geometry
rectangle
pigeonhole principle
analytic geometry
search
function
induction

Did any one use proofs other than the pigeon-hole one given in the math jam?

here's my proof:

We will prove that a $4 \times 19$ rectangle is the smallest rectangle that satisfies the condition of all four of its vertices being the same color.

Because we are only allowed to use three distinct colors (red, green, or blue), one sidelength of the rectangle must be atleast $4$ units. This is because--using the pigeonhole theorem--the fourth unit forces us to use two of the same colors.

There are $\binom{4}{2}=6$ ways we can choose two units from the sidelength to place two of the same colors, and $3$ colors to choose from, giving us a total of $3 \cdot \binom{4}{2}=18$ different permutations of the colors.

Using the pigeonhole theorem once again, we have that the other distinct sidelength of the rectangle must be at least $19$ units long before we are forced to have a repeated permutation.

Thus, because there are an infinite number of points in the plane, there exists a rectangle with all four vertices the same color---the smallest of which has dimensions $4 \times 19$.

$\mathbb{Q}\mathfrak{ED}$

I used pigeonhole but got $19$ instead of $82$, which I think is also acceptable

Since there are an infinite number of points in the plane, at least one of the three colors must be repeated an infinite number of times. Additionally, the distance from one vertex of a rectangle to the remaining three is a pythagorean triple which, of course, there are an infinite number of. The intersection of two such infinite sets is hard to define, given the rules of cardinal numbers, set theory etc. However, the intersection cannot be a null set (by definition of infinity) and as such, a rectangle with the same color vertices must exist.
I was just considering that during the math jam, any comments or critiques

Why didn't they actually ask for the smallest possible grid of lattice points?  As it is the problem seems almost trivial.

I did the same thing that Iversonfan2007 did, just I worded it poorly.

[quote="lotrgreengrapes7926"]Consider the intersection of all even numbers and all odd numbers.  :wink:[/quote]
haha thanks, i knew it couldnt be that easy but it had just occured to me--that wasnt the answer i sent in

[quote="oppenhejo"]Since there are an infinite number of points in the plane, at least one of the three colors must be repeated an infinite number of times. Additionally, the distance from one vertex of a rectangle to the remaining three is a pythagorean triple which, of course, there are an infinite number of. The intersection of two such infinite sets is hard to define, given the rules of cardinal numbers, set theory etc. However, the intersection cannot be a null set (by definition of infinity) and as such, a rectangle with the same color vertices must exist.
[/quote]

I used that answer (basically).

I used a $4\times28$ grid, basically the same way.

I just proved a generalization. I think it is different from the Pigeonhole one given in the math jam.

We will show that there exists a rectangle in the plane such that all four of its vertices are the same color for $n$ colors. This will prove the problem for 3 colors ($n$=3).

Define ($x,y$) as a set of lattice points such that 1$\leq x\leq n+1$ and 1$\leq y\leq n^{n+1}+1$. Also, let there be $n$ colors. By the Counting Principle, each row can be colored $n^{n+1}$ ways. Since, there are $n^{n+1}+1$ rows, by the Pigeonhole Principle, there are at least 2 rows that have the exact same coloring. Let the two rows have $y$-coordinates of $a$ and $b$. Since 1$\leq x\leq n+1$ for a $y$, and we have $n$ colors, by the Pigeonhole Principle, there are 2 colors that are the same in one row. So, there are 2 points that have the same color in row $a$ and in row $b$.

For each $i\in\{1,2,\hdots,n+1\}$, and $(i,a)$ and $(i,b)$ have the same color. We know that there are 2 colors that are in each row. Suppose, $(r,a)$ and $(s,a)$ have the same color in row $a$. We also know that $(r,a)$ has the same color as $(r,b)$, and $(s,a)$ has the same color as $(s,b)$. So, $(r,a)$, $(r,b)$, $(s,a)$, and $(s,b)$ have the same color. The four points form a rectangle. This shows that for $n$ colors, there exists a rectangle such that all 4 of its vertices are the same color.

[quote="renatzu"]I did a contradiction proof (I was never comfortable with pidgeon hole).

Basically, I set up a line, which had to have an infinite amount of some color, say blue.  Then I looked at a parallel line and all the points on the line perpindicular to a blue point.  There can only be one blue, otherwise a rectangle is formed, so there must be an infinite number of red or green. Continue adding parallel lines, and you continue to knock the number of possible colors down by 1.  Eventually you get a set which has to have an infinite number of points, but can't have more than 1 point of each color.[/quote] Somewhere in that strategy you get 4 lines, and that's basically equivalent to the pidgeonhole proofs on an infinite scale.

Some what off topic, but it's obvious that you can generalize this using similar proofs to $n$ colors in $n$ dimensions. However, a thought I had, if we stay in two dimensions (may be possible in more, but let's keep it simple for now), do you think that there exists a rectangle if their are a countably infinite number of colors? Just a thought I had, I'd like to hear other people's thoughts on this.

[quote="ehehheehee"]Every point then could be colored a different color[/quote]
No, because there are an uncountably infinite number of points on the plane, so if each were colored a different color, you'd get an uncountably infinite number of colors.

what does countably infinite number of colors mean.  You can prove that there exists a rectangle for any finite number of colors.

[quote="ehehheehee"]what does countably infinite number of colors mean.[/quote]Basically, it's equivalent to the size of the set of all natural numbers.
Uncountably infinite $\equiv$ number of complex numbers ($\equiv$ the number of points in a plane).

Are the number of points on a line uncountably infinite?
Is an uncountably infinite set infinitely greater than a countably infinite set (I have a feeling this is in the AOPS Book II, but as I am lending it I can't look it up)?

If a set of uncountably infinite set is infinitely greater than a countably infinite set, then the there would have to be an infinite number of colors from the countably infinite set to color at least on the the uncountably infinite set.  But that just sounds wierd.

If the number of points on a line is countably infinite, then you can just "shift" the colors by one to make every vertical line and horizontal line have all different colors.  That sounds wierd too.

The number of points on a line is uncountably infinite, since you can turn it into the real number line and assign exactly one real number to each point. The number of points on any finite line segment is also uncountably infinite.

[quote="draperp"]Uncountably many vs. countably many, it's sort of like "one dimensional infinity" and "two-dimensional infinity".

I have to admit that I had a tough time solving this problem...until I search the web and found a similar problem and solution (it was with two colors instead of three).
Do you think that extending the proof to (finite) $n$ colors would give more of a chance of commendation?[/quote]

But a "two-dimensional infinity" is still countable, if the one dimensional case is.  Consider the set of rational numbers.

By one lemma, if $A$ is countable, then $A \times A$ is also countable.  Well, you can also have an injective function from $\mathbb{Q}$ to $\mathbb{Z}\times \mathbb{N}$.

Also, there exists a bijection from $\mathbb{Z}$ to $\mathbb{N}$.  Consider $f(n) = 2n$, if $n > 0$, and $f(n) = 1-2n$, if $n \leq 0$.

Thus, $\mathbb{Q}$ is injective into $\mathbb{N}\times \mathbb{N}$, which we know is countable by the first lemma.

In fact, by induction, $A^{n}$ is countable, given that $A$ is countable, and $n \in \mathbb{N}$.  I'm not sure if it holds in the case $n = |\mathbb{N}|$.

[quote="PTynan89"]In fact, by induction, $A^{n}$ is countable, given that $A$ is countable, and $n \in \mathbb{N}$.  I'm not sure if it holds in the case $n = |\mathbb{N}|$.[/quote]
It does not.  (By the way, $|\mathbb{N}|$ is usually denoted as $\aleph_{0}$.)  In fact, a fairly small subset of the ordered $\aleph_{0}$-tuples of integers (specifically, those in which each of the integers is a 0 or a 1) is uncountable.  We call this set $\{0,1\}^{\aleph_{0}}$.  To prove this, suppose that there is a bijective function $\phi : \mathbb{N}\rightarrow \{0,1\}^{\aleph_{0}}$.  Now, we construct an element $\mathbf{x}$ of $\{0,1\}^{\aleph_{0}}$ such that the $n$th coordinate of $\mathbf{x}$ is the element of $\{0,1\}$ that is [i]not[/i] equal to $\phi(n)$.  Thus for all $n\in \mathbb{N}$, $\mathbf{x}\neq \phi(n)$.  But this is a contradiction, since we said that $\phi$ was bijective.  (This is just the Cantor diagonal argument, by the way.)


You can find the entire list on the AMC's homepage (its in a pdf file as for now)

Here are the 2009 Indiana Qualifiers, copied and pasted from the pdf file.

Lyndon Ji 9 Carmel HS Carmel IN USA
Shawn Qian 11 Carmel HS Carmel IN USA
David Liang 7 Creekside MS Carmel IN USA
David Ramsey 12 Homestead HS Ft Wayne IN USA
Richard Ni 10 Park Tudor School Indianapolis IN USA
Jeffrey Shen 10 Park Tudor School Indianapolis IN USA
Sachin Shinde 11 Terre Haute South Vigo HS Terre Haute IN USA
Nathan Benjamin 11 Harrison HS West Lafayette IN USA
Philip Thomas 12 Harrison HS West Lafayette IN USA
Mark Sellke 7 Klondike MS West Lafayette IN USA
Yixin Wang 12 West Lafayette HS West Lafayette IN USA
Justin Ahmann 9 Zionsville Community HS Zionsville IN USA

Congratulations to all of them and best of luck on the test!
EDIT: I flipped the names around since they had it last, first on their website.

California has like over 100 qualifiers!??!


logarithms
geometric sequence
arithmetic sequence
algebra unsolved
algebra

For each geometric progression $ a_{1},a_{1},...,a_{n}$ with positive terms and for each increasing arithmetic progression $ b_{1},b_{1},...,b_{n}$ we can choose a logarithmic base $ r$ such that
$ b_{1}\minus{} \log_{r}( a_{1})\equal{} b_{2}\minus{} \log_{r}( a_{2})\equal{}...\equal{} b_{k}\minus{} \log_{r}( a_{k})$. Prove it.

Let $ a_1\equal{}a$ and $ a_k\equal{}as^{k\minus{}1}$, $ b_1\equal{}b$ and $ b_k\equal{}b\plus{}(k\minus{}1)d$

We need to find $ r$ such that $ b\plus{}(k\minus{}1)d\minus{}\log_ras^{k\minus{}1}\equal{}b\plus{}(k\minus{}1)d\minus{}(k\minus{}1)\log_r s\minus{}\log_r a$ is independent of $ k$, and obviously $ r\equal{}s^{\frac{1}{d}}$ will do.


linear algebra
matrix
Gauss
linear algebra unsolved

For any non-singular matrix $A\in{\mathbb R}^{n\times n}$, a well known fact is that $AB=C$ where $C$ is a upper or lower triangular matrix.
However, could we explicitly written out the matrix $B\in{\mathbb R}^{n\times n}$ in the above?

[quote="yevon"][quote="cnbjy"]It is just Gauss-Jordan elimination.[/quote]

could you tell more detail what you mean?[/quote]

...You can find it in any book of linear algebra.
column transformation=multiplication by matrix on the right.
If you really want to some striking name, you may call it L-U decomposition.

Of course it is Jordan Elimination but what I mean is the explicit format of $B$ in terms of the elements of the $A_{i,j}$'s

You have already known the elimination method, why can't you get the form you want?
If you really can't, try a 3X3 matrix, you can generalized it to nXn easily.

Three comments:

1. If you're multiplying on the right, as $AB,$ then you're doing column elimination. The more usual row elimination would involve multiplying on the left, as $BA.$

2. $B^{-1}=E_{1}E_{2}\cdots E_{m}$ where the $E_{j}$ are the elementary matrices corresponding to the elementary column steps needed in the elimination process. The definition of the $E_{j}$ is algorithmic and conditional: if we've arrived at this state so far, then we do that next.

3. If we can do Gaussian elimination without row interchanges, then we can factor $A=LU$ where $L$ is lower triangular and $U$ is upper triangular. If that were the case, and if $U$ were invertible, then we would have $AU^{-1}=L.$ More generally, we would need row (or column) interchanges to complete the process; in that case, we would wind up with $A=PLU$ or $A=LUP,$ where $P$ is a permutation matrix.

For further reference, look up "LU factorization" in the index of either a linear algebra textbook or (possibly more to the point) a numerical analysis textbook.


trigonometry
floor function
modular arithmetic
complex numbers

For how many positive integers $n$ less than or equal to $1000$ is \[(\sin t + i \cos t)^n=\sin nt + i \cos nt\] true for all real $t$?

My answer seems a little weird... can anyone find anything I did wrong?

[hide="Answer"]$\cos t+i\sin t=e^{it}$, $\cos t=\sin (\frac{\pi}{2}-t)$, $i\sin t=i\cos (\frac{\pi}{2}-t)$, so $\sin t+i\cos t=e^{i(\frac{\pi}{2}-t)}$
$e^{in(\frac{\pi}{2}-t)}=e^{i(\frac{\pi}{2}-nt)}\Rightarrow e^{i\frac{n\pi}{2}-it}=e^{i\frac{\pi}{2}-int}$
$i\frac{n\pi}{2}-it=i\frac{\pi}{2}-int\Rightarrow \frac{n\pi}{2}-t=\frac{\pi}{2}-nt$
$n(\frac{\pi}{2}-t)=\frac{\pi}{2}-t\Rightarrow n=1+2k\pi$, so we have $1+\lfloor \frac{999}{2\pi}\rfloor=1+158=\boxed{159}$.[/hide]

Remember that you're working in radians.
$e^{im}=e^{in}$ implies $m=n+2k\pi$ where $k\in\mathbb{Z}$.

Okay, if I understood what you're saying, I think I have it now.

[hide="my solution"]we know that $cis(\theta)=cos\theta+isin\theta$ and $cis(-\theta)=cos\theta-isin\theta$
so the expressison $(sinX+icosX)$ is basically $(i)(cis(-X))$
and by De'Morevs therom $(cos\theta+isin\theta)^n=cos(n\theta)+isin(n\theta)$
so we apply this to $(i)(cis(-X))$ and get

$(i^n)(sinX+icosX)^n=sin(nX)+icos(nX)$
this is only true if $i^n=1$ which happens when $n$ is divisible by 4 so  we find how many multiples of 4 are less than 1000.

so our answer is $\frac{1000}{4}=250$[/hide]

i think i did this correctly.

Yeah... that's probably right... assuming it is, what did I do wrong in my solution?

[hide]factor out the i in both of them, (note that $-i=\frac{1}{i}$)

${i^n(\cos{t}-i\sin{t})=i(\cos{nt}-i\sin{nt}})$

then put that in exponential form...

$i^n*e^{-i*n*t}=i*e^{-i*n*t}$

this true when $n\equiv 1 \mod{4}$

and we find that there are $\boxed{250}$ answers[/hide]

i dont c how you get n congruent to 1 mod 4 :?:

[hide="Solution"]Multiplying by $(-i)^n$, we get $(\cos t -i\sin t)^n = \text {cis}^n \ t$.

By Euler's formula, we have $e^{in} = \text {cis}^n \ t$. Applying to both sides, we have \[ \sin nt + i\cos nt = \text {cis } nt = e^{int},  \] and \[ \frac{\text {cis}^n \ t}{(-i)^n} = \frac{(e^{it})^n}{(-i)^n} = \frac{e^{int}}{(-i)^n}. \]

Now we have \[ \frac{e^{int}}{(-i)^n}=e^{int}  \] so $(-1)^n=1$ and $n \equiv 0 \pmod {4}$. Thus there are $250$ solutions: $4, 8, \ldots , 996, 1000$.
[/hide]

[hide]gyan, 250 is correct, think that 4n+1<1000

n<249, but n=0 makes the power 1, so we have n is 0 to 249 inclusive, so 250

also to verify... [url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2005-aa/05answersaime2.html[/url] 250 is indeed the correct answer[/hide]

Altheman - Yes. Thanks. I have deleted the oriignal post to remove confusion.

Okay, now I have a more clear solution.

$\\\text{cis }t=\cos t+i\sin t\\\text{cis }(-t)=\cos t-i\sin t\\i\text{cis }(-t)=\sin t+i\cos t$
$\\(i\text{cis }(-t))^n=i^n(\cos nt-i\sin nt)=(\sin t+i\cos t)^n=(\sin nt+i\cos nt)\\\\ i^n=\frac{\sin nt+i\cos nt}{\cos nt-i\sin nt}=\frac{ie^{-itn}}{e^{-itn}}=i\implies i^{4n}=1\implies 4n\equiv 0\pmod 4\\\implies n\equiv 0\pmod 4\\ 4n=1000\\ n=\boxed{250}$

Is the reasoning correct?

Masoud Zargar

[quote="boxedexe"]Okay, I have been thinking about this.  After we reached
$i^n=i$, I raised it to them to the power of 4 and got $i^{4n}=1\rightarrow n\equiv 0\pmod 4$, which gives us 250.  However, Altheman divided it by $i\rightarrow n\equiv 1\pmod 4\rightarrow 4n+4=1000\rightarrow n=249$
0-249 inclusive gives us 250 numbers.  How can we tell whether it is 249 or 250, and how can we tell that our answer requires us to find all numbers from 0 to 249 inclusive?

Masoud Zargar[/quote]
I was wondering the same thing.  It says in the original problem that we are looking for "positive integers."  However, from my understanding, "positive integers" and "non-negative integers" are different in that non-negative integers include 0.

Yeah, $n = 0$ is not a valid solution because the problem asks for positive integers.

i think that there might be some confusion as to what i did

when i wrote 4n+1<1000, i probably should have used an extra variable, i should have written something like
n=4k+1<1000

also when k=0, n=1, so that value of k works since it makes $n$ a positive integer

[quote="boxedexe"][quote="chess64"][hide="Solution"]$(-1)^n=1$ and $n \equiv 0 \pmod {4}$[/hide][/quote]
Did you mean $i^n=1\rightarrow n\equiv 0\pmod{4}$?

Masoud Zargar[/quote]

I meant "and" as in the second statement ($n\equiv 0 \pmod {4}$) follows from the first ($i^n = 1$). Sorry, I should have been more clear.

[quote="boxedexe"]Sorry for being unclear.  You just had a $-1$ instead of $i$.  That's what I was referring to. :)[/quote]

Ohhh yeah, :blush: it's supposed to be $-i$. :D

Here is an interesting way to do it. The credit does not go to me, but a friend of my mine named Tyler.

[hide]$(\sin t + i \cos t)^n = (\sin nt + i\cos nt)$

So that looks like DeMoivre's Theorem mutated right?

So let's just multiply by DeMoivre's Theorem to see what happens...

$(\cos t + i \sin t)^n (\sin t + i \cos t)^n = (\cos nt + i \sin nt) (\sin nt + i\cos nt)$
$(1\cos t \sin t + i\cos^2t + i\sin^2t - 1\sin t \cos t)^n = (\cos nt \sin nt + i\cos^2 nt + i\sin^2 nt - \sin nt \cos nt)$
$\left(i(\cos^2t + \sin^2t)\right)^n = i(\cos^2 nt + \sin^2 nt)$
$i^n = i = \text{Tyler happy}$

Hence $n = 1 \mod(4)$ and $1 \le n \le 1000$, or $n = {1, 5, \cdots, 997}$; there are $\frac{991-1}{4} + 1 = \boxed{250}$ of possible $n$s.[/hide]

Uh...I'm reading the solution posted for this problem in the contests section, and it says that n is congruent to 0 mod 4. However, I think it should be 1 mod 4. The solution also says that cis(x)=cos(x)-i*sin(x), which i don't think is correct...

Huh? Last time I checked, $(-1)^1\not= 1$... and I don't see where I said that cis (x) = cos x - i*sin x...

Hmmm...I'm looking at the solution posted in the contests section:

[quote] Multiplying by (-i)^n, we get (\cos t -i\sin t)^n = \text {cis}^n \ t.[/quote]

Yeah, that's my solution. And that's what you get when you multiply the equation by $(-i)^n$. ;)

[quote="pkerichang"]I heard this irresponsible rsolution from a lecture:

[hide]
since it is true for all $t$, we can plug in $t = 0$...

we get $(\sin{0} + i\cos{0})^n = \sin{0} + i\cos{0}$
and therefore $i^n =i$

From now we see all $n = 1 (\bmod\ 4)$, and therefore $\fbox{250}$ is the answer
[/hide][/quote]
LOL, I remember that. :lol:

[quote="filletwho"]Except that's a false statement ;) . Last time I checked, cis^n(x)=(cos(x)+i*sin(x))^n.  ;) Get it?[/quote]

OK, I'll start from the beginning again.

We have the equation \[ (\sin t + i \cos t)^n=\sin nt + i \cos nt. \] Now, multiply this [entire equation] by $(-i)^n$. You should get \[ (\cos t - i\sin t)^n = \text {cis}^n \ t. \] No offense, but I think you might need to brush up on your complex numbers. And if you don't understand my solution, just read somebody else's.

well I come up with my own solution, and it's very easy...

[hide]
since it is so similar with the DeMoivre's formula, let's make it one.  Using the comp. angle property that $\sin{(90-t)} = \cos{t}$ and $\cos{(90-t)} = \sin{t}$, the original becomes $(cis(90-t))^n = \sin{nt} + i\cos{nt}$, or $\cos{(90n - nt)} + i\sin{(90n -nt)} = \sin{nt} + i\cos{nt}$.  Equating the reals and using the comp. angle thing again we see that $\cos{(90n-nt)} = \cos{(90-nt)}$, or $90n - nt \equiv 90 - nt (\bmod\ 360)$, since it is true for all angles $t$.  adding and dividing using the mod property we see that $n \equiv 1(\bmod\ 4)$ again and we get $\fbox{250}$ as the answer
[/hide]

[quote="Zellex"][quote="JesusFreak197"][quote="chess64"]\[ (\cos t - i\sin t)^n = \text {cis}^n \ t.  \][/quote]

But it's not. $\cos t +i\sin t=\cis t\not= \cos t-i\sin t$ (except when $t=k\pi$ :roll: ).[/quote]

JF197...I think he's saying that...after you multiply by $(-i)^n$, you get that equation, which you are supposed to solve for...[/quote]

YES! SOMEBODY CAN ACTUALLY READ! THANKYOU ZELLEX! ^_^

After you multiply what by (-i)^n? No offense, but I think you should make it clearer that you're multiplying both sides. And i'm not sure, since apparently, i need to brush up on my complex numbers, but i also think that you dropped a negative for the coefficient of i*sin(x) on the RHS. Either that or it's still not clear to me what you did.

I did this problem a little differently (well it's the same solution by a different method).

[hide="Solution"]
We recall the well known facts that $ \sin t\equal{}\frac{e^{it}\minus{}e^{\minus{}it}}{2i}$ and $ \cos t\equal{}\frac{e^{it}\plus{}e^{\minus{}it}}{2}$ (these are proven by expanding $ e^{it}$ using Euler's identity).

Therefore, $ \text{sic}\,t\equal{}\frac{e^{it}\minus{}e^{\minus{}it}}{2i}\plus{}i\cdot\frac{e^{it}\plus{}e^{\minus{}it}}{2}\cdot\frac{i}{i}\equal{}\frac{e^{it}\minus{}e^{\minus{}it}\minus{}e^{it}\minus{}e^{\minus{}it}}{2i}\equal{}ie^{\minus{}it}$

Now if $ \text{sic}^n\,t\equal{}\text{sic}\,nt$, we have $ i^ne^{\minus{}itn}\equal{}ie^{\minus{}itn}$, implying that $ i^n\equal{}i$.

This only happens when $ n\equiv 0\bmod 4\implies n\equal{}4k$, so there are $ \left\lfloor\frac{1000}{4}\right\rfloor\equal{}\boxed{250}$ solutions in the positive integers.
[/hide]

This was too easy to be a #9  :huh:

[quote="grn_trtle"] $ i^n \equal{} i$.

This only happens when $ n\equiv 0\bmod 4\implies n \equal{} 4k$

[/quote] Should this be $ i^n\equal{}i \implies i^{n\minus{}1}\equal{}1$ so $ n\minus{}1 \equiv 0\bmod 4$ and $ n \equiv 1 \bmod 4$?

Yeah, sorry, stupid mistake. Unfortunately, I can't edit my post anymore.

[hide="Solution"]

$(\sin t+i\cos t)^n=\sin(nt)+i\cos(nt)$
$\implies (-\cos t+i\sin t)^n=i^n(\sin(nt)+i\cos(nt))$.
$\implies (-1)^n(\cos nt-i\sin nt)=i^n(\sin(nt)+i\cos(nt))$.
$\implies (-1)^n=i^{n+1}$ and $(-1)^n*-i=i^n$.  The second equation gives $(-1)^n=-i^{n-1}=i^{n+1}$ which is the same as the first.
Therefore $n\equiv 1\pmod{4}$ or $n\equiv 3\pmod{4}$... however, testing $n\equiv 3\pmod{4}$ gives $-1=1$ and $n\equiv 1\pmod{4}$ gives $-1=-1$ therefore, $n\equiv 1\pmod{4}$ which gives $\boxed{250}$ solutions. [/hide]

Perhaps the biggest mistake in this problem is to pull out your hair for 10 minutes wondering why the above is not true for all $n$ by De Moivre's...truly a sad moment.

[b]Cheat Solution[/b]: We test $t=0$ to find $i^n = i$, so $n \equiv 1 \mod 4$, giving $\boxed{250}$ solutions
[b]Real Solution[/b]: Factor out $i$ from the RHS to get $$i(\cos(nt)-i \sin(nt)) \implies i(\cos(-nt)+ i \sin(-nt)) \implies i \text{cis}(-nt)$$ Do the same for the LHS to get $$i \text{cis} (-nt) = (i \text{cis}(-t))^n$$ Clearly $i$ has to cancel out, so $n \equiv 1 \mod 4$ or $n \equiv 3 \mod 4$. If it is $3 \mod 4$ then test $t=0$, contradiction. Therefore $n \equiv 1 \mod 4$, giving $\boxed{250}$ solutions

Fast way: Assume that $t=0$, the equation turns to $I^n=i$, leads to $250$ as answer

Factor out the $i$ in both of them, getting ${i^n(\cos{t}-i\sin{t})=i(\cos{nt}-i\sin{nt}})$. Putting this in exponential form, $i^n \cdot e^{-i \cdot n  \cdot
 t}=i  \cdot  e^{-i  \cdot  n  \cdot  t}$, which is true when this true when $n\equiv 1 \mod{4}$. Then, $250$.


[quote="Peter"]The Fibonacci sequence $\{F_{n}\}$ is defined by \[F_{1}=1, \; F_{2}=1, \; F_{n+2}=F_{n+1}+F_{n}.\] Show that $F_{mn}-F_{n+1}^{m}+F_{n-1}^{m}$ is divisible by $F_{n}^{3}$ for all $m \ge 1$ and $n>1$.[/quote]

Once again let's use induction for $m$.

1. For $m=1$ everything is obvious.

2. Let's suppose that we have proved the statement for $m=k$.

3. And finally let's prove it for $m=k+1$.

\[F_{nk+n}=F_{nk}F_{n+1}+F_{nk-1}F{n}\equiv F_{n+1}(F_{n+1}^{k}-F_{n-1}^{k})+F_{nk-1}F_{n}\]
=>
\[F_{n(k+1)}-F_{n+1}^{k+1}+F_{n-1}^{k+1}\equiv F_{nk-1}F_{n}-F_{n+1}F_{n-1}^{k}+F_{n-1}^{k+1}\]

As we have proved here:[url]http://www.problem-solving.be/pen/viewtopic.php?t=415[/url]
\[F_{nk-1}\equiv F_{n-1}^{k}(mode F_{n}^{2})\]
=>
\[F_{n}F_{nk-1}\equiv F_{n}F_{n-1}^{k}(mode F_{n}^{3})\]
=>
\[F_{nk-1}F_{n}-F_{n+1}F_{n-1}^{k}+F_{n-1}^{k+1}\equiv F_{n}F_{n-1}^{k}-F_{n+1}F_{n-1}^{k}+F_{n-1}^{k+1}=0\]

we are done! :wink:


AMC USA(J)MO USAMO inequalities function vector calculus

$0 \leq a, b, c \leq 1$. Prove:
\[\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1\]

P.S. Pre-Olympiad seems to have become a sub-forum of the Inquality Section.  :P

[quote="Rushil"]This is a  standard question to be done using Weistrass's theorm....[/quote]
then please post it.

The inequality is convex in $a,b,c$, so we need only calculate the value of the LHS over $a,b,c\in\{0,1\}$

This proof from "Problem Solving Through Problems" by L Larson may be helpful.It shows that we dont always need to use powerful techniques to get results.
Assume WLOG that a<= b<= c

Then we have a/(b+c+1) + b/(c+a+1) + c/(a+b+1) <=(a+b+c)/(a+b+1)

We will show that (a+b+c)/(a+b+1) +(1-a)(1-b)(1-c)  <= 1. Write L for the left side of the inequality we have to prove .

Then  L =(a+b+1)/(a+b+1) +(c-1)/(a+b+1) +(1-a)(1-b)(1-c)   = 1 -(1-c)/(a+b+1).[1-(1+a+b)(1-a)(1-b)]

The desired inequality follows from the last expression after noting that

(1+a+b)(1-a)(1-b) <= (1+a +b +ab)(1-a)(1-b) =( 1+a)(1+b)(1-a)(1-b)=(1-a^2)(1-b^2) <=1

[quote="blahblahblah"]The inequality is convex in $a,b,c$[/quote]

can u show that it is convex?

That's obvious. Fix $b$ and $c$, then the left-hand side is the sum of linear and homographic functions of $a$, all of which are convex, and thus the function itself is convex too.

can u argue that way?
i think $\phi(a,b,c)=a^2b^2c^2$ is concave and if u fix b and c it is convex as well  :?:

the only method to show that a function of a vector is concave is with the [url=http://mathworld.wolfram.com/Hessian.html]hessian[/url]

correct me if im wrong ...

In order to prove the inequality $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right) \leq 1$, we don't need the left hand side to be convex as a function from $\mathbb{R}^3$ to $\mathbb{R}$, but it is enough that if we fix two of the variables - for instance, b and c -, then the resulting one-variable function of a will be convex. This is what Blahblahblah meant with "convex in a, b, c". So, a function like $a^2b^2c^2$ would also work, although it is not convex as a function from $\mathbb{R}^3$ to $\mathbb{R}$. The underlying method of proving inequalities was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=15431 (particularly, Lemma 2 in post #3); you don't need Bolzano-Weierstraß or any kind of calculus to justify it.

The inequality in question is well-known, posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=11854 , http://www.mathlinks.ro/Forum/viewtopic.php?t=3753 and http://www.kalva.demon.co.uk/usa/usoln/usol805.html , and every time the same solution was given.

  darij

I meant that the LHS is convex in $a$ (and) $b$ (and) $c$, when any two variables are fixed, not that the LHS is itself convex as a function of three variables.  The advantage of my method is that we may conclude that $a,b,c\in\{0,1\}$, and then the result follows easily.

could somebody please give some links for weierstraß or bolzano-weierstraß, or maybe even explain it.

i looked quickly on mathworld didnt quite see the conclusion of that i was reading, especially in the case of a bounded vector and an inequality ...  :?:

We are just using the fact that a set in $\mathbb{R}^n$ is compact iff it is closed and bounded (both of which should be obvious here).  Also, continuous functions attain their maximum on compact sets.  As far as I remember, neither of these theorems go under the name Bolzano-Weierstrass, so I don't know what that has to do with anything.


I got stuck with this one .

OM,ON and OR are  three line segments.Three parallelograms OMAN,ONBP and OPCM are drawn.Prove that AP,BM and Cn bisect each other.

[hide="Hint1a"]Diagonals of parallelograms bisect each other. [/hide]
[hide="Hint1b"]Try to prove the quadrilateral $ MNBC$ is a parallelogram.[/hide]
[hide="Hint2"]A quadrilateral is a parallelogram [b]if and only if[/b] a pair of opposite sides are both parrallel and equal.[/hide]

Well I have started to work on it .Let's see...

Oh! What a fool I was (or I am). I troubled the helpful people of this forum unnecessarily.Sorry and of course, thank you


Let $ABC$ be a right triangle with $A=90$ and $\angle B>\angle C$ and let $D$ be a point on $BC$. The angle bisector of $\angle ADB$ and $\angle ADC$ intersect the sides $AB$ and $AC$ at $M$ and $N$ respectively. Let $MN$ intersect $BC$ at $P$. Show that if $\angle MPC = \frac{1}{2}(\angle B-\angle C)$, then $\angle ADC$ is a right angle.

Angle chasing!
$\angle{MPC}=180^{o}-\angle{C}-\angle{CNP}=180^{o}-\angle{C}-90^{o}-\angle{AMN}$

Hence: $\frac{1}{2}(\angle{B}-\angle{C})=90^{o}-\angle{C}-\angle{AMN}$
$\angle{B}-\angle{C}=180^{o}-2\angle{C}-2\angle{AMN}$
$\angle{B}+\angle{C}=180^{o}-2\angle{AMN}$
$2\angle{AMN}=90^{o}$
$\angle{AMN}=45^{o}$

Since $\angle{ADN}=\angle{CDN}$, $\angle{ADM}=\angle{BDM}$ and $\angle{ADN}+\angle{CDN}+\angle{ADM}+\angle{BDM}=180^{o}$.
Hence: $\angle{NDM}=\angle{ADN}+\angle{ADM}=90^{o}$

So $\angle{A}+\angle{NDM}=180{o}$. Now we see that $ANDM$ is cyclic. Hence:
\[\angle{AMN}=\angle{ADN}=45^{o}\]
Hence $\angle{ADC}=90^{o}$.

[quote="warut_suk"]You solved it even though I made a typo![/quote]
Did you? I didn't notice it....


Find all functional equation $ h:Z-->Z $ satisfying :
  $ h(x+y) + h(xy) = h(x)h(y) +1 $

Romano! You can use [code]\to[/code] for $-->$  ;)

Apologies in advance for the ugly non-LaTeX solution...

I'm supposed to be studying for Physics exam, but I assumed that if you can find h(1), then you could probably solve it..

SO here goes:

In the h(-1)=0 case,
Let x=y=-1
then h(-2) + h(1) = 1
so h(-2) = 1 - h(1)

Let x = -2, y=1
then h(-2) = h(1)h(-2) + 1

substituting:

1-h(1) = (1-h(1))h(1) + 1
Simplify:

-h(1) = h(1) - h(1)^2
Either h(1) = 0 or h(1) = 2

if h(1) = 0 then
h(x+1) + h(x) = 1, so the function would alternate since h(0) = 1..
so ...,0,1,0,1,0,1,...

Argh! Gotta go for now.. figure out the h(1) = 2 case for me.. somebody?

This problem is rather easy .
  I will solve in the case $ h(1)=2 $
  Let y=1 :
 $ h(x+1)=h(x)+1 $
 $ \to h(x+n)= h(x) +n $
  Let x=0 :
 $ \to h(n)=n+1 $
 We can prove for $ f:Q \to Q $

Thanks, my mom was ready to yell at me..  :D

I knew the answer though.. :P


[quote="SplashD"]how did you get the values of a^5+b^5 and a^6+b^6?[/quote]
[hide]$(a+b)(a^4+b^4)=a^5+b^5+ab(a^3+b^3)$[/hide]
[hide]$(a^3+b^3)^2=a^6+b^6+2a^3b^3$[/hide]
No answerers  :( . It isn't really hard as much as it is a lot of work.

I think you messed up your $a^4+b^4$
$(a^2+b^2)^2=a^4+b^4+2a^2b^2 \Longrightarrow a^4+b^4+2(ab)^2=4 \Longrightarrow a^4+b^4=\boxed{\frac{7}{2}}$

[quote="ragnarok23"][quote="pgpatel"]If $a+b=1$ and $a^2+b^2=2$ what is $a^3+b^3+a^4+b^4+a^5+b^5+a^6+b^6$?
The hints are in no order. This would be sprint 30 probably.
[hide="hint"]ab=-1/2[/hide]
[hide="hint"]$a^3+b^3=(a+b)(a^2+b^2-ab)$[/hide]
[hide="hint"]$a^3+b^3=5/2$[/hide]
[hide="hint"]$(a^2+b^2)^2=a^4+b^4+2a^2b^2$[/hide]
[hide="hint"]$a^4+b^4=15/4$[/hide]
[hide="hint"]$a^5+b^5=5$[/hide]
[hide="hint"]$a^6+b^6=13/2$[/hide]
[hide="answer"]$71/4$[/hide][/quote]
How did you get 5 for $a^5+b^5$?
[hide]What I did was $(a^3+b^3)(a^2+b^2)=5 \Longrightarrow a^5+b^5+a^3b^2+a^2b^3=5 \Longrightarrow a^5+b^5+a^2b^2(a+b)=5 \Longrightarrow a^5+b^5+(ab)^2(1)=5 \Longrightarrow a^5+b^5=5-1/4 \Longrightarrow a^5+b^5=\boxed{\frac{19}{4}}$[/hide][/quote]
I messed up $a^4+b^4$ which threw that one off. THanks to ragnorak for pointing that out.

okay I agree with that answer. I am too lazy to type up my full page of work but i will post what I got for each individual set...
[hide]
a^3+b^3=2.5
a^4+b^4=3.5
a^5+b^5=4.75
a^6+b^6=6.5
[/hide]
If I went wrong (which is highly likely), tell me what part.

[quote="ragnarok23"]Ok my final answer:
[hide]$\frac{69}{4}$[/hide][/quote]
This is my edited answer.


how do u get muted for saying retart
and secondly what is false abuse report?

First of all, you've posted in the wrong forum. This is the Mathcounts forum, which is meant to be a place to talk about Mathcounts, not about FTW issues with mutes/abuse reports/etc. This should be in the FTW Questions and Support forum.

Second of all, it's pretty clear that calling someone a retard is harassing other users. Read the Terms of Service before playing FTW. It's just common sense rules, no harassing/insulting/flaming other users.

A false abuse report is an abuse report being sent that falsely accuses someone of violating a rule when he/she has not actually violated a rule. Clearly you should not be sending false abuse reports because that's like slander, which damages one's reputation.

on my mute it says false abuse report but im still muted

Um.. Well insulting someone is not the best thing to do in FTW. FTW was made for math peoples? And this is suspicy. Cause nikeballa has the same avatar. I think.

3 months??? I don't think it's 3 months.. Usually it's like 1 week. But I cursed like the bad word. The really bad word. I deserved to be muted. So I guess I got muted for life or something.. Cause it's been 2 weeks. I think.  :maybe:

[quote="ACT36"]is it really possible to be muted for 3 months for false abuse report[/quote]
u haven't even been on aops for 3 months

ok, why did you say this topic was about mc nats?
if you want to talk about getting muted, pm levans or something if youre that worried...

[quote="ACT36"]is it really possible to be muted for 3 months for false abuse report[/quote]

wtf dude?
my avatar much?
delete it now, yuo already stole my signature (everyone go look at his signature, he didn't get any of those scores.)

Yeah, seriously, why is the topic named "mathcounts: nats"? :huh:


I need help!!!!!!!!!
I am trying to download latex but the diresttions on AOPS do not seem to match the things on the texnic center website. I keep following the AOPS direstions but cannot seem to find the right download client.
please help. :!:  :blush:  :what?:

Please give us the [i][b]full[/b][/i] details of what you tried and what went wrong.

ok,
first i clicked on the link in the latex section of aops diresting me to the texnic center web page. Then the directions tell me to click on the downloads of section the website. This is the URL: [url]http://www.texniccenter.org/resources/downloads/29[/url]
Then i cannot find the "the link next to Texnic center setup section next to top of the page" so i do not know what to click after that and how i should download this.

Please advise :(  :o  :thumbup:  :maybe:

The link doesn't seem to be very clear. Click on [i]TeXnicCenter Installer[/i] which is just after [i]1.[/i] and has the words [i]Download the latest Setup if you want to use TeXnicCenter[/i]. after it.

i downloaded it, but when i try and build the first document, an error occurs. I think one of the reasons is that in the bar that has latex => ..., it is supposed to say latex => PDF but mine says latex => DVi. Do you know how to fix this?
I understand if [i]you[/i] don't understand because my description of events is quite unclear.

Did you tell TexnicCenter where to find the LaTeX files when you first started it? From AoPSWiki
[quote]You need to tell TeXnicCenter where MiKTeX is. If you are using MiKTeX 2.7 it will probably be C:\Program Files\MiKTeX 2.7\miktex\bin (this folder contains lots of exe files including latex.exe). You do this by clicking on Build, Define Output Profiles and clicking on the Wizard button in the bottom left corner. [url=http://www.youtube.com/watch?v=-h8Oi1oCTQg]This video[/url] will show you what to do. [/quote]

yes i did but it cannot find the distributor in the file that i type.

What do you mean by 'distributor'? All you need is the directory which contains lots of files including latex.exe. Where did you install MiKTeX? In that directory there will be subfolder called miktex and nside that is a directory called bin. That's the one you want ie [i]where_MiKTeX_is_installed[/i]\miktex\bin

wait!!!!!!!!!!!!!!! :o
i figured out how to install it properly and make it work!!!!!!!!
thanks anyway for you patience and cooperation!
i know it is kind of hard to communicate with someone and explain to them how to install a program properly and make it work.

all i need to do is learn [i]how[/i] to use latex properly.

thanks


Welcome to the Joke Game!
Anyone can post as long as you say a joke.

Here's 1:
What happened to the peanuts as they crossed the road?
[hide]
[hide]
[hide]

They got a-salted!
[/hide] [/hide] [/hide]

Can we post a vid? If so:

[youtube]Ygy7UDADXDg[/youtube]

WARNING: some profanity

In a 5th grade class, the teacher asks: "Which president said 'Ask not what your country can do for you, ask what you can do for your country.' ?"

No one raises their hand. Then,

A Japanese boy in the class says, "John F Kennedy!"

Teacher says, "Correct! See, even a Japanese person knows, you americans should pay more attention to american history!"

A second boy whispers, "FCUK the Japanese!"

The teacher angrily asks, " WHO SAID THAT????"

The Japanese boy exclaims, "Franklin D. Roosevelt!!

LOL!

A first-grade class in Brooklyn comes in from recess. The teacher asks Sarah: What did you do at recess?

Sarah: I played in the sandbox.

Teacher: That's good. If you can go to the blackboard and write "sand" correctly, I'll give you a cookie.

She does, and gets a cookie.

The teacher asks Timmy what he did at recess.

Timmy: I played with Sarah in the sandbox.

Teacher: That's good. If you can go to the blackboard and write "box" correctly, I'll give you a cookie.

He does, and gets a cookie.

The teacher then asks Song-he what he did at recess.

Mohammed: I tried to play in the sandbox with Sarah and Timmy, but they threw rocks at me.

Teacher: They threw rocks at you? That sounds like blatant racial discrimination. If you can go to the blackboard and write "blatant racial discrimination" correctly, I'll give you a cookie.

Who is Mohammed to interrupt Song-he? He must be really rude!

My classmate Isaac has a ton of really bad puns.
It's his hobby to make bad puns.

Why did the chicken cross the street?

Because it was a game of Scrabble!

[hide="Havent you ever had the urge to loose control when ordering that burrito or burger and drink combo?"]

1. Ask for last months specials.

2. Place your order in three different languages if you don't know any, make them up.

3. When they repeat your order totally change it. Repeat as desired.

4. Order a whopper from McDonalds, when they say they don't have whoppers insist that they do. If they still argue demand to see a manager then when you talk to them order a normal meal and say i don't know what's up with kids these days.

5. Go to any burger joint and order Chinese.

6. When ordering in the drive through, ask if its happy hour on draft beer.

7. In summer turn stereo up full volume to Christmas music while ordering in drive through.

8. Drive in the drive through, park, then go inside and order.

9. Go through the drive through in reverse, again.

10. Wait for the busiest time of day, after paying get out of car, get jack out of trunk and proceed to rotate tires.

11. Check oil in drive through, possibly touch up the windscreen with some glass cleaner.

12. Walk up to drive through window with hands in the air like your holding onto a steering wheel.

13. Go through drive through and ask for directions to the place you're at.

14. Go through drive through naked, excluding the rabbit slippers of course!

15. Go inside and ask in a load voice if they got rid of the Ebola problems they've been having

16. Argue with your passenger (that's not there) and continue until you pay.

17. When they ask if that will be all tell them to hold on, your dog won't make up his mind.

18. After eating half your meal return it and say you got the wrong thing and ask for a refund.

19. As you're pulling away and they say "Have a nice day!" (with that retarded smile) put the car in reverse and ask them what the hell that's supposed to mean.

20. Order something from one fast food place then go to another drive through when they give you your food take one bite and say "No thank you
this tastes better (pointing to bag from other place), may i have my money back."

21. Play name that tune with person taking the order.

22. Pretend your deaf and order in sign language.

23. Go through Taco Bell order the 7 layer burrito. When they repeat your order ask if each layer of the burrito is sold separately.

24. When they ask how you are today proceed by telling them about your diarrhea problems, most likely due to the burger you bought from them yesterday and the day before that.

25. Go through a drive through after they tell you to pull to the window drive past it and go inside to get your order.[/hide][hide="Answering Service At The Mental Institute"]

"Hello, and welcome to the mental health hotline.

If you are obsessive-compulsive, press 1 repeatedly.

If you are co-dependent, please ask someone to press 2 for you.

If you have multiple personalities, press 3,4,5, and 6.

If you are paranoid, we know who you are and what you want. Stay on the line so we can trace your call.

If you are delusional, press 7 and your call will transferred to the mother ship.

If you are schizophrenic, listen carefully and a small voice will tell you which number to press.

If you are manic depressive, it doesn't matter which number you press, no one will answer.

If you have a nervous disorder, please fidget with the hash key until someone comes on the line.

If you are dyslexic, press 6969696969.

If you have amnesia, press 8 and state your name, address, phone number, date of birth, social security number, and your mother's maiden name.

If you have post-traumatic-stress disorder, slowly and carefully press 000.

If you have bipolar disorder, please leave a message after the beep, or before the beep, or after the beep. Please wait for the beep.

If you have short-term memory loss, press 9. If you have short term memory loss, press 9. If you have short term memory loss, press 9. If you have short term memory loss, press 9.

If you have low self esteem, please hang up. All our operators are too busy to talk to you."[/hide]

well, this is a somewhat cheesy math joke, but here i go:

I shall prove that 2+11-1=12.

two

tw[u]o[/u]el[u]e[/u]ve[u]n[/u], take away the o, the n, and an e, [/u]i.e. take away one, and what are you left with?

1=2's second joke reminded me of [url=http://www.rd.com/clean-jokes-and-laughs/mary-roach-my-planet----customer-service/article26862.html]this[/url].


trigonometry
real analysis
calculus
calculus computations

i Have the feeling that i need to use riemann sum for this question. but i can't seem to get it. can someone help?

Evaluate the sum:

$\sum^{\infty}_{k=1}{\arctan{(\frac{1}{1+k+k^{2}})}}$


What is the smallest whole number greater than $ \sqrt{30} \plus{} \sqrt{56} \plus{} \sqrt{110}$?

[quote="GameBot"]What is the smallest whole number greater than $ \sqrt {30} \plus{} \sqrt {56} \plus{} \sqrt {110}$?[/quote]

$ 7.5^2\equal{}56.25$
$ 5.5^2\equal{}30.25$
$ 10.5^2\equal{}110.25$

Thus, about $ 7.5\plus{}5.5\plus{}10.5\equal{}23.5$, or 24.


Of the following, which is the best buy, A or B?

A) 10 lbs of apples for $ \$3.95$ \\
B) 5Kgs of apples for $ \$3.85$

We know that kilograms is weighes slight more than $ 2$ lbs, so B) is the better buy.


A firm makes two kinds of shoes. Each shoe of the first type requires twice the labor to produce a shoe of the second type. If only shoes of the first type are produced, the firm can make $ 500$ shoes a day. The market limits sales to $ 150$ and $ 250$ shoes of the first and second types, respectively. Profits per shoe are $ \$8$ for type $ 1$ and $ \$5$ for type $ 2$. The firm wants to maximize profits. Solve this.

So let $ x_1 \equal{} \ \text{number of type 1 shoes produced}$ and $ x_{2} \equal{} \ \text{number of type 2 shoes produced}$. So $ \text{max} \ z \equal{} 8x_1 \plus{} 5x_2$ subject to $ x_1\leq 150$, $ x_2 \leq 250$ and $ x_i > 0$ for $ i \equal{} 1,2$. And the answer is $ (150,250)$?

Is this correct?

[quote="ragingbullrocky"]A firm makes two kinds of shoes. [color=red]Each shoe of the first type requires twice the labor to produce a shoe of the second type. If only shoes of the first type are produced, the firm can make $ \color{red}500$ shoes a day.[/color] The market limits sales to $ 150$ and $ 250$ shoes of the first and second types, respectively. Profits per shoe are $ \$8$ for type $ 1$ and $ \$5$ for type $ 2$. The firm wants to maximize profits. Solve this.

So let $ x_1 \equal{} \ \text{number of type 1 shoes produced}$ and $ x_{2} \equal{} \ \text{number of type 2 shoes produced}$. So $ \text{max} \ z \equal{} 8x_1 \plus{} 5x_2$ subject to $ x_1\leq 150$, $ x_2 \leq 250$ and $ x_i > 0$ for $ i \equal{} 1,2$. And the answer is $ (150,250)$?

Is this correct?[/quote]
Your solution takes no account of the information in red. However, the question is so badly worded that it's not obvious why that information is there.

I think the only way to interpret the question is to assume that all the information is about a single day's output. If that is the case, then you need to add another inequality $ x_1\plus{}2x_2\leqslant500$. The solution $ (150,250)$ is not then viable, because the firm lacks the capacity to produce that number of shoes in a day.

I actually read that as $ x_1\plus{}\frac12x_2\le 500,$ in which case $ (150,250)$ is feasible.

This is a linear programming problem, not a calculus problem. This post doesn't belong in the calculus section, but where does it belong? There's not a particularly good answer to that, but I think I'll go for the linear algebra section.

And to the O.P.: do look up the words "linear programming".


Find all continuous fuctions $ f$ defined on $ \mathbb{R}$such that $ f(x)\equal{}f(2x)$ for all $ x\in \mathbb{R}$.

Let $ g(x) \equal{} f(e^x)$.  Then $ g$ is continuous and has period $ \ln 2$.

Edit:  My apologies, this is only valid on $ \mathbb{R}_{>0}$.

For any $ x\ne0$, $ f(x)\equal{}f(2^{\minus{}n}x)$. Let $ n$ go to infinity, $ f(x)\equal{}f(0)$.


advanced fields
advanced fields unsolved

Solutions to problems on Langer's uniform approximation from Bender and Orszag book (Chapter 10 - WKB theory chapter)

What is going on¿  :huh:

What is this topic even about¿ Do you want the solutions¿ Then post the questions!

Deduce Langer's uniform approximation to the solution of Schrodinger equation directly from the differential equation


vector
superior algebra
superior algebra unsolved

Let $ A$ be a ring with property : for every $ a \in A$ , exist $ K \in N$ s.t $ a^{2k}\equal{}a$ . Prove that :
a) $ |A|$ is a power of 2
b) $ \sum_{\substack{
a \in A \\}} a \equal{}0$

Well, this is blatantly false if $ |A| \equal{} 1, 2, \infty$.  Otherwise:

Since every element is a square, $ z\mapsto z^2$ is surjective, hence injective.  But $ z^2 \equal{} ( \minus{} z)^2$, so $ z \equal{} \minus{} z$ for all $ z\in A$, hence every nonzero element of $ A$ has additive order $ 2$.

Then the additive group of $ A$ has the structure of an $ \mathbb{F}_2$-vector space of dimension at least $ 2$, from which a) and b) follow immediately.

It's not false for $ \left|A\right| \equal{} 1$, in which case the only element is 0 so that (a) $ \left|A\right| \equal{} 2^0$ and (b) $ \sum 0 \equal{} 0$.


geometry
superior algebra
superior algebra theorems

Hello,

I am sorry that this is more geometry than algebra.  But incidence geometry doesn't really seem to have a place here. :(

Consider a projective plane $P$.  Let $L$ be a line.  An elation is by definition a collineation( an automorphism of the projective plane) fixing each points on the line L, but not fixing any other points.

Now let $a$ and $b$ be two elations.  Prove that the composition $a b$ is again an elation, or trivial.

This is important, as it is the only difficult axiom that needs to be proven in order to see that they form a group, together with the trivial collineation.  Of course, the composition will still be a collineation fixing all points on the line $L$, but how can one prove it can't fix any other points as well?

Thank you,
fredbel6

Equivalently, we want to show that $ab$ fixes all points if it fixes one point not on $L$.

This is false for general collineations fixing $L$- we need to use that $ab$ is a product of elations which fix no points outside $L$.

I'll look at it.

I got it :

first one needs to know that given an elation with fixed axis $L$, the elation is completely determined by its action on just one other point

now let $a,b$ be elations with axis $L$ and assume that $a b=u$ is nontrivial and not an elation

It will still be a perspectivity with axis L, thus a homology.  Let its center be c.

now $c^{a}=c^{u b^{-1}}=c^{b^{-1}}$

This means that $a$ and $b^{-1}$ map c onto the same point, so they are equal.  Thus $a b$ is tirival, contradicting our assumption

Thanks jmerry for the efforts.


Έστω ότι συμμετέχουμε σε ένα τηλεπαιχνίδι, το οποίο συμπεριλαμβάνει το εξής μέρος:

Η παραγωγή έχει ετοιμάσει τρεις πόρτες. Πίσω από τις δύο πόρτες κρύβονται δύο κατσίκες αντίστοιχα. Πίσω από την τρίτη κρύβεται ένα αυτοκίνητο. Ο παρουσιαστής λοιπόν σας καλεί να διαλέξετε μία πόρτα, στην οποία νομίζετε πως βρίσκεται το αυτοκίνητο. Αφού διαλέξετε εσείς την πόρτα, ο παρουσιαστής ανοίγει μία από τις υπόλοιπες 2 πόρτες που κρύβει μία κατσίκα. Στη συνέχεια σας ρωτάει αν θέλετε να παραμείνετε στην επιλογή σας ή αν θέλετε να αλλάξετε απόφαση και έτσι να διαλέξετε την άλλη πόρτα.

Τι σας συμφέρει να διαλέξετε;

α) Να παραμείνετε πιστός στην αρχική επιλογή σας.
β) Να αλλάξετε την επιλογή σας και να διαλέξετε την πόρτα που έχει απομείνει.
γ) Δεν έχει σημασία, οι πιθανότητες είναι 50% για την κάθε πόρτα.

den sini8izo na pezo se tilepexnidia.....ala an pote to kano 8a exo meletisi poli kala proigumenos tin simperifora tu tileparusiasti se analoges periptosis....
apo dies8isi+mono den 8a alaza porta...

 :D

p.s mostel tlk ti 8a kerdiza...??
to aytokinito i 1 kilo fresko gala??... :P

Εγώ το πρόβλημα το ξέρω έτσι :

Eίστε σε ένα τηλεπαιχνίδι.. και έχετε να διαλέξετε ανάμεσα σε 3 κουτιά αβγ από τα οποία μονό το ένα έχει δώρο μια επιταγή των 100.000 ευρώ !!!!
κάνετε την επιλογή σας και διαλέγετε ένα κουτί , αφού έχετε κάνει την επιλογή σας ο τηλεπαρουσιαστής ο όποιος ξέρει που είναι το δώρο ανοίγει το ένα από τα αλλά δυο κουτιά και σας δείχνει ότι είναι άδειο και δεν είναι εκεί το δώρο και τώρα σας ερωτάει "θέλετε να επιμένετε στην αρχική σας επιλογή η προτιμάτε να διαλέξετε το άλλο κουτί?"

[hide]Έστω Α - Β - Γ τα κουτιά
Αν Ε η επιταγή και Κ το κενό τότε οι πιθανοί συνδυασμοί των κουτιών είναι οι ακόλουθοι

Α   Β   Γ

Ε   Κ   Κ
Κ   Ε   Κ
Κ   Κ   Ε

Έστω ότι επιλέγουμε το Α κουτί. Όπως βλέπουμε απότην 1η στήλη οι πιθανότητες να βρούμε την Επιταγή (Ε) είναι 1 στις 3, δηλ 33%

Ο παρουσιαστής ανοίγει ένα κουτί και μας δείχνει ότι είναι κενό. Έχουμε δηλαδή (αν Χ είναι το κενό):

Α   Β   Γ

Ε   Κ   Χ
Κ   Ε   Χ
Κ   Χ   Ε

Παρατηρούμε ότι αν κρατήσουμε το κουτί Α έχουμε πιθανότητα 33% (παραμένει η αρχική προφανώς), ενώ αν αλλάξουμε κουτί έχουμε πιθανότητα 66% να πετύχουμε την Επιταγή.
Επομένως, με την αλλαγή της αρχικής μας πρόβλεψης έχουμε διπλασιασμό των πιθανοτήτων να κερδίσουμε την Επιταγή .[/hide]

Σωστά giannis, η "αυστηρή" απόδειξη γίνεται με το θεώρημα του Bayes. Όταν βρω χρόνο θα τη βάλλω.

PS: Ποια τακτική ακολουθούμε στην περίπτωση που έχουμε $n$ πόρτες;  :wink:

[i]Απόδειξη με θεώρημα [u]Bayes[/u]:[/i]

Θα πρέπει να υπολογίσουμε την πιθανότητα για την οποία το αυτοκίνητο βρίσκεται πίσω από τις δύο κλειστές πόρτες. Για να το κάνουμε αυτό, μπορούμε να χρησιμοποιήσουμε τον κανόνα του Bayes. Αρχικά να εξηγηθεί το σύμβολο $P(A|B)$, το οποίο χαρακτηρίζει όλες τις πιθανές περιπτώσεις όπου το $B$ είναι αληθινό (ισχύει) έτσι ώστε και το $A$ να είναι αληθινό. Με άλλα λόγια, $P(A|B)$ είναι η αναλογία: $\frac{P(A\land B)}{P(B)}$, όπου $P(A\land B)$ σημαίνει ότι και το $A$ ισχύει και το $B$, και $P(B)$ σημαίνει ότι το $B$ ισχύει. Επίσης ισχύει για τον κανόνα του Bayes: $P(A\land B)=P(A|B)P(B)=P(B|A)P(A)$.

Έστω λοιπόν $A$ η περίπτωση όπου "το αυτοκίνητο βρίσκεται πίσω από την πόρτα $2$". Τότε πρέπει να υπολογίσουμε την πιθανότητα του $A$, έχοντας ως δεδομένο ότι ο παρουσιαστής έχει ανοίξει την πόρτα $3$. Έστω λοιπόν $B$ η περίπτωση όπου "ο παρουσιαστής έχει ανοίξει την πόρτα $3$". Αυτό που πρέπει να υπολογίσουμε είναι το: $P(A|B)$. Όμως από τον κανόνα του Bayes ισχύει: $P(A|B)=\frac{P(A\land B)}{P(B)}=\frac{P(B|A) P(A)}{P(B)}$

Το παραπάνω δηλαδή σημαίνει ότι ο παρουσιαστής θα ανοίξει την πόρτα $3$, έχοντας ως δεδομένο ότι το αυτοκίνητο βρίσκεται πίσω από την πόρτα $2$. Από τη στιγμή που ο παρουσιαστής δεν πρόκειται να ανοίξει την πόρτα που κρύβει το αυτοκίνητο, θα έχουμε: $P(B|A)=1$. Επίσης είναι εύκολος και ο υπολογισμός του $P(A)$. Πρόκειται για την "απόλυτη" πιθανότητα ότι το αυτοκίνητο βρίσκεται πίσω από την πόρτα $2$, χωρίς περαιτέρω υποθέσεις. Έτσι έχουμε λοιπόν $P(A)=\frac{1}{3}$. Έτσι παίρνουμε: $P(A|B)=\frac{\frac{1}{3}}{P(B)}$.

Για να υπολογίσουμε το $P(B)$, πρέπει να χρησιμοποιήσουμε ένα άλλο δεδομένο. "Αν όλες οι πιθανές περιπτώσεις χωριστούν σε ξέχωρα μεταξύ τους κομμάτια (χωρίς αυτές οι πιθανές περιπτώσεις να συμπίπτουν), τότε η πιθανότητα της κάθε περίπτωσης μπορεί να υπολογιστεί από το άθροισμα των πιθανοτήτων της συγκεκριμένης περίπτωσης". Πιο συγκεκριμένα στην προκειμένη περίπτωση έχουμε τις εξής περιπτώσεις: είτε το αυτοκίνητο βρίσκεται πίσω από την πόρτα $1$, είτε πίσω από την πόρτα $2$, είτε πίσω από την πόρτα $3$. Εάν θέλουμε να βρούμε την περίπτωση στην οποία ο παρουσιαστής ανοίγει την πόρτα $3$, τότε πρέπει να υπολογίσουμε τις περιπτώσεις:

$a$) ο παρουσιαστής ανοίγει την πόρτα $3$ και το αυτοκίνητο βρίσκεται πίσω από την πόρτα $1$,
$b$) ο παρουσιαστής ανοίγει την πόρτα $3$ και το αυτοκίνητο βρίσκεται πίσω από την πόρτα $2$,
$c$) ο παρουσιαστής ανοίγει την πόρτα $3$ και το αυτοκίνητο βρίσκεται πίσω από την πόρτα $3$.

Στη συνέχεια αρκεί να προσθέσουμε τις $a$, $b$, $c$.

Για να το εκφράσουμε συμβολικά, ας δώσουμε ονόματα σε κάθε μία από τις $3$ περιπτώσεις. Έστω $A_{1}$ ότι "το αυτοκίνητο βρίσκεται πίσω από την πόρτα $1$", $A_{2}$ ότι "το αυτοκίνητο βρίσκεται πίσω από την πόρτα $2$", $A_{3}$ ότι "το αυτοκίνητο βρίσκεται πίσω από την πόρτα $3$".

Τότε λοιπόν θα έχουμε: $P(B)=P(B\land A_{1})+P(B\land A_{2})+P(B\land A_{3})$. Εφαρμόζοντας και πάλι τον κανόνα του Bayes παίρνουμε: $P(B)=P(B|A_{1})P(A_{1})+P(B|A_{2})P(A_{2})+P(B|A_{3})P(A_{3})$.

Τώρα μπορούμε να υπολογίσουμε όλες τις παραπάνω περιπτώσεις. $P(B|A_{1})$ είναι η περίπτωση στην οποία ο παρουσιαστής ανοίγει την πόρτα $3$ έχοντας ως δεδομένο ότι το αυτοκίνητο βρίσκεται πίσω από την πόρτα $1$, δηλαδή $P(B|A_{1})=\frac{1}{2}$. $P(A_{1})$ είναι η περίπτωση στην οποία το αυτοκίνητο βρίσκεται πίσω από την πόρτα $1$, που είναι $\frac{1}{3}$. $P(B|A_{2})$ είναι η περίπτωση όπου ο παρουσιαστής ανοίγει την πόρτα $3$ έχοντας ως δεδομένο ότι το αυτοκίνητο βρίσκεται πίσω από την πόρτα $2$, που είναι ίσο με $1$. Τέλος, $P(B|A_{3})$ είναι η περίπτωση στην οποία ο παρουσιαστής ανοίγει την πόρτα $3$ έχοντας ως δεδομένο ότι το αυτοκίνητο βρίσκεται στην πόρτα $3$. Όπως είναι φυσιολογικό, κάτι τέτοιο δεν θα το έκανε ποτέ ο παρουσιαστής, άρα η συγκεκρίμενη περίπτωση είναι ίση με $0$.

Αν τα προσθέσουμε όλα αυτά λοιπόν θα πάρουμε: $P(B)=\frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{1}{3}+0=\frac{1}{2}$. Με τις κατάλληλες αντικαταστάσεις παίρνουμε λοιπόν: $P(A|B)=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$. Επομένως, οι πιθανότητες το αμάξι να βρίσκεται πίσω από την πόρτα $2$ είναι $\frac{2}{3}$ και οι πιθανότητες να βρίσκεται πίσω από την πόρτα $1$ είναι συνεπώς $\frac{1}{3}$. Έτσι, θα πρέπει να [i][b]αλλάξετε[/b][/i] πόρτα!

ενδιαφερον αποτελεσμα....χρονια πολλα σε ολους  :)

[quote="RAKAR"]poli endiaferousa apodiksi mostel..
apla stin apodiksi ayti ipe8eses oti o parusiastis ikseret tin sosti apantisi
kais skopima epelekse ali porta...
an ute o idios iksere tin sosti apantisi??(gt iparxun tilepexnidia sta opia ute o tileparusiastis kseri tin sosti apantisiaplos rotai ton diagonizomeno:ise siguros i 8a i8eles na to ksanaskeftis...)
an sew ekini tin periptosi otileparusiastis (exontas pliri agnia pu vriskete to amaksi)aniksi mia porta me mnia katsika,tote 8a prepi na alaksume porta i na paraminume stin rxiki mas epilogi..??
ego ixa katalavi oti enouses oti ute o parusiastis iksere tin sosti apantisi..(diko mu la8os..)gia ayto kai arxika skeftika 50-50%...+meta ipa na kratiso tin idia porta...

 :D[/quote]

Βασικά, νομίζω πως θεωρείται δεδομένο ότι ο παρουσιαστής γνωρίζει το πού βρίσκεται το αυτοκίνητο, αφού σε αντίθετη περίπτωση, θα υπήρχε και πιθανότητα να ανοίξει την πόρτα που κρύβει το αυτοκίνητο με αποτέλεσμα το παιχνίδι να μην έχει νόημα. Δηλαδή, αυτό που θέλω να πω είναι το εξής: το παιχνίδι δεν είναι τόσο ρεαλιστικό, αφού τίθενται οι παρακάτω περιορισμοί που προφανώς σε αντίστοιχες "πραγματικές" περιπτώσεις υπάρχει πιθανότητα να μην ισχύουν:

α) Ο παρουσιαστής γνωρίζει το πού βρίσκεται το αυτοκίνητο.
β) Δεν λαμβάνονται υπόψη (όπως ανέφερες και στην αρχή) οι αντιδράσεις και οι μορφασμοί του παρουσιαστή.

Πάντως, ναι, ίσως ήταν λάθος μου που δεν ανέφερα εξαρχής ότι ο παρουσιαστής γνωρίζει το πού βρίσκεται το αυτοκίνητο.

[color=blue]ΧΡΟΝΙΑ ΠΟΛΛΑ[/color] ΣΕ ΟΛΟΥΣ

Αυτό το παράδοξο είναι γνωστό σαν [u]Monty Hall[/u]. Μπορείτε να το δείτε και  [url=http://math.ucsd.edu/~crypto/Monty/monty.html]εδώ[/url] σε έφαρμογή

@giannis Σ'αυτό το γρίφο την είχα πατήσει (είχα πει ότι είναι ίδιες οι πιθανότητες)

@mostel Όμορφη η απόδειξη με τις δεσμευμένες πιθανότητες  :)

Λουκάς

Ναι Pontios, η απόδειξη αυτή όμως είναι καλή μόνο στην περίπτωση που έχουμε περιορισμένο αριθμό πορτών. Π.χ. αν είχαμε $n$ πόρτες όπως αναφέρω σε ένα παραπάνω post μου, αυτή η τακτική δεν θα απέφερε κανένα γρήγορο και έξυπνο αποτέλεσμα. Αν θέλει κάποιος ας πει τι κάνουμε σε περίπτωση που έχουμε $n$ πόρτες πάντως. Είναι ακόμη πιο παράδοξη η λογική  :wink:

Χρόνια πολλά και από μένα σε όλους!!!  :)

@mostel Για την περίπτωση με τις $n$ πόρτες νομίζω ότι σκεφτόμαστε ως εξής: Διαλέγοντας αρχικά μία πόρτα από τις $n$, η πιθανότητα να έχουμε πετύχει το αυτοκίνητο είναι $\frac{1}{n}$. Ο παρουσιαστής από τη στιγμή που ξέρει ποια είναι η σωστή πόρτα μπορεί πάντα να ανοίξει $n-2$ πόρτες που κρύβουν κατσίκες, χωρίς εμάς να μας προσθέσει κάποια επιπρόσθετη πληροφορία για την πόρτα μας. Συνεπώς όταν μένουν δύο πόρτες η πιθανότητα να είναι το αυτοκίνητο πίσω από αυτή που διαλέξαμε αρχικά είναι ίδια με την αρχική πιθανότητα που είχαμε, δηλαδή $\frac{1}{n}$. Άρα η άλλη πόρτα που μένει έχει πιθανότητα $\frac{n-1}{n}$ να είναι η σωστή και συνεπώς μας συμφέρει να αλλάξουμε.
Στην περίπτωση τώρα που ανοίγει μία μία τις πόρτες και μας ρωτάει συνέχεια για το αν θέλουμε να αλλάξουμε, με το ίδιο σκεπτικό μας συμφέρει να παραμείνουμε στην πόρτα μας μέχρι να μείνουν δύο πόρτες και τότε να αλλάξουμε επιλογή. Θα έχουμε αυξήσει έτσι τις πιθανότητές μας σε $\frac{n-1}{n}$.   :wink:

@pontios Όλοι έτσι την πατάμε στην αρχή.  :P

Στέλιος

Πολύ σωστά Stedes :first:

PS: Ελπίζω πάντως αν κάποιος βρεθεί σε ένα παρόμοιο τηλεπαιχνίδι και ακολουθήσει την προαναφερθείσα τακτική να μη τύχει και διαλέξει τη λάθος πόρτα, γιατί τότε ...κούνια που μας κούναγε :P


Can someone tell me how the weather is going to be at Hanoi during the IMO?(Google does not seem too helpful :P)

Thanks
Bhargav

The info used to be one the official website.

It is pretty hot, it's 30-35 degrees and sunny.

EDITED

25-30 degrees Celsius? I thought it would be much warmer.

in fact, the temperature is 30-35 the day (noon), and at night less than 25°C

The temperature is high as they said and the humidity is also high too (about 80%)


AMC AIME USA(J)MO USAMO inequalities MATHCOUNTS

How did the AOPSers prepare for the upcoming math competitions in spring break? (If u studied at all.........)

Just do math constantly.  Only my spring break is over.

My Spring Break is in two weeks. In fact, it's the week right before the USAMO. You know what that means.

The three things (in order) that I did most over Spring Break:
sleep
play xbox
prep for the aime

Didn't prepare at all in the past year, and I ended up being one AIME question short of USAMO this year.  Serves me right.  Next year, PREP PREP PREP PREP PREP PREP LIKE MAD.

Next year, oh, if only it had been this year.

I didn't do anything.....except sleep :D  :D

I think I read math books, thought about math, and took AIMEs over Spring Break, mostly just because I like the problems.  I will still take them after I am graduated.

[quote="MithsApprentice"][quote="bubala"]You lucky duckies :)  My spring break is during the USAMO!  I have no idea what to do!  I'm going to be in the Caribbean!!![/quote]

Psh. Skip USAMO. It's not everyday you can go to the Caribbean!!! Oh man...you lucky duck.[/quote]

And I suppose it is everyday that you go to the USAMO? :P

Actually, I'm working out a way with Prof. Dunbar that I can take the USAMO while i the Caribbean.  Now that certainly doesn't happen every day! :)

Oh yeah! Nothing like watching the sunset while doing math - notice that would be a GOOD sunset...

The Caribbean time zone is such that one couldn't watch a sunset while taking the USAMO, at least, but maybe you meant that he could prep while watching the sunset, which would indeed be a very comfortable way to prep.

Our homeschooling group includes kids who live in several school districts, so our "spring break" ends up spanning several weeks of various classes being on or off. If I had MY druthers, the guys on my math team would use all their extra time to do more math problems, but that isn't happening yet.

[quote="xxreddevilzxx"]mine starts next week.. hahah and USAMO is in 3 weeks....

my priorities:

1.USAMO-ing off my butt.
2.AoPS
3.IB reading logs for English
4.more MATH
5.then, finally some HalfLife[/quote]

Just curious: In my school there is not IB classes, only AP classes are provided. Why is it different? And i heard that IB has more credit, is it true?

in IB they give you a class in ninth grade called inquiry skills. taught by the most anti-math teacher they can find. then in eleventh grade you get theory of knowledge, a class not even the teacher cares about. lots and lots of homework, much less time for study. yes you do get higher gpa but you really have to earn it.

usually people apply in middle school, it's virtually impossible to transfer in.

[quote="tokenadult"]If I had MY druthers, the guys on my math team would use all their extra time to do more math problems, but that isn't happening yet.[/quote]

Some of the admins were of the opinion that those who had spring break before the AIME should not have used their time to study. It seems that relaxing and not doing hard math problems would be better than studying.

[quote="MithsApprentice"]AP is something only administered within the United States (correct me if I'm wrong...there may be a few international schools that also have APs). IB's are given throughout the world. They are graded differently, but the credits and difficulty level are comparable.[/quote]

That is right. IB is more Worldly standardized, because it began as a set program for sons and daughters of ambassadors, who constantly moved everywhere.

IB and AP are different. AP testings are way easier than IB. IB also requires you to take alot of junk classes, such as TOK (Theory of Knowledge as mentioned above) and 6th IB subject your junior year. If you plan to get an IB diploma, you MUST take 6 IB classes in Junior Year. That sux, because now i have no choice next yr but to take IB Psychology, which i heard was CRAPload of hmwk.

BTW, back to the spring break. it began today for me. I just finished a chapter on telescoping and trivial inequality of MO Challenges... whew... 5hours straight

Meh. I don't have Spring Break for another week... :(

[quote="henryyangrui"]Paladin, are you the jeffery that got second place in the nation in Mathcounts last year?[/quote]

Ha! I wish :P That was probably Jeffrey Chen, who is a freshman (I'm a soph) and got a 14 on the AIME. He's on this forum I believe. I went to national MC two years ago, but didn't do spectacularly (barely made top 25%).


From $ z \equal{}\minus{}(x\plus{}y)$ we derive:
-  $ x^{5}\plus{}y^{5}\plus{}z^{5}\equal{} 5xyz(x^{2}\plus{}xy\plus{}y^{2})$
-  $ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{} 3xyz$

So if $ xyz \equal{} 0$, it's trivial. Otherwise, we have to prove that $ x^{2}\plus{}xy\plus{}y^{2}\equal{}\frac{1}{2}(x^{2}\plus{}y^{2}\plus{}z^{2}) \equal{}\frac{1}{2}(x^{2}\plus{}y^{2}\plus{}(x\plus{}y)^{2}) \equal{} x^{2}\plus{}xy\plus{}y^{2}$, which is true, I guess.


trigonometry
geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

1. Find the tangent line to the curve at $ x\equal{}a$ and express as a linear equation:

a. $ y\equal{}\cos x;\ a\equal{}\frac{\pi}{2}$.

b. $ y\equal{}\csc x;\ a\equal{}\frac{\pi}{2}$.

2. Let $ f(x)\equal{}\left \{ \begin{array}{ll}\cos x, & 0 \le x \le \dfrac{\pi}{3} \\ ax\plus{}b, & \dfrac{\pi}{3}<x \le \pi \end{array} \right \}$.

Find $ a$ and $ b$ so that $ f$ is differentiable at $ x\equal{}\dfrac{\pi}{3}$.

For a), draw the graph of $ y \equal{} \cos x$ on the interval $ [\minus{}\frac{\pi}{2}, \frac{\pi}{2}]$. Draw a straight line crossing the x-axis at $ a \equal{} \frac{\pi}{2}$ and crossing the y-axis at another point. What is the equation of this line? It is the reflection of the line $ y \equal{} x$ shifted to the right by $ \frac{\pi}{2}$.

Let's calculate: $ y \equal{} f(x) \equal{} \cos x \implies f'(x) \equal{} \minus{} \sin x$. So $ f'(\frac{\pi}{2}) \equal{} \minus{}1$ and this is going to be the slope of the line you are after. What is the equation of the line: $ y \minus{} y(a) \equal{} m(x\minus{}a)$. Now go ahead and plug in the values.

For 1) you may be knowing the formula $ y\minus{}y_1\equal{}m(x\minus{}x_1)$ where $ m\equal{}\left(\frac{dy}{dx}\right)_{x\equal{}a}$ right?
2)
For the function to be differentiable $ x\equal{}\frac{\pi}{3}$ it has to be continuous at $ x\equal{}\frac{\pi}{3}$. Left limit at $ x\equal{}\frac{\pi}{3}$ is $ \cos(\frac{\pi}{3})$ and the right limit is $ a.\frac{\pi}{3}\plus{}b$. Using this one can find the value of $ a$.
Find the left and right derivatives at $ x\equal{}\frac{\pi}{3}$ and equate them to get $ b$.


[APH]
THEOREM

Let ABC, A#B#C# be two triangles and P a point.

Denote

Ac := AC# /\ CP, Ab := AB# /\ BP

A* := AcB /\ AbC. Similarly B*,C*.

ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective
for all Ps

Synthetic proof??

Antreas

------------------------------------------------------------
Dear Friends,

Why should this problem prove so intractable?
The algebra is absolutely a mess!

Jeff

-----

I need your help with both the analytic and synthetic solutions.

Sincerely, Jeff Brooks

Nice one...

[color=blue][b]Theorem.[/b] Let ABC and $A^{\#}B^{\#}C^{\#}$ be two triangles and P a point. Denote

$A_b=AB^{\#}\cap BP$, $A_c=AC^{\#}\cap CP$, $B_c=BC^{\#}\cap CP$, $B_a=BA^{\#}\cap AP$, $C_a=CA^{\#}\cap AP$, $C_b=CB^{\#}\cap BP$;
$A^{\ast}=BA_c\cap CA_b$, $B^{\ast}=CB_a\cap AB_c$, $C^{\ast}=AC_b\cap BC_a$.

Then, the lines $AA^{\ast}$, $BB^{\ast}$, $CC^{\ast}$ concur if and only if the lines $AA^{\#}$, $BB^{\#}$, $CC^{\#}$ concur.[/color]

[i]Proof.[/i] The main idea of the proof is to apply the Ceva theorem to the triangle ABC to every three lines that concur in a point. Let

$A^{\ast}_1=AA^{\ast}\cap BC$, $B^{\ast}_1=BB^{\ast}\cap CA$, $C^{\ast}_1=CC^{\ast}\cap AB$;
$A^{\#}_1=AA^{\#}\cap BC$, $B^{\#}_1=BB^{\#}\cap CA$, $C^{\#}_1=CC^{\#}\cap AB$;
$A_1=AP\cap BC$, $B_1=BP\cap CA$, $C_1=CP\cap AB$;
$A^{\#}_b=BA^{\#}\cap CA$; $A^{\#}_c=CA^{\#}\cap AB$, $B^{\#}_c=CB^{\#}\cap AB$, $B^{\#}_a=AB^{\#}\cap BC$, $C^{\#}_a=AC^{\#}\cap BC$, $C^{\#}_b=BC^{\#}\cap CA$;
$A_{b1}=CA_b\cap AB$, $A_{c1}=BA_c\cap CA$, $B_{c1}=AB_c\cap BC$, $B_{a1}=CB_a\cap AB$, $C_{a1}=BC_a\cap CA$, $C_{b1}=AC_b\cap BC$.

Since the lines $AA_1$, $BB_1$, $CC_1$ concur (namely, at the point P), the Ceva theorem yields

$\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}\cdot\frac{AC_1}{C_1B}=1$.

Since the lines $AB^{\#}_a$, $BB_1$ and $CA_{b1}$ concur (namely, at the point $A_b$), the Ceva theorem yields

$\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{CB_1}{B_1A}\cdot\frac{AA_{b1}}{A_{b1}B}=1$,

so that

$\frac{AA_{b1}}{A_{b1}B}=1: \left(\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{CB_1}{B_1A}\right)$.

Since the lines $AC^{\#}_a$, $BA_{c1}$ and $CC_1$ concur (namely, at the point $A_c$), the Ceva theorem yields

$\frac{BC^{\#}_a}{C^{\#}_aC}\cdot\frac{CA_{c1}}{A_{c1}A}\cdot\frac{AC_1}{C_1B}=1$,

so that

$\frac{CA_{c1}}{A_{c1}A}=1: \left(\frac{BC^{\#}_a}{C^{\#}_aC}\cdot\frac{AC_1}{C_1B}\right)$.

Since the lines $AA^{\ast}_1$, $BA_{c1}$, $CA_{b1}$ concur (namely, at the point $A^{\ast}$), the Ceva theorem yields

$\frac{BA^{\ast}_1}{A^{\ast}_1C}\cdot\frac{CA_{c1}}{A_{c1}A}\cdot\frac{AA_{b1}}{A_{b1}B}=1$,

so that

$\frac{BA^{\ast}_1}{A^{\ast}_1C}=1: \left(\frac{CA_{c1}}{A_{c1}A}\cdot\frac{AA_{b1}}{A_{b1}B}\right)=1: \left(1: \left(\frac{BC^{\#}_a}{C^{\#}_aC}\cdot\frac{AC_1}{C_1B}\right)\cdot 1: \left(\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{CB_1}{B_1A}\right)\right)$
$=\frac{BC^{\#}_a}{C^{\#}_aC}\cdot\frac{AC_1}{C_1B}\cdot\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{CB_1}{B_1A}$.

Similarly,

$\frac{CB^{\ast}_1}{B^{\ast}_1A}=\frac{CA^{\#}_b}{A^{\#}_bA}\cdot\frac{BA_1}{A_1C}\cdot\frac{CC^{\#}_b}{C^{\#}_bA}\cdot\frac{AC_1}{C_1B}$;
$\frac{AC^{\ast}_1}{C^{\ast}_1B}=\frac{AB^{\#}_c}{B^{\#}_cB}\cdot\frac{CB_1}{B_1A}\cdot\frac{AA^{\#}_c}{A^{\#}_cB}\cdot\frac{BA_1}{A_1C}$.

Since the lines $AB^{\#}_a$, $BB^{\#}_1$, $CB^{\#}_c$ concur (namely, at the point $B^{\#}$), the Ceva theorem yields

$\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{CB^{\#}_1}{B^{\#}_1A}\cdot\frac{AB^{\#}_c}{B^{\#}_cB}=1$,

so that

$\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{AB^{\#}_c}{B^{\#}_cB}=1: \frac{CB^{\#}_1}{B^{\#}_1A}$; similarly,
$\frac{CC^{\#}_b}{C^{\#}_bA}\cdot\frac{BC^{\#}_a}{C^{\#}_aC}=1: \frac{AC^{\#}_1}{C^{\#}_1B}$;
$\frac{AA^{\#}_c}{A^{\#}_cB}\cdot\frac{CA^{\#}_b}{A^{\#}_bA}=1: \frac{BA^{\#}_1}{A^{\#}_1C}$.

Thus,

$\frac{BA^{\ast}_1}{A^{\ast}_1C}\cdot\frac{CB^{\ast}_1}{B^{\ast}_1A}\cdot\frac{AC^{\ast}_1}{C^{\ast}_1B}=\left(\frac{BC^{\#}_a}{C^{\#}_aC}\cdot\frac{AC_1}{C_1B}\cdot\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{CB_1}{B_1A}\right)$
$\cdot\left(\frac{CA^{\#}_b}{A^{\#}_bA}\cdot\frac{BA_1}{A_1C}\cdot\frac{CC^{\#}_b}{C^{\#}_bA}\cdot\frac{AC_1}{C_1B}\right)\cdot\left(\frac{AB^{\#}_c}{B^{\#}_cB}\cdot\frac{CB_1}{B_1A}\cdot\frac{AA^{\#}_c}{A^{\#}_cB}\cdot\frac{BA_1}{A_1C}\right)$
$=\left(\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}\cdot\frac{AC_1}{C_1B}\right)^2\cdot\left(\frac{BB^{\#}_a}{B^{\#}_aC}\cdot\frac{AB^{\#}_c}{B^{\#}_cB}\right)\cdot\left(\frac{CC^{\#}_b}{C^{\#}_bA}\cdot\frac{BC^{\#}_a}{C^{\#}_aC}\right)\cdot\left(\frac{AA^{\#}_c}{A^{\#}_cB}\cdot\frac{CA^{\#}_b}{A^{\#}_bA}\right)$
$=1^2\cdot 1: \frac{CB^{\#}_1}{B^{\#}_1A}\cdot 1: \frac{AC^{\#}_1}{C^{\#}_1B}\cdot 1: \frac{BA^{\#}_1}{A^{\#}_1C}=1: \left(\frac{BA^{\#}_1}{A^{\#}_1C}\cdot\frac{CB^{\#}_1}{B^{\#}_1A}\cdot\frac{AC^{\#}_1}{C^{\#}_1B}\right)$.

Hence, $\frac{BA^{\ast}_1}{A^{\ast}_1C}\cdot\frac{CB^{\ast}_1}{B^{\ast}_1A}\cdot\frac{AC^{\ast}_1}{C^{\ast}_1B}=1$ holds if and only if $\frac{BA^{\#}_1}{A^{\#}_1C}\cdot\frac{CB^{\#}_1}{B^{\#}_1A}\cdot\frac{AC^{\#}_1}{C^{\#}_1B}=1$ holds.

Now, the lines $AA^{\ast}$, $BB^{\ast}$, $CC^{\ast}$ are the lines $AA^{\ast}_1$, $BB^{\ast}_1$, $CC^{\ast}_1$; thus, according to the Ceva theorem, these lines concur if and only if $\frac{BA^{\ast}_1}{A^{\ast}_1C}\cdot\frac{CB^{\ast}_1}{B^{\ast}_1A}\cdot\frac{AC^{\ast}_1}{C^{\ast}_1B}=1$. On the other hand, the lines $AA^{\#}$, $BB^{\#}$, $CC^{\#}$ are the lines $AA^{\#}_1$, $BB^{\#}_1$, $CC^{\#}_1$; thus, according to the Ceva theorem, these lines concur if and only if $\frac{BA^{\#}_1}{A^{\#}_1C}\cdot\frac{CB^{\#}_1}{B^{\#}_1A}\cdot\frac{AC^{\#}_1}{C^{\#}_1B}=1$. Hence, we conclude that the lines $AA^{\ast}$, $BB^{\ast}$, $CC^{\ast}$ concur if and only if the lines $AA^{\#}$, $BB^{\#}$, $CC^{\#}$ concur. The theorem is proven.

  Darij


Please specify your computer OS, as I have no knowledge about Linux and the others that are floating around.

I don't have a Linux distribution installed, but I plan to get Topoligilinux on an OLD laptop and possible SuSe or Debian on a sort of old desktop.

Right now, I'm trying out a lot of LiveCDs.  I like the Ubuntu livecd and its variants, Knoppix and its derivitaves, etc.


floor function
graph theory
combinatorics open
combinatorics

Let $ X$ be a set of $ 2n\plus{}1$ elements, and let $ N$ be the set of subsets of $ X$ with cardinality $ n$ or $ n\plus{}1$.
Let $ G$ be the bipartite unoriented graph whose nodes are the elements of $ N$ and where $ \left(A,B\right)$ is an edge iff $ A$ is a subset of $ B$, or vice-versa.
Prove that $ G$ has a perfect matching.

Your graph is regular, and every regular bipartite graph has a perfect matching:

Assume $G$ is an $r$-regular graph with bipartition $(A,B)$ (it follows that $|A|=|B|$), and take $\emptyset\ne X\subseteq A$. There are $r|X|$ edges with one endpoint in $X$, so if they had $<|X|$ endpoints in $B$, one of these would be an endpoint of more than $r$ edges, contradicting the regularity.

I think it's been discussed before.

Thanks a lot, and sorry if this problem has been posted before!

Notice that you can use this as a lemma to show that, given a set $X$ with $|X|=n$, there are at most $\displaystyle {n}\choose{\lfloor n/2 \rfloor}$ subsets of $X$, such that no subset includes another one.

I have already asked MindFlyer by PM some time ago, but it seems that he does not visit ML anymore, so let me ask the community:

[quote="MindFlyer"]Notice that you can use this as a lemma to show that, given a set $ X$ with $ |X| \equal{} n$, there are at most $ \displaystyle {n}\choose{\lfloor n/2 \rfloor}$ subsets of $ X$, such that no subset includes another one.[/quote]

How exactly do you conclude this assertion (which is, by the way, called Sperner's Theorem) from your problem?

  Darij

I believe the same argument shows that if $ k < \frac {n}{2}$, then there is a perfect matching from the $ k$ element subsets to the $ k \plus{} 1$ element subsets which satisfies this property.

Similarly, for $ k > \frac {n}{2}$, there is a perfect matching from the $ k$ element subsets to the $ k \minus{} 1$ element subsets.

By following the matchings upward/downward from the middle level, this enables us to divide the subsets of $ \{1, 2, \dots n \}$ into $ \binom{n}{\lfloor n/2 \rfloor}$ chains of the form $ A_1 \subseteq A_2 \subseteq A_3 \dots$.

Your collection of subsets can include at most one element from each chain.


calculus
integration
calculus computations

Is there any comparation between $m$ and $n$ i.e $m \geq n$ or $n \geq m$ ?

thx

since m and n are symmetrically placed in either integral, you may choose either to be the larger one (WLOG).

[quote="prithwijit"]Which of the following is bigger and why?

$\int_{0}^{1}(1-x^{n})^{\frac{1}{m}}dx$ ; $\int_{0}^{1}(1-x^{m})^{\frac{1}{n}}dx$. $m$ and $n$ are positive integers.[/quote]

If $m,n\geq 1$ then

$\int_{0}^{1}(1-x^{n})^{\frac{1}{m}}dx=\int_{0}^{1}(1-x^{m})^{\frac{1}{n}}dx= \left( \frac{1}{m}+1+\frac{1}{n}\right)\mbox{B}\left( \frac{1}{m}+1,\frac{1}{n}+1\right)$

where B(x,y) is the beta function.


I have 5 red plates and 4 blue plates. If I randomly select two plates to serve dinner on, what is the probability that they're both the same color?

It's either 2 red or 2 blue: 5*4/2+4*3/2=16 ways
Total: 9*8/2=36 ways
Probability: 16/36=4/9

I am bad at probability, so please correct me if I am wrong

Alternatively, we can count what we don't want - one red plate and one blue plate - and subtract that from the total number of possible ways. We see there are $ 5 \times 4 \equal{} 20$ ways to get one red and one blue plate, and  $ \binom{9}{2}$ ways in total. Thus the probability of what we don't want occurring is $ \frac{5}{9}$. We subtract this from 1 to get $ \frac{4}{9}$, confirming Bugi's answer.


geometry
geometric transformation
rotation
geometry proposed

In the figure below :

[color=red][b]Problem[/b][/color]

 Two circles meet at $ A,B$ and a line through $ A$ meets the first circle at $ C$ and the second at $ D$. Let $ K,L$ be the midpoints  of the arcs $ CB,DB$ which do not contain $ A$.If $ M , N$ are the midpoints of $ CD ,KL$ , prove that $ NM\equal{}NA$.

[b] Comment[/b]

[color=blue][i] I like this problem and I do not remember :(  if I have posted it again .[/i][/color]

It is suffice to show $ \angle KML \equal{} \angle KAL \equal{} 90$. Since $ KA,LA$ is internal bisector of $ \angle BAC,\angle BAD,$ $ \angle KAL \equal{} 90$. Let $ X$ be a symmetric point of $ K$ wrt $ M$. Since $ MX \equal{} MK$, It is suffice to show $ LX \equal{} LK$, or $ \triangle XDL$ and $ \triangle KBL$ are congruent. We have $ DX \equal{} KC \equal{} BK,DL \equal{} BL$.
$ \measuredangle XDL \equal{} \measuredangle XDA \plus{} \measuredangle ADL \equal{} \measuredangle KCA \plus{} \measuredangle{ABL} \plus{} 180 \equal{} \measuredangle KBA \plus{} 180 \plus{} \measuredangle{ABL} \plus{} 180 \equal{} \measuredangle KBL$. Hence $ \triangle XDL$ and $ \triangle KBL$ are congruent.
In fact, $ \angle KAL \equal{} 90$ and $ \measuredangle XDA \equal{} \measuredangle KCA$ hold iff $ A$ lies inside the segment $ CD$. if $ A$ lies outside the segment $ CD$(this case exist), then $ NM \equal{} NA$ does not hold(I drew by gsp in this case, and I notice $ NM\neq NA$). Hence the condition that $ A$ lies inside of the segment $ CD$ is necesary.

a rotarion with angle $ \angle DLB$ around L brings D to B and a rotation with angle $ \angle BKC$ around $ K$ brings B to C .the combination of these rotations is a rotarion with angle $ \angle DLB+\angle BKC=180^{\circ}$ that brings D to C.hence the center of this rotarion is $ M$.
hence $ \angle KML=90$ and also $ \angle KAL=90$ hence points $ K,M,A,L$ lie on a circle with center $ N$


[quote="Anonymous52111"]Dear users,

I need your help to answer and provide explanation (or show your work) each (5) question would be appreciated. Also, you don't have to answer each one. By the way, Merry Christmas to you all!

1. If Lynn can type a page in p minutes, what piece of the page can she do in 5 minutes?

2. If Sally can paint a house in 4 hours, and John can paint the same house in 6 hours, how long will it take for both of them to paint the house together?

3. The sales price of a car is $12,590, which is 20% off the original price. What is the original price? 4. Solve the following equation for A: 2A/3 = 8 + 4A 5. Alfred wants to invest
$4,000 at 6% simple interest rate for 5 years. How much interest will he receive?[/quote]

These are just some hints to help get you started.

1. To think about number one, just plug in easy number and think about what to do.
Once you think about it with the number just change it back to the variable.
Note--- this helps me if I get confused, but might not help you at all.

2. Think of how much each one will do in one hour.

3. If it is 20% OFF the total price, than what portion of the original price is it. Then write in terms of x and solve for x.

4. To start off multiply both sides of the equation by 3 to get rid of the fraction; from there it should be pretty simple.

5. Interest equ als principle*rate*time Just plug it in!

Hope I helped! :)

[hide="Here are all the answers"]1. Suppose $p=10$. Then the answer would be $\frac{5}{10}$. If $p=20$, the answer would be $\frac{5}{20}$. Thus the answer is $\frac{5}{p}$.

2. Sally can paint $\frac{1}{4}$ and John $\frac{1}{6}$ of the house in one hour. Thus $\frac{5}{12}$ of the house is painted in one hour if they work together. Thus, to get $\frac{12}{12}$, you need to multiply it by 2.4. 1 hour multiplied by 2.4 is [b]2 hours and 24 minutes[/b].

3. 12590 is 80% of the original price and we want 100%. So we have to take$\frac{100}{80}$ and multiply it with 12590. Thus $\frac{100}{80}\times 12590=15737,50$

4. $\frac{2A}{3}=8+4A$
    Multiply both sides with three to get:
    $2A=24+12A$
    Take the A's to the left-hand side to get:
    $-10A=24$
    Thus you divide both sides with $-10$ to get: $A=-\frac{12}{5}$

5. $\frac{6}{100}\times 4000\times 5=1200$

[/hide]

[hide]1. 5/p, because... yeah.  :P
2. Sally's rate: 1/4 per hr; Jhon's rate: 1/6. Together rate: 1/4 + 1/6 = 5/12 per hr. So it should be 2 2/5 hrs, or 2 hours 24 minutes.
3. Make a proportion. 12590/80 = x/100, 100*12590 = 80x, Solve for x. X = 15737.50
4. 2/3*A = 8 + 4A. Multiply each side by 3/2 to get rid of all the fractions. The equation becomes A = 12 + 6A. -5a = 12, a = -12/5
5. The formula for interest is I = prt (principal * rate * time). I = 4000 * .06 * 5 = 40*6*5, 1200

Um, yeah. :)[/hide]

[hide]
1. I think the formula rate*time=distance can be adapted to rate*time=pages typed. Call the rate r. If Lynn can type a page in p minutes, then r*p=1. At the same rate, we can say that r*5=x, where x is the number of pages Lynn can type in 5 minutes. Adding these two equations gets r(5+p)=1+x, or [b]x=r(5+p)-1[/b].
2. This question also implies r*t=d. Every hour, Sally can paint 1/4 of the house, while John can paint 1/6, so their combined rate is (1/4+1/6). Now we have (1/4+1/6)*t=1. This can be simplified to (5/12)*t=1, or [b]t=12/5[/b].
3. If 12950 is 20% of the original price, then we know 12950=.8n, where n is the original price. n=12950/.8, or [b]n=16,187.50[/b].
4.2A/3=8+4A (multiply both sides by 3) -> 2A=24+12A (subtract 2A from both sides) -> 0=24+10A (subtract 24 from both sides) -> -24=10A (divide both sides by 10) -> [b]A=-12/5[/b].
5.The formula for the amount of simple interest someone will receive is (r*t)(n), where n is the invested amount, r is the interest rate, and t is the number of years. Our rate is 6% or .06, our time is 5 years, and the invested sum is 4,000. (0.06*5)(4,000)=[b]1200[/b].[/hide]

[quote="-Terminator-"][quote="superwizard188"]5. The formula for interest is I = prt (principal * rate * time). I = 4000 * .06 * 5 = 40*6*5, 1200[/quote]

The formula is $I=pr^t$, not $I=prt$. The answer is $5352.9023104$.[/quote]

That's only when interest is compounded continously.

[quote="Anonymous52111"]Dear users,

I need your help to answer and provide explanation (or show your work) each (5) question would be appreciated. Also, you don't have to answer each one. By the way, Merry Christmas to you all!
1. If Lynn can type a page in p minutes, what piece of the page can she do in 5 minutes?

2. If Sally can paint a house in 4 hours, and John can paint the same house in 6 hours, how long will it take for both of them to paint the house together?

3. The sales price of a car is $12,590, which is 20% off the original price. What is the original price? 4. Solve the following equation for A: 2A/3 = 8 + 4A 5. Alfred wants to invest
$4,000 at 6% simple interest rate for 5 years. How much interest will he receive?[/quote]

[hide]1. 5/p amount. I kinda used guess and check.

2. For sally it takes 1/4 of house in 1 hour
John is 1/6 house in 1 hour
Together add them up so 5/12 of house in 1 hour
do 1 divided by 5/12
That is 12/5 hours to complete the house

3. (100-20)x=12590
80x=12590
x=1259/8

4. First move 4a over.
That would be -10a/3 after simplifying. Then multiply 3 on both sides.
-10a=24.
a=-12/5

5. 5000*1.3=6500. 1.3 was 1 was original. .3 was .05*6.
Not sure. IS it compounded yearly still even if it is simple interest?
[/hide]
Not sure on all because I did them all in my head

[hide]
1. Plug in a easy number in place of $\ p$(for example 10) then it would be $\ 5/10$. Then use another few numbers (for example 15, 20, or 25). The result is the same, so the answer is $\ 5/p$

2. John and Sally together can paint a house in $\ x/4+x/6=1$.
Then multiply that by 12 to cancel fractions and simplify; $\ 5x=12$
Thus the answer is $\12/5$...or $\ 2.2$...or 2 hours and 24 minutes...

3. Just multiply $\ 1.2$ by $\ 12590$ and you get $\ 153750$

4. $\ 2A/3=8+4A$...multiply 3 on both sides; $\ 2A=24+12A$
Now divide by 2...and you get $\A=24+6A$ and then simplify it to $\ A=-24/5$

5.$\ I=P*R*T$so...$\ I=4000*.06*5$; simplified to $\ I=1200$
[/hide]

1.  simple stuff- 5/p

2. 2.4 hours or 2 hours and 24 minutes. i used 1/4+1/6=1/x and then u add the fractions and then cross multiply to solve.

3. just do a proportion. 80/100=12950/x and solve. the answer is 15,737.50 dollars.

4. just solve for x. multiply both equations by 3, then move 12x (it becomes that) over and solve. answer is -2.4 or -12/5.

5. use the formula principal times rate times time= interest. 4000 times 0.06 times 5= 1200

 :welcomeani:


ratio
number theory
least common multiple

Three boys, Tom John and Ken, agreed to share some marbles in the ratio of 9 : 8 : 7 respectively.  John then suggested that they should share the marbles in the ratio 8 : 7 : 6 instead. Who would then get more marbles than before and who would get less than before if the ratio is changed?

Any ideas??

There are 168x marbles.

First ratio is equal to:  63x:56x:49x
Second ratio is equal to: 64x:56x:48x

Tom would get more marbles than before (64x>63x)
Ken would get less marbles than before(48x<49x)

 :wink:   :)

[quote="nick42"]yeah.

what he said. find the LCM.  :D[/quote]

Or, just try it with a number that is divisible by 6, 7, 8, and 9.

which is LCM without the L :P

anyways, LCM is easiest to work with (cuz its the smallest)

Wait!!! If one person gets more marbles, and another person gets less, then would the third person just keep what they have?

Not necessarily.  If one person gets the same number of marbles as the number of marbles another person loses, then yes.


When the $n's$ dice are thrown at one time, find the probability such that the sum of the cast is $n+2.$

Do you mean that $n$, six-sided dice are thrown one at a time?

each die must give at least 1 to the total, subtract that to make it dice with 0,1,2,3,4,5, on the sides, that gives a sum of 2 using the new die, the rest is trivial

To get a cast of $n+2$, you either need two of the $n$ dice to show 2, or one die to show 3. So the number of combinations of $n-2$ die showing 1 and $2$ die showing 2 would be $1+2+...+(n-1)$, or ${n(n-1)}/{2}$. Add this to $n$, the number of die that could show a 3, and divide by the total number of combinations of the 6-sided die, which is $6^n$, and you get:

$\frac{n^2 + n}{2*6^n}$

(edited, i forgot the 2)

Yes, your second answer is correct, cincodemayo5590.

kunnyさん、京都大学の試験をとりましたか。京都大学はいい大学ですね。高校生ですか。東大の試験もとりましたか。^_^


What is the largest even integer that cannot be written as the sum of two odd composite numbers?

No, because I think 10 works.  But the answer will be much harder to find (because this is an AIME #10)  Remember, 1 is neither prime nor composite.

This isn't a #10, I think it's a #14 from the mid 80's.

The answer is

[hide]38, and the solution deals with trying to express an even integer in terms of a sum of two products of two odd numbers each, then showing that the upper bound is 38 using BALZ N URNS formula...it's pretty messy,[/hide]

[quote="cosinator"][hide="is it..."]8??[/hide][/quote]

Well, 8 is one of them but you can check that 38, is larger, and proceed as elemnop wanted.

yeah...I just realized that both odd numbers are composite


function
real analysis
advanced fields
advanced fields unsolved

Let f be a real valued Leb meas function on R^k. Prove that there exist Borel functions g and h such that g(x)=h(x) a.e [m] and g(x)<=f(x)<=h(x) for all x in R^k

You can write $\mathbb{R}$ as a countable disjoint union of open and bounded intervals $\cup {I_{n}}$. Now you can write $f^{-1}(I_{1}=]a_{1},b_{1}[)=E \cup N$ where $E$ is a Borel set and $N$ is a null set (this comes from the fact that the Lebesgue $\sigma$-algebra is the completion of the Borel $\sigma$-algebra). You also have $N \subseteq A$ where $A$ is a Borel set with measure $0$. From here all you have to do is to define $\forall x \in E, g(x)=f(x)$ and $\forall x \in N, g(x) = y < a_{1}$. That means you will have $g^{-1}(I_{1})=E$ Borel set and $g \leq f$ everywhere. But now you iterate the process for $I_{2}$, etc ... eventually that will provide you with a suitable definition for $g$ (I leave it to you to write it down formally, i.e to verify that $g$ will be well defined and Borel-measurable).
By a similar process you can construct $h$.


algebra
polynomial
calculus
derivative
parameterization
linear algebra
matrix

Let $ n$ be a positive integer and let $ V$ denote the space of real polynomials of degree less than $ n$. We have $ \dim V\equal{}n$
Let $ A: V\to V$ be the linear transformation sending each polynomial $ p(x)$ to its derivative $ p'(x)$.
Find $ e^{tA}$

Is $ t$ a scalar parameter?

In any case, this problem isn't too bad:  $ A$ is nilpotent, so only a finite summation needs to be taken.

Note that the operator $ e^{tA}$ sends the polynomial $ p(x)$ to the polynomial

$ \sum_{k\ge0}\frac{p^{[k]}(x)t^k}{k!}$

As t0rajir0u indicated, this series is actually a finite sum, so there's no issue with convergence.

And we note that that's just a Taylor series. In particular, the operator $ e^{tA}$ maps $ p(x)$ to $ p(x\plus{}t).$

Which is a nice result.

If you don't trust that, let's walk through a concrete case. Let $ n\equal{}4$ so that we're talking about polynomials of degree no greater than $ 3.$ Using the basis $ \{1,x,x^2,x^3\},$ we have that

$ A\equal{}\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{bmatrix}.$ Then,

$ A^2\equal{}\begin{bmatrix}0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\end{bmatrix}$ and $ A^3\equal{}\begin{bmatrix}0&0&0&6\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}.$

$ e^{tA}\equal{}I\plus{}tA\plus{}\frac12t^2A^2\plus{}\frac16t^3A^3\equal{}
\begin{bmatrix}1&t&t^2&t^3\\0&1&2t&3t^3\\0&0&1&3t\\0&0&0&1\end{bmatrix}$

Then if $ p(x)\equal{}a\plus{}bx\plus{}cx^2\plus{}dx^3,$

$ e^{tA}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\equal{}
\begin{bmatrix}a\plus{}bt\plus{}ct^2\plus{}dt^3\\b\plus{}2ct\plus{}3dt^2\\c\plus{}3dt\\d\end{bmatrix}.$

That gives us the polynomial

$ (a\plus{}bt\plus{}ct^2\plus{}dt^3)\plus{}(b\plus{}2ct\plus{}3dt^2)x\plus{}(c\plus{}3ct)x^2\plus{}dx^3.$

Now, if we expand out $ a\plus{}b(x\plus{}t)\plus{}c(x\plus{}t)^2\plus{}d(x\plus{}t)^3,$ we get exactly the same thing.

Note in particular, for $ t\equal{}1,$ that our matrix for $ e^{tA}$ is simply a portion of Pascal's triangle.

We could do the same thing for the infinite-dimensional vector space of all polynomials; we would get in infinite matrix that is all of Pascal's triangle.


is it possible to construct a ring with n units where n is any natural we choose?

Someone should move this. Rings are not appropriate for MOEMS :wink:

or he thought this is the math olympiad section

[quote="mathnerd314"]I think there's a bit of confusion about the meaning of the word "algebra". Try the Olympiad Section for superior algebra (a.k.a. abstract algebra).

Here, algebra means stuff like $3x+5=9$.[/quote]
Ooh! $x=\frac{4}{3}$

[quote="daermon"]can a mod lock this?  :P[/quote] Or just post the original topic in a different forum... :wink:


Let $ X$ be a random variable which satisfies $ P(|X|>M)\equal{}0$ for some real $ M$. Prove that $ P(X \ne 0) \ge \frac{EX}{M}$.

Hint:
\[ E|X| \equal{} P(X\equal{}0) \cdot 0 \plus{} P(X\neq 0) \cdot E[|X|\, | X\neq 0].\]

maybe more explicitly

(using $ E(X;A) \triangleq \int_A XdP$)

\begin{eqnarray*}
EX &=& E(X;X=0) + E(X;0<|X|\leq M) + E(X;|X|>M) \\
&=&E(X;0<|X|\leq M) \\
&\leq& MP(0<|X|\leq M) \\
&=& MP(X\not=0)
\end{eqnarray*}


A board game spinner is divided into three regions labeled
$ A$, $ B$ and $ C$.  The probability of the arrow stopping on region $ A$
is $ \frac{1}{2}$ and on region $ B$ is $ \frac{1}{5}$.  What is the probability
of the arrow stopping on region $ C$?  Express your answer as a common fraction.

The probability that it stops on region $ C$ is the same as the probability that it DOESN'T stop on either region $ A$ or region $ B$.  Thus, this is simply $ 1\minus{}\left (\frac{1}{2}\plus{}\frac{1}{5} \right ) \equal{} 1\minus{}\frac{7}{10}\equal{}\boxed{\frac{3}{10}}$.


decode the following message. spaces are also encoded. There is one space in the encoded output.

RUOELTWKEINHXFEQHZEYTDJPEHVONERUOEBGCAEMHS


topology
real analysis
real analysis unsolved

How would I go about solving the following proofs?

1. Prove that the set {(x, y) ∈ R^2 : y > 0} is an open subset of R^2.

2. Prove that every finite subset of R^d is closed.

3. Let A be an open set and B a closed set. If B ⊂ A, prove that A \ B is open. If A ⊂ B, prove that B \ A is closed.

#1. if $(x,y)$ is a point of the set, then the open disk of radius $y$ centered at $(x,y)$ is contained in the set.

Your approach to #2 and #3 may depend on how closed sets are defined. Two common definitions are: (1) a set is closed if its complement is open; (2) a set $A$ is closed if for any convergent sequence $(x_{n})\subset A$ its limit belongs to $A$.

A closed set is defined as having a complement that is open.

3. Let $A=\{a_{1},\ldots,a_{n}\}$ -is our set. $CA=R^{d}\setminus A$.
$\forall x\in CA \quad \exists r=\min_{k=1,..,n}\parallel x-a_{k}\parallel $ (because $n<\infty$), so
$B(x,r/2)=\{z\in R^{d}: \parallel z-x\parallel <r/2\}\subseteq CA$. Thus $CA$-is open.
Then $A$ -is closed.

3. We knew than if $X,Y$ - is open sets in topological space, then $X\cap Y$ and $X\cup Y$ - is also open.
If $B\subseteq A$ then we have $A\setminus B = A\cap CB$ - is open because $A,CB$ -is open.
If $A\subseteq B$ then we have  $C(B\setminus A)=A\cup CB$. The last set is open, because $A,CB$- is open, so $B\setminus A$ - closed.