AMC
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Here is a slightly different solution to #10.  [See the Math Jam transcripts for the Web Site owners' excellent solutions.]

[quote]10. Let S be the set of integers between 1 and 2^40 whose binary expansions have exactly two 1's. If a number is chosen at random from S, the probability that it is divisible by 9 is p/q, where p and q are relatively prime positive integers. Find p + q.[/quote]

We are looking for two powers of 2 (2^0 through 2^39) whose sum is divisible by 9.  We'll do this with modular arithmetic.

2^0 =  1 == 1 (modulo 9)
2^1 =  2 == 2 (modulo 9)
2^2 =  4 == 4 (modulo 9)
2^3 =  8 == 8 (modulo 9)
2^4 = 16 == 7 (modulo 9)
2^5 = 32 == 5 (modulo 9)
2^6 = 64 == 1 (modulo 9)

The pattern repeats with a cycle length of 6.

[code]<Residue>  <#powers of 2>  <list of powers of 2>
   1     7   0,  6, 12, 18, 24, 30, 36
   2     7   1,  7, 13, 19, 25, 31, 37
   4     7   2,  8, 14, 20, 26, 32, 38
   8     7   3,  9, 15, 21, 27, 33, 39
   7     6   4, 10, 16, 22, 28, 34
   5     6   5, 11, 17, 23, 29, 35[/code]

Find all pairs of residue, which sum to 9:  1+8, 2+7, and 4+5.

There are 7*7 + 7*6 + 7*6 such pairs.  49+42+42=133.  To finish the problem, the total number of combinations of 2 bits is 40C2=780.  p/q=133/780.  p+q=913.

[quote]15. A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness.   Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?[/quote]
I attacked this as a "shuffling" problem.  The following "magic formula" was discovered after observing the patterns created by folding strips of length 8 and 16.  [I'll explain why it works, after a few days, if you folks don't explain it, first.]

1.  Number the unit squares from 0 to 1023.  We are looking for the position of the 942nd square, which is numbered 941.
2.  Convert to base 2.  941_10 = 1110101101_2
3.  Reverse the bits:  1011010111
4.  Remove the high-order bit, append a low-order zero-bit:  0110101110
5.  If the removed bit was "1", toggle all of the bits:  1001010001
6.  Convert to base 10:  1001010001_2 = 593_10

[quote]13. Let ABCDE be a convex pentagon with AB || CE, BC || AD, AC || DE, < ABC = 120 degrees, AB = 3, BC = 5, and DE = 15. Given that the ratio between the area of triangle ABC and the area of triangle EBD is m/n, where m and n are relatively prime positive integers, find m + n.
[/quote]

http://www.artofproblemsolving.com/Classes/AIME/Images/2004aimeB13diags.gif

Refer to the diagram posted in Math Jam above, extend BA and DE to meet at P and extend BC and ED to meet at Q, then both ACEP and ACQD are parallelograms. So PE=DQ=AC=7. PQ=7+15+7=29

The ratio of areas [ABC] / [PBQ] = (AC/PQ)^2 = (7/29)^2, since ABC is similar to PBQ.

The ratio of areas [EBD] / [PBQ] = ED/PQ = 15/29, since EBD and PBQ have the same heights.

Therefore, [ABC] / [EBD] = (7/29)^2 / (15/29)  = 49 / 435

Problem 7.  Let B'D be x.  AB' = 15, C'F = 3, DF= 22, and B'C' = x + 15.  Note that B'DC'F forms a cyclic quadrilateral, with diameter B'F.  Thus x^2 + 22^2 = (B'F)^2 and (x+15)^2 + 3^2 = (B'F)^2.  Solve these for x = 25/3 and the perimeter is 290/3.  m + n = 293

#7, a different way to do.

[hide]The crease EF should perpendicularly bisect BB'. We already know one dimension of the perimeter 17+8=25.  DF=22, and BF=FB'

and BE=EB'. Therefore, we automatically know EB'=17. Pythagorean Theorem we have AB=15.  Call AD x, BD=x-15 and BC=x.

Pythagorean Theorem

(x-15) :^2: +22 :^2: =(B'F) :^2:

x :^2: +9=BF :^2:

set them equal

x :^2: +9=(x-15) :^2: +22 :^2:

9=225+484-30x

30x=700

x=70/3

Perimeter=

(25+(70/3))*2= 290/3

the answer is 293[/hide]

#15.  Folding a strip of paper, 2n units wide.

I previously posted a "magic formula" for solving this question.  So far, there have been no takers for explaining why the formula works.  So here goes...

Let's tackle the first part of the problem.

Let L be the number of squares to the left of our target square.  (For this half of the problem, we won't concern ourselves with the vertical position of the target.)  Initially, L is a 10-bit number, 0 <= L < 1024.  After one fold, L is a 9-bit number, 0 <= L < 512.  After 9 folds, L is a 1-bit number, 0 <= L < 2.  After 10 folds, L=0.

For clarity, we will denote these successive values of L as L10, L9, L8, ..., L0.

At each stage of the folding, the new value of L depends solely on the old value of L and whether the target is in the left half or the right half of the horizontal strip.  We define the "Folding Sequence", F, as follows:

F(n-1) = 0, if Ln < 2(n-1)
F(n-1) = 1, if Ln >= 2(n-1)

Now, we can define L recursively, as follows:
L(n-1) = Ln, if F(n-1) = 0
L(n-1) = (2n-1)-Ln, if F(n-1) = 1

The top line of the definition essentially says we drop the high-order bit of L if the high-order bit was 0.
The bottom line of the definition essentially says we drop the high order bit of L and complement the remaining bits, if the high order bit was 1.

Clearly, after 10 operations, L has been reduced to zero.

The Folding Sequence bits can be combined into a single value as follows:
F = :Sigma:(i=0 to 9, Fi x 2i)

Just as there were 210 distinct possible values of L, there are 210 distinct possible values of F.  And just as L uniquely identified the target square, F also uniquely identifies the target square.

We'll consider the problem statement as an example.  The 942nd square has 941 squares to the left, so L=941.
[code]
Initially, L10=1110101101_2.
   F_9=1.  L_9= 001010010_2.
   F_8=0.  L_8=  01010010_2.
   F_7=0.  L_7=   1010010_2.
   F_6=1.  L_6=    101101_2.
   F_5=1.  L_5=     10010_2.
   F_4=1.  L_4=      1101_2.
   F_3=1.  L_3=       010_2.
   F_2=0.  L_2=        10_2.
   F_1=1.  L_1=         1_2.
   F_0=1.  L_0=[zero]
[/code]
F=10011110112 = 63510 = 512+64+32+16+8+2+1

Now, I refer you to any standard explanation of "Gray Code" (sometimes known as "Reversing Gray Code").  Note that 1110101101binary = 1001111011gray.

The folding sequence, F, is identical to the Gray Code equivalent of the initial position, L!  And the algorithm to obtain F from L is equivalent to the algorithm to obtain Gray Code from binary!

Now, let's tackle the second part of the problem.

Let's assume the paper has been completely folded, such that the stack is 1024 squares high and one square wide.  We will consider unfolding the paper.  Let B be the number of squares below our target square.  (For this half of the problem, we won't concern ourselves with the horizontal position of the target.)  Initially, B is a 10-bit number, 0 <= B < 1024.  After one unfold, B is a 9-bit number, 0 <= B < 512.  After 9 unfolds, B is a 1-bit number, 0 <= B < 2.  After 10 unfolds, B=0.

For clarity, we will denote these successive values of B as B10, B9, B8, ..., B0.

At each stage of the unfolding, the new value of B depends solely on the old value of B and whether the target is in the bottom half or the top half of the vertical strip.  We define the "Unfolding Sequence", U, as follows:

U(n-1) = 0, if Bn < 2(n-1)
U(n-1) = 1, if Bn >= 2(n-1)

Now, we can define B recursively, as follows:
B(n-1) = Bn,		if U(n-1) = 0
B(n-1) = (2n-1)-Bn,	if U(n-1) = 1

The top line of the definition essentially says we drop the high-order bit of B if the high-order bit was 0.
The bottom line of the definition essentially says we drop the high order bit of B and complement the remaining bits, if the high order bit was 1.

Clearly, after 10 operations, B has been reduced to zero.

The Unfolding Sequence bits can be combined into a single value as follows:
	U = :Sigma:(i=0 to 9, Bi x 2i)

Just as there were 210 distinct possible values of B, there are 210 distinct possible values of U.  And just as B uniquely identified the target square, U also uniquely identifies the target square.

[Does any of this look familiar?]

We'll consider the problem's answer as an example.  When the strip is completely folded, there are 593 squares below our target, so B=593.
[code]
Initially, B10=1001010001_2.
   U_9=1.  B_9= 110101110_2.
   U_8=1.  B_8=  01010001_2.
   U_7=0.  B_7=   1010001_2.
   U_6=1.  B_6=    101110_2.
   U_5=1.  B_5=     10001_2.
   U_4=1.  B_4=      1110_2.
   U_3=1.  B_3=       001_2.
   U_2=0.  B_2=        01_2.
   U_1=0.  B_1=         1_2.
   U_0=1.  B_0=[zero]
[/code]
U=11011110012 = 88910 = 512+256_64+32+16+8+1

Again, I refer you to any standard explanation of "Gray Code".  Note that 1001010001binary = 1101111001gray.

The unfolding sequence, U, is identical to the gray code equivalent of the final position, B !!!

Now we can tackle the final part of the problem.

What is the relationship between the the folding sequence and the unfolding sequence?  They are the reverse of each other.  [Using the problem as an example, F=10011110112.  U=11011110012.]

Now, we can apply three simple transformations to solve the problem:

	L-->F-->U-->B

	F = binarytogray(L)
	U = bitreverse(F)
	B = graytobinary(U)

The "magic formula" in my previous post was merely the combined effect of these 3 simple operations.

You made it look so simple the first time.


modular arithmetic
group theory
number theory unsolved
number theory

show that there are infinitely many composite $n$ such that for all $a$ with $\gcd(a,n)=1$, we have $a^{n-1}\equiv a\pmod{n}$.

Let $n=p_{1}^{k_{1}}...p_{m}^{k_{m}}$. Your condition $a^{n-1}\equiv a(mod \ n) \ \forall a: \ (a,n)=1$ equavalent to $lcm((p_{1}-1)p_{1}^{k_{1}-1},...,(p_{m}-1)p_{m}^{k_{m}-1})|n-2$. Therefore $n=2^{a}p_{1}...p_{m},a\le 2$ all $k_{i}=1$ for odd prime divisors $p_{i}$. Obviosly if n odd, then n=1.
Obviosly $n=2p$ work, because $(p-1)|2p-2$.

[quote="Rust"]. Your condition $a^{n-1}\equiv a(mod \ n) \ \forall a: \ (a,n)=1$ equavalent to $lcm((p_{1}-1)p_{1}^{k_{1}-1},...,(p_{m}-1)p_{m}^{k_{m}-1})|n-2$. [/quote]

i dont understand this. why is this true?

suppose that there is a finit set of as numbers then use Fermat's theorem.


quadratics
geometry
geometric transformation
rotation
dilation
search
vector

Let $ x$, $ y$, and $ z$ be real numbers such that $ 6x\minus{}9y\plus{}7z\equal{}2$ and $ 7x\plus{}2y\minus{}6z\equal{}9$. Evaluate $ x^2\minus{}y^2\plus{}z^2$.

I hope you remember [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170124&search_id=1840578115]this thread[/url].  :)

Surely! I had actually seen this problem before I wrote the comment that thread.

Is there an easy way to characterize all quadratic forms that are ruled surfaces? (I feel like there should be something with semidefinite matricies but I haven't thought about it.. xT A x form, etc)

You gave a good geometric intuition of how to interpret the answer in that thread.

[hide="Let me give another"] This system can be written as

$ \left[ \begin{array}{cc} 7 & \minus{} 6 \\
6 & 7 \end{array} \right] \left[ \begin{array}{c} x \\
z \end{array} \right] \equal{} \left[ \begin{array}{cc} 9 & \minus{} 2 \\
2 & 9 \end{array} \right] \left[ \begin{array}{c} 1 \\
y \end{array} \right]$.

The matrices in question are the composition of two different rotations and dilation by $ \sqrt {6^2 \plus{} 7^2} \equal{} \sqrt {2^2 \plus{} 9^2} \equal{} \sqrt {85}$, so it follows that $ (x, z)$ and $ (1, y)$ are at the same distance from the origin (and the exact same interpretation is possible in the other thread).

Of course, where there are rotation matrices there are complex numbers.  We can also write this system as $ (7 \plus{} 6i)(x \plus{} zi) \equal{} (9 \plus{} 2i)(1 \plus{} yi)$. [/hide]
As for your question, a quick search doesn't turn up anything nice.

[hide="thoughts"]Let $ x$ be a $ n$ by $ 1$ column vector with variables $ x_1$, $ x_2$, ..., $ x_n$. Let $ A$ be a $ n$ by $ n$ symmetric matrix. Then $ x^T A x$ gives us the quadratic forms in those $ n$ variables with no degree one or zero terms. Let $ x \equal{} v_1 \plus{} v_2 t$ be the line where $ t$ is a real parameter. Then
\[ (v_1 \plus{} v_2 t)^T A (v_1 \plus{} v_2 t) \equal{} v_1^T A v_1 \plus{} v_2^TAv_2 \plus{} t\cdot (x_1^T Ax_2 \plus{} x_2^TAx_1)
\]
So we'd like that second factor to be zero. I'm not sure how to see if a matrix has that property or not. I'm not really ontop of my linear algebra techniques.

Also I'm not even if we take a real ruled surface, does there exist complex vectors $ v_1$, $ v_2$ such that this works. I don't know. [/hide]

[hide="When you put it that way..."] $ v_1^T A v_2 \equal{} (v_2^T A v_1)^T$, and since both are scalars we require that $ v_1^T A v_2 \equal{} 0$.  In other words, $ v_1, v_2$ are orthogonal with respect to the [url=http://en.wikipedia.org/wiki/Inner_product]inner product[/url] $ \left < v_1, v_2 \right > \equal{} v_1^T A v_2$.

This is true of every pair of eigenvectors associated to distinct eigenvalues of $ A$.  Here's why:  if $ A v_1 \equal{} \lambda_1 v_1$ and $ A v_2 \equal{} \lambda_2 v_2$ then $ (A v_1)^T \equal{} v_1^T A \equal{} \lambda_1 v_1^T$ and

$ v_1^T A v_2 \equal{} v_1^T \lambda_2 v_2 \equal{} \lambda_1 v_1^T v_2$.

If $ \lambda_1 \neq \lambda_2$, the result follows. [/hide]
It's late at night, though, so I'm not sure that first expansion you've got is valid...

[hide="Wow yea"] it is

\[ (v_1 \plus{} v_2 t)^T A (v_1 \plus{} v_2 t) \equal{} v_1^T A v_1  \plus{} t\cdot (x_1^T Ax_2 \plus{} x_2^TAx_1)\plus{}t^2\cdot (x_2^T Ax_2)
\]

This is interesting then since obviously $ A$ cannot positive or negative definite.

Also $ A$ doesn't always define an inner product because the positivity axiom or whatever isn't necessarily satisfied. [/hide]

This problem is pretty much a copy of the old AMC 10 or 12 problem (can't remember which).

[hide]As I pointed out in the other thread, hyperboloids of one sheet are ruled surfaces...so it isn't like the AMC was the first one to come up with this type of problem. [/hide]

Since I'm not so good with those tools yet, is there a way to solve this using diophantine equations? I remember an AIME problem this year that was very similar, and that could be solved with diophantines (although I'm not sure that would make the proof rigorous though).
EDIT: Oh shoot it says "real numbers"...can someone post one anyways?

What do you mean?  This isn't a Diophantine problem.

[quote="1=2"]Could we possibly solve for x in terms of y and z in both equations, set them equal, find y in terms of z, then find x in terms of z, and square?[/quote]

This is what I did, and it worked.

You don't need to do anything advanced, but it is interesting to talk about the more advanced interpretations.

I like to think of talking about the advanced interpretations in terms of motivation and problem [i]writing[/i] rather than problem solving.  There is no reason here, [i]a priori[/i], to expect $ x^2 \minus{} y^2 \plus{} z^2$ to be a constant independent of what solution $ (x, y, z)$ we pick (which determine a line).


I have no clue why, but this thread is hilarious.

Umm, if you know where the nest is, you could pour boiling water in it, in hopes of killing the queen? Better run fast, though.

Or for fun, on a sunny day, you could get a magnifying glass and...
no, that's cruel.

[quote="helenamei"]any way to get rid of those d*** creatures without using chemical insectide?[/quote]

just get away from them

buy ant eaters

use a water hose to constantly spray on the ants

use flies, yes flies, i saw it on tv, ant killing flies that lay eggs in the ants and causes them to die  :D

i like the magnifying glass idea too, just get bigger ones, or...

FIRE!!!

pour any flammable liquids, alcohol, gasoline... or charcoal or just dry stick, and set it on fire.

fun.

You could bomb the whole place, assuming there's nothing you [b]don't[/b] want to get rid of...

hum.. i haven't though of that, but 20 pounds of tnt will work great.

You should try http://www.wikihow.com/Get-Rid-of-Ants. Apparently baby powder is one method.

Now if this were in G&FF, I would be ready with sarcastic ideas involving wormholes, quantum tunneling, and antimatter.

[quote="K81o7"][quote="mdk"]you guys are evil.[/quote]

Why do you say that?  I'd never hurt an ant.  As far as I'm concerned, my notion is entirely hypothetical: you have to admit, it would serve the asked-for purpose![/quote]

I guess it would, wouldn't it?

[quote="youdonotknowme"]

use a water hose to constantly spray on the ants

[/quote]

Actually, I tried that before. Ants' tunnels are designed to let them survive floods (it would kind of suck if you had to rebuild your home every time it rained). Also, once you spray them, all your giving them is giving them is a drink, and even more ants will come out.

i've tried putting them in water, but it always seems that surface tension keeps them from dying. soap takes care of that.

If they're invading your kitchen, another solution is simply to always clean up very well -- when they aren't able to find food, they'll stop bugging you.

funny, i never had problems with ants, but with roaches...
 :furious:  :cursing:  :mad:

[quote="mathnerd314"]You should try http://www.wikihow.com/Get-Rid-of-Ants.[/quote]

Hmm, I've tried tracking down all the holes in my backyard, but my backyard is really large, so I probably haven't even found all of them (I've found literally 40 already).

anyone has problem with fireants?
if you don't have fireants where you live, consider your self very lucky

why don't you just stomp on them?  :D  in my opinion that's the most effective way to do get rid of the ants, quickly and decisively  :)

not when:

1) There are crawling on a shelf or something.
2) There are swarms of them.
3) Stomping on the line of ants causes them all to scatter, and it becomes even messier

i dont believe in hurting wildlife :( , unless they annoy me to death, but in general, if they dont bother me, nobody gets hurt :lol:

Ants taste pretty good  :rotfl:

I spray them with orange spray

Actually, many people do eat ants

http://en.wikipedia.org/wiki/Entomophagy

also, see

http://en.wikipedia.org/wiki/Ants#Ants_as_food

Check this out:

http://content.answers.com/main/content/wp/en-commons/thumb/a/a2/360px-Insect_food_stall.JPG

MMMM! Delicious, deep fried insects!

Well It is quite expected. Many people eat food that is odd. Like in the rural part of Indonesia, the people consider Tarantulas to be a steak.

Yeah, I've heard of that before. Ever tried snails? Or pork blood? Pork blood kinda tastes weird, I don't like it. Chicken liver on the other hand is quite delicious. (I'm serious about this)

Snails, when prepared in the French manner ([i]escargots[/i]) are quite good -- the snail itself doesn't have much flavor, but it's nice and chewy and typically covered/sitting in a delicious sauce of some kind.

The first time I had escargot was with a friend while we were in France.  He ordered as his main cource beef tartar, which is a raw beef dish.  He didn't speak any French (and I speak only poorly), so the waitress had a somewhat difficult time making sure that he actually knew what he was ordering.  (He did.)

Oh well, this is getting off topic. This is about getting rid of ants, not about disgusting foods.

I'm sorry to say this, but i think pesticide is still the fastest and most efficient ant killer, although not the best for the environment.

I don't think fire would help. I mean it would get rid of the ants, but your house would be burned down, you may be accused of being an arsonist, and be sent to jail.

stop planning to kill the ants! they have feelings too, ya know! how would u feel if u were te ant & ppl stepped on u?

yr being like the ant bully!

actually, are you sure ants have feelings? as far as i know the only animal that has been proved to have be self - conscious are humans and chimps. Ants arent aware that they are living.

i dont care, if u were an ant, would u lke to be stepped on?

[quote="Nerd_of_the_Ages"]yeah, im glad alright. Once ants invaded the room, but i was too lazy get rid of them. I kept on playing gamecube, and by the time i noticed ants were all over my leg, i got bitten like 10 times. Good thing those werent fire ants....[/quote]

ouch! good thing those werent fireants!! :rotfl:

hmmmm...very interesting....a thread about ants...hehe :roll:

[quote="Nerd_of_the_Ages"]I wouldnt care, because if i were an ant, i wouldnt know that i was an ant.[/quote]

LOL.

The best way to get rid of ants is to instruct them that they are invading the privacy of your home (it may have been their home before your home was built, but who cares about that?) and tell them to please leave you alone. Or...you could step on them. Either one.

Well no one posted this one so:

Put butter on your back, and have the ants carry you to their home. Then you can attack them from the  inside.

I don't get the "attack them from the inside" part.

[quote="hunter34"]By that that means blow up the ant hill with dynamite and sell the charred remains on ebay.[/quote]

Don't forget to also sell the charred remains of your house  [hide=":roll:"] Congratulations on finding the secret message [/hide]. Also, good luck stuffing a dynamite into an ant tunnel  :roll: .

[color=white]Click the first roll[/color]


[color=darkred]I hads a trigonomtric usage . Can you resolve it by non-zero ? orientate demob of a Situated friend topic " new " error then
Let  $a,b,c$ be sides of a non-obtuse triangle  . Prove that
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/color]

[quote="nguyenphihung"]
Let  $a,b,c$ be sides of a non-obtuse triangle  . Prove that
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130812

But In coequal role those article, as for who dice me differ, it is there much more powerful!


Show that for any $a,b,c$ positive reals,

\[ (a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}  \]

[quote="Johann Peter Dirichlet"]Show that for any $a,b,c$ positive reals,

\[ (a+b)(a+c) \geq 2 \sqrt{2}{abc(a+b+c)}  \][/quote]
$a=b=c=1$ :lol:

yah this does not hold for a=b=c=1.
  u cited wrongly. see it again.

I think it's
\[ (a+b)(a+c)\geq 2 \cdot\sqrt{abc(a+b+c)}  \]

The problem indeed is easy
By AM-GM, we have
\[ (a+b)(a+c) = bc + a(a+b+c) \geq 2\sqrt{abc(a+b+c)} \]

Well, this is my solution:

We can suppose a=1 because the inequality is homogeneous. Then

$(1+b)(1+c) \geq 2 \sqrt{bc(1+b+c)} \Leftrightarrow \frac{(1+b+c)+(bc)}{2} \geq \sqrt{(1+b+c)(bc)}$
which follows by AM-GM!

By AM-GM,

\[(a+b)(a+c) = bc + a(a+b+c) \ge 2\sqrt{bc(a(a+b+c))}=2\sqrt{abc(a+b+c)}\]

[quote=MariusBocanu]What about squaring and writing like $c^2(a-b)^2+c(2a^3-2ab^2)+(a^4+a^2b^2+2a^3b)\ge0$. It is easy to prove that the discriminant is not positive,etc.[/quote]

Umm for $a \leq b$?

Construct a triangle $ABC$, $I$ its incircle, let $I$ intersect $AB$, $BC$ and $CA$ at $F$, $D$ and $E$, and $AF=a=AE$, $BF=b=BD$ and $CD=c=CE$. Then the area of $ABC$ is $\sqrt{abc(a+b+c)} \]=\frac { (a+b)(a+c) }{ 2 } \sin { \angle BAC } \le \frac { (a+b)(a+c) }{ 2 } $.
Equality case is trivial.
So, we´re done.

$a,b,c$ are positive real so by AM-GM we have:
$\frac { (a+b)(a+c) }{ 2 } =\frac { { a }^{ 2 }+ac+ab+bc }{ 2 } =\frac { a(a+b+c)+bc }{ 2 } =\sqrt { abc(a+b+c) }$


LaTeX
inequalities
inequalities unsolved

Nice proof! I tried to use that method to prove the obvious generalisation but didn’t succeed. Anyway, I’ve found another solution that proves the following generalisation:
If $n\ge 3$, $x_1,\ldots,x_n>0$ and $\sum_i x_i^{n-1}=n$ we have
$\sum_i x_i^{n-2}\geq \prod_i x_i^{n-2}\cdot \sum_i \frac1{x_i^{n-2}}$ :)
Proof: By the monotonicity of the generalized means, we have $\left(\frac{\sum x_i^{n-2}} n \right)^{\frac1{n-2}} \leq\left(\frac{\sum x_i^{n-1}} n \right)^{\frac1{n-1}}=1$, so $\frac{LHS}n\leq 1$.
By the “symmetric means ineq” ( ) putting $k=n-1$, $s=n-2$ and $r=\frac{n-2}{n-1}$, we have
$\left( \frac{\sum (x_1\cdots x_{n-1})^{n-2}}{n}\right)^{\frac 1{n-2}} \leq \left( \frac{\sum x_i^{n-2}}{n} \right)^{\frac{n-1}{n-2}}$, thus
$\left (\frac{RHS}n)^{\frac 1{n-2}}\leq (\frac{LHS}n \right) ^{1+\frac1{n-2}}\leq \left(\frac{LHS}n \right) ^{\frac1{n-2}}$ and the result follows :). I think it’s quite nice.

Note that according to this proof, the condition on the $x_i$ can be weakened to only require : “$\sum_i x_i^t=n$ for a certain $t\geq n-2$”. And the ineq remains still valid.
I wonder if there are other inequalities with this type of restriction where the same applies : that it is in fact sufficient to require $\sum_i x_i^t=n$ for “a certain” $t\geq t_0$ instead of requiring it for a given fixed $t$.
(Note this can only make sense if $x_1,\ldots,x_n$ are normalized in a way that we have $RHS=n$, not $1$ etc.)


For IE, I think you can use Google toolbar.

Also, you could use Firefox instead of IE, which has a built-in pop-up blocking ability.

I think all versions of Mozila have one built-in. (including firefox)

I had this problem a while ago, and installing 3-4 different pop-up blockers didn't to anything.  The best pop-up blocker there is (in my opinion) is called WebWasher.  If that doesn't work, there may be spyware on your computer.

[quote="nr1337"]there may be spyware on your computer.[/quote]
Indeed, that sounds like the case. Download, install, and run the latest version of Ad-Aware (it's free).

i would download spybot too. its a little different from ad-aware. also, do you have norton antivirus? that would check for trojans....

[quote="Rep123max"]wow, how does all this stuff get on computers.[/quote]

Sometimes free software that you download off the internet automatically installs unwanted programs without telling you.

and sometimes when you go to sites, you can get cookies. if you ran spybot and you got something from advertising.com, it was probably a cookie. its kinda scary, they can put your whole name and everything out there.

Rep, for the ones you can't get off, try restarting your computer and scanning right away (if you use Ad-Aware, it has an option that let's it scan at the moment you restart).  You can also try ctrl-alt-delete and then end the process that you suspect is a virus and then try to manually delete it (you can't delete something if it is running).

EDIT: and about getting that junk on your computer, it is seriously EVERYWHERE on the internet.  I don't even like to go to many websites except AoPS and Yahoo b/c of that junk.  It's not your fault, just scan as often as possible w/Ad-Aware and McAffee/Norton.

[quote="BHorseMath"]Places  like google  have bad stuff on the site that other  sites have hacked into google and  put the stuff there you can get in just by going to places like google.  I got it all on my computer and crashed it.[/quote]
[img]http://shogun.shafted.com.au/temp/bunny_with_pancake.jpg[/img]
As for Google having spyware, if that's what you're trying to say... no.

Going to basically to any site with advertising on it will download adware on your computer. Spyware isn't really on that many sites. If you find an advertisement saying you have more than 1000 different spyware on your computer and to download a spyware remover, don't believe it, it'll jsut download even more spyware. If you want to see which works best consult with the PC magazine website. Google is probably one of the most secure search engines ever. Yahoo just just full of adware.

How come noone has heard of Proxomitron? It's the best pop-up blocker I've used - it's lightweight, junk-free, and [b]scriptable[/b].

Wow..... I just realized i use Proxomitron also. I forgot completely about it. The only time I really use it, is to use other proxy servers so i can't be tracked.

Also, what you could do....my friend is the genius who got this idea, but you can  have a system restore.  You didn't have to make a CD on this certain date or anything, your computer just does it automatically makes a system checkpoint thing.  The only problem is that if you downloaded something, you have to download it again.  However, saved documents aren't affected.  So you just a pick a date when your computer wasn't infected with spyware, and click it.  And if that doesn't work, you can always put your computer back to where it was.  As Julie said, get Ad-Aware, and if that doesn't work, you can get Spybot - Search and Destroy (http://www.download.com - i think has both).

Here's the steps to do a system restore (Only if you got XP though):
1.  Click start and Control Panel

2.  Click Performance/Maintance

3.  In the left-hand corner it is going to say, "See Also" and below that "System Restore".

4.  Then click "Restore my computer to an earlier time" and just click a bolded date.   As you can see you can revert the process by going to the same place and clicking ,"Undo my last restoration."

It works well if you have like a severe virus and you can't delete it.

BHorseMath just means that website DO go on google and say if you type, "Cool stuff at AoPS" it links you to another search engine that says, "Find 'cool stuff at AoPS'  here at http://www.weirdowebsite.com."  I can't seem to find an example - maybe google fixed it.  Or well, maybe that's what she means.  I never clicked one of those links, but I can imagine the trashcans full of spyware you get though.

You shouldn't system restore for everything though... (if thats what you were trying to imply)

I usually choose not to accept any cookies or whatever those security licensing things because they're usually bad.  Also, if you install a free program, check their user agreement talking about installing third party software... Thats a big no no. I even think AOL has some spyware too! Hah.

[quote="Rep123max"]Whats better, firefox or mozilla 1.7?[/quote]
I like Mozilla a lot more, but Firefox runs marginally faster and takes up a bit less hard drive space (neither of those things should be an issue. Go for Mozilla's extra features)

[quote="JS1527"]Give her a break.  She's in 6th/7th grade, and you are 18.  Be nice.[/quote]

I'm going into 8th and I'm right.  Google might be secure but searching on it takes you to other sites that aren't secure.  those are the sites that have the spyware.  Thats what the people fixing my computer said.

That makes sense.  I think Confuted was being a little mean...

He doesn't like me what else it new?  But w/e back to the original topic.

mcAfree is good.

Mozilla has a very good password manager. unlike IE you can have multiple passwords remembered for a single site, and you choose which one to use.


calculus
derivative
analytic geometry
graphing lines
slope
parameterization
quadratics

Question: Find the equation of the circle centered at the point (-1,3) tanget to the line y = 0.5x - 4

Here is what i have been able to do:

 i formulated the general equation:
    (x+1)^2 + (y-3)^2 = r^2

i then expanded it out:
x^2 + 2x + 1 + y^2 -6y + 9 = r^2

next, i performed implicit differentiation:
2x + 2 + 2y(dy/dx) -6(dy/dx) = 2r(dr/dx)

but since the radius is a constant, this becomes:
2x + 2 + 2y(dy/dx) -6(dy/dx) = 0

they give you the equation of the tangent line, so we let dy/dx = 1/2 and plug it in:

2x + 2 + 2y(1/2) -6(1/2) = 0

now, i simplified:

2x + 2 + y -3 = 0

y = - 2x + 1

i THINK this equation relates the x and y coordinates of the tangent points to the circle, but i am not really sure. I am lost. What do i do to finish this problem? Thanks for your help.

I don't think you even need calculus for this.  A line tangent to a circle is perpendicular to the radius.  Using that, you can find the radius given the slope (perpendicular to the given line) and the center of the circle.

... Or you can just use the [url=http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html]distance formula[/url] $\boxed{r = \frac{ax_{o}+by_{o}+c}{\sqrt{a^{2}+b^{2}}}}$

You right again the equation of the line in the form $ax+by+c=0:$
$y = 0.5x-4 \Longrightarrow-0.5x+y+4=0$

and $(x_{o}, y_{o}) = (-1,3)$ are the coordinates of the center

So you have $r = \frac{(-0.5)\cdot(-1)+1 \cdot 3+4}{\sqrt{(-0.5)^{2}+1^{2}}}= \frac{7.5}{\sqrt{1.25}}= \sqrt{45}$

Finally, the equation of the circle is $(x+1)^{2}+(y-3)^{2}=45$

Here's another approach (Descartes would approve):

The line is $y=\frac{x}{2}-4.$ To get rid of fractions, I'll rewrite that as $x=2y+8.$

The circle is $(x+1)^{2}+(y-3)^{2}=r^{2}.$

Treat $r$ as an unkonwn parameter, and solve those two equations together algebraically. If we substitute $x$ from the linear equation into the quadratic, we have

$(2y+9)^{2}+(y-3)^{2}=r^{2}$

$5y^{2}+30y+90=r^{2}$

$y^{2}+6y+\left(18-\frac{r^{2}}5\right)=0$

If the line and the circle are to be tangent, this polynomial equation must have a double root. It must be a perfect square. That means that we must have

$18-\frac{r^{2}}5=9$

$r^{2}=45$

The answer is the same as the one pontios gave.


linear algebra
matrix
trigonometry
geometry
geometric transformation
rotation
LaTeX

There is a certain matrix $ A$ with order 2 such that $ A_{1,1}\equal{}\frac{\sqrt{3}}{2}$, $ A_{1,2}\equal{}\minus{}\frac{1}{2}$, $ A_{2,1}\equal{}\frac{1}{2}$, and $ A_{2,2}\equal{}\frac{\sqrt{3}}{2}$. Find $ A^{2008}$

Ordinarily, I'd say "diagonalize the matrix," but there's a simpler approach to this one.

The $ 2\times 2$ matrix $ \begin{pmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}$ acts on $ \mathbb{R}^2$ by rotating the plane counterclockwise through an angle $ \theta.$

So this particular matrix is rotation by what? And it takes how many steps to get back to where you started?

(By the way, you can write matrices in $ \text{\LaTeX}.$ Click on my formula above to see how.)

[quote="Kent Merryfield"]Ordinarily, I'd say "diagonalize the matrix," but there's a simpler approach to this one.

The $ 2\times 2$ matrix $ \begin{pmatrix} \cos \theta & - \sin\theta \\
\sin\theta & \cos \theta \end{pmatrix}$ acts on $ \mathbb{R}^2$ by rotating the plane counterclockwise through an angle $ \theta.$

So this particular matrix is rotation by what? And it takes how many steps to get back to where you started?

(By the way, you can write matrices in $ \text{\LaTeX}.$ Click on my formula above to see how.)[/quote]
So this particular matrix is a rotation of 60 degrees? And it would take 6 steps to get back to the original because 60*6=360. Is this right or am I totally off?
(thanks for the matrix in latex, didnt know how to :O)

[quote="Kent Merryfield"]Not quite. That's a $ 30^{\circ}$ angle, not a $ 60^{\circ}$ angle.[/quote]
Oh, sorry I got them mixed up D:
So with a $ 30^{\circ}$, it would go back every 12 times?

[quote="Kent Merryfield"]Yes, that's correct. Now, what is the final answer? (Intermediate question: $ 2008\equiv ?\pmod{12}.$)[/quote]
Since it repeats itself every 12 times, it would be the same as $ A^{4}$, since 2008 is equivalent to 4 in mod 12. Then doing that out I get...$ \begin{pmatrix} \minus{}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \minus{}\frac{1}{2} \end{pmatrix}$.

You have one sign error.

$ A^{2008}\equal{}\begin{pmatrix}\minus{}\frac12&\minus{}\frac{\sqrt{3}}2\\ \frac{\sqrt{3}}2&\minus{}\frac12\end{pmatrix}.$


hello, i have only found $ x\equal{}y\equal{}z\equal{}\minus{}\frac{1}{3}$, $ x\equal{}y\equal{}z\equal{}0$, $ x\equal{}y\equal{}z\equal{}1$.
Sonnhard.

[quote="julio"]Solve it:
x+2y^2=3z^3
y+2z^2=3x^3
z+2x^2=3y^3

x,y,z are real numbers.[/quote]
$ x \equal{} y \equal{} z$ gives the solution $ x\equal{}y\equal{}z\frac{\minus{}1}{3}$, $ x\equal{}y\equal{}z\equal{}0$ and $ x\equal{}y\equal{}z\equal{}1$. Assume $ x > y,z$. If $ x > 1$ then we have $ x^3 > z^2, y$ and thus $ 3x^3 > 2z^2 \plus{} y$. So that is not an option. So assume $ x < 1$. (It's easy to see that $ x \equal{} 1 \Rightarrow x\equal{}y\equal{}z\equal{}1$) We have $ x > z^3$ implying $ z > |y|^{2/3} > |y|$. But we know $ x^2 > y^3$ and $ z > y$ and thus $ z\plus{}2x^2 > 3y^3$.
Then we are left with the possibility $ x < y,z$. If $ x < 1$ we get $ x^3 < y$ and $ x^3 < z^2$ and thus $ 3x^3 < y \plus{} 2z^2$. So assume $ x > 1$. From $ x\plus{}2y^2 \equal{} 3z^3$ we obtain $ y > z$. But then $ y^3 > z$ and $ y^3 > x^2$ gives $ 3y^3 > 2x^2 \plus{} z$. So $ x\equal{}y\equal{}z$ giving the only solutions: $ x\equal{}y\equal{}z\frac{\minus{}1}{3}$, $ x\equal{}y\equal{}z\equal{}0$ and $ x\equal{}y\equal{}z\equal{}1$

hi! I already posted this system. It is a system invented by me when I was 9-th grade.

Yes! borislav_mirchev already posted this system. You see here http://www.mathlinks.ro/viewtopic.php?search_id=1206710720&t=163713


I would like a formula or idea of how to solve a generalized version of this problem:

What is the next exact time after 2:00 that a clock has its hands at an angle of 70 degrees?

So, basically I would like to know how to solve this problem even if the time, which in this case is 2:00, and the measure of the hands, which in this case is 70 degrees, was different.

Thanks for the help in advance.

Assuming the time is accurate to the minute and hour, the number of degrees between the hands is (where $ m$ is the minutes and $ h$ is the hours) $ \left|\frac{11}{2}m\minus{}30h\right|.$ In your example, we would like the hours to be as small as possible, so $ h\equal{}2.$ Now, $ \left|\frac{11}{2}m\minus{}60\right|\equal{}70.$ Now we separate this into two cases.
[hide="Case 1"]
$ \frac{11}{2}m\minus{}60\equal{}70.$ This gives the value of $ m\equal{}23\frac{7}{11}.$[/hide]

[hide="Case 2"]
$ \frac{11}{2}m\minus{}60\equal{}\minus{}70.$ This gives a negative value for $ m,$ but we can't have negative time, so there is no solution.[/hide]

Finally, the answer is $ 2: 23\frac{7}{11}.$

In general, the most important thing to remember is that the degree measure between the hands is $ \left|\frac{11}{2}h\minus{}30m\right|.$ Now just set this equal to the amount of degrees, and divide it into two cases.

[quote="RoFlLoLcOpT"]Assuming the time is accurate to the minute and hour, the number of degrees between the hands is (where $ m$ is the minutes and $ h$ is the hours) $ \left|\frac {11}{2}m \minus{} 30h\right|.$ In your example, we would like the hours to be as small as possible, so $ h \equal{} 2.$ Now, $ \left|\frac {11}{2}m \minus{} 60\right| \equal{} 70.$ Now we separate this into two cases.
[hide="Case 1"]
$ \frac {11}{2}m \minus{} 60 \equal{} 70.$ This gives the value of $ m \equal{} 23\frac {7}{11}.$[/hide]

[hide="Case 2"]
$ \frac {11}{2}m \minus{} 60 \equal{} \minus{} 70.$ This gives a negative value for $ m,$ but we can't have negative time, so there is no solution.[/hide]

Finally, the answer is $ 2: 23\frac {7}{11}.$

In general, the most important thing to remember is that the degree measure between the hands is $ \left|\frac {11}{2}h \minus{} 30m\right|.$ Now just set this equal to the amount of degrees, and divide it into two cases.[/quote]

Did you mean that its supposed to be $ \left|\frac {11}{2}h \minus{} 30m\right|$ or $ \left|\frac {11}{2}m \minus{} 30h\right|$ ?


[u]πως μπορουμε να αποδειξεουμε οτι αν 2 πολυωνυμα ειναι ισα (δηλ εχουν την ιδια τημη για καθε χ) τοτε ειναι του ιδιου βαθμου και οι αντιστοιχοι συντελεστες ειναι ισοι :?:  [/u]

μια σκεψη ειναι να αναγουμε το παραπανω προβλημα σε  ενα πιο απλο δηλ να δειξουμε οτι εαν ενα πολυωνυμο ειναι ισο με το μηδεν για καθε χ τοτε οι συντελεστες του θα ειναι μηδεν.Σε αυτο το προβλημα η αποδειξη προκυπτει απο το θεμελιωδες θεωρημα της αλγεβρας δηλ οτι καθε πολυωνυμο ν βαθμου εχει ν μιγαδικες ριζες (η το πολυ ν πραγματικες)
Ομως για την αποδειξη αφτου του θεωρηματος χρειαζεται μιγαδικη αναλυση και δεν ξερω εαν χρησιμοποιηται κ εκει η συνθηκη ισοτητας πολυωνυμων.

Μπορειτε να βρειτε μια απλη αποδειξη στο παρα πανω ερωτημα :?:

[quote="DIMITRIS T"]μια σκεψη ειναι να αναγουμε το παραπανω προβλημα σε  ενα πιο απλο δηλ να δειξουμε οτι εαν ενα πολυωνυμο ειναι ισο με το μηδεν για καθε χ τοτε οι συντελεστες του θα ειναι μηδεν.Σε αυτο το προβλημα η αποδειξη προκυπτει απο το θεμελιωδες θεωρημα της αλγεβρας δηλ οτι καθε πολυωνυμο ν βαθμου εχει ν μιγαδικες ριζες (η το πολυ ν πραγματικες)
Ομως για την αποδειξη αφτου του θεωρηματος χρειαζεται μιγαδικη αναλυση και δεν ξερω εαν χρησιμοποιηται κ εκει η συνθηκη ισοτητας πολυωνυμων.[/quote]

Έλεος, όχι και μιγαδική ανάλυση :D

To θεώρημα ότι για κάθε πολυώνυμο $ P$ υπάρχει πολ/μο $ Q$ ώστε $ P(x) \equal{} (x \minus{} a) Q(x) \plus{} P(a)$ είναι εντελώς στοιχειώδες (απλή Ευκλείδεια διαίρεση είναι, δεν έχει να κάνει με το θεμελιώδες θεώρημα της άλγεβρας).

Έστω $ P(x) \equal{} a_n x^n \plus{} \cdots \plus{} a_1 x \plus{} a_0$ με $ a_n\neq 0$ και $ P(x) \equal{} 0 \forall x\in\mathbb{R}$. Kάνε τη διαίρεση $ P(x)/\left((x \minus{} r_1)\cdots (x \minus{} r_n)\right)$ όπου $ r_{1\ldots n}$ διαφορετικοί πραγματικοί αριθμοί (κ' άρα $ P(r_i) \equal{} 0$).

Have fun με το άτοπο στο οποίο καταλήγεις :)

Cheerio,

Durandal 1707

Edit: ακόμα πιο απλά, assume WLOG that $ a_n > 0$. Then:
$ \lim_{x\to\infty} P(x) \equal{} \lim_{x\to\infty}a_n x^n (1 \plus{} \cdots \plus{} \frac {a_1}{x^{n \minus{} 1}} \plus{} \frac {a_0}{x^n}) \equal{} \plus{} \infty$, άτοπο αν $ P\equiv 0$.

Το μόνο που χρειάζεσαι είναι ότι κάθε πολυώνυμο ν βαθμού έχει το πόλυ ν πραγματικές ρίζες και για να το αποδείξεις αυτό χρειάζεσαι μόνο άλγεβρα. Την (μιγαδική) ανάλυση την χρειάζεσαι για να δείξεις ότι κάθε μη σταθερό πολυώνυμο έχει τουλάχιστον μία μιγαδική ρίζα.

Το αξίωμα Bezout δεν είναι και ό,τι χαρακτηριστικότρο από μιγαδική ανάλυση...  :D Τέσπα...

Ένα μικρούλι ευκολάκι στα πλαίσια αυτά αλλά είναι κομψό και μου άρεσε:

Να βρεθούν όλα τα πολυώνυμα τέτοια ώστε:

$ P(x^2\plus{}1)\equal{}[P(x)]^2\plus{}1$ και $ P(0)\equal{}0$

[quote="Durandal"]

Edit: ακόμα πιο απλά, assume WLOG that $ a_n > 0$. Then:
$ \lim_{x\to\infty} P(x) \equal{} \lim_{x\to\infty}a_n x^n (1 \plus{} \cdots \plus{} \frac {a_1}{x^{n \minus{} 1}} \plus{} \frac {a_0}{x^n}) \equal{} \plus{} \infty$, άτοπο αν $ P\equiv 0$.[/quote]

Ωραια σκεψη Durandal!!! δειχνεις δηλ οτι ο $ a_n$ ειναι αναγκαστικα μηδεν κ μετα κανεις την ιδια  διαδικασια για ολους τους ορους του πολυωνυμου.

Προχθες καθως κοιταγα το βιβλιο αλγεβρας του Τογκα πετυχα την αποδειξη αυτου που εψχνα (δηλ εαν ενα πολυωνυμο ειναι μηδεν για καθε χ τοτε ολοι οι συντελεστες του ειναι μηδεν)

Εστω $ P(x)\equal{}a_nx^n\plus{}..\plus{}a_1x\plus{}a_0\equal{}0     \forall x  \Leftrightarrow  x(a_nx^{n\minus{}1}\plus{}...\plus{}a_2x\plus{}a_1)\plus{}a_0\equal{}0     \forall x$
αρα το $ a_0\equal{}0$ γιατι αλλιως αν βαζαμε οπου χ το μηδεν θα ειχαμε $ P(x)\equal{}a_0\neq 0$ ατοπο απο την υποθεση.

Οποτε το πολυωνυμο μας γινεται $ P(x)\equal{}a_nx^n\plus{}...\plus{}a_1x\equal{}0  \forall x$  Aρα εχουμε $ P(x)\equal{}a_nx^n\plus{}...\plus{}a_1x\equal{}x(a_nx^{n\minus{}1}\plus{}...\plus{}a_2x\plus{}a_1)\equal{}0 \forall x  \Leftrightarrow a_nx^{n\minus{}1}\plus{}...\plus{}a_2x\plus{}a_1\equal{}0 \forall x$
 κ ομοια με πριν βγαζουμε οτι $ a_1\equal{}0$

κανοντας την ιδια διαδικασια με ολους τους συντελεστες τους βγαζουμε ολους μηδεν :D

Ετσι αυτο που θελαμε να αποδειξουμε δεν ισχυει μονο για πραγματικους,ουτε μονο για μιγαδικους αλλα [u]για καθε σωμα [/u]:lol:

Sorry, που σας κουρασα τζαμπα :blush:

[quote="DIMITRIS T"]
Ετσι αυτο που θελαμε να αποδειξουμε δεν ισχυει μονο για πραγματικους,ουτε μονο για μιγαδικους αλλα [u]για καθε σωμα [/u]:lol:

[/quote]

Δεν ισχύει για πεπερασμένα σώματα. Για παράδειγμα, στο $ \mathbb{F}_p$, όπου $ p$ πρώτος, έχουμε $ f(x) \equal{} x^p \minus{} x \equiv 0$.

[quote="Demetres"]Δεν ισχύει για πεπερασμένα σώματα. Για παράδειγμα, στο $ \mathbb{F}_p$, όπου $ p$ πρώτος, έχουμε $ f(x) \equal{} x^p \minus{} x \equiv 0$.[/quote]

Με πρόλαβες :D  Θα σε λέω Frobenius :ninja:

Τι ιδιότητες λοιπόν πρέπει να έχει ένα σώμα ώστε το μόνο μηδενικό πολυώνυμο να είναι αυτό με μηδενικούς συντελεστές?  ;)

Cheerio,

Durandal 1707

[quote="Protonios"]Το αξίωμα Bezout δεν είναι και ό,τι χαρακτηριστικότρο από μιγαδική ανάλυση...  :D Τέσπα...

Ένα μικρούλι ευκολάκι στα πλαίσια αυτά αλλά είναι κομψό και μου άρεσε:

Να βρεθούν όλα τα πολυώνυμα τέτοια ώστε:

$ P(x^2 \plus{} 1) \equal{} [P(x)]^2 \plus{} 1$ και $ P(0) \equal{} 0$[/quote]
Έστω $ a_{n\plus{}1}\equal{}(a_n)^2\plus{}1 , a_0\equal{}0$ και $ b_n\equal{}P(a_n)$ Εύκολα δείχνεις επαγωγικά ότι $ a_n\equal{}b_n\equal{}P(a_n)$ άρα το πολυώνυμο $ P(x)\minus{}x$ έχει απειρες διαφορετικές ρίζες όλους τους όρους της γνήσιας αύξουσας ακολουθίας $ a_n$ οπότε $ P(x)\equal{}x$

[quote="DIMITRIS T"][quote="DIMITRIS T"]

Προχθες καθως κοιταγα το βιβλιο αλγεβρας του Τογκα πετυχα την αποδειξη αυτου που εψχνα (δηλ εαν ενα πολυωνυμο ειναι μηδεν για καθε χ τοτε ολοι οι συντελεστες του ειναι μηδεν)

Εστω $ P(x) \equal{} a_nx^n \plus{} .. \plus{} a_1x \plus{} a_0 \equal{} 0 \forall x \Leftrightarrow x(a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1) \plus{} a_0 \equal{} 0 \forall x$
αρα το $ a_0 \equal{} 0$ γιατι αλλιως αν βαζαμε οπου χ το μηδεν θα ειχαμε $ P(x) \equal{} a_0\neq 0$ ατοπο απο την υποθεση.

Οποτε το πολυωνυμο μας γινεται $ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \equal{} 0 \forall x$  Aρα εχουμε $ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \equal{} x(a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1) \equal{} 0 \forall x \Leftrightarrow a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1 \equal{} 0 \forall x$
 κ ομοια με πριν βγαζουμε οτι $ a_1 \equal{} 0$

[/quote]
sorry pu xanafernw afto to 8ema alla xanaberdeftika... :blush:
[color=red]einai swsto afto pu  exw se quote??[/color][/quote]

Σίγουρα πρέπει να δικαιολογήσεις το

Aρα εχουμε

\[ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \equal{} x(a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1) \equal{} 0 \forall x \Leftrightarrow a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1 \equal{} 0 \forall x\]

Αν γνωρίζω πως $ P(x) \equal{} a_nx^n \plus{} ... \plus{} a_1x \equal{} x(a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1) \equal{} 0 \forall x$, τότε έπεται ότι $ Q(x) \equal{} a_nx^{n \minus{} 1} \plus{} ... \plus{} a_2x \plus{} a_1 \equal{} 0 \forall x \neq 0$. Πως λοιπόν μετά μπορείς να πεις ότι $ Q(0) \equal{} 0,$ άρα $ a_1 \equal{} 0$;

Οι μόνες δικαιολογήσεις που βλέπω είτε χρησιμοποιουν αυτό που θες να αποδείξεις, είτε χρησιμοποιούν ανάλυση που δεν χρειάζεται εδώ. Αν κάποιος βλεπει πως να το δικαιλογήσει (ίσως χάνω κάτι προφανές) ας μας πει.

Όπως είχε πει κάποιος χρήστης στην προηγούμενη συζήτηση, μπορεί να αποδειχτεί χρησιμοποιώντας τον Ευκλείδιο αλγόριθμο.


Find all positive integers $ n$ such that there exist real numbers $ a_1,a_2,...a_n$ such that

$ \{a_i\minus{}a_j| 1\leq j<i\leq n\}\equal{}\{1,2,...,\frac{n(n\minus{}1)}{2}\}$

Answer: 2, 3, 4

Obviously $n\ge2$ because for $n=1$ the difference set is empty. For $n=2,3,4$, the sequence $\{1,2\}$, $\{1,2,4\}$, $\{1,2,5,7\}$ satisfy the condition.

Assume $n\ge5$. Let $k=\frac{n(n-1)}2$. Note that each difference appears exactly once and obviously the sequence is increasing. The difference $k$ is the largest, so $k=a_n-a_1$, WLOG assume $a_n=k,a_1=0$. The difference $k-1$ must equal $a_n-a_2$ or $a_{n-1}-a_1$, WLOG it is $a_n-a_2$ (otherwise we can transform the sequence $a_i\rightarrow k-a_{n+1-i}$). Thus $a_2=1$.

Let $k-2=a_x-a_y$. Note that $a_{n-1}\ne k-1$ because the difference $a_n-(k-1)=a_2-a_1$, so $a_{n-1}\le k-2$. From this we get $x\ge n-1$. If $x=n$ then $a_y=2$, which implies $a_y-a_2=a_2-a_1$, a contradiction. Therefore $x=n-1$ and hence $a_{n-1}=k-2$.

Let $k-4=a_p-a_q$. Note that $a_{n-2}$ cannot be $k-3,k-4$ because the difference 1 and 2 already appears in the sequence. So $a_{n-2}\le k-5$. Therefore $p\ge n-1$. If $p=n-1$ then $a_q=2$, which is impossible. So $p=n$ and hence $a_q=4$. Since 2 and 3 cannot be in the sequence, we must have $q=3$. Our sequence is \[0,1,4,\ldots,k-2,k\]
Let $k-5=a_r-a_s$. We have $a_{n-2}<k-5$, so $r\ge n-1$. If $r=n-1$, then $a_s=3$, which is impossible. If $r=n$, then $a_s=5$, and we have $5-4=1-0$, a contradiction. Therefore for $n\ge5$ no such sequence exists.


integration
function
calculus
algebra
polynomial
calculus computations

[quote="kunny"]The equality holds when $b-\frac{3}{5}=0$ and $a+\frac{1}{3}c=0$ and $c=0\Longleftrightarrow a=0,\ b=\frac{3}{5},\ c=0.$
[/quote]
I agree this.
[quote="kunny"]Thus the desired minimum value is $\frac{2}{7}.$[/quote]
I think it is $\frac{8}{175}$.

Yes: kunny added a term of $\frac{2}{3}\cdot\frac{9}{25}$ in order to complete the square with $b$, but he didn't also subtract it.

Given that this is an orthogonal projection problem, an orthogonal basis would simplify the calculations. That is (knowing the Legendre polynomials), I'm suggesting that you make the replacement

$a+bt+ct^{2}=\alpha+\beta t+\gamma(3t^{2}-1).$

$\alpha,\beta,$ and $\gamma$ depend in some linear fashion on $a,b,c,$ but we really don't need the details of that. The span is the same. Minimizing over $\alpha,\beta,\gamma$ is the same problem as minimizing over $a,b,c.$

We determine $\alpha,\beta,\gamma$ just by taking the normalized inner product of $t^{3}$ with each of these basis functions. From that perspective, it is clear that $t^{3}$ is orthogonal to all the even functions and that only $\beta$ will be different from zero.

[quote="puuhikki"]
[quote="kunny"]Thus the desired minimum value is $\frac{2}{7}.$[/quote]
I think it is $\frac{8}{175}$.[/quote]

You are right. :blush:
I have just edited.

kunny


ABC  is a triangle  with a>=b>=c  and  P  is  a  point in  the  interior  of  the triangle . Prove  that  PA+PB+PC>b+c

[i]unknown source[/i]

this inequality is not good. Take a=b=c=\sqrt 3 and P=O the center of the triangle. Then PA+PB+PC = 3 > 2\sqrt 3 => 3 > 4 which is absurde.

ABC  is a triangle  with ${a}\geq{b}\geq{c}$  and  P  is  a  point in  the  interior  of  the triangle .
Prove  that  $PA+PB+PC<a+b$
author: Lev Radzivilovsky(around 1999)

Is it possible to find out which point in the interior of the triangle maximizes $PA+PB+PC$?

$P\notin AB\cup AC\cup BC.$ ;)
And enough that $c=min\{a,b,c\}.$


Use Cauchy for  1/ \sqrt (x+y),... and  \sqrt  (x+y),... for the left side. For the right side use the inequality 4/(x+y)\leq 1/x+1/y.


Can anyone help me with simple interest calculations? I can't seem to get it right.

All I know is the formula which is:

INTEREST= Principle amount * Interest rate * Time

To find the interest, you simply plug it into the formula. Now, it's important to understand how the formula works. Suppose I had 500 dollars, and there was a 7% interest rate. That means each year, the amount of money I have increases by 7%. So after the first year, I'll get $ 0.07 \times 500 \equal{} 35$ dollars interest. By simple interest, I'll have another 35 dollars the second year, and then another 35 the third year, so that after 3 years I'll have a total of $ 500 \plus{} 35 \plus{} 35 \plus{} 35 \equal{} 605$ dollars. This is why the formula:

interest = principle amount (what you started with, 500 dollars in the example) * interest rate (7% in the example) * time (3 years in the example)

works.

Let's do another example. Say I had 470 dollars, and there's 4.3% interest rate. How much will I have after twelve years? First we need to find the interest. The principle amount is 470 dollars, the interest rate is 4.3%, which we write as 0.043, and the time is 12 years. Therefore, we have $ 470 \times 0.043 \times 12 \equal{} 242.52$. That means the interest was 242.52 dollars, so in total I would have $ 470 \plus{} 242.52 \equal{} 712.52$ dollars.

Does that make sense?

That does make scene and that is what I was doing. Let me give you an example of some problems I couldn't do:

1)Jamie wants to earn 500 dollars in interest so she’ll have enough to buy a used car. She puts 2000 dollars into an account that earns  2 and a half percent interest. How long will she need to leave her money in the account to earn 500 dollars in interest?

2)A local bank is advertising that you can double your money in eight years if you invest with them. Suppose you have 1000 dollars to invest. What interest rate is the bank offering?

Just plug in the formula.

For the first one:

500 dollars = the interest, so we have:

500 = priniciple * interest rate * time

2000 dollars = the principle:

500 = 2000 * interest rate * time

Now we have 2.5% = the interest rate:

500  = 2000 * 0.025 * time

We need to figure out the time. This is simply an equation, so solving for time gives us $ \boxed{10}$ years.

And for number 2 the intrest has to be the same as the principle if you want it to double so:
$ \text{Interest}\equal{}\text{Principle}\times\text{Rate}\times\text{Time}$
$ 1000\equal{}1000\times\text{Rate}\times8$
$ 1000\equal{}8000\times\text{Rate}$
$ .125\equal{}\text{Rate}$
So the Interest Rate is $ \boxed{12.5\text{ Percent}}$

Could somebody please explain compound interest....I never really understood that :(

OK, so in simple interest, you do the interest on the ORIGINAL amount. So if I had 500 dollars, with 7% interest, after the first year I would have 535, and after the second, I would have 570. That's with simple interest. For compound interest (what's in real life), the second 7% goes on the 535, not the 500. So we have $ 535 \times 1.07 \equal{} 572.45$ after the second year. It's fairly simple to derive a formula for compound interest. Try doing it!  :)

Ok thanks so the formula that i came up with is for $ p$ dollars at a compound interest rate of $ i \text{ percent}$ over the span of $ t$ years the formula is: $ p\times(1\plus{}\frac{i}{100})^t$
So for your problem:

$ 500\times(1\plus{}\frac{7}{100})^2$

$ 500\times(1.07)^2$

$ 500\times1.1449$

$ \boxed{572.45}$

[quote="SuperNerd123"]And for number 2 the intrest has to be the same as the principle if you want it to double so:
$ \text{Interest} \equal{} \text{Principle}\times\text{Rate}\times\text{Time}$
$ 1000 \equal{} 1000\times\text{Rate}\times8$
$ 1000 \equal{} 8000\times\text{Rate}$
$ .125 \equal{} \text{Rate}$
So the Interest Rate is $ \boxed{12.5\text{ Percent}}$

Could somebody please explain compound interest....I never really understood that :([/quote]

okay but what I understood from the problem is that you can DOUBLE your money in 8 years so why would it not be:

$ 2000 \equal{} 1000\times\text{Rate}\times8$

Yes but the formula is
[b]interest[/b]$ \equal{}principle\times\text{rate}\times\text{time}$
So if the interest is equal to the principle than the principle will double


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

If $ A^{m}\equal{}I_{n}$ , $ m \in \mathbb{N},$ and  $ m \geq 1$

then can we get the conclusion: the elementary divisor of $ \lambda I_{n}\minus{}A$ is order $ 1$ $ ?$

In other words: if $ f_{1}(\lambda)$ is its elementary divisor, then $ deg(f_{1}(\lambda)) \equal{} 1$ $ ?$

 :D

Not if I understand you correctly.  Take $ A \equal{} I_n$.

Or you could do something like this:

$ A \equal{} \begin{bmatrix}0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0\end{bmatrix}.$

$ A^2 \equal{} I.$ Minimal polynomial $ (x \minus{} 1)(x \plus{} 1).$ Characteristic polynomial $ (x \minus{} 1)^2(x \plus{} 1)^2.$


probability
function
combinatorics unsolved
combinatorics

Are the people starting out in the same place?

It's pretty important - for instance, if they start out an odd number of squares apart, then either of the two probabilities is zero.

[quote="Diarmuid"]Are the people starting out in the same place?
[/quote]

Hmm... let's assume so.

[quote="warut_suk"]Two people are walking randomly on the number line. Each take a step of length $1$ every minute. The probability of going right and left are equal. After $K$ steps, find the probability that
(a) they're at the same places
(b) they're two units apart[/quote]

First, let us call our two random walkers [b]A[/b]lice and [b]B[/b]ob.

In order to get to an arbitrary position x, Alice walks n steps to the right and K-n steps to the left. $P(A = x) =P(n\; \text{right}|K-n\; \text{left})\cdot P(K-n\; \text{left}) =(\frac{1}{2})^n(\frac{1}{2})^{K-n} = (\frac{1}{2})^K$.

Likewise, in order to get to an arbitrary position y, Bob walks m steps to the right and K-m steps to the left. $P(B = y) =P(m\; \text{right}|K-m\; \text{left})\cdot P(K-m\; \text{left}) =(\frac{1}{2})^m(\frac{1}{2})^{K-m} = (\frac{1}{2})^K$.

Alice and Bob are statistically independent $\Rightarrow P(A =x,\; B =y) = P(A = x)P(B = y) =(\frac{1}{2})^K(\frac{1}{2})^K=(\frac{1}{2})^{2K}$.

Since x and y are arbitrary the distance between x and y is arbitrary; hence the answers to a) and b) are the same: $(\frac{1}{2})^{2K}$.

[quote="Kondr"]Well, I think that most of you wrote is wrong, because the probabilities of reaching diferent positions after  K steps are diferrent. After K steps they can reach K+1 different positions, let's number them from left to right with numbers 0 to K. Probability of reaching position p is $P(p)=\frac {\binom  {k} p} {2^k}$. Because of that, the probability a) is equal to
$\sum_{i=0}^k{P(i)^2}=\sum_{i=0}^k{\frac {\binom k i^2} {4^k}}=\frac {\binom  {2k} k} {4^k}$.[/quote]You are assuming that one can get to any given position from any other directly. This is not so. The proposed walk is that of a one dimensional Markov chain. From any given position one may only enter immediately adjacent positions, and regardless of which direction one chooses to move the transition probabilities for this chain are each 1/2.

[quote="Kondr"]I still think you are wrong. See http://mathworld.wolfram.com/RandomWalk1-Dimensional.html .[/quote]Yes, well the Mathworld page specifies the probability of taking "particular sequence of steps (to the right)" and a particular sequence of steps to the left. In the case given in the problem at hand we don't have to worry about how many steps we take in either direction because the probability of going left equals the probability of going right. In such a case, the probability of landing on any particular spot in K steps is $p = (\frac{1}{2})^K$.

[quote="Kondr"]After k steps can Bob be in one of k+1 positions. If you were right, $(k+1)\frac 1 {2^k}=1$ should be true.
If you want to know where is the mistake in your post, read the stuff at mathworld more carefuly.[/quote]Actually you're correct the probability of an arbitrary walker ending up at a position K-2n steps to the right of his starting point is binomially distributed.

$P(K-2n) =\binom{K}{K-n}p^{K-n}q^n$

Therefore,

$P(A = K-2n\; \text{right},\; B = K -2m\; \text{left})=(\binom{K}{K-n}p^{K-n}q^n)(\binom{K}{K-m}q^{K-m}p^m)$

[b]Caveat: In this problem: m is the number of rightward steps taken by Bob; n is the number of leftward steps taken by Alice.[/b]

In this case, $ p = q=\frac{1}{2}$,

$P(A = K-2n\; \text{right},\; B = K -2m\; \text{left})=\binom{K}{K-n}\binom{K}{K-m}p^{2K}$

The distance between Alice and Bob at the end of K steps for each is given by: $D =2(K-(m+n))$

Thus they are at the same point when $D = 0 \Rightarrow K = m + n$,

$P(D = 0)=\binom{K}{m}\binom{K}{n}p^{2K} = \binom{m+n}{m}\binom{m+n}{n}\frac{1}{4^{m+n}}$

They are two steps apart when D = 2,

$P(D = 2)=\binom{K}{m+1}\binom{K}{n+1}p^{2K}=\binom{m+n+1}{m+1}\binom{m+n+1}{n+1}\frac{1}{4^{m+n+1}}$

[edit]I removed a reaction to Dr. No's post, because after his edit it loose sense.

In the case b) we can split the probability int two cases, depending on who is more left. WLOG suppose it's Allice. Then Bob's position m can be from 2 to k, so probability of b) while Allice is more left then Bob is $\sum_{i=2}^k{P(i)P(i-2)}$. Because of the symetry, the total probability of b) is$2\sum_{i=2}^k{P(i)P(i-2)}=2\sum_{i=2}^k{\frac {\binom k i \binom k {i-2}} {4^k}}=2\frac {\binom{2k}{k-2}} {4^k}$.

A good way to think about this kind of problem is to imagine concatenating the two random walks.  More precisely, consider Alice's walk followed by the mirror image of Bob's moves.  That concatenation is a random walk with $2K$ steps.

For example, part (a) asks whether Alice and Bob end at the same place.  That is equivalent to the probability that a random walk on $2K$ steps (starting at the origin $0$) ends at $0$.  That formulation immediately leads to Kondr's formula.

Similarly, part (b) is equivalent to the probability that a random walk on $2K$ steps ends at ${-2}$ or $2$.  Again, we quickly get Kondr's second formula.

[quote="Kondr"]That seems better but look at the formulas you proved. There is a variable m. Because in the first case m can have arbitrary value from 0 to k, you have to sum these expresions for  these m and you wil get the sum I wrote above. In the second case we can split the probability int two cases, depending on who is more left. WLOG suppose it's Allice. Then Bob's position m can be from 2 to k, so probability of b) while Allice is more left then Bob is $\sum_{i=2}^k{P(i)P(i-2)}$. Because of the symetry, the total probability of b) is$2\sum_{i=2}^k{P(i)P(i-2)}=2\sum_{i=2}^k{\frac {\binom k i \binom k {i-2}} {4^k}}=2\frac {\binom{2k}{k-2}} {4^k}$.[/quote]Please review my edit of Post #10. No addition is necessary; this is where you go wrong. Summation in a probability statement implies a union of events. Here there is no such union. In this case one attempts to evaluate the intersection of Alice's walk and Bob's walk.

The journey of either walker is exactly K steps. If it is possible they are to wind up in the same position at their respective finishes, then K = m+n. Otherwise, winding up in the same position is impossible, and therefore it will occur with probability zero by definition.

Likewise, if it is possible for Alice and Bob to end up two steps apart then K = m+n+1. Else, it is impossible and occurs with probability zero.

[quote="Dr. No"]  No addition is necessary; this is where you go wrong. Summation in a probability statement implies a union of events. Here there is no such union. In this case one attempts to evaluate the intersection of Alice's walk and Bob's walk.
[/quote]
That's the point wher you go wrong :). Ravi B gave a nice and correct prove, which shows that. The union is there because I separete it into k+1 (respectively k-1) diferent  cases, depending on which position do they meet each other. The probability i denoted P(i) is probability of given walker being on the position i after k steps. Do you see it yet?

[quote="Ravi B"]Dr. No, I am afraid your formulas are off.  They involve variables $m$ and $n$.  If you look back at the original problem, there are no variables $m$ and $n$.  The final formulas you end up with should be purely functions of $K$.

You say that there is no union of events, but actually there is.  We have a union over all the possible values of $m$ and $n$.  We don't know in advance how many left or right steps Alice or Bob take, so we have to take the union.[/quote]I concede your point.

It is the intersection of two binomial distributions, but then I suppose this intersection could have any number of steps {0,1,2,...,K} left or right by either party implying that the total probability would be equal to the probability of the union of these individual walks.


superior algebra
superior algebra solved

:downto: Find the smallest field which contain  :sqrt: 2 and  :sqrt: 3

Well, it's clearly $\mathbb{Q} [\sqrt 2 , \sqrt3]$ when the roots are considered as elements of $\mathbb{Q}$.
Otherwise take $\mathbb{F}_2$   ;)


Doesn't Physics deserve, any more, a place in the new (and nice) format of the Portal?

seriously speaking it does but then the decision obviously rests with valentin


linear algebra
matrix
linear algebra unsolved

Let $A$ be a complex $n\times n$ matrix such that $A^{2}+A=O .$ If ${\mbox tr}(A)$ denotes the trace of $A$ , prove  ${\mbox tr}(A) \in \{-n,-n+1,...,-1,0\}.$

It's kind of obvious (unless I made a mistake, that is).

Let $\lambda$ be an eigenvalue of $A$. Then, $\lambda^{2}+\lambda = 0$, so $\lambda \in \left\{ 0,-1 \right\}$.

We have that $\textrm{tr}\, A = \lambda_{1}+\lambda_{2}+\ldots+\lambda_{n}= (-1) \cdot \, \textrm{the algebraic multiplicity of }-1$, so we're done.

[quote="perfect_radio"]It's kind of obvious ...[/quote]
"perfect_radio"="perfect-solution" ! Only a supplementary question : [b]what about $\mbox{rang}(A)$ ? [/b]

Ovious, Since every eigenvalue is in set $\{0,-1\}$ then $0\leq \mbox{Rank}(A)\leq n$ and all of number s between $0$ and $n$ could be taken.

[quote="Omid Hatami"]...then $0\leq \mbox{Rank}(A)\le n$... [/quote]
Are you sure ? It's not true that $\mbox{Rank}(A)=-\mbox{trace}(A)$ ? This is my opinion,but if you have... a counterexample
 please inform me .Regards,Alex

[quote="harazi"]It is true, since the rank of any projector is equal to its trace and here $-A$ is a projector.[/quote]
Very nice,Alex


Find all positive integer solution of the equation
$ (\frac{x}{y})^{2}\minus{}(y\minus{}1)^{2}\equal{}4x\minus{}1$


function
logarithms
inequalities
inequalities proposed

Let $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}n$. Prove that:
\[ (a_1^2\plus{}a_1\plus{}1)(a_2^2\plus{}a_2\plus{}1)...(a_n^2\plus{}a_n\plus{}1)\le 3^n\]

Happy birthday, my brother!

Your problem is nice but.. easy. Use the same trick we recently make use of  :lol:

happy birth to you
your problems are nice
your book is very interesting.

As a reference for Hoang's problem: For case $ n\equal{}3$ it is posted in Diendantoanhoc (someone proposed, but I dont remember who). For case $ n\equal{}4$, it is posted too, by me, recently.

General of Hoang can be proved by the same method.

Consider the function $ f(x)\equal{}\ln{(x^2\plus{}x\plus{}1)}\minus{}(x\minus{}1)$

It has maximum at x=1 so we get the desired inequality . :)

BTW , happy birthday Hoang  :)

[quote="silouan"]Consider the function $ f(x) \equal{} \ln{(x^2 \plus{} x \plus{} 1)} \minus{} (x \minus{} 1)$

It has maximum at x=1 so we get the desired inequality . :)

BTW , happy birthday Hoang  :)[/quote]

Why can you find out the function $ f(x) \equal{} \ln{(x^2 \plus{} x \plus{} 1)} \minus{} (x \minus{} 1)$?
Sorry, I only be the member of My website recently! Need help, please! Thanks!

thank everybody  and silouan's solution is my solution too  :wink:. A well-known solution used to mixing variables method but it is not really simple :P  An inequality is very nice and it is an easy one. The following inequality can be solved by this method but It is not correct for all n-variables.
Let $ a,b,c,d\ge 0$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that:
\[ (a^2 \minus{} a \plus{} 1)(b^2 \minus{} b \plus{} 1)(c^2 \minus{} c \plus{} 1)(d^2 \minus{} d \plus{} 1)\ge 1
\]

It is also true for 7-variables  :roll:
@anh Hùng: anh ổn định chỗ ở chưa, chắc anh đi học rồi nhở, chúc anh may mắn nhé  :P

[quote="FOURRIER"]The inequality is equivalent to:
$ \sum ln(a_{i}^2 \plus{} a_{i} \plus{} 1)\leq nln(3)$
but jensen doesnt help here  :mad:[/quote]

We prove it term by term, is your proof zaizai hoang the same?

$ ln(a_{i}^2 \plus{} a_{i} \plus{} 1)\leq ln(3)$

No, I don't think that it is a best way, and maybe the function is not correct $ \forall a,b,c\ge 0$. Here is my proof. I consider to the following function:
\[ \frac {\ln{3^n}}{n} \minus{} \ln{(a^2 \plus{} a \plus{} 1)} \plus{} a \minus{} 1\ge 0,\forall a,b,c\ge 0
\]
It is not hard to see that the minimum value of function hold on $ a \equal{} b \equal{} c \equal{} 1$. Multiplying them with each other, we have done. The base of this method is Fermat's theorem.

$ \sum_{k \equal{} 1}^n \log_3 (a_k^2 \plus{} a_k \plus{} 1)\leq \sum_{k\equal{}1}^n \left(\frac{1}{\ln 3}a_k\plus{}1\minus{}\frac{1}{\ln 3}\right)\equal{}n.$

[quote="zaizai-hoang"]The base of this method is Fermat's theorem.[/quote]

Wich one? can you give me a link  please?
Thanks  :bye:

[quote="nktp"][quote="silouan"]Consider the function $ f(x) = \ln{(x^2 + x + 1)} - (x - 1)$

It has maximum at x=1 so we get the desired inequality . :)

BTW , happy birthday Hoang  :)[/quote]

Why can you find out the function $ f(x) = \ln{(x^2 + x + 1)} - (x - 1)$?
Sorry, I only be the member of My website recently! Need help, please! Thanks![/quote]

we have the original inequality is equivalent to:
$ \sum \limits_{i = 1}^n \ln{(a_{1}^2 + a_{1} + 1)} \le n \ln{3}$
SO we will find the constant $ k$ such that $ f(x) = \ln{(x^2 + x + 1)} + a(x - 1) - \ln{3} \le 0$.
Let $ f^'(1) = 0$ we have $ a = - 1$.


On each lattice point $(a,b)$ with $a,b < 0$, there is a cute little mouse, and all the other lattice points are empty. Each move, you can let a mouse $(A)$ jump over one of his neighbours $(B)$ in horizontal or vertical direction, to an empty lattice. You remove the mouse $B$ from the lattice.

On which lattice points in $\mathbb{Z}^{2}$ can you get a mouse?

[hide="a suggestion?"]You could assign to each point in the lattice a value $f(x, y)$ s.t. $f(x, y)+f(x+1, y)=f(x+2, y)$, $f(x, y)+f(x, y+1)=f(x, y+2)$.  Then, the sum of all values in the plan can never increase.  So then let, WLOG, $f(-1,-1)=1$, and let $f(-1,-2)=f(-2,-1)=\frac{2}{1+\sqrt{5}}=\frac{1}{\phi}$.  We have $1+\frac{1}{\phi}= \phi$
\[\implies \phi^{n}+\phi^{n+1}=\phi^{n+2}\]

Then, \[f(x, y)=\phi^{-x-y+2}\]  So the sum of the values of all of the points on the plane is \[1+\frac{2}{\phi}+\frac{3}{\phi^{2}}+...=(1+\phi^{-1}+\phi^{-2}+...)^{2}=\frac{1}{(1-\frac{1}{\phi})^{2}}\]
\[=\frac{1}{(\frac{1}{\phi^{2}})^{2}}=\phi^{4}\]
We thus conjecture that we can have a mouse on any point $(x, y): x+y<2$ (in order to get to a point with value $\phi^{4}$, we'd have to use all of the mice).  But then, in order to get to points with very large magnitude x and y, we'd need to make downward jumps...so perhaps it whould just be the points with $x, y \le 1$, not including $(1, 1)$?[/hide]

[b]Solution.[/b]

Let $f(x)=x^{2}+x-1$ and let $\alpha$ be such that $f(\alpha)=0$, $\alpha>0$.  Consider the norm $n(x,y)=|x-a|+|y-3|, a\in\mathbb{Z}_{<0}$ assigned to each point $(x,y)$.

Assign to each point $(x,y)$ the weight $\alpha^{n(x,y)}$. Consider the set of points $(x,y)\in\mathbb{Z}^{2}_{x,y<0}$. The total weight of the board is therefore
\[\\ W=\sum_{(x,y)\in\mathbb{Z}^{2}_{x,y<0}}\alpha^{n(x,y)}=\sum_{m\geq 0}\sum_{x<0}\alpha^{|x-a|+4+m}\\ =\sum_{i>0}\frac{\alpha^{|-1-a|+4+i}}{1-\alpha}\\ =\frac{\alpha^{|-1-a|+4}}{1-\alpha}(1+\alpha+\alpha^{2}+\hdots)\\ =\alpha^{|-1-a|}\in(0,1] \]
Note that the total weight of the board is invariant upon a vertical jump upward or a horizontal jump towards the line $x=a$.  The weight of point $(a,3)$ is $\alpha^{n(a,3)}=\alpha^{0}=1$.

You can clearly reach lattice points $(x,2), x<0$. Since $W\in (0,1]$ and you have finitely many mouses, you can never reach points $(a,3)$ such that $a<0$.

Masoud Zargar

[quote="boxedexe"]You can clearly reach lattice points $(x,2), x<0$. Since $W\in (0,1]$ and you have finitely many mouses, you can never reach points $(a,3)$ such that $a<0$.[/quote]
Are you sure of your computations? I think that $(a,3)$ can be reached, say for $a <-8$... But it's a very long procedure, so I may have made some mistakes :blush:

Anyway, I'm $100 \%$ sure that you can't reach $(a,4)$ since I saw a similar problem before, in which you could place mice anywhere in the half-plane $y < 0$.

this problem is in acops, just with all lattice covered with nonegative y component,

And do they give the same solution in ACoPS as boxedexe's ?

That was my solution. I had seen a similar problem like this is Problem Solving Strategies, so I just used that method.  Just clarifying any misunderstanding about whether that was my solution or not.  By the way, is my conclusion correct?

Ok, no worries boxedexe  :)

I'm absolutely convinced that you found this solution yourself (And I'm quite sure it's a correct solution)

I thought that maybe in ACoPS, there was a different solution from yours, and I would find it most interesting to hear an alternative. That's the only reason why I asked this question to Altheman (and other people who read ACoPS).


topology
function
calculus
integration
real analysis
advanced fields

[b]Question[/b] Let $ (X,{\mathcal X},\mu)$ be a finite measure space. Construct a topology on $ L(X)$ (the space of measurable functions from $ X$ to $ \mathbb C$) such that a sequence $ f_n \in L(X)$ converges to $ f \in L(X)$ in this topology if and only if $ f_n$ converges to $ f$ pointwise almost everywhere.

One needs to find a set $ \mathcal U$ such that $ f_n$ converges to 0 almost everywhere if and only if for all $ U \in \mathcal{U}$ there is $ N$ such that for all $ n \ge N$ we have $ f_n \in U$. I can't find such $ \mathcal U$. There should be if the question has positive answer.

My question comes from Exercise 8 in the article [url=http://terrytao.wordpress.com/2009/02/21/245b-notes-11-the-strong-and-weak-topologies/]strong and weak topologies - Terence Tao[/url]

[quote][b]Exercise 8[/b] Let $ (X,{\mathcal X},\mu)$ be a $ \sigma$-finite measure space. Construct a topological vector space structure on $ L(X)$ (the space of measurable functions from $ X$ to $ {\mathbb C}$) such that a sequence $ f_n \in L(X)$ converges to $ f \in L(X)$ if and only if $ f_n$ converges to $ f$ pointwise almost everywhere. (Hint: use Egoroff’s theorem.) [I do not know if one can topologise pointwise almost everywhere convergence in the non-$ \sigma$-finite setting, unless one restricts attention to functions with $ \sigma$-finite support.] [/quote]

The hint to use Egoroff's theorem didn't help me but it may help you.

One may do this by constructing a set of metrics $ d_\alpha(\cdot,\cdot)$ such that $ f_n$ converges to 0 if and only if $ d_\alpha(f_n,0)$ converges to 0 for all $ \alpha$. One should not rely on integration alone when trying to create such metrics because many convergence theorems for pointwise convergence also hold for convergence in measure.

If you cannot prove some claim, it is a perfectly valid reason to spend some time trying to [i]disprove[/i] it regardless of who made that claim. ;) If you have trouble doing that either, here is a small
[hide="Hint"]
Take $ X\equal{}[0,1]$ with the usual Lebesgue measure and consider 2 cases:
1) For every open set $ U$ containing $ 0$ there exists $ \delta>0$ such that all functions supported on an interval of length less than $ \delta$ belong to $ U$.
2) There exists an open set $ U$ containing $ 0$ such that for all $ \delta>0$, there exists a function $ f$ supported on an interval of length less than $ \delta$ that does not belong to $ U$.
Show that in the first case your topology is too weak and in the second case too strong.
[/hide]

Whenever I made a topology in which every a.e. pointwise convergent sequence converges, verifying the converse direction failed and the reason was always same : all in-measure convergent sequence converge in that topology as well. This suggests the following :

Let $ \mathcal T$ be a topology on $ L[0,1]$. The followings are equivalent:
(i) If the sequence $ f_n \in L[0,1]$ converges pointwise a.e. to $ f$ then it converges topologically to $ f$.
(ii) If the sequence $ f_n \in L[0,1]$ converges in measure to $ f$ then it converges topologically to $ f$.

Proof : Use the fact that $ f_n$ converges to $ f$ in measure if and only if every subsequence has in turn a subsequence converging pointwise a.e. to $ f$.


Next saturday (tomorrow) the MIT regular action decisions will be released online!
Post here if you were accepted! your name, where are you from, and what you plan to major in!

I was.

Singapore, Math (of course :) ) and physics.

Mel. Citizen of Earth. Major: math?

*does happy dance*

[quote="manuel"]Post here if you were accepted![/quote]you mean.. this year?

Geehoon is teasing you because he was also accepted by MIT ... in 2005.

oh yeah.. ha i got it

Congrats Valowik!
hope to see u there!

Bah bah bah I got a shake of the head.   :noo:   :mad:

Wow, you're looking at the exact same three colleges I'm looking at.

[quote="joml88"][quote="mel"][quote="joml88"]Accepted :) :) - would probably major in math...still undecided on where I'm going though.[/quote]

Congrats! And if you don't mind sharing, where else are you looking at?[/quote]

University of Michigan and Princeton[/quote]

Those are great choices! Congrats again!

Just to swell aops's list of miters.... i got in early. Waiting on Princeton/Yale, but almost certainly going to MIT. Joml, have you looked at any of the dorms?

Simmons is tha besttttt

i hanged out in novmer for  a week in mit.. and simmons is really cool.. or baker

Bexley is pretty cool.  It's quite ghetto, but the location is good and the building is really cool.  Unfortunately I don't get to go to MIT.

I got waitlisted, so we'll see about that. Also looking at UC san Diego and Harvey Mudd (pending admission on HM).

I got waitlisted too.  But at least I made USAMo.

Hi I was also adMITted regular decision for class of 2010.
I am from Bulgaria.
Majoring i maybe physics and/or math(of course).
Hope to se ya soon :D

where are you planning to stay? im going to simmons !

hehehe, the metal waffle :-P i get to see it from my room now!

next year i'm moving off campus to the Women's Independant Living group, its an off campus independant living group about a block past random on Mass Ave, looking forward to it :)

You know, I was leaning toward Simmons, but then I found out that we had to clean our own bathrooms....   :o I'm not ready for that yet

I stayed in Next during CPW, though, and it was good  :lol:

[quote="daveed"]You know, I was leaning toward Simmons, but then I found out that we had to clean our own bathrooms....   :o I'm not ready for that yet

  [/quote]
FWIW -  If you play your cards right, the chances are that one of the other ones  ( who shares the bathroom) ) actually likes to do that.  My son has passed two years in Simmons without cleaning up duty :)

I am going to go a little bit of topic. Really, I should have strated to a new thread for this but I thought that this is the best place to get my questions answered since all of you are accepted into MIT.

Can anyone please tell me how I can get into MIT? What exactly do I need to get into MIT? Like what kind of marks, extra-curriclar activities, awards, etc. I want to become a quant, so I really want to get into MIT. I get excellent marks in math (99). I also won bronze level certificates and medals for Canadian Math Contests. You think I can get into MIT? If you need more info about me, please feel free to ask me.
Also, since I am planning to get a PhD. in mathematics, I can do my undergraduate degree at University of Toronto or University of Waterloo (I live in Canada so this will be cheaper for me) and then I can do my PhD at MIT. Is it possible to get into PhD at MIT with an undergraduate degree from Univesity of Toronto or University of Waterloo.

Any help will be appreciated.

Satyam - Like any other good school you have to apply to MIT and if you are selected  you are in. There is no single golden rule which makes you  in or out. Better a student you are (or better the admisson office at MIT think the student you are :) better are your chances.  You can look at statistics at MIT's web site to see the average SAT scores of students there but it may not mean much, as it is only a vauge guide to compare yourself with them.

Of course having good grades, and medals in math contests is somethng MIT looks as a positive.

There are plenty of Candian students in MIT, and  there have been many who do/did  PhD after graduating form U or T  or Waterloo.  Again your chances for graduate school depends on your transcripts etc.

Sorry if this is not very helpful but honestly  this is the most one can say..

Good luck and all the best..

The University of Toronto and the University of Waterloo are the two best schools in Canada for math, and probably both are in the top 10 in North America in terms of math (certainly if the Putnam is a good indicator, which is probably isn't).  So yes, they are good places to jump to MIT from, but MIT probably would accept one or two people from each school for Ph.D work at most, and probably none most years.  I think MIT only accepts about 20 people a year into their Ph.D program, so it is very competitive.  Doing well in your university coursework, on the Putnam, and doing research will help you (or hurt you in their absence) immensely.

You should probably detail your math contest results more, but most people need a combination of good test scores, good grades, and demonstrated leadership or excellence in extracurricular activities to get into MIT.


Prove that for all positive integers $n$ the following inequality takes place  \[ (2n^2+3n+1)^n \geq 6^n (n!)^2 . \]
[i]Cyprus[/i]

[hide="Solution"]By AM-GM, we have

$\frac{1^2+2^2+3^2+\cdots+n^2}{n} \ge \sqrt[n]{1^22^23^2\cdots n^2}$, or

$\frac{n(n+1)(2n+1)/6}{n}=\frac{(n+1)(2n+1)}{6} \ge \sqrt[n]{(n!)^2}$.

Raising both sides to the $n^{th}$ power and multiplying both sides by $6^n$ gives $((n+1)(2n+1))^n \ge 6^n \cdot (n!)^2 \Rightarrow (2n^2+3n+1)^n \ge 6^n \cdot (n!)^2$ as desired.[/hide]


arithmetic sequence
number theory
relatively prime

[quote="4everwise"]How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation)

$\text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 8$[/quote]

[hide]7,17,37,47,67,97 = 6 numbers (C)[/hide]

[hide]One possibility for the units digit which is 7. 6 possibilities for the tens digit which means that there are only 6 numbers less than 100 that are primes which end in 7.

You can use divisibility rules to check. Only one you really need to check for are the ones divisible by 3.[/hide]

[quote="shadysaysurspammed"]What if instead of $100$ it was $1000$?It wouldnt be so convinient to do each one manually(Don't say then this becomes a question for intermediate forum :P  )[/quote]
About that, do you know a way where you can do it a different way than counting and is quite time efficient? :(

[quote="shadysaysurspammed"]What if instead of $100$ it was $1000$?[/quote]

[hide="Then..."] We would eliminate the multiples of $3, 7, 11, 13, 17, 19, 23, 29, 31$ that ended in $7$ (these aren't so hard to find with modular arithmetic).  The numbers ending in $7$ left must be prime.

The annoying part is that you must be careful not to overcount, say, multiples of $33$. [/hide]

[quote="4everwise"]How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation)

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$[/quote]
Those numbers are $7, 17, 27, 37, 47, 57, 67, 77, 87, 97$
Since the multiple of 3 that is the closest one to 7 (but is greater) is 2 so when the tens digit is 2, 5 or 8, then it won't be prime. You are left with $7, 17, 37, 47, 67, 77, 97$ instantly eliminate 77 (divisible by 11, 7). and you are left with 6 numbers. (C)

[quote="shadysaysurspammed"]What if instead of $100$ it was $1000$?It wouldnt be so convinient to do each one manually(Don't say then this becomes a question for intermediate forum :P  )[/quote]

Isn't there some "primes in an arithmetic progression" theorem?

[quote="mathnerd314"]Isn't there some "primes in an arithmetic progression" theorem?[/quote]

You could be referring to a couple of different results, neither of which are particularly helpful.

[b]Theorem:[/b]  Given relatively prime integers $a, d$, there are an infinite number of primes of the form $ak+d$.  (Notice that this theorem says nothing about how to count them.)

[b]Theorem:[/b]  There exist arbitrarily long sequences of primes in arithmetic progression.

Edit:  There is a third theorem about the [b]asymptotic[/b] density of primes of the form $ak+d$ less than $N$, but I don't recall its exact form.


algebra
polynomial
superior algebra
superior algebra solved

If p > 2 is prime, show that $\Phi_{2p}(x)=x^{p-1}-x^{p-2}+...+x^2-x+1$.

Could anyone help me to proof it? Thanks a lot!

Let $1\neq\varepsilon$ be a $p$-th root of unity. Then $-\varepsilon$ is a primitive $2p$-th root of unity.
Since $\phi_p(\varepsilon)=0$, we have $f(-\varepsilon)=0$ where $f(X)=\phi_p(-X)$. $f$ is irreducible if and only if $\phi_p$ is irreducible, and $\phi_p$ is irreducible. Since $f$ is a monic irreducible polynomial and $f(-\varepsilon)=0$, $f$ must be the minimal polynomial of $-\varepsilon$ which is by definition $\phi_{2p}$.
But $f(X)=\phi_p(-X)=X^{p-1}-X^{p-2}+\ldots-X+1$


Let $ a$ be an integer.Prove that $ (a\plus{}1)(a\plus{}3)(a\plus{}7)\plus{}17$ isn't perfect square of integer $ ?$

[quote="Obel1x"]Let $ a$ be an integer.Prove that $ (a \plus{} 1)(a \plus{} 3)(a \plus{} 7) \plus{} 17$ isn't perfect square of integer $ ?$[/quote]
hello!
what for $ a\equal{}1$ !
$ (1\plus{}1)(1\plus{}3)(1\plus{}7)\plus{}17\equal{}81$ it's a perfect square.

[quote="mathematikos"][quote="Obel1x"]Let $ a$ be an integer.Prove that $ (a \plus{} 1)(a \plus{} 3)(a \plus{} 7) \plus{} 17$ isn't perfect square of integer $ ?$[/quote]
hello!
what for $ a \equal{} 1$ !
$ (1 \plus{} 1)(1 \plus{} 3)(1 \plus{} 7) \plus{} 17 \equal{} 81$ it's a perfect square.[/quote]

I need prove for general case $ ?$.

for $ a\equal{}1$ it's a perfect square and for $ 2$ isn't,so what we have to prove?

[quote="mathematikos"]for $ a \equal{} 1$ it's a perfect square and for $ 2$ isn't,so what we have to prove?[/quote]

??? we have to prove for $ a\in {\minus{}n,...,\minus{}3,\minus{}2,\minus{}1,0,1,2,3,...n}$ . Are u going to prove for all these numbers :)  ?

I'm sorry , I had problems with English  :oops:  :(

[hide]There is a typo in your problem
I think it should be : $ (a \plus{} 1)(a \plus{} 3)(a \plus{} 5)(a \plus{} 7) \plus{} 17$[/hide]

[quote="enndb0x"]There is a typo in your problem
I think it should be : $ (a \plus{} 1)(a \plus{} 3)(a \plus{} 5)(a \plus{} 7) \plus{} 17$[/quote]

Let $ a \equal{} b\minus{}4$

$ (b\minus{}3)(b\minus{}1)(b\plus{}1)(b\plus{}3) \plus{} 17 \equal{} k^2$

$ b^4 \minus{} 10b^4 \plus{} 26 \equal{} k^2$

Let $ c \equal{} b^2 \minus{} 5$

$ (c)^2 \plus{} 1 \equal{} k^2$

This clearly has no solutions

$ (a \plus{} 1)(a \plus{} 3)(a \plus{} 5)(a \plus{} 7) \plus{} 17\equiv 2(mod 3)$,which can never be a perfect square. By the same way,we ca prove that $ a(a \plus{} 2)(a \plus{} 4) \plus{} 17$ also can't be a perfect square.

You can also see that if $ a$ ends with one of the digits $ {2,3,4,5,7,8,9}$ then $ (a\plus{}1)(a\plus{}3)(a\plus{}5)(a\plus{}7) \plus{}17$ ends with $ 2$ or $ 7$,so thats not a perfect square.
Now we just use modular arithmetic for case $ a$ ends with $ 1$ or $ 6$  :blush:

[quote="geniusbliss"][quote="ocha"][quote="enndb0x"]There is a typo in your problem
I think it should be : $ (a \plus{} 1)(a \plus{} 3)(a \plus{} 5)(a \plus{} 7) \plus{} 17$[/quote]

Let $ a \equal{} b \minus{} 4$

$ (b \minus{} 3)(b \minus{} 1)(b \plus{} 1)(b \plus{} 3) \plus{} 17 \equal{} k^2$

$ b^4 \minus{} 10b^4 \plus{} 26 \equal{} k^2$

Let $ c \equal{} b^2 \minus{} 5$

$ (c)^2 \plus{} 1 \equal{} k^2$

This clearly has no solutions[/quote]
really nice how did u think of those substitutions? intuition? or logic?[/quote]The first one is standart(It's the AM of $ (a \plus{} 1),(a \plus{} 3),(a \plus{} 5),(a \plus{} 7)$). The second one is obvious ;).However I like that solution(not as much as mine, of course  :P  )

[quote="JBL"]The statement asks you to show that something is never a perfect square.  The other poster has shown that this is false by providing a counter-example, i.e., a specific value of $ a$ for which it [i]is[/i] a perfect square.  The statement is false, so there is nothing else left to be done.  Pay attention![/quote]

and you are a senior member of this forum...


Find the area of the plygon formed by all points (x,y) in the plane satisfying the inequality
||x|-2|+||y|-2|(less than or equal to) 4.

Anyone who knows how to solve such problem?

$ \left | |x|-2 \right |+\left | |y|-2 \right | \le 4$
Observe that if $ P1(u,v)$ satisfies this inequality $ P2(-u,v),P3(u;-v);P4(-u;-v)$ also satisfy it. So the polygon is symmetric with respect Ox and Oy and we can assume WLOG that $ x \ge 0$ and $ y \ge 0$ obtaining $ |x-2|+|y-2| \le 4.$
It is obvious that if $ M(u,v)$ is a solution so is $ N(v,u)$. The polygon is symmetric with respect $ y=x$ so we can assume $ y \ge x$

Case 1. Let $ 0 \le x \le 2, \& 0 \le y \le 2$ than $ 2-x + 2-y \le 4$ or $ x+y \ge 0$ with solutions (remember $ y \ge x$) the points from the blue triangle.

Case 2. $ 0 \le x \le 2, \&  y \ge 2$ we get $ 2-x+y-2 \le 4 \Rightarrow y\le x+4$. Points that satisfy all the restrictions are from the green trapezoid.

Case 3. $ x>3, \& 0\le y\le 2$ - this is impossible.

Case 4. $ x>4, \& y>2$ then $ x-2+y-2 \le 4 \Rightarrow x+y \le 8$. The red triangle.

It is easy to calculate the area of the figure and dont forget to multiply by 8.

thanks for providing the solution. I will try to verify of the answer is 96 square units using your solution. I only have the final answer but dont know hoe to get it.

Btw, what software did you use in drawing the graph?

Thanks!


Three circles of radius $ 1$ are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?

[asy]unitsize(0.8cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

real r = 1 + (2/3)*(sqrt(3));
pair A = dir(47.5)*(r - 1);
pair B = dir(167.5)*(r - 1);
pair C = dir(-72.5)*(r - 1);

draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(C,1));
draw(Circle(origin,r));[/asy]

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

[hide]
Join the centers of each circle to get an equilateral triangle with sidelength, $ 2$.

By pythag, the height of the triangle is $ \sqrt {2^2 \minus{} 1^2} \equal{} \sqrt {3}$
Since the height is also the median of the equilateral triange, the point where all three medians meet cuts each line into the ratio, $ 1: 2$.

So the radius of the larger circle is; $ 1 \plus{} \frac {2}{3}\cdot \sqrt {3} \equal{} \frac {3 \plus{} 2\sqrt {3}}{3} \longrightarrow \boxed{D}$[/hide]

How would you prove that the centroid of the equilateral triangle that connects the centers of the smaller circles is the same point as the center of the larger circle?

[hide="Alternative solution"]Three circles have curvature $ 1$.
The fourth has curvature $ c$.

By Descartes' Theorem, $ (3\minus{}c)^2\equal{}9\minus{}6c\plus{}c^2\equal{}2(3\plus{}c^2)\equal{}6\plus{}2c^2$
Therefore, $ c^2\plus{}6c\minus{}3\equal{}0\implies c\equal{}\frac{\minus{}6\pm4\sqrt{3}}{2}\equal{}\minus{}3\plus{}2\sqrt{3}$.

The radius must then be $ \frac{1}{\minus{}3\plus{}2\sqrt{3}}\cdot\frac{\minus{}3\minus{}2\sqrt{3}}{\minus{}3\minus{}2\sqrt{3}}\equal{}\frac{\minus{}3\minus{}2\sqrt{3}}{\minus{}3}\equal{}\boxed{\textbf{(D)}\ \frac{3\plus{}2\sqrt{3}}{3}}$

[url=http://en.wikipedia.org/wiki/Descartes'_theorem]Descartes' Theorem on Wikipedia[/url][/hide]

[quote="varunrocks"]ause do u use three because there are three circles with radius 1.[/quote]

And what does "ause" mean?


Show that the next integer above ${(\sqrt{3}+1)}^{2n}$ is divisible by $2^{n+1}$

Hint :
[hide]
$u_n = {(\sqrt{3}+1)}^{2n}$
$v_n = {(\sqrt{3}-1)}^{2n}$

Then  next integer above $u_n$ is $u_n+v_n$, then  ${(\sqrt{3}+1)} \cdot (\sqrt{3}-1) = 2$ and then induction[/hide]

[hide]Or also: we can see from the binomial rules and from the Tartaglia.Pascal triangle we can see that, $\displaystyle n=2m\to \sum_{i=0}^{m} \binom{n}{2i}=\sum_{i=0}^{m-1} \binom{n}{2i+1}=2^{n-1}$, so we can factorise and collecting the factor $(1+\frac{1}{\sqrt{3}})$[/hide]


[b]part-I[/b]

i was looking at a common right triangle the other day, and noticed that when i computed its perimeter i got the same number as when i computed its area though of course the units were differnt , being centimeters for perimeter and square centimeter for area.

(a) what right triangle might it have been ?
(b) given that the triangle had intger sides,what are all possible triangle?
(c)if you are not given that the sides were all integers,what relationcan you find among the sides?are there any limiting special cases?

[b][color=violet][size=150]part-II[/size][/color][/b]

after i  finished thinking about right triangle. i got to wondering about whether this property(of having perimeter in centimeters equalling area in square centimeters) was possible for other types of triangles.

(a)are there isosceles trianles with intger side that have this property ? if so find them. if not, explain why not.
(b)what if just the two legs areintegers or if just the base is an integers?
(c)what relation are there between the legs and the base?are there any special limiting case?

$ a \plus{} b \plus{} c \equal{} S \equal{} r\ \frac {a \plus{} b \plus{} c}{2}$ $ \Longrightarrow$ $ r \equal{} 2$

For right triangle with legs $ a$, $ b$, and hypotenuse $ c$, $ r \equal{} \frac {a \plus{} b \minus{} c}{2}$. Assuming $ a \equal{} m^2 \minus{} n^2$, $ b \equal{} 2mn$, $ c \equal{} m^2 \plus{} n^2$ are positive integers $ \Longrightarrow$

$ n(m \minus{} n) \equal{} 2$ $ \Longrightarrow$ $ (m, n) \equal{} (3, 1)$ or $ (3, 2)$ $ \Longrightarrow$ $ (a, b, c) \equal{} (8, 6, 10)$ or $ (5, 12, 13)$.

For isosceles triangle with base $ b$ and other sides $ a$,

$ \frac {b}{2} \sqrt {a^2 \minus{} \frac {b^2}{4}} \equal{} r\ \frac {b \plus{} 2a}{2}$

$ a^2 \minus{} \frac {4br^2}{b^2 \minus{} 4r^2}\ a \minus{} \frac {b^2}{4} \cdot \frac {b^2 \plus{} 4r^2}{b^2 \minus{} 4r^2} \equal{} 0$

$ a \equal{} \minus{} \frac {b}{2}$ or $ a \equal{} \frac {b}{2} \cdot \frac {b^2 \plus{} 4r^2}{b^2 \minus{} 4r^2}$

Discarding the negative root,

$ b^3 \minus{} 2ab^2 \plus{} 4r^2b \plus{} 8ar^2 \equal{} 0$

Assuming $ a \ge 2r \sqrt 3$ and $ r \ge 1$ are integers, $ b \equal{} 2n > 2r$ must be even,

$ n^3 \minus{} an^2 \plus{} r^2n \plus{} ar^2 \equal{} 0$

Possible positive integer roots of the last cubic are $ n \equal{} a$, $ ar$, $ ar^2$, $ r^2$, $ r$ or $ 1$. Last two are eliminated by $ n > r > 1$ and the remaining four lead to $ a \equal{} 0$, $ a^2 \equal{} \frac {1 \plus{} r}{1 \minus{} r}$, $ a^2 \equal{} \frac {1}{r^2} \cdot \frac {1 \plus{} r^2}{1 \minus{} r^2}$ and $ a \equal{} r^2 \cdot \frac {r^2 \plus{} 1}{r^2 \minus{} 1}$. First three are impossible for $ r > 1.$ The only remaining one (a), does not lead to the positive root $ n \equal{} r^2$ and (b), is not integer for any integer $ r > 1$, including $ r \equal{} 2$ $ \Longrightarrow$ $ a, b, r$ cannot be all integers.


Prove that in any triangle $ ABC$ exists the relation $ \sum\sqrt {\frac {b \plus{} c \minus{} a}{a}} \plus{} 2\cdot\sum\frac {a}{b \plus{} c}\ \ge \ 6\ .$

I think he meant for the radical to also be cyclic.

the inequality<=>$ \sum \sqrt{\frac{b\plus{}c\minus{}a}{a}} \geq 2\sum \frac{b\plus{}c\minus{}a}{b\plus{}c}$
Remember that:$ (b\plus{}c) \geq 2\sqrt{b\plus{}c\minus{}a}*\sqrt{a}$
so we get:$ \sqrt{\frac{b\plus{}c\minus{}a}{a}} \geq 2 \frac{b\plus{}c\minus{}a}{b\plus{}c}$
problem solved

[quote="tuandokim"][quote="Virgil Nicula"]Prove that in any triangle $ ABC$ exists the relation $ \sqrt {\frac {b \plus{} c \minus{} a}{a}} \plus{} 2\cdot\sum\frac {a}{b \plus{} c}\ \ge \ 6\ .$[/quote]
Try a=b=c;[/quote]
Thank you. I corrected.


How can we prove that if $A$ is diagonalizable then there exists $B$ such that $B^{2}=A$. Is the reciprocal true? :maybe:

I guess that [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=symmetric+root+square&t=24211]this[/url] is what you want.


trigonometry
AMC
AIME
geometry
geometric transformation
reflection
function

I had a question about the law of cosines. I thought the formula was
 $a^2 = b^2 + c^2 -2bccosa$

But on problem #9 on the 1987 AIME, according to the solutions it says it is
$a^2 = b^2 + c^2 +bc$

Heres a link, http://www.kalva.demon.co.uk/aime/aime87.html

[quote="Arne"]Probably because $\alpha = 120^{\circ}$ in that particular situation.[/quote]

right, $\cos120^{\circ}=-0.5$, so the formula the AIME gave was right only for the angle 120 degrees.

[quote="Calculus"]Oh ok, I used my calculator to check the the value of cos120, and it was on radians mode, so I got a different value.

Thanks for the quick reply.[/quote]

cos 120 is something you should memorize  :lol:

[quote]cos 120 is something you should memorize[/quote]
Absolutely! But "memorize" might not be exactly the right word.

There are two steps to the process:

1. You know, without spending any time thinking, what the sine and cosine are for the first quadrant angles $\frac{\pi}6,\frac{\pi}4,$ and $\frac{\pi}3$ ($30^{\circ},45^{\circ},$ and $60^{\circ}.$)

2. You know how to reflect and repeat, as necessary, to extend knowledge of these trigonometric functions to other quadrants.

Exercise to confirm what I mean: what is $\cot(210^{\circ})$? (Absolutely no calculators!!)

Tangent/cotangent are positive in Quadrant III. This is $\cot (30+180)$, so we have $\cot 210=\sqrt{3}$.

[quote="Kent Merryfield"][quote]1. You know, without spending any time thinking, what the sine and cosine are for the first quadrant angles $\frac{\pi}6,\frac{\pi}4,$ and $\frac{\pi}3$ ($30^{\circ},45^{\circ},$ and $60^{\circ}.$)
[/quote][/quote]

You can easily memorize that as following:

sin($0^{\circ}$) = $\frac{\sqrt 0}{2}$
sin($30^{\circ}$) = $\frac{\sqrt 1}{2}$
sin($45^{\circ}$) = $\frac{\sqrt 2}{2}$
sin($60^{\circ}$) = $\frac{\sqrt 3}{2}$
sin($90^{\circ}$) = $\frac{\sqrt 4}{2}$

The same for cosine in reverse direction