conics
ellipse
geometry
calculus
derivative
rhombus
analytic geometry

It's a tossup whether to classify this problem as geometry or algebra; based on the solution, I'm comfortable with calling it algebra.

Take a square, and choose an arbitrary point on the boundary of the square, not including the vertices. Prove that there is a unique ellipse (or circle) that is tangent to each of the four sides of the square and has one of those points of tangency at the selected point.

Furthermore, show that the major and minor axes of the ellipse (if it's not a circle) lie along the diagonals of the square.

Assume wlog the vertices of the square to be $ A\equiv(0,0)$, $ B\equiv(1,0)$, $ C\equiv(1,1)$ and $ D\equiv(0,1)$, and point $ P\equiv(p,0)$ ($ p\neq0$) the given point where the ellipse is tangent at side $ AB$.  Take the equation of the ellipse to be $ x^2 \plus{} ay^2 \plus{} bxy \plus{} cx \plus{} dy \plus{} e \equal{} 0$, where obviously $ a > 0$.  Taking derivative with respect to $ x$, we find $ 2x \plus{} 2ayy' \plus{} by \plus{} bxy' \plus{} c \plus{} dy' \equal{} 0$.

The ellipse being tangent to $ AB$ at $ P$ is given by the equation of the ellipse satisfying $ x \equal{} p$ and $ y \equal{} 0$, and simultaneously $ y' \equal{} 0$, and $ p^2 \plus{} cp \plus{} e \equal{} 0$ and $ 2p \plus{} c \equal{} 0$, yielding $ c \equal{} \minus{} 2p$ and $ e \equal{} p^2$.

Assume now that the ellipse is tangent to side $ DA$ at $ S\equiv(0,s)$.  Then, $ as^2 \plus{} ds \plus{} e \equal{} 0$ while simultaneously $ 2as \plus{} d \equal{} 0$, or $ d \equal{} \minus{} 2as$ and $ e \equal{} as^2$, or $ s \equal{} \frac {p}{\sqrt {a}}$, and $ d \equal{} \minus{} 2p\sqrt {a}$.

Assume now that the ellipse will be tangent to side $ CD$ at $ R\equiv(r,1)$.  Then, $ r^2 \plus{} a \plus{} br \plus{} cr \plus{} d \plus{} e \equal{} 0$ and simultaneously $ 2r \plus{} b \plus{} c \equal{} 0$.  Then, $ b \equal{} 2(p \minus{} r)$ and $ r^2 \equal{} (\sqrt {a} \minus{} p)^2$.  If $ r \equal{} p \minus{} \sqrt {a}$, then $ b \equal{} 2\sqrt {a}$, and $ 0 \equal{} p^2 \minus{} 2p\sqrt {a} \plus{} 2a \plus{} 2p\sqrt {a} \minus{} 4a \minus{} 2p^2 \plus{} 2p\sqrt {a} \minus{} 2p\sqrt {a} \plus{} p^2 \equal{} \minus{} 2p\sqrt {a} \minus{} 2a$, absurd since $ a,p$ are both positive.  Then $ r \equal{} \sqrt {a} \minus{} p$, and $ b \equal{} 4p \minus{} 2\sqrt {a}$.

Assume finally that the ellipse will be tangent to side $ BD$ at $ Q\equiv(1,q)$.  Then, $ 1 \plus{} aq^2 \plus{} bq \plus{} c \plus{} dq \plus{} e \equal{} 0$, and simultaneously $ 2aq \plus{} b \plus{} d \equal{} 0$, or $ aq \equal{} \sqrt {a} \minus{} 2p \plus{} p\sqrt {a}$, and $ (1 \minus{} p)^2 \equal{} 1 \plus{} c \plus{} e \equal{} aq^2$, or $ a(1 \minus{} p)^2 \equal{} a \plus{} 4p^2 \plus{} ap^2 \minus{} 4p\sqrt {a} \plus{} 2ap \minus{} 4p^2\sqrt {a}$, leading to $ 0 \equal{} (\sqrt {a} \minus{} 1)(\sqrt {a} \minus{} p)$.  If $ a \equal{} p^2$, then $ r \equal{} 0$ which is absurd, so $ a \equal{} 1$, $ q \equal{} r \equal{} 1 \minus{} p$, $ s \equal{} p$, and the equation for the ellipse is $ x^2 \plus{} y^2 \plus{} (4p \minus{} 2)xy \minus{} 2p(x \plus{} y) \plus{} p^2 \equal{} 0$, and since the equation is unchanged when we interchange $ x$ and $ y$, and when we make $ x\rightarrow1 \minus{} y$, $ y\rightarrow1 \minus{} x$, then lines $ y \equal{} x$ and $ x \plus{} y \equal{} 1$ contain the axes of the ellipse, qed.

[quote="Albanian Eagle"]do affine transformations count as algebra or geometry?  :)[/quote]
I think the answer to that question is "Yes." :wink:

A a procedural matter, I found it much nicer to work in the square $ [\minus{}1,1]\times[\minus{}1,1].$

I'll post my solution in another day or two.

[quote="Kent Merryfield"][quote="Albanian Eagle"]do affine transformations count as algebra or geometry?  :)[/quote]
I think the answer to that question is "Yes." :wink:

A a procedural matter, I found it much nicer to work in the square $ [ \minus{} 1,1]\times[ \minus{} 1,1].$

I'll post my solution in another day or two.[/quote]
 :lol:

I was thinking that translating the problem to a circumscribed rhombus of a circle, gives a somewhat geometrical solution right?

It is known that if an ellipse is inscribed in a quadrilateral $ ABCD$, then the centre of the ellipse lies on the line segment joining the midpoints of diagonals $ AC$ and $ BD$.  See this post, for example:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13984[/url]

In the case of a square, the midpoints of the diagonals coincide with the centre of the square, so the centre of the ellipse must also be the centre of the square.  This should give you most of the argument.

[quote="Kent Merryfield"]One word of caution - this was at the core of a debate I had with the colleague who showed me this problem - an affine transformation when applied to an ellipse does not in general transform the major axis to the major axis.[/quote]
We don't need that I guess...
Consider a square $ ABCD$ and a point $ P$ on $ AB$, let $ \frac{|AP|}{|PB|}\equal{}k$. Suppose an ellipse touches the sides at points $ P,Q,R,S$. an affine transformation will send the ellipse to a circle. the square will be sent to $ A'B'C'D'$ which is a parallelogram. now we have $ |A'P'|\equal{}k|B'P'|$ so the quadrilateral is completely determined from k and the ellipse is unique for each value of k. to see that it exists just use the inverse tranformation for each k.

Do the major and minor axes of an ellipse always lie along the diagonals of a tangential rhombus?

First, my long-delayed answer to the original question. Assume the ellipse is internally tangent to the square $ [\minus{}1,1]\times[\minus{}1,1].$ Since the ellipse is symmetric about its center and the greatest positive value of $ x$ on the ellipse balances the greatest negative value of $ x,$ we have that the $ x$ coordinate of the center of the ellipse is $ 0.$ A similar argument for the $ y$ coordinate shows that the center of the ellipse is the origin. Hence we can write the ellipse as $ ax^2\plus{}2bxy\plus{}cy^2\equal{}1$ for some $ a,c>0.$

Set $ x\equal{}1.$ Since the ellipse and the side of the square are tangent, the resulting equation for $ y$ must have a double root. That is,

$ cy^2\plus{}2by\plus{}a\minus{}1$ must be a perfect square, hence $ b^2\equal{}c(1\minus{}a).$

Repeat the argument setting $ y\equal{}1.$ We get $ b^2\equal{}a(c\minus{}1).$

Thus $ c(a\minus{}1)\equal{}a(c\minus{}1).$ Since neither $ a$ nor $ c$ can be zero, we conclude that $ c\equal{}a.$

Now suppose the ellipse is tangent to the circle at $ (1,u).$ Then $ cy^2\plus{}2by\plus{}a\minus{}1\equal{}0$ must have $ u$ as a double root, which means that $ \minus{}\frac{b}{c}\equal{}u$ Combine that with $ b^2\equal{}a(1\minus{}a)$ and solve to find that the equation for the unique such ellipse is
\[ x^2\minus{}2uxy\plus{}y^2\equal{}1\minus{}u^2.\]
Note that the major and minor axes of this ellipse (if not a circle) lie on the diagonals of the square.

Now to address geogau's question:

Place a coordinate system on the rhombus that places the center of the rhombus at the origin and one diagonal of the rhombus along the line $ x\equal{}y.$ But the diagonals of a rhombus are perpendicular, so the other diagonal lies along the line $ x\equal{}\minus{}y.$

Find the unique affine transformation that maps this rhombus to the square $ [\minus{}1,1]\times[\\minus{}1,1]$ while sending each vertex of the rhombus to the vertex of the square that lies in the same quadrant. This affine transformation maps any ellipse tangent to the rhombus to an ellipse tangent to the square. From above, the ellipse tangent to the square must have its axes along the diagonals of the square. Now, invert that affine transformation. Since it sends the origin to the origin, it is in fact a linear transformation. Furthermore, it must have $ (1,1)$ and $ (1,\minus{}1)$ as its eigenvectors, because of what it does to the vertices. Hence the the transformation stretches the ellipse along its axes, and in so doing preserves the direction of those axes.

Note that a rhombus is quite special, and very specific properties of a rhombus were used here.

if we project the ellipse to be a circle then the circumscribed rhombus becomes a circumscribed parallelogram which has to be a rhombus  :lol:
thus we projected along one of the diagonals which has to be the major axis of the ellipse.


Given triangle $ABC$ and $P$ is the point inside it ($P$ can lie on the side) fidn the maximum of the sum
\[PA^{3}+PB^{3}+PC^{3}\]

By $f(x)=x^{3}$ is convex over $[0,\infty)$ we have answer is $\max (AB^{3}+BC^{3},BC^{3}+CA^{3},CA^{3}+AB^{3})$.

Does your result have any relation with $f(x)=x^{3}$ is convex?

Denote $g(P)=PA^{3}+PB^{3}+PC^{3}$, we'll prove that $g(P)\leq \max (g(X),g(Y))\;\forall P\in [XY]$, then we're done!.

Exist $k\in [0,1]$ such that $PM\leq kXM+(1-k)YM\; \forall M$ , therefore by $f$ is convex : $g(P)\leq k g(X)+(1-k)g(Y)\leq \max (g(X),g(Y))$.

My problem is for triangle $ABC$ it isn't for segment $XY$

Assume that $AP\cap BC=X$ then by above $g(P)\leq \max (g(A),g(X))$ and $g(X)\leq \max (g(B),g(C))$ and we're done! :P


Carla spent two-fifths of her savings on a bracelet and had $ \$ 9$ left
over. How many dollars did Carla's bracelet cost?

Since Carla spent two-fifths of her savings and had $ \$9$ left over, those $ \$9$ were three-fifths of her total savings. Thus, her total savings was $ 9 \div \frac{3}{5} \equal{} 15$. However, we wanted to find the amount of money the bracelet cost, so our final answer is $ 15 \times \frac{2}{5} \equal{} \boxed{6}$.


Given that log_(4n)40 :sqrt: 3=log_(3n)45 find n^3

If I changed the base to n on both sides, could I solve it from there?


LaTeX trigonometry limit calculus derivative

[quote="pardesi"][hide="solution"]
first some $ \text{\LaTeX}$ tips...use \cos x rather than cos x to get the more beautiful $ \cos x$
so now
$ \lim_{x \to a} \frac {\cos x - \cos a}{\cot x - \cot a} = \frac {( \cos x )'_{a}}{(\cot x)'_{a}} = \sin^{3} a$[/hide][/quote]

If you've used L'Hospital then please give a soln without using it.

Hey cant you just apply the trig formulae n get it? :maybe:

no i simply used the definition of derivatives $ \lim_{x \to a} \frac {\cos x - \cos a}{\cot x - \cot a} = \lim_{x \to a}\frac {\frac {\cos x - \cos a}{x - a}}{\frac{\cot x - \cot a }{x - a}} = {\frac {( \cos x )'_{a}}{(\cot x)'_{a}} = \sin^{3} a}$

Oops. My bad! You missed the middle step and I thought you had just differntiated wrt x.  Yeah thats the method. :) I know this sum is "Chumma Jujubi" but I just wanted to illustrate to those who blindly use L'Hospital the idea of the rule and how its derived.

And why are you hidden da?

I bet he'll say
[hide]
[color=blue][b]CHUMMA JUJUBI[/b][/color] :P :D :D
[/hide]


How many of the twelve pentominoes pictured below have at least one line of symmetry?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
[asy]unitsize(5mm);
defaultpen(linewidth(1pt));

draw(shift(2,0)*unitsquare);
draw(shift(2,1)*unitsquare);
draw(shift(2,2)*unitsquare);
draw(shift(1,2)*unitsquare);
draw(shift(0,2)*unitsquare);

draw(shift(2,4)*unitsquare);
draw(shift(2,5)*unitsquare);
draw(shift(2,6)*unitsquare);
draw(shift(1,5)*unitsquare);
draw(shift(0,5)*unitsquare);

draw(shift(4,8)*unitsquare);
draw(shift(3,8)*unitsquare);
draw(shift(2,8)*unitsquare);
draw(shift(1,8)*unitsquare);
draw(shift(0,8)*unitsquare);

draw(shift(6,8)*unitsquare);
draw(shift(7,8)*unitsquare);
draw(shift(8,8)*unitsquare);
draw(shift(9,8)*unitsquare);
draw(shift(9,9)*unitsquare);

draw(shift(6,5)*unitsquare);
draw(shift(7,5)*unitsquare);
draw(shift(8,5)*unitsquare);
draw(shift(7,6)*unitsquare);
draw(shift(7,4)*unitsquare);

draw(shift(6,1)*unitsquare);
draw(shift(7,1)*unitsquare);
draw(shift(8,1)*unitsquare);
draw(shift(6,0)*unitsquare);
draw(shift(7,2)*unitsquare);

draw(shift(11,8)*unitsquare);
draw(shift(12,8)*unitsquare);
draw(shift(13,8)*unitsquare);
draw(shift(14,8)*unitsquare);
draw(shift(13,9)*unitsquare);

draw(shift(11,5)*unitsquare);
draw(shift(12,5)*unitsquare);
draw(shift(13,5)*unitsquare);
draw(shift(11,6)*unitsquare);
draw(shift(13,4)*unitsquare);

draw(shift(11,1)*unitsquare);
draw(shift(12,1)*unitsquare);
draw(shift(13,1)*unitsquare);
draw(shift(13,2)*unitsquare);
draw(shift(14,2)*unitsquare);

draw(shift(16,8)*unitsquare);
draw(shift(17,8)*unitsquare);
draw(shift(18,8)*unitsquare);
draw(shift(17,9)*unitsquare);
draw(shift(18,9)*unitsquare);

draw(shift(16,5)*unitsquare);
draw(shift(17,6)*unitsquare);
draw(shift(18,5)*unitsquare);
draw(shift(16,6)*unitsquare);
draw(shift(18,6)*unitsquare);

draw(shift(16,0)*unitsquare);
draw(shift(17,0)*unitsquare);
draw(shift(17,1)*unitsquare);
draw(shift(18,1)*unitsquare);
draw(shift(18,2)*unitsquare);[/asy]

[quote="13375P34K43V312"]I count 6.[/quote]

As do I. Some of the drawings seem strange though.

ya i got that too..ya whats with the diagram in the middle column and bottom row..did it not come up right? or is it like that...

I remember this problem, I believe it's 3 on the bottom, 2 on the top.
And it would help if your posts weren't filled with "ya"'s.

[quote="anirudh"]I think you gotta click on it.[/quote]

Shows up the same.

k click on it and after it goes to gallery_image..click on it again
and it will be clear..

[quote="ckck"]I tried it and its even worse.[/quote]
Zoom in.

[hide]I got 6 also.[/hide]
Note: Not all lines exact

[quote=proofbycalculator]Hey everyone! I apologize for the necropost, but can't the pentomino in the 3rd row, first column become symmetric if you split the middle square vertically? I think it can also be split symmetrically if you divide the three middle layer blocks horizontally.

This is probably a very obvious from my part, and I apologize if so, but can someone explain why this isn't symmetric?[/quote]

I don't understand how you think it is symmetric. Could you please draw a diagram as to draw the lines of symmetry that you see?

edit: wait nvm I thought it was the 4th row 1st column

[quote=proofbycalculator][quote=mahaler][quote=proofbycalculator]Hey everyone! I apologize for the necropost, but can't the pentomino in the 3rd row, first column become symmetric if you split the middle square vertically? I think it can also be split symmetrically if you divide the three middle layer blocks horizontally (Referring to the image with the lines of symmetry drawn above).

This is probably a very obvious mistake from my part, and I apologize if so, but can someone explain why this isn't symmetric?[/quote]

If you're referring to rotational symmetry, I believe that this particular problem wants you to find mirror symmetry, instead.[/quote]

Oh... that's right... I forgot about that. Thanks.
Sorry again for necroposting.[/quote]

No problem. :)


[quote="manlio"]This is a picture of me :blush:[/quote]
It's great, my friend  :lol:

[quote="zaizai-hoang"]I have thought that you are older than that  :D As I known you are a teacher who is teaching Physics in a hight school? Am I wrong?[/quote]

You are right. I am 52 years old but seeing all nice solutions posted in this forum makes me younger. :lol:

Ah...great! Here comes another theorem of Inequalities.  :D (Though I hardly suspect it to be a theorem  :P )

Can anyone find the general result?  :wink:


trigonometry
geometry
circumcircle
trig identities
Law of Sines
Law of Cosines

Can someone show me some of the trig formula's, and explain how they work?  I know what sine, cosine, and tangent are, but whenever I look at some of the formula's people use for the AMC's and stuff, I think , "How the heck does this work?"

Anyways, here is what I know so far -

definitions
--------------

sin = the length of the side oposite of the angle/the length of the hypotenuse

cos = the length of the side adgecent of the angle/the length of the hypotenuse

tan = the length of the side oposite of the angle/the length of the side adjecent of the angle

formulas
-----------

sin x = cos (90-x)

This is true becuase the angles x and 90-x, and 90 are the angles of a right trianlge.  The side adgecent to one angle is also the side opposite of the other angle, when both angles aren't 90.  Since the hypotenuse will always be the same, sin x = cos (90-x).

If y mod 90 = 0
sin x = sin y+x

I don't know why this works though.  It has something to do with the unit circle, I know that.

[quote="h_s_potter2002"]If y mod 90 = 0
sin x = sin y+x[/quote]

This is *not* true.

sin 60 = sqrt(3)/2
sin 150 = 1/2

It is true (working in degrees of course) that
sin (360+x) = sin(x)
sin (180+x) = -sin(x)
sin (90+x) = cos(x)

We are going to cover all of these topics, and many, many more, in the Intermediate Complex/Trig course starting next month!

For your last question, draw the unit circle, and apply your own definition of sines. You may be able to answer your own question, by looking at the opposite leg and hypotnuse for every angle.

Here are some: [tex]\sin{2x} = 2\sin{x}\cos{x} \\ \cos{2x} = \cos^2{x} - \sin^2{x} \\ \tan{2x} = \frac{2\tan{x}}{1 - \tan^2{x}} \\ \sin^2{x} + \cos^2{x} = 1[/tex]

The last one is probably the most useful.

Even though this is basic trig it should be in the intermediate.

I suggest finding a book to explain them to you.  As always AoPS Vol. 2 is at the top of my list.  There are probably others out there however.

i agree with White_Horse_King88. the best thing to do would be to draw a unit circle and play around with some of the values. However, the best thing to do would to get the AoPS vol. 1 or 2 books, or even a high school preCal textbook. Here are some additonal formulas:

sec 2 = tan 2 + 1;
csc 2 = cot 2 + 1;

the Law of Sines:
(A / sin(a) ) = (B /  sin(b) ) = (C / sin(c) )

the Law of Cosines:
A 2 = B 2 + C 2  - 2AB(cos(c))

[quote="htaanNERd"]the Law of Cosines:
A 2 = B 2 + C 2  - 2AB(cos(c))[/quote]

This is not correct. It should read, "For a, b, and c the side lengths of a triangle, and C the angle corresponding to side C, [tex]c^2 = a^2 + b^2 - 2ab\cos(C)[/tex]." The easiest way to remember this is to remember that the side length that is on the other side of the equation corresponds to the angle that is being "cosined".

[b]Extended Law Of Sines[/b]

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2R$

Where $R$ is the circumradius of the triangle, $a$, $b$, and $c$ are the side lengths, and $A$, $B$, and $C$ are the angle measures.


Does anyone know the formula for calculating the composite score in AP statistics, and the cutoff scores for a 5 in previous years?

I have a question for people who ask questions like this:  why do you need the answer?  Are you likely to do less well if you don't know the answer?  Why not just do as well as possible?

I'm pretty sure a google search can help you answer this question. Will probably take a lot less time than asking for responses, too.

I need the answer because I would like to know how I am doing right now.
For your second question, yes.
For your third question, I AM trying to do as well as possible.

according to my ap prep book, a 68 raw score will give you a 5 (out of 100).


inequalities
advanced fields
advanced fields unsolved

Let $ Lu \equal{} \minus{} \sum_{i,j \equal{} 1}^{n}(a^{ij}u_{x_i})_{x_j} \plus{} cu$. Prove that there exists a constant $ \mu > 0$ such that the corresponding bilinear form $ B[,]$ satisfies the hypothesis of the Lax-Milgram Theorem, provided $ c(x)\geq \minus{} \mu \hspace{3mm}(x\in U)$.

You need to specify the spaces that you are using (and perhaps hypotheses on a^{ij}).  Once you do, things should follow pretty much directly from the definitions.

Hello$ \$math\$$.
This problem is from Evans(PDE).
Just use ellipticity and Poincare inequality you asked in this forum. Then it is very easy.


How does one go about becoming fluent in a language, ie they already know how to say something in any tense, use all parts of speech and such, but basically just want to learn a lot of words in a systematic way?

The best way would be to go to that country since all the other alternatives are likely to fail but the best alternative would be to go to a community/place with tons of people speaking that language then gradually you'll learn all the words. Movies and plays will do also. The trick is to [b]not look up every word you don't know in the dictionary[/b], if you do that (like my mom) you might understand the conversation but it will help you less when trying to actually learn that word. Just think of looking words up as "cheating" in a test and try to stay away from it as much as you can. Why? Because if you do then you might understand the English translation but you won't be able to "feel" what it really means like the meanings behind the words. Also the next time you hear that unknown word you'll be able to put together the contexts and figure out the definition on your own. Words learned this way stays in the brain much longer than words which were read in the dictionary because you'll have something before and after the word which can help you remember what that word means. Just my two cents as a non-native English speaker...

Just out of curiosity, what is the language?

[quote="s0mp"]euh, i have been constantly bombarded by the language since my birth, so i do watch a lot of movies and songs. the trouble is, i almost always know all the words in the movies, so i never learn that much from them. thanks tare, i usually do all those things. does it pay off to memorize huge lists of words?[/quote]
Nope, because you'll never know if they're commonly used words or not. I mean who uses "behemoth"? You'd rather be better off memorizing "enormous" or "giant" but how would you know you'd be better off memorizing those?

I'm guessing Chinese since lots of people here are Chinese.

i grew up in china, and went through two grades, so i don't have a prob...

english... for the mechanics of it... i would just take classes...
and for pronounciation... 1st, you'd have to be young(sorta), then do what somp does(watch TV)

i'm guessing some injun language? sigh i feel guilty not knowing either my mom or my dad's language. ah well.

yeh, its Hindi. okay, maybe i should rephrase my question: are there any methods with which i can use to memorize large lists of words?

Use them in sentances, repeatedly.  Translate sentances (in both directions -- if you want to be able to recognize words, translate Hindi-to-English.  If you want to be able to use them, translate English-to-Hindi.)  Use them as much as you possibly can.  Or, go for the standard take-a-list-and-commit-it-to-memory rote memorization technique.

At least, that's how I would go about it.  Which maybe isn't a good recommendation.  Heh.

thanks guys, i've started to follow some of these.

For spanish, I just talk out loud.  A lot.  Y'know, it's kinda fun to talk to yourself in spanish when you're in the supermarket or whatnot.  Helps reinforce words you already know.

For instance,
Me: No hay queso en mi cabeza.
Me: Claro que si!  Si hay queso en tu cabeza, no podria caminar, hablar, comer, o vivir.
Me: Pero queso es muy importante!
Me: Callate!  Matematicas es mas importante.  Debes estudiar para USAMO.  No debes hablar.
Me: Hmph.

yeah, i understand hindi, but i cant speak it. blah, im an "abcd"-american born confused desi

where desi is slangish for injun which is slangish for indian, as in, from india


linear algebra
matrix
linear algebra unsolved

If T is a linear transformation from R3 to R3, then prove that, there always exists a line L such that, T(L)=L.

plz give the solution in detail
Thanks in advance :)

Intermediate step: A $ 3\times 3$ matrix has an eigenvalue.

No, I'm not going to write it out in detail. Put that together yourself.


[quote="mathnerd314"]Mathnerd314 makes N sandwiches for his school's Annual Math Party. For each sandwich, he uses B globs of peanut butter, at $4$ cents per glob, and J blobs of jam, at $5$ cents per blob.

The total cost of the peanut butter and jam is $\$2.53$. B, J, and N are positive integers.

What is the total cost of the jam?[/quote]

well n's gotta be 11, 23, or 253

253 and 23 obviously don't work

n = 11, b = 2, j = 3

so he spends 1.65 on jam.
funderful

[quote="8parks11"]elaborate why it's 11, 23, or 253[/quote]

There's going to be n sandwiches, so n has to be a factor of 253. If n is 253, $1 \neq 4B+5J$. If n is $23$, $11 \neq 4B+5J$. So $n=11$, etc etc.

[quote="8parks11"]elaborate why it's 11, 23, or 253[/quote]

Well, you spend $4B+5J$ on each sandwich. There are $N$ sandwiches, so the total is $N(4B+5J)=253$.


let $ a,b,c\geq\ 0$.Prove that
$ (ab \plus{} bc \plus{} ca)(\sum \frac{1}{(a \minus{} b)^2} \minus{} (a \minus{} b)(b \minus{} c)(c \minus{} a)) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\geq\ 4$
It is not nice,i see that

[quote="Allnames"]let $ a,b,c\geq\ 0$.Prove that
$ (ab \plus{} bc \plus{} ca)(\sum \frac {1}{(a \minus{} b)^2} \minus{} (a \minus{} b)(b \minus{} c)(c \minus{} a)) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\geq\ 4$
It is not nice,i see that[/quote]
Try $ a \equal{} 2;c \equal{} 1;b \rightarrow \plus{} \propto \rightarrow$   Wrong
I think U should put a;b;c be side lenghts of a triangle .
We 'll share Allnames's problem to 2 problem :
Problem 1 : $ (ab \plus{} bc \plus{} ca)(\sum \frac {1}{(a \minus{} b)^2}) \ge\ 4$ (TST VMO 2007)
Problem 2 : $ (ab \plus{} bc \plus{} ca)(a \minus{} b)(b \minus{} c)(a\minus{}c) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\ge 0$
With Problem1: We can easy prove by the entirely mixing variable method
And Problem 2 : To suppose : c = min{a;b;c} We have :
$ (ab \plus{} bc \plus{} ca)(a \minus{} b)(b \minus{} c)(a\minus{}c) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\ge 0$
$ \leftrightarrow c^3(a \minus{} b)^2 \plus{} (a^2b \plus{} abc \minus{} b^2c)(a \minus{} c)(b \minus{} c) \ge\ 0$
$ \rightarrow$  right

[quote="Allnames"][quote="onlylove_math"][quote="Allnames"]let $ a,b,c\geq\ 0$.Prove that
$ (ab \plus{} bc \plus{} ca)(\sum \frac {1}{(a \minus{} b)^2} \minus{} (a \minus{} b)(b \minus{} c)(c \minus{} a)) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\geq\ 4$
It is not nice,i see that[/quote]
Try $ a \equal{} 2;c \equal{} 1;b \rightarrow \plus{} \propto \rightarrow$   Wrong
I think U should put a;b;c be side lenghts of a triangle .
We 'll share Allnames's problem to 2 problem :
Problem 1 : $ (ab \plus{} bc \plus{} ca)(\sum \frac {1}{(a \minus{} b)^2}) \ge\ 4$ (TST VMO 2007)
Problem 2 : $ (ab \plus{} bc \plus{} ca)(a \minus{} b)(b \minus{} c)(a \minus{} c) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\ge 0$
With Problem1: We can easy prove by the entirely mixing variable method
And Problem 2 : To suppose : c = min{a;b;c} We have :
$ (ab \plus{} bc \plus{} ca)(a \minus{} b)(b \minus{} c)(a \minus{} c) \plus{} ab^3(a \minus{} c) \plus{} bc^3(b \minus{} a) \plus{} ca^3(c \minus{} b)\ge 0$
$ \leftrightarrow c^3(a \minus{} b)^2 \plus{} (a^2b \plus{} abc \minus{} b^2c)(a \minus{} c)(b \minus{} c) \ge\ 0$
$ \rightarrow$  right[/quote]
 thank your proof is very nice,,but the problem 1 is VNMO 2008 and the proof is quite hard,can you detail it[/quote]
Oh;my friend . If Prove this Ineq ; we only need prove $ f(a;b;0) \ge\ 0$ follow entirely mixing variable
[  with $ f(a;b;c) \equal{} (ab \plus{} bc \plus{} ca)(\sum \frac {1}{(a \minus{} b)^2}) \minus{} 4$ ]


floor function
induction
combinatorics unsolved
combinatorics

On a blackboard  a positive integer $n_0$ is written. Two players, $A$ and $B$ are playing a game, which respects  the following rules:

$-$ acting alternatively per turn, each player deletes the number written on the blackboard $n_k$ and writes instead one number denoted with $n_{k+1}$ from the set $\left\{n_k-1, \dsp \left\lfloor\frac {n_k}3\right\rfloor\right\}$;

$-$ player $A$ starts first deleting $n_0$ and replacing it with $n_1\in\left\{n_0-1, \dsp \left\lfloor\frac {n_0}3\right\rfloor\right\}$;

$-$ the game ends when the number on the table is 0 - and the player who wrote it is the winner.

Find which player has a winning strategy in each of the following cases:

a) $n_0=120$;

b) $n_0=\dsp \frac {3^{2002}-1}2$;

c) $n_0=\dsp \frac{3^{2002}+1}2$.

It's not so bad.
[hide]I claim that from $\frac{3^k - 3}2$ and $\frac{3^k +1}2$, B wins, while from $\frac{3^k - 1}2$ A wins for all $k \geq 2$.
Note that from 1 or 2 A wins, while from $3 = \frac{3^2 -3}2$ B wins (A can only move to 1 or 2), from $4 = \frac{3^2 -1}2$ A wins (moving to 3) and from $5 = \frac{3^2 +1}2$ B wins (A can only move to 1 or 4).  Now, suppose that our hypothesis is correct for all $k$ up to $n - 1$.

From $\frac{3^n - 3}2$, A may only move to (1) $\frac{3^{n - 1} - 1}2$ from which B wins (as B is moving first) by induction, or (2) $\frac{3^n - 5}2$.  From this position, B may move to $\left[\frac{3^n - 5}2/3\right] = \left[\frac{3^{n - 1} - 3}2 + \frac23\right] = \frac{3^{n - 1} - 3}2$ from which B wins by induction.  So B wins in this case.
From $\frac{3^n - 1}2$, A may move to $\frac{3^n - 3}2$ and win (by above), so A wins in this case.
From $\frac{3^n +1}2$, A may move to (1) $\frac{3^n - 1}2$ from which B will win (by above) or (2) $\left[\frac{3^n +1}2/3\right] = \left[\frac{3^{n - 1} - 1}2 + \frac23\right] = \frac{3^{n - 1} - 1}2$ from which B wins, by induction.  So B wins in this case.

With this, we're done: just note that $120 = \frac{3^5 - 3}2$.[/hide]


Call the sides of the two squares $ x$ and $ y$, with $ x$ the larger one.  Then $ x^{2}\minus{}y^{2}\equal{}100$ and $ 4x\minus{}4y\equal{}10$.  Solving the second equation, $ x\equal{}2.5\plus{}y$ and plugging into the first, $ \frac{25}{4}\plus{}5y\equal{}100\Rightarrow y\equal{}\frac{75}{4}$ (I leeft out the algebra).  So the perimeter is $ 4\times\frac{75}{4}\equal{}75$.

Also, $ x^2 \minus{} y^2 \equal{} 100$ simplifies into $ (x \plus{} y)(x \minus{} y) \equal{} 100$. We were given that $ 4x \minus{} 4y \equal{} 10$, so $ x \minus{} y \equal{} \frac {10}4 \equal{} \frac 52$. Then $ x \plus{} y \equal{} \frac {100}{\frac 52} \equal{} 40$. Then $ 2x \equal{} 40 \plus{} \frac 52 \equal{} \frac {85}2$ and $ x \equal{} \frac {85}4$. It then easily follows that $ y \equal{} \frac {75}4$, and its perimeter is then $ \boxed{75}$.


The sum of the lengths of the twelve edges of a rectangular box is 140, and the distance from one corner of the box to the farthest corner is 21. The total surface area of the box is

$\text{(A)} \ 776 \qquad \text{(B)} \ 784 \qquad \text{(C)} \ 798 \qquad \text{(D)} \ 800 \qquad \text{(E)} \ 812$


ceiling function
search
combinatorics unsolved
combinatorics

Where can I find a proof of dirac's theorem?

It states that if a graph has n vertices where n is at least 3, and if every vertex has degree $ \lceil \frac{n}{2} \rceil$, then there exists a hamiltonian cycle.

where? on this very forum, most likely, given that its most elegant proof is so well-known (to combinatorialists, at least). use the Search function.

[quote="BaBaK Ghalebi"]here is an "almost algebraic" proof,there is also a pure combinatorial proof... :)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159829[/url][/quote]For the purposes of easier Search, the theorem that BaBaK Ghalebi proved is called Ore's Theorem, with Dirac's Theorem as the corollary of it.

The link below is a story written by Rons Hossberger which is interestingly a good read related to this:
[url]http://www.math.uwaterloo.ca/navigation/ideas/articles/honsberger/index.shtml[/url]

At first we will prove that $ G$ is connected.Suppose for a contrary that it has some not connected parts.At least one of this part consists from less than $ \frac {n}{2}$ vertexes.But then each vertex in it has degree less than $ \frac {n}{2}$,contradiction.Hence $ G$ is connected.Now it is easy to understand that if we have some cycle with $ k$ vertexes,then we have some path with length $ k \plus{} 1$(because $ G$ is connected).
Now suppose for the contrary that the maximum length of cycle in $ G$ is $ m$,where $ m\leq n \minus{} 1$.Consider the path with maximum length.Let it be $ A_1A_2\dots A_k$(We know that $ k\geq m \plus{} 1$).Then $ A_1$ and $ A_k$ are connected only with $ A_i$,where $ i \equal{} 2,3,\dots,k \minus{} 1$.Now divide vertexes $ A_2,A_3,\dots A_{k \minus{} 1}$ into pairs $ \{A_1,A_2\},\{A_3,A_4\},\dots$(maybe $ A_{k \minus{} 1}$ will be without pair).Then since $ k\leq n$,and degrees of $ A_1$ and $ A_k$ are not less than $ \frac {n}{2}$,there exist pair,$ A_{2t},A_{2t \plus{} 1}$,such that $ A_1$ is connected with $ A_{2t}$ or with $ A_{2t \plus{} 1}$ and $ A_{k}$ with the other vertex from the pair.But it means that there is a cycle with length $ k\geq m \plus{} 1$,contradiction.


Source: AHSME 1973 #6

If $554$ is the base $b$ representation of the square of the number whose base $b$ representation is $24$, then $b$, when written in base $10$, equals
$\text{(A)} \ 6 \qquad \text{(B)} \ 8 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 14 \qquad \text{(E)} \ 16$

[hide]
$24_{b} \rightarrow (2b+4)_{10}$

$554_{b} \rightarrow (5b^2+5b+4)_{10}$

We get the equation $(2b+4)^2=5b^2+5b+4 \rightarrow (2^2)(b+2)^2=5b^2+5b+4 \rightarrow (4)(b^2+4b+4)=5b^2+5b+4 \rightarrow b^2-11b-12=0 \rightarrow (b-12)(b+1)=0$. Since $b>0$ and $b$ can not equal $1$, we rule out $b=-1$, making $b=12\rightarrow \boxed{C}$.
[/hide]

[hide]We can rewrite the numbers so that we have

$(2b+4)^2=4b^2+16b+16=5b^2+5b+4$. Solving this equation for the positive value of b (since bases are always positive) gives us b=12. [/hide]


group theory
abstract algebra
number theory
least common multiple
superior algebra
superior algebra unsolved

[size=150]A group is decomposable if it is the direct sum of two proper subgroups.

(i) For what values of $ n$ is the group $ \mathbb{Z}_n$ decomposable?

(ii) Is $ S_3$ decomposable?

(iii) Is $ A_4$ decomposable?[/size]

1. note that $ \mathbf{Z}_k$ is a subgroup of $ \mathbf{Z}_n$ iff $ k|n$. Proof: if $ k|n$, let $ n \equal{} km$ for integer $ m$. then the elements $ m, 2m, ..., km$ form a subgroup of $ \mathbf{Z}_n$ that's isomorphic to $ \mathbf{Z}_{k}$. the converse follows from Lagrange's Theorem.

if $ n$ is composite, we can write $ n \equal{} pq$, for $ \gcd(p,q) \equal{} 1$. then $ \mathbf{Z}_n \cong \mathbf{Z}_p\oplus \mathbf{Z}_q$ because $ (1,1)$ has order $ \mbox{lcm}(p,q) \equal{} n$ in $ \mathbf{Z}_p\oplus \mathbf{Z}_q$ and $ \mathbf{Z}_p$, $ \mathbf{Z}_q$ are subgroups of $ \mathbf{Z}_n$ by the above remarks.

if $ n$ is prime, then it's easy to show that $ \mathbf{Z}_n$ is not decomposable also by the above remarks. (propriety's important!)

2. edit: see Zetax's post below

No, $ S_n$ is not decomposable. Why should the map you gave be a homomorphism¿

certainly you're right, Zetax. something possessed me to forget that a transposition does not commute with everything

[hide="3"]
$ A_4$ isn't decomposable. If it were a direct sum of $ H$ and $ K$, then $ H,K$ are normal and one of them has order divisible by 3. WLOG let $ H$ be a normal subgroup of order divisible by 3. Then $ |A_4/H|$ divides 4, so for all $ \alpha \in A_4$, we have
\[ \alpha^4 H\equal{}\left( \alpha H\right) ^4\equal{}H\Rightarrow \alpha^4 \in H\]
In particular, all 3-cycles (and there are 8 of them) are in $ H$. So $ H$ is a subgroup of $ A_4$ of order at least 8 ,which is greater than 12/2, so we must have $ H\equal{}A_4$. This means any expression of $ A_4$ as a direct sum is trivial.
[/hide]


"Pick 5 random vertices from among vertices of regular dodecagon. Prove that some 3 of them are vertices of isosceles triangle"

I already know there are only 6 possible lengths to assign to all 10 segments in pentagon, but still I've no idea how to prove that one can find 2 segments of the same lengths connected with common vertice ...  :|

w00t, a reference to my solution.

This uses the Pigeon Hole Principle, which states that if there are $ n$ holes, and more than $ n$ pigeons, then there must be at least 2 pigeons in the same hole.  In this problem, we are applying this to say that because there are 10 sides, which is more than the number of different side lengths, two of the sides must be of the same length.  Thus, choose the 3 points used to determine those sides and you will have an isosceles triangle.

Try distributing out the distances between vertices; you won't be able to distribute all 10 without putting two of the same length.

I know the principle, but those two segments of the same length might be vertically disjoint.

Here U have one of such situations (hypothetical; with only 5 colours):
[img]http://img512.imageshack.us/img512/5005/pictiy9.png[/img]

where two segments of the same colour are supposed to be the same length.
I'm aware I couldn't do this with some real lengths, but HOW TO FORMALLY PROVE IT?

Oh, ok, I see what you mean.

Consider inscribing the dodecagon in a unit circle.  Assume that you have 4 points, $ A,B,C,D$, among the 5 points such that $ \overline{AB}\parallel\overline{CD}$, and $ AB \equal{} CD$, as shown in your picture.  Thus, $ ABCD$ is a parallelogram that is cyclic, so it must be a rectangle, implying that we have something like:

[asy]pair A,B,C,D,E; A=(0,0); B=(0,2); C=(4,2); D=(4,0); E=(2,1); draw(A--B--C--D--A); draw(A--C); draw(B--D);[/asy]

I think this can be done by considering that $ m\angle AEC\in\{30,60,90\}$, where $ E$ is the intersection of the diagonals.  Then, have a little casework with each of these possibilities to show that no matter which 5th point you chose, it would be possible to choose an isosceles triangle...

Thanks b-flat (for the ideas in both topics). I must think it over now.

Maybe there is some simpler solution? We are considering the problematic
situation, when there are no 3 same-length segments, thus there have to be
4 pairs of connections of the same length. But still the same problem... How
to use it - is it useful at all? - for the formal proof?


Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \sum_{cyc}\sqrt {a^2 \minus{} ab \plus{} b^2} \leq\frac {10(a^2 \plus{} b^2 \plus{} c^2) \minus{} ab \minus{} ac \minus{} bc}{3(a \plus{} b \plus{} c)}
\]
It seems true, but now I am not able to prove this inequality.

[quote="nguoivn"]See my solution here  :)[/quote]
Sorry, but your proof is wronge :( Try $ a\equal{}\frac{5}{8},b\equal{}c\equal{}1$ (this is also an equality case in the original inequality)
I proved it by the pqr technique. :)

[quote="arqady"][quote="can_hang2007"][quote="nguoivn"]See my solution here  :)[/quote]
Sorry, but your proof is wronge :( Try $ a \equal{} \frac {5}{8},b \equal{} c \equal{} 1$ (this is also an equality case in the original inequality)
[/quote]
Yes, can_hang2007! :lol:
I  proved it too. But my proof is very ugly:
$ \frac {\sum(10a^2 \minus{} ab)}{3(a \plus{} b \plus{} c)}\geq\sqrt {\sum_{cyc}\frac {a^2 \minus{} ab \plus{} b^2}{29a \plus{} 29b \plus{} 8c}\cdot\sum_{cyc}(29a \plus{} 29b \plus{} 8c)}\geq$
$ \geq\sum_{cyc}\sqrt {a^2 \minus{} ab \plus{} b^2}.$
What is your proof, can_hang2007? Thank you![/quote]
Haha, I used CYH the same as you, arqady. After that I use pqr theorem :)

Hi, today I also went to the way above. When I want to post, I see arqady posted it before  :lol:
That's so interesting!!! (althought I have't finished the proof because it has many ugly numbers in calculus).

Sorry, for asking this :

What is CYH and pqr theorem ?

Thanks..

When does the equality holds ? I don't think it holds if and only if $ a\equal{}b\equal{}c$
Can you help me?  :(

[quote="tronghieu"]When does the equality holds ? I don't think it holds if and only if $ a \equal{} b \equal{} c$
Can you help me?  :([/quote]
$ a\equal{}b\equal{}8$ and $ c\equal{}5.$


[quote="madness"][quote="pardesi"]actually there are number of ways...oxidative coupline,Edlington Reaction,Glacier Coupling[/quote]

Its Glaser coupling i believe and not Glacier coupling  :D   :P
And btw i would rather prefer to post all these problems in their resp forums than in indian forum bcos
[hide]
Ellarum ingai veen vambhu than pesaranga and response illa  :P
[/hide][/quote]

I don't think so... Personally if my chemistry was slightly better, I would take a personal effort to answer all the chem questions posed here but I least bother. So I prefer trying the math and physics questions here only. :D


This is an old question from IMO 77(probably). I saw a solution somewhere but it seems the author wrote it in haste and it went straight over my head. Can somebody give a proper solution from the basics not using some advanced thoery. I will really appreciate your spending time on it.

Anyways, the problem is :-
\[ Find\ all\ functions\ \mathbb{N} \mapsto \mathbb{N} \ s.t.\ \\
\\ f(\ n+1 )\ > f(\ f(\ n )\ )\ \]

Yes I have read that soln. but is not totally clear to me. Can you especially explain the 2nd para. i believe that is the crux of the argument

The place where I read a slight variant of the question wanted me to get the answer through a number of steps :-

1) prove that $ \exists \ some\ n \in \mathbb{N} $ s.t. F(n) = 1

The other steps can be obtained simply from this result. Think you can establish this.

There was the 3rd part which wanted me to prove the F is monotonically increasing. I determined that establishing this result woul make all others follow in one line. CAn you establish [u]this[/u]?


search
limit
real analysis
real analysis theorems

Just say i define the fibbonacci squence with:
$f(0)=1$
$f(1)=1$
and
$f(n) = f(n-1)+f(n-1)$ for $n>1$

how would I go about finding:
$lim_{n\to \infty}\frac{f(n+1)}{f(n)}$
or
$lim_{n\to \infty}\frac{f(n)}{f(n-1)}$

It's my understanding that the 2 answers are $\frac{5^{1/2}+1}{2}$ and $\frac{5^{1/2}-1}{2}$ respectively, and I'm racking my brain to come up with a starting point but I can't come up with anything.  So I'm not looking for a solution, I just want to know how to start it or any ideas on how to start it.

There was a thread on this subject recently in one of the high school fora (Pre-Olympiad or Intermediate).  You might also try a search of the forum -- this question and closely related questions have been discussed many times.

A starting point could be as follows:

assume f(n) takes the form f(n) = r^n for some r..

f(n) = f(n-1) + f(n-2)

=> r^n = r^(n-1) + r^(n-2)

Solve for r

I think you could go further...

Here's a quick formal argument (e.g., you need to show that the limits actually exist first).

We have

$f(n)=f(n-1)+f(n-2)$, hence

\begin{eqnarray*}1 &=& \frac{f(n-1)}{f(n)}+\frac{f(n-2)}{f(n)}\\ &=&\frac{f(n-1)}{f(n)}+\frac{f(n-2)}{f(n-1)+f(n-2)}\\ &=&\frac{f(n-1)}{f(n)}+\frac{1}{\frac{f(n-1)}{f(n-2)}+1}\\ \end{eqnarray*}

Let $L$ denote $\lim_{n\to\infty}\frac{f(n-1)}{f(n)}$ and send $n$ to $\infty$ in the last line of above to find that
\[1 = L+\frac{1}{\frac{1}{L}+1}\]
whence
\[L = \frac{\sqrt{5}-1}{2}\]
This reasoning is all valid assuming that you can show that the limit exists.

This is much like the previous proof (only now it's about the reciprocal), but it's based on that $\varphi = [1,\, 1,\, \ldots]$.

$\frac{f(n)}{f(n-1)}= \frac{f(n-1)+f(n-2)}{f(n-1)}= 1+\frac{1}{\frac{f(n-1)}{f(n-2)}}= \ldots = [\underbrace{1,\, 1\, \ldots ,\, 1}_{n}]$

Hence, as $n \rightarrow+\infty$ we see that $\frac{f(n)}{f(n-1)}\rightarrow [1,\, 1,\, \ldots] = \varphi$.

BTW: Here I'm assuming the more frequent definition of the fibonacci sequence, that is: $f(n) = n$ for $n \in \{0, 1\}$.


Self explanatory...

014

btw who had that last year?

I believe that was Noah Arbesfeld.

I have 66. Good thing it's not 666.

Dude, the point of the USAMO # is that it's supposed to make you anonymous.  Posting it misses the point.

[quote="The Zuton Force"]Oh  :cursing: the graders are going to discriminate against me based on my AoPS name  :(

My bad.[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1098332#1098332

I doubt it really matters, but just sayin'...

I guess if you don't care if anyone knows your score or info, you could post your number, but it seems like the AMC advises against this.

#421... It's only been used once and its a wittle baby for USAMO numbers. Last year it got an 18. I hope I will not fail you 421...  :)

Oh, wow, my number got a 13. I'm gonna fail...

from 2002 - 2007, my number got:

4,11,11,11,17,1

hopefully, l'll be able to match the 11's or get higher, though l seriously doubt that l could pull that off...

It's between #503 and #507

What happens if someone fakes being me by using it as their number?

Oh poop, you figured out the real motivation of my thread  :(

Anyway, nr1337, thanks for pointing out that thread -- I've just seen massive threads about this in previous years

My number from this year got half the score that I got last year last year.

Numberr got <10 in 2003-2007, and 34 in 2002.
Using this info, you can probably figure out my number...

Whoa my number had scores of 22, 8, 17, 11, 2, 2. Sweet. I hope i can live up to the 22!! xD

Last year, my number got a 12...I think. I hope I match up to that, since it might be enough to get into red mop...

My number is in 270s. The previous owners got the following scores:

07: 11
06: 3
05: -
04: -
03: -
02: 13

I only hope I can get that 13...

My number got an 11 last year. I would be surprised if I got that many.

My number has had an honorable mention AND a USAMO winner!  Hopefully I'll live up to those.  :D

wow my number is horrible: 10, 2, [b]0[/b], 8, 12, 1...

Since 2002, my number has gotten 12, 16, 30, 20, 2, and 14 in no particular order. That would be great if I could match that 30 or even the 14, 16, or 20.

Where do you check past scores by number?

Since 2004, my number has not scored less than 18... a streak likely to end this year. But I'd say it's a pretty good number to have. That is of course, if you think ID numbers have to do with anything. [size=9]And they obviously do =P[/size]

Well, I'd be thrilled to beat my number's average score since 2002.

@ Zuton Force, check the links next to the Individual Scoring tab here:

http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.shtml

Mine was a winner last year :) Pretty much failed every year before that though (maybe one other double digit score). Heh well let's see how I can serve it.

lol, this officially sucks  :mad: 3,2,12,11,0,2

i got number $ abc$ where $ 0/le a/le 6$, $ 0/le b/le 9$ and $ 0/le c/le 9$  :|

17,10,10,8,7,15. Last year I got a 12, but I think I should've gotten an 8; I'm hopeful but not at all confident that I'll beat last year's score. I'm **hopeful** I'll be the highest score ever with my number (divisible by 41), but I do greatly doubt it.

(Last year my number was 515.

[quote="nr1337"][quote="The Zuton Force"]Anyway, nr1337, thanks for pointing out that thread -- I've just seen massive threads about this in previous years[/quote]

Yeah, me too.  I wonder why they warn against it though.  I mean, are there people who submit correct solutions under someone else's number?  That'd be pretty dumb.  :D[/quote]

Actually, I think the point of this is to keep the ID numbers secret from the [i]graders[/i]. If the graders know who's papers they are grading, it will be much harder for them to grade things objectively.

[quote="Brut3Forc3"][quote="JAM"]Actually, I think the point of this is to keep the ID numbers secret from the [i]graders[/i]. If the graders know who's papers they are grading, it will be much harder for them to grade things objectively.[/quote]
I agree, but are graders really going to be actively trying to find out peoples' ID numbers?[/quote]

One would hope not, but it's still best if these things aren't publicly discussed [i]too[/i] much. The more it's discussed, the more likely it becomes that one of the graders will come across it somewhere.

People can talk all they want to [i]after[/i] the USAMO's been graded. :P

I'll say it again: it is strongly recommended that you not disclose your USAMO student number.  The purpose of the number is for students to be anonymous, both for grading and for score reporting.

Sorry if I sound cheeky or stupid, but this is an honest, better ask now then feel stupid later  question:
Where did you say it before? :blush:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199615&start=20

post #25

[quote="blackbelt14253"]in the triple digits
does your USAMO student number have anything to do with how you did?[/quote]

No. It's designed to be completely random.


If  $a,b,c,$ and $d$  are real numbers and $c^2 + d^2 = (a^2 + b^2)^3$, show that $\frac{a^3}{c} + \frac{b^3}{d} \geq1$

Maybe this is not the right place for this question since it's not that easy.

[hide]Anyway, $\left(\frac{a^3}{c} + \frac{b^3}{d}\right)\left(ac + bd\right) \geq \left(a^2 + b^2\right)^2$ by Cauchy, so it suffices to show that $\left(a^2 + b^2\right)^2 \geq ac + bd$, or $\left(a^2 + b^2\right)^4 \geq (ac + bd)^2$, or $\left(a^2 + b^2\right)\left(c^2 + d^2\right) \geq (ac + bd)^2$ which follows again from Cacuhy.[/hide]

[quote="shyong"]Are you sure it is [b]real[/b] or [b]positive real[/b] ? One counter example is by letting $a=0$ , $b<0$ while $c,d>0$ and you will see the inequality doesnt hold[/quote]

I'm not too sure, I got this while surfing the net, but I think it should be +ve real too.


I'm curious about how I got my username italicized. I don't know how it happened. Do you know how this phenomenon occurred?

It's italicized under the section that says, "Who is Online"

it means its YOU, silly. like, when im logged on as me, and i see "who is online"... my name is italicized... and it'd be the same for anyone else...

Oh, I thought it was because I was a moderator. *shrugs*

I know that when I'm a moderator and someone that is hidden is in my moderating forum, his or her hame is italicized.

[quote="4everwise"]:rotfl: Haha. I'm laughing at myself now.  :rotfl:

Would somebody like to banish this thread to the trash?[/quote]
lol no this is hilarious! :lol:
We're laughing WITH you, not AT you, by the way. :P Besides, it still technically could be considered a "phenomenon." ;)


This was a question in a local math competition that I didn't know how to do. (You're allowed to use a TI calculator under TI-89). I know the answer, but I don't know how to do it.

Let $B = 3^{10/3}-(3^{8/3})(6^{1/6})+(3^{2})(6^{1/3})-(3^{4/3})(6^{1/2})+(3^{2/3})(6^{2/3})-6^{5/6}$. Find the real number $K$ such that $KB = 75$. Give the answer in exact form. Do not approximate.

Is that just something that I'm supposed to memorize? Sorry, I'm a noob when it comes to competition math.

[quote="123456789"]Sorry, I'm a noob when it comes to competition math.[/quote]

Haha same here!

[quote="123456789"]Is that just something that I'm supposed to memorize?[/quote]

No but you should recognize when to use methods like the one t0rajir0u explained.

[quote="t0rajir0u"]Taking $x = 3^{ \frac{2}{3}}, y = 6^{ \frac{1}{6}}$, we have

$\begin{eqnarray*}B &=& x^{5}-x^{4}y+x^{3}y^{2}-x^{2}y^{3}+xy^{4}-y^{5}\\ &=& \frac{x^{6}-y^{6}}{x+y}$
[/quote]

How did you get the $\frac{x^{6}-y^{6}}{x+y}$?

Is it just some rule or law? If so, what is the name of it? Thanks.

One way to think about it is as the sum of a geometric series with common ratio $-\frac{y}{x}$.

Generally, it is good to know the identity

$a^{n}-b^{n}= (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+...+b^{n-1})$

Which can also be proven by sum of a geometric series, or just by expanding the RHS.  This is a very common factorization.

Let $n = 6, a = x, b =-y$ to give the desired result.


http://www.mathlinks.ro/viewtopic.php?t=206383

Perhaps the above question (linked to another subforum) is more of an intermediate level? Check it out! =)

[quote="CATboy22"]am getting the angles of triangle DEF as 270/7, 450/7 and 540/7. Hence it is not an iscosceles triangle. Please explain :([/quote]
I think that the triangle isn't isosceles.
I'm looking for errors in my solution, but I think it's correct! Somoene could help me?

[quote="gabriballets"][hide="Solution"]Well, you know that the three bisectors all come together in the incenter of  the triangle ABC, so

$ AF = AE$
$ FB = BD$
$ DC = CE$
(tangents' theorem)

and so the triangles AFE FBD and DEC are isosceles.

if you know that
$ \displaystyle \widehat{CAB} = {{180^{\circ}}\over{7}}$

$ \displaystyle \widehat{ABC} = {{360^{\circ}}\over{7}}$

$ \displaystyle \widehat{ACB} = {{720^{\circ}}\over{7}}$
you'll obtain all the angles of those 3 triangles.

Now by knowing that the sum of the anlges of a triangles is 180° you can find FED's angles![/hide]

I hope it's correct and clear!

Ciao![/quote]

You're assuming the E, F, and D are the feet of the perpendiculars from the incenter to each side.


Find the moment of inertia of a wire in the form of a rectangle with sides 2a, 2b about an axis perpendicular to its plane at one corner.

I have the ans; $ \frac{2}{3}M(2a^2\plus{}ab\plus{}2b^2)$.

Perhaps you could point out what you did instead of just throwing out an answer.  The method is usually more important than the answer itself.

Actually I don't know how to find the find the moment of inertia. I tried find it but still I couldn't. I am very poor with applied mathematics. Pls help me to solve the moment of inertia probs.

i have some more probs to solve, so it may help me to find the answers for other questions.

thanx.

[quote="JRav"]All you really need to do is find the moments of inertia of each piece individually and then add them together.[/quote]
See figure:

We have to find the Moment of Inertia about $ c$. Hence, using the [url=http://en.wikipedia.org/wiki/Parallel_axis_theorem]Parallel Axis Theorem[/url] find M.I of $ AB, AD$ and the M.I of a thin rod about an end is $ \frac {1}{3}ML^2$.
Although you did not mention the mass of the wire, which youreally should have, that being the most important part of the problem, I assume it to be $ M$. FInd individual masses of each of $ AB, BC, CD, AD$ and proceed.


If $ m\plus{}12 \equal{} p^{a}$ and $ m\minus{}12 \equal{} p^{b}$ where $ a, b, m$ are integers and $ p$ is a prime number. Find all possible values of $ p$.

This is equivalent to finding 2 prime powers which have a difference of 24.

Case $ b \equal{} 0$: $ p^{a}\equal{} 25$, and clearly $ 5^{2}$ is the only thing that works.

Case $ b > 0$: Then $ p^{b}\plus{}24 \equal{} p^{a}$. Since $ p\mid p^{a}\minus{}p^{b}$, we also have $ p\mid 24$, so then $ p \equal{} 2$ or $ p \equal{} 3$.

Subcase $ p \equal{} 2$:
$ 24 \equal{} 2^{a}\minus{}2^{b}\equal{} 2^{b}(2^{a\minus{}b}\minus{}1)$. Since the term in parentheses is odd, $ 2^{b}$ must be $ 2^{3}$, since $ 24$ has that in its prime factorization. So $ b \equal{} 3$, and then we find that $ a \equal{} 5$.

Subcase $ p \equal{} 3$:
$ 24 \equal{} 3^{a}\minus{}3^{b}\equal{} 3^{b}(3^{a\minus{}b}\minus{}1)$. Since the term in parentheses is not divisible by 3, $ 3^{b}$ must be $ 3^{1}$, since $ 24$ has that in its prime factorization. So $ b \equal{} 1$, and then we find that $ a \equal{} 3$.

I'm not going to bother proving $ b < 0$ gives nothing.

All $ (m,a,b,p)$: $ (13,2,0,5), (20,5,3,2), (15,3,1,3)$

from your two equalities we get that:
$ p^{b}\plus{}24 \equal{} p^{a}$
so first of all we get that $ p\mid 24$ so $ p \equal{} 2,3$...
now note that if $ 2^{b}\plus{}24 \equal{} 2^{a}$ then let $ 2^{b}(2^{a\minus{}b}\minus{}1) \equal{} 24$
so first of all $ b \equal{} 3$ and then $ 2^{a\minus{}3}\minus{}1 \equal{} 3$ so $ a\minus{}3 \equal{} 2$ so $ a \equal{} 5$...
so $ m\plus{}12 \equal{} 32,m\minus{}12 \equal{} 8$ so $ m \equal{} 20$ so $ p \equal{} 2$ is an answer...
and also if $ p \equal{} 3$ then we would have:
$ 3^{b}\plus{}24 \equal{} 3^{a}$ so we would have $ 3^{b}(3^{a\minus{}b}\minus{}1) \equal{} 24$ so first of all $ b \equal{} 1$ and also $ 3^{a\minus{}1}\minus{}1 \equal{} 8$ so $ a\minus{}1 \equal{} 2$ thus $ a \equal{} 3$
so $ m\plus{}12 \equal{} 27,m\minus{}12 \equal{} 3$ so $ m \equal{} 15$ so $ p \equal{} 3$ is an answer too
so we have $ p \equal{} 2,3$...
also we know that $ a>b$ so if $ b\equal{}0$ then we would have $ a\equal{}2,p\equal{}5$ so another answer is $ p\equal{}5$...
so for integer $ a,b,m$ we would have $ p\equal{}2,3,5$ but for natural $ a,b,m$ w would have $ p\equal{}2,3$


Let $I$ be the center of the incircle of a triangle $ABC$.The line $d$ tangent to $(I)$.The lines $d_{1},d_{2},d_{3}$ though $I$ and perpendicular to $IA,IB,IC$ intersect $d$ at $A_{1},B_{1},C_{2}$,respectively.
Show that $AA_{1},BB_{1},CC_{1}$ are cocurrent.

http://www.mathlinks.ro/Forum/viewtopic.php?t=18907

  darij


Take a good look at pi on [url=http://escape.com/~paulg53/math/pi/8192.html]this page[/url]. On the 25th column, on the 7th row, you can find the number 1776. Other than it being the year the Declaration of Independence was signed, what strange thing is there about it? There can be several answers. I've posted the weird things I've found out.

ANSWER 1 (this is the big one):
[hide]The number 1776 comes EXACTLY after the 888th decimal digit of pi. 888 * 2 = 1776.[/hide]

ANSWER 2:
[hide]There are two numbers nearby 1776. This is probably obvious just because the numbers are in groups of four, but you can see 0617 and 1159 pretty close by. 1776 - 1159 = 0617[/hide]

ANSWER 3 (probably a coincidence):
[hide]Two rows down, you can see 1712. 1776 and 1712 are 68 digits apart, and 1776 - 1712 = 64. 68 - 64 = 4, and 4*4*4=64.[/hide]

I'm still looking.  :lol:

I know somewhere there's the numbers 4999 and 9998 together...

I read somewhere that in the decimal expansions of $\pi$ and $\frac{1}{\pi}$, the sequence $042$ appears in the same place...

424242 is found at the 242424th position in pi (counting the decimal point)!!!!

...According to a search-pi website (some people have way too much time on their hands)

Not significant. Given an infinite string of random digits, you can make just about anything significant sounding appear.

Of course, the digits of pi may or may not be random but they're close enough for our purposes.

This is also why numerology is such nonsense: things like this are simply not meaningful at all.

Hey! Look! The number 8,345,703,897. This just happens to be the world's population on a given day when a four-leaf clover was found, representing luck. This means tomorrow must be a lucky day.

What nonsense. You can find order anywhere you look as long as you look hard enough. It's self-imposed order. It wasn't there until you came along and said it was.

[quote="perfectnumber628"]by the way, this is a base 10 representation of $\pi$, so in other bases you'd find other odd coincidences and stuff.[/quote]

Look, there are only 0s and 1s in binary! What amazing coincidence!!!  :D  :lol:

I wasn't trying to be a numerologist. I just though it was cool.  :P

There is nothing significant at all about patterns in $\pi$, nor any other seemingly random  numbers. Take $e$ for instance. $e$ has a lot of ties to Andrew Jackson
Our 7th president
He was elected in 1828

$2. 7 1828 1828\ldots$. It's like $e$ knew when our 7th president would take office

Big deal.

In numbers like $\pi$ or $e$ you'll find everything and anything you want. By the law of large numbers every possible string of digits occurs somewhere in $\pi$. So yes, if you go out far enough, you will find a string of numbers that starts 3141592653... And if you look far enough, you can get a longer and longer string. (This does not mean that if you look far enough you get $\pi\cdot10^{-something}$. That would imply that $\pi$ is rational.)
And guess what, that same string is somewhere in every other irrational number!

It's meaningless.

What isn't meaningless is the fact such these statements are true. In other words, the law of large numbers is not meaningless, but its consequences are.

I'll leave you with an [i]extremely[/i] meaningless question. If you take $\pi$ and reverse its digits, will this string be somewhere in $\pi$?

Actually, the only proof that you can present for numology is predicting a future event correctly, which might I point out, no one has been able to do without some degree of accuracy, whether in $\pi$, or the Bible, or whatever.


calculus
integration
function
trigonometry
calculus computations

Find the function $f(x)$ which satisfies the following equation.

\[ f(x)=(e^{-x}+e^x)\cos x -2x-\int_0^x (x-t)f'(t) dt.  \]


The modern age is rigged... And the difficulty curve screwed. it goes from
caveman- impossible to defend base- pro-comp evo
spartans- just fighting- slightly early evo if no specials were used
mideval- thriving-  way early evo
renaissance- owning the midveal dudes- evo while comp is still on mideval
modern- enemy turrets and units are useless- no evo needed b/c you will win by spamming (infantry infantry support)
futuristic age- never seen it yet

uh?
why did you create a thread? That post seems as an ongoing discussion...

I've never played Age of War, although I've played similar games (Age ofEmpires, Age of Mythology)

He's creating a thread to discuss the game.

It's interesting...the turrets in the futuristic age are quite strong.

If you get an egg cannon, the caveman stage is much easier.

^Proof dragon96 is an addicting games addict & also proof he can't spell addicting correctly^

It's on a whole bunch of websites... y8, maxgames, newgrounds...
( I love google (: )

this isn't a game right? a discussion?
i love this game. its really fun

by the way, meewhee, is your avatar holding a... gun?

Yeah sorry for the unclearness

I fail- the futuristic age turrets are too strong, so win in the modern age. or else its impossible.

It's not impossible; I won in the futuristic age on hard, no less. I was actually forced to win in said age.

I had to keep sending ACS ACS ACS ACS Cyborg Cyborg Cyborg Cyborg (Clicking order; ACSes were first)to whittle down the other guy slowly, and will everything maxed except infantry. I kept sending ACS Cyborg Cyborg Cyborg Cyborg to hold them off and gain money. Also, God's wrath is useless.

EDIT: oh, and I kept using special when my ACSes got close to their base to minimize the turret fire (his ACSes hold off mine for quite a while before my guys can fire on his base).

[quote="percy314"]This game is extremely easy on "easy mode". I beat the game in less than a minute.[/quote]

it is on normal, too- unless you let them get to futuristic.

[quote="abcak"]Uh lol.

Just get a futuristic turret, save your money and you get an easy win![/quote]

yeah? On which difficulty?

Oh sorry I was probably talking about Age of War I.

I just tried Insane mode. The first time I epically failed. The second time I held out until the CPU evolved to the futuristic age when I was stuck in the Renaissance.

Wow, another one? The first one was pretty good, played insane for the first time and survived to the renaissance.

But what were they thinking? Egypt came before Sparta if I remember correctly...

On hard level, I was caveman until Renaissance, and then I kept dying so I evolved to military directly. That's why I spent a lot of time last night being tied on future stage.

@james4l: Yeah...I just checked on Wikipedia.
@dragon96: what?

EDIT: I seem to have a good strategy for Insane, but I haven't used it successfully yet (last time I tried, Iaccidentally closed the browser). Focus on infantry and turrets, with full upgrades for infantry ASAP, and upgrade the damage and range of turrets, and upgrade anti-armor since armor pwns infantry (damage bonus). Get good turrets and use specials sparingly (good usage is like if they have lots of armor onscreen).

Why infantry? Support is way better than Infantry. Five slingers can take down a single assault dino before the first one dies.

They have about...no health.EDIT: I forgot to add: and you can't upgrade their health either.

I tried support; they got slaughtered by their infantry and support. Especially the infantry, which does extra damage.

[quote="piegraph"]by the way, meewhee, is your avatar holding a... gun?[/quote]
No. It is an very abstract video camera. So smile :).
Off topic.

I, forever and always will love the teh armored.
Infantary, support, anti-armored and special are fine...
yet I won, on hard... with my army of armored. :).
Once you get to the futuristic age, make sure you stock up on the range of the turret,
those infantary and armored got good aim.
Anyone have a screen of God's Wrath in use?
If so, can you show it to me?

I don't have a screenie, but trust me, they aren't that good. THey take too long to attack, both the first time and in refire, have no range. Thus, they get mowed down when they get to the turrets.

The good thing about ACS is that they have really good health, so you can use them as a spearhead to get your troops close to the enemy base.

I won a hard game using ONLY anti-armor.

And I won again using 80% support, 20% anti-armor.

Onto insane.

Hmm, Age of War? I've played it, but I don't remember spartans in there.

But I've seen the future level. Theres one glitch. The robots still bleed...

I've never played it on hard though, only on easy :P

yes, I'm a newbie. Don't rub it in...  :noo:

edit: WHOOPS: I WAS TALKING ABOUT AGE OF WAR I. AGH.

it's fail. 4x hp, almost same dpm as the ACS, and costs 40 times of the ACS.

YES "DEFEATED" INSANE MODE!
I say "defeated" as in "put the computer into a huge stalemate that causes him to lose like 53 hp every minute and I don't want to play further"

[hide="The Idea"]Anti-Armor + Support backup mass spam. Also, easier to defend than to attack[/hide]

[hide="Start"]AA atk +2, def +1, build 2 speedy dino, wait for them to use special and finish special, build 3 more dinos, once they kill 2 dinos use meteorite, build 1 dino, then 4 slingers, then 1dino/4slinger, etc, when troop + cash runs low use meteorite. After 2-3 meteors you should have enough money to sustain 1dino/4slinger, and xp up to sparta. (they should be middle sparta atm), up Support atk and range if possible[/hide]

[hide="Sparta/Egypt/Medi Strategy"]
SPARTA:
1 AA to 6 support-ish. when enemy comes REALLY CLOSE spam more supports, they stack
Ideal upgrades: AA (2/2) and Support (2/2)
EGYPT:
Beware of priests, they kill base HP when pretty far away. Use special to dispel them, and if they mass priests get a lot more priests
MEDI:
Archer spam, minimal use of mages (only as meatshield) Archer spam is awesome. Try not to even get mage unlocking tech if possible
Ideal: AA (3/2 or 2/3) and support (3/2 or 2/3)
[/hide]

[hide="Ren/Mod Strategy"]
REN:
Quickly get cannons (black powder tech)!!! ARMAGEDDON WITH AA (3/3)!!! YOU MUST ATTACK THE BASE LIKE [b]NOW[/b]. or you will want to kill yourself for not killing the base now. Future sucks. Really.
Yeah don't bother with riflemen, though possibly get support (3/3)

MOD:
You should be able to survive the enemy now pretty well with AA. Deal some damage while you can because Future is going to be a tough tedious battle.
[/hide]

[hide="Future "strategy"]Mass Mad Scientists like no tomorrow.[/hide]

wow this game is ridiculously easy now even on INSANE

Beat computer during the Early Renaissance (and had a 1/2 age tech headstart)

EDITS: And again Medieval vs Medieval, and yet AGAIN Egyptian vs Medieval


Functional Analysis
advanced fields
advanced fields unsolved

Let $H$ be a Hilbert space, $(T_{n})$ a sequence in $B(H)$ such that $(T_{n}x)$ is convergent for every $x \in H$. Prove that given a compact operator $K \colon H \to H$, the sequence $(T_{n}K)$ is convergent in $B(H)$.

// $B(H)$ is the Banach space of all bounded linear operators on $H$ with the norm $\|T\| = \sup_{\|x\|\le1}\|Tx\|$.

[b]P.S.[/b] I managed to prove that there exists a convergent in $B(H)$ subsequence of $(T_{n}K)$. I don't know how to prove the convergence of the whole $(T_{n}K)$ yet.

Yet, it seems that the statement will remain true, if we suppose $H$ to be any reflexive Banach space.

bump..?

i tried but got stuck.. define $ W=\{x\in H: \parallel{}x\parallel{}\le 1\}$ the unit ball in H

then $ K(W)$ must be precompact.. so if we try contradiction proof

assume $ \parallel{}T_nK-TK\parallel{}\to 0$ doesnt hold, then

$ \exists (T{}_{n{}_j}K)\subset (T_nK), (x_j)\subset W, \parallel{}(T{}_{n{}_j}K-TK)x_j\parallel{}\ge \delta >0$ for all j

but letting $ y_j=Kx_j$ we know $ \exists y{}_{j{}_m} \to y \in cl(K(W))$

i wanted to say that $ \parallel{}(T{}_{n{}_j{}_m}-T)y\parallel{}\to 0$ and $ \parallel{}(T{}_{n{}_j{}_m}K-TK)x{}_{j{}_m}\parallel{}\ge \delta$ give us the contradiction, but im pretty sure its not that easy?

any help appreciated

That's ok. By the Uniform Boundedness Principle, there exists a constant $ C$ such that $ \sup \|T_n\| \le C$. So,
\begin{align*} \|T_{n_j}Kx_{n_j} - T_{n_j}Kx_{n_j}\| \equiv \|T_{n_j}y_{n_j} - Ty_{n_j}\| & \le \|T_{n_j}y - Ty\| + \|Ty - Ty_{n_j}\| + \|T_{n_j}y_{n_j} - T_{n_j}y\|. \end{align*}
Obviously, $ \|T_{n_j}y - Ty\| \to0$; $ \|Ty - Ty_{n_j}\| \to0$ because $ y_{n_j}\to y$; $ \|T_{n_j}y_{n_j} - T_{n_j}y\|\to0$ because $ \|T_{n_j}y_{n_j} - T_{n_j}y\|\le C \|y_{n_j} - y\|$ and, again, $ y_{n_j}\to y$.

So, indeed, we have a contradiction.

thanks, i feel less guilty seeing the argument rigorously done =)


$ (210)^{210} \equal{} (2 * 3 * 5 * 7)^{210} \equal{} 2^{210} \cdot 3^{210} \cdot 5^{210} \cdot 7^{210}$
Each of these factors can be in any given factor up to 211 times (remember, $ 2^0$ is a factor as well) so the number of total factors is $ (210 \plus{} 1)(210 \plus{} 1)(210 \plus{} 1)(210 \plus{} 1) \equal{} 211^{4}$

[quote="Wickedestjr"][quote="xxxyyyy"]how many positive divisors (integers) are there for $ 210^{210}$? give a simple way to deduce.[/quote]

Do you know how to find the number of factors for smaller numbers?

Do you know how to find the number of factors of (x^u) times (y^v) times (z^w).

  - It is (u + 1)(v + 1)(w + 1), which gives something very ugly when you multiply it out, and it isn't that important to remember. You just have to know to add one fo each exponent, and then multiply the exponents.

Here's another problem;

If the number (25)(81)(2^x) has 60 factors, then what is 'x'?[/quote]

I got it now.
$ {5^2}{3^4}{2^x}$ implies $ (2\plus{}1)(4\plus{}1)(x\plus{}1) \equal{} 60$,
or $ x \equal{} 3$

I believe we need to partition the orginal number in lowest terms using powers of prime numbers...else it may not work?>

[quote="xxxyyyy"][quote="Wickedestjr"][quote="xxxyyyy"]how many positive divisors (integers) are there for $ 210^{210}$? give a simple way to deduce.[/quote]

Do you know how to find the number of factors for smaller numbers?

Do you know how to find the number of factors of (x^u) times (y^v) times (z^w).

  - It is (u + 1)(v + 1)(w + 1), which gives something very ugly when you multiply it out, and it isn't that important to remember. You just have to know to add one fo each exponent, and then multiply the exponents.

Here's another problem;

If the number (25)(81)(2^x) has 60 factors, then what is 'x'?[/quote]

I got it now.
$ {5^2}{3^4}{2^x}$ implies $ (2 \plus{} 1)(4 \plus{} 1)(x \plus{} 1) \equal{} 60$,
or $ x \equal{} 3$

I believe we need to partition the orginal number in lowest terms using powers of prime numbers...else it may not work?>[/quote]

Yes it only works when using prime numbers in the factorization.

Otherwise;

(25^1)(9^2)(2^x) has 60 factors
2 times 3 times (x+1) has 60 factors
6x + 6 is 60
6x + 6 = 60
54 = 6x
x = 9, which is a lot different from the real answer

*mod's job* There's never a good reason to quote the entire post above yours.

The reason that it does not work is because 5 is certainly a divisor of the number, and the formula passes over that factor. You [i]must[/i] prime factorize it first.

I'm pretty sure that this would be classified as combinatorics. Reason: Once you prime factorize, you need to choose how many of each prime in each divisor. This problem thus turns into a combinatorics problem.

Branch off: How many even factors are there in $ 349758$? You may use calculators.

1) There was 1 problem, and now it's getting a little bit like marathon topics- they have more than 1 problem in it
2) Formula: If $ n \equal{} p_1^{a_1}\cdot p_2^{a_2}\cdot ... \cdot p_k^{a_k}$ where $ p_1,p_2,...p_k$ where this is prime factorisation of $ n$, then number of divisors of $ n$ is: $ d_n \equal{} (a_1 \plus{} 1)(a_2 \plus{} 1)...(a_k \plus{} 1)$
3) The question is answered so lock this


If $d_nd_n-1...d_0$ is the decimal representation of a prime. Prove that
$f(x)=\sum\limits_{i=0}^{n}d_ix^i$ is irreduciple over $Z$

If it were reducible over $\mathbb{Z}$ then you would have $f(x)=P(x)Q(x)$ with $deg(P),deg(Q) \geq 1$. But then $p=f(10)=P(10)Q(10)$ is prime, meaning we can assume $P(10)=1$ without loss of generality. But this means $P=1$, which is absurd.

P.S: I used the fact that the $d_{i}$ are positive.

[quote="julien_santini"]But then $p=f(10)=P(10)Q(10)$ is prime, meaning we can assume $P(10)=1$ without loss of generality. But this means $P=1$, which is absurd.[/quote]
I read this paragraph carefully but don't understand. Why do you derive $P=1$