A sequence has 15 terms, each H or T.  There are 14 pairs of adjacent terms: 2 are HH, 3 are HT, 4 are TH, and 5 are TT.  How many sequences meet these criteria?

(AIME 1986, #13)

This is a fun problem that interesting_move and I solved earlier today, using some ridiculous notation (squiggly and straight lines, up and down arrows, and so on).

Meh, and still no one wants to solve it.  Here's a hint: [hide]how many times does the sequence switch from a series of one or more H's to a series of one or more T's, and vice versa?  Where must these 'switches' occur?[/hide]

ComplexZeta wrote:And to see how well you understand the solution, try this problem. (After you solve the other one, or maybe even before, it shouldn't take more than a few seconds.)

A sequence has 16 terms, each H or T. There are 15 pairs of adjacent terms: 2 are HH, 3 are HT, 5 are TH, and 5 are TT. (Note that only one number changed.) How many sequences meet these criteria?

[hide]Alright, there are five points where the sequence switches from a series of T's to a series of H's, and only three for the other way around.  But this can't be the case - there can only be at most one more switch from T to H than from H to T.  (Try listing THTH... to see.)  So there are no possible sequences.[/hide]

Good. The other problem should be easier now that you have figured that out. Then again, you already knew how to do it.

I will apologize in advance if I awe you with my profound stupidity, but I will do my best to try.

First, one sequence that works is:

T H H H T T T T T T H T H T H

In this example, there are 7 "switch" points.

At first I considered the possibility of isolating the T T T T, or considering it as one group and then determining the number of permutations possible from that arrangement. But then I realized that switching any of the T H with H T would violate the parameters of the sequence. So ...I'll have to get back to you. Meanwhile, are you asking for distinct sequences?

It seems that still no one has answered this question (DPopov, copying from Kalva doesn't count  :lol: ).  Bump...any takers?

I did this problem a while ago; the solution that has already been posted here illustrates the idea behind the only approach that I know about. Mathfanatic, is your solution fundamentally different?

Sadly, I don't have a fundamentally different solution; however, the original purpose of posting the problem (and the reason for bumping it, as well) was to encourage others to try it as it's interesting if you've never seen it before.

On a side note, I strongly discourage people from posting solutions copied from, say Kalva, regardless of whether they are acknowledged as such (and in this case, it wasn't).  It serves no purpose - if someone really desperately wants to see the solution, they know where to look it up - and it discourages others from trying it because they see it is already 'solved'.  I've seen this phenomenon occur several times, and hope it does not continue.

*attempt*

[hide]because we have 1 more TH than HT, the final thing will look somethign like this
TH...HT...TH....HT...TH...HT...TH
the 5 TT's can be placed between any of 3 occurrences of HT...TH and in front
thus it can be seen as 3 dividers to be placed between 5 things into 4 groups
yielding 8C3 or 56 ways of placing them
similarly, the 2 HH's can be placed between any of the 3 occurrences of TH...HT and in back
yielding 5C2 or 10 ways of placing them

56*10 = 560[/hide]


linear algebra
matrix
superior algebra
superior algebra unsolved

suppose $ L$ is simpl Li algebra.which is subalgebra of Matrix algebra $ Mat(n,C)$
(with Li Bracket $ [x,y] = xy - yx$ ) suppose that $ y \in L$ is nilpotent Matrix. Show that
$ \frac {(\ad y)^{k}}{k!}x = \sum_{i + j = k}\frac {y^{i}}{i!}x\frac {( - y)^{j}}{j!}$
2) prove that $ y^*$ is nilpotent and $ \exp(\ad y)x = (\exp( y))x(\exp y)^{ - 1}$

I changed the $ y$ in the second part to $ y^*$ (adjungated matrix) as this is what was probably meant.

Hi extremal; your post is very unclear for me.
The first result implies the second one.
The second is equivalent to $ x(e^{y}\minus{}id)\equal{}0$. If $ y^{k\plus{}1}\equal{}0$,  then the relation is: $ x(y\plus{}\cdots\plus{}y^{k}/k!)\equal{}0$. I don't know what is $ x$. If for example $ L\equal{}\{A: trace(A)\equal{}0\}$, then the result is obviously false; take for example $ x\equal{}y\equal{}\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$.


Let $a,b,c$ be the lengths of a triangle whose area is $S$. Prove that \[a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}S\]
In what case does equality hold?

It has been discussed many times in Inequality forum.
For example to look here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=geq&t=14388


Find the sum of the alternating sums of all subsets of {$1,2,3,4,5,6,7$}. For example, the alternating sum of {$1,2,3,4,5,6,7$} is $7-6+5-4+3-2+1.$

[hide]
Pick one of the numbers. Call it k.

k can be in 7 positions: 4 positive, 3 negative.

there are 6! ways to arrange the other 6 numbers, so all the k's added up are 6!(4-3) = 6! k's.

6!(1+2+3+4+5+6+7) = 6!*28 = 20160.

Sorry for the lack of LaTeX...

[/hide]

[quote="Ignite168"][hide="Hint"]The answer is between 0 and 1000 non-inclusive.[/hide][/quote]

[hide="Corrected version of the hint"]The answer is an integer between 0 and 999, inclusive.[/hide]

(any sequence with a seven)+(the corresponding sequence omitting the seven)=7 and there are exactly 2^6 of these pairs (because we have 2^6 choices for the other elements...include or not include for 6 numbers)
So the answer is 2^6*7=64*7=\boxed{448}

we did these sort of problems in my 6 grade class.... but they were much easier.. :)

NEVERMIND... WHAT WE DID WAS TOTALLY DIFFERENT.. :)


[quote="catalan"]Hello everyone,

I would like to know if there is another way to solve the following problem. In class, it was solved by finding the inverse equation of the original function.

Thanks!

---

1. An arrow is shot straight up in the air with an initial velocity of 49 m/s. The height of the arrow above the ground is a function of the (variable) time: $ h(t) \equal{} 49t \minus{} 4.9t^2$.

How long did it take the arrow to reach a height of 25 m? 40 m? h meters?[/quote]
[hide]
You could replace $ h(t)$ with each height and solve for $ t$.
$ 25 \equal{} 49t \minus{} 4.9t^2$
$ 4.9t^2 \minus{} 49t \plus{} 25 \equal{} 0$
$ t \equal{} \frac {49\pm \sqrt {2401 \minus{} 490}}{9.8}$
$ t\boxed{ \equal{} \frac {245\pm 5\sqrt {1911}}{49}}$
$ 40 \equal{} 49t \minus{} 4.9t^2$
$ 4.9t^2 \minus{} 49t \plus{} 40 \equal{} 0$
$ t \equal{} \frac {49\pm \sqrt {2401 \minus{} 784}}{9.8}$
$ t\boxed{ \equal{} \frac {35\pm 5\sqrt {33}}{7}}$
$ h \equal{} 49t \minus{} 4.9t^2$
$ 4.9t^2 \minus{} 49t \plus{} h \equal{} 0$
$ t \equal{} \frac {49\pm \sqrt {2401 \minus{} 19.6h}}{9.8}$
$ t\boxed{ \equal{} \frac {245\pm 5\sqrt {2401 \minus{} 19.6h}}{49}}$
[/hide]

Thank you ProtestanT for your help.

That was the exact kind of answer for which I was looking. Thanks again!


calculus
function
trigonometry
integration
calculus computations

I'm stuck on this problem, I can't find what I did wrong, but when I do the problem on the calculator, I get a different answer than working on my own.

Calculate the average value of the function $g(x)=\sin^{4}(2x-\frac\pi6) \cos(2x-\frac\pi6)$ over the interval $[0,\frac \pi4]$

Here's what I've got:

Let $u=2x-\frac\pi6\\du=2dx$

 AVG $=\frac4\pi(\frac12)\int^\frac\pi4_{0}\sin^{4}u \cos u du$

Let $w=\sin u\\dw=\cos u du$

 AVG $=\frac4\pi(\frac12)\int^\frac\pi4_{0}w^{4}dw$
 AVG $=\frac2\pi(\frac15)w^{5}$ over the interval $[0,\frac \pi4]$
 AVG $=\frac{9\sqrt3+1}{80\pi}$

Of course I've skipped a few steps here, but if you know calculus, you know what I'm doing.  Anyway, I'd appreciate some help on this, thank you.

What happens to the limits when you substitute?

Well, I didn't type it all out, but I reverse substituted $u$ and $w$ back into the equation, so I was working in terms of $x$ again, and solved for the limits the normal way; plugging the limits into the equation, subtracting $f(a)$ from $f(b)$.

[quote="datawerd"]
Calculate the average value of the function $g(x)=\sin^{4}(2x-\frac\pi6) \cos(2x-\frac\pi6)$ over the interval $[0,\frac \pi4]$

Here's what I've got:

Let $u=2x-\frac\pi6\\du=2dx$

 AVG $=\frac4\pi(\frac12)\int^\frac\pi4_{0}\sin^{4}u \cos u du$

Let $w=\sin u\\dw=\cos u du$

 AVG $=\frac4\pi(\frac12)\int^\frac\pi4_{0}w^{4}dw$
 AVG $=\frac2\pi(\frac15)w^{5}$ over the interval $[0,\frac \pi4]$
 AVG $=\frac{9\sqrt3+1}{80\pi}$

[/quote]

You're limits are wrong.

For example, you must plug the $x$-bounds into the expression for $u$ in order to get the $u$-bounds.

So:
1) As everyone else is pointing out, it's extremely bad form to re-write the integrals with a new variable without changing the limits of integration.

2) However, you did correctly back-substitute for the variable before plugging in your limits.  Because you followed such bad form, none of the other people reading figured out what you were doing.

3) You did, in fact, correctly compute the average value of that function.  You probably failed to correctly use your calculator.


Find $\sum_{n=1}^{infinity}\frac{x^{n}}{n(n+1)}$ such that the absolute value of  x is less than 1.

This probably requires calculus, since my calculus method gives transcendental functions. Of course, they vanish when $x=1$.

[hide]
Let $f(x)$ be the series, and $g(x)=xf(x)$. We see that $g^{\prime \prime}(x)=\frac{1}{1-x}$ with $g(0)=g^\prime(0)=0$. Now we integrate twice:
\[g^\prime (x)=\int_{0}^{x}\frac{1}{1-t}\ dt=-\ln (1-x)\]
\[g(x)=\int_{0}^{x}-\ln (1-t)\ dt=x-x\ln (1-x)+\ln (1-x)\]
\[f(x)=\boxed{1-\ln (1-x)+\frac{\ln (1-x)}{x}}\]
[/hide]

[hide="Calculus as well"]Remark that\[\mathcal{S}: = \sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)}= \sum_{n=1}^{\infty}\frac{x^{n}}{n}-\frac{1}{x}\sum_{n=2}^{\infty}\frac{x^{n}}{n}\]So let's compute $\mathcal{S}': = \sum_{n=1}^{\infty}\frac{x^{n}}{n}$:  \[\sum_{n=1}^{\infty}\frac{x^{n}}{n}= \sum_{n=1}^{\infty}\int_{0}^{x}t^{n-1}dt = \int_{0}^{x}\left(\sum_{n=1}^{\infty}t^{n-1}\right)dt = \int_{0}^{x}\frac{dt}{1-t}=-\ln(1-x)\]And we conclude that \[\mathcal{S}= \mathcal{S}'-\frac{1}{x}(\mathcal{S}'-x) = 1-\ln(1-x)+\frac{1}{x}\ln(1-x)\][/hide]


Jenny's grandmother has $ 24$ cats.  Seventeen of the cats do not catch mice.  Ten of the cats have black fur.  What is the least number of cats that do not catch mice that have black fur?

Since 17 catch mice and 10 have black fur, there should be $ 17 \plus{} 10 \equal{} 27$ cats in total. However, there are only 24, so that means some must overlap. Specifically, $ 27 \minus{} 24 \equal{} \boxed{3}$.

Well, I think the answer lies in the question if you read it carefully. You know that there are 24 cats total and that 17 of those cats don't catch mice, so subtract from the total cats by those who don't and you get those who do, which is 7.

Then you consider this, "Ten of the cats have black fur. What is the least number of cats that do not catch mice that have black fur?"

There are 10 cats that have black fur. Since the question is asking specifically for those who don't catch mice that have black fur, we can find that out by subtracting from those that we do know catch mice from the group of  10 black cats and have three left over for the black cats who don't catch mice.

This is at least how I arrived at the answer. May be wrong, may be a coincidental answer, but it made sense to me.

I cannot resist the temptation. I once had a cat with black fur (the current one is more like marmalade). She did not catch mice. Does that count?

The avatar is a memento of that momentous cat.

[hide]Sorry for the rating I gave, but it is based on how my 12 year old son solved it:
If you have 17 cats that do not catch rats AND 10 with black fur AND want to keep them 'as apart as possible' so their 'overlap' is minimized, add the two and 'trim the excess'.
So 17 + 10 - 24 = 3!
I stopped short of crying when I heard him say that![/hide]

[hide=Solution] We can create a Venn diagram with circles labeled "Black Fur" and "Catches Mice". We want to find the smallest number of cats that are black but do not catch mice, so we will label this part of our Venn diagram $x$. We will also let $y$ represent the number of cats with black fur that also catch mice, and $z$ represent the number of cats that do not have black fur but do catch mice.

[asy]
size(4cm);
import labelpath;
draw((0,-0.5)--(11,-0.5)--(11,8)--(0,8)--cycle,black+1);
fill(Circle((4,4),3),purple+darkblue+white);
fill(Circle((7,4),3),yellow+orange+white);
fill(Arc((4,4),3,-60,60)--Arc((7,4),3,120,240)--cycle,purple+yellow);
draw(Circle((4,4),3),black+1);
draw(Circle((7,4),3),black+1);
labelpath("Black Fur",Arc((4,4),4,180,315));
labelpath("Catches Mice",Arc((7,4),4,230,360));
label("${\bf y}$",(5.5,4));
label("${\bf z}$",(8.3,4));
label("${\bf x}$",(2.7,4));
[/asy]

We know that there are 17 cats that don't catch mice, so there are 24-17=7 cats that do catch mice. This tells us that $y+z=7.$ There are 10 cats with black fur, so $x+y=10$. The value of $x$ will be as small as possible when the value of $y$ is as large as possible. The largest value $y$ can be is $7$, since $y+z=7$. If $y=7$, then $x$ must equal $3$.

[asy]
size(4cm);
import labelpath;
draw((0,-0.5)--(11,-0.5)--(11,8)--(0,8)--cycle,black+1);
fill(Circle((4,4),3),purple+darkblue+white);
fill(Circle((7,4),3),yellow+orange+white);
fill(Arc((4,4),3,-60,60)--Arc((7,4),3,120,240)--cycle,purple+yellow);
draw(Circle((4,4),3),black+1);
draw(Circle((7,4),3),black+1);
labelpath("Black Fur",Arc((4,4),4,180,315));
labelpath("Catches Mice",Arc((7,4),4,230,360));
label("${\bf 7}$",(5.5,4));
label("${\bf 0}$",(8.3,4));
label("${\bf 3}$",(2.7,4));
[/asy]

The smallest possible number of cats that do not catch mice and have black fur is $\fbox{3}.$[/hide]


Functional Analysis
real analysis
real analysis unsolved

Show that for every infinite-dimensional Banach space there are linear subspaces of finite codimension that are not closed.(Hint:Use Zorn's Lemma)

Let $ (E,\parallel{}\cdot\parallel{})$ the infinite dimension-space, not necessarily Banach.

By Zorn lemma, $ E$ has a basis $ (e_n)_{n\in I}$. We can suppose $ \parallel{}e_n\parallel{} \equal{} 1$.

Let $ f$ linear such that $ f(e_n) \equal{} ne_n$.

It's clear that $ f$ is not continuous because $ \sup\limits_{x\neq 0}\frac {\parallel{}f(x)\parallel{}}{\parallel{}x\parallel{}} \equal{} \plus{} \infty$.

In fact $ \ker f$ will be codim 1 and not closed.


You see that  $\zeta ^3 =1 $  so $\frac {1}{\zeta} = \zeta^2$
and can also check that $\zeta +\zeta^2 = -1$
If you meant that  $x$ and  $\zeta$ are same than the answer is
[hide="zero"]$(x^2+x)+(x+x^2)+1+1 = -1-1+1+1 = 0$[/hide]

Also, DeMoivre's Theorem is very handy here.  Especially since $\zeta$ makes an "angle" of $\pm\frac{\pi}{3}$.

[quote="Silverfalcon"]Say $\zeta$ is a complex number such that $\zeta = \frac {-1 \pm i \sqrt 3}{2}$.

Then what is the numerical value of:

$x^2+ \frac {1}{x^2} + x^4 + \frac {1}{x^4} + x^9 + \frac {1}{x^9}$?[/quote]

[hide]Note that $\zeta$ is a root of the equation $x^2 +x+1=0$ divide by x we have

$x+\frac{1}{x}=-1$. Denote $S_n=x^n+\frac{1}{x^n}$

So $S_1 = -1$, our desired sum is $S_2+S_4+S_9$.

$S_2 = (x+\frac{1}{x})^2 - 2=S_1^2 - 2 = -1$

Consider $(x^n+\frac{1}{x^n})(x+\frac{1}{x}) = S_{n+1} + S_{n-1}=S_n(-1)$.
so $S_{n+1} = -S_n - S_{n-1}$
using this recursion
S_3 = -2,
S_4 = 3, S_5 = -1, S_6 = -2, S_7=3, S_8=-1, S_9=-2. So

-1 + 3 + -2 = 0[/hide]

Hi Silverfalcon

The fact that Gyan wrote $\zeta^3=1$ is that $\zeta$ is a root of $x^2+x+1=0$ which in turn is also a root of $x^3-1=0$. Hopefully this enlightened you a bit  :)  :cool:

[quote="Silverfalcon"]I'm still bit confused.

How do you get:

$\zeta^3 = 1$ and $\zeta^2 = \frac {1}{\zeta}$?

P.S. Yeah, $\zeta$ means $x$. :D[/quote]
Yes as 3X. said,   $x = (\frac {-1+\sqrt 3 i}{2})$  is  one of the cube-root of $1$  , You can directly see by calculating $x^3$ to see that indeed $x^3 = 1$ ( Go ahead and check it by doing actual calculation)
[hide]If   $x =   (\frac {-1+\sqrt 3 i}{2})$
      $x^2 =  (\frac {-1-\sqrt 3 i}{2})$    (If nothing else just check it by multiplying x by x)  (BTW it is also = ($1/x$)
     $x^3 = 1 $

(You can also check that $x+x^2 = -1 $   by direct substitution)
[/hide]
[hide](You may also  try to take the roots of  $x^3-1 = (x-1)(x^2+x+1) = 0 $ Check that above are two roots
of $x^2+x+1$  hence  $x+x^2 = -1$ and $x^3 = 1$ - This is what 3X was pointing out...)
[/hide]
  This $x$ (Normally people call it  $\omega$ ) is an expresson you may likely to  bump into  often,  so  you may like to know it  a little better...:).... (  For example if you want to know $x^7$ it is $x$ because  $x^3=1, x^6=1 $ etc..

Sorry for being stupid but what happened to $x^2+\frac {1}{x^2}$?

We're talking about $x^3$... Got that.. But what happened to those?? :(

Thanks a lot!

I get it now :D I should take Complex Number/Trig during the summer to learn it better :lol:

Even I take trig in my school now.. :)


[quote="mr. math"]If a team of four can do 10 problems in 15 minutes, how long does it take one person to do one problem in terms of seconds?[/quote]
[hide]Assume WLOTMG that the people have equal skill level.
1 person 10 problems 60 minutes.
1 person 1 problem 6 minutes.[/hide]

that is not correct.

Read the last word.

That's usually the reason that I miss questions on tests so I probably would have missed that too.
[hide="BTW,"]6 minutes = 360 seconds[/hide]

[hide="The longer solution"]$4$ $\textrm{people}$ $\rightarrow$ $10$ $\textrm{problems}$ $\rightarrow$ $15$ $\textrm{minutes}$
$4$ $\textrm{people}$ $\rightarrow$ $2$ $\textrm{problems}$ $\rightarrow$ $3$ $\textrm{minutes}$
$4$ $\textrm{people}$ $\rightarrow$ $2$ $\textrm{problems}$ $\rightarrow$ $180$ $\textrm{seconds}$
$4$ $\textrm{people}$ $\rightarrow$  $1$ $\textrm{problem}$ $\rightarrow$ $90$ $\textrm{seconds}$
$1$ $\textrm{person}$ $\rightarrow$ $1$ $\textrm{problem}$ $\rightarrow$ $360$ $\textrm{seconds}$[/hide]


Dear fellow LaTeXers,

I was wondering if it is possible to have a poll on which is more beautiful: the default LaTeX font, or horrible times new roman?? You might have guessed what I think on the subject.

My boss seems to think that 'times' is much, much better. I can only vehemently disagree.

The Times font was invented (by the newspaper of the same name) to squeeze as many words into one line as possible. Surely, there was a good reason why the original LaTeX/TeX opted for a far superior font???

Please help: I need ammo to turn the dark ways of my boss.

Cheers,
Yet another LaTeX lover.
 ;)

I have no preference in most cases, although I love Tempus Sans ITC.

Well, that depends on what you are doing. Any way, I usually use the default fonts!

Elaborating on what djimenez said, Times New Roman is useful when I want to fit as many words into a line as possible. Also, there's gotta be a reason why it's a default in several word processors.

There's nothing wrong with Times New Roman, it's a bit boring but I rarely see the point of changing default fonts in any application, whatever those fonts may be. When I do see the point I almost always head for Comic Sans. But the ugliest font in the universe can't obscure good content, providing it leaves said content legible.

I fully agree with you Complex Zeta! The beauty of the mathematics should be payed respect by the beauty of the presentation. Otherwise, we'd all be using Microsoft bloody Word (curse, spit, vomit!)!

If someone cannot be bothered to format their work elegantly, then the chances are that their treatise is just as bad.

The very fact that Microsoft favours times should be a good enough reason to avoid it!

YetAnotherLaTeXLover

What I don't understand is why some recent books have been written (in $\LaTeX$) in fonts that just look much worse than the default font.

Oh. Well, I've never seen a book like that, but I doubt the font used was Times New Roman. Times is 100% legible, not hard on the eyes, and not inelegant...I don't know where you got that idea LatexLover. I've never seen an application with a bad default font, so I stand by my statement that there's rarely a reason to change it. Especially when it's apparently possible to cause the opposite effect than the one intended. By the way...what [i]is[/i] the default font in LaTeX? And how do you change it?

I think the default font is called Computer Modern or maybe Computer Modern Roman. (I saw it abbreviated as CMR before.) You can change it by using some other font package. I wouldn't recommend it though. With a few exceptions (for example, I think that both $\epsilon$ and $\varepsilon$ leave much to be desired), the $\LaTeX$ default font is good.

Lynelle,

before you fall in love with the Times font, try it out and think again! On principal, I simply refuse to tell you how to switch to ugly Times. These 'other' packages, with Times as default that you speak of, can only imply that you are a Word lover, or at best, a Word user. In this case, you probably also light candles at midnight, sacrifice small defenceless mammals, and chant 'I love Bill Gates, death to all talented Mathematicians'.

Then again, you could simply be insane, drunk, or perhaps even insanely drunk....

;-)

eeee....

We're just talking about fonts, right?  This is getting a bit extrem.  As long as you are confortable with the font you use, thats's great.

Just don't tell people they're insane because this is what you think of the font they use....

Personnal opinion

ledurt

I wouldn't bother changing the LaTeX font anyway...I was just curious. And I never said I liked Times much. Like I said before, it's a bit boring. But there are much worse...and if the document's meant to be printed out, I [i]would[/i] prefer a font that's meant to be crammed into a small space. It saves paper. It saves trees.

A Theorem for Lynelle:-

Theorem 1.
Computer modern is better than crappy Times, any day of the bloody week.

Proof,

Recall Lynelle's previous reply. We proceed thus.

OK, Lynelle, well, maybe there's hope for you yet, then. Forget about Times, and if your a tree-hugger, simply use recyclable paper.

Note, the more recyclable paper you use, the stronger the recyclable paper companies become, and the more trees that are saved.

Ergo, computer modern is BETTER, ...

... and that, as they say, completes the proof.

Cheers,
LaTeXLover

[quote="ComplexZeta"]Sure it can. Have you ever opened an old math book in a really ugly font with all the math symbols in handwriting or something like that? (If you're interested, I can give you titles of a few on my shelf.) It makes me want to close them immediately. Occasionally I have sufficient willpower to get past the dreadful font and look at the content of the book, but it's really hard.[/quote]

I'm willing to bet that a lot of the reason ACOPS is so good is that it has an incredibly slick look and feel.

If you want something that looks pretty bad, by the first volume of Arbelos. It's typewritten, then horribly photocopied.

My dear, poor, uninformed Ledert,

Mathematics is not just The ultimate form of reasoning, it is also The language of greatest beauty. To render that beauty in a less than pristine fashion is akin to farting loudly in Church (although I am an atheist, I always try to respect this rule).

Just a question of fonts, you say. And I suppose next, you'll tell me that Da Vinci should have painted with inferior brushes, cheap paints, and spent most of his time watching 'American football' whilst drinking putrid, sickly sweet, American beer?

Strangely, the word 'philistine' comes forth unto ones mind...

It's one thing to have a favorite font, but it's quite another to be rude to people who disagree with your opinion. Please stop it.

Techy, techy! Hey, I thought I was being very restrained

I agree ComplexZeta. That's what I wanted to say in my previous post.

ledurt

I recall that Ahlfors is particularly badly formatted; every time I read a sigma sign I want to tear that page out of the book.

True, $\sum$ in Ahlfors is not so nice. But I think his $\wp$ is worse.


Fie $A \in M_2(R)$, astfel incat $det(A^4+4I_2)=0$.Sa se arate ca $|trA|=|detA|$.

I don't see how it could be :?
We can factorize $A^4+4I_4$ to $(A-\epsilon_1 I)(A-\epsilon_2 I)(A-\epsilon_3 I)(A-\epsilon_4 I)$, where $\epsilon_i$ are 4-th roots from $-4$. Therefore $\det (A-\epsilon_i I)=0$ for some $i$.

Si cum spuneam, aceeasi vesnica idee. Una din valorile proprii ale lui A trebuie sa fie unul din numerele $ 1+i, 1-i, -1-i, -1+i$ iar cealalta conjugatul ei (polinomul caracteristic are radacini reale). Si e gata!

I am continuing.
Note that $\epsilon=\pm 1\pm i$.
Write $\det(A-\epsilon I)=0$. We see that $ac-bd=(a+c)\epsilon-\epsilon^2$. Since $(a+c)\epsilon+\epsilon^2\in \mathbb{R}$ and $\epsilon^2=\pm 2i$ we conclude $|a+c|\cdot|Im\epsilon|=2$,  so $|a+c|\cdot 1=2$, i.e. $|a+c|=2$, moreover $|ac-bd|=|Re((a+c)\epsilon+\epsilon^2)|=|a+c|\cdot|Re \epsilon|=|a+c|=2$.

[quote="harazi"]Si cum spuneam, aceeasi vesnica idee:[/quote] :)
e tare faza asta.. dar.. parerea mea ca dak stii solutia instantaneu sa nu o pui una doua... ca nah... nu prea are rost... eventual dak nu primeste solutie dupa mai mult timp... in felul asta e mai usor pt cei care le deschid sa le incerce.. dak vad si solutia e f tentant sa te uiti peste ea.. si dupaia... nu mai are nici un farmec :| (cu toate ca.. pb nu are prea mult farmec in sine :) ) - just an opinion - sper sa nu te superi..

Nici o problema, de obicei nu prea intervin in acest subforum, dar genul ala de problema chiar ma enerveaza si simteam nevoia sa spun cuiva asta. Scuze si, intr-adevar, ai dreptate.


What math awards are available for smart, hard working mathmaticians? Why isnt there a math Noble Peace Prize?  I think mathmaticians should be paid more or rewarded better.  They work harder than anyone and there arent many high-paying career options for people with such talent and dedication for there work.  What kind of jobs do mathmaticians have?

The Fields Medal is a prestegious math award.  Heh, read the first sentence of this link -- its the first site that you get by typing "fields medal" into google: [url]http://www-gap.dcs.st-and.ac.uk/~history/Societies/FieldsMedal.html[/url]

[quote="SirErnest"]What math awards are available for smart, hard working mathmaticians? Why isnt there a math Noble Peace Prize?[/quote]

Um, it would be called the Nobel Prize in mathematics, if it existed. The famous prize for mathematicians is the [url=http://www-gap.dcs.st-and.ac.uk/~history/Societies/FieldsMedal.html]Fields Medal[/url], which has been [url=http://144.16.74.145/PEOPLE/halloffame.html]won by a few IMO medalists[/url]. A new mathematics prize is the [url=http://www.claymath.org/millennium/Rules_etc/]Clay Millennium Prize[/url],  which involves solving one of the famous unsolved problems that are mentioned here on the AoPS forum to name our level of posts (e.g, Riemann Hypothesis, Poincare Conjecture). It would be a BIG DEAL to solve one of those problems, and the person who solved one of them would, um, surely be able to make tenure at an accredited university somewhere, not to mention get a million dollar prize from the Clay Mathematics Institute.

The best reason to do math intensely is that it provides personal satisfaction. It can also pay the rent--it doesn't usually involve any heavy lifting or other risk of personal injury. There was a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9718]math careers thread[/url] earlier on the College Forum about what kinds of jobs math-likers end up with in adult real life.

Have fun. Good luck in achieving your dreams.

[quote]Why isnt there a math Noble Peace Prize? [/quote]

Well, it wouldn't be a PEACE prize anyways. But: Alfred Nobel's wife (or mistress or something) had an affair with a mathematician, Nobel was understandably pretty pissed about this.  So when he set up the whole Nobel Committee thing, he chose to leave out the field of mathematics.  I think i remember the story correctly.[/quote]

[quote="**********"]Well, it wouldn't be a PEACE prize anyways. But: Alfred Nobel's wife (or mistress or something) had an affair with a mathematician, Nobel was understandably pretty pissed about this.  So when he set up the whole Nobel Committee thing, he chose to leave out the field of mathematics.  I think i remember the story correctly.[/quote]

You remember the story correctly, but the story is an [url=http://www.snopes.com/science/nobel.htm]urban legend[/url] (that is, a false story). It's such a spicy story that one almost wishes it were true, and the story is certainly widely repeated, but it is most definitely false. The page I just referenced in my link has a good discussion of plausible reasons why Nobel didn't endow a prize in mathematics.

At any rate, I don't think Nobel and Mittag-Leffler got on very well.

The Clay Mathematics Prizes are the biggest of a largish number of prizes open for solving particular questions.  Usually, what happens is that a mathematician or someone else comes across a question which he would like solved or which he things is interesting, but which he for some reason can't or doesn't have the time to solve.  He then offers a small prize (usually less than $100) for the solution to the question.  Mathematicians who solve these problems often won't deposit the checks but will instead frame or otherwise save them.

Who did something like that other than Erdos? Interestingly, Erdos once offered $1010! (yes, ten to the ten factorial) for the solution to a problem that he thought would never be solved. Unfortunately for him, it was solved a few years later, but the person who solved it didn't ask for any money.


[quote="Interval"]The sides of a triangle are represented by the expressions
   2x2 + 4,    5x2 – 6x,    and    8x + 3.
What is the simplest expression for the perimeter of the triangle?[/quote]

I can't tell which is $x$ or $\cdot$. So I'll try my best. The side lengths are $8,10-6x,$ and $8x+3$. So adding up we have $21+2x$.

[quote="Interval"]By the way, the correct expression is:

2x^2 + 4, 5x^2 – 6x, and 8x + 3.

Now, what is the sum?[/quote]
Okay then.
[hide]Adding up like terms we get $7x^{2}+2x+7$.[/hide]


function
inequalities
inequalities solved

for x,y from (0,1) we need the following result:
x^y>x/(x+y-xy).
Let f : (0,1]-->R,f(x)=(x+y-xy)*x^(y-1)-1.We have:
f'(x)=(1-y)*x^(y-1)+(x+y-xy)*(y-1)*x^(y-2)=y*(x-1)*(1-y)*x^(y-2)<=0,
because x-1<=0. So we have that f strictly decreases on (0,1] so
x<1 implies that f(x)>f(1)=0.
So we wil have that:
x^y+y^x > x/(x+y-xy) + y/(x+y-xy)=(x+y)/(x+y-xy)>1.

cheers! :D

I can prove in the same conditions that x^y+y^x>=1+xy using Bernoulli.

[quote="alekk"]Prove that if  $0<x,y<1$ then $x^y+y^x>1$

[quote="cezar lupu"]I can prove in the same conditions that $x^y+y^x\ge 1+xy\/$ using Bernoulli.[/quote][/quote]
My have a solution using Bernoulli inequality, but I think not the same with cezar lupu. So just share the idea  :)  :)  :)
Bernoulli inequality state:
$(1+x)^n \ge 1+nx \/$ for $x\ge -1 \neq 0\/$ and $n>1\/$ an integer.
But in fact, the above inequality is also true for [b]real[/b] $n>1\/$.
We make use of this: $(1+x)^n \ge 1+nx > nx \/$ for $x\ge -1 \neq 0\/$ and real $n>1\/$.

Since $0<x,y<1$, we have $\left(1+\frac{x}{y} \right)^\frac{1}{x} > \frac{1}{y}\/$.
This equivalent to $\frac{x+y}{y} > \frac{1}{y^x}\/$, and hence $y^x > \frac{y}{x+y}\/$.
Similarly, $x^y > \frac{x}{x+y}\/$. Then it is easy to see that  $x^y+y^x>1$.


[quote="chien than"]For $x;y;z$ are positive real numbers satisfying $x+\sqrt{xy}+\sqrt[3]{xyz}=\frac{4}{3}$
Find the least value of the expression $A=a+y+z$[/quote]
What is $a$?


FTW LaTeX trigonometry Asymptote blogs Support

Hi Mathlinkers/AoPS'ers:

   Do any of you do web design or programming as a hobby or some kind of job here? If you are one, talk about your experience a little in web design. If you do programming, tell us what script language/platform you use.

I program a bit for fun.
I am very familiar with Visual Basic and a bit familiar with C.
I'm not learning any programming languages currently, and only program for fun when there is a purpose or a task.

I don't have trouble with HTML either...but it's not as fun.

I've been programming for quite a while. I use Java most often, but have used many other languages (Lisp/Scheme being my favourites). I'm not interested in web programming though.

HTML FTW

i know a bit of javascript as well. I don't really program stuff though.

I know

-HTML (getting rusty after lack of practice)
-BASIC (uhh I think I'm still okay with that)

Hi Californian. I am 16 and it's 9 years that I'm a professional user but until 2 years ago I didn't understand why people were being busy with programming or something like that and I cannot imagine I would be a programmist but I became! It's already one years that I'm fagging at ms.net. And almost I'm writing an applicative mathematics book on it. You can get advices whenever you want... turalinki@yahoo.com
From Tehran

I can manipulate the TI-84 BASIC program lang.

TI-BASIC -- fun for when in school
C++ -- use it a lot
HTML/CSS/JAVASCRIPT/PHP -- learning all these right now.

I can only do simple things with Javascript and PHP but I'm pretty knowledgeable about HTML and CSS.

Haha for BASIC I just program shorthands for long and tedious formulas and when my peers have to write it all out I just punch it into my calculator and I finish with 15 minutes of spare class time  :D .

I do a lot of html. but hey wassup hunter34, i haven't seen u in a long time. Ever since school started I've had no time to go on these forums

Meh, I can program in any non-really-old language you could possibly think of. (i.e. FORTRAN and COBOL qualify as really old) Except Python and ASP. I hate those.

The languages I'm most adept in are probably CSS, PHP, XHTML (web-related stuff in general, actually) and assembly languages, though C++ could be added to that list. I'm all right in LaTeX, I suppose, and not bad in Asymptote. I hate BASIC, sine it lets you fall into sloppy coding habits. TI-BASIC is pretty awesome, though.

Bleh...
I went to computer programming camp for a little while and learned C++

However, i've forgotten most of it now (already..) due to lack of practice :no:

anyways, i guess i could say beginner in C++ heh  :P

Eh, I'm 'fluent' as it were in C++, C# (awesome language, dare I say it), PHP/SQL (another one of those awesome languages), and XHTML/CSS/Javascript which, I suppose, are separate languages but they're all so closely tied I put them under one umbrella. I've also dabbled in LISP, which is an interesting take on things no matter who you are. All of these are self taught, the internet provides more than enough resources for the astute. And obviously I've got experience with $ \text{\LaTeX}$ and Asymptote, but, well, so what?

So in PHP I've worked with MySQL and other whizzes. Particular projects? Made a chat script for phpBB with a friend that ended up getting modified and used on Steve Vai's site, who is - I'm told - some famous guitarist bloke. So famous I've never heard of him, apparently! Did some stuff with the then up and-coming AJAX to make a kind of strange sticky system. To be honest nothing really that big. I've done a lot (too much) web design, however, wielding Photoshop like a mighty sword. I must've gone through twenty bloody blog redesigns before I got bored of it. Used to own [url=http://kaniaz.com]kaniaz.com[/url] to host these things, but I got bored, like I said, and some crappy spammer snapped it up when it expired.

As for C++ and C#, I did some Windows development once upon a time. XNA is a fantastic framework if you want to get into 3D game programming, and there's quite a lot of maths involved if you know where to look, which is always a nice thing to have. C++ is a much slower language (to program in, not to run) and not really that beneficial for my particular needs, so if you did want to get into these kind of things I would start with C# which has such a lovely object oriented architecture I sometimes want to marry it. In both C# and C++ I developed a rudimentary 3D game with DirectX. You could connect to the server (which I also coded), create an account with a password/username combo. I had got as far as the GUI. D'you know, I rather wish I had kept the source code to that.

I am now a Mac user, and much prefer the platform for a whole multitude of technical and usability reasons. Looking forward to the Leopard release, too. I've restrained myself from cracking open XCode and doing something again, although it might be fun to spend some time in that fashion. I'm a bit put off by the lack of decent C# support like Windows so openly touts, and is so much the better for.

As a result of this I've had extensive experience with subversion for code repository control. If you want to get serious about programming, you should look at SVN. Not CVS, unless you're a lunatic.

All that said, I'm quite bored of programming now. I don't think it's what I want to do. It's not half as intellectually challenging, I find, as mathematics, and to be honest I don't find it quite as rewarding either. Maybe in ten years you'll all be rolling in the dough and I'll be made to eat my hat, but, well, I'm willing to take that risk. I'm not here for the money anyway.

[b]EDIT:[/b] Holy crap, didn't mean to write an essay.

VB.NET  :D
Excellent in programming small games and applications
I attempted to program some "great" games by using C++ and DirectX, but in vain :(
It is a nightmare to trace which object is not released in C++. This job is, in most cases, automatically done when using VB.


Show that there is real $\alpha$  s.t
                         $\frac13\leq {\alpha^n}\leq \frac{2}{3}$
holds for all positive integers n. (where  {x} is fractional part of x)
to view all the Some problems from the Uzbekistan universities olympiads visit
[url]http://www.mathlinks.ro/Forum/topic-81099.html[/url]

already posted, try to search. It uses the fact that if $I_1$, $I_2$,..,$I_k$,.. is a decreasing sequence of compact subsets, then their intersection is non empty.

I think you meant
[quote="Orif"]Prove that there is a real number $\displaystyle \alpha$ such that \[ \displaystyle \frac13 \leq \left\{ \alpha^n \right\} \leq \frac23 , \, \forall n \in \mathbb N^\ast . \][/quote]

By the way, I found [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29550]the solution Alekk was talking about[/url].


Hello and congratulations to you all for your successful performance at your team selection test.

The list of guides and volunteers for this IMO has been announced; most of them are university students. Most of them are pretty and friendly female students. Each IMO team will own at least one of such nice girls  :lol:

And here is the hotel where contestants and Deputy leaders will be in.

http://www.hanoihotel.net/home/La_Thanh_Hotel_Hanoi_Vietnam.asp

I will be with you during this IMO.

For a great IMO.

I want to know if the organizers will make the photos and the names of the contestants like the last IMO in slovenia.

[quote="pvthuan"]

The list of guides and volunteers for this IMO has been announced;

[/quote]

Where could we find that list Thuan?

[quote="pvthuan"]Tiks, it is only in my university. See you all there. I will spend a lot of time with contestants and deputy leaders, and obersvers during your stay in Vietnam.[/quote]

OK :) ,see you there!!!! :wink:

[quote="hungkhtn"]He he, one of my girl friends in high School is the guider of IMO this year  :lol: So you may be happy to see her   :blush:[/quote]

You mean "girl friend" or just a female friend? :)
Anyway, you sound like a new Casanova :D (joke :) ).

[quote="hungkhtn"]Hi, female friend, of course, not lover  :blush: And Ticks, I am not joking. She is my old classmate. She is beautiful but very  shrewish  :blush:  unfortunate for who is guided by her ,[/quote]

I thought that you meant it that way,that's why I asked :wink: .
By the way, my nick-name is $Tiks$ and not "Ticks"!


solve in positive integers

aa+5bb=pp
aa-5bb=qq

as far as i know, there is at least one solution

[color=cyan]are we supposed to take p and q as primes, as well?

did you mean a^a, b^b, etc?[/color]

Hi,

This problem first appeared on imo-problem, then on mathlinks. I've worked on it a lot (without success  :( ).

I showed that :

. 12 divides b
. 7 divides q

I'm convinced that (a,b,p,q) = (41*k, 12*k, 31*k, 49*k) are the only solutions but failed to prove it for now  :evil:

ComplexZeta, yes
JBL, no they are not primes (or i would have stated it ;))

belenos, yes I was the one who posted on IMO mailing list then on mathLinks.  but I have gotten it from another message board (in a different language)

no one i know has been able to solve it thus far...  of course that includes myself


Prove that:
$ \frac{a}{\sqrt[]{b^3\plus{}1} }\plus{}\frac{b}{\sqrt[]{c^3\plus{}1} }\plus{}\frac{c}{\sqrt[]{a^3\plus{}1} } \geq 2$
With $ a,c,b > 0$ and $ a \plus{} b \plus{} c \equal{} 6$

[quote="mathz"]Prove that:
$ \frac {a}{\sqrt []{b^3 \plus{} 1} } \plus{} \frac {b}{\sqrt []{c^3 \plus{} 1} } \plus{} \frac {c}{\sqrt []{a^3 \plus{} 1} } \geq 2$
With $ a,c,b > 0$ and $ a \plus{} b \plus{} c \equal{} 6$[/quote]

This problem is in the mathematics and youth magagine,so i think we need to close this topic  :)


abstract algebra
vector
advanced fields
advanced fields unsolved

By using the Mayer-Vietoris sequence in cohomology, can we determine whether or not $ \mathbb{R}^2$ can be written as the union of two open connected sets, $ U$ and $ V$, such that $ U\cap V$ is disconnected?

Well, it seems that we are concerned with $ H^0$ stuff, since that tells us the number of connected components.

The sequence gives $ 0\to H^0(\mathbb{R}^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)\to 0$, a nice short exact sequence in which we mostly know they are connected, so

$ 0\to \mathbb{R}\to \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V)\to 0$, basically the question reduces to: Is it possible for the second to last term to have dimension greater than $ 1$, but the sequence to remain exact?

My guess is no, but I don't see why. Is it because $ \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V)$ would have trivial kernel (if dimension $ 2$), and thus the map previous would have trivial image?

Exactness on vector space gets us precise equalities of dimension- that last term must have dimension exactly one. It's the quotient of a dimension 2 space by a dimension one subspace.

[quote="HilbertThm90"]
My guess is no, but I don't see why. Is it because $ \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V)$ would have trivial kernel (if dimension $ 2$), and thus the map previous would have trivial image?[/quote]

you're right since $ 0\to\mathbb{R}\to\mathbb{R}\oplus \mathbb{R}$ is exact, it means that the second map is injective, so its image is one-dimentional and thus kernel of third map must be one-dimentional. Also from the exactness of the tail part you get that third map is surjective and thus by what jmerry mentioned the fourth real vector space must be one dimentional.

This is an interesting fact. Can we prove it without using cohomology or homology?

Sorry, I always forget that the third nonzero object in short exact sequences are so nice. i.e. that it will be isomorphic to the second object quotient the first.

Hmm...without cohomology. My first attempt was the infamous proof by picture? I came up with a $ U$ and $ V$ that give $ U\cap V$ a punctured part by picture (so some of the higher cohomologies are nontrivial), but never one that gave a disconnected $ U\cap V$.  :)

Also, I always feel like I'm trying to reinvent the wheel when working with exact sequences. Are there any good books with good exercises on this topic. I have Rotman's Advanced Modern Algebra, which was good for learning, but the exercises didn't teach me much, and it misses lots of major topics like Snake Lemma and Five Lemma. Any book with cohomology seems to assume knowledge of this stuff without building it, too.


I wonder if somebody could provide proof for these relations regarding nine-point circle and the Feuerbach point.
http://gogeometry.com/center/feuerbach_point_facts_triangle.html

Dear Mathlinkers,
for a synthetic proof see
Revistaoim (Espagne) 26 (2006) directed by F. Bellot ; /http://www.campus-oei.org/oim/revistaoim/
Sincerely
Jean-Louis

[quote="Mij"]Try these links:

[url]http://www.cut-the-knot.org/Curriculum/Geometry/SixPointCircle.shtml#explanation[/url]
[url]http://www.cut-the-knot.org/Curriculum/Geometry/FeuerbachProof.shtml#proof[/url][/quote]

Sorry, but I didn't find what I was looking for there.

Dear Grands,
if tou read german, look the book of Feuerbach on Google tape :
Feuerbach, Eigenschaften einiger merkwürdigen Punkte des geradlinigen Dreiecks, und mehrerer durch sie bestimmten Linien und Figuren
Sincerely
Jean-Louis

Thanks, jayme. It's pity I don't read German - only English (and Bulgarian). I still hope someone to find a way to arrive at these expressions.


Find all the function $f: R \to R$ satisfying:
   $f(x^2+f(y))=y+f^2(x)$

It's IMO 1992.

It's here : [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=366399#p366399]www.mathlinks.ro/Forum/viewtopic.php?p=366399#p366399[/url]


I think it use normal way:
$f(x)=(1+\frac{1}{x})^{x+0.5}$
We only need to prove:
$f''(x)\ge 0$
$f'''(x) \le 0$
Isn't it?


linear algebra
matrix
combinatorics proposed
combinatorics

In Mathlinks contest there are n problems, and every solver solves totally or doesnt solves a problem. Each problem gives a score determined by Valentin: The jury, and a contestant gets 0 points if he doesnt solve the problem and the number given by valentin if he does.  After the contest is over Valentin finds out that for every ordering of the solvers, he can give the problems some scores in such a way that the ranking of the contest matchs the given order.

determine the maximun number of contestants!

I remember solving this one before the crash of the site.

The answer was, I believe, $n$ (the maximal number of contestants).

Consider the matrix $(a_{ij}),\ i\in\overline{1,N},\ j\in\overline{1,n}$ (where $N$ is the number of contestants), s.t. $a_{ij}=1$ iff contestant $i$ has solved problem $j$, and $0$ otherwise.

Assume now $N>n$. The $N$ rows of the matrix (call them ${\bf v}_i$) are linearly dependent, and consider a null linear combination of them: $\sum_{i=1}^N \alpha_i{\bf v}_i={\bf 0}$, where $\bf{0}$ is the null vector. We thus get (after renumbering the $\alpha$'s, maybe) something like $\sum_{i=1}^p \alpha_i{\bf v}_i=\sum_{k=p+1}^N (-\alpha_k){\bf v}_k$, where $\alpha_i$ and $-\alpha_k$ are positive.

Note now that the exact same relation holds if we replace each ${\bf v}$ with the score of the respective contestant (I mean that we have the same coefficients $\alpha$).

Assume now that $\sum_{i=1}^p \alpha_i\le \sum_{k=p+1}^N(-\alpha_k)$. We cannot, in this case, arrange the contestants s.t. each contestant in the LHS is ranked lower than each contestant in the RHS, contradiction. This shows $N\le n$.

On the other hand, we can take each contestant to solve one problem, and each problem can be solved by precisely one contestant, so $N$ can be $n$.

These two show that $\max N=n$.

[quote="Valentin Vornicu"][quote="Myth"]Word "mean" has many distinct meanings. What meaning did you mean?
Hmm... mean-mean-mean-mean...[/quote]Me? I mean that I am mean :)[/quote]

He meant the "nasty" mean.  :lol:


On a plane surface, two circles with radaii of lengths 12 and 9 are x units apart where x is a positive integer and x < 19. If the length of a common external tangent segment of the two circles is $ 2\sqrt{w}$ when expressed in simplest radical form, find the sum of all possible distinct values of x.

[hide="Answer"]
16[/hide]

I attached the solution and I need help continuing on.
Thanks!

In your post, is $ 2w$ supposed to be $ 2\sqrt{w}$? Also, is $ x$ the closest distance between the circles, or the distance between their centers?

I fixed the square root. I thought I edited it before, but I guess not.

what I got in mind is that I think they mean the distance between the circles because the two radii added up to 21 and it cannot be <19.


geometry
circumcircle
incenter
symmetry
perpendicular bisector
angle bisector
geometry solved

The triangle ABC has CA = CB, circumcenter O and incenter I. The point D on BC is such that DO is perpendicular to BI. Show that DI is parallel to AC.  :blush:

Nice problem, but very simple:

We will use directed angles modulo 180°.

Since the triangle ABC is isosceles (CA = CB), its apex C, its circumcenter O and its incenter I all lie on its axis of symmetry, i. e. on the perpendicular bisector of its base AB. Thus, $OI\perp AB$, and consequently < (OI; AB) = 90°. Also, < (BI; DO) = 90° since $DO\perp BI$. Thus,

  < IOD = < (OI; DO) = < (OI; AB) + < (AB; BI) + < (BI; DO) = 90° + < (AB; BI) + 90° = 180° + < (AB; BI) = < (AB; BI) = < ABI.

But since the point I is the incenter of triangle ABC, the line BI is the angle bisector of the angle ABC, so that < ABI = < IBD. Thus, < IOD = < IBD, so that the points I, D, O and B lie on one circle. Consequently, < IDB = < IOB. But

  < IOB = < (OI; BO) = < (OI; AB) - < (BO; AB) = 90° - < (BO; AB) = 90° - < OBA.

Thus, < IDB = 90° - < OBA.

But since the point O is the circumcenter of triangle ABC, the points A, B, C lie on one circle centered at O, so that the central angle theorem yields < OBA = 90° - < ACB. Thus,

  < IDB = 90° - (90° - < ACB) = < ACB.

In other words, < (DI; BC) = < (AC; BC). Thus, DI || AC, and we are done.

  Darij


quadratics
ratio
geometry
3D geometry
algebra
quadratic formula

[hide]Let that crazy number be $x$. Then:
$x=\frac{1}{1+x}$
(Think about it)

Multiply both sides by $1+x$ and then subtract $1$ from both sides:
$x^{2}+x-1=0$

Applying the quadratic formula and ignoring the negative extraneous root (could it even be negative?), we get:
$x=\phi=\frac{-1+\sqrt{5}}{2}$[/hide]

[quote="DaReaper"]thanks...and why do you have $\phi$ there? what does that mean?[/quote]
That sign is phi. It equals $\frac{-1+\sqrt{5}}{2}$. It basically means the golden ratio.

[quote="DaReaper"]how do you know when something is going to equal $\phi$ ?[/quote]
if it equals $\frac{-1+\sqrt{5}}{2}$ then you know it equals $\phi$.

[quote="xscapezaer"]What the heck people? Am i just stupid? I was pretty sure $\phi$ = $\frac{1+\sqrt{5}}{2}$[/quote]
Yes, you're right.

$\phi$ is the positive solution to $x^{2}-x-1$, not $x^{2}+x-1$.

Technically, mathematicians generalize phi to be 1.618, which is related to all parts of the universe.  It's always the ratio of female bees to male bees in a honey come, the average of the spirals of a pine cone, etc.  Also, it is the redundant measure of your body, such as ratio of height TO hip to floor, shoulder to hand TO elbow to hand, and knuckle and spinal segments.  The ancients believed they had stumbled upon the building block of the world.


If (x + 1/x) = 2cos A , then prove that x^n + 1/(x^n) = 2cos(nA)

From $ x \plus{} \frac {1}{x} \equal{} 2\cos A$, we can set $ x \equal{} \cos A \plus{} i\sin A$, to get $ \frac {1}{x} \equal{} \cos A \minus{} i\sin A$, now de Moivre's theorem is coming.

X+1/X=2cosA
$ X^2$ + $ 1$ - 2cosA=0

x=$ /frac 2cosA +$/$ sqrt$($ 4cos^2A$ - 4)$ /2$

                      or

x=$ {/frac 2cosA -}$/sqrt($ 4cos^2$ - 4)${ /2}$

x=cosA+isinA

SO $ (cosA+isinA)^n$ = cos^nA+isin^nA
therefore x^n+x^-n=2cosnA SINCE  X+1/X=2cosA

to [b]sura33[/b]: Please make your reply typeset CLEARER, step by step solution. it seems you dropped other steps, i cann't follow the sequence. Thanks for the effort . . . .

to [b]Kunny[/b]: Thanks, you are always there to lead us to a better approach.


Express in simplest form: $ 576,000\div 800$.

576000/800=5760/8 applying hopefully not epicly failing division skills gives 720


search
combinatorics unsolved
combinatorics

Every cell of a 100X100 board is colored with one of 4 colors so that there are exactly 25 cells of each color in every column and in every row. Prove that one can choose  two columns and two lines so that the cells where they intersect are colored with diffenerent colors.


floor function
ceiling function
Pascal\u0027s Triangle

[hide]Divide by 2n and expand.  Notice that if n is even, there exists some 2k where n-2k=0.  If n is odd, the sum can be split into 2 equal halves since $\sum_{k=0}^{\frac{n-1}{2}}[(n-2k)\binom{n}{k}]^{2}=\sum_{k=\frac{n-1}{2}}^{n}[(n-2k)\binom{n}{k}]^{2}$.  This means that $\lfloor\frac{n-1}{2}\rfloor$ will cover both of these cases.  This means that $n+(n-2)^{2}n+(n-4)^{2}\frac{n(n-1)^{2}}{{2!}^{2}}+...+(n-2\lfloor\frac{n-1}{2}\rfloor)^{2}\frac{n(n-1)^{2}(n-2)^{2}...(n-\lfloor\frac{n-1}{2}\rfloor+1)^{2}}{(\lfloor\frac{n-1}{2}\rfloor!)^{2}}=1+(n-1)^{2}+\frac{(n-1)^{2}(n-2)^{2}}{(2!)^{2}}+...+1$
Now, subtract everything from the right hand side to the left, a pair at a time until all the pairs are removed, or there's only one number left on the right hand side.  Notice that $n-2+(n-2)^{2}n=(n-2)(n-1)^{2}$ and that $(n-2)(n-1)^{2}+(n-4)^{2}\frac{n(n-1)^{2}}{(2!)^{2}}-2(n-1)^{2}=\frac{(n-1)^{2}(n-2)^{2}(n-4)}{(2!)^{2}}$  It seems that if n=c, then n-c is always a factor in the numerator of the fraction that equals 0 and everything else are squared factors with the subtracted constants ranging from 1 to $\lceil\frac{n-1}{2}\rceil$ which is all divided by $(\lceil\frac{n-1}{2}\rceil!)^{2}$.  Now, let's show this pattern by induction.  The base n=1 is already known.  Assume it true for a given n.  Let's show it true for n+1.
There's 2 cases; n is even or n is odd.
Case 1: Even
Since it is true for n, one must realize that replacing n with n+1 is equally valid provided that one includes the parts not included in that sum as well.  Another thing to remember is that both the denominator and the terms being subtracted do not change.  (Think of the parts being subtracted in terms of k as it was done originally in the problem.)  Hence, $\frac{(n)^{2}(n-1)^{2}...(n/2+2)^{2}}{((\frac{n}{2}-1)!)^{2}}+\frac{(n+1)(n^{2})(n-1)^{2}...(n/2+2)^{2}}{(\frac{n}{2}!)^{2}}=\frac{(n)^{2}(n-1)^{2}...(n/2+1)^{2}}{(\frac{n}{2}!)^{2}}$  If one subtracts this and puts it under one denominator, $\frac{(n^{2})(n-1)^{2}...(n/2+2)^{2}}{((\frac{n}{2})!)^{2}}((n/2)^{2}+n+1-(n/2+1)^{2})$.  Obviously, the terms in the parenthesis equal 0.
Case 2: Odd
$\frac{(n)^{2}(n-1)^{2}...(\frac{n+1}{2}+1)^{2}}{(\frac{n-1}{2}!)^{2}}=\frac{n^{2}(n-1)^{2}...(\frac{n+1}{2}+1)^{2}}{(\frac{n-1}{2}!)^{2}}$  It's trivial because when odd advances to even, the change in the left hand side of the equation is nothing because one moves from n to $1^{2}$ to n+1 to $2^{2}$.  That can be accounted for just by shifting the already known equation up 1.  The only factor not accounted for is the right hand side's extra factor due to the fact that the n+1th row of the Pascal triangle has 1 more number than the nth row and that the numbers accounted for in the upwards shift are numbers that can be paired up.  For example, $\binom{5-1}{0}$, $\binom{5-1}{1}$, $\binom{5-1}{2}$, $\binom{5-1}{3}$, $\binom{5-1}{4}$.  To pair this up, we put the equal pairs together and find their counterparts on the fifth row.  This means $\binom{5}{n}$ if n is less than 2 and $\binom{5-1}{4-n}$ corresponds to $\binom{5}{5-n}$.  By doing so, we see that $\binom{5}{3}$ is not represented and is the number that's put on the left hand side.[/hide]


LaTeX
inequalities
inequalities unsolved

[quote="duj12358"][quote="tranthanhnam"]For $a,b,c,d \geq 0$ and$a^{2}+b^{2}+c^{2}+d^{2}+abcd=5$
[b][i]prove that[/i][/b] [b]:$a+b+c+d \leq 4$[/b]
 [u] PM:  can you write post  later by Latex [/u][/quote]
But where is Latex and how to use it?[/quote]

You can see how other people use Latex by looking at what you quoted...
Or you can also look around the forum...

[quote="duj12358"]For a,b,c,d>=0 and a^2+b^2+c^2+d^2+abcd=5
prove a+b+c+d<=4[/url][/list][/code][/u][/quote]
Is it true or false?

arqady
thank you for your nice solution :)

[quote="arqady"]
But easy to prove that $5\leq x^{2}+y^{2}+z^{2}+t^{2}+xyzt$
for all non-negative $x,$ $y,$ $z$ and $t$ such that $x+y+z+t=4.$ :wink:[/quote]
Here is my proof.
Let $x+y+z+t=4u$, $xy+xz+xt+yz+yt+zt=6v^2$ ($v\geq0$), $xyz+xyt+xzt+yzt=4w^3$ and $xyzt=p^4$ ($p\geq0$).
Since $(x-y)^2(z-t)^2+(x-z)^2(y-t)^2+(x-t)^2(y-z)^2\geq$ gives $p^4\geq4uw^3-3v^4$,
we need to prove that $(16u^2-12v^2)u^2+4uw^3-3v^4\geq5u^4$.
By Rolle's theorem it is obvious that there are non-negatives $a$, $b$ and $c$ for which $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Now we see that the last inequality is a linear inequality of $w^3$. Hence, by $uvw$ we need to check two cases:
1. $b=c=1$, which gives $(a-1)^2(11a^2+38a+5)\geq0$;
2. $b=1$, $c=0$, which gives $11a^4+8a^3-33a^2+8a+11\geq0$, which is obvious.

[b] Strengthening[/b]
For  nonnegative real numbers $a,b,c,d$ satisfying $a^{2}+b^{2}+c^{2}+d^{2}+\frac{4}{3}abcd =\frac{16}{3}$. Prove that \[a+b+c+d \leq 4.\]


Alright... So I have the equation $y=\frac{6x^2+3}{x-1}$

I tried to find the $x$ and $y$ intercepts to make graphing easier...

The $y$ intercept is obviously -3 as:

$y=\frac{6(0)^2+3}{(0)-1} \Rightarrow y=\frac{3}{-1}$

Yet the $x$ intercept is:

$0=\frac{6x^2+3}{x-1}$

Which is obviously unreal...or at least I think it is.

Yet my graphing calculator tends to disagree (I'm running Ti-89 Titanium). The numeric solver claims that there is a solution at 2, while the graph has an x intercept somewhere around 1!

What the heck is going on?  :?:

Your calculator is showing an asymptote at x=1.  I don't know about a solution at x=2, mine doesnt show one there, but it shows one around -0.5.

There are no x-intercepts.  It is factored as $\frac{3(x^2+1)}{x-1}$, which has zeroes when $3(x^2+1)=0$, and $x\neq1$.  The above clearly has no solutions in the reals, so there are no x-intercepts.

You probably made a typo when entering your solution into the numeric solver.  TI-89s don't lie. ;)  (By the way, I greatly prefer the solve() or csolve() functions over the numeric solver... they're a lot faster, and can return more than one answer)


Find solutions $(x, y)$ to the equation $3x^{2}-2y^{2}=1$.

Do you mean $x,y$ belong set $\mathbb{Z}$?

It is a generalized pelle equation. there are many sols.

3x^2 - 2y^2 = 1

minimum solution x_0 = 1 , y_0 = 1

root3 x_n + root2 y_n = (root3 x_0 + root2 y_0)^(2n-1)  ;
root3 x_n - root2 y_n = (root3 x_0 - root2 y_0)^(2n-1)

adding or subtracing.. we get x_n , y_n

If you are interesting, see the MOC (by T. Andreescu)


calculus
integration
calculus computations

This looks easy, but i can't believe i can't solve it! no one in my class can..... hope someone can help:

Evaluate:

$\displaystyle\int^{\frac{\pi}{2}}_0 \frac{dx}{xsinx}$

[quote="Kent Merryfield"]This is an improper integral, and is comparable near zero to $\int_0^{\delta}\frac{dx}{x^2},$ which diverges. Since this is a positive divergent integral, we can say that it computes to $+\infty.$[/quote]

arrrghhh!!! got it. so it IS easy. i forgot that we can use comparison : $\frac{1}{x} < \frac{1}{xsinx}$
 :fool:

thanks, kent

Another question:

let's say the integral i want to find is between on the interval [$\pi, 3\pi$] (or any interval that will yield a definite answer)
is it possible to get the answer?e.g. by substitution or trigo etc

Well, you don't want $[\pi,3\pi].$ On any interval that contains (or ends at) any integer multiple of $\pi,$ this integral will diverge.

As for an interval on which it does not diverge - say, $\left[\frac{\pi}4,\frac{3\pi}4\right]$ - there is no good reason to think that this could be done in closed form.


trigonometry
geometry
trig identities
Law of Sines

In any triangle $ ABC$ prove that :

$ \frac{a^2\minus{}b^2}{a^2\plus{}b^2}\equal{}\frac{\sin(A\minus{}B)\sin \ C}{1\plus{}\cos(A\minus{}B)\cos \ C}$

[quote="Pain rinnegan"]In any triangle $ ABC$ prove that :

$ \frac {a^2 \minus{} b^2}{a^2 \plus{} b^2} \equal{} \frac {\sin(A \minus{} B)\sin \ C}{1 \plus{} \cos(A \minus{} B)\cos \ C}$[/quote]

As $ A\plus{}B\plus{}C\equal{}\pi$ ,

$ R.H.S.\equal{}\frac {\cos (A\minus{}B\minus{}C) \minus{}\cos (A\minus{}B\plus{}C)}{2 \plus{} \cos(A \minus{} B\plus{}C)\plus{}\cos (A\minus{}B\minus{}C)}$
$ \equal{}\frac{\minus{}\cos 2A\plus{}\cos 2B}{2\minus{}\cos 2B\minus{}\cos 2A}$
$ \equal{}\frac{sin^2 A\minus{}sin^2 B}{sin^2 B\plus{}sin^2 A}$

By sine formula,

$ L.H.S.\equal{}\frac{(sin A)^2\minus{}(sin B)^2}{(sin A)^2\plus{}(sin B)^2}$

$ L.H.S.\equal{}R.H.S.$, done.

$ \frac{\sin(A\minus{}B)\sin\ C}{1\plus{}\cos(A\minus{}B)\cos\ C}\equal{}\frac{\sin(A\minus{}B)\sin\ (A\plus{}B)}{1\minus{}\cos(A\minus{}B)\cos\ (A\plus{}B)}$
$ \equal{}\frac{(\sin A \cos B \minus{} \cos A \sin B)(\sin A \cos B \plus{} \cos A \sin B)}{1\minus{}(\cos A \cos B \plus{} \sin A \sin B)(\cos A \cos B \minus{} \sin A \sin B)}$
$ \equal{}\frac{\sin^2 A \cos^2 B \minus{} \cos^2 A \sin^2 B}{1\minus{}(\cos^2 A \cos^2 B \minus{} \sin^2 A \sin^2 B)}$
$ \equal{}\frac{\sin^2 A (1\minus{}\sin^2 B) \minus{} (1\minus{}\sin^2 A) \sin^2 B}{1\minus{}((1\minus{}\sin^2 A) (1\minus{}\sin^2 B) \minus{} \sin^2 A \sin^2 B)}$
$ \equal{}\frac{\sin^2 A \minus{}\sin^2 A \sin^2 B \minus{} \sin^2 B\plus{}\sin^2 A \sin^2 B}{1\minus{}(1\minus{}\sin^2 A \minus{} \sin^2 B \plus{} \sin^2 A \sin^2 B \minus{} \sin^2 A \sin^2 B)}$
$ \equal{}\frac{\sin^2 A\minus{}\sin^2 B}{\sin^2 A \plus{} \sin^2 B}\equal{}\frac{\frac{\sin^2 A}{\sin^2 B}\minus{}1}{\frac{\sin^2 A}{\sin^2 B} \plus{} 1}$
By Law of Sines, $ \dfrac{a}{\sin A} \equal{} \dfrac{b}{\sin B} \rightarrow \dfrac{a}{b}\equal{}\dfrac{\sin A}{\sin B} \rightarrow \dfrac{a^2}{b^2}\equal{}\dfrac{\sin^2 A}{\sin^2 B}$ so
$ \frac{\sin(A\minus{}B)\sin\ C}{1\plus{}\cos(A\minus{}B)\cos\ C}\equal{}\frac{\frac{a^2}{b^2}\minus{}1}{\frac{a^2}{b^2} \plus{} 1}\equal{}\dfrac{a^2\minus{}b^2}{a^2\plus{}b^2}$


Why did you start a new topic then? Just try and check your own answers if no solutions are available and use the search function for problems you don't get. If they aren't there, feel free to start your own topic. But for some national things, there is an answer key.

[quote="bthings_2000"]I found the problems, now I need solutions from past tests. Anybody know where to find them?[/quote]

WHERE DID YOU FOUND THE PROBLEMS??? I WANT THEM TOO!!! :huh:

[quote="cat810930"]

WHERE DID YOU FOUND THE PROBLEMS??? I WANT THEM TOO!!! [/quote]

Maybe you could use some grammar problems and solutions too?! ;)

Answers to the 1993 sprint: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=73348#p73348]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=73348#p73348[/url]


Fie $k,m,n\in{\bf N}^*$ cu $k,m<n$. Sa se determine
\[ \displaystyle\sum\limits_{\sigma\in S_n}{\rm Card\{i\in\{1,2,\ldots,n\}\;|\;|\sigma(i+m)-\sigma(i)|=k\}  }\]
unde $\sigma(n+p): =\sigma(p)$, $\forall\;p\in\{1,2,\ldots,n\}$, $S_n$ este multimea permutarilor de ordin $n$, iar ${\rm Card}\{\ldots\}$ reprezinta cardinalul multimii $\{\ldots\}$.

[quote="Daniel Stanica"]Aratati ca \[ \displaystyle\sum\limits_{\sigma\in S_n}{\rm Card\{i\in\{1,2,\ldots,n\}\;|\;|\sigma(i+m)-\sigma(i)|=k\}=2n(n-k)(n-2)!  }\].[/quote]
    Cauta in cartea lui Gh.Rizescu de la editura Tehnologica Bucuresti.Se gaseste acolo!


algebra
function
domain
complex analysis
complex analysis theorems

Let $f$ be analytic on a starlike region $\Omega$.
Prove that $f$ has a primitive on $\Omega$.

Any starlike domain can be decomposed to a set of Triangles. The result then follows from the usual Cauchy theorem.


calculus
Harvard
college
MIT
integration
Princeton
linear algebra

Hi everyone.  I'm in ninth grade now, taking AP Calculus BC and AP Physics C.  I know I still have a while but I was wondering about placement when I get to college.  I hope to get to Harvard or MIT, and I know that they have strict placement policies.  If I get a 5 on the AP Calculus BC exam and a 5 on both sections of the AP Physics C exam, I will be able to place out.  But, I still have three years of high school left.  I was thinking about taking university level courses on EPGY but I am not sure of how this will work in college.  I know that it can not be taken as direct credit, but are you able to place out?  For example, if I took Multivariable Calculus (both differential and integral) via EPGY next year and got an A average, would I be able to take say linear algebra or differential equations or whatever the subsequent courise may be at college?  I hear about college placement exams as well that you can take at the school to place you in the appropriate course for you.  How does that work exactly?  I am insane and hope to have a triple major in math, physics, and economics.  The more I can get done now would help me a lot for my time in college.  If anyone has any knowledge or anything to say at all, it will be appreciated.

Whoa. Calm down. I had a similar decision to make.

Take Multivariate, except read the Title I in EPGY page. They tell you all about the credit transition and stuff. For the testing in Physics C and Calculus, dont worry. You will get 5 and place out.

Don't overpressure yourself!!!

Don't worry about placing out of Harvard and MIT courses. Some courses at each school are designed to provide interesting, challenging content for entering students who have already taken the "same" course somewhere else. Harvard's Math 55 course, for example, is approximately titled "multivariable calculus and linear algebra" (the full formal title is "Honors Advanced Calculus and Linear Algebra"), but that doesn't mean you know all of the course's content if you've had a multivariable calculus and linear algebra course before. The placement tests at the top schools (Caltech is included in this list too, as is Princeton) will put you in the right course. Anything you learned before you go to a college like that can only help you get through courses that are much more rigorous than courses with the same title at many other schools.


[u][b]The author of this posting is : kunny[/b][/u]
____________________________________________________________________

Let $a=\sqrt{x^2+xy+y^2}$, $b=p\sqrt{xy}$, $c=x+y$.
Determine the range of the value of $p$ such that there exist always a triangle with the side lengths of $a$, $b$, $c$ for any positive numbers $x$, $y$.

[u][b]The author of this posting is : kunny[/b][/u]
____________________________________________________________________

Thank you for your reply, paganinio.

You have not finished your answer yet.

  kunny

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

There are two necessary inequalities:

1) $p\sqrt{xy} > \left(x+y\right) - \sqrt{x^2+xy+y^2}$.
It is equivalent to $p\sqrt{xy} > \frac{xy}{\left(x+y\right)+sqrt{x^2+xy+y^2}}$, so for all $x,y$ we have
$p>\frac{\sqrt{xy}}{\left(x+y\right)+\sqrt{x^2+xy+y^2}}$.

But using AM-GM we obtain $\frac{\sqrt{xy}}{\left(x+y\right)+\sqrt{x^2+xy+y^2}} \leq \frac{\sqrt{xy}}{2\sqrt{xy}+\sqrt{2xy+xy}} = \frac{1}{2+\sqrt3} = 2-\sqrt3$ and for $x=y$ we will have equality, therefore $p>2-\sqrt3$.

2) $p\sqrt{xy} < \left(x+y\right) + \sqrt{x^2+xy+y^2}$.
And again $\left(x+y\right) + \sqrt{x^2+xy+y^2} \geq \left(2+\sqrt3\right)\sqrt{xy}$ and for $x=y$ we will have equality. So $p<2+\sqrt3$.

Answer: $2+\sqrt3 > p > 2-\sqrt3$.

[u][b]The author of this posting is : kunny[/b][/u]
____________________________________________________________________

Splendid! Here is my solution:

$c^2-a^2 = \left(x+y\right)^2-\left(x^2+xy+y^2\right) = xy>0$, so $c>a>0$. The condition for the existence of the triangle is given by $c-a<b<c+a \Longleftrightarrow x+y-\sqrt{x^2+xy+y^2}<p<\sqrt{xy}<x+y+\sqrt{x^2+xy+y^2}$. Dividing by $\sqrt{xy}$ of every side and setting $\sqrt{\frac{y}{x}} = t$, we have

$ \sqrt{t}+\frac{1}{\sqrt{t}}-\sqrt{t+\frac{1}{t}-1}<p<\sqrt{t}+\frac{1}{\sqrt{t}}+\sqrt{t+\frac{1}{t}+1} $.

Let $ f\left(t\right)=\sqrt{t}+\frac{1}{\sqrt{t}}+\sqrt{t+\frac{1}{t}+1}$, $g(t)=\sqrt{t}+\frac{1}{\sqrt{t}}-\sqrt{t+\frac{1}{t}-1} $,

in using AM-GM we obtain

${ f\left(t\right)\geqq 2\sqrt{\sqrt{t}\cdot\frac{1}{\sqrt{t}}}+\sqrt{{2\sqrt{t\cdot\frac{1}{t}}+1}}} \Longleftrightarrow f(t)\geqq 2+\sqrt{3} $.

On the other hand, since $f\left(t\right)g\left(t\right) = 1$, yielding $g\left(t\right) \leqq 2-\sqrt{3}$.

Therefore the range of the value of p which satisfies $g\left(t\right)<p<f\left(t\right)$ is $2-\sqrt{3} < p < 2+\sqrt{3}$.

  kunny


topology
complex analysis
real analysis
Functional Analysis

Anyone have info on 295/296/395/396?

Thx..

Course numbers, I suppose? Not that I know anything about it anyway.

samus - this collegeconfidential post has quite a bit of info on it:

http://talk.collegeconfidential.com/showthread.php?t=47757&page=1&pp=20

Wow! That [url=http://talk.collegeconfidential.com/showthread.php?t=47757]College Confidential thread[/url] is amazingly informative. I get the impression that most of those posters are AoPSers or immediate relatives of AoPSers.

It looks like U Michigan is not a bad choice at all. Tough courses...

For those who've read the college confidential thread, here's a slight correction: one poster gives a sample course sequence and says that the last 2 years are graduate courses.  That's not true.  Only the last year was standard grad courses.  The 59x courses at UMich are the first-year graduate courses, and courses like 512/513 are just advanced undergraduate courses for students on the grad-school track.  (Sometimes grad students who enter with a deficiency take one of them, but it's not the norm.)  There is a family of 500-level courses, those topics courses not in algebra/analysis/topology, which are considered grad/undergrad hybrid courses.  But these are graduate courses more in the sense that most grad students didn't take these topics as undergrads (or went to undergrad institutions that didn't offer them) rather than because the material is seriously graduate-level.

Indeed.  Whether a university chooses to give a 400 or 500 level label to a course can be purely arbitrary.  A more general rule of thumb would be that there is, for example, a two course undergrad complex analysis sequence, a three course undergrad real analysis sequence, and a three course algebra sequence.  Anything beyond this would then be a true graduate level course.  (still somewhat arbitrary, but hey, what can you do)

I'm exempting analysis on manifolds or other such things from such a real analysis sequence for simplicity's sake and considering more of a real analysis -> measure theory -> functional analysis sequence.


Happy New Year, [b]Abdek[/b]!
Your generalization is absolutely OK. And your solution is the solution to the problem.
Just a little remark. For my particular case, dedicated to 2010, the appropriate generalization is the following:

Let $ a_1,a_2,...,a_n$ be positive real numbers, and  $ n,k\in N, n,k\geq 2.$
Prove that
$ \frac {a^{k \plus{} 1}_1}{a^k_1 \plus{} (k \minus{} 1)a^k_2} \plus{} \frac {a^{k \plus{} 1}_2}{a^k_2 \plus{} (k \minus{} 1)a^k_3} \plus{} \cdots \plus{} \frac {a^{k \plus{} 1}_n}{a^k_n \plus{} (k \minus{} 1)a^k_1}\geq \frac {1}{k}\cdot (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)$

My example is given for: $ k \equal{} 2009$ and $ n \equal{} 2010$. [b]I just wanted to puzzle you, dear mathlinkers![/b] :lol:
But the essence of the proof is the same.
All the best to all of you! I really hope you enjoy my problems :roll:

[u][b]My proof[/b][/u]

Let's prove that:
$ \frac{a^{k+1}}{a^k+(k-1)b^k}\geq a-\frac{k-1}{k}\cdot b, \ (\forall)a,b>0, k\in N, k\geq2$
We have:
${ \frac{a^{k+1}}{a^k+(k-1)b^k}=\frac{a^{k+1}+(k-1)ab^k-(k-1)ab^k}{a^k+(k-1)b^k}=a-\frac{(k-1)ab^k}{a^k+(k-1)b^k}}\stackrel{(AM\geq GM) }{\geq}$
$ \geq a-\frac{(k-1)ab^k}{\sqrt[k]{a^k\cdot b^k\cdot...\cdot b^k}}=a-\frac{(k-1)ab^k}{ab^{k-1}}=a-\frac{k-1}{k}b$
Using this inequality, we get:
$ \frac{a^{k+1}_1}{a^k_1+(k-1)a^k_2}\geq a_1-\frac{k-1}{k}\cdot a_2$
$ \frac{a^{k+1}_2}{a^k_2+(k-1)a^k_3}\geq a_2-\frac{k-1}{k}\cdot a_3$
$ ......$
$ \frac{a^{k+1}_n}{a^k_n+(k-1)a^k_1}\geq a_n-\frac{k-1}{k}\cdot a_1$
Adding them up we get the desired inequality.


i need help with question G

1.  Consider the three reactions that occur in a car engine:
2 CO (g) + O2 (g) ---> 2 CO2 (g)					(1)
2 C8H18 (g) + 25 O2 (g) ---> 16 CO2 (g) + 18 H2O (g)		(2)
2 NO (g) ----> N2 (g) + O2 (g)					(3).
a.  Determine whether each reaction occurs spontaneously.  Give reasons.  For each reaction, determine whether the equilibrium constant is greater than 1 or less than 1.
b. According to Moore, et al., “The Chemical World”, 2nd ed., p. 548, “The largest growth in catalyst use is in emissions control for both automobiles and power plants.”  A catalytic converter in a car has two purposes.  What are the two purposes?
c.  In Reaction (1), CO is converted to CO2.  What is the source of CO?  Write a chemical equation that represents the reaction that produces CO.
d.  For each of the three reactions, speculate on the rate of the reaction without a catalyst.  In other words, do you expect the reaction to occur fast or slow?  Give reasons.
e. NO is a considered a serious air pollutant.  Look up the toxicity or safety of NO and describe briefly.  Cite the reference where you found this information.
f.  Describe the optimum temperature and pressure conditions for Reactions (1) and (3).
g.  Under the conditions of a car engine, is Reaction (1) likely to occur or is the reverse of Reaction (1) likely to occur?  Give reasons.  In a car engine, NO is produced when N2 reacts with O2.  Under the conditions of a car engine, is Reaction (3) likely to occur or is the reverse of Reaction (3) likely to occur?  Give reasons.

well the forward reactions are exothermic, and a cars engine has real high temperatures when its running, i said the forward reactions occur but am I right, if im wrong can u explain it to me

After reading the whole problem, I would say that reaction (1) occurs in the direct way and reaction (3) in the reverse way.


Let $a_1, a_2, \cdots$, be a sequence with the following properties.

I. $a_1 = 1$, and
II. $a_{2n}=n\cdot a_n$ for any positive integer $n$.

What is the value of $a_{2^{100}}$?

$ \textbf{(A)}\; 1\qquad
\textbf{(B)}\; 2^{99}\qquad
\textbf{(C)}\; 2^{100}\qquad
\textbf{(D)}\; 2^{4950}\qquad
\textbf{(E)}\; 2^{9999} $


ratio
geometry
trigonometry
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

[quote="math92"]Is a triangle whose side lengths are in the ratio 6 : 8 : 9 acute, right, or obtuse?

Also just a side question: Is it possible to find the area of a triangle given the three sides? Is there some sort of theoreom?[/quote]

[hide]Acute... think about it: a right triangle can be 6, 8, 10, and ur just taking 1 off the hypotenuse. So the right angle will be less a little bit which means its acute.[/hide]

[hide]Heron's Formula:

Given sides a, b and c and semiperimeter s, the area is

$\sqrt{s(s-a)(s-b)(s-c)}$[/hide]

[hide]Let the side lengths be 6x, 8x and 9x. Thus, sicne (6x)^2+(8x)^2>(9x)^2, the triangle is acute.

You can use Heron's to find the area.[/hide]

Drunken_math's solution is the easiest one. If you want a more mathematical/time-wasting solution, here it is. It might help on similar problems.
[hide]
Let $a=6, b=8, c=9$ We know the largest angle is opposite the $9$ so we can use the Law of cosines to determine the angle which we can call $C$. By the Law of cosines:
$c^{2}=a^{2}+b^{2}-2abcosC$ or
$81=36+64-96cosC$ or
$96cosC=19$ making $cosC=\frac{19}{96}$.
Since the cosine is positive then the angle must be, in degrees, between $0-90$ and $270-360$. Since in a triangle, all angles must be less than $180$ degrees, it is between $0$ and $90$. Furthermore since $0$ does not equal $\frac{19}{96}$, it is not a right angle, meaning that it's degree measure is less than 90 degrees, making it acute by definition.

sorry if its confusing, but it seems straightforward[/hide]

Well really all you do is take the sum of the squares of the two short sides, then if it's greater than the third, it's obtuse, less, then it's acute, equal, then it's right. As life=tennis+math pointed out, this is because of the law of cosines.

wow i know how to do this because you took the question right out of the chapter in volume 2 that I'm on :D
actually, this is easier than I thought :lol: wow why is this in volume 2 :maybe:
[hide]$(6x)^{2}+(8x)^{2}>(9x^{2})$ so acute[/hide]
for the side note, HERON'S FORMULA :D


Construct the circle that is tangent to three given circles.

This is the celebrated Apollonius' problem (CCC) that can be reduced to the case (LCC), transforming one of the circles into a line by an inversion, or the simpler case (PCC) by a Vieta's transformation: adding or substracting (according to the desired tangency) the smaller radius to the bigger radii.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=46945[/url]
[url]http://www2.washjeff.edu/users/mwoltermann/Dorrie/32.pdf[/url]
[url]http://www.cut-the-knot.org/pythagoras/Apollonius.shtml[/url]


Let be $ a,b,c\in \mathbb{R}$ such that $ abc\equal{}1$ and $ a^3>36$ . Show that :

$ \frac{2a^2}{3}<a^2\plus{}b^2\plus{}c^2\minus{}(ab\plus{}bc\plus{}ca)$

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}$ such that $ abc \equal{} 1$ and $ a^3 > 36$ . Show that :

$ \frac {2a^2}{3} < a^2 \plus{} b^2 \plus{} c^2 \minus{} (ab \plus{} bc \plus{} ca)$[/quote]

$ <\equal{}>(\frac{a}{2}\minus{}b\minus{}c)^2\plus{}\frac{a^3\minus{}36abc}{12a} \ge 0$
It's true  :lol:


[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=694296#694296]Problem 1[/url]
Find 4 primes less than 100 which divide $3^{32}-2^{32}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=694297#694297]Problem 2[/url]
Given a quadrilateral ABCD and points E, F on AB such that AE=EF=FB and points G,H on CD such that DG=GH=HC, show that $[CHAE]=[EFGH]=\frac{1}{3}[ABCD]$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=694298#694298]Problem 3[/url]
The positive integer 182634795 has the property that the digits 1 to 5 are in natural ascending order, but 1 to 6 are not, and each digit from 1 to 9 is used precisely once.

How many such positive integers exist?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=694301#694301]Problem 4[/url]
Let two circles S and T touch (once) at X. Let AB be a common tangent with A on S and B on T. Let P be the point on S diametrically opposite A. Draw a tangent to T through P, and call the point where it touches T, Q.

Show that $AP=PQ$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=694302#694302]Problem 5[/url]
Show that for all positive reals a,b,c
$(a^{2}+b^{2})^{2}\geq (a+b+c)(a+b-c)(b+c-a)(c+a-b)$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=694303#694303]Problem 6[/url]
Show that if $2+2\sqrt{12n^{2}+1}$ is an integer, then it is square.


Where are some of Martin Gardner's Puzzles , articles, problems etc. available free online? Are any scans of his books available free online??? Does anyone have an extensive list of ALL his books (that is written BY him, not EDITED by him!) ???

Hey guys, I'm bumping this after a log time? Does anybody know where I can get the above ONLINE???

[url=http://www.amazon.com/gp/search/ref=br_ss_hs/103-0187725-7453437?search-alias=aps&keywords=martin%20gardner]Amazon link[/url]

If you're that interested, why don't you try buying some of them?  If you feel that they are that worthwhile, then you may as well compensate the author for his work.

The only problem is... I Live in India!!!

Amazon's international shipping rates are not that bad.


combinatorial geometry
advanced fields
advanced fields unsolved

$K \subset \mathbb{R}^{d}$ a compact set of diameter $1$. Show that $K$ is contained in a ball of radius $\sqrt{\frac{d}{2d+2}}$

In other words, optimality is achieved for the regular simplex with edge equal to $1$.

Let $\mathcal B$ be a ball of minimal radius $r$ containing $K$. It's easy to see that center of $\mathcal B$ (which we may assume WLOG coincides with the origin of $\mathbb R^{d}$) is in the convex hull of $K$. By Caratheodory's Theorem, this means that it can be written as a convex combination $0=\sum_{i=1}^{d+1}\alpha_{i}v_{i}\ (*),\ v_{i}\in K$. It now suffices to prove that at least one edge of the simplex $\Delta$ having $v_{i}$ as vertices is not smaller than the edge of the regular simplex inscribed in $\mathcal B$. If $e_{i},\ i=\overline{1,d+1}$ are the vertices of a regular simplex inscribed in $\mathcal B$, what we want to prove boils down to showing that we can find $i\ne j$ such that $(v_{i},v_{j})\ge(e_{i},e_{j})\ (\#)$.

Taking scalar producs with $v_{j}$ in $(*)$ and summing after $j$, we get $\sum_{i,j}\alpha_{i}(v_{i},v_{j})=0\Rightarrow\sum_{i\ne j}\alpha_{i}(v_{i},v_{j})=-r$. Since the exact same thing holds with $e_{i},e_{j}$ instead of $v_{i},v_{j},\ (\#)$ follows (for some pair $i\ne j$).


Let A, B be square matrices satisfying: A 2 = A, B  2 = B, AB + BA = 0. Find det(A-B).

Namdung, thank you for posting the problems.
However, I belive that this problem statement is not complete. For example putting $B=0_n$ gives us $\det (A-B)= \det A \in \{0,1\}$. So no unique value.
Or should we find it as a function of $\det A$ and $\det B$ ? :)

Yes, I knew this question will be given. The correct question must be: "Which value can det(A-B) take?".

Namdung

Then it's very easy :) As I showed above, for $A=0_n$ and $B=I_n$ we obtain $(-1)^n$, for $A=I_n$ and $B=0_n$ we obtain 1, and for $A=B=0_n$ we obtain 0. We shall prove that only these 3 values can be obtained (if $n$ is even, then in fact only two values can be obtained :P).

We have $(A-B)^2 = A^2+B^2 - (AB+BA) = A^2+B^2$, but also $(A+B)^2 = A^2+B^2 + AB+BA = A^2+B^2 = A+B$, so $C=A+B$ is also idempotent (i.e. $C^2=C$) so it can only have $\det C \in \{ 0,1\}$. Thus $ (\det (A-B))^2 \in \{ 0, 1\}$ which leads us to the solution.


[quote="Jutaro"]Find all three-digit numbers $ abc$ (with $ a \neq 0$) such that $ a^{2}+b^{2}+c^{2}$ is a divisor of 26.[/quote]

[hide]We can do this systematically quite easily.
Case 1: $ a^{2}+b^{2}+c^{2}= 1$. Since $ a \neq 0$, 100 is the only valid number.
Case 2: $ a^{2}+b^{2}+c^{2}= 2$. a must be 1, so 110 and 101 are the only valid numbers
Case 3: $ a^{2}+b^{2}+c^{2}= 13$. Clearly, a,b,c must be 3 or less, and we see that one of the numbers must be 3. The only integral solution to $ x^{2}+y^{2}= 4$ is (2,0). So, the valid numbers are 320, 302, 230, 203.
Case 4: 26. Triplets (x,y,z) such that $ x^{2}+y^{2}+z^{2}= 26$ are what we want to find. If one is 5, then we only have (5,1,0). If one is 4, we only have (4,3,1). If none are 4, then at least one must be 3, and we find that there are no integral solutions to $ i^{2}+j^{2}= 17$ for i,j < 4. So, the valid numbers are 510, 501, 150, 105, and 431, 413, 314, 341, 143, 134.

And we are done.[/hide]


function
integration
real analysis
real analysis solved

Let f:[0,1]->R be a function of class C1. Prove that In=n[(1/n)\sum{k=1->n}f(k/n)-\int{0->1}f(x)dx]->(f(1)-f(0))/2 as n->00.

Are you sure about the limit (f(1)-f(0))/2 ?
On my homework sheet the limit is  (f(1)+f(0))/2
--------------------------------------------------
Your results is grobber
because the version I have, k goes until n-1, your version k untill n and
take the opposite.

This is strange... I thought I had a solution for (f(1)-f(0))/2, but I might have made a mistake. Can you prove it for (f(1)+f(0))/2? I mean, could you please post the solution? I'd like to see where I went wrong  :?

Oh, so it was Ok after all! :D

I used the fact that f' is continuous on [0,1], which is a compact interval, so it must be uniformly continuous (a well-known thm). I think it's pretty much the same as your solution, Moubinool. My idea was that if we take Jn=n[\int{0->1}f(x)dx-(1/n)\sum{k=1->n}f((k-1)/n)], then it's easy to see that In+Jn=f(1)-f(0) for all n \in N*, so in order to get to the conclusion, it's enough to show that In-Jn->0 as n->00.

Sorry,I mention it which is long ago :lol: .Here is my proof for the original problem.hope it is right.
Proof:Let $m_{k}=\inf \{f'(x): \frac{k-1}n \le x \le \frac{k}{ }n\},M_{k}=\sup \{f'(x): \frac{k-1}n \le x \le \frac{k}{ }n\}$ \[I_{n}=n\left(\frac{1}{ }n \sum_{k=1}^{n}f(\frac{k}{ }n)-\int_{0}^{1}f(x)dx\right)=n \left(\sum_{k=1}^{n}f(\frac{k}{ }n)-\int_{\frac{k-1}n}^{\frac{k}{ }n}f(x)dx \right)\] but we have: \[f(\frac{k}{ }n)-f(x)=(\frac{k}{ }n-x)f'(\xi_{x}),x \le \xi_{x}\le \frac{k}{ }n\] so we get: \[m_{k}\int_{\frac{k-1}n}^{\frac{k}{ }n}(\frac{k}{ }n-x) dx \le f(\frac{k}{ }n)-\int_{\frac{k-1}n}^{\frac{k}{ }n}f(x)dx \le M_{k}\int_{\frac{k-1}n}^{\frac{k}{ }n}(\frac{k}{ }n-x)dx\] which means \[\frac{m_{k}}{2n^{2}}\le f(\frac{k}{ }n)-\int_{\frac{k-1}n}^{\frac{k}{ }n}f(x)dx \le \frac{M_{k}}{2n^{2}}\] Thus we get: \[\[\sum_{k=1}^{n}\frac{m_{k}}{2n}\le I_{n}\le \sum_{k=1}^{n}\frac{M_{k}}{2n}\] But let $n \to \infty$,both side will go to $\frac{1}{ }2\int_{0}^{1}f'(x)dx=\frac{f(1)-f(0)}2$.
So we proved the problem. :)
[b]Edit[/b]
Thanks to ffrogg[/b]

zhaobin,I daresay you posted by mistake $f(x)$ for $f'(x)$ in the second line of your proof ,which you can possibly simplify by just referring to the definition of integral.


algebra
polynomial
calculus
integration
trigonometry
function
complex analysis

kool_dudy wrote:Find the sum

I'm thinking [hide]use the a^2 - b^2 = (a+b)(a-b) factorization to simplify things a bit.

Or if we had an explicit formula for the sum from 1^4 to 1/n^4, this would work:

[/hide]

If you go to infinity, i.e. \[\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^4},\] then there's a nice solution.

Let $\zeta(s) = \sum_{k=0}^{\infty}\frac1{k^s}$ wherever that converges. All the values at even integers: $\zeta(2), \zeta(4), \zeta(6), \dots$ are known exactly. For example, it is known that $\zeta(2) = \frac{\pi^2}6$.

The sum with all postive terms, the alternating sum, and the sum with only the odd terms are all related. If you know any one of those three, you can calculate the other two. Mathfanatic's first post shows how.

I know of three methods to do these calculations. One involves calculating the Fourier series of certain piecewise-polynomial functions. The second involves complex analysis: setting up a contour integral with a certain collection of poles. The third is Euler's original method: look at the infinite product $\frac{\sin x}x = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots$ and compare the coefficients you get multiplying that out with the coefficients of the power series for $\sin x$.

None of these methods is elementary (precalculus, or even calculus level), although Euler's probably comes closest. For each of these methods, $\zeta(4)$ takes more time and effort to calculate than $\zeta(2)$, $\zeta(6)$ more time than $\zeta(4)$, and so on.

The original problem wasn't the infinite series, it was a partial sum of that series. By the alternating series test, the distance between this partial sum and the sum of the infinite series is smaller than the next term.

kool_dudy wrote:Hmm...mathfanatic...i don't know...my friend found this problem in her book

The book had an answer in decimal and this forumla for this question...

[hide][/hide]

Eventually, I tried using my TI-89.  I got a very complicated fraction (perhaps 15 digits in the numerator and 15 in the denominator).  So I would say there is not a nice solution, at least not to the original problem.

Well how about this one. Its nicer.
$F_{1} = F_{2} = 1$. $F_{n} = F_{n-1} + F_{n-2}$ for all $n > 2$. Find $\displaystyle\sum_{i = 1}^{\infty}\frac{F_{i}}{3^i}$

There's a way to do jelyman's problem without thinking very hard: use the explicit formula for the Fibonacci numbers and break it into two geometric sums. But it's possible to do it with never a $\sqrt5$ in sight.

Let $S = \sum_{i=1}^{\infty}\frac{F_i}{3^i}$. Then $3S=\sum_{i=1}^{\infty}\frac{F_i}{3^{i-1}}=F_1+\sum_{i=1}^{\infty}\frac{F_{i+1}}{3^i}$.

Add these two formulas together to get

$4S = 1 + \sum_{i=1}^{\infty}\frac{F_i+F_{i+1}}{3^i} = 1 + \sum_{i=1}^{\infty}\frac{F_{i+2}}{3^i} = 1 + \sum_{i=3}^{\infty}\frac{F_i}{3^{i-2}} = 1 + 9\sum_{i=3}^{\infty}\frac{F_i}{3^i}$.

Add and subtract the two missing terms in the last sum to get:

$4S = 1-9\left(\frac{F_1}3+\frac{F_2}{3^2}\right) + 9\sum_{i=1}^{\infty}\frac{F_i}{3^i}$

$4S = 1 - 3 - 1 + 9S$, so $5S = 3$ or $S = \frac35$.

Alternatively and even more simply: this is just the evaluation of the generating function of the Fibonacci numbers, which is known to be $\frac x{1-x-x^2}$, at $x=\frac13$.

I can figure it out quickly, but I have an unfair advantage of knowing what $\zeta(4)$ is (and how to figure it out in two ways).

[quote="Kent Merryfield"]Back to something related to the original problem. No one on this thread has yet posted the exact value of $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^4}$. Who knows the answer to that?[/quote]

Well,
Sum(1/n^4) = pi^4/90

Essentially, we are looking for the sum of the odd terms minus the sum of the even terms.  But the even terms are equal to 1/16*odd terms.

So the odds = 15/16*pi^4/90 = pi^4/96

The even terms = 1/16*pi^4/90 = pi^4/1440

Hence odds - evens = 7pi^4/720

Wow, that was a really badly worded argument.


A concyclic quadrilateral $ ABCD$ , $ AB\equal{}3 ,BC\equal{}3 , CD\equal{}5 ,DA\equal{}8$.

Find the area of $ ABCD$

:trampoline:

Notice that if $ ABCD$ is a cyclic quadrilateral, put $ AB \equal{} a,BC \equal{} b,CD \equal{} c,DA \equal{} d,p \equal{} \frac {a \plus{} b \plus{} c \plus{} d}{2}$ and we have $ S(ABCD) \equal{} \sqrt {(p \minus{} a)(p \minus{} b)(p \minus{} c)(p \minus{} d)}$ .


geometry
circumcircle
trigonometry
angle bisector
geometry proposed

Let ABC be a triangle with median AM and angle
bisector AN. Draw the perpendicular to line NA through N, hitting
lines MA and BA at Q and P, respectively. Also let O be the point
where the perpendicular to line BA through P meets line AN. Prove
that QO and BC are perpendicular .

Let $ L$ be the midpoint of the arc $ BC$ of the circumcircle $(O).$ $ S \equiv OP \cap LB.$ Since $ \angle RPO \equal{} \angle CAL \equal{} \angle CBS,$ it follows that $ BNSP$ is cyclic $ \Longrightarrow \angle BNS \equal{} 90^{\circ}.$ Let $ R \equiv PN \cap AC.$ In the isosceles triangle $ \triangle ARP$ we have:

$ \frac {QR}{PQ} \equal{} \frac {c}{b} \Longrightarrow \frac {PR}{PQ} \equal{} \frac {b \plus{} c}{b}\Longrightarrow \frac {PN}{PQ} \equal{} \frac {b \plus{} c}{2b} \Longrightarrow \frac {PO \cos\frac {A}{2}}{PQ} \equal{} \frac {b \plus{} c}{2b}$

$ \Longrightarrow \frac {PO}{PQ} \equal{} \frac {b \plus{} c}{2b \cos \frac {A}{2}} \ . (1)$

Notice that, since $ \triangle ABN \sim \triangle PSN,$ we have $ \frac {PS}{PN} \equal{} \frac {c}{AN} \equal{} \frac {b \plus{} c}{2b \cos \frac {A}{2}} \ . (2)$

From $ (1)$ and $ (2), \ \frac {PS}{PN} \equal{} \frac {PO}{PQ} \Longrightarrow OQ \parallel NS$  $\Longrightarrow$ $ OQ \perp BC.$

[hide="Another solution"]Let $ L$ be the second intersection of $ AN$ and the circumcircle of triangle $ ABC$. Let $ D$ on $ AB$ and $ E$ on $ AC$ such that $ LD\perp AB$, $ LE\perp AC$. Easy to see that the Simson line $ (DMN)$ is perpendicular to $ AN$. So $ DM\parallel PQ$. Apply Thales's theorem to pairs of parallel lines $ DM\parallel PQ$, $ LD\parallel OP$, we have that $ \frac{AO}{AL}\equal{}\frac{AP}{AD}\equal{}\frac{AQ}{AM}$.
Hence $ OQ\parallel LM$. And $ OQ\perp BC$ (since $ LM\perp BC$).[/hide]


quadratics
algebra
polynomial
quadratic formula

:bomb:

It's hardly necessary to use such a technique on a factorable quadratic.  It breeds exactly the wrong kind of idea about math - that there is a formula that gives you the answer to every type of question you encounter, if only you know it.  If you tried to do the same to a third degree equation, you would run into [b]significant[/b] difficulties, although many are easily solvable with a few [b]problem-solving skills.[/b]

[hide="The general idea is this:"]Given a polynomial $x^{2}+ax+b$ with roots $r_{0}, r_{1}$ we can say that

$x^{2}+ax+b = (x-r_{0})(x-r_{1})$

[hide="Discussion"] [b]Lemma:[/b]  If $r$ is a root of a polynomial $P(x)$, then there exists a polynomial $Q(x)$ such that

$P(x) = (x-r) Q(x)$

This is a special case of [b]polynomial division:[/b]

[b]Lemma:[/b]  Given two polynomials $A(x), B(x)$, there exists polynomials $Q(x), R(x)$ such that

$A(x) = Q(x) B(x)+R(x)$

And such that $R(x) = 0$ or $deg(R) < deg(B)$.  Now, if $r$ is a root of $A(x)$ and we let $B(x) = x-r$ then it follows

$A(r) = 0 = Q(r) (r-r)+R(r) = 0+R(r)$

And so $R(r) = 0$.  Since $deg(B) = 1$ we have $R(x)$ is a constant polynomial, so $R(x) = 0$ everywhere. [/hide]

And therefore that

$a =-(r_{0}+r_{1}), b = r_{0}r_{1}$

Now, $b =-40 =-8 \cdot 5$.  Moreover, we have $a = 3 =-(-8+5)$, so we have

$x^{2}+3x-40 = (x+8)(x-5)$

Giving us our two solutions $x = 5,-8$. [/hide]

[quote="i_like_pie"][quote="anirudh"]move the forty to the other side :rotfl:

then factor
$(x+8)(x-5)$
solutions are 8, -5 :rotfl:[/quote]
Incorrect. :rotfl:[/quote]
[hide=";)"]If x+8=0 then 8+8 doesn't equal zero...[/hide]

You can shortly solve this by factorisation.
$x^{2}+3x-40═0$
$═(x+8)(x-5)$
so $x_{1}═-8$ and $x_{2}═5$

[quote="mary15"]how do you find x?
x^2+3x=40[/quote]
[hide="solution"]
$x^{2}+3x-40=0$
$(x+8)(x-5)=0$
Thus the roots are -8 and 5. anirudh, please do not laugh, it doesn't help much. Btw, your answer is wrong.[/hide]

[quote="i_like_pie"][quote="anirudh"]move the forty to the other side :rotfl:

then factor
$(x+8)(x-5)$
solutions are 8, -5 :rotfl:[/quote]
Incorrect. :rotfl:[/quote]

$-8, 5$; it is now right

[quote="mary15"]how do you find x?
x^2+3x=40[/quote]

Let me present a more detailed solution:

We know that because this is an equation with degree (highest exponent) 2, it will have 2 solutions.  How do we find these two solutions?  We first will need to use a unique propert of 0.  This property states that if $x \cdot y = 0$ then $x=0$ or $y=0$.  So let's make your equation equal to 0, as that will elp us later.

$x^{2}+3x=40$
$x^{2}+3x-40=0$

Now if we can make the left hand side look like $(x+a) \cdot (x+b)$ we will be able to solve, as you will soon see.  Take a look at the equation below to see what we need to do next:

$(x+a) \cdot (x+b) = x^{2}+ax+bx+ab = x^{2}+(a+b)x+ab$

Well, we want to make your equation look like my equation.  So let's let the your equation equal my equation:

$x^{2}+(a+b)x+ab = x^{2}+3x-40$
$(a+b)x+ab =3x-40$

Now we know that $a+b=3$.  Do you see why?  In your equation, the x is paired with 3, while in mine, the x is paired with a+b.  Since our equations are equal, those must be equal.
The same logic can be applied to give us $ab=-40$.  So now we have a system of 2 equation that we can solve:

$a+b=3$
$ab=-40$

We can see (by experimenting or using algebra) that:
$a=8$
$b=-5$

So let's plug these values back into our original form:

$(x+a) \cdot (x+b) = x^{2}+(a+b)x+ab$

$(x+8) \cdot (x-5) = x^{2}+(3)x-40$

Now, once we get it into this form, let's go back to that special property of 0 we mentioned earlier.  We know that $x+8=0$ or $x-5=0$.  Solve those equations, and get $x=-8$ or $x=5$.

This was a long process, but after a few practice problems, you can do these in your head.  Just ask yourself:

What numbers multiply together to get the constant (c in the below equation), and what numbers add together to get the coefficient of x (b in the below equation):

$x^{2}+bx+c=0$

Now if we have something like $3x^{2}$ instead of just $x^{2}$, it gets harder, but still bery doable...

[quote="Padmavathi"]you get x equal to +8 and -5 right?[/quote]
No, you get $(x+8)(x-5)=0.$ That's $x=\{-8,\ 5\}$

The roots of $(x+a)(x+b)=0$ are $-a$ and $-b$.
The roots of $(x-a)(x-b)=0$ are $a$ and $b$.

If you're careless, then you probably will mess up.

[quote="vishalarul"]Think of it like this:
$x^{2}+3x=40$
$x^{2}+3x-40=0$
$x^{2}-8x+5x-40=0$
$x(x-8)+5(x-8)=0$
$(x+5)(x-8)=0$

Now, if one (or more) of the factors is (are) 0, then the product is 0. Therefore, either $x+5=0$ or $x-8=0$. Solving is easy now. You get $x=-5$ and $x=8$.

Generalization:
$ax^{2}+bx+c=a(x-\frac{-b+\sqrt{b^{2}-4ac}}{2})(x-\frac{-b-\sqrt{b^{2}-4ac}}{2})$

Try to solve for $x$ now.[/quote]
I like your explanation, but you made a mistake: $x^{2}-8x+5x-40=0$ is supposed to be $x^{2}+8x-5x-40=0.$

You guys make it toooooooooooo complicate.

[hide=" :D "]$x(x+3)=40$The only possibilities are $-8$ and $5$[/hide]

[hide="nicer solution"]We can subtract 40 from both sides to get
$x^{2}-3x-40=0$
We can factor:

$(x-8)(x+5)=0$
So x is either
$8$
or
$-5$[/hide]