You are in a room with 10 people, all of whom have perfect senses of logic. Then, someone announces that they will be auctioning off a dollar. But, the person who comes in second will also have to pay what they bid. What will happen?

What about this scenario:

Someone is auctioning off a dolar. Everyone will (secretly) write down their bids. The winner gets the dollar, but everyone has to pay the amount they bid (ignore ties or something).

The first question to ask is: what happens when two bid the same¿

Did anyone ever try to sell a dollar bill on ebay?

yep [url]http://coins.search.ebay.com/dollar_Paper-Money-US_W0QQsacatZ3412[/url]

but they seem to be like "antique" "collectable" dollars.  not normal ones.

My friend bought a twenty dollar bill for 13 dollars.

"Can I just have 7 dollars transfered directly to my Paypal account?"

[quote="B-tone_Party"]I read about this once- more often than not, the 2 bidders will first bid to win the dollar, then, after the bidding has reached a dollar, they bid to minimize their losses (Would you rather pay a dollar and not get anything, or pay a dollar and 2 cents, and win the dollar?)  The bidding can supposedly get up to 3 or 4 dollars.[/quote]

Technically, because of this and the fact that each person has a perfect sense of logic, no one will bid.

Or 1 person will bid 1 cent and take the dollar.

this reminds me of that one problem where you buy a magic thing that gives you everything you want, but you have to sell it to someone for less than you bought it, or else you lose all the stuff and your life stinks.  and you have to sell it for a positive integer number of cents  :roll:

the question is, how much would you pay for it?  you don't want to pay 1 cent for it, because then you can't sell it.  but you don't want to pay 2 cents for it, because no one will buy it from you for 1 cent.  etc.

Once i auctionate a 50 pesos bill because I desperately needed change. I sold it on 41.50


calculus
derivative
function
algebra
domain
calculus computations

What is the difference between a global maximum and a local maximum?

and

Name two types of critical numbers and how they differ.

Thanks in advance  :lol:

The critical numbers are, according to the standard definition used by most math texts, either the solutions of the equation $f'(x)=0$ or the points where the first derivative of a function does not exists.
Ex: $f(x)=\frac{1}{x}$, then $f'(x)=\frac{-1}{x^{2}}$, hence the only critical point is 0 since at 0 the derivative is not well defined.
Now, if $f(x)=x^{2}-3x+2$ then $3/2$ is a critical number i.e. $f'(3/2)=0$.

Global max is the maximum value of a given function on a specific interval, however the local maximum is the maximum value of f near a  point (of course the point is a critical point), not on the whole interval.
Does it make any sense?

[quote]Ex: $f(x)=\frac{1}{x},$ then $f'(x)=\frac{-1}{x^{2}},$ hence the only critical point is 0 [/quote]
No, zero is not a critical point for the function; instead, it is a point not in the domain of the function. I would restrict use of the word "critical point" to something that actually is a point - hence to something in the domain of the function.

$f(x)=|x|,$ at $x=0,$ would be an example of a critical point where a derivative doesn't exist.


He means are we doing ladder (Denote a(n) as the nth place finisher, a(12) competes against a(11) and the winner is w(1). Denote w(n) as the winner when a(n) first competes. a(n) competes against w(n) n>1)

Or are we doing bracket (I am at a loss to express it mathematically, but it just pitches random people against each other, so that places 1 through 4 go less than the other places).

And yes, that is a good question: which one are you doing? Also, which are they doing this year at nats? Probably bracket...

[quote="jb05"]Also, which are they doing this year at nats? Probably bracket...[/quote]

I think it's ladder at Nats, but with my tendency to be wrong today, it is probably bracket. :(

Actually I think it is a bracket like last year with the top 12 written round.  Also, besides whether ladder or bracket for this, how does it work, how do we do it?

I like ladder better.
In 6th grade at chapter I beat 4 8th graders in a row but was eliminated
 before facing one of my team mates.

I have never done a ladder but it sounds cool

Bracket includes the top 12 instead of 10, so i'm for that ;)

Gives us a tiny bit better chance at nats...
1/22.4 : 1/19
.....

Well [i]some[/i] people don't randomly get their rank...

I used to like ladder (when I went to chapter and state and always got top two), but in the rare event that I make the countdown, I think I'll be hoping for bracket.

The thing is (unless you have byes, or double elimination) you have to have 1,2,4,8,16,32,or 64(etc.) people to do a bracket.  Because if you have 12 people, that would cut down to 6, then 3.  But with 3 people left, you could no longer continue the bracket (unless you give a bye to the highest rank, or do double elimination.  I can't see MC giving byes, but if it is a bracket, it will be with double elimination or byes.

Mathcounts divides 12 into 8 and 4. It brings the 8 down to 4, and then does an ordinary round robin tournament (I think; correct me if I'm wrong). How is this going to work tomorrow?

[quote="mysmartmouth"][quote="jb05"]Mathcounts divides 12 into 8[/quote]

I believe you, but how does that work?[/quote]

They take ranks 5-12, have 5 v 6, 7 v 8, etc... eliminate those 8 to 4, then do it until they have 1.

Actually, it's 8v9 7v10 6v11 5v12, and then against the other four

Like this:

this took 2 min for me to make.

I think it will be like that again this year.

You can read the ladder format rules and the bracket format rules on the [url=http://www.mathcounts.org/webarticles/articlefiles/510-C&SGenInstruct.pdf]MATHCOUNTS Web Site[/url].


cananyone explain what is a median of a trapezoid ?

let $ s$ be the median than

$ s \equal{} \frac {a \plus{} c}{2}$

this is a quote from AoPSWiki:

[quote]The median of a trapezoid is defined as the line connecting the midpoints of the two legs. Its length is the arithmetic mean of that of the two bases $ \dfrac{b_1 \plus{} b_2}{2}$. It is also parallel to the two bases.[/quote]


function
parameterization
induction
algebra
functional equation
strong induction
algebra unsolved

Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that \[ f(m+n)f(m-n) = f(m^2)  \] for $m,n \in \mathbb{N}$.

I like this - you really just need to calculate...

The functional equation $f\left(m-n\right)f\left(m+n\right)=f\left(m^2\right)$ allows us to recurrently compute all values of the function f in terms of the first three values, which we denote by f(1) = a, f(2) = b and f(3) = c. Note that these numbers a, b, c are naturals, so we can divide by them (this is important, since we will often do this in future). Then,

$f\left(4\right)=f\left(2^2\right)=f\left(2-1\right)f\left(2+1\right)=f\left(1\right)f\left(3\right)=ac$.

Now,

$af\left(5\right)=f\left(1\right)f\left(5\right)=f\left(3-2\right)f\left(3+2\right)=f\left(3^2\right)$
$=f\left(3-1\right)f\left(3+1\right)=f\left(2\right)f\left(4\right)=b\cdot ac$.

Division by a yields f(5) = bc. Furthermore,

$bf\left(6\right)=f\left(2\right)f\left(6\right)=f\left(4-2\right)f\left(4+2\right)=f\left(4^2\right)$
$=f\left(4-1\right)f\left(4+1\right)=f\left(3\right)f\left(5\right)=c\cdot bc=bc^2$.

Division by b yields $f\left(6\right)=c^2$. Continuing the game,

$cf\left(7\right)=f\left(3\right)f\left(7\right)=f\left(5-2\right)f\left(5+2\right)=f\left(5^2\right)$
$=f\left(5-1\right)f\left(5+1\right)=f\left(4\right)f\left(6\right)=ac\cdot c^2=ac^3$.

Division by c yields $f\left(7\right)=ac^2$. Hence,

$a^2c^2=a\cdot ac^2=f\left(1\right)f\left(7\right)=f\left(4-3\right)f\left(4+3\right)=f\left(4^2\right)$
$=f\left(4-1\right)f\left(4+1\right)=f\left(3\right)f\left(5\right)=c\cdot bc=bc^2$.

Division by $c^2$ yields $a^2=b$. Thus, f(2) = b becomes $f\left(2\right)=a^2$, and f(5) = bc becomes $f\left(5\right)=a^2c$.

So we have kicked out b from the parameters a, b, c - our first progress. Let's continue the computing spree:

$ac\cdot f\left(8\right)=f\left(4\right)f\left(8\right)=f\left(6-2\right)f\left(6+2\right)=f\left(6^2\right)$
$=f\left(6-1\right)f\left(6+1\right)=f\left(5\right)f\left(7\right)=a^2c\cdot ac^2=a^3c^3$.

Division by ac yields $f\left(8\right)=a^2c^2$. Thus,

$a^4c^2=a^2\cdot a^2c^2=f\left(2\right)f\left(8\right)=f\left(5-3\right)f\left(5+3\right)=f\left(5^2\right)$
$=f\left(5-1\right)f\left(5+1\right)=f\left(4\right)f\left(6\right)=ac\cdot c^2=ac^3$.

Division by $ac^2$ transforms this into $a^3=c$. Thus, the variable c is unnecessary and we are left with a only. Let's first combine:

$f\left(1\right)=a$;
$f\left(2\right)=a^2$;
$f\left(3\right)=c=a^3$;
$f\left(4\right)=ac=aa^3=a^4$;
$f\left(5\right)=a^2c=a^2a^3=a^5$;
$f\left(6\right)=c^2=\left(a^3\right)^2=a^6$;
$f\left(7\right)=ac^2=a\left(a^3\right)^2=a^7$;
$f\left(8\right)=a^2c^2=a^2\left(a^3\right)^2=a^8$.

Reminds you of something? Well, then you might be disappointed in a few moments. We note that

$af\left(9\right)=f\left(1\right)f\left(9\right)=f\left(5-4\right)f\left(5+4\right)=f\left(5^2\right)$
$=f\left(5-1\right)f\left(5+1\right)=f\left(4\right)f\left(6\right)=a^4\cdot a^6=a^{10}$,

so that, upon division by a, we get $f\left(9\right)=a^9$. But this leads to

$a^9=f\left(9\right)=f\left(3^2\right)=f\left(3-1\right)f\left(3+1\right)=f\left(2\right)f\left(4\right)=a^2a^4=a^6$,

and thus to $a^3=1$, so that a = 1. How boring! Anyway, the problem is now almost solved. We will show that f(n) = 1 holds for every natural n. This will be proven by induction over n. In fact, for n = 1, n = 2, n = 3, n = 4 and n = 5, this follows from the above formulas f(1) = a, $f\left(2\right)=a^2$, $f\left(3\right)=a^3$, $f\left(4\right)=a^4$ and $f\left(5\right)=a^5$ combined with the killer equation a = 1. Now, assume that N is a natural number $\geq 5$, and the identity f(n) = 1 holds for all naturals n < N. Then, particularly, f(N - 1) = 1, f(N - 3) = 1 and f(N - 4) = 1. Thus,

$f\left(N\right)=1\cdot f\left(N\right)=f\left(N-4\right)f\left(N\right)=f\left(\left(N-2\right)-2\right)f\left(\left(N-2\right)+2\right)$
$=f\left(\left(N-2\right)^2\right)=f\left(\left(N-2\right)-1\right)f\left(\left(N-2\right)+1\right)=f\left(N-3\right)f\left(N-1\right)$
$=1\cdot 1$,

so that f(N) = 1, and the induction step is done. Thus, f(n) = 1 holds for every natural n.

So the only solution of our functional equation is the function f(x) = 1. It is indeed a solution, as a reader could easily verify.

  Darij

$0 \not\in \mathbb{N}$ as long as nothing else is said.

why do you exlude 0

 it's a positive and negative integer

In most parts of the world $0$ is considered neither positive nor negative, but thats not important here (did anybody say anything about that¿).
$0 \not\in \mathbb{N}$ for me and a lot of other peoples, and it's the problem of the problem-poster when it's not well defined what he wants (and there are good reasons here that $0$ is excluded, like that then also $f(0)$ is defined and the problem gets a at least not harder).

Yes 0 is not natural number in Bulgaria, too :D

And I think that this problem is too easy to be in TST

Here is my approach:
By induction we obtain that :$f(m+2)=cf(m)$ for all $m\geq 3$ where $c$ is constant.So :
$F(2n)=c^{n-1}f(2)$ and $f(2n+1)=c^nf(1)$
This one is really classic !

[hide]Let $ g_p(n) \equal{} v_p(f(n))$ for all primes $ p$, where $ v_p(n)$ is the largest integer $ k$ such that $ p^k | n$. Our recurrence relation becomes $ g_p(m\plus{}n) \plus{} g_p(m \minus{} n) \equal{} g_p(m^2)$, and $ g_p$ is a function from the naturals to the nonnegative integers. We claim that $ g_p(n) \equal{} 0$ for all nonnegative $ n$.

Let $ g_p(1) \equal{} a_1$, $ g_p(2) \equal{} a_2, g_p(3) \equal{} a_3$. Note that if $ m > 2$, then $ g_p(m\plus{}1) \plus{} g_p(m\minus{}1) \equal{} g_p(m \plus{} 2) \plus{} g_p(m \minus{} 2) \equal{} g_p(m^2)$, so $ g_p(m\plus{}2) \equal{} g_p(m\plus{}1) \plus{} g_p(m\minus{}1) \minus{} g_p(m\minus{}2)$, so for all $ m > 4$, $ g_p(m) \equal{} g_p(m \minus{} 1) \plus{} g_p(m \minus{} 3) \minus{} g_p(m \minus{} 4)$.

Note that $ g_p(2^2) \equal{} g_p(2 \plus{} 1) \plus{} g_p(2 \minus{} 1) \equal{} a_3 \plus{} a_1$, so $ g_p(4) \equal{} a_3 \plus{} a_1$.  With this recurrence, we can easily compute that $ g_p(5) \equal{} a_2 \plus{} a_3$, $ g_p(6) \equal{} 2a_3$, $ g_p(7) \equal{} a_1 \plus{} 2a_3$, $ g_p(8) \equal{} a_2 \plus{} 2a_3$, and $ g_p(0) \equal{} 3a_3$. But $ g_p(3^2) \equal{} g_p(3 \plus{} 2) \plus{} g_p(3 \minus{} 2) \equal{} a_1 \plus{} a_2 \plus{} a_3$, so we see that $ 2a_3 \equal{} a_1 \plus{} a_2$. In addition, $ g_p(6^2) \equal{} g_p(6 \minus{} 1) \plus{} g_p(6 \plus{} 1) \equal{} g_p(6 \minus{} 3) \plus{} g_p(6 \plus{} 3)$, that is, $ (a_2 \plus{} a_3) \plus{} (a_1 \plus{} 2a_3) \equal{} a_3 \plus{} 3a_3$. Rearrange this to get $ a_1 \plus{} a_2 \equal{} a_3$. But have just found that $ 2a_3 \equal{} a_1 \plus{} a_2$, so $ a_1 \equal{} a_2 \equal{} a_3 \equal{} 0$. In other words, $ g_p(n) \equal{} 0$ for all positive integers $ n$.

It follows trivially that $ f(n) \equal{} 1$ for all nonnegative integers. [/hide]

[hide=Flawed Solution]Let $P(m, n)$ be the assertion $f(m + n)f(m - n) = f(m^2).$ Let $a = f(1).$

[b]Lemma.[/b] $f(4k + 1) = a$ and $f(4k + 3) = 1$ for all $k \in \mathbb{N}_0.$
[b]Proof:[/b] We go by induction. For the base case, $k = 0$, note that $f(1) = a$ trivially and $P(2, 1) \implies f(3) = 1.$ Now, for the inductive step, suppose that $f(4k - 3) = a$ and $f(4k - 1) = 1$ for $k \ge 1.$ Then by the inductive hypothesis, $P(4k , 1) \implies f(4k + 1) = a$ and $P(4k + 2, 1) \implies f(4k + 3) = 1$, completing the induction. $\blacksquare$
______________________________________________________________________________________________________________________________________________________
Back to the problem at hand, we will show that $a = 1.$ By applying the lemma, we make the following calculations:
\begin{align*}
P(3, 2) &\implies f(5)f(1) = f(4) \implies a^2 = f(4) \\
P(3, 1) &\implies a^2f(2) = a \implies af(2) = 1.
\end{align*}
Hence, $a \mid 1 \implies a = 1.$ From here, it follows quickly that $f(k) = 1$ for all $k \in \mathbb{N}.$ Indeed, choose an odd positive integer $m > k$, and let $n = m - k.$ Then as $m^2 \equiv 1 \pmod{4}$, the lemma implies that $f(m^2) = a = 1.$ Hence, $P(m, n) \implies f(2m - k)f(k) = 1$, whence it follows that $f(k) \mid 1.$ Therefore, $f(k) = 1$ for all $k \in \mathbb{N}$, as desired. It is immediate to check that this solution is indeed valid. $\square$[/hide]

EDIT: I mixed up $m$ and $n.$

Could someone check if my solution is correct? Thanks in advance! :D
[hide = "Solution"]We start by proving the following lemma:
[b]Lemma.[/b] $f(3) = 1$
[hide = "Proof of Lemma"]From our relation, we have:
1. $f(1)f(3)=f(4)$
2. $f(2)f(4)=f(9)$
3. $f(1)f(5)=f(9)$
4. $f(6)f(2)=f(7)f(1)$
5. $f(6)f(2)=f(5)f(3)$
6. $f(7)f(3)=f(6)f(4)$
7. $f(6)f(4)=f(1)f(9)$
8. $f(8)f(4)=f(5)f(7)$
9. $f(4)f(6)=f(2)f(8)$

From 2 and 3, we have $f(2)f(4)=f(1)f(5)$ so $f(5)=\frac{f(2)f(4)}{f(1)}=f(2)f(3)$

From 5, we have $f(6)f(2)=f(5)f(3)=f(2)f^2(3)$ so $f(6)=f^2(3)$.

From 6, $f(7)f(3)=f(6)f(4)$ so $\frac{f(7)}{f(6)}=\frac{f(4)}{f(3)}=f(1)$.

From 4, $f(7)f(1)=f(6)f(2)$ so $\frac{f(7)}{f(6)}=\frac{f(2)}{f(1)}=f(1)$. Thus, $f(2)=f^2(1)$.

From 7, $f(6)f(4)=f(1)f(9)=f(1)f(2)f(4)=f^3(1)f(4)$ so $f(6)=f^3(1)$.

From 2 and 3, $f(1)f(5)=f(2)f(4)=f^2(1)f(4)$ so $f(5)=f(1)f(4)$.

From 9, $f(6)f(4)=f(2)f(8)$ so $f^3(1)f(4)=f^2(1)f(8)$. Thus, $f(8)=f(1)f(4)=f(5)$.

From 8, $f(8)f(4)=f(5)f(7)=f(8)f(7)$ so $f(4)=f(7)$.

From 6, $f(7)f(3)=f(6)f(4)$ so $f(3)=f(6)=f^2(3)$. Thus, $f(3)=1.  \blacksquare$.[/hide]

Since $f(6)=f^2(3)=f^3(1)$ and $f^2(1)=f(2)$, we have $f(1)=f(2)=1$. Now, we will show, by strong induction, $f(n)=1 \forall n \in \mathbb{N}$ with $n \geq 3$.

For the base case, consider $n=3$. By the above lemma, $f(3)=1$ so our base case satisfies what we are trying to show!

For the strong inductive step, suppose $\exists k \in \mathbb{N}$ with $k \geq 3$ such that $f(n)=1$ for $n=1,2,\cdots,k$.

For the strong induction, consider $n=k+1$.
[u]Case 1:[/u] $k$ is odd
Then, we have $f(k)f(3)=f(k+1)f(2)$. However, we already know $f(2)=f(3)=f(k)=1$, implying $f(k+1)=1$.
[u]Case 2:[/u] $k$ is even
Then, we have $f(k)f(2)=f(k+1)f(1)$. However, we already know $f(1)=f(2)=f(k)=1$, implying $f(k+1)=1$.

Hence, regardless of the parity of $k$, we have $f(k+1)=1$, completing the induction.

Therefore, $f(x)=1  \forall x \in \mathbb{N}$ is the only solution.[/hide]

Note that for positive $x$ and $d$ such that all terms are positive, we have
\begin{align*}
f(x-3d)f(x-d)f(x+d)f(x+3d)=(f(x-3d)f(x-d))(f(x+d)f(x+3d))&=(f(x-3d)f(x+d))(f(x-d)f(x+3d)) \\
\implies f((x-2d)^2)f((x+2d)^2)&=f((x-d)^2)f((x+d)^2) \\
\implies f(x^2+4d^2)&=f(x^2+d^2)
\end{align*}
and
\begin{align*}
f(x-5d)f(x-d)f(x+d)f(x+5d)=(f(x-5d)f(x-d))(f(x+d)f(x+5d))&=(f(x-5d)f(x+d))(f(x-d)f(x+5d)) \\
\implies f((x-3d)^2)f((x+3d)^2)&=f((x-2d)^2)f((x+2d)^2) \\
\implies f(x^2+9d^2)&=f(x^2+4d^2)
\end{align*}

In particular, this means that $f(17)=f(20)$ and $f(45)=f(40)$, by setting $(x,d)$ to be $(4,1)$ and $(6, 1)$, respectively. Now $f(17)=f(20)\implies f(16)=f(19)$ because $f(18^2)=f(17)f(19)=f(20)f(16)$.

This actually implies for all $x$, we know $f(x)=f(x+3)$ because depending on whether $x$ is odd or even, we can multiply both sides by $f(16)$ or $f(17)$ to get an equality we know is true.

Similarly, from $f(45)=f(40)$, we prove that $\forall x, f(x)=f(x+5)$. Now, since $3$ and $5$ are relatively prime, this implies that $\forall x, y\in\mathbb{N}, f(x)=f(y)=k$, for some positive integer $k$. Then, the original equation becomes $k^2=k$, which implies $k=1$.

Thus, we've concluded that the only solution is $f(x)=1$.


Nesbit's inequality after substitution $x=a^2,y=b^2,z=c^2$ ;)


Let $a,b,c,d$ be four non-negative numbers such that $a^2+b^2+c^2+d^2=1$, find the minimum of
\[ f(a,b,c,d)= 5^{ab}+5^{bc}+5^{cd}+5^{da}.  \]

It seems very hard. I'm trying to solve your problem!

Thats trivial , because $\forall x\geq 0,\ 5^x \geq 1$.  So $f(a,b,c,d) \geq 4=f(1,0,0,0)$
 :cool:

I don't think so .because he said that $a,b,c,d > 0$ ;) .

$A = f(a,b,c,d)$

$A = 5^{ab+bc+ca+dc} ???$ :?

for me,i don't [u]think[/u] that it is possible to find the $MIN(f (a,b,c,d))$exception if $a,b,c,d \geq 0$,in this case the $MIN(f (a,b,c,d))$ is 1  :huh: .

[quote="toetoe"]I don't think so .because he said that $a,b,c,d > 0$ ;) [/quote]
You are wrong, he said non-negative numbers :D

 [quote="toetoe"]... exception if $a,b,c,d \geq 0$,in this case the $MIN(f (a,b,c,d))$ is 1  .[/quote]
Are you sur ?

 :cool:

for the $MIN(f(a,b,c,d))$ i am sure it will be (in the second

case ):rotfl: ,it was just a mistake in pressing the '4'.


Let $ C$ be a point on a fixed circle whose diameter is $ AB \equal{} 2R$ ($ C$ is different from $ A$ and $ B$). The incircle of $ ABC$ touches $ AB$ and $ AC$ at $ M$ and $ N$, respectively Find the maximum value of the length of $ MN$ when $ C$ moves on the given fixed circle.

This is a problem on M&Y in Vietnam. Pleases lock it!


The question goes as
Consider a circle of radius $ 5cm$ with centre $ O$. Let a chord $ BC$ be drawn whose length is $ 6cm$.
Let a point $ A$ be marked on the circumference of the circle nearer to $ B$ such that there exists [b]only one chord[/b] through $ A$ (say $ AD$) such that $ AD$ is bisected by $ BC$.
Let $ \angle{AOB}$ be $ \theta$ then $ sin\theta$ is $ \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}$

Let's call $ { E } AD \cap BC$. Since $ E$ is unique ( a single chord through A), it follows that $ BC$ is tangent at $ E$ to the circle of diameter $ OA$. But we know that $ D$ is the middle of the $ arc BC$; with $ E$ being the middle of $ AD$ we get $ OE \perp AD$, hence $ E$ belongs to the circle diameter $ OD$. Further, it seems to be a piece of cake ( not for diabetics, of course).

Best regards,
sunken rock

[quote="sunken rock"] But we know that $ D$ is the middle of the $ arc BC$; with $ E$ being the middle of $ AD$ we get $ OE \perp AD$, hence $ E$ belongs to the circle diameter $ OD$.[/quote]
Hey
How do you tell that $ D$ is middle of $ arc{AB}$??   :maybe:

I can't quite understand the question.
I don't think unique $ AD$ exists


inequalities
logarithms
function
inequalities unsolved

$a,b,c>0,\ abc=1$. For $n\geq1$ prove $\sum\frac{a^{3n}-1}{a^n}\geq n\sum\frac{a^3-1}a.$

ok,a nice problem.
I have draw the Graphics.
it tells me jensen does work :)
so I think maybe you will have a nice solution :)

Which graph did you draw? The one of $f(x)=\frac{x^{3n}-1}{x^n}-n\frac{x^3-1}x$?
Because for applying Jensen IMO we would need to first use logs to get an additive constraint,
Like putting $x=\ln a$ etc in the original ineq, so the condition becomes $x+y+z=0$. But then the function needed for Jensen would be $f(x)=\frac{e^{3nx}-1}{e^{nx}}-n\frac{e^{3x}-1}{e^x}$ and that is not convex, it behaves roughly like $x^3$. ???

I drawed $f(x)=\frac{e^{3nx}-1}{e^{nx}}-n\frac{e^{3x}-1}{e^x}$
and didn't you notice what I said?
I mean it is not convex,either

OK, finally I got it. If it is not all convex, there is still the Right-Convex Function Theorem ([url]http://www.mathlinks.ro/Forum/viewtopic.php?p=226013#p226013[/url]). And it works, as $f(x)=\frac{e^{3nx}-1}{e^{nx}}-n\frac{e^{3x}-1}{e^x}$ is convex for $x\geq0$. We can use it with three variables and $s=0$ as soon as we can show $2f(x)+f(-2x)\geq0$ for all $x\geq0$. Writing again $a$ for $e^x$ (for convenience of notation), this is equivalent to
$\frac{(2a^n-1)(a^{3n}-1)}{a^{2n}}+n\frac{a^3-1}{a^4}\geq0$, which is obviously true as $a\geq1$. :) :)

I found a result to this idea due to Titu Andreescu, also.
Let $m,n$  positive integers. Prove that  $\frac{x^{mn}-1}{m}\geq\frac{x^n-1}{x}$

for any positive real number $x$ .


Planet Ypsilon has a calendar similar to ours: A year consists of $365$ days, and every month has $28$, $30$ or $31$ days.
Prove that on Planet Ypsilon, a year must have $12$ months.

[hide]

i ono if theres a more general way to do it, so i just took it in cases

when there are 0 months with 28 days, we know there must be 5 months with 31 days (units digit of 365 is 5), and thus:

$365 - 5(31) = 210 = 7(30)$

so there will be 5+7 = 12 months

when there is 1 month with 28 days, we know there must be 7 months with 31 days (again units digit of 365 is 5), and we have:

$365 - 28 - 7(31) = 120 = 4(30)$

so there will be 1+7+4 = 12 months

similarly, when there are 2 months with 28 days, we will have 9 months with 31 days, giving us:

$365 - 2(28) - 9(31) = 1(30)$

so there will be 2+9+1 = 12 months

finally, we know the rest of the cases will be impossible, because:

$3(28) + 9(31) = 363 < 365$, and thus we will never be able to reach 365 days again after this case

[/hide]

[hide]If the year has at most 11 months, it has at most 31*11=341 days, not enough.
If the year has at least 13 months, the least number of days will be 28*13=364. Trying to increase the number of days by adding more months or making a month at least 30 days will create at least 366 days, too much.[/hide]


In an attempt to find the concentration of iron, which is the main component of the mineral hematite ($Fe_{2}O_{3}$), a chemist takes $1.25$ g of the mineral, dissolve it in $2$ mol/L, $0.025$ L of $H_{2}SO_{4}$, and put in zinc to reduce all into $Fe^{2+}$. Then the leftover zinc is taken out, and the remaining solution is added $H_{2}SO_{4}$, and titrate with $0.1$ mol/L of $KMnO_{4}$. At the equivalence point, it is found that $0.02$ g of $KMnO_{4}$ has been used. Find the percentage of iron in hematite. Given the half-reactions involved:
$MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$
$Fe^{2+}\rightarrow Fe^{3+}+e^{-}$

[quote="Cancer"]At the equivalence point[/quote]

You mean, "at the final point".

Well, let's start by writing the titration reaction:

$5Fe^{2+}+MnO_{4}^{-}+8H^{+}\longrightarrow 5Fe^{3+}+Mn^{2+}+4H_{2}O$.

Given the molar mass of potassium permanganate, 158.03 g/mol, then 0.02g of it correspond to $1.3 \times 10^{-4}$ mol. Then, the number of moles of $Fe^{2+}$ that reacted must be five times that: $n(Fe^{2+}) = 6.5 \times 10^{-4}$ mol, which corresponds to a mass of 0.036 g. But then the percentage of iron in hematite would be $\frac{0.036g}{1.25 g}\times 100 = 2.9per cent$, which is a too low percentage. So perhaps I am forgeting something, or some of your data is incorrect.

OK, it should be $0.02$ L instead of $0.02$ g

But what I don't understand is why is the equation $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$ used, when in the problem the $3+$ is converted to the $2+$...

In that case the number of moles of potassium permanganate that reacted were 0.002 mol, and so the number of moles of $Fe^{2+}$ that reacted were 0.01 mol, which correspond to a mass of 0.56 g and a percentage of iron of 45%, which still seems to be low but at least more reasonable.

[quote="Cancer"]But what I don't understand is why is the equation $Fe^{2+}\longrightarrow Fe^{3+}+e^{-}$ used, when in the problem the $3+$ is converted to the $2+$...[/quote]

The $Fe^{3+}$ ions present in the sample (and possibly other higher oxidation states) are previously converted to a reduced state because you are going to use a redox titration using an oxidising agent (potassium permanganate). Potassium permanganate is commonly used in redox titrations for two reasons: is a good oxidising agent, and its not necessary to use an indicator because the pink $MnO_{4}^{-}$ ion have a very low minimum detectable concentration.


linear algebra
matrix
vector
induction
combinatorics proposed
combinatorics

In a regular $n$-gon each side is colored blue or yellow.From such a coloring a new coloring is obtained in one step as follow:if the two neighbors of a side have different colors,the new color of that side will be blue,otherwise it will be yellow(the colors are modified simultaneously).Show that after finitely many steps all side will be color yellow.

This isn't true in general, for all $n$, if you start with any coloring whatsoever. Think about an equilateral triangle ($n=3$) having two blue sides and one yellow side.

I believe there is a proof on the forum that this is true (you end up with an all-yellow polygon no matter how you begin) iff $n$ is a power of $2$. Try searching in the 'Combinatorics' section (it won't be an easy task; the problem I'm referring to was posted quite a while ago, but I do remember the thread was started by orl).

Yes,grobber :)  :) .I conjectured this problem after i solve for $n=2^{k}(k \in n^{*}\geq 2)$ and all $4 \leq n \leq 20$,so i think the general may be true.
Have you already found a counter-example of $n \geq 4$ for which the conclusion doesn't take place? :blush:

Take $n=5$, for instance. We'll call our pentagon $ABCDE$, and let's say we color the vertices instead of the edges. It's easier to work that way. Start with $B,E$ blue, and the rest of the vertices yellow. After three iterations you're back where you started.

So I guess you didn't really prove it for $4\le n\le 20$, did you? :)

Edit:

Yes, Jose, it was a typo. I apologize.

I tried something with Linear Algebra, but it didn't work out.

Let $0 = \textrm{yellow}, \, 1 = \textrm{blue}$ and $A$ be the matrix having: $\left( 0, 1, 0, \ldots, 0, 1 \right)$ on the first line and cyclic permutations of that (one step to the right, then one more step to the right) on the next rows. Then $Ax$, where $x$ is the current state vector, is the state of the polygon after we apply our transformation.

$A$ should look like this $\left( \begin{matrix}0 & 1 & 0 & \ldots & 0 & 1 \\ 1 & 0 & 1 & \ldots & 0 & 0 \\ 0 & 1 & 0 & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & 0 & 1 \\ 1 & 0 & 0 & \ldots & 1 & 0 \end{matrix}\right)$.

The fact $A^{m}x = \left( \begin{matrix}0 \\ 0 \\ \vdots \\ 0 \end{matrix}\right)$ for all $x \in \mathbb F_{2}^{n}$, where $m$ is dependent of $x$, is equivalent to $A$ being nilpotent.

I computed $\det \left( A+\lambda I_{n}\right)$ with Mathematica for some small values. The results are as grobber predicted.

e.g.:
$n=7$: $P_{A}= \lambda+\lambda^{5}+\lambda^{7}$ - bad, has roots other than $0$;
$n=8$: $P_{A}= \lambda^{8}$ - good;
$n=9$: $P_{A}= \lambda+\lambda^{5}+\lambda^{7}+\lambda^{9}$ - bad, has roots other than $0$;
etc.

But I can't find a proof. I tried computing circulant determinants (in $\mathbb C$, then reducing to $\mathbb F_{2}$), but I kind of failed. :(

Is the idea any good?

PS:

[quote="José"][quote="grobber"]We'll call our pentagon $ABCDEF$[/quote]

Typo grobber[/quote]

No s**t, Sherlock! Viva la spamolucion. All must bow before you :D

How about something like the following:?

Working modulo 2, we can label the sides as $(x_{1}, \dots x_{n})$ where $x_{i}$ is 0 if side $i$ is yellow, 1 otherwise.  The recurrence we are given is effectively

$x_{i}(t+1)=x_{i-1}(t)+x_{i+1}(t)$, where addition in the subscripts is modulo $n$.

Instead of following this recurrence directly, however, let us next shift the vector over by 1 unit to the left.  This additional shift won't affect whether all sides eventually become yellow or not, but it allows us to think of the recurrence relation as

$x_{i}(t+1)=x_{i}(t)+x_{i+2}(t)$.

You can now show via induction on $k$ that $x_{i}(t+2^{k-1})=x_{i}(t)+x_{i+2^{k}}(t)$.

If $n=2^{k}$, this immediately gives that all the sides are yellow at $t=2^{k-1}$.

If on the other hand $n=2^{k}*j$, $j$ odd, we can use this to reduce from an $n-$gon to a $j-$gon, so I guess it's enough to just give a counterexample for any odd $j$.

I found the thread I was referring to:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1073[/url]

As you can see, it's not exactly a complete solution, just an outline.

Actually, I think the case $j$ odd isn't that bad either.

The only way we can reach all yellow is if in the prior step everything was the same.

The only way everything can be the same at time $t+1$ is if either

-Everything was the same at time $t$

or

-Each side is a different color than the side two after it at time $t$.  But this is impossible, since the polygon has an odd number of sides (We can "go right two edges" an odd number of times to get from an edge back to itself, so we can't have changed colors every time).

This implies that for an odd number of sides the only way we could ever have reached the all yellow state is if all sides started the same color in the first place.


Prove $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ for all $a,b,c$

This is known as Nesbitts inequality.  If you want an good proof  try a Google search on "[b]Nesbitts inequality" proof[/b]. Perhaps you will find a Ph. D's work.

Heh, i got this from Engel's book. :) n I guess it's a pretty popular one.

This is one of the nicer applications of rearrangement.

Let $a\geq b\geq c$, and so $\frac {1}{b+c}\geq \frac {1}{a+c}\geq \frac {1}{a+b}$.

Now,

$\sum \frac {a}{b+c}\geq \frac {1}{2}\sum\left( \frac {b}{b+c}+\frac {c}{b+c}\right)=\frac {3}{2}$

Jensen's: (this is making up for my mistake in the other thread)
It's homogeneous, so let $a+b+c=1$.  Then by the CONVEXITY of $\frac {1}{x}$, we get that:

$\sum \frac {a}{b+c}\geq \frac {1}{a(b+c)+b(c+a)+c(a+b)}$

But $2/3(a+b+c)^2\geq 2(ab+bc+ca)$, and the result follows.

Engel gave a good 10,000,000 proofs of this inequality.

Another way to apply Jensen's (from MO Treasures):

Let $s=a+b+c$. Consider the convexity of the function $f(x)=\frac{x}{s-x}$ on [0,s).

Then using equal weights, $f(a)+f(b)+f(c) \ge 3f(\frac{a+b+c}{3}) = 3f(\frac{s}{3}) = 3\cdot\frac{\frac{s}{3}}{s-\frac{s}{3}} = \frac{3}{2}$.

Well, I don't have the book yet, but here is the one I came up with:

\[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}\]

We rewrite this as

\[\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3 \geq \frac{3}{2}\]
\[(a+b+c)(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})-3  \geq \frac{3}{2}\]
\[\frac{1}{2} ( (a+b)+(b+c)+(a+c) ) (\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})-3  \geq \frac{3}{2}\]

Now let $x=a+b$, $y=b+c$, and $z=a+c$ to obtain
\[\frac{1}{2}(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})-3  \geq \frac{3}{2}\]
\[\frac{(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})-6}{2}  \geq \frac{3}{2}\]
\[\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}+3-6 \geq 3\]
\[{\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}}-3 \geq 3\]
\[{\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}} \geq 6\]

Since for all $m$ and $n$, $\frac{m}{n}+\frac{n}{m} \geq 2$ $($Note: $m^2+n^2 \geq 2mn  )$ the last inequality is clearly true, thus we are done.

[quote="DPopov"]
\[\frac{1}{2}(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})-3  \geq \frac{3}{2}\]
[/quote]

More simply, $\left(x+y+z\right)\left(\frac {1}{x}+\frac {1}{y}+\frac {1}{z}\right)\geq 9$ holds by Cauchy and by AM-HM

:(  I'm still working to learn to apply those

[quote="Peter VDD"][quote="paladin8"]There are two proofs of this in the Math Olympiad Treasures book, one by Cauchy-Schwarz and one by Jensen's. Both are pretty cool.[/quote]What is that Math Olympiad Treasures book you're talking about? :) I've been looking for a good treasure book for a while. What math level is it aimed at?[/quote]

It's called "Math Olympiad Treasures" by Titu Andreescu and Bogdan Enescu. It's level is probably like USAMO.

Yes, DPopov. The first statement is AM-HM. The second is also direct Cauchy-Schwarz.

LHS is homogeneous then you cam take $a+b+c=1$
It is easy to prove that
$\frac{a}{b+c}=\frac{a}{1-a}\geq \frac{9a-1}{4}$ then you can easily can get the result

It's worth noting that that last inequality is just a statement about convex functions; that is, a convex function is essentially a function which lies above its tangent at any point (the RHS is just the linear approximation at x=1/3)

This message is for DPopov: there is an extra term in your proof-step in the last 3 lines of your reasoning in the post #8. There should be 6 terms in it, not 7 terms


What are all the awards? I know that bronze is something like 60-74, silver like 75-95, and gold 96+(which, btw, I'm NOT going to get  :) ) But what about HM and the awards for getting 5's on certain problems?

I think the honorable mention range is from 40-59, but I could be wrong.

Bronze, silver, and gold all get key chains and I think gold and silver can choose two books or items (I'm not sure on how many so you have to check them out for sure). But I do know that gold choose items first then silver then bronze then HM. HM does NOT get any keychain. As for other people, I think you just get t-shirt. I'm not sure for that either. I just know that HM gets one shirt and one item because I was HM last year (could've done better but then time is something I do NOT have.. haha..).

http://www.usamts.org/About/U_AbPrizes.php The prize ranges are all correct. :D

So even Honorable Mention gets a book? That's cool :D . What kind of books are they?

Last year I chose 360 Problems for Mathematical Contests and Challenging Problems in Geometry as a Silver, but they also have more biographical or fun books if that's what you like. If I remember correctly, there were around 7 choices or so, not just books, either.

What kind of book is left for the honorable mentions? I think that  I'm getting that this year.

[quote="SplashD"][quote="13375P34K43V312"]Ugh...well I guess after my somewhat pitiful round 1, I'll be able to scrape out a bronze if I get around the same on the other rounds...[/quote]
Wait, did you already get your score?[/quote]
Probably an estimate. :D


quadratics
number theory
number theory unsolved

Could anyone help me on the following:

Suppose we have an equation:

y^2 = [ x^3 + 1 ]  mod m

Now I want to solve for x,
[y^2 - 1]^1/3  =  x   mod m

Here comes my question : Why is the above similar to:

[y^2 - 1] ^ (2m-1)/3 = x mod m

Or more explicitly: here is anothe equation where I want to find the quadratic residue

y^2 = a  mod m
This is equivalent to

[y^2]^ (2m-1) = a mod m
________________________

I could not think of any relationship between the above and Fermal Theorem. Perhaps there isn't...

Thanks

[quote="t0rajir0u"]Neither of those manipulations are valid unless $ m$ is prime; when they are valid, they follow from Fermat's Little Theorem, although they do not really tell you what the solutions are.[/quote]

Thanks alot that will do


logarithms
limit
integration
calculus
LaTeX
real analysis
real analysis unsolved

Let $s_n = \sum_{k=n}^{\infty}\frac{\ln k}{k^3}$. For which values of $a$ does $\sum_{n=1}^{\infty}s_n^a$ converge?

I find $s_n = \sum_{k=n}^{\infty}\frac{\ln k}{k^3}=O(\frac{\ln n}{n^2})$
is it useful?

Yes, from the comparison test you get that $a > 1/2$ immediately from that (if it is indeed true). You can compare $s_n^a$ with $\left(\frac{\ln n}{n}\right)^{2a}$ to get that  $\sum_{n=1}^{\infty}s_n^a$ is convergent if  $\sum_{n=1}^{\infty}\left(\frac{\ln n}{n}\right)^{2a}$  is convergent, which happens for $a > 1/2$. How did you find it?

edit: I've edited this post many times now. Maybe it is not as simple as I thought it would be at first. The current idea doesn't show what happens for $a \le 1/2$. But yeah, your findings must definitely help. Is it possible that you could say something directly about the limit of $\frac{n^2}{\ln n}\sum_{k=n}^{\infty}\frac{\ln k}{k^3}$ ?

Try the condensation test.

$\sum \left(\frac k {2^{3k}\right)^}{\alpha}$ is convergent iff $\sum \left(\frac{k^{\alpha} \ln 2^{\alpha}}{2^{(3\alpha-1)k}}\right)$ which is convergent for $\alpha > \frac 1 3$ use: $\sum_k \frac{k^t}{m^k}$ with $m>1$ is convergent; and diverget if $\alpha \le \frac 1 3$    use :$\sum_k k^t\cdot m^k$ with $m>1$ is divergent.

[quote="Kalle"] Is it possible that you could say something directly about the limit of $\frac{n^2}{\ln n}\sum_{k=n}^{\infty}\frac{\ln k}{k^3}$ ?[/quote]

Yes. it is $\infty$ Maybe you want the limit: $\frac{\ln n}{n^2}\sum_{k=n}^{\infty}\frac{\ln k}{k^3}$.  By Cesaro stolz method the limit is equal with:

$\lim_{n\to \infty } \frac{\frac{\ln n+1 }{(n+1)^3}} {\frac{(n+1)^2}{\ln n+1}-\frac{ n^2}{ \ln n}}=$
$\lim_{n \to \infty } \frac{\frac{(\ln n+1)^2 \ln n} {(n+1)^3}}{ \ln(n+1)n^2- \ln n (n+1)^2}=$
$\lim_{n\to \infty} \frac{\frac{(\ln n+1)^2 \ln n} {(n+1)^3}}{\ln n (2n+1)-ln(1+\frac 1 n) n^2}$
Dividing bpth numerator an denominator by $n$ we get that limit is $\frac 0 {\infty-1}=0$.

Corect me if I'm wrong.

I can't understand why this equation is true.
$\lim_{n \to \infty } \frac{\frac{(\ln n+1)^2 \ln n} {(n+1)^3}}{ \ln(n+1)n^2- \ln n (n+1)^2}=$
$\lim_{n\to \infty} \frac{\frac{(\ln n+1)^2 \ln n} {(n+1)^3}}{\ln n (2n+1)-ln(1+\frac 1 n) n^2}$?

Any way here my solution for:$s_n = \sum_{k=n}^{\infty}\frac{\ln k}{k^3}=O(\frac{\ln n}{n^2})$
notice $\frac{\ln x}{x^3}$ was reduce when x large enough.
So we get:
$\int_{n}^{+\infty}\frac{\ln x}{x^3} dx \le \sum_{k=n}^{\infty}\frac{\ln k}{k^3} \le \int_{n-1}^{+\infty}\frac{\ln x}{x^3} dx$
and then :)
maybe I am wrong,I am not sure.

Complete and correct solution:

For large enough n:
$\int_{n}^{\infty}\frac{\ln x}{x^3} dx \le \sum_{k=n}^{\infty}\frac{\ln k}{k^3} \le \int_{n-1}^{\infty}\frac{\ln x}{x^3} dx$

From primitively calculating the integrals it follows that
$\frac{n^2}{\ln n}\cdot\frac{2\ln(n) + 1 }{4n^2} \le \frac{n^2}{\ln n}\sum_{k=n}^{\infty}\frac{\ln k}{k^3} \le \frac{n^2}{\ln n}\cdot\frac{2\ln(n-1) + 1 }{4(n-1)^2}$

Taking limits, we find by the squeeze theorem that $\lim_{n\to\infty}\frac{n^2}{\ln n}\sum_{k=n}^{\infty}\frac{\ln k}{k^3} = \frac12$

Now, let $b_n = \frac{\ln n}{n^2}$.

$b_n^a$ and $s_n^a$ are both positive sequences and we have that that $\frac{s_n^a}{b_n^a} \to \frac{1}{2^a} > 0$ as $n\to\infty$. By the comparison test $\sum s_n^a$ is convergent iff $\sum b_n^a$ is convergent. The latter is convergent precisely when $a > \frac12$.

edit: You make so many mistakes when you type LaTeX on these forums :( I don't understand any of your posts, xirti  :P

I forgot a "-" sign :D

\[ \ln (n+1)n^2 -\ln n(n+1)^2=[\ln (n+1)-\ln n]n^2 -(\ln n)( 2n+1)= \]
\[ \ln (\frac{n+1}{n})n^2-(\ln n)(2 n+1)=\ln (1+\frac{1}{n})n^2-(\ln n)(2 n+1) \]

the limit remains $0$ however.

your method works even for $\sum_{k=n}^{\infty} \left(\frac{\ln k}{k^3}\right)^{\alpha}$. I dare to state that that $s_n = \sum_{k=n}^{\infty}\frac{\ln k}{k^3}=o(\frac{\ln n}{n^2})$. That's what I proved.


the following incredible formula was discovered by Srinivasa Ramanujan:
\[ \frac{1}{\pi}\equal{}
\sum_{n\equal{}0}^{\infty}{2n\choose n}^3\cdot\frac{42n\plus{}5}{2^{12n\plus{}4}}\]

have anyone seen the proof of this ??

if yes, please tell me where can I find it :)

There's another formula similar to that one that also yields $ \frac{1}{\pi}$.  It is somewhere on his wikipedia page.

Yes.  Unless that is just the same thing written differently.

By fastest, do you mean fastest converging?

[quote=ghaprameya]Do you Know that This is the fastest algorithm ever discovered to calculated the values of pie..[/quote]

The Chudnovsky brothers formula $$\frac{426880 \sqrt{10005}}{\pi} = \sum_{k=0}^\infty \frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}$$ is actually faster but as you can see it is based on Ramanujan's formula.


probability
symmetry
AMC
AIME
number theory
relatively prime

[hide="Solution"]The probability one rolls one's die for $ n$ times until the first six appears is
\[ \left(\frac56\right)^{n - 1} \cdot \frac16 = \frac {5^{n - 1}}{6^n}.
\]
Then the probaility that Linda will roll her die $ n$ times while Dave does for $ n - 1$, $ n$, or $ n + 1$ times (i.e. within one of the number of times Linda rolls her die) equals
\[ \frac {5^{n - 1}}{6^n} \left( \frac {5^{n - 2}}{6^{n - 1}} + \frac {5^{n - 1}}{6^n} + \frac {5^{n}}{6^{n + 1}} \right)
\]
for all $ n \ge 2$. For $ n = 1$, we must exclude the case when Dave rolls 0 times to obtain a probability of
\[ \frac16 \left( \frac16 + \frac56 \cdot \frac16 \right) = \frac {11}{216}.
\]
Now, summing for all natural numbers $ n$,
\begin{eqnarray*} & & \frac {11}{216} + \sum_{n = 2}^{\infty} \frac {5^{n - 1}}{6^n} \left( \frac {5^{n - 2}}{6^{n - 1}} + \frac {5^{n - 1}}{6^n} + \frac {5^{n}}{6^{n + 1}} \right) \\
& = & \frac {11}{216} + \sum_{n = 2}^{\infty} \frac {5^{n - 1}}{6^n} \left( (36 + 30 + 25) \cdot \frac {5^{n - 2}}{6^{n + 1}} \right) \\
& = & \frac {11}{216} + \sum_{n = 2}^{\infty} \frac {91 \cdot 5^{2n - 3}}{6^{2n + 1}} \\
& = & \frac {11}{216} + 91 \cdot \frac {5/6^5}{1 - 5^2/6^2} \\
& = & \frac {11}{216} + 91 \cdot \frac {5}{6^5} \cdot \frac {6^2}{11} \\
& = & \frac {11}{216} + \frac {455}{216 \cdot 11} \\
& = & \frac {121 + 455}{216 \cdot 11} \\
& = & \frac {8}{33} \\
& = & \frac {m}{n} \end{eqnarray*}
and therefore $ m + n = 8 + 33 = \boxed{41}$.[/hide]

[hide="Solution"]
Let $ p_n \equal{} \frac16\left(\frac56\right)^{n \minus{} 1}$ be the probability that it takes $ n$ rolls. Then
$ P \equal{} p_1(p_1 \plus{} p_2) \plus{} p_2(p_1 \plus{} p_2 \plus{} p_3) \plus{} p_3(p_2 \plus{} p_3 \plus{} p_4) \plus{} \cdots$
$ \equal{} (p_1^2 \plus{} p_2^2 \plus{} \cdots) \plus{} 2(p_1p_2 \plus{} p_2p_3 \plus{} \cdots)$
$ \equal{} \left(\left(\frac16\right)^2 \plus{} \left(\frac16\right)^2\left(\frac56\right)^2 \plus{} \left(\frac16\right)^2\left(\frac56\right)^4\right) \plus{} 2\left(\left(\frac16\right)^2\left(\frac56\right) \plus{} \left(\frac16\right)^2\left(\frac56\right)^3 \plus{} \left(\frac16\right)^2\left(\frac56\right)^5 \plus{} \cdots\right)$
$ \equal{} \left(\frac16\right)^2\left(\frac1{1 \minus{} \left(\frac56\right)^2} \plus{} 2\cdot\frac56\cdot\frac1{1 \minus{} \left(\frac56\right)^2}\right) \equal{} \left(\frac16\right)^2\left(1 \plus{} \frac53\right)\left(\frac {6^2}{11}\right)$
$ \equal{} \frac83\cdot\frac1{11} \equal{} \frac8{33}\implies8 \plus{} 33 \equal{} \boxed{41}$.
[/hide]

[hide="solution"]
Let $ p_0$ be the probability they end on same roll.
This is the chance they both roll 6, plus $ p_0$ times the chance they both roll non-6.
$ p_0 \equal{} \left(\frac {1}{6}\right)^2 \plus{} \left(\frac {5}{6}\right)^2 p_0$
$ p_0 \equal{} \frac {1}{11}$
Let $ p_D$ be probability that Dave ends 1 roll before Linda.
This is the chance of Dave rolling 6, Linda rolling a non-6, then Linda rolling a 6 the roll after, added to $ p_D$ times the chance both of them roll non-6.
$ p_D \equal{} \left(\frac {1}{6}\right)\left(\frac {5}{6}\right)\left(\frac {1}{6}\right) \plus{} \left(\frac {5}{6}\right)\left(\frac {5}{6}\right)p_D$
$ p_D \equal{} \frac {5}{66}$
The probability that Linda ends 1 roll before Dave is also equal to $ p_D$.
None of the three probabilities intersect (if one of them happens, the other two can't also happen). Thus the total probability is $ p_0 \plus{} p_D \plus{} p_D \equal{} p_0 \plus{} 2p_D \equal{} \frac {1}{11} \plus{} \frac {5}{33} \equal{} \frac {8}{33}$
[/hide]

[hide]
All the times when 6's are rolled at the same time can be written as ${ (\frac {5^{n-1}}{6^n}})^2=\frac {5^{2n - 2}}{6^{2n}}$, all the times when they aren't rolled at the same time can be written as ${{ (\frac {5^{n-1}}{6^n}})(\frac {5^{n}}{6^n+1}})=\frac {5^{2n - 1}}{6^{2n+1}}$, but either person may roll a 6 first, so it needs to be multiplied by two.

$ \sum_{i = 1}^{\infty} \frac {5^{2n - 2}}{6^{2n}} +  2 \sum_{i = 2}^{\infty} \frac {5^{2n - 1}}{6^{2n+1}}$

$ \frac {\frac{5^0}{6^2}}{1-\frac{5^2}{6^2}}+2\frac {\frac{5^1}{6^3}}{1-\frac{5^2}{6^2}}$
$ \frac{\frac{1}{36}}{\frac{11}{36}}+2\frac {\frac{5^1}{6^3}}{\frac{11}{6^2}}$

$ \frac{1}{11}+2\frac {5}{66}$

Would have gotten it right if I had remembered that I needed to multiply by 2.

$ \frac{16}{66}=\frac{8}{33}$[/hide]

Wouldn't it be more accurate to name this thread "Rolling Sixes Simultaneous-ish-ly?"  :)

Surprisingly, I thought this was the easiest question on the test. I solved it in about 3-5 minutes (I failed #1 lol)

strange, i got to 11/216+455/216*11 and epic failed :(

Isn't there symmetry in this problem? 3 cases: 1) Dave has same # of rolls as Linda. 2) Linda has more rolls than Dave. 3) Dave has more rolls than Linda.

Number 1 has 1/11 chance of happening, and 2 and 3 have an equal chance of happening, since they both roll under the same conditions, so why can't you just divide 1-1/11 = 10/11 by 2?
As in, what's wrong with the logic there?

That's not what the question is asking for.

It's asking for same number of rolls or within. So, why couldn't you just add the chance of equal rolls (1/11), with the chance that Linda has more rolls?

Georgiamathdude, I initially interpreted the problem exactly the same way as you did--and I got exactly your answer.

I kept reading the problem over and over again until I finally figured out my blind spot.  I had overlooked the word "one" in the last line of the problem.  "Within [i]one [/i]of the number of times" does not mean the same as "Within the number of times."

Oh, i see. I saw it as within "any one of the rolls" Like, if Linda rolled a 6, it could be any ONE of 1, 2, 3, 4, 5, 6... alrite. thanks.

I agree that the wording of the problem lends itself to some ambiguity of interpretation.  The English phrase "one of the number of times" is somewhat ambiguous, depending on the context.

For example, consider the use of the phrase in the sentence "During [i]one of the number of times[/i] he visited me in the hospital, he read me poetry."  In that context, the phrase "one of the number of times" means something entirely different than it does in the AIME problem.   It means "one of the numerous instances of times." Since "during" and "within" are somewhat similar, I can see how you and I might have been led to parse the problem statement incorrectly.

It really comes down to how your mind chops up the sentence into phrases.  A careful reading indicates that the correct parsing should immediately connect together the word "one" inside the phrase "within one of" rather than inside the phrase "one of the number of times."

I certainly don't think there's any grounds for protest, as a careful reading makes the intended interpretation evident, but it would have been even clearer, in my opinion, to have worded the problem as follows:

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Call the total number of times that Dave rolls $ D$, and call the total number of times that Linda rolls $ L$.  Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac {m}{n}$ is the probability that  $ |D \minus{} L| \le 1$. Find $ m \plus{} n$.

Yea, I was going for 7 and i got the first 7 so when I "got" the eigth one, I was like whoa that too easy... nah don't go back I'm done.

[quote=worthawholebean]Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m\plus{}n$.[/quote]

Oh man this problem was so confusing... I thought that if Linda rolled her dice $k$ then Dave could roll $1,2,3,\cdots,k$ times :(

Read the problem carefully.

Grinding [url=https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZS9iL2FjOWU0OTVjZDU0ZWMyM2I2MDMxYjU1OWEzM2RmNzg0NjljYjQyLnBkZg==&rn=QWltZUNvdW50aW5nUHJvYmxlbXMucGRm]@hurdler's AIME combinatorics handout[/url]

[hide=solution]Consider three cases. Let the requested probability be $p$.

If both get a six on their first roll, game over. This occurs with probability $\displaystyle\frac{1}{36}$.

If one gets a six on the first roll and the other doesn't (which occurs with probability $2\cdot\displaystyle\frac{5}{6}\cdot\displaystyle\frac{1}{6}=\displaystyle\frac{5}{18}$), the one who didn't get a six initially will need to get a six on the second roll (otherwise the difference won't be $\leq 1$). This occurs with probability $\displaystyle\frac{1}{6}$, so the entire situation has probability $\displaystyle\frac{5}{18}\cdot\displaystyle\frac{1}{6}=\displaystyle\frac{5}{108}$.

If no one gets a six on the first roll (which occurs with probability $\displaystyle\frac{25}{36}$), then we're back to square one.

Therefore $\displaystyle\frac{1}{36}+\displaystyle\frac{5}{108}+\displaystyle\frac{25}{36}p=p$, so $\displaystyle\frac{2}{27}=\displaystyle\frac{11}{36}p$ and $p=\displaystyle\frac{72}{297}=\displaystyle\frac{8}{33}$. Answer $8+33=\boxed{041}$.[/hide]

We take three cases.

Case 1: Dave = Linda

The probability is $\frac{1}{6} \cdot \frac{1}{6} + \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} + \ldots$, a geometric sequence with common ratio $\frac{25}{36}$ and first term $\frac{1}{36}$. This gives a sum of $\frac{\frac{1}{36}}{1-\frac{25}{36}} = \frac{1}{11}$.

Case 2: Dave = 1 + Linda

The probability is $\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} + \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} + \ldots$, a geometric sequence with common ratio $\frac{25}{36}$ and first term $\frac{5}{216}$. This gives a sum of $\frac{\frac{5}{216}}{1-\frac{25}{36}} = \frac{5}{66}$.

Case 3: Dave = -1 + Linda

This would be the same as the previous case by symmetry, so another $\frac{5}{66}$.

The total is $\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{8}{33} \implies \boxed{41}$.


Three tennis balls stacked one on top of the other, fit snuggly in a cylindrical container. What is the ratio of the volume of the tennis balls to the volume of the space inside the container not occupied by the tennis balls?

[hide="Solution."]
If the radius of one tennis ball is $ r$, then clearly the radius of the base of the container is $ r$ and the height of the can is $ 6r$. Then the volume of the three balls is $ 3\cdot \left(\frac {4\pi r^3}{3}\right) \equal{} 4\pi r^3$. Also, the volume of the can is $ (\pi r^2)(6r) \equal{} 6\pi r^3$. Thus the desired ratio is:

$ \frac {4\pi r^3}{6\pi r^3 \minus{} 4\pi r^3} \equal{} \boxed{2: 1}$
[/hide]

it is also useful to know that a sphere's volume is $ \frac23$ of the cylinder with same radius and height equal to diameter.
stacking 3 of these cylinders with spheres inside them, you get the structure described
since the occupied space is $ \frac23$ and the unoccupied space is $ \frac13$, the ratio is $ 2: 1$


geometry
number theory proposed
number theory

Given an equilateral triangle and a point in its interior. There is such a triangle with all 3 distances and the length of the triangle side being integers?

Which distances are you talking about?

If these are the distances to the sides of the triangle, the answer is no. By an area argument, the sum of the three distances is the altitude of the triangle, which is $\sqrt 3/2$ times the side, so it's impossible the three distances and the side to be all integers.

If these are the distances to the vertices, I don't know the answer right now, but I can try it later.

I was rushing  :blush:  The question was....

Given an equilateral triangle and a point in its interior.
There is such a point with all 3 distances to the vertices and the length of the triangle side being integers?


quadratics
algebra
polynomial
number theory
modular arithmetic

Find all $ (a ; b)$ such that $ a ; b$ are [b]interger[/b] and:
$ \frac {a \plus{} b}{a^2 \minus{} ab \plus{} b^2} \equal{} \frac {3}{7}$
or can write $ 7a \plus{} 7b \equal{} 3a^2 \plus{} 3b^2 \minus{} 3ab$
I found that $ a \equal{} 4, b \equal{} 5$ or $ a \equal{} 5, b \equal{} 4$ by an ugly proof
Help the new member please  :(

if we work in $ \mathbb{Z}/3\mathbb{Z}$ and $ \mathbb{Z}/7\mathbb{Z}$ we find that:
$ a \equal{} 21k \plus{} 4$ and $ b \equal{} 21k \plus{} 5$
so we get:

$ \frac {a \plus{} b}{a^{2} \minus{} ab \plus{} b^{2}} \equal{} \frac {3(14k \plus{} 3)}{7(63k^{2} \plus{} 27k \plus{} 3)}$
===> $ 14k \plus{} 3 \equal{} 63k^{2} \plus{} 27k \plus{} 3$
===>$ k \equal{} 0$ then:
$ (a,b) \equal{} (4,5) or (5,4)$

[quote="mathematikos"]if we work in $ \mathbb{Z}/3\mathbb{Z}$ and $ \mathbb{Z}/7\mathbb{Z}$ we find that:
$ a \equal{} 21k \plus{} 4$ and $ b \equal{} 21k \plus{} 5$[/quote]  Working mod 3 and 7, the most you could possibly get is that $ a \equal{} 21k \plus{} 4$, $ b \equal{} 21j \plus{} 5$.  There is no way modular arithmetic will tell you they differ by exactly 1, rather than by 1 modulo 21.

The equation you get is quadratic in $ a$, so the discriminant $ (3b \plus{} 7)^2 \minus{} 4\cdot 3 \cdot (3b^2 \minus{} 7b)$ must be nonnegative.  This implies $ 27b^2 \minus{} 126b \minus{} 49 \leq 0$.  This quadratic polynomial has reasonably nice roots, one in the interval $ ( \minus{} 1, 0)$ and the other in the interval $ (5, 6)$.  So we only need to check integer values of $ b$ from 0 to 5 and see if they have matching $ a$-values.  This is only moderately ugly.


inequalities
function
inequalities solved

by cauchy,
[aaa/(b+c+d)+bbb/(c+d+a)+ccc/(d+a+b)+ddd/(a+b+c)][ab+ac+ad+bc+bd+ba+cd+ca+cb+da+db+dc]>=(aa+bb+cc+dd)^2
2(ab+ac+ad+bc+bd+cd)<=3(aa+bb+cc+dd) (am-gm)

aaa/(b+c+d)+bbb/(c+d+a)+ccc/(d+a+b)+ddd/(a+b+c)
>=aa+bb+cc+dd/3
>=(ab+bc+cd+da)/3
=1/3

Just an idea ... couldn't this be done using Jensen ?

I mean, is the function f(x) = x/(a - x) convex on [0, a[ for a > 0 ?

If so, put a + b + c + d = S > 0. Put f(x) = x/(S - x).
Then LHS = f(a) + f(b) + f(c) + f(d) \geq 4 f(S/4).
So it suffices to prove that 1/3 \leq 4 f(S/4) = S/12.
So it suffices to prove that S \geq 4. Now
S = a + b + c + d + 2(ab + bc + cd + da) + 2ac + 2bd
= (a + c) + (b + d) + 2 \geq ((a + c) + (b + d))/2 + 2 = S/2 + 2.
So S \geq S/2 + 2, so indeed S \geq 4.

But i'm not sure whether f(x) = x/(a - x) is convex.
I'll calculate f"(x). (Boring !!!!!!)

To be continued ...

I think f is indeed convex.
Could someone verify my computations - e.g. with software ?
You know, I always make lots of mistakes in boring computations ...

A lemma Maverick posted before works fine here: for real ai and positive xi a1^2/x1+..+an^2/xn >= (a1+a2+..+an)^2/(x1+..+xn). We amplify each fraction by a, b, c, d respectively and we have to prove

 \sum a^4/ab+ac+ad >=1/3. We apply the lemma and we are left with (a^2+b^2+c^2+d^2)^2/2(ab+ac+ad+bc+bd+cd) >= 1/3. (a^2+b^2+c^2+d^2)/2(ab+ac+ad+bc+bd+cd) >= 1/3 is well known and a^2+b^2+c^2+d^2 >= 1= ab+bc+cd+da by rearrangement inequality.


function
integration
logarithms
limit
calculus
calculus computations

[quote="FieryHydra"]Why does the question read "Find the minimum?"[/quote]

oh, because i found the function to be strictly increasing. hence, the minimum of it is as x approaches 1

[quote="violetcraze"]Let $F: (1,\infty) \to \R$ be the function defined by

$F(x) = \int^{x^{2}}_{x}\frac{1}{\ln{t}}\,dt$
find $F(x)$ as $x\to 1$[/quote]
Nice problem, and though I got hold of the correct technique the first time I attempted this problem, for some inexplicable reason I didn't finish it off from there. It was only after I confirmed with Mathematica there was a solution that I went back and solved the problem.

Anyways, let me solve a more general problem: Find $\lim_{x \to 1}\int_{x}^{x^{n}}\frac{\,dt}{\ln t}.$

Since the function $\frac{1}{\ln t}$ is not integrable in the elemenary sense, we must somehow determine if we can use Taylor series expansion and then integrate term by term. To achieve this, we make the substitution $y = \ln t.$

So, we have $\int_{x}^{x^{n}}\frac{\,dt}{\ln t}$

$= \int_{\ln x}^{n \ln x}\frac{e^{y}\,dy}{y}$

$= \int_{\ln x}^{n \ln x}\frac1{y}(1+\frac{y}{1!}+\frac{y^{2}}{2!}+\frac{y^{3}}{3!}+\ldots) \,dy$

$= \int_{\ln x}^{n \ln x}\frac{\,dy}{y}+\int_{\ln x}^{n \ln x}(1+\frac{y}{2!}+\frac{y^{2}}{3!}+\ldots)\,dy$

$= \ln y \Big|_{\ln x}^{n \ln x}+(y+\frac{y^{2}}{2\cdot 2!}+\frac{y^{3}}{3 \cdot 3!}+\ldots)\Big|_{\ln x}^{n \ln x}$

$\Rightarrow \boxed{\lim_{x \to 1}\int_{x}^{x^{n}}\frac{\,dt}{\ln t}= \ln n}.$


abstract algebra
number theory unsolved
number theory

Let p be a prime. In Zp(the group with addition module p), let A and B be subsets of Zp and |A|+|B|-1<=p. Prove that
        |A+B|<=|A|+|B|-1, where A+B={a+b|a in A and b in B}

(Note: I am not sure whether the problem is right, that is, maybe there are some counterexamples. But I strongly believe there aren't any and I've write a program to test the cases p<30. No counterexample is found)

It's called Cauchy-Davenport-Theorem:

[url]http://planetmath.org/encyclopedia/CauchyDavenportTheorem.html[/url]
[url]http://www.mathlinks.ro/Forum/topic-43784[/url]

Thank you.

How sad it has been proved before, I think it is the best problem I've ever found by myself ;) .


A wall to wall carpet cost $ 15 per square yard. Rita paid$330 to carpet a room 6 yards long. What is the width of the room?

How do I find the answer?

First, we find the number of square yards.

=330/15
=22 sq. yds.

Then we divide 22 by 6 to get the width.

=11/3 yards.

[hide="Solution"]
The area of the carpet can be $ \text{length} \times \text{width}$. Dividing $ 330$ by $ 15$ yields $ 22$, which is the area of the carpet. Plugging it into our equation, we get $ 6 \times w \equal{} 22$, or $ \text{width} \approx \boxed{3. \overline{6}}$.
[/hide]

EDIT: Wow someone beat me...


Try this problem I made up:

Bob and Rob play toss an unfair coin 3 times.  It has a $\frac{2}{3}$ chance of landing heads and $\frac{1}{3}$ tails.  If exactly 2 out of 3 times the coin lands heads, Bob gets a point.  If otherwise, Rob gets a point.  They play until they have a total of 3 points.  What is the probability that Bob has exactly 2 points?

[hide]After 3 coin flips, Bob has a $\frac{2}{3}\cdot \frac{2}{3} \cdot \frac{1}{3} \cdot {3} = \frac{4}{9}$ chance of getting a point.  So after 3 points, Bob has a $\frac{4}{9} \cdot \frac{4}{9} \cdot \frac{5}{9} \cdot {3} =  \frac{80}{243}$[/hide]

[hide]The probablility that there is exactly two heads out of three is $(\frac{2}{3})^2\cdot\frac{1}{3}\cdot3 = \frac{4}{9}$. That means not getting 2/3  is 5/9.

$(\frac{4}{9})^2\cdot\frac{5}{9}\cdot3=\frac{80}{243}$. [/hide]


algebra
polynomial
complex numbers
algebra unsolved

Find all real polinomials $P(x)$ that $P(x)P(x+1)=P(x^2)$,for every $x\in{R}$

[quote="vietnamesegauss89"]$P(x)=0 \forall x$or$P(x)=x^n(x-1)^n$for any natural number n[/quote]
Please,then sent solution,and not only answer :mad:  :) .

I just want to mention that this problem was proposed in 2000 in the last round of the Bulgarian Mathematical Olympiad as Problem 1, Day 2 :)

Let $\omega$ be any root of $P$. Then, evaluating $P(x)P(x+1)=P(x^2)$ in $\omega$, we see that the LHS is $0$, so the RHS is $0$, and then $\omega^2$ is a root of $P$.

Suppose that $P$ is not $0$; then it has finitely many roots. If it has a root $\omega$ of modulus not $0$ or $1$, then $\omega^2, \omega^4, \omega^8, \ldots$ are all roots of $P$ with different modulus; so $P$ has infinitely many roots, contradiction. Then $P$ has roots of modulus only $0$ or $1$.

However, let $\omega$ be a root of $P$. Then the LHS is $0$ for $x=\omega-1$ also. Then the RHS is $0$, so $(\omega-1)^2$ is a root of $P$. So it has modulus $0$ or $1$, then $\omega-1$ has modulus $0$ or $1$. This allows for two possibilities: either $\omega$ is $0$ or $1$, or both $\omega, \omega-1$ are on the unit circle. This means that $\omega$ has angle $60$ or $300$ in the complex plane. However, if $\omega$ has angle $60$, then $\omega^2$ is a root and has angle $120$ which is forbidden; if it has angle $60$, then $\omega^2$ has angle $240$, which is forbidden. So all roots are $0$ or $1$.

Let $P(x)=x^m(x-1)^n$. Then $P(x+1)=(x+1)^mx^n$ and $P(x^2)=x^{2m}(x^2-1)^n$. Since $P$ has $m$ zero roots, $P(x+1)$ has $n$ and $P(x^2)$ has $2m$, we have $m+n=2m$ so $m=n$. Also, $x^n(x-1)^n$ works.

Whoa that's hard...I worked like a functional one, studyng x(x+1) and subtraing x and using series I had to modify the polyomial to x^2-x=x(x-1) but it's a terrifing method ;)

Can I reply? Find al polynomials for which $P(x)P((x-1))=P(x^2)$. And $P(x-1)P(x+1)=P(x^2)$

This also appeared on an Irish Olympiad.

Well, I had just discovered this technique and was trying to fit it into any problem I could... incidentally, it works in these  ;)

$P(x)P(x-1)=P(x^2)$ is similar to the first one: if $\omega$ is a root, then so are $\omega^2$ and $(\omega+1)^2$. Because of the first, all roots have modulus $0$ or $1$. Besides, $\omega+1$ must have modulus $0$ or $1$ If $1$ is a root, then $(1+1)^2$ is a root, contradiction. If $0$ is a root, then $(0+1)^2$ is a root, contradiction. If $-1$ is a root, then $(-1)^2=1$ is a root, contradiction. So, for $\omega, \omega+1$ to have both modulus $0$ or $1$, $\omega$ must be a third root of unity. Since they come in conjugate pairs, either $P=0$ or $P=1$, or $P$ is divisible by $x^2+x+1=(x-\omega)(x-\omega^2)$. If the latter happens, we divide by $x^2+x+1$ and repeat the argument. Solutions are $P=0, 1, (x^2+x+1)^n$.
[Edited because I had messed up the casework.]

For $P(x+1)P(x-1)=P(x^2)$, we see that if $\omega$ is a root then $(w-1)^2, (w+1)^2$ are too. Now note that at least one of $\omega+1, \omega-1$ has modulus greater than $\omega$ (+1 if it has positive real part, -1 otherwise). If $\omega$ has modulus at least $1$, and $\omega+1$ has greater modulus (the other case is analogous), then $(\omega+1)^2$ has it even greater modulus; so we can construct infinitely many roots this way. So all roots are inside the unit circle. However, for any nonzero $\omega$ in the unit circle, at least one of $\omega+1, \omega-1$ is outside, so its square is outside too, contradiction. Then only $\omega=0$ is possible. But then $(\omega+-1)^2=1$ which is nonzero, contradiction. There are no solutions except constants, $P=0, 1$.

[hide="another way"]and what about divisibility? checking 0 and 1 in the polinomial you see that the known term is 0. So $x+1|P(x+1)$ and $x+1|P(x^2)$ and $x|P(x+1)$ (and $x(x-1)|P(x^2)-P(x)$[/hide]


probability
AMC
AIME
USA(J)MO
USAMO
trigonometry
AMC 10

[quote="bookaholic"]Or 8th.

7. Though it should have been a 9...sigh.[/quote]

I'm sorry, but it clearly says 9th or 10th. You will be prosecuted to the fullest extent of the law. You have the right to remain silent; anything you say may be used against you in court.

But I'm a possible-floor-qualifier!

Actually, I'm merely a possible-floor-qualifier-providing-that-
solafidefarms's-and-mine-and-all-the-other-7-pointers'-wishful-
thinking-somehow-pays-off. But that's okay.

Ah, probability, she never said she voted, did she now? :D (Not that I'd be offended if she did!)

I think they'll make us have some sort of special qualifying process, since way too many of us got 7s. Or am I grasping at straws? :)

[quote="probability1.01"]..anything you say may be used against you in court.[/quote]

I believe it's "in a court of law". *pats*

Sophomore.  9.  I'll probably make USAMO.

[quote]The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.[/quote]

If they have too many qualifiers by the floor value then they'll look at your AMC score.

I'm fairly sure that 160 number is going to be more like 220-250 given the expansion.

So there's a chance of a floor value of 6? :D

I'm a freshman, and I got an 8. I hope I qualify.

6. I really really really really really hope that the floor will be 6 :P. And good job to all of you :)

[quote="inyoung"]I'm a seventh grade girl and I got an 8.. Do you think I'd qualify?[/quote]
They don't make a distinction between genders, but I think an 8 will probably qualify. Lucky duck.

Actually now I think of it the floor of 7 is very likely.

Here is the 2005 AOPS AIME statistics:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2005+AIME&t=29581]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2005+AIME&t=29581[/url]

41% of AoPSers last year scored the floor value(9) or above,

Now look at this year's AIME statistics

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77873&start=160]www.artofproblemsolving.com/Forum/viewtopic.php?t=77873&start=160[/url]
36% of AoPSers this year scored 9 or above, 47% scored 8 or above. Now we can expect AoPS to have gotten a little better since last year, so without expansion a floor value of 8 seems likely. Now we take account of the expansion, a floor of 7 seems likely.

I hope this makes some sense :lol:

[quote="filletwho"]AOPS has expanded dramatically over the last year also, so we'll probably have more of the qualifiers.[/quote]

Yeah the poll for this year is probably not done yet(but we already have more voters than last year, meaning aops probably expanded).

Thank you beta for actually taking the time to analyze the stats!  :D

[quote="K81o7"]10th grade.
Got a 7.  Though it should have been 12...I should have been a bit careful and checked more thoroughly.  Yes, I lost [b]5[/b] points out of utter carelessness  :roll:[/quote] Yeah, me too. Kind of. I got a 10, but I thought I had a 13, and the two I didn't think I had gotten I got after the test in like 5 minutes each. My mistakes were as follows:

8) Lack of geometric intuition...square is only max if side length is constant, off by factor of $\sqrt2$.
10) Just another stupid error, 4 and 5 are only one apart, not two, so we only need to shift 1/2 over...
11) Never thought of recursion until after the test.
12) Tried to overdo it and go the really long way about it (i.e. substituting trig identities for cos(ax) for a=3,4,5), drew a bad diagram and ended up being off by 9 degrees.
14) Mixed up Stewart's Theorem, b is on the same side as n, m on the same side as c. Yeah, that was just stupid, because I got everything else in that problem.

I also noticed that the three I thought I had but didn't (8, 10, and 14) were all geometry. Maybe that says something... :roll:

Sorry to anyone that feels envious of my 10, it was just really frustrating not to get the problems I thought I had gotten and end up with the same score as last year.

[quote="dhuang614"]If the floor is 7, i might be screwed over because of my relatively low index (roughly 214). Any opinions, or speculations on the cutoff Index for 10th and under?[/quote]

I don't think there's any such thing as an underclassman index. We qualify on the floor, and you'll qualify if the floor's 7.

sophomore: 7
what do u think the floor will be?

It could be influenced by the AIME IIs, but right now it most likely will be an 8 (7's still possible though).

OKay since it's quite possible aops on general is decreasing in math ability, we'll take it from another approach.
We still refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29581]this thread[/url].

 In 2005, 76 people scored 9 or above in that poll, out of 378 people last year on AIME I scoring 9 or above, that's approximately 20%.

But the % of people scoring the floor value or above that are frequent AoPS visitors cannot possibly decrease, so this year we expect at least 20% of the people scoring the floor or above to be AoPSers. Since there's expansion we expect 570ish people scoring the floor or above. So at least 115ish people will come from AoPS frequent visitors(by that I mean people who vote in the poll). Now 92 people scored 9 or above this year on AoPS, 120 scored 8 or above. So floor will be 8(well, at most an 8)

Freshman
AMC10:146.5
AIMEI:7

I'm also hoping for a floor of 7.

9th grade.
130 AMC 12B.
10 AIME.
230 index.

[quote="mathclass"]i predict an 8! i'm a 9th grader, got 11.[/quote]

Actually I don't the think the floor can possibly be 40320...


function
Euler
algebra
polynomial
logarithms
linear algebra
calculus

The functions $y_{1}=x^{3}$ and $y_{2}=x^{4}$satisfies some  linear homogeneous differential eq. of second order.

a)Show that the functions are linearly independent.

b)Determine the equation.

please help me on this.

One way to determine whether two proposed solutions are linearly independent is to consider their quotient and if their quotient is a constant then they are linearly dependent while if it is not a constant then they are linearly independent.

Another way is to consider Wronskian's. If the Wronskian's is NEVER zero then they are linearly independent.

Once you have established linearly independence, then the general equation is given by $y = c_{1}y_{1}+c_{2}y_{2}$ ( for some constants $c_{1}$, $c_{2}$)

I wasn't sure if your part b was asking to find the original differential equation which is satisfied by the given solutions.

[quote="akech"]

I wasn't sure if your part b was asking to find the original differential equation which is satisfied by the given solutions.[/quote]

[color=red]Yes how do I find that one ? [/color](the original differential equation which is satisfied by the given solutions.)

Thanks a lot for your reply.

Note that the general differential equation in question looks like this:
$y''+py'+qy = 0$
Now the question reduces to finding $p$ and $q$
Plugging in our two solutions we get two equations:
$qx^{2}+3px+6 = 0$
$qx^{2}+4px+12 = 0$

By elimination we get $px+6 = 0 \Rightarrow p = \frac{-6}{x}$
$q = \frac{12}{x^{2}}$ by substitution?

The original equation reads:
$y''-\frac{6}{x}y'+\frac{12}{x^{2}}y = 0$

also if we determine the wronskian $W$ of $y,x^{3},x^{4}$ and set $W=0$ we will have the eq.. Am I right?

That is right.
The method you are suggesting is somehow related to solving $AX = b$ in linear algebra (determinant of A and Wronskian).

In this case, the Wronskian is zero at $0$, but nowhere else. What does that mean? It means that our differential equation must be singular at zero, and we can't match arbitrary initial conditions there.

In general, equations with the Euler form $x^{n}y^{(n)}+a_{n-1}x^{n-1}y^{(n-1)}+\cdots+a_{1}xy'+a_{0}y=0$ (each $a_{i}$ a constant) have solutions of the form $y=x^{r}$, where the $r$ are roots of an associated polynomial equation. If that equation has multiple roots, we also get solutions of the form $y=x^{r}\ln^{k}x$.
Given that knowledge, we can reach the answer akech gave by simply matching coefficients.


algebra
polynomial
search
inequalities
number theory
number theory unsolved

Please help me with this one:

Let $P\left(x\right)$ be a polynomial with complex coefficients, and of degree $n$. All its roots $z_1$, $z_2$, $z_3$, ..., $z_n$ are different and are such that all $i$ satisfy $\left|z_i\right|=1$.

Prove that there exists a complex number $z$ with $\left|z\right|=1$ such that $\left|P(z)\right| \geq 2$.

I couldn't solve it after a lot of thinking.  :?

It's perfectly equivalent to this one:

[url]http://www.mathlinks.ro/viewtopic.php?t=5774[/url]

We just represent the roots of $P$ on the unit circle. Did you notice that they were the same? Im asking because it seems strange that you're the author of both topics :).

Yeah ...I'm not so stupid :D  There(in the algebra section), Myth said that it has been discussed before, but I couldnt find it in the forum, I wanted to put it in the number theory section, so i changed it a little bit, so that it looks more Number theoretical :D

By the way does any one know the solution, or knows where this problem has been discussed in the forum?

Please HELP  :D

But didn't Myth (in the Algebra section) give a link to the topic where he solved the problem?

My internet connection is too poor. So I can't search forum for appropriate link. In particular, such searching is difficult due to "crooked" topic titles. But I suppose my post was last in desired topic, thus you can try to find all topic which end with my post.

If z is the n-th root of unity one easily shows
p(1)+p(z)+p(z^2)+..p(z^(n-1))=2n from here the desired inequality follows
 Note: Use that 1^i+z^i+(z^2)^i+..+(z^(n-1))^i=0 if n doesn't divide i, and n otherwise

Let's assume $P(x)=ax^2+bx+c$. In this case $P(1)+P(-1)=2a+2c$, which isn't necessarily equal to $2n=4$. I think that sum you wrote depends on the coefficients of the polynomial, iura. Am I misunderstanding something?

is the polynomial monic? else we could multiply our polynomial by a constant and it would still satisfy the conditions, so in order to be always greater than 2 somewhere on the unit circle, there had to be a pole there, but polynomials don't like to have a pole...

[quote="Omid Hatami"]First it was not IMO selection test just an ordinary test.
Also there is a stronger problem from India 1998.
You can see it in book Problems and .... 1998-1999[/quote]
What is more stronger problem?

hi iura ,
i don't understand your hint .. i can just prove that sup(|P(x)|) >= 1+1/n .. how do you achieve 2 with your hint ?

I have found my previous post on similar problem:
[url]http://www.mathlinks.ro/viewtopic.php?t=4532[/url]
Though readability of this post is quite low :), I think you will be able extract its idea.


calculus
complex numbers
imaginary numbers
AMC

Let $ p(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$, where $ a$, $ b$, and $ c$ are complex numbers.  Suppose that
\[ p(2009 \plus{} 9002\pi i) \equal{} p(2009) \equal{} p(9002) \equal{} 0
\]What is the number of nonreal zeros of $ x^{12} \plus{} ax^8 \plus{} bx^4 \plus{} c$?

$ \textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$

[hide="solution"]Note that $ x^{12}\plus{}ax^8\plus{}bx^4\plus{}c\equal{}p(x^4)$. Therefore the roots are the fourth roots of $ p(x)$.

The square root of an imaginary number is an imaginary number, so we have that the fourth roots of $ 2009\plus{}9002\pi i$ are all imaginary.

Note that the fourth roots of 2009 are $ \sqrt[4]{2009}$, $ \minus{}\sqrt[4]{2009}$, $ i\sqrt[4]{2009}$, and $ \minus{}i\sqrt[4]{2009}$. There are similar fourth roots of 9002. Therefore there are two imaginary roots for each of 2009 and 9002.

Adding these all up we see that $ 4\plus{}2\plus{}2\equal{}8$.[/hide]

To clarify:

From the three zeroes, we have $ p(x) \equal{} (x \minus{} (2009 \plus{} 9002\pi i))(x \minus{} 2009)(x \minus{} 9002)$.

Then $ p(x^4) \equal{} (x^4 \minus{} (2009 \plus{} 9002\pi i))(x^4 \minus{} 2009)(x^4 \minus{} 9002)$.

Let's do each factor case by case:
$ x^4 \minus{} (2009 \plus{} 9002\pi i) \equal{} 0$: Clearly, all the fourth roots are going to be complex.
$ x^4 \minus{} 2009 \equal{} 0$: The real roots are $ \pm \sqrt [4]{2009}$, there are two complex roots.
$ x^4 \minus{} 9002 \equal{} 0$: Same.

So the answer is $ 4 \plus{} 2 \plus{} 2 \equal{} \boxed 8$.

I feel like I must point out a major flaw in high school mathematics here. I had no clue how to do this problem 2 weeks ago before my precalc teacher taught my class how to take the complex roots of a number (purely for the intention of calculus nonetheless) but now it's trivial (other than realizing that $ x^{12} \plus{} ax^8 \plus{} bx^4 \plus{} c \equal{} p(x^4)$). Why is the purpose of high school math calculus rather than algebra or discrete mathematics?

Lol this is a major revival, but I would like to say that this question does not require any calculus at all, just realize that in polynomials, if $r$, $s$, and $t$ are roots to $x^3 + ax^2 + bx + c$, then it factors as $(x-r)(x-s)(x-t)$. Then plug in $x=x^4$ and find the nonreal roots...


geometry
area of a triangle
Heron\u0027s formula
algebra
difference of squares
special factorizations

find the area of a triangle whose sides have lengths a, b, and c if 4a^2*b^2-(a^2+b^2-c^2)^2=16.  I got an answer, but im not sure if its right or if theres an easier way to solve it.  Thanks for any help.

This was a fun problem, although rather easier than I thought it would be.

The original equation was 4a^2(b^2) - (a^2 + b^2 - c^2)^2 = 16.
For my hand's sake, I wrote a^2 + b^2 - c^2 = x. So now we have
(2ab)^2 - x^2 = 16. Noting a difference of squares, we have
(2ab - x)(2ab + x) = 16. Reintroducing x's equivalent, we have
(2ab - a^2- b^2 + c^2)(2ab + a^2 + b^2 - c^2) = 16. Multiplying the first
expression by -1, we can then reduce down to
((a-b)^2 - c^2)((a+b)^2 - c^2) = -16. Once again, we have a diff. of squares. Now we get ((a-b+c)(a-b-c))((a+b-c)(a+b+c))=-16. Multiplying the second term by -1, we get (a+c-b)(b+c-a)(a+b-c)(P) = 16, where P is the triangle's perimeter. We can clearly see an unrefined version of heron's formula here. Using that 2s=P and 2s-a=b+c (which we shall do for each side length), we can get 2s(2s-2b)(2s-2c)(2s-2a)=16. Taking a 2 from each expression, and then dividing by 16, we get that s(s-b)(s-c)(s-a)=1. Square rooting both sides yields the desired 1 due to the left side then beign heron's formula.

It is often useful to know the alternate form of Heron's Formula:

A=1/4*sqrt((a+b+c)(a+b-c)(a-b+c)(-a+b+c)).

easier way

2abcos C = a^2+b^2-c^2

so the second term in the left hand side will be 4*a^2*b^2*(cos C)^2. Plugging in, you get 4*a^2*b^2*(sin C)^2=16. Solve that for absinC=2, K=1.


function
number theory
least common multiple
number theory theorems

What do you know about the function $ L(n) $ satisfying :
  Let $ n = p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k} $
$ L(n) = 1 $ for $ n=1 $
$ L(n) = lcm( (p_1^{\alpha_1-1}(p_1-1) , ..... , p_k^{\alpha_k-1}(p_k-1) ) $ for $ n>1 $

So if a is prime to p,
 a^(L(n)) == 1 (mod n ) ?  (and L(n) would be smaller than phi(n)) )

It looks like the Carmichael Function:
[url=http://mathworld.wolfram.com/CarmichaelFunction.html]mathworld.wolfram.com/CarmichaelFunction.html[/url]


function
geometry
geometric transformation
inequalities
advanced fields
advanced fields unsolved

Here is a piece of the proof in some paper I don't quite understand.

[quote]Suppose we know that $ M$ is a subspace of the Hilbert space $ H$, and the mapping
\[ X: M \rightarrow H \oplus \cdots \oplus H
\]
defined by
\[ X(a) \equal{} (X_1 (a) \oplus \cdots \oplus X_1(a))
\]
is an isometry. Since
\[ \dim M \equal{} \dim (XM),
\]
the map $ X \equal{} (X_1, \ldots, X_n)$ can be extended to an isometry $ A \equal{} (A_1, \ldots, A_n),$ defined on $ H$ and with values in the direct sum $ H \oplus \cdots \oplus H.$[/quote]

I do not know why
[hide]the map $ X$ can be extended to an isometry $ A,$ defined on $ H$.[/hide]

The professor wrote something like this:
[hide]\[ H \equal{} M \oplus (H \ominus M)
\]

\[ H\oplus \cdots \oplus H \equal{} XM \oplus (XM)^\perp
\]
Since
\[ \dim (H \ominus M) \equal{} \dim H \minus{} \dim M < n \dim H \minus{} \dim M \equal{} \dim (XM)^\perp,
\]

\[ \dim (H \ominus M) < \dim (XM)^\perp,
\]
we can construct an isometry from $ H\ominus M$ to $ (XM)^\perp.$
[/hide]
But he does not explicitly define the extension.

Can anyone provide the details for this argument showing how one can define the extension?

Also, in general, when can one say he can extend a linear operator from a subspace to the whole space?

In addition to this, I thought that the following might be helpful.

It is written in the paper that $ X\equal{}(X_1, \ldots, X_n)$ is an isometry on $ M \subset H$ because we can show
\[ \langle a,a \rangle \equal{} \sum_{j\equal{}1}^n \langle X_j^T X_j a, a \rangle, \qquad \text{for all } a \in M
\]

So I'm assuming that the image of $ X$ is all the elements of the form
\[ \sum_{j\equal{}1}^n X_j (a), \qquad a \in M
\]
and $ H \oplus \cdots \oplus H$ has the induced inner product
\[ \langle \sum_{j\equal{}1}^n a_j, \sum_{j\equal{}1}^n b_j \rangle \equal{} \sum \langle a_j, b_j \rangle.
\]
because then we can have
\[ \langle \sum_{j\equal{}1}^n X_j(a), \sum_{j\equal{}1}^n X_j(a) \rangle \equal{}
                          \sum_{j\equal{}1}^n \langle X_j^T X_j a, a \rangle.
\]

Is this correct?

are $ H$ infinite-dimesional, $ M$ of dimension $ n$?
it seems that these are reasonable hypotheses to make everything work.
in this case (changing $ n$ if necessary) we can suppose and $ X_i \equal{} \phi_i f_i$, where $ \phi_i$ is in $ H^*$ and $ f_i$ has norm 1.
in this case, simply extend the dual basis $ e_1,\dots,e_n$ relative to $ \phi_1,\dots,\phi_n$ to a basis (in the hilbert sense) $ \{e_\alpha\}_{\alpha\in A}$ of $ H$. hilbert spaces are classified up to isometry by cardinality of a basis, if we partition $ A$ in $ n$ sets of the same cardinality, with each of $ e_1,\dots,e_n$ belonging to a different subset, we obtain an isometry $ H\to H^{\oplus n}$, which we can choose to extend the given one.

anyway.. if $ M$ is not finite-dimensional, it doesn't seem to work: $ H$ is isometric to $ H_1\oplus H_2$ (where $ H_1,H_2$ are copies of $ H$, just recalled for sake of clarity). now choose $ M \equal{} H_1$ and an isometry $ H_1 \to H \equal{} H_1\oplus H_2$. of course you can't extend it to an isometry of the full $ H$, since it's already bijective...

hope this helps somehow..

Hi ma-go,

Thank you for replying. $ H$ should be finite-dimensional. [url=http://www.math.ucsd.edu/~helton/osiris/NONCOMMINEQ/sphereSubm.pdf]Here[/url] is the whole paper if you are interested. The part I'm mentioning is on page 5 to 7. What do you mean by $ H^*$??

$ H^*$ is the dual, the space of linear continuous functions $ H\to R$.

anyway, in the finite-dimensional case it's even easier: pick a (hilbert) basis $ B_i$ for $ H_i\cap {\rm Im}\, M$, extend it to a (hilbert) basis $ E_i$ of $ H_i$ (the i-th factor in the sum), take the counterimages of $ B_i$'s, extend it to a (hilbert) base $ E$ of $ H$ (you can do it because the map is an isometry, so take bases to bases), and choose distinct elements from $ \bigcup E_i\setminus B_i$ to creat your map.
you have room enough, since $ \dim H^{\oplus n} \minus{} \dim M \equal{} n\dim H \minus{} \dim M \ge \dim H \minus{} \dim M$.

Hi ma_go,

Thanks again. I don't quite follow what you said. Could you explain more?

1. It seems your are using the terms "basis" and "base" interchangeably.

2. What do you mean by the counter-images of $ B_i$'s? I'm assuming you mean the pre-images.

3. When you say "choose distinct elements from $ \bigcup E_i \backslash B_i$ to create your map," do you mean "choose the elements where $ E \backslash \bigcup \operatorname{pre\minus{}image}(B_i)$ can mapped to?"

Sorry, I'm not quite used to this material so I'd appreciate if you can elaborate it for me.

I have more questions.

1. Why do you pick a basis for $ H_i \cap \operatorname{Im} M$? Why don't you just pick a basis for $ \operatorname{Im} M$ and take the pre-image of that? and then extend it to a basis of $ H$?

2. When you say "you can do it because the map is an isometry," which part exactly are you referring to?

3. Are you assuming that the union of $ B_i$'s form a basis of $ \operatorname{Im} M$? And then you take the pre-image of this union?

4. Why is that the resulting map is also an isometry?

[quote="lenny"]Hi ma_go,

Thanks again. I don't quite follow what you said. Could you explain more?

1. It seems your are using the terms "basis" and "base" interchangeably.

2. What do you mean by the counter-images of $ B_i$'s? I'm assuming you mean the pre-images.

3. When you say "choose distinct elements from $ \bigcup E_i \backslash B_i$ to create your map," do you mean "choose the elements where $ E \backslash \bigcup \operatorname{pre - image}(B_i)$ can mapped to?"

Sorry, I'm not quite used to this material so I'd appreciate if you can elaborate it for me.[/quote]

1. i thought "bases" stood for the plural of "basis" (i've been checking on wikipedia): please correct me if it's not like this (in italian "basis" is called "base", and i think "base" (en) could be a good translation for "base" (it), so i'm quite messed up :p

2. pre-images, yes :) again, in italian you can use either "preimmagine" or "controimmagine" :)

3. yes.
[quote]
I have more questions.

1. Why do you pick a basis for $ H_i\cap \Im M$? Why don't you just pick a basis for $ \Im M$ and take the pre-image of that? and then extend it to a basis of $ H$?

2. When you say "you can do it because the map is an isometry," which part exactly are you referring to?

3. Are you assuming that the union of $ B_i$'s form a basis of $ \Im M$? And then you take the pre-image of this union?

4. Why is that the resulting map is also an isometry?[/quote]

1. it's not the same, of course. it's wrong as i wrote it. you need to take a basis for $ \Im M$: there are several "transversal" spaces that would made my proof totally wrong. your correction is ok.

2. let's name it: call $ \phi: M\to H^{\oplus n}$. since $ \phi$ is an isometry, it takes (hilbert) basis to (hilbert) basis for $ \Im \phi$. so you can pull back a basis for $ \Im \phi$ to a basis of $ M$.

3. yes. with "new" point 1. you don't need $ B_i$'s, but just $ B$.

4. a map between hilbert spaces is an isometry if it takes a basis into an orthonormal set.

by the way, this kind of problem could be easy generalized to a map $ M\to H'$, with $ \dim H'\ge \dim H$, even if $ H, H'$ are infinite-dimensional, and i think it works even for $ %Error. "codim" is a bad command.
_H M \le %Error. "codim" is a bad command.
_{H'} M$, without any further hypothesis.

It seems like you can start by choosing some orthonormal basis for $ M$ as well instead of $ \operatorname{Im} M.$ What do you think is the key in this proof? I'm still somehow not completely convinced by it.

the key is the relation on codimension: $ {\rm codim}_H\, M \le {\rm codim}_{H'}\, M$, with the notation of my last post.

Yeah but also what role does isometry play in here? I think the key is that one can extend orthonormal sets and any subset of an orthonormal set is still orthonormal.

Also is it correct that you can start by choosing some orthonormal basis for M?

[quote]and i think it works even for ....., without any further hypothesis.[/quote]

Here I don't understand ur notation in ur second last post.

yes for both.
same setting, just instead of $ H^{\oplus n}$ you put $ H'$, with the inequality on codimensions.