Angle $ B$ of triangle $ ABC$ is trisected by $ BD$ and $ BE$ which meet $ AC$ at $ D$ and $ E$ respectively. Then:

$ \textbf{(A)}\ \frac {AD}{EC} \equal{} \frac {AE}{DC} \qquad\textbf{(B)}\ \frac {AD}{EC} \equal{} \frac {AB}{BC} \qquad\textbf{(C)}\ \frac {AD}{EC} \equal{} \frac {BD}{BE}$
$ \textbf{(D)}\ \frac {AD}{EC} \equal{} \frac {AB\cdot BD}{BE\cdot BC} \qquad\textbf{(E)}\ \frac {AD}{EC} \equal{} \frac {AE\cdot BD}{DC\cdot BE}$

[hide="Click for solution"]
Since $ BD$ bisects $ \angle ABE$, we have $ \frac{AD}{DE}\equal{}\frac{AB}{BE}$.  Since $ BE$ bisects $ \angle DBC$, we have $ \frac{DE}{EC}\equal{}\frac{BD}{BC}$.  Thus, $ \frac{AD}{EC}\equal{}\frac{DE (\frac{AB}{BE})}{DE(\frac{BC}{BD})}\equal{}\frac{(AB)(BD)}{(BE)(BC)}$ or $ \boxed{\textbf{(D)}}$.
[/hide]


geometry
calculus
angle bisector
geometry solved

ABC triangle. Prove that:

(la+lb)^2<a+b+2c

Are you sure it's like that?
We have dimension 2 in the left term and dimension 1 in the second, so how do you compare an area ((la+lb)^2) with a length (a+b+2c) ?

Yeah sorry. The problem should be

(la+lb)^2<a^2+b^2+2c^2

In this case I have a solution
    We know from Cauchy that (la+lb)^2<=2(la^2+lb^2)
    Now we will use the lemma la^2<=p(p-a) and after some easy calculs we will have to prove that (a+b+c)*c<=a^2+b^2+2c^2
    Now after some other calculs we will have to prove that ac+bc<= \sum a^2 which is true

     Hope my solution is good!

I think the solution is good however I thought of another one. Could anybody find it?

I guess nobody found your solution! :(

so.. can you post it?

Well we know that:
la=(2b/(b+c)) *cos A/2<2bc/(b+c) \leq  \sqrt bc
the same  lb< \sqrt ca
So we now have: ( \sqrt bc+ \sqrt ca) 2 <a 2 +b 2 +2c 2
but ( \sqrt bc+ \sqrt ca) 2  \leq 2bc+2ca \leq b 2 +c 2 +c 2 +a 2 =a 2 +b 2 +2c 2

Have a nice day! :)

[quote="Maverick"]Let $ a,b,c$ are the sides of  $ \triangle ABC$,  $ w_b,w_c$ are the angle-bisectors. Prove that

$ (w_b \plus{} w_c)^2 < b^2 \plus{} c^2 \plus{} 2a^2$.[/quote]Note $ m_b,m_c$ are the medians, then

$ (w_b \plus{} w_c)^2\leq(m_b \plus{} m_c)^2\leq2(m_b^2 \plus{} m_c^2)$

$ \equal{} 2\left(\frac {2c^2 \plus{} 2a^2 \minus{} b^2}{4} \plus{} \frac {2a^2 \plus{} 2b^2 \minus{} c^2}{4}\right) \equal{} 2a^2 \plus{} \frac {b^2 \plus{} c^2}{2}$.

Actually,  if $ w_a$ is the angle-bisector of  $ \triangle ABC$, then

$ w_a\leq \frac {2b \plus{} 2c \minus{} a}{2\sqrt {3}}$,

with equality if and only if $ \triangle ABC$ is equilateral.

[hide="Proof"] $ \frac {(2b \plus{} 2c \minus{} a)^2}{12} \minus{} w_{a}^2 \equal{} \frac {(2b \plus{} 2c \minus{} a)^2}{12} \minus{} \frac {bc}{(b \plus{} c)^2}\left[(b \plus{} c)^2 \minus{} a^2\right]$

$ \equal{} \frac {(b \plus{} c \minus{} 2a)^2}{12} \plus{} \left(\frac {b \minus{} c}{b \plus{} c}\right)^2\frac {(b \plus{} c)^2 \minus{} a^2}{4}\geq 0$.[/hide]
In consequence, we have

$ w_b \plus{} w_c\leq \frac {4a \plus{} b \plus{} c}{2\sqrt {3}}\leq\sqrt {\frac {4a^2 \plus{} b^2 \plus{} c^2}{2}}$.


Sorry if there is another post like this, but in "the Big Picture" sections after chapters, I saw an error as they proclaimed the Mersenne numbers to be $2^{2^{n}}+1$ equals a prime number. But then 10000th User said that the Mersenne numbers were different and in fact, $2^{2^{n}}+1$ were called the Fermat Primes. Is it a typo?

you're right

mersenne primes are of the form $2^{n}-1$


I guess we should probably stop intruding the COMC discussion and set up a discussion area of our own.

so who has received an invitation so far?

[quote="soledad42"]All winter camp students last year that has not graduated plus Robin. Also, I think all the local students that went to the IMO training camp are invited. That's already around 12-13 people (I didn't count).[/quote]

13 people.

It is January 6-11., really late this year.  I don't think I will be able to come because of classes :(

When are the winter camp invitations generally sent out by?  :maybe:  :maybe:

I think the official one where you have to fill out a form is sent around mid-December. I remembered it came pretty late last year...

Here are some pictures from Winter Camp 2009. (Chen and Julian, protect your arms.  :rotfl: )

[img]http://photos-e.ak.fbcdn.net/photos-ak-snc1/v2063/12/11/500760782/n500760782_2420460_7895.jpg[/img]

[img]http://photos-f.ak.fbcdn.net/photos-ak-snc1/v2063/12/11/500760782/n500760782_2420461_8154.jpg[/img]

MAD PMS reunion:

[img]http://photos-b.ak.fbcdn.net/photos-ak-snc1/v2063/12/11/500760782/n500760782_2420465_9162.jpg[/img]

LMAO omg is that why the moose was not there when the group picture was taken? :rotfl: Yeah, the monster will definitely rip ur arms off.
Whoa MAD PMS is in order and actually sounds like something meaningful :D


If $ 2^x > 8000$, what is the least possible integer value of $ x$?

If $ 2^x > 8000$, then $ x$ is smallest for $ x \equal{} \boxed{13}$, as $ 2^{13} \equal{} 8192$ and $ 2^{12} \equal{} 4096$, and $ 2^x$ is monotonically increasing.


You have a 25-digit whole number A. Let's say that the abc 3-digit number can be read out of A if A contains the digits a, b, and c, so that b is to the right of a, and c is to the right of b. Prove that there is a 3-digit number (whose first digit is not zero, and it has 3 different digits), which cannot be read out of A.

Hes a neat trick. Consider the numbers $111,222,333,444,555,666,777,888,999$ For our number to have all of these, if must have at least 27 digits. Contradiction.

[quote="236factorial"]and it has 3 different digits[/quote]

Curses! I hate when i dont read the question!

I'm not sure if this is a solution yet.

---

there are two digits x,y that appear at most twice.  Let z and z' be the leftmost and rightmost of the x digits.

We need 15 digits to the right of z: this is because we have 8 digits left over, we place one, and we need 7 each to the left and right of this one, minimum.

We need 15 digits to the left of z'.

So suppose inbetween z and z' there are k digits.  Then we have 15 + 15 - (k+2) = 28-k digits being used.  But $k \geq 0$.  Uhoh.

I was thinking maybe, pick the digit as close to the right that dosnt appear any closer to the right, but that all the other digits are in there. This digit must be at least digit number nine. Then we have to be able to make all two digit numbers starting with this number with the remaining 16 digits of the number.But since each of these two digit numbers must occur in both orders, with any two digits, we must have at least three of these two. Therefore we need at least 15 digits to do this, so im one off...


For real numbers $a$ and $b$, define $a\diamond b=\sqrt{a^2+b^2}$. What is the value of $(5\diamond12)\diamond((-12)\diamond(-5))$?

$\text{(A)}\ 0 \qquad\text{(B)}\ \frac{17}{2} \qquad\text{(C)}\ 13 \qquad\text{(D)}\ 13\sqrt2 \qquad\text{(E)}\ 26$

[hide]Using the given imformation,
(supposing the given operation is shown as *)
5 * 12 would equal the square root of (5^2 + 12^2)
=squre root of 169
=13
12 * 5 would equal the square root of (12^2 + 5^2)
=square root of 169
=13
13*13 would equal the square root of (13^2 + 13^2)
=square root of 338
=13 multiply square root of 2
[b](D)[/b][/hide]

[hide]D,13sqrt2. Too lazy to include explanation.[/hide]

[hide]D 5,12,13 is a pythagorean triple. Use it twice. and you end up with 13^2 and 13^2. Add them up and you get D[/hide]

Gotta know your Pythagorean triplets.  :)

3,4,5
5,12,13
7,24,25
8,15,17
theres one for each odd number.
I figured that out sometime and the last two numbers of the odd numbers all are consecutive.

Yeah, there's a pattern for those triples.

This problem is fairly simple, just notice that, when squared, the first and second pairs of numbers are the same, and then notice that they're Pythagorean triples, and finally remember what happens when you take the square root of the sum of two equal squares, so you should get it in about 5 seconds. ;)


Let $a,b,c$ be length of the sides of triangle;and let $t \geq 1$ be real number.
Prove that: $\displaystyle (t+1)(\frac {a^2}{c}+\frac {b^2}{a}+\frac {c^2}{b})\geq \frac {b^2}{c}+\frac {c^2}{a}+\frac {a^2}{b}+t(a+b+c)$.

Of course we must prove it for $t=1$ since
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq a+b+c$.
So we must prove
$-\|(a-b)(b-c)(c-a)\|(a+b+c) \leq a^3b+b^3c+c^3a-abc(a+b+c)$
Now we can suppose $a>b>c$ (check!) and so the RHS is
$b(a^2-b^2)(a-c)+a(b-c)(b^2-c^2)$.
Now we can easily deduce the desired inequlality


em viet lau roi ma sao chang ai chiu giup vay

[quote="DuongChinh_K41SPHN"]cho tap hop A={+-a1,+-a2,.....,+-an}.xet so tu nhien 0<m<1024.CMR ton tai  tap S thuoc A sao cho neu a thuoc S  thi -a k0 thuoc S va tong tat ca cac phan tu cua S chia het cho m[/quote]
Nghe thằng Thế Anh nói thì mấy bài ông post thầy đều cho điểm cả,chơi xấu nhé :rotfl:
Mà ở trong kí túc bị dịch tiêu chảy thì phải,thằng Thế Anh nó bảo cả tiết học hôm qua ông cứ xin ra ngoài liên tục,lúc vào thì chưa kịp sơ vin,nếu bị thì cứ nói ra không bệnh này nguy hiểm lắm,5 tiếng là tử vong đó!


Hello everybody,
Lately, i've found a problem in proving that if :
x, y, z are from IN and x<y<z, and x >= 3,
(xyz)/x+y+z>=5

(sorry for not using Latex :roll: )

Ohh...Yeah! I didn't think it was that easy.
Anyway, great method  (have to remember it :oops: )


ratio
combinatorics unsolved
combinatorics

Given $n$ positive integers, consider all possible sums formed by one or more of them. Prove that there sums can be divided into $n$ groups such that in each group the ratio of the largest to the smallest does not exceed $2$.

[quote="N.T.TUAN"]see http://www.mathlinks.ro/Forum/viewtopic.php?p=353051#p353051  :wink:[/quote]
Thank you!


[color=darkred]Se considera un trapez isoscel $ABCD\ (AD\parallel BC)$ inscris in cercul $w$ si un punct fix $K\in w$ astfel incat dreapta $BC$ separa punctele $A$ si $K$. Pentru un paralelogram variabil $BCMN$ cu proprietatea $CM\parallel AK$ definim punctele $X\in AM\cap w$ si $Y\in AN\cap w$. Sa se determine locul geometric al intersectiei dreptelor $BX$ si $CY$.[/color]

super problema, locul geometric e dreapta $KD$, demonstratia aceasta este prea buna sa mai poata fi imbunatatita [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=138332[/url] [scuze de repetitie]


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

Suppose f is an extended real- valued measurable function and integrable on [0, $ \infty$) . Show that if f is
uniformly continuous on [a, $ \infty$) for some a $ \ge$ 0. Then show that f(x) goes to zero as x goes to infinity

If this fails, then for some $ \epsilon > 0 ,  |f(x_n)|  > \epsilon$ for a sequence $ x_n \rightarrow\infty$. Uniform continuity then implies there are intervals of the same length centered at these $ x_n$ where  $ |f(x_n)|  > \epsilon/2$.

But... f(x) = sin(x) is an extended real-valued measurable function and integrable on $ [0, \infty)$ and sine is uniformly continuous, but it's not going to 0???

$ \sin x$ is not integrable on $ [0,\infty).$ After all, $ \int_0^{\infty}\sin x\,dx$ is not defined.


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a^{4}\plus{}b^{4}\plus{}c^{4}\equal{} 3.$ Prove that:
$ \frac{a^{2}}{2b\plus{}c}\plus{}\frac{b^{2}}{2c\plus{}a}\plus{}\frac{c^{2}}{2a\plus{}b}\geq1.$
I have found the nice proof for it.

Very nice inequality Arqady .  :D
Here is my solution
By Holder we have $ (\sum\frac{a^{2}}{2b\plus{}c})(\sum\frac{a^{2}}{2b\plus{}c})(\sum\ a^{2}(2b\plus{}c)^{2})\geq (\sum a^{2})^{3}$

So it suffices to prove that $ (\sum a^{2})^{3}\geq\sum a^{2}(2b\plus{}c)^{2}$

But we know that $ (\sum a^{2})^{3}\geq 9\sum a^{2}b^{2}$   (why ? :wink: )

so it suffices to prove that
$ 9\sum a^{2}b^{2}\geq\sum a^{2}(2b\plus{}c)^{2}$  or $ a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2}\geq abc(a\plus{}b\plus{}c)$ which is true  :)

thanks arqady for the nice ineq

Beautiful proof, silouan! My congratulations! :lol:
[quote="silouan"]

But we know that $ (\sum a^{2})^{3}\geq 9\sum a^{2}b^{2}$   (why ? :wink: )

[/quote]
If $ a^{2}\equal{} x,$ $ b^{2}\equal{} y$ and $ c^{2}\equal{} z$ it's my old inequality:
$ (x\plus{}y\plus{}z)^{6}\geq27(x^{2}\plus{}y^{2}\plus{}z^{2})(xy\plus{}xz\plus{}yz)^{2}$ :D See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78236


function algebra polynomial trigonometry

alright i cant figure out how to get the inverse of this function. I cant seem to isolate y.

the function is f(x) = 2x + lnx

i need to find the inverse

the farthest ive gotten is e^2y + y = x..help me out here guys please

You aren't going to be able to invert that function using the "normal" functions you're used to (polynomials, exponentials, logs, trig functions, etc.).  It's inverse can't be expressed as a simple combination of them.  Sorry.


analytic geometry
AMC
AIME
ARML
geometry
similar triangles
algebra

The triangle is right-angled with sides $90, 120, 150$. The common tangents inside the triangle are parallel to the two sides Find the length of the dashed line joining the centers of the two small circles.
Simple.

[hide]

We know that the three triangles are 3-4-5 triangles.
We can find the largest circle's radius to be 30.

Then the bottom leg of the triangle is 60 + 60, so the radius of the lower right circle is 15. The radius of the upper left circle is 10.

We can right the coordinates of centers of the two smaller circles as (75, 15) and (10, 70).

Then the length of the segment is $\sqrt {(75 - 10)^2 + (70 - 15)^2} = \sqrt {65^2 + 55^2} = \sqrt {13^2 \cdot 5^2 + 11^2 \cdot 5^2} = \boxed {5\sqrt{290}}$

[/hide]

Yes, simple.

[hide]
I'll just outline the solution I made because I don't feel like making a diagram to explain things :P

Let C be the vertex of the triangle with a right angle and let A be the top vertex and B be the other vertex.  First, we find the radius of the largest circle.  The distance from C to the point where CA touches the circle can easily be found using the Power of a Point theorem and a system of equations, which I will leave out since I would need a diagram to explain it :P.  We find this value to be 30.  This is the same as the radius of the circle.

Now we use similar triangles to find the other radii.  The radius of fthe smallest circle comes out to be 10 and the radius of the last circle is 15.

Now we put this triangle on a coordinate plane with C = (0,0), A = (0,90) and B = (120,0).  We can easily find the center of the smallest circle to be (10, 70) and the vertex of the second smallest circle to be (75, 15).  Then the distance is sqrt((75-10)^2 + (70-15)^2) = sqrt(7250) = 5sqrt(290)
[/hide]

I put this in GS but it got moved up.... I really can't explain why.
oh yes your answers are correct.

This is a medium AIME problem, so it's perfectly justified. I think there was also something a little like this for the ARML tiebreaker this year too. In any case, both solvers used coordinate geometry, which is definitely not a getting started topic.

I would post my solution, but it's pretty much like what the others were...so i'll justsay that i got $5\sqrt{290}$ as well...


Usually, the AMC 12 cutoff for AIME qualification is 100, or in other words, top 5% of scorers.

probably in another week or so (at least it was around that for last year)

Frequently Asked Question AMC 10/12 #4
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


geometry
3D geometry
algebra unsolved
algebra

please which one is the greatest value?explain

They are the same.

They both satisfy the equation $ x^3 \plus{} 1 \equal{} \sqrt[3]{2}$ which has only one real root.

For the second number this can be verified directly by substitution.

For the first number just expand the cube and collect the terms to get the same result.

$ (x \plus{} y \plus{} z)^3 \equal{} x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \plus{} 3x^2y \plus{} 3x^2z \plus{} 3y^2x \plus{} 3y^2 z \plus{} 3z^2x \plus{} 3z^2y$

The rest I leave.


geometry
geometric transformation
rotation

On a 5 by 5 grid of unit squares, one unit square is colored blue, one unit square is colored red, and the rest of the unit squares are white. Grids are considered different if no rotation could turn one into the other. How many different grids are there.

There are 25 places for blue unit square and there are 24 places for red unit square. $ \frac{25*24}{4}\equal{}150.$

A harder problem:
On a 5 by 5 grid of unit squares, two unit squares are colored blue, two unit squares are colored red, and the rest of the unit squares are white. Grids are considered different if no rotation could turn one into the other. How many different grids are there.

[quote="AIME15"]I don't think that's right bachukas.  You didn't account for rotations.[/quote]

There are 4 rotations right?

I divided it by 4  :huh:

You're correct, Bachukas. Think about it for a second and it's common sense. For any selection of the blue and red ones, there are 4 ways that you could rotate it (around the center).

EDIT: the harder problem is $ \frac {\binom{25}{2}\binom{23}{2}}4 \equal{} 18975$. Same logic as above.

EDIT: nvm

much harder problem: There are two red squares, and the rest are white. How many are there [i]now[/i]?

[quote="math154"]
EDIT: the harder problem is $ \frac {\binom{25}{2}\binom{23}{2}}4 \equal{} 18975$. Same logic as above.[/quote]

Are you sure?

Here is example how we can get only 2 different grids by rotating 1 grid:

[img]http://www.part.lt/img/9ffd99191adea8ba39b5da2d557aaa72729.JPG[/img]

[quote="Wickedestjr"]So... do we have to divide by 2 instead?[/quote]

No, we can get only two different grids by rotating grid, also we can get 4 grids by rotating another grid.

[quote="math154"]I'm curious how you did it...I can think of dividing it into the "2 rotations" case and the "4 rotations" case.[/quote]

We can make only two rotations if coloured units are symetrical in point of grid's center.

I figured that out already. So did you just divide it into the cases and solve it (what I did)?

Yes, I divided it into "2 rotations" case and "4 rotations"  :)


Which of the following pair of electric and magnetic fields can represent a source of light-
(1) $B ={B_{0}}sin (wt-kx)$ and $E ={E_{0}}sin (wt-kz)$
(2) $B ={B_{0}}i+{B_{0}}j$ and $E ={E_{0}}i-{E_{0}}j$
(3) $B ={B_{0}}sin (wt-kx) i+{B_{0}}sin (wt-kx) j$ and $E ={E_{0}}sin (wt-kx) i-{E_{0}}sin (wt-kx) j$
(4)None of the above
Here $i$ and $j$ represent unit vectors along x and y axes respectively

well i suppose it has to be $3$ for the $\vec B$ and $\vec E$  are both perpendicular and also vary harmonically.


polynomial
algebra
complex analysis
complex analysis unsolved

[quote="hedgehuogis"]Is

$G_{2}(z) = \sum_{m =-\infty}^{\infty}\sum_{m^{2}+n^{2}\neq 0}(mz+n)^{-2}$

a modular form?[/quote]

No, it isn't, the series doesn't even converge [at any nonzero point, let alone uniformly on compacts...] You need an exponent strictly bigger than $2$ for the series to make sense. The Eisenstein series $G_{2k}(z)$ are only defined for $k \geq 2$ (but note that in most places you would encounter the notation $G_{k}$ in place of $G_{2k}$ for the exponent $2k$). The space of modular forms coincides with the polynomial algebra generated by $G_{4}(z) = \sum_{(m,n) \neq (0,0)}(mz+n)^{-4}$ and $G_{6}(z) = \sum_{(m,n) \neq (0,0)}(mz+n)^{-6}$.

Hope this helps :)


probability
geometry
symmetry
analytic geometry

A point $ P$ is chosen at random in the interior of equilateral triangle $ ABC$. What is the probability that $ \triangle ABP$ has a greater area than each of $ \triangle ACP$ and $ \triangle BCP$?

$ \textbf{(A)}\ \frac{1}{6} \qquad
\textbf{(B)}\ \frac{1}{4} \qquad
\textbf{(C)}\ \frac{1}{3} \qquad
\textbf{(D)}\ \frac{1}{2} \qquad
\textbf{(E)}\ \frac{2}{3}$

[hide]
By symmetry the answer is clearly $ \frac{1}{3}\equal{}C$.
[/hide]

[hide] Think of it this way- we draw the center of the equilateral triangle and draw the segments connecting the center to the vertices. This divides the triangle into 3 equal regions- if P lies anywhere in one region, it will be less than the other two. Thus, only 2/3 of the triangle if feasible, so the answer is E. [/hide]

I agree with the former solution.  Clearly each triangle has an equal chance of begin larger than the other two.  SO, the probability that it is a specific one is just $ 1/3$.

Yes, I agree with gfour84 and E^(pi*i) = -1

The answer is $ \frac{1}{3}$ due to symmetry.

mathdriven, er, using your method, it is true that $ ABP$ will be larger than at least one of the triangles, but it might also be smaller than the other triangle.

For example (counterexample):

 [geogebra]f576915aa3b7d5b3e62a8db84d3f81de572ea91b[/geogebra]

From the picture above, point $ P$ fits your constraints.

$ ABP$ is indeed larger than $ BCP$

However, clearly $ ACP$ is larger than $ ABP$

I hope this helped.

Geometric probability method. Call certain D center of triangle, so that AD, BD, CD are all same length. Now drop perpendiculars from D to closest side, see applet. For ABP is greater than BOTH ACP and BCP, P must be in shaded area, which is 1/3.

Or you could say each triangle has equal chance of being biggest so 1/3. This is more visual proof. Originally I thought it was 2/3 too I had the same reasoning as mathdriven then when I was double check my work I realized 2/3 is probability its bigger than at least 2 of triangles, not always both.

Applet:red/blue=1/3 answer.

 [geogebra]48267d932509f6134a73fcb2693be1b4d759b527[/geogebra]

Denote the barycentric coordinates to be $(x,y,z)$ with $x+y+z=1$. You want $x>y,z$. Clearly, this occurs with probability $\frac{1}{3}$.

Why does the [url=http://amc.maa.org/e-exams/e5-amc10/e5-1-10archive/2003-10a/03answers10AB.html]AMC answers[/url]have a way different answer?

And why is the AMC 10 problem way different? AMC 10A problem 19 is supposed to be finding the area of a lune...

Sorry for the revive, but just draw in the medial triangle and toss on the centroid, as a result the desired area is $\frac{1}{3}$ of the whole region.

[b]Solution:[/b] Firstly notice that the area of $\triangle{ABP}$ will have a greater area than each of $ \triangle ACP$ and $ \triangle BCP$? if and only if the distance from $P$ to $\overline{AB}$ is greater than that to each of $\overline{BC}$ and $\overline{AC}$. We can denote the centroid as $G$, and let the medians from $A, B, C$ be $X, Y, Z$ respectively. Then notice that the medians are only the angle bisectors (which give us insight into the distances from $P$ to one of the sides). Some analysis gives us $P$ must be in quadrilateral $XCYG$ to satisfy the problem condition. This gives us an answer of $\boxed{C}$. Furthermore, notice that we could theoretically switch up the vertices from $A, B, C$ into $B, C, A$ and $C, A, B$ and the same probability would hold in all cases, so one can get the answer by symmetry that way as well.


Prove that $ \displaystyle{\cos 27^{\circ}\equal{}\frac{1}{2}\sqrt{2\plus{}\sqrt{2\minus{}\varphi}}}$.

Note: I have already proven that $ \cos 72^{\circ}\equal{}0.5\varphi$ and $ \cos 36^{\circ}\equal{}0.5\phi$.

Thanks

Do you want this?
$ \cos 27^\circ \equal{} \cos\left( \frac{36^\circ \plus{} 18^\circ}{2} \right)$.

That really helped. I already have the expression for $ \cos 18^{\circ}$ and the half-angle formulas would help..

$ \cos 27^\circ \equal{} \sqrt {\frac {1 \plus{} \sin 36^\circ}{2}} \equal{} \sqrt {\frac {1 \plus{} \sqrt {1 \minus{} \cos ^ 2 36^\circ}}{2}}$.

By the triangle with $ 36^\circ,\ 36^\circ,\ 108^\circ$, or the ratios of 3 sides $ 1: 1: \frac {1 \plus{} \sqrt {5}}{2}$ gives $ \cos 36^\circ \equal{} \frac {1 \plus{} \sqrt {5}}{4}$


For $ 0\leq x\leq 1$, $ a\plus{}bx\leq \frac{e^{2x}\plus{}e^{\minus{}x}}{2}\leq a\plus{}cx$ always holds.

(1) Find the value of $ a.$

(2) Find the maxmumvalue of $ b$ and minimum value of $ c.$

hello, solution to your problem, for $x=0$ we get
$a\le 1\le a$ therefore $a=1$. The maximum value of $b$ is $b=0$ and the minimal value of $c$ we get if our straigthline goes through the points $(0,1)$ and $\left(1,\frac{e^{2}+e^{1}}{2}\right)$, therefore we get $c=\frac{e^{2}+e^{-1}}{2}-1$.
Sonnhard.

hello kunny, $b$ must be $1/2$, because if $f(x)=\frac{e^{2x}+e^{-x}}{2}-1-bx$, we get
$f'(x)=\frac{2e^{2x}-e^{-x}}{2}-b$ and $f'(x)\geq0$ for all $x$ with$ 0\le x \le 1$ if $\frac{1}{2}\geq b$ so we can take $b=\frac{1}{2}$.
Sonnhard.

Yes, $b=\frac 12$.

The problem is 2009 Hitotsubashi University entrance exam/Law, Economics e.t.c., 2nd exam


The students in Mrs. Reed's English class are reading the same
760-page novel. Three friends, Alice, Bob and Chandra, are in the
class. Alice reads a page in 20 seconds, Bob reads a page in 45
seconds and Chandra reads a  page in 30 seconds. \\ \\
Before Chandra and Bob start reading, Alice says she would like to
team read with them. If they divide the book into three sections so that each
reads for the same length of time, how many seconds will each have
to read?

Let's say how many pages each of them can read in 3 minutes.

Alice can read 9 pages in 3 minutes.

Bob can read 4 pages in 3 minutes.

Chandra can read 6 pages in 3 minutes.

So together, they read 19 pages in 3 minutes.

To get how many minutes each of them has to read, divide 760 by 19 and then multiply by 3.

760/19=40.  40*3=120.

Now multiply 120 minutes by 60 seconds/minute to get the number of seconds each of them needs to read.

120*60= 7200

Each of them needs to read 7200 seconds if they divide their book into sections that way.


graph theory
combinatorics proposed
combinatorics

There are 25 people. Every two of them are use some language to speak between. They use only one language even if they both know another one. Among every three of them there is one who speaking with two other on the same language. Prove that there exist one who speaking with 10 other on the same language.

Reformulate the problem:

[i]Consider a complete graph with $ 25$ vertices whose edges are colored with several colors such that in any triangle, at least two edges have the same color. Prove that there is a vertex with at least $ 10$ outgoing edges of the same color.[/i]

Suppose what we have to prove is false.
Call a triple $ (A,B,C)$ [i]good[/i] if $ AB$ and $ BC$ have the same color.
Count the number $ N$ of [i]good[/i] triples $ (A,B,C)$. From the condition, $ N$ is greater or equal to $ \displaystyle\binom{25}3$. Consider a vertex $ X$ and suppose $ d_1$ of the edges having $ X$ as an end are of the color $ 1$, $ d_2$ - color $ 2$ and so on.
Then $ d_1\plus{}d_2\plus{}\ldots\plus{}d_c\equal{}24$. The number of [i]good[/i] triples $ (A,B,C)$ with $ B\equal{}X$ is $ \displaystyle\sum_{i\equal{}1}^c\binom{d_i}2\equal{}\frac12\left(\minus{}24\plus{}\sum_{i\equal{}1}^cd_i^2\right)$. We have assumed that $ d_i\le9$. If there exist $ i\neq j$ such that $ d_i,d_j<9$ then we have $ d_i^2\plus{}d_j^2\le (d_i\plus{}d_j)^2$ if $ d_i\plus{}d_j\le9$, or $ d_i^2\plus{}d_j^2\le 9^2\plus{}(d_i\plus{}d_j\minus{}9)^2$, if $ d_i\plus{}d_j\ge9$. That's why $ d_1^2\plus{}d_2^2\plus{}\ldots\plus{}d_c^2\le 9^2\plus{}9^2\plus{}6^2$, or $ d_1^2\plus{}\ldots\plus{}d_c^2$ is maximal when $ c\equal{}3$ and $ d_1\equal{}d_2\equal{}9$, $ d_3\equal{}6$. So the number of [i]good[/i] triples $ (A,B,C)$ with $ X\equal{}B$ is at most $ \displaystyle2\binom92\plus{}\binom62\equal{}87$. Then $ N\le25\cdot87$ and from above we have $ 25\cdot 87\ge N\ge\displaystyle\binom{25}3$, or $ 87\ge92$. Contradiction. So, our assumption that no vertex has at least $ 10$ outgoing edges of the same color is false.

My congratulations to freemind.
 :first:


I think my teacher told me that I'll probably be taking AMC-12A at 30th, not 31st. Is this allowed?

oh and is it allowed for you to get your score graded by your teacher after you took the test? (I heard the answer keys come with the problems but I never got my score until very later and I like to see if I qualified or not :D )

Yeah, your teacher can grade your test right after everyone takes it.  But I don't know about taking it early.

i don't see how you could take the test on a different day...i am taking it on the 31st, maybe your teacher made a mistake or something

[quote="Silverfalcon"]I think my teacher told me that I'll probably be taking AMC-12A at 30th, not 31st. Is this allowed?[/quote]This is definitely [b]not[/b] allowed.

Here is a quote from the instructions that the AMC sends to schools:

[quote="AMC Instructions Packet"]If today is not yet Tuesday, January 31, 2006. STOP EVERYTHING. Under no circumstances is the contest to be given to anyone before the official day, nor should the contest package(s) be opened before that date.[/quote]and elsewhere it says

[quote="AMC Instructions Packet"]Early administration of the contests is never permitted, and will lead to disqualification.[/quote]Please double-check with your teacher that you are taking it on the [b]31st[/b], not the 30th.  Taking the test early (indeed, even your teacher looking at the test early) is grounds for disqualification.

Also, your teacher is not supposed to grade your answer sheet, but he/she is allowed to give you the answers *after* your answer sheet has been sent off to the AMC.


Please tell me what you would answer for the following question:

For which of the following molecules are resonance structures necessary to describe the bonding satisfactorily?

(A) $H_{2}S$
(B) $SO_{2}$
(C) $CO_{2}$
(D) $OF_{2}$
(E) $PF_{3}$

I would say (B), but my teacher says that that is incorrect... is (B) the correct answer or is another answer the correct one?

Thanks!!!

I think B is correct. I can come up with stable non-resonance structures for all others and since S and O are in the same group, SO2 is probably a lot like O3, which has resonance structures.

Yeah, that's what I was thinking...

Maybe one should think of this in terms of formal charge?  I have certainly come up with non-resonance structures for A, C, D, and E that each contain atoms with a formal charge of zero each...

The only one that I could possibly think of as having resonance structures is $SO_{2}$.


calculus
integration
function
calculus computations

Frankly, I'm too lazy to evaluate the integral as is.  So I'll use Fubini's Theorem and change the order of integration.  By drawing a picture, we see that the region of integration is $ x\equal{}0$ on the left, $ x\equal{}1$ on the right, $ y\equal{}0$ on the bottom, and $ y\equal{}x\sqrt{x}$ on top.  So we can rewrite it as
$ \int_{0}^{1}\int_{0}^{\sqrt{x}}\frac{ye^{x}}{x}\mbox{ }dy\mbox{ }dx$
Then we have
$ \frac{1}{2}\int_{0}^{2}e^{x}\mbox{ }dx$
which is really easy and equal to $ \frac{e\minus{}1}{2}$

How did you determine the y-bounds on the region of integration?

I just drew a picture based on what was already given and looked at the boundaries based on the picture.

Change "too lazy" to "can't be done". It's well-known that $ \frac {e^x}{x}$ doesn't have an elementary antiderivative.

[Edit: Error fixed, and I misread it anyway]

[quote="bthings_2000"]$ {\int_{0}^1}{\int_{y^2}^1}\frac {ye^x}{x}(dx)(dy)$[/quote]
Since $ y^2\le x\le1\implies y\le\sqrt x\le1$ and $ 0\le y\le1$ we can conclude that $ 0\le y\le\sqrt x\le1.$

From here, you can take easily the new integration limits when reversin' the integration order. It follows that $ 0\le y\le\sqrt x$ and $ 0\le x\le1.$ Now reverse the integration order:

$ \int_0^1 {\int_{y^2 }^1 {\frac{{ye^x }}{x}\,dx} \,dy}  \equal{} \int_0^1 {\int_0^{\sqrt x } {\frac{{ye^x }}{x}\,dy} \,dx}  \equal{} \int_0^1 {\frac{{e^x }}{x}\underbrace {\int_0^{\sqrt x } {y\,dy} }_{\frac{1}{2}x}\,dx} .$

I think you can do the rest from there.

Thanks all, I have it now. The picture method works well intuitively but the analytical method seems more rigorous.

The analytic method is based off of the picture method.  When you start dealing with functions of 2 and 3 variables, it becomes more difficult to draw the pictures so you'll be forced to use some annoying algebra, but as my professor constantly says, "seeing is believing".

The limits of integration on the inner integral are for $ x$ between $ y^2$ and $ 1$.  The limits of integration on the outer integral are for $ y$ between $ 0$ and $ 1$.  Thus $ y^2 \leq 1$ and so $ y^2 \leq x \leq 1$.  If you draw the picture of the region of integration, this will become abundently clear.


For each positive integer $ m > 1$, let $ P(m)$ represent the greatest prime factor of $ m$. (ex. $ m \equal{} 25$ $ P(m) \equal{} 5$, $ m \equal{} 21$ $ P(m) \equal{} 7$)

Find all positive integers $ m$ such that $ P(m) \equal{} \sqrt {m}$ & $ P(m \plus{} 48) \equal{} \sqrt {m \plus{} 48}$.

[hide="my solution"]
Say $ p \equal{} \sqrt {m}$ and $ q \equal{} \sqrt {m \plus{} 48}$. Clearly, since $ P(m) \equal{} p$, the prime factorization of $ m$ will be $ p^2$. And the one of $ m \plus{} 48$ will be $ q^2$. So it comes down to searching all couples of primes of which the difference of their squares is 48.

The difference between two subsequent squares is always odd, so $ q \equal{} p \plus{} 2n$ with $ n$ a positive integer:
$ q^2 \equal{} p^2 \plus{} 48$
$ (p \plus{} 2n)^2 \equal{} p^2 \plus{} 48$
$ 4pn \plus{} 4n^2 \equal{} 48$
$ n(p \plus{} n) \equal{} 12$

p and n are positive integers whose product is twelve so the only possible values for (p,n) are (11,1), (4,2) and (1,3). So for (p,q) we get the possible values (11,13), (4,8) and (1,7). Only in the first one p and q are primes so we can conclude:
$ p \equal{} 11$
$ m \equal{} 121$
[/hide]


A function is "flipping" if for certain numbers $a$ and $b$ $f(a) = b$ and $f(b) = a$ and for a given domain of $x$, $f(f(...f(x))) = a$ or $b$.  An example is the function $f(x) = 1-x^{2}$ over the domain $[0, 1]$.  Let's take $0.5$ for example:
$0.5$
$0.75$
$0.4375$
.
.
.
$0$
$1$
$0$
$1$
.
.
.

Find a flipping function that flips between $\frac{1}{2}$ and $\sqrt{3}$.

We can just follow your model and make a parabola that goes through $(1/2,\sqrt{3})$ with a vertex at $(\sqrt{3},1/2)$.

So, one such function is $f(x)=\frac{2}{-1+2 \sqrt{3}}(x-\sqrt{3})^{2}+1/2$

There are tons of functions with this property. It's really a concavity question, since you only force 2 points.

The second constraint is troublesome, since clearly you can find $x$ such that $f^{n}(x)$ cannot, say, be rational for finite $n$, so flipping between $0$ and $1$ cannot occur (except as limit behavior).

I think based on his example that he is in fact referring to limit behavior.  :maybe:


probability
quadratics
calculus
integration
absolute value
algebra
quadratic formula

not quite
i think what you proved is that it is not always an integer- just because n-1 and n-3 dont always divide $2n^{2}-10n-4$ they can sometimes.  x+1 is not a factor of x+3, but x+1|x+3 if x=1

[quote="daermon"] x+1|x+3 [/quote]

what does that mean? i thougt that symbol was only used in probability (bayes' theorem, etc.) and absolute value signs.

x+1 divides x+3.  The use of the symbol $|$ is very common in Number Theory.

[quote="The QuattoMaster 6000"][quote="max_tm"]The problem:

Given that $n$ is an integer, for how many values of $n$ is $\frac{2n^{2}-10n-4}{n^{2}-4n+3}$ an integer?

I thought to myself as I was writing this out that aha! I proved there were no integer solutions (obviously I was wrong). What was my mistake?:

[hide]
For $\frac{2n^{2}-10n-4}{n^{2}-4n+3}$ to be an integer, $n^{2}-4n+3\mid 2n^{2}-10n-4$, which means that $(n-1)(n-3)\mid 2n^{2}-10n-4$. So either $n-1\mid 2n^{2}-10n-4$ or $n-3\mid 2n^{2}-10n-4$ and by factor theorem, neither of these are factors of $2n^{2}-10n-4$, therefore the fraction can never be an integer.

Q.E.D. (NOT!)
[/hide]

So uh, what's the deal?

EDIT: Sorry if this should've been in the intermediate forum.[/quote]
[hide] Say that $\frac{2n^{2}-10n-4}{n^{2}-4n+3}=k$ where k is an integer.  Thus, $kn^{2}-4kn+3k=2n^{2}-10n-4\Rightarrow n^{2}(2-k)-n(10-4k)+(-3k-4)=0$.  Using the quadratic formula, we have that $n=\frac{(10-4k)\pm\sqrt{(10-4k)^{2}-(2-k)(-3k-4)4}}{2(2-k)}$
Expanding, we obtain that
\[n=\frac{(10-4k)\pm\sqrt{100-80k+16k^{2}-(12k^{2}-8k-32)}}{2(2-k)}=\frac{(10-4k)\pm\sqrt{4k^{2}-72k+132}}{2(2-k)}=\frac{(5-2k)\pm\sqrt{k^{2}-18k+33}}{2-k}\]
Thus, we have that $\sqrt{k^{2}-18k+33}$ is an integer, so $k^{2}-18k+33$ is a perfect square.  So, say that
\[k^{2}-18k+33=x^{2}\Rightarrow k^{2}-18k+(33-x^{2})=0\Rightarrow k=\frac{18\pm\sqrt{324-132+4x^{2}}}{2}=9\pm\sqrt{92+x^{2}}\]
Thus, $92+x^{2}$ is a perfect square.  So, say that $92+x^{2}=y^{2}\Rightarrow y^{2}-x^{2}=92\Rightarrow (y+x)(y-x)=2*46$.  The only way that makes x and y both integers is if $y=24, x=22$.  Since $x=22$, we have that $k^{2}-18k+33=22^{2}\Rightarrow k^{2}-18k-451=0$, but this makes k non-integral.  So, either I made a mistake or there are no possibities for n.  [/hide][/quote]

n=-1,2,4etc...

\[\frac{ 2n^{2}-10n-4}{n^{2}-4n+3}= 2-\frac{2n+10}{n^{2}-4n+3}\]
Limit the cases, then either test values or use a further argument.

[hide="Answer set"]
\[n \in \{-1, 2, 4, 7\}\][/hide]

UHH UGLY
[hide]
Long division first, and we get the expression to be:
$2-2\cdot \frac{n+5}{n^{2}-4n+3}$

Well it's an integer if the fraction is an integer.  Well the fraction is has even a SLIGHT possibility of being an integer if $n^{2}-4n+3 \le n+5$, or $n^{2}-5n-2 \le 0$.  BLAH too lazy to find the roots, so just check we find that we must have $0 \le n \le 5$.  six values, not too bad to check each of them.  only $n=2, 4$ worked, so 2 values.
[/hide]

Your work is so ugly I don't want to look...

[quote="pkerichang"]
[hide]if $n^{2}-4n+3 \le n+5$
[/hide][/quote]

that should have absolute value signs since n can be negative

Cincodemayo posted the Solution manual's method of solution and all the correct answers. Daermon is right in his answer to my question, all I proved is that the fraction is not [b]always[/b] an integer for all integral values of $n$. Thankee.

EDIT: actually everybody missed $-5$  :wink:

[hide=" rest of solution"]
The rational expression can only assume integer values when $2n+10 \ge n^{2}-4n+3$ or when $2n+10=0$ That is, $n^{2}-6n-7\le 0$ or $n=-5$. $-1\le n\le 7$ or $n=-5$. We consider all integers over these intervals (remembering the fact that $n \neq 1$ and $n \neq 3$ We conclude that $n\in\{-5,-1,2,4,7\}$

[/hide]

[quote="The QuattoMaster 6000"][quote="max_tm"]The problem:

Given that $n$ is an integer, for how many values of $n$ is $\frac{2n^{2}-10n-4}{n^{2}-4n+3}$ an integer?

I thought to myself as I was writing this out that aha! I proved there were no integer solutions (obviously I was wrong). What was my mistake?:

[hide]
For $\frac{2n^{2}-10n-4}{n^{2}-4n+3}$ to be an integer, $n^{2}-4n+3\mid 2n^{2}-10n-4$, which means that $(n-1)(n-3)\mid 2n^{2}-10n-4$. So either $n-1\mid 2n^{2}-10n-4$ or $n-3\mid 2n^{2}-10n-4$ and by factor theorem, neither of these are factors of $2n^{2}-10n-4$, therefore the fraction can never be an integer.

Q.E.D. (NOT!)
[/hide]

So uh, what's the deal?

EDIT: Sorry if this should've been in the intermediate forum.[/quote]
$k^{2}-18k+33=x^{2}\Rightarrow k^{2}-18k+(33-x^{2})=0\Rightarrow k=\frac{18\pm\sqrt{324-132+4x^{2}}}{2}=9\pm\sqrt{92+4x^{2}}$[/quote]
[hide="Oh, here is my mistake"] It should be $9\pm\sqrt{192+4x^{2}}$.  Dividing by 4, we obtain that $48+x^{2}$ should be a perfect square.  Now, our possibilities for $(x,y)$ are $(11, 13)$, $(4, 8)$, and $(1, 7)$.  Now, we plug these back in to get that $k\in\{-4, 1, 2, 16, 17, 22\}$  Now, we can plug these in to get what $n$ equals.  [/hide]


A truncated octahedron is a geometric solid with 14 faces (6 congruent squares and 8 congruent hexagons). In this particular solid, 2 hexagons and 1 square meet to form each forner. How many corners does the solid have?

[hide]I think that since each square is a part of 4 corners, that there are 4*6=24 corners, if there are no all-hexagon corners.[/hide]

Billy

[hide]Since there's a square at each of the six corners, you just have to multiply the number of corners by 4, giving the answer of 24.[/hide]

[quote="Farazy"]A [b]truncated [/b]octahedron is a geometric solid with 14 faces (6 congruent squares and 8 congruent hexagons). In this particular solid, 2 hexagons and 1 square meet to form each forner. How many corners does the solid have?[/quote]

 :D

[hide="my non sensible sol."]
square per side on the hexagon.
4*6=24

YAY!!!
[/hide]

Truncated - Having the apex cut off and replaced by a plane, especially one parallel to the base. Used of a cone or pyramid.

I know the answer...JUST THINK!  Check my blog out for a contest by the way!

I absolutely HATE these 3D geometric folding, edges, whatever. Yet at least 1 of these types of problems come up every competition... :(


It covers everything.  The assigned problems are from states and sometimes nationals.  I've taken the class and learned a lot.  AoPS wants you to feel like you are [u][i]always[/i][/u] being challenged, even if you won Mathcounts last year. :D
[b][i][u]-minicon[/u][/i][/b]

[quote="xpmath"]Sorry if this has been posted before, but around what level does the Mathcounts Problem Series cover? For example, does it cover the harder state problems and medium to hard national ones?
Thanks.[/quote]

Yes; for the most part, the course is aimed at more experienced MATHCOUNTS students.

Ok, thanks. I can't wait for it to start.


logarithms
function
number theory unsolved
number theory

Let $x, y, n$ denote positive integers. Find the number of digits of $x^{n}+y^{n}$ in base $10$.

$[\log_{10}(x^{n}+y^{n})]+1$
do you want smth more specific? :)

You're right. May be I should have told you before that I have tried to use elementary functions, but all of my attempts were in vain. Since the question was originally what you sent! I just used another interpretation. Can anyone put these another way?


inequalities
calculus
integration
parameterization
triangle inequality
complex analysis

We have just seen M-L formula in the university and i think the following problem directly relates to this formula :

Let C denote the line segment from z = 2i to z = 2. Prove that
 | Line Integral on C   ((z+2)/(z^4 + 1))dz | <= (8/3) sqrt 2 , without evaluating the integral.

In fact, i wonder what kind of strategies we can use to show that cont. func. f satisfies |f(z)|<= M for all z on a smooth curve.

The genreal formula is $\left|\int_{C}f(z) \ dz \right| \le L*M$, where $L$ is the length of the contour and $|f(z)| \le M \ \forall z\in C$.  Now, by the triangle inequality we have

$|z+2|\le|z|+2\le2+2 = 4$

$|z^{4}+1|\ge \parallel z^{4}|-1| \ge |4-1| = 3$

Thus $M=\frac{4}{3}$, and clearly $L = 2\sqrt{2}$, so we have $\left|\int_{C}f(z) \ dz\right| \le \frac{4}{3}* 2\sqrt{2}=\frac{8}{3}\sqrt{2}$.

[quote="cooley"]In fact, i wonder what kind of strategies we can use to show that cont. func. f satisfies |f(z)|<= M for all z on a smooth curve.[/quote]
There are various ways of proving inequalities. [url=http://books.google.com/books?vid=ISBN0521358809&id=u5fkpAJ5Rc0C&printsec=frontcover&dq=isbn:0521358809]This book[/url] (Hardy-Littlewood-Polya) has a bunch of them. But the only way I know that uses the smoothness of the curve is to maximize $|f(z(t))|$ over real parameter $t$, where $z(t)$ is a parametrization of the curve.


inequalities
calculus
inequalities unsolved

[quote="Aznboy256"]Let f(x,y,z)= x+2y+4z . Find the maximum value of f(x,y,z,) given that x^2+2y^2+z^2=1.[/quote]

Use Calculus of Variations,

$\nabla f = k \nabla g$ where $g(x,y,z)=x^{2}+2y^{2}+z^{2}$
$x^{2}+2y^{2}+z^{2}=1$

So,

$<1,2,4> = k <2x,4y,4z>$
$x^{2}+2y^{2}+z^{2}=1$

This produces the equations,
$\{ \begin{array}{c}1=2kx \\ 2 = 4ky \\ 4 = 4kz \\ x^{2}+2y^{2}+z^{2}=1 \end{array}$

[quote="RDeepMath91"]It just doesn't work since the given equality is $x^{2}+2y^{2}+z^{2}= 1$  :(
So, plug $a = 1, b = \sqrt2, c = 4$ to $(a^{2}+b^{2}+c^{2})(x^{2}+2y^{2}+z^{2}) \geq (ax+\sqrt2 \cdot by+cz)^{2}$.
We will obtain $x+2y+4z \leq \sqrt{19}$, equality holds if and only if $|x| = |y| = \frac{1}{\sqrt{19}}$ and $|z| = \frac{4}{\sqrt{19}}$.[/quote]

Your answer is correct, but the condition for which the equality holds is incorrect.

$x+2y+z=x+\sqrt{2}\cdot \sqrt{2}y+4z=(1,\ \sqrt{2},\ 4)\cdot (x,\ \sqrt{2}y,\ z)\leq \sqrt{19}$.

The equality holds when $(x,\ \sqrt{2}y,\ z)\parallel (1,\ \sqrt{2},\ 4)$ and ${x^{2}+2y^{2}+z^{2}=1,\ x>0,\ y>0,\ z>0\Longleftrightarrow \ (x,\ y,\ z) =(\frac{1}{\sqrt{19}},\ \frac{1}{\sqrt{19}},\ \frac{4}{\sqrt{19}}})$.

Well, maybe you can look the original problem once again, kunny..
$f(x,y,z) = x+2y+4z$, not $x+2y+z$..  :)

Oops! :blush: I just have edited. Thank you for your indication,  RDeepmath91. :)


Prove that a regular $2n$-gon ($n\ge 3$) can be divided into $\frac{n(n-1)}{2}$ rhombuses.
(The rhombuses don't have to be congruent.)

Can you prove that $\frac{n(n-1)}{2}$ is the minimum possible number of rhombuses?

[hide="Hint"]
$\frac{n(n-1)}{2}=\binom{n}{2}$

Observe how the rhombuses are placed in easier cases ($n=3,4,...$).

[hide="More hint?"]Parallel lines ;) [/hide]
[/hide]

(There was a mistake in the problem, so I fixed it.)


The circles $ w_{1}$ and $ w_{2}$ with centers $ O_{1}$ and $ O_{2}$ respectively intersect each other in points $ A$ and $ B$. On the same side as $ B$ there is a line which touches $ w_{1}$ and $ w_{2}$ circles at points $ C$ and $ D$ respectively. $ KD$ is diameter of circle $ w_{2}$. $ O_{1}O_{2}$ intersect $ KB$ in point $ P$. Prove that the lines $ DP$ and $ CB$ are perpendicular to each other.

The common external tangent CD meets the center line $ O_1O_2$ at E, the external similarity center of the circles $ w_1, w_2.$ Let $ CL \parallel DK$ be a diameter of $ w_1.$ Let EB, DP cut $ w_2$ again at F, Q, respectively. Intersections $ E \equiv BF \cap DD, P \equiv BK \cap DQ$ and center $O_2 \in DK$ are collinear By Pascal theorem for the degenerated cyclic hexagon DDBQKF, $O_2 \in FQ$ $ \Longrightarrow$ DQKF is a rectangle inscribed in $ w_2$. Triangles $ \triangle DFK \sim \triangle CBL$ are centrally similar with similarity center E $ \Longrightarrow$ $CB \perp (CL \parallel FK \parallel DQ)$ $ \Longrightarrow$ $CB \perp DP$.

Another Solution:

let $ AB \cap CD\equal{}M$
because $ BM \bot O_1 O_2$ and $ CD \bot KD$, $ \angle BMD\equal{}\angle PO_2 K$
and it is true that $ \angle BDM\equal{}\angle PKO_2$
So $ \triangle BMD \sim \triangle PO_2K$
On the other side, $ M$ is a mid-point of $ CD$ and $ O_2$ is a midpoint of $ KD$.
So we can conclude that $ \triangle BCD \sim \triangle PDK$ is also true.
from this, $ \angle BCD\equal{}\angle PDK$ is extracted, which implies the end of the proof.


Calculate the followning sums, for $x$ a real number.

(a) $S_{1}=\sum_{k=1}^{n}{kx^{k}}$

(b) $S_{2}=\sum_{k=1}^{n}{k^{2}x^{k}}$

(c)  $S_{3}=\sum_{k=1}^{n}{k^{3}x^{k}}$

[hide="..."]I'm curious how can the be calculated in different ways  :? [/hide]

well i assume that $|x|\le1$ and that $n$ tends towards $\infty$. i dont know about the last two but the first one can be done with some Gaussian paring.

maokid7: [hide]If n is finite, the results can be written in a  nice form, even when $x \ge 1$ .. Hint: Try subtracting  $x \cdot S$ from $S$ , term by term. ... It will work for $S_2$ and $S_3$ too .. you may have to repeate your steps. [/hide]

With some modifications, these can all be done with generating functions, although it's very tedious :D

by the theorem of finite difference, if you apply the subtracting couple times for number 2 and 3, eventually you get a constant coefficient for all x.  kind of nice

You may also use the Abel Summation Formula for the first one as far as I've observed the questions.

i think i got #2

[hide]
$S_2 = x + 4x^2 + 9x^3 + \ldots + n^2x^n$
$xS_2 = x^2 + 4x^3 + 9x^4 + \ldots + n^2x^{n+1}$
$S_2(1-x) = x + 3x^2 + 5x^3 + 7x^4 + \ldots + (2n-1)x^n - n^2x^{n+1}$

then we have:
$xS_2(1-x) = x^2 + 3x^3 + 5x^4 + \ldots + (2n-1)x^{n+1} - n^2x^{n+2}$

subtracting this from $S_2(1-x)$ we have:

$S_2(1-x) - x(1-x)S_2 = S_2 (1-x)^2 = x + 2x^2 + 2x^3 + \ldots + 2x^{n} - (n^2 +2n-1)x^{n+1} + n^2x^{n+2}$

this can then be simplified into:

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

and dividing both sides by $(1-x)^2$ gives you what $S_2$ is (yea im too lazy... to actually simplify that  :blush:

hopefully thats right

[/hide]


I see on this board the unhandy [tex] command has been replaced by the more common $ symbols. How exactly has this been done (in coding)? Or is that secret?

I found posts from Valentin asking it on phpbb.com but I haven't seen answers to it there, only people posting that it was not possible :P and some posting references that do not exist anymore at present...

Some hints or the full modification would be most welcomed :)

It was me arguing on the phpbb forum that it couldn't/shouldn't be done. Valentin persisted and so I came up with a method that proved he was right and I was wrong :blush:

If you'd like the code I sent him I will happily let you have it; I didn't include it in the standard LatexRender download because it could cause problems for those less skilled than Valentin; of course the code used by this forum has probably been updated and improved since then.

Roughly speaking, the code treats $ in the same way as [tex ] and for \ [ ... \ ] it adds \displaystyle to the formula (that has been changed to be the default on this forum) and an HTML center tag added to the image
A pair of double dollar signs are changed to \ [ ... \ ] before any latexing, so produces the same effect.


Prove that the deleted diameter plane is locally connected and path-connected.

I've no idea about this question...I'll be delighted to receive some guidance about it...

Thanks in advance

It would be nice to start with telling us what a "deleted diameter plane" is ;). Is it just a plane without an interval?

Well, let's define a "deleted diameter circle" to be a circle in R^2 without a finite number of diameters (can be 0... ) but it still contains its center...The deleted diameter plane is the topological space which for every x in R^2 the set of all deleted diameter circles that their center is in x is a neighbourhood base of x...

Hope you'll be able to help me now because I'm pretty desperate...

Thanks!

Path connected because any point can be connected to $ 0$ by straight line segment.

Locally connected because any non-zero point has a neighborhood that is a disk (i.e. is not crossed by any removed diameters). As for the zero point, its neighborhoods are connected because they are just like the space itself.

Well..

Your answer isn't that understandable...:
About the path connected thing- we need to find a function that for every two points x,y in this plane will do: f:[0,1]->Plane such as f(0)=x, f(1)=y and f is continous...How that fact that there's a line segmant from x to 0 and from y to 0 helps us?

About the locally connected thing-
We need to prove that for every point in the plane and for every neighbourhood around it, there is a sub-neighbourhood which is connected. It's true that every point has a neighbourhood that is a disk, but how is this helpful? It isn't a sub-neighbourhood of any other neighbourhood or something. And why did you seperate the case of (0,0) from the other cases?

Thanks!

First go from $ x$ to $ 0$, then go from $ 0$ to $ y$, both by straight line segments (so take $ f(\frac12) \equal{} 0$). Pasting lemma etc.

Take a neighbourhood $ N$ around some point not equal to zero. Then use that disk neighbourhood and fit it inside $ N$.

$ (0,0)$ was seperated from the other cases because that's where the deleted diameters "meet".

I still can't understand how to prove that it's path-connected...
You are saying we need to use a segment from x to 0 and from y to 0... But how is this helpful? :(

I'm becoming realy desperate in this...Hope you'll be able to help me

Thanks

Can you please state the definition of path-connectedness?

(I know it, but I am asking for didactic purposes)

A space is path-connected iff for every to points x,y in it there's a continous function f such as f:[0,1]->X where f(0)=x, f(1)=y...

Which function should I choose? The function that make use of a segment from 0 to x and from 0 to y isn't that correct because there might be a deleted diameter over that segment...

Hope you'll now be able to help me

Thanks a lot and sorry for my misunderstanding

I am assuming here that by deleting a 'diameter' (which is not standard terminology as far as I know) you mean deleting some straight line through the origin.

Call our deleted diameter plane $ D$.

Okay, take your points $ x, y \in D$. We are going to construct the function $ f$ we need.

Using that we are working in $ \mathbb{R}^2$ we can see that $ g: [0,\frac12] \to D$ defined by $ g(t) : \equal{} (1\minus{}2t) x$ is a "path" from $ x$ to $ 0$ (yes? is this continuous? you should show this).

Then define similarly $ h: [\frac12, 1] \to D$ by $ h(t): \equal{} (2t\minus{}1) y$. This is a "path" from $ 0$ to $ y$.

Now putting $ f(t) \equal{} g(t)$ on $ [0,\frac12]$ and $ f(t) \equal{} h(t)$ on $ [\frac12,1]$ gives our desired path.

The pasting lemma (wiki this) tells us that if we have two paths from points $ x$ to $ z$ and $ z$ to $ y$, we can paste them together to get a path from $ x$ to $ y$, which is exactly what we have done here, albeit a bit more explicit.

Is it clear that this works?

edit: wrote "'s as the OP insists on paths to be defined on the domain $ [0,1]$ instead of an interval. Also removed some silly dividing by norms.

Obviously this can be corrected?

Take $ g$ to be given by $ g(t) \equal{} (1 \minus{} t)x$ and $ h$ by $ h(t) \equal{} t y$. These are both paths, do you agree?

Then define $ f$ by $ f(t) \equal{} \begin{cases}g(2t) &\text{ if } t \in [0,\frac12)\\h(2t\minus{}1) &\text{ if } t\in[\frac12,1]\end{cases}$

Using the pasting lemma I mentioned earlier we can show that this is a path as well, from $ x$ to $ y$, as desired.

However, I will refrain from helping further if you continue to not show own ideas about the problems.

(In my other post I made the mistake of dividing by their norms; I defined $ g$ and $ h$ in the way I did to help you see the pasting better, as the construction of straight line segments  between points to and from the origin seemed obvious to me. Mlok's post basically gives you all you need to figure it out yourself.)


The recent cyclone in Burma and earthquake in China are both horribly tragic events.  The magnitude of human suffering makes both of these stories emotionally taxing to follow.

In previous threads there has been some discussion of (American) media coverage of these events.  JBL and others are correct that these are huge stories that have been extensively covered in respectable print media (e.g. the earthquake is still front page news on NYT).  But others have correctly pointed out that these stories have barely made a dent in the public consciousness, specifically noting the lack of coverage on 24-hour cable news (which is not nearly as pathetic as local TV news).  Someone suggested that this was the result of media bias (against Asians???) but that suggestion is way off base.

The real problem is that (most of) the media is primarily concerned with profit and it believes (rightly or wrongly) that world news does not sell.  It's that simple.

A smaller issue is that in print/internet media, important stories go on the front page, and everyone can see the headline.  On 24-hour cable news, you only see what's on when you tune in, so even if they "lead" with a story for 5 minutes, you might tune in during the next 55 minutes and see talking heads yapping about the '08 election.  (There's a lot of time to fill in 24 hours, and having "experts" yap about the election is a cheap and easy way to do it.)

Americans need to take a greater interest in world news, but I'm not sure how this can be achieved.  Most Americans barely know anything about domestic news.  Some argue that if you just offer people more world news, then they'll consume more of it, but I'm skeptical.  I think our best hope might be more emphasis on current events in high school.  As Xevarion pointed out, even bright and well-educated Princeton students will allow themselves to fall into ignorance without some intervention.

Finally, while the earthquake in China mostly makes me sad, the cyclone in Burma makes me mad.  The military junta's response to the cyclone is almost certain to be directly responsible for the death and suffering of many thousands of people, mostly because of its stubbornness in denying entry to foreign aid workers.  In contrast, China's response has been mostly good, but their early resistance to foreign aid was also counterproductive.  Moreover, China should have been using its influence in Burma to help open up the country to aid workers; it's hard to see how refusing aid workers itself helps to accomplish this.

(Aside to Quickster:  Why does having a natural disaster mean that you shouldn't criticize the government?  How is that "a sign of respect to those who have lost loved ones?")

I hope it was appropriate to create a new thread.  I didn't want to hijack either of the two previous ones, since this is sort of a tangent.

I think the media covered the disasters well enough for public awareness. How would you cover a disaster in a foreign country on a 24-hour TV show for a week? What would you say? Would you go over the rescue operation one body at a time? Or would you interview the people in charge at their reactions for 168 hours? Simply covering a story for 5 minutes is a lot of money. And face it, it might be a global disaster that affects everyone, but there isn't a lot of things to cover, which leaves the print media whatever they can get.

And remember, the cyclone that hit Myanmar was barely covered, since their government did not allow US reporters in for a while.

I agree with yenlee that the main reason is that it doesn't sell. When the VT shootings came, there was 24 hour coverage nonstop, almost. Next shooting? Almost nothing...

I was just observing that a thread originally designed to offer sympathy to the Chinese affected quickly turned into a discussion about the pros and cons of the Chinese Communist Government.  At this time, it's more important to focus on getting aid and support to the victims than the government.  Only sometimes the 2nd one stops the first from happening (in Myanmar).  Then, a problem arises


superior algebra
superior algebra solved

Would anybody be able to tell me what $\subseteqq$ means? I assume $\subsetneqq$ is simply the negation of this, but if not please let me know. Thank you.

$A\subseteqq B$ (or $A\subseteq B$) means that $A$ is a subset of $B$ -- each element of $A$ is also an element of $B$. Examples: $[3,4]\subseteqq [2,10]$ and $[3,4]\subseteqq [3,4]$.

$A\subsetneqq B$ means that $A$ is a proper subset of $B$, which is a subset that does not coincide with $B$ itself. It is true that $[3,4]\subsetneqq [2,10]$ but it is not true that $[3,4]\subsetneqq [3,4]$.

Some people (including me) use $\subset$ to mean $\subseteqq$; others use $\subset$ to mean $\subsetneqq$. It is analysis vs. algebra kind of thing.

Ok, thanks for your help.

I've always used $\subseteq$ for subset and $\subset$ for proper subset.


Mr. Bob has 46 animals, and 110 foods.  Animals with two legs need 2 foods, and animals with four legs need 3 foods.  He has a total number of 112 legs and 46 animals.  Does he have enough foods to feed them?

[hide]Let A be animal with 2 legs and B with 4 legs, so
A + B = 46 and
2A + 4B = 112
=>A = 36, B = 10.
2 x 36 + 3 x 10 = 72 +30 = 102 < 112.
He has enough food to feed them.[/hide]

[size=75][color=darkred]Please remember to use hide tags when posting solutions. Thanks.
-nebula42[/color][/size]

[hide]If all the animals were 2-legged, then there would be 92 legs, but since there are 112 legs, there must be ten 4-legged animals and thirty-six 2-legged animals.

36 X 2=72

10*3=30

110-(72+30)=8

YES
[/hide]


algebra
polynomial
abstract algebra
system of equations
number theory theorems
number theory

This is a famous theorem due to Warning...

Pierre.

More exactly: I know it as (special case of) the theorem of Chevalley-Warning.
There are a lot of different proves out there.

i know a theorem from Chevalley and it is that :
let $p$ be a prime and $f(x_1,x_2,...,x_n)$ be a polynomial with integer coefficents and degree of $m$ such that $m < n$ then if the equation
$f(x_1,...,x_n) \equiv ^p 0$ has a answer then its has another answer.
is this one usefull for proving the Chevalley-Warning?

If you know the proof of this variant, then you will certainly adapt it to obtain desired result. ;)

well , i find this harazi , any idea?
well, for proving the Chevalley theorem we use the fact that if $f(x_1,...,x_n) \equiv^p 0$ by degree of $d$ just has an answer like $(a_1,...,a_n)$ , a contradiction will appear by forming the polynomials
$1.H(x_1,...,x_n)=1-(f(x_1,...,x_n))^{p-1}$
$2.R(x_1,...,x_n)= \prod (1-(x_i-a_i)^{p-1})$ and using the fact that
these $2$ are equivalent for all set of $n$ integers by module $P$
but the degree of $H$ is $d(p-1)$ and degree of $R$ is $n(p-1)$ but we know that
$d < n$

well as harazi said , for adapting it for warning_Chevalley theorem we can form these polynomials (the same idea like above):
let the system of equations has $r$ answers showing them with $(a_{i,1},...,a_{i,n})$
$ 1 \leq i \leq n$, now changed the polynomial, $H,$ to
$G(x_1,...,x_n)= \prod_{i=1}^m ((1-(f_i(x_1,...,x_n))^{p-1})$ where m was the number of polynomails .
and $R$ to polynomail $S(x_1,...,x_n)=\sum_{i=1}^r \prod_{j=1}^n (1-(x_j-a_{i,j})^{p-1})$
these 2 are equivalent for the all sets of n inetegers by module $P$ and same  contardiction will happend , but what will we get(a new answer for system?)

The prove I know is very similar:

Let $p$ be a fixed prime from now on and lets work in the field $\mathbb{F}_p$ (because this prevents a lot of writing).

Lemma:
Let $k \in \mathbb{Z}, 0\leq k < p-1$, then $\sum_{x \in  \mathbb{F}_p} x^k  = 0 $ (here $x^k$ is seen as polynomial, so $x^0$ means the constant polynomial $1$)
Prove: already posted before, but still not complicted.

Chevalley-Warning:

Let be $f_1, f_2,...,f_m$ be $m$ polynomials in the $n$ variables $x=(x_1,x_2,...,x_n)$. If $\sum_{i=1}^m \deg (f_i)   < n $, then the number of solutions of
\[ f_1(x)=0 \\
f_2(x)=0 \\
...\\
f_m(x)=0\]
is divisible by $p$.

Prove:
Let be $s$ the number of solutions. We want to prove $s=0$ in $\mathbb{F}_p$ (be aware of that this just means that the number of solutions is divisible by $p$).

As lomos_lupin did, lets consider the polynomial $R(x) : = \prod_{i=1}^m ((1-(f_i(x_1,...,x_n))^{p-1})$.
Then $s= \sum_{x \in \mathbb{F}_p^n} R(x)$, because $R(x)=1$ iff $x$ is a solution.

Now lets look at the monoms $g(x)=\prod_{i=1}^n x_i^{v_i}$ that occure in $R$. For everyone of them there is an $i$, WLOG $i=1$,  with $v_i < p-1$, because otherwise the condition with the degrees would not be fulfilled. But then $\sum_{x \in \mathbb{F}_p^n} g(x) = \sum_{x_1 \in \mathbb{F}_p} \sum_{x_2 \in \mathbb{F}_p} ... \sum_{x_n \in \mathbb{F}_p} g(x) = \sum_{x_1 \in \mathbb{F}_p} x_1^{v_1} \cdot \sum_{x_2 \in \mathbb{F}_p} ... \sum_{x_n \in \mathbb{F}_p} x_2^{v_1} x_3^{v_1} ... x_n^{v_1}$ but the last one is $0$ because of $\sum_{x_1 \in \mathbb{F}_p} x_1^{v_1}$ after the lemma. So $s=0$ because we are summing up some monoms, which all sum up to $0$. But this is what we wanted to prove.

If we look back at the prove we see that there was never used some special property of $\mathbb{F}_p$, so the same properties hold for all fields with $p^k$ elements.

$ \textbf{Chevalley \minus{} Warning theorem}$. Consider polynomials $ f_1,...,f_m$ in variables $ x_1,...,x_n$ over a finite field $ \mathbb{F}_q$, where $ q \equal{} p^k$. Assume that $ n > \sum_{i \equal{} 1}^m \deg f_i$. Then the number of solutions $ (x_1,...,x_n)$ of the system of equations $ f_1(x_1,...,x_n) \equal{} 0$,...,$ f_m(x_1,...,x_n) \equal{} 0$ is divisible by $ p$.

$ \textbf{Proof}$. First, we find a nice way to count the number of solutions ($ \mod p$):

Let $ g(x_1,...,x_n) \equal{} \prod_{i \equal{} 1}^m (1 \minus{} f_i(x_1,...,x_n)^{q \minus{} 1})$.

Claim:
* If $ (x_1,...,x_n)$ is a solution, then $ g(x_1,...,x_n) \equal{} 1$.
* If $ (x_1,...,x_n)$ is not a solution, then $ g(x_1,...,x_n) \equal{} 0$.

The first claim is obvious. Now, let's prove the second. Since $ (x_1,...,x_n)$ is not a solution, we have $ f_i(x_1,...,x_n)\neq 0$ for some $ i$. Therefore $ f_i(x_1,...,x_n)^{q \minus{} 1} \equal{} 1$ and we see that one of the factors in the definition of $ g(x_1,...,x_n)$ is 0, thus we have proven the second claim.

Thus, counting solutions $ \mod p$ is the same as evaluating

$ \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} g(x_1,...,x_n)$.

Now, we can expand $ g$ into a sum of monomials:

$ \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} g(x_1,...,x_n) \equal{} \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} \sum_{(d_1,...,d_n)\in \mathbb{N}_0^n} c_{d_1,...,d_n} x_1^{d_1}\cdots x_n^{d_n} \equal{} \sum_{(d_1,...,d_n)\in \mathbb{N}_0^n} c_{d_1,...,d_n} \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} x_1^{d_1}\cdots x_n^{d_n}$.

Claim:
* If some $ d_i < q \minus{} 1$, then $ \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} x_1^{d_1}\cdots x_n^{d_n} \equal{} 0$.

Indeed, $ \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} x_1^{d_1}\cdots x_n^{d_n} \equal{} \prod_{i \equal{} 1}^n \sum_{x_i\in \mathbb{F}_q} x_i^{d_i}$. Thus it is enough to prove that $ \sum_{x\in \mathbb{F}_q} x^d \equal{} 0$ for $ d < q \minus{} 1$. But this is well-known; it can be deduced e.g. from the identity $ x^q \minus{} x \equal{} 0$ for $ x\in \mathbb{F}_q$ from which it follows that all elementary symmetric polynomials in the elements of $ \mathbb{F}_q$ of degree at most $ q \minus{} 2$ are 0 and therefore all symmetric polynomials of degree at most $ q \minus{} 2$.

Thus, in the sum $ \sum_{(d_1,...,d_n)\in \mathbb{N}_0^n} c_{d_1,...,d_n} \sum_{(x_1,...,x_n)\in \mathbb{F}_q^n} x_1^{d_1}\cdots x_n^{d_n}$, all terms vanish except those for which $ d_1,...,d_n\geq q \minus{} 1$. But those correspond to terms of degree at least $ n(q \minus{} 1)$. Now, the degree of $ g$ is at most $ \sum_{i \equal{} 1}^m (q \minus{} 1)\deg f_i < n(q \minus{} 1)$ by assumption. Therefore $ c_{d_1,...,d_n} \equal{} 0$ for those terms and we have proven the theorem. $ \hfill \Box$


linear algebra
matrix
algebra
polynomial
function

My dear friend, let's not try to make a mathematics forum the scene of a soap opera. One can express his marvelous and surely extremely passionate feelings in different settings, but I find it pretty stupid to do it here. So, think twice before posting such comments after a mathematical problem: many users (including me) will consider your post a spam and it will surely not have the desired attention.

To prove it: it is enough to use the fact that $ det$ is polynomial function from rows of matrix and besides we should conside that $ B, C$ 3x3 matrix.

$ P(x) \equal{} det(C \minus{} xB) \plus{} det(C \plus{} xB)$ is a polynomial of degree less or equal than  3

$ P$ is even function

$ P(x) \equal{} 2det(C) \plus{} ax^2$

with $ P(1)$ and $ P(2)$

you have easily the equality

You can use Lagarange Formula for $ f(x)\equal{}det(B\plus{}xC) \equal{}x^{3}det(C)\plus{}...$
To change ! Oke !! :roll:


inequalities
number theory
prime factorization
prime numbers

Show that for positive integers m,n :

(2m)!(2n)!/n!m!(m+n)! is always an integer.

I'll take a crack at it...

[[hide]Let M(x) denote the greatest integer less than or equal to x

Lemma:  M(2m)+M(2n)> or equal to M(n)+M(m)+M(m+n) for all positive reals m and n

Now we subtract 2M(n)+2M(m) from both sides to narrow our casework to positive reals between 0 and 1 as m and n (which cancels out the M(m)+M(n) terms immediately)

Without loss of generality let m> or equal to n

If the left side equals 0, since 2m>m+n, the left side equals zero as well.

If the left side is greater than 0, then the right side can only be at maximum one so the inequality still holds.

(2n)!(2m)!/((m)!n!(m+n)!) is not an integer if in the prime factorization of the bottom of the fraction features a prime with an exponent that is creater than the corresponding exponet of the prime in the top of the fraction.

The exponent of any prime p in the prime factorization of a! is:

M(a/p)+M(a/p^2)+M(a/p^3)+M(a/p^4)...

Since the Lemma proven above shows that M(2n/p^r)+M(2m/p^r)> or equal to M(n/p^r)+M(m/p^r)+M((m+n)/p^r) all the exponents of the prime numbers in the prime factorization of the numberater is greater or equal to the corresponding exponents of all the prime numbers in the prime factorizaion of the denominator.

Therefore (2n)!(2m)!/((m)!n!(m+n)!) is always an integer for positive integers m and n[/hide]


What is the value of the following expression: $ \frac{1}{3}\minus{}\frac{1}{9}\plus{}\frac{1}{27}\minus{}\frac{1}{81}\plus{}\frac{1}{243}$? Express your
answer as a common fraction.


Solve for t.

1000=500e^4t

I know how to do similar problems like this one:
(solve for x)
9^x+2 = 27^2x    (common base... blah blah)
3^2(x+2) = 3^3(2x)
2x+4=6x
4=4x
1=x

1000 = 500e^(4t)
First of all you can divide both sides by 500,
2 = e^4t now we use natural log which is denoted by ln,
ln 2 = 4t so
t = ln2/4


How many different prime factors does $20!$ have?

[quote="4everwise"]Full Solution:
[hide]$20!= 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20$
Now, we factor the products and the determine the number of different factors.
The factors are $1,2,3,5,7,11,13,17, 19$

The number of factors that we have are 9. Don't forget about 1!  ;)
[/hide][/quote]
hmm you may want to look at this topic:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40642]is 1 a prime number?[/url]

$1$ is not a prime number.....

EDIT: Palytoxin beat me to it :)

Oups, my bad. Completely forgot about that. :oops:

So, if I'm not mistaken, could the answer be 9-1 (taking away the 1) or 8?

[hide]the primes are 2,3,5,7,11,13,17,19 and there are 8[/hide]


How many moles of iodide are in 2.54g of CaI_2?
Iodine is 126.9 g/mole
Calcium is 40.078 g/mole

Thanks!

Formula for calculating moles: $ \text{\# of moles}\equal{}\frac{\text{mass of compound}}{\text{molar mass}}$

There is an ambiguity in your question: are you looking for the number of moles of [b]iodine[/b] or the number of moles of [b]calcium iodide[/b]? I ask this because you typed [b]iodide[/b], which causes a confusion.

I was looking for the mole of iodine, but iodine is called iodide in this particular compound.  Sorry for the confusion, but I still don't know how to solve it.

Iodi[b]d[/b]e is the chemical name given to compounds containing iodine; according to IUPAC rules, some simple compounds with nonmetals end with -ide, there are exceptions like larger ions with -ate and -ite and such.

First, you proceed to find the number of moles of $ \ce{CaI2}$ by using that formula (you are given the mass of $ \ce{CaI2}$ and you can find the molar mass of $ \ce{CaI2}$. Then, 1 mole of $ \ce{CaI2}$ contains two moles of $ I$ so ...  :idea:

[quote="10000th User"]Iodide is the chemical name given to compounds containing iodine;[/quote]

No. Iodide is the name given to compounds that contain the iodide anion $ I^ \minus{}$.

For each mole of calcium iodide there are two moles of iodide ion, this because each formula unit (which is the lowest possible neutral combination of ions for an aionic compound) contains two iodide ions for each calcium ion. Start by computing the molar mass of calcium iodide:

M(CaI2) = 40.078 + 2x126.9 = 293.9 g/mol.

No compute the chemical quantity of calcium iodide in the 2.54 g:

$ n(CaI_2) \equal{} \frac {2.54g}{293.9\, g/mol} \equal{} 8.64\, mmol$.

Therefore, the chemical quantity of iodide present in the sample is $ n(I^ \minus{} ) \equal{} 2 \times 8.64\, mmol \equal{} 17.3\, mmol$.

[quote="Cdymdcool"]But why does NAI_2 has two moles of Iodine?[/quote]

By the same reason calcium iodide does.

[quote="Cdymdcool"]How many grams of Potassium are in 23.543 moles of K_3P?[/quote]

Each formula unit of K3P contains three potassium atoms, so in 23.543 moles of K3P there are 3x23.543 = 70.629 moles of potassium. The molar mass of K is 39,0983 g/mol, and so the corresponding mass is

$ m(K) \equal{} 70.629\,mol \times 39,0983\, g/mol \equal{} 2.7615\, kg$.


algebra
polynomial
Vieta
function
number theory unsolved
number theory

Let $ a,b$ are positive integer.
$ a)$ if $ ab \minus{} 1|a^{2} \plus{} b^{2}$  $ \implies$  $ \frac {a^{2} \plus{} b^{2}}{ab \minus{} 1} \equal{} 5$  Prove that.
$ b)a^{2}\plus{}b^{2}\minus{}5ab\plus{}5\equal{}0$  find all $ (a,b)$

the first one is well-known,(we use vieta jumping as mentioned)
for the second one...
[hide="hint"] use the same idea used in solving this problem http://www.mathlinks.ro/viewtopic.php?t=150457, maybe this is not clear but think...[/hide]

A solution to 1:
I'm new to vieta jumping method,so please let me know if i made a mistake,because probably my solution is wrong.....

First of all let $ a \equal{} c \plus{} 1$.Then we have that $ (c \plus{} 1)b \minus{} 1|(c \plus{} 1)^2 \plus{} b^2$.
Let us consider the $ k \equal{} \frac {(c \plus{} 1)^2 \plus{} b^2}{(c \plus{} 1)b \minus{} 1}$.Fix $ k$ and consider all pairs of $ (b,c)$ satisfies the equation $ \frac {(c \plus{} 1)^2 \plus{} b^2}{(c \plus{} 1)b \minus{} 1} \equal{} k$.

Let $ (C,B)$ be a pair which minimizes the sum $ c \plus{} b$.We claim that $ B \equal{} C$.To prove this ,assume for the sake of contradiction that $ C > B$.

Let us now consider the equation
$ \frac {(x \plus{} 1)^2 \plus{} B^2}{(x \plus{} 1)B \minus{} 1} \equal{} k\Longleftrightarrow x^2 \plus{} (2 \minus{} kB)x \plus{} B^2 \plus{} 1 \minus{} kB \plus{} k$
We already know that $ x_1 \equal{} C$.From vieta formulas we have that $ x_2 \equal{} \frac {B^2 \plus{} 1 \minus{} kB \plus{} k}{C}$ which is smaller than $ C$.This contradicts our assumption that $ B \plus{} C$ minimizes $ b \plus{} c$.So $ B \equal{} C$,which leads to $ 2 \plus{} \frac {3}{B^2 \plus{} B \minus{} 1} \equal{} k$.So $ B^2 \plus{} B \minus{} 1|3 \Rightarrow B\equal{} 1$.So $ k \equal{} 5$.

[quote="cnyd"]Let $ (a,b) \in \mathbb{N}_0^2$ s.t. $ ab \minus{} 1 \mid a^{2} \plus{} b^{2}$. Show that $ \frac {a^{2} \plus{} b^{2}}{ab \minus{} 1} \equal{} 5$.[/quote]
Define the function $ f(a,b): \mathbb{N}_0^2 \to \mathbb{Q}: (a,b) \to \frac {a^2 \plus{} b^2}{ab \minus{} 1}$.Let $ k \in \mathbb{N}_0$ fixed and consider all pairs of $ (a,b) \in \mathbb{N}_0^2$ that satisfy $ f(a,b)\equal{}k$, and define $ S_k$ such set. Obviusly it must be $ ab\minus{}1 >0$, so the equation $ f(a,a)\equal{}2\plus{}\frac{2}{a^2\minus{}1} \in \mathbb{Z}$ has no solution in positive integers. If $ S_k \neq \{\}$, define $ (a_0,b_0)$ the couple of positive integers that minimize $ a\plus{}b$. Since $ f(a,b)\equal{}f(b,a)$ we can assume wlog that $ b>a>0$. We claim the if $ k \in \mathbb{N}_0$ is fixed and $ a>1$ is such that $ f(a,b)\equal{}0$ then $ f(a,ak\minus{}b)\equal{}k$ too, and $ 0<ka\minus{}b<b$ iff $ ba^{\minus{}1}<k<2ba^{\minus{}1}$. In fact the equation $ f(a,b)\equal{}k$ is a second degree polinomial in the variable $ a$, and the other solution $ a'$ must be positive (because the product is positive) and integer (because the sum is integer). Now, $ f(a,b)\equal{}k \ge 2ba^{\minus{}1}$ iff $ ab^2\le a^3\plus{}2b$ and since $ a^2 \le (b\minus{}1)^2$ (and $ b>1$) we have $ b^2 \ge a^2\plus{}2b\minus{}1$ that is $ ab^2 \ge a^3\plus{}2ab\minus{}a \ge a^3\plus{}2b$, that is a contradiction. On the other hand, if $ k \le ba^{\minus{}1}$ then $ a^2\plus{}b^2\plus{}k\equal{}a(a\plus{}a')\equal{}kab \le b^2$ that is clearly false. So we have concluded that if $ S_k \neq \{\}$ and $ (a_0,b_0)$ is its minimal-sum solution, then $ a_0\equal{}1$. Now, trivially $ f(1,b_0) \in \mathbb{Z}$ iff $ b_0\equal{}2$ (since it is minimal) and $ k\equal{}5$. Note that with this proof we have constructed also an algorith for finding all solution to the equation $ f(a,b)\equal{}5$ in positive integers.  :D

Ps. Thanks to DimitrisX for nothing me this nice question, that seems different from IMO88/6. What is the source?

[quote="bboypa"]

Ps. Thanks to DimitrisX for nothing me this nice question, that seems different from IMO88/6. What is the source?[/quote]

I have seen this problem on the [b]Englel's[/b] book:problem solving strategies dear friend :wink:


algebra
polynomial
Euler
LaTeX
calculus
calculus computations

Show using Eulers method that

$ \frac{\pi^2}{8}\equal{}\sum_{n\equal{}0}^{\propto}\frac{1}{(2n\plus{}1)^2}$

by,
a, finding the zeros of 1-sinx
b, wrting the taylor series of 1-sinx about x=0
c,let $ p(x) \equal{} a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{} ... \plus{} a_n x^n$
and $ \alpha_i$ be the zeros of p repeated acording to multiplicity. Derive the identity:

$ \sum_{i \equal{} 0}^{\propto} \frac {1}{\alpha^2_i} \equal{} \frac {a^2_1}{a^2_0} \minus{} 2\frac {a_2}{a_0}$

which leads to the final stage of the proof

Well for the taylor series, there's the maclaurin representation as follows:

$ f(x) \equal{} \frac{f(0)}{0!} \plus{} \frac{f'(0)x}{1!} \plus{} \frac{f''(0)x^{2}}{2!} \plus{} \cdots \plus{} \frac{f^{(n)}(0)x^{n} }{n!}$

Replace $ f(x)$ by $ 1\minus{}\sin(x)$ to get the desired result. The zeroes of $ 1\minus{}\sin(x)$ are trivial, so I'll leave the rest to you.

I put some hints for part (c) in the duplicate thread the O.P. created [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188028]here[/url].  Please don't double-post.

Also, the latex for $ \infty$ is \infty.

Aside: I think you can also use Fourier series  :maybe:

Don't have time to verify though


complex numbers
geometry unsolved
geometry

A solution strange with complex numbers :S

[quote="puuhikki"]While reading an another problem this came to my mind. Let $A=A_{1}A_{2}\cdots A_{n}$ be a regular polygon. Let $P$ be a point inside $A$ and $|PA_{i}|=d_{i}$ for every $1\leq i \leq n$. Find the side length of $A$.

Even harder. Generalize to higher dimensions by replacing a polygon with a standard $n$-simplex.[/quote]

Well I was looking for a solution by complex numbers but I've got something wich is very strange  :!:   :!:   :!:

well as we know, we can put $P_{i}=\rho \omega^{k-1}=\rho (cos((k-1)\frac{2\pi}{n})+isen((k-1)\frac{2\pi}{n})$ (it meas that the circunradii of the polygon is $\rho$) in the complex plane and the point $P=(x,y)$. So the distances $d_{i}=|P-P_{i}|$ so $d_{i}^{2}=|\rho (cos((k-1)\frac{2\pi}{n})-x,sen((k-1)\frac{2\pi}{n})-y|^{2}$ so $d_{i}^{2}=\rho^{2}+x^{2}+y^{2}-2x\rho cos((k-1)\frac{2\pi}{n})-2y\rho sen((k-1)\frac{2\pi}{n})$ and adding factors we get $\sum d_{k}^{2}=n\rho^{2}+n(x^{2}+y^{2})-2x\rho \sum cos((k-1)\frac{2\pi}{n}-2y\rho \sum sen((k-1)\frac{2\pi}{n}$. I have proved that $\sum sen((k-1)\frac{2\pi}{n}=0$ and $\sum cos((k-1)\frac{2\pi}{n}=0$ so $\sum d_{k}^{2}=n\rho^{2}+n(x^{2}+y^{2}) =n\rho^{2}+n(OP)^{2}$

If $n=2t$ the OP can be found byt median theorem. If $n=2t+1$, $OP$ can be found by stewart theorem

If you try by $n=2t$ you'll find a strange result  :!:   :!:   :!:


integration trigonometry limit algebra polynomial calculus derivative

Here's an outline of a proof, with three questions :

We assume by sake of contradiction that $\pi = \frac pq,$ with $p$ and $q$ positive coprime integers. We then consider $I_n = \int_0^{p/q} \frac {x^n(p-qx)^n}{n!} \sin x dx, n \in \mathbb{N}$ and show that $I_n$ satisfies
a) $I_n$ is an integer;
b) $I_n > 0;$
c) $\lim_{n \to +\infty} I_n = 0.$

[b]Historical remark(s) :[/b] Lambert has showed in 1761 that $\pi$ is irrational, Legendre has proved in 1794 that $\pi^2$ is irrational, Lindemann has proved in 1882 that $\pi$ is transcendental.

Okay, let's go.

(With the notations of my previous post,) at first, for $0 \leq x \leq \frac{p}{q},$ we have $0 \leq x(p-qx) = \frac p{2q}(p-\frac p{2q}q) = \frac{p^{2}}{4q},$ and so,
\[0 \leq I_{n}\leq \frac 1{n!}\int_{0}^{p/q}\left(\frac{p^{2}}{4q}\right)^{n}dx = \frac{\pi}{n!}\left( \frac{p^{2}}{4q}\right)^{n}.\]
Therefore, the sequence $(I_{n})$ converges and $\lim_{n \to+\infty}I_{n}= 0.$
Then, since for $x \in [0, \pi],$ we have $x^{n}(p-qx)^{n}\sin x \geq 0,$ for a given positive integer $n,$ we have
$I_{n}= \frac 1{n!}\int_{0}^{\pi}x^{n}(p-qx)^{x}\sin x dx \geq \frac 1{n!}\int_{\pi/4}^{3 \pi/4}x^{n}(p-qx)^{n}\sin x dx \geq \frac 1{n!}(3\pi/4-\pi/4)\left(\frac p{4q}(p-\frac p{4q}q) \right)^{n}\frac 1{\sqrt{2}}= \frac{\pi}{2n!}\left(\frac{3p^{2}}{16q}\right)^{n}> 0.$
Thus, $\forall n \in \mathbb{N}, I_{n}> 0.$
Let us finally check that, for any positive integer $n, I_{n}$ is an integer.
Let $P_{n}= x^{n}(p-qx)^{n}.$ $P_{n}$ is a polynomial with degree $2n$ and $0$ and $\frac{p}{q}$ are roots of order $n$ of $P_{n}$ and so, for $0 \leq k \leq n,$ roots of order $n-k$ of $P_{n}^{(k)}.$ Especially, $P_{n}^{(k)}(0)$ and $P_{n}^{(k)}(\frac{p}{q})$ are, for $0 \leq k \leq n,$ integers. Similarly, since $\text{deg}P_{n}= 2n,$ for $k \geq 2n+1, P_{n}^{(k)}\geq 0$ and especially, $P_{n}^{(k)}(0)$ and $P_{n}^{(k)}(\frac{p}{q})$ are, for $k \geq 2n+1,$ integers.
Let $k$ be an integer such that $n \leq k \leq 2n.$
$\frac 1{n!}x^{n}(p-qx)^{n}= \frac 1{n!}^{n}\sum_{i=0}^{n}\binom ni p^{n-i}(-1)^{i}q^{i}x^{i}= \sum_{i=0}^{n}\frac{\binom ni}{n!}p^{n-i}(-1)^{i}q^{i}x^{n+i}= \sum_{k=n}^{2n}\frac{\binom n{k-n}}{n!}p^{2n-k}(-1)^{k-n}q^{k-n}x^{k}.$
We then know that
\[P_{n}^{(k)}(0) = k! \times (\text{coefficient of }x^{k}) = (-1)^{k-n}\frac{k!}{n!}\binom n{k-n}p^{2n-k}q^{k-n},\]
which shows that $P_{n}^{(k)}(0)$ is an integer. Then, since $P_{n}(\frac{p}{q}-x) = P_{n}(x),$ we can deduce that $P_{n}^{(k)}(\frac{p}{q})$ is an integer.
We have thus showed that for any natural integer $k, P_{n}^{(k)}$ and $P_{n}^{(k)}(\frac{p}{q})$ are integers.
Let us then show that $I_{n}$ is an integer.
A first integration by parts yields : $I_{n}= [-P_{n}(x) \cos x]_{0}^{p/q}+\int_{0}^{p/q}P'_{n}(x) \cos x dx.$
$\cos$ takes integer values in $0$ and $\frac{p}{q}= \pi,$ like $P_{n}.$ Hence,
\[I_{n}\in \mathbb{Z}\iff \int_{0}^{p/q}P'_{n}(x) \cos x dx \in \mathbb{Z}.\]
A second integration by parts yields : $\int_{0}^{p/q}P'_{n}(x) \cos x dx = [P'_{n}(x) \sin x]_{0}^{p/q}-\int_{0}^{p/q}P''_{n}(x) \sin x dx.$
$\sin$ takes integer values in $0$ and $\frac{p}{q}= \pi,$ like $P'_{n}$ and
\[I_{n}\in \mathbb{Z}\iff \int_{0}^{p/q}P''_{n}(x) \sin x dx \in \mathbb{Z}.\]
Repeating the integration by parts and since $\sin$ and $\cos$ take integer values in $0$ and $\pi$ like the successive derivatives of $P_{n},$ we deduce that :
\[I_{n}\in \mathbb{Z}\iff \int_{0}^{p/q}P_{n}^{(2n)}(x) \sin x dx \in \mathbb{Z}.\]
But,
$\int_{0}^{p/q}P_{n}^{(2n)}(x) \sin x dx = \int_{0}^{p/q}\frac 1{n!}(-q)^{n}(2n)! \sin x dx = 2(-q)^{n}(2n)(2n-1)\cdots (n+1) \in \mathbb{Z}.$
So for any natural integer, $I_{n}$ is an integer, positive from what we've done above. We deduce that for any natural $n, I_{n}\geq 1.$ This last thing contradicts the fact that the sequence $(I_{n})$ converges towards $0.$
Therefore, $\pi$ is irrational.

[quote="Peter"]Show that $ \pi$ is irrational.[/quote]

The reference for this problem "CRUX, Problem A240, Mohammed Aassila" is not correct.


algebra
polynomial
algorithm
number theory
greatest common divisor
quadratics
ratio

I got the general form and just played with the coefficients for a few minutes and got:

[hide] (3x-2y+2)(2x+3y-1) [/hide]

I am not sure, but he asked for the factorization of an expression, not of an expression=0, so what we both said is not equivalent.

[quote="fuzzylogic"]Good job! But both of your answers have slight errors. Verify by multiplying out!

Now, can you see how to factor it using cross multiplying?[/quote]

what do you mean by [b]cross multiplying[/b]?

[quote="zabelman"]what do you mean by [b]cross multiplying[/b]?[/quote]

This is what I mean:

Assume that [tex]ax^2+bx+c = (a_1x+c_1)(a_2x+c_2)[/tex].

Find [tex]a_1,a_2,c_1,c_2[/tex] such that [tex]a=a_1a_2[/tex], [tex]c=c_1c_2[/tex] and [tex]b=a_1c_2+a_2c_1[/tex]

In my school textbook, the four numbers are putting in a [tex]2\times2[/tex] "matrix". Then cross multiply and add together.

sorry, i wrote the problem down wrong, heres the new factorization:

[hide] (3x+2y-2)(2x-3y+1) [/hide]

[quote="fuzzylogic"]Find the factorization of [tex]6x^2-5xy-6y^2-x+8y-2[/tex].

This problem is created on the fly after seeing [b]zabelman[/b]'s cool solution on another thread.[/quote]

Hmm...I did it and it was fairly straightforward.  Not to knock the problem, but this is a fairly common technique (as zabelman said when he posted his solution) known by most contest writers.

But treating the CONSTANT as a variable was ingenious - I've never seen that done, nor would I have thought of it off-hand.

[quote="fuzzylogic"][quote="blahblahblah"]Hmm...I did it and it was fairly straightforward.  Not to knock the problem, ...[/quote]

This is why it's in the [b]Intermediate Topics[/b].[/quote]

Point well taken :-)

In some sense [b]cross multiplying[/b] is kind of "try and error" method. In general, it works only for "small" polynomials.

Assume that a polynomial of degree n with integer coefficients has a rational root. The [b]Rational Root Theorem[/b] provides an algorithm for seaching the rational root by factoring the leading and constant coefficients. Unfortunately, integer factoring is a "hard" problem. On the other hand, polynomial factoring is "easy" in the sense that it is in [b]P[/b] (using the computer science language).

Another similar problem: people usually try to find the GCD by factoring the denominators when adding fractions instead of using Euclidean algorithm.

[quote="JS1527"]Can someone point out what I did wrong?  I know it does not work when I expand my factorization, but I am not sure why...[/quote]

After finding roots (3y-1)/2 or (-2y+2)/3,  the factorization should be:

6( x - (3y-1)/2 ) ( x - (-2y+2)/3 ),

which simplifies to:

(2x-3y+1)(3x+2y-2)

It's the leading coefficient.  In general, if we have a polynomial [tex]P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x+a_0, a_n \neq 0[/tex], with roots [tex]r_1, r_2, \ldots, r_n[/tex], then we can write it as [tex]P(x)=a_n(x - r_1)(x-r_2)\cdots(x-r_n)[/tex].  You just left off the leading coefficient there.

I know what a leading coefficent is...and I know that I left it out...but it being there does not come out of the method in which I solved the problem.  If you go back to my original work, I counted it as a=6 when I used the quadratic formula, so I don't see why I have to add it again.

Yes, you correctly solved for the roots.  But that isn't hte only effect the leading coefficient has.  THere are infinitely many polynomials with any given set of roots -- what differentiates them is the leading coefficient.  You simply picked one at random (well, not quite at random -- you picked the one with leading coefficient 1), but we knew that it needs to have leading coefficient 6.

Edited for interesting typo.  Thanks TripleM.

[quote="JBL"]You simply picked one at random (well, not quite at random -- you picked the one with leading coefficient 0), but we knew that it needs to have leading coefficient 6.[/quote]

Well, leading coefficient 1, not 0 :)

If we let the polynomial to be factored (call it P) be

P = (ax + by + c)(dx + ey + f)
= adx^2 + bey^2 + (ae + bd)xy + (bf + ce)y + (af + cd)x + cf,

then we come up with a system of equations:

ad = 6
be = -6
ae + bd = -5
bf + ce = 8
af + cd = -1
cf = -2.

The equation cf = -2 gives us four possible ordered pairs for (c, f):

(-2, 1)
(2, -1)
(-1, 2)
(1, -2).

Letting c = -2 and f = 1  we get

b = 2 and e = -3, or b = 6 and e = -1
d = -3/2 and a = -4, or d = 2 and a = 3.

The case where a = 3, b = 2, c = -2, d = 2, e = -3, and f = 1 gives a satisfactory result. The other case is where a = -3, b = -2, c = 2, d = -2, e = 3, and f = -1, which is obviously equivalent to the first case. So (3x + 2y - 2)(2x - 3y + 1) is the factorization of P.

[quote="JS1527"]I understand it.  Since using zableman's method you use the quadratic formula (which only finds roots), you have to add the correct leading coefficent, even though I already counted a=6.  Thanks!

Zenny's method is way more common, but I would say zableman's is cooler and a little easier.[/quote]

I suppose that you could say that the 'extended' form of the quadratic formula is the identity:

ax^2+bx+c = a(x - (b+sqrt(b^2-4ac))/2a)(x - (b-sqrt(b^2-4ac))/2a)

[quote="scm007"]So what is this common technique that zabelman used? I'd like to know it, or was it already said earlier? Personally, I usually use Zenny's.

Also, can all polynomials be factored the way that Zenny did so? And if not, how can you tell which can be factored that way. Meaning into (ax +by +c)(dx +ey +f) obviously that is a quadratic, but I mean how do you know what to write for something ilke 6x^4 + 9x^3y etc etc?[/quote]

Not all polynomials can be factored over the rationals, which is how we refer to such a factorization.  In fact, for anything higher than a first degree equation (obviously with rational coefficients), we cannot guarantee a rational factorization.

An obvious example is the golden ratio, which is the positive solution to the quadratic x^2 - x - 1 = 0.  This particular polynomial does not factor over Q.


Please look over this proof, tell me if it's clear enough.

Prove that :sqrt: xy <= (x+y)/2  for all nonnegative x, y

:sqrt: xy <= (x+y)/2
xy <= (x^2+y^2)/4 + xy/2   (since x and y are nonnegative, squaring both sides works)
2xy <= x^2+y^2
Let y=x-k for some 0<=k<=x (making y<=x does not lose anything since x and y can be switched in the original inequality)
2x^2 - 2xk <= x^2 + x^2 - 2xk + k^2
0<=k^2, which is true by the previous inequality.  Equality when k=0 i.e. x=y

When writing up a proof, though, it's probably better to write the proof forwards, not backwards. For instance, since you know how to get from your inequality to [tex](x-y)^2 \geq 0[/tex], start from there and work towards your inequality. There's a section of the article "How to Write a Solution" on thinking backwards and writing forwards: [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php[/url]. It's a good idea to do this because you never know when a step you took might not be reversible.

$(\sqrt{x}-\sqrt{y})^2 \geq 0$, expand it, $x+y-2\sqrt{xy} \geq 0$. the conclusion follows easily...


Farz Konid  $ f(x)=a_{1}\sin x+a_{2}\sin2x+\cdots+a_{n}\sin nx$ ke inja $ a_{1},a_{2},\cdots,a_{n}$ adadhaye Haghighiye Sabet hastand va $ n\in\mathbb{N}$. Midaanim ke baraye har adade haghighiye $ x$, $ \left|f(x)\right|\leq\left|\sin x\right|$. Sabet Konid:

$ \left|a_{1}+2a_{2}+3a_{3}+\cdots+n a_{n}\right|\leq1$

man ziad az kar kardane ba $ \lim$ khosham nemiado asanam vared nistam,vase haminam mokene ke rahe zir ghalat bashe,dar har soorat age ghalate maskharei dasht bebakhshid...  :oops:
khob darim:

$ |f(x)|\leq |\sin x|$

$ \Rightarrow |\frac{f(x)}{x}|\leq |\frac{\sin x}{x}|$

pas:

$ \lim_{x \to 0}|\frac{f(x)}{x}|\leq \lim_{x \to 0}|\frac{\sin x}{x}|=1$

$ \Rightarrow \lim_{x \to 0}|\frac{f(x)}{x}|\leq 1 (*)$

amma midoonim ke:

$ \lim_{x \to 0}\frac{f(x)}{x}=\lim_{x \to 0}\frac{a_{1}\sin x+a_{2}\sin 2x+...+a_{n}\sin nx}{x}$

$ =\lim_{x \to 0}\frac{a_{1}\sin x}{x}+...+\lim_{x \to 0}\frac{a_{n}\sin nx}{x}$

$ =a_{1}+2a_{2}+...+na_{n}$

pas tebghe $ (*)$ natije mishe ke:

$ |a_{1}+2a_{2}+...+na_{n}|\leq 1$

eyval kamelan doroste va rahehalle ghashangi ham bood :)


I will compile some results on resistors here, especially infinite grids.

$\line(1,0){200}$
Some webpages:
[url=http://www.geocities.com/frooha/grid/node2.html]Infinite 2D square grid of $1\Omega$ resistors[/url]
[url=http://www.geocities.com/frooha/grid/node1.html]Infinite ladder of $1\Omega$ resistors[/url]
$\line(1,0){200}$

Problems:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=360847785&t=150192]Grid of resistors[/url]

Please contribute but do not make this a messy marathon thread!!  :mad:

Those website links you gave are really cool man. Could you give some more? :D  :D


In the addition shown, different letters stand for different digits. If $ O$ stands for 7, what digit does $ F$ represent?
\begin{tabular}{r}
     TWO &
+   TWO &
\hline
     FOUR
\end{tabular}

two three digit numbers can add up to a maximum of 1998, thus F=1.


Can I start a Classroom Math Marathon???   :lol:

[quote="MCrawford"]For now, we are locking marathon threads and asking that students discontinue creating them. The reasons are that (1) many of the problems are growing too difficult for the level of the forum, (2) the depth of discussion of the solutions is low, and (3) it's hard to search/identify the problems in any way. Please keep posts to one problem per thread (except where another problem relates to the original problem discussed).
[/quote]

So [b][u][i]no[/i][/u][/b].

useless posts= :spam:
plz stop spamming!!


Two grasshoppers sit at opposite ends of the interval $ [0, 1]$. A finite number of points (greater than zero) in the interval are marked. In a move, a grasshopper selects a marked point and jump over it to the equidistant point the other side. This point must lie in the interval for the move to be allowed, but it does not have to be marked. Or, alternatively, in that move, the grasshopper may choose to stand still. Prove that in two moves, the grasshoppers can end up with no marked points between them.

Question: What about if in a move, each grasshopper must jump?

[hide="nevermind this is wrong"]WLOG, assume that the endpoints are not marked. If there are n marked spots, there are n+1 regions bounded by the spots and the end points. Since each spot cna be jumped over by at least one grasshopper and there are also the 2 options for each of them to stay put, there are a total of n+2 possible options. By the pgieonhole pricnicipe, two of the options must lead to the same region.
[/hide]

As for the for the second question, wouldn't a single marked point at .4 be a counter example? The grasshopper sitting at 1 is unable to jump.

[quote="0714446459923"]

As for the for the second question, wouldn't a single marked point at .4 be a counter example? The grasshopper sitting at 1 is unable to jump.[/quote]

I agree... so try the first question  :D