Can someone please explain summation notation (the big Greek "S")?  Also, I need a clarification on subscripts.  If I have s(sub)1 and s(sub)2, does that just mean the first and second terms of a sequence?  Is there neccesarily a relation?  Any help would be appreciated.  Thanks!

I don't think summation would really show up in MathCounts. I couldn't help you with it, unfortunately; my teacher explained it a little bit to my class but we never used it so I really don't know about it.

Sometimes terms in a sequence are donated with a(sub)0, a(sub)1, etc., right?

Okay, I'm sure other books do a much better job of describing $\sum$, but here is the basic way to "read" it.

For example:
$\displaystyle\sum_{k=1}^{10}k = 1+2+3+....+10$.
Using the formula $\frac{n(n+1)}{2}$, we find that $1+2+3...+10=55$.
Basically the variable on the bottom is a dummy variable.  It is substituted in the expression on the right, which in this case is just $k$.  The sum of the values when $k$ is substituted is added up until the number on top, which in this case is $10$.

Another more complicated example:
$\displaystyle\sum_{k=3}^{13}\frac{k}{k+1} = \frac{3}{3+1}+\frac{4}{4+1}+\frac{5}{5+1}...\frac{13}{13+1}$
Because in this case $k=3$ is on the bottom, we start summing it by plugging in $3$ for $k$ in $\frac{k}{k+1}$.  We add up all these plugging in $k=4, k=5$ all the way to $k=13$.  The variable on the bottom, which in my examples is $k$, can actually be any letter, such as $i$.  It doesn't have to be $k$.

See if you can figure out what this is:
$\displaystyle\sum_{k=0}^{10}2^{k}$

[quote]If I have s(sub)1 and s(sub)2, does that just mean the first and second terms of a sequence?[/quote]

It just denotes two numbers.  They are two different variables, as different from each other as $s$ and $t$.  But using subscripts indicates more clearly what's related to what.

Sometimes, $s_1$ and $s_2$ denotes the first two terms of a sequence that continues $s_3, s_4, \ldots$.  In that case, you have to use subscripts because if you write $a,b,c,\ldots$ soon you run out of letters :) :) :)

But sometimes you only have two numbers, but you still call them $s_1$ and $s_2$ because it helps you keep track of all the quantities in the problem and how they are related to each other.

For example, you might also have $t_1$ and $t_2$ in the same problem.  To give a really silly example (since I cannot think of a better one right now), maybe $s_1$ means the number of apples person 1 has, $s_2$ the number of apples person 2 has, and $t_1$ and $t_2$ the same but with bananas. Instead of writing $s_1$,$s_2$,$t_1$,$t_2$ you could use $s$,$t$,$u$,$v$ and nothing in principle would change, but it would be more difficult to remember what variable stood for what.  (Of course, $a$ and $b$ would be better choices than $s$ and $t$ if I was doing a problem with apples and bananas :) )

Yakko


function
topology
real analysis
real analysis unsolved

$K \in R^{n}$ compact, $V$ open set in metric space $E$. $f$ is continuous real-valued function on $K \times V$.
$z_{n}\to z_{0}$ in $V$. Clearly $f(x,z_{n})$ is a sequence of fucntions on $K$, pointwisely converges to $f(x,z_{0})$.
Why this convergence is uniform?

Fix $\epsilon>0$. By continuity of $f$, for any $x\in K$ there exist open sets $W_{x}\subset K$ and $U_{x}\subset V$ such that: (i) $(x,z_{0})\in W_{x}\times U_{x}$; (ii) $|f(x',z')-f(x,z_{0})|<\epsilon$ whenever $(x',z')\in W_{x}\times U_{x}$.
The set $K$ is covered by a finite family $W_{x_{i}}$. The set $U: =\bigcap U_{x_{i}}$ is open in $V$ and contains $z_{0}$. By its construction, $\sup_{x\in K}|f(x,z)-f(x,z_{0})|\le\epsilon$ for any $z\in U$.


A plane passes through $ (1,\minus{}2, 1)$ and is perpendicular to two planes $ 2x\minus{}2y\plus{}z \equal{} 0$ and $ x\minus{}y\plus{}2z \equal{} 4$. What is the distance of the plane from the point $ (1, 2, 2)$?

P.S. Where could i learn more about 3D geometry (equations of planes, etc) and how i could learn to solve problems like this? Any advice is appreciated. Thanks..

Well, I just learned some stuff about 3D geometry. I have no idea if this is right or wrong though, since I just learned this...

[hide]

Let's call $ 2x\minus{}2y\plus{}z\equal{}0$ plane $ A$, and $ x\minus{}y\plus{}2z\equal{}4$ plane $ B$.

So, if I read correctly, if a plane is perpendicular to another plane, the normal vectors must be orthogonal (dot product = $ 0$). So, since the plane is perpendicular to both plane $ A$ and $ B$, if we take the cross product of the normal vectors of $ A$ and $ B$, we should get the normal vector of the plane we are looking for...

I have no idea how to latex vectors, so... here goes...

$ <2,\minus{}2,1>\times <1,\minus{}1,2> \equal{} <\minus{}3,\minus{}3,0>$. So, the normal vector is $ <\minus{}3,\minus{}3,0>$.

So, we have a plane in the form $ \minus{}3x\minus{}3y\plus{}0z\plus{}d \equal{} 0$, plugging in $ (1,\minus{}2,1)$ gives us...$ d\equal{}\minus{}3$, so the equation of the plane is $ \minus{}3x\minus{}3y\minus{}3\equal{}0$

So, using the distance formula is $ \frac{\left|(\minus{}3)(1)\plus{}(\minus{}3)(2)\minus{}3\right|}{\sqrt{(\minus{}3)^{2}\plus{}(\minus{}3)^{2}}}\equal{}\boxed{2\sqrt{2}}$[/hide]

That should be correct since it was one of the options in the contest. Btw, what is the physical meaning of an expression with $ x$ and $ y$ variables but no $ z$ variable? E.g. $ 3x\plus{}3y \equal{} 0$. When you plot it in 3D, is it a plane, a line, or what?

Thanks  :)

An expression that is "missing" a variable allows that variable to take all values. Compare to $ x \equal{} c$ or $ y \equal{} c$ in the plane (where $ c$ is a constant). In this case, you have a plane parallel to the $ z$-axis.


Where can I  get previous year papers for INAO ? I am not able to find any on the internet :(   <except for the ones in mathlinks>Can anybody, who has  written previously post the questions please?

Last years paper had 20 obj.(obviously dn't remember them)
subjective
1.find th eequation to the durve that passes through ..........(some three points were given)
2.Find the amount of radioactive material left....half life etc. were given
3.Find an arrangemnet with min number of mirrors inside a closed box such that incident ray moves out of the box undeviated.
4.Find the amount of Tea consumed by People of Mumbai in a single day
5.Then there was a question on range of signal supply by a power station to a satellite(i suppose)...the on from communication chapter in case u study NCERT.
and if there were 6 questions i forgot one :D


How do you typeset the weird "C" letter in $\LaTeX$??? Thanks.


LaTeX
real analysis
real analysis unsolved

Let ai with i=1,2,3.....n be numbers such that the condition:

(x+1)*(x+a1+1)*(x+a2+1)*.....*(x+an+1)-x*(x+a1)*(x+a2)*......(x+an)-(n+1)*(x+1)^n=0

applies for any integer x.

Show that j*(j+a1)*(j+a2)*......(j+an)/(n+1)=1^n+2^n+......+j^n.

j an integer.

^ means power

It is not difficult at all, just trivial induction... By the way, try to learn $\LaTeX$ :)


algebra
polynomial
Vieta
quadratics
complex numbers

Show that z = 3 + 3i is a root of the equation z^3 - 18z + 108 = 0 and obtain the remaining roots of the equation.

Too lazy to show work, but basically, you plug it into the equation and if it does equal 0 then it is a root (or do division) and since thats a root, 3-3i must be a root.
Then from Vietas theorm 6+z=0 so z=-6 is also a root.

correct...except I don't know what vietas thorem is. I would have just multiplied (z - (3 + 3i)) by (z - (3 - 3i)) to get a quadratic factor. I would divide the cubic by this factor to get a linear factor (x + 6) which when set to 0 gives the required real root.

I thought it was the conjugate pair theorem not Vietas theorem.

da ban man's logic is actually fairly simple.  Because we know that $(z-(3+3i))(z-(3-3i))$ is a factor, we can write

$(z^{2}-6z+18)(z-a) = z^{3}-(6+a) z^{2}+(18-6a) z-18a = z^{3}-18z+108$

It follows from equating the quadratic coefficients that $6+a = 0 \implies a =-6$.  "Vieta's theorem" is a fancy name some people give for equating a polynomial with the expansion of its factored form.

Alternately, equating the constant coefficients would give $-18a = 108$ which implies the same thing.

Just to follow up the problem, prove that $3-3i$ must be a root of the polynomial if $3+3i$ is (don't use direct algebra).
Hint: $\overline{z^{n}}={\bar{z}}^{n}$ and $\overline{z+w}=\bar{z}+\bar{w}$ where $z$ and $w$ are complex numbers, and $\bar{z}$ denotes the complex conjugate of $z$.


a,b,c>0 and abc=1. prove that
 \sqrt a+ \sqrt b+ \sqrt c \geq 1+ ^3  \sqrt ((a+1)(b+1)(c+1))
i think it's quite hard. :D

I think noone can solve it, since it is not true. Take $ a=x^2, c=b=\frac {1}{x} $ with x very close to 0.


[quote="EFuzzy"]Are Local moderator's supposed to be able to delete other people's posts? I can add new user's, but I can't delete posts.[/quote]In short local moderator at this point means moderator for the local group (ie lets you decide which people are in your local group), so at this moment the answer is no, you are not supposed to delete other people's posts.

We are not yet sure how everything will work in it's final stage, because these forums and their groups are still not very populated and young.

[quote="SirErnest"]If you can not delete posts, then how can you make sure nothing inappropriate is posted?[/quote]At this point there isn't much posting at all. Admins can be announced of any in-appropriate behaviour, but it is supposed that local moderators choose users carefully before letting them in, right? :)

We are thinking now of temporarily allowing local mods to become mods of their forums, untill the population growth is enough to support "elections".

Val is working on a way to give you the moderator powers within your forum without cluttering AoPS with the green moderator bolds on the usernames.  This should be set up soon and then you can make sticky/announcement posts (like would be good for introductions and competition schedule posts) and also delete rare bad posts.

[quote="mathmom"][quote="Valentin Vornicu"] it is supposed that local moderators choose users carefully before letting them in, right? :) [/quote]

I'm not really sure how this could be done with any degree of success.  If someone is already a poster on the main boards, then you can check their postings, but otherwise, how would you know if the person is going to post appropriately or not until after you let them in?[/quote]

I don't think this is an issue in general.  Nearly all of our users behave themselves and those who don't are quickly put in line.


Let $ w,x,y,z$ be real numbers, Prove that

$ (w\plus{}ix)^2\plus{}(w\plus{}iy)^2\plus{}(w\plus{}iz)^2\minus{}(w\plus{}iy)(w\plus{}iz)\minus{}(w\plus{}iz)(w\plus{}iy)\minus{}(w\plus{}ix)(w\plus{}iy)\le 0$

$ 2iwx\plus{}2iwy\plus{}2iwz\minus{}2iyw\minus{}2iwx\minus{}2iwz\plus{}3w^2\minus{}3w^2\plus{}yz\plus{}wz\plus{}wx\minus{}x^2\plus{}y^2\plus{}z^2$

$ \minus{}(x^2\plus{}y^2\plus{}z^2\minus{}yz\minus{}wz\minus{}xy)$

Which is trivial by SOS or rearrangement.

It doesn't necessarily follow from rearrangment because these are real numbers. But it is trivial by SOS...

rearrangement does work for arbitrary REAL numbers.


Putnam
combinatorics proposed
combinatorics

Let $ n,k$ be positive integers. We call a permutation $ \sigma =${$ a_1;a_2,..;a_n$} of {1;2;3...;n} is a k-intercepted permutation if $ |a_i - i|\le k$ for every $ i\in$ {1;2;...;n}. Prove that the number of k-intercepted permutations is odd integer iff ${ n\equiv 0(mod \{2k + 1})$ or ${ n\equiv 1(mod \{2k + 1})$


probability
floor function
arithmetic sequence
combinatorics unsolved
combinatorics

[i]Prove that each element of the set $ M \equal{} (1,2,...,1988)$ can be colored with one of four colors such that no arithmetic sequence with 10 elements of $ M$ is monochromatic (Note: A sequence will be called monochromatic if all its elements are colored with the same color).[/i]

Could anyone help me this problem? Thanks so much

If my calculations are right, it turns out that a random coloring actually works here.

For any particular arithmetic sequence, a random coloring will make that sequence monochromatic with probability $ \frac{1}{4^9}$, so the probability that a random coloring fails is at most $ \frac{N}{4^9}$, where $ N$ is the number of arithmetic sequences of $ 10$ elements in $ M$.    It remains to find $ N$.

If a sequence begins with $ k$, the next term can be anything from $ k+1$ to $ k+\lfloor \frac{1988-k}{9} \rfloor$, meaning there are ${ \lfloor\frac{1988-k}{9} \rfloor}$ choices.  Adding up over all $ k$ gives $ N=9(1+2+\dots+219)+8(220)=218,570$.

Since $ \frac{218,570}{4^9} <1$, a random coloring works with positive probability, so there must be a good coloring.


Let $p$ be odd prime $p>3$.
a) Show that:There exist positive integers $x;y$ such that $x;y\in (0;\frac{p}{2})$ for which . $k=\frac{x^2+y^2+1}{p}$ is positive integer and not divisible by $4$
b) Show that:For each positive integer $n$ there exist pair$(x;y)$ for which $x^2+y^2+1$ is divisible by $p^n$

I think you can't exclude $xy=0$ in part a), e.g. take $p=5$.

Define $P : = \{ 0,1,2,..., \frac{p^n-1}{2}\} , S : = \{ h^2 | h \in P \} , T : = \{ -1-h^2 | h \in P \}$.
Then $|S|=|T|=|P|=\frac{p^n+1}{2}$.
Now there are different $a,b \in S \cup T$ such that $a \equiv b \mod p^n$ since $|S \cup T|=|S|+|T|=p^n+1$.
But then $a$ and $b$ can't both be from $S$ since that would give different $x,y \in P$ with $x^2 \equiv y^2 \mod p^n$, but the last one is equivalent to $x \equiv \pm 1 \mod p^n$, which cannot happen. By the same, also not both are from $T$.
Thus, we have (WLOG) $x,y \in P$ with $a=x^2 , b=-1-y^2$ and by this $x^2 \equiv -1-y^2 \mod p^n \iff p^n|(x^2+y^2+1)$.
Setting $n=1$ then also solves part a) since $x^2+y^2+1$ is not divisible by four for any integers $x,y$.

This problem i proposed in our Test for National Olympiad, fact it's not new. I took it from the following result:
$4\nmid n$ if and if there are $x,y$ which $n\mid x^2+y^2+1$


induction
USAMTS
modular arithmetic
ratio
number theory
greatest common divisor
function

Here is my solution to question #4 on round 3:

  Let $ a_1 \equal{} 2007$, $ a_2 \equal{} 20072007$ etc. It's easy to see that
    $ a_{k \plus{} 1} \equal{} a_k \plus{} 2007*10^{4k}$. Since $ 10^4 \equiv 1 (\mod 101)$, we have
    $ a_{k \plus{} 1} \equiv (a_k \plus{} 2007) (\mod 101) \equiv (a_k \plus{} 88)(\mod 101)$.
    Since $ a_1 \equiv 88 (\mod 101)$,
    by induction, we have $ a_k \equiv 88k (\mod 101)$.

    We need to show that there are infinitely many values of $ a_k$ divisible by $ 101$.
    If $ 101$ divides $ k$ then $ a_k \equiv 88 \times 0 \equiv 0 (\mod 101)$, or $ a_k$ is divisible by $ 101$.
    Since there are infinitely many values of $ k$ divisible by 101, there are infinitely many $ a_k$
    that are divisible by 101.$ \square$

I got 3 points for this submission. Can anyone tell me what's the problem with it?
From what I understand from the Grading Rubric this is precisely the most common type of correct solution that gets me 5 points.

Thanks,
--Artur

[quote="ar2rd"]Since $ a_1 \equiv 88 (\mod 101)$,
    by induction, we have $ a_k \equiv 88k (\mod 101)$.[/quote]

The only problem I see is that this step is out of order.  Your inductive step is the line above it; reorganize this proof the way you want it to be read.  In other words, start with your inductive claim and [b]then[/b] do your inductive step.

ar2rd, I don't see any problems with your solution above. And I am the person who wrote the Grading Rubric for problem 4. You followed the most common solution for Problem 4 without skipping any vital steps. What comment did the grader write?

[quote="t0rajir0u"]The only problem I see is that this step is out of order.  Your inductive step is the line above it; reorganize this proof the way you want it to be read.  In other words, start with your inductive claim and [b]then[/b] do your inductive step.[/quote]
We graders aren't that fussy. Or to be more precise, the few new graders who start out that fussy are lectured that we should not demand writing of math-journal quality in a high school contest.

Erin Schram
USAMTS grader

My grader wrote : "Good Start". That's all.

--Artur

[quote="ar2rd"]My grader wrote : "Good Start". That's all.[/quote]
You had the line, "$ a_k \equiv 88k \pmod{101}$," so you were past the start of the problem. I worry that part of your solution might have been missing. That sometimes happens with faxed submissions. If that is not the case, then somehow the grader misread your submission.

Your next step is to make a formal request for re-evaluation of the grading in the Grading Issues subforum. I see that you have done that already.

Erin Schram
USAMTS Grader

I also had a question about my solution to problem number 4.  My grader feedback just said "Good" and I was wondering what I got points off on (I got a 4)

To prove that 101 divides infinitely many members of the set {$ 2007,20072007,200720072007 \ldots$} We start by noticing that each member of the set is the sum of a geometric progression.  If 2007 is the first member of the set, 20072007 is the second, etc, then the nth member of the set can be represented by:

$ 2007(1\plus{}10^4\plus{}10^8\plus{}\ldots \plus{}10^{4(n\minus{}1)})$.

Rewrite this using the formula for the sum of a geometric sequence (common ratio is $ 10^4$, starting term is 2007, and there are n terms):

$ \frac{2007(10^{4n}\minus{}1)}{9999}\equal{}\frac{223(10^{4n}\minus{}1)}{1111}$

To prove that that infinitely many members of the set are divisible by 101, we find infinitely many n so that the expression is congruent to 0 modulo 101.

$ \frac{223(10^{4n}\minus{}1)}{1111}\equiv 0 (mod 101)$

$ 223(10^{4n}\minus{}1)\equiv 0 (mod 101)$

223 and 101 are relatively prime, so we can divide each side of the congruence by 223:

$ 10^{4n}\minus{}1\equiv 0 (mod 101)$

$ 10^{4n}\equiv 1 (mod 101)$

101 is prime, by Fermat's little theorem for prime p and $ a<p$, then $ a^{p\minus{}1}\equiv 1 (mod p)$, thus, $ a^{100}\equiv 1 (mod 101)$

If a=10, then we have $ 10^{100}\equiv 1 (mod 101)$.  Looking back to our original congruence, we seek to find n such that $ 10^{4n}\equiv 1 (mod 101)$.  Clearly, if n is a multiple of 25, then 4n is a multiple of 100, and all multiples of 100 make the congruence 0.  There are an infinite number of multiples of 25, so 101 divides infinitely many members of the given set.

End of problem 4.

[quote="flipandtuck"]

$ \frac {223(10^{4n} \minus{} 1)}{1111}\equiv 0 (mod 101)$

$ 223(10^{4n} \minus{} 1)\equiv 0 (mod 101)$
[/quote]

Don't forget that $ \text{gcd}(1111,101) \neq 1$, so the second line does not follow from the first.

I also had a question about my solution to #4. I know this solution has few "minor" mistakes and is a little bit ambiguous but does it deserve only a score of 1?

We can write successive numbers in the set as follows:

Number______________Factorization_____________________Number Y
1.  2007______________2007 x (1)_______________________1
2.  20072007__________2007 x (1 x 10,000 + 1)___________ 10001
3.  200720072007______2007 x (10,001 x 10,000 + 1)_______100010001
4.  2007200720072007__2007 x (10,0010001 x 10,000 + 1)___1000100010001

F(X)= (Y x 10,000 + 1)

In the above table each successive number is obtained by multiplying 2007 by X, where Y in the function X is obtained from the previous number, and the first Y=1.

In each successive ‘Y’, number since the multiplication by 10,000 doesn’t change the remainder, with each consecutive Y our remainder increases by one as we add Y x 1 (or +1 to the remainder) in the function f(x). Therefore remainder of first Y is 1, second 2, third 3 and so on. As a result we can conclude that the 101st Y will not have any remainder.  This proves that 101 divides infinitely many of the numbers in the given set.

[quote="flipandtuck"]I also had a question about my solution to problem number 4.  My grader feedback just said "Good" and I was wondering what I got points off on (I got a 4) [/quote]
My guess is that your proof has many elements of a correct proof, but it is not arranged to follow a clear and valid line of reasoning, so the grader decided that you had a correct proof in your head and were a disorganized writer. You lost a point because a proof needs a valid line of reasoning.

Your grader was wrong, because your proof is wrong. I would have given your submission only 2 points. Notice that the correct result is that every 101th element in the sequence is divisible by 101, but your result is that every 25th element is divisible by 101.

The erroneous step is the one that azpjs commented on. You said:
[quote]To prove that that infinitely many members of the set are divisible by 101, we find infinitely many n so that the expression is congruent to 0 modulo 101.

$ \frac{223(10^{4n}\minus{}1)}{1111}\equiv 0 \pmod{101}$
[/quote]
In rigorous language you are saying, "We wish to find all positive integers n such that the integer $ 223(10^{4n}\minus{}1)/1111$ is congruent to 0 when viewed modulo 101." That statement is fine so long as the division is integer division rather than modulus division. There is a great deal of trouble with division by 1111 when working mod 101. $ 1111 \equal{} 11\times 101$. So $ 1111\equiv 0 \pmod{101}$.

On the next line, when you multiply both sides by 1111 to remove the denominator, you really multiply both sides by zero mod 101. The equation
$ 223(10^{4n}\minus{}1)\equiv 0 \pmod{101}$
is true for all positive integers n and has no relation to the values of n for which
$ \frac{223(10^{4n}\minus{}1)}{1111}\equiv 0 \pmod{101}$
is true.

Algebra modulo a prime number is perfectly workable, but it does have slightly different rules than algebra over the rational numbers. You got caught by one of those different rules. (Algebra modulo a composite number is sometimes workable, but its rules are even stranger.)

Erin Schram
USAMTS grader

[quote="luke5dm"]I also had a question about my solution to #4. I know this solution has few "minor" mistakes and is a little bit ambiguous but does it deserve only a score of 1?[/quote]

I hope you have a durable ego, because I am going to enumerate most of the flaws in your proof.

1. In your table with columns named "Number", "Factorization", and "Number Y", you give a pattern. You spend a whole table on the pattern. However, you do not spend a single sentence of explanation on that pattern, so we graders don't know what you want to say about it.

2. You are implicitly saying that rather than looking at the sequence 2007, 20072007, 200720072007,..., you want to look at the sequence 1, 10001, 100010001,.... Looking at the second sequence is fine; however, you should say that explicitly.

3. In the statement "F(X)= (Y x 10,000 + 1)" you throw in an undefined variable X, an undefined function F, and a variable Y that appeared only as the label on a column in the table above. How is the reader supposed to find any meaning in this statement? Since I know the proof inside and out, I deduce that Y is an arbitrary element in the sequence, and F is the recursive relation that calculates the next element in the sequence. But most sequences don't have recursive relations, so you have to explicitly say that this sequence has a recursive relation to give any meaning to the function F. As for X, I think you meant Y.

4. Oh wait, in the next sentence you call X both an number and a function. So what is X? You are being very confusing.

5. [quote]In each successive ‘Y’, number since the multiplication by 10,000 doesn’t change the remainder, with each consecutive Y our remainder increases by one as we add Y x 1 (or +1 to the remainder) in the function f(x).[/quote]
The beginning of this sentence does not parse. Maybe that comma after the 'Y' is misplaced. But even without that confusing comma, and dropping the intervening "since" clause, you are saying, "In each successive ‘Y’ number with each consecutive Y our remainder increases...." You are repeating yourself, adding confusing to bad grammar.

6. The statement "the multiplication by 10,000 doesn’t change the remainder" is usually false. Multiplying a dividend by 10000 does change the remainder unless the divisor is a factor of 10000-1 = 9999. Now 101 happens to be a factor of 9999, so this is one of the rare special cases where the statement is true, but really you should explain that.

7. [quote]Therefore remainder of first Y is 1, second 2, third 3 and so on. As a result we can conclude that the 101st Y will not have any remainder.[/quote]
Not having any remainder is elementary school language for having a remainder of zero, so I can understand what you are saying. But it still looks like skewed reasoning to say that the sequence 1, 2, 3, and so on eventually reaches 0. A full line of reasoning would say, "As a result, the 101st element would have remainder 101, except that since 101 is our divisor, that reduces to remainder 0."

8. Your next sentence is, "This proves that 101 divides infinitely many of the numbers in the given set." Wait, you found one element that is divisible by 101: the 101st element. You have not shown anything about any other elements being divisible by 101. Sure, a smart reader could continue the line of reasoning: "And after the remainder 0 at the 101st element, the remainders increase again to 1, 2, 3, and so on. Thus, the 202nd element, 303rd element, and so on are all divisible by 101, too." However, you left that unsaid. You did not give that critical step.

I suspect that the correct proof existed in your head. But you failed to put that proof down on paper. We graders see only what you write, and you left too many steps unwritten or unexplained.

Erin Schram
USAMTS grader

Erin, is it possible for you to look at my solution for this question and tell me why I had a point docked? (I'm the guy who had a point docked for now stating the answer correctly.)

[quote="le00327146"]Erin, is it possible for you to look at my solution for this question and tell me why I had a point docked? [/quote]
I've looked over it. Your proof has two minor weaknesses where you failed to explain your reasoning. Your grader could have decided that two minor weaknesses merited one point off.

The first weakness is that you did not explain of why it is okay to divide the sequence by 2007. You simply divide and start looking at the sequence 1, 10001, 100010001, etc. The best explanation would have been to refer back to the original sequence at the end of the proof, "Multiplying this sequence by 2007 cannot decrease the number of elements that are divisible by 101, so the original sequence has infinitely many elements divisible by 101, too."

The second weakness is when you replace 10000 by 1 in your algebra mod 101. A word of explanation that $ 10000\equiv 1 \pmod{101}$ would have been nice.

Erin Schram
USAMTS grader

[quote]I suspect that the correct proof existed in your head. But you failed to put that proof down on paper. We graders see only what you write, and you left too many steps unwritten or unexplained.[/quote]

Mr. Erin J. Schram, thanks for a thorough dissection of my proof. I'll examine my work more closely and make it more explicit for round 4.

[quote="luke5dm"]Mr. Erin J. Schram, thanks for a thorough dissection of my proof. I'll examine my work more closely and make it more explicit for round 4.[/quote]

Uh...just a quick thing...isn't Erin a female spelling?
(Sorry if it isn't)

Also, thanks for looking over my proof. Do you know what the grader could have meant by not stating my solution correctly?

[quote="le00327146"]Uh...just a quick thing...isn't Erin a female spelling?[/quote]
The web site http://www.namestatistics.com claims that around 172,725 U.S. females are named Erin and around 8575 U.S. males are named Erin. Thus, around 95% of American people named Erin are female. Therefore, it is a female name. I, however, am in the 5% minority.

[quote]Also, thanks for looking over my proof. Do you know what the grader could have meant by not stating my solution correctly?[/quote]
Since the grader put an emphasis on "stating", I presume it means that your method of proof was valid, but your written proof was invalid. For example, it could have had expositional gaps or typographical errors. In your case, it was expositional gaps because of two weak explanations.

It also means that the grader needs a refresher on writing informative comments. Maybe I will conduct such a refresher at the beginning of the next grading session.

Erin Schram
USAMTS grader

Ok, thanks.
And sorry about the name thing...

Can anyone explain why I got a 4? The grader's comment was "good"

Given a number with $ n\plus{}1$ many 2007's, it's easy to verify that we can write this as
\[ 2007\times10^{4n}\plus{}2007\times10^{4n\minus{}4}\plus{}\ldots\plus{}2007\times10^4\plus{}2007\]
Now we need to show that the above number is divisible by 101 for infinitely many $ n$. First note that
\[ 10^4\minus{}1\equal{}(10^2\plus{}1)(10^2\minus{}1)\equal{}101\times99\]
So $ 10^4\minus{}1$ is divisible by 101, or equivalently, $ 10^4\equiv1\mod101$. Furthermore,
\[ 10^{4k}\equiv(10^4)^k\equiv1^k\equiv1\mod101\]
for any nonnegative integer $ k$. Consequently,
\[ 2007\times10^{4n}\plus{}\ldots\plus{}2007\equiv2007\plus{}\ldots\plus{}2007\equiv2007(n\plus{}1)\mod101\]
So the number with $ n\plus{}1$ many 2007's will be divisible by 101 if $ n\plus{}1$ is divisible by 101. Since our set obviously has infinitely many of those numbers, 101 divides infinitely many of the numbers in the set.

You're probably in the same boat as the rest of us...correct reasoning but bad proof-writing.

[quote="Jiang"]Can anyone explain why I got a 4? The grader's comment was "good"[/quote]
No, I cannot figure out why your grader gave you a 4. Your proof above covered all the critical steps.

Erin Schram
USAMTS grader


number theory
greatest common divisor
modular arithmetic
number theory unsolved

let $ d\equal{}\gcd(m,n)$ and write $ m\equal{}du,n\equal{}dv$. clearly $ a^{d}\minus{}1$ divides both $ a^{du}\minus{}1,a^{dv}\minus{}1$. now let $ t$ be any integer such that $ t|a^{du}\minus{}1$ and $ t|a^{dv}\minus{}1$, and let $ r$ be the order of $ a$ $ \pmod{t}$. then $ r|du$ and $ r|dv$, from which $ r|d$. then $ t|a^r\minus{}1|a^d\minus{}1$, which proves that $ \gcd(a^m\minus{}1,a^n\minus{}1)|a^{\gcd(m,n)}\minus{}1$, and we're done.


calculus
integration
vector
real analysis
real analysis unsolved

Let S1 be the part of the paraboloid z = 1 - x - y that lies above the xy-plane, and let S2 be the part of the cylinder x + y = 1 for which -1 = z = 0. Let S = S1 U S2, oriented with the outward unit normal. (Note that this surface is open on the bottom.) Let G = (0, 0, y ). Find G·dS in one of the following two ways:

a) use Stokes’s theorem to trade the surface integral for a line integral (by finding a
vector field F whose curl is G),

or

b) use the divergence theorem to trade the integral over S for an integral over a flat
surface.

Now I was able to get it using the a.  Or at least I think I was.  I got pi/4.  However, I wasn't able to do it using the divergence theorem, I can't quite see how to apply it as it has to be a closed surface.

So what I'm basically asking for is:

a. Check the answer of pi/4
b. How to set it up using method b, the divergence theorem.

My work for method a:

$F(x, y, z) = (0, xy^2, 0)$

So $\bigtriangledown \times F = (0, y^2, 0)$

Then:

$\int G \cdot dS = \int \bigtriangledown \times F \cdot dS = \int_a^b F(c(t)) \cdot c'(t) dt$

$c(t) = (cos(t), sin(t), -1)$
$F(c(t)) = (0, cos(t)sin^2(t), 0)$
$c'(t) = (-sin(t), cos(t), 0)$

So that's:

$\int_0^{2\pi} cos^2(t)sin^2(t)dt$

Which upon integration I get $\frac{\pi}{4}$


If $ m$ and $ b$ are real numbers and $ mb > 0$, then the line whose equation is $ y \equal{} mx \plus{} b$ [u]cannot[/u] contain the point

$ \textbf{(A)}\ (0,1997)\qquad
\textbf{(B)}\ (0,\minus{}1997)\qquad
\textbf{(C)}\ (19,97)\qquad
\textbf{(D)}\ (19,\minus{}97)\qquad
\textbf{(E)}\ (1997,0)$

[hide]

$ m$ and $ b$ are either both positive or both negative. If you consider this graphically, the line $ y\equal{}mx\plus{}b$ will always intersect the x-axis with a negative x-intercept. Hence, $ \mathrm{ (E) \ }$ cannot be on the line.

[/hide]

[hide="Cheap"]Randomly plugging in answer choices shows that with choice $ \boxed{\textbf{(E)}\ (1997,0)}$, we would get $ 0\equal{}1997m\plus{}b$. Therefore, $ m$ must be positive and $ b$ must be negative, or vice versa, which is impossible if $ mb>0$.[/hide]


man film va bazigara ro kheili doost daram,va be nazare man cinama zibatarin honare.
hadaf az poste in topic ine ke kami ham raje be cinama va HOLLYWOOD sohbat konim,albate age shoma ham doost darin.
az jadidtarin film haye HOLLYWOOD che khabar?

[quote="amirhtlusa"]Legend of Zorro with Antonio Banderas.[/quote]
Kathrine Zeta Jones ham bazy mikone,man khodam faghat tike hayee az in film ro didam vali be nazaram filme khoobi oomad,halla be nazare to filme khoobi hast ya na?

bayad khoob basheh. (film e avelish ke khoob bood).

also:

Harry Potter IV: Nov. 18th :P

ok i will see it,thanks.
did you see STAR WARS EPISON 3?

ina ro velesh konid faghat bechasbid be matrix,,,

kheily film e toopie :D

are filme kheilikhoobi bood bahat movafegham.

MATRIX ;)

hmm... badak nabood (ye zare Chert o pert bood!)

ta hodoody bahat movafegham vali dar majmoo be nazare man khoob bood.

emshab (ta 45 daghigheye dige), Harry Potter IV dar saat e 11:59 pm baaz mishe...


Problem (Proposed by Khang Tran) Let a,b,c be positive real numbers satisfying $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ \frac{a^5}{a^3\plus{}b^2}\plus{}\frac{b^5}{b^3\plus{}c^2}\plus{}\frac{c^5}{c^3\plus{}a^2}\geq\frac{3}{2}$


calculus
derivative
function
real analysis
real analysis unsolved

How can we find $f'(x)$ using limits?

$f(x)=sin(cos(cotgx^2))$

You don't. In practice, you only use the formal definition of the derivative to calculate very simple things, and to prove the rules we use to calculate everything else. This function is far too complicated for a grubby epsilon-delta approach.


I was just curious if anyone is into stock trading and if so are there any good stocks you've found?

I dunno much about stocks but my brother is a day trader.. you want me to ask him for advice?  :lol:

Also, its kinda hard to ask for a 'good stock' as they change all the time.. one stock thats worth 20 bucks one day can likely be worth 10 bucks the next.. and not gain much for a while.

mutual fund as well. could've gotten a whole boatload of money investing in a certain company who had some certain scandals but is doing much better now but the bad guy walked free and i am mad, but my mother wouldn't let me. "too risky" she said. FOURTEEN CENTS A SHARE!!! and then it went up to like, four bucks. big money.

[quote="furious"]i have mutual funds. i let a company decide what to invest in cuz i don't have a clue. i might do something stupid and invest all my money in an ice cream company or something.[/quote]

 :rotfl: lol! I would invest in an ice cream company!! haagan daaz (is that even spelled right??) but then again I don't even know what mutual funds ARE.

really nice if your momma works (or worked) for a public school and you can use TIAA-CREF. basically, you pick your risk factor (lowest being the most local stocks, and highest being national/international) and stick your money in there. they invest it, you get more money than you had, and they get a percentage for doing the investing.


The ratio of the interior angles of two regular polygons is $3: 2$. How many such pairs are thare?

[hide]
dont's the measure of angles = (sides-2)180

Then I can only think of square to pentagon.

1 pair
[/hide]

[hide]well, if you're just talking about the sum of all the interior angles, i'm pretty sure there's an infinite number of pairs... for example if you have a polygon with $a$ sides and another polygon with $n$ sides, the ratio of the sum of their interior angles would be
\[\frac{\left(a-2\right)180}{\left(n-2\right)180}=\frac32\]

We can see immediately that one pair, $(a,n)$ would be $\left(\left(3+2\right),\left(2+2\right)\right)$ but then you should consider all equivalent fractions of $\frac32$ (i.e. $\frac64$, $\frac96$, $\frac{12}{8}$ etc... which would generate pairs $(8,6)$, $(11,8)$, $(14,10)$ etc...) Thus, there are an infinite number of pairs that fit this condition[/hide]

That's strange. I got three as an answer.

Yeah, I agree that the answer is three because I have solved this problem before. My solution should be somewhere in the old GS articles, so I'll search it later.

dangvh, the answers you gave were not correct. For instance, $(8,6)$ does not give $3: 2$ ratio...

EDIT- Ok, now I see why dangvh got "infinitely many" answers. He interpretted the "ratio of the interior angles" as "ratio of the [b]sum[/b] of the interior angles".

The interior angle of a regular n-gon is $\frac{180(n-2)}{n}$ ;)

ah ok, my mistake, i misinterpreted the problem  :blush: sorry about that

[hide]equilateral triangle interior angle is 60 degrees, times 1.5 is 90 degrees, which is a square, and it's possible.
a square is 90 degrees measuring each interior angle, times 1.5 to get 135, which is an octagon, again possible.
A regular pentagon's interior angle is 108 degrees, multiply by 1.5 to get 162, some regular polygon, and possible.
A regular hexgon has interior angle of 120 degrees, multiply by 1.5 to get 180, but no regular polygon has each inteior angle of 180.
there are only three pairs. [/hide]

Not to mention, the ratio of 2 sides to three sides works as well.

Except thereare no "biangles".  ;)

[quote="4everwise"]Except thereare no "biangles".  ;)[/quote]

Did I put that? Why don't I see that? (did you edit it?)

No, I'm referring to how I solved the problem. I came up with an equation and then found the number of ways.

The number of ways came out to be 4, but a polygon with side 2 was included, which was invalid.

That's how I got the answer as 3.

I think biangle sounds really funny.   :D


hey guys, u kno how there was a boy / girl indicator before

where has it gone?

The new server initiated back in June got rid of it.

really,

didnt kno that

can a admin fix that?

Why?
Does that really matter?
I never look at it.

[quote="akalra1"]its kinda useful at times

for example, u say:

banana123 said [b]he[/b] liked mathcounts better than amc
banana123 said [b]she[/b] liked mathcounts better than amc

which is true?

it might be embarassing to banana123 to see a "she" if banana123 is a "he"[/quote]
Well, when in doubt, you can always go with "he/she"


algebra
binomial theorem
algebra unsolved

Thanks to Dr Sonnhard Graubner for the BIG calculating!  :D
I've been thinking that there are some identities I didn't know that can be used but till now I can't find it.
So, there doesn't exist any identities helping us at this point, does there?  :(

Well, you could use the binomial theorem to reduce the expression to $ 1 \minus{} \frac{\sum_{i\equal{}0}^8 \binom{20}{i} \cdot 3^i}{4^{20}}$.

But even so, the numerator 1028473918 of that fraction isn't a small number. Can it be reduced further?


Let n be an odd integer greater than 1.  Let permutations p of the set${1,2,...,n}$ so we have $|p(1)-1|+|p(2)-2|+...+|p(n)-n)|=\pm1\pm1\pm2\pm2\pm...\pm n\pm n.$.  Why is this true?  Each p(i) is a unique element of 1,2...,n Is the way I interpret the problem.

First of all, if you want the symbol $\pm$, it's \pm.  Second of all, it's just the fact that $|x| = \pm x$ together with a rearrangement of the terms by order.  The fact that the image of $p$ is $\{1, \ldots, n\}$, each appearing exactly once, explains why on the right we have each number appearing exactly twice.  And $n$ need not be odd.  And, of course, we can say further that there need to be an equal number of + and - signs among those additions and subtractions, as well as other limitations.

If you want to save yourself time editing, you can use the "preview" button before you post.

The answer to that question is contained within my observations in the previous post :)

Actually, my previous comment shows that your expression there is an upper bound -- so show that it's the maximum, you need to exhibit an actual permutation which achieves that bound.

so there are $n[((n-1)/2)!]^{2}$ such  permutations


This problem was extra credit for my math class and I haven't gotten it back yet but I'm assuming
a.) Everyone handed it in
and
b.) None of you here goes/takes/will go/take my math class

Anyways:
Suppose two of the zeroes of the following fourth-degree equation are the same and the other two zeroes are the recipricals of each other. Find a and b.

x:^4:+ax:^3:+bx:^2:+4x+4=0

It's not at all hard as it looks...a lot of work though, so I suggest orginizing as you go along.

Tare wrote:x:^4:+ax:^3:+bx:^2:+4x+4=0

Here's a shorter way:

[hide]Say the four roots are c, c, d, and 1/d.  Then the product of the four roots is the constant term of the polynomial, so c^2=4.  Then c= :pm: 2.  Similarly, from the linear term, c^2d+c^2/d+c+c=-4.  If we plug in c=2, we get d=-1,  so the roots are 2, 2, -1, -1.  So a = -(2+2-1-1)=-2 and b = 2*2+2(-1)+2(-1)+2(-1)+2(-1)+(-1)(-1)=-3.  If we plug in c=-2, we get 4d+4/d=0, so d+1/d=0.  Then a = -(-2-2+0)=4 and b=(-2)(-2)+(-2)(0)+(-2)0+1=5.  So either a=-2, b=-3 or a=4,b=5.

[/hide]

Thanks Tare for catching the mistakes.

--Dan

Well...it didn't specify that the solution is real and also you were supposed to get a and b...


[img]http://i102.photobucket.com/albums/m86/nguyenminhtrang_2006/T9-Inequality.jpg[/img]

Is this inequality very hard?[/img]

[quote="nmt_2006"]
Is this inequality very hard?[/quote]
I think  it's nice. It's a problem #106 from Kvant. :wink:

[b]Please stop using Photobucket or Imageshack for your graphics (particularly if they are no graphics)!

You can attach graphics in this forum![/b]

[quote="arqady"][quote="me@home"]what is Kvant?[/quote]

It's the Russian magazine about mathematics and physics. See here:
http://kvant.mccme.ru/[/quote]
Is there an  English version  of the magzine?

It has been discussed on the forum before. :D I think the English version, Quantum, didn't have all the same material either.
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=english+kvant&t=26677

[quote="perfect_radio"]Nobody knows what the original problem was?[/quote]
The following  is an original problem from Sharygin:
Given that $(q_{1}-q_{2})^{2}+(p_{1}-p_{2})(p_{1}q_{2}-p_{2}q_{1})<0.$
 Prove that each of the following equations
$x^{2}+p_{1}x+q_{1}=0$ and $x^{2}+p_{2}x+q_{2}=0$ has two distinct real roots
and  they alternate.


abstract algebra
algebra
function
domain
superior algebra
superior algebra unsolved

Give an egzample of a module of finite uniform dimension that is neither noetherian nor artinian.

[quote="wiosna"]

Give an example of a module of finite uniform dimension that is neither noetherian nor artinian.[/quote]

Any non-noetherian commutative domain $ R$ as $ R$ module, e.g. $ R \equal{} \mathbb{Z}[X_1,X_2, \cdots ],$ is an example. That's because the intersection of any two non-zero ideals of a commutative domain $ R$

is non-zero. Thus every non-zero ideal of $ R$ is essential and so $ \text{u} \cdot \dim_R R \equal{} 1.$


1. 50 % of 12 is the same as 25% 0f____.

2.  A rectangle is 6" long and 4" wide. Yoland has square pieces of cloth 2 " wide. How many pieces of cloth does she need to cover the rectangle?

3. A large pizza is made in shape of a rectange 24" by 9". Twelve people have to get and equal rectangular slice that is twice as long as it is wide. Each of these pieces should be ____ " long.

4. If 3 people can make 3 widgets in 3 hours then 25 people can make ___ widgets in 3 hours.

5. Burt and George set their watches correctly at 6 pm Wednesday. Burt's watch is fast and gains 1 min every 6 hrs. George's watch is slow and loses 1 min every 9 hours. At 9 A.M. actual time Friday morning, Burt's watch is ___min_____secs ahead of George's watch.

6. Marcy gave Juanita as many dimes as Juanita had. The Juanita gave Marcy as many dimes as Marcy had left. Juanita now had 26 dimes and Marcy had 20 dimes. Originally, Marcy had _____ dimes.

[hide]
1. 24
2. not enough information for an answer
3. um... not sure.
4. 25
5. 10, 50
6. 28
[/hide]

[quote="Klebian"][hide]
2. not enough information for an answer
[/quote]
Sorry, for # 2 I forgot to type they were SQUARE pieces.

What's Cml stand for? I havent seen this mentioned before.


[quote="sami.d"]Hello
Prove this:
[img]http://images3.hiboox.com/images/0508/8x4zwji4.gif[/img][/quote]
hi $ suppose f(x)\equal{}x\minus{}sin(x) and g(x)\equal{}x\minus{}tan(x) then calcul f'(x) and g'(x) ..................$[/img]

Thanks
but,without derivative if possible ^^


Sir i dont know how many goiitians are here..............
but i am one and hv a very high altitude
i am not bragginfg abt myself....... but me in 9 and i can solve iit physics sum 80% my id  http://www.goiit.com/user/profile/42885.htm
will prove it ............  now i want to go to the next level in physics thats.............advanced physics pls tell me where to dig in

pls pls

i dont think i understand what youre trying to say...

if you want books for ipho prep you should check out the feynman lectures. beyond that just solve problems.

sir i think that the theory for iit and that for ipho is basicall the same (assumption)  but where should we start solving which buk

You see varun thats one of the misconceptions in the mainstream today...
All of physics(and mathematics) teaching is focused on problemsolving... But the thing is thats just not the  point. The objective of physics, and science overall is not for that to solve the thousandth stupid problem... but to build theories, and perform experiments to verify the theories and understand nature itself.
To be honest i had the same misconception at your age, i thought physics was just solving problems, but then later my eyes opened n looked at the deep and beautiful theory behind it.
Thats a big problem in university too, that all of it is just focused on solving problems, and doing your exams.. but you wont get anywhere with this mentality..

My personal experience, is that i just got burned out of solving problems.. just became sick of it... no challenge at all.. even on uni level... you just build the solution up from the basics and you are done. You didnt learn anything new, or didnt broaden your knowledge.. you just used up what you knew already and thats it... just plain boredom...
So the real challenge is to build a new theory and just learn more theory...

And my advice is just about the same.. dont focus on the itty-bitty stupid problems..(of course you should know how to solve them because unfortunately that what counts at the uni entrance exam..) but just learn the theory behind it.

So good luck to you Varun !!!

Cheers


function
LaTeX
calculus
derivative
limit
complex analysis
algebra

Hello,

I am trying to understand test functions, I will restrict to the one dimensional case

consider a function $0\neq f\in C^{\infty}(\mathbb{R})$ that is nonzero in a bounded subset of the reals

what are the easiest examples?

my professor told us :

let f be $e^{\frac{1}{x^2-1}}^$ in $]-1,1[$ and zero elsewhere, or thus :
$f(x)=\{\begin{array}{c}0 \ |x|\leq 1\ e^{\frac{1}{x^2-1}}\end{array}$
(why does this look so bad in latex? :huh:  :blush: )

now with functions defined in this way, isn't it terribly hard to check that this really is infinitely differentiable, it's like everyone says it's obvious but nobody bothers to demonstrate  :roll:

And by the way : is this one of those weird infinitely differentiable functions that do not have a power series at each point, the kind that doesn't exist in complex analysis?

Use & to separate items on the same line of an array. Also, "cases" instead of "array" is nice for this purpose.

If you're checking differentiability: The $n$-th derivative is $q(x)\exp(\frac1{x^2-1})$ for some rational function $q$ inside the interval $(-1,1)$. Since $\lim_{x\to 1^-}\frac{\exp(\frac{-1}{(1-x)(1+x)})}{(1-x)^n}=0$ for all $n$, this tends to zero at $1$ and all derivatives are zero at 1. A similar argument gives us differentiability at $-1$.

This function does fail to be analytic at those two points; its Taylor series does not converge to it on an open interval. That failure is necessary, since a real analytic function which was zero in any interval would be zero on the whole real line.


Let $ p$ be the greatest prime factor of 9991. Then, the sum of the digits of $ p$ is

$ \text{(A)}\ 4 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 28$

91 x 109 is not equal to 9991 (just see the units digit).

lol he dropped the square root. Its 103X97, 103 is greatest, so the answer is A, 4.

[hide]
\[ 9991 \equal{} 10000 \minus{} 9 \equal{} 100^2 \minus{} 3^2 \\
\equal{} (100 \plus{} 3)(100 \minus{} 3) \\
\equal{} (103)(97)
\]
Both $ 103$ and $ 97$ are prime. Thus, the greatest prime factor of $ 9991$ is $ 103$, so $ p \equal{} 103$, and $ S(p) \equal{} 4 \Longrightarrow \boxed{A}$.[/hide]

[color=blue] $9991=7.13.103$ and no combination can generate a factor $>103$ prime so it is largest
$S(n)=1+0+3=4$[/color]

9991=10000-9=100^2-3^2=(100+3)(100-3)=103*97
Now both of 103 & 97 are prime.
So there are no other prime factors.
Largest Prime divisor of 9991=103.
Sum of digits=1+0+3=4.
Ans. (A) 4.


how do you do this problem?

my numer is greater than 2000, but less than 5000. it's unit's digit is 9 and it has exactly 3 factors. find the sum of all the numbers that could be my number.

please include an explanation

 :pilot:  :starwars:  :spider:

[hide]OK. One thing you just HAVE to know, is that if a number has an odd number of factors, its a perfect square (try it out, you'll see, and it's easy to prove why)

Also, the only way to get a perfect square to end in 9 is if the number you squared had a units digit of 7 or 3.
First we try to get the range of squares.
2000 = a bit more than $44^2$
5000 = a bit less than $71^2$
So our choices are:
$49^2, 53^2, 59^2, 63^2, 69^2$
Since "my number" has 3 factors, the number we're squaring must be prime.
That leaves $53^2 and 59^2$

Just square both, 2809 and 3481
Add together to get 6290.[/hide]

[hide="solution"]Since this number has 3 factors, it must be a perfect square. As it ends with 9, its square root must end in 3 or 7. By trial and error, we see that all the numbers between 45 and 70 have squares between 2000 and 5000. All the numbers inbetween that end in 3 or 7 are: 47, 53,57, 63, and 67. Now, we must check to see that their squares have only 3 factors. This will happen if the number isn't composite, so we delete 57 and 63. $47^2+53^2+67^2=9507$.[/hide]
Klebian, do 49, 59, and 69 end in 3 or 7?

[hide]Well, odd number of factors-->square of a prime number
$\sqrt{2000}$~$44.72$
$\sqrt{5000}$~$70.71$
To end in 9, the number must end in 7 or 3
That leaves 47, 53, 57, 63, 67
Or course, the number must be prime, so eliminate 57 and 63, both multiples of 3
$47^2+53^2+67^2=\boxed{9507}$[/hide]

[hide]
First thing you notice is that it has to be a square of a prime.
This is true because the prime factorization must be in the form $a^2$, where $a$ is prime, so add one to 2 and get 3 factors.
For the square of the number to end in 9, it has to end in 7 or 3.
The boundaries of the squares is
$\sqrt{2000}\leq x\leq \sqrt{5000}\approx 45\leq x\leq 70$
For the ones ending in 3, 53 is prime
For the ones ending in 7, 47,67 are prime.
Summing the squares, we get $53^2+47^2+67^2= 9507$
[/hide]


How to understand the line denoted by red line ?

Thanks

I agree the conclusion does not at all follow immediately. See below its detailed proof.

The proof of point (i) showed that the set of cluster points of the sequence $ (x_n)$ is the closed interval $ [\ell_{ \minus{} }, \ell_{ \plus{} }]$. For $ a \in (\ell_{ \minus{} }, \ell_{ \plus{} })$, this means there exists a sub-sequence $ (x_{k_n})$ converging to $ a$. But then, by the continuity of $ f$, the sequence $ (f(x_{k_n}))$ converges to $ f(a)$. According to the definition, $ f(x_{k_n}) \equal{} x_{k_n \plus{} 1}$. Now, the sequence $ (x_{k_n \plus{} 1} \minus{} x_{k_n})$ is a sub-sequence of the sequence $ (x_{n \plus{} 1} \minus{} x_n)$, which is given to converge to $ 0$. Since any sub-sequence of a convergent sequence is convergent to that same limit, it follows that $ (x_{k_n \plus{} 1} \minus{} x_{k_n})$ converges to $ 0$. But $ (x_{k_n})$ converges to $ a$, so $ (x_{k_n \plus{} 1})$ also converges to $ a$. Only that we proved just before that $ (x_{k_n \plus{} 1})$  converges to $ f(a)$. It follows $ f(a) \equal{} a$.

[quote="mavropnevma"]I agree the conclusion does not at all follow immediately. See below its detailed proof.

The proof of point (i) showed that the set of cluster points of the sequence $ (x_n)$ is the closed interval $ [\ell_{ \minus{} }, \ell_{ \plus{} }]$. For $ a \in (\ell_{ \minus{} }, \ell_{ \plus{} })$, this means there exists a sub-sequence $ (x_{k_n})$ converging to $ a$. But then, by the continuity of $ f$, the sequence $ (f(x_{k_n}))$ converges to $ f(a)$. According to the definition, $ f(x_{k_n}) \equal{} x_{k_n \plus{} 1}$. Now, the sequence $ (x_{k_n \plus{} 1} \minus{} x_{k_n})$ is a sub-sequence of the sequence $ (x_{n \plus{} 1} \minus{} x_n)$, which is given to converge to $ 0$. Since any sub-sequence of a convergent sequence is convergent to that same limit, it follows that $ (x_{k_n \plus{} 1} \minus{} x_{k_n})$ converges to $ 0$. But $ (x_{k_n})$ converges to $ a$, so $ (x_{k_n \plus{} 1})$ also converges to $ a$. Only that we proved just before that $ f(x_{k_n })$  converges to $ f(a)$. It follows $ f(a) \equal{} a$.[/quote]
Thanks very much!


superior algebra
superior algebra unsolved

Let $ R\equal{}\mathbb{Z}[\sqrt{3}]$. Prove/disprove: $ 1\plus{}\sqrt{3}$ is reducible in $ R$.


[b]Είναι μάλλον άσκηση επιπέδου ΙΜΟ.[/b]

Σε ένα τετράπλευρο $ ABCD$ η διαγώνιος $ AC$ διχοτομεί τη γωνία $ BAD$.Πάνω στην πλευρά $ CD$ παίρνουμε τυχαίο σημείο $ E$.Η  $ AC$ τέμνει την ευθεία  $ BE$  στο $ F$ και η  ευθεία $ DF$ τέμνει την $ BC$ στο σημείο $ G$.Να αποδείξετε ότι

  $ \angle CAE \equal{}\angle CAG$.

[u] Μπάμπης[/u]

[color=red] Edited.[/color]

[quote="stergiu"][b]Είναι μάλλον άσκηση επιπέδου ΙΜΟ.[/b]

Σε ένα τετράπλευρο $ ABCD$ η διαγώνιος $ AC$ διχοτομεί τη γωνία $ BAD$.Πάνω στην πλευρά $ CD$ παίρνουμε τυχαίο σημείο $ E$.Η  $ AC$ τέμνει την ευθεία  $ BE$  στο $ F$ και η  ευθεία $ DF$ τέμνει την $ BC$ στο σημείο $ G$.Να αποδείξετε ότι

  $ \angle CAE \equal{}\angle CAG$.

[u] Μπάμπης[/u][/quote]

κ.Στεργίου μήπως εννοείτε ότι  $ \angle CAE \equal{}\angle CAG$ ?   :huh:

[quote="bilot4"][quote="stergiu"][b]Είναι μάλλον άσκηση επιπέδου ΙΜΟ.[/b]

Σε ένα τετράπλευρο $ ABCD$ η διαγώνιος $ AC$ διχοτομεί τη γωνία $ BAD$.Πάνω στην πλευρά $ CD$ παίρνουμε τυχαίο σημείο $ E$.Η  $ AC$ τέμνει την ευθεία  $ BE$  στο $ F$ και η  ευθεία $ DF$ τέμνει την $ BC$ στο σημείο $ G$.Να αποδείξετε ότι

  $ \angle CAE \equal{}\angle CAG$.

[u] Μπάμπης[/u][/quote]

κ.Στεργίου μήπως εννοείτε ότι  $ \angle CAE \equal{}\angle GAB$ ?   :huh:[/quote]

[b]ΑΠΑΝΤΗΣΗ[/b]

 Είδες ; Αυτά παθαίνει κανείς , αν πάει από τα ... χαράματα στο κομπιούτερ :( .

  Το σωστό είναι    $ \angle CAE \equal{}\angle CAG$. Σε ευχαριστώ και συγνώμη για την ταλαιπωρία.

Ευτυχώς γιατί προσπαθούσα απεγνωσμένα να βρω κάποιο λάθος στη λύση μου που με οδηγούσε σε αυτό το συμπέρασμα :P

Θα δουλέψουμε αντίστροφα. Θα φέρουμε δηλαδή μια ευθεία KA όπου Κ επί της $ BC$ τέτοιο ώστε $ \angle KAC\equal{}CAE$ και θα αποδείξουμε ότι ισχύει $ D, F, K$ συνευθειακά κι οπότε το Κ ταυτίζεται με το G.
Οπότε έχουμε $ \angle KAC\equal{}CAE\equal{}y$ και $ \angle KAB\equal{}DAE\equal{}x$

Εχουμε:

$ \frac{DE}{DC}\equal{}\frac{(DAE)}{(DAC)}\equal{}\frac{sin{x}*AD*AE}{sin(x\plus{}y)*AD*AC}$, (όπου $ (XYZ)$ εννοούμε το εμβαδόν του τριγώνου ΧΥΖ.)

Eφαρμόζουμε το  αντίστροφο του θεωρήματος  Μενελάου στο τρίγωνο $ EBC$ για τα σημεία $ G,F,D$ και χρησιμοποίουμε την παραπάνω διαπίστωση για τους τρεις λόγους που δημιουργούνται.

Ετσι:

$ \frac{DE}{DC}*\frac{CG}{GB}*\frac{BF}{FE}\equal{}$

$ \equal{}\frac{sin{x}*AD*AE}{sin(x\plus{}y)*AD*AC}*\frac{sin(x\plus{}y)*AB*AF}{sin{y}*AF*AE}*\frac{sin{y}*AK*AC}{sin{x}*AB*AK}\equal{}1$

 Οπότε αποδείχθηκε το ζητούμενο

[quote="bilot4"]Ευτυχώς γιατί προσπαθούσα απεγνωσμένα να βρω κάποιο λάθος στη λύση μου που με οδηγούσε σε αυτό το συμπέρασμα :P

Θα δουλέψουμε αντίστροφα. Θα φέρουμε δηλαδή μια ευθεία KA όπου Κ επί της $ BC$ τέτοιο ώστε $ \angle KAC \equal{} CAE$ και θα αποδείξουμε ότι ισχύει $ D, F, K$ συνευθειακά κι οπότε το Κ ταυτίζεται με το G.
Οπότε έχουμε $ \angle KAC \equal{} CAE \equal{} y$ και $ \angle KAB \equal{} DAE \equal{} x$

Εχουμε:

$ \frac{DE}{DC}\equal{}\frac{(DAE)}{(DAC)}\equal{}\frac{sin{x}*AD*AE}{sin(x\plus{}y)*AD*AC}$, (όπου $ (XYZ)$ εννοούμε το εμβαδόν του τριγώνου ΧΥΖ.)

Eφαρμόζουμε το  αντίστροφο του θεωρήματος  Μενελάου στο τρίγωνο $ EBC$ για τα σημεία $ G,F,D$ και χρησιμοποίουμε την παραπάνω διαπίστωση για τους τρεις λόγους που δημιουργούνται.

Ετσι:

$ \frac{DE}{DC}*\frac{CG}{GB}*\frac{BF}{FE}\equal{}$

$ \equal{}\frac{sin{x}*AD*AE}{sin(x\plus{}y)*AD*AC}*\frac{sin(x\plus{}y)*AB*AF}{sin{y}*AF*AE}*\frac{sin{y}*AK*AC}{sin{x}*AB*AK}\equal{} 1$

 Οπότε αποδείχθηκε το ζητούμενο[/quote]

[b] Πολύ ωραία ! [/b]

 Προφανώς εννοείς :

$ \frac{DE}{DC}*\frac{BF}{FE}*\frac{CK}{KB}$ = $ \equal{}\frac{sin{x}*AD*AE}{sin(x\plus{}y)*AD*AC}*\frac{sin(x\plus{}y)*AB*AF}{sin{y}*AF*AE}*\frac{sin{y}*AK*AC}{sin{x}*AB*AK}\equal{} 1$

δηλαδή  στο δεύτερο κλάσμα έβαλα όπου G to K.Είναι άνευ ουσίας typo .

[u] Μπάμπης[/u]

Το θεώρημα αυτό είναι πολύ χρήσιμο, αφού μας επιτρέπει να κατασκευάσουμε συμμετρική ημιευθείας ως προς ευθεία, χρησιμοποιώντας [u]μόνο[/u] κανόνα

[size=150]Πρόβλημα[/size]
Έστω γωνία $ BAD$ και $ Ax$ η διχοτόμος της. Δίνεται σημείο $ E$ εκτός της ευθείας $ BD$. Να κατασκευαστεί η συμμετρική ημιευθεία της $ AE$ ως προς την $ Ax$

Με τα δεδομένα που έχουμε, μπορούμε μόνο με τη χρήση κανόνα να προσδιορίσουμε τα σημεία $ C,F,G:$
Η $ DE$ τέμνει την $ Ax$ στο $ C$
Η $ BE$ τέμνει την $ Ax$ στο $ F$
Η $ DF$ τέμνει την $ BC$ στο $ G$

Οι $ AE,AG$ είναι συμμετρικές ως προς την $ Ax$

Δια του $ A$ φέρνουμε την κάθετη ευθεία επί την $ AC,$ η οποία τέμνει την ευθεία $ BD$ στο σημείο $ P$ και έστω $ Q,$ $ Q',$ τα σημεία τομής της $ AC,$ από τις ευθείες $ EG,$ $ BD,$ αντιστοίχως.

Στο τρίγωνο $ \bigtriangleup ABD,$ τα σημεία $ P,$ $ Q',$ είναι αρμονικά συζυγή των $ B,$ $ D,$ ως ίχνη των διχοτόμων ( εξωτερικής και εσωτερικής αντιστοίχως ), της γωνίας $ \angle BAD.$

Στο τρίγωνο $ \bigtriangleup CBD,$ επειδή οι ευθείες $ BE,$ $ CQ',$ $ DG$ είναι συντρέχουσες, συμπεραίνουμε ως γνωστό, ότι η ευθεία $ EG,$ διέρχεται δια του $ P$ $ ($ το αρμονικό συζυγές του $ Q',$ ως προς τα $ B,$ $ D$ $ ).$

Έτσι τα σημεία $ P,$ $ G,$ $ Q,$ $ E,$ αποτελούν αρμονική τετράδα ( αρμονική σημειοσειρά ), ως προβολές από το σημείο $ C,$ της αρμονικής σημειοσειράς $ P,$ $ B,$ $ Q',$ $ D,$ επί την ευθεία $ PE.$

Έχουμε επομένως, το ότι η δέσμη $ A.PGQE,$ είναι αρμονική.

Επειδή τώρα, δύο ακτίνες αυτής της δέσμης είναι κάθετες μεταξύ τους $ ($ $ AP\perp AQ$ $ ),$ συμπεραίνουμε ότι είναι και διχοτόμοι ( εσωτερική και εξωτερική ), της γωνίας που σχηματίζουν οι άλλες δύο ακτίνες της $ ($ $ \angle EAG$ $ )$ και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.


The first practice of the Western Mass ARML team is tomorrow at Hudson High School, from 9 AM - 12:30 PM.

lol,

how come eastern mass arml team starts practicing so late (around end of April?)
we should begin practice now also to topple exeter (idk if i can use those words in the same sentence like that) :)

the practice yesterday was fun thank you


geometry
rectangle
ratio
trigonometry
perpendicular bisector

M is a point and ABCD is a rectangle, if $\angle AMB = \angle DMC$, then  triangles MAB and MDC are similar.

Ill assume it is $MAB~MDC$ ;)
$\angle{AMB}=\angle{DMC}$
and $AB=CD$
so the ratio:
$\frac{\sin{\angle{AMB}}}{AB}=\frac{\sin{\angle{DMC}}}{CD}$
 is constant
So, the triangles are similiar. :D

How does that ratio prove similarity? I can easily draw two triangles with a common angle and opposite side that are not similar.
I doubt that Mamat's original assertion is true in the first place.

Look at this,
$\displaystyle \frac{\sin{\angle{AMB}}}{AB}=\frac{\sin{\angle{DMC}}}{CD}$
$\frac{\sin{MCD}}{MD}$ is same as above.
So,
$MD=\frac{\sin{MCD}*AB}{\sin{AMB}}$
Similiarily in the different triangle,
$AM=\frac{\sin{ABM}*AB}{\sin{AMB}}$
So,
dividing the 2,
$\frac{MD}{AM}=\frac{\sin{MCD}}{\sin{ABM}}$
well it is not working but I believe there is a definate proof.

I think that frt is right, moreover triangles are congruent. You can prove it by using sine law or vectors.

[quote="frt"]If the only case to satisfy $\angle AMB=\angle DMC$ is when $M$ is on the perpendicular bisector of $AD$, then Mamat's statement is probably true.

Is there any other possible location for $M$? :?[/quote]
M must lie on the perpendicular bisector of AD. Check the geometry unsolved section. There is a problem similar to that.

How do you prove no other set of points will suffice????

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46346]http://www.mathlinks.ro/Forum/viewtopic.php?t=46346[/url]
I think this should help.

[quote="Rushil"]So the basic method is to make some cases and disprove them??[/quote]
Yes.

This problem isn't correct.   If  A,B,C,D,M lie on the same circle then that triangles aren't always congruent.
If M doesn't lie on that circle then  triangles are congruent.

   Sorry for my English ( It's very bad  :oops: )


geometry
rotation
trigonometry
angle bisector

OMG your back, even if I don't know you!

[quote="CircleSquared"]With a problem!

You have a triangle ABC.  Draw equilateral triangles ABD and ACE such that DA and EA are on the outside of ABC.  Then draw DC which intersects AB at F and EB which intersects AC at G.  DC and EB intersect at P.
Prove the area of AFPG = the area of PBC.[/quote]
Are you sure that's right?

[asy]size(200); pair a,b,c,d,e; a=(0,0); b=(3,0); c=a+5*dir(110); draw(a--b--c--cycle); d=rotate(-60,a)*b; draw(a--b--d--cycle); e=rotate(60,a)*c; dot(a); dot(b); dot(c); dot(d); dot(e); draw(a--c--e--cycle); draw(b--e); draw(c--d); label("$A$",a,E); label("$B$",b,E); label("$C$",c,E); label("$D$",d,E); label("$E$",e,E);[/asy]

Sorry, forgot to put that A=60.   :rotfl:

[quote="CircleSquared"]Sorry, forgot to put that A=60.   :rotfl:[/quote]
wow thanks

[quote="CircleSquared"]Sorry, forgot to put that A=60.   :rotfl:[/quote]

does that even help? A is a point...

*cough*degreemeasure*cough*

I hope that helps?

[hide="Solution"]
Okay, so AC=AE, AC=AD, and angles ABE and DAC are both equal since they are both 120.  So, [DAC]=[BAE] because the two triangles are congruent.  Now, we have that AF bisects angle DAC and AC bisects angle BAE.   Call AB=x and AC=y.  So, by the Angle Bisector Theorem,
\[ \frac {[AFD]}{[AFC]} \equal{} \frac {DF}{FC} \equal{} \frac {AD}{AC} \equal{} \frac {AB}{AC} \equal{} \frac {x}{y}
\]
So,
\[ [AFD] \equal{} (\frac {x}{x \plus{} y})([ADC])
\]

\[ [AFC] \equal{} (\frac {y}{x \plus{} y})([ADC])
\]
Also,
\[ \frac {[ABG]}{[AGE]} \equal{} \frac {BG}{GE} \equal{} \frac {AB}{AE} \equal{} \frac {AB}{AC} \equal{} \frac {x}{y}
\]

\[ [AGB] \equal{} (\frac {x}{x \plus{} y})([BAE]) \equal{} [AFD]
\]

\[ [AGE] \equal{} (\frac {y}{x \plus{} y})([BAE]) \equal{} [AFC]
\]
So, we can say that
\[ [AFC] \plus{} [AGB] \equal{} [AFD] \plus{} [AGE]
\]

\[ 2[AGPF] \plus{} [PFB] \plus{} [GPC] \equal{} [AFD] \plus{} [AGE]
\]

\[ [AGPF] \plus{} ([APGF] \plus{} [PFB] \plus{} [GPC]) \equal{} [AFD] \plus{} [AGE]
\]

\[ [APGF] \plus{} ([ABC] \minus{} [PBC]) \equal{} [AFD] \plus{} [AGE]
\]

\[ [APGF] \equal{} ([AFD] \plus{} [AGE] \minus{} [ABC]) \plus{} [PBC]
\]

\[ [AFD] \plus{} [AGE] \equal{} (\frac {x}{x \plus{} y})[BAE] \plus{} (\frac {y}{x \plus{} y})[BAE] \equal{} [BAE] \equal{} \frac {(AB)(AE)\sin 120}{2} \equal{} \frac {xy\sin 120}{2} \equal{} \frac {xy\sin 60}{2} \equal{} [ABC]
\]

\[ [AFD] \plus{} [AGE] \equal{} [ABC]
\]

\[ [ABC] \minus{} [AFD] \minus{} [AGE] \equal{} 0
\]

\[ [APGF] \equal{} 0 \plus{} [ABC] \equal{} [ABC]
\]
[/hide]

Does anyone want me to post a really long, complicated solution?

[quote="ProtestanT"]What is the angle bisector theorem?[/quote]
[hide="Angle Bisector Theorem and Its Proof"]
Take a triangle ABC and take angle bisector AD so that D is on BC.  Now, drop perpindiculars DX to AB and DY to AC.  Now, notice the following: $ \angle XAD \equal{} \angle YAD$, $ AD \equal{} AD$, and $ \angle AXD \equal{} 90 \equal{} \angle AYD$.  So, we have that triangles AXD and AYD are congruent and thus that $ DX \equal{} DY$ and say that these lengths are both equal to $ x$.

Now, we have that $ [ABD] \equal{} \frac {DX(AB)}{2} \equal{} \frac {ABx}{2}$.  Also, we have that $ [ACD] \equal{} \frac {DY(AC)}{2} \equal{} \frac {ACx}{2}$ (I use $ [ACD]$ to mean the area of triangle ACD).  So, $ \frac {[ABD]}{[ACD]} \equal{} \frac {\frac {ABx}{2}}{\frac {ACx}{2}} \equal{} \frac {AB}{AC}$.

Now, also notice that $ \frac {[ABD]}{[ACD]} \equal{} \frac {BD}{CD}$.  Thus, we have that $ \frac {BD}{CD} \equal{} \frac {[ABD]}{[ACD]} \equal{} \frac {AB}{AC}$, so $ \frac {BD}{CD} \equal{} \frac {AB}{AC}$.  This is known as the Angle Bisector Theorem.
[/hide]

Dude, I hope you realize that ProtestanT is a genius. He definitely already knows the Angle Bisector Thm.


Let $a_{n}$ be the last nonzero digit in the decimal representation of the number $n!$. Does the sequence $a_{1}$, $a_{2}$, $a_{3}$, $\cdots$ become periodic after a finite number of terms?

$ f(n)$ is last digit different from 0 of $ n!$
Suppose this sequence period T
Thwn exist $ m,k$ such that :
$ 10^m\equiv 10^k(\mod T)$
We has $ f(10^k) \equal{} f(10^{k \minus{} 1})$
$ f(10^k \plus{} 10^k \minus{} 10^m) \equal{} f(10^k \plus{} 10^k \minus{} 10^m \minus{} 1)$
So $ f(2.10^k \minus{} 10^m) \equal{} f(2.10^k \minus{} 10^m \minus{} 1)$
Other:
$ (2.10^k \minus{} 10^m)! \equal{} (2.10^k \minus{} 10^m \minus{} 1)!(2.10^k \minus{} 10^m)$
The last digit different from 0 of $ 2.10^k \minus{} 10^m$ is 9
So $ f(2.10^k \minus{} 10^m) \equal{} 9f(2.10^k \minus{} 10^m)$
So $ f(2.10^k \minus{} 10^m \minus{} 1)\equiv 9f(2.10^k \minus{} 10^m \minus{} 1)(\mod 10)$ so $ f(2.10^k \minus{} 10^m \minus{} 1) \equal{} 5$
But $ v_{2}(2.10^k \minus{} 10^m \minus{} 1) > v_{5}(2.10^k \minus{} 10^m \minus{} 1)$
So $ f(2.10^k \minus{} 10^m \minus{} 1)$ can not be 5 (tradition).
So the sequence can not period.


Evaluate $ \sum_{n\equal{}1}^{\infty}\frac{F_n}{5^n}$ where $ F_n$ are the terms in the Fibonacci sequence $ (F_0\equal{}F_1\equal{}1, \text{ and } F_n\equal{}F_{n\minus{}1}\plus{}F_{n\minus{}2} \text{ for } n \geq 2).$

[hide="approach 1"] Let the sum be S. Consider S, 5S, and 5S-S=4S.[/hide]

[hide="approach 2"] Use Binet to split it into two geometric series.[/hide]

[quote="unimpossible"][hide]im assuming thast a 5^n.

Using the formula for a geometric series, would it be 1/(1-phi/5)?[/hide][/quote]

The Fibonacci series is not a geometric series, although that approach can be salvaged (see Approach 2 in CatalystOfNostalgia's post).  I would rephrase the problem in the following way:

What is the generating function $ F(x) \equal{} \sum_{k\equal{}0}^{\infty} F_k x^k$ (with $ F_0 \equal{} 0, F_1 \equal{} 1$)?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178530&search_id=825310110]my proof of Binet's formula by generating functions[/url].  (This is the reverse implication of Approach 2.)


geometry
3D geometry
sphere
topology
function
group theory
abstract algebra

Can anyone clarify, a bit, what exactly CP^2 is. I am reading about it on some online lectures and am not real clear about it...

Formally, $ \mathbb{C}P^2$ is the set of equivalence classes on ordered triplets $ \mathbf{x} \equal{} (x_1, x_2, x_3) \in \mathbb{C}^3$ under the equivalence relation $ \mathbf{x} \sim \mathbf{y} \Leftrightarrow \exists \lambda \in \mathbb{C}^{*} : \mathbf{x} \equal{} \lambda \mathbf{y}$.  Intuitively, it is the set of lines through the origin in complex 3-space, which can also be identified with the complex 2-sphere (not to be confused with the Riemann sphere!).

I suppose that would be because the Riemann Sphere has a hole in the north pole... So you are saying that CP^2 is homeomorphic to the complex 2-sphere...or just any 2 sphere in general?

The Riemann sphere can be identified with $ \mathbb{C} P^1$, not with $ \mathbb{C} P^2$, and it doesn't have a hole anywhere.  As for your second question, I can't say for certain.

You think the Riemann Sphere is defined at (0,0,1)? But anyway...Can anyone say if CP^2 is homeomorphic to a 2 sphere?...S^2 I believe.

What do you mean by a complex $ 2$-sphere¿
The real $ 2$-sphere has dimension $ 2$, $ \mathbb CP^2$ has dimension $ 4$ (as real manifolds). Now it isn't the real $ 4$-sphere, too, as $ H_2(S^4)\equal{}0, H_2(\mathbb CP^2)\equal{}\mathbb Z$.

There are tons of interesting ways to interpret $ \mathbb{C}P^2$. If you really like topological groups you can identify it as either one of the orbit spaces $ S^5/S^1$ or $ \left(\mathbb{C}\setminus\{0\}\right)/\mathbb{C}^\times$.

I didn't say a complex 2 sphere..I said S^2 ...and the definition I've seen is (C^2\{0})\(C\{0}), but honestly, I have no idea what that is supposed to mean...or how I should visualize that..

$ S^2$ means the real $ 2$-sphere (the normal sphere everyone knows). $ (\mathbb C^2 \backslash{0}) / (\mathbb C \backslash{0})$ is by definition $ \mathbb CP$ (and $ (\mathbb C^2 \backslash{0}) \backslash (\mathbb C \backslash{0})$ is something I wouldn't call sphere as it is just $ \mathbb C^2 \backslash \mathbb C$).

PS:
[quote="Frank1987"]So you are saying that CP^2 is homeomorphic to the complex 2-sphere...[/quote]
You said complex $ 2$-sphere ;)

Ahh..I think I said that because the previous post used it...I just assumed it would be a unit 2 sphere on the complex plane...which is clearly homeomorphic to S^2...But still the question remains? Is CP^2 homeomorphic to either? And how am I supposed to visualize C^2\C ? What is that? and what is the difference between \ and /...geez...now of this is in any text of mine

[quote="Frank1987"]You think the Riemann Sphere is defined at (0,0,1)?[/quote]
I'm not sure you understand the [url=http://en.wikipedia.org/wiki/Riemann_sphere]definition[/url] of the Riemann sphere.  The Riemann sphere is not the image of the inverse of the stereographic projection.

[quote="Frank1987"]Ahh..I think I said that because the previous post used it...I just assumed it would be a unit 2 sphere on the complex plane...which is clearly homeomorphic to S^2...[/quote]
ZetaX has already stated that this is not true (unless you mean something different by "unit 2-sphere in the complex plane":  I mean $ \{ (x, y, z) \in \mathbb{C}^3 : x^2 \plus{} y^2 \plus{} z^2 \equal{} 1 \}$, which is homeomorphic to $ \mathbb{C}P^2$ unless I have misunderstood the definition of homeomorphic.)

Homeomorphic just means that there exists a function from one space to another that is 1-1, onto, and has continuous inverse.... Can anyone explain a little more how it is homeomorphic.. If that was done, I am sure that would help with trying to see the connection between the two.

$ \mathbb{C}P^2$ is a set of equivalence classes of points in $ \mathbb{C}^3$.  The set of points on the sphere $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ consists of exactly one representative from each of these equivalence classes.  In other words, given a point $ (X : Y : Z) \in \mathbb{C}P^2$ we map it to the point $ \left( \frac{X}{\sqrt{X^2 \plus{} Y^2 \plus{} Z^2} }, \frac{Y}{\sqrt{X^2 \plus{} Y^2 \plus{} Z^2} }, \frac{Z}{ \sqrt{X^2 \plus{} Y^2 \plus{} Z^2} }\right)$ on the complex $ 2$-sphere.

[quote="Frank1987"]And how am I supposed to visualize C^2\C ? What is that? and what is the difference between \ and /...geez...now of this is in any text of mine[/quote]

I'm pretty sure that $ X\setminus Y$ is the typical removing $ Y$ from the set $ X$ (that's how I was using it anyway). The quotient I was using is $ X/Y$ which is formed by looking at the cosets. $ X/Y\equal{}\{x\plus{}Y: x\in X\}$. Cosets partition the space $ X$ and so we have an equivalence relation $ x\sim y$ if $ x\in y\plus{}Y$. Now the topology on the space is the quotient topology $ X/\sim $ which means the opens sets $ U\in X/\sim $ are precisely the ones such that $ q^{\minus{}1}(U)\in X$ are open (where $ q$ is the natural mapping sending a point to its equivalence class $ x\mapsto [x]$).

Oh, I see. You are talking about the quotient group....the set of all left cosets.. Well, yeah... I suppose some additional questions come up...If the group operation is addition, why do we need to remove the {0} from C^2 and C in (C^2\{0})/(C\{0})..but I suppose that isn't a big deal...My bigger question would be... Doesn't C need to be a subgroup of C^2 before you talk about (C^2\{0})/(C\{0})...?

You remove the $ 0$ because the group is that of multiplication (not addition). You are not building the quotient of two groups (but you could do soto get it), but you are considering the group action (=scalar multiplication) of $ \mathbb C \backslash \{0\}$ on $ \mathbb C^2 \backslash \{0\}$ by coordinatewise multiplication. The quotient then is, like for groups, the set of equivalence classes for ths action.

t0rajir0u: $ \mathbb CP^n$ is compact, but your sphere isn't, thus they can't be homeomorphic.
You use square roots of complex numbers, which can't be continuous everywhere, or you need to define your map up to multiplication yb $ \pm 1$. Additionally, $ x^2\plus{}y^2\plus{}z^2$ can be $ 0$ without all of $ x,y,z$ being $ 0$. You seem to have thought to much in the reals^^

My apologies - I was indeed thinking too much in the reals.  I suppose I really meant the set of complex numbers satisfying $ |x|^2 \plus{} |y|^2 \plus{} |z|^2 \equal{} 1$, but possibly even this is inaccurate.

Then it's a $ 5$-dimensional real manifold (more exactly, it's $ S^5$) and is lacking a structure as a complex manifold. But its projection to $ \mathbb CP^2$ is continuous and surjective (the fibres being essentially $ S^1$).

ZetaX,

That makes a little since... But given any vector, or point in C^2, multiplying it by all elements in C would span a plan in C^2...right? So CP^2 can be viewed as the collection of all planes in C^2? Each conjugacy classes being each unique plane. Is this what you are saying?

Yes, $ \mathbb CP^2$ is the union of these conjucy classes, each of the being a plane (but note that the origin is missing from each plane).

well, then yeah...CP^2 isn't homeomorphic to S^2 then..

[i]Real[/i] planes - from an algebraic point of view, they are complex lines (that is, they have complex dimension $ 1$).

ahh yes... That is a good point also.. Well, thx.. That was an interesting little trinket to learn about.


I'd like to quantify "charging by induction," since the version presented in most high-school E&M courses in America is so elementary:

Suppose we have two conducting spheres, one of radius $ R$ and carrying net charge $ \plus{}Q$, and the other of radius $ r$ and with no net charge.  The spheres are separated by a distance $ d$ at the nearest points on their surfaces (so that the distance between their centers is $ d \plus{} R \plus{} r$).  What is the net charge $ q$ induced on the sphere of radius $ r$ by the sphere of radius $ R$?

Of course, the sphere of radius $ r$ will have two regions of charge $ \minus{}q$ and $ \plus{}q$.  However, I don't know the exact continuous distribution of charge that will be produced (although I suspect that it varies linearly); the charge distribution on the sphere of radius $ R$ ought to be affected as well.  My approach was to assume that the hemisphere of the sphere of radius $ r$ closer to the other sphere would have a net negative charge varying linearly as it got closer to the center, and the other hemisphere would have a positive charge distribution varying in exactly the same way.  Then, we could integrate and conclude that the "equilibrium condition" would occur when the electrostatic forces exerted on the hemisphere of charge $ \minus{}q$ by the hemisphere of charge $ \plus{}q$ and the sphere of radius $ R$ are equal.  But I don't know if this is right...

erm... won't the large sphere of charge Q change its charge distribution in response to the induced electric field from the small sphere?
if what you're saying is right, then there will be an electric field approximately due to a dipole inside the large sphere, and its charge will create no electric field inside the sphere, so there will be a non-zero electric field inside a conductor, which is bad.
so we have to compensate.
annoying.
so now we have two continuous charge distributions to figure out.

Exactly, that's what I meant when I said, "the charge distribution on the sphere of radius $ R$ ought to be affected as well".


topology
geometry
3D geometry
sphere
trigonometry
advanced fields
advanced fields theorems

Im doing some physics research and I was wondering if anyone could help me with some basic topology and or non-euclidean geometry. I need to find out about triangles on a sphere and possibly formulas for finding the sum of the angles of a triangle on a sphere. However anything you can tell me about these topics will help.

http://en.wikipedia.org/wiki/Spherical_trigonometry (see also the links there)


How come "Lord of the Rings" is "LoTR" and not "LotR"? :)

*sigh*...just BECAUSE...Lord of "The Ring"..."The Ring" in this case is important since it's basically what the entire story revolves around. Notice that in the title, its always Lord of The Ring not Lord of the Ring...b/c The Ring is referring to The One Ring...which, if it wasnt capitalized...just wouldnt be right

Then Aragorn deserves to have the "The" too? :)

no...Aragon is just Aragon Elendil (check my spelling on that last name Jen, im not sure its right)...Elessar means "royal heir" i believe...thats why he's also the Elessar...there's no "The" b/c there doesnt need to be one..."The" in "The Ring" means "one" or "most powerful"...thats y its there...

I'm talking about "RoTK", which translates too "Return of [b]The [/b]King"

in that case, "The King" is important...look in ur ELA textbook dangit...

:lol: I'll ask my English teacher Monday *imagines her going  :shock: *... :lol:

I thought "Elessar" meant "Elfstone". I know they're both his names.

Oh man! we found a round table post from MithsApprentice/MathFiend/Tare days! :'(

You shouldn't resurrect old posts. You especially shouldn't resurrect anything MA/MF/T posted, because it brings back memories of the good old days for some of us who've been here 5 or 6 or 7 years.

yes we all loved what they posted it unfailingly brings neverending joy to our hearts

and unfailing tears to our eyes, like right now and when I was posting before.

Hmm...I haven't been around long enough to know what you're talking about. Very interesting topic, though. I'm not sure I buy it. The original title of [i]The Return of the King[/i] definitely has "the" lowercase.

But, if you want to discuss something where no one posted since 5 years ago, I think that the best is create a new topic (maybe with a link to this).


It is an idea from me but i am not sure if it is true..
For positive reals a,b,c,
a^3+b^3+c^3+(15/4)abc >= 9/8 (ab(a+b)+bc(b+c)+ca(c+a))

Sorry siuhochung,but if $c=0$ and $a=b$,your ineq is fall.

Also. When a=b=1 and c is very near to 0 it falls.


trigonometry
trig identities
Law of Cosines
SFFT
special factorizations

Given a triangle $ ABC$ with side lengths $ a,\ b,\ c$ such that $ \angle{B}\equal{}60^\circ,\ b$ is an integer, both of $ a,c$ are primes.
Prove that $ \triangle{ABC}$ is an equilateral triangle.

[hide="Fixed"]

Use law of cosines to get $ a^2 \plus{} c^2 \minus{} ac \equal{} b^2$
b obviously has to be odd, unless a=2 and c=2, in which case b=2
Otherwise:
Subtract ac from both sides
$ a^2 \minus{} 2ac \plus{} c^2 \equal{} b^2 \minus{} ac$
$ (a \minus{} c)^2 \equal{} b^2 \minus{} ac$
$ (a \minus{} c \minus{} b)(a \minus{} c \plus{} b) \equal{} \minus{} ac$
Case by case analysis from here, where $ a \minus{} c \minus{} b$ will always be negative due to the triangle side inequal:
$ a \minus{} c \minus{} b \equal{} \minus{} 1, a \minus{} c \plus{} b \equal{} ac$
Simon's favorite factoring trick on the second, substitute the first to get a contradiction
$ a \minus{} c \minus{} b \equal{} \minus{} a, a \minus{} c \plus{} b \equal{} c$
$ 2c \plus{} 2b \equal{} 4a$, $ 2c \equal{} a \plus{} b$ to get $ a \equal{} b$ and the conclusion follows
$ a \minus{} c \minus{} b \equal{} \minus{} c, a \minus{} c \plus{} b \equal{} a$
$ a \equal{} b$ direct from the first equation and the conclusion follows
$ a \minus{} c \minus{} b \equal{} \minus{} ac, a \minus{} c \plus{} b \equal{} 1$
Same as the first equation
[/hide]

EDIT: Proof added.  Commence w00ting


geometry
geometric transformation
rotation
analytic geometry
calculus
calculus computations

I'm not sure if this has been posted before, but can someone give me the formula for the volume of a rotation in polar coordinates?

What do you mean by "volume of a rotation"? Only a solid can have volume.

If your intended question is what I suspect, polar coordinates are not well suited to the problem. There would be multiple approaches leading to very different expressions, none of which are simple.


function
number theory unsolved
number theory

consider function $ \phi(n,m)$ which means amount of numbers k such that $ k<\equal{}m$ and $ gcd(k,n)\equal{}1$.

find $ \phi(n,m)$.

This ist the Legendre Totient function (see e.g. http://www.emis.de/journals/JIPAM/images/017_02_JIPAM/017_02_www.pdf)


So anyone going to both?
Just curious if there was someone I'd be spending the whole 6 weeks with as I'm going to both.
I'm taking all 4 intermediate classes.

My goal is to go from my AMC10/AIME/USAMO of 120/3/unqualified to 140+ (AMC 12)/8+/more than 1 right

Hopefully with the two camps and hard work I can achieve this.

Eh with a 140+ on AMC 12 I think a 8+ AIME won't be too tough.

I asked my parents if I could go to both, but they think I'm too immature as a 7th grade to spend "6 WHOLE weeks away from home"...

ME TOO!!
I'm going to both camp and I'll probably take all 4 intermediate classes.

Lol AIME15. I also applied for COSMOS and got in, another camp at UC Santa Cruz. Basically what's going to happen is

1) Finish COSMOS
2) Walk across campus
3) Sign in for AwesomeMath and find my room

This puts me away from home for about seven weeks. Unfortunately, "away from home" is like a half-hour drive, so I'm probably going to get unexpected parent visits. Darn.

Awesome, so there will be at least one other person going to both camps...
At camp I'll probably be one of the tallest indians who walks around asking whether anyone else is going to both camps.


Find all positive integers n such that the set
$A = 1, 3, 5, ... , 2n - 1$
can be partitioned into 12 subsets, the sum of elements in each subset being the
same.

[quote="Beat"]Find all positive integers n such that the set
$A = 1, 3, 5; ... , 2n - 1$
can be partitioned into 12 subsets, the sum of elements in each subset being the
same.[/quote]
S(A)=1+3++...+(2n-1)=n^2. Therefore 12|n^2 or 6|n. if {1,3,...,(2n-1)} can be partitioned into 12 subsets then {1,3,...(2n-1),(2n+1),...,(2n+47)} can be partitioned, because 2n+1+2n+47=2n+3+2n+45=2n+5+2n+43=...=2n+23+2n+25. Therefore if 24|n A can be partitioned. If A=A(1)+A(2)+...A(12), and n(i) is number of members A(i) we have n(i)=n(j)(m0d 2), becouse all members is odd. Therefore 12|n=n(1)+....+n(12), and 12|S(A(i))=S(A)/12.
If n=12 can not be partitioned. If n=36 we have S(A)/12=108, and 108=71+37=69+39=67+41=65+43=63+45=61+47=59+49=57+51=55+53=35+33+31+9=29+27+25+23+1+3=
=21+19+17+15+13+11+7+5. Therefore for all n, with 12|n,n>=36 A can be partitioned into 12 subsets.

your solution is not true
the answer is all $n=6k;k\geq\ 4;k\neq\ 5$

"Rust" has shown that if $A_{6k}$ = {1,3,5,...,12k-1} can be partitioned into 12 subset where the sum of elements of each subset is the same, then so can the set $A_{6(k+2)}$. This means that if we can show that $A_{36}$ (as Rust has done) and $A_{42}$ (as Rust has not done) has this property, then so does $A_{6k}$ for every natural number $k \geq 6$. The set $A_{42} = \{1,3,5,\ldots,83\}$ can be partitioned into the following 12 subset where the sum of elements of each subset is 147:

{31,33,83}
{29,37,81}
{27,41,79}
{25,45,77}
{23,41,75}
{21,45,73}
{19,49,71}
{17,51,69}
{15,55,67}
{1,11,13,59,63}
{3,7,39,47,51}
{5,9,35,43,55}

The sum of the elements of each of the 12 subsets of $A_{6k}$ must be $S_k = 3k^2$. The largest element i $A_{6k}$ is 12k-1, hence $3k^2 \, \geq \, 12k - 1$, i.e. $3k(k - 4) \, \geq \, -1$. Therefore $k \geq 4$.

For k=4 the set $A_{6k} \,=\, A_{24}$ = {1,3,5,...,47} can be partitioned into 12 subsets {2k-1,49-2k} where $1 \leq k \leq 12$ and the sum of elements is 48.

In the case where k=5 we have $A_{30}$ = {1,3,5,...,59} and $S_k = 3\cdot 5^2 = 3 \cdot 25 = 75$. Since the subsets of $A_{30}$ consist of odd integers, the number of element in each subset must be odd. The integers 59, 57, 55, 53, 51 and 49 must be in 6 different subsets because 49 + 51 = 100 > 75. By adding the elements of these 6 subsets which consist of at least 3*6 = 18 integers, we get the sum 6*75 = 450. Hence the sum of the elements remaining after the 6 integers 59, 57, 55, 53, 51 and 49 has been removed, is 450 - (59 + 57 + 55 + 53 + 51 + 49) = 450 - 324 = 126. But this sum must be $\geq$ to the sum of the first 18-6 = 12 odd natural numbers which is $12^2 = 144$. This contradiction implies that the set $A_{30}$ does not have the mentioned property.

These calculations correspond with the answer given by the signature "Beat".


function
number theory
modular arithmetic

You know how there's a pattern in the units digits of powers of number? For example, for powers of 2s, they come in cycles of 4 (2, 4, 8, 6). This is also seen in 3's, 4's, 5's, and so on. So, are there any patterns in the number of digits per cycle as the base gets bigger and bigger?

Yep.  The three standard theorems about results of this type, in increasing order of generality and difficulty:

http://en.wikipedia.org/wiki/Fermat%27s_little_theorem
http://en.wikipedia.org/wiki/Euler%27s_theorem
http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem

(Actually, Carmichael's theorem isn't standard in most treatments, but I'm trying to raise awareness here.)

I think I get them, except for maybe the carmichael thing. thanks. :D

any number to the power of five will have a units digit that is the same as the units digit of the original number.
that was awkwardly phrased. in other words, $ a^5$ will have the same units digit as $ a$.

In fact, $ a^{4k\plus{}1}$ has the same units digit as $ a$ for all integers $ k$.  Additionally, we can write that there exists an integer $ n$ such that $ a^{nk\plus{}1}$ has the same units digit as $ a$ and specifically, $ n|4$ - i.e. $ n\equal{}1,2,4$.

This result and results like it are found using modular arithmetic, usually with the theorem's t0r presented above.  Also relevant here is the concept of the order of an element.  Specifically, the smallest $ n$ that works in my formula for a specific $ a$ is called the order of $ a$ in mod $ 10$ (mod 10 is refering to the fact that we are looking at the remainders when a number is divided by 10)


linear algebra
matrix
search
induction
algebra
polynomial
calculus

Give $ A \in \text{M}_n\left(\mathbb{C}\right)$. Prove that $ \det\left(A\cdot A^{\prime} \plus{} E\right) \geq 0$, where $ E$ is the identity matrix, and $ A^{\prime}$ is the conjugate matrix of $ A$ (that is, the matrix whose entries are the complex conjugates of the respective entries of $ A$).

OK I'll give the sketch of a proof (or hints) then
1. You can suppose $ A\in \mathrm{GL}(\mathbb{C})$, then use a density argument.
2. Show that $ \det(AA'\plus{}E)\in \mathbb{R}$ (that's not difficult, and there are half a dozen ways to show it).
3. Consider $ \lambda<0$ an eigenvalue of $ AA'$ of multiplicity $ \alpha$ and show by induction that $ \ker (AA' \minus{}\lambda E)^{\alpha}$ is of even dimension.

HTH

It's a very ,very,..... very difficult problem ; it appeared 10 (15 ?) years ago in the Amer. Math. Monthly, as an exercise. Nobody (I repeate: nobody in the mathematical world) sent a complete solution; the proposer's solution was published monthes later.

loup blanc,
I think I have a solution (summary in my post above). Was the solution in the AMM going a different road?
It is not easy, but it surprises me that no one submitted a solution to the AMM. Do you have a reference?

[quote="JC_math"]
1. You can suppose $ A\in \mathrm{GL}(\mathbb{C})$, then use a density argument.
2. Show that $ \det(AA' \plus{} E)\in \mathbb{R}$ (that's not difficult, and there are half a dozen ways to show it).
3. Consider $ \lambda < 0$ an eigenvalue of $ AA'$ of multiplicity $ \alpha$ and show by induction that $ \ker (AA' \minus{} \lambda E)^{\alpha}$ is of even dimension.
[/quote]
Ok, I worked out a proof with your hints, (with some difficulty  :oops: ).
Though I don't like this density argument.
Please, let me know if there is an essentially shorter proof.

First I need the following Lemma.
[b]Lemma[/b]
$ AB$ and $ BA$ have the same characteristic polynomial
for all $ n\times n$ matrices $ A,B$ over any commutative ring $ R$.
[b]Proof[/b]
We need to show $ \det (XI_n \minus{} AB) \equal{} \det (XI_n \minus{} BA)$ in $ R[X]$.
For that write $ A \equal{} (a_{ij})_{i,j \equal{} 1,\ldots,n},B \equal{} (b_{ij})_{i,j \equal{} 1,\ldots,n}$
and take $ 2n^2$ algebraically independent variables $ A_{ij},B_{ij},i,j \equal{} 1,\ldots,n$ over $ \mathbb{Z}$.
Take the integral domain $ S: \equal{} \mathbb{Z}\left [\{A_{ij},B_{ij}: i,j \equal{} 1,\ldots, n\}\right]$ with quotient field $ K$
and canonical $ \sigma: S\to R,A_{ij}\mapsto a_{ij},B_{ij}\mapsto b_{ij}$.
Then the $ n\times n$ matrices $ C: \equal{} (A_{ij})_{i,j \equal{} 1,\ldots,n},D: \equal{} (B_{ij})_{i,j \equal{} 1,\ldots,n}$ over $ S$
are mapped to $ A,B$ under $ \sigma$,
and $ \det (XI_n \minus{} CD) \equal{} \det (XI_n \minus{} DC)$ in $ S[X]$ would give
$ \det (XI_n \minus{} AB) \equal{} \det (XI_n \minus{} BA)$ in $ R[X]$ using $ \sigma$.
But now take $ \tau: S\to \mathbb{Z}$ canonical with
$ A_{ii}\mapsto 1,A_{ij}\mapsto 0$ for $ i\not \equal{} j$ and $ B_{ij}\mapsto 0$ for all $ i,j \equal{} 1,\ldots,n$.
Then $ C$ is mapped to $ I_n$ under $ \tau$ and $ \det C$ to $ 1$ under $ \tau$.
So $ \det C\not \equal{} 0$ in $ S$, which gives $ C$ invertible over $ K$.
And now
$ \det (XI_n \minus{} CD) \equal{} \det [C(XI_n \minus{} DC)C^{ \minus{} 1}] \equal{} \det C\cdot \det (XI_n \minus{} DC) \cdot\det C^{ \minus{} 1} \\
\hspace*{0.95in} \equal{} \det (XI_n \minus{} DC)$
in $ K[X]$ and $ S[X]$.
[b]q.e.d.[/b]

This Lemma now gives
$ \chi (X): \equal{} \det (XI_n \minus{} A\overline{A}) \\
\hspace*{0.37in} \equal{} \det (XI_n \minus{} \overline {\overline{A}A}) \\
\hspace*{0.37in} \equal{} \det (\overline{XI_n \minus{} \overline{A}A}) \\
\hspace*{0.37in} \equal{} \overline {\det (XI_n \minus{} \overline{A}A)} \\
\hspace*{0.37in} \equal{} \overline {\chi (X)}$
and so $ \chi (X)\in\mathbb{R}[X]$ of degree $ n$ and leading coefficient $ 1$
and we find $ \lambda_1,\ldots,\lambda_r\in\mathbb{R},\mu_1,\ldots,\mu_s\in\mathbb{C}\setminus\mathbb{R}$ with $ r \plus{} 2s \equal{} n$ and
$ \chi(X) \equal{} \prod_{k \equal{} 1}^r(X \minus{} \lambda_k)\prod_{k \equal{} 1}^s(X \minus{} \mu_k)(X \minus{} \overline{\mu_k})$.
This gives
$ \det (I_n \plus{} A\overline{A}) \equal{} ( \minus{} 1)^n\chi( \minus{} 1) \equal{} \prod_{k \equal{} 1}^r(1 \plus{} \lambda_k)\prod_{k \equal{} 1}^s(1 \plus{} \mu_k)(1 \plus{} \overline{\mu_k}) \\
\hspace*{1.73in} \equal{} \prod_{k \equal{} 1}^r(1 \plus{} \lambda_k)\prod_{k \equal{} 1}^s|1 \plus{} \mu_k|^2$
and we need to show the number of $ \lambda_k < \minus{} 1$ is even.

For that I'll show, that the multiplicity of all $ \lambda\in\mathbb{R},\lambda < 0$ as root of $ \chi(X)$ is even.
But this multiplicity is $ \dim_{\mathbb{C}}\text{Ker}(A\overline{A} \minus{} \lambda I_n)^k$ for $ k\in\mathbb{N}$ large enough using the Jordan normal form of $ A\overline{A}$.
So it's enough to show $ \dim_{\mathbb{C}}\text{Ker}(A\overline{A} \minus{} \lambda I_n)^k/\text{Ker}(A\overline{A} \minus{} \lambda I_n)^{k \minus{} 1}$ is even for all $ k\in\mathbb{N}$.

Now $ f: \mathbb{C}^n\to \mathbb{C}^n,v\mapsto A\overline{v}$ is $ \mathbb{R}$-linear with $ f(\alpha v) \equal{} \overline {\alpha}f(v)$ ($ f$ is $ \mathbb{C}$-semilinear)
and $ f^2(v) \equal{} A\overline{A}v$ for all $ \alpha\in\mathbb{C}, v\in\mathbb{C}^n$.
So $ \text{Ker}(A\overline{A} \minus{} \lambda I_n)^k \equal{} \text{Ker}(f^2 \minus{} \lambda \text{id}_{\mathbb{C}^n})^k$ is an $ f$-invariant $ \mathbb{C}$-subspace of $ \mathbb{C}^n$ for all $ k\ge 0$. (here $ \lambda\in\mathbb{R}$!)

Take $ V_1\subsetneq \cdots \subsetneq V_l$ a maximal chain of $ f$-invariant $ \mathbb{C}$-subspaces
with $ V_1 \equal{} \text{Ker}(f^2 \minus{} \lambda \text{id}_{\mathbb{C}^n})^{k \minus{} 1}$ and $ V_l \equal{} \text{Ker}(f^2 \minus{} \lambda \text{id}_{\mathbb{C}^n})^k$ for some $ k\in\mathbb{N}$.

I'll show $ v \plus{} V_m,f(v) \plus{} V_m$ form a $ \mathbb{C}$-basis of $ V_{m \plus{} 1}/V_m$ for all $ m \equal{} 1,\ldots,l \minus{} 1$ and $ v\in V_{m \plus{} 1}\setminus V_m$.
Then $ \dim_{\mathbb{C}}V_{m \plus{} 1}/V_m \equal{} 2$ for all $ m \equal{} 1,\ldots,l \minus{} 1$ and we are done.
We need to show $ V_{m \plus{} 1} \equal{} \mathbb{C}v \plus{} \mathbb{C}f(v) \plus{} V_m$
and $ \alpha v \plus{} \beta f(v)\in V_m$ gives $ \alpha \equal{} \beta \equal{} 0$ for all $ \alpha,\beta\in\mathbb{C}$.

Clearly $ \mathbb{C}v \plus{} \mathbb{C}f(v) \plus{} V_m$ is a $ \mathbb{C}$-subspace of $ V_{m \plus{} 1}$. ($ V_{m \plus{} 1}$ is $ f$-invariant)
Also $ f(\mathbb{C}v \plus{} \mathbb{C}f(v) \plus{} V_m)\subseteq \mathbb{C}f(v) \plus{} \mathbb{C}f^2(v) \plus{} V_m$ ($ f$ is $ \mathbb{C}$-semilinear and $ V_{m}$ is $ f$-invariant)
and $ f^2(v) \minus{} \lambda v\in (f^2 \minus{} \lambda \text{id}_{\mathbb{C}^n})\text{Ker}(f^2 \minus{} \lambda \text{id}_{\mathbb{C}^n})^k\subseteq \text{Ker}(f^2 \minus{} \lambda \text{id}_{\mathbb{C}^n})^{k \minus{} 1}\subseteq V_m$
$ \Rightarrow f^2(v)\in \mathbb{C}v \plus{} V_m$.
So $ \mathbb{C}v \plus{} \mathbb{C}f(v) \plus{} V_m$ is $ f$-invariant,
and by maximal choice of the chain we get $ V_{m \plus{} 1} \equal{} \mathbb{C}v \plus{} \mathbb{C}f(v) \plus{} V_m$.

Assume $ \alpha v \plus{} \beta f(v)\in V_m$ with $ \alpha,\beta\in\mathbb{C}$.
Then since $ V_m$ is $ f$-invariant we get
$ V_m\ni f(\alpha v \plus{} \beta f(v)) \equal{} \overline{\alpha}f(v) \plus{} \overline{\beta}f^2(v)$ and then
$ V_m\ni\overline{\alpha}(\alpha v \plus{} \beta f(v)) \minus{} \beta f(\alpha v \plus{} \beta f(v)) \\
\hspace*{0.2in} \equal{} |\alpha|^2 v \minus{} |\beta|^2f^2(v) \\
\hspace*{0.2in} \equal{} (|\alpha|^2 \minus{} \lambda |\beta|^2)v \minus{} |\beta|^2(f^2(v) \minus{} \lambda v)$
with $ f^2(v) \minus{} \lambda v\in V_m$ and $ v\not\in V_m$.
So $ |\alpha|^2 \minus{} \lambda |\beta|^2 \equal{} 0$ and $ \lambda < 0$ gives $ \alpha \equal{} \beta \equal{} 0$.
This completes the proof.

[quote="olorin"]Ok, I worked out a proof with your hints, (with some difficulty  :oops: ).
Though I don't like this density argument.
Please, let me know if there is an essentially shorter proof.
[/quote]
I used the density argument to simplify my proof, but it is not essential.
My proof is not shorter than the one above (actually it is slightly longer)

The density argument is short and alright.
But it's hard to use beyond $ \mathbb{C}$.
(unless you use formal power series or some other valuation/localisation?. :maybe: )
Also you need $ \chi(X)\in \mathbb{R}[X]$, I think, not just $ \det(A\overline{A} \plus{} I_n)\in\mathbb{R}$.
(using the density argument on the coefficients of $ \chi(X)$)
The argument in the Lemma can be used to extend all kinds of known formulas to commutative rings.
So I like that better. :oops:

With "essential" I meant a different approach, not just rewritten.
(I tend to complicate my arguments with basics.)
Please take no offense.

In the final argument I basically reworked the construction of the Jordan basis.
That seems too much. So I was wondering if there is something easier.

For example my first idea was the following.
$ \det (A\overline{A} \plus{} I_n) \equal{} \sum_{J\subseteq \{1,\ldots, n\}}\det B_J$,
where $ B_J$ is the $ J\times J$-minor of $ A\overline{A}$ for all $ J\subseteq\{1,\ldots,n\}$.
(This follows from $ \det$ being multilinear and repeated Laplace expansion)
Then using [url=http://en.wikipedia.org/wiki/Minor_%28linear_algebra%29]this[/url] generalization of the [url=http://en.wikipedia.org/wiki/Cauchy-Binet_formula]Cauchy-Binet formula[/url] we get
$ \det (A\overline{A} \plus{} I_n) \equal{} \sum_{J,K\subseteq \{1,\ldots, n\},|J| \equal{} |K|}\det A_{J,K}\det \overline{A}_{K,J} \\
\hspace*{0.8in} \equal{} \sum_{J,K\subseteq \{1,\ldots, n\},|J| \equal{} |K|}\det A_{J,K}\overline{\det A_{K,J}}$

where $ A_{J,K}$ is the $ J\times K$-minor of $ A$ for all $ J,K\subseteq\{1,\ldots,n\}$.
Maybe that's another way to prove $ \det (A\overline{A} \plus{} I_n)\ge 0$.
I don't see it. :huh: (if $ A$ is not symmetric)

I don't remember the reference; but it's before 1986 because this result is in the book (in french)  "Introduction à la théorie des groupes de Lie classiques" by Mneimné and Testard, (dated 1986) in the form of a long exercise..

A funny anecdote: this question was asked during the oral examination of "polytechnique" (a top french school) before the publication of the proof. One will guess under heard.

Do you know the former student who got this question?

Yes I knew him.
 Do you want to know if this exercise depressed this student ?

If he found a solution as a student, I was just curious to know how 20+ years later this person turned out as a mathematician (or not).

He did not find and the teachers either.

If $ A \in GL_n{(\mathbb{C})}$ You can give a solution implies ? :huh:
If You have a solution with $ A \in GL_n{(\mathbb{C})}$ I will give a solution implies  with $ A \neq GL_n{(\mathbb{C})}$!!!  :roll: