Let $ a,b,c\geq0$ such that $ abc\equal{}1$ Prove that:

$ a(a^3\plus{}1)\plus{}b(b^3\plus{}1)\plus{}c(c^3\plus{}1)\geq \frac{7}{4}(a^2\plus{}b^2\plus{}c^2)$

it is trivial to show that

$ 4(a^4\plus{}a) \ge 7a^2$

Moreover, we could prove the stronger inequality with the equality holds clearly

[i]Let a,b, c be positive reals satisfying $ abc\equal{}1$. Prove that

$ a(a^3\plus{}1) \plus{}b(b^3\plus{}1)\plus{}c(c^3\plus{}1) \ge 2(a^2\plus{}b^2\plus{}c^2)$[/i].

[quote="great math"]it is trivial to show that

$ 4(a^4 \plus{} a) \ge 7a^2$

Moreover, we could prove the stronger inequality with the equality holds clearly

[i]Let a,b, c be positive reals satisfying $ abc \equal{} 1$. Prove that

$ a(a^3 \plus{} 1) \plus{} b(b^3 \plus{} 1) \plus{} c(c^3 \plus{} 1) \ge 2(a^2 \plus{} b^2 \plus{} c^2)$[/i].[/quote]
It is also very weak:
It comes from:
$ a^4\plus{}1\ge 2a^2$.

I'm really suprised that my inequality is very easy !!! :blush:  :oops:

In fact I began with a long equivalence !!!

$ a(a^3\plus{}1)\plus{}b(b^3\plus{}1)\plus{}c(c^3\plus{}1)\equal{}a^4\plus{}b^4\plus{}c^4\plus{}a\plus{}b\plus{}c \geq a^4\plus{}b^4\plus{}c^4\plus{}3 \geq 2(a^2\plus{}b^2\plus{}c^2)$


Yet when you plug in $x=50$, it's correct.  :rotfl:
With graphing, there are 2 solutions, both rational.

Sorry, I don't get these replies. I can only see 1 solution:

x/20 = (5/2) log x 50
2x/100 = log x 50
x/50 = log x 50
x = 50 log x 50
x = log x (50^50)
x^x = 50^50
x = 50

I cannot for the life of me see where you get this other solution of 2/5.
When you plug in 2/5 for x into the original equation it blatantly does not work! Am I missing something?

jimi

Oops; it seems I did miss something. Thanks for pointing that out greengrapes.

[b]ashrafmod[/b], why do you make double posting! :mad:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=714763#714763[/url]

Moderators,\ you should warn him of his conduct.

Japanese Communities moderator

kunny


Se da triunghiul $\ ABC$ in care $\ P$ este mijlocul laturii $\ BC$.Fie $\ M\in (AB),N\in (AC)$ astfel incat dreptele $\ MN$ si $\ BC$ sa fie paralele.Vom nota $\ Q$ intersectia dreptelor $\ MP$ si $\ BN$.Perpendiculara din $\ Q$ pe dreapta $\ AC$ intersecteaza pe $\ AC$ in $\ R$ si paralela prin $\ B$ la $\ AC$ in $\ T$.Aratati ca:
$\ a)$ $\ TP\parallel MR$
$\ b)$ $\angle MRQ\equiv\angle PRQ$
$\text{Virgil Nicula si Mircea Fianu}$

Probabil ca li se pare cunoscuta fiind data la o judeteana in 2001 sau.2002


probability
complementary counting
pifinity is ze best

Two different positive integers less than 100 are randomly chosen and multiplied together. What is the probability that the resulting product is a multiple of 3?

It seemed easier for me to compute the probability of selecting two integers that are not multiples of 3 $ \frac{66}{99}\cdot\frac{65}{98}\equal{}\frac{65}{147}$, subtract this from 1 to get the probability of getting a multiple of 3.

They probably want to introduce casework in the solution.

OK sorry for reviving this, but I don't understand.

When doing complementary counting, it goes $1-\frac{67}{100}*\frac{66}{99}=\frac{83}{150}$. The reason is that for the number to not be divisible by 3, we first need to select 1 of the 67 and then one of the 66 out of the 99. I don't understand what is wrong with that.

EDIT: Thanks prezcoin! (I don't want to post because I want to save my 1000th post for something epic, LOL.)

[hide]Note that the question specifically mentioned "[b]less[/b] than 100". Thus, we may not include the integer 100 in the set of numbers not divisible by 3. Otherwise, your solution is correct:

$1-\frac{66}{99}\times\frac{65}{98}=1-\frac{2}{3}\times\frac{65}{98}= 1-\frac{65}{147}=\boxed{\frac{82}{147}}$.[/hide]

I know that $\binom{99}{2}=4851$ is in the denominator. But for the numerator I calculated it as $\frac{33 \cdot 98}{2}$ (since at least one of the numbers must be a multiple of 3 and there are 33 multiples of 3 less than 100; after that, we can choose any number of the remaining 98 numbers) to get my answer was $1/3$. What did I do wrong?

You overcounted the pairs in which both of the numbers are multiples of 3.

Wait, how are the pairs being over counted in Mathaddict11's solution?

So lets say we pick $3$ as the first number. We could then pick $6$ as another option. However, we could have first picked $6$ and then picked $3$. This is still the pair, so we cannot overcount this.

But isn't that why it's divided by 2 to make it (33 x 98)/2 in the numerator? So it doesn't overcount?

I did it by finding the number of ways to have the resulting product not a multiple of two. The numerator would then be 66x65, and you have to divide it by 2 to remove repetitions. The denominator will be 99C2, as we have to choose 2 numbers out of the 99 positive integers less than 100 without any repetitions. This fraction will result in 65/147, and we subtract it from 1 to get the answer [b]82/147[/b].

@above you should divide it first into two cases, because you can't divide it by 2 if on of the numbers is divisible by 3 and the other isn't because they will never be the same. You can have 1 case where they are both divisible by 3, and another where they aren't. I think it would be easier to find the opposite case and then subtract it from 1 to get the answer.

First, I misread "less than" as less than or equal to. Then I missed "different"...
 :wallbash_red:

[quote=sosiaops]I got this problem wrong on my first try since I thought "less than" was "less than or equal to", then I got it correct on my second try[/quote]

bruhhhh same

[hide=easy sol]complementary counting, 66/99*65/98=1/3*65/49=65/147 chance unsuccesful, the answer is (147-65)/147=82/147[/hide]

A moment of silence in chat for those who included 100 in first try.

But, this is just complementary counting.

However, to be kewl, use casework

One of the numbers is a multiple of 3 which means we have 33 choices for one of the numbers, we don't care which so we don't multiply it by two. After that we need to get one of the 66 ones, so we 33*66

Both are multiples of 3, that is 33C2, out of 33 multiples pick 2, which means the answer is (33*66 + 33C2)/(99C2)

omg i got 12 rating up on overall c+p for this

I was stupid and thought there were 100 numbers


Let $a,b,c$ be three positive real numbers such that $a + b +c =1$ . prove that among the three numbers $a-ab, b - bc, c-ca$ there is one which is at most $\frac{1}{4}$ and there is one which is at least $\frac{2}{9}$.

[hide]Since $\sqrt{ab} \leq \sqrt{\frac{a^2 + b^2}{2}}$ we have $ab + bc + ca \leq a^2 + b^2 + c^2$.

Since $a + b + c = 1$ we have $ab + bc + ca \leq \frac{1}{3}$.

$x = a - ab$, $y = b - bc$, $z = c - ca$.

$a - ab + b - bc + c - ca = 1 - (ab + bc + ca) = x + y + z$.

Then $x + y + z \geq \frac{2}{3}$.

So there is a number between $x, y, z$ that is at least $\frac{2}{9}$.[/hide]

How would you prove that "there is one which is at most 1/4"?

[hide]  Suppose the contrary:
$ a\minus{}ab>\frac{1}{4}$
$ b\minus{}bc>\frac{1}{4}$
$ c\minus{}ca>\frac{1}{4}$

Then we have

$ 4a>\frac{1}{1\minus{}b}$
$ 4b>\frac{1}{1\minus{}c}$
$ 4c>\frac{1}{1\minus{}a}$

Adding up we have

$ 4>\frac{1}{1\minus{}b}\plus{}\frac{1}{1\minus{}c}\plus{}\frac{1}{1\minus{}a}$

But,

$ \frac{1}{1\minus{}b}\plus{}\frac{1}{1\minus{}c}\plus{}\frac{1}{1\minus{}a}\ge\frac{9}{3\minus{}a\minus{}b\minus{}c}\equal{}\frac{9}{2}>4$

by AM-HM.

[/hide]

[quote="Andreas"]Since $ \sqrt {ab} \leq \sqrt {\frac {a^2 \plus{} b^2}{2}}$ we have $ ab \plus{} bc \plus{} ca \leq a^2 \plus{} b^2 \plus{} c^2$.

Since $ a \plus{} b \plus{} c \equal{} 1$ we have $ ab \plus{} bc \plus{} ca \leq \frac {1}{3}$.

$ x \equal{} a \minus{} ab$, $ y \equal{} b \minus{} bc$, $ z \equal{} c \minus{} ca$.

$ a \minus{} ab \plus{} b \minus{} bc \plus{} c \minus{} ca \equal{} 1 \minus{} (ab \plus{} bc \plus{} ca) \equal{} x \plus{} y \plus{} z$.

Then $ x \plus{} y \plus{} z \geq \frac {2}{3}$.

So there is a number between $ x, y, z$ that is at least $ \frac {2}{9}$.[/quote]

How would you justify that  $ \sqrt {ab} \leq \sqrt {\frac {a^2 \plus{} b^2}{2}} \implies ab \plus{} bc \plus{} ca \leq a^2 \plus{} b^2 \plus{} c^2$?  I do see any connection between the two since the first inequality does not have the term $ c$.

$ a^2\plus{}b^2 \geq 2ab$
$ b^2\plus{}c^2 \geq 2bc$
$ c^2\plus{}a^2 \geq 2ca$
adding up   :)

We can actually prove a sharper inequality: $abc(1-a)(1-b)(1-c) \le \left(\frac{2}{9}\right)^3$
$abc \le \left(\frac{a+b+c}{3}\right)^3 = \frac{1}{27}$ and
$(1-a)(1-b)(1-c) \le \left(\frac{3-a-b-c}{3}\right)^3 = \frac{8}{27}$ by AM-GM
Therefore multiplying we get $abc(1-a)(1-b)(1-c) \le \left(\frac{2}{9}\right)^3$
We notice that the previous inequality could not be any equality since equality only would hold if $a = 1-a$, $b = 1-b$ and $c=1-c$ which cannot happen.

Are you sure that it is a true solution to that problem?


Is there a way to place infinitely many points on a plane in general position so that any finite subset of points defines a convex polygon ? (Of course, two different subsets (of points) need not necessarily define two distinct polygons)

Take a circle?  Or any countable subset of a circle.


Notice that the left side is increasing.
this is not a hard one I think,because we can solve it by having a trial.

hallo,
are there any solutions ?
i think there aren't !  :)


function
trigonometry
number theory
greatest common divisor

if a function f(x) = ax^3 + bx^2 + cx +d, where a,b,c,d are integers, and a>0 is such that f(sin10) = 0,then the smallest possible value of f(1) is what?[b] please note it is 10 degrees NOT10 radians[/b]

THANKS A LOT FOR THE SOLUTION IN ADVANCE!

As far as I can get: They are looking for the sum of the coeficciants. :blush:

[hide="hint"]express $\sin 3\theta$ in terms of $\sin\theta$[/hide]

[quote="Bictor717"][quote="tjhance"][hide]
$\sin3\theta=\sin2\theta\cos\theta+\sin\theta\cos2\theta$
$=2\sin\theta\cos^{2}\theta+\sin\theta(cos^{2}\theta-\sin^{2}\theta)$
$=\sin\theta(2\cos^{2}\theta+\cos^{2}\theta-sin^{2}\theta$
$=\sin\theta(3\cos^{2}\theta-sin^{2}\theta) =\sin\theta(3-3\sin^{2}\theta-\sin^{2}\theta)$
$=\sin\theta(3-4\sin^{2}\theta)$
$=3\sin\theta-4\sin^{3}\theta$

$\sin30=3\sin10-4\sin^{3}10$
$\frac{1}{2}=3\sin10-4\sin^{3}10$
$0=3\sin10-4\sin^{3}10-\frac{1}{2}$

Since it has have integer coeficients, we multiply by the GCD

$0=\6\sin10-8\sin^{3}10-1$

Then if we let $f(x)=-8x^{3}+6x-1$, then $f(\sin10)=0$. Then $f(1)=\boxed{-3}$

[/hide][/quote]
Very small problem: a>0 (given). Otherwise, multiply your f(x) by arbitrarily large integers to get more negative f(1).[/quote]

[hide="so then it would be this?"]
$f(x)=-8x^{3}+6x-1$
$f(x)=8x^{3}-6+1$
giving $f(1)=\boxed{3}$
[/hide]


Find all answers of:

$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$

in $\mathbb{N}$

$x^{2}+y^{2} | x^{2}y^{2}$
Let (x, y) = d. then $(x/d)^{2}+ (y/d)^{2} = d^{2}$.  So it looks like we can convert pythagorean triples to (any and all) triples which satisfy this equation:

(3,4,5)->(15, 20, 25)
(a, b, c)->(ac, bc, cc)


Let $a,b,c$ positive integers, we have:

$(\frac{a^4+b^4+c^4}{a+b+c})^\frac{a+b+c}{n} \geq a^ab^bc^c$

Determine the maximun value of $n$.

[hide]$a,b,c=1 \to 1^{\frac{3}{n}}\ge1$
$n$ can be any finitely large number[/hide]


calculus
integration
algebra
function
domain
advanced fields
advanced fields unsolved

[quote="bchui"]Is there a way to transform or solve the 2nd order PDE

$L(f)(x)=\sum_{i,j=1}^na_{ij}(x){{\partial^2 f(x)}\over {\partial x_i\partial x_j}} +\sum_{i=1}^nb_i(x){{\partial f(x)}\over {\partial x_i}} +a(x)f(x) =\varphi (x)$

Do we have a table, for which type of integral transform should be in use for a given type of $c_{i,j}(x),b_i(x),a(x)$ are $x-$dependent?  :oops:[/quote]
In general (for x-dependent coefficients, in a general domain) this cannot be done explicitly or with transform methods. Even the case  $\sum_{i=1}^n {{\partial^2 f(x)}\over {\partial x_i^2}} = 0$ gives rise to an entire mathematical field (potential theory).

I am looking for a list of existing results. Including for example the well-known Heat diffusion equation. ${{\partial f}\over {\partial t}} = a(x,t){{\partial^2f(x,t)}\over {\partial x^2}}$
It seems certain type of integral transform (Laplace/ Fourier for example) always work for certain type of $a(x),b_i(x),c_{i,j}(x)$ (say all equals to constants). Could we possibly draw a list for our ease to solve such problems?


videos
LaTeX
Support
geometry
3D geometry

Let's hear it. Which is the best? Is it DDR: :ddr: or star wars (which be, since it's over popular these days :D :D) :starwars: or is it the juggling trick :juggle:. Maybe some of you can showoff :showoff: that would be  :wow:

Bye for now :bye: :P

My vote's on :pilot: . It's just too darn CUTE!

Hm...I'd say either  :wow:,  :ddr: , or  :starwars:

:starwars:  is the best....no doubt.....

:spider: or :juggle:
they're both too cool to decide between...

:starwars: of course.

I'm in a Star Wars mood now.

:wallbash:  :wallbash_red:
these two are pretty good.  Wow I never realized there are new ones until I read this thread.

:oops: heehee....that's so me!

but this one:    :wallbash:        kinda fits in with an inside joke w/ a friend of mine.

:moose:  it's a moose!

i luv 'em all! they're really cool!

Haha...Valentin these are most kool. ^_^

 :spam: HAHAHAHAHA. Dude I feel like I need one of those signs.

 :gleam: <--- That is so me. Hahaha. Well, most of my sunglasses are blue. Oh well...

 :starwars:  :ewpu:

[quote="Xantos C. Guin"][quote="paladin8"]But wait! There's more!  :play_ball: <-- I think this guy automatically wins 'coolest emoticon ever' award. He's my idol.[/quote]

In your signature, you say ":play_ball: <---- That's me"
Are you reffering to the smile, the ball, or the wall?

:starwars: is my favorite.[/quote]

The ball. Duh. :rotfl:

*blink blink* where did you all find the rubik smiley and the aops smiley?

We found them while travelling lands far away.. full of special rare emoticons. but these emoticons are tricksy little fellas, so we had to be sneaky and careful when we tried to catch them.. but with our patience and hard work, we were able to catch the special emoticons and aquire them for our avatars.

no really...
[hide]we made (edited) them with photoshop or something like that[/hide]

:weightlift:  $\sqrt{24!}$  :weightlift:
math weights

I won't make a poll anymore. Clearly SW will win, so no point making a poll for that :)

[url=http://artofproblemsolving.com/community/c48h1106726_poll_for_emoticon]the poll[/url]
how many people like it?


linear algebra
matrix
inequalities
function
triangle inequality

Supposed $ A$ is a $ n\times n$ matrix with positive real entries and the sum of all its entries is $ n$, prove that $ \left|\det \left(A\right)\right| \leq 1$.

This is Vojtech Jarnik Competition 2006, Category I, Problem 4 [b](a)[/b]. Let's extend it a bit, dropping the positivity of the entries and instead letting $ n$ be the sum of their [i]absolute values[/i]:

[color=blue][b]Problem[/b]. Let $ A\equal{}\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb{R}^{n\times n}$ be a matrix. Assume that $ \sum_{i\equal{}1}^n\sum_{j\equal{}1}^n\left|a_{i,j}\right|\equal{}n$ (that is, the sum of the absolute values of all entries of $ A$ is $ n$). Prove that $ \left|\det A\right|\leq 1$.[/color]

[i]Solution.[/i] We have $ \det A\equal{}\sum_{\pi\in S_n}\text{sign }\pi\prod_{i\equal{}1}^na_{i,\pi\left(i\right)}$, so that

$ \left|\det A\right|\equal{}\left|\sum_{\pi\in S_n}\text{sign }\pi\prod_{i\equal{}1}^na_{i,\pi\left(i\right)}\right|\leq\sum_{\pi\in S_n}\left|\text{sign }\pi\prod_{i\equal{}1}^na_{i,\pi\left(i\right)}\right|$       (by the triangle inequality)
$ \equal{}\sum_{\pi\in S_n}\underbrace{\left|\text{sign }\pi\right|}_{\equal{}1}\cdot \prod_{i\equal{}1}^n\left|a_{i,\pi\left(i\right)}\right|\equal{}\sum_{\pi\in S_n}\prod_{i\equal{}1}^n\left|a_{i,\pi\left(i\right)}\right|$.

The following inequality:

$ \sum_{\pi\in S_n}\prod_{i\equal{}1}^n\left|a_{i,\pi\left(i\right)}\right|\leq\sum_{f: \left\{1,2,...,n\right\}\to\left\{1,2,...,n\right\}}\prod_{i\equal{}1}^n\left|a_{i,f\left(i\right)}\right|$

holds because the sum on the right hand side contains all the terms that the sum on the left hand side contains (since every permutation in $ S_n$ is a function from $ \left\{1,2,...,n\right\}$ to $ \left\{1,2,...,n\right\}$), and the remaining terms are all nonnegative.

Now, by the product rule,

$ \sum_{f: \left\{1,2,...,n\right\}\to\left\{1,2,...,n\right\}}\prod_{i\equal{}1}^n\left|a_{i,f\left(i\right)}\right|\equal{}\prod_{i\equal{}1}^n\sum_{j\equal{}1}^n\left|a_{i,j}\right|$.

Finally, by the AM-GM inequality (applied to the $ n$ nonnegative reals $ \sum_{j\equal{}1}^n\left|a_{i,j}\right|$ for all $ i\in\left\{1,2,...,n\right\}$), we have

$ \prod_{i\equal{}1}^n\sum_{j\equal{}1}^n\left|a_{i,j}\right|\leq\left(\frac1n\cdot\sum_{i\equal{}1}^n\sum_{j\equal{}1}^n\left|a_{i,j}\right|\right)^n\equal{}\left(\frac1n\cdot n\right)^n\equal{}1^n\equal{}1$.

Combining this all, we get

$ \left|\det A\right|\leq\sum_{\pi\in S_n}\prod_{i\equal{}1}^n\left|a_{i,\pi\left(i\right)}\right|\leq\sum_{f: \left\{1,2,...,n\right\}\to\left\{1,2,...,n\right\}}\prod_{i\equal{}1}^n\left|a_{i,f\left(i\right)}\right|$
$ \equal{}\prod_{i\equal{}1}^n\sum_{j\equal{}1}^n\left|a_{i,j}\right|\leq 1$,

qed.

  darij


Angle $ BCD$ is an exterior angle of triangle $ ABC$ and has a measure of 55 degrees. Angle $ BAC$ has a measure of 32 degrees. How many degrees are in teh measure of angle $ ABC$?

The sum of the interior angles of a triangle is 180 degrees. Since angle BCD is an exterior angle, we must subtract it from 180 to get the interior angle. Therefore, the three interior angles of Triangle ABC is 125, 32, and x.
180 - 125 - 32 = 23, so our answer is 23.

Kiseki's answer is right, and his work contains the proof of a more general fact that is a bit simpler to remember:

In a triangle, the exterior angle at one corner is equal to the sum of the interior angles of the other two corners.

Use the same picture this problem has, but call
angle BAC $ \alpha$
angle ABC $ \beta$
exterior angle DCB $ \epsilon$

Angle BCA would have to be $ 180\minus{}\alpha \minus{}\beta$.  By definition of exterior angle, $ \epsilon\plus{}(180\minus{}\alpha \minus{}\beta)\equal{}180$
Simplifying:
$ \epsilon\plus{} 180\minus{}\alpha \minus{}\beta\equal{}180$
$ \epsilon\minus{}\alpha \minus{}\beta\equal{}0$
$ \epsilon\equal{}\alpha \plus{}\beta$

If you know this, then the problem is solved at a glance :)


Nonnegative real numbers $x,y,z$ satisfy $x+y+z=1$.  Prove that $7(xy+yz+zx) \leq  2+9xyz$.

Using the rearranjun inequality:
$xy+yz+zx\leq x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+xz)$ and
$3( xy+yz+zx)\leq (x+y+z)^{2}$ adding this 2 times:
$6( xy+yz+zx)\leq 2(x+y+z)^{2}+$ and adding one more in the first form:
$7( xy+yz+zx)\leq 2(x+y+z)^{2}+x^{2}+y^{2}+z^{2}$ or
$7( xy+yz+zx)\leq 2+x^{2}+y^{2}+z^{2}$ we just have to prove now that $x^{2}+y^{2}+z^{2}\leq 9xyz$ using chebychev:
$(x^{2}+y^{2}+z^{2})(yz/x+xz/y+xy/z)\leq 9xyz$ well... hehehe i wasn´t able to prove the last part, but i think that this is all right, since i tested some values and the last inequality works.

Your idea was looking good, I agree...

However, I get $x^{2}+y^{2}+z^{2}\geq 9xyz$.  I do not know what numbers you were testing (just 1/3,1/3,1/3?).

** Sorry! I didn't mean to erase my old post.  It just said that if you let x be 0 or approach 0, the claimed inequality fails.

yes you are right. About if 1 term is zero, we can just prove trivially that the niequality holds, thats why i didn´t care about this case, but i should have explicty said that, but....in the case when x approachs zero, you are totally right. I just was seduced by this because it sims to build the inequality perfectly with all its coefficientes, what was a hard task to me until i´ve seen this relation.

We can solve it with Lagrange multipliers.

Set $f(x,y,z) = 7(xy+yz+zx)-2-9xyz$. Then

$\nabla f = (7y+7z-9yz, 7x+7z-9xz, 7x+7y-9xy).$

So any extremum in the interior of the region satisfies

$7y+7z-9yz = 7x+7z-9xz = 7x+7y-9xy$
$(9y-7)(9z-7) = (9x-7)(9z-7) = (9x-7)(9y-7)$

If any one of $x,y,z$ equals $\frac{7}{9}$, then at least two must, which is impossible.  So none of them equal $\frac{7}{9}$, and we have $x = y = z = \frac{1}{3}$.  In that case, $f(1/3,1/3,1/3) = \frac{7}{9}-2-\frac{1}{3}=-\frac{14}{9}.$

The other possibility is that $x,y,z$ lie on the boundary of the region, i.e. at least one of $x,y,z$ is zero.  WLOG let $x = 0$. Then $f = 7yz-2 \leq-\frac{1}{4}$.  So this is the maximum value of $f$.

I have a more straightforward solution .
By Schur $7(ab+bc+ca)\leq\frac{7}{4}+\frac{63abc}{4}\leq 2+9xyz$
by AM-GM  :)


Let $ k$ is a natural number. We denote with $ f(k)$ the biggest natural number for which there exists a set $ M$ from natural numbers with $ f(k)$ elements, such that:

(i) Each element from $ M$ is a divisor of $ k$.

(ii) There are no element from $ M$ that divides some other element from $ M$.

Prove that if $ m$ and $ n$ are coprime numbers then:

\[ f(n\cdot 2^n)\cdot f(m\cdot 2^m)\equal{}f(mn\cdot 2^{m\plus{}n})\]

I guess you need also to require that $ m$ and $ n$ are odd numbers.
Then one simply needs to show that $ f(n\cdot 2^n) \equal{} \tau(n)$, the number of divisors of $ n$, and use multiplicativity of this function.

First, it is clear that all elements of the set $ M$ corresponding to $ f(n\cdot 2^n)$ are of the form $ d\cdot 2^k$ where $ d$ is divisor of $ n$. However, all $ d$ must be distinct since for two numbers $ d\cdot 2^{k_1}$ and $ d\cdot 2^{k_2}$, one must divide the other. Therefore, $ f(n\cdot 2^n)\leq \tau(n)$.

Second, it is easy to see that $ \tau(n)\leq n$.

Third, let $ 1 \equal{} d_1 < d_2 < \dots < d_{\tau(n)} \equal{} n$ be all divisors of $ n$. Then we can take
\[ M \equal{} \{ d_1\cdot 2^{\tau(n) \minus{} 1},\; d_2\cdot 2^{\tau(n) \minus{} 2},\; \dots,\; d_{\tau(n)}\cdot 2^0 \}.\]
This $ M$ proves that $ f(n\cdot 2^n) \equal{} \tau(n).$

On the other hand, for $ m,n>1$, the set $ M$ for $ f(mn\cdot 2^{m\plus{}n})$ can be taken as the pairwise products of the elements of the corresponding set for $ f(n\cdot 2^n)$ and $ f(m\cdot 2^m)$ so that
\[ \tau(m)\tau(n)\leq f(mn\cdot 2^{m\plus{}n})\leq f(mn\cdot 2^{m\plus{}n}) \leq f(mn\cdot 2^{mn}) \equal{} \tau(mn),\]
implying that $ f(mn\cdot 2^{m\plus{}n}) \equal{} \tau(m)\tau(n) \equal{} f(n\cdot 2^n)\cdot f(m\cdot 2^m).$ For $ m\equal{}1$ or $ n\equal{}1$, the statement is trivial.


On a number line, find the coordinate of a point $ \frac{1}{5}$ of the distance from $ \minus{}3$ to $ 17$.

The distance is $ 17\minus{}(\minus{}3)\equal{}20$. $ \frac{20}{5}\equal{}4$. However, this must be $ 4$ away from $ \minus{}3$, so $ \minus{}3\plus{}4\equal{}\boxed{1}$.


Let $a$,$b$,$c$ be positive real numbers with $abc=1$.Prove that the following inequality occurs: \[\frac{3a^{4}+1}{3ab+c^{3}}+\frac{3b^{4}+1}{3bc+a^{3}}+\frac{3c^{4}+1}{3ac+b^{3}}\geq 3\]

Come on,guys..no solutions? :(
Ok,I give a hint:Use AM-GM on the denominator in the right way :)

[quote="andyciup"]
Ok,I give a hint:Use AM-GM on the denominator in the right way :)[/quote]
Nice! :lol:

Ok,I agree that it`s not a big hint,but if I say more,the problem will be too easy.Maybe this hint will help:Use the "puzzle method" and the fact that ${\sum a^{4}\geq\sum a^{3}}$

You are right.that was also my main idea:
You must observe that $\sum\frac{3a^{4}+1}{3ab+c^{3}}\geq\frac{3(a^{4}+b^{4}+c^{4}+1)}{a^{3}+b^{3}+c^{3}+1}$ and use the fact that $\sum a^{4}\geq\sum a^{3}$
I wonder if there are other solutions..

[quote="andyciup"]Actually there is a mistake there :wink: you reversed a sign..[/quote]
What's the mistake :?:

"Mea culpa" :) ..I hurried and said that it was wrong without much investigation..I thought there was a mistake in line 2,but it`s ok.


on those problems like "if i can use 4,5,6,7 for _ _ * _ _, what is the smallest/greatest possible product?" or "if i can use 4,5,6,7 for _ / _ + _ /_ what is sum closest to 1.5?" or stuff like that? what's the general idea? thx!

follow up question:

What is the largest number you can make using only the four digits 1-4, exactly one time each, and only using the symbols + - * / ^(exponent) . ?

[quote="HiDN428"]general idea:

For multiplying:
 *The GREATEST number is produced by the greatest number and least number combined times second largest number and second least number combined. Ex: for 1,2,3,4 its 41*32 because its the largest and smallest times....u know it.

 *The SMALLEST number is produced by the least number and largest number combined times the second least number and the second largest number combined. Reversed basically. Ex: 1,2,3,4 its 13*24[/quote]

o, i think u made a typo on smallest. so smallest is 13*24 or 14*23? wut about the fractions problem i posted? thx!!!!!!!!!

Hey math king is it ((3^4)^2)+1??
Which is 6562?
Not sure if you have to use each symbol once though......

[quote]HiDN428 wrote:
general idea:

For multiplying:
*The GREATEST number is produced by the greatest number and least number combined times second largest number and second least number combined. Ex: for 1,2,3,4 its 41*32 because its the largest and smallest times....u know it.

*The SMALLEST number is produced by the least number and largest number combined times the second least number and the second largest number combined. Reversed basically. Ex: 1,2,3,4 its 13*24

o, i think u made a typo on smallest. so smallest is 13*24 or 14*23? wut about the fractions problem i posted? thx!!!!!!!!![/quote]

Oh, yes, i did make a mistake saying it, it's actually 13*24 because the least sum produced is the first number and the second least number combined times the second least number and the largest number combined.
And for the fraction one, i'm not that sure if i know how to do it, i usually do an educated guess and check. :huh:

o. k, wut about for 5+ numbers? (one two digit, one three digit----one three digit, one three digit-----one three digit, one four digit)... thanks in advance!  :)


[size=150]$ a_{0} \equal{} u, a_{1} \equal{} v,$
$ a_{n \plus{} 1} \equal{} \frac {a_{n}a_{n \minus{} 1} \plus{} 1}{a_{n} \plus{} a_{n \minus{} 1}}$
$ n \equal{} 1,2,....$
find  $ a_{n}$

Can anybody use generatingfunctionology to sovle this?
I have tried for 1 hour to handle,but no evolvement.
Any method can use if you it works.[/size] :)

If $ a_n\equal{}cth(b_n)$, then $ b_{n\plus{}1}\equal{}b_n\plus{}b_{n\minus{}1}$ or $ b_n\equal{}aF_n\plus{}bL_n$ and $ a_n\equal{}cth(aF_n\plus{}bL_n)$. Constants $ a$ and $ b$ you can find from u and v.

Generating functions are almost never of use for nonlinear recurrences.

$ cth$ may be called $ \coth$ in other places.  He probably came up with that solution by recognizing that the recurrence looks like the addition formula for the tangent, then checking more closely concluded that it is exactly the addition formula for the hyperbolic cotangent.


Would the following hypothetical situation be possible? Has anything similar ever been arranged?

A local company, Springfield Financial Services Inc., agrees to sponsor the math team of East Springfield High School. Each year, they will make a $1,500 tax-deductible donation to the Art of Problem Solving Foundation, which will store the money in an private account to be used by the East Springfield High School math team to pay for books and courses over the year.

We can't set up our foundation to be used for the express purpose of soliciting funds to buy our products - the IRS would then take away our tax-exempt status and possibly our non-profit status in the Foundation.

Other non-profit vehicles have been used this way (ICAE and the Charlotte Math Club are examples).

How do you set-up a non-profit where corporations could make tax-deductable donations?   I have made announcements in all the math classes and used the faculty as an outlet to reach students for my math club, and interest seems to be much higher than my optimistic projection.  I may end up with a membership of nearly 100.  I'm not kidding.  There is no way 100 people are going to survive on 1000 dollars.

[quote="schulmannerism"]A local company... will make a $1,500 tax-deductible donation to the Art of Problem Solving Foundation, which will store the money in an private account to be used by the East Springfield High School math team to pay for books and courses over the year.[/quote]

The donor could offer to re-imburse the school for up to USD 1500/yr of purchases from AoPS.
At the end of the year, it can donate the balance to the Foundation.  This is equally tax-deductible, but the money doesn't change hands up-front and so some deductions might be from the following year's taxes.


\[\displaystyle Cate \ perechi \ de \ submultimi \ (A , B) \ ale \ multimi \ X \ cu \ n \ elemente \ astfel \ incat \ A\subseteq B\subseteq X \ exista ? \]

Cred ca ar fi cam $3^n$.
Pentru $k$, avem $\binom{n}{k}$ posibilitati pentru a alege $B$ cu $|B|=k$. Dar atunci pentru $A$ avem $\sum_{i=0}^{k}\binom{k}{i} = 2^k$ posibilitati.
Deci, in total ar fi $\sum_{i=1}^n \binom{n}{k}\times 2^k = 3^k$. Cred ca este simplu! :D


Let $n$ be a positive integer. Prove that $n$ is prime if and only if \[{{n-1}\choose k}\equiv (-1)^{k}\pmod{n}\] for all $k \in \{ 0, 1, \cdots, n-1 \}$.

We will regardless use that we can calculate with fractions $\mod n$ as long as the denominator is coprime to $n$. It's not hard to show that we always can do that.

If $n$ is not prime, let $q<n$ be the smallest prime dividing $n=aq$:

$\binom{n-1}{q}= \frac{(n-1) \cdot (n-1) \cdot ... \cdot (n-q)}{1 \cdot 2 \cdot ... \cdot q} =$
$= \frac{n-q}{q} \cdot \frac{n-1}{1}  \cdot \frac{n-2}{2}  \cdot ...  \cdot \frac{n-q+1}{q-1} \equiv (a-1) \cdot (-1)^{q-1} \mod n$.
But $a-1 \not \equiv -1 \mod n$ (this would yield $a \equiv 0 \mod n$, so $n|a|n$, wrong).

If $n$ is prime, we do the same calculation to get
$\binom{n-1}{q}= \frac{(n-1) \cdot (n-1) \cdot ... \cdot (n-q)}{1 \cdot 2 \cdot ... \cdot q} =$
$= \frac{n-1}{1}  \cdot \frac{n-2}{2}  \cdot ...  \cdot \frac{n-q+1}{q-1} \cdot \frac{n-q}{q} \equiv (-1)^q \mod n$.

For a given $ n$, the set of congruences given are equivalent to $ 1\equal{}C(n\minus{}1,0)\equiv(\minus{}1)^0\equal{}1$ and $ C(n,i\plus{}1)\equal{}C(n\minus{}1.i)\plus{}C(n\minus{}1,i\plus{}1)\equiv(\minus{}1)^i\plus{}(\minus{}1)^(i\plus{}1)\equal{}0$ for $ 0\le i<n\minus{}1$. Thus the condition given is equivalent to $ C(n,i)$ being a multiple of $ n$ for all $ i$ in the range  $ 1\le i\le n\minus{}1$.

If $ n$ is prime, then for $ 1\le i\le n\minus{}1$, $ C(n,i)\equal{}n!/i!(n\minus{}i)!$ has a multple of $ p$ on the numerator but not on the denominator, so is divisible by $ p$.

Now suppose $ p$ is a prime factor of $ n$ but $ n$ is not equal to $ p$. Then $ C(n,p)/n\equal{}(n\minus{}1)(n\minus{}2)...(n\minus{}(p\minus{}1))/p!$ has a multiple of $ p$ on the denominator but not on the numerator, so is not an integer. Thus $ C(n,p)$ is not divisible by $ n$.

Luke
See my puzzle blog at [url]http://bozzball.blogspot.com[/url]

ZetaX very nice solution
.man ham shunday ishladim bosing brat.


calculus
derivative
analytic geometry
graphing lines
slope
conics
parabola

how do you find the vertex of y=-5x^2 + 4x - 6?

[quote="Altheman"][quote="cincodemayo5590"]Or if you're feeling lazy you can use [hide="a bulldozer"]
\[y' =0 \]
[/hide][/quote]

a bulldozer is not an adequate metaphor...i think a nuclear bomb would be more accurate...[/quote]

Rofl, Alex, meet me halfway.  :rotfl:

[quote="cincodemayo5590"]Or if you're feeling lazy you can use [hide="a bulldozer"]
\[y' =0 \]
[/hide][/quote]
how do you mean? y' means the inverse of y right?

Although we shouldn't be getting into calculus, if the derivative is equal to 0 the slope of the tangent at that point is 0. Turning points are perfect examples of places where the slope of the tangent is 0 (although this is not the only case, in a parabola the only critical point is the vertex so this is all we need to do.)

[quote="7h3.D3m0n.117"][quote="Treething"]derivative, might not have learned that yet lol[/quote]

haha i didn't know derivatives can be used to find vertex  :maybe:

just use $-\frac{b}{2a}$[/quote]I know we shouldn't be getting into calculus, but that formula can be derived quickly by using derivatives.

Given a general parabola $ax^{2}+bx+c$ the derviative is $2ax+b = 0$

Solving for x gets $\frac{-b}{2a}$

[hide="generalization method 1"]Consider the equation $ax^{2}+bx+c=y$
$a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}=y$
By the trivial inequality, the thing inside the square is minimized when it's 0, so the vertex is when
$x+\frac{b}{2a}=0$
$x=-\frac{b}{2a}$[/hide]

[hide="generalization method 2 (I like this one better :) ) "]
Okay, consider the axis of symmetry.
All points with constant y values are equidistant from it.
This means that the roots are equidistant from it.
So the midpoint of the roots is the vertex.
Using vieta's sums, we know that the sum of the roots is $\frac{-b}{a}$ so our answer is $\frac{-b}{2a}$[/hide]

Key point: All these are talking about the x-coordinate of the vertex. The y-coordinate is $f(-\frac{b}{2a})$.


[quote="Ubemaya"]Okay I don't understanding this, so some geo beast needs to explain this solution to me:

Problem:
A point D lies inside the triangle ABC. Let A_1, B_1, C_1 be the second intersection points of the lines AD, BD, and CD with the circumcircles of BDC, CDA, and ADB, respectively.
Prove that:
$ \frac {AD}{AA_1} \plus{} \frac {BD}{BB_1} \plus{} \frac {CD}{CC_1} \equal{} 1$

Solution:
Let $ O_1$ be the circumcircle of BDC, $ O_2$ be the circumcircle of CDA, and $ O_3$ be the circumcircle of ADB. Invert the plane about D with arbitrary radius $ r$, and let $ X'$ denote the inverse of $ X$. Let $ A'$ be the inverse of $ A$, $ B'$ be the inverse of $ B$, and $ C'$ be the inverse of $ C$. Since circle $ O_1$ passes through $ B$ and $ C$, the $ O_1'$ is line $ B'C'$; similarly, the $ O_2'$ is line $ C'A'$, and $ O_3'$ is line $ A'B'$. $ A_1$ is the intersection of $ AA_1$ and $ O_1$, so $ A'_1$ is the intersection of $ AA_1$ and $ B'C'$. Similarly, $ B_1'$ is the intersection of $ BB_1$ and $ A'C'$, and $ C_1'$ is the intersection of $ CC_1$ and $ A'B'$. Since $ \dfrac{AD}{AA_1} \equal{} \dfrac{A_1'D}{A_1'A'}$, the problem has been reduced to proving that $ \dfrac{A_1'D}{A_1'A'} \plus{} \dfrac{B_1'D}{B_1'B'} \plus{} \dfrac{C_1'D}{C_1'C'} \equal{} 1$, which is true.

So I really don't understand any of it, but can someone at least prove the last sentence: If $ XD$, $ YE$, and $ ZF$ are cevians of triangle $ XYZ$ which meet at $ M$, prove that $ \dfrac{DM}{DX} \plus{} \dfrac{EM}{EY} \plus{} \dfrac{FM}{FZ} \equal{} 1$. Should the proof of this be omitted from a formal solution and be able to receive full points?[/quote]
ROAR
I WAS THINKING OF DOING EXACTLY THAT
INVERSION WITH ARBITRARY RADIUS AROUND $ D$
ASDFKJHSADKJLFH
but I didn't do it
because I thought it would be too random and wierd and stuff
anyways, for the last sentence:
[hide="Proof"]
Let the cevians of a triangle $ ABC$ be $ AD$, $ BE$, and $ CF$ that meet at $ P$.  Notice that $ \frac {[PBC]}{[ABC]} \equal{} \frac {PD}{AD}$.  We do the same thing for the rest, so $ \frac {PE}{BE} \plus{} \frac {PD}{AD} \plus{} \frac {PF}{AF} \equal{} \frac {[PBC]}{[ABC]} \plus{} \frac {[APB]}{[ACB]} \plus{} \frac {[APC]}{[ABC]} \equal{} \frac {[PBC] \plus{} [PBA] \plus{} [PCA]}{[ABC]} \equal{} 1$ [/hide]
by the way
I talked to junggi, Ubey...
he didn't spend three hours on number 6...

[quote="Ubemaya"][quote="CircleSquared"][quote="Ubemaya"]Okay I don't understanding this, so some geo beast needs to explain this solution to me:

Problem:
A point D lies inside the triangle ABC. Let A_1, B_1, C_1 be the second intersection points of the lines AD, BD, and CD with the circumcircles of BDC, CDA, and ADB, respectively.
Prove that:
$ \frac {AD}{AA_1} \plus{} \frac {BD}{BB_1} \plus{} \frac {CD}{CC_1} \equal{} 1$

Solution:
Let $ O_1$ be the circumcircle of BDC, $ O_2$ be the circumcircle of CDA, and $ O_3$ be the circumcircle of ADB. Invert the plane about D with arbitrary radius $ r$, and let $ X'$ denote the inverse of $ X$. Let $ A'$ be the inverse of $ A$, $ B'$ be the inverse of $ B$, and $ C'$ be the inverse of $ C$. Since circle $ O_1$ passes through $ B$ and $ C$, the $ O_1'$ is line $ B'C'$; similarly, the $ O_2'$ is line $ C'A'$, and $ O_3'$ is line $ A'B'$. $ A_1$ is the intersection of $ AA_1$ and $ O_1$, so $ A'_1$ is the intersection of $ AA_1$ and $ B'C'$. Similarly, $ B_1'$ is the intersection of $ BB_1$ and $ A'C'$, and $ C_1'$ is the intersection of $ CC_1$ and $ A'B'$. Since $ \dfrac{AD}{AA_1} \equal{} \dfrac{A_1'D}{A_1'A'}$, the problem has been reduced to proving that $ \dfrac{A_1'D}{A_1'A'} \plus{} \dfrac{B_1'D}{B_1'B'} \plus{} \dfrac{C_1'D}{C_1'C'} \equal{} 1$, which is true.

So I really don't understand any of it, but can someone at least prove the last sentence: If $ XD$, $ YE$, and $ ZF$ are cevians of triangle $ XYZ$ which meet at $ M$, prove that $ \dfrac{DM}{DX} \plus{} \dfrac{EM}{EY} \plus{} \dfrac{FM}{FZ} \equal{} 1$. Should the proof of this be omitted from a formal solution and be able to receive full points?[/quote]

[hide="alternatively"]
obviously im too sucky at math to do something like this
but couldn't you try to prove ACD and A1C1D are similar? (they sure look like it to me)
[/hide][/quote]

By my diagram, they're not even close to similar. But I dunno.[/quote]

Now if i could just prove AC1A1C is cyclic....then it would work
im not too sure how it would help.

and my diagrams are usually really bad.

Don't complain about bad diagrams, say that to Quatto's face.

[quote="Ubemaya"]Let $ X'$ denote the inverse of $ X$.

Since circle $ O_1$ passes through $ B$ and $ C$, the $ O_1'$ is line $ B'C'$

$ A_1$ is the intersection of $ AA_1$ and $ O_1$, so $ A'_1$ is the intersection of $ AA_1$ and $ B'C'$.

$ \dfrac{AD}{AA_1} \equal{} \dfrac{A_1'D}{A_1'A'}$[/quote]

what is $ X$, and can you please prove those things that you stated without proving?

vishalarul has trouble with geometry, especially w/ hard things like inversions. [time to hit Geometry Revisited again]
[url=http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=12770]WARNING! Big (bad) diagram[/url]

Uh... this was someone else's solution.

I don't even know what inversion is.

And that's a sweet diagram!

Oh, and I think by $ X \equal{} X'$, he means for any point/shape/line $ X$, the inversion is denoted by $ X'$. Just some notation.

[quote=Ubemaya]Okay I don't understanding this, so some geo beast needs to explain this solution to me:

Problem:
A point D lies inside the triangle ABC. Let A_1, B_1, C_1 be the second intersection points of the lines AD, BD, and CD with the circumcircles of BDC, CDA, and ADB, respectively.
Prove that:
$ \frac {AD}{AA_1} \plus{} \frac {BD}{BB_1} \plus{} \frac {CD}{CC_1} \equal{} 1$

Solution:
Let $ O_1$ be the circumcircle of BDC, $ O_2$ be the circumcircle of CDA, and $ O_3$ be the circumcircle of ADB. Invert the plane about D with arbitrary radius $ r$, and let $ X'$ denote the inverse of $ X$. Let $ A'$ be the inverse of $ A$, $ B'$ be the inverse of $ B$, and $ C'$ be the inverse of $ C$. Since circle $ O_1$ passes through $ B$ and $ C$, the $ O_1'$ is line $ B'C'$; similarly, the $ O_2'$ is line $ C'A'$, and $ O_3'$ is line $ A'B'$. $ A_1$ is the intersection of $ AA_1$ and $ O_1$, so $ A'_1$ is the intersection of $ AA_1$ and $ B'C'$. Similarly, $ B_1'$ is the intersection of $ BB_1$ and $ A'C'$, and $ C_1'$ is the intersection of $ CC_1$ and $ A'B'$. Since $ \dfrac{AD}{AA_1} \equal{} \dfrac{A_1'D}{A_1'A'}$, the problem has been reduced to proving that $ \dfrac{A_1'D}{A_1'A'} \plus{} \dfrac{B_1'D}{B_1'B'} \plus{} \dfrac{C_1'D}{C_1'C'} \equal{} 1$, which is true.

So I really don't understand any of it, but can someone at least prove the last sentence: If $ XD$, $ YE$, and $ ZF$ are cevians of triangle $ XYZ$ which meet at $ M$, prove that $ \dfrac{DM}{DX} \plus{} \dfrac{EM}{EY} \plus{} \dfrac{FM}{FZ} \equal{} 1$. Should the proof of this be omitted from a formal solution and be able to receive full points?[/quote]

$ \dfrac{DM}{DX} \equal{} \dfrac{[ZMY]}{[XYZ]}$ (using area ratios and componendo dividendo)
Adding similar ratios gives you the desired result.

#CG


I dont know where to post this problem so I have posted it in a few sub forums!

" For any given real numbers $ a,b,c$ and satisfying $ a^{k}\plus{}b^{k}\equal{}c$, where $ k$ need not be an integer, find a general rule for finding $ k$!"

In the future, please do not post multiple copies of the same thread / question in different forums.

Sorry I am still a novice so please forgive me this time. I shall not repeat it another time!


Heres a problem I made while reading through a biography-book of mathematicians.

  Show that any number in the form $x^2+1$ can be factored using $i$.

($i=\sqrt{-1}$)

nice one, i beleive theres another solution ( if not an infinite)....

Another way to look at it:
[hide] Let $f(x)=x^2+1.$  Then $f(x)=\frac{x^4-1}{x^2-1}.$  Set the top equal to 0.  It has the fourth roots of unity as roots. The roots of the denominator are undefined so we have to consider them.  They are are the second roots of unity.  So $f(x)$ has as roots the fourth roots of unity except those that are also second roots of unity, i.e. $\pm i.$  Therefore \[ x^2+1=(x+i)(x-i).  \][/hide]

[quote="LoboSolitario"]This is funny  :)

$\left(1-\frac{x}{i}\right)\left(1+\frac{x}{i}\right)$[/quote]

That is similar to G UNITs.  This is because $i^{-1}=i^3=i^7=\cdots$.  I think you could go many places with this problem :P


inequalities
function
inequalities proposed

[quote="epitomy01"]If a_1+a_2 .. +a_n = 1
prove that:
[(1- a_1^k)/a_1^k]...[(1-a_n^k)/(a_n)^k] >= (n^k - 1)^n[/quote]
I think, it's wrong for $ k\equal{}\minus{}1$ and $ n\equal{}2.$ :wink:

[quote="Albanian Eagle"]It is true for k a positive integer :)[/quote]
If so then it's true by Jensen. :D

Arqady, how are you applying Jensen? It didn't work for me.
If we apply Jensen to the function f(x) = In [(1-x^k)/x^k] = In [x^(-k) - 1], then according to my calculations, f''(x) >=0 only when, 1/(k+1) > x^k, which clearly isn't always true.

You are right, epitomy01. My mistake.  :(  Thank you. But by AM-GM and Holder we obtain:
$ \prod_{i \equal{} 1}^n\frac {1 \minus{} x_i^k}{x_i^k}\geq\prod_{i \equal{} 1}^n\frac {1 \minus{} x_i}{x_i}\cdot\prod_{i \equal{} 1}^n\left(1 \plus{} \frac {1}{x_i} \plus{} \frac {1}{x_i^2} \plus{} ... \plus{} \frac {1}{x_i^{k \minus{} 1}}\right)\geq$
$ \geq(n \minus{} 1)^n\left(1 \plus{} \frac {1}{\sqrt [n]{x_1x_2...x_n}} \plus{} \frac {1}{\sqrt [n]{x_1^2x_2^2...x_n^2}} \plus{} ... \plus{} \frac {1}{\sqrt [n]{x_1^{k \minus{} 1}x_2^{k \minus{} 1}...x_n^{k \minus{} 1}}}\right)^n\geq$
$ \geq(n \minus{} 1)^n(1 \plus{} n \plus{} n^2 \plus{} ... \plus{} n^{k \minus{} 1})^n \equal{} (n^k \minus{} 1)^n.$

Beautiful solution, arqady.
Mine was a little different, a bit hard to explain.
If f(a1,a2 ... a_n) = [(1- a_1^k)/a_1^k]...[(1-a_n^k)/(a_n)^k],
we wish to prove f(a1,a2 ... a_n) >= f(1/n, 1/n ... 1/n).
Note f(a1 .. a_n) = [(1-a1)(1+a1 +a1^2 ... +a1^(k-1)] ... [(1-a_n)(1+a_n + ... + a_n^(k-1)] / [a1*a2 ... *a_n]^k = (a2+a3.. + a_n)... (a_1+a_2... +a_(n-1)) * (1+a1+ .. +a1^(k-1)) ... (1+a_n .. + a_n^(k-1) / [a1*a2 .. *a_n)^k.
By expanding the product of 2n terms in the numerator, since it is cyclic, we can right as,
the sum of cyclic sum (a1^i1*a2^i2* .. *a_n^i_n), where the set (i1 .. i_n) belong to a set S of all such n-tuples that occur in the expansion. It's easy to see that for all in set S, i1+i2 .. + i_n <= nk.
Clearly by AM-GM, cyclic sum (a1^i1 .. *a_n^i_n) >= C*(a1*a2 .. *a_n)^j, where j = (i1+ .. + i_n) / n, and C is the number of terms in the cyclic sum.
therefore, cyclic sum (a1^i1 ... *a_n^i_n)/(a1* ... *a_n)^k >= C*(a1*a2 .. *a_n)^(j-k) >= C*(1/n^n)^(j-k) (since j<=k, since i1+i2 .. + i_n <= nk).
Adding up all the inequalities of this kind, ranging over all sets in S,
gives us that f(a1 ... a_n) >= f(1/n ... 1/n), as required.


what do all those "poincare conjecture" and hodge conjecture" and so on and on mean? and what do they stand for? i dunno whether it's how long you've been an aopser or how many posts u have.

[quote="Dojo"]Well since you couldn't find it, I found it for you.  :maybe:
0 - 19 New Member (Zero Stars)
20 - 49 P versus NP (Half Star)
50 - 99 Hodge Conjecture (One Star)
100 - 249 Poincare Conjecture (Two Stars)
250 - 499 Riemann's Hypothesis (Two and Half stars)
500 - 999 Yang Mills Theory (Three Stars)
1000 - 2499 Navier-Stokes Equation (Four Stars)
2500 - Birch & Swinnerton Dyer. (Five Stars)[/quote]

How Many Posts you make (other than G&FF). It is also the name of the Millenium Problems.

See [url=http://www.claymath.org/millennium/]here[/url].  Unfortunately, to even state several of the Millennium problems requires years of background and experience.

[quote="1=2"]Shouldn't we replace Poincare with Goldbach? Hasn't the Poincare Conjecture been proved?[/quote]

The Goldbach Conjecture isn't a millenium problem.

Why are those millenium problems so special?

no one has proven them before. Their are like >1 million dollar prizes for each one, for the person who proves it.

[quote="Wickedestjr"]Is one of them the proof that every even number greater than 2 is the sum of two prime numbers, because that has never been proved before.

BTW, I just became a Reimann's Hypothesis about 20 minutes ago.[/quote]

No. That is the goldbach conjecture.

View the link that t0rajir0u gave for what the conjectures are.

Riemann Hypothesiss,
Poincare Conjcture (proven)
Hodge Conjecture
Navier-Stokes Equations
Birch-Swinterton Dryer Conjecture
P=NP?
...And i forgot the 7th one.

I did, but I didn't want to read them all. I will later though.

[quote="pythag011"]
Riemann Hypothesiss,
Poincare Conjcture (proven)
Hodge Conjecture
Navier-Stokes Equations
Birch-Swinterton Dryer Conjecture
P=NP?
...And i forgot the 7th one.[/quote]

[url=http://en.wikipedia.org/wiki/Yang%E2%80%93Mills_existence_and_mass_gap]Yang–Mills existence and mass gap[/url] - Wikipedia

Oh, they have these topics in the AoPSWiki?


algebra
polynomial
AMC
USA(J)MO
USAMO
number theory unsolved
number theory

[quote="melkor"]Let N be a positive integer. Prove that there is a positive integer n such that n² + 3 is divisible by at least N distinct primes.[/quote]
An old problem . More general
1 )If $ P(x)$ is a polynomial with integer coefficient and $ \deg P\geq 1$ then exist an integer m such that $ P(m)$ has at least N distant prime divisor .
2 ) $ a\in N$ and $ P(x)$ is a polynomial with integer coefficient .
$ G(x) \equal{} a^x \plus{} P(x)$
Prove that for each M there exist $ n_0$ such that $ G(n_0)$ has at least prime divisor .

See the discussion around [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202909]USAMO 2008 #1[/url] (especially the third [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2008-s.pdf]official solution[/url]).

It 's easy to prove that there exists N different primes p1,.., pN and N integers n1,.., nN where

pk | nk^2 +3

Applying the Chineese remainder theorem, there exist n where

n= nk (mod pk ) for all k


geometry
topology
algebra
polynomial
linear algebra
matrix
linear algebra unsolved

If $X$ is a map, we define $\exp{(X)}=\sum_{i=0}^{\infty}\frac{X^{i}}{i!}$

Let $A$ and $B$ be two maps, such that $[A,[A,B]]=[B,[A,B]]=0$ where $[A,B]=AB-BA$ is the commutator of the maps $A$ and $B$.

Prove that
\[\exp{(A)}.\exp{(B)}=\exp{\left(A+B+\frac{1}{2}[A,B]\right)}. \]

It's a particular case of a wellknown result in non commutative geometry (Baker-Campbell-Hausdorff theorem/cf.Mathworld or Wikipedia):
$log(e^{x}e^{y})=x+y+\frac{1}{2}[x,y]+\frac{1}{12}([x,[x,y]]-[y,[x,y]])+\frac{1}{48}H_{4}+\cdots$
where $H_{4}$ is the homogeneous part of degree 4.
Here $[x,[x,y]]=[y,[x,y]]=0,H_{4}=0,\cdots$.
   Remark: $x$ and $z=[x,y]$ commutate then $z$ is nilpotent ($\forall{k},trace(z^{k})=0$).
There exists a direct proof without the BCH theorem if $n\leq3$:
$n=2$: necessarily $xy=yx$ (easy).
$n=3$: if $rank(z)=2$ then we may assume that: $z=J_{3}$ the jordan block of dimension 3. Thus $x,y$ are polynomials in $J_{3}$ and commutate, a contradiction.
If $rank(z)=0$ then $x,y$ commutate.
If $rank(z)=1$ then we may assume that $z=diag(J_{2},0)$; $x,y$ commutate with $z$ therefore they are in the form \[\begin{pmatrix}a&b&c\\0&a&0\\0&d&e\end{pmatrix}\]
Moreover $z=xy-yx$ thus \[x=\begin{pmatrix}a&b&c\\0&a&0\\0&d&a\end{pmatrix},y=\begin{pmatrix}u&v&w\\0&u&0\\0&t&u\end{pmatrix}with : ct-wd=1\]
$x=aI+N,y=uI+P$ where $P,N$ are nilpotent matrices. $rank(z)=1$ implies that $x,y$ are simultaneously triangularizable over $\mathbb{C}$ and we may assume that $a=u=0$ and $N,P$ are strictly upper triangular matrices. Then the required conditions are automatically satisfied. Now it's easy to prove that $e^{x}e^{y}=e^{x+y+\frac{1}{2}[x,y]}={I+N+P+NP+\frac{1}{2}(N^{2}+P^{2})}$.


Express .$ \overline{28}$ as a common fraction.

We see that $ .\overline{28}\equal{}\frac{28}{99}$. You can't simplify any more so the answer is $ \frac{28}{99}$.


find the value of AI.BI.CI on thr basis of r and R.

in triangle $ ABC$ let $ D$ be the tangency of incircle and side $ AB$.so $ AI\equal{}\frac{p\minus{}a}{cos(A)}$ (2p=a+b+c) (let angles of triangle $ 2A,2B,2C$ :P ).so $ AI.BI.CI\equal{}\frac{(p\minus{}a)(p\minus{}b)(p\minus{}c)}{cos(A)cos(B)cos(C)} and we know that$(cos(A))^2=\frac{p(p-a)}{bc} and use that $ S^2\equal{}p(p\minus{}a)(p\minus{}b)(p\minus{}c)$ and $ S\equal{}rp$ and $ 4RS\equal{}abc$ ........$ AI.BI,CI\equal{}\frac{S.(abc)}{p^2}\equal{}4R(r^2)$


we are given a 2000-sided convex polygon in wich no 3 diagonals are concurrent. Considering any set of 1999 diagonals of this polygon, prove there exist a triangle whose sides lie entirely on these diagonals.

I think it is true but i can't show it.

thanks !!

Hint:  Let $ y$ be the number of intersections between pairs of the $ 1999$ diagonals inside your polygon.  Now consider the graph consisting of the $ 2000$ vertices of the original polygon plus the $ y$ vertices inside the polygon (with the edges consisting of the diagonals already drawn).  How many edges, vertices, and faces does this graph have?

If there were no triangles consisting entirely of diagonals, then what do we know about the number of edges on each face, and how can you reach a contradiction from that?

Sorry but I can't go any further with your hint ; could you develop please ?   :D


The Sagebrush student council has 6 boys and 6 girls as class representatives. Two different subcommittees, each consisting of 2 boys and 2 girls, are to be created. If no student can serve on both subcommittees, how many different combinations of subcommittees are possible?

A little too easy at first sight for number 28 nats?

[quote="Treething"]it's a States question I thought...[/quote]

it's on my national competition  :P

[hide="I am not sure..."]Well, in the first committee, there are ${6\choose2}=15$ ways to choose the girls and the same for boys.  For the second committee, there are only 4 choices for the boys and girls.  That makes the total number of choices ${6\choose2}{6\choose2}{4\choose2}{4\choose2}=15\times15\times6\times6=90^2=\boxed{8100}$[/hide]

[quote="Treething"]it's a States question I thought...[/quote]

Hmm...for some reason I also remember a problem like this from states 2005. :? Maybe it was just a similar one. *shrugs*

actually what I think you're remembering is the one from chapter 2005
It was similar
I don't remember anything from state like that


algebra
polynomial
quadratics
binomial theorem
Pythagorean Theorem
geometry

1. In how many distinct ways can you walk up a stairway which has 7 steps if you can take 1 or 2 steps up the stairs at a time?

2. Given that $ a_0\equal{}1, a_1\equal{}3, a_2\equal{}7$, and $ a_{n\plus{}3}\equal{}3a_{n\plus{}2}\minus{}3a_{n\plus{}1}\plus{}a_n$ for $ n \ge 0$, we can find a general polynomial expression for $ f(n)\equal{}a_n$.  What is the sum of the degree and the coefficients of f(n)?

3. Let a, b, and c be the roots of the polynomial $ x^3\minus{}17x\plus{}19$.  What is the value of $ a^3\plus{}b^3\plus{}c^3$?

4. What is the remainder when $ 6^{51} \plus{} 8^{51}$ is divided by 49?

5. A quadrilateral has perpendicular diagonals and three of its sides have lengths of 2, 3, and 4.  There are two possible lengths for the fourth side: $ \sqrt{m}$ or $ \sqrt{n}$.  What is m+n?

Some hints:

[hide="1"]
Let $ F_n$ be the number of ways to reach the $ n^{th}$ step. Note that $ F_1\equal{}1, F_2\equal{}2$, and $ F_{n\plus{}1}\equal{}F_{n\minus{}1}\plus{}F_{n}$ because you can reach the $ n\plus{}1^{st}$ step from either the $ n\minus{}1^{st}$ or $ n^{th}$ steps.
[/hide]

[hide="2"]
The characteristic polynomial for this recursion is $ x^3\minus{}3x^2\plus{}3x\minus{}1\equal{}0$, or $ (x\minus{}1)^3\equal{}0$. Therefore, $ a_n\equal{}c_1\plus{}c_2n\plus{}c_3n^2$ for some $ c_1,c_2,c_3$. Use the initial conditions to find the constants.

(See [url]http://en.wikipedia.org/wiki/Recurrence_relation[/url], and go down to "Solving generally" and "Theorem.")
[/hide]

[hide="3"]
We know $ s_1\equal{}a\plus{}b\plus{}c, s_2\equal{}ab\plus{}bc\plus{}ca, s_3\equal{}abc$ from Vieta. Also, $ (a\plus{}b\plus{}c)^3 \equal{} a^3\plus{}b^3\plus{}c^3\plus{}3(s_1s_2\minus{}s_3)$, so we can find $ a^3\plus{}b^3\plus{}c^3$ in terms of the three symmetric polynomials.
[/hide]

[hide="4"]
Use the binomial theorem: $ (7\minus{}1)^{51}\plus{}(7\plus{}1)^{51}$. Only the last two terms of each expansion are not 0 modulo 49.
[/hide]

[hide="5"]
I can't do this one off the top of my head... but I would try using the Pythagorean theorem a bunch of times on the right triangles formed by the diagonals.
[/hide]

[hide="Even easier for 2"]The degree of $ f$ is how far back the terms go in the recurrence relation ($ 3$ here since $ a_n$ is in terms of $ a_{n\minus{}3}$). The sum of the coefficients of $ f$ is $ f(1)$.[/hide]

[hide="5 hint"]If $ a,b,c,d$ are sides of a quadrilateral with perpendicular diagonals with $ a,c$ opposite, then $ a^2 \plus{} c^2 \equal{} b^2 \plus{} d^2$ by the Pythagorean Theorem on the diagonals.[/hide]

[hide="Number 3 Solution"]By Newton's Sums, we have:
$ S_1\plus{}0\equal{}0$, so $ S_1\equal{}0$
$ S_2\plus{}0S_1\minus{}34\equal{}0$, so $ S_2\equal{}34$
$ S_3\plus{}0S_2\minus{}17S_1\plus{}57\equal{}0$, so $ S_3\equal{}\boxed{\minus{}57}$[/hide]
[hide="Number 4 Solution"]Use the binomial theorem: $ (7 \minus{} 1)^{51} \plus{} (7 \plus{} 1)^{51}$. Only the last two terms of each expansion are not $ 0$ modulo $ 49$.

The last two terms of $ (7 \minus{} 1)^{51}$ are $ \binom{51}{50}(7^1)(( \minus{} 1)^{50}) \plus{} \binom{51}{51}(7^0)(( \minus{} 1)^{51}) \equal{} 51 \cdot 7 \cdot 1 \plus{} 1 \cdot 1 \cdot ( \minus{} 1) \equal{} 356$
The last two terms of $ (7 \plus{} 1)^{51}$ are $ \binom{51}{50}(7^1)(1^{50}) \plus{} \binom{51}{51}(7^0)(1^{51}) \equal{} 51 \cdot 7\cdot 1 \plus{} 1\cdot 1\cdot 1 \equal{} 358$

We see that $ 356 \plus{} 358 \equal{} 714$.
The remainder when $ 714$ is divided by $ 49$ is $ \boxed{28}$.[/hide]
[hide="Number 5 Solution"]Because the diagonals are perpendicular, we can dissect the quadrilateral into four triangles and get the equation $ a^2 \plus{} c^2 \equal{} b^2 \plus{} d^2$, where $ a, b, c, d$ are the lengths of consecutive adjacent sides.

The only two equations that yield positive solutions are:
$ 2^2 \plus{} 4^2 \equal{} 3^2 \plus{} x^2 \Rightarrow x \equal{} \sqrt {21} \\
3^2 \plus{} 4^2 \equal{} 2^2 \plus{} x^2 \Rightarrow x \equal{} \sqrt {11}$

Therefore, $ m \plus{} n \equal{} 21 \plus{} 11 \equal{} \boxed{32}$.[/hide]

[quote="MellowMelon"]The degree of $ f$ is how far back the terms go in the recurrence relation ($ 3$ here since $ a_n$ is in terms of $ a_{n \minus{} 3}$). [/quote]

...except that it isn't.  The degree of $ f$ is actually $ 2$.

Just for the record

[hide="Solution to 2"]Rewrite as $ (a_{n\plus{}3}\minus{}a_{n\plus{}2})\minus{}2(a_{n\plus{}2}\minus{}a_{n\plus{}1})\plus{}(a_{n\plus{}1}\minus{}a_n)\equal{}0$
Then use $ b_n : \equal{} a_{n\plus{}1} \minus{} a_n \qquad b_{n\plus{}2} \minus{} 2 b_{n\plus{}1} \plus{} b_n \equal{} 0$
And finally $ c_n : \equal{} b_{n\plus{}1} \minus{} b_n \qquad c_{n\plus{}1} \minus{} c_n \equal{} 0 \implies c_{n\plus{}1} \equal{} c_n$

And it's clear that since the second differences are constant, we have a quadratic, and the three initial conditions are enough to determine the polynomial as $ f(n) \equal{} n^2 \plus{} n \plus{} 1$
[/hide]

Qn #:
If a,b and c are the roots of the equation, then a+b+c = 0

Also a^3 - 17a + 19 = 0
       b^3 - 17b + 19 = 0
       c^3 - 17c + 19 = 0

Adding, we get (a^3+b^3+c^3) - 17(a+b+c)+57 = 0

Hence a^3+b^3+c^3 = -57

Number 1:
This is simply the n+2th term of the fibonacci sequence (with F_1=0, F_2=1). n=7, F_9=21 :)

I know this is a bad approach to number 1 because we were given small values of how many steps you could take and how many steps there are, but I did it this way:

You will either make 7, 6, 5, or 4 moves (a move consists of 1 or 2 steps)
if you make 7 moves there is 1 way.
for 6: there are 6c1 ways = 6
for 5 there are 5c2 ways = 10
for 4 there are 4c3 ways = 4

4+10+6+1 = 21


MATHCOUNTS
LaTeX
email
geometry
probability
number theory

Does anyone have any National Level Math Counts? I would really appreciate it if anyone could say what they have and maybe zip it and PM it to me.

I might have a few. PM me and tell me what you need.

Can you please list what you have?

Also, I need answers for a lot

i'm trying to prepare for mathcounts, and i have a lot of old contests, but not the answers. do you know where i can get them so i can actually figure out how many i get right? some of my national rounds are as old as 1990.

Talk to chess64 I think

Edit: my $12^{3}$th post

Um, national mathcounts is one of the most coveted previous tests, it's practically impossible to find the recent ones 2000-2006 unless you know preet patel, he copied his hardcopies for me.

Wait. I thought you were preet patel. :maybe:  :maybe:  :maybe:

Maybe scans?

No, I just know preet patel really well, even though he is an enemy of mine.

Could you give me some scans (the PM I sent you)

[quote="#H34N1"]Hi there :bye: Does anyone have any state competitions from 2000?

@bmps: I'm talking about Nationals before 2001[/quote]
1. You spelled my name wrong, four letters no symbols, it isn't that bad.
2. The nationals before 2001 are marginally easier to get, but they are still coveted.

Where can you get them?

Actually, here's my problem:

I have the questions, but no ansewrs :!:

I can get you some answer keys for older nationals if you can get me some more recent nationals.

Never mind. I found the answer keys for the older nationals  :)

(maybe we could work it out if we both make states)

I have all of 1996-2006, and 1990. Have any old AMC tests?

[quote="#H34N1"]Does anyone have the [b]2006 National Round[/b]?

If you do, please scan it and PM it to me. Thanks.[/quote]
i got some 2006 nationals, but i dont have time to scan it nor pm it to u sry

[quote="#H34N1"]I already have the 2006 Nat's. Does anyone have answers?[/quote]
I do, but you shouldn't openly request these things because mathcounts sells them.

[quote="#H34N1"]Okay. Does anyone have any answer keys? I need answer keys for 2006, 2001, 2002, 2004 National Round Questions[/quote]
I have 2004 Nationals Answers. It's a scanned pdf file and it's not exactly good-looking, but it's legible.

OOh. I meant 2005. Sorry. I have 2004 that's horribly scanned but readable. with answer

Can anyone post a link, then?
So it would be easier for other interested people to get to...

[quote="#H34N1"]Okay. Is there any way that I can prepare besides doing Saab, the handbook, old nationals and old states and old districts for district math coutns? Thansk.[/quote]

Work through the Art of Problem Solving textbooks (Vol. 1 and the new ones on Number Theory, Geometry, Counting & Probability), work through problems here on AoPS, identify your areas of weakness and spend extra time working on those types of problems.

Good luck!

[quote="mathmom"][quote="bpms"]
2006 nationals is the most coveted nationals there is, my pdf copy of it costed me an arm and a leg, I'll say that much.[/quote]

You can buy it from Mathcounts for $\$$25 (along with Chapter and State, which you probably already have).  That's expensive, but cheaper than selling off your body parts. ;-)[/quote]
But mathcounts doesn't sell it in pdf format.
About the body parts, I almost lost an eye too -_-

[quote="pianoforte"]how do you expect to compete w/o a leg, arm or eye?[/quote]
Well I just write with my write hand, and use it for countdown as well.
[quote="spiced hammer"]
He still has one leg = wheelchair. He uses his other arm. He uses his other eye. :P[/quote]
I don't get what's so funny about my predicament :mad:

I hate to be a partypooper,

but off topic folks :(.


I'm too lazy to check mine but mine sux
Its like 2.2 Gigahertz
I forgot the rest

AMD 2400 Sempron (around 2 GHz)
768 MB RAM
128 MB GeForce FX gracphics card
160 GB Hard Disk
Samsung DVD-R
LG DVD-RAM (writer)

But my UPS is almost six years old. I must get a new one soon.

And this most definitely belongs to G&FF. Someone do the needful.

Dell Laptop:
Dell Latitude
Intel Pentium III Processor
20 GB Hard Disk (11.3GB free)
256 MB RAM
CD and DVD reader (no DVD writer)
Windows XP Professional SP2

Northstar Laptop:
Intel Celeron
10 GB Hard Disk ($\approx$ 6GB free)
256 MB RAM
CD and DVD reader (no CD or DVD writer)
Ubuntu 6.06 Dapper Drake LTS

AMD64 2.2Ghz Westminster core (OC 2.6)
ATI X800XT Platinum
MSI Neo4 Platinum Motherboard
Seagate Barracuda 200GB HDD
Viewsonic G90fb 19" CRT Monitor
2x512MB RAM Corsair VS

INTEL CORE 2 DUO OVERCLOCKED TO LIKE ROUND 2.9GHTZ
2 GIGS OF RAM
UHHHHHH CANT FIND MY VID CARD BUT IM GUESSING ITS AVERAGE
AND 750 GIG OF HARDDRIVE

[quote="anirudh"]This is a really messed up version of a Dell XPS: :P (I modified it soo much)

Okay. Here we go:

Intel Pentium D (2x3.6 Ghz) Extreme
1.5 Terabytes of HD Space
Dual Layered DVD Burner
4 gigs of DDRSD2 Ram
Dell Ultrasharp 24" Monitor (Its oversized)
and some otehr stuff i don't remember[/quote]

WOW :o

i only have Intel Celeron 3.33 GHz and 1 GB of RAM  :(   [but i don't play PC games  :P ]

[quote="anirudh"]This is a really messed up version of a Dell XPS: :P (I modified it soo much)

Okay. Here we go:

Intel Pentium D (2x3.6 Ghz) Extreme
1.5 Terabytes of HD Space
Dual Layered DVD Burner
4 gigs of DDRSD2 Ram
Dell Ultrasharp 24" Monitor (Its oversized)
and some otehr stuff i don't remember[/quote]

daaaaaaaaaaaaaaaang that's sooo nice! *jacks ur comp and runs*

I have an okay system:

Manufacturer: HP Pavilion 061
Processor: AMD Athlon(tm) 64 X2 Dual Core Processor 4200+, MMX, 3DNow (2 CPUs), ~2.2GHz
Memory: 1982MB RAM
Hard Drive: 290 GB
Video Card: ATI RADEON XPRESS 200 Series
Monitor: SyncMaster 920BW(Analog)
Sound Card: Realtek AC97 Audio
Keyboard: USB Root Hub
Mouse: USB Root Hub
Operating System: Windows XP Professional (5.1, Build 2600) Service Pack 2 (2600.xpsp_sp2_gdr.050301-1519)

Personal computer:

Manufacturer: Me
Case: Lian Li with side panel
Proc: Athlon64 X2 4600+ Manchester I think (dualcore, 64bit, 2.4GHz)
HD space: 500GB
RAM: 2GB (used to be 2.5, before I took out the smaller stick for the server)
Vidcard: Geforce FX5500 256MB (Nvidia is nice)
Monitors: 21" Rosewill LCD, 16" Dell CRT (yes, that's confusing :D; I have a separate set of virtual desktops for each, but I can drag windows between them.)

Shared computer (i.e. server, made of used parts except the PSU and case)
Manufacturer: Me again
Case: Nasty no-name white box
Proc: AthlonXP 3000+ @1.8GHz (underclocked)
HD space: 200GB
RAM: 512MB
Vidcard: Servers don't have video cards. They have Ethernet, power, and printer cords emanating from them; nothing else.

OS on both is Gentoo Linux; WM on my computer is E17 ([url]http://www3.get-e.org[/url])

[quote="Thu7"][quote="anirudh"]This is a really messed up version of a Dell XPS: :P (I modified it soo much)

Okay. Here we go:

Intel Pentium D (2x3.6 Ghz) Extreme
1.5 Terabytes of HD Space
Dual Layered DVD Burner
4 gigs of DDRSD2 Ram
Dell Ultrasharp 24" Monitor (Its oversized)
and some otehr stuff i don't remember[/quote]
HOLY COW HOW MUCH DID THOSE COST
*drools* :o  :o  :o  :o[/quote]

All that together = $\$$ 4000


ratio
geometry
incenter
geometry unsolved

Let $ D$ be a point on the side $ BC$ of $ \triangle{ABC}$,and let $ E$ and $ F$ be the incenters of triangles $ ABD$ and $ ACD$,respectively.Suppose that $ B,C,E,F$ are concyclic.Prove that $ \frac{AD\plus{}BD}{AD\plus{}CD}\equal{}\frac{AB}{AC}$

[hide="A short hint (no detail)"]d is the line through A and parallel to EF. EF cuts AB,AC at P,Q. DE cuts AB,d at M,R, DF cuts AC,d at N,S. After that, we prove that AM=AN.[/hide]


Find the common ratio in the following sequences:
1,3,9,27...

general term an =a1r^n-1

[hide]The common ratio, r, is the multiple by which the terms are increased/decreased.

We can finding this by dividing $U_n$ by $U_(n-1)$

So the answer is $27/9 = 3$  or  $9/3 = 3$ Therefore the answer is: $3$[/hide]

[hide="hint"]

the formula is

$N/(1-R)$ when R's absolute value is smaller than 1 :D [/hide]

[hide="For the bonus questions"]
Let $S=1+\frac13+\frac19+\frac{1}{27}+...$
$3S=3+1+\frac13+\frac19+...=3+S$

So $2S=3\Leftrightarrow S=\frac32$.

You can do the same thing to prove the general formula for the sum of an infinite geometric series.[/hide]

[quote="mathnerd314"]Bonus problems from mathnerd314:

1. Sum the infinite geometric series $1+\frac 13+\frac 19+\frac{1}{27}+...$.

2. Prove the formula for the sum of an infinite geometric series.[/quote]

[hide="The second one"]Let the series be $a+ar+ar^2+ar^3...=S$.

$Sr=ar+ar^2... \Longrightarrow Sr=S-a \Longrightarrow Sr-S=-a \Rightarrow S(r-1)=-a \Rightarrow S=\frac{-a}{r-1}=\frac{a}{1-r}$[/hide]

 :D

[hide="Answer to the second bonus"]Take geometric series (and sum) $a+ar+ar^2+...+ar^{n-1} = S$ for example. Then multiply the series by $r$, getting $ar+ar^2+ar^3+...+ar^n = Sr$. Subtract the new series from the original, getting:

$S-Sr = a-ar^n$

Do some factoring out:

$S(1-r) = a (1-r^n)$

Divide both sides by $(1-r)$:

$S = \frac{a (1-r^n)} {(1-r)}$[/hide]


geometry
trigonometry
inequalities
inequalities unsolved

These problem appeared in geometry section in solving the problem
proposed by treegoner. Could anybody help to solve it.

√(1- sina*sinb) + √(1- sinb*sinj) + √(1- sina*sinj) ≥ 3/2 where a+b+j =180.

It is not so difficult as I thought first.

\[1-\sin a\sin b=1-1/2(\cos (a-b)-\cos (a+b))\]
\[\geq 1-1/2(1+\cos c)=\frac{1-\cos c}{2}=\sin^2(c/2).\]

$\sin(a/2)+\sin(b/2)+\sin(c/2)\geq\frac{3}{2}$ with $a/2+b/2+c/2=\frac{\pi}{2}$ follows with Jensen for sinus.

Misha

Misha are you sure in your solution.I think Jensen inequality gives us that
$\sum{\sin{\frac{\alpha}{2}}}\leq\frac{3}{2}$

of course you are right :( . sorry for the wrong solution. :blush:

Misha


Let be $ n\in N^*$. If $ a \equal{} \underbrace {11...1}_{n \plus{} 1}\underbrace {22...2}_{n \plus{} 1} \minus{} \underbrace {33...3}_{n \plus{} 1}$ and $ b \equal{} 1\underbrace {77...7}_n9\underbrace {55...5}_n2 \minus{} 5\underbrace {33...3}_{n \minus{} 1}28$ , compute $ \sqrt {\frac{{\sqrt a }}{{\sqrt b }}}$
[i] R.M.G. 29/2007, Mîndrilă Claudiu[/i]

The problem was translated from the Romanian Forum...

[quote="Mîndrilă Claudiu"]Let be $ n\in N^*$. If $ a \equal{} \underbrace {11...1}_{n \plus{} 1}\underbrace {22...2}_{n \plus{} 1} \minus{} \underbrace {33...3}_{n \plus{} 1}$ and $ b \equal{} 1\underbrace {77...7}_n9\underbrace {55...5}_n2 \minus{} 5\underbrace {33...3}_{n \minus{} 1}28$ , compute $ \sqrt {\frac {{\sqrt a }}{{\sqrt b }}}$
[i] R.M.G. 29/2007, Mîndrilă Claudiu[/i]

The problem was translated from the Romanian Forum...[/quote]

$ \underbrace {11...1}_{n \plus{} 1}\equal{}m$  --> $ 9m\equal{}\underbrace {99...9}_{n \plus{} 1}\equal{}10^{n\plus{}1}\minus{}1$ ---> $ 10^{n\plus{}1}\equal{}9m\plus{}1$ --->

$ a\equal{}\underbrace {11...1}_{n \plus{} 1}\underbrace {22...2}_{n \plus{} 1} \minus{} \underbrace {33...3}_{n \plus{} 1}\equal{}10^{n\plus{}1}m\plus{}2m\minus{}3m\equal{}(10^{n\plus{}1}\minus{}1)m\equal{}9m^{2}$

$ b\equal{}1\underbrace {77...7}_n9\underbrace {55...5}_n2 \minus{} 5\underbrace {33...3}_{n \minus{} 1}28\equal{}(10^{n\plus{}1}\plus{}7m\plus{}2)*10^{n\plus{}1}\plus{}5m\minus{}3\minus{}(5*10^{n\plus{}1}\plus{}3m\minus{}5)\equal{}(9m\plus{}1\plus{}7m\plus{}2)(9m\plus{}1)\plus{}5m\minus{}3\minus{}(5(9m\plus{}1)\plus{}3m\minus{}5)\equal{}(16m\plus{}3)(9m\plus{}1)\plus{}5m\minus{}3\minus{}48m\equal{}144m^{2}\plus{}27m\plus{}16m\plus{}3\plus{}5m\minus{}3\minus{}48m\equal{}144m^{2}$ ----->

$ \sqrt {\frac {{\sqrt a }}{{\sqrt b }}}\equal{}\sqrt[4]{\frac{a}{b}}\equal{}\sqrt[4]{\frac{9m^{2}}{144m^{2}}}\equal{}\sqrt[4]{\frac{1}{16}}\equal{}\frac{1}{2}$

Let $ c\equal{}\frac{10^{n\plus{}1}}{9}$, then $ 10^{n\plus{}1}\equal{}9c\plus{}1$ $ a\equal{}c(9c\plus{}1)\plus{}2c\minus{}c\equal{}9c^2$, $ b\equal{}(9c\plus{}1)^2\plus{}7c(9c\plus{}1)\plus{}2(9c\plus{}1)\plus{}5c\minus{}3\minus{}5(9c\plus{}1)\minus{}3c\plus{}2\equal{}144c^2\minus{}3$. I think may be $ b\equal{}144c^2$, then your expression is $ \frac 12$.

I like your proofs... The answer is correct.


algebra
polynomial
linear algebra
matrix

How would I  solve the following problem:
If $ A$ and $ B$ are square matrices of the same dimension and if $ \lambda$ is an eigenvalue of $ AB$ then $ \lambda$ is also an eigenvalue of $ BA$

First suppose that one of the matrices, say $ A$, is invertible. Then $ \det(A)\det(A^{\minus{}1})\equal{}1$. Now $ |AB\minus{}\lambda I|\equal{}|A^{\minus{}1}\parallel{}AB\minus{}\lambda I\parallel{}A|\equal{}0\equal{}|BA\minus{}\lambda I|$ hence $ \lambda$ is an eigenvalue of $ BA$ too. Now prove your statement in the general case by a continuity argument.

[i]PS[/i] Another method for proving this theorem is by cooking up two block-matrices with the same determinant.
[i]Remark[/i] We have shown that the two characteristic polynomials coincide.

thanks!
I thought up:
Since we have eigenvalue $ \lambda$, then consider arbitrary eigenvector $ v$.
Then
$ ABv \equal{} \lambda v$
$ (BA)Bv \equal{} \lambda Bv$
Denote, $ u \equal{} Bv$
So $ BA$ has eigenvalue $ \lambda$

Would this be sufficient?

Obviously, if $ u \equal{} Bv \equal{} 0$, you have found that $ BAu \equal{} BA0 \equal{} 0 \equal{} \lambda0\equal{}\lambda u$, which does not guarantee that $ \lambda$ is an eigenvalue of $ BA$.  However, in that case, independently on whether the $ \lambda$ you picked is zero or not, we can guarantee that $ B$ has eigenvalue 0, or $ B$ has rank less than its size, and so do $ AB$ and $ BA$, and 0 is eigenvalue of both $ AB$ and $ BA$, qed.

[quote="daniel73"]Obviously, if $ u \equal{} Bv \equal{} 0$, you have found that $ BAu \equal{} BA0 \equal{} 0 \equal{} \lambda0 \equal{} \lambda u$, which does not guarantee that $ \lambda$ is an eigenvalue of $ BA$.  However, in that case, independently on whether the $ \lambda$ you picked is zero or not, we can guarantee that $ B$ has eigenvalue 0, or $ B$ has rank less than its size, and so do $ AB$ and $ BA$, and 0 is eigenvalue of both $ AB$ and $ BA$, qed.[/quote]

 $ Bv\equal{}0\Longrightarrow 0\equal{}ABv\equal{}\lambda v \Longrightarrow \lambda \equal{} 0 \Longrightarrow 0\equal{}det(AB)\equal{}det(BA)$ .. and son on ...
 :cool:

[quote="The Captain"]First suppose that one of the matrices, say $ A$, is invertible. Then $ \det(A)\det(A^{ \minus{} 1}) \equal{} 1$. Now $ |AB \minus{} \lambda I| \equal{} |A^{ \minus{} 1}\parallel{}AB \minus{} \lambda I\parallel{}A| \equal{} 0 \equal{} |BA \minus{} \lambda I|$ hence $ \lambda$ is an eigenvalue of $ BA$ too. Now prove your statement in the general case by a continuity argument.[/quote]
This sounds like that the underlying field $ K$ is $ \mathbb R$ or $ \mathbb C$. But what is the final argument here if $ K$ is an arbitrary field?  :P

Yeah, I've implicitly assumed that the underlying field is $ \mathbb{C}$, just as Prasolov in his book has assumed.
This is the method for the general case that I've mentioned in my PS:
[b]Proposition.[/b] Let $ B, A$ be $ m\times n$ and $ n\times m$ matrices, respectively. Then we have $ \lambda^n|\lambda I_m \minus{} BA| \equal{} \lambda^m|\lambda I_n \minus{} AB|$.
[i]Proof.[/i] Let $ C \equal{} \begin{bmatrix}xI_n & A \\
B & I_m\end{bmatrix},\ D \equal{} \begin{bmatrix}I_n & 0 \\
\minus{} B & xI_m\end{bmatrix}.$ Now compute $ \det(CD) \equal{} \det(DC)$ to obtain the result.


ratio
integration
trigonometry
calculus
calculus computations

wassup fine people ?  :D

can you prove

$ \sum_{n\equal{}0}^\infty \; \dfrac{(n!)^2 } {(2n\plus{}1)! \cdot\phi ^{2n\plus{}1}  }\;\equal{}\; \boxed{\dfrac{2\pi}{5\cdot \sqrt{2\plus{}\phi }} }$

where $ \phi\;\equal{}\;\dfrac{1}{2}\left(1\plus{}\sqrt 5\right)$ (i.e. golden ratio)

[quote="misan"]...

can you prove ...

[/quote]

Sure,  :lol: .
it's not important about golden ratio, silver ratio or bronze ratio (If somebody don't no, I note that in math we have also silver and bronze ratio.)  :blush: .
more generaly:
find:
$ \sum_{n\geq 0}\frac {(n!)^{2}z^{n}}{(2n \plus{} 1)!}$
where $ |z| < 4$
hint:
$ \int_{0}^{\frac {\pi}{2}}\cos^{2n \plus{} 1}tdt \equal{} \frac {(2n)!!}{(2n \plus{} 1)!!}$ :)


linear algebra
matrix
vector
algorithm
linear algebra unsolved

[quote="DK"]Hello,

I am wondering if there is a famous result for the following problem I have:
I have a complex NxN nonsingular matrix H where we can arbitrary permute the rows and the columns in it. After such a permutation, we perform a QR decomposition of it. Now I'm interested in the QR decomposition that maximally concentrates the energy in the R matrix in the southeast part of the matrix.
Energy is the sum of the squares of the absolute values of the matrix elements. For comparison, there is a similar result for vectors, minimum phase FIR filters.

Thanks in advance.[/quote]Row permutations don't really matter: If $A = QR$ and $P$ is a permutation matrix, then $PA = PQR = \tilde Q R$, and $\tilde Q$ is also orthogonal (or unitary in the complex case).

One can find column permutations (say a permutation matrix P) such that $QR = AP$ and the northwest entries of $R$ are large, in the sense that $|r_{ii}|^{2}\ge \sum_{j= i}^{k}|r_{jk}|^{2}$ for all $k > i$. In particular, the diagonal entries of $R$ are decreasing in magnitude. This is done with "QR-decomposition with column pivoting". The algorithm is described in any textbook on numerical linear algebra, e.g. in the book by Golub and Van Loan (algorithm 5.4.1). The method is essentially the greedy algorithm for determinants of certain submatrices, so it's not guaranteed to be optimal.

So there is also a way to do the same but where |r_{ii}| >= sum(j=i:n)|r_{kj}|^2 for all k < i, that is, a similar result but where the energy is focused on the southeast part and on the rows?


trigonometry
quadratics
geometry
3D geometry
algebra
complex analysis

If I want to solve the equation $ 8x^3\minus{}6x\equal{}1$ then Maple outputs
$ x\equal{}1/4(4\plus{}4i\sqrt{3})^{1/3}\plus{}1/(4\plus{}4i\sqrt{3})^{1/3}$. But if I have understood correctly, $ (4\plus{}4i\sqrt{3})^{1/3}$ has three values $ x_1,x_2,x_3$. If I want to evaluate the value of $ x$, do I have to evaluate all the expressions $ 1/4x_k\plus{}1/x_l$ for all combinations $ k,l\in \{1,2,3\}$.

I was told that the correct way to compute those is to fix $ k\in \{1,2,3\}$ and the result is $ \frac {x_k}{4} \plus{} \frac {1}{x_k}$. The value is independent on choosing $ k$.

Obviously, the value [b]does[/b] depend on choosing $ k$ (and you get all 3 roots this way) but otherwise you are right: the cubic root should be chosen the same in both fractions. If you recall how the cubic equation formula is derived, you find that it starts with representing the root as $ u\plus{}v$, then you get the pair of equation giving you the value of $ uv$ and $ u^3\plus{}v^3$. You raise the first one to the third power to get the quadratic equation for $ u^3$ and $ v^3$. Then you solve this quadratic equation and get $ u\equal{}\sqrt[3]\dots$ with 3 possible values for the cubic root but still keep the relation $ uv\equal{}\dots$, not just the weaker relation $ u^3v^3\equal{}\dots$.

Okay. I thought the value does not depend on $ k$ as Maple outputs three different roots. It looks like Maple uses the principal branch of the cube root when outputting answers.

One more question on the roots. How do we compute for example $ \sqrt {2 \plus{} 3i} \plus{} \sqrt {4 \plus{} i}$ or $ \sqrt {2 \plus{} 3i} \plus{} \sqrt [3]{4 \plus{} i}$? Here we have different roots and different branches.

You will need to give the branch, or you will have to write down all possibilities.


I have till now not been able to be convinced about newton's laws of motion...

What does it define..??
force or mass...(or maybe both.. :P )

Replies awaited...

i would rather take mass definition as an axiom and then give newton's law as a definition of force...
i feel they are exceptional equations but would rather prefer Lagrangian Mechanics to newton's laws.
 I would rather like to know others opinion befor e procceding

What I am afraid of is [b]circular reasoning[/b]..

ok, I will tell you what i have understood..
[u]Mass[/u]
An arbitrary body was chosen and assigned the value 1...
It was put on a balance...
(though we till now don't have idea about gravity, but we can define from our observations...)
Any other body which balanced this body (by observation without understanding of gravity..) was said to have mass 1...

A balance can be used to check whether two bodies have equal mass...(by definition...)

If two mass 1 objects balance another object, the bigger object is said to have mass 2 and similarly... (But this breaks down when the objects become too large, by facts we know now though we wouldn't be aware of this fact then...)
But lets confine ourselves to primitive people who thought about these things first...

So now we have some idea of what 'mass' is..
We observe a body of mass 'm' travelling with acceleration 'a'..
we found that to move a body with more 'a' we had to strain ourselves more..
and also to move a body with more 'm' we had to strain more...

In order to quantify it, we defined sth called force and wrote it as m*a...
and then we do blah blah blah mathemtics and reach conclusions...

But now let's see the situation...

Somebody asks you,
can you explain why two bodies of the same mass kept on either side of the weighing scale balance each other..?
Ans 1: the torque due to the force called gravity is equal on both sides, so they balance each other...(vectorially, net torque is zero and so is the net force...)
Ans 2: Hadn't we defined mass that way...
So by definition they should balance...

Are we reaching a point of 'circular reasoning'...

BTW
I am waiting for people to post their views on how they 'feel' and 'understand' the concept of 'mass'...

how about [url=http://en.wikipedia.org/wiki/Mass]this[/url] article

I would feel really happy if sb could tell me things without 'circular logic'...
i.e. without implicitly assuming what he is trying to prove...

I don't see the need. The other post was one which required lots of views, lets see how this progresses na??

see.. if we begin to define mass to be inertial, i would be led to think that if there was no force on the body, there would technically be no mass.. so i was not very convinced by newton's laws.. now im looking at it differently -

Claim 1. every matter is made of atoms

Claim 2: atoms are further divided into smaller components.

some cool guy put these things in a neat little table (the standard table) and said that there are 2 types of tiny little things based on whether they constituted types of force or matter itself (are we coming to the fork in the road?) which depended on (seemingly) absurd properties like spin et al into bosons (integer spins) and fermions (half integer spins) which are further sub divided into other stuff. then, one day some guy started discovering these elements in particle accelerators by crashing particles at high energy with each other. amongst all these particles, there's one boson ( - force carrier(tada!). so where was the independence in the first case?) called the higgs boson which has been predicted by this table but not found yet. if found it would quite convincingly solve our problems. if you're wondering why, its because this thing is something like the fundamental unit of mas. moreover, as a fringe benefit, it even tells us about the reason of existence of "massless" particles like photons. so that would kill our argument.. i think they're looking for it in CERN (that swiss research place). not sure though..

i had another nice doubt.. about quantisation of time and how it maybe defined as the time for the smallest unit of increase in entropy (which is quantised!) and second law etc etc.. if u guys can think about it and come up with something, ill put forth my viewpoints. we can have a nice discussion..

[quote="Cyber-Shreyas"]Hmmm.....Good question.
http://en.wikipedia.org/wiki/Newton%27s_Laws

Anyways I think Newton's Laws define force and then go on to define mass.

"Mass is an inherent property of a body determined by virtue of amount of matter it contains, which inturn gives us a measure of the inertia of a body". The inertia is its resistance to change in state.

just like V=IR (in elec)
F=ma

Till now things are okay but then we realize that he has indeed used mass to define momentum. So we go around in circles.

So as pardesi said if we assume mass is defined, then we can define force but defn force first makes it easier for understanding except for the glitch.[/quote]

No no...
You are not getting my question...

don't assume the mass to be defined ... beacause my whole question now depends on how you 'understand' the concept of mass..

you needn't tell me a logical definition...
I am o.k. with an intuitive approach also..

just like I told you about a possible definition using scales
(and the definition turned out to be good because then mass becomes additive, and thus obeys a law of conservation ...)

Tell me the story of definition of 'mass'..
how it all started in the earlier times??

what do u think of that boson?

anyway, i presume it must be relative.. because of difficulty while attempting to lift objects of different masses? is that not the most natural thing? and hence the inertial idea of its being defined as opposition to force. right? thats natural progression isint it?


John flips a fair coin 3 times. Find the probability that exactly 2 of the flips come up the same.

there are 8 possible put comes.
hht tth thh htt
those are the combos that work

1/2 is the answer.

what about "hth";  "tht"  any other solutions?

[quote="G-UNIT"][hide]im not sure if this is right......    since you flip 3 coins you can get a maximum of 3  to a minimum of 0 tails ( head vice versa),  this eqauls 3 + 1= 4 different combinations.  Now you can get either 2 heads and a tail or 2 tails and a head.  There are two choices altogether, Thus the answer is 1/2. [/hide][/quote]

nvm its 6/8 just list out the possibilities

the total outcomes is 2^3
3 number of ways to arrange two heads
3 number of ways to arrange two tails
so 6/8 or 3/4

You guys made it way too hard. You can either list out the possibilities or

[hide]
Total possible outcomes are $2^3=8$. The probability that there are exactly 2 of the same, which is the same as the probability that one is "the same",  is opposite the probability that 3 are the same. The probability that 3 are the same is $\frac{2 \cdot 1 \cdot 1}{2^3}=\frac{1}{4}$. The opposite is $1-\frac{1}{4}=\frac{1}{4}$. That seems complicated, but you should learn to use complements - finding the opposite and substacting it from 1 - in probability.
[/hide]


Solve this equation.
 :lol: This is easy to solve!!! :(

$\frac{(x-1)(x+3)}{3x}+7=0$


Let a triagle $ABC :\angle A=20^0,\angle B=\angle C=80^0,M\in BC$ such that:$AM=BC$.
Find the value $\angle BMC$.

$M \in BC $ such that : AM=BC
Could it be???

I'm sure it's $M\in AB$.

Take point $P$ on the same side of $AB$ as $C$ s.t. $PAM$ and $ABC$ are congruent. In this case we have $AC=AP=AB=MP=\ell$ and $\angle CAP=60^{\circ}$, so $ACP$ is equilateral, meaning that $CP=\ell$. We now have $PM=PC=\ell$ and $\angle MPC=\angle APC-\angle APM=40^{\circ}$, so $\angle PMC=70^{\circ}$, and since $PM\|BC$, we get $\angle BCM=70^{\circ}$, so $\angle BMC=30^{\circ}$.

I supposed M is on AC but it doesn't really matter.
Let <ABM=x  then from triangle ABM and ABC : sinx*sin80=sin20*sin(20+x) so
cos(80-x)-cos(80+x)=cosx-cos(40+x)
cos(80-x)-cos(x)=cos(80+x)-cos(40+x)
sin40*sin(x-40)=-sin(60+x)*sin20
2cos20sin(x-40)=-sin(60+x)
sin(x-20)+sin(x-60)=-sin(60+x) but sin(x-60)+sin(60+x)=sinx so
sin(x-20)=-sinx=sin(-x) hence x-20=-x so x=10.We get <BMC=20+x=30.
That's all. :D


There are $5$ red balls and $7$ yellow ones. Picking three balls, what is the chance that a man will get two yellows and a red one.

[quote="liu"]$ \Omega = \{ A \subset \{r_1, r_2, \ldots, r_5, y_1, y_2, \ldots, y_7\}: |A|=3 \}$ ($|A|$ is number of elements in $A$).
$ |\Omega| = {12 \choose 3} $
$ |K| = {7 \choose 2}\cdot 5$ (because we have to choose 2 yellow balls from 7 and one red from 5 red.

$P(K) = \frac{|K|}{|\Omega|}$

Please do calculations because it's too difficult for me ;)[/quote]

I don't think all those symbols are necessary, just use the counting principle for it  ;)

I'm assuming it's without replacement.

[quote="liu"]But those symbols are funny ;)[/quote]
well, those symbols are a little complicated for middle school ;)

Thsi problem isn't hard, I dont' know what's this outside chatter...

[hide]
The number of choices is $ {12 \choose 3} $ and the chance to get 2 yellows is $ {7 \choose 2} $ and the chance for the red is $ {5 \choose 1} $
So the answer is $ \frac{105}{440}=\boxed{\frac{21}{88}} $[/hide]


At a meeting of 12k people  ,each person exchanges greetings with exactly other 3k+6 people.
For any 2 people , the number of people who exchanged greetings with both of them  is the same.
How many people are there in the meeting? :lol:

no one interested in this one? is it too hard ? it is not that hard...  :P


function
superior algebra
superior algebra unsolved

Let F be a field of characteristic 2.
a) Prove that the function Φ:F →F given by Φ(x)=x^2 is a ring homomorphism.
b) Prove that it is also one-to-one.

$ \phi(ab) \equal{} (ab)^2 \equal{} a^2 b^2 \equal{} \phi(a) \phi(b)$
$ \phi(a \plus{} b) \equal{} (a \plus{} b)^2 \equal{} a^2 \plus{} 2ab \plus{} b^2 \equal{} a^2 \plus{} b^2 \equal{} \phi(a) \plus{} \phi(b)$

Moreover, if $ \phi(a) \equal{} \phi(b)$ then $ a^2 \equal{} b^2 \Leftrightarrow a^2 \minus{} b^2 \equal{} 0 \Leftrightarrow (a \plus{} b)(a \minus{} b) \equal{} 0$, so $ a \equal{} b$ or $ a \equal{} \minus{} b$, but in a field of characteristic $ 2$ clearly $ b \equal{} \minus{} b$.

In general, given a ring of characteristic $ p$ the map $ \phi(x) \equal{} x^p$ is the [url=http://en.wikipedia.org/wiki/Frobenius_endomorphism]Frobenius map[/url].


Show that $\forall x,y,z>0$ we have:
$ \sum \sqrt{ \frac{x}{y+z}}>2$.

cheers! :D :D :cool:

We have sqrt(x/(y+z))>=2x/(x+y+z) and so on. Add them up. It is trivial to show that we cannot have equality.


[hide]it's clearly true cause $ A_1B_1C_1$ and ABC are orthologic tringle.[/hide]

[quote="balan razvan"]On the sides of the triangle $ ABC$ construct isoscelles triangles $ AB_1C,BA_1C,AC_1B$.Prove that the projections from $ A,B,C$ to $ B_1C_1,C_1A_1$ and $ A_1B_1$ concur in the same point.[/quote]
[hide="Solution"]
Let the circles with center $ B_1$ and passing through $ A$ and $ C$ be named $ \omega_B$.  Similarly, $ \omega_A$ is the circle with center $ A_1$ and passing through $ B$ and $ C$ and $ \omega_C$ is the circle with center $ C_1$, passing through $ A$ and $ B$.  Now, let $ \omega_C$ and $ \omega_A$ meet at $ B$ and $ B'$ and let $ BB'$ meet $ A_1C_1$ at $ K$.  It follows that $ BB'$ is a radical axis of the two circles, so $ BB'\perp A_1C_1\implies BK\perp A_1C_1$, which means that $ BK$ is the perpindicular from $ B$ to $ A_1C_1$.  Similarly, the perpindicular from $ C$ to $ A_1B_1$ is the radical axis of $ \omega_A$ and $ \omega_B$ and perpindicular from $ A$ to $ C_1B_1$ is the radical axis between $ \omega_B$ and $ \omega_C$.  It follows that all of these perpindiculars are radical axes of three circles and are thus concurrent. [/hide]


Given Triangle ABC with legs of length a and b, and a hypotenuse of length c. What is the measure of the altitude from point B of the vertex to the hypotenuse in terms of a, b and c.

[hide]c/b=a/x where x is the unknown length.  I'm quite sure that's right because of the similar triangles.[/hide]


calculus
integration
calculus computations


Construct a function $f: \mathbf{R}\rightarrow \mathbf{R}$ such that f is not injective on any interval ,but $f=g+h$,where also $g$ and $h$ are real valued functions,both injective.

Is there any problem with $g(x)=x, h(x)=-x$¿

OOps I forgot to mention that f should have at least two different  real values  :blush:

Set everything like above except that $g(n)=n+1$ for integers $n$¿

It's ok Zetax.But what about f not being monotone on any interva,l but can be wriiten as a sum of  two strictly monotone functions.


I am doing Solving Systems of equations using the substitution method, here is my problem

2x-y=11
6x-3y=-24

I ran off into (3/4, 12 1/2) which is not the answer, haha. Can someone help me out please, this book just doesnt explain it step-by-step  :?:

Okay, solve for $y$ in the first equation since it has a coefficient of 1... $y=2x-11$. Divide the second equation by 3 to get $3x-y = -8$. Substitute $y$ into this: \[ 3x - (2x-11) = -8, \] or \[ x+11 = -8, \qquad x=-19. \] Now $y=2(-19)-11=-38-11=-49$. $\boxed {}$

isn't it just

[hide]$-24=6x-3y=3(2x-y)=3(11)=33$ which clearly has no solution[/hide]

Multiply the first equation by three, and you'll get the same expression equal to two different numbers. ;)

AHHH no homework problems in this forum.

Slopes = Equal, and since their diff y intercepts, they never intersect.


algebra
polynomial
algebra proposed
Kazakhstan

Find all polynomials $ P(x)$ with real coefficients such that for every positive integer $ n$ there exists a rational $ r$ with $ P(r)=n$.

Clearly $ P(x)$ can't be constant, so $ d: \equal{} \deg P\ge 1$.
For all $ n\in\mathbb{N}$ take $ r_n\in\mathbb{Q}$ with $ P(r_n) \equal{} n$.
$ P(r_1) \equal{} 1,P(r_2) \equal{} 2,\ldots,P(r_{d \plus{} 1}) \equal{} d \plus{} 1$ gives $ P(x)\in\mathbb{Q}[x]$ using the [url=http://en.wikipedia.org/wiki/Lagrange_polynomial]Lagrange interpolation[/url].
Then for some $ t\in\mathbb{N}$ we have $ tP(x)\in\mathbb{Z}[x]$ with leading coefficient $ m\in\mathbb{Z}\setminus\{0\}$.
But then $ tP(x) \minus{} tn\in\mathbb{Z}[x]$ has rational root $ r_n$ and also leading coefficient $ m$.
So the denominator of $ r_n$ divides $ m$, i.e. $ r_n\in{1\over m}\mathbb{Z}$ for all $ n\in\mathbb{N}$.

Now assume $ d \equal{} \deg P > \equal{} 2$.
Then $ \left |{P(x)\over x}\right |\to \infty$ for $ |x|\to \infty$,
and we find $ N\in\mathbb{N}$ large enough with $ \left |{P(x)\over x}\right | > 2|m|$ for all $ |x|\ge N$.
But $ {1\over m}\mathbb{Z}\cap ( \minus{} N,N)$ contains exactly $ 2|m|N \minus{} 1$ elements.
So among the $ 2|m|N$ different numbers $ r_1,r_2,\ldots,r_{2|m|N}\in{1\over m}\mathbb{Z}$
we must find $ |r_k|\ge N$ for some $ k \equal{} 1,\ldots ,2|m|N$.
This gives $ 2|m| < \left |{P(r_k)\over r_k}\right | \equal{} {k\over |r_k|}\le {2|m|N\over N} \equal{} 2|m|$.
Contradiction!

So we must have $ P(x)\in\mathbb{Q}[x]$ with $ \deg P \equal{} 1$, which is indeed a solution for the problem.


hallo,
let $a, b, c$ be positive real numbers such that $ab+bc+ac=1$.
Show:
$\sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a}\leq \frac{1}{abc}$

No reason to post this spam over and over...
http://www.mathlinks.ro/Forum/viewtopic.php?t=30865

  darij


Four dogs are in the corners of a square ABCD of side 10 metres. They start runnigng all with the same constant speed s.t. A follows B, B follows C, C follows D and D follows A. They eventually meet in the center of the square. What distance has one dog covered?

[i]unknown source[/i]

well, the dog's trace is Equiangular Spiral

a common name of that is "logarithmic spiral"
and the four dogs meet together at the center of the square at last.

That's true, so I guess you could use calculus to find the answr, but there's a much nicer way.

notice that the running directions of two adjacent dogs are always perpendicular... :D

so the length of the path is equal to the side of the square..?

[quote="grobber"][i]unknown source[/i][/quote]
The source of this problem is
[b]Steinhaus, H.[/b] [i]One Hundred Problems in Elementary Mathematics.[/i] New York: Dover, 1979.


Is there any easy way to solve this?

X*A=0
X*B=m
X*X=1

Where A and B are 3 dimensional unit vectors and m is a  real scalar

I tried brute forcing it, but I got a really ugly equation

Is X a three-dimensional vector as well?
You've got X . A = 0 which means that X and A are orthogonal.
You've also got X.X = 1 which tells you what?

If the dot product of two vectors is 0, then the vectors are perpendicular.  If the dot product is 1, they are parallel.  From what you have, $ \vec{x}\perp\vec{a}$.

Basically, the first two equations are both planes, and they intersect in a line. If that line intersects the unit sphere, which will happen 0 (line misses), 1 (line is tangent), or 2 (line passes through) times, then those intersections are the possible values of x.

I can't conceive of a nice way of doing this problem for the general case. It seems pretty gross.

I guess I just have to write it out in terms of the x y and z components and do it the brute force method then

Was this problem given to you with particular vectors?

I can tell you right now you're going to have a nightmare brute-forcing it with components.

No, it came up as part of a program I am writing (in Visual Basic for Applications)

[quote="JRav"]If the dot product is 1, they are parallel.[/quote]
You mean if the dot product is equal to the product of their magnitudes.

How easy this problem is to do with components is going to depend a lot on what basis vectors (that is, axes) you pick.  The obvious choice is $ A, B, A \times B$.  (Edit:)  $ A$ and $ B$ aren't orthogonal, but writing $ X \equal{} pA \plus{} qB \plus{} r(A \times B)$ is going to make your life a lot easier.  Can you solve for $ p, q, r$?  Alternately, replace the second equation by $ X \cdot (B \minus{} \text{Proj}_{A} B) \equal{} m$ and then you've got an orthogonal (not orthonormal) basis.

[quote="TZF"]Basically, the first two equations are both planes, and they intersect in a line. If that line intersects the unit sphere, which will happen 0 (line misses), 1 (line is tangent), or 2 (line passes through) times, then those intersections are the possible values of x.[/quote]
In terms of what I just said, the cross product of the normal vectors of two planes in $ \mathbb{R}^3$ tells you what direction their intersection is pointing (why?).


AMC
USA(J)MO
USAMO
inequalities
geometry
algebra
functional equation

Can anyone provide insight on how to practice for the USAMO for this year AND the future?

I look at past USAMOs at http://www.unl.edu/amc . :D

I do not think that helps very much since similar problems RARELY show up again. However, I was wondering what theorems and ideas are important to know (of course one must also have a creative mind).

[quote]I do not think that helps very much since similar problems RARELY show up again.[/quote]

Pretty much every inequality that uses a certain theorem or technique somewhere in the proof is similar.

Also look over the last 10 years and you'll see several problems that ask you to either create a set S that satisfies certain properties (98 #5) or prove that a set consists of all integers (04 #2, 01 #5). While different techniques are used in each case, truth is the general feel of a situation makes having the practice very useful.

Along similar lines, just about every functional equation uses the same certain techniques.

Ah, thanks! I just wanted to review somethings before I actually have to take it. Although 20 days may not seem like a lot of time, miracles can happen, and also studying wouldn't hurt :P  :lol:

[quote]miracles can happen, and also studying wouldn't hurt Razz Very Happy[/quote]

Only takes two problems to get to MOP usually. Even a semi-miracle would work eh?

I work on AoPS book, especially on the chapters about proof, since i don't know anything about proof.
BTW, Whats the most frequent topic  that appears in USAMO?  Geometry or algebra?

How many points is in USAMO and what's the qualifying score for IMO?

42 points possible for the USAMO.  To get on the US IMO team there is  a Team Selection Test (TST) at MOP for that.  The score depends on how everyone does on that.  But basically you have to be top 6 to compete at the IMO and #7 is the alternate.


give an example of a compact set with countable no of limit points.

ps: another simliar qn- construct a set whose limit points are exactly the integers

Oh, I think that $ \left\{ {n \plus{} \frac{1}{m}} \right\}_{n,m \in \mathbb{N}}$ will be okey in your similar question...

For the compact version, take the reciprocals $ \frac1{n\plus{}\frac1m}$.