radical axis
geometry
circumcircle
trigonometry
AMC
USA(J)MO
USAMO

Let $ABC$ be a triangle. Take points $D$, $E$, $F$ on the perpendicular bisectors of $BC$, $CA$, $AB$ respectively. Show that the lines through $A$, $B$, $C$ perpendicular to $EF$, $FD$, $DE$ respectively are concurrent.

No, I'm quite sure you can invoke that without proof.

But, are you sure that they always are the altitudes?  I've drawn a diagram, and I'm pretty sure that this is not necessarily true...

[quote="probability1.01"]I looked at it and I think you can use an induction-like method and reduce it to the altitudes.[/quote]

Okay.  All I did was draw a rough diagram and see that it wasn't necessarily the altitudes...

D, E, and F are all midpoints, so DE, EF, and DF must be parallel to their respective bases. If the line through A is perpendicular to DE, then it must also be perpendicular to AB. Same reason with the others, so you have 3 altitudes of ABC (if you extend the lines to the sides of ABC).

Or am I doing something wrong?

They aren't midpoints.  They lie on the perpendicular bisectors.  However, I think you can sort of "induct" by starting them out at the midpoints.

So it wasn't as easy as I thought. Ah well, back to the scratch paper!

Here's a solution:
Let $G$, $H$, $I$ be the feet of the perpendiculars from $A,B,C$ to $EF,FD,DE$ respectively. Let $O$ be the circumcenter of $ABC$. Using the two right angles, we have that $\angle ACI=\angle OFD$ (as directed angles). Similarly, $\angle ICB=\angle FDO$, $\angle CBH=\angle ODE$, $\angle HBA=\angle DEO$, $\angle BAG=\angle OEF$, and $\angle GAC=\angle EFO$.

By Trig Ceva in $DEF$, $\frac{\sin\angle OFD\sin\angle ODE\sin\angle OEF}{\sin\angle EFO\sin\angle FDO\sin\angle DEO}=1=\frac{\sin\angle ACI\sin\angle CBH\sin\angle BAG}{\sin\angle GAC\sin\angle ICB\sin\angle HBA}$
$1=\frac{\sin\angle ACI\sin\angle CBH\sin\angle BAG}{\sin\angle ICB\sin\angle HBA\sin\angle GAC}.$
By Trig Ceva in $ABC$, this last equation tells us that $AG$, $BH$, $CI$ are concurrent, and we are done.

I was a freshman in 1997, and a first-time USAMO qualifier. I didn't know enough geometry to handle this question, but I did solve the other one using Power of a Point.

This problem can be proved using the fact which I mentioned in http://www.artofproblemsolving.com/Forum/topic-12761.html

the diagonals of a convex quadrilateral are perpendicular to each other if and only if the sums of squares of the correponding opposite sides are equal.

You can make use of the fact that the perpendiculars in triangle $DEF$ to the sides at pts $M,N,P$ on $DE,EF,DF$ are concurrent iff $DM^{2}-ME^{2}+EN^{2}-NP^{2}+FP^{2}-PD^{2}=0$. Then you just use Pythagorean theorem two times to get this result.

[quote]I found a pretty easy solution...but I'm not sure if its right...it looked TOO easy, in fact...

Solution: Since D,E,F lie on the perpendicular bisectors, then they must be the centers of three circles passing through two of A,B,C. Each pair of circles also intersect at one of D,E,F. Also, note that DE,EF, and FD are the lines joining the centers of two circles. But any line perpendicualr to the line joining the center of two circles and also passing through one of their intersection points must be the radical axis of the two circles. It is then well known that the radical axes of three pairwise intersecting circles are concurrent. QED[/quote]

This is correct (and a great solution!), and follows from the construction of the circumcenter of a triangle with a ruler and compass- it's basically the same process of having the radical axes intersect, with the centers of the circles being the vertices of the given triangle.  Then, as an extension you know that the point of concurrency is also the circumcenter of triangle DEF when the three circles happen to have the same radius.

Sorry to be reviving an old topic. But I think this question is basically a one liner(probably I may be wrong).

Consider triangle $ ABC$ with points $ D,E,F$ on the perpendicular bisectors of $ BC,CA,AB$ respectively. Let $ D \in l , E\in m, F\in n$ respectively. We know that $ l,m,n$ are concurrent at the circumcenter $ O$ of triangle $ ABC$.

Now consider the perpendiculars dropped from $ A$ to $ EF$, $ B$ to $ DF$, $ C$ to $ DE$ as $ p,q,r$ respectively.

[color=red]Orthologic Triangles Theorem :  Given two triangles[/color] $ ABC, A'B'C'$ [color=red]respectively. if the perpendiculars from[/color] $ A$ to $ B'C'$, $ B$  to $ C'A'$, $ C$ to $ A'B'$  [color=red]are concurrent if and only if the perpendiculars from[/color] $ A'$ to $ BC$, $ B'$ to $ CA$, $ C'$ to $ AB$ [color=red]respectively are concurrent.[/color]
(Darij has written about it many times in geometry forums)

So here, since $ l,m,n$ are concurrent, it implies obviously that $ p,q,r$ are concurrent . (since $ l$ perpendicular to $ BC$, etc).
$ \text{QED}$.

Let W1 W2 W3 be circles with centers D,E,F . perpendicular from C to DE lies on their radical axis and we can say the same thing for perpendiculars from A and B so by radical axis theorem we know they meet at a single point.
we're Done.

Consider the circles centered at $D,E,F$ with radii $DB, EC, FA$ respectively. Since they lie on the perpendicular bisectors, we know that $DB=DC, EC = EA,$ and $FA = FB$. Then the radical axes of these circles concur at the point desired, the radical center.

Denote by $\odot D$ the circle centered at $D$ with radius $DC=DB$ and define $\odot E$ and $\odot F$ similarly. The radical axis of  $\odot E,\odot F$ and $\odot F,\odot D$ and $\odot D,\odot E$ are the lines through $A$, $B$, $C$ perpendicular to $EF$, $FD$, $DE$ so they must concur at the radical center of $\odot D,\odot E,\odot F$ as required.

It definitely existed but may have not been well-known (no idea if it was citable). In any case, one can just prove the theorem (it is not so hard) and then use it to kill the problem. :)

The three lines are the radical axis of the circles centered at D,E,F so they will concur.

My figure might look strange, but try to decode it. The proof becomes too easy. [b]Tell me if I am wrong, because I'm not sure.[/b]

draw circle to prove AG as radical axis and GH as radical axis. Subsequently we get AH as radical axis. same with others

Let $\omega_A$ be the circle centered at $D$ passing through $B$ and $C$, and define $\omega_B$ and $\omega_C$ similarly. Then, the three lines in question are the pairwise radical axes of $\omega_A,\omega_B,\omega_C$, so we are done.

It is just orthological two triangles and this is trivial....

Another cool problem solved simply by radical center. There isn't much information for us to use angle chasing or "synthetic" techniques like Ceva. Construct circles centered at D passing through B (and consequently C), etc. Then their pairwise radical axes pass through A and perp. to EF, thru B and perp. DF, thru C perp. DE, and by radical center they concur. $\blacksquare$

Let the foot of the perpendicular from $A$ to $EF$ be $X$, the foot of the perpendicular from $B$ to $DF$ be $Y$, and the foot of the perpendicular from $C$ to $DE$ be $Z$. Let $P_1=AX \cap BY$, $P_2=BY \cap CZ$, $P_3=CZ \cap AX$. Since $FXP_1Y, DYP_2Z, EXP_3Z$ are cyclic, their circumcircles concur at the Miquel Point of $\triangle DEF$. Then, we can conclude that $P_1=P_2=P_3$, as desired. $\blacksquare$

^ >>>

The desired concurrency point is the radical center of the circle centered at $D$ through $B,C$, at $E$ through $C,A$, and at $F$ through $A,B$. $\square$

slightly different approach
[hide=motivation]
Lot of right angles $\Rightarrow{}$ circumcircles $\Rightarrow{}$ radical axis!
[/hide]
[hide=solution]
Can the midpoint of $BC$ as $M_A$, and do so similarity for $CA$ and $AB$. Let the $(AFM_C)$ be $O_1$, $(AEM_B)$ be $O_2$, $(CEM_B)$ be $O_3$, and $(CDM_A)$ be $O_4$. Let the radical axis of $O_1$ and $O_2$ be $l_1$, and the radical axis of $O_3$ and $O_4$ be $l_2$. Then, let $l_1$ and $l_2$ intersect at point $P$. By the radical axis theorem, $Pow_{O_1}(P)=Pow_{O_2}(P)=Pow_{O_3}(P)=Pow_{O_4}(P)$. Using $Pow_{O_2}(P)=Pow_{O_3}(P)$, we see that $P$ lies on $l_1$, $l_2$, and the perpendicular bisector of $AC$ (recall that $\triangle{AEC}$ is isosceles). Hence, $APC$ is and isosceles triangle.

Doing the same for sides $AB$ and $BC$, we figure that we essentially need to fit three isosceles triangles into a triangle. This can only be done when the common point of intersection is the circumcenter, which always exists, so we are done. $\blacksquare{}$
[/hide]

Dear Mathlinkers,

a long discussion...about two orthologic triangles

[url=http://jl.ayme.pagesperso-orange.fr/vol6.html] here [/url]

then

A propos de deux triangles orthologiques

Sincerely
Jean-Louis


what is mean by

level curve of f(xy)=1/sin(xy)  ?

See [url]http://en.wikipedia.org/wiki/Level_set[/url]
or Google on "level curve".


quadratics
function
algebra
algebra unsolved

given a functuon $ {y\equal{}\frac {ax^2\plus{}8x\plus{}b}{x^2\plus{}1}}$ have max value 9 and mininum value 1 . find the range of $ {y\equal{}\sqrt{ax^2\plus{}8x\plus{}b}}$ .

If $ \frac {ax^2 \plus{} 8x \plus{} b}{x^2 \plus{} 1}$ has some real number $ k$, there exists real number $ x$ such that $ (k \minus{} a)x^2 \minus{} 8x \plus{} k \minus{} b \equal{} 0\ \cdots [*]$.

Case 1: $ k \equal{} a$, $ [*]$ gives $ x \equal{} \frac {k \minus{} b}{8}$, which is a real number.

Case 2: $ k\neq a$, $ [*]$ is a quadratic equation in $ x$, so the condition for which there exists real number $ x$ satisfying $ [*]$,

we have $ D/4 \equal{} 16 \minus{} (k \minus{} a)(k \minus{} b)\geq 0\Longleftrightarrow k^2 \minus{} (a \plus{} b)k \plus{} ab \minus{} 16\leq 0\ \cdots [1].$

Now the given assumption : $ 1\leq k\leq 9$ gives $ (k \minus{} 1)(k \minus{} 9)\leq 0\Longleftrightarrow k^2 \minus{} 10k \plus{} 9\leq 0\ \cdots [2]$.

Since the left sides of $ [1],\ [2]$ are coinside, we have $ a \plus{} b \equal{} 10,\ ab \minus{} 16 \equal{} 9$, yielding $ a \equal{} b \equal{} 5$.

Conversely from  $ [1]$, $ a \equal{} b \equal{} 5$ leads $ (k \minus{} 5)^2\leq 16\Longleftrightarrow |k \minus{} 5| \leq 4\Longleftrightarrow 1\leq k\leq 9$. The equality holds , from $ [*]$

$ k \equal{} 1\Longleftrightarrow \minus{} 4(x \plus{} 1)^2 \equal{} 0$, yielding $ x \equal{} \minus{} 1$,

$ k \equal{} 9\Longleftrightarrow 4(x \minus{} 1)^2 \equal{}$, yielding $ x \equal{} 1$.

Therefore the desired range is $ y \equal{} \sqrt {5x^2 \plus{} 8x \plus{} 5} \equal{} \sqrt {5\left(x \plus{} \frac 45\right)^2 \plus{} \frac {9}{5}}$, yielding $ y\geq \frac {3}{\sqrt {5}}$

Thanks a lot kunny  :) ! I should now this .we put the first function into an equation about "x" we get "$ {\Delta >\equal{} 0}$" then we get an equation system abou $ {a,b }$ Just solve it ,we get $ {a,b }$ THEN easy to find the range.


[quote="worthawholebean"]How would old problems on (pre-1996) USAMO correspond to current difficulty? Would it be something like 3, 4, 5 correspond (roughly) to 1, 2, 3, respectively?[/quote]

It varies wildly from year to year.

The first year I took the test (1987), the highest score was in the mid 50s out of 100, only 6 students cleared 40 (I think DPatrick was one), and around 30 cleared 20.  The last year I took the test (1989), there were around a dozen students at 70+, and one perfect score.

[quote="rrusczyk"]The first year I took the test (1987), the highest score was in the mid 50s out of 100, only 6 students cleared 40 (I think DPatrick was one), and around 30 cleared 20...[/quote]I was not.  I was honorable mention in 1987 (I think I was about 15th or so oervall) and I was tied for 7th in 1988.  I'm amazed that you remember the actual scores because I have no recollection whatsoever.  I don't even remember my AIME scores anymore, and I only remember one AHSME score (you can guess which one).


What can be tyhe best program(study) for a 9 grade math olympiad student for preparing for IMO and JBMO

Reading math olympiad books and solving problems on mathlinks.

So may I ask you some questions?

what is the hardest part in Geometry? Is it harnomious line?
How can I find the best Mathematics books for training?
Yours sincerely,

I guess different people find different things difficult. You can order books right here on [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]AOPS[/url] or you may go to [url=http://www.amazon.com/]Amazon[/url] for example.


algebra
polynomial
Putnam
symmetry
number theory
relatively prime
system of equations

I'm assuming that the following questions can be solved using polynomial symmetric sums (somebody gave me a sheet of problems with the title "polynomial symmetric sums" on the top.)

I can't do the following problems. Anyone care to help?

[list]
1. The solutions to the following system of equations can be denoted as: $\{a_{1},b_{1},c_{1}\},\{a_{2},b_{2},c_{2}\},...,\{a_{n},b_{n},c_{n}\}$ where $n$ is a positive integer and $a,b,c$ are real numbers that satisfy:
\[\left\{ \begin{array}{ c c }abc =-6 \\ 4a+b-3c=-7 \\ 4ab-3bc-12ac=-26 \end{array}\right\}\]
If $\sum_{i=1}^{n}(a_{i}+b_{i}+c_{i}) = \frac{-p}{q}$, where $p$ and $q$ are relatively prime positive integers, compute the remainder when $p+q$ is divided by 1000.

2. Evaluate:
\[\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-...}}}\]
Express your answer in the form: $\frac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are positive integers and $c$ is not divisible by the square of any prime. (Putnam 1995/B-4)
[/list]

Likely force of habit.  His method is correct, since $47^{2}= 2209$.

I never realized there were two of you  :|

[quote="t0rajir0u"][hide="Big hint for 1"] Consider the polynomial $(x-4a)(x-b)(x+3c)$.  [/hide]
[hide="Big hint for 2"] Let the desired expression be $x$.  Then

$x = \sqrt[8]{ 2207-\frac{1}{x^{8}}}$ [/hide][/quote]

Well, for the first question, I realize that in the polynomial you provided me, the sum of the three roots add up to -7. (Polynomial Symmetric Sums), and if I expand your polynomial, I get a long string that seems to have some reminisce of the three equations in the system of equations:

$x^{3}+3cx^{2}-bx^{2}-4ax^{2}\boxed{-3bcx-12acx+4abx}\boxed{+12abc}$

But where do I go from there? Do I factor out the $x$?

[quote="t0rajir0u"]Yes, yes, good.  Keep going.

[hide="Your next step should be..."] To realize that $(x-4a)(x-b)(x+3c) = x^{3}+7x^{2}-26x-72$ (this is merely substituting the given equations into the expressions of the coefficients).

Now, solve this cubic equation.  What are its roots?  How many ways can you assign these roots to $4a, b,-3c$? [/hide][/quote]

Oh, while you were responding, I had realized that $(x-4a)(x-b)(x+3c) = x^{3}+7x^{2}-26x-72$, but how did you first [b]realize[/b] the hint to tell me about $(x-4a)(x-b)(x+3c)$?

[quote="t0rajir0u"]The system is clearly a set of symmetric sums.  I was first introduced to this idea in Engel.[/quote]

What is Engel? And I realized that is was a set of symmetric sums as well, but how did you know that the polynomial would be able to be used for substitution by the sums in the brackets?

And sorry again, but how do I solve the cubic?

[quote="123456789"]What is Engel? And I realized that is was a set of symmetric sums as well, but how did you know that the polynomial would be able to be used for substitution by the sums in the brackets?

And sorry again, but how do I solve the cubic?[/quote]

http://www.amazon.com/Problem-Solving-Strategies-Problem-Books-Mathematics/dp/0387982191

The coefficients of a polynomial are the symmetric sums of its roots.  Given values for some symmetric sums, one can construct a polynomial to find the roots.

You're best off guessing at rational roots by the Rational Root Theorem.  Engel has this exact problem, and the roots are nice.

(I call this the [b]Problems-have-nice-solutions Principle.[/b]  It doesn't always hold true, especially for Vietnam TST problems and the like, but it is a good guiding principle that can often lead you to nice solutions you would not have otherwise noticed if you hadn't assumed they were there.)

Well, I get $(x-4)(x+9)(x+2)$, so the roots are $4,-9,-2$.

I get a=1, b=-2, c=3. Where do I go from there? Are there multiple solutions?

Yes.

You seem to have assumed that $4 = 4a,-9 =-3c,-2 = b$.  But that isn't the only way to assign values.

For example, you could assign $4 = b,-9 = 4a,-2 =-3c$ to get the solution $(a, b, c) = \left(-\frac{9}{4}, 4, \frac{2}{3}\right)$.

So, [i]how many ways are there to assign values?  What happens when all of those ways are added together as described in the problem?[/i]

[quote="t0rajir0u"]His method is correct, since $47^{2}= 2209$.  [/quote]

Some contest-solvers are just very used to using years in problems.   :P

[quote="Glen Mann"]123456789 could you please explain these three lines from your answer of number 2?

[hide="question"]
$\Rightarrow (x^{4}+\frac{1}{x^{4}})^{2}-2=2007$

$\Rightarrow (x^{4}+\frac{1}{x^{4}})^{2}-2=2009$

$\Rightarrow x^{4}+\frac{1}{x^{4}}=47$

I assume in the second line the -2 in there was an error? but sqaure root of 2009 isnt 47 either..

thanks
[/hide][/quote]

He meant:
$\Rightarrow (x^{4}+\frac{1}{x^{4}})^{2}-2=2207$

$\Rightarrow (x^{4}+\frac{1}{x^{4}})^{2}=2209$

$\Rightarrow x^{4}+\frac{1}{x^{4}}=47$

blah the first problem is contrived...this is the kind of problem that you have seen and you can solve; i suppose the thinking that you would do to think of that would be: "how can I increase symmetry?" this is done by the substitutions:

x=4a, y=b, z=-3c

I saw this first in AoPS volume 2.

But man, I need to get engel soon...like I am doing problems out of math olympiad challenges, and a russian problem book, but I want something that has more exposition.

Well, I added the 6 possible solutions, and I got $\frac{-77}{6}$, so p is 77 and q is 6. $77+6=83$, so is the answer 83?

Also, is there a way to add up all 6 possible solutions without finding them manually and adding them up? I feel so stupid.

[quote="123456789"]is there a way to add up all 6 possible solutions without finding them manually and adding them up? [/quote]

Yes.

[hide="Specifically..."] Consider the expression

$2 \cdot (4-9-2) \cdot \left( \frac{1}{4}+1-\frac{1}{3}\right) =-\frac{77}{6}$

I claim that this expansion gives the sum of all six possible solutions.

To see why:  Every solution is of the form

$\{ 4x,-9y,-2z \}$

Where $\{ x, y, z \}= \{ \frac{1}{4}, 1,-\frac{1}{3}\}$ in some permutation.  There are six of these permutations, and when adding these six permutations up, each possible term (for example, $4 \cdot \frac{1}{4}$ or $-9 \cdot 1$) appears twice (once in two different permutations each).  Another way to see this is that in all six permutations there are a total of $18$ terms, but because we have $9$ possible terms each one must appear twice.

Expanding the first expression I gave gives every possible term once, and multiplying by two completes the sum. [/hide]


function
real analysis
real analysis unsolved

Function $f: \mathbb R \to \mathbb R$ is such that $ |f(a)\minus{}f(b)|<|a\minus{}b|$ for all reals $ a\ne b$. Prove that if $ f(f(f(0)))\equal{}0$ then $ f(0)\equal{}0$.

First I would consider the following cases :
If $ f \circ f(0) \equal{} f(0)$ $ \Rightarrow$ $ 0 \equal{} f \circ f\circ f(0) \equal{} f \circ f(0) \equal{} f(0)$
If $ f \circ f \circ f(0) \equal{} f \circ f(0)$ $ \Rightarrow$ $ 0 \equal{} f \circ f \circ f(0) \equal{} f\circ f(0)$ $ \Rightarrow$ $ 0 \equal{} f \circ f \circ f(0) \equal{} f(0)$.
So in both situations we get that $ f(0) \equal{} 0$

Now, in order to prove that $ f(0) \equal{} 0$ I will use the contradiction method.Let's suppose that $ f(0) \neq 0$. So from the relations from the hypothesis we get :
$ |f \circ f(0) \minus{} f(0)| < |f(0) \minus{} 0| \equal{} |f(0)|$.Furthermore we can suppose that $ f \circ f \neq f(0)$ and $ f \circ f \circ f(0) \neq f \circ f(0)$.So from the relation from the hypothesis we have :
$ |f \circ f \circ f \circ f(0) \minus{} f \circ f \circ f(0)| < |f \circ f \circ f(0) \minus{} f \circ f(0)| < |f \circ f(0) \minus{} f(0)| < |f(0)|$ $ \Rightarrow$ $ |f(0)| < |f(0)|$ contradiction.

Proposition)  Define $ f^n \equal{} \underbrace{f \circ \cdots \circ f}_{n \text{ tuples}}$. If $ |f(x) \minus{} f(y)| < |x \minus{} y|$ for all $ x \neq y$ and $ f^n(0) \equal{} 0$ for some positive integer $ n$, then $ f(0) \equal{} 0$.

proof)  By hypothesis, $ | f^n(x) \minus{} f^n(y)| < |f^{n\minus{}1}(x) \minus{} f^{n\minus{}1}(y)| < \cdots < |x \minus{} y|$ for $ x \neq y$. In particular, $ |f^n(x)| \equal{} |f^n(x) \minus{} f^n(0)| < |x\minus{}0| \equal{} |x|$ for $ x \neq 0$.

Now assume $ f^n$ has a fixed point other than $ 0$, namely $ x_0$. Then $ |x_0| \equal{} |f^n(x_0)| < |x_0|$, a contradiction. Thus $ f^n$ has a unique fixed point.

Then $ f^n(f(0)) \equal{} f^{n\plus{}1}(0) \equal{} f(f^n(0)) \equal{} f(0)$, hence $ f(0) \equal{} 0$.

[quote="Svejk"]sos440  , I believe you are wrong when saying that
"$ |f^{k \plus{} 1}(x) \minus{} f^{k \plus{} 1}(y)| < |f^k(x) \minus{} f^k(y)|$ for $ x\neq y$ "since if $ x\neq y$ it is not necessary that $ f^k(x) \neq f^k(y)$ so you cannot apply the property from our hypothesis.[/quote]

Ah, you're right. Hmm... maybe this would work:

... Exploiting hypothesis, $ |f(x) \minus{} f(y)| \leq |x \minus{} y|$ for all $ x, y$. Then for $ x \neq y$, $ |f^{n}(x) \minus{} f^{n}(y)| \leq |f^{n \minus{} 1}(x) \minus{} f^{n \minus{} 1}(y)| \leq \cdots \leq |f(x) \minus{} f(y)| < |x \minus{} y|$.

Let $ f(0)\equal{}x,\; f(x)\equal{}y$. Then $ f(y)\equal{}0$. Now using our condition:
$ |x\minus{}0|\geq|f(x)\minus{}f(0)|\equal{}|y\minus{}x|\geq|f(y)\minus{}f(x)|\equal{}|0\minus{}y|\geq|f(0)\minus{}f(y)|\equal{}|x\minus{}0|$.
And this means $ x\equal{}y\equal{}0$, or $ f(0)\equal{}0$.


function
inequalities
real analysis
real analysis solved

Find all functions $f:R ->R$ verifying that $f(x^2)=f(x)^2$ and $f(x+1)=f(x)+1$.

From $f(x+1)=f(x)+1$ it follows: $f(x+k)=f(x)+k$. Now we substitute $f(x^2)$ by
$f(x)+x^2-x$ and get $f(x)+x^2-x=f(x)^2$ and $f(x)^2-f(x)+x-x^2=0$. It follows: $f(x)=\frac{1\pm\sqrt{1-4x+4x^2}}{2}=x,1-x$ for all $x$. Assume that there exists an $x_0$ with $f(x_0)=1-x_0$ then $f(x_0+1)=2-x_0=3-(x_0+1)$ which must be $x_0+1$ because of the formula above. We get $x_0=\frac{1}{2}=f(x_0)$. So we get the only solution $f(x)=x$.

Misha

Can u explain a bit more clear why can we substitute $f(x^2)$ by $f(x)+x^2-x$ ? Personally I didnt get it.

Hey, killer, do you have a solution for this one?  ;)  I see it in the proposed section.Personally, I understood exactly nothing from Misha's post. Here are the results of the combined efforts of adyo and me:
  First of all, it is easy to see that $f(x)$ is nonnegative for nonnegative x. Next, using $ f(x+1)=f(x)+1$ we find ${ f(x)=[x]+f(\{x})$, thus $ f(x)\geq [x]$ for all positive $x\geq 1$. Suppose we have proved that $ f(x)\geq ([x^{2^n}])^{\frac{1}{2^n}}$ for all  $x\geq 1$. Then for all $x\geq 1$ we have $ f^2(x)=f(x^2)\geq ([(x^2)^{2^n}])^{\frac{1}{2^n}}$ and the inductive step is proved. Now, for a fixed $x\geq 1$ we make n very large in the inequality $ f(x)\geq ([x^{2^n}])^{\frac{1}{2^n}}$ and we get $f(x)\geq x$. Taking into account that $f(x+1)=f(x)+1$, we get $f(x)\geq x$ for all nonnegative x. Then, from ${ f(x)=[x]+f(\{x})$ we get $f(x)\geq x$ for all real x. I mean by ${f(\{x})$ the value of f in the fractional part of x.
   Next, we obviously have $ f^2(x)=f^2(-x)$. Suppose that there exists $x$ non zero such that $ f(x)=f(-x)$. Then we have $ f(x)-1=f(-x)-1=f(-x-1)$, thus $ (f(x)-1)^2=f^2(-x-1)=f^2(x+1)=(f(x)+1)^2$, from where we find $f(x)=0$ and also $f(-x)=0$. But this implies that $0$ is greater than x and -x, false. Thus $f(x)=-f(-x)$ for all real number x (this is true for 0, because it is clear that f(0) is 0 or 1. Suppose that $ f(0)=1$. Then $ f(1)=2, f(2)=3, f(3)=4, 9=f^2(2)=f(4)=5$ ). Thus $ x\leq f(x)=-f(-x)\leq -(-x)=x$. That's all!

It seems to me that we have the same ideas sometimes!


I think $x_{1},x_{2},..,x_{k}$ are real roots, $g(x)$ has not real roots.
It is obviosly.

yeah sorry! I mean $x_{1},x_{2},...,x_{k}$ are real and $g(x)$ has no real root
But how can the problem be obvious ?

Number of roots not in (0,$\infty$) not greater than $sign(c_{0},c_{1},...,c_{n})$ and number of roots in ($-\infty ,0$) not greater than $sign(c_{0},(-1)^{m_{1}}c_{1},...,(-1)^{m_{n}}c_{n})$. It give number of roots in ($-\infty,\infty$) not greater than n. All this is known theorems.


Find all continous functions st:

$ f(x)\equal{}f(x^2\plus{}\frac{1}{4})$

[hide="An uncomplete solution:"]As $ f$ is odd, we study it just in $ \mathbb{R^\plus{}}$
Consider : $ g(x)\equal{}x^2\plus{}\frac{1}{4}$

and set $ x_n$ st
$ x_0\equal{}x$
$ x_{n\plus{}1}\equal{}g(x_n)$,
$ x_n$ is increasing

* If $ x \in [0,\frac{1}{2}]$
Then we can prove that

$ x_n \le \frac{1}{2}$
and $ x_n$ is increasing
so $ \lim x_n \equal{} \frac{1}{2}$
and $ \lim f(x_n)\equal{}f( \frac{1}{2})$
So $ f(x)\equal{}f( \frac{1}{2})$ for all $ x \in [0,\frac{1}{2}]$

* If $ x \in [\frac{1}{2},\infty]$
Then $ x_n$ is still increasing
[hide](On a book, I found that it's decreasing in this case but this is obviously false)
[/hide]
And I can't go from here ...[/hide]

[quote="delegat"]Please look at: http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf

You can find solution on page 20 (Bulgaria, problem 15)[/quote]

Thanks,
But Im talkin about this book, Its solution is wrong in case $ x\geq \frac {1}{2}$
 :lol:

[size=150][u]the sequence is increasing in both cases[/u][/size] since:

$ g(x) \minus{} x \equal{} (x \minus{} \frac {1}{2})^2$

ie
$ u_{n \plus{} 1} \minus{} u_{n} \equal{} (u_{n} \minus{} \frac {1}{2})^2$

Oh sorry I haven't looked at your [color=red]click to reveal hidden content[/color] part.

[quote="FOURRIER"]Find all continous functions st:

$ f(x) \equal{} f(x^2 \plus{} \frac {1}{4})$
[/quote]

For $ x\in[0,\frac{1}{2})$, the sequence $ a_0\equal{}x$ and $ a_{n\plus{}1}\equal{}a_n^2\plus{}\frac{1}{4}$ is such that $ f(x)\equal{}f(a_0)\equal{}f(a_n)$ $ \forall n$. Then, since $ \lim_{n\rightarrow \plus{}\infty}a_n\equal{}\frac{1}{2}$ and since $ f(x)$ is continuous, $ f(x)\equal{}f(\frac{1}{2})$

For $ x>\frac{1}{2}$, the sequence $ a_0\equal{}x$ and $ a_{n\plus{}1}\equal{}\sqrt{a_n\minus{}\frac{1}{4}}$ is such that $ f(x)\equal{}f(a_0)\equal{}f(a_n)$ $ \forall n$. Then, since $ \lim_{n\rightarrow \plus{}\infty}a_n\equal{}\frac{1}{2}$ and since $ f(x)$ is continuous, $ f(x)\equal{}f(\frac{1}{2})$

So $ f(x)\equal{}f(\frac{1}{2})$ $ \forall x\geq 0$

Then, for $ x<0$, $ f(x)\equal{} f(x^2 \plus{} \frac {1}{4})\equal{} f((\minus{}x)^2 \plus{} \frac {1}{4})\equal{}f(\minus{}x)\equal{}f(\frac{1}{2})$

And the solutions are all constant functions $ f(x)\equal{}a$


trigonometry
function
geometry
circumcircle
incenter
ratio
angle bisector

Let $ABC$ be a triangle. Let $P$, $Q$, $R$ be the points of intersection of the angle bisectors of its angles $\angle A$, $\angle B$, $\angle C$ with the sides $\left[BC\right]$, $\left[CA\right]$, $\left[AB\right]$ respectively. Prove that $\measuredangle RPQ = 90^{\circ}$ if and only if $\measuredangle CAB = 120^{\circ}$.

i have not completed the calculations, but a simple algebraic aproach seems to kill the problem.

we express everything in terms of a,b,c the side lengths of the triangle ABC.

for easier computations we can denote x=BP, y=CQ, z=AR.

then we need to prove that

a2 = b2+c2 + bc \sqrt 3  <=>
RQ2 = RP2 + PQ 2 .

But from Cosine law in triangles ARQ, BRP and CPQ we obtain the expressions of RQ2, RP2 and PQ2 as a function of x,y,z,a,b,c.

We also know that x= ac / (b+c) , y = ba / (c+a) and z= cb / (a+b) and we have obtained a relation only in a,b,c (no \sqrt -s involved) and we just group, cancel, and factorize the terms to obtain the LHS.

I've tried desperately to find a geometric approach, but I couldn't come up with anything really different from Valentin's approach. The calculations look too ugly to me, though. I don't think I've got the patience to finish them...  :D

Let's considere the equilateral triangle BCD (D is opposite to A whit respect to BC).
Given that BACD is cyclic, is easy to see that D is on the line AP. It results that BPD and APC are similar. Then BP/BD=AP/AC from which we can write  BP/AP=BD/AC=BC/AC=BR/AR then PR is the bisector of angle(BPA).

the rest follows ...

Well ...

This obviously proves <A = 120° => <RPQ = 90°.

Probably I'm missing something but how do you prove <RPQ = 90° => <A = 120° ?

:?:?:?

So the wonderful geometric proof above works only for

<CAB = 120° => <RPQ = 90°

and not for

<RPQ = 90° => <CAB = 120° ?

Is there nothing similar that can be done for

<RPQ = 90° => <CAB = 120° ?

extend PR and PQ to meet a line parallel to BC and passes thru A at R' and P' respectively

R'AR and BPR, AP'Q and PQC are similar

so we have:
R'A=BP*AR/RB=BP*AC/BC
AP'=PC*AQ/QC=PC*AB/BC

notice that AB/BP=AC/PC
BP=AB*PC/AC
sub that into R'A we get
R'A=AB*PC/BC which is equal to AP'
it follows that A is the midpoint of R'P' since R'PP' is right triangle A is the center of the circumcircle R'P'P and it follows that AP=AR'=PP' and APR'=AR'P=R'PB, AP'P=APP'=P'PC
hence APB is bisected by RP and APC is bisected by QP
continue as  sprmnt21

Trigo shows it one way nicely, with an interesting lemma that the intersection of CR and PQ is the incentre of APC, and the intersection of BQ and PR is the incentre of APB.

This is because: $\frac{AQ}{CQ}=\frac{AB}{BC}=\frac{AP}{PC}$. Thus by angle bisector thm, it is the incentre. (It shows that it is 90 only one way though... because it proves a stronger theorem.)

We have $R,Q$ are excenters of triangles $APC,ABP$ so $PR,PQ$ are bisectors of $\angle BPA$ and $\angle APC$, hence $\angle RPQ=90^0$.

-shrugs- that was exactly the same as my solution!

Let ABC be a triangle. Let P, Q, R be the intersections of the angle bisectors of <A, <B, <C with [BC], [CA], [AB] respectively. Prove that <RPQ = 90° if and only if <CAB = 120°.

My solution is as follows:
Drop perpendiculars $QD$, $QE$ and $QF$ on $AB$ (possibly extended), $AP$ and $BC$ respectively as well as perpendiculars $RD_1$, $RE_1$ and $RF_1$ on $AC$, $AP$ and $BC$ respectively. Note that $RD_1=RF_1$ and  $QD=QF$. We also have $\angle RAD_1=\angle QAD$ and $\angle BAP=\angle CAP$, so that triangles $ADQ \sim ARD_1$ and $AQE \sim ARE_1$ from which, since triangles $ADQ$ and $AQE$ share the side $AQ$ and triangles $ARD_1$ and $ARE_1$ share the side $AR$, we get $\frac {DQ}{QE} = \frac {RD_1}{RE_1}$. Hence we have $\frac {QF}{QE} = \frac {RF_1}{RE_1}$, if this ratio exceeds $1$ then (from the triangles $QPF$, $QPE$ and $PRE_1$, $PRF_1$) we would get $\angle RPF_1 > \angle RPE_1$ and $\angle QPF  > \angle QPE$ so that $\frac{\angle PC}{2}=\frac{\pi}{2}>  \angle QPE + \angle RPE_1 =\angle RPQ$. If the ratio is less than $1$, from the same triangles we would get $\angle RPQ= \angle QPE + \angle RPE_1 > \frac{\angle BPC}{2}=\frac{\pi}{2}$. Hence $\angle RPQ=\frac{\pi}{2}$ only if $QF=QD=QE$ which can happen only if $\angle BAC$ is obtuse and $AQ$ bisects $\angle PAD$, so that $\angle BAP = \angle PAC = \angle CAD$. Since those angles add up to $180$° each of them is $60$°, hence $\angle CAB = 120$°


FTW AMC USA(J)MO USAMO MATHCOUNTS USAMTS

Anyone from MA wishing to hone their countdown skills and have some fun is invited to play FTW with our team tomorrow at 3:00 pm.
I believe we will have some normal games to determine the starting rankings and do countdowns in ladder or bracket style.
Hopefully we will have 8-10 ppl.  :D

Hmmmm should I play for the first time in like 4 months?

how was my test

Oh it was a breeze.
We all got 46s.
Well, I got 45 because of the stupid mistake on #1, but otherwise, it was too easy.
Please make the next one harder.

Anyways, the password to the game will be CatNos' name (first name only, non-capitalized).
See you there.

you might as well say 'the password is carl' because uh, my first name is in my sig. for that matter, now that i've said what the password is, there isn't much of a point of a password.

Can I come too? :P
Lol, our state team is only going to have one practice...and it's on April 26th. We're going to suck pretty badly.

Wow...
Ok, lets change the password to your last name then.

Sure, xpmath, you can come if you want.
As long as you know carl's last name.

[quote="alkjash"]Wow...
Ok, lets change the password to your last name then.[/quote]

There aren't a whole lot of people that list their location as lexington, ma, do math, went to nationals, made the usamo twice, named carl,. i mean, screw the named carl part. it's pretty obvious that my last name is lian.

also, see the shoutbox in my blog.

Well, we're only in middle school...
and I am not as beast as everyone else :(

[quote="shentang"]then lets change the password to the first two names of the school that Carl attended last year.
*Remember that the second word in the school ends with an e and not a k.[/quote]

yeah, it really irritates me that people always say jonas clark instead of jonas clarke

and when they got it right in the MATHCOUNTS YEARBOOK and the MATHCOUNTS WEBSITE last year, they took out "middle school," for some reason. I wonder if looked at my USAMTS USERNAME or my very first AoPS USERNAME. Or the THREE USERNAMES I MADE BY ACCIDENT, which happen to be next to 1337P34K43V312, which was formerly jclarkemathL314159.

Omg, we're only using a password so that random people won't join.
People won't try Carl's last name as soon as they see there is a pw to the game.
Seriously, it's not like a top secret event or nething.
Carl's last name is fine.

arg, my internet was down for the whole day yesterday :(

I checked in later and someone I didn't know begged me to play....I finally won one :lol:

Any thoughts on another one of these?  Did it go well for those who did show up?

It didn't work out terribly well....
We only had 5 people, so we played like 3 games and then left..
I think we should leave more notice next time.
Maybe Sunday?

It would probably be better on Saturday because on Sunday, people probably will have to do a bunch of things like
going to church, preparing for school (finishing their hw), going to chinese school, sports, and other things..


calculus
integration
trigonometry
calculus computations

$ \sin ^ 4 x \equal{} (\sin ^ 2 x)^2$, so you can imagine a trianglular formula.


superior algebra
superior algebra solved

Does anyone know how to show the following:

[color=green]Suppose that R is a ring with identity, and suppose that 1R = 0R. Show that
R = { 0R }.[/color]

Thx guys.  :oops:

Say that there is another element $x$.  Then $x=1x=0x=0$, a contradiction.


Could somebody post the partial results of imotc 2008(especially soumya and vihang)
 :roll:

the results aren't out yet...but check out the discussions [url=http://iit-redefined.zzl.org/viewtopic.php?f=34&t=134]here[/url]


real analysis
topology
calculus
calculus computations

Prove that $ f: (M,d)\rightarrow (N,\rho)$ is uniformly continuous if and only if $ \rho((x_{n}),f(y_{n}))\rightarrow 0$ for any pair of sequences $ (x_{n})$ and $ (y_{n})$ such that $ d(x_{n},y_{n})\rightarrow 0$.

[quote]$ \rho((x_{n}),f(y_{n}))\rightarrow 0$[/quote]

Do you mean $ \rho(f(x_{n}),f(y_{n}))\rightarrow 0$?

If so, suppose that $ f$ is uniformly continuous on $ X$ then given $ \epsilon > 0$, there exists $ \delta > 0$ st $ \rho(f(x),f(y)) < \epsilon$ when $ d(x,y) < \delta$

Since $ d(x_{n},y_{n})\rightarrow 0$, then for the $ \delta$ above, there exists $ N$ st :  $ d(x_{n},y_{n}) < \delta$ for $ n > N$. Hence  $ \rho(f(x_{n}),f(y_{n})) < \epsilon$ for $ n > N$

For the other direction -  suppose that $ f$ is not uniformly continuous the there is $ \epsilon > 0$ and for all $ \delta > 0$ there are $ x_{\delta}, y_{\delta}\in X$ and $ d(x_{\delta},y_{\delta}) < \delta$ but $ \rho(f(x_{\delta}),f(y_{\delta}))\geq \epsilon$

Now let $ \delta \equal{} 1/n$ we can find $ (x_{n})_{n\geq 1}, (y_{n})_{n\geq 1}\in X$ st $ d(x_{n},y_{n}) < 1/n$ but $ \rho(f(x_{n}),f(y_{n}))\geq\epsilon$

and from the hypothesis, we have a contradiction

------------------------------------------------------------------------------------------------------

[b]Follow up question (from Rudin's book) [/b]

Let $ E$ be a dense subset of a metric space $ X$ and $ f : E\rightarrow \mathbb{R}$ is uniformly continuous , prove that $ f$ has a continuous extension from $ E$ to $ X$. Could the range $ \mathbb{R}$ be replaced by $ \mathbb{R}^{k}$? by any compact metric space? by any complete metric space? by any metric space?

Yes, I did mean $ \rho(f(x_{n},f(y_{n}))$.  Thanks for the proof.

That follow-up problem is actually a theorem in another analysis book I have (and in many I believe).


$ \left\{ a_1,a_2,a_3,\cdots,a_n\right\} \in \mathbb{N}$ and all distinct.

If $ a_1^2\plus{}a_2^2\plus{}a_3^2\plus{}...\plus{}a_n^2\equal{}2002$ , find the max. value of $ n$.

[quote="stephencheng"]We have $ 2002 \equal{} \sum_{i \equal{} 1}^{n} a_i^2 \ge 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$

So $ n \le 17$. For $ n \equal{} 17$, we can easily find

$ 1^2 \plus{} 2^2 \plus{} .... \plus{} 11^2 \plus{} 13^2 \plus{} 14^2 \plus{} ... \plus{} 17^2 \plus{} 19^2 \equal{} 2002$

, so max. value of $ n$ is $ 17$[/quote]

How can you easily find out  when replacing  $ 12$th  by $ 19$th term from  $ 1^2 \plus{} 2^2 \plus{} .... \plus{} 11^2 \plus{} 12^2\plus{} 13^2 \plus{} 14^2 \plus{} ... \plus{} 17^2$  we could get $ 2002$?

[quote="PKMathew"][quote="stephencheng"]We have $ 2002 \equal{} \sum_{i \equal{} 1}^{n} a_i^2 \ge 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$

So $ n \le 17$. For $ n \equal{} 17$, we can easily find

$ 1^2 \plus{} 2^2 \plus{} .... \plus{} 11^2 \plus{} 13^2 \plus{} 14^2 \plus{} ... \plus{} 17^2 \plus{} 19^2 \equal{} 2002$

, so max. value of $ n$ is $ 17$[/quote]

How can you easily find out  when replacing  $ 12$th  by $ 19$th term from  $ 1^2 \plus{} 2^2 \plus{} .... \plus{} 11^2 \plus{} 12^2 \plus{} 13^2 \plus{} 14^2 \plus{} ... \plus{} 17^2$  we could get $ 2002$?[/quote]

$ 1^2\plus{}2^2\plus{}...\plus{}17^2\plus{}(x^2\minus{}y^2)\equal{}2002  \Leftrightarrow (x\minus{}y)(x\plus{}y)\equal{}7 \times 31$.


(a) Consider a damped, driven harmonic oscillator (in one dimension) with equation of motion
$ m\ddot{x}\equal{}\minus{}m\omega_{0}^{2}x\minus{}\gamma\dot{x}\plus{}A\cos(\omega t)$
What is the time averaged rate of energy dissipation?

(b) Consider an anharmonic oscillator with equation of motion
$ m\ddot{x}\equal{}\minus{}m\omega_{0}^{2}x\plus{}\alpha x^{2}\plus{}A\cos(\omega t)$
where $ \alpha$ is a small constant.  At time $ t\equal{}0$, $ x\equal{}0$ and $ \dot{x}\equal{}0$.  Solve for the subsequent equation motion including terms of first order in $ \alpha$.

The second part in particular is very ugly.  Good luck.

Both of these are just Brute-force problems.... , nothing extra :(

Post some problems like the solenoid one, those are much better :roll:

[quote="JRav"]While I have two whole books of problems that I could give you guys to solve, actually looking through them to try to find "tricky" problems might take a while.  But I'll try to find some nice ones.  I have a really nice E&M problem from MIT that I'll post in a few days (messy, but cool).[/quote]

I think that they're Yung Kuo Lim's books :)

[quote="JRav"]Indeed they are.  Do you like his books?  I find that they provide very nice practice, even at the undergraduate level.[/quote]

Of course and I've got whole 8 volumes of his course :)

I have only the book on Mechanics and E&M.  They are rather expensive, but they're really good.


Stabiliti daca $\{x_{n}\}$ are limita, $x_{n}=1+\frac{1}{2}+...+\frac{1}{n}$, unde $\{x_{n}\}$ reprezinta partea fractionara a lui $x_{n}$.E cumva $e-2$ limita?

Nu, n-are limită. În general, avem:

Fie $\left( a_{n}\right)_{n \geq 1}$ un șir de numere reale pozitive astfel încât $\lim_{n \to \infty}a_{n}= 0$ și $\sum_{k=1}^\infty a_{k}= \infty$. Definim șirul $\left( x_{n}\right)_{n \geq 1}$ prin \[x_{n}= \left\{ \sum_{k=1}^{n}a_{k}\right\}.\]
Atunci oricare ar fi $\alpha \in \left[ 0, 1 \right]$ există un subșir $\left( x_{n_{k}}\right)_{k \geq 1}$ cu limita $\alpha$.

Fie sirul de numere reale $ \left(x_{n}\right)_{n \geq 1}$ cu $ x_{n}=1+\frac{1}{2}+...+\frac{1}{n}$, pentru orice n natural. Atunci $ \left(x_{n}\right)_{n \geq 1}$ este crescator ($ x_{n+1}-x_{n}=\frac{1}{n+1}\geq 0$) si $ \lim_{n \to \infty}x_{n}=+\infty$.
Fie $ \left(y_{n}\right)_{n \geq 1}$ astfel incat oricare ar fi n natural, $ y_{n}={x_{n}}$.
Sirul $ \left(y_{n}\right)_{n \geq 1}$ este un sir alternant deoarece $ \lim_{n \to \infty}x_{n}=+\infty$, deci exista o infinitate de numere n naturale astfel incat $ y_{n}\to 1$ si $ y_{n+1}\to 0$. Cele doua subsiruri vor avea limite diferite, deci sirul initial nu are limita.

Să presupunem că $ \underset{n\rightarrow\infty}{\lim}\left\{ x_{n}\right\}=L.$ Atunci $ \underset{n\rightarrow\infty}{\lim}\left( \left\{ x_{n+1}\right\}-\left\{ x_{n}\right\}\right) =L-L=0.$

Dar $ \left\{ x_{n+1}\right\}-\left\{ x_{n}\right\}=x_{n+1}-\left[ x_{n+1}\right]-x_{n}+\left[ x_{n}\right] =\frac{1}{n+1}-\left( \left[ x_{n+1}\right]-\left[ x_{n}\right] \right) ,$ de unde $ \underset{n\rightarrow\infty}{\lim}\left( \left[ x_{n+1}\right]-\left[ x_{n}\right] \right) =0.$

Cum $ \left( \left[ x_{n+1}\right]-\left[ x_{n}\right] \right)$ e şir de numere naturale, rezultă că, de la un rang, este constant egal cu zero. Dar atunci obţinem că şirul $ \left[ x_{n}\right]$
este constant (de la un rang încolo), ceea ce contrazice faptul că $ \underset{n\rightarrow\infty }{\lim}x_{n}=\infty.$

Corect. Si sa demostrezi ca $ \lim_{n \to \infty}x_{n}=+\infty$ e foarte usor. Deci sirul $ y_{n}=$ {$ x_{n}$} nu are limita.

[quote="BlueSy1v3r"]Corect...etc[/quote]
normal că e corect..doar e soluţia domnului Enescu.. :yup:

Mi-a cerut posabogdan o demonstrație a acestui rezultat:
[quote="perfect_radio"]Fie $ \left( a_{n}\right)_{n \geq 1}$ un șir de numere reale pozitive astfel încât $ \lim_{n \to \infty}a_{n}= 0$ și $ \sum_{k=1}^\infty a_{k}= \infty$. Definim șirul $ \left( x_{n}\right)_{n \geq 1}$ prin
\[ x_{n}= \left\{ \sum_{k=1}^{n}a_{k}\right\}. \]
Atunci oricare ar fi $ \alpha \in \left[ 0, 1 \right]$ există un subșir $ \left( x_{n_{k}}\right)_{k \geq 1}$ cu limita $ \alpha$.[/quote]
Am cam uitat (n-aveam "pregătită" o soluție riguroasă atunci când am fost întrebat) :blush: da' mi-am amintit acuma...

[hide="Soluție"]
Fie $ m$ un număr natural și $ N$ astfel încat $ a_{n}< \frac1{m}, \, \forall n \geq N$.
Acuma împărțim $ \mathbb R^{+}$ în intervale de forma $ I_{k}= \left[ \frac{k}{m}, \frac{k+1}{m}\right)$.
Să presupunem că $ S_{N}= \sum_{n=1}^{N}a_{n}$ se află în $ I_{u}$ pentru $ u \geq 0$.
Cum $ S_{n}\to \infty$, există un indice $ q$ astfel încât $ S_{q}\in I_{u}$ și $ S_{q+1}\not\in I_{u}$.
Dar $ a_{q+1}< \frac1{m}$, deci $ S_{q+1}\in I_{u+1}$.
Prin inducție, demonstrăm că pentru oricare $ v \geq u$ există un $ S_{r}\in I_{v}$.
Așadar, oricare ar fi $ k \in \overline{0,m-1}$, există o infinitate de $ S_{w}$-uri a căror parte fracționară e între $ \frac{k}{m}$ și $ \frac{k+1}{m}$.
[/hide]


superior algebra
superior algebra unsolved

Suppose R is a field, then R has no nontrivial ideals.

[hide="Hint"]What are the ideal(s) containing a unit?[/hide]


What is the smallest possible number of whole 2-by-3 non-overlapping rectangles needed to cover a square region exactly, without extra over-hangs and without gaps?


symmetry
rotation
geometry
geometric transformation

Classify each of the following sentences as true or false:

(a) All compounds having a carbon atom with four different groups attached to it are chiral.

(b) Stereoisomers with different configurations are enantiomers.

(c) Meso compounds have chiral carbons and are therefore chiral.

(d) Enantiomers are stereoisomers of a chiral compound.

(e) Diastereomers are compounds with the same molecular formula, but that are not mirror images of each other.

(f) All tertiary amines with three different groups and the pair of electrons as a fourth group are chiral.

(g) Cis and trans isomers are always diastereomers.

(h) Molecules composed by two chiral fragments are necessarily chiral.

(i) Enantiomers of the same compound have different physical properties.

(j) Compounds with several chiral centers are always chiral.

(k) Meso compounds are achiral.

(l) If a molecule have a plane of symmetry then it is necessarily achiral.

(m) All structures that are mirror images of each other are enantiomers.

(n) If a Fisher projection is rotated in the plane of the paper by 90º the same molecule is obtained.

(o) A pair of enantiomers have the same solubility in ethanol.

(p) An optically active compound must have at least one stereoisomer.

(q) Meso compounds are achiral because they all have a center of simmetry.

(r) Two diastereomers can be distinguished by their melting points or other physical property.

(s) The designations "erythro/threo" refer to diasteromers.

(t) The compound $ CH_{3}CH_{2}CH_{2}As(CH_{3})CH_{2}CH_{3}$ is chiral.

(u) A pair of enantiomers have slightly different NMR spectra.

(v) Two enantiomers have different pKa's.

(w) If two enantiomers are volatile, they will have the same aroma.

(x) A pair of enantiomers have identical infrared spectra.

(y) Two enantiomers may not have the same vapor pressure.

(z) Two enantiomers have different solubilities in chiral solvents.

a.F
b.F
c.F
d.F
e.F
f.T
g.
h.F
i.F(except rotation of plane polarised light)
j.F
k.T
l.T
m.F
n.
o.T
p.T
q.F
r.T
s.T(but with specific properties)
t.T
u.F
v.F
w.T
x.T
y.F
z.T

[hide="Comment"]Your answers for (d), (e), (f), and (w) are wrong.[/hide]

[hide="my reasoning"]
d.also diastereomers are stereoisomers of a chiral compound.
so although the statement is true but i wrote false because it was ambiguous w.r.t the definition of enantaiomers

e.consider two compounds which show say chain isomerism clearly one of them is not the mirror image of other but they are not diastereomers

f.was a guess  :D

w.was also a guess :rotfl:

would like an explaination on those from u [/hide]

[hide="Answer for (d)"]The statement doesn't say that enantiomers are the only stereoisomers for a chiral compound. It can also have diastereomers, but the sentence doesn't exclude that possibility. So the statement is true.[/hide]

[hide="Comment for (e)"]You are right, because I forgot one word. What I wanted to write was "Diastereomers are stereoisomeric compounds with the same molecular formula...". [/hide]

[hide="Explanation of (f)"]All open chain tertiary amines (i. e., amines in which the nitrogen doesn't belong to a ring) with three different groups, $ R_{1}R_{2}R_{3}N$, are not chiral due to the fenomenon of [i]pyramidal inversion[/i], which consists in a rapid oscillation of the unshared pair of electrons from one side of the $ R_{1}R_{2}R_{3}$ plane to the other, thus rapidly converting the molecule to its enantiomer. However, there are two main types of nitrogen for which the inversion is prevented: a nitrogen belonging to a three membered ring and simultaneously connected to at least one atom having an unshaired pair of electrons, and a nitrogen located at a bridgehead position in a bicyclic system. So the statement is false, because only those molecules having one of those nitrogens with three different groups are in fact chiral.[/hide]

[hide="Explanation of (w)"]The statement is false, because the receptor sites in the nose are chiral. One known example is of (+)-Carvone (responsible for the odor of caraway seed oil) and (-)-Carvone (which gives spearmint oil its characteristic odor).[/hide]


search
combinatorics proposed
combinatorics

There are n + 1 fixed positions in a row, labeled 0 to n in increasing order from right to
left. Cards numbered 0 to n are shed and dealt, one in each position. The object of the game is to
have the card i in the ith position for 0  i  n. If this has not been achieved, the following move is
eqecuted. Determine the smallest k such that the kth position is occupied by a card l > k. Remove
this card, slide all cards from the (k+1)st position to the lth position one place to the right, and place
the card l in the lth position.
(a) Prove that the game lasts at most 2n  1 moves.
(b) Prove that there exists a unique initial conguration for which the game lasts exactly 2n 1
moves.

This one has been posted before by Pascual2005 in the "Combinatorics Proposed and Own Problems" section. There should be enough good key words here for a successful search, for anyone who's interested :).


vector
function
integration
topology
real analysis
real analysis unsolved

Let $ X$ be the vector space of all continuous functions from [0,1] to $ \mathbb{R}$.

Let $ X_1$ be this vector space with the $ L^1$ norm, $ \| f \|_1 \equal{} \int_0^1 |f(x)|dx$.

Let C be the same vector space with the C norm, $ \| f \|_C \equal{} \sup_{x\in [0,1]} |f(x)|$.

 Let $ \displaystyle E \equal{} \{ f\in X \mid f(x) > 0 \hspace{8 pt} \forall x\in [0,1]\}.$

Show $ E$ is an open set in $ C$

Show $ E$ is not an open set in $ X_1$.

Use the Weierstrass maximum theorem, for instance. ;)

If the minimum value of $ f$ is $ \alpha$ and $ \|f\minus{}g\|_C\leq\alpha/2$, then the minimum value of $ g$ is at least $ \alpha/2$.

There's no such analogue in the $ 1\minus{}$norm -- for example, consider the function $ f_n$ which is equal to $ \minus{}n$ at $ x\equal{}1/2$, $ 0$ at $ 1/2\pm 2^{\minus{}n}$, and is linear on $ [1/2\minus{}2^{\minus{}n},1/2]$ and $ [1/2,1/2\plus{}2^{\minus{}n}]$ and $ 0$ everywhere else.  If $ f$ is an arbitrary continuous function, then for large enough $ N$ $ f\plus{}f_N$ will be negative at $ x\equal{}1/2$.  Compute $ \|(f\plus{}f_N)\minus{}f\|_1$ and you should be almost done.

so you show that g is positive $ \in E$ and that implies that $ E$ is open, because g was an arbitrary function with the given distance restriction?

also, I have been going over the justification for why E is not open in X1, but I cannot exactly figure it out for the formulation that was given. blahblahblah or anyone else, do you think that you could elaborate on why E is not open in X1.  I tried using that function, but I couldn't get that it wasn't open.  maybe a different approach would be more clear, or maybe I am just missing something.

THanks so much for all the help and happy holidays

Sure.  Given a continuous function $ f$ which is everywhere positive on $ [0,1]$, for every $ \varepsilon > 0$ you can find a function $ f_\varepsilon$ such that $ \|f \minus{} f_\varepsilon\|_1 < \varepsilon$ but $ f_\varepsilon$ is nonpositive somewhere in $ [0,1]$.

This means exactly that $ E$ (as you define it) is not open in $ X_1$, since then for every $ \varepsilon > 0$ the ball $ B(f,\varepsilon)$ is not contained in $ X_1$.  The construction of $ f_\varepsilon$ can be found in my post.

intuitively: a long 'spike' with a narrow base can have a large sup norm but an arbitrarily small 1 norm.


[quote="balan razvan"]Prove that
\[ (\sum_{i \equal{} 1}^n p_i x_i^{\beta})^{\alpha \minus{} \gamma} < \equal{} (\sum_{i \equal{} 1}^n p_i x_i^{\alpha})^{\beta \minus{} \gamma} (\sum_{i \equal{} 1}^n p_i x_i ^ {\gamma})^{\alpha \minus{} \beta} ,
\]

\[ \alpha > \beta > \gamma > 0
\]

\[ p_1,p_2,...,p_n,x_1,...,x_n > 0,n > \equal{} 2
\]
[/quote]
\[ \sum_{i\equal{}1}^n p_ix_i^{\beta} <\equal{}(\sum_{i\equal{}1}^n p_ix_i^{\alpha})^{\frac{\beta\minus{}\gamma}{\alpha\minus{}\gamma}} \cdot (\sum_{i\equal{}1}^n p_ix_i^\gamma)^{\frac{\alpha\minus{}\beta}{\alpha\minus{}\gamma}}\] and we observe that is similar to Holder because the numbers $ \frac{\alpha\minus{}\beta}{\alpha\minus{}\gamma}$ and $ \frac{\beta\minus{}\gamma}{\alpha\minus{}\gamma}$ have the sum 1.But to be in the case where we should aply Holder we have to have $ (x_i^{\alpha})^{\frac{\beta\minus{}\gamma}{\alpha\minus{}\gamma}} (x_i^{\gamma})^{\frac{\alpha\minus{}\beta}{\alpha\minus{}\gamma}} \equal{} x_i^{\beta}$ wich is true.
The equality takes place when $ x_1\equal{}...\equal{}x_n$


Write the ideals of the ring $ A\equal{}Z_2[x,y]/(x^2,y^2,xy)$. Use $ A$ to show that
"at most two non-prime" hypothesis in the Prime Avoidance Lemma is essential.

It would probably help if the people reading this post without access to a text knew what the Prime Avoidance Lemma was.

[b]Lemma:[/b]  Suppose that $ I_1, ... I_n, J$ are ideals of a ring $ R$ and suppose that $ J \in \bigcup I_j$.  If $ R$ contains an infinite field or if at most two of the $ I_j$ are not prime, then $ J$ is contained in one of the $ I_j$.


Let S1, S2, S3, ... be a sequence of non-empty closed sets in Rn satisfying

S1 \supset S2 \supset S3 \supset ... \supset Sk \supset ...

and

lim diam S_k = 0

where diam S = sup{ |x-y| | x,y\in S}. |x-y| means the distance between x and y.

Prove that there is a unique point in  \cap_{k=1}^\infty S_k

What does "\supset" mean? does "A \supset B" mean that B is included in A?

Yes.
it's a standard tex command, isn't it?

It might be, I don't know Tex :). And since it's not implemented here.. Anyway, thanls for making it clear.

Pick a point in each set. The condition about the diameters going to zero now makes this sequence of points into a Cauchy sequence. Follow the consequences of that.

You can view this either as a completeness result or as a compactness result. It's also very close to the Heine-Borel Theorem, or you could choose to use the Bolzano-Weierstrass Theorem.


Six members of the team of Fatalia for the International Mathematical Olympiad are selected from $13$ candidates. At the TST the candidates got $a_1,a_2, \ldots, a_{13}$ points with $a_i \neq a_j$ if $i \neq j$.

The team leader has already $6$ candidates and now wants to see them and nobody other in the team. With that end in view he constructs a polynomial $P(x)$ and finds the creative potential of each candidate by the formula $c_i = P(a_i)$.

For what minimum $n$ can he always find a polynomial $P(x)$ of degree not exceeding $n$ such that the creative potential of all $6$ candidates is strictly more than that of the $7$ others?

[i]Proposed by F. Petrov, K. Sukhov[/i]

WLOG, $a_1 < a_2 < \cdots < a_{13}$. Suppose the set of people the leader wants to see in the team is those $a_i$ for which $i$ is even. Then the graph of $P(x)$ must have at least 11 turning points, and so $P(x)$ has degree at least 12. On the other hand, we can determine the creative potential of all the students exactly with a polynomial of degree 12, by Lagrange interpolation. So $n=12$.

Out of curiousity, what is Tuymaada?

tuymaada is an international olympiad. students (from romania at least) who don't make it to IMO go there


Find the integer $ n$ such that $ n\times3^{4} \times 7^{5} \equal{} 21^{6}$.

i just want to clarify you got the 6-4 is from the 21^6 right  ?


Driving in your car with a constant speed of v = 13.2 m/s, you encounter a bump in the road that has a circular cross section. If the radius of curvature of the bump is r = 33.6 m, calculate the apparent weight of a 63.5 kg person in your car as you pass over the top of the bump.

Hint: Using Newton's second law

[hide]
The apparent weight of the person at the top of the bump is $m(g-\frac{v^{2}}{r})$[/hide]


find all real number solutions of x^2-yz=1, y^2-xz=2, z^2-xy=3 Thank you

[quote="huangxiahui"]find all real number solutions of
 $ x^2\minus{}yz\equal{}1$,
$ y^2\minus{}xz\equal{}2$,
 $ z^2\minus{}xy\equal{}3$
Thank you[/quote]
LaTeXed.

Adding them all together, doubling, and factoring gives $ (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2\equal{}12$, which sadly doesn't help too much.

I swear I used to remember a factorization for this exact symmetric pattern :(

Multiply the first equation by y+z, multiply the second by x+z, multiply the third by x+y.  Add them all together to get 3x+4y+5z=0, I think.  This may or may not be accurate or helpful.

Well, Thanks, at least it's a little more developed.

Can someone give me a solution to the problem please? since(x-y)^2+(y-z)^2+(z-x)^2=12, the variables are not integers. But can you guys help me with it?

Solution:

$ \begin{cases}x^2\minus{}yz\equal{}1.........(1) \\
y^2\minus{}xz\equal{}2...........(2) \\
z^2\minus{}xy\equal{}3............(3) \end{cases}$

(1)-(2), $ (x\minus{}y)(x\plus{}y\plus{}z)\equal{}\minus{}1$
(2)-(3), $ (y\minus{}z)(x\plus{}y\plus{}z)\equal{}\minus{}1$

So $ x\minus{}y\equal{}y\minus{}z\equal{}k$, where $ k$ is a constant, and $ x\plus{}y\plus{}z\equal{}\minus{}\frac{1}{k} \Leftrightarrow y\equal{}\minus{}\frac{1}{3k}$

Now (2) becomes $ (\minus{}\frac{1}{3k})^2\minus{}(\minus{}\frac{1}{3k}\plus{}k)(\minus{}\frac{1}{3k}\minus{}k)\equal{}2$
$ \Leftrightarrow k^2\equal{}2, k\equal{}\pm \sqrt{2}$

So $ (x,y,z)\equal{}(\minus{}\frac{1}{3\sqrt{2}}\plus{}\sqrt{2}, \minus{}\frac{1}{3\sqrt{2}}, \minus{}\frac{1}{3\sqrt{2}}\minus{}\sqrt{2})$
or $ (x,y,z)\equal{}(\frac{1}{3\sqrt{2}}\minus{}\sqrt{2}, \frac{1}{3\sqrt{2}}, \frac{1}{3\sqrt{2}}\plus{}\sqrt{2})$

After checking, both of them work.

thank you very much. you are smart, man.

But how did you find the factorization formula? By testing? Or it looks familar to you?


Lemma: Let I be a minimal left ideal. Either I^2 = 0 or I = Ae, where A is
the ring and e is an idenpotent.

Proof: Suppose that I^2 is different from 0. Choose a in I such
that Ia is different from 0. Then it is clear that I is contained in Ia,
but I do not understand that why is Ia contained in I. Hence I = Ia.

Then let us consider the left ideal
a^{-1}(0) = {r in A: ra =0}.

Then why is true that a^{-1}(0) intersect I is different from I and
by minimality of I we have a^{-1}(0) intersect I is 0?

Next choose e in I such that a = ea. Why is it true that e-e^2 belong to
a^{-1}(0) intersect I =0. because we have to show that e is idempotent,
and we can not use this fact in the proof.
And Ae is different from 0 and why is it true that I = Ae?
And the other thing is that where have we shown that Either I^2 = 0 or I = Ae

[quote="peteryellow"]Then it is clear that I is contained in Ia,
but I do not understand that why is Ia contained in I. Hence I = Ia.
[/quote]
$ a\in I$ and $ I$ left ideal gives $ Ia\subseteq Aa\subseteq AI\subseteq I$.
Then $ 0\not \equal{} Ia\subseteq I$ with $ Ia$ also left ideal and $ I$ minimal gives $ I \equal{} Ia$.
Otherwise $ I\subseteq Ia$ is not clear at all.

Then $ ea \equal{} a\not \equal{} 0$ gives $ e\in I\setminus a^{ \minus{} 1}(0)$ by definition.
So the left ideal $ I\cap a^{ \minus{} 1}(0)\subsetneq I$ must be $ 0$ by minimality of $ I$.
Now $ e^2\in Ae\subseteq AI \subseteq I$, and then also $ e \minus{} e^2\in I$.
Also $ (e \minus{} e^2)a \equal{} ea \minus{} e^2a \equal{} a \minus{} ea \equal{} a \minus{} a \equal{} 0$, so $ e \minus{} e^2\in a^{ \minus{} 1}(0)$ by definition.
This gives $ e \minus{} e^2\in I\cap a^{ \minus{} 1}(0) \equal{} 0$ and $ e^2 \equal{} e$ idempotent.

Finally $ ea \equal{} a\not \equal{} 0$ gives $ 0\not \equal{} e \equal{} e^2\in Ae$ a left ideal.
So $ 0\not \equal{} Ae\subseteq AI\subseteq I$ and $ I$ minimal gives $ I \equal{} Ae$.

Sorry, but I can't make it more clear than that.

Thankyou very much for your reply, the only question I have left is that have you used that e is idempotent in
$ (e \minus{} e^2)a \equal{} ea \minus{} e^2a \equal{} a \minus{} ea \equal{} a \minus{} a \equal{} 0.$ But is it something we have to show in the proof that e is idempotent or we can just use it, thanks

Pay attention to how $ e$ was chosen.  $ e \in I$ so that $ ea\equal{}a$.  Then $ e^2a\equal{}e(ea)\equal{}e(a)\equal{}a$.  Consequently,
$ (e\minus{}e^2)(a)\equal{}0$ and then $ e\minus{}e^2 \equal{}0$.


A Rubix Cube comes in sizes of 2 by 2 by 2, 3 by 3 by 3, 4 by 4 by 4, and 5 by 5 by 5. What is the difference in value of the volume of the biggest and smallest cubes and the middle size cubes?

[hide]
80[/hide]

[hide]Ok the hardest part of this problem is figuring out what it means.  After like 10 minutes i got it.  What it means is

$ (5\times5\times5 \minus{} 2\times2\times2) \minus{} (4\times4\times4 \minus{} 3\times3\times3)$

this is the same as $ (125 \minus{} 8) \minus{} (64 \minus{} 27) \equal{} (117) \minus{} (37) \equal{} \boxed{80}$[/hide]

EDIT:  Changed + to - (thanks stevenmeow! thats what i meant)

@ zolojetto
Notice that You put 5*5*5[b]+[/b]2*2*2 that confused me for a while.

and @ MrRubix, the problem was very ambiguously worded

[quote="stevenmeow"]and @ MrRubix, the problem was very ambiguously worded[/quote]

i will agree with that, i had to try around with different meanings of the problem untill i got the answer of 80

well answer is obviously 80...i can solve a 2x2x2, 3x3x3 and a 4x4x4 rubiks cube  :lol:

what the?? why rubics cubes? and shouldnt answer be 117?

1-2=341(8*67)d(67=54)

albert einstein's scientific equation of the world's astolarynx side theory

mathnerdworm, this is an old thread. And what the heck are you asking us to do with your post???


algebra
function
domain
polynomial
Ring Theory
vector
superior algebra

Could you help me with this problem (need help with (ii)):

Describe all ideals of the following quotient rings (R represents the real numbers):

(i) $ R[x]/ < x^2 \minus{} 1 >$

(ii) $ R[x]/ < x^2 \plus{} 1 >$

Now, (ii) is isomorphic to C (the complex numbers). As C is a field, its only ideals are itself and {0}.

But what is (i) isomorphic to? How would one describe its ideals?

Since $ \mathbb{R}[x]$ is a principal ideal domain, all ideals will be of the form $ (f)$ for some polynomial $ f$. Modding out by $ (x^2\minus{}1)$, we obtain that the ideal $ (f) \subset \mathbb{R}[x]$ is mapped to the ideal $ (\gcd(f,x^2\minus{}1)) \subset \mathbb{R}[x]/(x^2\minus{}1)$. We have that $ x^2 \minus{} 1 \equal{} (x\minus{}1)(x\plus{}1)$, so the only proper ideals are $ (x\minus{}1)$ and $ (x\plus{}1)$.

$ \mathbb{R}[x]/(x^2 \minus{} 1)$ is the algebra of functions on the points $ 1, \minus{} 1$, so it is isomorphic to $ \mathbb{R}^2$.  (You can prove this using the Chinese remainder theorem.)  More generally, we have the correspondence theorem:

[b]Theorem:[/b]  Let $ R$ be a commutative ring and $ J$ an ideal.  The ideals of $ R/J$ are naturally in bijection with the ideals of $ R$ containing $ J$.  (The correspondence is as follows: ideals of $ R$ containing $ J$ are precisely those ideals $ I$ with the property that the map $ R \to R/I$ factors through $ R/J$.)

Thank you both for the answers - they were very helpful.

However, I'm still confused about something. On one hand, we have that:

[quote="t0rajir0u"][b]Theorem:[/b]  Let $ R$ be a commutative ring and $ J$ an ideal.  The ideals of $ R/J$ are naturally in bijection with the ideals of $ R$ containing $ J$.  (The correspondence is as follows: ideals of $ R$ containing $ J$ are precisely those ideals $ I$ with the property that the map $ R \to R/I$ factors through $ R/J$.)[/quote]

which implies what Nukular said:

[quote="Nukular"] We have that x^2 - 1 = (x-1)(x+1), so the only proper ideals are (x-1) and (x+1).[/quote]

On the other hand, we have

[b]Theorem: [/b] If $ f: R \to S$ is an isomorphism of the rings $ R,S$, then $ I \subseteq R$ is an ideal of R if and only if $ f(I)$ is an ideal of $ S$.

Then, if indeed

[quote="t0rajir0u"]$ \mathbb{R}[x]/(x^2 \minus{} 1)$ is the algebra of functions on the points $ 1, \minus{} 1$, so it is isomorphic to $ \mathbb{R}^2$.  (You can prove this using the Chinese remainder theorem.)  [/quote]

we would have that the ideals of $ \mathbb{R}[x]/(x^2 \minus{} 1)$ would be isomorphic to the ideals of $ \mathbb{R}^2$, (with an appropriate definition of multiplication)... doesn't $ \mathbb{R}^2$ have no proper ideals? And hence wouldn't this imply that $ \mathbb{R}[x]/(x^2 \minus{} 1)$  doesn't have any proper ideals?

I think I'm going wrong somewhere in the second argument, but I can't quite figure out where (maybe that $ \mathbb{R}^2$ has no proper ideals)?

$ \mathbb{R}^2$ has non-trivial proper ideals $ \mathbb{R}\times\{0\}$ and $ \{0\}\times\mathbb{R}$, corresponding to $ (x\minus{}1)$ and $ (x\plus{}1)$.

Steve

[quote="SD5225272"]$ \mathbb{R}^2$ has non-trivial proper ideals $ \mathbb{R}\times\{0\}$ and $ \{0\}\times\mathbb{R}$, corresponding to $ (x \minus{} 1)$ and $ (x \plus{} 1)$.

Steve[/quote]

Thank you for the clarification. I found my mistake now: what was misleading me was that $ \mathbb{R}^2$ is usually said to be isomorphic to $ \mathbb{C}$, which is a field, and hence has no proper subrings. I forgot that in order to do this, we needed to define a different type of multiplication on $ \mathbb{R}^2$ :
\[ \left( {a,b} \right).\left( {c,d} \right) \to \left( {ac \minus{} bd,ad \plus{} bc} \right)\]
And while $ \mathbb{R}^2$ with this type of multiplication has no proper subrings, $ \mathbb{R}^2$ with the usual multiplication does (as you pointed out). Finally, it is the latter ring-with-operations to which we constructed the isomorphism, and hence we really have no problem...

Thanks again to everyone!

I agree the notation is confusing; I maybe should've said $ \mathbb{R} \times \mathbb{R}$, since otherwise it's not clear whether $ \mathbb{R}^2$ refers to the affine space or the vector space or what.


could anyone explain what is reverse hyperconjugation??? :D

the phenomenon of reverse hyp is observed in X-ch-c=c where x is halogen.i.e halogen atoms are in conjugation with a pi bond. its similar to hyp, just that in this case halogen atom has the lone pair of electrons in the resonating structures and a carbocation is formedwhich is in conjugaion with double bond

I have never heard of "reverse hyperconjugation". VARUNARASU, regarding "and a carbocation is formedwhich is in conjugaion with double bond", could you explain how exactly a carbocation is formed and in what carbon?

in hyperconjugation, you know that in the resonating structures, there is a h+ ion. and a carbanion which is conjugation with the double bond, similarly in case of reverse hyp in the resonating structures, there is a X- ion, and a carbocation on the carbon alpha to the double bond which is resonance with the double bond. the req of reverse hyp is that a halogen atom should be present on the carbon alpha to the double bond

An halogen can already stabilize an adjacent positive charge by ressonance, so postulating a ressonance form with an halide ion seems odd.

but there is no positive charge adjacent to the halogen


geometry
similar triangles
geometry open

I am trying to find all triangles that can be dissected into 5 triangles, that are similar to each other (but not necessary to the original triangle). If 2 or more of the parts are of the same size, that's OK.

1) Prove, that any right-angled triangle can be dissected into 5 similar triangles.

2) Prove, that any isosceles triangle (including equilateral) can be dissected into 5 similar triangles.

3*) Are there any other triangles that can be dissected into 5 similar triangles?

I do not know the solution for point 3. If anybody knows, please, write here or contact me.

Thank you!
Sergey Markelov
markelov@mccme.ru

(1) Starting from the point which has the right angle, you can draw perpendicular line to the hypotenuse - Through 4 steps, we can get 5 triangles
(2) Divide the triangle into two same right-angle triangles - Same with (1)
(3) Well, what I can prove is that
     if there is stronger qualifiaction that all the 5 similar triangles are similar, then there are only two types of triangles.
     A triangle with 90 degree angle, 120 degree angle.
     (Titu Andreescu, Mathematical Olympiad Challenge)

Exemple :
[img]http://img179.imageshack.us/img179/1356/premiercas.jpg[/img]
See also [url=http://www.les-mathematiques.net/phorum/read.php?8,593720,page=1//url]Les mathématiques.net[/url] , in french !!!

Domi


Does there exist an integer k which can be expressed as the sum of two factorials ${k = m! + n!}$ (with $m\leq{n}$) in two different ways?

[hide="Solution"]
$2 = 1! + 1! = 0! + 0! = 0! + 1!$
$1! + n! = 0! + n!$
 :D
[/hide]

[hide="Proof"]
For $m \le n$ and $p \le q$, with all of them $> 1$, let us denote $k = m! + n! = p! + q!$.  WLOG, let $(n-m) < (q-p)$.  Then $p < m, q > n$.

We know that $m! | k$.  We therefore know that $m! | (p! + q!)$, which can only be true if $\frac{m!}{p!} | \frac{q!}{p!} + 1$.  If $m > p+1$, then $\frac{m!}{p!}$ is even.  Then the equation can only hold true if $\frac{q!}{p!}$ is odd, which only occurs when $q=p$ or $q=p+1$, both of which contradict our assumption that $p < m, q > n$.

This means $m \le p+1$, and because $p < m$, we have $m = p+1$ and $p+1 | \frac{q!}{p!} + 1$.  Because we have assumed $p > 1$ in order to prevent a trivial solution, $p+1 \ge 3$.

For $p+1 \ge 3$, then, we write our original equation as $k = (p+1) p! + n! = p! + q!$.  Subtracting yields $p(p!) = q! - n! = n! ( \frac{q!}{n!} - 1 )$.  This requires that $\frac{p(p!)}{n!}$ be an integer because $q > n$, but because we have $p < m \le n$, the fraction becomes $\frac{p}{(p+1)(p+2)...(n)}$ which is clearly never an integer.

Hence, only the trivial solutions work.
[/hide]

i was also thinking [hide]$3=0!+2!=1!+2!$[/hide]

edit-[hide]now that i think about it, if you have any $n$ and you take $n!$, you can express $n!+1=0!+n!=1!+n!$ so there's an infinite number of numbers $k$ that will satisfy this equality[/hide]


[color=blue]Buenos dias,

En esta seccion, podremos postear algunos tips teoricos q nos ayuden a entender mejor las Inecuciones, a resolverlas y a construir nuevas  :lol: .

Saludos,

Carlos bravo:)
Lima - PERU[/color]

Hace unos dias, recorde una desigualdad muy util q se utiliza en prob de Ineqs:

Dado: $x>0$,

$\frac{1}{1+x^{2}}\geq \frac{2-x}{2}$,
lo que nos lleva a:
$x(x^{2}-2x+1) \geq 0$, lo cual es obviamente cierto.

Esto se utiliza para expresiones como esta:

---
Dados $a,b,c>0$, tal que: $a+b+c = 4.$

Probar que:
$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}\geq 1.$
---

La idea original de esta tecnica es obtener una expresion algebraica que se cumpla para algun intervalo en particular (en este caso $x>0$) y generar expresion con 3 o mas variables para una ineq,
ademas, debemos considerar una condicion q nos ayude a reducir expresiones y acotar....

pero podemos generalizar esta primera ineq:

Dado: $x>0$, $a>0$,

$\frac{1}{1+x^{2}}\geq \frac{a-x}{a}$,

lo cual nos lleva a:

$x(x^{2}-ax+1) \geq 0$
Para q esto sea cierto, extiste un grupo de valores para $a$, como:
-2,
-1,
1,
2,
etc

intenten construir inecuaciones con esta desigualdad...

Saludos,

Navegando por la seccion Inequalities, encontre algo curioso:

[quote="nayel"]Prove that for $x,y,z>0$,
\[\sum{\frac{x^{2}}{y^{2}+z^{2}}\ge \sum\frac{x}{y+z}}\]
[/quote]

y parece ser q esto es asi:

[quote="eagerkill"][img]http://img157.imageshack.us/img157/4913/untitled7xy.jpg[/img][/quote]

alguna idea??

Saludos,

Carlos bravo :)
Lima -PERU

Si tomamos $y=z=1$, $x<1$ y $\alpha>1$ vemos que algo anda mal con la desigualdad expuesta.
Por otro lado, derivando la función $f$, vemos que $f'(\alpha)\geq 0$ si y sólo si
$(b^{\alpha}+c^{\alpha}).ln(a)\geq b^{\alpha}.ln(b)+c^{\alpha}.ln(c)$
Un ejemplo simple en el que esta desigualdad es válida para todo $\alpha\in R$, es cuando $a\geq max\{b,c\}$


Hi,

I have a problem. An electron enters a magnetic field as shown in the image. Draw its path.

I am tempted to think that it would spiral, but I am not entirely sure.

[img]http://img246.imageshack.us/img246/9973/electronaz3.jpg[/img]

Any help would be appreciated.

Cheers.

Yes it would spiral, because the horizontal component would go around as if it were traveling in a circle, while the vertical component of velocity remains constant.

The direction of the force is perpendicular to the electron at all times, and hence, due to the initial velocity, it is almost like centripetal force.

[quote="leovinus"]Hi,

I have a problem. An electron enters a magnetic field as shown in the image. Draw its path.

I am tempted to think that it would spiral, but I am not entirely sure.[/quote]

[hide]
The electron has a helical trajectory with a pitch angle $\alpha =\sin^{-1}\left(\frac{v_{z}}{v}\right),$ and axis parallel to the B-field.

The Lorentz force is given by: $\vec{F}=-e \vec{v}\times \vec{B}$

$\vec{v}= v_{\theta}%Error. "mathbold" is a bad command.
{\hat{\theta}}+v_{z}%Error. "mathbold" is a bad command.
{\hat{z}}$

$\vec{B}= B%Error. "mathbold" is a bad command.
{\hat{z}}$

$\Rightarrow \vec{F}=-e \left[v_{\theta}%Error. "mathbold" is a bad command.
{\hat{\theta}}+v_{z}%Error. "mathbold" is a bad command.
{\hat{z}}\right] \times B%Error. "mathbold" is a bad command.
{\hat{z}}$

$\Rightarrow \vec{F}=-e v_{\theta}B %Error. "mathbold" is a bad command.
{\hat{r}}$

The Lorentz force is a centripetal force.

$\vec{F}=-e v_{\theta}B %Error. "mathbold" is a bad command.
{\hat{r}}=-m\frac{v_{\theta}^{2}}{r}%Error. "mathbold" is a bad command.
{\hat{r}}$

$\Rightarrow \frac{v_{\theta}}{r}= \frac{e}{m}B$

$\frac{v_{\theta}}{r}\equiv \omega.$

$\therefore \omega = \frac{e}{m}B$

$\alpha = \sin^{-1}\left(\frac{v_{z}}{v}\right) \text{ where }v = \parallel \vec{v}\parallel$
[/hide]


$a,b,c\in\mathbb{R}$.
suppose that $|ax^2+bx+c|\le 1$ if $|x|\le 1$. prove that $|cx^2+bx+a|\le 2$ if $|x|\le 1$.

This is a problem for the SIMUW 2006 application, please do not answer, mods lock this.

A summer camp in Washington (as in the state) that requires you to solve 8 problems as part of the application.


A number is called "cute" if it has exactly 4 (positive) integer divisors. What percent of the first 10 positive integers are cute?


[quote="pascal_1623"]Convert $\frac{13}{20}$ into percent and decimal.[/quote]

[hide="Decimal"]
Decimal: 0.65[/hide]

[hide="Percent"]Percent: 65%[/hide]

Please don't give titles to problems that suggest that their level of difficult is "easy" or "hard" or anything like that.  It may be that problems in Classroom Math are easy for some students and more challenging for others.  The goal is to present and solve problems, not judge their difficulty.  Judging their difficulty might turn some students off to the forums and that's not good for building a community of math students.


Express the sum as a common fraction:
\[ \frac{1}{10}\plus{}\frac{1}{100} \plus{} \frac{1}{1,000} \plus{} \frac{1}{10,000}.\]


inequalities
logarithms
ratio
calculus
integration
inequalities proposed

I was not thinking for very long on this one, but it may be hard and interesting (and it may be not ;)):
Let $a,b>1$ be positive integers and:

$T=\log{a}$
$D=-\log{\left|\frac{a}{b}-\pi\right|}$

Find maximum of $\frac{D}{T}$ or prove that the maximum doesn't exists. It seems that it is unbounded, but then you can try maximize $\frac{D}{T^2}$ or something like that, I don't know.

Good luck.

The ratio $\frac{D}{T}$ is bounded above.  It follows from [url=http://mathworld.wolfram.com/IrrationalityMeasure.html]Hata's theorem[/url].

I think the first bound of the form
$| \pi - p/q| > 1/q^C$    was by Mahler, with $C = 42$.  (Or maybe Siegel??)

Hata's result is the latest and greatest in this genre, with $C$ around 8.
If I remember correctly, he actually finds such a bound for $\pi^2$ (or $\zeta (2)$) and from this gets the result on $\pi$.

Yes, it was Mahler who was first.  I should have credited him.

Hata does have some better bounds for $\pi^2$, but I'm pretty sure that his bound on $\pi$ was directly for $\pi$, not $\pi^2$.  I studied his technique in [url=http://www.math.leidenuniv.nl/~naw/serie5/deel01/dec2000/pdf/beukers.pdf]this paper[/url].  It's pretty interesting stuff.

Interesting, thanks.   Apparently I'm referring to the wrong article by Hata, i.e. not the one with the current strongest bound for $\pi$.   Hata actually has several articles in which integrals are written down whose positivity gives irrationality bounds on $\pi, \zeta(2), \zeta(3), log(2), ...$, and rational linear combinations of these.   I am fairly certain that $\pi^2 / 6$ is easier to handle by this method and that in the article I saw, Hata produces a very good irrationality measure for $\zeta(2)$ and gives the corresponding bound for $\pi$ as corollary.


Let $A,B$ 2 points and $e,f$ 2 lines(they have a common point $O$) in the plane such that neither $A$ nor $B$ belongs to any of the lines.We take $M$ variable on $e$ and $N$ variable on $f$.Find their position such that $AM+MN+NB$ is minimal.

In this problem I think we can give only the way how to find the minimum.
The basic idea is that if we construct the symmetric of A wrt to [b]e[/b],
symmetric of B wrt to [b] f[/b], or both together - the minimum doesnt change
as sum of the distances AM+MN+NB doesnt change.

Lemma: In the triangle if  60 then 2a > (b+c)
Proof:     a 2 > b 2 + c 2 - bc > (b+c) 2/4

Here is a sketch. Let lines [b]e [/b]and [b]f[/b] be lines XOX= e and YOY= f which divide
our plane in 4 cadrans. We will devide our problem in 3 cases.

1.    A and B are in 1 and 2 cadran respectively and WLOG suppose they lie in the
  same semiplane wrt to the line [b]e[/b]. Then if we construct A symmetric to A  wrt to line [b]e [/b]
then the line AB intersects lines [b]e[/b] and [b]f[/b] in points M and N that satisfy the minimum.
Proof:  If we have another variant M, N then
          AM+MN+BN = AM+MN+NB > AM+MN+NB = AM+MN+NB
          because the shortest distance is a line.

2.   If A and B are in the same cadran XOY.
      Let A and B be the images of A and B wrt to lines e and f respectively.
a)	< XOY ≤ 60
     Then it is easy to see that AB intersect sides of XOY in M and N.
     Then (M, N) satisfy the minimum by the same logic as in (1case)
b)	<XOY > 60
      In the case if AB intersect XOY - we use same way.

      Suppose AB doesnt intersect  XOY.
      If symmetric are in neighbour cadrans - (1 case) we have solution.
      Hence A, B are in opposite cadrans (1,3) or (2,4).

      If ray AB first intersect line e after that f  - then this points are M, N
      which we are finding ( line is the shotest distance)

      So  AB intersect firstly [b] f[/b] then [b]e[/b]. In this case I want to prove  maximum is AO+BO.
      Suppose we have M and N that give minimum.
  If M, N   (XO) and (OY) then obviously AO+BO < AM+MN+NB.
  If M, N   (XO) and (OY) or (OY) and (OX)
       AM+MN+NB = AM+MO+ON+NB > AO+BO.
  If M, N  (OX) and (OY) then
       AM+MN+NB = AN+BM+2MN > AN+ON+OM+BM >AO+BO (by lemma) q.e.d.


ratio
real analysis
real analysis solved

[quote="silviu"]Having $(x_n)_{n>0}$ with $x_1,x_2,x_3>0$ and $x_{n + 3}  = \frac{{x_{n + 2}^2  + 5x_{n + 1}^2  + x_n^2 }}{{x_{n + 2}  + 5x_{n + 1}  + x_n }}\forall n \ge 1$
Prove that $(x_n)_{n>0}$ converge.[/quote]

$x_{n+3}$  is a weighted average of the $x_n, x_{n+1}, x_{n+2}$  with weights $x_n, 5x_{n+1}, x_{n+2}$.  This means that the (minimal) interval $I_n$ containing $x_n, x_{n+1}, x_{n+2}$ is shrinking, and that the ratios $x_i / x_{i+1}$ are bounded away from zero.  If $I_{\infty} = [a,b]$, with $a < b$,  and $I = [a - \epsilon, b+ \epsilon]$, and we take a weighted average of 3 points in $I$ (at least one near $a$, at least one near $b$) with any ratio of 2 weights larger than $C > 0$, then for sufficiently small $\epsilon$ (depending on $C, a, b$)  this weighted average must lie in the interior of $(a,b)$.  From this we conclude that $a=b$, i.e. $x_n \to a$.

A better (more constructive) version of the same argument:
there exists some positive constant $c < 1$  so that
$|I_{n+1}| \leq c|I_n|$.

yes fleeting guest its called contractive sequences.They are Cauchy sequence & are hence convergent(R is complete)

O sorry I am horribly wrong.The contractive sequences are those:

mod(xn+1 -xn) <= c*mod(xn-xn-1) where 0<c<1

Sorry for the mistake..

The argument above does not show that $d_n = |x_n - x_{n-1}|$ is contracting, it shows $d_{n+3} < c d_n$.  This certainly implies convergence, but in the above argument it is enough that $|I_n| \to 0$ (not necessarily geometrically).   It would be interesting to prove directly that $d_{n+1} < c d_n$, if that is true.


Let $x,y,z$ be positive real numbers. Prove that:
$\frac{xyz[(x+y+z)+\sqrt{x^2+y^2+z^2}]}{(x^2+y^2+z^2)(xy+yz+xz)}\leq \frac{3+\sqrt{3}}{9}$

From Cauchy-Schwarz : $ x+y+z \leq \sqrt{3(x^2+y^2+z^2)} $
  From AM-GM : $ x^2+y^2+z^2 \geq 3.(xyz)^{\frac{2}{3}} $ and $ xy+yz+xz \geq 3.(xyz)^{\frac{2}{3}} $ .
  The rest is for you , Gohan  :)


Find 4 prime numbers less than 100 which are factors of $ 3^{32}\minus{}2^{32}$.


Find all continuous functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x)^3 = -\frac x{12}\cdot\left(x^2+7x\cdot f(x)+16\cdot f(x)^2\right),\ \forall x \in \mathbb{R}.\]

Moved to pre-olympiad. Anyone who wants to try? :)

[quote="Peter VDD"]Find all continuous functions $f: \mathbb{R}\rightarrow\mathbb{R}$ such that \[ f(x)^3 = -\frac x{12}\cdot\left(x^2+7x\cdot f(x)+16\cdot f(x)^2\right),\ \forall x \in \mathbb{R}. \][/quote]

 the epsilon R means is a member of the set of real numbers... , but an someone help me out and tell me what the little upside down A thing means?  I'm taking the USAMO and feel like i should know this stuff  :blush: ...

$\forall x \in \mathbb{R}$ means "for all real numbers $x$".

I don't really know how to approach this problem but plugging in values for x could get you ideas of what f(0) f(1) f(2) etc. would be by solving the resulting cubics.

$f(x)^3 = -\frac x{12}\cdot\left(x^2+7x\cdot f(x)+16\cdot f(x)^2\right)$
$\Rightarrow 12[f(x)]^3 + x \cdot \left(x^2+7x \cdot [f(x)]+16 \cdot [f(x)]^2\right) = 0$
$\Rightarrow 12[f(x)]^3 + 16[f(x)]^2 \cdot x + 7[f(x)] \cdot x^2 + x^3 = 0$
$\Rightarrow (3 f(x) + x)(2 f(x) + x)^2 = 0$
$\Rightarrow f(x) = \frac{-x}{3} \text{or} \frac{-x}{2}$

Edit: I was in a rush. All we can deduce from the factored form to the next step is that one of the factors must be zero for all $x$, not that the function can be entirely linear the whole way through, my apologies.

[quote="Pork_Chop8"]$f(x)^3 = -\frac x{12}\cdot\left(x^2+7x\cdot f(x)+16\cdot f(x)^2\right)$
$\Rightarrow 12[f(x)]^3 + x \cdot \left(x^2+7x \cdot [f(x)]+16 \cdot [f(x)]^2\right) = 0$
$\Rightarrow 12[f(x)]^3 + 16[f(x)]^2 \cdot x + 7[f(x)] \cdot x^2 + x^3 = 0$
$\Rightarrow (3 f(x) + x)(2 f(x) + x)^2 = 0$
$\Rightarrow f(x) = \frac{-x}{3} \text{or} \frac{-x}{2}$[/quote]

Wow, from the source I assume you just cite the solution.  But can anyone explain in more detail how one go from the second step to the third???

[quote="Pork_Chop8"]
$(3 f(x) + x)(2 f(x) + x)^2 = 0$
$\Rightarrow f(x) = \frac{-x}{3} \text{or} \frac{-x}{2}$[/quote]

You should explain that step better. It's not that "immediate".

Who says we can't have $f(x) = -x/3$ for some values of $x$ and $f(x) = -x/2$ for all other values of $x$?

There are two more solution , but based on these two, allready found(for x<0 , x>0 ), since f has only one root ect.A problem with my PC does not allow me to write the functions.

Babis

hi, i think that f(x) = 0 is also an acceptable solution !

may be ,the third step is false .

f(x)=0 is not an acceptable solution. You end up with $0=-\frac{x^3}{12}$.

The point is that we need $f(x) = -x/2$ for some values of $x$ and $f(x) = -x/3$ for all other values of $x$, but of course, $f$ must be a continuous function, so there are in fact four solutions.


parameterization
algebra unsolved
algebra

Solve the following system equations with parameter a
$ \left\{\begin{array}{l} 2z(x^2 \plus{} a^2 ) \equal{} x(x^2 \plus{} 9a^2 ) \\
2y(z^2 \plus{} a^2 ) \equal{} z(z^2 \plus{} 9a^2 ) \\
2x(y^2 \plus{} a^2 ) \equal{} y(y^2 \plus{} 9a^2 ) \\
\end{array} \right.$

i am getting
$ (0,0,0),(\sqrt{7}a,\sqrt{7}a,\sqrt{7}a),(\minus{}\sqrt{7}a,\minus{}\sqrt{7}a,\minus{}\sqrt{7}a)$


A rectangle is dissected into eleven congruent pieces.Prove or disprove that each piece is also a rectangle.

Isn't it obvious? We can only make 11 congruent if we cut them either horizontally or vertically, which makes the divide pieces rectangles to themselves.

[quote="xxreddevilzxx"]Isn't it obvious? We can only make 11 congruent if we cut them either horizontally or vertically...[/quote]

That's not obvious at all to me.  I think it's intuitive, but I don't think it's obvious at all.  How do you know it couldn't involve triangles with angles of 45 degrees, say?  Also, there are some rectangles which can be divided into rectangles in other ways than simply 11 horizontal or vertical slices.

well, it couldn't be triangles, since 2 triangles make a rectangle, so you need an even number of triangles.. if they're congruent. And 11's not even.

The fact that two right triangles can be put together to make a rectangle certainly does not preclude the possibility that some number other than 2 triangles might be put together to make a rectangle.  I am interested if the problem poser has any hints for us?


geometry
incenter
circumcircle
geometry unsolved

[quote="ks1458"]ABC is a triangle with angle C = 90'  The two tangents to the incircle perpendicular to AB meet AB at P and Q. Find angle PCQ[/quote]

Let's $ O$ is incenter of of $ \triangle ABC$. Let two tangents to the incircle perpendicular to AB meet incircle of $ \triangle ABC$ at M and N(so MP and QN is tangents). Let incircle of $ \triangle ABC$ meet $ BC,AC,AB$ at $ D,E,F$ respectively.
We have that Squares $ MPFO$, $ OFQN$, $ CEOD$ is equel. So we have $ PO\equal{}QO\equal{}CO$ and $ O$ is circumcenter of $ \triangle PQC$ and $ \angle POQ\equal{}90^o$
So $ \angle PCQ\equal{} \frac{1}{2} \cdot \angle POQ\equal{}45^o$


[b]Hi

How can I get hungerford exercises of abstract algebra ??

Yara[/b]

I have the book

But I want the solutions of its exercises...

I believe problems like that are discussed here:
[url=http://www.mathlinks.ro/Forum/index.php?f=61]www.mathlinks.ro/Forum/index.php?f=61[/url]
You can try posting them.


function
algorithm
counting
distinguishability
algebra unsolved
algebra

I'll need to give some background about what computability means, so if you already know, skip to the last paragraph.

A function $f: \mathbb{N}\rightarrow \mathbb{N}$ is called "computable" if there exists a Turing machine algorithm that can take $n \in \mathbb{N}$ as an input and return $f(n)$ as an output. You can think of this most easily as that there is an algorithm (think of any modern programming language) which calculates $f(n)$ exactly in a finite number of steps (not necessarily bounded in $n$ though).

A real number $r$ is called "computable" (sorry for the repeated use of terminology, it wasn't my idea) if there exist four computable functions $S(n), N(n), D(n), E(n)$ (and $E(n) \to \infty$) so that
$\left\vert r-(-1)^{S(n)}\frac{N(n)}{D(n)}\right\vert \le 2^{-E(n)}$
for all $n$.

Important things to note:
- The above definition uniquely determines $r$ (that is, $\{S, N, D, E\}$ determines exactly one real. That's not to say another $\{S', N', D', E'\}$ can't produce the same $r$).
- There are only countably many computable functions.
- Therefore, there are only countably many computable reals.

It is also well-known that there are only countably many algebraic reals. So I ask the following question: [b]Are all algebraic numbers computable?[/b]

[quote="Xevarion"] [b]Are all algebraic numbers computable?[/b][/quote]

As soon as there is a definition of what it means to be given a specific (unique, in particular, distinguishable from its conjugates) algebraic number you will see that the answer is "yes".


Let f(x) and g(x) be two single-variable polynomials. I want to prove that there is a non-zero two-variable polynomial h(y,z) such that h(f(x),g(x))=0.

Any help?


Prove that for all $ a,b\in\mathbb{R}$ polynomial
\[ a\cdot P(x) \plus{} b\cdot P'(x)
\]
has less or equal number of complex roots than $ P(x)$.

[quote="Nemion"]Didn't you mean less or equal number of complex roots? Is a complex root one that has an imaginary component? I mean, is $ \minus{} 1$ a complex root?

PS. I know my last question may be stupid, but it's just to clarify.
My first question comes from:
What if $ b \equal{} 0$?[/quote]
Yes,sorry,It should be less or equal...
And yes to the second question.


What is the $ y$-coordinate of the image when $ (5, 3)$ is reflected over the line $ y \equal{} x$?

The image is $ (3,5)$ so the $ y$-coordinate is $ \boxed{5}$.


One liter of paint will exactly cover all sides of a cube. How many
liters are needed to cover all sides of a second cube with an edge that
is twice as long as an edge of the first cube?

If the side of a cube is doubled, the surface area is multiplied by four (because each square face is multiplied by 4). Therefore the answer is 4.


A boy has as many sisters as brothers , but each sister has only half as many sisters as  brothers .
How many brothers and sisters are there in the family ?

[hide="not really clear cgyao15..."]
Let $ b\equal{}\text{the number of boys}, g\equal{}\text{the number of girls}.$ Each boy has $ b\minus{}1$ brothers (do you see why?) and $ g$ girls, while each girl has $ b$ brothers and $ g\minus{}1$ girls.
We are also given $ b\minus{}1\equal{}g$ (because the boy has as many brothers and sisters) and $ \frac{b}{2}\equal{}g\minus{}1$ (because the girl has half as many sisters as brothers. Rearranging gives \[ b\equal{}g\plus{}1\newline b\equal{}2g\minus{}2.\]
Equating by the transitive property, we have \[ g\plus{}1\equal{}2g\minus{}2\Rightarrow g\equal{}3.\] Now we plug $ g\equal{}3$ back into either one of the equations and we get \[ b\minus{}1\equal{}3\Rightarrow b\equal{}4.\]
\[ \boxed{\text{There are 4 boys and 3 girls in the family}}.\][/hide]

oh w/e RoflLoLcOpT
but thanks for clarifying it  :)

[quote="cgyao15"]if the boy has the same # of bros as sis then
the sis will have twice as many bros as sis
we can solve as $ \ 4 bros$ and $ \ 2 sis$
but we need to add one sis so $ \ 4 bros$, $ \ 2 sis$

Hope this helped  :)[/quote]but the boy is a brother too

They are referring to how many brothers the original boy has.


Consider a $\triangle {ABC}$, with $AC \perp BC$. Consider a point $D$ on $AB$ such that $CD=k$, and the radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals. Prove that the area of $\triangle {ABC}$ is equal to $k^2$.

[quote=Satyaprakash2009rta][quote=Bourne]
radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals
$\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$
[/quote]

How did you find that??[/quote]

this follows as $r=\frac{[ABC]}{s}$ where $s$ is semiperimeter and $r$ is inradius.
apply this for both triangles and put them equal to get above relation

[quote=Bourne]My solution :
Let $AB=c,AC=b,BC=a,AD=u,BD=v$
$u+v=\sqrt{a^2+b^2}$
radius of the inscribe circles on $\triangle {ADC}$ and $\triangle {CDB}$ are equals
$\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$
$\frac{b+k}{u}=\frac{a+k}{v}$
$\frac{b^2v+a^2u}{u+v}-uv=k^2$
$b^2\frac{a+b}{a+b+2k}+a^2\frac{b+k}{a+b+2k}=k^2+\frac{(a^2+b^2)(a+k)(b+k)}{(a+b+2k)^2}$
$4k^4+4ak^3+4bk^3+2k^2ab-2a^2b^2-2kab^2-2kba^2=0$
$2(2k^2-ab)(a+k)(b+k)=0$
$\frac{ab}{2}=k^2$[/quote]

How to get $\frac{AC+CD}{AD}=\frac{BC+CD}{BD}$?


How can i prove this inequality
$ \frac{a}{\sqrt(b\plus{}c)}$+$ \frac{b}{\sqrt(a\plus{}c)}$+$ \frac{c}{\sqrt(a\plus{}b)}>2$

hello, what numbers are $ a,b,c$?
Sonnhard.

$ {a,b,c}$ are real numbers greater than zero

But Alex , there is a cas of equality ,just try for $ a \equal{} b \equal{} c \equal{} \frac {8}{9}$ for exemple


Japanese, yes. (short sticky rice) Whatever else, no. (including but not limited to any Italian long rice, Chinese dry long rice, or South-East Asian rice)

yay for rice! if you put sugar in rice, and then warm it up. its really really good

I dont think Asians are allowed to not like rice...

Well...rice pudding is kinda still rice...

a.k.a. child abuse :lol:

Rice pudding does not count, obviously, neither does rice cake but that's a whole 'nother story :lol:

Mmm, that's a bit stereotypical since you can't really grow rice in North-Eastern Russia but I'll let it fly :)

Is everyone who's posting here Asian?

me not Asian...rice is good tho' especially seasoned wild rice or that 'Lipton Rice Sides' stuff, but not the spanish kind...it's too spicy...

they cultivate rice in other regions of the world too. i have to admit i like rice, but not the steamed variety. chicken fried is preffered.

Rice grows wildly in many parts of the world, too. It's in some swamps and next to a few rivers around here (Michigan). I've never tried eating it, though. Mostly because I think it would be a lot of work to harvest it from the plants, and I'm not sure how legal it would be.

I wonder how the responses here compare to those of a "normal" school-age population.  Of course, it's biased here because a lot of people here are asian.  [hide](Like me).[/hide]

I like white rice, brown rice, rice pilaf, rice cakes, and rice pudding.  I can't think of any rice that I don't like.  Oh, and rice cakes don't really count.

i like all kinds of rice... my big sis LOVEs  rice, especially the sticky jpn kind.  we eat rice almost every dinner time.  and, im asian, so i guess thats why (mandarin chinese)

[quote name="FaThEr-RuDeSpAm" url="/community/p29022056"]
[quote=tigeryong][quote name="justJen" url="/community/p29014116"]
(This probably doesn't belong in TRT).
[/quote]

This is exactly what TRT needs! Enough with politics, abortion, lgbtq, and gun debates. We need more lighthearted stuff xD[/quote]

Then go to AtAC
[/quote]

Then go to [url=https://artofproblemsolving.com/community/c6184t29597f6184h1381086_random_spam_posts]Frivolous Forum[/url][/quote]

Your arguments never make sense.

[quote name="FaThEr-SqUiRrEl" url="/community/p29033865"]
[quote=tigeryong]Neither do yours.[/quote]

No they did actually, because someone made a post about how light hearted stuff should be in TRT and I said they shouldn't and should be in AtAC. You said I should go to the frivolous forum, which doesn't make any sense.

Also I like how AG05 is just liking every comment against me because I said celebrating her birthday doesn't belong in trt.
[/quote]

#1: Users should not spend time in forums where they do not belong, and you clearly are somebody that belongs in Frivolous Forum or Fun Factory. Note the specific thread that I sent you to in FF. [url=https://artofproblemsolving.com/community/c6184t29597f6184h1381086_random_spam_posts](^人^)[/url]

#2: If AoPS Sheriff didn't have a problem with the report that was sent about that thread, then it's fine.

#3: Reporting AG's below post is a very childish and immature way to disagree. Why not bring up the problem with her personally? The only reason I see to report instead of discuss would be because the reporter is a mentally underdeveloped {female animal of the subfamily bovidae} -ard. If the post broke AoPS rules, then it can be reported, but again: That post didn't breach ToS, nor TRT rules. So it doesn't warrant a report.

[img]https://i.ibb.co/Hgg2tZG/Screenshot-2023-10-26-155847.png[/img]

[quote name="AlienGirl05" url="/community/p29033968"]
Tigeryong, what types of rice are the most suitable for ducks? :)
[/quote]

Ducks like all types of rice, wild rice especially. While cooked rice is not harmful to them, fried or seasoned rice in large amounts should be avoided. However, small amounts like what goes into the compost would be ok.

[quote=King tigeryong][quote name="AlienGirl05" url="/community/p29033968"]
Tigeryong, what types of rice are the most suitable for ducks? :)
[/quote]

Ducks like all types of rice, wild rice especially. While cooked rice is not harmful to them, fried or seasoned rice in large amounts should be avoided. However, small amounts like what goes into the compost would be ok.[/quote]

Ah, thank you for letting me know. I think that I have a lot of wild rice in my pantry. :P

[quote=ujulee]i love the microwavable rice
or sticky rice those are soo good[/quote]

I don't think that I have tried microwavable rice before. I don't have a microwave.

[quote name="FaThEr-SqUiRrEl" url="/community/p29034741"]
[quote=tigeryong]

#1: Users should not spend time in forums where they do not belong, and you clearly are somebody that belongs in Frivolous Forum or Fun Factory. Note the specific thread that I sent you to in FF. [url=https://artofproblemsolving.com/community/c6184t29597f6184h1381086_random_spam_posts](^人^)[/url][/quote]

Source?

[quote=tigeryong] #2: If AoPS Sheriff didn't have a problem with the report that was sent about that thread, then it's fine. [/quote]

I'm not sure how that's relevant.

[quote=tigeryong] #3: Reporting AG's below post is a very childish and immature way to disagree. Why not bring up the problem with her personally? The only reason I see to report instead of discuss would be because the reporter is a mentally underdeveloped {female animal of the subfamily bovidae} -ard. If the post broke AoPS rules, then it can be reported, but again: That post didn't breach ToS, nor TRT rules. So it doesn't warrant a report. [/quote]

Very classy, calling me coward.

Also it was a post farm as the original post has to do with one's opinion on rice, not how ducks should consume rice, which should be left to a PM. That's why I reported the post.
[/quote]

Source: Common sense

How it's relevant: If you didn't notice, AG's birthday thread was reported but no action was taken. Therefore, it's fine.

Thanks for the compliment

Well, it's related to rice. Additionally, anyone with a sense of humor could see that that was a joke meant to 'stoke the fire,' as I believe the saying goes. Also, since when has AG been known to postfarm?

Either way, back to the topic

Rice:
I love sushi rice and wild rice in particular. Sometimes, I eat plain rice with salt.

i like rice
i've been eating it my entire life and its good

this thread is more on the AtAC side

[quote name="AlienGirl05" url="/community/p29034902"]
Tigeryong, stop arguing with Father-Squirrel.

Sushi rice is so delicious!!! I swear, sushi in general is so delicious!!!

Have you ever tried bomba rice? It is divine!!!
[/quote]

Nope. But I will try some soon!