trigonometry
logarithms
irrational number
algebra proposed
algebra

Could the power n in de moivre's theorem be irrational number? Can any prove it if the answer is yes? Many thanks.

This boils down to the question of defining $z^{p}$ for complex $z$ and arbitrary $p.$ If $p=n$ is an integer, then $z^{n}$ has a single unambiguous definition. Note that De Moivre's theorem is usually stated for integer powers. You want to deal with irrational powers but even non-integer rational numbers will cause trouble.

Any attempt at a definition must rely on the complex logarithm.

$\log(z)=\ln|z|+i\arg(z)$

The argument of $z,$ abbreviated as $\arg(z),$ is the angle needed to represent $z$ in polar coordinates. In other words, if you write $z=re^{i\theta}$ (or in alternative notation that means the same thing, $z=r\text{cis}\theta=r(\cos\theta+i\sin\theta)),$ then $\arg(z)=\theta.$ But the argument is ambiguous and multiple-valued. One cannot define an argument that is continuous on $\mathbb{C}\setminus\{0\}.$ In particular, since De Moivre focuses on the unit circle, you cannot define an argument that is continuous all the way around the circle.

If we have a logarithm defined on some set, which in practice means that we have defined an argument (we call that "picking a branch of the logarithm"), then we can define $z^{p}=e^{p\log(z)}=e^{p\ln|z|}e^{ip\arg(z)}.$

The factor $e^{p\ln|z|}$ causes no trouble; in fact, it is equal to $1$ for $z$ on the unit circle. The trouble lies in the second factor. You would like for that to read that $(e^{i\theta})^{p}=e^{ip\theta}.$ But, as I said, we can't pick a branch of the logarithm that is continuous all the way around the circle.

If $p=n$ is an integer, we're fine. In that case, if we replace $\theta$ by $\theta+2\pi k,$ then $e^{in\theta}$ gets replaced by $e^{in\theta}e^{2\pi kni}.$ But $e^{2\pi kni}=1,$ so nothing has changed.

But if $p$ is not an integer, then by replacing $\theta$ by $\theta+2\pi k,$ then we multiply our original result by $e^{2\pi kpi}$ If $p$ is a fraction in lowest terms with denominator $q,$ then $e^{2\pi kpi}$ takes on any of $q$ different values, depending on what $k$ is. And if $p$ is a real irrational number, then $e^{2\pi kpi}$ takes on infinitely many different possible values - in fact, those values are densely distributed around the circle.

Here's another way to look at this.

How many solutions does the equation $z^{n}=1$ have for a given positive integer $n?$ That answer is that there are $n$ solutions (the "roots of unity") and we can write those solutions by writing $z=e^{2\pi ik/n}$ for $k=0,1,\dots,n-1.$

How many solutions does the equation $z^{p}=1$ have for a given real irrational number $p?$. It has infinitely many solutions. We can write these solutions as $e^{2\pi ik/p}$ for $k\in\mathbb{Z}.$ Any interval of the unit circle, no matter how small, has infinitely many of these solutions.


AOPS Volume 1 has a question that asks if you need if, only if, or if and only if for:

A number ends with a 5...?...it is divisible by 5.

The answer is "only if," (solution manual).  Is this implying that a number that ends with a 5 isn't divisible by anything else?

In addition

I understand how "only if" provides more specifity to "if," but I don't understand how "if" does that to "only if."

Using the example in the book, it says a number is even if and only if it is divisible by 2.  If it said merely "only if," then that would imply that anything not divisible by 2 is not an even number.  (The book says that "we don't say that all multiple of two are even").  But how are we not saying that?  Doesn't saying that a number is even only if it is divisble by 2 imply that the only numbers that are even are the ones that are divisible by two and no other divisibility rule?  If a number is divisible by two, isn't it a multiple of two by default?

I would greatly appreciate it if someone could explain the three possibilities in more detail.

Thanks for all the help!!  :)

[quote="catcurio"][quote="brady00n"]A number ends with a 5...?...it is divisible by 5.

The answer is "only if," (solution manual). Is this implying that a number that ends with a 5 isn't divisible by anything else? [/quote]

No, it simply says that any number that ends with a 5 must be divisible by 5. It does not say that it can only be divisible by 5.

[quote="brady00n"]Using the example in the book, it says a number is even if and only if it is divisible by 2. If it said merely "only if," then that would imply that anything not divisible by 2 is not an even number.[/quote]

"If and only if" means the statement can be taken both ways; you can say "a number is even if it is divisible by 2" or "a number is divisible by 2 if it is even."[/quote]

Gosh this is at the tip of my tongue...

What's the difference between "if" and "only if"?  If "only if" means that any number that ends with a 5 must be divisible by 5, then doesn't "if" say the same thing? (a number ends with 5 if it is divisble by 5...)

I'm not sure if this is a good way to explain it, but it's a try.

"A number ends with 5 [b]if[/b] it is divisible by 5" would be equivalent to "If a number is divisible by 5, then it ends in 5."

"A number ends with 5 [b]only if[/b] it is divisible by 5" would be equivalent to "If a number ends in 5, then it is divisible by 5."

Clearly, the first example is incorrect, because the number could end in 0 as well.

This way, "if and only if" would mean both ways are true, both "if" and "only if."

I hope this makes sense and helps. :)

[i]Only if[/i] means the same thing as [i]then[/i].

I know that seems counterintuitive, but think of it this way.  Consider the statement:  "[i]I will sleep only if you leave[/i]."  In that statement, notice that [i]only[/i] if means that the only reason I can sleep is if you leave.  Suppose I am sleeping; because the only reason I can sleep is if you leave, it follows logically that you left.  Therefore, I am sleeping implies that you left, so the statement "[i]I will sleep only if you leave[/i]" is equivalent to "[i]If I am sleeping, then you left[/i]."

Now do you see why [i]only if[/i] means the same thing as [i]then[/i]?

So in your problem, "[i]A number ends with a 5 only if it is divisible by 5[/i]" can be read as "[i]If a number ends with a 5, then it is divisible by 5[/i]."  And since we know the converse is not true (a number divisible by 5 does not have to end with 5), we cannot make it into a biconditional by using "[i]if, and only if[/i]."

Weiyan, that's what I was trying to explain. You did it better though. :D

Somebody hasn't been paying attention in their geomety class. :D

If and only if can also be written "iff," so, a general explanation,

P if Q does not imply Q if P
P iff Q implies Q iff P

Or, in terms of a Venn diagram,

P if Q means P is inside Q, but other elements not in P could also be in Q
P iff Q means P is inside Q and no other elements are inside Q


geometry
combinatorics proposed
combinatorics

[i]This is, to the best of my knowledge, not an original problem - it was probably used some time in the past in a competition or in a book of some kind. This was Problem 2 (Combinatorics) for 10th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]

A natural number is written in each cell of $ 10 \times 10$ table. It is known that, no matter which $ 5$ columns and $ 5$ rows of this table we choose, the sum of numbers in their $ 25$ intersection cells is even. Prove that all the numbers in the table are even.

Let us fix column numbers $ C1$,...$ C5$. Denote $ S(r)$ as the sum of five numbers at row $ r$ and column $ C1$ up to $ C5$. Let $ r1,r2,...r6$ be distinct integers in $ [1,10]$. Since $ S(r1)\plus{}S(r3)\plus{}S(r4)\plus{}S(r5)\plus{}S(r6)$ have same parity as $ S(r2)\plus{}S(r3)\plus{}S(r4)\plus{}S(r5)\plus{}S(r6)$, $ S(r1)$ has same parity as $ S(r2)$. In other words, $ S(r)$ are all evens or all odds for any $ r\in[1,10]$. $ S(r2)\plus{}S(r3)\plus{}S(r4)\plus{}S(r5)\plus{}S(r6)$ being even implies all $ S(r)$ are evens.

Now we know any five number in one row has even sum. Using similar trick again, we know each number is even.


geometry
probability
articles
ARML
MATHCOUNTS
3D geometry
sphere

Does anyone know any good mathematical problem solving formula sheets that contains almost all types of formulas in all categories of mathematics?

It's because I keep seeing a lot of random formulas that pop up in these forums and I don't know what and where these formulas do and come from.

The Noah Sheets are something a lot of IL people that I know have memorized. It's not really worth memorizing, but it is worth looking it. If you can derive or prove everything on the Noah Sheets you should be good. They are from a former IMSA student , a guy named Noah. http://www.payton.cps.k12.il.us/karafiolp/ARMLD/NOAH_Sheets_03.pdf

Thank you for the website.

I'm looking at it right now and I suggest others to take a look too.  It should be really helpful in geometry contests.

I'm trying to find some formula sheets right now...especially in probability and algebra.

Don't memorize formulas.  If you don't understand something ask for help or try to derive the formula yourself.

Gah!  Formula sheets!  [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]Read this article![/url]

The formula sheet that was previously referenced is at a (very high) high school level (ARML) -- not at a Getting Started level.

The [url=http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html]Nick Diaz MATHCOUNTS Bible[/url] is universally recommended by successful MATHCOUNTS coaches.

The [url=http://www.mathcounts.org/Program/toolbox.pdf]MATHCOUNTS Toolbox[/url] is also available.

Formula sheets are not the Holy Grail.  You really should be able to derive the total surface area of a cube, for example.  On the other hand, some formulas must be memorized -- the volume of a sphere, for example.

I think of formula sheets sort of like multiplication tables.  Do you *really* need to memorize the 12x tables?  You can manage "six dozen", even if you've only learned up to the 10x tables.  But it's sure faster and more accurate if you've memorized a few more.

Likewise with formulas.  There are a few basic formulas you *have* to memorize.  You ought to be able to derive everything else.  But it's sure faster and more accurate if you've memorized some more formulas.


function
inequalities
geometry
complex analysis
complex analysis unsolved

Suppose that $f: \mathbb{D}\to \mathbb{C}$ is an injective function defined on the open unit disk such that:
$f(0)=0$
$f'(0)=1$

Show that
$|f(z)| \leq \frac{|z|}{(1-|z|)^{2}}$
for $|z|<1$

This in the famous Koebe distortion theorem.
Didi

http://planetmath.org/encyclopedia/KoebeDistortionTheorem.html

No proof though. :P

Also known as [i]Verzerrungssatz[/i], this theorem is usually stated  as a two-sided inequality: \[\frac{|z|}{(1+|z|)^{2}}\le |f(z)|\le \frac{|z|}{(1-|z|)^{2}}\] Recently it has been popularized as [i]the egg yolk principle[/i]. Think of the unit disk $\mathbb D$ as an egg containing yolk of radius $0<r<1$. You can break an egg into an irregular shape (by the Riemann mapping theorem), but its yolk will remain somewhat round by the Koebe distortion theorem.

Eggs are mostly three-dimensional, but the egg-yolk principle holds in higher dimensions as well, and for maps that are merely [url=http://planetmath.org/encyclopedia/QuasiconformalMapping.html]quasiconformal[/url] rather than conformal.

I don't know of a straightforward proof of Koebe's theorem. It can be derived from Bieberbach's $|c_{2}|\le 2$ estimate, which itself follows from the Gronwall's area theorem. Duren's book "Univalent functions" has all this.

I can show using normal family tricks that the maximums and minimums exist. I'm still working on how to show that $\frac{z}{(1+z)^{2}}$ gives these maxima and minima.


trigonometry
geometry
algebra unsolved
algebra

Let $ x,y,z \in R_ \plus{}$ and

$ x^2 \plus{} y^2 \plus{} xy \equal{} 1\\
x^2 \plus{} z^2 \plus{} xz \equal{} 2\\
y^2 \plus{} z^2 \plus{} yz \equal{} 3$

Find the value of $ xy \plus{} xz \plus{} yz$

[quote="NikolayKaz"]Let $ x,y,z \in R_ \plus{}$ and

$ x^2 \plus{} y^2 \plus{} xy \equal{} 1 \\
x^2 \plus{} z^2 \plus{} xz \equal{} 2 \\
y^2 \plus{} z^2 \plus{} yz \equal{} 3$

Find the value of $ A\equal{}xy \plus{} xz \plus{} yz$[/quote]
 one way is to solve this system and find (x,y,z) so we will take the value of this A , but this way is the long way but now i cant find another way , but surly you can solve this system and find (x,y,z)  :oops:

[quote="NikolayKaz"]Let $ x,y,z \in R_ \plus{}$ and

$ x^2 \plus{} y^2 \plus{} xy \equal{} 1 \\
x^2 \plus{} z^2 \plus{} xz \equal{} 2 \\
y^2 \plus{} z^2 \plus{} yz \equal{} 3$

Find the value of $ xy \plus{} xz \plus{} yz$[/quote]

Subtracting first line from second, we get $ (z \minus{} y)(x \plus{} y \plus{} z) \equal{} 1$
Subtracting second line from third, we get $ (y \minus{} x)(x \plus{} y \plus{} z) \equal{} 1$

So $ z \minus{} y \equal{} y \minus{} x \equal{} t$ and so $ z \equal{} y \plus{} t$ and $ x \equal{} y \minus{} t$ which implies $ x \plus{} y \plus{} z \equal{} 3y$ and so $ 3ty \equal{} 1$ and $ 9t^2y^2 \equal{} 1$

The first line becomes then $ 3y^2 \minus{} 3yt \plus{} t^2 \equal{} 1$ and so $ 3y^2 \plus{} t^2 \equal{} 2$

So $ 3y^2$ and $ t^2$ are solution of the equation $ X^2 \minus{} 2X \plus{} \frac 13$ and so are $ 1 \minus{} \sqrt {\frac 23}$ and $ 1 \plus{} \sqrt {\frac 23}$

But $ A \equal{} 3y^2 \minus{} t^2$ and so is either $ 2\sqrt {\frac 23}$, either $ \minus{} 2\sqrt {\frac 23}$ and so, since $ x,y,z > 0$ $ \boxed{A \equal{} 2\sqrt {\frac 23}}$

$ 1^2\equal{}x^2\plus{}y^2\minus{}2xy\cos 120^\circ$ etc

Not a soloution!!!
very bad stupid mistake.
sorry :blush:

ok,Can we determine minimum value of $ x^2\plus{}y^2\plus{}z^2$?

[quote="amparvardi"]
My soloution and answer are not like pco.
Soloution:
$ x^2 \plus{} y^2 \plus{} xy \equal{} 1(*)$
$ x^2 \plus{} z^2 \plus{} xz \equal{} 2(**)$
$ y^2 \plus{} z^2 \plus{} yz \equal{} 3(***)$
Ok,pco showed that $ x \plus{} z \equal{} 2y$.we use this.:

$ (*) \Longrightarrow x^2 \plus{} \frac {{x \plus{} z}^2}{4} \plus{} \frac {x(x \plus{} z)}{2} \equal{} 1$

$ \Longrightarrow 4x^2 \plus{} x^2 \plus{} z^2 \plus{} 2xz \plus{} 2x^2 \plus{} 2xz \equal{} 4$

$ \Longrightarrow 7x^2 \plus{} z^2 \plus{} 4xz \equal{} 4$

$ (***) \Longrightarrow 7z^2 \plus{} x^2 \plus{} 4xz \equal{} 12$   $ \triangle$

And $ (**): x^2 \plus{} z^2 \plus{} xz \equal{} 2$   $ \star$

Subtracting  $ \star$ and $ \triangle$ $ \Longrightarrow z^2 \minus{} x^2 \equal{} \frac 43$

$ \Longrightarrow (z \minus{} x)(z \plus{} x) \equal{} \frac 43 \Longrightarrow yz \minus{} zx \equal{} \frac 23$.[/quote]

I'm affraid there is an error here : $ (z \minus{} x)(z \plus{} x) \equal{} \frac 43 \Longrightarrow yz \minus{} yx \equal{} \frac 23$ and not $ yz \minus{} zx \equal{} \frac 23$

Sorry

I made a mistake here too,sorry. :blush:  :|

[quote="amparvardi"]:)  mistake in mistake. :)
I wrote that $ yz \minus{} zx \equal{} \frac 23$ that was wrong and the right answer was $ yz \minus{} yx \equal{} \frac 23$ and you had a mistake on writing,so the original soloution is right,because:
 $ \minus{} \times \minus{} \equal{} \plus{}$

(the error corrected,thank you)

amparvardi[/quote]

I begin to be tired.
You made a mistake and I pointed to you this mistake.
So the remainder of you proof is wrong.
And your result is wrong.

And, sorry, but my result is OK :

Choose $ (x,y,z)\equal{}(\sqrt{\frac{1\plus{}\sqrt{ \frac 23}}3}\minus{}\sqrt{1\minus{}\sqrt{\frac 23}},$ $ \sqrt{\frac{1\plus{}\sqrt{ \frac 23}}3},$ $ \sqrt{\frac{1\plus{}\sqrt {\frac 23}}3}\plus{}\sqrt{1\minus{}\sqrt{\frac 23}})$

You can check that these values fit the system equations and are such that $ xy\plus{}yz\plus{}zx\equal{}2\sqrt{\frac 23}$

[quote="amparvardi"][quote="pco"][quote="amparvardi"]:)  mistake in mistake. :)
I wrote that $ yz \minus{} zx \equal{} \frac 23$ that was wrong and the right answer was $ yz \minus{} yx \equal{} \frac 23$ and you had a mistake on writing,so the original soloution is right,because:
 $ \minus{} \times \minus{} \equal{} \plus{}$

(the error corrected,thank you)

amparvardi[/quote]

I begin to be tired.
You made a mistake and I pointed to you this mistake.
So the remainder of you proof is wrong.
And your result is wrong.

And, sorry, but my result is OK :

Choose $ (x,y,z) \equal{} (\sqrt {\frac {1 \plus{} \sqrt {\frac 23}}3} \minus{} \sqrt {1 \minus{} \sqrt {\frac 23}},$ $ \sqrt {\frac {1 \plus{} \sqrt {\frac 23}}3},$ $ \sqrt {\frac {1 \plus{} \sqrt {\frac 23}}3} \plus{} \sqrt {1 \minus{} \sqrt {\frac 23}})$

You can check that these values fit the system equations and are such that $ xy \plus{} yz \plus{} zx \equal{} 2\sqrt {\frac 23}$[/quote]

I edited the soloution,please tell me where is wrong.
and note that $ 2\sqrt {\frac 23} \equal{} \sqrt {\frac 83}$.
now it's better.
((one of the proofs is wrong sure.))

amparvardi[/quote]

Maybe you could do a little bit of work !!!!

Your solution says $ xy\equal{}\frac 23$, $ xz\equal{}\frac 23$ and $ yz\equal{}\frac 43$ and so $ (x,y,z)\equal{}(\frac 1{\sqrt 3},\frac 2{\sqrt 3},\frac 2{\sqrt 3})$

And so :

$ x^2\plus{}y^2\plus{}xy\equal{}\frac 73\ne 1$

$ x^2\plus{}z^2\plus{}xz\equal{}\frac 73\ne 2$

$ y^2\plus{}z^2\plus{}yz\equal{}4\ne 3$
Check your result before claiming other results are wrong ! :(

[quote="pco"][/quote][quote="amparvardi"][quote="pco"][quote="amparvardi"]:)  mistake in mistake. :)
I wrote that $ yz - zx = \frac 23$ that was wrong and the right answer was $ yz - yx = \frac 23$ and you had a mistake on writing,so the original soloution is right,because:
 $ - \times - = +$

(the error corrected,thank you)

amparvardi[/quote]

I begin to be tired.
You made a mistake and I pointed to you this mistake.
So the remainder of you proof is wrong.
And your result is wrong.

And, sorry, but my result is OK :

Choose $ (x,y,z) = (\sqrt {\frac {1 + \sqrt {\frac 23}}3} - \sqrt {1 - \sqrt {\frac 23}},$ $ \sqrt {\frac {1 + \sqrt {\frac 23}}3},$ $ \sqrt {\frac {1 + \sqrt {\frac 23}}3} + \sqrt {1 - \sqrt {\frac 23}})$

You can check that these values fit the system equations and are such that $ xy + yz + zx = 2\sqrt {\frac 23}$[/quote]

I edited the soloution,please tell me where is wrong.
and note that $ 2\sqrt {\frac 23} = \sqrt {\frac 83}$.
now it's better.
((one of the proofs is wrong sure.))

amparvardi[/quote][quote="pco"]

Maybe you could do a little bit of work !!!!

Your solution says $ xy = \frac 23$, $ xz = \frac 23$ and $ yz = \frac 43$ and so $ (x,y,z) = (\frac 1{\sqrt 3},\frac 2{\sqrt 3},\frac 2{\sqrt 3})$

And so :

$ x^2 + y^2 + xy = \frac 73\ne 1$

$ x^2 + z^2 + xz = \frac 73\ne 2$

$ y^2 + z^2 + yz = 4\ne 3$
Check your result before claiming other results are wrong ! :([/quote]

Oh,I'm so so so sorry,a stupid mistake,( $ yz-xz-xy=2x^2$ not $ yz-xz-xy=0$)

Always check the solution before posting!!!
THANK you for guidance.

[quote="pco"][quote="kunny"]$ 1^2 \equal{} x^2 \plus{} y^2 \minus{} 2xy\cos 120^\circ$ etc[/quote]

Very sorry, kunny, but I dont understand how does this equality help solving the problem. Could you tell us some more (instead of etc), thanks ?[/quote]

First we draw a triangle with sides $ 1$, $ \sqrt{2}$ and $ \sqrt{3}$. Now consider the Fermat's point. Let the distance from the the Fermat's point to the three vertices be $ x$, $ y$ and $ z$ respectively. It can be seen clearly that
$ x^{2}\plus{}y^{2}\plus{}xy \equal{} 1\\ x^{2}\plus{}z^{2}\plus{}xz \equal{} 2\\ y^{2}\plus{}z^{2}\plus{}yz\equal{}3$
by cosine formula.

So $ \frac{1}{2}(xy\plus{}xz\plus{}yz)\sin 120^\circ\equal{}[\triangle]$, and we have $ xy\plus{}yz\plus{}zx\equal{}\frac{1}{\sqrt{3}}\sqrt{(\sqrt{1}\plus{}\sqrt{2}\plus{}\sqrt{3})(\sqrt{1}\plus{}\sqrt{2}\minus{}\sqrt{3})(\sqrt{1}\minus{}\sqrt{2}\plus{}\sqrt{3})(\minus{}\sqrt{1}\plus{}\sqrt{2}\plus{}\sqrt{3})}$.

[quote="KenHungKK"] ...$ xy \plus{} yz \plus{} zx \equal{} \frac {1}{\sqrt {3}}\sqrt {(\sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})(\sqrt {1} \plus{} \sqrt {2} \minus{} \sqrt {3})(\sqrt {1} \minus{} \sqrt {2} \plus{} \sqrt {3})( \minus{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})}$.[/quote]

And just notice that $ (\sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})(\sqrt {1} \plus{} \sqrt {2} \minus{} \sqrt {3})$ $ (\sqrt {1} \minus{} \sqrt {2} \plus{} \sqrt {3})$ $ ( \minus{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})\equal{}8$

And so $ A\equal{}2\sqrt{\frac 23}$  :)

Congrats (and kunny too) for your geometrical solution (I'm completely unable to find such solutions as I have no talent at all in geometry; for example, I dont know what is Fermat point  :blush: )

[quote="pco"][quote="KenHungKK"] ...$ xy \plus{} yz \plus{} zx \equal{} \frac {1}{\sqrt {3}}\sqrt {(\sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})(\sqrt {1} \plus{} \sqrt {2} \minus{} \sqrt {3})(\sqrt {1} \minus{} \sqrt {2} \plus{} \sqrt {3})( \minus{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})}$.[/quote]

And just notice that $ (\sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3})(\sqrt {1} \plus{} \sqrt {2} \minus{} \sqrt {3})$ $ (\sqrt {1} \minus{} \sqrt {2} \plus{} \sqrt {3})$ $ ( \minus{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3}) \equal{} 8$

And so $ A \equal{} 2\sqrt {\frac 23}$  :)

Congrats (and kunny too) for your geometrical solution (I'm completely unable to find such solutions as I have no talent at all in geometry; for example, I dont know what is Fermat point  :blush: )[/quote]

just like me!!!
don't like geometry.


I was just wondering why in the U.S. there are 2 liter bottles of soda sold, but everything else is sold in standard units(gallons, pounds,etc)? Just wondering.


function
linear algebra
matrix
algebra unsolved
algebra

i) Find all functions $ f$ continuous on $ \mathbb{R}$ such that :

\[ f ( f ( f (x))) \plus{} f (x) \equal{} 2x, \;\forall x \in \mathbb{R}.\]
ii) Find all continuous functions $ f : \mathbb{R} \to \mathbb{R}$ such that :
\[ 9 f (8x) \minus{} 9 f (4x) \plus{} 2 f (2x) \equal{} 100x, \;\forall x \in \mathbb{R}.\]
iii) Find all the functions $ f : \mathbb{Z} \to \mathbb{Z}$ satisfying the following conditions:
[color=white]....[/color][b]-[/b] $ f \big( f (m \minus{} n)\big) \equal{} f (m^2) \plus{} f (n) \minus{} 2n f (m)$,[color=white]... [/color]$ \forall m, n \in \mathbb{Z}$.
[color=white]....[/color][b]-[/b] $ f (1) > 0$.

[quote="linda2005"]  Find all functions $ f$ continuous on $ \mathbb{R}$ such that :
\[ f ( f ( f (x))) \plus{} f (x) \equal{} 2x, \;\forall x \in \mathbb{R}.
\][/quote]

$ f(x_1)\equal{}f(x_2)$ $ \implies$ $ f(f(f(x_1)))\plus{}f(x_1)\equal{}f(f(f(x_2)))\plus{}f(x_2)$ and so $ x_1\equal{}x_2$.  So $ f(x)$ is injective, so monotonous, since continuous.

If $ f(x)$ is decreasing, $ f(f(x))$ is increasing, $ f(f(f(x)))$ decreasing and $ f(f(f(x)))\plus{}f(x)$ decreasing, which is impossible ($ 2x$ is increasing). So $ f(x)$ is increasing.

Then, if $ f(a)<a$, so $ f(f(a))<f(a)<a$ and then $ f(f(f(a)))<f(f(a))<a$ and $ f(f(f(a)))\plus{}f(a)<2a$, which is impossible.
Then, if $ f(a)>a$, so $ f(f(a))>f(a)>a$ and then $ f(f(f(a)))>f(f(a))>a$ and $ f(f(f(a)))\plus{}f(a)>2a$, which is impossible.

So $ f(a)\equal{}a$ and the only solution is $ f(x)\equal{}x$. And we can immediatly verify that this solution is OK.

[quote="linda2005"] ii) Find all continuous functions $ f : \mathbb{R} \to \mathbb{R}$ such that :
\[ 9 f (8x) \minus{} 9 f (4x) \plus{} 2 f (2x) \equal{} 100x, \;\forall x \in \mathbb{R}.
\][/quote]

The equation may be written $ 9(f(8x)\minus{}\frac{5}{2}(8x))\minus{}9(f(4x)\minus{}\frac{5}{2}(4x))\plus{}2(f(2x)\minus{}\frac{5}{2}(2x))\equal{}0$

We can then write $ g(x)\equal{}f(x)\minus{}\frac{5x}{2}$ and we have $ 9g(8x)\minus{}9g(4x)\plus{}2g(2x)\equal{}0$, or also :

$ g(2x)\equal{}\frac{9}{2}(g(4x)\minus{}g(8x))$

So, we can create the sequence :
$ u_0\equal{}g(a)$
$ u_1\equal{}g(\frac{a}{2})$
$ u_{n\plus{}2}\equal{}\frac{9}{2}(u_{n\plus{}1}\minus{}u_n)$

Obviously, $ u_n\equal{}g(\frac{a}{2^n})$

And we have $ u_n\equal{}\alpha 3^n \plus{} \beta (\frac{3}{2})^n$ for some $ \alpha$ and $ \beta$

But, when $ n\longrightarrow \plus{}\infty$, this expression may have a finite limite ($ g(0)$, since $ f(x)$ and $ g(x)$ are continuous) only if $ \alpha\equal{}\beta\equal{}0$

So $ g(x)\equal{}0$ and $ f(x)\equal{}\frac{5x}{2}$ and we can easily verify that this solution works.

[quote="linda2005"] iii) Find all the functions $ f : \mathbb{Z} \to \mathbb{Z}$ satisfying the following conditions:
[color=white]....[/color][b]-[/b] $ f \big( f (m \minus{} n)\big) \equal{} f (m^2) \plus{} f (n) \minus{} 2n f (m)$,[color=white]... [/color]$ \forall m, n \in \mathbb{Z}$.
[color=white]....[/color][b]-[/b] $ f (1) > 0$.[/quote]

According to me, there are no solutions to this equation :

Let $ P(m,n)$ be $ f(f(m\minus{}n))\equal{}f(m^2)\plus{}f(n)\minus{}2nf(m)$

Let $ f(0)\equal{}a$. $ P(0,0)$ gives $ f(a)\equal{}2a$

$ P(n,0)$ $ \implies$ $ f(f(n))\equal{}f(n^2)\plus{}a$
$ P(0,\minus{}n)$ $ \implies$ $ f(f(n))\equal{}a\plus{}f(\minus{}n)\plus{}2an$

And so $ f(n^2)\equal{}f(\minus{}n)\plus{}2an$. Putting this in $ P(x,y)$, we get :

$ Q(m,n)$ : $ f(f(m\minus{}n))\equal{}f(\minus{}m)\plus{}2am\plus{}f(n)\minus{}2nf(m)$

$ Q(n,n)$ $ \implies$ $ 2a\equal{}f(\minus{}n)\plus{}2an\minus{}(2n\minus{}1)f(n)$ and so $ f(\minus{}n)\equal{}\minus{}2a(n\minus{}1)\plus{}(2n\minus{}1)f(n)$. Putting this in $ Q(m,n)$, we get :

$ R(m,n)$ : $ f(f(m\minus{}n))\equal{}2a\plus{}(2m\minus{}2n\minus{}1)f(m)\plus{}f(n)$

Then $ R(n\plus{}1,n)$ $ \implies$ $ f(f(1))\equal{}2a\plus{}f(n\plus{}1)\plus{}f(n)$ and so $ f(n\plus{}1)\equal{}k\minus{}f(n)$ and $ f(n\plus{}2)\equal{}f(n)$
So $ f(n)$ would take only two values but $ P(1,n)$ would be $ f(f(1\minus{}n))\equal{}f(1)\plus{}f(n)\minus{}2nf(1)$, so $ 2nf(1)\equal{}\minus{}f(f(1\minus{}n))\plus{}f(1)\plus{}f(n)$

And this last equation is impossible since left part is unbounded ($ f(1)>0$) and right part is bounded (since $ f(x)$ can take only two values).

[hide="Problem 3"]Putting $ m = 0, n = 0$ into the equation, we get $ f(f(0)) = 2f(0)$

Putting $ m = 1, n = 1$ we get $ f(f(0)) = 0$

Hence $ 0 = f(f(0)) = 2f(0)\implies f(0) = 0$

Putting $ n = m$ we get $ 0 = f(m^2) + f(m) - 2mf(m)\quad (1)$

Putting $ n = m - 1$ we get $ f(f(1)) = f(m^2) + f(m - 1) - 2(m - 1)f(m)\quad(2)$

Subtracting $ (2)$ from $ (1)$, with notation $ D: = f(f(1))$, we get

$ - D = f(m) - f(m - 1) - 2f(m)\iff f(m) + f(m - 1) - D = 0$

Hence $ f(m - 1) + f(m - 2) - D = 0$, so $ f(m) = f(m - 2)$

Therefore $ f(m) = \{\begin{matrix}0 & \mathrm{m\,\, even} \\
C & \mathrm{m\,\, odd}\end{matrix}$ for some integer $ C$.

But if we put $ m$ odd, $ n$ odd in the initial equation, we get $ 0 = 2C(1 - n)$ which is impossible as the condition $ f(1) > 0$ yields $ C>0$.

Therefore no such function exists.[/hide]


1. Give $ A\in M_{n}K$. Prove that:
$ det(I_n\plus{}A^2) >\equal{}0$
2. Give $ A\in M_{n}K$. Prove that exist matrix $ B,C$ with $ AA^{\minus{}1}\equal{}I$ and $ BB^{\minus{}1}\equal{}I$ such as $ A\equal{}B\plus{}C$

Hi vipCD_A1,
rewrite your post.
1) $ K\equal{}\mathbb{R}$ and $ det(I\plus{}A^2)\equal{}det(I\plus{}iA)det(I\minus{}iA)$.
2) Is 2) a joke ?

Perhaps 2) is supposed to be "Show that for any non-singular matrix $ A \in M_n(\mathbb{K})$ there exist matrices $ B,C$, with $ B$ non-singular, such that $ A \equal{} B \plus{} C$."

If this is the case, then it is true. In fact, it can be shown that, given any matrix $ A \in M_n(\mathbb{K})$, this matrix can be written as the sum of:

1) Two non-singular matrices;
2) Two singular matrices;
3) One singular matrix and one non-singular matrix.

[quote="loup blanc"]Hi vipCD_A1,
rewrite your post.
1) $ K \equal{} \mathbb{R}$ and $ det(I \plus{} A^2) \equal{} det(I \plus{} iA)det(I \minus{} iA)$.
[/quote]
I think so but I don't make finish it. Please, help me!!

$ \det(I \plus{} A^2) \equal{} \det (I \plus{} iA) \det(I \minus{} iA) \equal{} \det (I \plus{} iA) \det [(I \plus{} iA)^*]$
$ \equal{} [\det (I \plus{} iA)][\det (I \plus{} iA)]^* \equal{} |\det (I \plus{} iA)|^2 \geq 0$,

where I have used * to denote complex conjugate.


geometry
rectangle
combinatorics unsolved
combinatorics

Hi,

Who can proof that 6X6 rectangle cannot be tiled with 1X4 rectangles? Or otherwise who can show that 6X6 rectangle cannot be tiled with nine 1X4 rectangles?

Thanks in advance.

It is fairly easy just to place the first rectangle down in one corner and reason where the other recatangles would be and find a contradiction.

Or break the 6X6 into 2X2's consider the upper left square in eac 2X2. if a rectangle touches an upper left square, then it also touches another upper left square. So each tile tiles either 0 or 2 upper left squares. But there are 9 upper left squares so theres a contradiction.

It has been shown before on this forum that an $m\times n$ rectangle can be tiled with $1\times a$ rectangles iff $a$ divides one of the naturals $m,n$. Try it, it's a nice problem :).


[i]Trying to solve a problem from a Bulgarian olympiad, I arrived to the following result. I have a trigonometrical approach, but I'd like to have, if possible, a pure synthetic one. The problem is this:[/i]

[b]PROBLEM[/b]

Let a triangle $ ABC$. The excircle to angle $ C$ touches side $ AB$ at $ E$ and to the line $ CA$ at $ M$. The excircle to angle $ A$ touches side $ BC$ at $ D$ and to the line $ AC$ at $ L$. If line $ LD$ meets line $ ME$ at $ P$, prove that $ BP$ is perpendicular to $ AC$.

[u]Babis[/u]

See at the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=164092]Three excircles and three points[/url], of [b][size=100]Parkdoosung[/size][/b].

Best regards, Kostas Vittas.

Does [url=http://www.mathlinks.ro/viewtopic.php?t=222221]Vector equality[/url] serve as a reference, too?


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$. Show that:

\[ \frac{a}{11a^2\plus{}4}\plus{}\frac{b}{11b^2\plus{}4}\plus{}\frac{c}{11c^2\plus{}4}\le \frac{1}{5}\]

The given hypothesis implies $ \sum_{cyc}(1 - a) = 0$ .Using  AM-GM we have :
\[ \sum_{cyc}{\frac {a}{11a^2 + 4}} \leq \sum_{cyc}{\frac {1}{7a + 8}}\]
We need to prove that :
\[ \sum_{cyc}{\frac {1}{7a + 8} \leq \frac {1}{5}}\]

\[ \iff \sum_{cyc}{\left( \frac {1}{7a + 8} - \frac {1}{15}\right)} \leq 0\]

\[ \iff \sum_{cyc}{\frac {7(1 - a)}{15(7a + 8)}\leq 0}\]
Let $ a \geq b \geq c$ ,and observe that $ 7(1 - a)$ and $ 15(7a + 8)$ are not similarly sorted .Conclusion follows by Chebyshev

EDIT : Ignore this post

[quote="enndb0x"]The given hypothesis implies $ \sum_{cyc}(1 - a) = 0$ .Using  AM-GM we have :
\[ \sum_{cyc}{\frac {a}{11a^2 + 4}} \leq \sum_{cyc}{\frac {1}{7a + 8}}\]
We need to prove that :
\[ \sum_{cyc}{\frac {1}{7a + 8} \leq \frac {1}{5}}\]

\[ \iff \sum_{cyc}{\left( \frac {1}{7a + 8} - \frac {1}{15}\right)} \leq 0\]

\[ \iff \sum_{cyc}{\frac {7(1 - a)}{15(7a + 8)}\leq 0}\]
Let $ a \geq b \geq c$ ,and observe that $ 7(1 - a)$ and $ 15(7a + 8)$ are not similarly sorted .Conclusion follows by Chebyshev[/quote]

I think this prove is wrong...

$ 7(1 - a)$ and $ 15(7a + 8)$ are not similarly sorted, so it should have

$ \sum{\frac {7(1 - a)}{15(7a + 8)}\geq \frac13 \sum{7(1-a)}\sum{15(7a+8)} = 0.}$

... :wink:

I think it is incorrect since $ 15(7a + 8)$ is in the denominator. But
$ a\geq b\implies \{\begin{array}{c}77(1 - a)\leq 7(1 - b) \
15(7a + 8)\geq 15(7b + 8)\end{array}$
So sequences $ \{7(1 - a)\}$ and $ \{\frac 1 {15(7a + 8)}\}$ are [b]similarly sorted[/b].
Please rectify me if I am wrong. :oops:


[quote="nguyentrang"]Let $AA_1$, $BB_1$, $CC_1$ be angles bisectors in $\triangle ABC$, and let $G_1,G_2,G_3$ be the medians of triangles $AB_1C_1, BA_1C_1, CA_1B_1$ respectively. Prove that $AG_1,BG_2,CG_3$ are concurrent.[/quote]

...let $G_1,G_2,G_3$ be [b]the intersection points of[/b] the medians of...

Let $M$ be the midpoint of $B_1C_1$. Then:
\[ \frac{\sin{\angle{G_1AC_1}}}{\sin{\angle{AMC_1}}}=\frac{C_1M}{AC_1}  \]
and
\[ \frac{\sin{\angle{G_1AB_1}}}{\sin{\angle{AMB_1}}}=\frac{B_1M}{AB_1}  \]
Therefore we get $\frac{\sin{\angle{G_1AC_1}}}{\sin{\angle{G_1AB_1}}}=\frac{AB_1}{AC_1}$. Hence multiply them togather we get:
\[ \prod \frac{\sin{\angle{G_1AC_1}}}{\sin{\angle{G_1AB_1}}} = \prod \frac{B_1M}{AB_1} = 1  \]
Thus by trig Ceva they are concurrent.

[b]Generalization[/b] Let $ABC$ be a triangle and $A'B'C'$ be the Cevian triangle of a point $P$ wrt $ABC$. Call $Q$ a point and $A_1B_1C_1$ the Cevian triangle of $Q$ wrt $A'B'C'$. Then $ABC$ and $A_1B_1C_1$ are perspective.


Solve in the set of real numbers the equation \[ 3x^3 \minus{} [x] \equal{} 3,\] where $ [x]$ denotes the integer part of $ x.$

[hide="One Solution:"]Let $ h$ equal the decimal part of x, i.e. $ h \equal{} x \minus{} [x]$. Then the equation reads $ 3x^3 \minus{} (x \minus{} h) \equal{} 3$, where $ 0 \leq h < 1$.

This gives $ 3x^3 \minus{} x \plus{} h \minus{} 3 \equal{} 0$. Let $ f(x) \equal{} 3x^3 \minus{} x \plus{} h \minus{} 3$ and note that $ f'(x) \equal{} 9x^2 \minus{} 1$, giving $ f$ a local maximum at $ x \equal{} \minus{} \frac {1}{3}$. Since $ f\left( \minus{} \frac {1}{3}\right) \equal{} \frac { \minus{} 25}{9} \plus{} h < 0$, f has exactly one real zero.

Because $ f(1) \equal{} h \minus{} 1 < 0$ and $ f(2) \equal{} 19 \plus{} h > 0$, by the intermediate value theorem this zero must be strictly between 1 and 2. Therefore the integer part of x, [x], equals 1.

So the equation now reads, $ 3x^3 \minus{} 1 \equal{} 3$, which gives $ x \equal{} \sqrt [3]{\frac {4}{3}}$ as the only real solution.[/hide]

[quote="pelao_malo"][hide]if $ x > 2$ then $ 3x^3 \minus{} [x] > 3$ .[/hide][/quote]

why ??  :?:

$ a>b\to 3a^2\minus{}1>3b^2\minus{}1\to a(3a^2\minus{}1)>b(3b^2\minus{}1)$ then $ 3a^3\minus{}a>3b^3\minus{}b$ if $ a>b$ and if $ a>2$ then $ 3a^3\minus{}[a]\ge 3a^3\minus{}a>3$

This seems really easy for an IMO problem.. even if it IS longlist.
Cool solutions though :)

[hide="Solution"]Since $ 3x^{3}\equal{}\lfloor x\rfloor \plus{}3$, $ 3x^{3}$ is an integer $ k$.  Then $ x\equal{}\sqrt[3]{k/3}$.  Substituting gives $ k\minus{}\lfloor\sqrt[3]{k/3}\rfloor\equal{}3$.  Let $ f(k)\equal{}k\minus{}\lfloor\sqrt[3]{k/3}\rfloor$ for integers $ k$.  Then $ f(k)$ is increasing, and we have $ f(3)\equal{}2,f(4)\equal{}3,f(5)\equal{}4$, so the only solution is $ k\equal{}4$, so $ x\equal{}\sqrt[3]{4/3}$.[/hide]

[quote="matt276eagles"][hide="Solution"]Since $ 3x^{3} \equal{} \lfloor x\rfloor \plus{} 3$, $ 3x^{3}$ is an integer $ k$.  Then $ x \equal{} \sqrt [3]{k/3}$.  Substituting gives $ k \minus{} \lfloor\sqrt [3]{k/3}\rfloor \equal{} 3$.  Let $ f(k) \equal{} k \minus{} \lfloor\sqrt [3]{k/3}\rfloor$ for integers $ k$.  Then $ f(k)$ is increasing, and we have $ f(3) \equal{} 2,f(4) \equal{} 3,f(5) \equal{} 4$, so the only solution is $ k \equal{} 4$, so $ x \equal{} \sqrt [3]{4/3}$.[/hide][/quote]
nice solution

[quote="matt276eagles"]Then $ f(k)$ is increasing[/quote] $ f(2.9) \equal{} 2.9 > 2.1 \equal{} f(3.1)$.


Hey, it's odd to talk about school selection when everyone is talking about going nats :D

But I am still curious...

Can someone tell me how many tests did your school offer to you when selecting school team? Only one test or a few?

Our coach gives us many school rounds and selects the best performers of the collective total for the school team.

12 tests

1 was nat, 2 were state, 8 were chapter, 1 was school

i powned on the nat

with a 25/14 :lol: (dan got 11/6)

-jorian

Really...? Could it be that we have the same coach?

Wow, so many tests? In which timeframe? from Dec - January? or from Oct - January right before chapter?

Our coach starts practicing from the summer.

Start selection test from summer? Don't you burn out then?

nonono, september/october - december/january

have fun!

jorian

[quote="jhredsox"]team is decided in december, indis in january

jorian

omg, we had like 12 meetings, and you got a null score if you didn't come to a certain meeting for that test

i missed 3 meetings. dan missed 1.

and he comes in first and i come in second (more amazed at me coming in second)

i guess my nat really helped me

anyways, meet on weekends or something and do tests and have fun...something like that[/quote]

I think you were talking to me: No, we don't burn out, neither literally nor metaphorically.

[b]jhredsox write:[/b]
[quote]i missed 3 meetings. dan missed 1.

and he comes in first and i come in second (more amazed at me coming in second) [/quote]

Wow, incredible! missing 3 meetings, you still got into 2nd!

With that many selection tests, those students who have done this kind of selection tests in previous years, will definitely have advantage, correct?

Also, after the tests, will the teacher return the test paper to you or teacher just kept by him/herself?

When I was in Mathcounts (way, way, back...), we had a statewide math league during what you might call the "regular season", and after that was over, we just did the school competition.  Then the coaches somehow picked the team; I think it was partly based on the school test, partly on league scores, and partly on just who they thought would do well.

when my school first began practice (AKA right when basketball started) i missed like the first 3 practices.  then during the season, i missed a total of 5 or 6.  it didnt matter after january though.  my teacher bases our placing by points, like the total right for warm-ups and workouts.  i was 40 points ahead f the 2nd place kid in january, and i finished close to 70 points up.

We had four tests to find the 4 team members and 11 other students. Then, we had another 4 team rounds to find 4 individuals from those 11 students

If I remember rightly, I got screwed on the first few tests and then got a 40 on the last one and got 6th overall.  I have the same coach as Peijin.  Wait, that was what the team rounds were for?

[quote="magixter"]WoW! Did you make nationals? If so, what was your state score?[/quote]

Yes, i made nationals, and i also think that the 2nd place kid might next year, but he didnt get to go downstate this year because of some band contest thing

Somehow
My coach always chooses the right people for the team (top 4 ppl)
The Past Two Years (including this year):
Last Year:

Team Members:
1st on team got 2nd at Chapter and 1st at State and 19th at Nationals
2nd on team got 3rd at Chapter and 4th at State and 100 something place at Nationals
3rd on team got 8th at Chapter and 11th at State (thats me)
4th on team got 21st at Chapter and 12th at State
Individuals:
1st Individual got 6th at Chapter and 39th at State
2nd Individual got 10th at Chapter and 14th at State
3rd Individual got 14th at Chapter and I think 60 something place at State
4th Individual got 16th at Chapter

This Year:

Team Members:
1st on team got 3rd at Chapter and 10th at State (thats me)
2nd on team got 1st at Chapter and 14th at State (1st Individual from last year)
3rd on team got 9th at Chapter and 15th at State (3rd Individual from last year)
4th on team got 5th at Chapter and 5th at State
Individuals:
1st Individual got 4th at Chapter and 25th at State
2nd Individual got 10th at Chapter and 20 something at State
3rd Individual got 14th at Chapter
4th Individual got 20th at Chapter

So apparently all the team members do good at state but mediocre at chapter

My coach just gave us the school test the past two years and chose the top 4 for team and 5-8 for individual

[quote="magixter"]Your coach only gave you guys 2 tests? What if somebody did really bad one day? Isn't that prone to too many risks?[/quote]

Thats why idk how my coach always seems to pick the right ppl

and its just one test
not two

Oh I thought your coach gave you the school tests OF the last two years not IN the last two years. Sorry.

You guys are so lucky you get to do states and nats. Our coach just gave us the school sprint round, and then took the top 8 (which I totally screwed up on)

Our coach gives us 1 national, 1 state, 2 Chapter, 4 Schools

We meet on weekdays, and I was individual because I missed a whole lot of meetings for basketball, but then I got perfect scores on the 5 last tests (4 schools, 1 chapter), so it got me in as the 8th man.

Our coach usually gives 2 or 3 chapter or state rounds (sprint + target) throughout the early school year (before chapter, of course). This year, he only gave one chapter round, and I fortunately didn't make too many mistakes...

Wow. Our coach gives more than 10 competitions easily...

you know...it really doesn't matter, as long as you're not an alternate and you made the team or indis.

i'd rather be an individual (as i was last year and i missed about 5 meetings)...less pressure

oh...did i mention dan was ranked 3rd on our team last year? (the other two kids pown him)

jorian

My coach has all of the math club take the school round test that he writes. We have over 60 people. He takes the top 12 and we practice every Saturday for I think 8 weeks. Each week we take 1 Sprint, 1 Target, 1 Team, and do Countdown Practice. After those 8 weeks, on the day of before chapter, he announces the 4 team members, the 4 individuals, and the 4 alternatives. It's pretty even but he picks people I think more based on growth and consistency so he knows they'll do good at the competitions.


In a certain school, there are three times as many boys as girls and nine times as many girls as teachers. Using the letters $b,g,t$ to represent the number of boys, girls, and teachers, respectively, then the total number of boys, girls, and teachers can be represented by the expression

$\textbf{(A) }31b\qquad
\textbf{(B) }\frac{37b}{27}\qquad
\textbf{(C) }13g\qquad
\textbf{(D) }\frac{37g}{27}\qquad
\textbf{(E) }\frac{37t}{27}$

Here is my [hide="answer"]
Let there be:
$81 boys$
$27 girls$
$3 teachers$

So
$b+g+t = 111$

Now the answers:
A) $2511$
B) $111$
C) $351$
D) $37$
E) $4.111$

So $b$ is the answer.
[/hide]

Probably not the easiest way to do it...

Ooops...
But my way still works...just takes a bit longer...

Yeah, it's still the right approach, just a little more complicated than you needed. Just remember to always go simplest possible case for future problems. ;)

Yeah, I was working from top to bottom, instead up from bottom to top...

er..

never mind

From the given information,can't we find the number of boys and teachers in terms of th number of girls and then add them up?In that case the answer would come as $C$?

No, because look at it in terms of G.

Number of boys is 3g. Number of girls is g. Now read it carefully. It says there are nine times as many girls as teachers.
That means that if there are 9 girls, there is one teacher. Therefore, number of teachers is not 9G, but $\frac{1}{9}$ G.
Add it up, and you get 4$\frac{1}{9}G$, not 13.

[hide]

the answer would be $B$

let $g = 9t$

let $b = 27t$

the total number of people in terms of $t$ is $37t$

$t$ can be expressed in two ways:

$t = \frac{b}{27}$ or $t = \frac{g}{9}$

so the total number of people can be expressed either as $\frac{37b}{27}$ or $\frac{37g}{9}$

by looking at the choices available, the obvious answer would be $B$

[/hide]

[quote="lund"]I did the same as ritchp. However, I found it faster and easier to comprehend in terms of [b][i]b[/i][/b].[/quote]

true, the number of boys would be $b$, girls $\frac{1}{3}b$, and teachers $\frac{1}{27}b$. Add them up and it's a direct answer...


Katie wants to buy a CD player for $ \$150$. She has $ \$45$ that she received as a birthday gift and plans to earn the rest of the money by babysitting for which she makes $ \$3.75$ per hour. How many hours will she have to babysit to earn exactly what she needs?

She already has $ \$45$.  Thus, she needs to make $ \$150 \minus{} \$45 \equal{} \$105$ from babysitting.
Note that she makes $ \$3.75 \equal{} \$3\plus{}\$\frac{3}{4}\equal{}\$\frac{12}{4}\plus{}\$\frac{3}{4}\equal{}\$\frac{15}{4}$ per hour.  Thus, she needs to babysit for a total of $ \frac {\cancel{\$}105}{\cancel{\$}\frac {15}{4}} \equal{} \frac {105\cdot4}{15} \equal{} 7\cdot 4 \equal{} \boxed{28}$ hours.


Hmm, for $a = b = c > \sqrt{2}$ it seems to be false. Since then we have

$2(a^{2}+1) \ge 3a^{2}\Rightarrow a^{2}\le 2$.

Thank you, paladin8. I have corrected it.

Another nice one is the following
$\sqrt[3]{4(a^{6}+1)(b^{6}+1)(c^{6}+1)}\geq ab+ac+bc.$

[quote="arqady"]Let's $a,$ $b$ and $c$ are non-negative numbers. Prove that
$\sqrt[3]{4(a^{3}+1)(b^{3}+1)(c^{3}+1)}\geq ab+ac+bc.$[/quote]

Now it follows directly from Holder:

$\left(\frac{a^{3}}{2}+\frac{a^{3}}{2}+1\right) \left(\frac{b^{3}}{2}+1+\frac{b^{3}}{2}\right) \left(1+\frac{c^{3}}{2}+\frac{c^{3}}{2}\right) \ge \frac{1}{4^{\frac{1}{3}}}(ab+bc+ca)$.


in normal casual speech it rymes, and it can be considered ryhming in poems, but it doesnt really ryme since they dont end with the same letter. (not like people usually pronounce out the "t" in breakfast anyways...)
p.s. who the heck is soulja boy?
p.s.s. how do you spell "ryme"? (i suk at spelling)

[quote="dreambold22"]p.s. who the heck is soulja boy?[/quote]
YouTube it.

[quote="dreambold22"]p.s.s. how do you spell "ryme"? (i suk at spelling)[/quote]
Rhyme. Also, It's P.P.S. (post post script), not P.S.S.

ok. i see. so it's not "post sub script"?

Hahaha. post sub script? I think it rhymes...

I dont think it rhymes, but i is not no goodly at writings, so I'm not really sure

Nope. According to rhymezone.com, there are no rhymes for breakfast. Then again, I think their rules for "rhyming" are quite strict. On the other hand, websters-online-dictionary.com gives quite a few rhymes (but no Texas). However, no online dictionary that I looked up even mentioned the colloquial pronounciation of "breakfast" without the "t". So no.

 :lol:  :lol:   :lol:

I don't think there is a verb "to rhymes", so no.

[quote="i_like_pie"][quote="dreambold22"]p.s.s. how do you spell "ryme"? (i suk at spelling)[/quote]
Rhyme. Also, It's P.P.S. (post post script), not P.S.S.[/quote]

Yes. The prefix post- means after, which is why the PS is always at the end: AFTER the main message. So PSS would mean post script script, which doesnt makes sense. PPS means post post script, which would be a passage after the PS.

It does not rhyme!

BTW, Soulja boy of in dat ***
        Watch me crank dat watch me roll

roll doesn't rhyme with ***, so if it fits in souljaboy, it doesn't necessaruly rhyme. :D

[quote="ra5249"]ding....it.does.rhyme.
and i posed the question to my entire french class
the vote was 20 in favor of rhyming, 4 in favor of not rhyming[/quote]

But you live in Texas...

but the french are just weird...and it makes people who take french weird too

Just because it fits in soulja boy doesn't mean it rhymes!

What, the TX team copied our idea?  I wouldn't know cuz I left early.  When did this happen?

And Kevin, you know Arvind?  Wow, I missed out on a lot...

PS: It doesn't [i]strictly [/i]rhyme.  If we take off the first consecutive consonants, breakfast becomes eakfast and Texas becomes exas.  "ea" and "e" sound the same, so we examine kfast and xas.  If we say them the same way we'd say them in the two words, we immediately see that -kfast has a minor "t" sound at the end while xas doesn't.  Thus, they do not strictly rhyme, though in a looser sense, in writing songs like Soulja Boy, they could because they sound similar.

PPS: Who won the spirit stick?  Colorado?  With their "do you like waffles" and amazing guitar solo?

What?  That's really ridiculous, no offense to any NYers.  Their cheer was pretty bad, and according to Arvind, they only won b/c of their hats...

Ppl, you seriously need to study poetry.

Texas and breakfast are an ASSONANCE not a rhyme.

An assonance is where the vowels sounds near the end are the same.

And they're also a...whatchamacallit...............like relative rhyme or something.

[quote="ra5249"]absolutely man
btw this is a joke from nats
apparently they claim that texas does not rhyme with breakfast
but seriously (and arvind agrees with me) if you can use it in soulja boy it should work right?
see
soulja boy up in dat texas
watch me crank dat watch me breakfast

...

okay so fine it doesn't exactly work
and sure breakfast has a "t" at the end
but still if you take off the t it sort of works doesn't it

...[/quote]

YEAH DUDE OF COURSE IT WORKS!!! Here's another example.

Mary had a little lamb, little lamb, little lamb

Mary had a little lamb whose fleece was white as Texas

And everywhere that Mary went, Mary went, Mary went

Everywhere that Mary went that lamb was sure to breakfast. PWNED!!!

But like it would be better if breakfast was a verb. rite?

Dude. I asked this to like 30 ppl in my school and they were all like DUH of course...btw i cant belive dreambold din know who soulja boy was. Colorado shud hav definitely won spirit stick. And Ding, IT RHYMEZ!!! Just like cheese and beef...:D

[quote="Math Geek"]I know this is spam, but:

Do you like waffles?
Yes I like waffles...
Do you like pancakes?
Yes I like pancakes...
Do you like french toast?
Yes I like french toast...

NaNaNaNaNaNa,NaNaNaNaNaNa!

Do you like waffles?
I wanna get a mouthful!

Beastly guitar solo:

NaNaNaNaNaNa,NaNaNaNaNaNa,NaNaNaNaNaNa,NaNaNaNaNaNa,NaNaNaNaNaNa,NaNaNaNaNaNa!

And so forth...

PS: hadasah, it's funny how you say we need to study poetry when you got the definition of assonance wrong.  It's the repetition of vowel sounds to create internal rhyming within phrases or sentences, not "where the vowels sounds near the end are the same."  rotflmao :P Yeah, you just got pwnd.  No offense or anything though. :thumbup:[/quote]

DOOD I LUV DAT SONG!!!!!!

the guitar solo guy refused to perform it again, tho... :(


Let $f: \mathbb{R}\to\mathbb{R}$ and $a\in(0,\frac{1}{2})$ such that

                            \[ f(\frac{x+y}{2})\leq af(x)+(1-a)f(y)  \] and \[ f(x)\leq x,  \]
 for all $x, y>0.$

     Prove that $f=0.$ ;)

[quote="cezar lupu"] $f(x)\leq x$, Prove that $f=0.$[/quote]
 :?:

Is-it for $x, y > 0$ only ?


O generalizare a inegalitatilor lui Cebisev

O corectie
   In articolul “O generalizare a inegalitatilor lui Cebîşev”, la pagina 5, teorema 9 a fost preluată eronat din [2]. Varianta corectă este:
Teorema 9. Aplicaţia
$F\left( t \right)=\frac{{{M}^{[t]}}\left( \vec{p},\vec{a} \right)}{{{M}^{[t]}}\left( \vec{p},\vec{b} \right)},$
este crescătoare dacă \[\vec{b}\succ \prec \frac{{\vec{a}}}{{\vec{b}}}\]  şi descrescătoare dacă \[\vec{a}\succ \prec \frac{{\vec{a}}}{{\vec{b}}}.\]
Prin urmare, afirmaţia că inegalitatea (15) este mai “tare” decat inegalitatea lui Holder este greşită, întrucat demonstraţia acestui fapt se baza pe teorema 9 în variantă eronată.                                               Se poate însă afirma altceva în cazul inegalităţii (15), şi anume faptul că dacă inegalitatea nu este mai “tare” dacat inegalitatea lui Holder, atunci schimband rolurile lui $r$  şi $s$  obtinem o variantă mai “tare”. Intradevăr, dacă notăm
$f\left( r,s \right)={{\left( \frac{\sum{\vec{p}\vec{b}}}{\sum{\vec{p}{{{\vec{a}}}^{r-1}}}} \right)}^{\frac{1}{r}}}{{\left( \frac{\sum{\vec{p}\vec{a}}}{\sum{\vec{p}{{{\vec{b}}}^{s-1}}}} \right)}^{\frac{1}{s}}}\,,$
atunci   este suficient să demonstrăm că $f\left( r,s \right)\cdot f\left( s,r \right)\le 1.$
Aceasta se reduce însă la inegalitatea
$\left( \sum{\vec{p}\vec{a}} \right)\left( \sum{\vec{p}\vec{b}} \right)\le\left[ {{\left( {{\sum{\vec{p}\vec{a}}}^{{r}/{s}\;}} \right)}^{{1}/{r}\;}}{{\left( {{\sum{\vec{p}\vec{a}}}^{{s}/{r}\;}} \right)}^{{1}/{s}\;}} \right]\cdot \left[ {{\left( {{\sum{\vec{p}\vec{b}}}^{{r}/{s}\;}} \right)}^{{1}/{r}\;}}{{\left( {{\sum{\vec{p}\vec{b}}}^{{s}/{r}\;}} \right)}^{{1}/{s}\;}} \right],$
care este o consecinţă a inegalităţii lui Holder.                                                                                Intradevăr, avem
$\sum{\vec{p}\vec{a}}\le {{\left( {{\sum{\vec{p}\vec{a}}}^{{r}/{s}\;}} \right)}^{{1}/{r}\;}}{{\left( {{\sum{\vec{p}\vec{a}}}^{{s}/{r}\;}} \right)}^{{1}/{s}\;}}$
si $\sum{\vec{p}\vec{b}}\le {{\left( {{\sum{\vec{p}\vec{b}}}^{{r}/{s}\;}} \right)}^{{1}/{r}\;}}{{\left( {{\sum{\vec{p}\vec{b}}}^{{s}/{r}\;}} \right)}^{{1}/{s}\;}}$
care prin inmulţire conduc la inegalitatea dorită.
Are loc  egalitatea $f\left( r,s \right)\cdot f\left( s,r \right)=1$ numai daca $r=s=2$ sau ${{a}_{1}}={{a}_{2}}=...={{a}_{n}}$ si ${{b}_{1}}={{b}_{2}}=...={{b}_{n}}$.
Rezultatul se extinde şi pentru varianta integrala.


complex analysis
complex analysis unsolved

How to construct an analytic bijective map from $ \mathbb{C}$\[ - 1,1] to B(0,1)\{o}

Use Zhukovski map to map the complement of the unit interval to the complement of the unit disk and than apply $ \frac 1z$.

Well, I never remember which one is Zhukovski and which one is the inverse Zhukovski (if $ z\mapsto \frac12\left(z\plus{}\frac 1z\right)$ is Zhukovski, then, clearly, you need the inverse to get the complement of the disk from the complement of the interval), but otherwise what I said was correct :).

Yes! Thanks!

The map $ z\mapsto \frac12\left(z\plus{}\frac 1z\right)$ is from C\[- 1,1] to B(0,1)\{o} analytic bijective map!

But I have some small problems to prove it is surjective


superior algebra
superior algebra unsolved

Is there a finite(infinite) extension of $ \mathbb{R}$ not isomorphic to $ \mathbb{C}$ ?

What type of extension¿ There definitely is no finite algebraic extension as $ \mathbb C$ is already the algebraic closure. There are a lot of infinite ones.

Sorry for not specifying. Let's say infinite Galois extensions ? Are there plenty of them ? No way to classify ?

What is a Galois extension here¿ As said, there is no algebraic one.

Yes, yes. After asking the first question about extension of reals I was already thinking about infinite Galois extension over $ \mathbb{Q}$, because there is indeed no sense to consider any even algebaric extensions over reals.


In how many ways can we color $ n$ points in a line with 5 colors in such a way that for any 3 consecutive points, 2 are the same color?

I really like problems like this :), but Open Problems is not the correct place for them.

[hide="Solution"]Change 5 to $ k$ (because why not?).  Let $ a_n$ be the number of such sequences that begin with two points of the same color and let $ b_n$ be the number that begin with two points of different colors.  The number we want is $ c_n = a_n + b_n$.  Consider coloring the first three points: they can be colored $ rrr$, $ rrb$, $ rbr$, or $ rbb$ for two different colors $ r, b$.  Those colored $ rrr$ and $ rrb$ are counted by $ a_n$; those colored by $ rrr$ are counted by $ a_{n - 1}$, while those colored $ rrb$ are counted by $ b_{n - 1}$.  Those colored $ rbr$ and $ rbb$ are counted by $ b_n$; those colored $ rbr$ are counted by $ b_{n - 1}$ while those colored $ rbb$ are counted by $ (k - 1)a_{n - 1}$.  Thus $ a_n = a_{n - 1} + b_{n - 1}$ and $ b_n = (k - 1)a_{n - 1} + b_{n - 1}$.  Then do a bit of algebra: $ a_n = c_{n - 1}$ and so $ b_n = (k - 2)a_{n - 1} + c_{n - 1} = (k - 2)c_{n - 2} + c_{n - 1}$, so $ c_n = a_n + b_n = 2c_{n - 1} + (k - 2)c_{n - 2}$.  We have initial values $ c_1 = k, c_2 = k^2$, so we can recursively compute as many values as we like.  We can also use our favorite method for solving linear recurrence relations to get that $ c_n = \frac {k}{2} \cdot \left((1 + \sqrt {k - 1})^n + (1 - \sqrt {k - 1})^n\right)$.  The "nice" particular cases are $ k = 1$, giving $ c_n = 1$, $ k = 2$, giving $ c_n = 2^n$ and the given problem $ k = 5$, giving $ c_n = \frac {5(3^n + ( - 1)^n)}{2}$.[/hide]


Could somebody understand the problem and help me draw the figure for this following problem (which appeared in AIME2004, problem 14)?

 A unicorn is tethered by a 20 foot silver rope to the base of a bagician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower and the length of the rope that is touching the tower is $\frac{a-\sqrt{b}}{c}$ feet, where $a$,$b$, and $c$ are positive integers, and $c$ is prime. Find $a+b+c$

Many thanks.

Am I correct in thinking that the rope must be coiled around the tower?

There was another thread on this problem...
[quote="scorpius119"][hide]Let $U$ be the point representing the unicorn, $P$ be the point where the rope touches the tower, $Q$ be the point directly above $P$ at the height of the unicorn, $C$ be at the center of the cylinder at the height of the unicorn.

Note that $CQ\perp QU$. $UC=12,\ CQ=8\Rightarrow UQ=4\sqrt{5}$ from Pythagorean Theorem.

Let $x$ be the length $UP$. Since the length of the rope is 20 and the height of the unicorn is 4, the rope rises 1/5 for each foot of rope, so that $PQ=\frac{x}{5}$.

Using Pythagorean Theorem, $x^2=\left(4\sqrt{5}\right)^2+\left( \frac{x}{5}\right)^2=80+\frac{x^2}{25}\Rightarrow \frac{24x^2}{25}=80\Rightarrow x^2=\frac{250}{3}\Rightarrow x=\frac{\sqrt{750}}{3}$.

The length of the rope touching the tower is $20-x=\frac{60-\sqrt{750}}{3}\Rightarrow \boxed{813}$.[/hide][/quote]
Side view:
[img]http://img162.imageshack.us/img162/8002/18it.png[/img]
Top view:
[img]http://img113.imageshack.us/img113/7180/21gr.png[/img]

:mad: The thing that got me was for some reason I thought it was a circular rope, and thought the answer would have to be $20 - 2x$ instead, and kept getting answers that didn't fit under 1000....

[quote="123s"]I don't think thats a very good picture, though I solved it the same way and got the same answer. The place where the rope leaves the circle in a straight light is a tangent. Thus, the radius to that point should form a right angle with that line. However, in the picture, the right angle is between the radius and the line connecting the center of the circle to the end of the rope.[/quote]

Do you mean that angle UPC in the top view is right?
I think either diagram is correct, actually.


abstract algebra
superior algebra
superior algebra unsolved

If $ R$ has zero divisors show that every nonzero $ R\minus{}$module has nonzero torsion elements.

Of course you're going to be using the fact that there exists $ a,b \in R$ such that $ ab \equal{} 0$ when neither $ a$ nor $ b$ is.  Now since we have a nonzero $ R$-module, we certainly have a nonzero element, let it be $ x$.  What happens if $ bx \equal{} 0$?  What do we do when $ bx \neq 0$?


Has anyone tried this before?

I gave it ago today and got problems 1-11 and 20.

I mostly used PHP and Perl, but used a tiny bit of C++ when I couldn't get it in the first 2.

http://www.projecteuler.net

Give it a try.  :)

I've been doing it for a while, its good for practicing new languages on it

I got into the 40s in one day, most of them aren't too hard.

Yeah, not too difficult so far, I did about 50 yesterday. Still, fun though, thanks for mentioning it :)

[quote="TripleM"]Heh, they do start to get a little harder. Currently at 86/168 problems and going strong..[/quote]

I know the feeling. I detest the problems which involve primes. Not because I have anything against primes, but because I still can't get a sieve of atkins right and have to resort to erathantos

I'm stuck on 12.

[code]
<?php
	function factors($x)
	{
		$testnum = 1;
		$counter = 0;
		while ($testnum <= $x)
		{
			if ($x % $testnum == 0)
			{
				$counter++;
			}
			$testnum++;
		}
		return $counter;
	}

for ($i = 10002; $i < 10100; $i++)
	{
		echo ($i*($i+1))/2 . "  " . factors(($i*($i+1))/2) . "<br />";
	}
?>[/code]

The factors function returns how many factors the number has.  This works for small numbers but it goes REALLY slow for larger triangle numbers. Can anybody help me make it faster or suggest a different method?

You can either find a way to calculate the number of factors of $ n$ in $ \mbox{O}(\sqrt{n})$ time, or you can just notice that you're factoring $ \frac{i(i \plus{} 1)}{2}$ each time, so you can factor $ i$ and $ i \plus{} 1$ separately and combine the results.

Or... if you want a really fast program, you can do both of those together.  :D

[quote="math92"][quote="Temperal"]#14 was tricky; my algorithm was very inefficient and brute-forcy.

I'm to #15 right now.[/quote]

What was your method for 14?[/quote]

Something VERY close to testing just about every sequence under 1,000,000.

[quote="math92"]The factors function returns how many factors the number has.  This works for small numbers but it goes REALLY slow for larger triangle numbers. Can anybody help me make it faster or suggest a different method?[/quote]

Your current method is $ O(n)$ while a [i]very[/i] simple modification speeds it up to $ O(\sqrt {n})$:  once you find a divisor that is greater than or equal to $ \sqrt {n}$, stop.  If $ \sqrt {n}$ is not a divisor, double the number of divisors you have (because for each divisor $ d$ that is less than $ \sqrt {n}$, the divisor $ \frac {n}{d}$ is greater than $ \sqrt {n}$, so you can pair them up).  Otherwise, double the number of divisors you have and add $ 1$ (for $ \sqrt {n}$).

Once you split $ n$ up into $ i, i \plus{} 1$ you can speed up even further to $ O(\sqrt [4]{n})$.

Additionally:  you might want to divide out small prime factors first (say, $ 2, 3, 5, 7, 11$) and then count what's left.  Odds are for "most" triangular numbers this will speed up your calculation a lot.

Edit:  [b]Temperal:[/b]  I haven't thought about it much, but it seems like what you should be doing is a sort of sieve.  That is, every number that appears in a sequence has a path length strictly less than that of the number you started with, so don't start with any of those numbers.  Once you hit numbers with longer paths, it should start to speed up your calculation a little.  Of course, memory might be an issue.

Thanks!

It runs a LOT faster now.

But, I'm still not getting the correct answer. I'm getting: 842161320 and its wrong.

 :(

[quote="t0rajir0u"]Did you forget to take the division by two into account?[/quote]
Oh, yes I forgot about that.  Thanks.

I got the right answer now.   :lol:

Thanks, t0r, that tripled the speed of my calculations... my computer's memory is suitably large.

129/168.. and going much much slower :)

Can somebody else read problem 149 and tell me what the intended problem is, because what is written there makes no sense whatsoever. Any ideas, anyone?

[edit]Nevermind, for some weird reason they made the - an image?! And I had them off.

I've done problems 1-139. I just haven't really started with the newer problems this year.  :D

Whew. Just 6 to go: 141, 154, 155, 156, 165, 167; I'm one spot off moving onto the first page of scores. The problems definitely get harder and harder as they go along.. but I'm not stopping til I get 100%.

[quote="TripleM"]Yay :D 100%. Boy, those last ones were tough.[/quote]

I think that must be a record for coding through them all!

Heh, heh.. for possibly the first time ever, theres a programming contest style thing which actually fits in my (NZ) time zone perfectly. Which lets me jump straight to #1 after problem 170 came out :D

They do change it around each release. You must have got lucky for it to be your first new problem in your time zone.


Let a,1,0 be non-equal and be algebraic and beta irrational. is a to the square root  of beta power  is then transcendental?

It is closely related to the 7th [url=http://mathworld.wolfram.com/HilbertsProblems.html]Hilbert's Problem[/url].


Find the greatest positive integer $ n$ such that we can choose $ 2007$ different positive integers from $ [2.10^{n \minus{} 1},10^{n}]$ such that for each two $ 1\leq i < j\leq n$ there exists a positive integer $ \overline{a_{1}a_{2}...a_{n}}$ from the chosen for which $ a_{j}\geq a_{i} \plus{} 2.$

Given a problem that has gone unsolved in the Olympiad forums, I hardly see how it is appropriate to repost it to the Preolympiad forums, especially without providing some sort of note that you are double-posting (or at least naming the source of the problem).

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1984436241&t=149152


There is a hill that is two miles from the base to the top on the north side and one mile from the top to the bottom on the south side. Jack has an old car that can only go up the hill at an average speed of 40 miles per hour, but he can race down the hill as fast as he desires. What will Jack's average speed have to be going down the south side of the hill to average 60 miles per hour over the entire hill?

[hide]$\infty$!! :D  60 mph is one mile per minute, so 3 miles should take you 3 minutes.  But going two miles at 40 mph will take all of the three minutes, so you need to go down in no time at all, at infinite speed![/hide]


What is the smallest whole number that has a remainder of 1 when divided by 4, a remainder of 1 when divided by 3, and a remainder of 2 when divided by 5?

[hide=" what I think"]This number has to be one more than a number that is both divisible by 4 and 3. It also has to be 2 more than a multiple of 5. I just did guess and check starting at 13 since that's one more than a multiple of 3 and 4. That didn't work, and neither did 25, but $ 37 \div 4$ is 9 R 1, $ 37 \div 3$ is 12 R 1, and $ 37 \div 5$ is 7 R 2, so 37 is the smallest number that works[/hide]

[hide] Since the number gives a remainder of $ 1$ when divided by both $ 3$ and $ 4$, it will also give a remainder of $ 1$ when divided by $ 12$. Trying out the values of $ 12x\plus{}1$ (where $ x$ is a positive integer), we see that $ \boxed{37}$ is the first which leaves a remainder of $ 2$ when divided by $ 5$.[/hide]


I have been allowed to use the letters A–Z, symbols [, ], - and ^ (group of characters), ?, * and + (repetition of symbols), | (alternative) and ( and ) (grouping). Construct as small regular expression as possible which accepts those words having KORONKORKO as a substring but reject any other words.

Well, that problem is trivial. I read the problem too fast, sorry. It should be:
Construct a regular expression that accepts those strings that have no KORONKORKO as a substring and rejects other strings. The shorter expression the better.

I don't know how short it can be made (if for example you could take advantage of the fact that K is always followed by OR), but a possible solution is to go character by character:

([^K] | K[^O] | KO[^R] | .. KORONKORK[^O])*

ffao's solution doesn't work though, it will accept for example the word KKORONKORKO. There exists an algorithm for creating a regular expression if you have the DFA for your language (which is in this case pretty simple to construct) but the result would be a rather huge and incomprehensible expression. A sample solution for a simplified version of your problem with "AB" instead of "KORONKORKO" would be:

([^A] | A+[^AB])*A*

[quote="t0rajir0u"]I believe you can translate the way the [url=http://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm]KMP algorithm[/url] works to a feasible solution.[/quote]
That's how you make the DFA, but I'm not sure how to translate it to efficient regular expression...

[quote="bus"]ffao's solution doesn't work though, it will accept for example the word KKORONKORKO.[/quote]

I overlooked the string overlap... It would work perfectly if we were allowed to use lookahead, but I guess that would make it too easy. Good luck anyway!


\[ \sqrt {x^2  \minus{} 1}  \cdot \left( {ax^2  \minus{} x} \right) \geqslant 0{\text{ with a }} \in {\text{ }}\left] {{\text{0}}{\text{,1}}} \right[{\text{ }}
\]

then $ ax\minus{}1 \ge 0$ just try to understand why ... :rotfl:

[quote="othman"]then $ ax \minus{} 1 \ge 0$ just try to understand why ... :rotfl:[/quote]
Sorry, I edited.

try a=1,x=0,5.You shouldn't post two topics,


Given a circle centered at point P with radius r, and two A, B within the circle.
How to find a point C on the cirlce (on the boundary of the cirlce) such that

angle ACB = alpha. where alpha is a given costant within range [0, pi]

I listed the equation for such problem, but seems there is no way I can solve these equations. Is there any solution for finding the points C?

[quote="SpongeBob"]Of course there is, but not with calulations.

You should construct a locus of such point $C$, then wanted point will be intersection of that locus and given circle.
Construction of that locus really isn't olympiad - like problem, so I guess you can do it yourself :)

Bye[/quote]

Yes, I think you are right. But the problem is that, even if I listed all of the equations, I cannot solve it, and therefore cannot find an close-form solution. I am wondering if there is any better way to calculate it.

could anyone here at least try it and give me some suggestions?

I'll give it a try.

From A, and from B construct an angle (90 - alpha) ...construct a perpendicular at the point then copy the given angle to subtract.  You will form an isosceles triangle with an angle of 2 times alpha at the apex.  That will be the center of a circle containing chord AB and sector angle 2alpha. Draw the circle.  Where it cuts the original circle you'll have the subtended [within the second drawn circle] angle required.


geometry
incenter
exterior angle
geometry open

[quote="ertanrock"][size=150][color=darkred][b]Let O be the intersection point of the interior angle bisectors then show that O is incenter.
Show for exterior angle bisectors (excenter) [/b][/color][/size][/quote]
[quote"ertanrock"]This is easy to prove. Suppose O is incenter. Then write all the angles and add them. you wil find the sum of the interior angles in triangle is the same as the sum of the angles at O.[/quote]
????????????? What do you want to say ????????? I don't understand at all.

Dear Mathlinkers,
a interesting proof thaht O is the incenter can be obtained by considering the Morley's circle:
http://perso.orange.fr/jl.ayme  vol. 2  Les cercles de Morley... p. 1
Sincerely
Jean-Louis


For $ x \neq \frac{3}{2}$, what is the value of $ \frac{3}{2x\minus{}3} \minus{} \frac{2x}{2x\minus{}3}$?

Since the denominators are the same, we can subtract the numerators, so $ \frac{3\minus{}2x}{2x\minus{}3}\equal{} \frac{\minus{}(2x\minus{}3)}{2x\minus{}3}\equal{}\minus{}1$.

The denominators are the same so combine the numerators to get $ \frac{3\minus{}2x}{2x\minus{}3}\equal{}\minus{}\frac{2x\minus{}3}{2x\minus{}3}\equal{}\minus{}1$.


trigonometry
real analysis
real analysis unsolved

$\µ(x)=\tan(\sum_{k=0}^{1000}\frac{(-1)^k}{2k+1}x^{2k+1})$

Give an asymptotic expansion of order $2005$ at $x=0$

You must find $a_k$ such that

$\µ(x)=\sum_{k=0}^{2005}a_kx^k+o(x^{2005})$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ be a prime number and $ (1~2)\in H<S_p$. Prove that if $ H$ acts transitively on the set $ \{1,2,\ldots ,p\}$, then $ H\equal{}S_p$.

Consider the subgroup $ H_p\equal{}\{h\in H\,|\,h(p)\equal{}p\}$ of $ H$. Since $ H$ operates transitively, we have $ p\equal{}|H: H_p|$ from the orbit formula, in particular, $ H$ must contain an element of order p. Since $ |S_p|\equal{}p!$, all Sylow-p-groups of $ S_p$ have order p and are generated by some corresponding p-cycle, i.e. $ H$ contains an involution and a p-cycle. Since $ S_n$ (for arbitrary n) is generated by an involution and an n-cycle, the proof is complete.

but not by arbitrary choices: Sn is generated by (ij) and (12...n) if and only if i-j is coprime to n. but in the exercise, you just need a weak form of this.


Let a,b,c>0 such that ab+bc+ca+abc=4 Prove that
$ a^2\plus{}b^2\plus{}c^2\plus{}2(a\plus{}b\plus{}c)\plus{}3abc \geq 4(ab\plus{}bc\plus{}ca)$

[quote="tran300491"]Let a,b,c>0 such that ab+bc+ca+abc=4 Prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(a \plus{} b \plus{} c) \plus{} 3abc \geq 4(ab \plus{} bc \plus{} ca)$[/quote]
$ LHS\equal{}f (abc)$.Because$ f'' (abc)\equal{}0$.So we need to prove it in 2 cases:
Case 1: \[ c \equal{} 0 \to ab \equal{} 4 \to LHS \equal{} a^2  \plus{} b^2  \plus{} 2(a \plus{} b) \ge 4
\]
Case 2:\[ c \equal{} b \to a \equal{} \frac{{4 \minus{} b^2 }}{{b^2  \plus{} 2b}} \to LHS \minus{} RHS \equal{} {\rm{(b \minus{} 2)(b \plus{} 2)(b}}^{\rm{2}} {\rm{ \plus{} 2b}}^{\rm{3}} {\rm{ \minus{} b}}^{\rm{2}} {\rm{ \plus{} 4b \minus{} 4)}} \ge {\rm{0}}
\]
Done :)


a) Find all function $f: \mathbb R \to \mathbb R$ satisfying \[ f(x\plus{}f(xy))\equal{}f(x)\plus{}xf(y)\] for all $x,y \in \mathbb R$.
b) Find all function $f: \mathbb R \to \mathbb R$ satisfying \[f(x\plus{}f(x)f(y))\equal{}f(x)\plus{}xf(y)\] for all $x,y \in \mathbb R$.

[quote="SUPERMAN2"]Problem:
a)Find all function $ f: R \to R$ satisfying $ f(x \plus{} f(xy)) \equal{} f(x) \plus{} xf(y)$
[/quote]

Let $ P(x,y)$ be the assertion $ f(x\plus{}f(xy))\equal{}f(x)\plus{}xf(y)$

We obviously have a solution $ f(x)\equal{}0$ $ \forall x$. So we'll now consider only solutions non all-zero.

1) $ f(0)\equal{}0$, $ f(1)\equal{}1$ and $ f(x)\neq 0$ $ \forall x\neq 0$
If $ f(x)\equal{}0$ $ \forall x\neq 0$, Then $ P(0,0)$ implies $ f(f(0))\equal{}f(0)$ and so $ f(0)\equal{}0$. So $ \exists v\neq 0$ such that $ f(v)\neq 0$
Suppose now that $ \exists u$ such that $ f(u)\equal{}0$. Then $ P(\frac uv,v)$ $ \implies$ $ f(\frac uv\plus{}f(u))\equal{}f(\frac uv) \plus{}\frac uvf(v)$ $ \implies$ $ uf(v)\equal{}0$ $ \implies$ $ u\equal{}0$
But $ P(\minus{}1,\minus{}1)$ $ \implies$ $ f(\minus{}1\plus{}f(1))\equal{}0$ and so $ f(1)\equal{}1$ and $ f(0)\equal{}0$
Q.E.D.

2) $ f(x)$ is injective
Suppose $ f(a)\equal{}f(b)$ with $ a,b\neq 0$ and $ a\neq b$ (we already know that $ f(a)\equal{}0$ $ \implies$ $ a\equal{}0$)
Let $ x\neq 0$. Comparing $ P(x,\frac ax)$ and $ P(x,\frac bx)$, we get that $ f(\frac ax)\equal{}f(\frac bx)$ $ \forall x\neq 0$
Let then $ u\neq 0$ (and so $ f(u)\neq 0$) and let $ x\equal{}\frac {a\minus{}b}{f(u)}\neq 0$  so that $ \frac ax\equal{}\frac bx \plus{} f(u)$ :
$ f(\frac ax)\equal{}f(\frac bx)$ $ \implies$ $ f(\frac bx)\equal{}f(\frac bx\plus{}f(u))$
Then $ P(\frac bx,\frac{ux}b)$ $ \implies$ $ f(\frac bx)\equal{}f(\frac bx\plus{}f(u))$ $ \equal{}f(\frac bx)\plus{}\frac bxf(\frac {ux}{b})$ and so contradiction since $ a,b,x,u\neq 0$
So $ a\equal{}b$
Q.E.D.

3) $ f(x)\equal{}x$ $ \forall x$
$ P(1,x\plus{}f(xy))$ $ \implies$ $ f(1\plus{}f(x)\plus{}xf(y))\equal{}1\plus{}f(x)\plus{}xf(y)$
Setting $ x\equal{}\minus{}1$ in this equation and writing $ f(\minus{}1)\plus{}1\equal{}c$, we get $ f(c\minus{}f(y))\equal{}c\minus{}f(y)$
Setting then $ y\equal{}c\minus{}f(z)$ in this last equation, we get $ f(f(z))\equal{}f(z)$
And, since $ f(x)$ is injective, this implies $ f(z)\equal{}z$
Q.E.D.

Hence the two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$

Thanks a lot.
Could you please send me the links for downloading the problems and solutions from AMM,nguyenvuthanhha?Thanks.


function
trigonometry
algebra
domain
absolute value
complex analysis

I am having a difficult time figuring this out :(
Let $ g$ be an analytic function on the set shown below, such that $ g(z)^6=z$ and $ g(-1)=-i$. Find $ g(i)$ and ${ g(\frac{3i}{4}})$ in Cartesian form.
[img]http://photos-f.ak.fbcdn.net/hphotos-ak-snc3/hs157.snc3/18441_290419425660_509745660_4981830_6822668_n.jpg[/img]

I dont really know where to start. I tried this:
since $ g(-1) = -i = \cos (\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})$
and we want g(i), I tried "tracing" an angle from $ -1$ to $ i$ but staying within the set. I think it takes $ -3\pi + \frac{\pi}{4}$ radians so I added this value to the RHS of $ g(-1)$. I am no idea if its right or what I really am doing :P Thanks

You have a right idea: you need a continuous branch of the argument of $ g$. Note that its increment is $ 1/6$ times the increment of the argument of $ z$. Now, just find the increment of the continuous branch of the argument of $ z$ from $ \minus{}1$ to $ i$ along the domain, divide it by $ 6$ and add the result to the argument of $ g(\minus{}1)$. You'll get the argument of $ g(i)$. The absolute value shouldn't be a problem.

To be honest that was sort of confusing. To find the increment of the continuous branch of the argument of $ z$ from $ \minus{} 1$ to $ i$ along the domain... I am not sure what you mean. Do you mean $ \sqrt {2}$?

Edit : Ah, scratch that. Argument = angle XD
Well the argument that I thought was from $ \minus{} 1$ to $ i$ along the domain was $ \minus{} 3\pi \plus{} \frac {\pi}{4}$ because starting from the $ \minus{} 1$, we go around to the left of it counter clockwise to get $ \minus{} 2\pi$ then a half circle right, then down, then up (to cancel the "down) and then up $ \frac {\pi}{4}$ radians. Am I going for the completely wrong thing?

That is correct except the right angle is $ \frac\pi 2$, not $ \frac\pi 4$ :P.

O no! Okay here was my really big problem during the long while I as working on it.
How come it is $ \frac{\pi}{2}$? I have to start counting from the origin?
I was verifying the actual trig. values so that it equals what I want it to (ie i needed $ \frac{\pi}{2}$) but I can't actually figure it out mentally. I thought I had to start tracing the angles starting from $ \minus{}1$ instead of the origin :(.

So my actual answer will be $ g(i) \equal{} \cos \left( \frac{\frac{3 \pi}{2} \plus{} \frac{\pi}{2} \minus{}3\pi }{6} \right) \plus{} i\sin \left( \frac{\frac{3 \pi}{2} \plus{} \frac{\pi}{2} \minus{}3\pi }{6} \right )$ ??

I am not sure how to finish ^_^

[quote]How come it is $ \frac{\pi}{2}$? I have to start counting from the origin? [/quote]
How came that it is $ \frac \pi 4$? :? Can you show me the angle that you counted as $ \frac\pi 4$? My $ \frac\pi 2$ is the angle from the direction of the positive semi-axis to $ i$.

As to your "answer", the angle should be $ \frac{3\pi}2\plus{}\frac{\minus{}3\pi\plus{}\frac\pi 2}{6}$ (argument of $ \minus{}i$ plus $ 1/6$ of the increment, not $ 1/6$ of the sum of the argument of $ \minus{}i$ and the increment).

As to "how to finish?", I don't think anybody really wants you to compute $ \sin$ and $ \cos$ of $ \frac\pi{12}$, though it is not terribly hard from the double angle formula.

Ah my angle comes from me considering $ \minus{} 1$ as the origin and from the positive x-axis going up to $ i$. I know its not the origin :P but I was asking why is it we start from the origin and not subtend the angle from the point we're starting at (ie. $ \minus{} 1$).
Thanks a bunch by the way :)

So for the next part is it just the angle $ \frac {3\pi}{2} \plus{} \frac { \minus{} 4\pi \minus{} \frac {\pi}{2}}{6}$ (and scaled) ? So $ g(\frac {3i}{4}) \equal{} (\frac {3}{4})^{\frac {1}{6}}cis \left( \frac {3\pi}{2} \plus{} \frac { \minus{} 4\pi \minus{} \frac {\pi}{2}}{6} \right)$?

Also that is where the $ \sqrt{2}$ came from in my earlier post. $ \minus{}1$ and $ i$ differ by that length so I thought I had to scale by that too.

Well, we count the argument of $ z$ from $ 0$ because it is the argument counted from the origin that is multiplied by the power and not the argument from any other point.

Your answer to the second part is correct.


Congruent segments are used to form equilateral triangles in
this sequence so that each figure contains one more triangle
than the preceding figure. Figure 3 of the sequence is made
with seven congruent segments. How many congruent
segments must be used to make Figure 25 of the sequence?

[asy]draw((0,0)--(10,0)--(5,8.7)--cycle);
draw((15,0)--(25,0)--(30,8.7)--(20,8.7)--cycle);
draw((20,8.7)--(25,0));
draw((35,0)--(55,0)--(50,8.7)--(40,8.7)--cycle);
draw((40,8.7)--(45,0)--(50,8.7));
draw((60,0)--(80,0)--(85,8.7)--(65,8.7)--cycle);
draw((65,8.7)--(70,0)--(75,8.7)--(80,0));
label("Figure 1",(2,0),S);
label("Figure 2",(20,-10),S);
label("Figure 3",(45,0),S);
label("Figure 4",(76,-10),S);[/asy]

This is actually an [url=https://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219917]arithmetic sequence[/url], with a common difference of 2. Therefore, the 25th figure has $ 2 \times 24 \plus{} 3 \equal{} \boxed{51}$ congruent segments.


[quote="amparvardi"]Download following link and $ enjoy$ it!!!
it's 2007 inequality problems.
some of the inequalities are very hard that make you cry!
[hide="pp"]I hope that you don't say:posted before!! :D [/hide][/quote]

This is definitely paradise my friend :D
P.S As i see you have some of Vasc's inequalities. But they have been posted before :P  here http://www.mathlinks.ro/viewtopic.php?t=100399

[quote="aakansh92"]" posted before!! "
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191690[/url] :P[/quote]

so cool!!!

thanks to both amparvadi for the file and aakansh for the link
both are awesome...


rotation
geometry
geometric transformation

(i) 15 chairs are equally placed around a circular table on which are name cards for 15 guests. The guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in the correct seat. Prove that the table can be rotated so that at least two of the guests are simultaneously correctly seated.

(ii) Give an example of an arrangement in which just one of the 15 guests is correctly seated and for which no rotation correctly places more than one person.

[hide](i) For each guest, there is exactly one rotation that seats them correctly. Thus, the sum of the number of guests seated correctly for all $ 15$ rotations is $ 15$. If none of the other rotations have more than $ 1$ guest correctly seated, then the sum of the number of guests seated correctly for all $ 15$ rotations would be less than or equal to $ 14$. Therefore, at least one other rotation must have $ 2$ or more guests correctly seated.

(ii) One such arrangement occurs when the cards are for guests $ 1,2,3,\ldots,15$ going clockwise and the guest sit in that order, but counter-clockwise. [/hide]


geometry
abstract algebra
topology
superior algebra
superior algebra unsolved

Let $ R$ be a sheaf of (commutative unitary) ring and $ P, Q$ are two sheaves of $ R$-modules (over a space topological D). They have $ \Gamma(U,P\otimes_R Q)$not in $ \Gamma(U,P)\otimes_R \Gamma(U, Q) ?$

Let $ R$ be a sheaf of (commutative unitary) ring and $ P, Q$ are two sheaves of $ R$-modules (over a space topological $ D$). we have  $ \Gamma(P\otimes_R Q, U) \equal{} \Gamma(U,P)\otimes_{R_U} \Gamma(U, Q)$ (with $ \Gamma (U, P)$ is presheaf of sections of sheaf $ P$, $ U\subset D$ is an open set)? Thanks


integration
limit
Putnam
calculus
calculus computations

Let $ F_0(x) \equal{} ln x$. For $ n \geq 0$ and $ x > 0$, let $ F_{n\plus{}1}(x) \equal{} \int^x_0F_n(t) dt$. Evaluate

$ \lim_{n\minus{}>\infty} \frac{n!F_n(1)}{ln n}$

This was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244051]Putnam 2008 B2[/url].


Hi, I'm working with a math submission that I have to be finished with tomorrow. I have done the equations, but I'm not completely sure if I've done them right. It would be nice to know that I have. So, can you please help me with these equations?

[img]http://home.no.net/atllars/images/matte2.gif[/img]

Uh...try graphing? I don't know how much that will help...

Could you show me the calculations? I don't understand all the english terms of the words you are saying. What does square and Pythagorean Identity mean, and why should I subtract 1 from each side?

[quote="Altheman"][quote="knexpert"]Notice how (a) is a quadratic equation that is factorable. For (b), use the double angle formula for sine and cosine and add 2sin^2(x) to each side. Then, factor the left side and take the square root of both sides. For part (c), square both sides and use the Pythagorean Identity. Subtract 1 from each side and solve the resulting quadratic equation.[/quote]

for part c, you are adding many solutions, you have to check back that your solutions work[/quote]

Thank you, Altheman, for pointing that out. Squared means a number multiplied by itself once and is represented by a superscript (exponent) 2 next to the number. For example, $2^{2}=4$. The Pythagorean Identity is $sin^{2}x+cos^{2}x=1$ for all values of $x$. The Double Angle Formulas are here:
[url]http://mathworld.wolfram.com/Double-AngleFormulas.html[/url]

I'll post the solutions later, probably tommorow evening.

[quote="MatiasL"]You don't have to do the equations, because I'm going to hand it in today, but thanks anyway. On task c) I took (2 cosx)^2 = (1- sinx)^2 and got a "second degree" function. Is this what you meant with squaring both sides?[/quote]
That's what I mean by squaring both sides. However, I solved this by squaring both sides as given. Then, I used the Pythagorean Identity once and subtracted one from each side to get a quadratic equation. Make sure to check the solutions you receive from this second-degree equation in the original equation because one answer should not work.


[quote="ffdbzathf"]does anyone know of a good book for someone with a strong AP Calculus BC background to learn the necessary mathematics for research in theoretical physics in a field like quantum gravity or string theory?  a few books i've seen on the internet are "Mathematics of Classical and Quantum Physics" by Byron, "Essential Mathematical Methods for Physicists" by Afken and Weber, and "Mathematics for Physicists" by Dennery and Krzywicki.  does anyone know anything about these or any others?  thanks a lot.[/quote]You might try [i]Mathematical Methods in Physics and Engineering[/i] by John W. Dettman, ISBN 0-486-65649-7.  I've found it a useful alternative to Arfken.

[quote="ffdbzathf"]does anyone know of a good book for someone with a strong AP Calculus BC background to learn the necessary mathematics for research in theoretical physics in a field like quantum gravity or string theory?  a few books i've seen on the internet are "Mathematics of Classical and Quantum Physics" by Byron, "Essential Mathematical Methods for Physicists" by Afken and Weber, and "Mathematics for Physicists" by Dennery and Krzywicki.  does anyone know anything about these or any others?  thanks a lot.[/quote]

The mathematics background subsumes what is taught for a first degree in pure (or applied) mathematics.  After that, there is more geometric and topological material needed for string theory or QG.

another question relating to these books,

why is it that the books by Dettman or Byron are only about 10 to 20 dollars while the book by Afken is 100 dollars?  there's no way that it is only because the Afken book is hardcover.  also as i browse these books' contents online, there seems to be nothing about topology.  any thoughts or suggestions?

[quote="ffdbzathf"]would it be at all feasible for me to learn enough of this background in order to do research in string theory or quantum gravity next summer?  i would be working on a specific research project, so wouldn't i not really need the full background that a graduate student would, but rather a more narrow range of topics in depth as well as just a brief understanding of all of the main ideas leading up to the fact?  i thought that was why they had books of mathematics for physicists because all of the math isn't really entirely necessary for an individual research project.[/quote]

It is unclear why it's necessary to (a) do research, (b) only in those fields, (c) by next summer.  If there is a project the lab can set you on where the time spent instructing you is less than the time they save by having you perform the work (and if you do this for free as opposed to undergrads whom they have to pay) there might be something.  But if you never heard of, say, a Riemann surface, or quantum field theory, until last Thursday, then it is a waste of time for a grad student to supervise you to the point that you can code up usable and reliable computer calculations.   If the actual goal is to state "I have done research", or to increase college-admission credentials, or whatever else, there are probably better ways to go about it.

it's really not just about improving college admissions credentials.  i did research this past summer at MIT in geophysics which was mostly experimental, and although it was a great experience, i didn't love it.  i am much more interested in quantum gravity and string theory based on solely qualitative books like "The Elegant Universe" and my physics classes in school.  i think i would enjoy and get a lot out of doing research in this field, but the problem is that i only know single variable calculus.  that is why i am asking about ways to learn the mathematical foundations necessary to pursue my goal.

Witten's advice for prospective string theory researchers was to first master quantum field theory.   QFT doesn't require immense mathematics and physics background if you are willing to fill things in or take them on faith as you go along, but it requires quite a bit of mathematical and physical sophistication to use that material reliably.  It is also unfinished mathematically (largely heuristic and nonrigorous), and the standard QFT graduate course takes at two semesters or more.  I knew people who learned it in high school but they had to re-learn it "for real" later on.   There was a girl about 15-16 years old who learned QFT to the level of doing a Westinghouse project on weak interactions a few years ago, and she subsequently went to MIT.    String theory is a whole additional level of mathematical sophistication.

There are popularizations about exciting fields other than quantum gravity and strings, such as Laughlin's memoir on his career in condensed matter physics, or books and articles about the recent flurry of activity in quantum computing.  Some of those might interest you as well.

thanks for the advice!  i'll take a look at some things about quantum field theory.  i was thinking about quantum computing as well but i wasn't really sure how to go about looking into that.  i'll try and find some articles or books on that as well.


Do you mean $ m^3 \plus{} n^3 \plus{} 99mn \equal{} 33^3$?

ARML Individual (at least, I think so).  So what do you expect  :P

not 33^3, it's 33^2

m, n doesn't need to be integers. i've found two (m,n) and actually the answer is 2,

but i can't proove why there is only two of them.. help me

[quote="Seo-Yeon, Sohn"]
m, n doesn't need to be integers. [/quote]

Unless I'm missing something, then there clearly are an infinite number of solutions (just consider any small $ m$, we get a cubic in $ n$ which will have a positive root) ..

This (but with a $ 33^3$) was a #30 on an AHSME, which used the factorization \[ (m)^3 \plus{} (n)^3 \plus{} (\minus{}33)^3 \minus{} (m)(n)(\minus{}33)\] \[ \equal{} \frac 12 (m\plus{}n\minus{}33)((m\minus{}n)^2\plus{}(n\plus{}33)^2\plus{}(m\plus{}33)^2)\], but which doesn't help here.

$ \frac 12 (m \plus{} n \minus{} 33)((m \minus{} n)^2 \plus{} (n \plus{} 33)^2 \plus{} (m \plus{} 33)^2) \equal{} 33^2 \minus{} 33^3$

$ \iff (33 \minus{} m \minus{} n)((m \minus{} n)^2 \plus{} (n \plus{} 33)^2 \plus{} (m \plus{} 33)^2) \equal{} 64 \cdot 33^2$

Let $ a \equal{} n \plus{} 33$, $ b \equal{} n \plus{} 33$

$ (99 \minus{} a \minus{} b)((a \minus{} b)^2 \plus{} a^2 \plus{} b^2) \equal{} 2^6 \cdot 3^2 \cdot 11^2$.

$ (99 \minus{} (a \plus{} b))(2(a \plus{} b)^2 \minus{} 6ab) \equal{} 2^6 \cdot 3^2 \cdot 11^2$.

$ r \equal{} ab$, $ s \equal{} a \plus{} b$

$ (99 \minus{} s)(s \minus{} 3r) \equal{} 2^5 \cdot 3^2 \cdot 11^2$, clearly $ 3|s \implies s \equal{} 3t$

$ (33 \minus{} t)(t \minus{} r) \equal{} (t \minus{} 33)(r \minus{} t) \equal{} 2^5 \cdot 3 \cdot 11^2$

Which yields $ (6\plus{}1)(1\plus{}1)(2\plus{}1)\equal{}36$ solutions in $ (r,t)$, and thus $ 36$ solutions in $ (r,s)$

This gives you 36 quadratics to solve and hope for integral solutions.  :rotfl:


Objects $ A$ and $ B$ move simultaneously in the coordinate plane via a sequence of steps,
each of length one. Object $ A$ starts at $ (0,0)$ and each of its steps is either right or up,
both equally likely. Object $ B$ starts at $ (5,7)$ and each of its steps is either left or down,
both equally likely. Which of the following is closest to the probability that the objects meet?
\\
A. 0.10
\\
B. 0.15
\\
C. 0.20
\\
D. 0.25
\\
E.0.30

Say that they meet at the point $ (x,y)$. Then $ A$ has moved $ x \plus{} y$ steps, and $ B$ has moved $ (5 \minus{} x) \plus{} (7 \minus{} y) \equal{} 12 \minus{} x \minus{} y$ steps. These must be equal (since they move simultaneously), so we have $ x \plus{} y \equal{} 6$.
Now, we can have (diagram may help here, and I won't explain that much)
$ (0,6)$: a total of $ \dbinom60\cdot\dbinom61 \equal{} 6$ ways.
$ (1,5)$: a total of $ \dbinom61\cdot\dbinom62 \equal{} 90$ ways.
$ (2,4)$: a total of $ \dbinom62\cdot\dbinom63 \equal{} 300$ ways.
$ (3,3)$: a total of $ \dbinom63\cdot\dbinom64 \equal{} 300$ ways.
$ (4,2)$: a total of $ \dbinom64\cdot\dbinom65 \equal{} 90$ ways.
$ (5,1)$: a total of $ \dbinom65\cdot\dbinom66 \equal{} 6$ ways.
Thus, there are a total of $ 792$ ways, so our probability is $ \frac {792}{2^{12}} \equal{} \frac {99}{512}\approx19.3\%\implies\boxed C$.

Edit: Ugh, being stupid.

Better solution: as shown above, $ A$ and $ B$ must each move $ 6$ to meet. In each path from $ (0,0)$ to $ (5,7)$, the first $ 6$ moves uniquely determine the path for $ A$, and the last $ 6$ moves uniquely determine the reverse path of $ B$. Thus, the probability is $ \frac1{2^{12}}\dbinom{12}5\equal{}\frac{99}{512}$.

where did the 12 choose 5 come from in the solution?

I think this is where 12 choose 5 came from:

It takes 12 steps to move from Point A (0,0) to Point B (5,7). In those 12 steps, you move to the right 5 times, and up 7 times. So 12 choose 5 is choosing which of those 12 steps are steps to the right. The rest of the steps all go up. Keep in mind you could also find which of those 12 steps are steps up, which is 12 choose 7 (which produces the same answer as 12 choose 5).

So now, take any one of those 12 choose 5 paths. Let Point A start walking on the path to its destination, Point B. Point B will start at the end of the path and walk 'backwards' on the path towards Point A. They each meet after six steps. So on any of those 12 choose 5 paths, point A can walk the first six steps and point B can just walk the last six steps 'backwards' and they would each meet after 6 steps. Since those 12 choose 5 paths cover all the ways for Point A to reach Point B, they also include all the ways for B and A to meet, with A walking the path forwards and B walking the path 'backwards' at the same time.

Probability is the number of successful outcomes over total outcomes, and 12 choose 5 is the amount of successful outcomes in which each point moves six steps and meet. The total outcome, then, is the number of outcomes in which each point moves six steps, [i]regardless[/i] of whether the two points meet or not.

The number of ways for point A to move six steps is 2^6, and the number of ways for point B to move six steps is 2^6. For each of the 2^6 ways Point A can take six steps, there are 2^6 ways point B can take six steps, so the total number of ways for the two points to move six steps is 2^6 times 2^6, which is 2^12. Therefore, 2^12 is the number of total outcomes.

Now divide the number of successful outcomes (12 choose 5) by the number of total outcomes (2^12) to get the answer. That simplifies to 99/512, which is about 0.193, closest to C)0.20.

Why is the denominator $2^6 * 2^6$ what if Point A went to the right all 6 times?

Because you also haveta account for point B

[hide]Alcumus solutions is overly complicated, we present a simple solution.

First, we determine the total paths from (0,0) to (5,7). In total, we take 12 steps, from which 5 must be right and 7 must be up. So we have 12C5 = 12C7 total paths.

Now we must determine the successful paths, Since as discussed above, there are 12 total paths, each object will have to travel 6 paths to intersect. Since there are 2 options for each object, there are 2^6 ways for object A to travel, and by similar logic, 2^6 ways for object b to travel. Since these moves are not dependent on each other we have 2^6*2^6 = 2^12 as the number of successful outcomes.

So in total, we have 2^12/12C5 which approximately = 0.2 :)[/hide]

I think what @Jwilson1990J meant was why is the denominator $2^6 \cdot 2^6$ instead of $(2^6-1)(2^6-1)$.

I have the same question, as point A cannot go six steps to the right, nor could point B go six steps to the left, so shouldn't it be $(2^6-1)(2^6-1)$?

This should be really just be a "think about it"...

[hide=Solution]
If $A$ and $B$ meet, they form one of the $\binom{12}{5}=792$ paths from $(0,0)$ to $(5,7)$. In total they each have to move $6$ times out of the $12$ step path, so the total is $2^6 \cdot 2^6 = 4096$. The answer is $\frac{792}{4096} \approx 0.20 \Rightarrow \boxed{C}$.

~ :gleam:
[/hide]


It's given a triangle ABC.
$ < A \equal{} 20^\circ , < B \equal{} 130^\circ$
M is a point in the triangle ABC,such that $ < MAC \equal{}< MCA \equal{} 10^\circ$
AL=AB,where $ L \in AC$
And $ K \equal{} AM\cap BC$
Find <AMB=?

Well,I don't know  :lol: and i don't know what's the role of K ?
But it's interesting problem.

Well,here's the problem now.(without the points L and K)

It's given a triangle ABC.
$ < A \equal{} 20^\circ , < B \equal{} 130^\circ$
M is a point in the triangle ABC,such that $ < MAC \equal{}< MCA \equal{} 10^\circ$
Find <AMB=?
Everything is clear now.

[quote="Yavor Tsvetanov 52"]Well,here's the problem now.(without the points L and K)
It's given a triangle ABC.
$ < A \equal{} 20^\circ , < B \equal{} 130^\circ$
M is a point in the triangle ABC,such that $ < MAC \equal{} < MCA \equal{} 10^\circ$
Find <AMB=?
Everything is clear now.[/quote]
u didn't see my post?!!!
[quote="shoki"]i used geometer's sketchpad for your problem and i saw that $ \angle AMB$ can be $ 141$ and $ 142$(i'm sure of it :D )
there is obviously sth wrong with your problem...check it again...[/quote]

[b]Answer[/b] : $ m(\angle AMB) \equal{} 140^{\circ}$ .

[b]Hint.[/b] If you consider the second intersection $ D$ of the line $ BC$ with the circle $ C(M, MA)$ , then prove easily that $ m(\angle AMD)\equal{}60^{\circ}$ ,
the triangle $ ADM$ is equilateral and the triangles $ BDM$ , $ CMD$ are isosceles, i.e. $ MC\equal{}MA\equal{}MD\equal{}AD\equal{}DB$ a.s.o.

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension of the proposed problem[/u].[/b] Let $ ABC$ be a triangle for which  exists an interior point $ M$ so

that $ MA\equal{}MC$ and $ m(\angle MCB)\equal{}2\cdot m(\angle MAB)$ . Prove that[/color] $ C\equal{}30^{\circ}\ \Longleftrightarrow\ m(\angle ABM)\equal{}30^{\circ}$ .[/quote]

[quote="Virgil Nicula"][color=darkred][b][u]A similar problem[/u].[/b] Let $ ABC$ be a triangle for which exists an interior point  $ M$ be an interior point for which $ MA\equal{}MC$

and $ m(\angle MBA)\equal{}2\cdot m(\angle MAB)$ . Prove that $ m(\angle MCA)\plus{}m(\angle MBA)\equal{}30^{\circ}\ \Longleftrightarrow\ m(\angle MCB)\equal{}30^{\circ}$ .[/color][/quote]