geometry
parallelogram
analytic geometry
vector
AMC

Parallelogram $ ABCD$ has area $ 1,\!000,\!000$. Vertex $ A$ is at $ (0,0)$ and all other vertices are in the first quadrant. Vertices $ B$ and $ D$ are lattice points on the lines $ y\equal{}x$ and $ y\equal{}kx$ for some integer $ k>1$, respectively. How many such parallelograms are there?

$ \textbf{(A)}\ 49\qquad
\textbf{(B)}\ 720\qquad
\textbf{(C)}\ 784\qquad
\textbf{(D)}\ 2009\qquad
\textbf{(E)}\ 2048$

[hide]
Let B = (a,a) and D=(b,kb). Then, C = (a+b, a+kb), which must be a lattice point, so we don't have to worry about C not being a lattice point. Use the parallelogram area formula to get that the area = ab(k-1). Since everything is in the first quadrant, a and b are positive, and so is k-1 because k > 1. These are also all integers. We have ab(k-1)=$ 2^65^6.$ The sum of the exponents of 2 in a,b, and k-1 = 6 therefore. Balls in urns gives us that there are 8 choose 2 = 28 choices for the exponents of a,b, and k-1. Also, there are 28 choices for the exponent of 5 by symmetry. Therefore, the answer is 28*28 = 784 (C).[/hide]

How come you don't have to consider that $ k\minus{}1 \ge 1$ and that some of the variables can't be $ 0$?

[quote="123456789"]How come you don't have to consider that $ k \minus{} 1 \ge 1$ and that some of the variables can't be $ 0$?[/quote]None of the variables will ever be zero in that count. No factorization of a nonzero quantity will contain a zero.

[quote="123456789"]But when you use balls and dividers, aren't you allowing the possibility that two dividers are adjacent to each other?[/quote]In which case the variable in question will be the empty product, or $ 1$.

[quote=skyhog][quote="cdymdcool"][quote="skyhog"][hide="Another way to find the area"]
Let point $ B$ be $ (b,b)$ and point $ D$ be $ (d,dk)$
Also, let vector $ \bf b \equal{} (b,b)$ and vector $ \bf d \equal{} (d,dk)$
Vector $ \bf d$ is an "edge" vector of the parallelogram, and to find the other we take:
$ \bf {b} \minus{} \bf {d} \equal{} (b \minus{} d,b \minus{} dk)$
The area of the parallelogram, $ 1000000$, will be equal to the magnitude of the cross product of these two vectors.  I don't know how to display matricies in TeX, but you can easily find the cross product as being equal to:
$ bd(k \minus{} 1)\bf k$ ($ \bf k$ referring to the basis vector, not the variable in this problem), the magnitude of which is just $ bd(k \minus{} 1)$
[/hide][/quote]

Did you use vectors to calculate the area of the parallelogram?  Is that portion covered in AoPS V2?[/quote]

Yes, if you have two vectors spanning a parallelogram, the length (also referred to as "magnitude" or "norm") of the cross product is the area of the parallelogram.  I am fairly certain this is included in the vectors chapter in AoPS volume 2, where I think you'd find a better explanation than mine.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Vector#Cross_.28Vector.29_Product]The AoPS Wiki has a page on this, too[/url][/quote]
The page doesn't exist.

Ouch. Apparently I thought $2kab-2ab=ab(2k-1)$ and this cost me to not break my streak of scoring in the $120-130$ range on an AMC 12 for the fourth time in a row.


trigonometry
induction
limit
algebra proposed
algebra

[quote="skytin"]prove that: $ \pi\equal{}\frac{180\sin(x)}{x*\cos\frac x2\cos \frac x4\cos\frac x8\cos\frac x{16}...}$[/quote]

That's wrong.

We obviously have $ \sin x\equal{}2^n\sin\frac x{2^n}\prod_{k\equal{}1}^n\cos \frac x{2^k}$

So $ \frac {\sin x}{x\prod_{k\equal{}1}^n\cos \frac x{2^k}}\equal{}\frac{\sin\frac x{2^n}}{\frac x{2^n}}$

Setting $ n\to\plus{}\infty$ in this equality, we get $ \frac {\sin x}{x\prod_{k\equal{}1}^{\plus{}\infty}\cos \frac x{2^k}}\equal{}1$

And so $ \frac {180\sin x}{x\prod_{k\equal{}1}^{\plus{}\infty}\cos \frac x{2^k}}\equal{}180\ne\pi$

[quote="skytin"]to pco
if i am not wrong[/quote]

Huhh ?

How could you conclude that limit is $ 0$ ? You have a product of :

1) a quantity whose limit is $ \plus{}\infty$ : 2^n
2) a quantity whose limit is $ 0$ : $ \sin\frac x{2^n}$
3) a product of infinitely many numbers whose limit (numbers, not product) is $ 1$

And the proof of my equality is immediate using induction and the fact that $ \sin 2x\equal{}2\sin x\cos x$

[quote="skytin"]i cant anderstend wy if n is going to infty than sin(x/2^n)/(x/2^n)=1[/quote]

Basic standard limit : $ \lim_{x\to 0}\frac{\sin x}x\equal{}1$

Simply because as $ x$ tends to zero $ \sin{x}$ also tends to zero. We can 'intuitively' think that they both reach the same value (infinitesimally small) simultaneously, thereby giving the limit $ \lim_{x\to 0}\frac{\sin x}{x}\equal{}1$
The limit perhaps appears within the first 3 pages of any standard book of elementary analysis.


What is the smallest positive integer greater than $ \frac{3}{2} \plus{} \frac{3}{4} \plus{} \frac{3}{8} \plus{} \frac{3}{16} \plus{} \frac{3}{32} \plus{} \cdots \plus{} \frac{3}{2^n}$, for a finite value of $ n$?


algebra
polynomial
binomial theorem
complex numbers

Sorry I need help with these, please post full solution.

1. Prove that there is a N such that $ \left(\sqrt {2007} \minus{} \sqrt {2006}\right)^{2008} \equal{} \sqrt {n} \minus{} \sqrt {n \minus{} 1}$
[hide="hint"]Use the binomial theorem[/hide]

2. Make a polinomial $ p(x)$ with real coefficients such that

$ (x \plus{} 10)P(2x) \equal{} (8x \minus{} 32)P(x \plus{} 6)$

Edit: Really sorry there's more to the problem:
For all real $ x$ and $ p(1)\equal{}210$.
Prove that the polinomial found satisfies the conditions.

3. Let $ a$ and $ b$ be integers. Demonstrate that the equation.

$ (x \minus{} a)(x \minus{} b)(x \minus{} 3) \plus{} 1 \equal{} 0$

Gives at most one solution. (Im not sure if it's at most or atleast one, I had to translate it from spanish. The original sentence is "Admite a lo sumo una solucion entera.)

For the first one, take reciprocals and rationalize to get

$ (\sqrt{2007}\plus{}\sqrt{2006})^{2008}\equal{}\sqrt{n}\plus{}\sqrt{n\minus{}1}$

Then add the two equations to get $ \sqrt{n}$, subtract the two equations to get $ \sqrt{n\minus{}1}$ and then show that is consistent (the difference of the squares of those sums and differences is equal to 1 or whatever)

"Admite a lo sumo una solución entera" means it has at most one solution in intergers
=D

2. take $ x\equal{}\minus{}10$ to get $ 0\equal{}\minus{}32p(\minus{}4)$...so $ p(x)\equal{}(x\plus{}4)p_1(x)$

$ (x\plus{}10)(2x\plus{}4)p_1(2x)\equal{}(8x\minus{}32)(x\plus{}10)p_1(x\plus{}6)$

$ (x\plus{}2)p_1(2x)\equal{}4(x\minus{}4)p_1(x\plus{}6)$

Take $ x\equal{}\minus{}2$ to get $ 0\equal{}\minus{}24p(4)$ so $ p_1(x)\equal{}(x\minus{}4)p_2(x)$

$ (x\plus{}2)(2x\minus{}4)p_2(2x)\equal{}4(x\minus{}4)(x\plus{}2)p_2(x\plus{}6)$

$ (x\minus{}2)p_2(2x)\equal{}2(x\minus{}4)p_2(x\plus{}6)$

Take $ x\equal{}2$ to get $ 0\equal{}\minus{}4p_2(8)$ so $ p_2(x)\equal{}(x\minus{}8)p_3(x)$

$ (x\minus{}2)(2x\minus{}8)p_3(2x)\equal{}2(x\minus{}4)(x\minus{}2)p_3(x\plus{}6)$

$ p_3(2x)\equal{}p_3(x\plus{}6)$

Looking at the lead coefficient of $ p_3(x)$ we see that it must be constant...

Therefore, $ P(x)$ is necessarily in the form $ k(x\minus{}8)(x\minus{}4)(x\plus{}4)$

It is easy to see that this works.

Thanks @Altheman sorry I forgot to put a part of the problem 2.
If you read up top you'll see it, and now you can find k.
$ p(1) \equal{} k(1 \minus{} 8)(1 \minus{} 4)(1 \plus{} 4) \equal{} 210$
$ k( \minus{} 7)( \minus{} 3)(5) \equal{} 210$
$ k \equal{} 2$

"Looking at the lead coefficient of $ P_3(x)$  we see that it must be constant... " Can someone explain this?

Let $ p_3(x)\equal{}a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_0$ where $ a_n\ne 0$.

We have that $ p_3(2x)\equal{}p_3(x\plus{}6)$

The coefficients of $ x^n$ should be equal on both sides, so we have that $ a_n(2)^n\equal{}a_n$, so we conclude that $ n\equal{}0$ because $ a_n\ne 0$.

Is there a way someone can answer the first question using the binomial theorem?

For #3, I think you should specify if $ x$ is in the set of all reals or complex numbers, or just the integers... if it's the set of complex numbers, I doubt there'd be only one solution.

Shoot I keep making mistakes on my post :(
In the third problem it's " Gives atleast only one INTEGER solution" , When I originally wrote it I didn't notice Integer on the other page Sorry.

It won't let me edit my original post, to change the problem number 3.

As $ x$, $ a$, and $ b$ are all integers, $ x \minus{} a$, $ x \minus{} b$, and $ x \minus{} 3$ are all integers. Since their product is -1, we have that either one of the three factors is equal to -1 and the others are 1, or that all three of them equal -1.

In the second case, $ x \equal{} 2$, and $ a \equal{} b \equal{} 3$ for obvious reasons.

Now we have three cases:
[list]
[*]$ x \minus{} 3 \equal{} \minus{} 1$: again, $ x \equal{} 2$, and $ a \equal{} b \equal{} 1$
[*]$ x \minus{} a \equal{} \minus{} 1$ or $ x \minus{} a \equal{} \minus{} 1$; hence, $ x \minus{} 3 \equal{} 1$, so $ x \equal{} 4$. [/list]

In each case, $ x$ has at most one solution. Since these are the only cases which yield solutions--if $ a$ and $ b$ are anything else, there are clearly no solutions--we are finished.

I think my solution is correct, at least... =/


What means to say that two normal vectors N1 and N2 are linearly independent ?

Thanks

Linearly independent means the following:
A set of vectors, $a_1, a_2, \ldots a_n$ are linearly independent if and only if $c_1\cdot a_1 + \ldots c_n \cdot a_n = 0$ only when $c_1 = c_2 = \ldots = c_n = 0$.  In $\mathbb{R}^n$, two vectors orthogonal to each other are always linearly independent.  (In $\mathbb{R}^1$, there do not exist 2 orthogonal vectors.  Also, any 2 vectors are linearly dependent.)

All vectors in R^1 are linearly dependent ? Or you wanted to say independent ?

Thanks

No - dependent is right (in $\mathbb{R}^1$, any vector $(a)$ is a multiple of any other vector $(b)$, thus linearly dependent.)

Speaking as a three dimensional creature:
(And an[b] expert [/b] in vector calculus - How else would I be an expert in catching Frisbies :)  .. not to mention that an expert  guide dog needs   to be excellent in vectoring )

A vector is some thing which has magnitude and direction ... (Eg Force, Velocity etc.) and can be represented as an "arrow"  (of the right maginitude and in the  right direction)

Two vectors, if they  have same direction then they are linearly  dependent, if they are not in the same direction then they are linearly  independent.

Three vectors are linearly dependent if they are in the same plane..

More than three vectors are  always linearly dependent, (Or course true, [b]only [/b] if you are working with a 'physical' word of three dimensions where we deal with regular vectors such as force etc..)  Of course, if your  vector space has more than 3 dimension you may have more than three independent vectors....). In fact , the dimension of a  vector space ( if it is finite) is same as the maximum number of linearly independent vectors you can have.

Two $normal$  vecotrs are always linearly independent..
Basically [b]this means that two vectors which are perpendicular to each oter can not be in the same direction.[/b]
For crying out loud, if the vectors are  'normal ' (= perpendicular to each other) they can't be in the same direction. Can they?
(BTW - You can always tell how "snobish" a fellow is by noticing what noun (s)he uses -  "perpendicular" or "normal" or "orthogonal" .. They mean the samething :D)

(Also that's why you can not have more than 3 vecotors - in three dimensonal space $normal$  to each other).

Now this is some  thing easy to visuaize  .. so naturally .. our mathematican friends are going to  spend all their  time to prove all these  so called "important"  aka silly  theorems :)   whoever said "mathematics is nothing but physics without a purpose" wansn't  too much of the mark..:)

Anyway I hope this has been helpful.


One digit of the decimal representation of 5/7 is randomly selected. What is the probability that the digit is a 4? Express your answer as a common fraction.

[hide]the probability is 1/6. any number of the form 1/7 has six repeating distinct integers, one of which is 4.[/hide]


Even though I do keep a TI (since we're allowed to have two calculators) for graphing problems or problems that can be solved or just checked easily with a program or something,

I usually use an RPN calculator cuz they're really fast at computation. Especially since we have this fun but weird competition in Texas called TMSCA and one part is dedicated to 80 long expressions that we have to evaluate with a calculator in a short time period. Ya it might not be very useful in later life but it kind of makes my speed on my RPN calculator really fast so....

I haven't been in Mathcounts for a long time, but when I was in 7th grade, I just used a standard TI scientific calculator.  Then in 8th grade, after I got my TI-89 from nats, I took both to competitions.  I still have that 89, and I find it pretty amusing that I've had it for 7 years now, and it's still pretty much top-of-the-line.

yeah that's all cool
my coach says that the only reason that you should
use a calculator is if it takes longer to do it on paper, like adding, subtracting, multiplying, or dividing.
he says you should memorize how to do the rest, not put it on programs, etc.
Well then again he took calculus in 7th grade

[quote="happyme"]I use a Bank of America free calculator...

It can calculate monthly payments for loans in seconds!  :huh:[/quote]

certainly helpful in MATHCOUNTS  :rotfl:

I use TI-83+ only...i think the silvers lag just a little bit with all those apps and my parents won't get me one...either way most of them support BASIC...

[quote="happyme"]I use a Bank of America free calculator...

It can calculate monthly payments for loans in seconds!  :huh:[/quote]
hey you ever went to Bank of America museum in Charlotte? it is the coolest place to buy Bank of America calculators.

hmmm...other not going to specify.

I use a TI-83+ Silver Editions
and
a TI-89 Titanium

TI-83 has the programs
TI-89 has some of the same programs but since TI-89's can calculate exact numbers, the TI-89 is much better.
Like I can calculate the area of an octagon given its sides and my  TI-89 will leave it in radical form instead of decimal form
very helpful
Plus it has an equations solver (haha...for lazy americans like me who sometimes don't feel like solving equations)

Well my calculator gets me by just fine I would say.

If I ever NEED to use trig (which I never do), I borrow my sister's TI-30A, other than that, I think that MathCounts only requires basic functions...

Maybe thats just me  :ninja:

who wants to do math when a calculator can practically think for them?

I use MathCounts as a learning opportunity. I couldn't care less if I  make it first in nationals, or if I get a zero on my school test (well, I do care more than that but...).

Using a pre-programmed graphing calculator that can use the Shoelace Method and all that fancy stuff doesn't help my cause.

I use a TI-84 Silver Edition that I bought at a WalMart.  I'm going to use the free one with Mathcounts on the back for tests where they clear your RAM..but then I have OmniCalc.

cool dudes
i personally think that
u need more brains than a good calculator
to win nationals.
my states are march 23rd wish me luck

the TI-89 was allowed on the 2007 AMC.. and seriously.. being able to browse through the history is sooo helpful (unlike the Z80 calculators: 82, 83, 84, and 86)

It's also allowed on the SAT, though I didn't find the math sections too hard.. breezed through  :lol: and I thought if you got a good score on the SAT, the ACT isn't required..?

However, probably because of its CAS, it was not allowed on one of my local competitions (ICTM) so I had to borrow a TI-84 Silver from a classmate.. but usually when I can't use my TI-89T i use my TI-36

I don't know if an algebra class really requires a graphing calculator..

I still have my handy dandy TI-83+. I've had it for who knows how long at it has about 4 million programs.

[quote="undefined117"]
There are no such things as TI-83+ Silver Editions, but I think I remember seeing a "Blue Edition" way back when..[/quote]

Then what are those 2 calculators I have that I got at Nationals that say TI-83+ Silver Editions on them :lol:

I use the first calculator I can find at my house. :P

[quote="Moldytape"][quote="undefined117"][quote="erdogankerem123"]I used my sister's TI-83+ Silver Edition (why just "some other I don't want to list on the poll?) last year, then I got the free 84+ silver from Nats, and I use it now. (but I still voted for 83 other on the poll). However, this is only because my sister wanted her calc back after a little chip off of the battery flap thing broke off. Otherwise I would still be using it now because its faster (and I think better) than the bulky 84+ silvers, though they aren't bad.[/quote]

There are no such things as TI-83+ Silver Editions, but I think I remember seeing a "Blue Edition" way back when..[/quote]

There are TI-83+ Silver Editions, search it in google. and what's a Blue edition?  :huh:[/quote]

Ehh sorry. For some reason I didn't see that (released in 2001?) when I bought my TI-89T last year.. but the Texas Instruments page says it's discontinued..

and about the TI-83 Blue Edition click here: [url]http://www.ticalc.org/images/calcs/83plus-blue-big.gif[/url]

I saw one for sale when I bought my TI-89T from OfficeMax, but I think the technical capabilities are the same as the TI-83 Plus.

The TI-83+ Cool Blue Edition was released in 2001 at Target stores...
[img]http://www.ticalc.org/images/calcs/83plus-blue-big.gif[/img]

In 2002 they made a TI-83+ Ultra Violet:
[img]http://img.photobucket.com/albums/v475/footballrocks41237/Image1.jpg[/img]

They're the same as an 83+ (in terms of memory, speed, etc.)

Has lots of Graphing Calculator Info:
http://www.albion.edu/mathcs/MBollman/GraphCalc.html

EDIT:I resized the 2nd picture

I use one that looks like this:
[img]http://i88.photobucket.com/albums/k168/calc84maniac/pic_00076.png[/img]
But mine isn't really made out of virtual plastic bricks. :D

I use a TI-86, the calculator I had since I was 7 (I wanted it for my birthday, a classic nerd story :) ).

Sometimes I use a TI-89 Titanium but that has too many different functions for me.

[quote="frost13"]The blue and violet editions are just the TI83+ in a different color.  It's not a different Calc.  The Silver Edition is a different calc in terms of memory/speed(not sure of functions).[/quote]

I said in my post that they are the same as a TI-83 Plus...  I don't think there's any other functions.  The only upgrades is the speed and memory i think.

[quote="philB"]a bit of topic, but how did you put a picture in your post, footballrocks41237?? :huh:[/quote]

Click the Image tag (it's above the font size and color tools), paste the url of the image (make sure it ends in a image file like .jpg; you can always use a hosting site like imageshack), then click the image tag again to close it.

I use this TI-84+ Silver Edition (which sometimes doesn't accept files from my computer :huh: ).  It's doing what I normally do with it, play games  :lol:  :D

[img]http://img.photobucket.com/albums/v475/footballrocks41237/mycalc.jpg[/img]

like i actually no what kind of calculator i use, my teacher just gives us random ones.  and she has two kinds, which we were allowed to take both into r state and chapter comps with


calculus
integration
inequalities
algebra proposed
algebra

F(x) = integr[0 to x] e^(t^3).dt

how to show whether or not this inequality is true

e < F(e) < e^(e^3 +1)

I am new in this forum, i don't know how to get the notations in here.

integr is integral from 0 to x (lower limit to upper limit)

Use inequality (f is continuous)
\[(b-a)\min_{x\in[a;b]}f(x)\leqslant\int_{a}^{b}f(x)\,dx\leqslant(b-a)\max_{x\in[a;b]}f(x) \]
If $f$ is not constant both inequalities are strict

Thanks for your reply, but i don't understand how your expression can be used to solve my inequality. What is F(e) ?

$a=0$, $b=e$, $f(x)=e^{x^{3}}$. And you have already defined $F(x) = \int_{0}^{x}e^{t^{3}}\,dt$, so isn't it clear what is $F(e)$?


algebra
polynomial
inequalities
algebra unsolved

Let $ \varphi(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ be a polynomial with real coefficients. Given that $ \varphi(x)$ has three positive real roots and that $ \varphi(0) < 0$, prove that
\[ 2b^3 \plus{} 9a^2d \minus{} 7abc \leq 0.
\]

Let the roots be $ r_1,r_2,r_3$.  Then, since $ \frac{\minus{}d}{a}\equal{}r_1r_2r_3>0$ and $ d<0$, we have $ a>0$.  Thus, we can divide the inequality by $ a^3$ to obtain an equivalent inequality:

$ 2\frac{b^3}{a^3}\plus{}9\frac{d}{a}\minus{}7\frac{b}{a}\frac{c}{a}\le 0$

$ \Leftrightarrow \minus{}2(r_1\plus{}r_2\plus{}r_3)^3\minus{}9(r_1r_2r_3)\plus{}7(r_1\plus{}r_2\plus{}r_3)(r_1r_2\plus{}r_2r_3\plus{}r_3r_1)\le0$

Since this inequality is homogeneous, we can assume $ r_1\plus{}r_2\plus{}r_3\equal{}1$ to obtain

$ \minus{}2\minus{}9r_1r_2r_3\plus{}7(r_1r_2\plus{}r_2r_3\plus{}r_3r_1)\le0$

$ \Leftrightarrow \minus{}\frac{2}{9}\minus{}\frac{7^3}{9^3}\plus{}\frac{49}{81}(r_1\plus{}r_2\plus{}r_3)\minus{}r_1r_2r_3\plus{}\frac{7}{9}(r_1r_2\plus{}r_2r_3\plus{}r_3r_1)\le \minus{}\frac{7^3}{9^3}\plus{}\frac{49}{81}(r_1\plus{}r_2\plus{}r_3)$

$ \Leftrightarrow \prod\left(\frac{7}{9}\minus{}r_i\right)\le \frac{64}{729}$

Now, if the right hand side is negative, we're done.  If not, either all or one factor is positive.  Obviously, there can't be two terms negative since then there would be two roots greater than $ \frac{7}{9}$ and so the sum of the roots is greater than one.  If all factors are positive then by AM-GM,

$ \prod\left(\frac{7}{9}\minus{}r_i\right)\le \left(\frac{1}{3}\sum\frac{7}{9}\minus{}r_i\right)^3\equal{}\frac{64}{729}$

as desired.

[quote="cosinator"]for $ r_1\plus{}r_2\plus{}r_3 \equal{} 1$: $ \minus{} 2 \minus{} 9r_1r_2r_3 \plus{} 7(r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1)\le0$[/quote]

This is British MO 1999, Round 2, Question 3.

[quote="minsoens"]You can just expand from here and cancel stuff. Substitute $ - \frac {b}{a} = x_1 + x_2 + x_3$, $ \frac {c}{a} = x{}_1x{}_2 + x{}_2{}_3 + x{}_3x{}_1$, $ - \frac {d}{a} = x{}_1x{}_2x{}_3$, and the inequality is equivalent to $ x_1^2x_2 + x{}_1x{}_2^2 + x_2^2x_3 + x{}_2x{}_3^2 + x_3^2x_1 + x{}_3x{}_1^2\leq 2(x_1^3 + x_2^3 + x_3^3)$, true by Schur.[/quote]This is also trivial by Muirhead (and equivalently true by AM-GM).

$ \sum_\text{sym}r_1^2r_2\le\sum_\text{sym}\frac{r_1^3+r_2^3+r_2^3}{3}$

how can we write r1+r2+r3=1????? :maybe:  :maybe:

We can always normalize homogeneous inequalities.


Well Basically... I quit math at school which I kinf of regret now since I have come to this problem...

I never listened to my teacher when he said that quadratic formulas and stuff would help me in life but I have finally come to a point where I just can't find the answer.

Ok here it goes...

I play this game called kingsofchaos... Basically you play the game and buy upgrades when you get enough money to buy them. You need to buy these upgrades to get more units for your army in the game. Now I think they use some kind of exponential formula to find this out. Here is the list of the upgrades...

from 0 unit production to 5: 47 000 gold
from 5 unit production to 10: 64 222 gold
from 10 unit production to 20: 184 316 gold
from 20 unit production to 40: 560 463 gold
from 40 unit production to 80: 1 822 330 gold
from 80 unit production to 160: 6 404 000 gold
from 160 unit production to 320: 24 632 756 gold

Now I want some kind of formula so that I can simply plug in to tell me the cost of the next upgrade... I am pretty sure u could work this out...

Thx in advance...

So, (at the beginning at least), a doubling in unit production leads to a tripling in gold required.  This is just a linear progression (roughly).

i think there's no gurantee the solution will be pretty, or if there even is a formula.

[quote="Blurgish_Blade"]I don't want to disaagree with u but that isn't how it works...

The multiple increases every time you make an upgrade...

I want to basically know how much this multiple increases each time...

I have tried to figure it out but I just can't find a logical answer!!!

I remember that in math there was a way of figuring this out but I just can't do it...[/quote]

As I said, with these numbers, it's roughly linear.  It does appear to be 'fairly' exponential though, and plugging it into Excel and doing a regression will give you a better answer.  It doesn't look like it's quadratic though; or if it is, it has a very, very small leading coefficient.


If the polynomial $ x^8\plus{}5x^6\plus{}13x^4\plus{}20x^2\plus{}36$ is evenly divisible by the trinomial $ x^2\minus{}x\plus{}a$, then determine the positive interger $ a$.

[hide="Hint"]$ a$ has to divide 36.
Also, since there exist no odd powers of $ x$ in the expanded sum, this means that it must also have a factor of $ (x^2 \plus{} x \plus{} a)$, which implies that $ a^2$ has to divide 36 as well, or $ a|6$. This leaves the possibilities to $ a \equal{} 1, 2, 3, 6$.[/hide]

Oh, I didn't see that. The a^2  divisible by 36, not a.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

[quote="Extremal"][quote="ashrafmod"]$\cos^{2}\frac{n\pi \cos x}{2}$
from 0 to 2 pi[/quote]
you need find this integral$\cos(n\pi \cos x)$ can do it?[/quote]This is related to [url=http://mathworld.wolfram.com/BesselsFirstIntegral.html]Bessel's First Integral[/url], $\int_{0}^{2\pi}\cos(x \cos \theta) d \theta = 2 \pi J_{0}(x)$, where $J_{0}$ is a Bessel function of the first kind. Asymptotically, $2 \pi J_{0}(n \pi) \approx \frac{\pm 1}{\sqrt{n}}$.

[quote="ashrafmod"]$\cos^{2}\frac{n\pi \cos x}{2}$
from 0 to 2 pi[/quote]
Ok  and $\int_{0}^{\pi}\cos^{2}\frac{n\pi \cos x}{2}$~$\pi/2+\frac{\pm 1}{2\sqrt{n}}$


find all integers $ m,n$ such that:
\[ n^2 \plus{} 2 \equal{} m^3
\]
Edited  :D

[quote="funtwo"]n=5, m=3
simple brute force[/quote]

I edited my problem .. look again  :D

This problem is special case of the [url=http://mathworld.wolfram.com/BachetEquation.html]Bachet Equation[/url].
Furthermore, this problem should have been solved by Fermat.
[url=http://mathworld.wolfram.com/FermatEllipticCurveTheorem.html]Fermat Elliptic Curve Theorem[/url] - MathWorld

Let, $ m=t-1$ and $ n=t+1$.
So,
\begin{eqnarray*}
(t+1)^2+2=(t-1)^3
& \iff & t^2+2t+3=t^3-3t^2+3t-1 \\
& \iff & t(t^2+1)=4(t^2+1) \\
& \iff & t=4
\end{eqnarray*}
Thus, $ t=4 \iff (m, n)=(3, 5)$.

Please examine the other information. :)

Is there an elementary proof, that $ |m \minus{} n| \equal{} 2$ ?

Yeah... how do you know
that m can be expressed
as t-1 and n as t+1?


Suppose that $ S_n$ is the sum of first $ n$ terms  of A.P., $ S_{50}\equal{}a ,S_{100}\equal{}b$, find $ S_{150}\plus{}S_{250}\equal{}$?

[quote="vinskman"]
Let $ t$ be the first term of the A.P. , $ d$ be common difference
$ T_1,T_2,T_3,T_4,T_5$ be the sum of the first $ 50$ terms, the second $ 50$ terms, and then the third, fourth and fifth respectively.

$ \left\{\begin{array}{c}{1} T_1 = t + (t + d) + (t + 2d) + \cdots + (t + 49d) \
T_2 = (t + 50d) + (t + 51d) + \cdots + (t + 99d) \
\end{array}\right.$

Hence, the first term is >> $ T_1 < < = a$ , common difference is >> $ b - 2a$ <<
[/quote]
First, ignore my other post.

vinskman,

the first term of the series is not $ T_1$ as you have it.  It is t.
$ T_1$ = a.

$ T_1$ = 50t + 1225d = a  . . (*)
$ T_2$ = 50t + 3725d

$ T_1$ + $ T_2$ = b

b - (50t + 3725d) = 50t + 1225d
b - 50t - 3725t = 50t + 1225d
b = 100t + 4950d  . . (**)

Multiply (*) by 2 on each side
and place (**) above it, reversed:

100t + 4950d =  b
100t + 2450d = 2a
-------------------------
2500d = b - 2a

d = (b - 2a)/2500

The common difference is (b - 2a)/2500.

Let's solve for the first term of the series, t.

a = 50t + 1225d

50t = a - 1225d

50t = a - 1225(b - 2a)/2500

50t = a - (49/100)(b - 2a)

50t = (100a/100) - (49)(b - 2a)/100

50t = (100a - 49b + 98a)/100

50t = (198a - 49b)/100

t = (198a - 49b)/5000 . . . This is the first term of the series.

[hide]this can also be solved by simple linear equations
$ s_{50} \equal{} a$ let the first term be m and common difference be n
$ m \plus{} 49n \equal{} a\ \ \ .............(1) \\
m \plus{} 149n \equal{} b\ \ \ .............(2) \\
(2) \minus{} (1)\ we\ have \\
n \equal{} \frac {b \minus{} a}{100}$ putting this value in any of the previous equations we can obtain velue of m and the susequent answer[/hide]

[quote="rahul1993"]
$ s_{50} \equal{} a$ let the first term be m and common difference be n
$ m \plus{} 49n \equal{} a\ \ \ .............(1) \\
m \plus{} 149n \equal{} b\ \ \ .............(2) \\
(2) \minus{} (1)\ we\ have \\
n \equal{} \frac {b \minus{} a}{100}$ putting this value in any of the previous equations we can obtain velue of m and the susequent answer[/quote]
No, I already posted a corrected version of a set of linear
equations for it:

100t + 4950d = b
100t + 2450d = 2a
-------------------------
2500d = b - 2a

d = (b - 2a)/2500, not (b - a)/100

And b does not equal the sum of the terms of $ T_2$.
It is equal to the sum of the first 100 terms, or
put another way, b = $ T_1$ + $ T_2$.

hey how can i write the way rahul1993 did
how to use such symbols
like vinskman  wrote T1,T2
how 2 write like that

i still prefer vinksman's solution as the more likely solution. ARRANNE YOUR TAN, please reread his message well . . .  :roll:  :)  :)


Let $ a,b,c$ be non negative real numbers. Prove that

1/  $ (\frac a{b\plus{}c})^2\plus{}(\frac b{c\plus{}a})^2\plus{}(\frac c{a\plus{}b})^2\plus{}\frac{10abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge2$

2/  $ (\frac a{b\plus{}c})^2\plus{}(\frac b{c\plus{}a})^2\plus{}(\frac c{a\plus{}b})^2\plus{}\frac{15\sqrt[3]{abc}}{4(a\plus{}b\plus{}c)}\ge2$

Equality holds for $ (a,b,c)\sim (1,1,1)$ or $ (1,1,0)$  :)

[quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that

1/  $ (\frac a{b \plus{} c})^2 \plus{} (\frac b{c \plus{} a})^2 \plus{} (\frac c{a \plus{} b})^2 \plus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge2$

[/quote]
It's equivalent to $ \sum_{cyc}(a^6\plus{}2a^5b\plus{}2a^5c\minus{}a^4b^2\minus{}a^4c^2\minus{}4a^3b^3\plus{}a^2b^2c^2)\geq0,$ which obviously.

[quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that

1/  $ (\frac a{b \plus{} c})^2 \plus{} (\frac b{c \plus{} a})^2 \plus{} (\frac c{a \plus{} b})^2 \plus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge2$

2/  $ (\frac a{b \plus{} c})^2 \plus{} (\frac b{c \plus{} a})^2 \plus{} (\frac c{a \plus{} b})^2 \plus{} \frac {15\sqrt [3]{abc}}{4(a \plus{} b \plus{} c)}\ge2$

Equality holds for $ (a,b,c)\sim (1,1,1)$ or $ (1,1,0)$  :)[/quote]
The second follows from the first since
\[ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge \frac{8}{9}(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c) \ge \frac{8}{3}\sqrt[3]{a^2b^2c^2}(a\plus{}b\plus{}c)\]
We will show the first
Setting $ x\equal{}\frac{2a}{b\plus{}c}$ and $ y,z$ are similar, we have
\[ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4\]
The inequality becomes
\[ x^2\plus{}y^2\plus{}z^2\plus{}5xyz \ge 8\]
Setting $ p\equal{}x\plus{}y\plus{}z,q\equal{}xy\plus{}yz\plus{}zx,r\equal{}xyz$, then
\[ q\plus{}r\equal{}4\]
the inequality is equivalent to
\[ p^2\minus{}2q\plus{}5r \ge 8\]
\[ \Leftrightarrow p^2\minus{}7q\plus{}12 \ge 0\]
If $ 4 \ge p$, By Schur Inequality for third degree, we have
\[ r \ge \frac{p(4q\minus{}p^2)}{9}\]
\[ \Rightarrow 4 \ge q\plus{}\frac{p(4q\minus{}p^2)}{9}\]
\[ \Leftrightarrow q \le \frac{p^3\plus{}36}{4p\plus{}9}\]
It suffices to show
\[ p^2\minus{}\frac{7(p^3\plus{}36)}{4p\plus{}9} \plus{}12 \ge 0\]
\[ \Leftrightarrow (p\minus{}3)(p^2\minus{}16) \le 0\]
which is true since $ 4 \ge p \ge \sqrt{3q} \ge 3$.
If $ p \ge 4$, then $ p^2 \ge 16 \ge 4q$, hence
\[ p^2\minus{}2q\plus{}5r \ge p^2\minus{}2q \ge \frac{p^2}{2} \ge 8\]
We have done. :)
Equality holds for $ x\equal{}y\equal{}z\equal{}1$ or $ x\equal{}y\equal{}2,z\equal{}0$ or any cyclic permutations.

[quote="can_hang2007"][quote="manlio"][quote="can_hang2007"]

.....If $ 4 \ge p$, By Schur Inequality for third degree, we have
\[ r \ge \frac {p(4q \minus{} p^2)}{9}
\]
[/quote]

Sorry, can_hang2007, but I don't understand why if $ 4 \ge p$ then $ 4q \ge p^2$

Thank you very much.[/quote]
Manlio, I didn't say "if $ 4 \ge p$ then $ 4q \ge p^2$", please check my solution again. Thanks :)[/quote]

You are right, but I asked myself  if it is correct to use

\[ r \ge \frac {p(4q \minus{} p^2)}{9}
\]

when $ 4q \minus{} p^2 \leq 0$

Sorry if it is a stupid question.  :blush:

[quote="manlio"][/quote][quote="can_hang2007"][quote="manlio"][quote="can_hang2007"]

.....If $ 4 \ge p$, By Schur Inequality for third degree, we have
\[ r \ge \frac {p(4q - p^2)}{9}
\]
[/quote]

Sorry, can_hang2007, but I don't understand why if $ 4 \ge p$ then $ 4q \ge p^2$

Thank you very much.[/quote]
Manlio, I didn't say "if $ 4 \ge p$ then $ 4q \ge p^2$", please check my solution again. Thanks :)[/quote][quote="manlio"]

You are right, but I asked myself  if it is correct to use

\[ r \ge \frac {p(4q - p^2)}{9}
\]

when $ 4q - p^2 \leq 0$

Sorry if it is a stupid question.  :blush:[/quote]

In this case [b]manlio[/b] you do not need to seperate it in to two other cases $ 4q \ge p^2$ and $ 4q \le p^2$

In this you do not do any multiplication or any division so do not need to care about that, you just change the member of an equation and reduce to the same denominator...

We seperate the proof into 2 case $ p \ge 4$ and $ p \le 4$ because with $ p \ge 4$ the ineq is trivial, it does not relate to $ 4q \ge p^2$

You just need to reread Can's Proof carefully. Hope you will see that

Hix, it is really nice but not hard :D I will prove it by Mixing Varible method, maybe it will help manlio to understand another solution for this. :wink:
We put:
\[ f(a,b,c)\equal{}a^2\plus{}b^2\plus{}c^2\plus{}5abc\minus{}8\]
We will prove that:
\[ f(a,b,c)\ge f\left(a,\frac{2}{a\plus{}1},\frac{2}{a\plus{}1}\right)\]
In fact we have:
\[ f(a,b,c)\minus{} f\left(a,\frac{2}{1\plus{}a},\frac{2}{1\plus{}a}\right)\]
\[ \equal{}(b\minus{}c)^2(5abc\plus{}b^2\plus{}c^2\plus{}4bc\minus{}20a\plus{}8b\plus{}8c\plus{}8)\ge 0\]
It is always true because if we assume that $ a\equal{}min(a,b,c)$ then:
\[ 5abc\plus{}b^2\plus{}c^2\plus{}4bc\minus{}20a\plus{}8b\plus{}8c\plus{}8\ge (b^2\plus{}4)\plus{}(c^2\plus{}4)\plus{}8(b\plus{}c)\minus{}20a\ge\]
\[ 12(b\plus{}c)\minus{}20a\ge 4a\ge 0\]
We are done because:
 \[ f\left(a,\frac{2}{a\plus{}1},\frac{2}{a\plus{}1}\right)\equal{}\frac{a(a\plus{}4)(a\minus{}1)^2}{(a\plus{}1)^2}\ge 0,\forall a\ge 0\]
I hope I haven't made mistake in calculating  :wink:

[quote="zaizai-hoang"]
In fact we have:
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{1 \plus{} a},\frac {2}{1 \plus{} a}\right)
\]

\[ \equal{} (b \minus{} c)^2(5abc \plus{} b^2 \plus{} c^2 \plus{} 4bc \minus{} 20a \plus{} 8b \plus{} 8c \plus{} 8)\ge 0
\]
 [/quote]

I don't understand for this,zaizai-hoang  :(

I think that
$ f(a,b,c) \minus{} f\left(a,\frac {2}{1 \plus{} a},\frac {2}{1 \plus{} a}\right)\equal{}a^2b^2\plus{}a^2c^2\plus{}5a^3bc\plus{}2ab^2\plus{}2ac^2\plus{}10a^2bc\plus{}b^2\plus{}c^2\plus{}5abc\minus{}20a\minus{}8$

[quote="akai"][quote="zaizai-hoang"]
In fact we have:
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{1 \plus{} a},\frac {2}{1 \plus{} a}\right)
\]

\[ \equal{} (b \minus{} c)^2(5abc \plus{} b^2 \plus{} c^2 \plus{} 4bc \minus{} 20a \plus{} 8b \plus{} 8c \plus{} 8)\ge 0
\]
[/quote]

I don't understand for this,zaizai-hoang  :(

I think that
$ f(a,b,c) \minus{} f\left(a,\frac {2}{1 \plus{} a},\frac {2}{1 \plus{} a}\right) \equal{} a^2b^2 \plus{} a^2c^2 \plus{} 5a^3bc \plus{} 2ab^2 \plus{} 2ac^2 \plus{} 10a^2bc \plus{} b^2 \plus{} c^2 \plus{} 5abc \minus{} 20a \minus{} 8$[/quote]
I think that
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{1 \plus{} a},\frac {2}{1 \plus{} a}\right)
\]

\[ \equal{} (b \minus{} c)^2(5abc \plus{} b^2 \plus{} c^2 \plus{} 4bc \minus{} 20a \plus{} 8b \plus{} 8c \plus{} 8)\ge 0
\]
which is really obvious  :wink:

[quote="ye109"][quote="dduclam"]Let $ a,b,c$ be non negative real numbers. Prove that

1/  $ (\frac a{b \plus{} c})^2 \plus{} (\frac b{c \plus{} a})^2 \plus{} (\frac c{a \plus{} b})^2 \plus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge2$

2/  $ (\frac a{b \plus{} c})^2 \plus{} (\frac b{c \plus{} a})^2 \plus{} (\frac c{a \plus{} b})^2 \plus{} \frac {15\sqrt [3]{abc}}{4(a \plus{} b \plus{} c)}\ge2$

Equality holds for $ (a,b,c)\sim (1,1,1)$ or $ (1,1,0)$  :)[/quote]

1/  $ (\frac a{b \plus{} c})^2 \plus{} (\frac b{c \plus{} a})^2 \plus{} (\frac c{a \plus{} b})^2 \plus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge2$
 :)[/quote]

Hi i think ye109 proved wrong: $ \sum cos^{2}B cos^{2}C \geq \dfrac{(1 \minus{} 2cosAcosBcosC)^{2} \minus{} \frac {3}{16}}{2}$
This we have $ \sum cos^{4}A \leq \dfrac{3}{16}$
But i proved that $ \sum cos^{4}A \geq \dfrac{3}{16}$
Because $ \sqrt {3(\sum cos^{4}A)} \geq cos^{2}A \equal{} 1 \minus{} 2cosAcosBcosC \geq 1 \minus{} 2\sqrt [4]{(\dfrac{\sum cos^{4}A}{3})^{3}}$
We put $ t \equal{} \sum cos^{4}A$ and see that $ t \geq \dfrac{3}{16}$ :roll:

[quote="zaizai-hoang"]Hix, it is really nice but not hard :D I will prove it by Mixing Varible method, maybe it will help manlio to understand another solution for this. :wink:
We put:
\[ f(a,b,c) \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 5abc \minus{} 8
\]
We will prove that:
\[ f(a,b,c)\ge f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)
\]
In fact we have:
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{1 \plus{} a},\frac {2}{1 \plus{} a}\right)
\]

\[ \equal{} (b \minus{} c)^2(5abc \plus{} b^2 \plus{} c^2 \plus{} 4bc \minus{} 20a \plus{} 8b \plus{} 8c \plus{} 8)\ge 0
\]
It is always true because if we assume that $ a \equal{} min(a,b,c)$ then:
\[ 5abc \plus{} b^2 \plus{} c^2 \plus{} 4bc \minus{} 20a \plus{} 8b \plus{} 8c \plus{} 8\ge (b^2 \plus{} 4) \plus{} (c^2 \plus{} 4) \plus{} 8(b \plus{} c) \minus{} 20a\ge
\]

\[ 12(b \plus{} c) \minus{} 20a\ge 4a\ge 0
\]
We are done because:
\[ f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right) \equal{} \frac {a(a \plus{} 4)(a \minus{} 1)^2}{(a \plus{} 1)^2}\ge 0,\forall a\ge 0
\]
I hope I haven't made mistake in calculating  :wink:[/quote]

I don't think your proof is correct, Hoang. Please check it again.  :)

General problem:

$ \sum\left(\frac a{b\plus{}c}\right)^k\plus{}\frac{\left(2^{k\plus{}1}\minus{}3\right)abc}{2^{k\minus{}3}(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge2$

I think it's hold for all $ k\ge log_23\minus{}1.$  :)

[quote="desmrhai"]dduclam! do you think ye109'proof is right????????[/quote]

[quote="desmrhai"]

Hi i think ye109 proved wrong: $ \sum cos^{2}B cos^{2}C \geq \dfrac{(1 \minus{} 2cosAcosBcosC)^{2} \minus{} \frac {3}{16}}{2}$
[/quote]

You're right, desmrhai. Ye109'proof is incorrect. Infact, we have $ \sum cos^{2}B cos^{2}C \leq \dfrac{(1 \minus{} 2cosAcosBcosC)^{2} \minus{} \frac {3}{16}}{2}$
and Ye109 has solved a weaker inequality my one.  :)


Let $g$ be a Fibonacci primitive root $\pmod{p}$. i.e. $g$ is a primitive root $\pmod{p}$ satisfying $g^2  \equiv g+1\; \pmod{p}$. Prove that [list=a] [*] $g-1$ is also a primitive root $\pmod{p}$. [*] if $p=4k+3$ then $(g-1)^{2k+3} \equiv g-2 \pmod{p}$, and deduce that $g-2$ is also a primitive root $\pmod{p}$. [/list]

a.:
We have $(g-1)g=g^2-g \equiv 1 \mod p$, thus $g-1$ is the inverse of $g \mod p$, thus primitive root, too.

b.:
$g-1$ is a nonsquare $\mod p$ since it's a primitive root. Thus $(g-1)^\frac{p-1}{2} \equiv -1 \mod p$.
This gives $(g-1)^{2k+3} = (g-1)^2 \cdot (g-1)^\frac{p-1}{2} \equiv -(g^2-2g+1) \equiv g-2 \mod p$.

All we are left to show is $\gcd(2k+3,p-1)=1$, which follows via $\gcd(2k+3, 4k+3) = \gcd(2k+3, -3)=1$, the latter because otherwise $3|p$, impossible.

let $g$ be a primitive root of $p$ then we shall assert that $g^{-1}$ is also a primitive root,

assume that $(g^{-1})^x \equiv 1 \mod p$ that $x<p-1$

$(g)^x(g^{-1})^x \equiv g^x \equiv 1 \mod p$ which is a contradiction so $x=p-1$ and $g^{-1}$

now $g^2 \equiv g+1 \mod p \implies g(g-1) \equiv 1 \mod p$ so $g-1=g^{-1} \implies$ $g-1$ is a primitive root as well.

now we know if $x^2 \equiv 1 \mod p \implies p \mid x+1$ or $p \mid x-1$ now part $b$ we can see that $g^{\frac{p-1}{2}} \equiv -1 \mod p$

$g^{2k+1} \equiv -1 \mod p$ and because $g^{-1}=g-1$ then $(g-1)^{2k+3}=g^{-(2k+3)} =g^{-(2k+1)} \times g^{-2}$

now it's straightforward to prove $-g^{-2} \equiv g-2 \iff -1 \equiv g^3-2g^2 \iff g^3 - 2g^2 +1 = (g-1)(g^2-g-1) \equiv 0$


A sequence of integer satisfying $\frac{1}{1998}\leq\frac{|a_m-a_n|}{|m-n|}\leq\1998$ and every integer number appear in it once time
Prove that $|a_n-n|\leq\ 2000000$


inequalities
function
inequalities proposed

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that:
\[ \frac {a}{1 \plus{} 12a^2} \plus{} \frac {b}{1 \plus{} 12b^2} \plus{} \frac {c}{1 \plus{} 12c^2}\leq\frac {3}{7}
\]
[hide]For all non-negative numbers $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} 1$
and for all $ k\geq\frac {13 \minus{} 4\sqrt {10}}{9}$ the following inequality holds:
\[ \frac {a}{k \plus{} a^2} \plus{} \frac {b}{k \plus{} b^2} \plus{} \frac {c}{k \plus{} c^2}\leq\frac {9}{1 \plus{} 9k}
\]
For $ k \equal{} k_{best} \equal{} \frac {13 \minus{} 4\sqrt {10}}{9} \equal{} 0.0389877...$ the equality also holds when $ a \equal{} b \equal{} \frac {4 \minus{} \sqrt {10}}{3}$ and $ c \equal{} \frac {2\sqrt {10} \minus{} 5}{3}.$
For $ k \equal{} \frac {1}{12}$ we obtain:
\[ \frac {a}{1 \plus{} 12a^2} \plus{} \frac {b}{1 \plus{} 12b^2} \plus{} \frac {c}{1 \plus{} 12c^2}\leq\frac {3}{7}
\]
[/hide]

Nice one  :)

We have to prove that $ \sum\frac{a}{1\plus{}12a^2}\leq\frac{3}{7}$ or equivalently

$ \sum (\frac{a}{1\plus{}12a^2}\plus{}\frac{3}{49}(a\minus{}\frac{1}{3})\minus{}\frac{1}{7})\leq 0$ or equivalently

$ \sum (\frac{4(a\minus{}2)(3a\minus{}1)^2}{49(1\plus{}12a^2)})\leq 0$

which is true ... :)

By the way for all $ a,b,c>0$ such that $ a \plus{} b \plus{} c \equal{} 1$ and for all $ k\leq 15$.The Ineq is hold
  $ \frac {a}{1 \plus{} ka^2} \plus{} \frac {b}{1 \plus{} kb^2} \plus{} \frac {c}{1 \plus{} kc^2}\leq\frac {9}{9 \plus{} k}$
This is my solution:
 we will prove: $ \frac {1}{1 \plus{} ka^2}\leq\frac {3}{9 \plus{} k} \plus{} \frac {(27 \minus{} 3k)}{(9 \plus{} k)^2}(3a \minus{} 1)$
                      $ \Leftrightarrow (3a \minus{} 1)^2(6 \plus{} a(9 \minus{} k))\geq0$ (Right for all $ k\leq 15$ because  $ a\leq1$ )
done! :blush:

Very well, silouan and toanIneq!
The following inequality true too:
for all non-negative numbers $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} 1$ prove that:
\[ \frac {a}{1 \plus{} 18a^2} \plus{} \frac {b}{1 \plus{} 18b^2} \plus{} \frac {c}{1 \plus{} 18c^2}\leq\frac {1}{3}\]

We can rewrite the last inequality in a nicer form:
for all non-negative numbers $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} 3$ prove that:
$ \frac {a}{1 \plus{} 2a^2} \plus{} \frac {b}{1 \plus{} 2b^2} \plus{} \frac {c}{1 \plus{} 2c^2}\leq1.$

I will use mixing variables Assume that a is the maximum .We have

$ f(a,b,c)\leq f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2})$  <=>

$ \frac{(b\minus{}c)^2(b\plus{}c)(2bc\minus{}3)}{(1\plus{}2b^2)(2\plus{}(c\plus{}b)^2)(1\plus{}2c^2)}\leq 0$  which is true .

So it remains to prove that

$ f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2})\leq 1$ which is equivalent to

$ \frac{\minus{}(a\minus{}1)^2(2a^2\minus{}5a\plus{}5)}{(a^2\minus{}6a\plus{}11)(1\plus{}2a^2)}\leq 0$

Nice proof, silouan!
I have meant the following reasoning:
Let $ f(a) = \frac {a}{1 + 2a^2}$ and $ a = \max\{a,b,c\}.$
Then $ f'(a) = \frac {1 - 2a^2}{(1 + 2a^2)^2}.$
Hence, ${ f(a)\leq f(\frac {1}{\sqrt2}}) = \frac {1}{2\sqrt2}.$
We obtain: $ \sum_{cyc}f(a)\leq1\Leftrightarrow\sum_{cyc}(\frac {1}{3} - \frac {a}{1 + 2a^2} - \frac {1}{9}\cdot(a - 1))\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a - 1)^2(2 - a)}{1 + 2a^2}\geq0.$
If $ a\leq2$ it's done.
If $ a > 2$ then
$ \sum_{cyc}f(a) < f(2) + 2f(\frac {1}{\sqrt2}) = \frac {2}{9} + \frac {1}{\sqrt2} < 1.$ :)

For the following inequality my last method not helps.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a+b+c=3.$ Prove that:

\[ \frac {a}{3 + 8a^2} + \frac {b}{3 + 8b^2} + \frac {c}{3 + 8c^2}\leq\frac {3}{11}\]

[quote="arqady"].
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {a}{3 \plus{} 8a^2} \plus{} \frac {b}{3 \plus{} 8b^2} \plus{} \frac {c}{3 \plus{} 8c^2}\leq\frac {3}{11}
\]
[/quote]
let $ x \equal{} \frac {a}{3},y \equal{} \frac {b}{3},z \equal{} \frac {c}{3}$
the given inequality then becomes
$ \sum_{cyc} xf( \minus{} (1 \plus{} 24x^{2})) \leq f( \minus{} \frac {11}{3})$
where $ f(x) \equal{} \frac { \minus{} 1}{x}$
which is clearly concave and increasing so
$ \leftrightarrow f(\sum_{cyc} \minus{} x(1 \plus{} 24x^{2})) \leq f( \minus{} \frac {11}{3})$
$ \leftrightarrow \sum_{cyc} x \plus{} \sum_{cyc} 24x^{3} \geq \frac {11}{3}$
which is obviously true since $ \sum_{cyc} x^{3} \geq  \frac {(\sum_{cyc}x)^{3}}{9}$

pardesi, I think your proof is wrong:
$ \minus{} (1 \plus{} 24x^2) < 0.$ But $ f$ is convex function on $ (\minus{}\infty,0).$ :wink:

[quote="arqady"]
I have meant the following reasoning:
...
[/quote]
It was close! Nice proof.
[quote="arqady"]
For the following inequality my last method not helps.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {a}{3 \plus{} 8a^2} \plus{} \frac {b}{3 \plus{} 8b^2} \plus{} \frac {c}{3 \plus{} 8c^2}\leq\frac {3}{11}
\]
[/quote]
Indeed, it doesn't work here for all cases.
But I think we can prove it using silouan's idea - by Mixing Variables:
Assume $ a \equal{} max\{a, b, c\}$
1) $ f(a, b, c) \minus{} f(a, \frac {b \plus{} c}{2}, \frac {b \plus{} c}{2}) \equal{} \frac {2(b \plus{} c)(b \minus{} c)^2 (8bc \minus{} 9)}{(3 \plus{} 8b^2)(3 \plus{} 8c^2)(3 \plus{} 2(b \plus{} c)^2)} \leq 0$
and the last one is true.

2)$ f(a, \frac {b \plus{} c}{2}, \frac {b \plus{} c}{2}) \leq \frac {3}{11} \Leftrightarrow$
$ \Leftrightarrow \frac { \minus{} 2a^3 \plus{} 4a^2 \plus{} 6a \plus{} 3}{(3 \plus{} 8a^2)(3 \plus{} 2(3 \minus{} a)^2)} \leq \frac {3}{11} \Leftrightarrow$
$ \Leftrightarrow (a \minus{} 1)^2(8a^2 \minus{} 21a \plus{} 15) \geq 0$,
which is true.

[quote="arqady"]
For the following inequality my last method not helps.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a + b + c = 3.$ Prove that:
\[ \frac {a}{3 + 8a^2} + \frac {b}{3 + 8b^2} + \frac {c}{3 + 8c^2}\leq\frac {3}{11}
\]
[/quote]
For this inequality with $ a,b$ and $ c$ which are non-negative real numbers then we can proved by the same way here
http://www.mathlinks.ro/viewtopic.php?t=175288
And I'm so sorry because my solution in that topic is wrong,  :( . It's only true with $ a,b$ and $ c$ are positive real numbers.
\[ \sum {\frac{a}{{3 + 8a^2 }}}  \le \frac{3}{{11}}
\]
\[ \Leftrightarrow f( r ) = 81r^2  - ( {99q + 504} )r + 216q^2  - 1944q + 4608 \ge 0
\]
We have $ f'( r ) = 162r - 99q - 504 \le 0$ so it suffices us to prove the inequality in the case $ a=b$, then
\[ \Leftrightarrow \frac{{2a}}{{3 + 8a^2 }} + \frac{{3 - 2a}}{{3 + 8( {3 - 2a} )^2 }} \le \frac{3}{{11}}
\]
\[ \Leftrightarrow ( {a - 1} )^2 ( {768a^2  - 1296a + 576} ) \ge 0
\]
Equality holds if and only if $ a=b=c=1$, :oops: .

But with this
[quote="Ji Chen"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are real numbers such that $ a + b + c = 3.$ Prove that:

$ \frac {a}{3 + 8a^2} + \frac {b}{3 + 8b^2} + \frac {c}{3 + 8c^2}\leq\frac {3}{11}.$[/quote]${ \frac {3}{11} - \sum{\frac {a(a + b + c)}{(a + b + c)^2 + 24a^2}}\equiv\frac {16F(a,b,c)}{11\prod{[(a + b + c)^2 + 24a^2}]}},$[/quote]

Actually the above simple proof has been failed,  :( .

The following more general statement holds:

Let $ a_1,a_2,...,a_n$ ($ n \ge 3$) be nonnegative real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$.
If $ 1\minus{}\frac 2{n} \minus{}\frac 5{2n^2} \le p\le 1 \plus{} \frac 3{n} \plus{} \frac {15}{2n^2}$, then

$ \frac {a_1}{1 \plus{} pa_1^2} \plus{} \frac {a_2}{1 \plus{} pa_2^2} \plus{} ... \plus{} \frac {a_n}{1 \plus{} pa_n^2} \le \frac n{1 \plus{} p}$.


modular arithmetic
number theory
relatively prime

[quote="Peter VDD"]$\phi(7)=6$ ;)[/quote]

What does that mean??? I'm really not familiar with that symbol.

Another hint: it's worth remembering that 1001 is divisible by 7 (and by 11, and by 13). So if you want to figure out what 2222 or 5555 are mod 7, subtract off 2002 or 5005 to work with a much smaller number.

[quote="Rep123max"]$\phi{n}$ denotes the number of positive integers less than $n$ that are relatively prime to $n$. It is obvious that if $p$ is prime, then $\phi{p}=p-1$.[/quote]

very funny rep123max :)

It also means something to questions like this: if a,p are relatively prime (which is clearly true for 7,2222 and 7,5555) then we know that $a^{\phi(p)} = 1 \pmod{p}$

So knowing this, it's easy to conclude that the sum of them will be 0 mod 6 :)

[quote="Rep123max"]$\phi{n}$ denotes the number of positive integers less than $n$ that are relatively prime to $n$. It is obvious that if $p$ is prime, then $\phi{p}=p-1$.[/quote]

actually it is less than or equal to, [tex]\phi{1} = 1[/tex]

[quote="Peter VDD"]
It also means something to questions like this: if a,p are relatively prime (which is clearly true for 7,2222 and 7,5555) then we know that $a^{\phi(p)} = 1 \pmod{p}$

So knowing this, it's easy to conclude that the sum of them will be 0 mod 6 :)[/quote]

That's cool. I didn't know that, but I definitely will remember that concept, and the purpose of $\phi$. Thanks guys!


Help! I posted this problem, but wrong section  :( . Solve it!
http://www.mathlinks.ro/Forum/viewtopic.php?p=469964#

Let $x_1 = \ln a,\ x_n = \ln(a-x_{n-1})$. Prove that:

\[ \lim_{n \to \infty} x_n = \ln a  \]


Let a triangle $ABC$  and its altitudes $AD,BE,CZ$ and the orthocenter $H$. Let  $G$ the point of intersection of $EZ$ with $BC$.Let a parallel from $D$ to $EZ$  which intersects $AB,AC$ at $K,L$ respectively and let $R,P$  the midpoints of  $AH,ZE$ respectively. Prove that the lines $RP, BC$  and the circumcircle of  the triangle $GKL$
pass through the same point.

Use http://www.mathlinks.ro/Forum/viewtopic.php?t=29208 and the conclusion follows.


Let a in [0,1] and C the Cantor Set, proof that exists x,y in C such that x-y=a.

Maybe there are better approaches, but it looks like writing everything in ternary expansion and using the fact that the Cantor set is the set of points which are wriotten only with digits $0$ and $2$ should work. It's not obvious, of course, but maybe it's a good start. I don't have time to look into that right now.. :(

Really, it gives the result, but it is so boring to describe... :)

I think I know what you mean, but anyway describing is boring.

I don't know.. It does have a very nice geometric interpretation. Basicly, I'm sure you can see why it's equivalent to the following:

On any segment parallel to a diagonal running through the unit square, there is a point of [url=http://mathworld.wolfram.com/CantorDust.html]Cantor's dust[/url]. I think this shouldn't be too hard to show :).


Ring Theory
superior algebra
superior algebra unsolved

I am trying to prove the following result.

Let $\phi : A \to B$ a ring homomorphism, of commutative rings with unit. Then $\phi$ satisfies the conclusion of Going Down, whenever the associated map ${^{a}\phi}: \text{Spec}(B) \to \text{Spec}(A)$ is an open mapping.

I know (i.e. have been told) this is in EGA, but I do not know how to read French.

Thank you for any help!

The argument in EGA (IV, 1.9.1/2 and 1.10.3/4) is as follows: consider primes $\mathfrak p$ in $A$ and $\mathfrak P$ in $B$ such that ${}^{a}\phi(\mathfrak P)=\mathfrak p$. To show going-down, it suffices to show that $\mathrm{Spec}\,B_\mathfrak P\to\mathrm{Spec}\,A_\mathfrak p$ is surjective. For every $f\in B\setminus\mathfrak P$, ${}^{a}\phi(\mathrm{Spec}\,B_{f})$ is a neighbourhood of $\mathfrak p$, hence it suffices to show that ${}^{a}\phi(\mathrm{Spec}\,B_\mathfrak P)\supseteq\bigcap_{f}{}^{a}\phi(\mathrm{Spec}\,B_{f})$. Take $\mathfrak q\notin{}^{a}\phi(\mathrm{Spec}\,B_\mathfrak P)$. This means $0=B_\mathfrak P\otimes_{A}\kappa(\mathfrak q)=\varinjlim_{f}B_{f}\otimes_{A}\kappa(\mathfrak q)$, but a direct limit of rings is zero iff it is zero already at some finite stage ($1=0$ has to be true at some finite stage). For this $f$, we have $B_{f}\otimes\kappa(\mathfrak q)=0$, hence $\mathfrak q\notin{}^{a}\phi(\mathrm{Spec}\,B_{f})$. owk


How can it be short, rigourous, and correct and inelegant?  :)

[quote="ZennyK"]If I have a solution that is short, rigorous, and correct, but extremely boring and inelegant, will I probably get a 5?[/quote]

Of course.

What if it's rigorous, but very long, boring, and tedious? Would you still get a 5?

ditto to tekno10m, and is longer necessarily better on USAMTS?

If it is rigorous and correct, you'll get a 5.

Most graders would emphatically tell you that longer is not necessarily better.  That is not a green light to skip steps, of course.  It just means that you shouldn't be trying to make your solution longer than it needs to be to fill up space.

[quote="Dark_Shadow"]So, would I get a five by (in a "Find all solutions in integers less than two thousand" problem) checking all possibilities?[/quote]

Yes, though you'd make one grader really really unhappy.

[quote="ZennyK"][quote="Dark_Shadow"]So, would I get a five by (in a "Find all solutions in integers less than two thousand" problem) checking all possibilities?[/quote]

It could take a while to search all integers less than 2000 . . .[/quote]

Remember that you could submit a computer program solution.  Searching those integers would take a fraction of a second.

[quote="ZennyK"]Note that there are an infinite number of integers less than 2000.[/quote]
True, but that's not the point.  :)
I'm pretty sure there could be some way that you could put a lower bound on the problem.  Once you do, there is only a finite amount of time before a computer checks all the possibilities.

Yes, yes, of course. It was probably supposed to be POSITIVE integers less than 2000. But we get the idea.  ;)


I'm sad...

RIP Bonkura, a pen spinning legend. You will be remembered.

5 December, 1985 - 13 September, 2007

are you a pen spinning pro? I know a few people that have every one of them tricks on pentrix down


There are infinitely many chords of length 2 in a circle of radius 8. Find the area of the region consisting of all such chords.

[hide="probably not right..."]
We can see that the chords create a circe within a circle. By the Pythagorean thereom, the radius of the inner circle is $3\sqrt{7}$. Thus, the area of the region between the circles is $64\pi-63\pi=\boxed{\pi}$ [/hide]

I would like to note that the area determined by all chords of length 2 does not depend on the radius of the big circle (well, it has to be greater than 1)

[quote="Klebian"]Hm? Are you sure? I'd think that a circle with radius 200 would have a larger area determined by chords of length 2...

I'd also like explained how you get the radius of the inner circle as $3\sqrt7$[/quote]But think about this.... the annulus is thinner and thinner as the radius of the circle increases. So in a circle of radius 200, you would end up having an annulus whose thickness is a little bit larger than 0.0025, it's so thin, it looks like a line.

[quote="Klebian"]Hm? Are you sure? I'd think that a circle with radius 200 would have a larger area determined by chords of length 2...

I'd also like explained how you get the radius of the inner circle as $3\sqrt7$[/quote]
the hypotneuse is 8 (the radius), one leg is 1 (half of the chord), and the other leg would be $3\sqrt7$ to complete the right triangle.

Yo[hide]u can think of this as the pythagorean theorem.

there be one chord of length 2.

Then let there be two concentric circles of radius r and R, where R>r.

Then, by Pythagorean theorem one can check that R^{2}-r^{2}=1.

Now notice that, you can multiply both sides by pi, and get

pi*R^{2}-pi*r^{2}=pi.

And so that is why pi is the answer.[/hide]

Ah, ok. I also see what you're saying, 1000th user. I was actually thinking of it the other way around, looking at the smaller circle instead of the annulus.


advanced fields
advanced fields unsolved

Can you make a string (infinite in both directions) of A's and B's with an automorphism (in the category of partially-ordered sets) that switches the A's and B's?

[hide="This arose from the following discussion, which might make the question more clear."]To get rid of the ninja pirate/pirate ninja (or even ninja ninja pirate, etc) kinds of disagreements, why don't we make an infinite chain (in both directions) in which pirates and ninjas are interchangeable. This actually raises an interesting question: Can you make a string (infinite in both directions) of A's and B's with an automorphism (in the category of partially-ordered sets) that switches the A's and B's?[/hide]

This is sort of combinatorial, but I'm thinking it could get into the axiom of choice or something advanced like that.

I'm not sure I understand either.  In the category of [b]totally[/b] ordered sets, aren't shifts the only automorphisms?

[quote="fedja"]Frankly speaking, I fail to understand what is asked: what is wrong with the alternating string ...ABABABABA... and the shift by one position? Apparently, that's not what you had in mind :?.[/quote]  Oh good ... I had exactly this thought also, and it always pleases me to be no more confused than the people around me :)

t0r -- yes if we're only talking about totally ordered sets with some local finiteness condition (but I don't know if "locally finite" is actually the right condition, partly because I don't remember what it means), but $ \bf Q$ has lots of non-shift (poset-)automorphisms.

[quote="fedja"]Frankly speaking, I fail to understand what is asked: what is wrong with the alternating string ...ABABABABA... and the shift by one position? Apparently, that's not what you had in mind :?.[/quote]

Actually, that is what I had in mind. I didn't notice how trivial it was ;)


Does there exist a natural number $N$ which is a power of$2$, such that one
can permute its decimal digits to obtain a different power of $2$?

Let's assume there is two powers of 2 $2^m$ and $2^n(m<n)$ such that one can permute decimal digits of the 1st to obtain 2nd. Then $n-m\leq 3$, else $2^n\geq 2^{m+4}>10*2^m$ and $2^n$ has more decimal digits, than $2^m$.
Hence $2^n-2^m=2^m(2^{n-m}-1)=2^m$, $3*2^m$ or $7*2^m$ is not divisible by 9, and $2^m\not\equiv2^n(mod9)$. And since $S(2^m)\equiv2^m(mod9)$, $S(2^n)\equiv2^n(mod9)$ we have $S(2^m)\not\equiv{S(2^n)}(mod9)$ :arrow: $S(2^m)\not=S(2^n)$ :arrow: we can't permute decimal digits of $2^m$ to obtain $2^n$.

P.S. $S(x)$ is the sum of decimal digits of $x$.

What if N has zeros in its decimal form and those zeros when permuted are placed in front so the result of permutation can decrease more than 10 times?


induction
LaTeX
Pascal\u0027s Triangle
binomial coefficients
algebra
binomial theorem
strong induction

Proofs are highly intimidating.  Usually, beginners to proofs just lock up and don't have a clue as to how to approach the problem. However, with practice the problems become much more feasible.  I'll try to post new problems and techniques as much as possible.  Others are encouraged to contribute as much as possible!  If you've seen something before, allow a day for others to try their hand before posting a solution.

1. Prove that \[ \sum_{j=0}^n \binom{n}{j}^2 = \binom{2n}{n}.  \]

[quote="Iversonfan2005"]AHHHHHHH!!!!!!!! PROOFS!!!!!!!!!!!

btw, you might wanna sticky this![/quote]

Done.

could somebody give me a good explaination of everything that's going on with the sigma notation. i know the general idea (summations) and how to use them in statistics, but what do the values above and below it stand for and how is it used?

it's combinatorics

$\binom{n}{j}-\frac{n!}{j!(n-j)!}$

Here are some examples that should be helpful with becoming familiar with sigma notation:

\[ \sum_{i=1}^{10} i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10  \]

\[ \sum_{j=0}^n \binom{n}{j}^2 = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2+\cdots+\binom{n}{n}^2  \]

Below the sigma symbol is the lower limit of the sum (where you start summing) and the upper limit is written above the sigma (where you stop summing).

thanks! i knew they kind of had something to do with how you were adding up the terms. this made it clear for me

eh i havent studied this block walking technique very thoroughly yet..

\[ \text{Block Walking}  \]

Let me explain what 'block walking' refers to.  Most people recognize Pascal's triangle:

\[ \begin{array}{ccccccc} & & & 1 & & &\\ & & 1 & & 1 & &\\ & 1 & & 2 & & 1 &\\ 1 & & 3 & & 3 & & 1\\ & \vdots & & \vdots & & \vdots & \end{array}  \]

But less people are familiar with Pascal's triangle presented in the following form:

\[ \begin{array}{ccccccc} & & & \binom{0}{0} & & &\\ & & \binom{1}{0} & & \binom{1}{1} & &\\ & \binom{2}{0} & & \binom{2}{1} & & \binom{2}{2} &\\ \binom{3}{0} & & \binom{3}{1} & & \binom{3}{2} & & \binom{3}{3}\\ & \vdots & & \vdots & & \vdots & \end{array}  \]

where the $k$th entry of the $n$th row is $\binom{n}{k}$.

The first natural question one should ask is why is this true?  The entries of Pascal's triangle are defined by the rule that each element equals the sum of the two elements directly above it starting with a row with a single 1 (blank spaces are 0's).  The binomial coefficient, $\binom{n}{k}$, is combinatorially understood to be the number of ways to choose $k$ objects from a set of $n$ (all objects distinct) where order does not matter.  It turns out that the binomial numbers tell us how many ways you can 'walk' from the 1 at the very top to the $k$th entry of the $n$th row if you can only go from each entry to either the lower left or lower right entry below it.  If you show that both the entries of Pascal's triangle and the binomial coefficients give the number of paths to the $k$th entry of the $n$th row you will have shown their equivalence!  (I'll leave this as an exercise).

OK, so how does 'block walking' help with proving combinatorial identities?  Block walking provides a clean, efficient way to cominatorially think about the identities (as opposed to looking at them algebraically or some other way).

\[ \text{Problems}  \]

For the problems, you don't have to use just block walking.  The more methods posted, the better.  You may assume the validity of the binomial theorem except on #6 where I ask you to show it's validity :D

1. Show that the coefficients of Pascal's triangle are equivalent to the binomial coefficients by showing that they both count the number of ways to 'walk' to the $k$th entry of row $n$.

2. Show that $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.  Remember that $\binom{n}{k}$ is defined to be the number of ways to choose a subset of $k$ elements from a set of $n$.

3.  Prove that \[ \binom{n}{k}=\binom{n}{n-k}.  \]

4. Prove that \[ \binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n.  \]

5.  Prove Pascal's Identity:  \[ \binom{n+1}{k+1} = \binom{n}{k+1} + \binom{n}{k}.  \]

6.  Demonstrate that the binomial theorem works for positive integers $n$: \[ (x+y)^n = \sum_{i=0}^n\binom{n}{i}x^iy^{n-i}.  \]

7.  Prove the Christmas stocking theorem (a.k.a. the Hockey Stick Theorem): \[ \binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots+\binom{n+k}{k} = \binom{n+k+1}{k}.  \]

8.  Prove Vandermonde's Identity: \[ \sum_{k=0}^r \binom{n}{k}\binom{m}{r-k}=\binom{n+m}{r}.  \]

Here's a link to [url=http://binomial.csuhayward.edu/index.html]more than you'd probably ever need or want to know about Pascal's triangle and binomial coefficients[/url].

I think we need some hints Joe.  Proofs are tedious; I think I could prove a few by induction, but with the added pain of Latex, I really don't feel up to it.  How about you solve a few of your own problems to get us started.

Here are 3 different methods to solve problem 3:

[hide="Algebra"]
By definition, we have that \[ \binom{n}{n-k}=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{(n-k)!k!}.  \]  But that is just $\binom{n}{k}$.[/hide]

[hide="Block Walking"]
To 'walk' to the $k$th entry of the $n$th row you have to take $k$ right turns (the rest being left).  On the other hand, you can look at is as taking $n-k$ left turns (and the rest being right).

(remember that in block walking we're going down to the row below so 'left' and 'right' might be better understood as downward left and downward right...)[/hide]

[hide="Committee Forming"]
There are $\binom{n}{k}$ ways to form a committee of $k$ people from $n$ members.  Alternatively, you could choose who is NOT in the committee which gives $\binom{n}{n-k}$.  There is a 1-1 correspondence (aka bijection or isomorphism) between each committee of $k$ people and each non-committee of $n-k$ people.[/hide]

If you find it easier, you could even think about the [hide="binomial theorem"]
In $(x+y)^n$ the coefficient of $x^k$ is $\binom{n}{k}$.  But in $(y+x)^n$ the coefficient of $x^k$ is $\binom{n}{n-k}$.  However, these two expressions are equivalent.  Therefore each $x^k$ term must be equivalent.[/hide]

\[ \text{Induction}  \]

Induction usually only works if you know what you're trying to prove.  It's not so helpful for deriving formulas.  Nonetheless, it is an extremely powerful tool which every math person should know.

Here are the steps for induction:

1) Show that the base case is true.
2) Assume that the statement is true for $n=k$.
3) Show that the statement must then hold for $n=k+1$.

Note:
There is something called strong induction which essential uses the same logic.  However, you assume that the statement is true for every value below $k+1$, i.e. $1,2,\ldots,k$, and then show that it holds for $k+1$.

A common way to think of induction is with dominoes.  Once you set up a line of dominoes, it only takes knocking down the first one for all the rest to fall down.  Induction encorporates the same idea.  You set up that if the statement holds for some value of $n$, then it holds for the next value of $n$.  Then, you have to show that it works for $n=1$ (otherwise the dominoes would still be standing!).

\[ \text{Problems}  \]

1.  Prove that \[ 1+2+\cdots+n=\frac{n(n+1)}2. \]

2.  Prove that \[ 1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6. \]

3.  Prove that \[ 1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2. \]

(more should be coming)

Once again, de-stickyed due to overwhelming lack of participation.

I was looking around ICCSD and saw [url=http://www.iccsd.k12.ia.us/schools/west/ClubsOrg/Math/fcubed/f^3-Induction.pdf]this PDF on Induction[/url].  Thought it would help everyone, particularly Shobhit.  :P

I've seen MCrawford post a whole list of Fibonacci related problems that are excellent induction practice.

I think someone should sticky this, this thread is dying out before it even started.

[quote="joml88"]1. Prove that \[ \sum_{j=0}^n \binom{n}{j}^2 = \binom{2n}{n}.  \][/quote]

How do you do this?

Maybe use Binomial Theorem general cases....maybe?  I'm not good with summation notation....

Summation notation is not a problem:

\[ \text {Prove that } \binom{n}{0}^2+\binom{n}{1}^2+\cdots+\binom{n}{n}^2=\binom{2n}{n}. \]

[hide="Hint"]
Is there anyway you can convert this into a statement about Pascal's triangle?
[/hide]

[hide="Different hint"]
What about choosing a committee? Can you convert that formula into a statement about choosing some kind of committee?[/hide]

can non-SCers post in this thread? If not delete this:

[hide]
so by not writing the squares, we have:

$\binom{n}{0}\binom{n}{0} + \binom{n}{1}\binom{n}{1} + \dots + \binom{n}{n}\binom{n}{n} = \binom{2n}{n}$

Define the first term of the product to be the number of ways to walk from the top of the pascal triangle to the place where you take $i$ steps to the left in $n$ moves, and the second term the number of ways to walk from the position you are at to the place where you take $i$ steps to the RIGHT in $n$ moves.  Immediately we see that we end at the place $\binom{2n}{n}$.  To walk to the $2n$th row, you must be at the spots $\binom{n}{0}, \binom{n}{1}, \dots \binom{n}{n}$on the $n$th row.  Therefore the sum on the left completely covers all the ways to walk from the top of the triangle to the place $\binom{2n}{n}$
[/hide]


Warren buys limes at a cost of three for 50 cents. He then sells
them at a price of five for 90 cents. How many limes must he sell
to make a total profit of $ \$$2?

He buys 15 for 2.50 and sells 15 for 2.70. He makes 0.20 every 15, meaning he must sell $ 15*10\equal{}\boxed{150}$ to make a 2 dollar profit.


limit
real analysis
real analysis unsolved

Consider the set of all points in R^2 that lie outside the unit ball in $ l_n ^1(\mathbb{R})$ and inside the unit ball in $ l_n ^{\infty}(\mathbb{R}).$ Prove that every point in this region lies on the boundary of the unit ball in $ l_n ^{p}(\mathbb{R})$ for some p between 1 and $ \infty$.

I think I might have an analytic argument, but is there another way to do this?

When you say you have an analytic argument, you probably mean this same simple thing, but just in case:

Essentially (to be perfectly formal, we'd have to prove the following thing using the definition of a boundary and then we'd have to find closures of an open ball and its complement, but these are all easy problems), we want to prove that $ (|x|^p \plus{} |y|^p)^{1/p} \equal{} 1$ for some $ p > 1$, which is, of course, equivalent to $ |x|^p \plus{} |y|^p \equal{} 1$. We know that $ |x| < 1$ and $ |y| < 1$ and that $ |x| \plus{} |y| > 1$. Take $ f(p) \equal{} |x|^p \plus{} |y|^p \minus{} 1$. This is a one variable continuous function. We know that $ f(1) > 0$ and $ \displaystyle\lim_{x\rightarrow \plus{} \infty}f(x) \equal{} \minus{} 1$. (Since $ |x| < 1$ and $ |y| < 1$). So, there exists $ t > 1$ such that $ f(t) < \frac {1}{2}$. Therefore, by intermediate value theorem, there exists $ p > 1$ such that $ f(p) \equal{} 0$.


Fie A,B doua matrici patratice de ordinul n, n numar natural diferit de zero cu elemente complexe. Sa se demonstreze ca daca (AB-BA)^2=I_n atunci n este par. (I_n=matricea patratica unitate de ordinul n).

Valoriile proprii ale matricei AB-BA verifica ecuatia x^2=1, deci sunt in multimea {-1,1}. Cum Tr(AB-BA)=0 si Tr(AB-BA)=suma valorilor proprii, rezulta ca acestea sunt in numar par. Deci n este par.
Unde ai gasit problema?

A fost propusa de un profesor de la Facultate.


search
function
integration
real analysis
real analysis unsolved

$f$ is differentiable on $[a,b]$, with $f'(a)=f'(b)$.  Prove that there exists $c\in(a,b)$ such that

$f'(c)=\frac {f(c)-f(a)}{c-a}$

I am sure that I posted it before. You can make a search on the forum

It's easy to prove that since $f'(a)=f'(b)$, we can find some $t\in(a,b)$ s.t. $\frac{f(t)-f(a)}{t-a}=\frac{f(b)-f(a)}{b-a}$, and now just look at the function $g:\left[\frac{a+t}2,b\right],\ g(x)=\frac{f(x)-f(a)}{x-a}$. $g(t)=g(b)$ implies that there is some $c\in(t,b)\subset(a,b)$ s.t. $g'(c)=0$, and $g'(c)=\frac{f'(c)(c-a)-[f(c)-f(a)]}{(c-a)^2}$, so we're done (this is the $c$ we were looking for).

This is Fleet's mean value theorem. Anyway,Grobber, can you prove what you said at the begenning of the paragraph? ;)

I can, but there's a little mistake. It's not always true. For instance, we could have $f'(a)=f'(b)=\frac{f(b)-f(a)}{b-a}$, with the entire graph of $f$ being above the segment with endpoints $(a,f(a)),(b,f(b))$, but then we can just take $c=b$. In all other cases, what I said is true, and since it's almost obvious, I'll let other people prove it.

I have a proof  for a more general result with Fermat's and Cauchy's theorems. ;)
But your proof,except that case, is shorter and also very nice. ;)


Given that $ a,b,c$ are non-negative real numbers that satisfy $ a \plus{} b \plus{} c \equal{} 3$, find the maximum value of;
\[ a \plus{} ab \plus{} abc
\]
with the same condition ($ a \plus{} b \plus{} c \equal{} 3$ and $ a,b,c \geq 0$) , find the minimum and maximum of
\[ a^2 \plus{} b \plus{} bc
\]
Thanks.

hello, we have $ a\plus{}ab\plus{}abc\le4$ and the equal sign holds for $ a\equal{}2,b\equal{}1,c\equal{}0$.
Sonnhard.

hello, we have $ 0\le a^2\plus{}b\plus{}bc\le9$.
Sonnhard.

Thank you , Dr Sonnhard Graubner  :)

Can you solve this problem too ?

Give that $ a,b,c \geq 0$ and $ a\plus{}b\plus{}c\equal{}3$, and find the maximum of:

\[ \frac{a}{b}\plus{}\frac{a\plus{}b}{(a\plus{}c)(b\plus{}c)}
\]

Thanks.

hello, i think in this case exists no maximum.
Sonnhard.

yes sorry. you are right!  :blush:

because as $ b$ becomes smaller and smaller, the expression becomes bigger and bigger.
What about the following;

$ a\plus{}b\plus{}c\equal{}3$, find the [u]maximum and minimum[/u] of

\[ a^3\plus{}b^2\plus{}c
\]

I think maximum is 27, attained in $ a\equal{}3, b\equal{}0, c\equal{}0$

Am i right?

Please also tell me minimum value. Thanks.

[quote="shaaam"]yes sorry. you are right!  :blush:

because as $ b$ becomes smaller and smaller, the expression becomes bigger and bigger.
What about the following;

$ a \plus{} b \plus{} c \equal{} 3$, find the [u]maximum and minimum[/u] of
\[ a^3 \plus{} b^2 \plus{} c
\]
I think maximum is 27, attained in $ a \equal{} 3, b \equal{} 0, c \equal{} 0$

Am i right?

Please also tell me minimum value. Thanks.[/quote]
Your result is right, shaam (but I think you should give the addition condition is $ a,b,c \ge 0$). :) And we can prove it as follow:
$ a^3 \plus{} b^2 \plus{} c \le 9a \plus{} 3b \plus{} c \le 9a \plus{} 9b \plus{} 9c \equal{} 27.$
:)
For the minimum, we can use AM-GM balancing technique as follow:
For all $ u,v \ge 0,$ we have
$ a^3 \plus{} u^3 \plus{} u^3 \ge 3u^2a,$ with equality holds when $ a \equal{} u,$
$ b^2 \plus{} v^2 \ge 2vb,$ with equality holds when $ b \equal{} v$.
Therefore
$ a^3 \plus{} b^2 \plus{} c \ge 3u^2a \plus{} 2vb \plus{} c \minus{} 2u^3 \minus{} v^2.$
Now, choosing $ u \equal{} \frac {1}{\sqrt {3}},v \equal{} \frac {1}{2},$ we will obtain
$ a^3 \plus{} b^2 \plus{} c \ge a \plus{} b \plus{} c \minus{} \frac {2}{3\sqrt {3}} \minus{} \frac {1}{4} \equal{} 3 \minus{} \frac {2}{3\sqrt {3}} \minus{} \frac {1}{4} \equal{} \frac {11}{4} \minus{} \frac {2}{3\sqrt {3}}.$
We have equality holds when $ a \equal{} \frac {1}{\sqrt {3}},b \equal{} \frac {1}{2},$ and $ c \equal{} 3 \minus{} a \minus{} b \equal{} \frac {5}{2} \minus{} \frac {1}{\sqrt {3}}.$
:)

WOW Can_Hang!  :clap2:

Your solution is beautiful. Thank you!

So from your solution, i think we can find generalization. if, $ a,b,c\geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$, and $ m, n$ and $ k$ (lets just assume that $ m \geq n\geq k$) are positive integers. Then:
\[ a^m \plus{} b^n \plus{} c^k
\]
the maximum will be attained when $ a \equal{} 3$, $ b \equal{} 0$ , $ c \equal{} 0$

and minimum will be attained when $ a \equal{} \frac {1}{\sqrt [m \minus{} 1]{m}}$ and $ b \equal{} \frac {1}{\sqrt [n \minus{} 1] {n}}$

(because we can write

$ a^m \plus{} u^m \plus{} u^m \plus{} \cdots \plus{} u^m \geq mau^{m \minus{} 1}$ and

$ b^n \plus{} v^n \plus{} v^n \plus{} \cdots \plus{} v^n \geq nbv^{n \minus{} 1}$ and

$ c^k \plus{} w^k \plus{} w^k \plus{} \cdots \plus{} w^k \geq kcw^{k \minus{} 1}$ and then we should put

$ a \equal{} u \equal{} \frac {1}{\sqrt [m \minus{} 1]{m}}$ and $ b \equal{} v \equal{} \frac {1}{\sqrt [n \minus{} 1] {n}}$ )

can you please verify it to me?

sorry for double posting. Can_hang, i have some difficulties in understanding Am-Gm balancing method.
for the problem $ a^3 \plus{} b^2 \plus{} c$, i understood your solution completely  :)  it was very nice and easy.
but lets say $ a,b,c \geq 0$ , and $ a \plus{} b \plus{} c \equal{} 3$, i want to find minimum of
\[ a^4 \plus{} b^3 \plus{} c^2
\]
using Am-Gm balancing technique,

$ a^4 \plus{} u^4 \plus{} u^4 \plus{} u^4 \geq 4au^3$

$ b^3 \plus{} v^3 \plus{} v^3 \geq 3av^2$

$ c^2 \plus{} w^2 \geq 2cw$

so then, $ a^4 \plus{} b^3 \plus{} c^2 \geq 4au^3 \plus{} 3av^2 \plus{} 2cw \minus{} 3u^4 \minus{} 2v^3 \minus{} w^2$

therefore, we are going to put $ u \equal{} \frac {1}{\sqrt [3]{4}}$ , $ v \equal{} \frac {1}{\sqrt {3}}$, $ w \equal{} \frac {1}{2}$

but then $ a \equal{} u$, $ b \equal{} v$, $ c \equal{} w$ . however, the condition $ a \plus{} b \plus{} c \equal{} 3$ is not satisfied anymore.  :(

Can_Hang, please explain how to find minimum of $ a^4 \plus{} b^3 \plus{} c^2$

Thank you!

[quote="shaaam"]
so then, $ a^4 \plus{} b^3 \plus{} c^2 \geq 4au^3 \plus{} 3av^2 \plus{} 2cw \minus{} 3u^4 \minus{} 2v^3 \minus{} w^2$

therefore, we are going to put $ u \equal{} \frac {1}{\sqrt [3]{4}}$ , $ v \equal{} \frac {1}{\sqrt {3}}$, $ w \equal{} \frac {1}{2}$

but then $ a \equal{} u$, $ b \equal{} v$, $ c \equal{} w$ . however, the condition $ a \plus{} b \plus{} c \equal{} 3$ is not satisfied anymore.  :(

Can_Hang, please explain how to find minimum of $ a^4 \plus{} b^3 \plus{} c^2$

Thank you![/quote]
In this case, we will choose $ u,v,w$ so that $ 4u^3\equal{}3v^2\equal{}2w,$ my friend. ;) And we must choose $ u,v,w$ so that the equality holds, i.e.
$ 4u^3\equal{}3v^2\equal{}2w$ and $ u\plus{}v\plus{}w\equal{}3.$
Solving this system equalition, we will easily get the complete solution. :) I hope you understand the method, now. :)

[quote="can_hang2007"][quote="shaaam"]
so then, $ a^4 \plus{} b^3 \plus{} c^2 \geq 4au^3 \plus{} 3av^2 \plus{} 2cw \minus{} 3u^4 \minus{} 2v^3 \minus{} w^2$

therefore, we are going to put $ u \equal{} \frac {1}{\sqrt [3]{4}}$ , $ v \equal{} \frac {1}{\sqrt {3}}$, $ w \equal{} \frac {1}{2}$

but then $ a \equal{} u$, $ b \equal{} v$, $ c \equal{} w$ . however, the condition $ a \plus{} b \plus{} c \equal{} 3$ is not satisfied anymore.  :(

Can_Hang, please explain how to find minimum of $ a^4 \plus{} b^3 \plus{} c^2$

Thank you![/quote]
In this case, we will choose $ u,v,w$ so that $ 4u^3 \equal{} 3v^2 \equal{} 2w,$ my friend. ;) And we must choose $ u,v,w$ so that the equality holds, i.e.
$ 4u^3 \equal{} 3v^2 \equal{} 2w$ and $ u \plus{} v \plus{} w \equal{} 3.$
Solving this system equalition, we will easily get the complete solution. :) I hope you understand the method, now. :)[/quote]

Then we will have

$ w\equal{}2u^3$ and $ v\equal{}\sqrt{\frac{4u^3}{3}}$,

direct substitution yields:

$ u\plus{}\sqrt{\frac{4u^3}{3}}\plus{}2u^3\equal{}3$

which is equivalent to :

$ 12u^6\plus{}12u^4\minus{}40u^3\plus{}3u^2\minus{}18u\plus{}27\equal{}0$

now, how are we going to solve this?  :huh:

Thanks for attention.


I thought Christmas was the answer just like the guy (this question was on Russian Roulette in GSN)
I mean why New Year's? Christmas is much more depressing (well maybe not to people with tons of friends/lovers/family/etc. but those people don't suicide anyways)

So depressing are these times...*sigh*

Why not New Year's? There's far more significance there relating to beginnings or endings...if that made sense.

Well yeah...still why not New Year's Eve? :)

confuted I apologize for the inconvenience but can you split this and name it "When to suicide"? Thanks.

[quote="MysticTerminator"][quote="Tare"]Did you know that most people suicide on New Year's Day [/quote]

well it makes sense doesn't it?[/quote]

I rather doubt that the statement is true. In the United States, the peak months for suicide are in the spring and autumn. What is the source for the statement?

See

http://learninfreedom.org/suicide.html

for some surprising facts about suicide, with some citations of scholarly research on the subject.

i dont know why you would say new year's day, but winter makes some sense to me.  It gets dark a lot earlier and its the holidays, so it could be a tough time for people w/out friends or family.  Also people tend to be less active during the winter (colder, darker earlier) so they have less to do, gain weight, etc. in general more depressed.

also, in winter, it's dull and drearyand cold and all out depressing...I mean, there's no leaves on trees, and nothing's growing...

[quote="Tare"]Thank you.

I would personally choose the start of a school year if I have to and not on a holiday...I mean holidays are cheery and hopeful and stuff, some more than others I guess but still you get a day or two off which makes most people feel better.[/quote]

You wouldn't be feeling cheery and hopeful at all if you want to commit suicide.

Well for other people, and just like depression is "contagious" in a sense it can travel from one person to another happiness can too.

[quote="MysticTerminator"]now that you have fully contemplated the wonderful correlation between the seasons and that oh so exciting adventure of suicide, I feel compelled to ask wtf you're discussing it for...[/quote]
Well someone reminded us of Thanksgiving being almost over and I thought "hmm when's our next break?" and I answered "Christmas!" and then I got kinda depressed so I thought "yeah this reminds me of that [b]question on Russian Roulette on Game Show Network[/b] (answers tokenadult's question) about when most people suicide, I still think more people ought to :twisted: suicide on Christmas than New Year's Day" and here we are.

Mmm I'm not so sure about spring but autumn could provoke suicide, but I wouldn't die until I eat all the scrumtiliumtuous season foods of autumn, mackerals and chestnuts and...oh that just reminded me of food which I haven't eaten since last night...oooh...

[quote] winter makes some sense to me[/quote]

Valentine's Day is in winter.  :roll:

Of course, that could be a positive or a negative thing in terms of suicide depending on the person...

Ya, it's kind of like Christmas in a sense that non-suicidal people gets more non-suicidal and vice-versa unless you're in elementary school or something but then most of people there aren't suicidal...or are they? :?:

suicide isn't very common in elementary...

Fierytycoon wrote:Quote: winter makes some sense to me

Valentine's Day is in winter.

Of course, that could be a positive or a negative thing in terms of suicide depending on the person...

Fiery, don't tell me you're feeling suicidal over a girl.  I mean, [hide]writing sonnets about a certain someone was bad enough...[/hide]

Anyway, haven't yall heard about suicidal first graders?  They get sick of storytime, go outside, and immerse themselves in algebra.


Prove that for any possitive real numbers $a, b, c, d$ the following inequality holds
$\frac{bcd}{a^{3}+3bcd}+\frac{cda}{b^{3}+3cda}+\frac{dab}{c^{3}+3dab}+\frac{abc}{d^{3}+3abc}\leq  1$

Lemma:  For pos. reals x,y; Given xy=1, lets prove 1/(x+3) + 1/(y+3) <= 1/2.

Expand.  You get x+y >= 2 which is true.  The steps are reversible.

---

Now, take a^4 = w, b^4 = x, etc.  We get sum 1/(w+3) <= 1  after making each term solely in one variable.  By lemma, done.

Let ${a^{3}\over bcd}=x,\ldots$ then we prove $\sum{1 \over 3+x}\le 1$ for $xyze=1$. Just expand.

i have i nice solution, can you check out it?
The following ineq is equivalent to
$\sum_{}(\frac{1}{3}-\frac{bcd}{a^{3}+3bcd})\geq  \frac{1}{3}$
or $\sum_{}(\frac{a^{3}}{a^{3}+3bcd})\geq  1$
Now, by AM-GM inequality, we have
$\sum_{}(\frac{a^{3}}{a^{3}+3bcd})\geq  \sum_{}(\frac{a^{3}}{\sum_{}a^{3}})= 1$ :D


When I use the \usepackage[pdftex]{graphicx} it works, but the words
[pdftex]graphicx also show up. Is this normal? How can I eliminate it?
Also, is there anyway to copy their images from their pdf?

[quote="bpms"][quote="joml88"]Are you compiling as PDF?

To copy images you can use Prnt Scrn to take a picture of your screen, paste this into MS Paint, and crop what you want.[/quote]
In what file do you do this, and how? Thanks. Yes I compile as PDF but the created by go2pdf bar is a pain in the ...[/quote]

omg.... why do you need to use go2pdf?? are you using TeXnicCenter?

[hide="click this"]
[img]http://www.toolscenter.org/fileadmin/images/full.png[/img]
[/hide]

do you see the little drop-down menu at the top that says "LaTeX=>PDF"? :wink:

and also, do you see a button at the top of your keyboard, after F12 and before Scroll Lock that says "Print Screen"? Find the picture you want in Adobe Acrobat, click "Print Screen", click Start, click Run, type "mspaint", and press Ctrl+V (or edit->paste). then click the fill button and then click the select button and then select the part of the screen that you want. :wink:

Thanks for the help, but how do I get rid of the words [pdftex]graphicx? Thanks,

-BPMS

[quote="bpms"]Also, I read somewhere that we had to have complete solutions to at least two problems if we wanted to send them in. I just started them and I am unsure if I can completely solve two in this much time... Thanks.[/quote]Delete the word "complete" from what you just said.

We ask that you send work on at least 2 problems.  They don't have to be full solutions.  If you've made some progress on at least 2 problems, then by all means please submit whatever you've got.

[quote="chess64"][quote="bpms"][quote="DPatrick"]There must be an error message that LaTeX is spitting out that goes with this...without knowing that message, it's hard to tell what's going wrong.

Is your \usepackage command before \begin{document}?  It's not allowed to be after.[/quote]
Oh, I thought that we weren't allowed to alter the solutions document. I made it the first line in my solutions text. Thanks a bunch.

-BPMS
PS. Also, I read somewhere that we had to have complete solutions to at least two problems if we wanted to send them in. I just started them and I am unsure if I can completely solve two in this much time... Thanks.[/quote]

is it outputting as a pdf?[/quote]
Yeah. The only thing I was using go2pdf for was to insert pictures, but now I don't need it.

Just for reference (because I hate the line in go2pdf too although you can remove it by setting the latex boundary to the perfectly right place): you do NOT need pdf pics, JPG works just as well and there's no difference between coding except the name of the file becomes "name.JPG".

I hope it works with png too, because that's what I normally use.

And I just sent in my solutions with a png file.


$ m$ and $ n$ are positive integers $ \neq0$
$ x$ and $ y$ are strictly positive numbers.
what is the minimum of this expression
$ x^n\plus{}\frac{1}{x^ny^m}\plus{}y^{2m}$  when $ x$ an $ y$ change .

[quote="ghostchungho"]$ m$ and $ n$ are positive integers $ \neq0$
$ x$ and $ y$ are strictly positive numbers.
what is the minimum of this expression
$ x^n \plus{} \frac {1}{x^ny^m} \plus{} y^{2m}$  when $ x$ an $ y$ change .[/quote]

$ x^n \plus{} \frac {1}{x^ny^m} \plus{} y^{2m}\equal{}\frac{1}{2}x^n\plus{}\frac{1}{2}x^n\plus{}\frac {1}{2x^ny^m}\plus{}\frac {1}{2x^ny^m}\plus{}y^{2m}\geq5\sqrt[5]{\frac{1}{2}x^n*\frac{1}{2}x^n*\frac {1}{2x^ny^m}*\frac {1}{2x^ny^m}*y^{2m}}\equal{}5\sqrt[5]{\frac{1}{16}}$  ----->

$ x^{n}\equal{}\sqrt[5]{2}$ ,$ y\equal{}\sqrt[10]{\frac{1}{16}}$


let $a_1=a>0$ if $a_{n+1}=n\sqrt{a_n}+1$ Prove that $\lim_{n\to\infty}\frac{a_n}{n^2}=1$.

Denote $b_n=\sqrt{a_n}$, then you'll have $b_{n+1}^2=nb_n+1 (*)$. The growth of the sequence $b_n$ is not less then that of $a_{n+1}^2=na_n$ for which we have $a_n \sim n$, $n \to \infty$. On the other hand since $b_n >1$ for sufficiently large $n$, the growth of the sequence $b_n$ is not greater then that of $c_{n+1}=nc_n+c_n=(n+1)c_n$, for which we also have $c_n \sim n$, $n \to \infty$. Thus we have that $\lim_{n \to \infty} \frac{b_n}{n}=1$ which solves our problem.


4. Find all pairs of integers (x,y) such that : $ x^2 \plus{} y^9 \equal{} 33$
The problem in RWMO2 , x,y,z is a real numbers

I have a solution, but if you're submitting this, is this a contest problem in an ongoing contest?

If this is something due, then you should try to solve them on your own.  In fact, by doing so, you'll improve your problem solving skills, and so you would not encounter problems in the future.  It is very interesting that these questions were recently posted and solved.


Prove that F=mv^2/r where m-mass in kg, v-velocity in m/s, r-radius of circle motionin m, F is force on mass m in Newtons.

Bomb

[hide="hint"]
Firt prove that $a_n=\frac{v^2}{r}$[/hide]

Shobber isn't that asking the exact same thing? :\

When the mass in the circle turns in a very small angle $d\phi$, its velocity vector turns the same angle, with magnitude of velocity remaining constant. The magnitude of the vector of velocity change is then $dv = v.d\phi$.

Since force needed to accelerate is $F = m.a$,

$F = m\frac{dv}{dt} = mv\frac{d\phi}{dt} = mv\omega = m\frac{v^2}{r}$

PS: is there a special character for the d's denoting differentials in TEX? How do you normally write something like "a = dv/dt" in TEX?

[quote="kubus"]PS: is there a special character for the d's denoting differentials in TEX? How do you normally write something like "a = dv/dt" in TEX?[/quote]

There is a special character "\partial" for partial derivatives, $\frac{\partial F}{\partial x}$. For dv/dt. I usually use the "\text" character in front of d's, $a = \frac{\text d v}{\text d t}$. It can also be used for units: 10\ \text{kg} looks like $10\ \text{kg}$. For multiplication, the "\cdot" character looks better than the period, $F = m \cdot a$.

Ahoj, Yetti.


Let $ a$, $ b$, and $ c$ be positive integers such that $ ab|c(c^2 \minus{} c \plus{} 1)$ and $ c^2 \plus{} 1|a\plus{}b$.  Show that $ \{a,b \} \equal{} \{ c, c^2 \minus{} c \plus{} 1\}$.

This problem has been posted before
http://www.mathlinks.ro/viewtopic.php?p=233038#233038 .
And I read  an other solution ,using following lemma :
If x,y,n are three natural numbers such that $ \frac {xy}{x \plus{} y} > n$ then $ \frac {xy}{x \plus{} y}\geq n \plus{} \frac {1}{n^2 \plus{} 2n \plus{} 2}$ with the equality iff $ \{x,y\} \equal{} \{n \plus{} 1,n^2 \plus{} n \plus{} 1\}$


the equation of a circle is $ 9x^2\plus{}y^2\minus{}12x\plus{}15y\equal{}16$

is the radius 4 and the center at (12,-15). is it even a circle?

[quote="abacadaea"]the equation of a circle is $ 9x^2 \plus{} y^2 \minus{} 12x \plus{} 15y \equal{} 16$

is the radius 4 and the center at (12,-15). is it even a circle?[/quote]

$ \displaystyle 9\left(x^2 \minus{} \frac {4}{3}x\right) \plus{} (y^2 \plus{} 15y) \equal{} 16$
$ \displaystyle \left(x \minus{} \frac {2}{3}\right)^2 \plus{} \frac {\left(y \minus{} \frac {15}{2}\right)^2}{9} \equal{} \frac {305}{36}$

I believe that is an ellipse.

[quote="abacadaea"]is it an ellipse because the coeficient for x^2 is different from y^2?[/quote]
Yup. See [url]http://www.artofproblemsolving.com/Wiki/index.php/Ellipse[/url].

In that case, you would divide through by 9 and then complete the square to get a circle. An ellipse is just a distortion of a circle, where the coefficients of the $ x^2$ and $ y^2$ terms determine the minor/major axes of the ellipse.

That is a degenerate ellipse and has one or no solutions.

An ellipse is in the form $ \frac{(x \minus{} h)^{2}}{a^{2}} \plus{} \frac{(y \minus{} k)^{2}}{b^{2}} \equal{} 1$

so how should it look? and what is the answer?

[quote="abacadaea"][quote="abacadaea"]the equation of a circle is $ 9x^2 \plus{} y^2 \minus{} 12x \plus{} 15y \equal{} 16$

is the radius 4 and the center at (12,-15). is it even a circle?[/quote]
so if it were instead $ 9x^2 \plus{} 9y^2 \minus{} 12x \plus{} 15y \equal{} 16$, then what would it be? why do the coeficients matter? or how, rather.[/quote]
The equation would be $ \left(x\minus{}\frac{2}{3}\right)^2\plus{}\left(y\plus{}\frac{5}{6}\right)^2\equal{}\frac{105}{36}$.
Center at (2/3,-5/6) with $ r^2\equal{}\frac{105}{36}$.


Let $ ABCD$ be an harmonical quadrilateral which is inscribed in the circle  $ w$ . For a point  $ M$ which belongs to the line  $ AB$ define

the second intersection $ N$ of the line $ DM$ with the circle $ w$ and the intersections $ \left\|\begin{array}{c} P\in DM\cap BC \\
 \\
R\in CM\cap BN\end{array}\right\|$ . Prove that  $ P\in AR$ .

Denote E=DM∩AC, I=BN∩AC. The quadri ABCD is harmonic, and N lies on circle (ABCD), so the pencil N(ACBD)=-1, it follows that (ACIE)=-1, then (BRIN)=-1. Thus, AR,BC,EN are concurrent. hence result.


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advanced fields
advanced fields unsolved

If $ f_1(x): \equal{}|x|    (|x|\le \frac{1}{2} )$.  $ f_1$ will periodically continued $ f_1(x\plus{}1): \equal{}f_1(x\plus{}n): \equal{}f_1(x)$ for all $ n\in \mathbb{Z}$

If $ f_{1}(x\plus{}n): \equal{}f_{1}(x)$  for all $ n\in \mathbb{Z}$
If $ f_{2}(x): \equal{}f_{1}(4x)$,  thus $ f_{2}(x\plus{}\frac{1}{4})\equal{}f_{2}(x).$
If $ f_n(x): \equal{}f_1(4^nx)$,  thus $ f_n(x\plus{} \frac{1}{4^{n\minus{}1}})\equal{}f_n(x)$. Sei $ f(x): \equal{}\sum4^{\minus{}n}f_n(x)$. Then is $ f$ on $ \mathbb{R}$ continuing, but $ f$ is in no single point differentiable.

Fitim


ceiling function
symmetry
combinatorics unsolved
combinatorics

In a $m\times{n}$ grid are there are token. Every  token [i]dominates [/i] every square on its same row ($\leftrightarrow$), its same column ($\updownarrow$), and diagonal ($\searrow\hspace{-4.45mm}\nwarrow$)(Note that the token does not \emph{dominate} the diagonal ($\nearrow\hspace{-4.45mm}\swarrow$), determine the lowest number of tokens that must be on the board to [i]dominate [/i] all the squares on the board.

I can solve only the square case. The board can be dominated with $\left\lceil \frac{2n-1}{3} \right\rceil$ and this is the minimum.

For the rectangular case, I have only a criterion that gives a lower bound:

You need at least $\min\left(m, n, \frac{m+n-1}{3}\right)$ pegs.

However this is not optimal: 4x6 needs 4 pegs, but the criterion gives only 3.

This problem baffles me because I would think that cmmmrc's bound should be correct.  But for example, like cmmmrc says, a $4\times6$ board needs $4$ tokens, while a $5\times8$ only needs $4$.  Does anyone see a pattern in the types of boards that don't obey cmmmrc's bound?

Me, of course!!

I made a little step further, solving the cases when n-1 pegs are sufficient (n is the minor dimension).

It turns out that n x 2n-2 can be covered with n-1 pegs iff n is not 0 (mod 4). n x 2n-1 can never be covered with n-1 pegs.

So 4x6, 8x14, 12x22 etc... are cases when my lower bound is not optimal.

Moreover some other cases are there when things go bad. 7x9 should be covered with 5 pegs, but this is indeed impossible.

My guess is (but is just an impression, I may well be very wrong) that when 3 | n+m-1 and we are not in the square case or in the (n-1)-pegs case, the lower bound is not optimal, and probably one more peg does the trick. Just my guess, beware.

M.

I don't why I didn't look before, but Kalva's homepage states that we only need solve the $n\times n$ board.  Thus cmmmrc's bound solves the problem.

It is still interesting to see if a solution exists for an $m\times n$ board.  Maybe this string should be moved to the open problems section.

Assume that $ k$ pegs dominate the entire board, and call "naughty" the squares that are not in a row or column where a peg lies (ie, those that are not dominated horizontally or vertically, but are dominated diagonally).  Now, clearly there are $ (m \minus{} k)(n \minus{} k)$ naughty squares, that we distribute in diagonals $ \searrow\hspace{ \minus{} 4.45mm}\nwarrow$, which in turn we number starting from the lower left corner of the board.  Clearly, the first and last such diagonals contain at most one square; if the first diagonal contained two squares, the square that is in the intersection of the row of one and the column of the other, lying lower and left, would also be naughty, and its diagonal $ \searrow\hspace{ \minus{} 4.45mm}\nwarrow$ would be lower and left from the first one, contradiction - similarly for the last diagonal.  The second and next to the last diagonals contain at most two naughty squares, in the intersection of their row and column lie the squares in the first and last diagonals, and so on.  Hence naughty squares are covered diagonally in bunches of $ 1,1,2,2,3,3,\dots$, ie, with the first $ 2n \minus{} 2k \minus{} 2$ pegs we dominate at most $ (n \minus{} k)(n \minus{} k \minus{} 1)$ naught squares.  Hence, assuming $ n\leq m$ since otherwise we perform a symmetry around diagonal $ \searrow\hspace{ \minus{} 4.45mm}\nwarrow$ leaving the problem unchanged except for reversing $ m,n$, we find that $ (n \minus{} k)(n \minus{} k \minus{} 1) < (m \minus{} k)(n \minus{} k)$, and we need additionally $ (m \minus{} k) \minus{} (n \minus{} k \minus{} 1) \equal{} m \minus{} n \plus{} 1$ pegs to dominate each $ n \minus{} k$ naughty squares (since each diagonal contains $ k$ squares dominated horizontally, hence not naughty).  Hence the number of pegs satisfies $ k\geq2n \minus{} 2k \minus{} 2 \plus{} m \minus{} n \plus{} 1$, or $ 3k\geq m \plus{} n \minus{} 1$ and $ k\geq\lceil\frac {m \plus{} n \minus{} 1}{3}\rceil$, where $ \lceil x\rceil$ is the least integer larger than or equal to $ x$.  Note that this is NOT necessary if $ m$ or $ n$ are less than this value, since then we need just one peg per row or column to dominate the entire board, and no naughty squares exist.

The original problem only asks for you to consider the square case ($m \times m$).


let $XOY$ be a constant angle and $A$ a constant point on $OX$ . the line $L$ passing throw $A$ meets the $OY$
in $B$.the circle $C$, with the center on $B$ and radius$BO$, intercepts the $L$ in $D,C$.

find the locus of the point $M$ by the moving of line $L$ such that:
1.$M$ be on the line$L$
2.$D$ be between the $M$ and $A$
3.$\frac{MD}{MC}=\frac{DA}{AC}$( points $M,A,D,C$ are called harmonic points)

Try to prove this (it should be a very simple computation):

If $X,Y,Z,T$ form a harmonic cross-ratio and are in this order on a line, and $U$ is the midpoint of $XZ$, then $UX^2=UZ^2=UY\cdot UT$.

We apply this to our configuration: $B$ is the midpoint of $CD$, so $BO^2=BC^2=BA\cdot BM$. This shows that $M$ lies on the circle through $A$ and $O$ which is tangent in $O$ to $OB$.


Let $a_1, a_2, . . . , a_{1999}$ be a sequence of nonnegative integers such that for
any $i, j$ with $i + j \leq 1999$,
$a_i + a_j \leq a_{i+j} \leq a_i + a_j + 1$.
Prove that there exists a real number $x$ such that $a_n = [nx]$ for each $n$.

Iura solved it long time ago in the forum! He proved that the intervals of working $x$ were not disjoint

Can you give link? I don't know what to look for exactly  :?


Problems solving this:

An object traveling around a circle with a radius of 5 inches.  If in 50 seconds a central angle of 1/2 radian is swept out, what is the linear speed of the object?

Recall that linear velocity $ v$ is given by $ v\equal{}r\omega$, where $ r$ is the radius and $ \omega$ is angular velocity.  Now you have all the information you need.

So I would take .5/50 seconds x 5 inches

could i leave the answer in in/sec or would i have to put it into mph

I have no idea what the problem says so if it doesn't specify units, you could leave it in inches/second or convert to miles/hour or meters/sec or angstroms/picosecond (or whatever you want).