vector
linear algebra
linear algebra unsolved

Let $ V$ and $ W$ be vector spaces over the same field, and let $ T\colon V \rightarrow W$ be a linear transformation. For any $ H \in \mathcal{B}(W)$, where $ \mathcal{B}(W)$ denotes the set of all bilinear forms on $ W$, define $ \widehat{T}(H)\colon V \times V \rightarrow F$ by $ \widehat{T}(H)(x,y) \equal{} H(T(x),T(y))$ for all $ x,y \in V$. Prove the following result.

If $ T$ is an isomorphism, then so is $ \widehat{T}$.

I think I have to show that $ \widehat{T}$ is surjective, but this has me stumped. Thanks for any help.

Given linear map $ T\colon V\to W$, you have constructed a linear map $ \widehat{T}\colon \mathcal B(W)\to \mathcal B(V)$. (Note the reversal of direction: $ T$ pulls bilinear forms from $ W$ to $ V$).

If $ T$ has an inverse $ T^{\minus{}1}\colon W\to V$, then you also have $ \widehat{T^{\minus{}1}}\colon \mathcal B(V)\to \mathcal B(W)$. This looks like a good candidate for the inverse of $ \widehat{T}$.

Sorry, but how would I explicitly definte $ \widehat{T^{\minus{}1}}$?

I've been fiddling away, and I can only think of showing that $ \widehat{T}$ is a bijection, and hence has the inverse $ \widehat{T^{\minus{}1}}$.

In a previous part of the exercise I showed that $ \widehat{T}$ is linear.
Also, to show $ \widehat{T}$ is an injection, I have:
Assume $ \widehat{T}(H)\equal{}O$, where $ O$ is the zero bilinear form on $ V$, so $ O(x,y)\equal{}0, \forall x,y \in V$. Then for any x,y in V
$ \widehat{T}(H)(x,y) \equal{} H(T(x),T(y)) \equal{} O(x,y) \equal{} 0$
so $ H(T(x),T(y)) \equal{} 0$ for all x,y in V.
However, since T is an isomorphism, and hence onto, it follows that T(x) and T(y) map V onto W, and so if I let $ T(x)\equal{}w_1$ and $ T(y)\equal{}w_2$ for $ w_1, w_2 \in W$, then $ H(T(x),T(y)) \equal{} H(w_1,w_2) \equal{} 0$ for all $ w_1,w_2 \in W$, and hence H is the zero bilinear form on W. So $ \widehat{T}$ is one-to-one.

I'm getting stuck showing that it's onto. I started out with:
Let $ J \in \mathcal{B}(V)$, and I want to show $ \exists H \in \mathcal{B}(W)$ such that $ \widehat{T}(H)(x,y)\equal{}H(T(x),T(y))\equal{}J(x,y)$ for all $ x,y \in V$, but I'm stumped as to where to go from there.

Thanks so far for your tip mlok.

For any linear transformation $ T$, you have $ \widehat T$ (as defined in the original post).

$ T^{\minus{}1}$ is just that - a linear transformation. If you find $ \minus{}1$ in the superscript distracting, give it another name: say, $ S\equal{}T^{\minus{}1}$. Go back to the definition of $ \widehat T$  and replace $ T$ with $ S$ everywhere.

You can show (only slightly more) generally that if $ X,Y,Z$ are vector spaces, and if $ f : X \to Y$ and $ g : Y \to Z$ then $ \widehat{gf} \equal{} (\widehat{f})(\widehat{g})$. Then show that $ \widehat{id_{X}} \equal{} id_{\mathcal{B}(X)}$. That should be enough to do the problem; you don't have to worry separately about injectivity or surjectivity.

Ah, I think I have it now.

We have $ \widehat{T}\colon\mathcal{B}(W)\to\mathcal{B}(V)$ and $ \widehat{T^{ \minus{} 1}}\colon\mathcal{B}(V)\to\mathcal{B}(W)$.

So for all $ J \in \mathcal{B}(V)$ define $ \widehat{T^{ \minus{} 1}}(J)\colon W \times W \to F$ by $ \widehat{T^{ \minus{} 1}}(J)(w,z) \equal{} J(T^{ \minus{} 1}(w),T^{ \minus{} 1}(z))$ for all $ w,v \in W$.

So let $ H \in \mathcal{B}(W), J \in \mathcal{B}(V)$ and $ T(x) \equal{} w, T(y) \equal{} z$ for arbitrary $ x,y \in V$.

Hence, for any $ x,y \in V$,
$ \widehat{T}\widehat{T^{ \minus{} 1}}(J)(x,y) \equal{} \widehat{T^{ \minus{} 1}}(J)(T(x),T(y)) \equal{} \widehat{T^{ \minus{} 1}}(J)(w,z) \equal{} J(T^{ \minus{} 1}(w),T^{ \minus{} 1}(z)) \equal{} J(x,y)$.

So $ \widehat{T}\widehat{T^{ \minus{} 1}}(J)(x,y) \equal{} J(x,y)$ for all $ x,y\in V$, so it follows that $ \widehat{T}\widehat{T^{ \minus{} 1}}(J) \equal{} J$.

Similarly, $ \widehat{T^{ \minus{} 1}}\widehat{T}(H)(w,z) \equal{} H(w,z)$ for all $ w,z\in W$, so it follows that $ \widehat{T^{ \minus{} 1}}\widehat{T}(H) \equal{} H$. Hence $ \widehat{T}$ is linear and invertible, so $ \widehat{T}$ is an isomorphism from $ \mathcal{B}(W) \to \mathcal{B}(V)$.

Again, much gratitude mlok.

Thanks also Nukular, I tried using your tip in conjunction with with mlok's suggested transformation.


HI, as part of a math project i'm doing, I have to research and prove various circle related geometry theorems. I've done thing's like Ptolemy's theorem, SImson's line, and stuff but i was wondering whether anybody knows any sites that explain and prove the theorem really simply? I've been to mathworld and cut the knot but they don't provide all the proofs and go through all the steps..

Thanks,
nic ox

Go to the Geometry theorem section, or use the search function. There must be proofs of these two famous theorem.
(I remember Yetti posted incredibly long article talking about Simson line...)


Maybe this is too easy for this section, if so, please forgive me and move this to the correct place

Prove the following formula

$\sum_{k=0}^{n}{\binom{n}{k}}^{2}= \binom{2n}{n}$

[b]Solution.[/b]

[b]Lemma (Vandermonde's Identity).[/b]  $\binom{m+n}{k}=\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i}$.

[b]Proof.[/b] Let there be a set with cardinality $m+n$, $m$ and $n$ of the objects colored with two colors.  We choose a total of $k$ elements from the set with $i$ of $n$ objects and $k-i$ of $m$ objects in $\binom{n}{i}\binom{m}{k-i}$.  Sum over all values of $i=\overline{0,k}$.  And the conclusion follows.$\blacksquare$

For $m=n=k$, we have
\[\\ \binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}\\ =\sum_{k=0}^{n}\binom{n}{k}^{2}, \]
as required.  $\blacksquare$

Is there another way to solve this?

I don't really understand that proof.

I am new to combinatorial problems like this, so I am not familiar with many theorems and such.....

Thanks for your help so much.

Ah, thanks

You are quick

I had just edited that out of my post as I saw the calculuation....

in other words, consider the coefficient of $x^{n}$ in the expansion of: $(x+1)^{n}*(x+1)^{n}=(x+1)^{2n}$

And isn't the generalized Vandermonde's just considering coeffs of $(x+1)^{m}(x+1)^{n}=(x+1)^{m+n}$

One proof uses generating functions, which is pretty standard (above), and the other uses a combinatorial argument shown above in my post.


[quote="smoothcriminal005"]Evaluate [2003!/(2002!+2001!+2000!+...+3!+2!+1!] where brackets denote the greatest integer function.
I think the answer is 2001[/quote]
[hide] We have that
\[\frac{2003!}{2002!+2001!+2000!...+3!+2!+1!}=\frac{2003!}{2003(2001!)+2000!+1999!...+3!+2!+1!}\]
We know that the denominator is more than $2003(2001!)$, so we know that the amount is less than $\frac{2003!}{2003(2001!)}=\frac{2002!}{2001!}=2002$.  Now that we have found an upper-bound, let's try to find a lower bound.  Imagine that the fraction was instead
\[\frac{2003!}{2002!+2001!+2000!(2001)}=\frac{2003!}{2002(2001!)+2001!(2)}=\frac{2003!}{2004(2001)!}=\frac{2002(2003)}{2004}\]
This is greater than 2001 (since $(2003+1)(2002-1)<2003(2002)$).  We also know that the denominator is more than the given denominator, so this is less than the actual sum.  Thus, if we call our sum $x$, then $2001<x<2002$.  So the greatest integer less than or equal to x is $\boxed{2001}$. [/hide]


The function $ f(x)$ is defined for all real values of $ x$ by $ f(x) \equal{} 2 \minus{} (x\plus{}1)^{\frac{1}{3}}$.

Evaluate $ ff(\minus{}126)$

Find the set of values of $ x$ for which $ f(x) \equal{} \mid f(x) \mid$.

Find an expression for $ f^{\minus{}1}(x)$.

To find $ f^{\minus{}1}$, write $ x\equal{}2\minus{}(f^{\minus{}1}(x)\plus{}1)^{1/3}\Rightarrow (f^{\minus{}1}(x)\plus{}1)^{1/3}\equal{}2\minus{}x\Rightarrow f^{\minus{}1}(x)\equal{}(2\minus{}x)^{3}\minus{}1$.

[quote="AndrewTom"]Evaluate $ f(f(\minus{}126))$[/quote]
We have $ 2\minus{}(\minus{}126\plus{}1)^\frac13\equal{}2\minus{}(\minus{}5)\equal{}7$.

Plugging this in, $ f(7)\equal{}2\minus{}(7\plus{}1)^\frac13\equal{}2\minus{}2\equal{}\boxed{0}$.

Also, the answer for the second one (don't have time to explain right now) is $ (\minus{}\infty,7]$.

Solution for the second one:
In order for $ f(x)\equal{}|f(x)|$ to be true, we msut have $ f(x)\ge0$. Thus, we shall first find the case where $ x\equal{}0$.
We have $ \minus{}2\equal{}\minus{}(x\plus{}1)^\frac13 \implies x\plus{}1\equal{}8 \implies x\equal{}7$ (which agrees with the statement of the previous problem). Next, we note that as $ x$ increases, $ f(x)$ decreases. Hence, the set of values is $ (\minus{}\infty,7]$.


Let A be a subset of {$1,2,3,\cdot\cdot\cdot,2n-1,2n$} containing n+1 elements. Show that
a.	Some two elements of $A$ are relatively prime.
b.	Some two elements of $A$ have the property that one divides the other.

This problem was solved by me using Pigeon Hole Principle.
But, the book in which it was there the solution told to try solving using induction also. :?:

[b]MY QUESTION IS HOW DO WE SOLVE IT USING INDUCTION[/b]

Suppose a and b are true for n. Then for n+1, the set A or whatever must contain both 2n+2 and 2n+1. Clearly (2n+2, 2n+1) = 1. Also 2n+2 is for all intents and purposes equal to n+1 in terms of numbers dividing one another.

[quote="probability1.01"]Also 2n+2 is for all intents and purposes equal to n+1 in terms of numbers dividing one another.[/quote]

What do you mean by this? :?:


calculus
derivative
function
calculus computations

I need help with this question.....Why does the function y=e^x+c have a derivative that is itself for all values of c.

thanks in advance for your help.

It does not except when $ c\equal{}0$.  :huh:
Did you mean $ y\equal{}ce^x$?


Is it possible to download an Alcumus video?
I heard you can lose your videos so I wanted to download them before I lose them.

Unfortunately, I'm not sure about the downloading question.

However, even if you lose videos, the complete list is available here:
[url]http://www.artofproblemsolving.com/Alcumus/Videos/VideoList.php[/url]

This list should be updated as new videos come and essentially includes all the videos.


Let be $ x,y,z\in \mathbb{R}$ such that $ xyz \equal{} 1$ . Find the values of $ E$ :

$ E \equal{} \frac {x \plus{} 1}{xy \plus{} x \plus{} 1} \plus{} \frac {y \plus{} 1}{yz \plus{} y \plus{} 1} \plus{} \frac {z \plus{} 1}{zx \plus{} z \plus{} 1}$

Are u sure x,y,z are real numbers? Since 2 will be the minimum value if x,y,z are positive. And I think u made a typo on the 2nd fraction. It should be $ yz\plus{}z\plus{}1$

Yes , $ x,y,z\in \mathbb{R}$ , but i changed  :D

With the condition xyz=1 we easy to prove
$ \sum_{cyc}\frac {1}{xy \plus{} x \plus{} 1} \equal{} 1$
and
$ \sum_{cyc}\frac {x}{xy \plus{} x \plus{} 1} \equal{} \sum_{cyc}\frac {1}{xy \plus{} x \plus{} 1} \equal{} 1$
Then E=2


Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ R$ its circumradius.
Prove that $ R\geq\frac {a^2 \plus{} b^2}{2\sqrt{2a^2 \plus{} 2b^2 \minus{} c^2}}$, and determine when equality occurs.

[quote="socrates"]Let $a,b,c$ the sides of a triangle and $R$ its circumradius.
Prove that $R\geq\frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}$
and determine when equality occurs.[/quote]

By a well-known formula, the length $m_c$ of the median of triangle ABC issuing from the vertex C is $m_c=\frac12\sqrt{2a^2+2b^2-c^2}$. Thus, $2\sqrt{2a^2+2b^2-c^2}=4m_c$, and hence your inequality becomes $R\geq\frac{a^2+b^2}{4m_c}$. This is equivalent to $m_c\geq\frac{a^2+b^2}{4R}$. But this was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=16023 .

  darij

ok. thank you darij !!! :lol:  :lol:  :)


ARML AMC AIME USA(J)MO USAMO vector geometry

I did 11 chapters (Ch. 1-10, 18)in AoPS Vol. 2 in two weeks; I want to finish the book before the AMC tests start. Am I working to fast? However, I know that when school starts, I could only do about $\approx$1 chapter per week.

Did you completely understand it? Have you been exposed to any of the material before? The book says to go fast when you can, so it probably isn't bad unless you aren't getting it. as for getting it done in time for the AMC tests, you have plenty of time.

I don't think 11 chapters in 14 days is too fast, as long as you're spreading them out and not doing all 11 in the same day. Keep in mind that the AMC's are still 6 months away, so you have plenty of time.

Remeinds me: How long did it take you all to get therough aops1? aops2?

When I started aops 1, it took me forever since I had no idea how to attack problems. I then came back to finish 3/4 of it recently, after learning some more basic things, and I blew through 2/3 of it in a week.

For me I just flip thru volume I (it's too easy for me at that time), and I spend some time (like a month...??) on volume II then I just go to ACOPS

I'll prob try to get vol 2 done by Christmas time and peruse vol 1 if I have time between end of December and AMC 12.

why would you do vol 2 before vol 1? :maybe:

As long as you do all or most of the problems, 11 chapters in 2 weeks is valuable.

@SplashD: Maybe AoPS 1 was too much review for them, so they decided to do AoPS 2, but then later they decided they might as well pick up any details they might have missed in AoPS 1?

[quote="JSteinhardt"]Does this include working through all of the problem sets without looking at the solutions (until after you've solved the problems)?[/quote]

For about 98% of the problems, I can do straightforward without looking at the solutions. For the ones at the very end of the problem sets (the threaded needle ones), I would sometimes look at the picture in the solutions manual or some hint, and try to solve it from there.

@k8reindeer: Yeah, I completely understand it. I tried to do the book in 8th grade and I found it too hard, but after doing an ARML problem book, AoPS Vol. 2 seemed not too hard.

Another Question: I want to get above 6 on the AIME (preferably qualify for USAMO--10th grade), is doing just AoPS Vol. 2 enough? I plan to do some recent AMC12s/AIMEs sometime before the competition.

6 is pretty easy...AoPS II is definitely enough for it.  I'll try for 2 digits.

I got 8 last year (then I try the alternate one and got 12...), I'll say ACOPS helps a lot for AIME level.  Then just do problems.

SplashD, I feel confident with most of the material in vol. 1, and wish to learn as much as possible by AMC 12. So I think working through vol 2 then checking anything I forgot/missed in vol 1 in the spare will be better. Oh yeah, and doing a lot of practice problems.

I think a good indicator of whether you're going too fast and really learning the material is to look back at it later. Give it a month or so, and if you can go back to that section and still do all of the problems (hopefully not from memory), then I'd say you have a pretty good grasp of the concepts in that section.

[quote="funcia"]vol 1 is bugging me b/c i have no idea what they're saying about imaginary numbers...i don't look at the solutions..[/quote]

You should always look at their solution after you solve a problem to see if there is a better way.

[quote="funcia"]vol 1 is bugging me b/c i have no idea what they're saying about imaginary numbers...i don't look at the solutions..[/quote]

I think ACoPS's chapter on imaginary number really grasps the main concept of imaginery numbers (in competition, at least), that basically everytime you see imaginary think geometry.  This way you usually find a nice solutions.  (like if you give me $i+1$ I think the vector sum between $<0, 1>$ and $<1, 0>$)

[quote="pkerichang"] if you give me $i+1$ I think the vector sum between $<0, 1>$ and $<1, 0>$ [/quote]

I'm glad I'm not the only one that is analytic-geometry-dependent after reading that book. I'm not too sure how much ACoPS helped with AIME, but I'm pretty sure it is helping in Olympid problems.


What are the points of intersection of the graph with equations x^2 + x and -2x^2 - x + 1?

Set up a two variable equation and subtract to obtain $3x^2+2x-1=0$.

x = 1/3, -1

Substituting, we get y = 4/9, 0

(1/3, 4/9), (-1, 0)


calculus
derivative
calculus computations

I have a question when a problem says:

Find ∂z/∂x and dz/dx if:
z=xy, where y=f(x)

If I am going to find ∂z/∂x, I will find it with ∂y/∂x !?
Is that right? and it will be ∂z/∂x=x*(∂y/∂x) + y

and how can I find dz/dx if I have to variables which are x & y, but if I am going to put f(x) instead of y and make the equation as
z=x*f(x), then I will find it with f'(x), is that right?

By the way, I am new to Partial Derivatives, so please don't get mad of me if I am asking alot of questions.   :oops:

Thanks

This belongs in the calculus section, not "Advanced fields".

Partial derivatives don't work that way. When taking a partial derivative with respect to $x$, treat $y$ (or any other variable) as a constant. The partial derivative $\frac{\partial z}{\partial x}$ in this problem is just $y$, independent of the relation between $x$ and $y$. Equivalently, $\frac{\partial y}{\partial x}$ is zero. Similarly, the partial derivative $\frac{\partial z}{\partial y}$ is $x$.

Now, we introduce a relation: Let $x=t,y=f(t)$. The multivariable chain rule tells us that $\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt}=y\cdot 1+t\cdot f'(t)=f(t)+tf'(t)$.
This can also be done by simply writing $x$ and $y$ in terms of $t$ and using the ordinary product rule.

Thank you very much jmerry for your great help!
and by the way the answer that I put in the message was totally wrong since I got it from another problem in my notebook, so I am sorry About that.
Once again, Thank You! :lol:


modular arithmetic
quadratics
number theory
relatively prime

Let $ a,\ b,\ c$ and $ d$ are postive integers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} d^2$.

(1) If $ d$ is not divisible by 3, then prove that there exists exactly two numbers which are divisible by 3 among $ a,\ b,\ c$.

(2) If $ d$ is not divisible by 2 or 3, then prove that there exists more than one number which is divisible by 6 among $ a,\ b,\ c$.

I'll have to work on my wording, but the meat of the solution is here.  :blush:

[hide="Solution for #1"]
=====
If $ d$ is not divisible by $ 3$, then it is congruent to either $ 1$ or $ 2 \pmod{3}$, and so $ d^2\equiv 1 \pmod{3}$. The quadratic residues $ \pmod{3}$ are $ \{0,1\}$, and so $ a^2,b^2,c^2$ are congruent to either of $ \{0,1\} \pmod{3}$. If there are zero, one, or three of $ a^2,b^2,c^2$ congruent to $ 0 \pmod{3}$, then the sum $ a^2 \plus{} b^2 \plus{} c^2$ will be congruent to $ 0, 2, 0$, respectively, whereas if there are exactly two of these congruent to $ 0 \pmod{3}$, then their sum will be congruent to $ 1 \pmod{3}$, as desired. Thus exactly two of $ a^2,b^2,c^2$ are divisible by $ 3$, and therefore exactly two of $ a,b,c$ are divisible by $ 3$.
=====
[/hide]

[quote="kunny"]Let $ a,\ b,\ c$ and $ d$ are postive integers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} d^2$.

(2) If $ d$ is not divisible by 2 or 3, then prove that there exists more than one number which is divisible by 6 among $ a,\ b,\ c$.[/quote]

I think you mean "at least", because $ 6^2\plus{}3^2\plus{}2^2 \equal{} 7^2$  :wink:

[hide="2"]
Since d is relatively prime to 6, and phi(6) = 2, then $ d^2$ is congruent to 1. Assume that none of a, b, or c are multiples of 6. then their squares must all be either 1, 4, or 3 modulo 6. testing out combinations, the only combinations that work are 3,3, and 1. But then a,b,c are odd, and we take the equation modulo 12 to get that $ d^2$ is 3 modulo 4, which is impossible.
[/hide]

Prove that if d is not divisible by 3 then exactly two of a,b,c are multiples of 3.
[hide="#1"]If d is not divisible by 3, then $ d \equal{} 1,2 (mod{3}) \implies 2^2(mod{3}) \equal{} 1(mod{3}), 1^2(mod{3}) \equal{} 1(mod{3})$, so $ d^2 \equal{} 1(mod{3})$. (0,1) are the only quadratic residues modulo 3, so if none of a,b,c are divisible by 3, that implies that $ d^2 \equal{} 0(mod3)$, which is false by the problem statement. If 1 of a,b,c is divisble by 3, then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2 (mod{3})$, which is not true. If all three are divisible by 3, that implies that $ d^2 \equal{} 0(mod{3})$, which is false. Therefore 2 of a,b,c are divisible by 3 because $ 0^2 \plus{} 0^2 \plus{} 1^2 \equal{} 1(mod 3)$ which is correct.[/hide]

Hi, everyone, what do you think how difficult the problem is?

[url=http://www.hit-u.ac.jp/eng/education/undergraduate.html]Hitotsubashi University[/url] entrance exam starts the problem related to number theory in Problem 1.

The applicants have trouble to solve the Problem 1.

I don't have much experience with writing solutions to number theory problems, but I think this problem is a fairly straightforward application of quadratic residues.

I agree... @jvenezuela where did you learn your number theory--I just picked it up on aops

Anything I know about contest math has come from reading/skimming AoPS 1/2, Zeitz, and Engel, as well as just reading posts/solutions here in the AoPS community. Similar to you, I kinda pick some of this stuff up... that explains why I have trouble sitting down and formally learning something.  :(

you own v1/v2 and zeitz and engel but havent read them?

I discovered contest math very late in my high school career; I am now a rising senior and have only had these books a few months. I have finished AoPS 1, am a few chapters through AoPS 2, and have read selections of Zeitz and Engel. I work very slowly but thoroughly; I do every example/exercise/problem, to the point of marking typos (I'm thanked in AoPS 1!). Don't get me wrong... I make an effort to contribute many hours a day to studying this stuff on top of my job and schoolwork.


AMC
AIME
modular arithmetic
USA(J)MO
USAMO
ARML
AMC 10

For those students who are planning to take the AMC 10 or 12, and are not sure whether they will qualify for AIME, here are my recommendations on the number of questions to answer.

If your only goal is to get a 120 or higher (for example, to qualify for AIME via the AMC 10), then you should answer 21 or 25 questions.  (Any other choice is dominated by these two choices.)  If you answer 21 questions, you can get 2 questions wrong and still get a 120 (exactly).  If you answer all 25 questions, you can get 5 wrong and still get a 120.

If your only goal is to get a 100 or higher (for example, to qualify for AIME via the AMC 12), then you should answer 14, 18, 22, or 25 questions.  (Any other choice is dominated by these four choices.)  If you answer 14 questions, you have to get them all right to get a 100.5.  If you answer 18 questions, you can get 3 wrong and still get a 100.5.  If you answer 22 questions, you can get 6 wrong and still get a 100.5.  If you answer all 25 questions, you can get 8 wrong and get a 102.

Note:  If the AMC decides to lower the qualifying scores (120 on the AMC 10, or 100 on the AMC 12), then the recommendations above may not apply.  That's hard to plan for in advance.

I don't think they will lower the qualifying scores.  Personally, I usually answer all 25 questions on the AMC 10, guessing if I can eliminate 2 answer choices.

If you are careful and don't make many careless mistakes, perhaps don't answer them all, and take the sure points.  But if you know you'll make a few careless errors, then answer a few extra questions to make up for it.  Or that's what has worked for me, anyway...(bare minimum 120.5 on the AMC 10 WOOT!)

Could you explain how answering 21 dominates answering 22? For example, say I've already answered 21 questions and I get to number 22, and I totally pwn this question. Why would I not want to answer it?

if your absolutely sure about 22, there is no reason to not answer it.

However, if your unsure, remember that getting it wrong could make you lose a point and could possibly keep u from passing.

Personally, I would probably make enough mistakes in the first 21 questions that I would attempt to move on. So I gotta be super duper extra careful.

If your only goal is to get a 120 or higher, then under the new rules it doesn't make sense to answer exactly 22 questions.  Let's say you have answered 22 questions, and there are 30 seconds to go, not enough time to solve another question.  Then you might as well guess randomly on the remaining three questions.  Whether you answer 22 or 25 questions, you still need to get 20 right.  So you might as well guess.

Even erasing one of your answers makes more sense than leaving 22 answers.  (In that case, completely erase the answer you are least sure of.)  That's because whether you answer 21 or 22 questions, you can afford to get only 2 wrong.  But in general, my recommendation is to add answers, not erase them, until you get to one of my target numbers.

just so i know

how many points do you get for a blank?
for a wrong answer?
correct answer?

Under the new rules taking effect this year, a correct answer is worth 6 points, a blank is worth 1.5 points, and a wrong answer is worth nothing.

Well suppose I want to get 130. What is the magic number of questions to answer?

I think, that you should answer either 22 or 25 questions if you want to get a 130, using Ravi's method.

It is not possible to get a 130 exactly. The general formula for a score $s$ is $6x+1.5y$. Therefore, any score must be divisible by 1.5.

Let $s=1.5m$. From the scoring guidelines, $s=6c+1.5b+0w$ and $c+b+w=25$. $c$ is the number of questions correct, $b$ is the number of questions blank, and $w$ is the number wrong.

Since, $s=6c+1.5b+0w$, we have $s=6c+1.5(25-c-w)+0w$ which simplifies down to $1.5m=4.5c-1.5w+37.5$.  Further, this simplifies, $m-25=3c-w$. So, $w\equiv 25-m\pmod 3$. So $w\equiv 25-\frac{s}{1.5}\pmod 3$. For $c$ to be an integer, $\frac{1}{3}(\frac{s}{1.5}-25+3k)$ must be an integer.

When $s=132$, m=88. So, $w\equiv 0\pmod 3$. $w=3k$. So, $c=21+k$ and $b=4-4k$. We also know that $c+w\leq25$, so $k\leq 1$. So, we know $k=0$ or $1$.

The possible ways to get 132 in the form $(c,b,w)$, there are two possibilities, $(21,4,0)$ and $(22,0,3)$.

Or you can do $(21, 3, 1)$ for a $130.5$. That is alright. Then you need a 10 on AIME for USAMO qualification.

This is a very interesting analysis. In my state, because the top ten AMC 12 scorers are invited automatically onto the state ARML team, the goal here for the top forty or so math students is to maximize points on the AMC 12 compared to other students in the state. (By historical data, clearing 140 points on the AMC 12 ought to be enough to lock in ARML, but each year the students get better here, in large part thanks to AoPS.) So given that goal, is there any more refined advice than simply writing down as many correct answers as possible?

In Minnesota last year, it looks like the 9th best student on the AMC 12 got a 130.  (There was a four-way tie for 10th best at 129.)  In 2005, the 10th best student there got a 127.5.

So, can the AMC people change the point values or the qualification scores at the test site?

The AMC has already announced the new point values listed above, and won't change them again for the 2007 contests.  The AMC has the option of lowering the qualifying scores, but has promised not to raise the qualifying scores.

For example, the qualifying score on the 2001 AMC 12 was 84 (!).  The qualifying score on the 2004 AMC 10 A was 110.  The qualifying score on the 2003 AMC 10 B was 121, but that was before the AMC guaranteed that 120 would qualify for AIME.

So if it was really hard and I got like a 115, I mgiht still have a chance?

[quote="Boy Soprano II"]Um, guys, the AMC director has posted numerous times saying that the point of the contest is not to find out exactly how many questions you need to answer to qualify for the AIME and then answering exactly that number of questions, but to have a good mathematical experience.  That's why the scoring system was changed in the first place.

(And if you can qualify for the AIME, you can probably figure out the scoring system on your own, anyhow. :maybe: )[/quote]

The fact of the matter is that the AMC is scored.  Getting to take the AIME depends on this score.  Thus it makes sense that people would wonder how many problems they need to get right, leave blank, etc.  So it naturally makes for a discussion topic.  If you don't want to discuss it, then don't.


Gauss
algebra
polynomial
number theory
greatest common divisor
superior algebra
superior algebra unsolved

1)Give an example of an f in Z[x] which is irreducible in Z[x] but not in Q[x]
2) Give an example of an f in Z[x] which is irreducible in Q[x] but not in Z[x]
------------

since Q is the quotient field of Z....we can use Gauss' lemma.
1) by Gauss's lemma, if a polynomial is reducible in Q[x], f(x)=a(x)b(x), where both of a(x) and b(x) in Q[x], then there exist r and s in Q such that A(x)=r times a(x), and B(x)=s times b(x), where f(x)=A(x)B(x), and A B belongs in Z[x].
SO THERE EXIST NO SUCH F

2) f(x)=4, it is irreducible in Q[x]  since it is a unit in Q[x] // 4 times 1/4 =1
  it is reducible in Z[x] since f(x) = 2 times 2
--------------

And I got zero on both questions.......so..can anyone help???

I'm new to this but doesn't Gauss's lemma apply only for primitive polynomials? It doesn't say anywhere there that the polynomial must be primitive.

We can just extract the gcd of the coefficients first.

I have no idea what the question's writer was looking for. You have a valid complaint- ask what's going on.


Are cineva baremele de la barajele date la Scoala cu ceas in 28 martie 2009?
Multumesc.

Pe net nu au aparut nici [b]subiectele[/b] de la probele de baraj. Daca le are cineva as fi recunoscator daca le-ar posta!

Vreau sa le vad si eu . Le poate posta cineva ?  :lol:

[quote="mihai miculita"][b]Adresa solicitata este: [/b] http://www.scoalacuceas.ro/matematica2.html[/quote]

Aici sunt doar subiectele si baremele la proba pe clase (IV-VIII).

la adresa prcedenta sunt probele de clasa. ma refeream la barajele de tip oim si obmj care s-au dat sambata.
la acestea nu sunt afisate nici subiectele nici baremele.

Le-am postat eu pe mateforum la "Alte concursuri" pe cele de la seniori...    :P

Multumim. Subiectele de la seniori au fost destul de dificile ...


vector
geometry
3D geometry
sphere
geometry unsolved
vector geometry

Given $n$ unit vectors in the plane whose sum has length less than one, show that you can arrange them so that the sum of the first $k$ has length less than $2$ for every $1 < k < n$.

Do an algorithmic solution. Prove at each step we can choose one or two of the vectors that the sum of selected vectors is less than 1.
This is obvious :D  :D

sorry,
I just dont understand your solution at all.
Could you expain it  clearer?

chao

Start with a random vector and say it number 1.
Suppose we have selected i vectors with numbers 1,2,...,i
We can easily prove each time we can selectone vector and name it (i+1) that sum of selected vectors has length less than 1 or we can find two vectors and name them (i+1) and (i+2) that sum of selected vectors has length less than 1. So for each k<n the sum of first k vectors is at most 2.

You may find related informations by looking for the Steinitz' constants.

More generally :
Let a,b be nonegative real numbers, and S = {v_1,...,v_n, u, w} be a set of vectors in an m-dimensional normed real space satisfying :
|u|  \leq  a, |w|  \leq  b, |v_i|  \leq  1 for each i, and
u + w  \sum {1 \leq i \leq n} v_i = 0.
For each permutation  \pi  of {1,2,...,n} let's define :
F( \pi , n, S) = max{1 \leq k \leq n} |u +  \sum {1 \leq i \leq k} v_ \pi (i) |.
That is F is the radius of the smallest sphere with center O circumscribing the polygon with sides u, v_ \pi (1),...,v \pi (n). Of the vector orderings determined by all possible  \pi n there is one which minimizes the spherical radius. Define K_m(a,b) be the least number for which
|u +  \sum {1 \leq i \leq k} v_ \pi (i)|  \leq  K_m(a,b) for some permutation  \pi , for all integer n and sets S (that is K_m(a,b) = max{over n,S}(min{over  \pi } F( \pi , n, S) ), which is the m-dimensional Steinitz constant.

The best known upper bound for the m-dimensional Steinitz constant, with no restriction on the nom, is K_m(0,0)  \leq  m - 1 + 1/m.

If we restrict ourselves to the euclidean norm in the plane, it is known that :
K_2(1,0) = K_2(1,1) =  \leq 2 (which is the case of the problem, but with no exact value for the bound), and K_2(0,0) = ( \sqrt 5)/2.
More generally K_2(a,b) =  \leq (1 + max(a 2 , b 2  , 1/4))

Pierre.


There is an old and famous problem that consist to obtain as many consecutive non-negative integer numbers as possible using four fours and the usual arithmetic operations, Addition (+), Substraction (-), Multiplication (*), Division (/) and powers (^). You can concatenate, but you cannot use any other operation or digit. Example:

$0=\frac{4-4}{4+4}$

$1=4^{\frac{4-4}{4}}$

$2=\frac{4\cdot4}{4+4}$

How long can you continue?

$3=4-\left(\frac44\right)^4$
$4=4\cdot\left(\frac44\right)^4$
$5=4+\left(\frac44\right)^4$
We were just doing this yesterday in math. Are any of negatives, decimal points/repeating bars ($99=\frac{44}{.\overline{44}}$), factorials or roots of any kind allowed?

[quote="Bictor717"]We were just doing this yesterday in math. Are any of negatives, decimal points/repeating bars ($99=\frac{44}{.\overline{44}}$), factorials or roots of any kind allowed?[/quote]Negatives, well, we already said that substraction is allowed, then say you are using that one. Decimal Points and Repeating bars... let's try not to use them, to keep the tools very simple.

[quote="jimli"]how do you get 11?[/quote]That is a good question... Does anyone has an answer... I need either the square (42 ) function or five fours!  :(

As I said, At least I can't do it with less than five 4s or need to use the square, but in that case we were using a 2 in the expression. My solutions (that fail with the rules) are below

[hide="My "Solutions""]$4^2-4-\frac44=11$
$4\cdot4-4-\frac44=11$[/hide]

[quote="Ravi B"]If you allow either decimal points or square roots, then you can do 11.[/quote]Really, how?

[quote="jmerry"]You can do 11 with less than five fours- it only takes three: $\frac{44}4$.
That doesn't mean you can do it with exactly four.[/quote]Yep, you are right.. Any way it still fails to the original rules.  ;)

[quote="djimenez"][quote="Ravi B"]If you allow either decimal points or square roots, then you can do 11.[/quote]Really, how?[/quote]I'll leave both problems as a challenge for others!  If no one responds in a few days, ask me again.

The square root would effectively burn a 4.

How about we continue going and see how many we can get in a row like the problem says

Here's the next one:

4(4-(4/4))=12

I don't think 13 is possible unless we twist the rules in which case 4! - 44/4=13, but prove me wrong if you can.

I would like to share some notes on four [b]n[/b]'s in my attachment. With no restrictions of mathematical operations it is supposed to express any number.
However I don't understand the proof of that theorem(please view the attachment). Can anyone enlighten me?

[quote="probability1.01"]Basically it's $\log_2 2^k$.[/quote]

[quote="djimenez"]Basically what they are doing is just to find an expression for k with four or five n's and after that telling you that you can add as many as you want by adding pairs (n-n). But in that cas you need square roots (implicitely using 2 :P ) and logaritms that are out of our scheme so far. But is nice anyway.[/quote]

Thanks! :)


geometry
incenter
vector
analytic geometry
geometry proposed

Im looking for an elegant solution of a following problem:

In a triangle $ ABC$ let incircle touches $ CA$ in $ K$. If $ S$ is an incenter, prove that a line through $ PS$ intersects $ BK$ in its middle, where $ P$ is the middle point of $ CA$.

I think $ P$ on $ AC$.
We can prove this problem easily by vector or barycentric coordinate or complex numbers.

CKAB is a (degenerate) tangential quadrilateral with diagonals CA, BK and incircle (S). If P, Q are midpoints of these diagonals, then Newton line PQ is of the tangential quadrilateral CKAB goes through its incenter S.

Dear Mathlinkers,
this redult has been posted severals times... but where?
For a synthetic proof see
http://perso.orange.fr/jl.ayme    vol. 3  Cinq théorèmes de Nagel scolie 3 p. 7
Sincerely
Jean-Louis

Do $ Homo(B,k)$, Just that $ ω - > ω'$,$ ω'$ is incircle,
ω is othercircle(also tangents $ BC,CA,AB$,and it tangents $ CA$ at point $ L$,and then $ PL = PK$).
Let $ CA->C'A',L->L'$
then $ ω$ tangants $ C'A'$,Easy to prove that $ SK = SL'$,and $ S,K,L'$ on a line.
So ......
prove.


function
calculus
integration
conics
parabola
calculus computations

How come I'm getting a negative volume? I think I chose the outer curve and the inner curve correctly, or am I missing something here?
[img]http://grimbb.phpnet.us/work6.gif[/img]

I didn't mindlessly apply the formula JBL I did in fact draw the graph which is why I chose one curve for the disc and the other for the washer (I just didn't post the graph here). If we take a look at a value between the two limits of the functions, for instance x = -1.5 then the corresponding y-value for the disc curve is 0.75 and the corresponding y-value for the hole curve is 0.125 which is why I chose the curves in that way. Unless I'm drawing the wrong graph, which doesn't seem likely as an x^2 curve should be situated above an x^3 curve along the range from 0 to 1.

The washer method is much uglier here than the other method (shell method?) would be.  And your error stems from choosing to go that ugly route.  In particular, you need to more carefully consider the following question: how many square roots has $1-y$?

(The comment which I deleted to which you are responding was because I initially assumed you made a far sillier error that what you actually did.)

Yes it's come to my attention that I neglected to account for a second region that intersects at (0, 0) and (-2, -8). However, shouldn't this integral produce a positive volume? The volume of an exterior shape minus the volume of an interior shape?

What you just mentioned is an entirely different problem, and not one that has anything to do with why you're getting negative volume.  Notice that the parabola lies on both sides of the cubic curve.  It's a function of $y$ in terms of $x$.  However, its inverse isn't a function: there are many values of $y$ for which there are two possible values of $x$.  You made a choice of one of them, $\sqrt{1-y}-1$.  But there is another possible value of $x$.

Try actually graphing the curve $x = \sqrt{1-y}-1$ and you'll see what I mean.  (And I do mean to graph exactly that curve; don't first solve for $x$ in terms of $y$.)


calculus integration trigonometry function

I have the following solution to a Fourier Transform problem

$U(x,t)=\frac{2}{\pi}\int_0^\infty \frac{\cos\lambda}{\lambda^2}\left(1-e^{-\lambda^2 t \right)\cos(\lambda x) \,d\lambda}$

I need to plot it for t =0,1,2,3,4,5  and 0 < x< 5 in Maple/Mathematica, but neither program can evaluate the integral

Is there a way to numerically get the program to plot it??

Use Plot[NIntegrate (YOUR FUNCTION TO BE INTEGRATED HERE), {x, min, max}] in mathematica and, after many warning messages, it will plot it.  No guarantees that the plot is accurate, though :)

Also, note that the case t = 0 you actually don't want: that seemed to be causing mathematica problems.

Here's the picture I get out of Mathematica.  I had to do the graphs seperately or it wouldn't take them (too many errors?).

Thanks! I guess I wasnt patient enough with Mathematica!

However...I just found a way to calculate the integral explictly using erf functions...should circumvent any troubles I am having


trigonometry
geometry
circumcircle
geometry unsolved

In a $\triangle {ABC}$, consider a point $E$ in $BC$ such that $AE \perp BC$. Prove that $AE=\frac{bc}{2r}$, where $r$ is the radio of the circle circumscripte, $b=AC$ and $c=AB$.

$S=\frac 12.bc\sin A=\frac 12.AE.a\ ,\frac{a}{\sin A}=2R$then $bc=AE.2R$

Let $w=C(O,R)$ be the circumcircle of the triangle $ABC$. Define the point $A'\in w\cap (AO$. Then $\widehat {ABE}\equiv\widehat {AA'C}\ ,$ $AE\perp BE$ and $AC\perp A'C$ $\Longrightarrow$ $\triangle ABE\sim \triangle AA'C$ $\Longrightarrow \frac{AB}{AA'}=\frac{BE}{A'C}=\frac{EA}{CA}$ $\Longrightarrow$ $\frac{c}{2R}=\frac{h_a}{b}$ $\Longrightarrow$ $\boxed {\ bc=2Rh_a\ }\ .$[u]Remark.[/u] $S=\frac 12ah_a=$ $\frac 12 a\cdot \frac{bc}{2R}$ $\Longrightarrow$ $\boxed {\ 4RS=abc\ }\ .$

The problem tells us to prove the following well-known relation in a trinagle i.e \[ h_{a}=\frac{bc}{2R}, \]
where $h_{a}$ is the altitude of the triangle $ABC$ and $a, b, c$ are the side lenghts of the same triangle, $ABC$.
Well, here is one-lin solution: We know that \[ S=\frac{abc}{4R} \] and \[ a\cdot h_{a}=2S. \]
The conclusion follows immediately. ;)


If you do miserably on Round 1, will you be prohibited from continuing onward to Round 2, etc.?

If so, may I please know the cut-off line if it can be disclosed?

Thanks,  :)

There are no cutoffs to get to the next round or anything, but there are cutoffs for HM, Bronze, Silver, and Gold:

96-100 - Gold
75-95 - Silver
60-74 - Bronze
40-59 - Honorable Mention

These are the overall scores. There are 4 rounds, 5 problems per round, 5 points per problem, giving a maximum total of 100 points. Each group gets some prizes and the top 30 male and top 30 female participants get a partial scholarship to MathCamp.

More info [url=http://www.usamts.org/Rules/U_RulesPrizes.php]here[/url].

No. You can do submit solutions to any rounds you wish, all of them, none of them, skip some, start partway through the year, whatever.

As a matter of statistical correlation, I would assume that those who get low scores on the first round are less likely to submit answers for the next round than those who get high scores. But that's human behavior and statistical likelihood, not the result of any rule. And there will be many individual exceptions.

Can you see a way to improve? Will you still appreciate the challenge even if you find it difficult? Will you learn from this? There are many good motivations for continuing.

wait - just wondering....

do you get anything for being an honorable mention????

Yes, you recieve a USAMTS T-shirt, and 1 selection from the USAMTS prizes for 2009.

[quote="abacadaea"]Yeah. I know this guy who got 7 on round one(partially because of laziness), and ended up getting bronze or something..[/quote]

dang that guy sounds tricky.

[quote="hurdler"][quote="themorninglighttt"]eh also, 75+ gets auto aime qualification (with a 100 amc score i think).[/quote]

really? I recall last year round 4 was not finished before aime qualification had to be announced. Last was 70/75 for first 3 rounds, which is a lot tougher than 75/100[/quote]

wasn't it just 45/50 for rounds 1 & 3? but still yeah, that's much harder than 75/100 o.0


linear algebra
matrix
algebra
polynomial
induction
inequalities
trigonometry

[img]http://photos.piry.net/images/8745matrixprob.gif[/img]
 I can't solve it. Can someone help me please?  :|

Then we must prove that [img]http://photos.piry.net/images/67758aa.gif[/img].  :help:

First some basics.
For $ C\in M_k(\mathbb{C})$ and $ P(x) \equal{} a_0 \plus{} a_1x \plus{} \cdots \plus{} a_nx^n\in\mathbb{C}[x]$,
we always set $ P(C): \equal{} a_01_k \plus{} a_1C \plus{} \cdots \plus{} a_nC^n\in M_k(\mathbb{C})$.

If $ U\in GL_k(\mathbb{C})$ is any invertible matrix (i.e. $ \det U\not \equal{} 0$),
then it's easy to see, that $ \det (UCU^{ \minus{} 1}) \equal{} \det U\cdot\det C\cdot (\det U)^{ \minus{} 1} \equal{} \det C$
and $ P(UCU^{ \minus{} 1}) \equal{} UP(C)U^{ \minus{} 1}$ for any $ C\in M_k(\mathbb{C})$ and $ P(x)\in\mathbb{C}[x]$.
(like $ (UCU^{ \minus{} 1})^3 \equal{} (UCU^{ \minus{} 1})(UCU^{ \minus{} 1})(UCU^{ \minus{} 1}) \equal{} UC\cdot C\cdot CU^{ \minus{} 1} \equal{} UC^3U^{ \minus{} 1}$)

If $ C \equal{} (c_{ij})$ is an upper triangular matrix, i.e. $ c_{ij} \equal{} 0$ for $ i > j$, and $ P(x)\in\mathbb{C}[x]$,
then $ P(C)$ is also upper triangular with diagonal entries $ P(c_{11}),\ldots,P(c_{kk})$.
(The sum and product of upper triangular matrices are again upper triangular
with diagonal entries being added and multiplied, resp.)

And now define the sequence of polynomials $ P_0,P_1,P_2,\ldots\in\mathbb{Z}[x]$ by
$ P_0\equiv 0, P_1\equiv 1, P_{n \plus{} 1} \equal{} xP_n \plus{} P_{n \minus{} 1}$ for all $ n\in\mathbb{N}$.
Then by induction $ P_n(A) \equal{} X_n$ for all $ n\in\mathbb{N}_0$.
Also by induction $ P_n$ has degree $ n \minus{} 1$ and only non-negative coefficients
and $ P_{2n}$ is odd and $ P_{2n \plus{} 1}$ is even for all $ n\in\mathbb{N}_0$.
(i.e. $ P_{2n}( \minus{} x) \equal{} \minus{} P_{2n}(x)$ and $ P_{2n \plus{} 1}( \minus{} x) \equal{} P_{2n \plus{} 1}(x)$)

So we can write $ P_{2n \plus{} 1}(x) \equal{} a_0 \plus{} a_1x^2 \plus{} \cdots \plus{} a_nx^{2n}$ with $ a_0,a_1,\ldots,a_n\in\mathbb{N}_0$,
which gives $ P_{2n \plus{} 1}(\lambda)\ge 0$ for all $ \lambda\in\mathbb{R},n\in\mathbb{N}_0$.
Also $ P_{2n \plus{} 1}(\mu)P_{2n \plus{} 1}(\overline{\mu}) \equal{} P_{2n \plus{} 1}(\mu)\overline{P_{2n \plus{} 1}(\mu)} \equal{} |P_{2n \plus{} 1}(\mu)|^2\ge 0$ for all $ \mu\in\mathbb{C}$.

Now using the [url=http://en.wikipedia.org/wiki/Jordan_normal_form]Jordan normal form[/url] we can find some $ U\in GL_k(\mathbb{C})$
with $ B: \equal{} U^{ \minus{} 1}AU\in M_k(\mathbb{C})$ upper triangular.
If $ \lambda_1,\ldots,\lambda_k\in\mathbb{C}$ are the diagonal entries of $ B$, then $ x1_k \minus{} A\in M_k (\mathbb{R}[x])$
and $ x1_k \minus{} B\in M_k (\mathbb{C}[x])$ upper triangular with diagonal entries $ x \minus{} \lambda_1,\ldots,x \minus{} \lambda_k\in\mathbb{C}[x]$.
This gives
$ \chi : \equal{} (x \minus{} \lambda_1)\cdots (x \minus{} \lambda_k) \equal{} \det (x1_k \minus{} B) \equal{} \det (xU^{ \minus{} 1}U \minus{} U^{ \minus{} 1}AU) \\
\hspace*{1.54in} \equal{} \det [U^{ \minus{} 1}(x1_k \minus{} A)U] \equal{} \det (x1_k \minus{} A)\in\mathbb{R}[x]$.
(The last part follows by the [url=http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants]Leibniz formula[/url] of the determinant)

Therefore the diagonal entries of $ B$ are of the form $ \lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s}$ (in some order)
with $ \lambda_1,\ldots,\lambda_r\in\mathbb{R}$ and $ \mu_1,\ldots,\mu_s\in\mathbb{C}\setminus\mathbb{R}$. (and $ k \equal{} r \plus{} 2s$)
And finally we get
$ \det X_{2n \plus{} 1} \equal{} \det P_{2n \plus{} 1}(A) \equal{} \det P_{2n \plus{} 1}(UBU^{ \minus{} 1}) \\
\hspace*{0.6in} \equal{} \det [UP_{2n \plus{} 1}(B)U^{ \minus{} 1}] \equal{} \det P_{2n \plus{} 1}(B) \\
\hspace*{0.6in} \equal{} P_{2n \plus{} 1}(\lambda_1)\cdots P_{2n \plus{} 1}(\lambda_r)P_{2n \plus{} 1}(\mu_1)P_{2n \plus{} 1}(\overline{\mu_1})\cdots P_{2n \plus{} 1}(\mu_s)P_{2n \plus{} 1}(\overline{\mu_s}) \\
\hspace*{0.6in}\ge 0$.

P.S. I tried to avoid definitions like eigenvalue or characteristic polynomial,
but i guess there is no escaping Jordan's normal form now.  :lol:

This is very useful, but I'm only in the eleven degree. I haven't studied so much about matrix. I think that it can be done easily.  :stink: (the problem is from a contest for the eleven degree)

OK, here is another proof. :P

$ P_{2n \plus{} 1}(x) \equal{} a_0 \plus{} a_1x^2 \plus{} \cdots \plus{} a_nx^{2n}$ with $ a_0,a_1,\ldots,a_n\in\mathbb{N}_0$,
then $ P_{2n\plus{}1}(\lambda)\ge P_{2n\plus{}1}(0)$ and by induction $ P_{2n\plus{}1}(0)\equal{}P_1(0)\equal{}1$ for all $ \lambda\in\mathbb{R},n\in\mathbb{N}_0$.
So $ P_{2n\plus{}1}$ has no real roots and we can find $ \lambda_1,\ldots,\lambda_n\in\mathbb{C}\setminus\mathbb{R}$ with
$ P_{2n\plus{}1}(x)\equal{}(x\minus{}\lambda_1)(x\minus{}\overline{\lambda_1})\cdots (x\minus{}\lambda_n)(x\minus{}\overline{\lambda_n})$.
(by induction $ P_{n}$ has degree $ n\minus{}1$ and leading coefficient $ 1$ for all $ n\in\mathbb{N}$)
So $ X_{2n\plus{}1}\equal{}P_{2n\plus{}1}(A)\equal{}(A\minus{}\lambda_11_k)(A\minus{}\overline{\lambda_1}1_k)\cdots (A\minus{}\lambda_n1_k)(A\minus{}\overline{\lambda_n}1_k)$
and $ \det X_{2n\plus{}1}\equal{}\det (A\minus{}\lambda_11_k)\det (A\minus{}\overline{\lambda_1}1_k)\cdots \det (A\minus{}\lambda_n1_k)\det (A\minus{}\overline{\lambda_n}1_k)$.
(btw the determinant is multiplicative, i.e. $ \det (AB)\equal{}\det A\det B$)
But using $ A\in M_k(\mathbb{R})$ and the [url=http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants]Leibniz formula[/url] we have
$ \det (A\minus{}\overline{\lambda}1_k)\equal{}\overline {\det(A\minus{}\lambda 1_k)}$ and                                  $ \det (A\minus{}\lambda 1_k)\det (A\minus{}\overline{\lambda}1_k)\equal{}|\det(A\minus{}\lambda 1_k)|^2\ge 0$.
So $ \det X_{2n\plus{}1}\ge 0$ follows.

Thanks.. but this is the second part of the problem. First, we have to find Xn. Any idea? :)

Finding $ X_n \equal{} P_n(A)$ is the same as finding $ P_n(x)$, which is similar to the [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]Chebyshev polynomials[/url].

In fact by induction
$ P_n(x) \equal{} {1\over\sqrt {x^2 \plus{} 4}}\left [ \left ({x \plus{} \sqrt {x^2 \plus{} 4} \over 2}\right )^n \minus{} \left ({x \minus{} \sqrt {x^2 \plus{} 4} \over 2}\right )^n \right ] \\
\hspace*{0.34in} \equal{} {1\over 2^{n \minus{} 1}}\sum_{i \equal{} 0}^{\left [{n \minus{} 1\over 2}\right ]}\binom{n}{2i \plus{} 1}x^{n \minus{} 2i \minus{} 1}(x^2 \plus{} 4)^i$
and
$ X_n \equal{} {1\over 2^{n \minus{} 1}}\sum_{i \equal{} 0}^{\left [{n \minus{} 1\over 2}\right ]}\binom{n}{2i \plus{} 1}A^{n \minus{} 2i \minus{} 1}(A^2 \plus{} 4\cdot 1_k)^i$
for all $ n\ge 0$.

Also by induction $ P_n(x) \equal{} i^{n \minus{} 1}U_{n \minus{} 1}( \minus{} { ix\over 2})$ for all $ n\ge 0$, where
$ U_n(x)\in\mathbb{Z}[x]$ is the Chebyshev polynomial of the second kind satisfying
$ \sin (n \plus{} 1)x \equal{} U_n(\cos x)\sin x$ for all $ n\ge \minus{} 1$.
Then $ P_n( 2i\cos \varphi) \equal{} i^{n \minus{} 1}U_{n \minus{} 1}(\cos \varphi) \equal{} i^{n \minus{} 1}{\sin n\varphi\over \sin \varphi} \equal{} 0$ for $ \varphi \equal{} {m\over n}\pi\; ,m \equal{} 1,\ldots ,n \minus{} 1$.
So $ P_n$ has the $ n \minus{} 1$ different roots $ 2i\cos {1\over n}\pi,\ldots , 2i\cos {n \minus{} 1\over n}\pi$,
and since it has degree $ n \minus{} 1$ and leading coefficient $ 1$, we get
$ P_n(x) \equal{} (x \minus{} 2i\cos {1\over n}\pi)\cdots (x \minus{} 2i\cos {n \minus{} 1\over n}\pi)$
and using $ \cos {m\over n}\pi \equal{} \minus{} \cos {n \minus{} m\over n}\pi$ we get

$ P_{2n \plus{} 1}(x) \equal{} (x^2 \plus{} 4\cos^2 {1\over 2n \plus{} 1}\pi )\cdots (x^2 \plus{} 4\cos^2 {n\over 2n \plus{} 1}\pi )$
and
$ P_{2n}(x) \equal{} x(x^2 \plus{} 4\cos^2 {1\over 2n}\pi )\cdots (x^2 \plus{} 4\cos^2 {n \minus{} 1\over 2n}\pi )$.

So
$ X_{2n \plus{} 1} \equal{} (A^2 \plus{} 4\cos^2 {1\over 2n \plus{} 1}\pi \cdot 1_k)\cdots (A^2 \plus{} 4\cos^2 {n\over 2n \plus{} 1}\pi \cdot 1_k)$
and
$ X_{2n} \equal{} A(A^2 \plus{} 4\cos^2 {1\over 2n}\pi \cdot 1_k)\cdots (A^2 \plus{} 4\cos^2 {n \minus{} 1\over 2n}\pi \cdot 1_k)$.

What else do you need?  :D

Edit: fixed a little $ \pm$ error

Thank you!  :D .
By the way, I have found an easier solution :) . We can prove using the principle of induction that [img]http://photos.piry.net/images/11258matrixproblem.gif[/img].

Ok, this is a complete shock!  :|

$ P_n(x) \equal{} \sum_{k \equal{} 0}^{\left[{n \minus{} 1\over 2}\right]}\binom{n \minus{} 1 \minus{} k}{k}x^{n \minus{} 2k \minus{} 1}$
gives
$ U_n(x) \equal{} ( \minus{} i)^nP_{n \plus{} 1}(2ix) \equal{} \sum_{k \equal{} 0}^{\left[{n\over 2}\right]}( \minus{} 1)^k2^{n \minus{} 2k}\binom{n \minus{} k}{k}x^{n \minus{} 2k}$
and
$ T_n(x) \equal{} U_n(x) \minus{} xU_{n \minus{} 1}(x) \equal{} \sum_{k \equal{} 0}^{\left[{n\over 2}\right]}( \minus{} 1)^k2^{n \minus{} 1 \minus{} 2k}{n\over n \minus{} k}\binom{n \minus{} k}{k}x^{n \minus{} 2k}$.

So
$ \cos nx \equal{} \sum_{k \equal{} 0}^{\left[{n\over 2}\right]}( \minus{} 1)^k2^{n \minus{} 1 \minus{} 2k}{n\over n \minus{} k}\binom{n \minus{} k}{k}\cos^{n \minus{} 2k}x \\
\hspace* {0.37in} \equal{} 2^{n \minus{} 1}\cos^n x \minus{} 2^{n \minus{} 3}n\cos^{n \minus{} 2}x \plus{} 2^{n \minus{} 6}n(n \minus{} 3)\cos^{n \minus{} 4}x\pm\cdots$
for all $ n \ge 1$ and $ x\in\mathbb{R}$.

I didn't know the coefficients of the Chebyshev polynomials were that explicit.
Could anyone confirm that?  :maybe:

What means here: "find $ X_n$" ? Is it find a compact form for $ X_n$ or find an expression which can be calculated quicker than using the recursive formula ? (we assume that $ n$ is big). The cost of the recursive formula is essentially $ n$ matricial multiplications. The cost of St3faN' formula is essentially $ n/2$ matricial multiplcations.
But we can also use the Olorin's formula: $ X_n \equal{} (\sqrt {x^2 \plus{} 4})^{\minus{}1}\left [ \left ({x \plus{} \sqrt {x^2 \plus{} 4} \over 2}\right )^n \minus{} \left ({x \minus{} \sqrt {x^2 \plus{} 4} \over 2}\right )^n \right ]$ with $ x\equal{}A$ and $ \sqrt {x^2 \plus{} 4}\equal{}U$ an invertible matrix s.t. $ U^2\equal{}A^2\plus{}4I$ and $ UA\equal{}AU$. Now the cost of the calculus of $ X_n$ is $ O(ln(n))$ matricial multiplications and we have a compact form. The condition can be reduced to: $ B\equal{}A^2\plus{}4I$ is invertible because in this case $ U$ exists in $ \mathcal{M}_n(\mathbb{R})$:
proof: $ B$ is a square in $ \mathcal{M}_n(\mathbb{R})$ iff each negative $ B$-eigenvalue has an even multiplicity. This is true because each negative $ A^2$-eigenvalue has an even multiplicity. $ a$ $ priori$ $ U$ commutes with $ A^2$ but we can choose it s.t. it commutes also with $ A$.
Obviously the method doesn't work if $ A^2\plus{}4I$ isn't invertible.

[quote="loup blanc"]$ B$ is a square in $ \mathcal{M}_n(\mathbb{R})$ iff each negative $ B$-eigenvalue has an even multiplicity.[/quote]
I'm sorry, but this only works for $ B$ diagonalizable. There is no $ C\in M_{2}(\mathbb{R})$ with $ C^2= (\begin{matrix}-1&1\\0&-1\end{matrix})$.
However I managed to prove, that for all $ A\in \mathcal{M}_n(\mathbb{R})$ with no eigenvalues $ \pm 2i$
we have $ B^2=A^2+4I_n$ and $ AB=BA$ for some $ B\in\mathcal{M}_n(\mathbb{R})$.
This is actually very long, but interesting.

[b]1) [/b]
If $ N\in \mathcal{M}_n(\mathbb{R})$ nilpotent (i.e. $ N^m=0$ for some $ m\in\mathbb{N}$)
then we can find $ q(x)\in\mathbb{Q}[x]$ with $ [q(N)]^2=I_n+N$.
[b][u]Pf:[/u][/b]
This follows from [url=http://en.wikipedia.org/wiki/Binomial_theorem]Newton's generalized binomial theorem[/url].
In fact we can take $ q(x)=\sum_{k=0}^{m-1}\binom{1/2}{k}x^k\in\mathbb{Q}[x]$,
where $ \binom{1/2}{0}: =1$ and $ \binom{1/2}{k}={{1\over 2}({1\over 2}-1)\cdots ({1\over 2}-(k-1))\over k!}$ for $ k\in\mathbb{N}$.

Then with $ N^0: =I_n$ and $ N^m=0$ we get
$ q(N)^2=(\sum_{k=0}^{m-1}\binom{1/2}{k}N^k)(\sum_{l=0}^{m-1}\binom{1/2}{l}N^l)\\
\hspace*{0.35in}=\sum_{0\le k,l\le m-1}\binom{1/2}{k}\binom{1/2}{l}N^{k+l}\\
\hspace*{0.35in}=\sum_{n=0}^{m-1}(\sum_{k,l\ge 0, k+l=n}\binom{1/2}{k}\binom{1/2}{l})N^{n}$
and we need to show
$ \sum_{k,l\ge 0, k+l=n}\binom{1/2}{k}\binom{1/2}{l}= \{\begin{array}{cl}1&\text{ for }n=0,1\\ 0&\text{ for }n\ge 2\end{array}$.

But for all $ p\in\mathbb{N}$ we have $ (1+x)^{2p}=(1+x)^p(1+x)^p$ in $ \mathbb{Q}[x]$
with $ (1+x)^{2p}=\sum_{n\ge 0}\binom{2p}{n}x^n$ and $ (1+x)^{p}=\sum_{n\ge 0}\binom{p}{n}x^n$ (here $ \binom{p}{n}=0$ for $ n>p$)
and then
$ \sum_{n\ge 0}\binom{2p}{n}x^n=(\sum_{k\ge 0}\binom{p}{k}x^k)(\sum_{l\ge 0}\binom{p}{l}x^l)\\
\hspace*{0.73in}= \sum_{n\ge 0}(\sum_{k,l\ge 0, k+l=n}\binom{p}{k}\binom{p}{l})x^{n}$
and the $ n$-th coefficient is
$ \binom{2p}{n}=\sum_{k,l\ge 0, k+l=n}\binom{p}{k}\binom{p}{l}$ for all $ p\in\mathbb{N}$ and $ n\in\mathbb{N}_0$.
So for every $ n\in\mathbb{N}_0$ the polynomial
$ P_n(x): =\binom{2x}{n}-\sum_{k,l\ge 0, k+l=n}\binom{x}{k}\binom{x}{l}\in\mathbb{Q}[x]$
has infinitely may roots $ p\in\mathbb{N}$, which gives $ P_n\equiv 0$.
And then $ x={1\over 2}$ gives the result
$ \sum_{k,l\ge 0, k+l=n}\binom{1/2}{k}\binom{1/2}{l}=\binom{1}{n}$ for all $ n\in\mathbb{N}_0$.
(Here $ \binom{x}{n}: ={x(x-1)\cdots (x-(n-1))\over n!}\in\mathbb{Q}[x]$ for all $ n\in\mathbb{N}$ and $ \binom{x}{0}\equiv 1$)

[b]2) [/b]
If $ J=(\begin{matrix}\lambda&1&&\\ &\ddots &\ddots &\\ &&\ddots &1\\ &&&\lambda\end{matrix})\in \mathcal{M}_n(\mathbb{C})$ is a [url=http://en.wikipedia.org/wiki/Jordan_block]Jordan block[/url] with $ \lambda\in\mathbb{C}\setminus \{\pm 2i\}$,
then we can find $ K\in \mathcal{M}_n(\mathbb{C})$ with $ K^2=J^2+4I_n$ and $ JK=KJ$,
and if $ \lambda\in\mathbb{R}$ we can take $ K\in \mathcal{M}_n(\mathbb{R})$.
[b][u]Pf:[/u][/b]
$ J=\lambda I_n+N$ with $ N=(\begin{matrix}0&1&&\\ &\ddots &\ddots &\\ &&\ddots &1\\ &&& 0\end{matrix})\in \mathcal{M}_n(\mathbb{Q})$ nilpotent. ($ N^n=0$)
Then $ \lambda^2+4\not =0$ and
$ J^2+4I_n=(\lambda I_n+N)^2+4I_n=(\lambda^2+4)I_n+2\lambda N+N^2=(\lambda^2+4)(I_n+\tilde{N})$

with $ \tilde{N}=N{2\lambda I_n +N\over \lambda^2+4}\in \mathcal{M}_n(\mathbb{C})$ nilpotent.
($ N$ and $ {2\lambda I_n +N\over \lambda^2+4}$ commute, so $ \tilde{N}^n=N^n({2\lambda I_n +N\over \lambda^2+4})^n=0$)
And now using 1) we get $ q(x)\in\mathbb{Q}[x]$ with $ q(\tilde{N})^2=I_n+\tilde{N}$, and we also find $ \mu\in\mathbb{C}$ with $ \mu^2=\lambda^2+4$.
Then $ K: =\mu q(\tilde{N})\in\mathcal{M}_n(\mathbb{C})$ satisfies $ K^2=J^2+4I_n$,
and since $ J=p_1(N)$ and $ K=p_2(N)$ for some polynomials
$ p_1(x)=x+\lambda\in\mathbb{C}[x]$ and $ p_2(x)=\mu q({x(x+2\lambda)\over \lambda^2+4})\in\mathbb{C}[x]$,
we therefore get $ JK=KJ=(p_1\cdot p_2)(N)$.
Finally if $ \lambda\in\mathbb{R}$, then $ \tilde{N}\in\mathcal{M}_n(\mathbb{R})$ and $ \mu=\pm\sqrt{\lambda^2+4}\in\mathbb{R}$,
and therefore $ K=\mu q(\tilde{N})\in\mathcal{M}_n(\mathbb{R})$.
(Here $ q(x)\in\mathbb{Q}[x]$)

[b]3) [/b]
If $ J=(\begin{matrix}J_1&&&\\ &J_2 & &\\ &&\ddots &\\ &&& J_k\end{matrix})\in \mathcal{M}_n(\mathbb{C})$ is a [url=http://en.wikipedia.org/wiki/Jordan_block]Jordan matrix[/url] with no eigenvalues $ \pm 2i$
and Jordan blocks $ J_1\in\mathcal{M}_{n_1}(\mathbb{C}),J_2\in\mathcal{M}_{n_2}(\mathbb{C}),\ldots,J_k\in\mathcal{M}_{n_k}(\mathbb{C})$,
then we can find $ K\in \mathcal{M}_n(\mathbb{C})$ with $ K^2=J^2+4I_n$ and $ JK=KJ$,
and if $ J\in \mathcal{M}_n(\mathbb{R})$ we can take $ K\in \mathcal{M}_n(\mathbb{R})$.
[b][u]Pf:[/u][/b]
Using 2) we find complex matrices $ K_1\in\mathcal{M}_{n_1}(\mathbb{C}),K_2\in\mathcal{M}_{n_2}(\mathbb{C}),\ldots,K_k\in\mathcal{M}_{n_k}(\mathbb{C})$
with $ K_i^2=J_i^2+4I_{n_i}$ and $ J_iK_i=K_iJ_i$ for $ i=1,\ldots ,k$.
Then take $ K=(\begin{matrix}K_1&&&\\ &K_2 & &\\ &&\ddots &\\ &&& K_k\end{matrix})\in \mathcal{M}_n(\mathbb{C})$ and we are done.
(by 2) $ K\in\mathcal{M}_n(\mathbb{R})$ for $ J\in\mathcal{M}_n(\mathbb{R})$)

And finally
[b]4) [/b]
If $ A\in\mathcal{M}_n(\mathbb{R})$ with no eigenvalues $ \pm 2i$,
then $ B^2=A^2+4I_n$ and $ AB=BA$ for some $ B\in\mathcal{M}_n(\mathbb{R})$.
[b][u]Pf:[/u][/b]
Using the [url=http://en.wikipedia.org/wiki/Jordan_normal_form]Jordan normal form[/url] and $ A=\overline{A}$ we get
$ U\in GL_n(\mathbb{C})$ and Jordan matrices $ J_1\in\mathcal{M}_r(\mathbb{R}),J_2\in\mathcal{M}_s(\mathbb{C}),n=r+2s$ with
$ J: =U^{-1}AU=(\begin{matrix}J_1&&\\ &J_2 & \\ &&\overline{J_2} \end{matrix})$.
(Here if $ C=(c_{ij})\in \mathcal{M}_n(\mathbb{C})$ then $ \overline{C}: =(\overline{c_{ij}})$)
Then using 3) we find $ K_1\in\mathcal{M}_r(\mathbb{R}),K_2\in\mathcal{M}_s(\mathbb{C})$ with $ K_1^2=J_1^2+4I_r,K_2^2=J_2^2+4I_s$
and $ J_1K_1=K_1J_1$ and $ J_2K_2=K_2J_2$.
But then for $ K: =(\begin{matrix}K_1&&\\ &K_2 & \\ &&\overline{K_2} \end{matrix})\in\mathcal{M}_n(\mathbb{C})$ we have $ K^2=J^2+4I_n$ and $ JK=KJ$.
Let's write $ J_2=P+iQ$ and $ K_2=R+iS$ with $ P,Q,R,S\in\mathcal{M}_s(\mathbb{R})$.
Then $ V: =(\begin{matrix}I_r&&\\ &I_s &iI_s \\ &iI_s&I_s \end{matrix})\in GL_n(\mathbb{C})$ with $ V^{-1}={1\over 2}(\begin{matrix}I_r&&\\ &I_s &-iI_s \\ &-iI_s&I_s \end{matrix})$,
and $ \tilde{J}: =VJV^{-1}=(\begin{matrix}J_1&&\\ &P &Q \\ &-Q&P \end{matrix})\in\mathcal{M}_n(\mathbb{R})$
and $ \tilde{K}: =VKV^{-1}=(\begin{matrix}K_1&&\\ &R &S \\ &-S&R \end{matrix})\in\mathcal{M}_n(\mathbb{R})$.
Then $ \tilde{K}^2=VK^2V^{-1}=V(J^2+4I_n)V^{-1}=(VJV^{-1})^2+4I_n=\tilde{J}^2+4I_n$
and $ \tilde{J}\tilde{K}=VJKV^{-1}=VKJV^{-1}=\tilde{K}\tilde{J}$.

Also $ A=UJU^{-1}=UV^{-1}\tilde{J}VU^{-1}=W\tilde{J}W^{-1}$ for $ W: =UV^{-1}\in GL_n(\mathbb{C})$ and $ A,\tilde{J}\in\mathcal{M}_n(\mathbb{R})$.
Then if we write $ W=G+iH$ with $ G,H \in\mathcal{M}_n(\mathbb{R})$ we get
$ AG+iAH=AW=W\tilde{J}=G\tilde{J}+iH\tilde{J}$ with $ AG,AH,G\tilde{J},H\tilde{J}\in\mathcal{M}_n(\mathbb{R})$.
This gives $ AG=G\tilde{J}$ and $ AH=H\tilde{J}$.
Since $ \det (G+iH)=\det W\not =0$ we have the polynomial $ p(x)=\det(G+xH)\in\mathbb{R}[x]$ not constant $ 0$.
Then $ p(x)$ has only finitely many roots, and we find $ \alpha\in\mathbb{R}$ with $ 0\not =p(\alpha)=\det(G+\alpha H)$.
But now $ \tilde{W}: =G+\alpha H\in GL_n(\mathbb{R})$ with
$ A\tilde{W}=AG+\alpha AH=G\tilde{J}+\alpha H\tilde{J}=\tilde{W}\tilde{J}$.
So $ A=\tilde{W}\tilde{J}\tilde{W}^{-1}$ and $ B: =\tilde{W}\tilde{K}\tilde{W}^{-1}\in\mathcal{M}_n(\mathbb{R})$ satisfies $ B^2=A^2+4I_n$ and $ AB=BA$.
Done! :D

Well done olorin.
I was wrong; in fact the set of the matrices $ B$ s.t. each negative $ B$-eigenvalue has an even multiplicity is the closure in $ \mathcal{M}_n(\mathbb{R})$ of the set of the invertible matrices which are square of real matrices.

Hi Olorin,
I read now your proof and there exists an important gap in the part 2): you haven't solved the case $ \lambda^2\plus{}4<0$. In fact the 2) with a real  $ \lambda$  is false in this case (as your example with the $ C$-matrix  -cf. top page-  proves it!). You must work with several Jordan blocks in this hot case.
Example with $ \lambda^2\plus{}4<0$: $ a\equal{}\begin{pmatrix}0&\minus{}3\\3&0\end{pmatrix}$. $ a^2\plus{}4I\equal{}\minus{}5I$.
Moreover let $ J$ be the nilpotent Jordan block of dimension 2:
$ \minus{}I\plus{}J$ isn't the square of a real matrix but $ diag(\minus{}I\plus{}J,\minus{}I\plus{}J)$ is.

You misunderstood my proof.
In 2) I never say $ \lambda,\mu\in\mathbb{R}$.
Just $ \lambda\in\mathbb{C}\setminus\{\pm 2i\}$
and $ \mu\in\mathbb{C}$ is some square root of $ \lambda^2\plus{}4\in\mathbb{C}$.
($ \mathbb{C}$ is algebraically closed, so this always exists.)
If $ \lambda\in\mathbb{R}$ in 2), then $ \lambda^2\plus{}4\ge 4>0$ and $ \mu\equal{}\pm\sqrt{\lambda^2\plus{}4}\in\mathbb{R}$.

The step from $ \mathbb{C}$ to $ \mathbb{R}$ is first done in 4),
where I "work with several Jordan blocks" just as you say.

[quote="loup blanc"]
Example with $ \lambda^2 \plus{} 4 < 0$: $ a \equal{} \begin{pmatrix}0 & \minus{} 3 \\
3 & 0\end{pmatrix}$. $ a^2 \plus{} 4I \equal{} \minus{} 5I$.[/quote]
Now I don't understand.
This $ a$ has eigenvalues $ \pm 3i\not\in\mathbb{R}$.
So how is this an example with $ \lambda\in\mathbb{R}$ and $ \lambda^2\plus{}4<0$?
(such $ \lambda$ don't exist anyway)

[quote="loup blanc"]
$ \minus{} I \plus{} J$ isn't the square of a real matrix but $ diag( \minus{} I \plus{} J, \minus{} I \plus{} J)$ is.[/quote]
In 1) it's suposed to be $ N\in\mathcal{M}_n(\mathbb{C})$, not just real $ N$. I'm sorry.
Then $ I_n\plus{}N$ is a square of a complex matrix. (even of a rational matrix, if $ N$ is rational.)
However it doesn't say anything about $ \minus{}I_n\plus{}N$ over $ \mathbb{R}$.
This again is done first in 4).

Hi olorin, once more I read too quickly.
 Yet your proof could be much better;you can prove easily with your method the following result:
Let $ U$ be a real invertible $ (n,n)$ matrix; then $ U$ is the square of a real matrix iff, in the $ U$-Jordan form, for all $ k$, the number of $ (k,k)$ Jordan blocks associated to each negative $ U$-eigenvalues is even.
Proof: ($ \rightarrow$)    $ A$, a real invertible matrix has the Jordan form: $ diag({\lambda}I_r\plus{}J_r,\cdots,{\mu}I_s\plus{}J_s,{\bar{\mu}}I_s\plus{}J_s,\cdots)$ with $ \lambda\in\mathbb{R}^*$ and $ \mu\notin\mathbb{R}$; then $ A^2$ has the Jordan form: $ diag({\lambda}^2I_r\plus{}J_r,\cdots,{\mu}^2I_s\plus{}J_s,{\bar{\mu}}^2I_s\plus{}J_s,\cdots)$. The case $ \mu\in{i\mathbb{R}^*}$ gives the result.
($ \leftarrow$)     a candidate $ U$ has the Jordan form: $ diag({\lambda}I_r\plus{}J_r,\cdots,{\mu}I_s\plus{}J_s,{\bar{\mu}}I_s\plus{}J_s,\cdots,
{\nu}I_k\plus{}J_k,{\nu}I_k\plus{}J_k,\cdots)$ with
$ \lambda>0$, $ \mu\notin{\mathbb{R}}$ and $ \nu<0$.
$ {\lambda}I_r\plus{}J_r$ is a square (binomial formula).
$ diag({\mu}I_s\plus{}J_s,{\bar{\mu}}I_s\plus{}J_s)$ is a square  (your proof).
$ diag({\nu}I_k\plus{}J_k,{\nu}I_k\plus{}J_k)$ is a square  (your proof).    QED

Now it's obvious that if $ A$ is a real matrix and $ \alpha>0$ s.t. $ A^2\plus{}\alpha{I}$ is invertible, then $ A^2\plus{}\alpha{I}$ is a square. The result is false if $ \lambda<0$ except if we add a condition about the positive $ A^2$-eigenvalues.

I have forgotten the condition about commutation.
With the notations of  $ (\rightarrow)$, the $ A^2\plus{}\alpha{I}$ Jordan form is:
$ diag(({\lambda}^2\plus{}\alpha)I_r\plus{}J_r,\cdots,({\mu}^2\plus{}\alpha)I_s\plus{}J_s,
({\bar{\mu}}^2\plus{}\alpha)I_s\plus{}J_s,\cdots,({\nu}^2\plus{}\alpha)I_k\plus{}J_k,({\nu}^2\plus{}\alpha)I_k\plus{}J_k,\cdots)$

with $ \lambda\in\mathbb{R}, \nu^2\plus{}\alpha<0$ and $ \mu$ the other eigenvalues.
Remark: I don't write, as in my precedent post, the matrix which changes the basis. It isn't important because if 2 real matrices are similar over $ \mathbb{C}$ then they are similar over $ \mathbb{R}$.
$ ({\lambda}^2\plus{}\alpha)I_r\plus{}J_r\equal{}B^2$ with $ BJ_r\equal{}J_rB$ then $ B$ and $ {\lambda}I_r\plus{}J_r$ commute. Same argument for the 2 other blocks; for example:
$ diag(({\nu}^2\plus{}\alpha)I_k\plus{}J_k,({\nu}^2\plus{}\alpha)I_k\plus{}J_k)\equal{}B^2$ with $ Bdiag(J_k,J_k)\equal{}diag(J_k,J_k)B$ then $ B$ and $ diag({\nu}I_k\plus{}J_k,{\nu}I_k\plus{}J_k)$ commute.
Conclusion: $ A^2\plus{}\alpha{I}\equal{}B^2$ where $ B$ and  $ A$ commute.
Beware: $ B$ isn't necessarily a polynomial in $ A$.

[quote="loup blanc"]for all $ k$, the number of $ (k,k)$ Jordan blocks associated to each negative $ U$-eigenvalues is even.[/quote]
This is equivalent to $ \dim \text{Ker}(A\plus{}\lambda I_n)^k$ even for all $ \lambda\in\mathbb{R}_\plus{}$ and $ k\in\mathbb{N}$.

could the moderator(s) move this topics to College playground / Matrix Algebra / Unsolved and proposed problems ?
[color=red][[b]Moderator edit:[/b] done - darij][/color]

I think that the problem was:
Let $A\in\mathcal{M}_k(\mathbb{R})$ be a given matrix and let $(X_i)_{i\geq{0}}\in\mathcal{M}_k(\mathbb{R})$ be a sequence defined as follows:
$X_0=0_k,X_1=I_k,X_{n+2}=AX_{n+1}+X_n$.
Find $X_n$.


[quote="CCMath1"]let $ {{a[n]}}$ be a sequence satisfying a1=1,$ {a[n \plus{} 1]}$=2$ {a[n]}$+n*(1+$ {2^n}$), n=1,2,3,...... then find $ {a[n]}$ .[/quote]

We are to determine the values $ p,\ q,\ r,\ s$ such that the given reccursive formula as follows.

$ a_{n\plus{}1}\plus{}p(n\plus{}1)\plus{}q\plus{}\{r(n\plus{}1)\plus{}s\}2^{n\plus{}1}\equal{}2(a_{n}\plus{}pn\plus{}q\plus{}(rn\plus{}s)2^{n})$

[quote="CCMath1"]let $ {{a[n]}}$ be a sequence satisfying a1=1,$ {a[n \plus{} 1]}$=2$ {a[n]}$+n*(1+$ {2^n}$), n=1,2,3,...... then find $ {a[n]}$ .[/quote]

[color=blue]
[hide="My Solution"]
Given : $ a_{n\plus{}1}\equal{}2a_n \plus{} n(1\plus{}2^n)$

$ \implies \frac{a_{n\plus{}1}}{2^{n\plus{}1}}\equal{}\frac {a_n}{2^{n}} \plus{} n\left(\frac{1}{2^{n\plus{}1}}\plus{}\frac 12\right)$

Let $ \  \frac{a_{n}}{2^{n}} \equal{} b_n$

$ \implies b_n \minus{}b_{n\minus{}1} \equal{} (n\minus{}1)\left(\frac{1}{2^n}\plus{}\frac 12\right)$

$ \implies b_n \minus{} b_{1}\equal{} \sum_{r\equal{}1}^{n\minus{}1}\frac{r}{2^{r\plus{}1}}\plus{}\sum_{r\equal{}1}^{n\minus{}1}\frac{r}{2}\equal{}1\minus{}\frac {n\plus{}1}{2^n}\plus{}\frac {n(n\minus{}1)}{4}$

$ \implies\frac{a_{n}}{2^{n}}\minus{}\frac{1}{2}\equal{}1\minus{}\frac {n\plus{}1}{2^n}\plus{}\frac {n(n\minus{}1)}{4}$

$ \implies \boxed{\boxed{a_{n} \equal{}(n^2\minus{}n\plus{}6)2^{n\minus{}2} \minus{}(n\plus{}1)}}$
[/hide]
[/color]

[quote="kunny"][quote="CCMath1"]let $ {{a[n]}}$ be a sequence satisfying a1=1,$ {a[n \plus{} 1]}$=2$ {a[n]}$+n*(1+$ {2^n}$), n=1,2,3,...... then find $ {a[n]}$ .[/quote]

We are to determine the values $ p,\ q,\ r,\ s$ such that the given reccursive formula as follows.

$ a_{n \plus{} 1} \plus{} p(n \plus{} 1) \plus{} q \plus{} \{r(n \plus{} 1) \plus{} s\}2^{n \plus{} 1} \equal{} 2(a_{n} \plus{} pn \plus{} q \plus{} (rn \plus{} s)2^{n})$[/quote]

Sorry, this approach aren't useful in this case.

Determine the values $ p,\ q,\ r,\ s$ such that the given reccursive formula as follows.

$ a_{n \plus{} 1} \plus{}(n \plus{} 1) \plus{} f(n\plus{}1) \equal{} 2(a_{n} \plus{} n \plus{} f(n))$ for some function $ f(n)$.


what is the remainder when $ (1!\plus{}2!\plus{}3!\plus{}4!\plus{}...\plus{}2006!\plus{}2007!)$ is divided by 7?

(can some1 explain the solution both with and without using mods)

Well simply all of these are multiples of $ 7$ $ 7!,8!,9!,\cdots,2007!$ and congruent to $ 0$ mod 7. So all we need to check is the remainder of $ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!$ when divided by $ 7$:

$ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!\equiv 1 \plus{} 2 \plus{} 6 \plus{} 24 \plus{} 120 \plus{} 720\pmod{7}$
$ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!\equiv (1\plus{}6) \plus{} 2 \plus{} 3 \plus{} 1 \plus{} 6\pmod{7}$
$ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!\equiv 7 \plus{} 7 \plus{} 5\pmod{7}$
$ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!\equiv \boxed{5}\pmod{7}$

The non-modulus way:
We can use the fact that for all the factorials except the factorials of $ 1,2,3,4,5,6$ are divisible by $ 7$, so we only need to check the remainder when $ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!$ is divided by $ 7$. $ 1!\plus{}2!\plus{}3!\plus{}4!\plus{}5!\plus{}6!\equal{}873$ which is the same as $ 124*7 \plus{} 5$ and $ \therefore$ the remainder is $ 5$.


geometry
geometric transformation
rotation
number theory
prime numbers

The problems:

1.  How many such 5-digit numbers are there in which the sum of the digits is even?

2.   Find the prime numbers a, b, c, and d, if a + b = c, and a - b = d. (a minus b = d)

3.  A red line is going upwards in a spiral from the bottom of a cylinder to the top, rising by the same rate all along, making 6 complete rotations around the cylinder. The radius of the cylinder is 12 centimeter, and its height is 1 meter. How long is the line. (Ignore the width of the line.)

For number 2,
[hide]
a=5
b=2
c=7
d=3[/hide]
[/hide]

[hide]6 right triangles with legs of $\frac {50}3$ and $24 \pi$.

$6\sqrt{\left(\frac {50}3\right)^2 + (24 \pi)^2} \approx 463.09$

Unless I messed up putting it in my calculator.[/hide]

[hide="1."]
The number can have 5 even digits.
For this, we have $4*5^4$.
The number can also have 3 even digits and 2 odd digits.
These can be arranged in 10 ways, with 6 having an even at the front.
So, with an odd at the front we have:
$4\left(5^5\right)$
And for an even at the front, we have:
$6\left(4*5^4\right)$
Finally, we can have 1 even and 4 odds.
There is one way to put the even in front and 4 to put an odd in front so:
$4*5^4$ and $4*5^5$
Summing all of these up gives us:
$8*5^5+32*5^4=8*5^5+6(5)5^4+2*5^4=14*5^5+2*5^4=45000$
[/hide]

[hide="2."]
Since all primes are positive integers, d is a positive integer.  Therefore a>b.  Since a>b and a+b=c, c>a>b.  So, c cannot be 2, the only even prime.  Since c is odd, either a or b must be even.  b must be even and 2, since it is smaller than a. Thus, we have:
b=2
a+2=c
a-2=d
So d,a,c must be a set of triple primes, namely 3,5,7, with d<a<c.

Finally,
a=5
b=2
c=7
d=3
[/hide]

[hide="3."]
Since the radius is 12, the circumference is $24\pi$.
Since the line wraps around 6 times, and is going upwards, we have a right triangle with leg $144\pi$.
Our other leg is the height, 100.
Using Pythagoras, we get $463.310$cm, or $4.633 m$
[/hide]

basically the number of 5 digit numbers is 90000
half have even digital roots


On a True/False test, Amy answered the first three questions wrong but answered the rest of the questions correctly. On the same test, Scott answered exactly two questions wrong. He answered the last question wrong, the 26 questions before it correctly, and the question before that wrong. If Amy and Scott were both incorrect on the same question exactly once, what is the greatest possible percent of the total number of questions that Amy could have answered correctly?

[hide]
If Scott answered the last question wrong, the 26 before that right, and the one before that wrong, and only missed two, than Amy missed one of those two.  We know that Amy couldn't have missed the last question, so she missed the 28th to last question (which Scott also missed). We can now see that there are 30 questions on the test, and that Amy missed 3 of them, so she had 27/30=90 percent accuracy.
[/hide]


Solve the following equation in $ Z$: $ 1\plus{}4^{x}\plus{}4^{y}\equal{}z^{2}$

[quote="Rust"]Let $ x < y$, then $ z\ge 2^{y}\plus{}1$. $ (2^{y}\plus{}1)^{2}\equal{} 4^{y}\plus{}2^{y\plus{}1}\plus{}1 \equal{} 4^{y}\plus{}24^{x}\plus{}1$ if y is odd $ x \equal{}\frac{y\plus{}1}{2}$.
If $ x < x_{0}\equal{}\frac{y\plus{}1}{2}$, then $ 2^{y}< z < 2^{y}\plus{}1$ is not integer.
If $ x > x_{0}$, then $ z \equal{} 2^{y}\plus{}k,k > 1, k(k\plus{}2^{y\plus{}1}) \equal{} 4^{x}\plus{}1,k \equal{}\pm 1\mod 2^{y}$. It can not be.[/quote]

I think there must be something wrong. The Equation have at least 1 solution $ (x,y,z)\equal{}(2,3,9)$

[quote="Mashimaru"]I think there must be something wrong. The Equation have at least 1 solution $ (x,y,z) \equal{} (2,3,9)$[/quote]
These solution I find $ z\equal{}2^{y}\plus{}1,x\equal{}\frac{y\plus{}1}{2}$, were y is odd. For example y=1 give (1,1,3), y=3 give (2,3,9),...

[quote="Rust"]Let $ x < y$, then $ z\ge 2^{y}\plus{}1$. $ (2^{y}\plus{}1)^{2}\equal{} 4^{y}\plus{}2^{y\plus{}1}\plus{}1 \equal{} 4^{y}\plus{}24^{x}\plus{}1$ if y is odd $ x \equal{}\frac{y\plus{}1}{2}$.
If $ x < x_{0}\equal{}\frac{y\plus{}1}{2}$, then $ 2^{y}< z < 2^{y}\plus{}1$ is not integer.
If $ x > x_{0}$, then $ z \equal{} 2^{y}\plus{}k,k > 1, k(k\plus{}2^{y\plus{}1}) \equal{} 4^{x}\plus{}1,k \equal{}\pm 1\mod 2^{y}$. It can not be.[/quote]

I agree that a triple $ (x,2x\minus{}1,2^{2x\minus{}1}\plus{}1)$ for $ (x,y,z)$ will construct a solution for every positive integer $ x$

but i do not see why there are no other solutions. because from $ k(k\plus{}2^{y\plus{}1}) \equal{} 4^{x}\plus{}1$ u can only conclude that $ k^{2}\equal{} 1(mod2^{y})$ not that $ k\equal{}\pm1(mod 2^{y})$ what about for example $ k \equal{} 3(mod8)$ or $ k \equal{} 5(mod8)$? and i also dont get why $ k \equal{}\pm1(mod 2^{y})$ can't be true :huh:

[quote="Mashimaru"]Solve the following equation in $ Z$: $ 1\plus{}4^{x}\plus{}4^{y}\equal{} z^{2}$[/quote]
Once more? :huh: It seems to be a conspiracy (click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160202]here[/url])...

Let $ x < y$ ,
Then
$ 1\plus{}4^{x}(1\plus{}4^{y\minus{}x}) \equal{} z^{2}$
$ 4^{x}(1\plus{}4^{y\minus{}x}) \equal{} (z\plus{}1)(z\minus{}1)$
Obviously,$ z$ is odd
so let $ z \equal{} 2a\plus{}1$
$ 4a(a\plus{}1) \equal{} 4^{x}(1\plus{}4^{y\minus{}x})$
$ a(a\plus{}1) \equal{} 4^{x\minus{}1}(1\plus{}4^{y\minus{}x})$

If $ a \equal{} 4^{x\minus{}1}\equal{} 4^{y\minus{}x}$, we get $ (x,y,z) \equal{} (x,2x\minus{}1,2^{x\minus{}1}\plus{}1)$

but I can't see why there isn't other solutions.

I'm sorry but I really can not understand the solution of Rust. Could you explain your proof easier?

If $ z^{2}\equal{} 1\plus{}4^{x}\plus{}4^{y}$ and $ \frac{y\plus{}1}{2}< x\le y$, then $ z \equal{}\pm 1\plus{}\mod 2^{2x\minus{}1}$ and $ 2x\minus{}1 > y$,
 therefore for minimal value of z: $ z_{0}\equal{}\minus{}1\plus{}2^{2x\minus{}1}, z_{0}^{2}\equal{} 1\minus{}4^{x}\plus{}4^{2x\minus{}1}> 1\plus{}4^{x}\plus{}4^{y}$.

[quote="Mashimaru"]Solve the following equation in $ Z$: $ 1\plus{}4^{x}\plus{}4^{y}\equal{} z^{2}$[/quote]
have a look here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=160202[/url]
 :)


trigonometry
geometry
geometric transformation
dilation
complex numbers

Compute arctan(4/3) - arctan(1/7)

I'm not really sure how to start this problem.  Maybe drawing a picture of the triangles that these each term would correspond to and/or using identities could help, but I'm not exactly sure how at the moment.

$ tan(\tan^{\minus{}1}{(4/3)} \minus{} \tan^{\minus{}1}{(1/7)})\equal{}x$
Use the angle add formula for tangent, and then arctan your x.
Should be fairly simple, since
$ tan(\tan^{\minus{}1}{x)}\equal{}x$

[hide="solution"]
The trick was to set up an equation and take the tangent of both sides. After using the angle difference identity to get a numerical value, we just need to take arctan of both sides to isolate x.
\begin{eqnarray*} \arctan^{-1}(4/3) - \arctan(1/7) & = & x \\
\tan[\arctan(4/3)-\arctan(1/7)] & = & \tan(x) \\
\frac{\tan(\arctan(4/3))-\tan(\arctan(1/7))}{1+\tan(\arctan(4/3))\tan(\arctan(1/7))}&=& \\
\frac{(4/3)-(1/7)}{1+(4/3)(1/7)}&=& \\
\frac{(25/21)}{(25/21)}&=&\tan(x)\\
\arctan(1)&=&x \\
x&=&\boxed{\pi/4}
\end{eqnarray*}
[/hide]

[hide="Cool solution"]
Notice that $ \arctan\frac ba \equal{} \text{arg}\{a \plus{} bi\}$.

$ \text{arg}\{3 \plus{} 4i\} \minus{} \text{arg}\{7 \plus{} i\} \equal{} \text{arg}\left\{\frac {3 \plus{} 4i}{7 \plus{} i}\right\}$, because the product of two complex numbers multiplies the magnitudes and adds the arguments.

$ \frac {3 \plus{} 4i}{7 \plus{} i}\cdot \overline{7 \plus{} i}\cdot (\overline{7 \plus{} i})^{ \minus{} 1} \equal{} \frac {(3 \plus{} 4i)(7 \minus{} i)}{|7 \plus{} i|} \equal{} \frac {25 \plus{} 25i}{|7 \plus{} i|}$.

Dilate by $ |7 \plus{} i|$ (dilation preserves angles) to get $ 25 \plus{} 25i$, which is clearly an isosceles right triangle in the complex plane, making the angle $ \boxed{\frac {\pi}{4}}$.
[/hide]

The [url=http://en.wikipedia.org/wiki/Arg_%28mathematics%29]argument[/url] of a complex number.


In aTriangle ABC, draw AD, where D is the midpoint of BC.

If <ACB = 30° and <ADB = 45°, find <ABC.

Let $E$ be a point on $AB$ such that $BE\bot AC$, let $BD=x$
Because $D$ is a center of circumcircle of $\triangle BCE$,
$DE=x$ but $\angle CBE=60^\circ$ so $BE=x$.
Because $\angle EAD=\angle EDA=15^\circ$, $AE=x$ so $AE=BE$.
$\angle EAB=\angle EBA=45^\circ$. We get $\angle ABC=105^\circ$


This is how this topic will work..
I will start by posting a puzzle (which i will by the end)
and if you get the answer, you pm me.
if i pm you back, saying that you got it right, you have the right to post another puzzle.

do not post until you have received a pm from me...

okay!
now to the first puzzle...

[b]water!![/b]

A man walks into a bar, asking for a glass of water. The bartender instead gets out a gun and points it at the man. The man says, "Thank you" and walks out.
Why did the man ask for the glass of water?

oh, and also, you can post yes/no questions here, but they will be replied to as soon as possible (probably the next day)

Thanks!!

[hide="Silly riddle"]The man had the hiccups, and the waiter was trying to cure him by scaring him.[/hide]

At the end of World War II the few scattered members of a once-rich and important Central European family who had survived concentration camps managed to come together. The family fortune, although much diminished, was still of considerable size. In due course these few known survivors died and the courts began a search for a possible heir to the estate. Two candidates showed up, both claiming to be the same person - the only son of the last head of the family. Their stories were similar (each claimed to have been in slave-labor camps); their information on the family history was much the same; there was a superficial resemblance between them so that old acquaintances could not decide which was the true heir. The magistrate in charge of the case decided to interview them separately. To the first candidate the magistrate proposed a simple identity test. However, when the man readily agreed, the magistrate immediately said, "Arrest this man: he is an impostor. The other is the true heir." How come?

Copied shamelessly from a website, word-for-word.

i heard of this..
[hide]blood type test..[/hide]
i don't know any off the top of my head right now...

heres one: One day, my dad decided to do a little traveling around Asia. He left me in charge of collecting his mail. Then, two days after he left, he called me, asking how much mail he had gotten. i told him "none, because i dont have the key to the mailbox(who has keys to mailboxes anyways?). He said sorry and promised to mail it to me. When he came back, he again asked how much mail he had gotten. I once again replied "none." and then explained to him why he had gotten no mail.

WHY?

(very simple, if you think about it, and if you read it carefully)

Um, if you don't have the key to the mailbox, how do you get the key that he mails to you?

My cousin Sheng Wu can walk faster than the speed of sound. How?

[quote="Temperal"]Um, if you don't have the key to the mailbox, how do you get the key that he mails to you?[/quote]
yeah. exactly. told you it was easy

as for yours...my only guess is (s)he was on a whip, which is highly unlikely

In space, sound cannot travel.

Okay, yeah. Also, very thin air might do the trick (although it'll pretty much have to be in space).

new problem...

this is really hard to get, and the solution is stupid...

A man goes into a restaurant and orders albatross soup. after he tastes a little, he walks outside and commits suicide.

why did he do that?

and those of you who know this, don't spoil it..
you can ask random yes/no questions..

I know the answer, so I'll guess a wrong answer, and the most common wrong one:
He was crazy.
Just so that no one tries to guess that....

I think this one is too well known.

yeah..
well..
I didn't know this one until mid-July 2007 and I've been looking at lateral thinking puzzles since 05..
xD


Dr.Boring got bored of the class, so he decided to go duck hunting

Dr.Boring bought a rifle and some super gigantic bullets

the total cost was 30.00 dollars

Dr.Boring thought it was too expensive, so he begged for discounts

when he explained the clerk about his duck hunting situation, the clerk discounted 50 percent of
the rifle...

if the total cost including the discount was 23 dollars and Dr.Boring bought 8 bullets, how does a single bullet costs?

( I think I gave you more than enough information to solve this problem..)

[color=darkred]Fixed[/color]

[size=75]Edited by rcv[/size]

[hide="solution"]
Well, if the 50% discount on the gun took away 7 dollars, that means the gun cost 14 dollars without the discount. 30 - 14 = 16 dollars for the 8 bullets, meaning each bullet costs two dollars :what?: .
[/hide]

Yup, it costs 2 bucks each because they are the Mr.GiganticBullet brand

well, what did you do with my post? :?

I didn't touch it. I think it might have to do with the dollar signs...

oooo....

I forgot that LaTeX works with dollar sign... :D  :blush:

Quite gory... ;)

[hide]
The difference between the original price and the discounted gun price is 7 dollars. Since the gun was 50% off, then it must have originally been 14 dollars. If you subtract 14 from 30, you'll get 16, which is the cost of the 8 bullets. 8 bullets cost 16 dollars, so 1 bullet costs 2 dollars.
[/hide]


(1) Let $ a,$ $ b,$ $ c$ be real numbers in the interval $ [\minus{}1,\; 1]$ such that $ a\plus{}b\plus{}c\equal{}0.$ If $ k \ge \frac{3\minus{}\sqrt{2}}{2}\approx 0.79289\ldots,$ then
$ \sqrt{1\plus{}a\plus{}kb^2}\plus{}\sqrt{1\plus{}b\plus{}kc^2}\plus{}\sqrt{1\plus{}c\plus{}ka^2} \ge 3.$

(2) Let $ a,$ $ b,$ $ c,$ $ d$ be real numbers in the interval $ [\minus{}1,\; 1]$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}0.$ If $ k \ge \frac{5\minus{}2\sqrt{2}}{2} \approx 1.08578\ldots,$ then
$ \sqrt{1\plus{}a\plus{}kb^2}\plus{}\sqrt{1\plus{}b\plus{}kc^2}\plus{}\sqrt{1\plus{}c\plus{}kd^2}\plus{}\sqrt{1\plus{}d\plus{}ka^2} \ge 4.$

I will post my solution for (1) first. I am typing the solution for (2), please wait for some moment...

For (1), see my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-30-p-t-nam-v-q-b-can.html

For (2), see my proof here (please check it carefully): http://can-hang2007.blogspot.com/2010/01/inequality-31-p-t-nam-v-q-b-can.html

I think that the following inequalities hold too but I cannot prove them...

(1) Let $ a,$ $ b,$ $ c$ be real numbers in the interval $ [ \minus{} 1,\; 1]$ such that $ a \plus{} b \plus{} c \equal{} 0.$ If $ k \ge \frac {7}{9},$ then
$ \sqrt {1 \plus{} a \plus{} kb^2} \plus{} \sqrt {1 \plus{} b \plus{} kc^2} \plus{} \sqrt {1 \plus{} c \plus{} ka^2} \ge 3.$

(2) Let $ a,$ $ b,$ $ c,$ $ d$ be real numbers in the interval $ [ \minus{} 1,\; 1]$ such that $ a \plus{} b \plus{} c \plus{} d \equal{} 0.$ If $ k \ge 1,$ then
$ \sqrt {1 \plus{} a \plus{} kb^2} \plus{} \sqrt {1 \plus{} b \plus{} kc^2} \plus{} \sqrt {1 \plus{} c \plus{} kd^2} \plus{} \sqrt {1 \plus{} d \plus{} ka^2} \ge 4.$


email MATHCOUNTS ARML AMC USA(J)MO USAMO Mafia

JV state is on the last day of this month at Woodward Academy. Who's going?

I'm not, I'm going to the state science fair instead. Pwn for me, will you? ;)

They give plaques to the top 20, so I, of course, got 21st last year  :lol:, but I'm much better at math this year. Also, I've been learning math from the teacher who organizes the tournament from a book that he wrote for the past two years, so fortunately, I know how he writes problems at this point. I usually only miss one or two questions when practicing with the old tests (almost always careless mistakes), so I think that top five is a realistic goal. That being said, I don't want to get my hopes up, so I'm going to set my goal at top ten.

[quote="Boy Soprano II"]I'm going.  I'm hoping to beat my score from last year. :P[/quote]

 :rotfl:

0th place then?

Walton will be there. (Assuming our damn coach can get off his butt and do the paperwork!) I don't think that we will be able to repeat our relay performance from last year however. That was just amazing to watch. We'll be bringing some of the Dickerson Middle School kids also.

[quote="Boy Soprano II"]I'm going.  I'm hoping to beat my score from last year. :P[/quote]

So you're going, how about Santosh? He's a sophomore too. Hmmh, sorry Billy, but I can't pwn against these guys *-_-  :D

At this moment, no. I don't know how or where to register as an individual. If I can figure that out, I'll go.

No Wheeler JV???!!! You guys were WAY too strong not to go. The website is:

http://www.woodward.edu/academics/us/math/mathteam_sjv_tournament.asp

Send an email to Jeff Floyd at Woodward Academy:

jeff.floyd@woodward.edu

If for some reason you end up going as an individual, I can probably help with transportation.

does anyone have a copy of past tests? pm me  :)

[quote="gtf"]No Wheeler JV???!!! You guys were WAY too strong not to go.[/quote]

Agreed. Regardless, I am now going, but I will be at a severe disadvantage during team ciphering.

[quote="gtf"]Need a ride? Personally, I'm waiting to see you during the relay!  :D[/quote]

He'll probably do similarly to me last year :D. [I held up a wild guess about 2 seconds after the start, which got everyone wonderfully psyched out when my runner yelled out "FIRST!". Though he sort of was confused, because my answer wasn't right :D.]

Ahhh, so that's what happened. My guys were moving so fast, I was amazed that anyone could beat them. We were first on the first two relays and got a right answer for the third. One little mistake or we might have been first on all three. It was one of the most amazing performances I've ever seen. (Maybe it proves that my guys work best when I don't let them talk! :-)

Thinking about it, I've now coached a JV State Championship team and a MATHCOUNTS State Championship team. Just need a varsity to go with it!  :D

[quote="Karth"]whats non manipulative calculator? can I take a TI89 Titanium?[/quote]

No, if it can solve equations, factor, etc., it isn't allowed.

Side note: is there such thing as a TI85, 86, 87, or 88? I've never heard of any of them. Did they just decide to skip four numbers?

Edit: To answer my own question, I found a TI85 and 86 online; still no sign of an 87 or 88...

(Oh boy! This will be post #50!)

There were TI-85 and 86's the 85 was the predecessor to the 86. The TI-86 is even older than the 83's. I've never much liked it. It has a lousy user interface. It's mostly as functional as the 83's and 84's.

Never heard of an 87 or 88.

The other part of the question (Karth's), it's talking about a symbol manipulator. For TI, no 89's or 92's.

Fulton

($25$ answered) - (one incorrect guess) - (two extremely careless mistakes) = A score of $88$ = $6th$ place = Division AAAA champion = probably won't help toward ARML very much = oh well, I can still try to do well at GCTM.

My first careless mistake was computing $A+B$ instead of $A-B$. My other was something equally as careless, but I can't remember exactly what.

On the relay ciphering, on the last problem, I was in the back, and someone was having trouble, so in the last ten seconds, having not received any answer to work with, I randomly wrote $6.25$ and got it right. You should have seen my face when the person who took my paper told me I was right. This has to be the luckiest thing that has ever happened to me. :rotfl:

Tied for 2nd with 92.
4th place team ciephering 150.

I think I did fairly well by my self.

Congratulations to the Northview team. Very impressive today. For those of you that weren't there:

Team
1st - Northview - 744.3
2nd - Cedar Shoals - 557.3
3rd - Alpharetta - 523.4
4th - Pace Academy - 465.7
5th - Woodward Academy - 451.7
6th - Lassiter - 449.7
7th - Walton - 424.9
8th - Rockdale - 405.2
9th - Sculley Academy Math Club - 382.1 (Who is this?)
10th - Dickerson Middle School -375.6

Individual
1st - Miles - 96.0
2nd - Santosh, Heng Tong and Ben Hu (Northview) - 92.0
5th - Alvin Yan (Northview) - 91.3
6th - Eric (Twiz) and Tony Huang (Northview) - 88
8th - Jimmy Dong (Northview) - 81.1
9th - Eric Chen (Brookwood) - 80.4
10th - Clara Starkweather (Cedar Shoals) - 77.5
11th - Maja Wichrowska (Alpharetta) - 74.2
12th - Coby Lu (Alpharetta) - 72.4
13th - Arvind and Bellen Shen (Northview) - 67.7
15th - Catherine Cheng (Walton) - 67

Fulton

The Sculley Academy is a homeschool math club up in Atlanta. Of course, the star student(s) are mostly the children of the founders. Anyway, I think it was Peter Sculley who got 3rd in JV at GT this year and 2nd in JV [to me] last year there. They have some good younger students too.

Yes, we played mafia with the Sculley people while we were waiting for the awards to begin.  It was lots of fun.  First I was the narrator, and then, when everybody knew how to play properly, I was a townsperson who voted to kill two innocent people (but for very good reasons) and was subsequently killed by the mafia. :o

There were lots of ties this year.  Too bad we bombed relays.  The test was nice—there were a lot of problems that looked ugly, but turned out to have nice solutions.  And there were no mistakes, although I thought at first that there was one on the written test.

[quote="gtf"]Congratulations to the Northview team. Very impressive today. [/quote]

Getting 7 of the top 14 and all 8 participants in the top 20 is tough.

I also tied for second. The plaque reads "Ben Hu" instead of "Benjamin Hu." :mad: *sigh

Since Northview did well, are we allowed to send more than four people to GCTM Varsity State, assuming that we're invited?

No team is allowed to send more than 4. The ARML coaches will choose various "persons of interest" and invite them to compete as individuals. I would guess (and it really is STRICTLY a guess) that there is a good chance Northview may have a couple invited as individuals. I say a couple, but that's not a specific number. The team must still specify which four individuals are competing for the team.

Fulton

Five questions:
1) When are the individual invitations going to be sent out (I assume sometime after the USAMO indexes come out)?
2) Do you send them to people who are going on a team anyways?
3) Do you send them to the individual or the coach?
4) About how many do you usually send out?
5) If you aren't invited as an individual, is it theoretically possible to make ARML anyways (if you go on a team)?

The best resource for understanding selection to the ARML team is Steve Sigur's website.

http://www.paideiaschool.org/Teacherpages/Steve_Sigur/resources/GAARML/GEORGIAARML/Team%20Selection.html

I'll try to answer a few of the questions.

1. I'm not sure when that happens. I suspect that we do need to see the USAMO indexes to help make those decisions.
2. We basically make sure that people are coming if we are interested. We'll let a coach know to bring that person even if they aren't on their official team.
3. I think that they mostly go to the coach, but I don't believe that there are "official" invitations. We just let people know we are interested.
4. Don't know.
5. The majority of ARML members ARE on a team to begin with.

Here's a great example from last year. Miles Edwards from Lassiter (a freshman at the time) qualified for the USAMO and was very interesting to the ARML coaches. It wasn't a done deal that he would be part of Lassiter's official team. Miles would have been invited to come as a 5th member from Lassiter if he hadn't been chosen as part of Lassiter's team. Turns out that he was part of their team, but I still think the example is useful.

Now understand, I am one of the ARML coaches, but I have not been involved as much in choosing the "extra" individuals that come to state. It's quite possible that I've made a mistake in my statements. I don't think so, but you never know! :-) I'm putting this out as a method of information dissemination.

Fulton

[quote="gtf"]6th - Eric (Twiz) and Tony Huang (Northview) - 88 [/quote]

What school is Eric go to?


number theory
greatest common divisor
prime factorization

Hi, just wondering if someone could help me solve this question please. I need to find the GCD of the following:

gcd(360^450, 450^360)

Thanks in advance.

A nice systematic way to do this is to count powers of primes.

$360^{450}= 2^{1350}\cdot 45^{450}$
$450^{360}= 2^{360}\cdot 225^{360}$

Hence the gcd's prime factorization contains $2^{min(1350, 360)}= 2^{360}$, and continue with $3, 5$.

Sorry i didn't quite follow that, could you please explain it to me again.

gcf?
Question i got here states gcd.

http://members.optusnet.com.au/gujju_style/maths1.JPG

I thought GCF = Greatest Common Factor, while GCD = Greatest Common Denominator.

[quote="vishalarul"]GCD = Greatest Common Denominator.[/quote]

You are getting the term confused with Least Common Denominator / Least Common Multiple.

GCD and GCF are equivalent terms.  In the natural numbers, $gcd(a, b) = gcf(a, b)$ refers to the largest number $d$ such that $d | a, d | b$.

Oh, now I get it.  :idea: Gollum000, this is the main idea of the solution:
[hide="If you are still stuck, click here"](1)Let the numbers be $n$ and $m$. Find the prime factorizations of the numbers:
$n=p_{1}^{k_{1}}\cdot p_{2}^{k_{2}}\cdot p_{3}^{k_{3}}\cdot\cdot\cdot$

$m=q_{1}^{l_{1}}\cdot q_{2}^{l_{2}}\cdot q_{3}^{l_{3}}\cdot\cdot\cdot$

(2)Now in the prime factorization, suppose there is a prime $p$ that is common to both $m$ and $n$. WLOG, let $m$ have a factor of $p^{k}$, while $n$ has a factor of $p^{l}$, where k<l. You can infer that the GCD has a factor of $p^{l}$.
(3)Repeat this process to find the prime factorization of the GCD.[/hide]