algebra
polynomial
function
Diophantine equation
number theory

[quote="Scrambled"][quote="t0rajir0u"]Let $f(x) \in \mathbb{Z}[x]$ be such that $\deg f \ge 2$ (edit: and $\deg f \equiv 0 \bmod 2$).  Prove or disprove:

If $f(k)$ is a square for infinitely many integers $k$, then $f(x)$ is the square of a polynomial.

(Actually, I may or may not have seen this problem somewhere on AoPS.)[/quote]

False of course.
2(x^2+x)
is a square infinitely often. In fact, every for every pell equation solution, theres , theres a number that satisfies the property that 2(x^2+x) is a square[/quote]
[hide]You posted $2x(x+1)=4\left(\frac{x(x+1)}{2}\right)$, which is 4 times a triangular number, so if there are an infinite number of triangular numbers that are perfect squares, then this polynomial has an infinite number of perfect square values.
(Are there an infinite number of perfect square triangular numbers? A few examples I can think of are $\frac{8\cdot9}{2}$ and $\frac{49\cdot50}{2}$, but there doesn't seem to be any pattern.)

Also, could you explain why these perfect square values are related to the Pell equation? Thanks.[/hide]

The Pell equation $m^{2}-2n^{2}= 1$ has infinitely many solutions. Of course, if $(m,n)$ is a solution, $m$ must be odd: $m = 2k+1$. So $(2k+1)^{2}-2n^{2}= 1$ which gives $k(k+1)/2 = (n/2)^{2}$. Clear?

[quote](Are there an infinite number of perfect square triangular numbers? A few examples I can think of are $\frac{8\cdot9}{2}$ and $\frac{49\cdot50}{2}$, but there doesn't seem to be any pattern.)[/quote]

Pattern is: (If $u_{n}^{2}$  is a triangular number)

$u_{n}= 6 u_{n-1}-u_{n-2}$
For example first two are ($1,6$  )
(That is $1^{2}= (1 \cdot 2)/2$ and $6^{2}= (8\cdot 9)/2)$

So next  one is $6.6-1 = 35$ and you indeed see $35^{2}= (49\cdot 50 )/2$
And the next one would be $6 \cdot 35-6 = 204$ and $204^{2}$ is a  triangular number.  Next one would be $204*6-35$ and so on.

Another formula (which is little less known)

$u_{n+1}= 3u_{n}+\sqrt{ 8 u_{n}^{2}+1}$

I'm pretty certain that that t0rajir0u's conjecture/theorem thing is true for monic polynomials, because then a number would be squeezed in between two notnecessarily integer square polynomials. Which means that there would be a finite set of numbers it has to equal to be a square. Since it has degree n and a polynomial of degree n has at most n integer solutions, then there would be at most n numbers such that it would take on a square.\

[quote="paladin8"]What if $f(k)$ is square for ALL integers $k$?[/quote]

How about $|f(k)|$ is a square for all integers and $\deg f$ is not necessarily even?  :D

[quote="B23"]Well,
$P(x)=2x^{2}+3x+4$
is a non-square polynomial which nevertheless takes infinitely many squares as values.
Some examples: P(1),P(36),P(65),...

See e.g. http://mathforum.org/library/drmath/view/65773.html
(the bit of number theory is actually the solving of a Pell equation)[/quote]

Yes, this point has been brought up before.  How about if we require that our polynomial takes square values at every integer?  :)

DUH

as for rigor
Assume the polynomial f(x) has leading coefficient a and degree k. Then there exists an n such that a(P(n))^2<f(x)<a(P(n+1))^2 (P(n) being a monic polynomial)
so we can square root both of them.
since a is constant, there exists a finite set of polynomials that sqrt(f(x)) must lie under.
Since the polynomials have a finite degree k, but they are the same infinitely often (as k+1 points define a unique polynoimal of degree k), then sqrt(f(x)) must be some polynomial (not necessarily rational).

EDIT--- this method only works for evens. Im assuming negative values don't count in the odd analysis

What do you mean by "there exists a finite set of polynomials that sqrt(f(x)) must lie under. " Which polynomials are "the same infinitely often." ENGLISH 101!!!!!

More complaints: Why do you throw P(n) and P(n+1) at us all of a sudden. Introduce new functions when you use them.


Two girls and three boys sat in a five-seat row at the movie theater. What is the probability that the two people at each end of the row were both boys or both girls? Express your answer as a common fraction.

Case 1: We want both as boys.

Suppose the row looked like this: __ __ __ __ __ . The ends have to be boys, so: $ B_1$ __ __ __ $ B_2$. For $ B_1$, there are $ 3$ choices, and for $ B_2$, there are $ 2$ choices. Thus, we have $ 6$ choices so far. Now for the middle seats - this is simply $ 3!  \equal{} 6$. Thus, there is a total of $ 6 \times 6 \equal{} 36$ ways for boys to be on the ends.

Case 2: We want both as girls.

This is very similar to Case 1 - the answer is $ 2 \times 1 \times 3! \equal{} 12$.

In total, we have $ 36 \plus{} 12 \equal{} 48$ favorable ways, out of a total of $ 120$ ways. Our final probability is then $ \frac{2}{5}$.

We could also use combinations. 5C3 for different arrangements. 5C3 = 10.

Then 3C2 for boys at ends. 3C2 = 3.

Finally 2C2 for girls at end. 2C2 = 1

3+1 = 4

4/10 =[b] 2/5[/b]


I could'nt understand what the author of these problems want me to find out because of my poor English,so I want some suggests/solutions for all problems below.
[img]http://i212.photobucket.com/albums/cc112/le0thanh/deee.jpg[/img]
[img]http://i212.photobucket.com/albums/cc112/le0thanh/eeeeee.jpg[/img]

Thanks all in advance

Why do you need to post the same posts 4 times? :mad:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=253878[/url]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=253879[/url]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=253880[/url]

Japanese Communities Moderator

kunny

It's wave interference, or ondulatory optics. Do you know the subject?
Just for testing, does (i) gives 2,27º ?


How thick should a coin be to have a $ \frac{1}{3}$ chance of landing on its edge when thrown onto a flat surface?

Do you have any assumptions on how the coin is moving when it lands?


$AB$ and $CD$ are two diameters of a circle with center $O$. $M$ is a point on $AC$ such that $ADOM$ is cyclic. The line passing through $B$ and parallel to $OM$ meets the circle at $F$. Prove that the points $D, M, F$ are on a straight line.
(maybe its completely obvious :D )

$\angle ADM = \angle AOM = \angle ABF = \angle ADF$, so M, F and D are collinear. :D


THe first one is [hide]6 feet[/hide]

THe second one is [hide]27um....:pie sign:[/hide]

[quote="k8reindeer"]I was raking leaves last fall. My rake is 6 feet long, and my waist is 3 feet off the ground. I am a funny raker: I only rake with the end of the rake at my waist or higher. I don't extend my arms or anything to make my range longer. (must be really inneficient!)
First of all, how far away from my feet can my rake reach?
I decide to stand in one spot. What's the area of leaves I can rake?
I decide that I don't want to be a lazy raker as well as a funny raker, so I go somewhere completely different in the yard. I will only walk along a 10 foot line. Now what's the area I can rake?
Going to yeat another area of the yard where I won't overlap naything I've already raked, I decide to walk in a circle of radius 7 feet. I can't go inside the circle, just on it. What's the area I can rake?
Biggest question of all...when my mom looks out her bedroom window will she think aliens have landed in the backyard and began making wierd leaf patterns? And who on earth is going to rake the rest of the leaves in the backyard?[/quote]
 :rotfl:  :lol:  :rotfl:
The rake can reach $3\sqrt3$ feet away. The area is $27\pi$. Errr...10 feet? $14\pi$ .Your mom looks out the window and magically transports me to your yard to rake the rest of the leaves. (Isn't it odd how mothers do that?)

[quote="k8reindeer"]On the 10-foot line and the 7 foot circle I can put the rake on eaither side of the line I'm walking on but I just can't move my feet from the line. I'd draw a diagram but
a) part of the question is knowing what it should look like (i didn't make the question very clear the first time)
b) I have no clue how to put pictures on the forum anyway![/quote]

When you reply to a post it should say 'Add an Attachment' on a button by options. Click on that and put the name of your file in.


So I've made a new thread for the second round cuz Andrei requested it.

How'd you all do?

$ 5/5/n/5/5$ such that $ n$ is any positive integer much less than $ 5$ for me...


This problem that kyyuanmathcount created is similar to this problem from the 2000 Mathcounts State Competition.

Of the 400 eighth-graders at Pascal Middle School, 117 take algebra, 109 take advanced computer, 114 take industrial technology.  Furthermore, 70 take both algebra and advanced computer, 34 take both algebra and industrial technology, and 29 take both advanced computer and industrial technology.  Finally, 164 students take none of these courses.  How many students take all three courses?

So, of course, it has a similar solution  :D

[hide]First, define some variables.  The people who take algebra are $A$, computer, $C$, technology, $T$, and combinations $A\cap C$, etc.
By PIE, we know
$\#(A\cup C\cup T)=\#(A)+\#(C)+\#(T)-\#(A\cap C)-\#(A\cap T)-\#(C\cap T)+\#(A\cap C\cap T)$.
Substitute using the informtation we know to get:
$400-164=117+109+14-70-34-29+\#(A\cap C\cap T)$
$\#(A\cap C\cap T)=129$[/hide]

Why do more people take algebra and industrial technology than take technology?  My solution ignores this fact.

What round and what number was this problem?

[quote="mathgeniuse^ln(x)"]It was 2000 State Question number 21 I believe.[/quote]

Well, you got it correct, except for the part that [b]114 people take industrial technology instead of 14 people[/b]

[hide]well i got 29 from 117+109+114-70-34-29+x=400-164 from PIE, u just solve for x and its 29[/hide]


inequalities
floor function
calculus
derivative
inequalities proposed

If $ x\ge y \ge z\ge 0$, then

$ \frac {(x\minus{}z)^2}{4(x\plus{}y\plus{}z)} \le \frac {x\plus{}y\plus{}z}{3} \minus{}\sqrt[3]{xyz} \le \frac {4(x\minus{}z)^2}{3(x\plus{}y\plus{}z)}$.

Very nice inequality, Mr Vasile. Please take a look at my proof at: http://canhang2007.wordpress.com/2009/11/05/inequality-48-v-cirtoaje/

What about this ?

If $ x\ge y \ge z\ge 0$, then

$ \frac {(x \minus{} z)^2}{2(x \plus{} z)} \le x\plus{}y\plus{}z \minus{} 3\sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{x\plus{} z}$.

Notice that the right inequality is sharper that the one above.

[quote="Vasc"]What about this ?

If $ x\ge y \ge z\ge 0$, then

$ \frac {(x \minus{} z)^2}{2(x \plus{} z)} \le x \plus{} y \plus{} z \minus{} 3\sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{x \plus{} z}$.

Notice that the right inequality is sharper that the one above.[/quote]
See my proof here: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/

[quote="Vasc"]Nice solution, Can_hang.  :lol:
Note that for the right inequality there is at least another one nice solution (only with AM-GM inequality).[/quote]
Dear Mr Vasile,

Please take a look again the link: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/ to see my second proof for (b).

[quote="Vasc"][quote="Vasc"][quote="can_hang2007"]
Dear Mr Vasile,

Please take a look again the link: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/ to see my second proof for (b).[/quote]
Nice proof, Can_hang.  :lol:
There is still at least another one nice solution.[/quote]
[b]My solution. [/b]
Using twice AM-GM inequality, we show that
$ x \plus{} y \plus{} z \minus{} 3\sqrt [3]{xyz} \le 2x \plus{} z \minus{} 3\sqrt [3]{x^2z}\le \frac {2(x \minus{} z)^2}{x \plus{} z}$.
The left inequality reduces to
 $ \sqrt [3]{x^2} \plus{} \sqrt [3]{xy} \plus{} \sqrt [3]{y^2}\ge 3\sqrt [3]{xz}$,
and the right to
$ z^2 \plus{} 3\sqrt [3]{x^5z} \plus{} \sqrt [3]{x^2z^3}\ge 7xz.$

[b]Generalization.[/b] If $ x_1\ge x_2\ge ...\ge x_n\ge 0$, then

$ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \minus{} n\sqrt [n]{x_1x_2...x_n}\le \frac {(n \minus{} 1)(x_1 \minus{} x_n)^2}{x_1 \plus{} x_n}$.[/quote]
Your solution is really nice, Mr Vasile.

For your generalization, I found now that you have in your book the following problem:

"Let $ 0 \le a < b$ and let $ a_1,a_2, \ldots, a_n \in [a,b]$. Prove that
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \minus{} n\sqrt [n]{a_1a_2\cdots a_n} \le (n \minus{} 1)\left( \sqrt {b} \minus{} \sqrt {a}\right)^2.$"

According to the result of this problem, we can see that the above generalization is just a direct consequence of it.

[quote="can_hang2007"] Your solution is really nice, Mr Vasile.

For your generalization, I found now that you have in your book the following problem:

"Let $ 0 \le a < b$ and let $ a_1,a_2, \ldots, a_n \in [a,b]$. Prove that
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \minus{} n\sqrt [n]{a_1a_2\cdots a_n} \le (n \minus{} 1)\left( \sqrt {b} \minus{} \sqrt {a}\right)^2.$"

According to the result of this problem, we can see that the above generalization is just a direct consequence of it.[/quote]
Indeed, this is true. :lol:

[quote="Vasc"][quote="can_hang2007"][quote="Vasc"]The following inequalities are a little harder.

If $ x\ge y\ge z\ge 0$, then

$ (a) \ \ \ \ \ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \ge \frac {2(y \minus{} z)^2}{9(y \plus{} z)}$;

$ (b) \ \ \ \ \ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \ge \frac {(x \minus{} y)^2}{5(x \plus{} y)}$.[/quote]
Please take a look at my proof here, Mr Vasile: http://canhang2007.wordpress.com/2009/11/08/inequality-57-v-cirtoaje/[/quote]
Nice proofs, Can-Hang. :lol:
The following inequalities hold, too.

If $ x\ge y\ge z\ge 0$, then

$ (a) \ \ \ \ \ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \le \frac {(x \minus{} z)^2}{3z}$;

$ (b) \ \ \ \ \ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{x \plus{} 5z}$.[/quote]
I think (a) follows directly from (b). For (b), we can also prove that the stronger inequality holds
$ x\plus{}y\plus{}z\minus{}3\sqrt[3]{xyz} \le \frac{2(x\minus{}z)^2}{x\plus{}5z}.$

P.s: Actually, I think Mr Vasile meant (b) is my stronger one, but he got somemistakes in his typing.

[quote="can_hang2007"] [quote="Vasc"]
The following inequalities hold, too.

If $ x\ge y\ge z\ge 0$, then

$ (a) \ \ \ \ \ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \le \frac {(x \minus{} z)^2}{3z}$;

$ (b) \ \ \ \ \ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{x \plus{} 5z}$.[/quote]
I think (a) follows directly from (b). For (b), we can also prove that the stronger inequality holds
$ x \plus{} y \plus{} z \minus{} 3\sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{x \plus{} 5z}.$
P.s: Actually, I think Mr Vasile meant (b) is my stronger one, but he got somemistakes in his typing.[/quote]
You are right, Can, I wanted to post
$ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{3(x \plus{} 5z)},$
but I made a print-mistake.
We can generalize this interesting inequality to $ n$ numbers. For instant, if $ n \equal{} 4$, then

$ x \plus{} y \plus{} z \plus{} w \minus{} 4\sqrt [4]{xyzw} \le \frac {3(x \minus{} w)^2}{x \plus{} 5w}$

for $ x\ge y\ge z\ge w\ge 0$.

Can someone find the best $ k_n$ such that

$ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \minus{} n\sqrt [n]{x_1x_2...x_n}\le \frac {(n \minus{} 1)(x_1 \minus{} x_n)^2}{x_1 \plus{} k_nx_n}$

for $ x_1\ge x_2\ge ...\ge x_n\ge 0$?

or equivalently,

If $ 0\le a < b$ and $ x_1, x_2, ..., x_n\in [a,b]$, find the best $ k_n$ such that

$ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \minus{} n\sqrt [n]{x_1x_2...x_n}\le \frac {(n \minus{} 1)(b \minus{} a)^2}{b \plus{} k_na}$?

[hide]$ k_3 \equal{} k_4 \equal{} 5$; $ k_5 \equal{} 17/2$; $ k_7 \equal{} 6$.[/hide]

[quote="can_hang2007"]
$ x \plus{} y \plus{} z \minus{} 3\sqrt [3]{xyz} \le \frac {2(x \minus{} z)^2}{x \plus{} 5z}.$
[/quote]
EDIT: Flipped $ a,b,c$ and $ x,y,z$, sorry. :oops:
[b]Problem[/b]
$ a \plus{} b \plus{} c \minus{} 3\sqrt [3]{abc} \le \frac {2(a \minus{} c)^2}{a \plus{} 5c}.$
[b]Solution[/b]
We replace $ a \equal{} x^3; b \equal{} y^3; c \equal{} z^3 \Longrightarrow x\geq y\geq z\geq 0$, so we are only required to prove that
$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz\leq \frac {2(x^3 \minus{} z^3)^2}{x^3 \plus{} 5z^3}$
$ \Longrightarrow \frac 12 (x \plus{} y \plus{} z)[(x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2]\leq \frac {2(z \minus{} x)^2 (z^2 \plus{} x^2 \plus{} zx)^2}{x^3 \plus{} 5z^3}$
Since $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2\leq (z \minus{} x)^2$, so it suffices to show that
$ \frac 12 (x \plus{} y \plus{} z)(x^3 \plus{} 5z^3)\leq (z^2 \plus{} zx \plus{} x^2)^2$
Using $ x\geq y$, this equivalents
$ \iff (2x \plus{} z)(x^3 \plus{} 5z^3)\leq 2z^4 \plus{} 2z^2x^2 \plus{} 2x^4 \plus{} 4zx^3 \plus{} 4z^3x \plus{} 4z^2x^2$
$ \iff 2x^4 \plus{} zx^3 \plus{} 10z^3x \plus{} 5z^4\leq 2x^4 \plus{} 2z^4 \plus{} 6z^2x^2 \plus{} 4zx^3 \plus{} 4z^3x$
$ \iff 6z^3x \plus{} 3z^4\leq 6z^2x^2 \plus{} 3zx^3$
$ \iff 6z^2x(x \minus{} z) \plus{} 3z(x^3 \minus{} z^3)\geq 0$
This is perfectly true, since $ x\geq y\geq z$.
(Edited to fix a huge mistake)

Nice, Potla. :lol:
I wait for a solution in the case  $ n\equal{}4$.
In addition, the best $ k_n$ has a nice formula. We have $ 5\le k_n<7$ for $ n\ge 3$.

[quote="Vasc"]In addition, the best $ k_n$ has a nice formula. We have $ 5\le k_n < 7$ for $ n\ge 3$.[/quote]
In my result, I found that $ k_n$ is $ \frac {2n(n \minus{} 1)}{\left\lfloor \frac {n}{2}\right\rfloor \left\lfloor \frac {n \plus{} 1}{2}\right\rfloor} \minus{} 1.$ But my proof was too complicated and I am afraid of getting something wrong in my computations. So, I dont wanna post it here. I hope Mr Vasile has an easier proof than mine.

By the way, the case $ n \equal{} 4$ is easy and it has a nice proof. I will post it in some days. Now I am a little busy with my school test.

[quote="can_hang2007"] In my result, I found that $ k_n$ is $ \frac {2n(n \minus{} 1)}{\left\lfloor \frac {n}{2}\right\rfloor \left\lfloor \frac {n \plus{} 1}{2}\right\rfloor} \minus{} 1.$ But my proof was too complicated and I am afraid of getting something wrong in my computations. So, I dont wanna post it here. I hope Mr Vasile has an easier proof than mine.[/quote]
Indeed, the best $ k_n$ for $ n\ge 2$ is $ k_{i\minus{}1}\equal{}k_i\equal{}7\minus{}\frac 8{i}$ for even $ i$. My proof uses derivatives, but is not complicated.

[quote="Vasc"][quote="can_hang2007"] In my result, I found that $ k_n$ is $ \frac {2n(n \minus{} 1)}{\left\lfloor \frac {n}{2}\right\rfloor \left\lfloor \frac {n \plus{} 1}{2}\right\rfloor} \minus{} 1.$ But my proof was too complicated and I am afraid of getting something wrong in my computations. So, I dont wanna post it here. I hope Mr Vasile has an easier proof than mine.[/quote]
Indeed, the best $ k_n$ for $ n\ge 2$ is $ k_{i \minus{} 1} \equal{} k_i \equal{} 7 \minus{} \frac 8{i}$ for even $ i$. My proof uses derivatives, but is not complicated.[/quote]
Could you please post it, Mr Vasile? I am interested in seeing your nice solution. Thank you very much.

Thank you very much, Mr Vasile. Your solution and mine are similar (they are only different at the first step), and I think they are not simple at all.  :(


induction
algorithm
number theory
relatively prime
Euclidean algorithm

Prove that for $F_0 = F_1 = 1$ and $F_n = F_{n-1}+F_{n-2}, n>1$

For all $n$, $(F_n , F_{n+1}) = 1 $ (adjacent Fibonacci numbers are relatively prime)

[hide]you could also use eulclid algorithm[/hide]
edit whoops i meant euclid

hmm i tried using induction
so i got for $n=1$
$ (F_1,F_2) = (1,2) = 1 $
and then i tried to prove it was true for $n=N+1$
so i got
$(F_{N+1}, F_{N+2}) = 1$ but how do i get that to equal $(F_{N},F_{N+1}) = 1$?

First, you should make sure you're doing your induction properly -- the way you have stated it there is not correct.  Second, you haven't yet used the fact that $F_n$ isn't just [i]any[/i] number, it's a Fibonacci number.

and im not very familiar with the special characteristics of fibonnaci numbers... so i kinda need help on that too

This is induction -- you should never have to use anything particularly special.  (yes, I know, I'm being deliberately vague.  Think about what kinds of things you need to do induction and what kinds of things you know about the Fibonacci sequence.)

The Euclidean algorithm is a classic here.

You can do it in your head!

It telescopes to $F_1$ easily, and then we're done.

$F_n=F_{n-1}+F_{n-2}$

....

$F_3=F_2+F_1$

errr im not familiar with the euclidean algorithm either... and i tried to mathworld it but its kinda confusing there
can anyone explain it to me like step by step?

[hide="Idea"]
Also, note that $F_n=F_{n+2}-F_{n+1}$ and also $\gcd(m,n)=\gcd(m-n,n)$, hence

[tex]

\begin{eqnarray*}

\gcd(F_{n+2},F_{n+1}) & = & \gcd(F_{n+2}-F_{n+1},F_{n+1}) \\

 & = & \gcd(F_n,F_{n+1}) \\

 & \vdots & \\

 & = & \gcd(F_1,F_2)=1

\end{eqnarray*}

[/tex]

And that's it!

[/hide]


$A, B, C, D$ are four natural numbers such that

$A^{2}+B^{2}+C^{2}+D^{2}= A \times B \times C \times D$

Obviously, $A=B=C=D=2$ is a solution.

Can you find a solution where $A, B, C, D$ are different integers?

And then can you find another solution?

How many can you find?

[hide]
There are infinitely many.

To see this, let $n=a^{2}+b^{2}+c^{2}+d^{2}=abcd$ and suppose there are a finite number of solutions. Then there exist finitely many $n$, so we can choose the largest $n$.

WLOG let $d$ be the smallest of $a,b,c,d$. (We can make this assumption since the equation is symmetric, so the variables can be rearranged in any way.) Then $a^{2}+b^{2}+c^{2}$ is divisible by $d$. If we let
\[x=\frac{a^{2}+b^{2}+c^{2}}{d}\]
then $x\geq 3d>d$. On the other hand, multiplying both sides of the equation
\[a^{2}+b^{2}+c^{2}+d^{2}=abcd\]
by $x^{2}$ and dividing by $a^{2}+b^{2}+c^{2}$ gives
\[a^{2}+b^{2}+c^{2}+x^{2}=abcx\]
and since $x>d$, $abcx>abcd=n$, a contradiction! Therefore, infinitely many solutions.

(Given $(2,2,2,2)$, this process gives $(6,2,2,2)$, $(22,6,2,2)$, $(262,22,6,2)$, $(34582,262,22,6)$, etc.)
[/hide]

I believe this is an application of Reid Barton's technique of root-flipping.  Very nice  :)

In essence:  the equation is a quadratic in a given variable, say $A$; given any solution $(A, B, C, D)$ this corresponds to a quadratic equation of which $A$ is a root.  It has another root $A'$ which is positive, and, as it turns out, integer, which gives another solution $(A', B, C, D)$.


greatest common divisor
number theory theorems
number theory

[b]Lemma[/b]
For four nature numbers $m,n,p,q$ we have $\gcd(m,n) = 1, \gcd(p,q) = 1$
Find the value of $\gcd(p^m+q^m, p^n+q^n)$ where $\gcd() \text{ means \"the greatest common divisor\" }$

Answer: (i) when $p, q$ are both odd numbers $\gcd(p^m+q^m, p^n+q^n) = p+q \text{ while } m-n \text{ is even }$
and $\gcd(p^m+q^m, p^n+q^n) = 2 \text{ while } m-n \text{ is odd }$
(ii) when $p,q$ contains one even number, appearently another cannot be even.
$\gcd(p^m+q^m, p^n+q^n) = p+q \text{ while } m-n \text{ is even }$
$\gcd(p^m+q^m, p^n+q^n) = 1 \text{ while } m-n \text{ is odd }$

[u][b]The proof part[/b][/u]
Assume that $m \ge n$

while $2n-m > 0$
    $p^m+q^m = (p^n + q^n)(p^{m-n}+q^{m-n}) - p^{m-n}q^n - p^nq^{m-n} \\ \Rightarrow p^m+q^n = (p^n + q^n)(p^{m-n}+q^{m-n}) - p^{m-n}q^{m-n}(p^{2n-m}+q^{2n-m})$
    $\Rightarrow \gcd ( p^m+q^m, q^n+q^n) = \gcd(p^n+q^n, p^{m-n}q^{m-n}(p^{2n-m}+q^{2n-m})$

Since we can easily see that $\gcd (p^n+q^n, p^{m-n}q^{m-n}) = 1 \rightarrow \\ \gcd (p^m+q^m, q^n+q^n) = \gcd (p^n+q^n, p^{2n-m}+q^{2n-m})$
Appearently $2n-m < m$ since $m>n$
so that we can always decrease the line $\gcd (p^m+q^n)$ into an equilvalent line $\gcd (p^n+q^n, p^{2n-m}+q^{2n-m})$, then we can get $\gcd (p^{2n-m}+q^{2n-m}, p^{2m-2n}, q^{2m-2n}), ...$
Until we get:
    (i) $(m,n) = (1,0) \rightarrow \gcd (p^m+q^m, q^n+q^n) = (p+q, 2)= 2 \text{or} 1$depending on value of $p,q$
    (ii)$(m,n) = (1,1) \rightarrow \gcd (p^m+q^m, q^n+q^n) = (p+q, p+q) = p+q$
(sorry, somebody can help to write a proof of this)

while $2n<m$
we can adjust $p^m+q^m = (p^n + q^n)(p^{m-n}+q^{m-n}) - p^{m-n}q^{m-n}(p^{2n-m}+q^{2n-m})$
into $p^m+q^n = (p^n + q^n)(p^{m-n}+q^{m-n}) - p^{n}q^{n}(p^{m-2n}+q^{m-2n})$
and i think the rest should be the same (using the same method)

Frank


[quote="shadysaysurspammed"]Evaluate  $\frac{1^4}{1\cdot3}+\frac{2^4}{3\cdot5}+\frac{3^4}{5\cdot7} \cdots$[/quote]
[hide]Clearly
\[ \sum_{n=1}^{\infty}\frac{n^{4}}{\left(2n-1\right)\left(2n+1\right)}=\infty  \]
because $\lim_{n\rightarrow\infty}\frac{n^{4}}{\left(2n-1\right)\left(2n+1\right)}=\infty$.[/hide]

[quote="lotrgreengrapes7926"][quote="t0rajir0u"]Then give a closed form in terms of $n$ instead.   :)[/quote]

$\displaystyle\sum_{n=1}^{n}{\frac{n^4}{(2n-1)(2n+1)}}=\displaystyle\sum_{n=1}^{n}{\frac{1}{(2n-1)(2n+1)}}*\displaystyle\sum_{n=1}^{n}{n^4}$
Using partial sums and the power sums from MathWorld :lol: , we get: $\frac{n}{2n+1}*\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}=\frac{n^2(n+1)(3n^2+3n-1)}{30}=\boxed{\frac{3n^5+n^4+2n^3-n^2}{30}}$ I think.  :D[/quote]
You made a mistake there:

\[ \sum_{k=1}^{n}\frac{k^{4}}{\left(2k-1\right)\left(2k+1\right)}\neq\left(\sum_{k=1}^{n}\frac{1}{\left(2k-1\right)\left(2k+1\right)}\right)\left(\sum_{k=1}^{n}k^{4}\right) \]


Let $a_n$ be a decreasing sequence and $x_n$ a sequence of numbers each being equal to $1$ or $-1$ and such that the series $x_1a_1+x_2a_2+...$ converges. Then prove that $ (x_1+x_2+...+x_n)a_n$ converges to 0.

Actually,if i'm not wrong the result is true for any sequence $(x_n)$.Let $S_m=\sum_{1}^{m}x_na_n$.Denote $S_m \to A$, $S=max_{1 \leq n \leq M}S_n$, so $|S_m-A|<\epsilon$ for all $m>M$. It is left to notice that from the series convergence it follows that $a_n \to 0$ as $n\to \infty$. From now we can write $a_m|\sum_{1}^{m}x_n|=a_m|\sum_{1}^{m} \frac{S_n-S_{n-1}}{a_n}|=a_m|\frac{S_m}{a_m} + \sum_{1}^{m} S_n(\frac{1}{a_n}-\frac{1}{a_{n+1}} )| \leq |S_m|+a_mS \sum_{1}^{M} (\frac{1}{a_{n+1}}-\frac{1}{a_n})+|A+\epsilon|a_m\sum_{M}^{m} (\frac{1}{a_{n+1}}-\frac{1}{a_n}) = |S_m|+S\frac{a_m}{a_M}-|A+\epsilon|<\epsilon$
since $a_m \to 0$ and $|S_m|-|A+\epsilon|<0$


geometry
rectangle
algebra unsolved
algebra

The equation $ 5x\plus{}7y\equal{}c$ has exactly three solutions $ (x,y)$ in positive integers.Find the greatest possible value of $ c$.

I must be missing something.  Why does that equation have only three positive integer solutions?  If $ c$ is just some number, then why aren't there infinite solutions.  And similarly, why does $ c$ have a maximum?  Couldn't I make $ c$ as large as I want and probably find integer solutions?

yes, it is very very simple
if you fix a lattice point in the plane then (by all lines y=-5/7x+c/7) all lattice points are determinated. if (a,b) is a solution also (a \pm 7, b \mp 5), and the diagonal rectangle with vertex (0,0),(7,5) does not contain any lattice point. so we can put three of them in the space x,y>0, and other two at max on two lines x=-1 ad y=-1. a simple scheme shows that $ c_{max}\equal{}128$

bye
paolo

@ olorin, you're right
if $ x,y \in N$ then $ c_{max}=  128$
if ${ x,y \in N-\{0}$ then $ c_{max}= 140$

 :wink:

[quote="bboypa"]yes, it is very very simple
if you fix a lattice point in the plane then (by all lines y=-5/7x+c/7) all lattice points are determinated. if (a,b) is a solution also (a \pm 7, b \mp 5), and the diagonal rectangle with vertex (0,0),(7,5) does not contain any lattice point. so we can put three of them in the space x,y>0, and other two at max on two lines x=-1 ad y=-1. a simple scheme shows that $ c_{max} \equal{} 128$

bye
paolo[/quote]
any algebric reasonning :?: .

because the next lattice point that not holds in the equation must be out of semiplane $ x,y \ge 0$ (in my case) and at the intersection of two lines || to axis, one of whichis constant <0, so <=-1. if both of extremal point satisfy this condition than we have a minimum. (however the example can be easily find).


geometry
rectangle
combinatorics unsolved
combinatorics

A $ 2007*2008$ rectangle is divided into squares. We number these squares by $ (u,v)$ is the square at row $ u$ and column $ v$ $ (1\leq u\leq 2007$ and $ 1\leq v\leq 2008)$. At first, we colour the 3 squares $ (u,v), (u\plus{}1,v\plus{}1)$ and $ (u\plus{}2,v\plus{}2)$. Then, for each move, we colour 3 squares on a row or a column that has not been coloured before. With this rule of colouring, can we colour all the squares?

This is a well-known trick.

[hide="hint"]Number the squares of the board with 1's, 2's and 3's such that any $ 3\times 1$ tile will cover exactly one 1, one 2, and one 3.  It is possible to do this in such a way that the three removed squares are all the same number.
[/hide]


trigonometry
function
complex numbers
geometric series
calculus
calculus computations

Please give me an idea how to solve this problem :
$S_{n} = sin (x)+sin (2x)+sin(3x)+ . . . +sin (nx)$
if x is from $Z$ ( excluding  $x=2\pi k$) then what is the general formula of this sum ?

Answer is

\[ \[ \sum_{k=1}^{n} \sin(kx)=\frac{\cos(\frac{x}{2})-\cos(\frac{2n+1}{2}x)}{2\sin(x/2)} \]

Solution:
[hide]
it is well known that $\sin\alpha\sin\beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))$.
Then $\sum_{k=1}^{n} \sin(x/2)\sin(kx) =\frac{1}{2} \sum_{k=1}^{n} \cos(\frac{2k-1}{2}x)-\cos(\frac{2k+1}{2}x) = \frac{1}{2}\left(\cos(\frac{x}{2})-\cos(\frac{2n+1}{2}x)\right).$
And we obtain \[ \sum_{k=1}^{n} \sin(kx)=\frac{\cos(\frac{x}{2})-\cos(\frac{2n+1}{2}x)}{2\sin(x/2)} \]
[/hide]

Another method, with complex numbers:
$\sin x$ is the imaginary part of $e^{ix}$ (for real $x$), so $\sum_{k=1}^{n}\sin kx=\text{Im}\sum_{k=1}^n e^{ikx}.$
That's a geometric series; we can apply the general formula for a sum. We get
$\sum_{k=1}^{n}\sin kx=\text{Im}\frac{e^{ix}-e^{i(n+1)x}}{1-e^{ix}}.$
$=\text{Im}\frac{e^{i\frac{n+2}{2}x}\left(e^{-i\frac n2x}-e^{i\frac n2x}\right)} {e^{i\frac12x}\left(e^{-i\frac12x}-e^{i\frac12x}\right)}$
$=\text{Im }e^{i\frac{n+1}2 x}\cdot\frac{-2i\sin\frac{nx}2}{-2i\sin\frac x2}$
$=\sin(\frac{n+1}2x)\cdot\frac{\sin\frac{nx}2}{\sin\frac x2}=\frac{\sin\frac{nx}2\sin\frac{(n+1)x}2}{\sin\frac x2}$
If you apply a product-to sum identity, you should get the form that anik_andrew got.


limit
trigonometry
calculus
derivative
function
calculus computations

evaluating algebraically $ \displaystyle\lim_{x\to 0}\frac{\sin{x}}{x}$ algebraically

Start by showing, with geometrical arguments, that, for $ x \in \,\,]0, \frac {\pi}{2}[$, $ \sin x < x < \tan x$ (1). Then note that $ f(x) \equal{} \frac {\sin x}{x}$ is an even function, and so it is enough to compute the lateral limit from the right of x = 0, for example. Then, use relation (1) (and taking into account that cos(x) is a continuous function) and the squeeze theorem.

[quote="Dvagner"]$ \displaystyle\lim_{x\to 0}\frac {\sin{x}}{x}$ without using anything with derivatives (definition of derivative, l'hopital's rule, taylor series etc), because this limit appears in the definition of the derivative of sin(x) itself and using those would involve circular reasoning[/quote]

like carcul said, see

http://www.ies.co.jp/math/java/calc/LimSinX/LimSinX.html


For $ a_1,a_2,...,a_n>0,$ and $ s\equal{}a_1\plus{}a_2\plus{}\cdots\plus{}a_n,$ prove that

$ \left.\frac{n}{n\minus{}1}\sqrt[n\minus{}1]{(s\minus{}a_1)(s\minus{}a_2)\cdots(s\minus{}a_n)}\geq a_1\sqrt[n\minus{}1]{s\minus{}a_1}\plus{}a_2\sqrt[n\minus{}1]{s\minus{}a_2}\plus{}\cdots\plus{}a_n\sqrt[n\minus{}1]{s\minus{}a_n}\right.$

I hope this would help

We modify the original inequality as follows
\[ a_1\frac{1}{\sqrt[n\minus{}1]{(s\minus{}a_2)(s\minus{}a_3)\ldots(s\minus{}a_n)}} \le \frac{a_1}{n\minus{}1} \left( \sum^{n}_{i\equal{}2}\frac{1}{s\minus{}a_i} \right)\]

Similarly, we continue this procedure and adding them up, we essentially get the desired result.

P/s. Just a flashing moment in my head so I haven't had enough time to check it. Forgive me if I  made some mistakes.

Dear [b]great math[/b],
Your idea is absolutely OK. I would like to see a complete solution to this inequality. It's easy!

that's pretty much a complete solution. AM-GM to get $ n$ inequalities of the form greatmath presented. then adding them up gives the result. (the terms on the right sum to $ \frac {1}{n \minus{} 1}\sum_{i \equal{} 1}^n\frac {1}{s \minus{} a_i}\sum_{j\neq i}a_j \equal{} \frac {1}{n \minus{} 1}\sum_{i \equal{} 1}^n 1 \equal{} \frac {n}{n \minus{} 1}$.)


when should one start thinking about what they are going to be when they grow up or what college they are going to?  im in eigth grade and im worried about what will happen if i procratinate. ive procrastinated on homework but this is a decision that will affect the rest of my life.

Ah my dear friend. . . finding out what you want to be, or where you want to go isn't something you should start worrying about. As you grow older you'll encounter new experiences and new people, through this you will find out what you want to do. It's not something that you merely think about and it's just clicks. One day it might just come to you. . . you might see a commercial, or talk to someone interesting. So don't worry about that, I didn't really care when I was in 8th grade, but as I've talked to people, and just grown up I figured out what my calling was, and you will too. The point? Worry more about getting a broad sample of knowledge than picking an occupation and focusing on that subject.

do u have any i dea what you will be doing in the future?

lets just say i've been obsessed with MIT since about 6th grade? =) but dats just me personally.. *shrugs*

[quote="JBL"]I would say that the earliest it could possibly be healthy to start thinking about college in any serious way would be 10th grade.[/quote]

I largely agree, but I would modify that thought with the thought that it depends on what one's family experience base is with college. I actually had a big change in career aspirations between eleventh and twelfth grade, which excluded from my college list some colleges that I had earlier (casually) considered. But I had REALLY LOUSY high school teachers and counselors, and my parents knew of but two colleges: my dad's and paternal grandmother's alma mater, and my paternal grandfather's and great-uncle's alma mater, both in the same small town that is our family home, and there simply wasn't a knowledge base in my family or in my school about colleges out of state. For the inquirer in this thread, and for MANY of the people who participate in this forum, whose parents perhaps received undergraduate degrees outside the United States in countries with very different systems of education, the time to start thinking about how the rules work here is EIGHTH GRADE, at the latest.

I wouldn't advise anyone to pick a top choice school any earlier than tenth grade, for the reasons already eloquently mentioned by i.w.o.a.v., but I would definitely advise middle-schoolers to think ahead to what kind of preparation they will need to show a college admissions officer if they perchance desire to attend a "top" school. Some high schools do that thinking for their students, by how their curriculum is designed, but most high schools in the United States only educate students to be mediocre, and I think most participants on AoPS want to be better than mediocre, and need to think about how to engage in self-help to improve their academic preparedness before they apply to college.

I'd say to at least wait until high school to worry too much about it...When I was in 6th grade one of my teachers told me that its never too early to choose your future and I disagree with that. As you get older your interests change (I never liked math much until last year and was going to be anything that didn't involve math, but now I absolutely love it...) and you will find out as you get closer to the end of high school and graduation that many things will cross your mind that might sound interesting. I think you should be more concerned with doing well at the present time and giving yourself some time and sooner or later your future will start to become more clear.

I definitely agree that it's not worth it to look too seriously at college in middle school.  Heck, when I was in middle school, I went through a phase where I decided I didn't want to go to college: I'd do something creative and nontraditional (although I didn't know what).  In eighth grade, I was determined to be a writer... and now here I am trying to decide between good math schools.

It's probably a good idea to keep your ears open and hear what other people have to say, though.

[quote="SirErnest"]what are some good medical schools?[/quote]

John Hopkins.

Also Washington University in St. Louis, UPenn, and Harvard. As for college, the decisions you make, even in 8th grade, can mold what happens with your high school experience in terms of courseload or activities. It's not a good idea to stress about it in middle school, but be careful where your decisions will take you and have a vague idea of what direction you want to go.

MIT is engineering and science, right?  what kind of college is syracuse? whats the difference between a college and university?

most colleges are usually 2-yr programs.. while universities are 4
MIT rocks!! hehe.. well...dats cuz im sciency and mathy and engineering-y.. *shrugs*
Harvard is only grad school good for medical..

[quote="lunmu"]most colleges are usually 2-yr programs.. while universities are 4.[/quote]

That's not really true.  It's really just a naming issue.  It is true that community colleges, which call themselves colleges and not universities, are by and large 2-year programs.  However, there are plenty of "colleges" that are 4-year programs.  Generally, university is used by larger institutions, but there are no "rules" about it -- it's just whatever they wanted to name themselves.

[quote="JBL"]Generally, university is used by larger institutions, but there are no "rules" about it -- it's just whatever they wanted to name themselves.[/quote]

JBL is correct. There is no [b]sure[/b] distinction between a "college" and a "university" in the United States. That is very confusing to people from other countries (both English-speaking and non-English-speaking) in which there is a distinction, sometimes defined by national law, between different kinds of schools of higher education. Since I learned that some people from Canada (especially) or Australia might think that a "university" is a higher-ranked academic institution than a "college," I have taken care to use phrases such as "when I was in university" (rather than "when I was in college," the normal American idiom) to keep the whole honor of my alma mater, which in fact was a major research university that has always had the name "University." In the United States of America such a terminological distinction is completely illusory.

Most universities consist of more than one school, and usually have graduate programs. Most colleges only have one undergraduate school.


Two circles of radius $r_{1}$ and $r_{2}$ intersect at points $A$ and $B$.

A common tangent is drawn to the two circles, touching them at points $C$ and $D$ respectively.

Under what circumstances will the two triangles $\triangle ABC$ and $\triangle ABD$ be equal in area?

The answer is
[hide]Always as triangles have a common side AB and M is the middle of CD as $MC^{2}=MA.MB=MD^{2}$ so C and D are equidistant to AB.[/hide]


modular arithmetic
Euler
number theory
relatively prime

Find all natural numbers x,y which satisfy  xy=(2^x)-1 (13.Korean M.O)

Obviously $ x \equal{} 0$ is a solution since both sides of the equation are 0 regardless of the value of $ y$.  Assume that $ x \equal{} 1$.  then, the equation becomes $ y \equal{} 1$, yielding another solution.  We will next show that the only two possible solutions are indeed $ x \equal{} y \equal{} 1$, and $ x \equal{} 0$ and $ y$ any (assuming that 0 is considered natural).

Obviously, $ x > 1$ cannot be even, since it must divide $ 2^x \minus{} 1$, which is odd.  Consider the least odd prime $ p$ that divides $ x$.  Obviously, $ 2^{x}\equiv1\pmod p$ if the given equality is true, while $ 2^{p \minus{} 1}\equiv1\pmod p$ because of Euler-Fermat's theorem.  Now, $ p \minus{} 1$ and $ x$ are relatively prime, since $ p \minus{} 1$ is an even number smaller than the smallest prime factor of $ x$, or integers $ a,b$ exist such that $ 1 \equal{} ax \plus{} b\left(p \minus{} 1\right)$, and consequently $ 2^1\equiv1\pmod p$, which is absurd.  There can therefore be no solution for $ x > 1$.


For the life of us we can't figure out this pattern, tried prime numbers and many other approaches.
1,8,27,64,125...

[url=http://www.research.att.com/~njas/sequences/]The On-Line Encyclopedia of Integer Sequences[/url] can be very helpful with such problems.

YAY, that may have looked ridiculously easy but I think we were stuck on the fact that the three prior to that problem were REALLY ridiculously easy and we may have started going way too in depth to solve it. Thanks so much!! :lol:


algebra
polynomial
function
domain
Ring Theory
superior algebra
superior algebra unsolved

If $K$ is a field, can the number of irreducible polynomials in $K[X]$ be finite ?

Since $\mathbb{K}[X]$ is a euclidean domain,ca'nt one adapt the proof of euclid for existence of infinitely many primes in $\mathbb{Z}$ here and conclude that its not possible to have only finitely many primes? :maybe:

Yes, but what is happening if our field is supposed to be non commutative ? In this case it seems that we cannot keep the Euclid-like proof, since the $P_{i}$ in $1+\prod uct_{i=1}^{n}{P_{i}}$ cannot commute.

It seems like it can be handled this way.

Assume the result is false.

First, $K$ must be assumed finite since every $X+\alpha$ ($\alpha \in K$) is irreducible.

Now, since there are only finitely many irreducible polynomials in $K[X]$, one of them must have a maximal degree $d$. Then consider the polynomials of $K[X]$ with degree $d+1$ and leading coefficient $1$. Since every such polynomial can be factored into a finite product of irreducible polynomials (the factorization is not necessarily unique, but it doesn't matter), we could count the maximal number of such distinct factorizations and show that there are less than $\mid K\mid ^{d}$, which would solve the question.

It seems like I don't know how to get a decent upper bound for that (if we work with simple roots polynomials only, this is easy; otherwise, it seems like we have to give bound on certain partitions). I'm not even sure the result will be true eventually, by the way.

Yep, the worst is that I thought about it yesterday ...  :D thanks

[quote="jmerry"](Use \prod, not \product)

If it's not commutative, we don't call it a field. We would call the object a skew field or a division ring.

Also, since every finite division ring is actually a field, we don't have a problem. If the division ring has infinitely many elements, we can exhibit infinitely many irreducibles of the form $X-a$.[/quote]

On second thought, conjuring Wedderburn theorem is far less elementary than trying to build up a more elementary solution (in the case of the division ring). Anybody can go further with my idea ?


I want to learn how to solve problems like these:

1. An alien has 5 legs. There are 3 different colored socks in the drawer, colored red, white, and blue. What is the least number of socks the alien must pull out so that it has 5 socks of the same color?

2. There are 12 socks in a drawer, 6 of them are colored red, and the rest are colored green. What is the least number of socks that Bob must pull out so that he gets 2 socks of the same color?

Please help!! I know I'm pretty bad at math.

You have to think of the worst case scenario. For #1, you didn't state how many socks does he have.

#1) Assuming he had 15 socks.Worst case scenario:
First four socks:red

Next four socks: white

Next four socks: blue

The next sock he picks out, regardless of what color, will assure him 5 same colored socks. Therefore he only needs to pick out $ 4\plus{}4\plus{}4\plus{}1 \equal{} \boxed{13}$.

#2) Worst case scenario:

First sock = red

Second sock = green

Third sock = red or green

Therefore he only needs to pull out $ \boxed{3}$.

Look up the [url=http://www.artofproblemsolving.com/Wiki/index.php/Pigeonhole]Pigeonhole Principal[/url]. It can help with these types of problems if you understand it.

In these types of problems, always consider the worst possible scenario. If there are n red socks, n white socks, and n blue socks, you pull n-1 socks out of each because if you pull out n socks then the pair is already complete. Then you simply pick any sock, so you just pick 1. Then you just add and you get 3(n-1)+1. <= That's pretty much like a formula I came up with, but I don't really use it, I just solve the problem.

Hope this helped! :lol:

For me, logic is the way to do it. I really wouldnt recommend memorizing a formula, because these problems are way too easy for a formula, if you just think about it. Lets say you have 5 red socks, 5 blue sock, and 5 green socks. How many would you need to take to be sure to get 2 of the same color? Drawing a picture

[color=red]R R R R R[/color]
[color=blue]B B B B B[/color]
[color=green]G G G G G[/color]

Now, we pick one sock. Say its red (it honestly doesnt matter). Now we have:

[color=red]R R R R [/color]
[color=blue]B B B B B[/color]
[color=green]G G G G G[/color]
I have:[color=red]R[/color]

Now we pick another sock. It could easily be red and we would be done. But does it have to be red? No. Remember, we want to be SURE that we have 2 of the same color, and the probability for picking two of the same colors in a row is 1/3. That does not assure us of picking two of the same color.

Now lets say, to be on the safe side, that we draw a blue. Now we have:
[color=red]R R R R [/color]
[color=blue]B B B B [/color]
[color=green]G G G G G[/color]
I have:[color=red] R[/color], [color=blue]B[/color]

Again, we could easily pick either a blue or a red and have 2 of the same color socks, but the probability of that happening is only 2/3. That doesnt assure us completely [b](imagine you were in the dark doing this and had no idea what colors you were picking. You want to be as safe as possible).[/b]

To be as safe as possible, say we pick a green now.
[color=red]R R R R [/color]
[color=blue]B B B B [/color]
[color=green]G G G G [/color]
I have:[color=red] R[/color], [color=blue]B[/color], [color=green]G[/color]

Ok, now we covered the worst case scenario, which is picking 3 different color socks after 3 different draws. Now heres to cool part. We already have a green, red, and blue, right? So no matter what color we pick, we will always have a sock of the same color with us. So to be absolutely as safe as possible, we need to pick $ 4$ socks.

 We could easily pick all red socks, but that is not very likely, and does not cover the worst case scenario, which we did.

[i][b]One Important thing:[/b] [/i]In this example,we used 5 socks of each color. That 5 doesnt really matter. We could have just as easily had 2 socks of each color, or even 20,000 socks of each color, and the answer would have been the same. (do you see why?). The number of socks of each color we start out with will not affect our worst case scenario at all. (well, it may sometimes, but you dont need to worry about that most of the time)

Try this with 10135 Red socks, 1337 Green socks, 58 Blue, and 3 Orange socks in a drawer. We are trying to pick 3 of the same color. Remember, cover the worst case scenario of picking different color socks each time.
[hide="Answer"]Did you get $ 9$?[/hide]

Stephen has four blue socks and six gray socks in a drawer. What is the least number of socks he must select, at random, to guarantee having at least two socks of the same color?

Idk what to do now.

The worst possible thing that could happen is that we get a blue sock, then a gray sock (or the other way around.) So suppose that happens. Stephen gets a blue sock, and then he gets a gray sock. But no matter what color he gets on the third try, he will then have two socks of the same color. If it's blue, he will have two blue socks; if it's gray, he will have two gray socks. Therefore, the answer is $ \boxed{3}$.

Cliu, can you tell us what specifically confuses you?

Basic Pigeonhole is basically common sense, but the Wickipedia article's Intermediate and Olympiad problems looked quite challenging.


Let $ a,b,c\geq0$ s.t. $ ab\plus{}bc\plus{}ca\equal{}1$.Prove that:
$ \dfrac{1}{\sqrt{a\plus{}b}}\plus{}\dfrac{1}{\sqrt{b\plus{}c}}\plus{}\dfrac{1}{\sqrt{c\plus{}a}}\geq2\plus{}\dfrac{1}{\sqrt{2}}$

This is a very known ineq.
Please post someone a solution.


SMO 2007, Open #5

Let $A = \{1,2,3,4,\cdots,1000\}$. Let $m$ be the number of 2-element subsets ${a, b}$ of $A$ such that $a \times b$ is divisible by 6. Find the value of $\lfloor\frac{m}{10}$.

[hide]1000/6=166+2/3 so there's 166 multiples of 6 from 1-1000.
1000/2=500 so there's 500 multiples of 2 from 1-1000, but we don't want to overcount so there's 500-166=334 multiples of 2 that are relatively prime to 3.
1000/3=333 so there's 333 multiples of 3 from 1-1000, but we don't want to overcount so there's 333-166=167 multiples of 3 that are odd.
First consider the set with all the multiples of 6.  Since the first number already is a multiple of 6, the product must be a multiple of six and there's no requirements in terms of divsibility on the second term.
Thus, there's 166*(1000-1)=166*999 possibilities.
Next, consider the set with all the multiples of 2 that weren't previously considered.  There's 334 of those.  These must be multiplied by the multiples of 3 that weren't previously considered.  Thus, there's 334*167 of those.
Thus, m/10=(166*(999+334)+334)/10=(166*1333+334)/10=22161.2[/hide]

Apparently, the answers to this contest are supposed to be integers.  You do a lot of overcounting; for example, in the first step, you count all the sets with two multiples of sixes twice, but not the others.

Edit: Apologies, the question is supposed to ask you to find the floor of $\frac{m}{10}$.

But anyhow, yes i think you overcounted, you have to subtract ${166 \choose 2}$ from $166 \cdot 999$ right?

[hide="Hope I did this right"]We have 166 multiples of 6, 334 multiples of 2 not multiples of 6, and 167 multiples of 3 not multiples of 6.

First of all, we can pick 2 multiples of 6, giving $166 \cdot 165 / 2 = 13695$.

Second, we can pick a multiple of 6 and a non-multiple of 6, giving $166 \cdot 834 = 138444$.

Finally, we can pick a multiple of 2 and a multiple of 3, giving $334 \cdot 167 = 55778$.

$13695+138444+55778 = 207917$ so the requested answer should be $\fbox{20791}$.[/hide]

EDIT: Thanks to the below poster for pointing out a typo.

MellowMelon, you inverted the 9 and 1.
Adding your numbers or subtracting $\binom{166}{2}/10$ from mine gives 20791.7 which is converted to 20791.


The task is:

Participants of math competition are solving six tasks. For each task you can get one of marks - 6,5, 3 or 0 .
It transpired that for each pair of participants we can indicate two tasks , that in each of them participant A got different mark from participant B.
Delimit the highest number of participants for which this situation is possible. :)

Thank you for help in solving it

This one is from Polish Mathematical Olympiad which is still in progress - locked till 11th of October.


I think we discussed in open section problems from inequalities this problem of Ozeki.

:blush: Sorry, I'm new here, I haven't seen that :blush:


Hi everybody,

I have created a forum (phpbb) and I want to install LaTeX.
Is there any one who have an idea how to do it?

Thanks :)

You need to have LaTeX, ImageMagick and GhostScript installed first on your server - you may need to consult your host. Then install LatexRender from http://www.mayer.dial.pipex.com/tex.htm You'll find the necessary files and instructions in the phpBB directory of the LatexRender download.

If you can't install LaTeX etc then, assuming you can run cgi programs, an alternative is to install mimeTeX. Instructions are also in the same download.

Thanks a lot!  :)  In fact I have created a Vbulletin forum. I will try to install it.

Thanks again stevem :)


ARML
geometry
3D geometry
sphere
ratio
email
tetrahedron

1.) $f(x) = x+\sqrt{x}$ and $g(x) = x+\frac{1}{4}$. $g(f(g(f(g(f(g(f(a)))))))) = 5$, a > 0. Find the value of a.

2.) How many ways can you paint the faces of a cube such that each face is a different color, and no coloring can be obtained by rotating any other coloring?

3.) A sphere is inscribed in a regular tetrahedron. Find the ratio of the volume of the sphere to the volume of the tetrahedron.

4.) Find the sum of the series: $1+2\Big(\frac{1}{2}\Big)+3\Big(\frac{1}{4}\Big)+4\Big(\frac{1}{8}\Big)+\ldots$

Marcus,
I will send out an email within the week to let students know that we are starting to post problems for ARML.  Thank you for putting them up.  I am also sending out some problems with acceptance letters.  I ran out of ink in my printer, and I will get them out as soon as I get ink.

They seemed to get easier, so I started with number 4  :P ...

[hide="Number 4"]
Let the sum equal S.

$S = 1+2\Big(\frac{1}{2}\Big)+3\Big(\frac{1}{4}\Big)+4\Big(\frac{1}{8}\Big)+\ldots$

Now find 2S:

$2S = 2+2\Big(1\Big)+3\Big(\frac{1}{2}\Big)+4\Big(\frac{1}{4}\Big)+\ldots$

Line up the two sums:

$2S = 2+2\Big(1\Big)+3\Big(\frac{1}{2}\Big)+4\Big(\frac{1}{4}\Big)+\ldots$
$S = 0+1(1)+2\Big(\frac{1}{2}\Big)+3\Big(\frac{1}{4}\Big)+4\Big(\frac{1}{8}\Big)+\ldots$

Subtract S from 2S:

$2S-S = S = 2+1+\frac{1}{2}+\frac{1}{4}+\ldots$

Now we can use the formula for the sum of an infinite series:

$S = \frac{a}{1-r}$ where a is the first term, and r is the common ratio.

$S= \frac{2}{1-\frac{1}{2}}= \boxed{4}$
[/hide]
That was a nice problem.  I liked it.  Unlike number 3...

Number 3 looked brute force-able with a lot of 30-60-90 right triangles, so that's what I did.  I think you could also coordinatize it, although that seemed even harder.  I bet there's an easier way to do it though.

[hide="Outline for Number 3"]
If we call the side length of the tetrahedron $a$, and then find the length of the inscribed radius in terms of $a$, we have everything we need to solve the problem.

In order to find the radius, we draw a right triangle.  Leg 1 will be from the center of one of the equilateral faces to the center of the sphere.  Leg 2 will be from the center of one of the equilateral faces to a vertex of the tetrahedron.  And the hypotenuse will be from the center of the sphere to a vertex of the tetrahedron.  Note that the radius of the sphere is Leg 1.

(Here it becomes an outline)

Then I drew in 30-60-90 right triangles all over the place, and came up with the radius being $\frac{a\sqrt{6}}{12}$.  This would give the sphere a volume of $\pi\frac{a^{3}\sqrt{6}}{216}$.

Then we find the volume of a tetrahedron with side length a, which (after either memorization, or a few more 30-60-90 right triangles) comes out to be $\frac{a^{3}\sqrt{2}}{12}$.

So the ratio of the sphere to the tetrahedron is $\boxed{\frac{\pi\sqrt{3}}{18}}$.

But there is quite probably an arithmetic error in there somewhere, so it would definitely be better approached a different way.

[/hide]

[hide="Guess for Number 2"]
We have 6 possible colors for 6 sides.  Choose a color, paint one face that color, and then glue that face to the table, so it is permanently the bottom face.  This can be done in $\binom{6}{1}= 6$ ways.

Then we have 5 colors left, and we pick 4 of them to go on the four faces adjacent to the bottom face.  These 4 colors can be arranged in a circle in $(4-1)!$  ways.  So this gives us a total of $\binom{5}{1}\cdot 3! = 30$ ways.

Then of course, the color of the top is the color not yet used, and there is only 1 choice.

So multiply these together:
$6 \cdot 30 \cdot 1 = \boxed{180}$

But I'm not very sure at all on that one.
[/hide]

[quote="generating"]Your problem 1 needs to be tweaked. The mega-function has a minimum value of 4 on the range $x\geq 0$. Note that $f$ and $g$ are increasing on that interval, so the minimum of the composition occurs at $x=0$.

I don't want to spoil the intended solution -- I think the problem is pretty cool. I recommend switching the '3' to a '5' or a '7' or a '2007'  :)[/quote]

You're right. You should change the 3 in problem 1 to a 5.

And Sean, your answers to numbers 3 and 4 are correct, but your answer to number 2 is wrong.

Note:
Problem 3 was inspired from a problem I worked on when I was in high school. The problem was to find the limit of the ratio of the total volume of several spheres stacked in a "pyramid formation" to the volume of a circumscribed tetrahedron as the number of spheres tends to infinity.

Then I tried to generalize the problem to find the limit of such a ratio for "spheres" and "tetrahedrons" of higher dimensions. It was an interesting project (at least to me). It forced me to really think about where the formulas for volume and area come from, and the meaning of dimension. It is also tangentially related to some areas of [url=http://en.wikipedia.org/wiki/Sphere_packing]Sphere Packing[/url].

sean's solution for #2 is right, except that when you fix a face at the beginning you don't multiply by 6...

#1 is also really slick once you get it.

[quote="generating"]mysmartmouth, that's a pretty neat way of approaching #4. I'd never seen that before. Here's how I would do it  :)

[hide="Different #4"]
For the kinda-geometric series $S=1+2r+3r^{2}+4r^{3}+\cdots$, write it like:
\begin{eqnarray*}S & = & 1+r+r^{2}+r^{3}+r^{4}+\cdots \\ & &+\,r+r^{2}+r^{3}+r^{4}+\cdots \\ & &+\,r^{2}+r^{3}+r^{4}+\cdots \\ & &+\,\cdots \\ & = & \frac{1}{1-r}+\frac{r}{1-r}+\frac{r^{2}}{1-r}+\frac{r^{3}}{1-r}+\cdots \\ & = & \frac{1}{1-r}\left(1+r+r^{2}+r^{3}+\cdots\right) \\ & = & \frac{1}{(1-r)^{2}}\end{eqnarray*}

So just plug in $r=1/2$ and you've got it  :)
[/hide][/quote]

Interesting... I've never seen your way, even though it appears to be the easiest. I've always seen Sean's (that and calculus, which is even more annoying).


Problem 10 on the AwesomeMath Admission Test B:
The result of the addition

AWESOME + MATH + SUMMER

is a seven-digit number whose digits are all equal.  Different letters stand for different decimals.  There are only two possible numbers AWESOME can represent.  What are they?

I was looking through the solution, and all they gave was the answer!  I understand how to get the answer through deductive trial and error and just pure logic, but how does one approach this type of problem with rigor?  I tried applying mod arithmetic to it, but I'm not good enough to figure out the numbers that are carried over.

i used excel and i got one of the answers

im not sure how to do it without guess and check

Note that some of these problems involve a certain degree of trial and error; it probably is even better than a non-guess solution.

And how do you use excel to find solutions to these problems?

i set up this thing where when you put in a value for A, it puts it in for all the A's

also, it added the corresponding collumns and carried over to the next column when it was needed.

There is a very similar problem on the test C so i think thats why they didnt post the full solution


B and C is a 3 x 3 matrices and C is invertible.

A is the matrix below
| 0 1 0|
| 1 0 1|
|-2 0 4|

Det(A) is -6.

Det of (3CA(C^-1)B) = 54.  *I don't know how to make small -1 invertable sign... it's [b]not[/b] to the power of -1

And i'm suppose to find Det B.

i got down to 18 after dividing by 3... but i'm just lost after that.

The key theorem: the determinant of a product is the product of the determinants. That is, $ \det (AB)\equal{}\det(A)\det(B)$ for all square matrices $ A$ and $ B.$

As a corollary, if $ C$ is invertible, then $ \det(C^{\minus{}1})\equal{}\frac1{\det(C)}.$

What about $ \det(rA)$ for some scalar $ r?$ It's important to note that the determinant is [b]not[/b] a linear function. But we can write $ rA\equal{}(rI)A$ and it's easy to see that for an $ n\times n$ identity matrix, $ \det(rI)\equal{}r^n.$ Therefore $ \det(rA)\equal{}r^nA.$ (When you said you got down to 18 after dividing by 3: you were wrong. You have to divide by $ 3^3,$ not $ 3.$)

So:

$ \det(3CAC^{\minus{}1}B)\equal{}3^3\det(C)\det(A)\det(C^{\minus{}1})\det(B)\equal{}54.$

The multiplication on the right hand side of that is the ordinary multiplication of (real) numbers, and that multiplication, unlike the multiplication of matrices, is commutative. So we get to use that $ \det(C)\det(C^{\minus{}1})\equal{}1.$ And, as you have calculated, $ \det(A)\equal{}6.$ Oops! Your calculation is incorrect. In fact, $ \det(A)\equal{}\minus{}6.$ (Use minors of the first row, but don't forget the alternating signs that are part of that definition.) Hence:

$ 27\cdot(\minus{}6)\cdot\det B\equal{}54,$ from which $ \det B\equal{}\minus{}\frac13.$

Ah that clears a lot. Thank you kent. I appreciate your time answering my question. and putting symbols o_o(how you do that !! lol)


Given a triangle $ ABC$ such that $ AB\equal{}1,\ AC\equal{}2, \angle{BAC}\equal{}120^\circ$. Draw the perpendicular $ AH$ fronm $ A$ to $ BC$. Express $ \overrightarrow{AH}$ in terms of $ \overrightarrow{AB},\ \overrightarrow{AC}.$

Outline:
find angle $ CAH$ and $ BAH$; express AH as the orthogonal projection of CA and BA

See the attached diagram.

By Cosine Law, $ BC^2 = 1^2 + 2^2 - 2\cdot 1\cdot 2\cdot\left( - {1\over 2}\right) = 7\iff BC = \sqrt {7}$

Expressing the area of $ \triangle ABC$, $ AB\cdot AC\cdot\sin 120^\circ = BC\cdot AH\implies AH = {1\cdot 2\cdot{\sqrt {3}\over 2}\over\sqrt {7}} = \sqrt {3\over 7}$

By Pythagoras, $ BH = \sqrt {1 - {3\over 7}} = {2\over\sqrt {7}}, CH = \sqrt {4 - {3\over 7}} = {5\over\sqrt {7}}$, hence $ BH: HC=2: 5$

If $ PH\parallel AB$ and $ QH\parallel AC$, then obviously $ AP = {2\over 7}AC$ and $ AQ = {5\over 7}AB$, so we get $ \overrightarrow{AH} = {5\over 7}\overrightarrow{AB} + {2\over 7}\overrightarrow{AC}$

That's correct.  :)

Alternative Approach:

In the attched diagram 3 points $ B,\ C,\ H$ are collinrar, so we can set $ BC: CH\equal{}t: 1\minus{}t\ (0<t<1)$.


There is say, a 5x3 square, made of 15 equal unit squares. A line starts on one vertex and goes to the vertex that is diagonal from the original vertex (Basically 15 unit squares in 5x3 with a diagonally line through it touching 2 vertexes). How many squares (not of any size) would the line touch?

Now, I remember that I originally learned this formula as, if the length and width are A and B, then the diagonal would pass A+B-1. However, I remember learning somewhere that if its 7x4 or 15x8 or a very tall or wide rectangle, then the formula is different (more complicated version of the first formula I stated)

What's the formula? And no this isn't like a trick question that you see on a mathcounts test, I'm just trying to re-learn a formula I SHOULD KNOW!!! Especially since I plan on nabbing the 4th spot on the Indiana team (not going to happen, I know :()  :blush:

Thanks to all responses and for the formula tarquin!

[quote="Xantos C. Guin"]Whatever you do, don't draw out all squares and draw a line.[/quote]

Well, I did that once, sort of... and didn't find the right answer. :blush:


superior algebra
superior algebra unsolved

On page 9, it says:

[quote]If [b]F[/b] is any field, there is a canonical map $\phi$ from the integers to [b]F[/b] determined by $\phi(1) = 1$.  If char([b]F[/b]) = 0, then $\phi$ is an injection; using $\phi$ we regard the integers as a subring of [b]F[/b], and then we also regard $F_0 = \Re$ as a subfield of [b]F[/b].  If char([b]F[/b]) = p, then $\phi$ gives an injection from $F_p$ into [b]F[/b] and we regard $F_0 = F_p$ as a subfield of [b]F[/b].  In either case, $F_0$ is called the prime field.[/quote]

I'm a bit confused by the entire paragraph and not quite sure where to start.

1. What are F_0 and F_p?  Does the 0 or p just mean the characteristic of the field?
2. In the book, it uses an equals sign as I wrote it: $F_0 = \Re$.  Does this mean isomorphism?  I've just always seen isomorphism written as $\approx$.
3. In the first sentence, does the author state $\phi(1) = 1$ just to indicate that $\phi$ can not be the zero map?  Or is there something more I'm missing?

I have more questions, but they'll probably be based off of the above.

1. $\mathbb F_p$ is probably $\mathbb Z/\mathbb Z=\mathbb Z_p$ with $p$ prime.
2. yes. It says that there exists a subfield of your field(the big one) which is isomorphic (the sign $=$ is there just to make it easier to work with and not use $\phi(...)$ all the time) with $\mathbb Q$ when $char F=0$ and with $\mathbb Z_p$ when $char F=p$ ($p$ will be prime).
3. No. It defines $\phi(1)=1$ because if $\phi$ is an isomorphism of fields in this case it is perfectly determined by $\phi(1)$.

$1=\phi(1)=\phi\left(\displaystyle n\cdot\frac1n\right)=n\phi\left(\frac1n\right)$ thus $\displaystyle\phi\left(\frac1n\right)=\frac1n$.
and of course $\phi(n)=n$ where the first $n$ is $1_F+1_F+...+1_F$ $n$ times and the second $n$ is from the field with which you want to make it isomorphic (i.e. $\mathbb Q$ if $char F=0$ and $\mathbb Z_p$ when $char F=p$ ). It is not perfectly rigurous what I wrote but I hope you understand...

I don't think it works with $\mathbb R$ instead of $\mathbb Q$ in the general case. For example $\mathbb Q$ is a field with $char\mathbb Q=0$ but it does not have a subfield isomorphic with $\mathbb R$. You may need some other hypothesis to use $\mathbb R$ instead of $\mathbb Q$.
If it still unclear or I missed (or did smth wrong) smth pls tell me.


inequalities
combinatorics unsolved
combinatorics

For $\{a_i\}$ such that $\prod a_i=a^n$.prove that:
(1+a_1)(1+a_2)...(1+a_n)≥(1+a)^n
Please find the combinatorics solution for this inequality


Find a formula for the area of a cyclic $n$-gon in terms of its side lengths $s_1, s_2, s_3, ... s_n$.

This is a research-level problem for which no complete solution is known.  The state of the art can be found on the Internet in recent (last 2-3 years) articles by Igor Pak and Maxim Fedorchuk.


Not use calculator, compare $ \log_{2}{3}$ and $ \log_{3}{2}$ [size=150]?[/size]

both are reciprocal.
[hide="proof"]
expand as ln 3 / ln 2. U can also use some other base.[/hide]
Thank u.


geometry
circumcircle
incenter
trigonometry
inequalities
inradius
geometric transformation

let $ O , I , H $ be center of circumcircle , incentre , and orthocentre of a triangle

prove that $ OH \geq (IH) \sqrt2$

[color=red][Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=2950 .][/color]

Denote the triangle sides $a = BC, b = CA, c = AB$, angles $\angle A = \alpha, \angle B = \beta, \angle C = \gamma$ and circumradius $R$. WLOG, assume that the angle $\alpha = \angle A < 90^o$ is acute. Then

$AH = 2R \cos \alpha$

If the angle $\alpha$ was not acute, we would have $AH = -2R \cos \alpha$ for $\alpha > 90^o$ or $AH = 0$ for $\alpha = 90^o$, which is what we want to avoid.

$AI = \sqrt{bc \tan{\frac \beta 2} \tan{\frac \gamma 2}} = 2R \sqrt{\sin \beta \sin \gamma \tan{\frac \beta 2} \tan{\frac \gamma 2}} = 4R \sin{\frac \beta 2} \sin{\frac \gamma 2}$

If $\beta \le \gamma$ resp. $\beta > \gamma$,

$\angle IAH = \frac \alpha 2 + 90^o - \beta = \frac{\gamma - \beta}{2}$ [color=white].[/color] resp. [color=white].[/color] $\angle IAH = \frac \alpha 2 + 90^o - \gamma = \frac{\beta - \gamma}{2}$

Hence, $\cos{\widehat{IAH}} = \cos{\frac{\gamma - \beta}{2}} = \cos{\frac{\beta - \gamma}{2}}$ is the same in both cases.

Using the cosine theorem for the triangle $\triangle IAH$,

$IH^2 = AI^2 + AH^2 - 2 AI \cdot AH \cos{\widehat{IAH}} =$

$= 4R^2\left(4\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2} + \cos^2{\alpha} - 4 \sin{\frac \beta 2} \sin{\frac \gamma 2} \cos \alpha \cos{\frac{\gamma - \beta}{2}}\right) =$

$= 4R^2\left(4\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2} (1 - \cos \alpha) + \cos^2{\alpha} - \sin \beta \sin \gamma \cos \alpha\right) =$

$= 4R^2\left(8 \sin^2{\frac \alpha 2}\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2} + \cos^2{\alpha} - \sin \beta \sin \gamma \cos \alpha\right)$

According to Coxeter & Greitzer, Geometry Revisited (chapter 1, problem 1.7.2),

$OH^2 = 9R^2 - (a^2 + b^2 + c^2) = R^2\left(9 - 4(\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma})\right)$

Also, compare the 2 solutions of the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25726]Distance from a vertex to ninepoint center[/url]. One uses the above relation and the other does not. Hence, their combinantion yields this relation. We have to show $OH^2 \ge 2 IH^2$, i.e.,

$(?)\ \ R^2\left(9 - 4(\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma})\right) \ge 8R^2\left(8 \sin^2{\frac \alpha 2}\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2} + \cos^2{\alpha} - \sin \beta \sin \gamma \cos \alpha\right)$

$(?)\ \ 9 \ge 4(\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma}) + 8\left(8 \sin^2{\frac \alpha 2}\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2} + \cos^2{\alpha} - \sin \beta \sin \gamma \cos \alpha\right)$

Using $\cos \alpha = \cos{\left(180^o - (\beta + \gamma)\right)} = -\cos{(\beta + \gamma)}$,

$\sin^2{\beta} + \sin^2{\gamma} - \sin \beta \sin \gamma \cos \alpha = \sin^2{\beta} + \sin^2{\gamma} + \sin \beta \sin \gamma \cos(\beta + \gamma) =$

$= \sin^2{\beta} + \sin^2{\gamma} + \sin \beta \sin \gamma \cos \beta \cos \gamma - \sin^2{\beta} \sin^2{\gamma} =$

$= \frac 1 2(\sin^2{\beta} + \sin^2{\gamma} + \sin^2{\beta} (1 - \sin^2{\gamma}) + \sin^2{\gamma} (1 - \sin^2{\beta})+ 2\sin \beta \sin \gamma \cos \beta \cos \gamma) =$

$= \frac 1 2 \left(\sin^2{\beta} + \sin^2{\gamma} + (\sin \beta \cos \gamma + \sin \gamma \cos \beta)^2 \right)  =$

$= \frac 1 2 (\sin^2{\alpha} + \sin^2{\beta} + \sin^2{\gamma})$

Substituting this to the above inequality, it remains to show

$(?)\ \ 9 \ge 8(\sin^2{\alpha} + \cos^2{\alpha}) + 8\left(8 \sin^2{\frac \alpha 2}\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2}\right)$

$(?)\ \ 1 \ge 64 \sin^2{\frac \alpha 2}\sin^2{\frac \beta 2} \sin^2{\frac \gamma 2}$

Sines of all triangle angles are always positive, not to mention sines of half-angles, we can take the square root of the last inequality.

$(?)\ \ 1 \ge 8 \sin{\frac \alpha 2}\sin{\frac \beta 2} \sin{\frac \gamma 2}$

This inequality has been proved more than one time on this forum, for example, see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24803]Triangle inequality[/url] for the use of Lagrange multipliers. Let's do something different:

$(?)\ \ 8 \sin{\frac \alpha 2}\sin{\frac \beta 2} \sin{\frac \gamma 2} \le 1$

$(?)\ \ 8 \sin{\frac \alpha 2}\sin{\frac \beta 2} \cos{\frac{\alpha + \beta}{2}} \le 1$

Cosines of triangle half-angles are always positive, hence we can multiply the last inequality by $2 \cos{\frac \alpha 2}\cos{\frac \beta 2}$.

$(?)\ \ 16 \sin{\frac \alpha 2} \cos{\frac \alpha 2} \sin{\frac \beta 2} \cos{\frac \beta 2} \cos{\frac{\alpha + \beta}{2}} \le 2 \cos{\frac \alpha 2}\cos{\frac \beta 2}$

$(?)\ \ 4 \sin \alpha \sin \beta \cos{\frac{\alpha + \beta}{2}} \le \cos{\frac{\alpha - \beta}{2}} + \cos{\frac{\alpha + \beta}{2}}$

$(?)\ \ 8 \sin \alpha \sin \beta \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha + \beta}{2}} \le 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}} + 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha + \beta}{2}} $

$(?)\ \ 4 \sin \alpha \sin \beta \sin{(\alpha + \beta)} \le \sin \alpha + \sin \beta + \sin{(\alpha + \beta)}$

${(?)\ \ 4 \sin \alpha \sin \beta \sin \gamma \le \sin \alpha} + \sin \beta + \sin \gamma$

${(?)\ \ \frac{8R^3 \sin \alpha \sin \beta \sin \gamma}{R^2} \le 2R \sin \alpha} + 2R \sin \beta + 2R \sin \gamma$

Let $r, s, \triangle$ be the inradius, semiperimeter and area of the triangle $\triangle ABC$.

$(?)\ \ \frac{abc}{R^2} \le a + b + c = 2s$

$(?)\ \ \frac{4 \triangle}{R} \le 2s$

$(?)\ \ r = \frac{\triangle}{s} \le \frac R 2$

But the last inequality follows from Euler's relation $R^2 - d^2 = 2rR$, where $d = OI \ge 0$ is the circumcenter to incenter distance:

$r = \frac{R^2 - d^2}{2R} \le \frac{R^2}{2R} = \frac R 2$

with equality only for an equilateral triangle.

OH^2 - 2 IH^2 = R^2 - 4 r^2
How can we prove that without too much computation?
Kind regards. Jean-Pierre

Let U = reflection of O in I.
OH^2 - 2 IH^2 = OU^2 - 2 IU^2 - UH^2 (Leibniz) = 2 OI^2 - UH^2
IF N is the NP-center, we have UH = 2 IN = R - 2r (Feuerbach)
As OI^2= R(R-2r), it follows that OH^2 - 2 IH^2 = R^2 - 4 r^2
Kind regards. Jean-Pierre

Dear JPE,

    You are being extremely unfair to our Yetti  in your reply. You know my position on
    the subject,   but in this particular case 'nobles oblige' s  me to take his well
    represented  side.

     Maj.  Pestich

Dear Pestich
I don't see what you mean. If you think that I've been unfair, be sure that it was not my intention. I was just trying to answer to the original problem; so, I've suggested a relation and, later, I've found how to prove this relation.
Kind regards. Jean-Pierre

I was trying to emphasize that Yetti here singlehandedly is building a temple,
   a kind of Babylonian tower, a monument  to math, and you, obviously having the best
   intentions in mind, are trivializing  his royal attempts.
   In my book it's unfair, and I realized it myself only recently.

    ML should strictly enforce a lower boundary on number of pages in replies.

    Col.  Pestich.

[quote="pestich"]... ML should strictly enforce a lower boundary on number of pages in replies.

    Col.  Pestich.[/quote]

The lower boundary is already being enforced by Mathlinks and the value of this boundary is zero. I do not see how could they go lower than this.

Yetti

[quote="lomos_lupin"]let $ O , I , H$ be center of circumcircle , incentre , and orthocentre of a triangle

prove that $ OH \geq (IH) \sqrt2$[/quote]

Actually, the inequality $ OH\geq\frac32\cdot IH$ holds for every triangle ABC with circumcenter O, orthocenter H and incenter I. Since $ \frac32 > \sqrt2$, this inequality is stronger than your $ OH\geq \sqrt2\cdot IH$.

How to prove the inequality $ OH\geq\frac32\cdot IH$ ? The proof is surprisingly simple. We assume the triangle ABC to be non-equilateral (else, everything is trivial, since O = H = I and thus OH = IH = 0). If G is the centroid of triangle ABC, then, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=16552 , the incenter I of triangle ABC lies inside the circle with diameter GH. Thus, the segment IH must be smaller than the diameter GH of this circle. Thus, GH > IH. On the other hand, by the Euler line theorem, the point G lies on the segment HO and divides it in the ratio $ \frac {GH}{OG} \equal{} 2$. Thus, $ OG \equal{} \frac {GH}{2}$. Hence,

$ OH \equal{} OG \plus{} GH \equal{} \frac {GH}{2} \plus{} GH \equal{} \frac32\cdot GH > \frac32\cdot IH$.

Hence, the inequality $ OH\geq\frac32\cdot IH$ is proven for all triangles ABC.

[i]Remark.[/i] The same configuration proves another inequality, namely $ OH\geq OI$:

Since the point I lies inside the circle with diameter GH, we have $ \measuredangle GIH>90^{\circ}$, so that $ \measuredangle OIH\geq \measuredangle GIH>90^{\circ}$ (here, $ \measuredangle OIH\geq \measuredangle GIH$ follows from the fact that the point G lies inside the segment OH). Hence, the triangle OIH is obtuse-angled at I. The longest side of an obtuse-angled triangle is the side opposite to the obtuse angle; hence, the longest side of triangle OIH is OH, so that $ OH\geq OI$. Proof complete.

  Darij

Edit:This post is edited to the following :

Can we have both of these inequalities: :!:

$OH \frac32 HI $ and $OH \sqrt2 HI$.

Answer :See the darij`s post.

[quote="lomos_lupin"]Darij proved $ OH \ge \frac {3}{2} IO$ and yetti proved $ OH \ge \sqrt2{OI}$
But it is cleat we cant have point of them :?  :?[/quote]

At first, we are talking about IH, not IO (and $ OH\geq\sqrt2 OI$ cannot be true; the best such inequality is $ OH\geq OI$). So the inequalities should be $ OH\geq\frac32\cdot IH$ and $ OH\geq\sqrt2\cdot IH$.

And then, indeed both of these inequalities are true. It sounds strange, but both inequalities are sharp (i. e. they can become equalities) and both are correct. The reason: Both inequalities become equalities if triangle ABC is equilateral, what is equivalent to OH = IH = 0, and in this case, the coefficient in front of IH doesn't matter.

However, "my" inequality $ OH\geq\frac32\cdot IH$ is the sharpest possible, what you can easily see if you note that the point I can lie everywhere inside the circle with diameter GH, particularly it can lie very near to the point G.

  Darij


search
number theory
calculus
calculus computations

Let $ [a_n, b_n]$, $ n \equal{} 1, 2, \ldots$ be closed intervals with $ [a_n, b_n] \cap [a_m, b_m] \neq \emptyset$ for all $ n,m$. Prove that $ \bigcap_{n \equal{} 1}^{\infty} [a_n, b_n] \neq \emptyset$.

Since any pair of intervals have some element in common, isn't it obvious that the intersection of all intervals has some element in common? Or am I missing something  :maybe:

[quote="aidan"]Since any pair of intervals have some element in common, isn't it obvious that the intersection of all intervals has some element in common? Or am I missing something  :maybe:[/quote]  Consider the sets $ C_n \equal{} (0, \frac {1}{n})$ and let me know how far that logic gets you.

Also: you and this other poster have now dumped a large number of the same problems into the forum.  Obviously they come from some homework, contest, exam or the like.  It would be polite to include the source with your post, e.g. in the "Subject description" field.  Perhaps you should look through all the other poster's posts, as well, before you go posting any more of this block of questions.  As a general rule, when posting a problem that "comes from somewhere" it's a good idea to do a quick search for some obvious keywords to see if it's been posted before.

[quote="JBL"]Also: you and this other poster have now dumped a large number of the same problems into the forum.  Obviously they come from some homework, contest, exam or the like.  It would be polite to include the source with your post, e.g. in the "Subject description" field.  Perhaps you should look through all the other poster's posts, as well, before you go posting any more of this block of questions.  As a general rule, when posting a problem that "comes from somewhere" it's a good idea to do a quick search for some obvious keywords to see if it's been posted before.[/quote]

Noted. Will do that from now on - for the record, these problems come from my first-year university math course Numbers and Sets (a mixture of number theory and set theory). Apparently this is covered in Analysis as well.

[quote="JBL"]Consider the sets $ C_n \equal{} (0, \frac {1}{n})$ and let me know how far that logic gets you. [/quote]

Wasn't this used to prove the last time I posted a similar problem that the intersection of an infinite number of subsets is empty? But in this case we require closed intervals, so it should be $ C_n \equal{} [0, \frac{1}{n}$. An infinite intersection would still contain $ 0$, so it is certainly not empty?


I typed my solutions already, but each has a different header, so therefore, I cannot combine them into one Word document... I have to keep them separate.

How could i combine the PDF files so it makes one final PDF rather then 5 individual PDF files?

and sorry about the last board I made... it was a mistake.

There's a utility called [url=http://www.brothersoft.com/pdf-merge-25708.html]pdfmerge[/url] that you could try to install.  (We use it on our Linux machines and it works pretty well, but I don't know anything about how well it works in Windows.)  Or if you have access to a scanner, you can print them out and rescan them into one file.

And, of course, remember that you can also just print out and mail your solutions.


function
calculus
real analysis
probability
search
complex analysis

Ahh, the almighty generatingfunctionology. How can you pass up a free math book?  ;)

Having worked my way through most of this book, I must say that it is very readable and extremely well-written. Some familiarity with generating functions would be helpful, as would some calculus knowledge, but neither are necessary to get alot from it. I suggest looking at even a small portion for anyone curious enough to open this thread.

[quote="zabelman"]How can you pass up a free math book?  ;)
[/quote]

Wait till you get to university (in age, not ability, I mean  :) )

I just paid 165 CDN for Rudin (which is a hair over 200 pages), and if I take a 400-level complex analysis course that I've been looking at, my textbook cost this semester will run to nearly 1000 CDN

[quote="blahblahblah"][quote="zabelman"]How can you pass up a free math book?  ;)
[/quote]
Wait till you get to university (in age, not ability, I mean  :) )
[/quote]Why should he? Tsimmerman didn't wait :P :)

There are certain stigmas associated with going to university early, unfortunately.  College is an experience, not just a place where you learn stuff, and I think going to university when you're younger than everyone else can have very negative consequences.

Mind you, I'm the youngest person in my classes (for the most part), so I guess that's kind of the same thing.

So they say. I don't do much other than learn stuff at university. But everyone else here thinks I'm a madman. I find being the youngest person in my classes fun for the most part. I don't think the graduate students hate me.

[quote="Ravi B"]Is this the start of a trend?[/quote]

I hope so!  :D

As for going to college early, I was pretty seriously considering skippping my senior year. I ultimately decided against it. The only reason I wanted to go is to learn lots of math. But as y'all mentioned earlier, college is not just a place for learning; the same goes for high school. I found a greater importance in finishing my high school experience because of my friends, my other activities (sports, etc), self-taught mathematics, and of course competitive math. There's no way I was gonna miss out on all of that.

[quote="blahblahblah"]Wait till you get to university (in age, not ability, I mean  :) )
[/quote]

It's great to be in high school and have my parents buy me all the books I want :p

Still, I feel bad on occasion, which is why this is awesome.  I sure hope that this is the start of a trend.

I read about the first 20 pages of generatingfunctionology a couple weeks or so ago and it is very interesting so far.  I hope to find more time to read more of it.  I am also very glad it is free.

I also found [url=http://www-math.mit.edu/~rstan/papers/comb.pdf]this[/url] a little while back.  I have not read very much of it.  It appears to get pretty advanced.  It's on combinatorics.

I sat in on Herbert Wilf's class on generatingfunctionology back probably in 1990. He says books cost way too much; he's generated enough money selling textbooks, and WAS trying to make a statement by selling his books for $6.00
Since the download site is his own page on the math.upenn.edu web page, I guess he retained his rights to distribute it for free. He did have a class policy of giving out tiny cash rewards for anyone who found typo's in his manuscript (like Donald Knuth for _The Art of Computer Programming_) so that's why the quality is high.
-DK

This is a GREAT probability book.  I use it for everything.  And since y'all like free stuff so much........[url=http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/amsbook.mac.pdf]The Dartmouth Probability Book[/url]

[quote="Ravi B"]I think it's great that [i]Generatingfunctionology[/i] is available as a free download.  But I'm wondering why the publisher would allow it.  What is their incentive?  Is this the start of a trend?[/quote]

I'm seeing many many many free books, usually about computers, but some about math.  Just search something like 'number theory' or 'abstract algebra' on Google and you'll come up with some free, online books from tons of universities.  Some are better than others.

It's the book that's linked up there.  Or it's a word meaning the study of generating functions.  But I'm pretty sure that it became a word because of the book.

Wow, my finger rapidly twitched to "right-click, save target as."  I didn't even know what I was doing.  I envy anybody whose parents buy them textbooks.  They come for a high price.  I get my books for Christmas and birthday.  Unfortunately, they tend to overlap eachother and treat the audience like an average high school graduate.  Textbooks get to the point.  Can't wait for college.

I have found that the popular science books are only good for history, because the math is very basic (maybe why it can be popular).  Not having access to textbooks, my philosophical books help me more because they explain the current standards for science.  Like, any time we use the word "infinity", it can be replaced by "there is no number sufficiently large".  That really helps in understanding what exactly you are saying and what exactly is being said whenever math is the topic.  These books include "Godel, Escher, Bach" and "A Compact History of Infinity"

If you know Russian (which, sadly, I don't), there's gobs and gobs of free books over at http://www.mccme.ru/free-books/.

oooooo i cannot understand a single word!
I am a Chinese so we've been studying english therefore i have no time to learn Russian...
I'll try google.

[quote="shobber"]My god, generatingfunctionology and probabilit...that's amazing!
But i still want to get more... :blush:  :P
Who can give other links of free books about high school math?[/quote]

Go to [url=http://goodlymaths.0catch.com/books/online_book_links.htm]Goodlymaths[/url] for a lot of free ebooks.  Also, check out [url=http://www.math.gatech.edu/~cain/textbooks/onlinebooks.html]this[/url] and [url=http://us.geocities.com/alex_stef/mylist.html]definitely this![/url]


induction
number theory unsolved
number theory

Let $p$ be an odd prime. Prove that\[1^i+2^i+...+(p-1)^i\equiv 0(\bmod p)\]for all $i,0\leq i\leq p-2$

[edited]

Perhaps the right question is:
$1^i+2^i+...+(p-1)^i\equiv 0(\bmod p)$ ?
[hide]Classical. For example by induction using $\sum_{k=0}^{p-1} \left( (k+1)^{i+2}-k^{i+2} \right) \equiv0[p]$[/hide]


limit
calculus
trigonometry
logarithms
calculus computations

The limit (as x approaches 0) of  $ \frac {ln(1 \minus{} x)sin(x)}{1 \minus{} cos^2x}$

In the movie the math competition director was wrong because he called
this a limit of an equation.

More to the point:

The girl (who went first) opposite Lindsay Lohan said it was "negative one,"
but she was told that was incorrect by the head of the competition.

Then Lindsay Lohan's character said the answer was "does not exist,"
and she was told that answer was correct.

On the Internet Movie Database site under goofs, that response from Lohan
is listed as a  "factual error."
It states that the answer is "-+ infinity," and that the answer is not
"does not exist."

When I worked it out, I changed it to the

limit (as approaches 0) of $ \frac {ln(1 \minus{} x)sin(x)}{sin^2(x)}$.

Then I applied L'Hopital's Rule to the fraction twice to get $ \minus{}1$.

Making a point:  The portion limit (as x approaches) of $ ln(1 \minus{} x) \equal{} 0$.

Also, when I graphed the original expression on a calculator on the
interval $ [\minus{}1, 1)$, for example, it passes through the point $ (0, \minus{}1)$.

I feel that the first girl (character in the math competion) was/is correct
about her answer of $ \minus{}1$.  Also, it was a sudden death question deciding
which school's team won.

Which is the answer to this movie math problem, for sure?

Well, the limit that you describe does approach $ \minus{}1$ as $ x\rightarrow 0$.

Investigating further, though, I found that the limit you describe is not the limit in the movie. I looked at a clip titled "Math in Mean Girls" on youtube and found the limit on the chalkboard to be
\[ \lim_{x\rightarrow 0}\frac{ln(1\minus{}x)\minus{}sin(x))}{1\minus{}cos^{2}x}\]

In fact, this limit is undefined...

If the limit is as mismatchtea wrote, "does not exist" is correct, by the usual standards of a calculus class. $ \lim_{x\to 0^ \plus{} }\frac {\ln(1 \minus{} x) \minus{} \sin x}{1 \minus{} \cos^2 x} \equal{} \minus{} \infty$ and $ \lim_{x\to 0^ \minus{} }\frac {\ln(1 \minus{} x) \minus{} \sin x}{1 \minus{} \cos^2 x} \equal{} \plus{} \infty$; combining the two sides leaves us with no possible limit even in the extended reals.
Of course, either multiplication or addition in the numerator makes for a more interesting question.

[quote="mismatchtea"]Well, the limit that you describe does approach $ \minus{} 1$ as $ x\rightarrow 0$.

Investigating further, though, I found that the limit you describe is not the limit in the movie. I looked at a clip titled "Math in Mean Girls" on youtube and found the limit on the chalkboard to be
\[ \lim_{x\rightarrow 0}\frac {ln(1 \minus{} x) \minus{} sin(x))}{1 \minus{} cos^{2}x}\]
In fact, this limit is undefined...[/quote]

I looked again.  I have it on videotape.  Up there in the numerator it shows as a
multiplication dot between the logarithm and the sine, and this character
(the dot) is visually shorter than the subtraction signs seen elsewhere in the
expression.  And there are no parentheses around the x-variable where the
sine and the cosine are concerned in the problem in the movie.

On screen, I see it as $ \lim_{x\rightarrow 0}\frac {ln(1 \minus{} x)\cdot sin x}{1 \minus{} cos^{2}x}$

It doesn't look like a multiplication dot to me...

http://pokit.etf.ba/upload/?pokit0b229b5ea84cc5f79dcaf2984efb7e45.jpg

hello, but this limit
$ \lim_{x \to 0}\frac{\ln(1\minus{}x)\sin(x)}{1\minus{}\cos(x)^2}$
does exist, it is $ \minus{}1$.
Sonnhard.

yes, of course it is -1, but they were talking about
$ \lim_{x\to 0}\frac{\ln(1\minus{}x)\minus{}\sin(x)}{1\minus{}\cos(x)^{2}}$
(only in the first post they were talking about $ \lim_{x\to 0}\frac{\ln(1\minus{}x)\sin(x)}{1\minus{}\cos(x)^{2}}$ , but then they figured out that it is a minus, not a mulitplication dot (is that the correct english expression for this sign?^^)

on youtube it also looks like a minus

[url]http://www.youtube.com/watch?v=QIMSC-RWvF8[/url] (7:42 and 8:34)


In a plane, what is the locus of points 8 centimeters from a given line?

(A)two parallel lines             (C)a line
(B)two intersecting lines      (D) a circle

My initial intuition would think to say that they are two parallel lines.

Think about it.. the locus of points 8 units away from a given line.  If it were a point, then it'd be a circle (the locus would be).  But the points 8 units away from a line would be 8 on each side of the line and they'd be parallel to each other because they're parallel to the same line.

$\boxed{A}$


Here is the book about number theory by Naoki Sato, maybe he is nsato admin of mathlinks !

I know it, nice but it isn't a text book on number theory ( rank Olympiad).


10 points are plotted on a circle. How many different convex polygons of 3+ sides can be drawn using a number of the points as vertices?

[hide="probably wrong"]
$ \binom{10}{3}\plus{}\binom{10}{4}\plus{}\cdots\plus{}\binom{10}{10}$

[/hide]

Actually, that might be right. Specifically, the answer is 1024-56=968.

Correct, xpmath. I'll check yours later, Circle. Or maybe I won't.

This was an 1989 AIME problem 2.

Circle's is right. His expression is equal to $ (1 \plus{} 1)^{10} \minus{} \left(\binom{10}{0} \plus{} \binom{10}{1} \plus{} \binom{10}{2}\right) \equal{} 1024 \minus{} 56 \equal{} 968$.


[quote="tz"]A spiral staircase turns 270 degrees as it rises 10 feet. The radius of the staircase is 3 feet. What is the number of feet in the length of the handrail?[/quote]
[hide]
Um..I'm going to make a half random guess...
$15\pi$?[/hide]

didn't we have this problem here before? oh well, I'm going to answer.

[hide]I think you have to roll out the cylinder's outside face to form a rectangle. The top edge of the rectangle will be 3*2*pi*(270/360) which is 4.5 pi. The other edge is 10 feet. we just find the diagonal of the rectangle.
**gets out calculator.**
calculator says it's 17.316451400967217568331088115519[/hide]

[hide]Baa i got 12pi with a crazy calculation

anyone care to post a solution[/hide][/hide]


induction
combinatorics proposed
combinatorics

We are given three machines which act on pairs of naturals as follows: the first one gets $(a+1,b+1)$ when given $(a,b)$, the second one gets $(\frac a2,\frac b2)$ when given $(a,b),\ a,b\in 2\mathbb N$ and nothing if $a,b$ are not both even, and the third one gets $(a,c)$ when given $(a,b),\ (b,c)$. Assume that after you have a new pair obtained by a machine, you also have the old pair. Initially, we have $(5,19)$. Can we get

a) $(1,50)$?
b) $(1,100)$?
c) If we have $(a,b)$ ($a<b$) and we want to get $(1,n)$, for which $n$ is this possible?

a) Yes, see c).

b) No.
Suppose that we start with $(a,b)$ where $b>a$ and that $p$ is an odd integer which divides $b-a$.
Now assume that after some steps (possibly none) we only have pairs $(x,y)$ where $p$ divides $y-x$. Then, it is easy to verify that using any machine once again we only obtain a pair $(x,y)$ such that $p$ divides $y-x$ too.
It easily follows by induction that after any finite number of steps we only obtain pairs $(x,y)$ such that $p$ divides $y-x$.
Thus, since $7$ divides $19-5$ and not $100-1$, we deduce that we cannot reach $(1,100)$ from $(5,19)$.

c) Let $b-a = k \cdot 2^t$ where $k$ is odd.
We will prove that $(1,n)$ can be reached if and only if $n-1$ is divisible by $k$.

From b), we easily see that if $(1,n)$ can be reached then $n$ has to be divisible by $k$.

Conversely, assume that $n=1+kp$ for some positive integer $p$.
We will prove that we may obtain $(1,n)$ by induction on $a$.

First, assume that $a=1$, so that $b=1+k$.
Note that only using the first machine, we may obtain $(1+k,1+2k),(1+2k,1+3k), \ldots, (1+k(p-1), 1+kp)$ from $(1,1+k)$.
Then using the third machine, we may obtain $(1,1+kp) = (1,n)$.

Now, let $a \geq 1$ be given. Assume that for all $x \in \{1,2, \cdots, a \}$, we may obtain $(1,n)$ from $(x,x+k)$.
Then, we start from $(a+1,a+1+k)$.
Using machines 1 and 3, we may obtain $(a+1+k,a+1+2k)$ and $(a+1,a+1+2k)$.
- If $a+1$ is even, using machine 2, we may obtain $(\frac {a+1} 2 , \frac {a+1} 2 + k)$ and $\frac {a+1} 2 \leq a$ since $a \geq 1$.
- If $a+1$ is odd, then $a \geq 2$. We first use machine 3 to get $(a+2,a+2+2k)$. Then, as above, we obtain $(\frac {a+2} 2 , \frac {a+2} 2 + k)$ and $\frac {a+2} 2 \leq a$ since $a \geq 2$.

In each case, using the induction hypothesis, we may obtain $(1,n)$.
This ends the induction step and the proof.

Pierre.


calculus
integration
function
algebra
domain
limit
algebra unsolved

A problem that very much resembles [url=https://artofproblemsolving.com/community/c6h4556]the "masterpiece"[/url].

Let $f(z)$ and $g(z)$ be integral functions (i.e., analytic at all finite points of the complex plane $\mathbb{C}$) such that for every fixed positive integer $n,$ the function $\frac{f(z)^{n}-1}{g(z)^{n}-1}$ is integral as well.
Prove that there exist positive integer $k$ such that $f(z) = g(z)^{k}.$

I have two corrections to the statement above. First: $g(z)$ must not be constant (consider $g(z)=0$). Second: the existence of POSITIVE integer $k$ does not generally hold ($f(z)=e^{kz}$, $g(z)=e^{z}$ give examples for ANY integer $k$).

So it needs to be proven that, for entire functions $f(z)$ and (nonconstant) $g(z)$ satisfying the condition in the post above, there is an integer $k$ such that $f(z)=g(z)^{k}$. My proof is based on the following lemma:

Let $U(z)$ be a function analytic on some (open) domain $D$ having a common point with the circle $|z| = 1$. Suppose that, for any $z \in D$ and any positive integer $n$, $z^{n}= 1$ implies $U(z)^{n}= 1$. Then $U(z) = C z^{k}$ for some integer $k$ and some constant $C$.

(Proof.) Let's denote $E(z)=e^{2zi\pi}$ for brevity. Now take any $z=E(p/q)$ in $D$ (with $q>1$ and $p$ coprime to $q$); we must have $U(z)=E(r/q)$ for some integer $r$. Moreover, for all sufficiently large integers $n$ there must exist integers $a_{n}$ such that $U(E(p/q+1/q^{n})) = E(r/q+a_{n}/q^{n})$; we may assume (using the continuity of $U$) that $\lim_{n\to\infty}(a_{n}/q^{n}) = 0$. Now compute $U'(z)$:
\[U'(z)=\lim_{n\to\infty}\frac{E(r/q+a_{n}/q^{n})-E(r/q)}{E(p/q+1/q^{n})-E(p/q)}=E((r-p)/q)\lim_{n\to\infty}a_{n}\frac{(E(a_{n}/q^{n})-1)/(a_{n}/q^{n})}{(E(1/q^{n})-1)/(1/q^{n})}=E((r-p)/q)\lim_{n\to\infty}a_{n}. \]
Thus $a_{n}$ has a limit $a$, and therefore $a_{n}= a$ for all sufficiently large $n$ (recall that $a_{n}$ are integers). So the equation
\[U(z) = C z^{a}\qquad(C = E((r-ap)/q)) \]
holds for $z=E(p/q)$ and for $z=E(p/q+1/q^{n})$ for all sufficiently large integers $n$. The lemma follows by applying the [url=http://mathworld.wolfram.com/PrincipleofPermanence.html]principle of permanence[/url].

Coming back to the solution; applying, for instance, [url=http://mathworld.wolfram.com/PicardsLittleTheorem.html]Picard's theorem[/url] to $g(z)$, we have a point $t$ such that $|g(t)|=1$ and $g'(t) \neq 0$. By the [url=http://mathworld.wolfram.com/InverseFunctionTheorem.html]inverse function theorem[/url] there is an open neighborhood $O$ of $t$ such that $g$ has an inverse function $h$ analytic on $g(O)$. As for any positive integer $n$ and any $z \in O$, $g(z)^{n}= 1$ implies $f(z)^{n}= 1$,  the function $U(z) = f(h(z))$ meets the conditions of the above lemma (with $D=g(O)$), so $f(z) = C g(z)^{k}$ for $C$, $k$ from the lemma and for any $z \in O$ (hence for any complex $z$). Now $g(z)^{n}= 1$ implies $C^{n}= 1$ (for any positive integer $n$ and any complex $z$); applying Picard's theorem once again gives $C^{n}= 1$ for any positive integer $n$ except at most one. This means that $C = 1$.