MATHCOUNTS
function
modular arithmetic
probability

Could you please post the hardest math problem you have ever seen from mathcounts? It has to be from an old year or the handbook. I just need some practice with really hard problems. I'm sure everyone here could use some practicing of hard problems. :ninja:

here's what I got...
[hide]
60 for the first one----might have messed up :?

2nd one : 26*1*25*24*12*10*9=16848000
the 12 is from the ways you can arrange the numbers: AABC - (4*3*2*1)/(2*1)

[/hide]

26*25*24/6*3*12*10*9

Biff is right, I think, I tried this problem in practice but forgot the *3 part,  I know the answer starts with an 8 and looks like biffs or the number you would get if you multiplied all this stuff above...

I think a hard problem is something that I can't solve (I don't mean on the spot, timed, and under pressure...)

[quote="pgpatel"][quote="Ignite168"]Pgpatel wants to choose a license plate with exactly 4 letters and then a hyphen followed by two numbers. If only one letter is repeated and both numbers are different, how many different license plates can Pgpatel choose?
(You might want to use a calculator for this one.)[/quote]
Binonuniquiniest (sp?) that is the pedro target problem isn't it. [hide]26*1*25*24*24*10*9=33,696,000[/hide][/quote]
Incorrect.
[hide="One Solution"]
Keep combinations of letters separate from ways to arrange them.
26 ways to pick the letter repeated, then (25C2) to choose the next two letters as 25*24 would be arranging them. Finally, 4! for arrangements but divide by 2 to arrange the repeated letter. Finally, multiply by 90 to get your answer.[/hide]

I have seen the answer key

and to select three letters

you need 26*25*24/6

because selecting abc and acb is the same

multiply by 3 because you need to select the double

multiply by 12 to order them

multiply by 90 for the numbers

I would not suggest trying to solve the first posted problem in 3 minutes.

During a mathjam, Mr.Crawford commented
[quote]That's the kind of problem I would skip.[/quote]

Of course, you should try to solve it, it uses some important geometric skills(z), but I really doubt we'll see another one of those.

It is just a bunch of triangles...the pattern repeats each time...I know that I posted the answer in the problem solving marathon, but I don't want to dig through all that

[quote="binonunquineist"]Here's a hard problem. Actually, it isn't that hard if you use listing but I want to know how to do it mathematically and efficiently.

For how many positive integers n is n^2 - 19n + 99 a perfect square?[/quote]

Are you allowed to use a calculator? If so, I'd use the table function ;)

[quote="binonunquineist"]Here's a hard problem. Actually, it isn't that hard if you use listing but I want to know how to do it mathematically and efficiently.

For how many positive integers n is n^2 - 19n + 99 a perfect square?[/quote]
[hide] n^2-19n+99=p^2
(n+p)(n-p)=19n-99 is the answer infinitely many?[/hide]

[quote="binonunquineist"]Here's a hard problem. Actually, it isn't that hard if you use listing but I want to know how to do it mathematically and efficiently.

For how many positive integers n is n^2 - 19n + 99 a perfect square?[/quote]

[hide="Now I am positive that this is correct"]

We can rearrange terms to get (n-19/2)^2+35/4.  Assume that k^2 is the perfect square
Now we equate this to get k^2-(n-19/2)^2=35/4.
So we can factor this to get (k+n-19/2)(k-n+19/2)=35/4
So the possibilities are k+n-19/2=35/2, k-n+19/2=1/2, vice versa and k+n-19/2=7/2, k-n+19/2=5/2, and vice versa.  Also we could have k+n-19/2=-35/2, k-n+19/2=-1/2, vice versa and k+n-19/2=-7/2, k-n+19/2=-5/2, and vice versa.
So the ordered pairs possible is (9,18), (9,-18), (3,10), (3,-10), (-9,1), (-9,-1), (-6,9), and (-6,-9).  So there are only four possible positive values of n which give n^2-19n+99 a perfect square
18, 10, 9, and 1.

This should be correct.
Thanks pgpatel
$4$
[/hide]

[quote="mathgeniuse^ln(x)"][quote="binonunquineist"]Here's a hard problem. Actually, it isn't that hard if you use listing but I want to know how to do it mathematically and efficiently.

For how many positive integers n is n^2 - 19n + 99 a perfect square?[/quote]

[hide="I am positive that this is correct"]

We can rearrange terms to get (n-19/2)^2+35/4.  Assume that k^2 is the perfect square
Now we equate this to get k^2-(n-19/2)^2=35/4.
So we can factor this to get (k+n-19/2)(k-n+19/2)=35/4
So the possibilities are k+n-19/2=35/2, k-n+19/2=1/2, vice versa and k+n-19/2=7/2, k-n+19/2=5/2, and vice versa.
So the ordered pairs possible is (9,18), (9,-18), (3,10), and (3,-10).  So there are only two possible positive values of n which give n^2-19n+99 a perfect square
18, 10

$2$[/hide][/quote][hide]What about 1 and 9.[/hide]

I'll edit that, I forgot that you can consider negative solutions for products as well.   :oops_sign:

for Pgpatel I got this
[hide]8424000 (26x25x24x6x10x9)
the 6 is for the 4-letter word's repeated letter
aabc
abbc
abcc
abac
abcb
abca
everything else is pretty obvious  ;)[/hide]

[quote="daermon"]for Pgpatel I got 4212000 (26x25x24x3x10x9)
the 3 is for the 4-letter word's repeated letter
aabc
abbc
abcc

did i get it right?  :?[/quote]

No.

[quote="binonunquineist"]Here's an okay problem that is absurdly messy. (You can strengthen your computation skills by working it out without a calculator):

Pgpatel starts in the center of his circular garden with radius 120 miles. He walks (ever so slowly) north, stopping at the point halfway between the center and the northern tip. Then he goes east, stopping halfway to the end of the garden. Then he goes south and west in the same fashion. After the end of his useless hike, how many miles away is he from his starting point?

Note: I changed the problem a bit but the overall solution is still the same.[/quote]

Umm...I keep getting 40.7 miles. Doesn't sound right. Do you have the answer, binonunquineist? :)

[quote="spx2"]$0$[/quote]
Are you refering to my answer?

[quote="spx2"]what else ?[/quote]
He doesn't walk 60 miles every time, so he doesn't get back to where he started. The first time, he walks 60 miles. The second time he walks $30\sqrt{3}$ miles. It's a quarter of the length of a chord 60 away from the center. You can use Pythagorean and see half the chord is $\frac{1}{2}\sqrt{120^2-60^2}=30\sqrt{3}$ It is also a 30-60-90 triangle.

[quote="biffanddoc"]For that problem I get
[hide]
$26*6*25*24*10*9=8,424,000$
The 26 is the repeated digit, and the 6 is the ways to arrange where the repeated letteres are. Then the 25 and 24 and 9 and 10 are obvious[/hide][/quote]

I think thats right  :lol:

[quote="lotrgreengrapes7926"][quote="Ignite168"]Somehow I think that's wrong :P[/quote]
 :dry: Well, it matches my answer at the top, which I just did with a calculator.

P(ignite gets a 47 at nationals)=0 Haha. It's true.[/quote]

Actually, $47\equiv1\pmod{46}$ so the probability would be quite high.

40.7 IS the correct answer, the real problem said nearest mile so it IS $\boxed{41}$

[quote="lotrgreengrapes7926"][quote="kyyuanmathcount"]
40.7 IS the correct answer, the real problem said nearest mile so it IS $\boxed{41}$[/quote]
Yay!! :wow:

[quote="kyyuanmathcount"]Actually, $47\equiv1\pmod{46}$ so the probability would be quite high.[/quote]
 :huh: Are you saying it's likely that Ignite will get a 1??[/quote]

 :P More than the chance he'll get 47 XD.

When you see problems like that on the target, you do the other one, and MAKE SURE the other one is right.  Then, try to do this one really fast, but don't get bummed and mess up the rest of the target if you don't get it, I mean, you pretty much have a 1/50 chance of guessing it.  Its probably between 30 and 80.

I don't see why this problem seems to be so infamous...you just store the distance it is to the right and above the center as a and b respectively.  Then you make triangles and repeat the process 4 times...

I probably would have a panic attack and have problems though...


I am by no means a expert in this kind of subject, Computer Vision. But after I saw the project “Camera Kombat” http://camerakombat.googlepages.com/ (Warning, is in Portuguese.) I began to wonder some things.

To have a fast recognition we need bad quality?
I'm making (debugging on my mind, some non-programmers call this brainstorm) a strategy to conciliate quality with speed.

First of all I need to present the method of the recognition.

A camera is fixed and record one frame. This is the base frame. After that a person, or a object is inserted in the room and the camera start to record each frame.

A frame is just a picture, so we can reproduce it with a pixel matrix, we can use gray scale (0 for black, 1 for white, and the other are between 0 and 1), since we just want to know if the pixel of the new frame S is equal to the old pixel in the same position of the original frame, say O.

We want to test if $ |S_{i,j}\minus{}O_{i,j}| \ge \rho$, where $ \rho$ is a number near zero (if it where zero-zero, you would notice that some pixels are missing, etc...)

If the image is 1024x768 we have 786432 pixels. So we will need to make 786432 subtractions, apply the function Abs 786432 times and 786432 logical tests.

If the image is 20 frames per second we would end up making 47185920 operations per second.

Not smart. But there is a little tricky.

If we resize the images to a small size, for example 100x75, we would end up with 450000 operations per second. Very good.

But the final image is bad quality. But we know where the pixel of S originated. For example.

Let the original image have width $ w$, and height $ h$. The new image have width $ w'$ and height $ h'$. Let $ n_w \equal{} \frac{w}{w'}$ and $ n_h \equal{} \frac{h}{h'}$.

The pixel of $ S_{i',j'}$ originated from $ M_{i n_h \plus{} k_1,j n_w \plus{} k_2}$ for $ 0 \le k_1 \le n_h\minus{}1$ and $ 0 \ge k_2 \le n_w\minus{}1$.

This last result map a pixel to a pixel matrix.

If this pixel is black and all it's neighbors are black too, we are sure that all elements of this sub-pixel matrix are black too. (black? White? What? If the test $ |S_{i,j}\minus{}O_{i,j}| \ge \rho$ is true, the pixel is changed to black, if not, white. The important think is to know where the pixel is, and not it's color)

So, through relations like this we can reconstruct the original image. Note that we didn't calculate the test of the sub-pixel matrix, we assigned it a value, because we where sure that if we tested it, it would come out this value.

Since the test function cost more time than assigning a value to a matrix, we are done. With some extra calculations that are less time consuming we could end up with the same result.

Now the questions:

All solid things (Real world) projected in a plane are closed (some there are holes, but the central idea here if the recognition of a person. Unless you where shot with a magnum you don't net to pay attention in this.)

So to determine a Object we don't need to give all pixel location of this object. Just the interface is ok, because everything else is in the interior of the object.

So we can create a new rule for the first step, eliminating additional and futile calculations.

For example. We calculate and a cell turns to be black, but if neighbor up it and on the up left and up right and the left cell are black too we don't need to search the neighbor on it's right, because it's probably black. So we are construction the object from outside to inside. For some patterns the next cell can be surely black, probably black or maybe white, so we can jump a lot of cells, depending upon a very long database of patterns and etc.
Some cells could be flagged to “Test” and others.

The problem is, this new implementation is just a piece of junk? It will not improve drastically with speed or will just make the program bigger? It's really useful to spend time on this? NOTE: The process, patterns, etc, are a little bit more sophisticated than I mentioned, but is just a draft of the end product.

My power to resume things is really astonishing , sorry for being too long on this small question

NOTE: The text is confusing, the part I said about sub pixel matrix, the neighbords etc. Just consider the Idea of using a low quality image to determine the pixels on the good quality image.

[quote="EulerIntegral"]Just consider the Idea of using a low quality image to determine the pixels on the good quality image.[/quote]
This idea can and has been taken to extremes in the literature - see this excellent post on the [url=http://terrytao.wordpress.com/2007/04/13/compressed-sensing-and-single-pixel-cameras/]one-pixel camera[/url].


The length of a rectangle is decreased by 7 inches to create a square that has an area of 81 square inches. How many inches are in the positive difference between the perimeters of the two quadrilaterals?

Obviously, we have $ l\equal{}w\plus{}7$, so we see that

$ |2(l\plus{}w)\minus{}4w|\equal{}|2(2w\plus{}7)\minus{}4w|\equal{}\boxed{14}$ (the $ 81$ square inches part is irrelevant).

it might be easier to understand this way
2l+2w-old perimeter
if length is decreased by 7
then it is 2x7 less than the old perimeter
14!

Or we could find the numbers.

Square=9x9 so the rectangle is (9+7)x9=16x9

2(16+9)-2(9+9)=50-36=14


Find a closed form for the sum:
$1\cdot 1!+2\cdot 2!+3\cdot 3!+4\cdot 4!+...+n\cdot n!$

[hide="Hint for method 1"] Induction ;) [/hide] [hide="Hint for method 2"] Telescoping sum ;) [/hide]

The problem is, with method 1, you have to actually know what the sum equals BEFORE you use induction to prove it.  :noo:

How do you do it with a telescoping sum?

So is the answer $(n+1)!-1$, which you can then prove with induction?

let $P(n) = 1.1!+2.2!+..... n.n!$ then $P(n) = (n+1)!-1$ hence we can say that P(1) is true.
let us suppose $P(n)$ will true then we have to prove that $P(n+1) = 1.1!+2.2!+..... (n+1).(n+1)!$ hence $(n+2)!-1 = (n+1)!-1+(n+1).(n+1)!$ taking
LHS we can write $(n+2)! = (n+1)(n+1)!+(n+1)!$ hence we can say that it is proved by mathmatical induction

Are there other ways to do it than telescoping sums or induction?

[hide]I don't think so, but yes, telescoping series would help.

You get the answer $(n+1)!-1$[/hide]

Maybe you could express it combinatorically (as a series of $a \binom{n}{m}$ terms), then try to apply some kind of combinatoric identity. Just a thought, I haven't actually found anything good with this myself.

Add a $1!$ at the start of the sequence and you should notice something interesting.

This is one of our problems this week for our Intermediate Counting & Probability course.  Here's a solution by counting argument:

[hide]Write the first $n+1$ natural numbers in increasing order.  Consider the permutations of these numbers such that not all numbers are in the same position they started in.  There are clearly $(n+1)!-1$ of these permutations.

Suppose $k$ is the greatest number not in the same position it started in, where $2\le k\le n+1$ (If $k=1$, then all of the other numbers are in the same position they started in, and then 1 is in the same position it started in as well, a contradiction).  The number $k$ can go in any of the first $k-1$ places, and then the $k-1$ numbers less than $k$ can be placed in $(k-1)!$ ways.  Therefore, $(n+1)!-1=\sum_{k=2}^{n+1}(k-1)\times (k-1)!=1\times 1!+2\times 2!+3\times 3!+\cdots+n\times n!$.[/hide]


What do you think the AMC10A cutoff for AIME will be?  Post your predictions.  Then, when the official cutoff is posted, the person closest to the score gets a fabulous prize: the URL of my *secret* website.

P.S. Don't post a range - that's just for the poll because I can't put thirty options up there...

i'd say 115 or so because it definitely got harder, but i think there are a lot of smart people this year and the competition's gonna be tough still.

[quote="Tare"]Or the top 1%.

I haven't got a good feeling about this "prize" of yours...something tells me it involves grammar lessons and bunnies :lol:[/quote]

Nope, no grammar lessons or bunnies on my website.  Just math.  (Don't call it a sucky prize either.  It's not.  Ask confuted.)

112.5, it seems that majority of people didn't do well on it (me included)

[quote="tAsDoOm"]It's fixed at 120 this year.[/quote]

wat makes u say that?

it said on the test; it's a new rule.  They still take the lower score, though.

[quote="xxreddevilzxx"]121[/quote]

Psst...that would be impossible, since the qualifying score is [b]at most [/b]120.

Hmm...in my opinion the AMC 10 was quite easy, I just went too slow on it.  I'd still say that the cutoff would be 114, it seems like a lot of people thought it was quite hard.

I think the cutoff will be 113 (that might have been taken already but oh well).
 :wink:

-interesting_move

[quote="TheSouthpaw"]I'm gonna guess it's gonna be around a 112.[/quote]

As the official AIME cutoff for the 2004 AMC10A is 110, our contest winner is TheSouthpaw.  Congratulations!  I'll be PMing you your reward.


We solve the case where $m$ is even. For $m$ odd, the proof can be adapted with no difficulty (leading to a strict inequality).

Since the $a_{i}$'s and the $b_{i}$'s are playing symmetic parts, we may assume that among the $m$ couples $(a_{i},b_{i})$ where $i=1,\cdots ,m$ at least $\frac{m}{2}$ are such that $a_{i}\geq b_{i}.$ (1)
For these $i$ satisfying (1), since the $\frac{m}{2}$ greatest (possible) values of the $a_{i}$'s are $1,\frac{1}{2}, \cdots, \frac{1}{\frac{m}{2}}= \frac{2}{m}$, it follows that $a_{i}\leq \frac{2}{m}$ for some $i \leq m$ satisfying (1).
Thus $a_{m}+b_{m}\leq a_{i}+b_{i}\leq 2a_{i}\leq \frac{4}{m}$, as desired.

Pierre.


Solve in real numbers the following system:

$ \{ \begin{array}{cccc}
(x_1+x_2+x_3)x_4  =  a\\
 \\
(x_1+x_2+x_4)x_3  =  a\\
 \\
(x_1+x_3+x_4)x_2  =  a\\
 \\
(x_2+x_3+x_4)x_1  =  a\end{array}$

Denote $ s \equal{} x_1\plus{}x_2\plus{}x_3\plus{}x_4$. Then the system writes as $ sx_i \minus{}x_i^2 \equal{} a$, for $ i\equal{}1,2,3,4$, hence $ (x_i\minus{}x_j)(x_k\plus{}x_{\ell}) \equal{} 0$, for $ \{ i,j,k,\ell \} \equal{} \{ 1,2,3,4 \}$.
Now a tedious discussion leads to $ x_1 \equal{} x_2 \equal{} x_3 \equal{} x_4 \equal{} \sqrt{\frac {a} {3}}$ when $ a\geq 0$; $ x_i \equal{} x_j \equal{} x_k \equal{} 0$, $ x_{\ell}$ arbitrary when $ a\equal{}0$; $ x_i \equal{} x_j \equal{} \sqrt{\minus{}a}$, $ x_k \equal{} x_{\ell} \equal{} \minus{}\sqrt{\minus{}a}$ when $ a \leq 0$.


$ a,b,c$ are the sides of a triangle. Prove that if $ \frac{a\minus{}b}{c}\plus{}\frac{b\minus{}c}{a}\plus{}\frac{c\minus{}a}{b}\equal{}0$ then the triangle is isosceles.


For the two functions, f(x) and g(x), tables of values are shown below.  What is the value of g(f(3))?

$ \begin{tabular}{c|cccc}
x&f(x)\\\hline
\minus{}5&7\\
\minus{}2&\minus{}5\\
1&3\\
3&2\\
\end{tabular}$

$ \begin{tabular}{c|cccc}
x&g(x)\\\hline
\minus{}2&3\\
1&\minus{}1\\
2&\minus{}3\\
3&\minus{}5\\
\end{tabular}$

[quote="varunrocks"]Are these SAT questions.[/quote]

No. These are practice questions for a community college entrance exam called COMPASS (http://www.act.org/compass/sample/math.html).  I'm trying to help my son study, and I'm posting the questions I don't remember how to solve.


calculus derivative function inequalities

Prove that if $ x,y,z \in [0,1]$ then $ x(1\minus{}y)\plus{}y(1\minus{}z)\plus{}z(1\minus{}x)\leq 1$

Taking the second partial derivative of the LHS with respect to any of $ x,y,z$ gets $ 0$, so the function is convex and it suffices to check the endpoints (i.e. $ x,y,z \in \{0,1\}$). The conclusion is apparent.

(cf. example 6 [url=http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf]here[/url])

No need to take the partials; the function is clearly linear in any one variable.

[quote="balan razvan"]$ \{x,y,z\} \subset [0,1]\ \implies\ x(1 \minus{} y) \plus{} y(1 \minus{} z) \plus{} z(1 \minus{} x)\leq 1$ .[/quote]
$ \minus{}xyz\le 0\le (1\minus{}x)(1\minus{}y)(1\minus{}z)\ \implies\ 0\le 1\minus{}$ $ \sum x\plus{}\sum yz\ \implies\ \sum x\le 1\plus{}\sum yz\ \implies$

$ x(1 \minus{} y) \plus{} y(1 \minus{} z) \plus{} z(1 \minus{} x)\leq 1$ with equality iff $ \{0,1\}\ \subset\ \{x,y,z\}$ .

[b]Another example :[/b] $ \{x,y,z\}\subset [0,1]\ \implies\ x(1\minus{}y)(1\minus{}z)\plus{}y(1\minus{}z)(1\minus{}x)\plus{}z(1\minus{}x)(1\minus{}y)\plus{}xyz\le 1$ .

[quote="balan razvan"]$ \{x,y,z\} \subset [0,1]\ \implies\ x(1 - y) + y(1 - z) + z(1 - x)\leq 1$ .[/quote]
Here is a gemetrycal proof for this inequality;
Consider an equilateral triangle with the sides $ 1$.
Let $ M,N,P$ on the sides of the triangle such as $ AM=x$, $ NC=y$,$ BP=z$
${ [AMN}+[NPC]+[BMP]\leq [ABC]$ $ \implies$ $ x(1-y)\frac{sin60}{2}+y(1-z)\frac{sin60}{2}+z(1-x)\frac{sin60}{2} \leq 1\cdot 1 \frac{sin60}{2}$

[hide="Without partial derivatives or geo"]
$ x \minus{} xy \plus{} y \minus{} yz \plus{} x \minus{} xz \le 1$
$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \le 2 \plus{} xyz$, which is clearly true, because LHS $ \le 0$
[/hide]


complex analysis
function
integration
topology
complex analysis unsolved

Let $E=\{z\in \mathbb{C}\ | 1\leq |z|\leq 2\}.$ Prove that there doesn't exist the holomorphic function $f(z)$ on the closed set including $E$ such that
\[ Re f(z)>0\ \ \text{when}\ \ |z|=1.  \]
\[ Re f(z)<0\ \ \text{when}\ \ |z|=2.  \]

A function holomorphic on this annulus is given on the interior of the annulus by a Laurent series:

$f(z)=\sum_{n=-\infty}^{\infty}a_nz^n.$

WLOG, we can assume that this series converges absolutely and uniformly on the circles $|z|=1$ and $|z|=2.$

But we now have two formulas (Fourier coefficient formulas) for the constant term $a_0.$

1. $a_0=\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\,d\theta$

2. $a_0=\frac1{2\pi}\int_0^{2\pi}f(2e^{i\theta})\,d\theta$

But from formula #1, $\text{Re}(a_0)>0$ and from formula #2, $\text{Re}(a_0)<0.$ These can't both be true.


function
inequalities
inequalities proposed

Prove that for any $a,b,c \ge 0$ we have
\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+3\sqrt{3}\cdot\sqrt{\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}}\ge \frac{7}{\sqrt{2}}\]
:)

Oh , it's very beautiful but hard , too ! Could you please to give me a hint ?  :blush:
The following is easier but beautiful , too !

Let $a,b,c \geq\ 0$ , prove that :

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{5}{2}\sqrt{\frac{10(ab+bc+ca)}{a^{2}+b^{2}+c^{2}}}\geq\ \frac{15}{2}$
 :lol:

:blush: It is easer than you thought, tanpham. It is much easier than problems you sent me. Just a little mixing variable or using the statement  I posted here

http://www.mathlinks.ro/Forum/viewtopic.php?t=144697

Yes, that's my solution, Hung. :)
Can you show me the harder problem?

[quote="tanpham90"]Cool !  :blush:
Very nice , hungkhtn and toanhocmuonmau !  :lol:
May be this one is harder !
Let $a,b,c \geq\ 0$ Prove that :
$\frac{a^{4}}{b^{2}+c^{2}}+\frac{b^{4}}{c^{2}+a^{2}}+\frac{c^{4}}{a^{2}+b^{2}}+12\sqrt{6}(ab+bc+ca) \geq\ 15(a^{2}+b^{2}+c^{2})$[/quote]
Note that $ab+bc+ca \ge \sqrt{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}$, hence it suffices to prove
\[\sum\frac{x^{2}}{y+z}+12\sqrt{6(xy+yz+zx)}\ge 15(x+y+z) \]
or equivalently
\[(x+y+z)\sum\frac{x}{y+z}+12\sqrt{6(xy+yz+zx)}\ge 16(x+y+z) \]
Setting $p=x+y+z,q=xy+yz+zx,r=xyz$, then we have to prove
\[\frac{p(p^{3}-2pq+3r)}{pq-r}+12\sqrt{6q}\ge 16p \]
It's easy to check that this is an increasing function in terms of $r$, hence if suffices to show that
\[\frac{p^{3}-2pq}{q}+12\sqrt{6q}\ge 16p \]
Without loss of generality, suppose that $q=1$, hence $p \ge \sqrt{3}$, we have to prove
\[p^{3}-18p+12\sqrt{6}\ge 0 \]
By AM-GM inequality, we have
\[p^{3}+12\sqrt{6}=p^{3}+6\sqrt{6}+6\sqrt{6}\ge 18p \]
We are done. :)

5 minutes ! Very Nice toanhocmuonmau !  :D
 :first:


A rancher owns land by a river bend. The radius of the river bend at the water's edge is $ 300$ feet. He has $ 1600$ feet of fencing. He wants to fence off a lot at the water's edge, so that it has one side on the front and two equal sides ending at the river. The corners are to be right angles. The river bends into the lot, so that the resulting lot will be a rectangle with one concave side. What should the 3 sides be for the area of the lot to be the maximum? I'd like an exact integer answer, please?

PS: How to create a diagram?

How do you know the answer is an integer?  And have you set up equations yet?

Well, the question asks for an integral answer.  I know the equations and it was solved using calculus (with aid from softwares).  I'm looking for a non-calculus way.

By the way, why aren't I receiving topic notification after you posted?  That's strange.


Nowadays I'm disappointed by the relatively low number of people at my school carrying around tracks containing the words "concerto", "symphony", "sonata", etc. on their i-pods and cd's.  Classical music most definitely needs a new era of rejuvination (is that spelled correctly?) with young people, like the majority of us.

I'm sure the percentage of classical music lovers on AoPs is quite a bit higher.  So if you like the [i]true[/i] "oldies", what are some of your favorites?  And what are some good sites for free (and legal) classical music downloads?  I know classiccat.net is one.  But there must be others.

Some of my all-time favorites (usually things I have played) are Dvorak's 9th Symphony (New world), which I haven't played.  I also am really starting to like a number of overtures, including Ruslan and Ludmilla, Oberon, and even the 1812 one.  For piano music, I really like 19th century-early 20th century music, mainly because of how complicated it is.

And you?

63 out of 263 songs are classical
I like canon

I mostly like symphonies. Especially Dvorak 1, 4, 7, 8, 9; Beethoven 7; Shostakovich, Mahler. I also like some violin and piano concertos, like Tchaikovsky (especially his violin concerto), Korngold, Brahms double concerto, Dvorak violin and cello, blah blah so many.

Heifetz was good.

I also listen to a lot of industrial music and folk though, so I had to vote for a lower percentage.

[quote="akalra1"]
new music is so much better, i havent seen any of these beethoven guys on the top ten

[/quote]

What r u talking about ?  That Beethoven guy is dead from 1827

:D

XD

IM TALKING ABOT ANY NEW CLASSICAL SONGS

Anyway...I love classic music  :)

Rachmaninov and Liszt  still ROCK    :D

Just because songs are popular doesn't make them better than those that aren't. Being in the top 100 just means that the average person likes a song.

Well, it's hard to give a percentage, it's not zero for me, but it won't be much either.

I once read a quote "Classical music is music during which you keep hoping something catchy will still come along eventually..." :)
It is true that during some classical pieces, I just sit there thinking "this is going nowhere"
But in fact there are many very famous classical "catchy tunes" around.

...maybe it wouldn't be so bad if there was a list or something like that, so people could rediscover how many fine things classical music has given us. :|

[quote="Ramanujan"][quote="Ignite168"]I haven't seen any new classical songs in the top 100.[/quote]

Do you guys listen only song that are in the top 100 or 1000 or 276 ???

And of course you will not find since classical music is pretty much over.[/quote]

I never said I did. I'm just pointing out akalra1's point was wrong.

[quote="akalra1"]no u were defending it

top 10 is inside top 100[/quote]

Attacking a crappy attack is not defending

Yeah but no "new" classical music is in top 100 meaning it's not in the top 10 either

Dvorak wrote his first symphony for a competition. He submitted the only copy of the music, and they didn't give it back to him. Years later, he told some of his students about this, and they asked him what he did about it. He replied, "Nothing. I sat down and wrote another one."

I do like Brahms as well (especially the double concerto).

I recently did a study(Meaning I went out and gained access to the scores of the music(usually meaning a guitar tab with a melody over it) and analyzed it by the methods of music theory and the musicologists) of a great amount of the modern music, ignoring rap though. What I discovered about it is that a good amount of it is simply bad classical to late romantic music on different instruments. Now there are of course many exceptions, I listen to ACDC, Paul McCartney, Elton John, etc. Now for a lot of the others... Evanescence, Blink 182, etc. That's just bad music.

If the music is actually interesting I'll probably give it a try, but unfortunately all most of it seems to be is one or two lame progressions that are repeated ten times on overdriven instruments. I wont' even begin on the drummers... Particularly for punk music...

I've often thought that truly good music can be played in many different genres, and be considered good. An example: I have heard Queen's Bohemian Rhapsody played beautifuly by a string quartet, and I play it on saxophone.

Take the tune of the song "I Don't Know How To Love Him" from Andrew Lloyd Webber's musical "Jesus Christ Superstar," change the meter from 4/4 back to 3/4 or 6/8, take away the bit of bombast at the end of the phrase ... and surely the violinists here would know where that melody comes from.

Yeah there's songs like that.

I could have sworn there was some Brahms in this other song I was listening to...Good Charlotte or something.

Also, I was listening to some Led Zeppelin, and I realized that they were playing a twelve bar blues :D , one of the most common jazz styles.

I think the Howard Shore Lord of the Rings music soundtrack is pretty much the best Western music I've listened to.

[quote="thatin"]is nightwish classic?
and why this topic isn't at music&art forum?[/quote]

no, [url=http://en.wikipedia.org/wiki/Nightwish]nightwish[/url] isn't classic.

[quote="kyyuanmathcount"][quote="akalra1"]I HATE CLASSICAL, NO OFFENSE

new music is so much better, i havent seen any of these beethoven guys on the top ten

the only way i listen to classical music is if i have to
for example at skool wen i have to sing in chorus and the background is violins, etc[/quote]

Ah em, have you ever really tried the good things in classical music?

100% definitely.  Half of the pop stars can't sing for their lives, so I just don't listen.[/quote]

Can you show me all the pop singers in the top 10? Thanks.

Just because they are in the top 10 doesn't mean that they are good singers.

[quote="Treething"][quote="thatin"]is nightwish classic?
and why this topic isn't at music&art forum?[/quote]

that's like asking if dragonforce is classic...[/quote]

Certainly I know that there are not classic if we want to be accurate. I was talking of some songs like sleeping sun and imo this piece is more classic than metal or opera metal etc
BTW I haven't listened dragonforce


Can anyone post links? For books like an excursion in mathematics etc. Basically I want books with more solved examples and theory to build up a good base in the 5 days I'll have for preparing from 13-18.  :huh:
And I thought it stupid to purchase books only for 5 days :p
Any help would be appreciated.  :lol:

[quote="hell_ever"]I have some packets on geometry and inequalities. You can give me your email id and i can mail it if you so wish...[/quote]
Thanks! Sending you a mail.


are all consecutive interior angles supplementary? no matter if lines parallel or not?

Hi. Unless I've misunderstood your question, the answer is that they are only supplementary when the lines are parallel.

To add to what AndrewTom said, no, they're not supplementary unless the lines are parallel.

Many of those kinds of theorems, like the Corresponding Angle or the Alternate Exterior Angles, are only true if the lines intersected by the transversal are parallel.

For an example in your particular case, extend the sides of a polygon, like a regular hexagon. The sum of any two (consecutive or not) interior angles is 240 degrees.

The lines must be parallel. On the other hand, if you have supplementary consecutive interior angles, then the lines must be parallel.

Yes, you can use the converse to prove that the lines are parallel.

They are only supplementary if the interior angles are made by parrallel lines.

Don't post the SAME answer OVER AND OVER again.


Let $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}1$
 Find max of $ \frac{1}{1\minus{}ab}\plus{}\frac{1}{1\minus{}bc}\plus{}\frac{1}{1\minus{}ca}$

[quote="Allnames"][quote="Algadin"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
 Find max of $ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}$[/quote]
It is old,I remember zhaobin's proof is one of the nicest one for it
Hint
Use AmGM in the first and CS then
The result is $ \frac {9}{2}$
Let try[/quote]
Can you post the proof?

It 's here
http://www.mathlinks.ro/viewtopic.php?t=225743


take a look here, I found this article very interesting (maybe because I know most of the people involved, but never heard the story) :)

http://articles.findarticles.com/p/articles/mi_m1511/is_6_21/ai_62277745

[quote="PiDeltaPhi"]Dominating,Majorizing  :D  :D  :D  this is so so so funny,i cant believe it is true[/quote] it is! :) what the writer didn't mention, was that also the coach of the USA team was Romanian :P

His name was mentioned,so you can figure it out  :)

[quote="Valentin Vornicu"]take a look here, I found this article very interesting (maybe because I know most of the people involved, but never heard the story) :)
[/quote]

Hi Valentin,

You say that you know most of the people involved in the story about Melanie Wood.  You may have lost touch with Melanie in the last few years (have you?).  She still makes news.  She won the 2004 AMS-MAA-SIAM Frank and Brennie Morgan Prize for Outstanding Research in Mathematics by an Undergraduate Student.  See  [url]http://www.maa.org/news/morganprize04.html[/url] for a news release from the Mathematical Assocation of Amercia.

The person who was awarded Honorable Mention, Karen Yeats, I'm proud to say, is Canadian.  I actually met Karen briefly at the Canadian Undergraduate Mathematics Conference in Calgery in 2002.

Yours truly,

mathfreak


function
algebra
polynomial
algebra solved
Polynomials

Given non-constant linear functions $p_1(x), p_2(x), \dots p_n(x)$. Prove that at least $n-2$ of polynomials $p_1p_2\dots p_{n-1}+p_n,  p_1p_2\dots p_{n-2} p_n + p_{n-1},\dots p_2p_3\dots p_n+p_1$ have a real root.

Let $q_i$ be the $i$'th polynomial in the list in the previous post, and let $x_i$ be the root of $p_i$. Assume there are three $q$'s with no real roots. We may assume WLOG that $q_1,q_2,q_3$ have no real roots. Now let $P=\prod_{i\ne 1,2,3}p_i$.

We thus know that $p_1p_2P+p_3,p_2p_3P+p_1,p_3p_1P+p_2$ have no real roots. By plugging the roots of $P$ in these polynomials, we find that $P(x_1),P(x_2),P(x_3)$ all have the same sign. Now plug $x_i$ in $q_i$ to get a contradiction: we find $(x_1-x_2)(x_1-x_3),(x_2-x_1)(x_2-x_3),(x_3-x_1)(x_3-x_2)$ to have the same sign, which is impossible.

I will assume that the coefficients of $p_i$ are real. Let us sort $p_i$ by the size of its root $r_i$. Let $q_i = \prod_{j = 1, j \neq i}^n p_j +p_i$. Clearly there is a sign change between $q_i(r_{i-1})$ and $q_i(r_{i+1})$ for all $2 \le i \le n-1$ and that is n-2 of the polynomials.


geometry
incenter
geometric transformation
homothety
geometry solved

Let I be the incenter of ABC.  let incircle meet BC at D.  Let M be the midpoint of BC and DD' be the diameter of incircle.  show that AD' is parallel to IM

Let D'' be the pt where the exinscribed circle corresponding to side BC touches this side. It's well-known that MD=MD'' .

The tangent to the incircle at D' is parallel to BC, which means that this tangent is obtained from BC by a homothety of center A. We can see that by this homothety the exinscribed circle becomes the incircle, so D'' becomes D', so A, D'', D' are collinear.

I is the middle of DD' and M is the middle of DD'', so MI || D'D''=AD'


http://www.nytimes.com/2007/04/16/us/16cnd-shooting.html?em&ex=1176868800&en=e24ccc9c3571672f&ei=5087%0A
So, so ...   :mad:

32 according to foxnews. http://www.foxnews.com/story/0,2933,266310,00.html

It's just horrible...

Several people in my neighborhood go there...

Wow.  My dad, my mom, three of my uncles, and an aunt went there.  :o

i like how most of the campus was like "lol gunfire" even two hours into the shootings.

I have just seen the news on television. I condole with them on their death.

Holy crap....

My condolences to all those affected...

Does anyone go to or has anyone gone to Virginia Tech?  That was scary...

[quote="Ignite168"][quote="Art of Owna"]omg rumor has it that the shooter was asain... :o and not that it should start anything, but I bet someone will say something about that fact.[/quote]

why must you bring race into this[/quote]
Dude Im not bringing race.  I personally am Asian, if you read my thing, you would note that I dont care what race the person was, it will always be a dispicable act.  I was just pointing out that the person was Asian and that might cause some controversy.

[quote="Thu7"]deaths up to 33 now
i give my respects to those that died[/quote]

I might be completely wrong, but I think the 33 includes one of the shooters who shot himself after the shootings.

[quote="Ignite168"][quote="Art of Owna"]omg rumor has it that the shooter was asain... :o and not that it should start anything, but I bet someone will say something about that fact.[/quote]

why must you bring race into this[/quote]
I agree with Ignite. Most likely the motivation behind the shooting involved some sort of aberrant psychology on the part of the shooter and unfortunate social circumstances. Even if the shooter was in fact Asian (witness testimony is often wildly wrong), it's rather presumptive to claim that race is relevant. It wouldn't be a controversy at all if it weren't for people like you bringing it to attention.

One searches for reasons why it couldn't happen "here." Isn't there something about VPI that would make it uniquely vulnerable, some reason special to that university and that that campus? But I can't think of anything that would apply to VPI that wouldn't apply equally well to a hundred or a thousand other campuses. Whatever you think might have or could have been done - from other students packing heat, to a group of people suicidally charging the shooter, to a quicker or more effective police response - it boils down to this: no one went to class expecting that anything like this was even possible. It's very, very difficult to react effectively to something so unexpected, so alien to all of your experiences.

[quote="Xevarion"]Didn't some prominent blogger claim that this would not have happened / been much less likely if VA allowed concealed weapons? Personally, I think WAY fewer people should own weapons and it should be MUCH more difficult to obtain them. The 2nd Amendment clearly states that the purpose of the right to bear arms is to ensure that the populace would be capable of staging another revolution in case the government went bad. Obviously having a concealed handgun (or any other small weapon for that matter) is not going to help in this situation nowadays, so I see no Constitutional justification for having deadly weapons in the first place. (I suppose you could argue for "pursuit of happiness" -> "ppl should be allowed to hunt" but that doesn't have anything to do with carrying concealed or visible weapons around when you're walking about town. Also, just because we have the right to do something doesn't mean we have the right for it to be easy or lightly regulated.)[/quote]
[urlhttp://www.cnsnews.com/ViewNation.asp?Page=%5CNation%5Carchive%5C200209%5CNAT20020917a.html]The thing is, it already did happen.[/url]

And we see now why we should not be so quick to assume things right after the innocent.

We've gone from a 30 year old Chinese immigrant that wasn't even a VT student to a 23 year old Korean US Resident who was a VT student.

I read that he has a sister at Princeton. I really feel sorry for her -- people are probably going to fear/hate her now because of her brother.

Very sad.

(Among victims is a prof who was from   IIT Kanpur  who died.  Also One of my son's friend was shot)

The shooter's family won't necessarily be feared and shunned by others.  I hope that the Princeton community does not allow that to happen to his sister.  A few kind and encouraging words from professors, classmates, and friends can have a huge impact.

My campus has had a few recent events where our students have been charged with murder.  A police officer was killed at an off-campus party that got out of hand a few weeks ago.  The arrested student's mother happens to also be a student in a class that I'm teaching this term.  Our campus community encouraged her to return to classes and get on with her life, despite looming capital murder charges against her son.  She's taken our advice and returned.  I've not sensed any fear or hate towards her by anyone on our campus.

I'd like to suggest that those of you who want to discuss policy issues (e.g. gun laws) go do it in another thread.  Thanks.

Wow... these people who died didn't get to enjoy life after all of that stress and hard school work... I guess where we go to college can also be a matter of life and death too  :(

wow. just wow. i wonder how he (shooter person) became like this.

[quote="jeez123"]I guess where we go to college can also be a matter of life and death too[/quote]
I don't see anything in the story that could not happen in my current or former university, or any other. Last year I seriously considered applying for a job at Virginia Tech. (They have a good math department, and some of their research is in my area.) Will definitely apply this fall if they are hiring again.

I can't blame the university administration or anyone other than the killer. Perhaps they could introduce a more efficient emergency communication system (e.g., texting to students' cell phones instead of sending emails.) But nothing will change the simple fact that none of us are guaranteed to live another day.

mlok, it appears that some aspects of Virginia Tech's history are extremely similar to Texas A&M - a time in the past when it was all-male and every student belonged to a Corps of Cadets and did military training, but now a much larger a much more general university, with the Corps of Cadets still in existence but representing a minority of the students.

Unfortunately, the VPI-TAMU similarity goes further: Texas A&M had a [url=http://bonfire.tamu.edu/]deadly accident[/url] on its campus before I got here. Aggies can feel the pain of Virginia Tech.
[img]http://dmc-news.tamu.edu/thumbnails/4496.gif[/img]

[quote="Treething"]Hmmm.

Have you read his Richard McBeef play?

http://news.aol.com/virginia-tech-shootings/cho-seung-hui/_a/richard-mcbeef-cover-page/20070417134109990001[/quote]

wow, that's intense... no wonder his English teacher was worried. All of his resentments seem to be pouring into his writing. That's a very sad play.

[quote="Treething"]Hmmm.

Have you read his Richard McBeef play?

http://news.aol.com/virginia-tech-shootings/cho-seung-hui/_a/richard-mcbeef-cover-page/20070417134109990001[/quote]I know. That's just...it seems as if he had uncontrollable anger building inside of him...for one reason or another...and he was quiet and never let it out...until last Monday...far too violently...

Ya that was a pretty disturbing play...I can see why his english teacher made him transfer :(

i havent read his play
and dont want to
after all ive heard about him i hope i never hear anything about that again
its just too depressing

[quote="Stgmihnea"]Guns should be made illegal in USA. Without the possibility of buying a gun legally in a competitive market, this guy (and other killers) would not habe been able to get weapons and massive amounts of ammunition (this guy fired 50-70 bullets by police estimates). (the prices for guns and bullets would be MUCH MUCH higher since you would have to get them illegally and they would not be easily available)[/quote]

No. Guns should be more widely available. If a lot of people have guns, then people will think twice about committing crimes.

[quote="Treething"]No. Guns should be more widely available. If a lot of people have guns, then people will think twice about committing crimes.[/quote]
Will you carry your gun loaded all day if you have one? Especially when you are in school?
Furthermore, concealed weapons should be allowed as well if you want to scare the potential shooter, otherwise the shooter can decide whether to begin his/her job by simply looking around.
What if the shooter decide to kill as many as he/she can before comitting a suicide?

Stgminhea, Treething, Buffalo: There's another thread for discussion of gun control. It's like the second thread in this forum. Or were you too afraid to enter into a discussion where people like JBL and PenguinIntegral were defending the two sides of this issue with actual well-thought-out arguments and references?

[quote="Xevarion"]Or were you too afraid to enter into a discussion where people like JBL and PenguinIntegral were defending the two sides of this issue with actual well-thought-out arguments and references?[/quote]

How rude and elitist.

Anyways, I'm wondering how you guys got the feeling of "anger" out of his Richard McBeef play, it might've been "anger" if the kid's rage was building inside him and he ended up killing the dad, but it was quite the opposite.

Those that died are in my prayers. May they rest in peace.

I wonder what could have driven someone to such drastic action.  :|

you have to admit though, those kids love their schools. I only heard one person say that they are considering not attending next year.

I hear college life is one of the truly exciting times in a person's lifetime; well maybe all of school accounts for that.

This incident probably could have happened to any of various colleges in the world, so I don't see how this makes Virginia Tech better or worse to go to or stay than to go to another school.


How can I express an absolute value function as a piecewise function?

For example:

How can i rewrite

f(x) = |x-2| - |x + 3| as a new piecewise function so it is easier to work with.

Any help is appreciated.

You know that $|x+a|$ is

$x+a$ if $ -a \leq x$

$-x-a$ if $x \leq -a$
--------------------------

For your problem consider the cases
$x \leq -3$
$-3\leq x \leq 2$
$2\leq x$


[hide="Think so"]
$ a \minus{} b \equal{} r$
$ b \minus{} c \equal{} s$
$ c \minus{} a \equal{} t$

$ \equal{} r^3 \plus{} s^3 \plus{} t^3$
$ \equal{} (r \plus{} s)(r^2 \minus{} rs \plus{} s^2) \plus{} t^3$
$ \equal{} (a \minus{} c)[(a \minus{} b)^2 \minus{} (a \minus{} b)(b \minus{} c) \plus{} (b \minus{} c)^2 \minus{} (a \minus{} c)^2]$
$ \equal{} \boxed{3(a \minus{} c)(b \minus{} a)(b \minus{} c)}$

$ \equal{}3rst$[/hide]

Notify me of any errors thanks.  :lol:

Yeah. I think that's right. That's how I did it.  :)

Does anybody know if there's some other way to do this? Like a sum of three cubes formula or something?

Well, $ x^3\plus{}y^3\plus{}z^3\equal{}3xyz$ if $ x\plus{}y\plus{}z\equal{}0$. :)

i directly used formula fiven by Brut3Forc3

[hide="The actual formula is"]
$ x^3\plus{}y^3\plus{}z^3\minus{}3xyz \equal{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx)$

by putting $ x\plus{}y\plus{}z\equal{}0$,LHS becomes $ 0$ and thus we have the formula as given by Brut3Forc3
LHS of the formula is often written in different ways as per requirement
[/hide]

Yes, the formula is given [url]http://www.mathlinks.ro/viewtopic.php?t=221470[/url] and can also be used to prove AM-GM for three variables.

I had solved a problem which said: Find all integers $ x,y,z$ such that $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}p$, where $ p$ is a prime. Now with the factorization which skand provided, it's easy!

Let $ P \equal{} (a\minus{}b)^3 \plus{} (b\minus{}c)^3 \plus{} (c\minus{}a)^3$.
$ P$ is cubic alternating polynomial of $ a$, $ b$, $ c$.
So, $ (a\minus{}b)(b\minus{}c)(c\minus{}a)|P$.
$ \therefore P\equal{}k(a\minus{}b)(b\minus{}c)(c\minus{}a)$  ($ k$:const)
$ P_{(a, b, c)\equal{}(0, 1, 2)} \equal{} 6$, $ k(a\minus{}b)(b\minus{}c)(c\minus{}a)|_{(a, b, c)\equal{}(0, 1, 2)} \equal{} 2k$
$ \therefore k\equal{}3$

Thus, $ (a\minus{}b)^3 \plus{} (b\minus{}c)^3 \plus{} (c\minus{}a)^3 \equal{} 3(a\minus{}b)(b\minus{}c)(c\minus{}a)$

Let $ a\minus{}b\equal{}x,\ b\minus{}c\equal{}y$, then $ c\minus{}a\equal{}\minus{}(x\plus{}y)$.
$ \therefore (a\minus{}b)^3\plus{}(b\minus{}c)^3\plus{}(c\minus{}a)^3\equal{}x^3\plus{}y^3\minus{}(x\plus{}y)^3$
$ \equal{}\minus{}3xy(x\plus{}y)\equal{}3(a\minus{}b)(b\minus{}c)(c\minus{}a)$.


for a rectangle ABCD (BC=2AB), take an arbitrary point X on BC.
draw altitude BE on AX, altitude CF on DX.
let M be the midpoint of AD.

prove that $ \square{MEXF}$ are cyclic.

[hide="Nice!"]
Let $ AB \equal{} CD \equal{} n$, $ AD \equal{} BC \equal{} 2n$, so $ AM \equal{} MD \equal{} n$. $ \triangle CDF \sim \triangle CDX \implies DF \cdot DX \equal{} n^2 \implies \triangle DMF\sim \triangle DMX \implies \angle DMF \equal{} \angle DXM.$
Similarly $ \angle AME \equal{} \angle AXM$. So $ 180 \minus{} \angle EMF \equal{} \angle AME \plus{} \angle DMF \equal{} \angle EXM \plus{} \angle FXM \equal{} \angle EXF \implies MEXF$ is cyclic. [/hide]

By the way, does anyone know how to fix the spacing in the above solution?

Use the align* environment instead of that one ridiculously long equation.


[quote="onlymath"]Consider $ \frac {1}{x \plus{} y \plus{} 1}$ $ \plus{}$ $ \frac {1}{x \plus{} z \plus{} 1}$ $ \plus{}$ $ \frac {1}{z \plus{} y \plus{} 1}$ $ \geq$ $ 1$
prove that:

$ 3$ $ \geq$ $ xy \plus{} xz \plus{} zy$[/quote]
It's wrong. Try $ x\equal{}2,$ $ y\equal{}\minus{}2.9$ and $ z\equal{}\minus{}10.$ :wink:

Dear Arqady $ x,y,z$ $ \in$ $ R^\plus{}$

[quote="onlymath"]Dear Arqady $ x,y,z$ $ \in$ $ R^ \plus{}$[/quote]
Now it's true.
$ \sum_{cyc}\frac {1}{1 \plus{} x \plus{} y}\geq1\Leftrightarrow2(x \plus{} y \plus{} z \plus{} 1)\geq(x \plus{} y)(x \plus{} z)(y \plus{} z).$
But $ (x \plus{} y)(x \plus{} z)(y \plus{} z)\geq\frac {8}{9}(x \plus{} y \plus{} z)(xy \plus{} xz \plus{} yz)$
and $ x \plus{} y \plus{} z\geq\sqrt {3(xy \plus{} xz \plus{} yz)}.$
Hence, $ 2(x \plus{} y \plus{} z \plus{} 1)\geq\frac {8}{9}(x \plus{} y \plus{} z)(xy \plus{} xz \plus{} yz).$
Let $ xy \plus{} xz \plus{} yz \equal{} 3v^2,$ where $ v\geq0.$
Hence, $ \frac {9(x \plus{} y \plus{} z \plus{} 1)}{4(x \plus{} y \plus{} z)}\geq xy \plus{} xz \plus{} yz$ gives $ \frac {9(3v \plus{} 1)}{12v}\geq3v^2.$
But $ \frac {9(3v \plus{} 1)}{12v}\geq3v^2\Leftrightarrow4v^3 \minus{} 3v \minus{} 1\leq0\Leftrightarrow v\leq1.$ Done!


Here are the rules:

1. Answer the problem that the person who posted above you wrote, with a full written solution.
2. Post your own problem. The problem can be anywhere from MATHCOUNTS to AIME level, but please don't make insanely hard olympiad problems (that might reduce the popularity of this, and not many Illinoisians might come here).
3. You can only post here if you live in Illinois. Outsiders are welcome to view the problems, but if they want to use them, only in the public marathons, not here.
4. Have fun! :D

Ok let's begin! :D

Ruzan and Rubin are playing a countdown round. Ruzan has a 1/3 chance of getting a problem right. Rubin has a 1/2 chance of getting a problem right, and there's a 1/6 chance that neither player gets a problem right. The person who has the most correct problems after 3 problems wins. If there's a tie, tiebreak problems are imposed such that the first person to answer a tiebreak problem correctly wins. What's the probability that Rubin wins?


Solve the following equation:
$ x-\sqrt{x^{2}-16}=\left(\sqrt{x-3}-\sqrt{x-5}\right)^{4}.$

Multiply the left hand side by : x+sqrt(x^2-16)/(x+sqrt(x^2-16)), also multiply the right hand side by : sqrt(x-3)+sqrt(x-5)/(sqrt(x-3)-sqrt(x-5)), we will find that : x+sqrt(x^2-16)=sqret(x-23)-sqrt(x-5). May it can help you.
God bless you.

The hint: the root is $ \frac{65+\sqrt{257}}{16}.$ :wink:

K81o7, my solution is the same! See here number 47:
http://hydepark.hevre.co.il/topic.asp?whichpage=24&topic_id=2250792


The logarithm of $ 27\sqrt[4]{9}\sqrt[3]{9}$ to the base $ 3$ is:
$ \textbf{(A)}\ 8\frac{1}{2} \qquad\textbf{(B)}\ 4\frac{1}{6} \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these}$


MATHCOUNTS function Support geometry trigonometry videos probability

To the best of my memory from my kids and I getting calcs, here is hte last 9 calculators given out at Nationals.

1999 TI-92 (not allowed in MC due to QWERTY keyboard)
2000 TI-89
2001 TI-89
2002 TI-89
2003 TI-83+Silver
2004 TI-83+Silver
2005 TI-84+Silver
2006 TI-84+Silver
2007 TI-84+Silver
2008 ??????????  Probably TI-84+ silver, hopefully TI-Titanium......time for a new calc in my bag :lol:

[quote="bleuwiki"]I'm a 7th grader, and this is my first year going to Nationals, so I don't know much about this stuff..... , but do they give calculators to all the people who qualified to Nationals or do you have to be a certain place?[/quote]

They give it to everybody.

[quote="Lawrence Wu"][quote="happyme"][quote="karatemagic7"]does anyone have any idea what kind they are giving, or if they decided not to give one this year?[/quote]

They usually give TI-84 Silvers.

Unless Texas Instruments somehow stopped sponsoring MathCounts (which I highly doubt) all Nationals Kids will be getting their hands on one of those.[/quote]
YAY!!!
As a Texan, I'm extremely proud of Texas Instruments (note my sig and avatar)
lol, when compaq was still in houston, we got free laptops at chapter, but now all we get are these stuffed animal thingies :wallbash_red:[/quote]

Did they really give free laptops?  I mean the cost of a average bad laptop is $ \$$400, so multiply that by 200, thats a lot of money compaq wasted.

[quote="levans"]One thing I've always wondered about these calculator prizes...

Don't most contestants already have a good calculator?  I guess the kids could always donate them to their school :)[/quote]

When I went in 06, I had a year and a half old 83. So, I guess there are worse calcs, but I needed a new one.

[quote="levans"]One thing I've always wondered about these calculator prizes...

Don't most contestants already have a good calculator?  I guess the kids could always donate them to their school :)[/quote]
I still have a scientific one.

I have an 89-Titanium for competitions, but I still have my trusty TI-30 for the ACT...  :lol:
Though, when I was in MC, I guess I kinda, sorta, failed miserably...

Does it matter what kind of calculator they give at Nats? I thought people did math competitions purely for sport and math, not for calculators.

It's always exciting to get a (Free) prize!

[quote="gauss1181"]Will the Nationals ever give TI-89 Titanium calculators?[/quote]

Once they become a second-hand thing, I guess. Maybe when the TI-100-something comes into popular demand.

/me hates blockdude also    :D
But anyways, my goal is also to get top 57. My extremely hopeful is to get 20, my near impossible is countdown, but as you said, who knows, there are some people who do horrible at states and great at nats and vice versa. (I had a cold at states and only got a 33  :mad:) I bet you will do very well ulvidad
 [size=59] backwards [/size]

Back on topic, about the prizes, when not including calculators, they might have some grand prizes for the winners, considering they are in their 25th year.

I have a cold almost every day of my life. :mad:
So I really hope I don't get a cold on the 7th of May through the 9th of May (May 10th and May 11th are OK days for a cold - no competitions!)

I hope to get in the top 228 - not a terribly hard goal to get.
My really hopeful goal is to get in the top 227 - only infinitesimally harder...
.
.
.
My $ \text{hopeful}^{228}$ goal is to be in first place - with a probability of $ \frac {1}{\infty}$.

I remember seeing that the trophy for the National champion is made of glass with a marble base, whose height divided by the State trophy's height equals State trophy's height divided by Chapter trophy's height.

[quote="gauss1181"][quote="CatalystOfNostalgia"]It's not real glass; it's hard plastic. Well I mean, it has been for at least the past ten years, so uh, I would assume they're not going to get real glass this year.

but if Bobby wins this year they're probably going to have to help him carry it[/quote]
really why? is he serously so weak he couldn't carry that????? :huh:

andyeah, real glass is simply not logical for a trophy. it'd be wayyy to fragile[/quote]

I like the chapter & state trophies because almost all  other regular MT tournament trophies fall apart on the spot...

If I make cd (1/57846395786347956435798688754500111222 times yongyi's chances) everyone will know because I will squeal and fall to the ground and all will hear the squeal and the entirety of the earth shaking from my fall......

[quote="Math Geek"][quote="xpmath"][quote="CatalystOfNostalgia"]It's not real glass; it's hard plastic. Well I mean, it has been for at least the past ten years, so uh, I would assume they're not going to get real glass this year.

but if Bobby wins this year they're probably going to have to help him carry it[/quote]
Don't make fun of bs like that.

Uh, I just want to make top 57, which probably won't happen.[/quote]

lol, bs.  profanity...

Anyways, what does the trophy look like?   I missed the awards ceremony last year and will do so this year as well. :([/quote]
I think it's something transparent or plastic for the top teams. The individual trophies are actually marble and wood like the chapter and state trophies.

[quote="gauss1181"]
Anyways, what does the trophy look like?   I missed the awards ceremony last year and will do so this year as well. :([/quote]
I think it's something transparent or plastic for the top teams. The individual trophies are actually marble and wood like the chapter and state trophies.[/quote]

It is about two times the height and width of a state trophy, the top is made of glass/plastic (but I think the champion trophy at our school is glass) The bottom out of marble. Ironically, it's actually not too difficult to carry. Yeah we have a national champion trophy at our school :) not to brag about it...

I bet they're gonna do something really special this year at nats since it's the 5² Anniversary. :D

Yep, and that's 25 years, which means a QUARTER of a CENTURY!!! :D

YIPPEE! :lol:

yeah, and it a FORTIETH OF A MILLENIUM, which sounds like a million years lolz

[quote="PI-Dimension"][quote="uldivad9"]It's also a perfect square, and the first composite number not divisible by the first two primes!
But we're getting off-topic. I don't really think we're going to get anything special at Nats this year.[/quote]

So depressing.

I bet we will all get calculators with MATHCOUNTS 25 years on the back.[/quote]
Yeah. DEFINITELY.


calculus
integration
calculus computations

I can't seem to figure out how to do these integrals. I've tried the substitution rule, but to no avail. :(

1) [b]∫[/b] (t^2 + 1/t^2 + 2)^1/2 dt

2) Almost identical to the 1st one:
    [b]∫ [/b](4e^2 + 1/e^2 + 4)^1/2 dt

any idea/suggestions/solutions would be greatly appreciated!  :lol:

but how'd u make it the denominator? it's not even to the power of -0.5 :S

EDIT: sorry nvm..i got the hint.  :blush:

okiyyyy...cant figure the 2nd one out :(

Yes, it's easier,

$\int \sqrt{t^{2}+\frac{1}{t^{2}}+2}\cdot dt \, = \, \int \frac{1}{t}\sqrt{t^{4}+2t^{2}+1}\cdot dt \, = \int \frac{1}{t}\left( 1+t^{2}\right) \cdot dt$

 :bye:

[quote="eagle"]yeaa..thnx...i did get that one..confused abt:

∫ (4e^2 + 1/e^2 + 4)^1/2 dt ![/quote]

See that $\sqrt{4 e^{2}+e^{-2}+4}$ is a constant


inequalities
number theory proposed
number theory

Find all $x,y,z,k\in Z^+$ such that
$(x+y+z)^2=kxyz$

very easy $k <= 9$ and k different to 7
and for each case we have infinite solutions. :D  :D

it is the answer but it is not very easy ! it is from Turkey 2000 i think, problem 5, my solution is pick up the smallest triple $(x,y,z)$ and verify they satisfy they are the sides of possibly a degenerated triangle. Now do some inequalities, to get $k<10$. to prove $k=7$ is not ateinable, prove at least one of the numbers is $1$, and that the other 2 numbers are consecutive or equal, and deal with some cases.

I have read the general problem
x1^2+...+xn^2=k.x1...xn
which can solved same as Pell equation.


[quote="DPatrick"]Round 4 problems should be posted later this month (probably the last week of January) on the USAMTS web site.[/quote]

Am I missing something, because I have not seen the problems up yet. Is there just a delay in releasing them or is there somewhere else they are being posted? I only asked this because I have not seen rubrics for round 3 problems either and i am starting to wonder if something is going on that i am unaware of...

i hate to sound rushy but i just don't want to not say anything and then miss out on this round.

Thanks

The problems will be released Wednesday.  The scores for Round 3 will hopefully be released early next week.


If a and b are positive real numbers and each of the equations $x^{2}+ax+2b=0$ and $x^{2}+2bx+a=0$ has real roots, then find the smallest possible value of a+b.

[hide]by taking the discriminant of each side, and knowing that the discriminant must be greater than or equal to 0, we get
$a^{2}-8b\ge0$
$4b^{2}-4a\ge0$
We then add $4a$ to each side of the second inequality to get $4b^{2}\ge4a$, or $b^{2}\ge a$.  Square each side to get $b^{4}\ge a^{2}$
Substitute this into the first inequality, so we have $b^{4}-8b\ge a^{2}-8b\ge0$.
Therefore, we have $b^{4}-8b\ge0$, and if we solve this for the positives, we get $b\ge2$.  When that is substituted into our first inequality, we can get $a\ge4.$  We then check to see that the solutions work with our original inequalities (they do; we get $x^{2}+4x+4$, which of course has a double root of $x=-2$  Finally, we add $a$ and $b$ to get $6$.[/hide]

couldn't $a$ and $b$ both be $0$, making the answer $0$?

[quote="daermon"]:oops_sign:
man, am I stupid[/quote]
not stupid, just need to pay more attention. :)
I do too.


just use the well  known inequality
$ 2 \sqrt {n} \minus{} 2 < \sum_{i \equal{} 1}^{n}\frac {1}{\sqrt {i}} < 2\sqrt {n} \minus{} 1$
here $ n \equal{} 10^{4}$
so answer is c) $ 197$

can you plz post the inequality again? The latex seems to be not proper
And can you give a method of derivin gthat whatever ineq?
One cant expect to rem things  :maybe:

and by your method, the ans comes to be 198  :huh:

i think it's pretty clear with $ \text{\LaTeX}$.... :huh:
also u missed out $ 1$ in ur sum u posted [url=http://www.mathlinks.ro/viewtopic.php?p=1063225#1063225]here[/url]
where the sum starts from $ \frac {1}{\sqrt {2}}$ and not 1 :wink:

ok here is the proof of the inequality
$ 2\sqrt {m + 1} - 2 \sqrt {m} = \frac {2}{\sqrt {m + 1} + \sqrt {m}} < \frac {2}{2 \sqrt {m}} = \frac {1}{\sqrt {m}}$
also $ 2 \sqrt {m} - 2\sqrt {m - 1} = \frac {2}{\sqrt {m} + \sqrt {m - 1}} > \frac {2}{\sqrt {m}} = \frac {1}{\sqrt {m}}$
so we get
$ 2(\sqrt {m + 1} - \sqrt {m}) < \frac {1}{\sqrt {m}} < 2(\sqrt {m} - \sqrt {m - 1})$
thus adding this telescoping series we get the required result

programmed mind? One wasnt missed out so the answer is 197 yeah!  :D

i pretty much sure this is some really old and brilliant imo problem (or something very closely related) or maybe fom the bmo

i dunno abt that man, i found it in a magazine as a practise sum for ISI  :D
N the original inequality that mythili posted was derived in a different manner  :D

Go Shreyas!!!!!!!!!!!!!!!!!!!  :first:

And Mythili, that was a very good proof :)

[quote="shruthibaskaran"][u][b]Go Shreyas!!!!!!!!!!!!!!!!!!!  [/b][/u]:first:

And Mythili, that was a very good proof :)[/quote]

Ahem i quoting to highlight the controversy  :P   :rotfl:

Lol..... As Vikku would put it, Controversy creates cash. LOL. Sorry, couldnt help myself!

Ahem i am a partial layman and i dont understand why u are saying go shreyas and awarding him the first prize seeing that shreyas hasnt posted any soln here. May i ask why?  :D   :huh:

The inequality is a very simple inequality to state and prove but is quite powerful at times.

Oh no no! Shreyas, please do remember, I solved it when we were in UKG and gave the credit to you, because you were crying very badly.... :D

n believe me when i say this, i think chappli n sanjith n ashwath n soumya n pardesi n me have never heard abt that inequality before (we are very much interested in math) n among us atleast someone shd have heard if the user of the ineq had heard abt it. So how come only limited people know it? What's the defect in me argument?  :huh:   :P   :rotfl:

I don't think pardesi couldn't have not heard of it. Do you?

well i just happened to phone them all n sanjith is here so ask him himself  :D

Ah well. You're forgetting one cruel fact - coincidence. Maybe they all just decided to skip over it, but of course, Shre knew it in UKG because of me. :D

And Shre, we all know that you are the lean mean math machine man. Dont show off :D (55kg apparently, did we ask for that? Forgive my poor memory, of course)

You people are veering of the topic too much. So lets get into the discussion of the inequality shall we?

I wont do that bcos its
[hide]
Chumma jujubi  :P   :rotfl:   :D
[/hide]

its bcos we all love the moderator (not in a wrong sense  :P ) and want him to suffer with as many mod edits as possible so we try to trouble him a lot thus showing him we care  :rotfl:   :D

[quote="mythili"]why is it that all the threads here get into irrelevant discussions  so often(no offence) :D[/quote]

Most people would rather die than think; in fact, they do so.
Bertrand Russell

where is he/she ???
there are lots of irrelevant discussion here :huh:

yeah right i was waiting fer that  :D  i knew that was coming but as they say, always put the blame on others eh?  :D  Well as long as its the problem i posted and the soln is done, i dont mind spamming even if necessary but some people just pick on up frm the spamming and they later deny liking the spamming just like the person cyber shre  :rotfl:

[quote="madness"]yeah right i was waiting fer that  :D  i knew that was coming but as they say, always put the blame on others eh?  :D  Well as long as its the problem i posted and the soln is done, i dont mind spamming even if necessary but some people just pick on up frm the spamming and they later deny liking the spamming just like the person above me  :rotfl:[/quote]
where did i spam :huh:

Hey mythili - No offense, but seriously :D We just like playing around with it a bit :D I mean, as madness said, chumma jujubi - I do hope you understand that!!!!

@Shre: I will not pick on Shrids, because he, unlike you, doesnt bother. And he is well.. Shridhar, and you are well.... YOU!!!

sorry i didnt see that you had posted before i had done so, that was a satire on cyber shre  :D   :rotfl:


geometry
circumcircle
geometric transformation
reflection
trigonometry
analytic geometry
angle bisector

[quote="manlio"]Let $PQ$ be a diameter of a circle, and let $A$ and $B$ be two distinct points on the circle lying on the same side of $PQ$. Let $C$ be the point of intersection of the tangents to the circle at $A$ and $B$. Let the tangent to the circle at $Q$ meet the lines $PA$, $PB$, and $PC$ at the points $A'$, $B'$, and $C'$, respectively. Prove that the point $C'$ is the midpoint of $A'B'$.[/quote]

Thanks, Manlio, a nice problem! Yes, I have an elementary solution which may be of some interest (although it is not BEAUTIFUL and by far not INGENIOUS  :) ).

My solution uses a bit of theory. We begin heuristically: At the very first sight, the problem above tells us nothing. What do we know about some weird circles with certain diameters and about some strange intersections of tangents? Not really much. We want to have something familiar. But what is familiar? Well, triangles are familiar. Thousands of theorems known about triangles give us a good chance to solve the theorem once it is translated into triangle geometry language. In fact, inquiring which triangle would be most appropriate for using as the "base triangle", we expect the triangle A'PB' to be the best choice.

Now, the line A'B' touches the circle with diameter PQ at the point Q; therefore, the line PQ is perpendicular to the line A'B'. Hence, in the triangle A'PB', the line PQ is an altitude. The point Q is the foot of this altitude. Now, as < PAQ = 90° (since the point A lies on the circle with diameter PQ), the point A turns out to be the orthogonal projection of the point Q on the side PA' of our triangle A'PB'. Similarly, the point B is the orthogonal projection of the point Q on the side PB'. The point C is the point where the tangents to the circumcircle of triangle APB at the points A and B meet, and the point C' is the point of intersection of the lines PC and A'B'. We must show that the point C' is the midpoint of the segment A'B', or, equivalently, that the line PC is a median of triangle A'PB'.

Thus, replacing the letters P, A', B' by A, B, C, we can rewrite our problem as follows:

[i]Let ABC be an arbitrary triangle, and let X be the foot of the altitude from A in this triangle. Further denote by Y and Z the orthogonal projections of the point X on the sides AB and AC of the triangle ABC. The tangents to the circumcircle of triangle YAZ at the points Y and Z meet at a point T. Then prove that the line AT is a median of triangle ABC.[/i]

Okay, now we have transformed the problem into a form in which it can be easily attacked by triangle geometry. In the following, the [i]A-median[/i] of triangle ABC will mean the median issuing from the vertex A, and similarly, the [i]A-bisector[/i] will mean the angle bisector issuing from the vertex A (i. e., the angle bisector of the angle CAB). If you reflect the A-median of triangle ABC in the A-bisector, you get a new line, which is called the [i]A-symmedian[/i] of triangle ABC. Now there is the following well-known fact:

[i]If the tangents to the circumcircle of a triangle ABC at the vertices B and C meet at a point D, then the point D lies on the A-symmedian of triangle ABC.[/i]

Apply this, but not to triangle ABC, but to triangle YAZ. Then you see that the point T, defined as the point of intersection of the tangents to the circumcircle of triangle YAZ at the points Y and Z, lies on the A-symmedian of triangle YAZ. In other words, the line AT is the A-symmedian of triangle YAZ, i. e. the reflection of the A-median of triangle YAZ in the A-bisector of the triangle YAZ. If you denote by Q and Q' the midpoints of the segments BC and YZ, then the A-median of triangle YAZ is the line AQ'; hence, the line AT must be the reflection of the line AQ' in the A-bisector of triangle YAZ. This yields < TAZ = < Q'AY.

Now, since < AYX = 90° and < AZX = 90°, the points Y and Z lie on the circle with diameter AX. This means that the four points A, X, Y and Z lie on one circle, and hence < AYZ = < AXZ. But using the right-angled triangles AZX and AXC, we have < AXZ = 90° - < XAZ = 90° - < XAC = < ACX = < ACB. Hence < AYZ = < ACB. Similarly, < AZY = < ABC. Thus, the triangles AYZ and ACB are similar. (Actually, they are oppositely similar; we could see this if we would work with directed angles, but we don't.)

Now, the triangles AYZ and ACB are similar, and the points Q' and Q are corresponding points in these two triangles (since the point Q' is the midpoint of the side YZ of the former triangle, and the point Q is the midpoint of the side CB of the latter triangle). In similar triangles, corresponding points make equal angles; hence, < Q'AY = < QAC. Combining with < TAZ = < Q'AY, this yields < TAZ = < QAC, or, in other words, < TAC = < QAC. Therefore, the point T lies on the line AQ, i. e. on the A-median of triangle ABC. Hence, the line AT is a median of triangle ABC. This completes the proof.

  Darij

IF R=PA.QB and H = PB.QA the circle c(ABC) is the nine-points circle of PQR. From this we have that C is The midpoint of RH. The rest follows.

Here's another proof, although it's not elementary (it uses some projective geometry):

Let $P_c$ be the intersection between the circle and $PC$. Then the quadrilateral $PAP_cB$ is a harmonic quadrilateral, so the tangent through $P$ to the circle, $PA,PC,PB$ form a harmonic quadruple, and by intersecting this quadruple with the tangent through $Q$ to the circle, we must get a harmonic quadruple of points on the tangent through $Q$, but one of the points is $\infty$, so $(A',B';C',\infty)=-1\Rightarrow C'$ is the midpoint of $A'B'$.

OK thank you very much . Now I will study deeply all three solutions.
However I want to post two other solutions:

A solution by inversion:

If we invert the figure in the circle with center at $P$ and radius $PQ$, we see that the circle with diameter $PQ$ inverts to the line tangent to the circle at $Q$. The circle through $A$ and $B$ with center $C$, which is orthogonal to the circle with diameter $PQ$, inverts to a circle with center $C'$ and radius $C'A'=C'B'$ because orthogonality must be preserved and because the center of the inverse must lie on the line $PC$

A solution by analytic geometry.
If you use analytic geometry you can solve it in a easy way but I don't like to solve geometric problems using algebra.

It suffices to prove the statement for the unit circle with $P=(0,1)$. WLOG we may assume $A=(\cos(\theta-\alpha),\sin(\theta-\alpha))$and $B=(\cos(\theta+\alpha),\sin(\theta+\alpha))$ with $\mid\theta+\alpha\mid < \frac{\pi}{2}$ and $\mid\theta-\alpha\mid < \frac{\pi}{2}$ , so that $C=(\sec\alpha\cos\theta,\sec\alpha\sin\theta)$.
Because the line trough $P$ and a point$(r,s)$ , $s$ not equal to  $1$ intersects $y=-1$, the tangent at $Q$ at the point$(\frac{2r}{1-s},-1)$we have

$A'=(\frac{2\cos(\theta-\alpha)}{1-\sin(\theta-\alpha)},-1)=(\frac{2(1+\sin(\theta-\alpha)}{\cos(\theta-\alpha},-1)$

$B'=(\frac{2\cos(\theta+\alpha)}{1-\sin(\theta+\alpha)},-1)=(\frac{2(1+\sin(\theta+\alpha)}{\cos(\theta+\alpha},-1)$

$C'=(\frac{2\sec\alpha\cos\theta}{1-\sec\alpha\sin\theta},-1)=(\frac{2\cos\theta}{\cos\alpha-\sin\theta},-1)$

Beginning with the expression for the  x-coordinate of the midpoint derived from $A'$ and $B'$

$\frac{1+\sin(\theta-\alpha)}{\cos(\theta-\alpha)}+\frac{1+\sin(\theta+\alpha)}{\cos(\theta+\alpha)}= \frac{2\cos\theta}{\cos\alpha-\sin\theta}$

which shows that $C'$ is the midpoint of $A'$ and $B'$

So we have 5 solutions !!

$PC$ is symmedian in $\Delta PAQ$, so it is a median in $\Delta PQX$, where $X$ is the intersection of $AP$ with tangent at $Q$ to circle, done.

Best regards,
sunken rock


I need some drastic help with these problems

1. A steamboat travels at the rate of 30 Km/h in still water. If it can travel 45 Km upstream in the same amount of time that it takes to go 63 Km downstream, what is the rate of the current?

2. A plane has a speed of 50 mph in still air. It makes a round trip, flying with and against a wind of 30 mph. Find its average speed for the trip.

3. If the distance between the lines 5x+12y-1=0 and 10x+24y+k=0 is 2, then k is:
a. 50 or 54
b. -50 or -54
c.50 or -54
d. -50 or 54

4. What is the area of the portion of the circle $ x^2\plus{}y^2\minus{}4y\equal{}0$ lying below the x axis?

[hide="1"]
Suppose the current's rate is $ c$.

Upstream, it's going at $ 30 \minus{} c$ kmph. Downstream, $ 30 \plus{} c$ kmph.

To go $ 45$ km at $ 30 \minus{} c$ kmph requires $ \frac {45}{30 \minus{} c}$ hours.

To go $ 63$ km at $ 30 \plus{} c$ kmph takes $ \frac {63}{30 \plus{} c}$ hours.

Equating and solving gives $ c \equal{} \boxed{5\text{ kmph}}$.
[/hide]

[hide="2"]
Suppose it travels from A to B, and the distance between them is $ d$ miles.

From A to B, the plane goes at $ 80$ mph, taking $ \frac {d}{80}$ hours.

From B to A, it takes $ \frac {d}{20}$ hours, so in total it spends $ \frac {d}{16}$ hours to travel $ 2d$ miles.

Then its average speed is $ \frac {2d}{\frac {d}{16}} \equal{} \boxed{32\text{ mph}}$.
[/hide]

[hide="3 (this is a terrible solution)"]
The first one is $ y \equal{} \minus{} \frac {5}{12}x \plus{} \frac {1}{12}$.

The second is $ y \equal{} \minus{} \frac {5}{12}x \minus{} \frac {k}{24}$.

In the first, let $ x \equal{} 0$ to get $ y \equal{} \frac {1}{12}$. Then the perpendicular line through this is $ y \equal{} \frac {12}{5}x \plus{} \frac {1}{12}$.

The perpendicular intersects the other line when $ x \equal{} \minus{} \frac {5(k \plus{} 2)}{338}$ and $ y \equal{} \frac {25 \minus{} 72k}{2028}$.

The distance between the two points ends up being $ \pm\frac {k \plus{} 2}{26}$ after bashing it out.

$ \frac {k \plus{} 2}{26} \equal{} 2\implies \boxed{k \equal{} 50\text{ or } \minus{} 54}$.
[/hide]

[hide="4"]
$ x^2 \plus{} (y \minus{} 2)^2 \equal{} 2^2$, so our circle is translated up $ 2$ units and has a radius of $ 2$. This means it's tangent to the $ x$ axis on top, so the answer is $ \boxed{0}$.
[/hide]

I got the first one right and the second one wrong

[quote="batteredbutnotdefeated"]I need some drastic help with these problems

1. A steamboat travels at the rate of 30 Km/h in still water. If it can travel 45 Km upstream in the same amount of time that it takes to go 63 Km downstream, what is the rate of the current?

2. A plane has a speed of 50 mph in still air. It makes a round trip, flying with and against a wind of 30 mph. Find its average speed for the trip.

3. If the distance between the lines 5x+12y-1=0 and 10x+24y+k=0 is 2, then k is:
a. 50 or 54
b. -50 or -54
c.50 or -54
d. -50 or 54

4. What is the area of the portion of the circle $ x^2 \plus{} y^2 \minus{} 4y \equal{} 0$ lying below the x axis?[/quote]
IMHO, these are not geometry problems, they're rate problems. They're not that hard once you get the hang of them.

[hide="Solution for Number one"]
So firstly, we know that distance is rate multiplied by time ($ d \equal{} rt$, if you don't understand this post saying so)
Let r=rate of the river's current for the solution.
Since when the boat goes upstream it goes against the flow of the river we get the equation:

$ \text{Time taken to go upstream} \equal{} \frac {45}{30 \minus{} r}$

similarly they go with the current when they go downstream so,
$ \text{Time taken to go downstream} \equal{} \frac {63}{30 \plus{} r}$

Since the time taken to go upstream and downstream are the same, we get the equation:

$ \frac {45}{30 \minus{} r} \equal{} \frac {63}{30 \plus{} r}$
Solving this gives:

$ r \equal{} 5km/h$

[/hide]

:)

[quote="grn_trtle"]
[hide="3 (this is a terrible solution)"]
The first one is $ y \equal{} \minus{} \frac {5}{12}x \plus{} \frac {1}{12}$.

The second is $ y \equal{} \minus{} \frac {5}{12}x \minus{} \frac {k}{24}$.

In the first, let $ x \equal{} 0$ to get $ y \equal{} \frac {1}{12}$. Then the perpendicular line through this is $ y \equal{} \frac {12}{5}x \plus{} \frac {1}{12}$.

The perpendicular intersects the other line when $ x \equal{} \minus{} \frac {5(k \plus{} 2)}{338}$ and $ y \equal{} \frac {25 \minus{} 72k}{2028}$.

The distance between the two points ends up being $ \pm\frac {k \plus{} 2}{26}$ after bashing it out.

$ \frac {k \plus{} 2}{26} \equal{} 2\implies \boxed{k \equal{} 50\text{ or } \minus{} 54}$.
[/hide]
[/quote]
I had a different approach for this one, otherwise using similar methods:
[hide="3"]
The distance from a point to a line can be found as $ \dfrac{|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$

Where A, B, and C are the variables in a line's standard form Ax+By+C=0. X and Y are the coordinates of the point you are measuring the perpendicular distance from.

So now I found a point on 5x+12y-1=0. I wanted nice whole numbers so I chose (5, -2) Essentially the problem becomes the distance from (5, -2) to the line 10x+24y+k=0. This distance the problems gives as two, so we now have an equation to work from:

$ \dfrac{10(5) \minus{} 2(24) \plus{} k}{\sqrt {24^2 \plus{} 10^2}} \equal{} 2$

$ \dfrac{k \plus{} 2}{\pm26} \equal{} 2$

$ k \equal{} 50, \minus{} 54$
[/hide]


Hi,

How would I go about graphing the hyperbola $ \frac{(x\plus{}2)^2}{4}\minus{}\frac{(y\plus{}3)^2}{9}\equal{}1$?  I found that the foci are $ (\minus{}2\minus{}\sqrt{13},\minus{}3)$ and $ (\minus{}2\plus{}\sqrt{13},\minus{}3)$, but I don't know what to do next.

You could always get it into $ y\equal{}f(x)$ form, pick out $ x$-values, and solve for $ y$ so you can plot points.

Setting a common denominator gives $ 9(x\plus{}2)^{2}\minus{}4(y\plus{}3)^{2}\equal{}36$.  Therefore, we get
$ 9(x\plus{}2)^{2}\minus{}36\equal{}4(y\plus{}3)^{2}\Rightarrow\sqrt{\frac{9}{4}(x\plus{}2)^{2}\minus{}9}\minus{}3\equal{}y$.
So maybe get $ x\equal{}0,\pm 1,\pm 2$ and go from there.

The graph has a center at $ ( \minus{} 2, \minus{} 3)$ and asymptotes at $ y \equal{} \frac {3x}{2}$ and $ y \equal{} \frac { \minus{} 3x}{2} \minus{} 6$.
Graph the center, the asymptotes, and a box going b and -b units in the y direction and a and -a units in the x direction from the center. The asymptotes should be the diagonals of the box.

The hyperbola opens up in the positive and negative x directions and the vertices are on the box at $ ( \minus{} 4, \minus{} 3)$ and $ (0, \minus{} 3)$. Now, graph the hyperbola with those vertices and asymptotes.

And make sure the hyperbola does not cross the two asymptotes.


Hi, I just had my final school maths exam today, there was a question on it that was a bit confusing, dealing with masses.

Ok, imagine a conveyor belt AB, which is inclined at 30 degrees to the horizontal. When turned on, the conveyor belt moves in such a way that it is able to drag objects from its base up to its peak.

Now, a mass of 5kg is placed onto the conveyor belt. The belt is switched on, and the mass begins to move up the belt. When the acceleration of the belt is 0.8m/s^2, the mass is on the point of slipping down the plane. When the acceleration of the belt is greater than 0.8m/s^2, the mass begins to slip. Find the coefficient of friction of the surface of the belt correct to 2 decimal places, assuming g (acceleration due to gravity) is equal to 9.8m/s^2.

your question is pretty easy, just draw the diagram and the forces ;)

also FYI, there is a Physics Forum, that you can post your problems in.. ;)

Oh sorry :(. Well, this is more of a quick check-thing, because I have an answer, but I don't know if it's right. I worked coefficient out to be 0.67, and I hope that's right.


function
calculus
integration
real analysis
calculus computations

Can anybody help me answer the following question:

Is the function f(x) = x, x is an element of Q, f(x) = 0, x is an element of Q^c integrable on [1, 2]?

What definition of integrability would you like to start from?

What Joel is saying: it matters whether you're using the Riemann integral or the Lebesgue integral. Of course, I'm about 99.9% sure you're asking this question from a Riemann integral viewpoint.

The Lebesgue viewpoint: of course the function is integrable. It's zero almost everywhere, and thus might as well be the zero function. Why are you bothering me with the zero function?

Which is why I assume you're using the Riemann integral. Tell me - what estimates can you make for upper and lower sums?


Hi! I am a math student who has only very recently discovered his passion for mathematics. It was only in mid October of 2008 when I found this site did I realize how much I really love mathematics. I am currently in 10th grade. I always loved my school mathematics but I never really went .. further, you know?
Ever since I've been working really hard at becoming a better problem solver. I got a 114 on my AMC 10B( maybe a 108, I think I made a scantron mistake). This was a huge improvement over my last score, which I do not want to reveal.
I think if I keep on working at the pace I'm going I'll be a good problem solver. I was able to solve most of the problems for the test A of the admission form. If I wasn't able to come up with a solution I was able to come up with a strong outline as to how to solve the problem.
I really want to advance my skills because I love math so much.  One problem I definitely face is the fact that I probably do not have 2 mathematics teachers to give me a letter of recommendation. I may have one, but the gifted program at my school is relatively small.
Anyway, can some former students speak to me about my concerns?

Hey cipos2! :)

That's really cool that you have an interest in math xD
and you definitely should apply to amsp, since you said you were able to come up solution for most of the test A problems, you're probably in good shape for admission.

Btw,
you don't need 2 math teachers for rec. letters.
you only need "one from a mathematics teacher or mentor and one from someone who is familiar with the student’s activities and accomplishments outside the world of mathematics."
so you can have one math teacher write a rec. letter & then someone else [maybe a piano teacher or a soccer coach] something like that. the 2nd person should know you from doing an activity other than math.

but good luck on ur apps.
if u can solve most of the test A problems, you probably can be admitted. xP
and if u do go, have fun! its amazing!!

hello! :]

i think that it's never too late to go to a math camp..i got invovled with math stuff at my school during my 10th grade too.

regarding admission to amsp,
i dont think how many questions you get right on your admission test matters. well i guess it matters to certain extent, but i think the most important thing is to show your passion for math and let Dr. Andreescu see that through your application! :]

I mean..amsp doesnt care about your school GPA, or anything unlike other camps out there,
so don't think about the results, and just apply! :]


linear algebra
matrix
linear algebra unsolved

When this matrix:
$
\left( \begin{array}{cccccc}
a_1 & c_1 & 0 & 0 & 0 & 0 \\
b_2 & a_2 & c_2 & 0 & 0 & 0 \\
0 & 0 & \dots down & \dots down & \dots down & 0 \\
0 & 0 & 0 & b_{n-1} & a_{n-1} & c_{n-1} \\
0 & 0 & 0 & 0 & b_n & a_n \end{array} \right)
$
is irreducible and when it is reducible?

[quote="Kent Merryfield"]Gaussian elimination is an efficient process with such a matrix. In fact, it would be possible (although somewhat tedious) to do Guassian elimiation on the general case and determine a formula for the determinant, which should answer your question. I don't see any insights that are any better than that.[/quote]
I found a solution.. I think all the b_i and c_i should be different from 0.
This way the graph induced by the matrix is strongly connected and it means A is irreducible...

could you explain a little bit more what you mean ? The relation between matrix and graphs seem very interesting !
thanx  :)

I don't have the proof for that ...
But the graph is built this way.
if $a_{ij}$ is not 0 then there is an edge (path) between node i and j.
This graph should be strongly connected for the matrix to be irreducible... can't tell you more then that :)

This is all new to me too, but I believe it's not at all hard to prove, if reducible means what I think it means (i.e. [url=http://mathworld.wolfram.com/ReducibleMatrix.html]this[/url]), and if I'm not misunderstanding the whole thing :).

A digraph is strongly connected iff given any bipartition of its vertices $A,B$, there is at least one arc from $A$ to $B$. I think that with this characterization of strong connectedness, it becomes rather obvious.


How many AP courses do people usually take during high school.    What are the advantages of taking AP computer science over AP chemistry, and vice versa?   Is it better to take a course during the summer so that you have less courses during the year and more time to study for the AMCs?  or should you study for the AMCs during the summer and work on school work during the year?  Do colleges look at how many AP classes you take?  AP courses are college classes, so why do people want to take so many?

This isn't really very appropriate for Getting Started, there are other forums to discuss this sort of thing, but I think I can help a little:

1)  If you do well on AP tests, some colleges will give you college credit for those courses so you won't have to take them in college - it saves time and tuition money.

2)  The number of AP classes that people take depends a lot on what their goals are and what kind of colleges they want to go to.  I know crazy people who've taken 12 or more AP classes in their high school career (I'll be one of those people :P), but plenty of people do fine with 1 or none.  If you want to get into really top schools, AP classes tend to be the most challenging and rigorous courses you can take, and more selective colleges will be looking at the rigor of your coursework in addition to just your grades.  I mean, if you have a 4.0 but you spent the whole time taking classes like Wood Shop that won't be anywhere near as impressive to a good college as a 3.5 student who took classes like AP Physics.

3)  If you want to work in science, take AP Chemistry, and if you want to work with computers, take AP Computer Science.  (I took both :P )

4)  Courses over the summer won't give you college credit like AP courses, and if you were really good you'd study for AMCs and AP tests at the same time :P

I'll be taking like.... seven, but a bunch of aditional post APs (Multi variable calculus, Linear Algebra, AI, etc.)

Colleges ask for AP scores and what courses you took. A few also ask for AMC and AIME scores, but I'm sure you can stick those in somewhere on your application even if they don't ask. Taking AP courses show that you can manage your time and handle the workload of college level courses and that you are willing to do something extra.
I took 5 total, but I also had multivariable calculus and some college classes.
You shouldn't devote a whole summer to studying. I think colleges would rather see that you attended a program or found a job or volunteered. If you were an admissions officer, would you prefer a candidate who spent two months inside studying for a two hour test or someone who was outside, interacting with people and contributing to his community?


Problem:

M leaves a remainder of 5 when divided by 6.

N leaves a remainder of 3 when divided by 6.

What is the remainder when M+N is divided by 6?

[hide="Answer"]226[/hide]

Obviously something is wrong... you can't have a remainder of

226 when dividing by 6. I reported this about a month ago, but I

saw it again yesterday, still accepting 226.

When the problem comes up, I don't know whether to answer

226 (the bug) or 2 (the right answer) because I don't know if its

been fixed...

You should probably contact one of the administrators about this, if you reported it and it still hasn't been fixed.


i am a sudent from bangladesh.....want to attend imo next time..i need a good note for combinatorics for this purpose....please help
ishfaq :rotfl:


function
limit
inequalities
calculus
derivative
linear algebra
matrix

I know this is rather easy, but it's a cute problem :).

Let $f:\mathbb R\to\mathbb R$ be a three times differentiable function s.t. $\lim_{x\to\infty}f(x)=l\in\mathbb R,\ \lim_{x\to\infty f^{(3)}(x)=0}$. Show that $f',f''$ also tend to $0$ as $x\to\infty$.

I'm not sure about the proof but... here is what I did:
If $\lim\limits_{x\to\infty}f(x)=l$ results that for any
$\varepsilon>0$ there is a $\delta_\varepsilon>0$ s.t. $-\varepsilon<f(x)-l<\varepsilon$ for any $x>\delta_\epsilon$.
Let $(x_n)_{n\ge1}$ be an arbitrary positive sequence of real numbers s.t. $x_n\to 0$.
We have that for every $n$ there are $\delta_{n_1}>0$ s.t. $-x_n^2<f(x)-l<x_n^2$ for all $x\ge\delta_{n_1}$ and
$\delta_{n_2}$ s.t. $-x_n^2<f(x+x_n)-l<x_n^2$ for all $x>\delta_{n_2}$. Now for $\delta_n=\max(\delta_{n_1},\delta_{n_2})$ by multiplying the first ineq by -1 and sum them up:
$-2x_n^2<f(x+x_n)-f(x)<2x_n^2\Leftrightarrow -2x_n<\displaystyle \frac{f(x+x_n)-f(x)}{x_n}<2x_n$ for all $x>\delta_n$, and because $x_n$ was chosen arbitrary results that $\lim\limits_{x\to\infty}f'(x)=0$.

is it corect?

or... another approach, maybe this is better :)
Let $x_n,y_n$ be two arbitrary sequences, strictly increasing and not bounded s.t. $y_n-x_n\to\infty $ and $y_n>x_n,\forall n\in N^*$(eg. $y_n\sim \sqrt n,x_n\sim\sqrt n$)

Now by Lagrange Theorem we get that $\displaystyle\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(z_n)$, with $z_n\in(x_n,y_n)$. Because $x_n$  and $y_n$ were chose arbitrary results that $f'(z_n)\to\infty\Rightarrow f'(x)\to\infty$ as $x\to\infty$.

I was thinking at smth else... since $f(x)\to l$ as $x\to\infty$ means that the line $y=l$ is a horizontal asympthot for $f$'s graphic and because $f'(x)$ is interpreted as the tangent in the point $(x,f(x))$ to the $f$'s graphic results that $f'(x)=0$ when $x\to\infty$.

Here's another approach:

WLOG, $l=0$; we can freely add a constant to $f$.
Suppose that $|f(x)|<\epsilon_1$ and $|f^{\prime\prime\prime}x)|<\epsilon_2$ for all $x>A$.
Then, by Taylor's theorem, $f(x+t)=f(x)+tf^{\prime}(x) + \frac{t^2}{2}f^{\prime\prime}(x) + \frac{t^3}{6}f^{\prime\prime\prime}(x+u)$ for some $u$ between $0$ and $t$. We rearrange to
$tf^{\prime}(x) + \frac{t^2}{2}f^{\prime\prime}(x) = f(x+t)-f(x)-\frac{t^3}{6}f^{\prime\prime\prime}(x+u)$
and estimate using the triangle inequality that
$\left|tf^{\prime}(x) + \frac{t^2}{2}f^{\prime\prime}(x)\right| < \epsilon_1+\epsilon_1+ \frac{t^3}{6}\epsilon_2$
Using $t=1$ and $t=2$, we have
$\left|f^{\prime}(x) + \frac12f^{\prime\prime}(x)\right| < 2\epsilon_1+\frac16\epsilon_2$
$\left|2f^{\prime}(x) + 2f^{\prime\prime}(x)\right| < 2\epsilon_1+\frac43\epsilon_2$
Again by the triangle inequality,
$|f^{\prime}(x)| < 2(2\epsilon_1+\frac16\epsilon_2)+ \frac12(2\epsilon_1+\frac43\epsilon_2) =5\epsilon_1+\epsilon_2$ and
$|f^{\prime\prime}(x)| < 2(2\epsilon_1+\frac16\epsilon_2) +1\cdot(2\epsilon_1+\frac43\epsilon_2) =6\epsilon_1+\frac53\epsilon_2$

For every pair $(\epsilon_1,\epsilon_2)$, there is an $A$ such that these inequalities hold for all $x>A$. Therefore we have that
$\lim_{x\to\infty}f^{\prime}(x)=\lim_{x\to\infty}f^{\prime\prime}(x)=0$

A closely related argument would work for the more general situation
$\lim_{x\to\infty}f(x)=l$ and $\lim_{x\to\infty}f^{(n)}(x)=0$.
We would need to evaluate the Taylor formula at $n-1$ places, but since the coefficients of the derivatives form an invertible matrix, each derivative would be bounded by a linear combination of the bounds for $f$ and $f^{(n)}$.

By the way, grobber, you mistyped the problem. You obviously meant that $\lim_{x\to\infty}f^{(3)}(x)=0$, but you left off a curly brace.
Also to Kuba: Your solutions are wrong, and they can't be right unless they use the condition on the third derivative. Without that condition, the problem is false, as demonstrated by $f(x)=\frac1x \sin(x^3)$.

This is almost what I did. I didn't read it all, but I also used Taylor expansion.

The same conclusion also holds in the more non-restrictive case when $ f(x+1)-f(x)$ goes to 0 when x goes to infinity.


modular arithmetic
algebra unsolved
algebra


function
real analysis
real analysis unsolved

Let $ f(x,y)$ be a differentiable function on $ \mathbb{R}^2$ such that $ \frac {\partial f}{\partial x}|_{(0,0)} > 0,\frac {\partial f}{\partial x}|_{(0,1)} < 0$.Prove that there is a point $ (x,y)$ such that $ \frac {\partial f}{\partial x}|_{(x,y)} \equal{} 0$.

Is $ f$ continuously differentiable?  If so, you can use the intermediate value theorem.  (If not, there is no need to say $ f$ is continuous when you've already said it's differentiable.)

Sorry,$ f$ is differentiable.
I have edited it.Thank you.

Ok,I have a solution for this problem(From a friend of me),but does the stronger conclusion is still true?
Prove the is a point $ (0,y)$ with $ 0<y<1$ such that $ \frac {\partial f}{\partial x}|_{(0,y)} \equal{} 0$.
Can anybody help me?

Surprisingly enough, the answer to the last question is "No". Take any $ C^1$-function $ g(x)$ such that $ g(x)\equal{}x$ when $ |x|>1$ and $ g'(x)<0$. Now put
\[ f(x,y)\equal{}\begin{cases}
x& y\le\frac 12
\\
(y\minus{}\tfrac12)^2g(x/(y\minus{}\tfrac12)^2)& y>\frac12
\end{cases}\]
This function is clearly continuously differentiable at every point except $ (0,\tfrac 12)$ and it is not hard to see that it is differentiable at that exceptional point as well (though the differential is no longer continuous there). On the other hand, we have $ f_x(0,\cdot)\equal{}1>0$ on $ [0,\tfrac12]$ and $ f_x(0,\cdot)\equal{}g'(0)<0$ on $ (\tfrac12,1]$.

[quote="zhaobin"]Thank you,fedja.But sorry,this is just my "conjecture". :wink: ,and also surprise to me for your conterexample.In fact,I want to the follow theorem which appeared in an exam of pku.The problem is:
Let $ f,g$ be differentiable function on $ \mathbb{R}^2$ such that there exist some points $ (x_1,y_1),(x_2,y_2)$,$ \frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}|_{(x_1,y_1)}>0,\frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}|_{(x_2,y_2)}<0$,Prove that there is a point $ (x,y)$ such that $ \frac{\partial f}{\partial x}+\frac{\partial g}{\partial y}|_{(x,y)}=0$.[/quote]
Can you post a solution please?
Thank you very much. :oops:


how can we know:
if the remainder of f(x)/(x-4) is 3 and the remainder of f(x)/(x-3) is 4 what is the remainder of f(x)/[(x-3)(x-4)]

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311516[/url]


In ham yek soale ziba

Farz konid $ 0\le x\le 1$ maximome ebarate zir ra bedast avarid
\[ 13\sqrt{x^2\minus{}x^4}\plus{}9\sqrt{x^2\plus{}x^4}\]

Kasi nabood

rahnamayi:
[hide]AM-GM[/hide]


thank you so much N.T.TUAN for your help.


Two concentric circles are such that a chord of the larger, 36 inches long, is trisected by the smaller.  The sum of the radii of the two circles is also 36 inches.  Find the inch-length of the radius of the larger circle.

[hide]Refer to the attached diagram.  The length marked with the x is, by the Pythagorean theorem, sqrt(r^2 - 36).  Then, using the Pythagorean theorem on the larger right triangle yields

sqrt(r^2 - 36)^2 + 18^2 = R^2

r^2 - 36 + 324 = R^2

R^2 - r^2 = 288

(R+r)(R-r) = 288

We also are given that R + r = 36, so that

36(R-r) = 288

R-r = 8

Adding this last equation to R + r = 36 yields 2R = 44, or R = 22.[/hide]

[hide]Draw another chord from one of the tri-section point (chord segments 12 one side 24  on other side) passing through the center.

so 24*12 = 36* (R-r) which gives  (R-r)= 8   and R=22.[/hide]


hmm one of the questions on the rsi app is

list below your technical skills level and academic background as of June 2006

wat does that mean exactly....

That (I presume) refers to the little checkboxes at the bottom of the page (for math/science classes and programming ability).