since $ y|(x^2 + 1)$ and $ x^2|(y^3 + 1)$, there exists $ k, t$ such that $ x^2 + 1 = ky$ and $ y^3 + 1 = tx^2$.
This leads to $ y^3 + 1 = t(ky - 1)$ which is equivalent to $ y^3 - (kt)y + (t + 1) = 0 \quad (1)$. From this, we have $ y$ must be a divisor of $ t + 1$ or $ t + 1 = my$.

Substituting this into $ (1)$, we then have $ y^2 - k(my - 1) + m = 0 \Leftrightarrow y^2 - kmy + (k + m) = 0$.
Hence, due to Viete Theorem, if there is a solution $ y_1 \in \mathbb{N}$ then there is another solution $ y_2 \in \mathbb{N}$ such that:
\[ \{ \begin{array}{c} y_1 + y_2 = km \\
y_1 y_2 = k + m \end{array}
\]
From this, we check that if $ y_1, y_2 \geqslant 2$ then $ \dfrac{km}{k + m} = \dfrac{1}{y_1} + \dfrac{1}{y_2} \leqslant 1$. This means $ km \leqslant k + m \Rightarrow k = 1 \vee m = 1$. If $ k = 1 \Rightarrow x^2 + 1 = y$ and $ y - 1|2 \Rightarrow y = 2 \vee y = 3 \Rightarrow x = 1$ and $ y = 2$. Similarly, $ m = 1$ leads to $ t + 1 = y$ and since $ t|y^3 + 1$, we also end up with $ x = 1$ and $ y = 2$.

If $ y_1<2$ or $ y_2 < 2$ then $ x^2 | 2$. This means $ x = y = 1$ is the only solution.

Hence, we have only 2 solutions of $ (x,y) \in \mathbb{N}^2$: $ (1,1)$ and $ (1,2)$.

P.S. I checked my solution at [url]http://www.mathlinks.ro/resources.php?c=1&cid=69&year=2002&p=1066654[/url] and found out that it 's incorrect. Could somebody please help me change it ? :(


function
limit
quadratics
algebra unsolved
algebra

Let f:R->R be a function for which : 1) f(1)=6 , 2) f(5)=30 , 3) f(x+y)-f(x)=kxy + y^2
a)find k
b)find f'(7)
c)find the type of the function

a) set $x=1$ and
\[\displaystyle f(y+1)-f(1)=f(y+1)-6=y^2+ky\]
\[\displaystyle f(y+1)=y^2+ky+6\]
\[\displaystyle 30=f(5)=f(4+1)=4^2+4k+6\]
\[k=2\]
b)
\[\displaystyle f'(x)=\lim_{y\rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim_{y\rightarrow 0}\frac{2xy+y^2}{y}=\lim_{y\rightarrow 0} 2x+y=2x\]
\[\displaystyle f'(7)=14\]
c)
\[\displaystyle f(y+1)=y^2+2y+6=(y+1)^2+5\]
\[\displaystyle f(x)=x^2+5\]
quadratic

thanks..i suppose i should use more frequently the definition of differentiation..


In the spirit of making a strong Alabama mathematics community I encourage students to post here with your any information you like about yourself including your name, school, and grade.  This will help you all know each other at the competitions around Alabama and at ARML tryouts and practices.

Ryan Muller
Vestavia Hills High School
10th grade

I was with the Alabama ARML team last year, so I've met John and others. I'll be at MAO and probably all of the Alabama tournaments.

Anything else?

Seva Tchernov
Vestavia Hills High School
10th grade

I went to MAO 2005, placed in most of the topics but not as good as I would've liked; I definitely plan on working more on math and one of my goals is to at least get in top 50 of USAMO.

Evan Zhao
10th grade
Vestavia Hills High School

Parker Robbins
10th Grade
Vestavia Hills

All these Vestavia people.

Richard Rusczyk
Class of 1989
Austin High School

John Hu
10th Grade
Grissom High School

I hope to see all of you again at UAT on Saturday.

David Sampson
Grissom (just for math)
8th grade

A lot of Vestavia people around here ;) Who went to the Vestavia tournament Saturday (besides obvious Vestavia people.)?

[quote="bleumoose"][quote="MCrawford"]Fan, what are you doing now that you have graduated?[/quote]

he went to MIT. actually we had two grads go to MIT last year. Lance went too.[/quote]

Good!  I'm jealous.  I wish I'd applied there.

Hyun Woo
Grissom Highschool
10th Grade

I went to Vestavia this year; 2005

I am a homeschooler in Mobile, I think I'm the ONLY one.

Smartie alert. Do you do MathCounts, lemur?
Oh, sweet. I'm no longer the only white person here!

lol i just found out about MATHCOUNTS this year and i hope to compete next year (ill be in 8th grade.) It sounds pretty cool, but i'll have to study a bunch.....   :P

are u tired of being around chinese people, dynamo?

cuz we rock

everyone else< CHINESE PEOPLE

I'm Rickie Jang
6th Grade
Baldwin Arts and Academics

Made Nationals this year (beginners luck)! :laugh:

I am Asutosh Nanda (ash is good enough)
Challenger Middle School
made nats 2009 :)  except i failed
7th grade

Well...
Jamie Zhang
Pizitz Middle School
8th Grade

Hi guys. My name is Eric Wu. I go to Davidson High School and I'm in 9th grade.

/bump
My name is Sam, and I go to the New Century Technology Virtual School (I take my classes online at home), and I'm a 9th grader. In case you don't me, I'm the tall kid who is very introverted in public settings (such as math tournaments :P).

[quote=MCrawford][quote="rrusczyk"]All these Vestavia people.
[/quote]

And then some...

Mathew Crawford
Vestavia Hills High School
Class of 1995[/quote]
 WOW ! VHHS has some DISTINGUISHED Alumini.
Tina Gao
Pizitz Middle
No biggie achievements

Allan Yuan
Simmons Middle
<insert flex here>

Eric Wang
Pizitz Middle

Did nothing significant.


[quote="silouan"]Let two circles $C_{1},\ C_{2}$ externally tangent at $M$ .From
a point $A$ which belongs to $C_{1}$ we draw the tangents $AB,\ AC$ to the other circle .Now let $BM,\ CM$ meet $C_{1}$ at the points $E,\ D$ respectively .Prove that $ED$ passes through the midpoint of $AC$ .[/quote]

Please see :

 http://www.mathlinks.ro/Forum/viewtopic.php?t=131661

But I please the proposer of the above link to send his solution. I think we must see one more solution to this nice problem.

[u] Babis    [/u]

Here is my solution
Let intersection points of $AB$ and $AC$ with $ED$ be $K$ and $L$ respectively.Let $l$ be common tangent to circles through $M$ and let $U$ and $V$ be points on $l$ on the same side with $B$ and $C$ respectively.Then
$\angle MCB=\angle UMB=\angle VMD=\angle DEM$
$\Longrightarrow BC \parallel ED$
Let $\angle CBM=\alpha$ and $\angle BCM=\beta$Then
$\angle ACM=\angle DBC=\angle BDE=\angle MAE$
$\Longrightarrow \bigtriangleup AEM \sim \bigtriangleup CEA$
Let $\angle AED =\gamma$Then from law of Sine we have
$\frac{AL}{\sin\gamma}=\frac{AE}{sin \angle ALE}$
$\frac{LC}{sin\beta}=\frac{EC}{sin\angle ELC}$
Dividing side by side gets
$\frac{AL}{LC}.\frac{sin\beta}{sin\gamma}=\frac{AE}{EC}$
On other hand from similarity
$\frac{AE}{EC}=\frac{AM}{AC}=\frac{AM}{AB}=\frac{\sin \angle ABM}{sin \angle AMB}=\frac{sin\beta}{sin\gamma}$
So $AL=LC$


convex is the opposite of concave.  convex is sloped outward and to see aroung the corner the drawing would look like a big lens around a corner that is curved outward.

Some general spelling tips:

1) convex
2) lens

Yeah, a convex mirror would like kind of like a bubble =)

[quote]convext is the opposite of concave. convext is sloped outward and to seel aroung the corner the drawing would look like a big lense around a corner that is curved outward.[/quote]

BHorseMath, I hope you don't take offense, but your sentence has been added to a list of "correct this sentence" exercises for my school's english class. This one in particular may have the worst grammar ever. Congratulations.

w/e I was in a hurry I typed that before a class. I'm fixing it. (would u believe that I have the highest grade in english?)

Don't forget to change aroung to around too :D


Denote by $ f(m)$ sum of digits of an integer m. Is it true that for each positive integer $ n$ there exist integer $ m$ such that $ \frac{m}{f(m)}\equal{}n$

Thank you for these useful links. Now it is easy to prove that for examle equation $ \frac{m}{f(m)}\equal{}75$ does not have solution is set of positive integers.


geometry
geometric series
geometric sequence

Well I decided to find the area of [url=http://en.wikipedia.org/wiki/Koch_snowflake]Koch Snowflake[/url].
[hide="Derivation"]
Let us first find the sum for a non-infinite series.

Let $ 1$ be the first term of the sequence

\begin{align} & \text{Let }s = 1 + r + r^2 + r^3 + \cdots + r^{n - 1}. \\
& \text{Then }rs = r + r^2 + r^3 + r^4 + \cdots + r^n. \\
& \text{Then }s - rs = s(1 - r) = 1 - r^n,\text{ so }s = \frac {1 - r^n}{1 - r}. \end{align}

However, If we let $ a_1$ be the first term, we have
$ \begin{align} & \text{Let }s = a_1 + r(a_1) + (r^2)(a_1) + \cdots + (r^{n - 1})(a_1). \\
& \text{Then} rs = r(a_1) + (r^2)(a_1) + (r^3)(a_1) + \cdots + (r^n)(a_1). \\
& \text{Therefore} s - rs = s(1 - r) = a_1 - (r^n)(a_1) = a_1(1 - r^n), \text{so,} s = \frac {a_1(1 - r^n)}{1 - r}$

Now let us assume that $ - 1 < r < 1$,
If we take a fraction to a greater power, the fraction will keep getting smaller and smaller and smaller.

So when take a fraction to the infinite power, we can just let the final answer be equal to zero!!

So that makes,
$ \frac {a_1(1 - r^n)}{1 - r} = \frac {a_1(1 - 0)}{1 - r} = \boxed{\boxed{\boxed{\frac {a_1}{1 - r}}}}$

Now, we shall find the area of the first stage of Koch's Snowflake.  Let us assume that the side has a length of $ s$.  The area would be $ \frac {s^2\sqrt {3}}{4}$.  Now, in the second stage, we add three triangles with side length of $ \frac {s}{3}$.  This would make the area of one of those triangles $ \frac {\frac {s^2}{9}*\sqrt {3}}{4}$.  The area of 3 of these triangles would be $ \frac {\frac {s^2}{3}*\sqrt {3}}{4} = \frac {s^2\sqrt {3}}{12}$.  These are the first two terms of an infinite geometric sequence.  If we let $ a_1 = \frac {s^2\sqrt {3}}{4}$ and let $ r(a_1) = \frac {s^2\sqrt {3}}{12}$
Therefore, $ r = \frac {1}{3}$.  We know have all the parts of our formula and can now plug it into the formula we found for an infinite geometric series.  The formula is $ \frac {a_1}{1 - r}$
Plugging in our information gives us ${{ \frac {\frac {s^2\sqrt {3}}{4}}{1 - \frac {1}{3}}} = \frac {\frac {s^2\sqrt {3}}{4}}{\frac {2}{3}}} = \boxed{\boxed{\boxed{\boxed{\frac {3s^2\sqrt {3}}{8}}}}}$

[/hide]
This is wrong.....what did I mess up on?
I seriously don't know, what I messed up on

Why do you assume the areas form a geometric sequence?

Wait, I thought that it did.....Does it not?

There is certainly a geometric series involved, but I think you've really misunderstood the construction of the snowflake.

For example, the first step, you add three triangles to the initial triangle. But in the next step, do you really add 9 triangles? Absolutely not --- you add 12. The thing you should realize is where the new triangles are coming from.

Create a sequence $ c_n$ that counts the number of segments that make up the stage-n snowflake, where $ c_0 \equal{} 3$. Notice that each stage, the number of segments multiply by four: you go from ___ to _/\_. Also, note that the number of new triangles in stage $ n\plus{}1$ is equal to the number of segments in stage $ n$.

The area of the triangles in each step, as you noticed, decreases by a factor of 9, since the sidelength decreases by a factor of 3. Thus, construct a geometric sequence $ a_n$ representing the area of each new triangle used to generate the nth snowflake.

Let $ A_n$ be the running area. We know $ A_n \minus{} A_{n\minus{}1} \equal{} a_n c_{n\minus{}1}$. Now crank it out, using your initial conditions!

Sorry, the way my post would be interpreted is not what I meant. TZF covered my objection in his post. After adding the first three triangles, the real pattern emerges, so you need to look at later stages.

The correct answer is $\frac{2\sqrt{3}}{5}$.


function
combinatorics unsolved
combinatorics

how many surjective functions can you create from A to B, where cardA = m, cardB = n; m, n pozitive integers, m>=n

yes, it is quite classical and related to the Stirling numbers of second kind.

Pierre.


The second one:

A left block of a natural number N is any block of consecutive digits in the decimal representation of N, which starts from the leftmost digit. For example, 137 is a left block of 13729, but 1373 and 729 are not. An infinite sequence of natural numbers n1 < n2 < ... is called special if no ni is a left block of nj for different indices i, j. Prove that for any special sequence we have
\sum 1/ni \leq 1 + 1/2 + 1/3 + ... + 1/9,
where the summation ranges over all terms (i.e. infinitely many) of the sequence.

for the sequence 1,2,3,4,5,6,7,8,9 we have the equality of course

pick one of the items in the row. let's take X here as an example.
so for each number that goes, even after doing this trick a thousand times

we get these new options 10X, 10X+1, 10X+2, ... 10X+9  taking its place

and I guess it's kind of logical that X \leq (1/10X + 1/(10X+1) + ...)

(as it can never outrise (10/10).X )

proved :D

I like these kind of questions!


modular arithmetic
number theory proposed
number theory

Find all natural solutions of the equation:     $ 7^{x}\minus{}2\cdot3^{y}\equal{} 1$.

For x>2, 7^x -1 has a prime divisor other than 2,3.
So the only solution is x=1.

My solution:

First it's not difficult to see that $ (1,1)$ is a solution. So assume that $ x,y\ge 2$. If we consider the equation $ \mod 9$ we find that $ 7^{x}\equiv 1\pmod 9$. But as $ \text{ord}_{9}(7)\equal{}3$, we conclude that $ 3\mid x$.

Let us rewrite the equation as follows: $ 1\plus{}7\plus{}7^{2}\plus{}\dots\plus{}7^{x\minus{}1}\equal{}3^{y\minus{}1}$. Using $ 1\plus{}7\plus{}7^{2}\equal{}3\cdot 19$, this can be written as $ 19(1\plus{}7^{3}\plus{}7^{6}\plus{}\dots\plus{}7^{x\minus{}3})\equal{}3^{y\minus{}2}$, which implies $ 19\mid 3^{y\minus{}2}$, contradiction. Thus the only solution is $ (1,1)$.

In general, if we replace 7 and 3 by any 2 odd nos, we have at most one solution.

[quote="nktp"]If $ x \equal{} 1$ then $ y \equal{} 1$.
If $ x\geq 2,y\geq 2$ we have:
$ 7(7^{x\minus{}1}\minus{}1) \equal{} 6(3^{y\minus{}1}\minus{}1)\Rightarrow 3^{y\minus{}1}\minus{}1\vdots 7\Rightarrow y\minus{}1 \equal{} 6m$
Then $ 3^{y\minus{}1}\minus{}1\vdots 3^{6}\minus{}1 \equal{} 728\vdots 13\Rightarrow 7^{x\minus{}1}\minus{}1\vdots 13\Rightarrow x\minus{}1 \equal{} 12n$
Now $ 7(7^{x\minus{}1}\minus{}1)\vdots 7^{12}\minus{}1\vdots 9\Leftrightarrow 6(3^{y\minus{}1}\minus{}1)\vdots 9$,contradiction![/quote]

The technique to solve this problem also is the technique to solve problem :

Find all natural solutions of the equation:$ 3^{x}\plus{}4\equal{}7^{y}$

And it was posted before : http://www.mathlinks.ro/Forum/viewtopic.php?t=164434

guys using zsigmondy's theorem in such particular cases is not very nescessary ( to Abel )
As you can see we can just give simple solutions as nayel and nktp did!  :wink:


superior algebra
superior algebra solved

I am trying to show that in a commutative ring with 1 has at most one non-zero nilpotent if -1 and 1 are only units of this ring.  I am trying to show this by contra positive (2 nilpotent implies there is nontrivial unit), but I am having hard time showing this.  (Should I try different method?)

More generally:

In a ring with 1 (not necessarily commutative), the number of units is greater or equal to the number of nilpotent elements (including 0).

In order to prove this, it will be enough to prove that for every nilpotent element a of the ring, the element 1 - a is a unit.

In fact, since a is nilpotent, there exists some natural number n such that $a^{n}=0$. Then,

$\left(1-a\right)\left(1+a+a^{2}+...+a^{n-1}\right)=\left(1+a+a^{2}+...+a^{n-1}\right)-\left(a+a^{2}+a^{3}+...+a^{n}\right)$
$=1-a^{n}=1-0=1$,

so that 1 - a is a unit, qed.

  Darij


Sean made a one-time deposit of $145 into a bank account. If the bank pays 3.35% simple interest, how much will the deposit be worth in 4 years?


modular arithmetic
number theory
relatively prime
algebra
binomial theorem

Whats the remainder when $ 7^{571}\plus{}17^{150}$ is divided by 100?

As for this problem, do u split them up separately and do $ mod 100$ on each one, or is there another way to do this?

[quote="Mathalete"]Whats the remainder when $ 7^{571} \plus{} 17^{150}$ is divided by 100?

As for this problem, do u split them up separately and do $ mod 100$ on each one, or is there another way to do this?[/quote]
Well, that is probably the quickest way to solve the problem, so here is the solution:
[hide="Solution"]
$ 7^{4} \equal{} 2401\equiv 1\bmod 100$ and $ 17^{40}\equiv 1\bmod 100$ by Euler's Totient Theorem.  Thus,
\[ (7^{4})^{142}\equiv 1\bmod 100
\]\[ 7^{568}\equiv 1\bmod 100
\]\[ 7^{571} \equal{} 7^3(7^{568})\equiv 7^{3}\equiv 43\bmod 100
\]\[ 17^{120} \equal{} (17^{40})^3\equiv 1\bmod 100
\]\[ 17^{150} \equal{} (17^{40})^3(17^{30})\equiv 17^{30}\equiv (17^2)^{15}\equiv 89^{15}\equiv ( \minus{} 11)^{15}\equiv ( \minus{} 11^{5})^3\bmod 100
\]\[ 17^{150}\equiv ( \minus{} 16051)^3\equiv 49^3 \equal{} 49^2(49) \equal{} 2401(49)\equiv (1)(49)\equiv 49\bmod 100\] Thus, $ 17^{150} \plus{} 7^{571}\equiv 49 \plus{} 43\equiv \boxed{92\bmod 100}$ [/hide]

As far as $ 17^{150} \bmod 100$ goes, you can do better:  by CRT, $ 17^{20} \equiv 1 \bmod 100$.  (This is true when $ 17$ is replaced by any integer relatively prime to $ 100$.)  The generalization is given by http://en.wikipedia.org/wiki/Carmichael's_theorem .

(Just a shortcut I saw.)
You can use the Binomial Theorem on $ 17^{150}$:
$ (10\plus{}7)^{150}\equiv 7^{150} \pmod 100$.
Then you are looking for the remainder of $ 7^{571}\plus{}7^{150}$, and you can use similar methods as above...


group theory
abstract algebra
symmetry
superior algebra
superior algebra unsolved

Let $G$ be a finite group and $H, L, K$ be proper subgroups of $G$ such that $G=H\cup K\cup L$. Prove that index of these subgroups in $G$ is equal to $2$. Then consider the case in which $G$ is infinite.

I don’t really know what happens in the infinite case, but here’s the finite one:

First of all, a group cannot be a union of two proper subgroups. Assume the contrary, i.e. that $G=A\cup B$ for two proper subgroups $A,B$ of the group $G$. If $a\in A\setminus B,\ b\in B\setminus A$, then $ab$ cannot belong to either $A$ or $B$, contradicting $G=A\cup B$. I’ll use this below.

For every $k\in K\setminus L,\ l\in L\setminus K$ the element $kl$ belongs to $H\setminus(K\cup L)$, by an argument identical to the one above. Suppose now that $l\in (L\setminus K)\cap H$. Then for every $k\in K\setminus L$ we have $kl\in H\Rightarrow k\in H$. This means that $K$ is the union of $K\cap L$ and $K\cap H$, so by the previous paragraph one of these two groups must be the entire $K$, or, equivalently, $K$ is a subset of either $L$ or $H$. Again, this contradicts the result above (because then $G$ would be the union of the proper subgroups $H,L$). This means that $L\cap H\subseteq L\cap K$. By symmetry, we see that the sets $H\cap K, K\cap L, L\cap H$ are all equal (and equal to $H\cap K\cap L$).  In other words, the group $G$ is partitioned into the four subsets $A=H\cap K\cap L, H\setminus A, K\setminus A, L\setminus A$. All of this holds regardless of whether $G$ is finite or not.

Now in the finite case, we can assume WLOG that $|H|\ge |K|,|L|$ (actually, it’s easy to prove that their cardinalities are equal). It is then obvious from the above that $|G|<3|H|$, so, since $|H|$ is a divisor of $|G|$ strictly less than $|G|$, we must have $|G|=2|H|$, and the same holds for $K,L$.

The infinite case is just as easy, actually.

If $h\in H,\ k\in K,\ l\in L$ lie outside of $A=H\cap K\cap L$, then on the one hand we have $hk\in L$, and on the other hand we have $kl\in H$. This means that $hkl\in H\cap L=A$, i.e. $h\in H$ belongs to the coset $A(kl)^{-1}$ of $A$ in $H$ whenever it does not belong to $A$ itself (and similarly for $K$ and $L$). It follows that $A$ is of index two in each of $H,\ K,\ L$, and this, in turn, implies $[G: A]=4$. Clearly, this means that $[G: H]=[G: K]=[G: L]=2$, as desired.


imparatul verde, foarte bogat, a constatat ca are 1000 de sticle de vin (:D). din pacate, cineva nu il are la suflet si una din sticle este otravita. din fericire, paharnicul sef al imparatiei este un tip destept si bun la matematica, si de aceea imparatul a decis sa ii ceara ajutorul. el are 40 de zile la dispozitie, si N condamnati la moarte care sa testeze sticlele de vin. de la o picatura de otrava se moare dupa 30 de zile.
paharnicul va cere sa ii ziceti numarul N minim de condamnati la moarte astfel incat sa poata sa`si indeplineasca sarcina.

(nota : includeti o demonstratie completa:D)

pt ca n-avem decat 40 de zile in care sa indeplinim sarcina, si nu stim efectele bauturii decat peste 30 de zile, inseamna ca nu are rost sa facem doar cateva teste si apoi sa asteptam rezultatele lor ca sa decidem cum sa testam mai departe. rezulta ca toate testele se pot face intr-o singura zi, si sticla otravita se determina pur si simplu prin a observa care submultime de condamnati moare dupa 30 de zile.

daca sunt mai putini de 10 condamnati, numarul de submultimi din multimea condamnatiilor este cel mult $2^{9}= 512 < 1000$, si rezulta ca sunt doua sticle care trebuie baute de aceeasi submultime. in cazul in care una din astea doua sticle este cea otravita, submultimea care a baut din ele va murii dar nu vom putea stii care din ele este cauza. dar daca sunt 10 condamnati, atunci sunt $2^{10}-1 = 1023 > 1000$ submultimi nevide, si fiecare sticla poate fi bauta de o submultime diferita de condamnati. observand care submultime de condamnati moare, deducem ca sticla (unica) din care a baut aceasta submultime este aia otravita.

bai nu prea ai neaparat dreptate, la faza ca sub 10 nu se poate.

alte pareri ?
rezolvarea mea se afla mai jos, insa v`as ruga mult sa nu o cititi decat dupa ce aveti o rezolvare proprie, deoarece am auzit foarte multe raspunsuri la aceasta problema.
nu stiu daca eu am gresit sau nu

[hide="parerea mea"]
raspuns : $N=3$ :)
demonstratie :
raspunsul va rezulta in ultimul rand al solutiei. paharnicul va gasi sticla otravita in functie de ce se va intampla dupa 40 de zile. cum abia dupa 30 de zile se afla informatii, rezulta ca zilele care ar trebui studiate sunt 31,32,...,40.
voi presupune ca sunt $N$ condamnati. atunci o situatie finala o voi "codifica" sub forma unui n`uplet ordonat : $(z_{1},z_{2},\ldots,z_{N})$.  semnificatie $z_{k}$ reprezinta ziua in care a murit condamnatul $k$, mai exact : $z_{k}=0$ daca el ramane in viata, $z_{k}=j$ daca el a murit in ziua $30+j$. atunci fiecare componenta a acestui n`uplet poate lua 11 valori, iar orice doua n`uplete distincte dau o anumita situatie finala, si orice situatie finala se reprezinta unic in acest fel. cu alte cuvinte, situatiile finale sunt in bijectie cu n`upletele de tipul de mai sus. multimea n`upletelor este chiar $\mathbb{Z}_{11}^{N}$.
sa zicem ca paharnicul a gasit o modalitate "desteapta" de a ii face sa bea. inseamna ca, in functie de situatia finala, el va putea determina sticla. inseamna ca fiecarei sticle ii corespunde o unica situatie finala, deoarece altfel paharnicul n`ar putea sti sigur. deci exista o aplicatie injectiva de la multimea $S=\left\{1,2,\ldots,1000\right\}$ la multimea $M$ a situatiilor finale. deci $|S|\geq |M|$, si deci $N\geq 3$, deoarece $|M|=11^{N}$.
pe de alta parte, fie $P$ o submultime a lui $M$ cu exact 1000 de elemente si $f$ o bijectie de la $P$ la $S$. atunci $f^{-1}: S\rightarrow M$ este o aplicatie injectiva de care ziceam. apoi, fie $z=(z_{1},z_{2},\ldots,z_{N})$ un n`uplet din $P$, si fie $k=f(z)$. atunci condamnatul $j$ va bea din sticla $k$ in ziua $z_{j}$, daca $z_{j}\neq 0$, si nu va bea deloc, daca $z_{j}=0$. daca sticla $k$ este cea otravita, in final se obtine exact configuratia corespunzatoare n`upletului $z$. deci sticla otravita este $f(z_{final})$. cum pt $N=3$ are sens sa vorbesc de cele de mai sus, rezulta ca $N=3$ este minimul cautat.
practic : fie sticlele etichetate $0,1,\ldots,999$. voi scrie fiecare numar $i$ dintre acestea in baza $11$. cum $11^{3}>999$, rezulta ca am nevoie de trei cifre. cele trei cifre din baza 11 vor determina cum va fi bauta respectiva sticla de condamnati : fie $k=\overline{z_{1k}z_{2k}z_{3k}}$. observatie : toate cifrele pot fi $0$, de exemplu $0=\overline{000}$. atunci, din sticla $k$ va bea condamnatul $1$ in ziua $z_{1k}$ daca $z_{1k}\neq 0$, etc. la final, daca sticla $k$ este cea otravita, atunci situatia finala corespunde tripletului $z_{1k}z_{2k}z_{3k}$. reciproc, stiind situatia finala $(a,b,c)$, fie numarul $k=121a+11b+c$. el este scris in baza 11 astfel : $k=\overline{abc}$. atunci sticla $k$ este cea otravita, si acest fapt este verificat.

[/hide]

edit : typing errors.

m`am mai gandit la ceva

[hide="in legatura cu parerea mea"]
pt a fi inteles mai usor fenomenul pe care il descriu eu acolo, voi lucra doar cu baza 10 (facand problema putin mai slaba, insa rezultatul se mentine exact aici)
sticlele le numerotez de la 0 la 999.
acum, unei sticle $\overline{abc}$ ii aplic urmatorul tratament :
condamnatul 1 va bea din aceasta sticla in ziua $c+1$.
condamnatul 2 va bea din aceasta sticla in ziua $b+1$.
condamnatul 3 va bea din aceasta sticla in ziua $a+1$.
la sfarsit, daca aceasta sticla era otravita, atunci condamnatul 1 va muri in ziua $c+31$, condamnatul 2 va muri in ziua $b+31$, condamnatul 3 va muri in ziua $a+31$.
deci stiind zilele $x,y,z$ in care mor cei trei condamnati (orice condamnat bea din toate sticlele - evident), sticla cautata va fi $100\cdot(z-31)+10\cdot(y-31)+(x-31)$.
in acest fel solutia e mai usor de citit, insa se poate observa ca am renuntat situatiile finale in care cel putin un condamnat ramane viu.
de aceea am folosit baza 11 initial, deoarece folosind baza 11 obtin toate configuratiile finale, si pot "ghici" pana la 1331 sticle.

[/hide]


calculus
integration
trigonometry
induction
function
analytic geometry
algebra

the problem is like this:
prove that

$ \int_0^{\frac{\pi}{2}}\sin^{2n}dx\equal{}\frac{2n\minus{}1}{2n}\cdot\frac{2n\minus{}3}{2n\minus{}2}\cdot......\cdot\frac34\cdot\frac12\cdot\frac{\pi}{2}$

$ n\in N$

actually the problem in my book asks to prove by induction, but anybody knows to prove without induction..?
thanks for your help..:)

Consider the equivalence of your integral to it's beta function representation, and then consider how wallis's product is related to the gamma function.

could you please tell me more detail about beta and gamma function..
I've never learned before, I think..
thx

Or let's take this from the ground up.

Let $ I_n\equal{}\int_0^{\frac{\pi}2}\sin^{2n}x\,dx.$

Assume $ n\ge 1$ and integrate by parts, letting $ dv\equal{}\sin x\,dx$ and $ u\equal{}\sin^{2n\minus{}1}x.$

$ I_n\equal{}\left.\minus{}\sin^{2n\minus{}1}x\cos x\right|_0^{\frac{\pi}2}\plus{}(2n\minus{}1)\int_0^{\frac{\pi}2}\sin^{2n\minus{}2}x\cos^2x\,dx$

The boundary terms are zero at both ends. Writing $ \cos^2x\equal{}1\minus{}\sin^2x:$

$ I_n\equal{}(2n\minus{}1)\int_0^{\frac{\pi}2}\sin^{2n\minus{}2}x\minus{}\sin^{2n}x\,dx$

$ I_n\equal{}(2n\minus{}1)I_{n\minus{}1}\plus{}(2n\minus{}1)I_n$ so $ (2n)I_n\equal{}(2n\minus{}1)I_{n\minus{}1}$

$ I_n\equal{}\frac{2n\minus{}1}{2n}I_{n\minus{}1}.$

Since we easily have that $ I_0\equal{}\frac{\pi}2,$ this recursion is exactly what we need to derive your expression.

By exactly the same process, we have that

$ \int_0^{\frac{\pi}2}\sin^{2n\plus{}1}x\,dx\equal{}\frac{2n}{2n\plus{}1}\cdot \frac{2n\minus{}2}{2n\minus{}1}\cdot\,\cdots\,\cdot\frac35\cdot\frac23\cdot 1.$

[quote="mhar_teens"]actually the problem in my book asks to prove by induction, but anybody knows to prove without induction..?[/quote]
My argument was a proof by induction, of course. Given the recursive nature of even the statement of the result, I don't see any way to avoid induction or recursion - that much is built in.

My alternate computation would be to compute $ \int_0^{\infty}\int_0^{\infty}y^{2n}e^{\minus{}x^2\minus{}y^2}\,dx\,dy$ in two different ways: in rectangular coordinates, and in polar coordinates. We will have to deal with the Gamma function in doing that, and the induction/recursion would then be buried in facts about the Gamma function.

I'd like to see that solution Kent, but, can you first tell me how did you turn the original integral into a double one?

We can use the Euler's formula  $ \sin x \equal{} \frac{{e^{ix}  \minus{} e^{ \minus{} ix} }}{{2i}}$ and binomial theorem $ \left( {a \plus{} b} \right)^n  \equal{} \sum\limits_{k \equal{} 0}^n {C_n^k a^{n \minus{} k} b^k }$ where $ C_n^k  \equal{} \frac{{n!}}{{k!\left( {n \minus{} k} \right)!}}$.

$ \begin{gathered}
  \int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern\minus{}\nulldelimiterspace} 2}} {\sin ^{2n} \left( x \right)dx}  \equal{} \frac{1}
{2}\int\limits_0^\pi  {\sin ^{2n} \left( x \right)dx}  \equal{} \frac{1}
{2}\int\limits_0^\pi  {\left( {\frac{{e^{ix}  \minus{} e^{ \minus{} ix} }}
{{2i}}} \right)^{2n} } dx \equal{} \frac{1}
{{2^{2n \plus{} 1} \left( { \minus{} 1} \right)^n }}\int\limits_0^\pi  {\left( {e^{ix}  \minus{} e^{ \minus{} ix} } \right)^{2n} dx}  \equal{}  \hfill \\
   \equal{} \frac{1}
{{2^{2n \plus{} 1} \left( { \minus{} 1} \right)^n }}\int\limits_0^\pi  {\left( {e^{ix}  \plus{} \left( { \minus{} e^{ \minus{} ix} } \right)} \right)^{2n} dx}  \equal{} \frac{1}
{{2^{2n \plus{} 1} \left( { \minus{} 1} \right)^n }}\int\limits_0^\pi  {\left[ {\sum\limits_{k \equal{} 0}^{2n} {C_{2n}^k \left( {e^{ix} } \right)^{2n \minus{} k} \left( { \minus{} e^{ \minus{} ix} } \right)^k } } \right]} dx \equal{}  \hfill \\
   \equal{} \frac{1}
{{2^{2n \plus{} 1} \left( { \minus{} 1} \right)^n }}\sum\limits_{k \equal{} 0}^{2n} {\left[ {C_{2n}^k \int\limits_0^\pi  {\left( {e^{ix} } \right)^{2n \minus{} k} \left( { \minus{} e^{ \minus{} ix} } \right)^k } } \right]} dx \equal{} \frac{1}
{{2^{2n \plus{} 1} \left( { \minus{} 1} \right)^n }}\sum\limits_{k \equal{} 0}^{2n} {\left[ {\left( { \minus{} 1} \right)^k C_{2n}^k \int\limits_0^\pi  {e^{2i\left( {n \minus{} k} \right)x} dx} } \right]} . \hfill \\
\end{gathered}$

Now we will consider the integral $ \int\limits_0^\pi  {e^{2i\left( {n \minus{} k} \right)x} dx}$ and two cases:

$ \begin{gathered}
  {\text{1}}{\text{.  }}n \minus{} k \equal{} 0 \hfill \\
  \left. {\int\limits_0^\pi  {e^{2i\left( {n \minus{} k} \right)x} dx}  \equal{} \int\limits_0^\pi  {e^{2i0x} dx}  \equal{} \int\limits_0^\pi  {dx}  \equal{} x} \right|_0^\pi   \equal{} \pi . \hfill \\
  2.{\text{  }}n \minus{} k \ne 0 \hfill \\
  \left. {\int\limits_0^\pi  {e^{2i\left( {n \minus{} k} \right)x} dx}  \equal{} \frac{1}
{{2i\left( {n \minus{} k} \right)}}\int\limits_0^\pi  {d\left( {e^{2i\left( {n \minus{} k} \right)x} } \right)}  \equal{} \frac{{e^{2i\left( {n \minus{} k} \right)x} }}
{{2i\left( {n \minus{} k} \right)}}} \right|_0^\pi   \equal{} \frac{{e^{2\pi i\left( {n \minus{} k} \right)}  \minus{} 1}}
{{2i\left( {n \minus{} k} \right)}} \equal{} 0. \hfill \\
  \left[ {e^{2\pi i}  \equal{} \cos \left( {2\pi } \right) \plus{} i\sin \left( {2\pi } \right) \equal{} 1} \right] \hfill \\
\end{gathered}$

So we have

$ \begin{gathered}
\int\limits_0^{{\pi  \mathord{\left/
 {\vphantom {\pi  2}} \right.
 \kern\minus{}\nulldelimiterspace} 2}} {\sin ^{2n} \left( x \right)dx}  \equal{} \frac{1}
{{2^{2n \plus{} 1} \left( { \minus{} 1} \right)^n }}\sum\limits_{k \equal{} 0}^{2n} {\left[ {\left( { \minus{} 1} \right)^k C_{2n}^k \int\limits_0^\pi  {e^{2i\left( {n \minus{} k} \right)x} dx} } \right]}  \equal{} \frac{\pi }
{{2^{2n \plus{} 1} }}C_{2n}^n  \equal{} \frac{{\pi \left( {2n} \right)!}}
{{2^{2n \plus{} 1} \left( {n!} \right)^2 }}.
\end{gathered}$

Yes, that's the computation I was going to suggest, but I wasn't sure how nice it would be from $ 0$ to $ \frac{\pi}{2}$.  Note that we are essentially computing the [url=http://en.wikipedia.org/wiki/Fourier_series]Fourier series[/url] of $ \sin^{2n} x$, which behaves nicely with respect to integrals of this type.


Given $R, r > 0$. Two circles are drawn radius $R$, $r$ which meet in two points. The line joining the two points is a distance $D$ from the center of one circle and a distance $d$ from the center of the other. What is the smallest possible value for $D+d$?

Really easy one $\sqrt{\max\{R,r\}^2-\min\{R,r\}^2}$.

yes, but try to do the other 5 problems of 97-OBM, and after it, imagine how it´s difficult today!


First note that the function $f(x) = a^{x}$ is strictly decreasing in $x$ for $0 < a < 1$.
So $g(x) = (3/5)^{x}+(4/5)^{x}$ is strictly decreasing in $x$. Note that $g(2) = 1$.
So $g(x) < 1$ for $x>2$ and $g(x)>1$ for $x<2$.

If $x>2$, then $g(x) < 1= (3/5)^{x}+4^{y}/5^{x}$. This implies $4^{x}< 4^{y}$, hence $x<y$.
As a consequence $y>2$, and the same reasoning gives $y<z$.
As a consequence $z>2$, and the same reasoning gives $z<x$.
So $x<y<z<x$. Contradiction.

If $x<2$ we similarly get the contradiction $x>y>z>x$.

So $x=2$, which easily implies $y=2$ and $z=2$. :)


the first isnt even a quadratic really..
very itnerested to see how that one is done

[quote="linzk3"]solve for x.

a. (x+3)^(2/3) - 3(3+3)^(1/3) - 40 = 0

b. $1/((x+3)^{2})-3(1/(x+3))-40 = 0$

im having some trouble with problems like these.[/quote]

first of all, since you didn't put it in latex form, i had a hard time deciphering the problem. But this is what i think u meant:

${1}/{3}(x+3)^{2}-3({1}/(x+3))-40=0$

when i say divided 1/ ...., i mean 1/ by the entire thing. Anyhow, [hide] First, you would set it up like so:

$1$                               $3$
____                   $-$                   $-$  40  = 0
 $(x^{2}+6x+9)$(x+3)    [/hide]

[quote="tridecalogism_lover"][quote="linzk3"]solve for x.

a. (x+3)^(2/3) - 3(3+3)^(1/3) - 40 = 0

b. $1/((x+3)^{2})-3(1/(x+3))-40 = 0$

im having some trouble with problems like these.[/quote]

when i say divided, i mean the entire thing divided.

first of all, since you didn't put it in latex form, i had a hard time deciphering the problem. But this is what i think u meant:

${1}/{3}(x+3)^{2}-3({1}/(x+3))-40=0$[/quote]

yeah thats right

[quote="linzk3"]solve for x.

a. (x+3)^(2/3) - 3(x+3)^(1/3) - 40 = 0

b. 1/((x+3)^2) - 3(1/(x+3)) - 40 = 0

im having some trouble with problems like these.[/quote]
[hide="B"] Say that $\frac{1}{x+3}=y$.  Thus, we have that $y^{2}-3y-40=0\Rightarrow (y-8)(y+5)=0\Rightarrow y=8, y=-5$.  So, we have that $\frac{1}{x+3}=8\Rightarrow 8x+24=1\Rightarrow x=\frac{-23}{8}$ or $\frac{1}{x+3}=-5\Rightarrow-5x-15=1\Rightarrow 16=-5x\Rightarrow x=\frac{-16}{5}$.  So, the solutions for $x$ are $\boxed{\frac{-16}{5},\frac{-23}{8}}$ [/hide]


Let x,y,z in R satisfy that:
                     $x^{2}$+$z^{2}$=1
                     $y^{2}$+2y(z+x)=6
Prove that y(z-x) $\leq$ 4

That you shall be patient please!
That means not bumping this thread every day, too.


geometry
3D geometry
IMO Shortlist
ellipse
Volume
sphere

A plane cuts a right circular cone of volume $ V$ into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the volume of the smaller part.

[i]Original formulation:[/i]

A plane cuts a right circular cone into two parts. The plane is tangent to the circumference of the base of the cone and passes through the midpoint of the altitude. Find the ratio of the volume of the smaller part to the volume of the whole cone.

The plane cuts the conic surface in an ellipse with (unknown) semimajor and semiminor axes p, q. We have to calculate the ellipse area $S = \pi p q$ and the distance d of the cone apex A from the plane. Then the volume of the cone part containing the cone apex A will be $V_A = \frac{S d}{3}$. Let O be the center of the cone circular base with radius r, h  = OA the cone height, M the midpoint of OA and B the tangency point of the cutting plane with the cone circular base. Put the xyz-coordinate origin at the center O of the cone base, the positive z-axis in the direction of the ray OA and negative y-axis in the direction the ray OB. This makes the x- axis parallel and the yz-plane perpendicular to the cutting plane. The conic surface cuts the yz-plane in 2 cone elements AB, AB', with the points B, B' on the y-axis, and the cutting plane cuts the yz-plane in the line BM. Let the line BM cut the line AB' at a point C. The segment $BC = a = 2p$ is the ellipse major axis and the ellipse focal points $F_1, F_2$ are on this segment. The ellipse focal points $F_1, F_2$ are the tangency points of 2 spheres inscribed in the conic surface with the cutting plane. These 2 spheres cut the yz-plane in the incircle (I) and excircle (E) of the triangle $\triangle ABC$ touching the side BC. Let N be the midpoint of the segment BC (the ellipse center). The distances of the tangency points from the ellipse center are $ep = NF_1 = NF_2$, where e < 1 is the ellipse excentricity. The semininor axis q is then given by Pythagorean theorem for the right angle triangle $\triangle F_1NQ$, where Q, Q' are the intersections of a line in the cutting plane through the ellipse center N perpendicular to the ellipse major axis BC. That's because $QF_1 + QF_2 = 2p$ for a point on the ellipse and $QF_1 = QF_2 = p$ because of symmetry. Hence, $q = \sqrt{QF_1^2 - NF_1^2} = \sqrt{p^2 - e^2p^2}$.

Working in the yz-plane, the points A, B, B', M have the yz-coordinates A = (0, h), B = (-r, 0), B' = (r, 0), M = (0, h/2). The equations of the lines BM, AB' intersecting at the point C are

$BM: \ \ \ \frac{2z}{h} - \frac{y}{r} = 1$

$AB': \ \ \ \frac{z}{h} + \frac{y}{r} = 1$

Solving, we get the coordinates of the point $C = \left(\frac{r}{3}, \frac{2h}{3}\right)$ and the distance BC

$BC = \sqrt{(y_C - y_B)^2 + (z_C - z_B)^2} = \sqrt{\frac{16r^2}{9} + \frac{4h^2}{9}} = \frac 2 3 \sqrt{4r^2 + h^2}$

Thus the ellipse semimajor axis is equal to $p = \frac{BC}{2} = \frac{\sqrt{4r^2 + h^2}}{3}$. Labeling the sides of the triangle $\triangle ABC$ as a = BC, b = CA, c = AB and its semiperimeter s, the distance of the tangency point $F_1$ of the incircle (I) with the side BC from the vertex C is equal to $CF_1 = s - c$ and the distance of this tangency point from the midpoint N of the side BC is equal to $NF_1 = \frac a 2 - (s - c) = \frac{c - b}{2}$. Since $z_C = \frac{2h}{3} = \frac{2z_A}{3}$ and the z-axis bisects the angle $\angle B'AB \equiv \angle CAB$, it is clear that $b = CA = \frac{AB}{3} = \frac{\sqrt{r^2 + h^2}}{3}$ and $NF_1 = \frac{c - b}{2} = \frac{\sqrt{r^2 + h^2}}{3} = b$. The ellipse semiminor axis is then equal to

$q = \sqrt{QF_1^2 - NF_1^2} = \sqrt{\frac{a^2}{4} - b^2} = \sqrt{\frac{4r^2 + h^2}{9} - \frac{r^2 + h^2}{9}} = \frac{r}{\sqrt 3}$

and the ellipse area is $S = \pi pq = \frac{\pi r \sqrt{4r^2 + h^2}}{3 \sqrt 3}$. The distance d of the cone apex A from the cutting plane is equal to the A-altitude AD of the triangle $\triangle ABC$. The altitude passes through the point A = (0, h) and has the slope $m_{AD} = -\frac{1}{m_{BM}} = -\frac{2r}{h}$, being perpendicular to the line $BC \equiv BM$. Hence, the altitude equation is

$AD: \ \ \ z = -\frac{2r}{h}\ y + h$

Solving the equations of the lines BM, AD, we get the coordinates of their intersection $D = \left(\frac{rh^2}{4r^2 + h^2}, \frac{2r^2h}{4r^2 + h^2} + h\right)$ and the distance d = AD

$d = AD = \sqrt{(y_D - y_A)^2 + (z_D - z_A)^2} = \frac{\sqrt{r^2h^4 + 4r^4h^2}}{4r^2 + h^2} = \frac{rh}{\sqrt{4r^2 + h^2}}$

As a result, the volume $V_A$ is equal to $V_A = \frac{S d}{3} = \frac 1 3 \cdot \frac{\pi r \sqrt{4r^2 + h^2}}{3 \sqrt 3} \cdot \frac{rh}{\sqrt{4r^2 + h^2}} = \frac{\pi r^2 h}{9 \sqrt 3}$. On the other hand, the volume of the entire cone is equal to $V = \frac{\pi r^2 h}{3}$, so that $\frac{V_A}{V} = \frac{1}{3\sqrt 3}$. Clearly, $V_A = \frac{V}{3 \sqrt 3} < \frac V 2$ is the smaller volume.


In a certain kingdom, the king has decided to build $25$ new towns on $13$ uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections.

Have I understood the problem correctly? If so, then...

Let the islands have $t_1, ..., t_{13}$ towns respectively, where $t_1+...+t_{13}=25$. For a town on island $i$, it must be joined to $25-t_i$ towns. So the amount of connections given by island $i$ is $t_i(25-t_i)$ (each connection is counted twice here). So we want to minimize the total, that is $ \frac{t_1(25-t_1)+...+t_{13}(25-t_{13})}{2}$ which is the same as minimizing $t_1(25-t_1)+...+t_{13}(25-t_{13})$. Then the problem restates as:

If $t_1, ..., t_{13}$ are positive integers such that $t_1+...+t_{13}=25$, find the minimum of $t_1(25-t_1)+...+t_{13}(25-t_{13})$.

Now

\[t_1(25-t_1)+...+t_{13}(25-t_{13})=-(t_1^2+...+t_{13}^2)+25(t_1+...+t_{13}) \\
-(t_1^2+...+t_{13}^2)+25(t_1+...+t_{13})=-(t_1^2+...+t_{13}^2)+25^2 \]

Since $25^2$ is a constant, we want to minimize $-(t_1^2+...+t_{13}^2)$, that is, to maximize $t_1^2+...+t_{13}^2$. Now we use an optimization argument. Suppose $a \ge b$ and consider $a^2+b^2$. Now note that $(a+1)^2+(b-1)^2=a^2+2a+1+b^2-2b+1=(a^2+b^2)+2(a-b+1)>a^2+b^2$. Thus we can't have two numbers greater than $1$ in an optimal case, because otherwise we would apply that transformation and increase the sum. So WLOG $t_1=...=t_{12}=1, t_{13}=13$. Then the least possible number of connections is $\frac{-181+625}{2}=222$.

The most edges are when the partition is quasi-balanced; the least - when the partition is the most unbalanced.
By the way, the problem shuns the more economic way of just establishing $\binom {13} {2} = 78$ ferry connections between the islands - as nothing is done for towns on a same island, they must be considered on en equality foot :)


function
limit
real analysis
real analysis solved

Let $ A$, $ B$, $ C$ be three nonempty subsets of the space $ \mathbb R^n$ such that $ A$ is bounded, and such that $ C$ is closed and convex. Suppose that $ A \plus{} B\subseteq A \plus{} C$. Show that $ B\subseteq C$.

If we suppose $\exists b\ :\ b\in B,\, b\not\in C$, then $b$ is isolated from $C$ by mean of hyperplane, i.e. there is linear function $f$ s.t. $f(b)\leq f(x)-\epsilon$ $\forall x\in C$. Since $A$ is bounded there exists finite $\inf\limits_{x\in A}f(x)=a$. It implies that $f(x)\geq a+f(b)+\epsilon$ $\forall x\in A+C$ $\Rightarrow$ $\inf\limits_{x\in A+C}\geq a+f(b)+\epsilon$. But $\inf\limits_{x\in A+B}f(x)=\inf\limits_{x\in A}f(x)+\inf\limits_{x\in B}f(x) \leq a+f(b)$ -- contradiction with condition $A+B\subseteq A+C$.

Is it true that Vojtec Jarnik competition's problems are quite simple? And does anybody knows results and problems from Vojtec Jarnik'04?

so so but some of the prob from vojtec are really chalenging one ,up to now in its official site ,the result isnt announced

[quote="sam-n"]Let $ A$, $ B$, $ C$ be three nonempty subsets of the space $ \mathbb R^n$ such that $ A$ is bounded, and such that $ C$ is closed and convex. Suppose that $ A + B\subseteq A + C$. Show that $ B\subseteq C$.[/quote]

Let $ b$ be an element of $ B$ . Let $ A_n$ denote $ A_n = \{\frac a{2^n}\ | \ a\in A \}$.
For all $ a\in A$ there are $ a_1\in A, c_1\in C$  and $ a_2\in A,c_2\in C$ such that

$ b + a = a_1 + c_1 \ ,\ b + a_1 = a_2 + c_2$ , so $ \ b + \frac a2 = \frac {a_2}2 + \frac {c_1 + c_2}2$ , so $ b + A_1 \subset A_1 + C$

In the same way $ \ \forall n,\ b + A_n \subset A_n + C$ , so ${ b + \frac a{2^n} = \frac {\alpha_n}{2^n}} + \gamma_n$ , where $ \alpha_n\in A ,\gamma_n\in C$

Now I can write : $ (A \ bounded) \Longrightarrow \lim_{n\rightarrow\infty} \frac a{2^n} = \lim_{n\rightarrow\infty} \frac {\alpha_n}{2^n} = 0 \Longrightarrow \lim_{n\rightarrow\infty} \gamma_n = b$

But $ C$ is closed, so $ b\in C$ (q.e.d)

 :cool:


a man has fallen into a ditch of width d.his friends are pulling him out from opposite sides of the ditch with equal force using a rope that is connected to a pulley at both the ends of the ditch.find the force acting on the man when he is at a depth h.

[hide="Suggestion"]Consider all of the forces acting on the man (gravity, the pulling forces, the tensions), and then consider the angle that the rope makes depending on the height.  The force will be a function of angle.[/hide]

Why are you trying to eliminate mass?  If the man is bigger, he will be harder to pull up so mass (of the man) should be in your solution.

[quote="aadil"]
jrav,i am getting the force as a function of theta but i am unable to eliminate the mass.please post a detailed solution

[/quote]

You should eliminate the $ \theta$, not the mass!

$ 2T\cos\theta \equal{} mg \\
 \\
\tan\theta \equal{} \frac {d}{2h} \\
 \\
\cos\theta \equal{} \frac {2h}{\sqrt {4h^2 \plus{} d^2}}$

Tension in the rope is the force applied by the two men.

$ \therefore T \equal{} \frac {mg\sqrt {4h^2 \plus{} d^2}}{4h}$


[youtube]FLn45-7Pn2Y[/youtube]

 :rotfl:

SOME of us don't HAVE sound cards. Inconsiderate -!

While surfing on youtube, I found this guy:
http://www.youtube.com/profile?user=Backwurdz

He joined 2 weeks ago and has like 80 suscribers and each of his videos have over 5000 views. He's also pretty funny.


ARML AMC USA(J)MO USAMO algebra function domain

The list of qualifiers for the SC all-state team is now available online.  Click [url=http://scall-statemathteam.com/default.aspx]here[/url] to visit the website for the team, and [url=http://scall-statemathteam.com/Documents/2007%20All-State%20Qualifier%20List.xls]here[/url] to view the list of qualifiers.

Btw, congrats to Sean, Harry, and Jim on USAMO!

[quote="Iversonfan2007"]nice site, did you just make it?[/quote]

No, Mr. R made it.  And Shobhit, you need to come to ARML.  It's a once in a lifetime thing, whereas you can go to India any time.  Trust me, you don't want to miss it!

lol, I was wondering why you wouldn't use Google for it.

Eh, well it's a registered domain, so http://www.whois.net/


For those who took the putnam, how did u do?

Well, we don't get official results for a while (like March or something).  I think I solved 6, but I don't think I had full solutions for all six.  Definitely I had one error on one question and wasn't able to write up a proof in full for another one (ran out of time).  So I'm hoping to get in the 50-60 range, possibly with a little extra credit on a 7th problem.  I don't really know if that is realistic, though, but I feel like I have a 40 nailed down and then the rest is at the hands of the graders.

I think I nailed three problems, but that's it. I'd be surprised if my score turns out to be more than a point or two away from 30.