A circle has a radius of $ \log_{10}(a^2)$ and a circumference of $ \log_{10}(b^4)$. What is $ \log_ab$?

$ \textbf{(A)}\ \frac {1}{4\pi} \qquad \textbf{(B)}\ \frac {1}{\pi} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ 2\pi \qquad \textbf{(E)}\ 10^{2\pi}$

[hide] So we have $ 4\pi\log_{10}{a} \equal{} 4\log_{10}{b}$. Now, we need to use the fact that $ \log_a{b} \equal{} \frac {\log_{10}{b}}{\log_{10}{a}}$. So, it is $ \boxed{\textbf{(C)} \; \pi}$
.[/hide]

EDIT: Fixed my Latex.

[quote="xpmath"][hide] So we have $ 4\pi\log_{10}{a} \equal{} 4\log_{10}{b}$. Now, we need to use the fact that $ \log_a{b} \equal{} \frac {\log_{10}{b}}{\log_{10}{a}}$. So, it is $ \boxed{\textbf{(C)} \; \pi}$
.[/hide][/quote]
I think you've made a few mistakes in the typing. I corrected them in the quote.

[quote="Altheman"]We know that $ 2\pi r \equal{} C$, so
\[ 2\pi \log_{10}(a^2) \equal{} \log_{10}(b^4)
\]
By the power rule,
\[ 2\pi \cdot 2\log_{10}(a) \equal{} 4\cdot \log_{10}(b)
\]
by the power rule. A bit of simplification gives us
\[ \pi \equal{} \frac {\log_{10}(b)}{\log_{10}(a)}
\]
Now we can apply the change of base formula to get
\[ \frac {\log_{10}(b)}{\log_{10}(a)} \equal{} \log_a(b) \equal{} \pi
\]
so the answer is $ \boxed{C}$.[/quote]

What is the change of base formula, and how do we use it?  Thanks!

Change of base formula: $ log_{b_2} (x) \equal{} \frac {\log_{b_1} (x)}{\log_{b_1}(b_2)}$.

How do you use it? When you have the need to switch on another base (i.e. $ b_2$) divide the original logarithm $ \log_{b_1} (x)$ with the logarithm of the new base (base of the logarithm is the old one, though), and you get the logarithm in new base. This is useful in this kind of transforms, but also in calculating on standard calculators, where only bases available are $ e$ and $ 10$, so you have the need for this type of conversion (quick and easy) if you're working in other bases (i.e. $ 2$).


function
calculus
derivative
topology
real analysis
inequalities
probability

1) Let $ (X, B, \mu)$ be a $ \sigma$-finite measure space and let $ \{ N_i \}_{i \equal{} 1}^{\infty}$ be a family of subsets of $ B$ such that:
i) Each $ N_i$ is a countable partition of $ X$.
ii) Every element of each $ N_i$ has finite measure
iii) If $ E\in N_{i \plus{} 1}$, $ F\in N_i$ and $ E\cap F \not \equal{} \emptyset$, then $ E\subseteq F$.

Let $ \lambda$ be a real finite measure over $ B$.
Define
\[ f_i : \equal{} \sum_{E\in N_i} \frac {\lambda (E)}{\mu (E)}\chi_E\]
(where $ \chi_E$ is the characteristic function of $ E$).

(i) Prove that $ f_i$ converges almost everywhere
(ii) Does $ f_i$ converge to the derivative of some measure with respect to $ \mu$?

2) Let $ X$ be a locally compact Hausdorff space and let $ M$ be a non-zero multiplicative linear functional over $ C_0(X)$. Prove that there exists $ x\in X$ such that $ M(f) \equal{} f(x)$ for every $ f\in C_0(X)$.

3) Hardy- Littlewood´s maximal theorem states the following result

If $ \mu$ is a complex Borel measure on

$ \mathbb {R}^k$  and $ \lambda$ is a positive real number, then
\[ m(\{x\in \mathbb {R}^k | M\mu(x) > \lambda\}) \le \frac {3^k|\mu|(R^k)}{\lambda}\]
where $ m$ is the Lebesgue measure and  $ M\mu(x) \equal{} \sup_{r\in \mathbb{R^ \plus{} }} \dfrac {|\mu|(B_r(X))}{m(B_r(X))}$

Is the constant $ 3^k$ sharp in these inequality?

[b]To the Problem 1[/b]. At first, it is sufficient to consider the case of finite measure $ \mu$, because we can consider the problem on each element of $ N_1$ separately. Multiplying $ \mu$ by a constant we reduce the problem to the case when $ (X, B, \mu)$ is a probability space (i.e. $ \mu(X) \equal{} 1$). In this case the sequence $ (f_n)$ is a typical example of [i]martingale[/i], and the problem is a consequence of the classical theorem that an $ L_1$-bounded martingale converge almost everywhere. I don't see how your particular case can be proved simpler than the general theorem itself. A reference to that martingale pointwise convergence theorem that I know (a version for Banach space valued martingales) is: J. Diestel and J.J. Uhl ``Vector measures'': AMS Mathematical surveys no. 15 (page 130, Theorem 9). I am sure you can find the original version (for scalar-valued martingales) in many other manuals.

[b]To the Problem 2[/b]. The locally compact case can be reduced to the compact one by one-point compactification $ \bar X$ of $ X$ (i.e. by adding formally to $ X$ the point $ \infty$, and taking complements to compact subsets of $ X$ as neighborhoods of  $ \infty$). Then $ C_0(X)$ can be considered as the collection of those $ f \in \bar X$ that $ f(\infty) \equal{} 0$. A non-zero multiplicative linear functional over $ C_0(X)$ can be extended to a multiplicative linear functional over $ C(\bar X)$ (put $ M(1) \equal{} 1$).

The solution of the problem for compact case can be found for example in Rudin's ``Functional Analysis'' (see 11.13: examples of multiplicative linear functionals over commutative Banach algebras).


conics ellipse counting distinguishability algebra polynomial ratio

E,  F  and  G  are three mutually intersecting ellipses in the same plane  with respective foci , A and B,  B and C, and  C and A ,which are not all three collinear

Prove that the common chords of each pair of ellipses are concurrent.

Yeah... It was my doubt too. But I don't know how the condition about coinciding focii can affect it.

[quote="Myth"]Ops... Fleeting_guest have already removed his remark. :lol:[/quote]

For those who missed it --- we are discussing whether the ellipses have 2 or 4 intersection points.  (a nice exercise:  intersection at 3 points is impossible, even allowing for tangencies as multiple intersections).    I removed the remark after noticing that the problem requires the ellipses to share foci.

So a few questions.

If 2 ellipses share one focus (e.g. 2 Kepler orbits of planets around a star) is it possible that they intersect at 4 points?   I think the answer is YES, because one can take a very eccentric ellipse close to a line segment (foci are near the endpoints) and draw a circle-like ellipse (foci close together) sharing one focus and intersecting at 4 points.  So maybe for the IMO proposal we simply [i]assume[/i] 2 points of intersection.

Suppose that 2 ellipses intersects in 2 points.  Is there a nice algebraic description of the line through the intersection points, as in the case of intersection of 2 circles (I am thinking of the "linear combination of equations of circles = equation of line" or something similar with determinants, etc)?  I think that this is not possible because the complex intersection points may be indistinguishable algebraically from the real ones (i.e. same Galois orbit; degree 4 polynomial for the intersection irreducible over the field of definition of the ellipses).   In particular, some idea like "radical axis of a pair of ellipses" probably does not exist.   But maybe sharing a focus improves the bad Galois situation...

Anyway, a beautiful problem, if it is true.

[quote="fleeting_guest"]If 2 ellipses share one focus (e.g. 2 Kepler orbits of planets around a star) is it possible that they intersect at 4 points?   I think the answer is YES, because one can take a very eccentric ellipse close to a line segment (foci are near the endpoints) and draw a circle-like ellipse (foci close together) sharing one focus and intersecting at 4 points.[/quote]
Hmm... The example seems to be incorrect. I was thinking about it in the beginning  too.

Indeed, for Imo Proposal 1985 it is too specific. Many school students don't know what is ellips.

I believe the answer to fleeting_guest's first question to be no. I think that some time ago I proved this (I'll try to do it again; as I recall, it was a nice, short and simple proof): if $\ell$ is a line formed by two points of intersection of two such ellipses, and $\ell_1,\ell_2$ are the directrices of the ellipses corresponding to the common focus, then $\ell,\ell_1,\ell_2$ are concurrent. This should show that it's impossible to have four points of inetrsection, because all $6$ lines determined by them should be concurrent.

Let $A,B$ be two points of inetrsection, and $F$ the common focus. Then $\frac{AF}{d(A,\ell_1)}$ is the eccentricity of the first ellipse, while $\frac{AF}{d(A,\ell_2)}$ is the eccentricity of the second one. The same holds for $B$, so we find that $\frac{d(A,\ell_1)}{d(A,\ell_2)}=\frac{d(B,\ell_1)}{d(B,\ell_2)}$.
However, this doesn't show that the three lines are concurrent, as we might have $A$ inside one of the angles formed by the directrices, while $B$ is in another angle. I'll see if I can fix it.

Actually, come to think of it, it seems clear that all the inetrsection points of the ellipses lie in the same angle formed by the directrices of the ellipses corresponding to $F$. That's because all the points of the first ellipse lie in a certain half-plane determined by $\ell_1$, and all the points of the second ellipse lie in a certain half-plane determined by $\ell_2$. This means that among the four combinations of two half-planes, one determined by $\ell_1$, and one determined by $\ell_2$, all the common points of the two ellipses lie in only one such combination.

Together with what I wrote above, I believe this shows that there cannot be $4$ points of intersection. Is it Ok? :? Will anybody read this and answer? :)

grobber, your observation about the directrix lines seems to solve the IMO proposed problem.
(I agree with your proof of 2 intersection points.)
Namely, suppose the directrices intersect in points $A', B', C'$. (Directrix of AB ellipse and AC ellipse intersect at $A'$, etc).   Then the common secant of AB ellipse and AC ellipse divide $B'C'$ in ratio $e_{AB} : e_{AC}$, and the same for the other secants; Ceva's theorem applies immediately.

[quote="fleeting_guest"]Namely, suppose the directrices intersect in points $A', B', C'$. (Directrix of AB ellipse and AC ellipse intersect at $A'$, etc).   Then the common secant of AB ellipse and AC ellipse divide $B'C'$ in ratio $e_{AB} : e_{AC}$, and the same for the other secants; Ceva's theorem applies immediately.[/quote]

Can you explain where this ratio comes from?  I don't see it.

You are right, the statement about ratios is complete nonsense.

So the progress remains at grobber's lemma about the directrices.

Also, in using that lemma at each focus, all 6 directrices will appear, not 3 as implied in my posting.


Ring Theory
superior algebra
superior algebra unsolved

What is an example of a commutative ring $ A$ with unit and its maximal ideal $ \mathfrak{m}$ such that $ \mathfrak{m}[x]$ is not a maximal ideal of $ A[x]$?

Every example of a ring $ A$ and a maximal ideal $ \mathfrak m$¿

[hide]the reason is that A[x]/m[x]= (A/m)[x] is not a field. [/hide]


FTW
MATHCOUNTS
summer program
MathPath
AMC
AIME
geometry

Hello everyone. As you all know, there are more tournaments than I can count (and I can count past 10 thank you very much!) I've decided it is finally time to take action. I am now founding The Comitee for Tournament Creation, CTC for short. This is a group that will create tournaments for FTW. THe only people allowed to join these tournaments are CTC members. Joining CTC is easy! All you need to do is apply by pming me and agreeing to follow these rules:

1. DOn't join any non-CTC tournaments
2. Don't make any tournaments unless told to by a high-ranked member of CTC

Any questions? Comments? Concerns? Things the voice in your head told you to say? Just post them here!

hey mewto! umm can i sign up later? like can i help host a few tourney and then make my own cuz i haven't made tournys b4   :blush: i'd luv to help though! :D

Joined people:

Skaldjfhrulez
Violin321
Mewto55555
Kl2836
(^_^)
Mathpassion101
Isabella2296
Xoangieexo
Lotsofmath
Ernie
FantasyLover
Yoh2241
Vallon22
AIME15
Jongao
Dojo
Pacman2812

Please read first post. I have an idea now. Positions within CTC!

President(1)
Vice President(1)
Board Member (8 of these)
Normal members

Everyone is elected. Board members every month, president vice president every 3 months. Inm making this up as i go along btw  :oops:

hi!! mewto..can i join!?

mathpassion =]

Clarification: YOu do not have to quit any tournaments you are currently in, but don't sign up for any more. There may be up to a 2 week wait before any ne wtournies appear (that are CTC certified)

I'd just like to say -

The Butterscotch Tourney is now CTC-Certified! ( :) )The following monthlies are questionable, however. ( :( )

I told mewto I wanted to join, yesterday.  :(  :(  :(

[quote="mewto55555"]All decisions made by mewto55555 are final. He is supreme dictator of CTC
[/quote]

 :rotfl:

If you mean implementing the nonimplementation of guests in tournaments, then read the first post of this thread, and you will see that this is the correct place for this discussion to take place.

I don't see the problem with guest accounts. Is it so bad if we want to protect our rating? The point of a tourney is not to see how our ratings are affected - it's to see who beats whom, who wins, etc., etc.

Well, if you were like Federer (who lost his No. 1 seed to Nadal), you can't play with guest accounts.  Wow, that made no sense at all.  I mean, did Federer scurry away into a guest account?

Well, FTW is just a game anyways.

actually, i agree with iin now
the point of the rating system is to show your level compared to others
if you only play unrated, that doesn't show your level of playing
likewise, if you get free rating, that doesn't show your level of playing either

however, how is it possible to get everyone to stop using guest accounts?
there is no way to guarantee that nobody will use guest accounts to save their ratings, but you can decide not to play anyone using a guest account (although i'm not sure how well that'll work)

with that, i'll bring this up in the board member discussion thread and see where that discussion will take us

[quote="vallon22"]actually, i agree with iin now
the point of the rating system is to show your level compared to others
if you only play unrated, that doesn't show your level of playing
likewise, if you get free rating, that doesn't show your level of playing either
[/quote]

That is the point of the CTC rating system.

[quote="iin77"]You also might argue that there's the CTC rating, but no one cares about that. [/quote]
That is very true, that is just used for seeding purposes in future tournaments. It is based on best 3 out of 5 cds, as opposed to taking each countdown individally. The main problem is that people (myself included) see their rating sort of as a "score" and wish to make it higher. With the high score system rating is made more of a contest, which gets to the point where people are unwilling to play others. Also, when people tank, it makes their rating lower than it should be. Tanking really screws up the rating process because most people with high ratings are unwilling to play a 1200 who should actually be a 1600.

Everyone knows that rating means absolutely nothing. Even if everyone played rated, it wouldn't mean much because of other factors like memming and tanking.

@iin: No one claims their rating means anything. Most people just try and raise their rating, even if it doesn't mean you're a good player.

@mewto: I agree. My rating would be way lower if I wanted it to reflect my ability. I just want to see how high I can get it. It's a way I challenge myself.

@superllamel: You are absolutely right. Rating doesn't mean anything especially now when people memorise half of the problems on FTW and then answer in less than 2 seconds.
@ernie: FTW is just a game, but you can't explain it to many people whose rating on FTW is their no.1 in life...

wow
all of these responses to other's statements
next time, i think using the pms would work better

anyway, let's drop this guest and rating stuff
we've talked about it enough and it's obvious that people want to continue using guests for tournies
therefore, we'll keep the guests in, and this discussion will end
if you want to talk to a certain person about this some more, then pm them

Errr, mewto? I sent you a PM but you haven't read it. [Sent 8:08 pm August 26, PST]

Maybe he's has not been on AoPS since that time.

Could someone post a list of all CTC members?

Ignore teh funny numbers.

Skaldjfhrulez - 1200
Violin321 -1175
Mewto55555 -1223
Kl2836 -1223
Isabella2296 -1225
Xoangieexo -1200
Lotsofmath -1200
Ernie -1231
FantasyLover-1202
Yoh2241 -1200
Vallon22 -1248
AIME15 -1171
Jongao -1200
Dojo -1200
Pacman2812-1200
Myyellowducky82 -1175
Vamathletes -1200
Guineapiggies -1175
Pinkpiglet -1200
Superllamel -1252
Jain_harshi -1150
Splatyango -1200
Mathblitz -1200
Cherry_pi -1150
RomanianGenius-1298
Weird888 -1200
Dragonlaird7 -1150
Dragon96 -1200
Dannyhamtx -1225
Kookamunga -1200
Nikeballa96 -1175
Pi guy -1200
27r -1200
Yclarinet -1200
Chenhsi -1200
Textangle -1200
Jjx1 -1175
Aime_is_hard - 1200
Pi-dimension-1200
Mind Storm-1200
Bisbmard9-1200
Mathnerd711-1177
Emilylambkin-1200
Johnnybeard-1200
Discrete_math-1200
AsianPascal-1200
Bugi-1200
Irsmart22-1200
Sup3rsolv3r-1200
James4l-1200

it's never too l8 to join CTC
just pm mewto and ask to join (meaning, plz don't use this thread)


advanced fields
advanced fields unsolved

let $ X$ be a scheme over a field $ k$ of characteristic $ p>0$. clearly $ X$ can be covered by (spectra of) rings of characteristic $ p$, and these have a frobenius endomorphismus. but why do they glue to a "frobenius" endomorphism of $ X$? I can't see why they should be the same on intersections ...

Maybe it is easier to just consider this ("absolute") Frobenius as the morphism of schemes defined as the identity on the underlying topological space, and the $ p$-th power map on the entire structure sheaf. Once you have shown that this is a morphism of locally ringed spaces, it is clear that this construction is compatible with passage to open subsets. But for affine schemes, morphisms as locally ringed spaces correspond bijectively to the ring homomorphisms between the rings of global sections. owk


geometry
rectangle
number theory
Diophantine equation

[quote="frt"]1. How many ways can 4 women and 6 men be divided into 2 five-person teams if there is at least one women on each team?

2. A 4 by 3 checkerboard has 60 rectangles. Find the sum of the areas of the 60 rectangles.

3. Find the ordered pair of integers $(x,y)$ such that $7x+11y=1$ and $|x+y|$ is the minimum.[/quote]

[hide="1"]  There are 5 teams.  By the conditions given by the problem, there must be at least one woman on each team.  Thus, the minimum number of woman is 5.  However, there are only 4 women.  A contradiction, and consequently, no such teams can be made. [/hide]

  Im completely confused by the wording of question 2.  Could you please re-phrase the question so it is more understandable, frt?

[hide="3"]  All we need to do is a little manupulation in terms of $y$ and then solving equations to get:

(-3, 2)  Which equals: $|(-3) + 2|=\boxed{1}$ [/hide]

Is the second one 12? :blush:

G-UNIT,the first one says there are 2 five person teams,not 5 teams :lol:

Is the first answer 2 ways?

it sounds pretty lame :blush:  , but hey i got 12 too , is it correct??

all i did was 4 x 3 , the checker board area = the sum of the areas of the 60 squares

:rotfl:  :rotfl: Thats the same thing  did...first i went into complicated things findin the no. of rectamgles and the side of each an i realised that 4*3=area of chessboard=sum of aras of rectangles.

heheheh ,  :lol:  :P  ;) first i too thought it wont so easy and started doing rot like finding  area of each rectangle then multiplying by 60, but i realised then....... , haha :oops:

1) If we consider men and women as distinct people it is $\left(\stackrel{8}{4}\right) = 70$
    If we consider only the sex is $2$

2) $12$

3) The solution is the particular solution of the diophantine equation , so it is $(-3,2)$

#2 isn't 12 :P

[hide="solution"]
There are 12 1x1 rectangles
there are 8+9=17 1x2 rectangles
there are 4+6=10 1x3 rectangles
there are 3 1x4 rectanges
there are 6 2x2 rectangles
there are 3+4=7 2x3 rectangles
there are 2 2x4 rectanges
there are 2 3x3 rectangles
and there is 1 4x3 rectangle

This means there are 12+17+10+3+6+7+2+2+1 = 60 rectangles which matches the problem.

So the total area is:
$12(1)+17(2)+10(3)+3(4)+6(4)+7(6)+2(8)+2(9)+1(12) = \boxed{200}$

You guys didn't count the overlap :P
[/hide]

Sorry that the wording of #2 was unclear (although it was directly taken from a math contest problem). Yes, 200 is the correct answer :)

The question should have mentionesd rectangles of diff dimensions....could someone explain me hamster's sol,i cant get it ;)

Well, how was i supposed to know about the overlap?  :(

How big is each rectangle in number 2?

By the phrase "4 by 3 checkerboard", the problem meant that 12 $1\times 1$ squares are tiled to form a 4 by 3 checkerboard. (m by n checkerboard is made up of m*n squares, right?)

So if you disregard the overlap, you have only 12 squares, but if you count all the overlaps, you get 60 rectangles (including squares).

Again, sorry for the confusion regarding the wording of the problem.

1.[hide]
$C(8,4)=\frac{8!}{4!4!}=70
[/hide]

2.[hide]
12?

[color=green][size=75]mod edit: Read the previous posts...[/size][/color]
[/hide]

[hide]
no idea
[/hide]


inequalities
geometry
inradius
trigonometry
geometry proposed

[quote="Virgil Nicula"]Prove that in an acute triangle $ ABC$ there is the inequality $ \boxed {\sum\frac {m_a}{s \minus{} a}\ \le\ \frac sr}\ ,$ where denoted

$ 2s \equal{} a \plus{} b \plus{} c$ , the length $ m_a$ of the $ A$-median and the length $ r$ of the inradius for the triangle $ ABC\ .$[/quote]

Let $ A'$ - midpoint of $ BC$ and $ A''$ - midpoint of arc $ BC$ , then $ \boxed{m_a*A'A'' \le \frac{a^2}{4}}$ ,  ( It is true for acute triangle )

$ A'A'' \equal{} \frac{a}{2}\tan(\frac{\alpha}{2}) \Rightarrow m_a \le \frac{a}{2}cot(\frac{\alpha}{2})$

$ s\minus{}a \equal{}  r cot(\frac{\alpha}{2}) \Rightarrow \frac {m_a}{s \minus{} a} \le \frac{a}{2r} \Rightarrow \boxed{\frac{m_a}{s \minus{} a} \plus{}\frac{m_b}{s \minus{} b}\plus{}\frac{m_c}{s \minus{} c} \le \frac{s}{r}}$


When I saw solutions of some problems in inequalities.

So many of them suppose that abc=1 ,a+b+c=1

For a+b+c=1 sometimes they let a=x/(x+y+z) b= y/(x+y+z)

c= z/(x+y+z) or a = xy/(xy+yz+xz) b = yz/(xy+yz+xz)

c = xz/(xy+yz+xz) . But I really not understand why did they

can do that. So please help me by answer these questions.

1) What conditions of problems we can suppose that abc=1

(and better for give me the reason.)

2) What conditions of problems we can suppose that a+b+c=1

(and better for give me the reason.)

Thank you ver much for answering

Sorry for my English. :D


For which values of $ x$, where $ x\in \mathbb{R}$, does the inequality $ 2\log_{2}\left ( \frac{a\plus{}b}{2} \right )\leq \log_{x}a\plus{} \log_{x}b$ hold for $ \forall a,b\in \mathbb{R}^\plus{}$.

[quote="The guy above me"]For which values of $ x$, where $ x\in \mathbb{R}$, does the inequality $ 2\log_{2}\left ( \frac {a \plus{} b}{2} \right )\leq \log_{x}a \plus{} \log_{x}b$ hold for $ \forall a,b\in \mathbb{R}^ \plus{}$.[/quote]

I'll give it a shot.

Lets say $ 2\log_{2}\left ( \frac {a \plus{} b}{2} \right ) \equal{} k$

$ \log_{2}\left ( \frac {a \plus{} b}{2} \right )^2 \equal{} k$

$ \frac {a \plus{} b}{2} \equal{} 2^{k/2}$

$ {a \plus{} b} \equal{} 2^{(k\plus{}2)/2}$

brb

Please keep in mind that only the most elementary proofs should be posted in High School Basics, so moved to Intermediate.  In this case, the facts that the problem could involve log base things other than integers, there are many variables are involved, and the problem does not reduce in about 2 to 3 steps show that it is too difficult for HSB.

Thanks!

i apologize if I posted the question in this subforum...It's just that,  I thought the question is most suitable for High School Intermediate Topics appropriate as well for students aging 13 and above...I believe students of this age are well equipped enough to attack questions like this...Moreover, isn't it that topics such as logarithm an intermediate topic for high school math?

If we change the base to 2 in second term we get log(2)x  <= [log(2)ab]/{log(2)[(a+b)/(2)]^2}
Using AM-GM we get  log(2)1 <= log(2)x <= log(2)2 so x  is between 1 and 2 .... since the inequality holds for every a,b we get x=1,hope i'm rigth(i didn't got to exponential inequality at school) and sorry because i don't know latex(i'm new)

$ 2\log_{2}\left (\frac{a-b}{2}\right )\leq\log_{x}a+\log_{x}b$
$ \log_{2}{{\frac{a-b}{2}}^2\right }\leq\frac{\log_{x}{ab}}{log_{x}2}*{log_{x}2}$
then do 2^ of everything
$ {(\frac{a-b}{2})}^2\leq (ab)^{\log_{x}2}$

Strange, everything seems going quite well until now. I'm sure im missing just a step

Here's my solution:
Actually the inequality may be written as $ \log_{x}\left ( \frac{a\plus{}b}{2} \right )\leq \log_{x}\sqrt{ab}$. Based from the fact that $ \frac{a\plus{}b}{2} \geq \sqrt{ab}$, then it follows that the $ \log_{x}$ function is decreasing only at the interval $ 0<x<1$.


about the left part of the inequality , it's equivalent to
$\sum \frac{abc}{a^{3}+b^{3}+abc}\leq 1$
but we see that $\sum \frac{abc}{a^{3}+b^{3}+abc}\leq \sum \frac{abc}{ab(a+b+c)}=\sum \frac{c}{a+b+c}=1$ which is completed our proof

[hide]i think SOS may able to do the rest of the problem [/hide]

[quote="PHONGVAN"]Let $a,b,c$ be positive real numbers .Prove that:
$\ 2 \leq\frac{a^{3}+b^{3}}{a^{3}+b^{3}+abc}+\frac{b^{3}+c^{3}}{b^{3}+c^{3}+abc}+\frac{c^{3}+a^{3}}{c^{3}+a^{3}+abc}\leq\frac{2(a^{2}+b^{2}+c^{2})}{ab+bc+ca}$[/quote]
The first is easy, let I post my solution for the right, it's equivalent to
\[abc\sum\frac{1}{a^{3}+b^{3}+abc}+\frac{2(a^{2}+b^{2}+c^{2})}{ab+bc+ca}\ge 3 \]
By Cauchy Schwarz inequality, we have
\[\sum\frac{1}{a^{3}+b^{3}+abc}\ge \frac{9}{2(a^{3}+b^{3}+c^{3})+3abc}\]
It suffices to prove that
\[\frac{9abc}{2(a^{3}+b^{3}+c^{3})+3abc}+\frac{2(a^{2}+b^{2}+c^{2})}{ab+bc+ca}\ge 3 \]
Suppose that $a+b+c=1$, setting $x=ab+bc+ca,y=abc$, then the inequality is equivalent to
\[\frac{9y}{9y+2(1-3x)}+\frac{2(1-2x)}{x}\ge 3 \]
or
\[(1-3x)(9y+2-5x) \ge 0 \]
which is true, since $x \le \frac{1}{3}(a+b+c)^{2}=\frac{1}{3}$. :)

I think that
 $\sum{cyc}\frac{a^{3}}{a^{3}+abc+b^{3}}\leq\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}$
And you?


fek mikonam soale marufi basheo ghablan post shode bashe!vali man peydash nakardam...
$ P$ yek chand jomleie ba zarayebe guya ast,ke baraye har adade tabi'e $ n$,$ P(n)$ barabare morabae yek adade sahih ast.sabet ya rad konid chand jomleie $ Q$ ba zarayeb guya vojud darad ke:
$ P\equal{}Q^2$

niazi be hame aadade tabii nis,age ye chan jomlei be ezae tavanhaye 2 morabae kamel besheam halle...

khob che juri mishe sabetsh kard?
dar morede tamimesh chi?
age ye chand jomleie baraye hameye adade tabi'e tavane d ome kamel bashe,mishe goft barabare tavane d ome ye chanjomleie digast?
merc babak jan... :D

are doroste,soalesh too ghesmate paine poste inja matrah shode:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=326560#326560[/url]
javabesham 2ta post paintar dade shode :D


I could'nt find a "Hi" kinda forum so i decided to post this here.
Hope its ok.

Anyways.
Hi guys, im new at the forums, and im here to broaden my knowledge on math.
I want math to be my major in College, but im abit slow learner for unknown reasons.

I was hoping you guys n' girls could help me solve some problems now and then if thats allowed in here.
I read the latest news on the site and was bedazzled on the advanced level of maths that exist.
Anyways, hi and i hope you guys can help me.

i doubt i could help u with math as i'm a freshman in hs and my math isn't very good, but if you want to say hi and introduce yourself, i'd suggest u go to the Games and Fun Factory and Click on "Please Introduce Yourself" and introduce yourself.

Welcome to AoPS!!!


invariant
trigonometry
number theory
trig identities
Law of Cosines
algebra proposed
algebra

[i]Find all positive real numbers [b]m,n,k,l,s,t[/b] such that the equalities hold:[/i]

[b](i)[/b]$m+n=s+t=k+l$
[b](ii)[/b]$m^2+l^2=k^2+s^2=t^2+n^2$

Davron Latipov

What does the title mean¿ Is it Diophantine¿ When yes, then the statment that it shall be solved in the reals makes no sense...
Please correct it.

[u] Zetax: i stupidly confused the paragraph name, now it's ok.[/u]

[color=red]Why don't you solve this problem.[/color]

Davron

I changed the title, I think it's better so (better to refind etc.).
And reals are not really number theory (at least not here) $\implies$ off it goes to Algebra.

[b][i][u]Is there anyone who will solve this problem[/u][/i][/b]

[hide="Do you need a need a hint or you will solve this yourselves?"]Just answer me[/hide]

Davron

:rotfl: Nice bluff, Davron. Nobody should worry about integers, Diophantine etc. :D
If you assume the variables are all different, it's easy to see there is [b]no [/b]solution. :P
The system is invariant under a simultaneous cyclic change $m\to t\to k\to m$ and $n\to s\to l\to n$. So we can assume wlog either $m>t>k$ or $m<t<k$.
In the first case, we have $n<s<l$, so $m^2+l^2$ is strictly bigger than the two other sums.
In the second case, we have $n>s>l$, so $m^2+l^2$ is strictly smaller than the two other sums.

[b]Thanks for your reply: now can you help me with this one, the idea of this systematic equation is coming from the IMO 2005 problem1.[/b][i]

[b](i)[/b]$m+n=s+t=k+l$
[b](ii)[/b]$m^2+l^2-ml=k^2+s^2-ks=t^2+n^2-tn$

Davron Latipov[/i]

That equation is observed after the application of the Law of Cosines to the angles with the degree of 60 and the first one is coming since ABC is an equilateral triangle.

Davron Latipov

[i][color=darkred]Is there anyone who can help me with this problem ?[/color][/i]

Davron Latipov


Let $x_{1},x_{2},...,x_{5}\in R$. Find the least positive integer $n$ such that if some $n$ distinct sums of the form $x_{p}+x_{q}+x_{r}$ are equal to $0$, then $x_{1}=x_{2}=...=x_{5}=0$.

Solution n=5

n=4 has a set different from all zero like:
0,1,-1,1,-1

for n=5
let k= x1+x2+x3+x4+x5
then there are five different sums of xm+xn = k
which gives a system of 5 independent equation with only one solution the trivial one.

[b]mszew[/b], I don't think you are right. I think the answer is 7. For n = 6 the example is 1,1,1,1,-2. Which gives $\binom{4}{2}=6$ zero sums.
For n=7. By pigeon-hole principle we have that one of $x_{i}$ is in at least 5 of zero sums(7.3=21>4.5).
Wlog this is $x_{1}$. Since $\binom{4}{2}=6$ at most one sum of the form $x_{1}+x_{p}+x_{q}$ is not zero. Let this sum is $x_{1}+x_{4}+x_{5}$ this means all other sums $x_{1}+x_{p}+x_{q}$ are 0. From here easily follows that $x_{2}=x_{3}=x_{4}=x_{5}$ and since there is one zero-sum without $x_{1}$ all numbers are 0.


linear algebra
matrix
induction
superior algebra
superior algebra solved

grobber wrote: (It was on the old forum)

(3)
A is in M_4(C), A=/=0_4.

a) if rank A=r<4 show that there exist U, V in M_4(C), invertible, s.t.
UAV=
(I_r 0), where I_r is the unit matrix of order r.
(0 0)
b) Prove that if A and A^2 have the same rank k, then the matrix A ^n has rank k for all n>=3.

-----------------------------------------------------------------------------------

Moubinool wrote
a) Can be find in text book in the  rank
b)
If rgA=rgA^2 with
dim(_C)C^4 = rgA + dimkerA=rgA^2 + dimkerA^2 => dimkerA=dimKerA^2 but kerA is included in kerA^2 so
kerA=kerA^2

(*) let x\in  kerA^3=> A^3(x)=0 => A(x) \in kerA^2=kerA=>A^2(x)=0
this prove kerA^3 is included in kerA^2 but kerA^2 is included in kerA^3
so kerA^2=kerA^3 => rgA^2=rgA^3

We apply the same idea with rgA^2=rgA^3 => kerA^2=kerA^3=>
working like in (*) with x \in kerA^4 => A(x) \in  kerA^3=kerA^2=>
A(x)=0 so kerA^4=kerA^3 we deduce ImA^4=ImA^3 and
rgA^3=rgA^4

By induction if rgA=rgA^2 then for any n \in N n  \geq  2, rgA^n=rgA^2.

for a) we have an article in GM 7-8/2003 written by Cristinel Mortici

You can still find the old questions, Moubi. I think they're somewhere in the "Algebra" section. You could ask Valentin.


Define $ x_0\equal{}0, x_1\equal{}1, x_{n\plus{}1}\equal{}4x_n\minus{}x_{n\minus{}1}$, and $ y_0\equal{}1, y_1\equal{}2, y_{n\plus{}1}\equal{}4y_n\minus{}y_{n\minus{}1}$. Prove that: $ y_n^2\equal{}3x_n^2\plus{}1$ for all $ n \ge 0$.


For how many two-digit positive integers is the product of the digits 12?

I answered 4:

62, 26, 43, and 34

When I reviewed it, it said the answer is 16.

Is that a system error?

If not, please explain.

I'm not sure the the previous poster is doing... Your answer is correct. Next time, make sure you get the problem number. Then, pm levans about the problem, so he can change the answer.

What happens after you click "Report Error"? I've clicked it multiply times for wrong problems, but most of them were not fixed....

[quote="poultry"]you forgot 82
because 8+2=10 and 10/2=5 and 5*8=40 and 40+8=48 and 48*10=480 and 8^2 is 64[/quote]
Which universe?

Seriously, are we still continuing this pointless, spammy, misplaced thread? (Sorry for this post, but I really had to point this out.) And also, as if the above wasn't bad enough, please actually post only if you know what you're talking about, you're on topic, or if you actually have something to contribute to the mathematical discussion here (no, a two-word question pointing out the already known that a question does not make sense is definitely far from contributive).

[quote="isabella2296"]Or you can click on "Report Error", which is next to "Ask On Forum" on the Review.[/quote]

Wow, I blanked out (i didnt think of that?!)..... Yah, do that.  :)


Does anybody know  where the  IMC 2008 will be held?

I'm not sure at all, but I think once I heard it will be in Hungary.

Nima Ahmadi Pour!You are right!
Imc 2008 will be held july 26-31 in Budapest, Hungary. :)

15th IMC 2008 will be again in Blagoevgrad, Bulgaria during 25-31 July 2008

http://www.imc-math.org.uk/

First Day
25th July
Arriving and Accommodation
Registration

Second Day
26th July
Opening Ceremony
Additional Registration
Meeting of the Jury

Third Day
27th July
First Exam Day

Fourth Day
28th July
Second Exam Day

Fifth Day
29th July
Meeting of the Jury

Sixth Day
30th July
Closing Ceremony
Final Dinner

Seventh Day
31st July
Departure


If $ 2x \plus{} 1 \equal{} 8$, then $ 4x \plus{} 1 \equal{}$

$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 19$

Well we have that $ 2x\equal{}7$, so $ 4x\equal{}14$, meaning $ 4x\plus{}1\equal{}15$.

Multiply the first equation by 2:
$ 2(2x\plus{}1)\equal{}2(8)$
$ 4x\plus{}2\equal{}16$
Subtract 1 to get the desired result:
$ (4x\plus{}2)\minus{}1\equal{}(16)\minus{}1$
$ 4x\plus{}1\equal{}15$
$ \fbox{\text{A}}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $ H$ is a subgroup of $ G$,and both $ K$,$ S$ are subgroups of $ H$.If $ K$,$ S$ are conjugate,does it follow that $ K$,$ S$ are also conjugate in $ H$,i.e.$ \exists x \in H$,s.t.$ K\equal{}x^{\minus{}1}Sx$. :)


Prove by induction:
$\sum\limits_{m = 1}^n {(2m - 1) = n^2 }$

I think you only need a [hide="hint"]$n^2+(2n+1)=(n+1)^2$[/hide]. ;)


#1help me plssssss

--------------------------------------------------------------------------------
pls help me to answer this im gettin crazy but i really dunno how to convert it

NUTRITION FACTS
serving size 2 tsp (5g)
serving per container about 45
--------------------------
Amount Per Serving
_______________________
calories 20 Calories from fat 15
______________________________
% Daily Value*
total fat 1.5g 2%
saturated fat !g 5%
cholesterol 5mg 2%
sodium 85mg 4%
total carbohydrates 0g 0%
dietary fiber 0g 0%
sugar 0g
Protien 2g
------------------------------
Vitamin A 0%
Calcium 6%
-----------------------------
*per dailyvalues are based on a 2,000 aclorie
diet.Your daily values may be higher or lower
depending on your calorie needs:
Calorie: 2,000 2,500
-----------------------------------------
total fat less than 65g 80g
sat. fat less than 20g 25g
cholest less than 300mg 300mg
soduim less than 2,400mg 2,400mg
total carb 300mg 375g
fiber 25g 30g

--------------------------------------------
this is a convertion and i really dunno this plss help me or its ok if you will give me the formula please.......that is the nutrition fact chart the question is
how may milligrams of total fat are present in each serving?
how many grams of cholesterol?
how many grams of soduim?

It looks to me like one serving has $ 1.5$ grams of fat.

$ 1.5 grams \times \frac{1000 milligrams}{1 gram} \equal{} 1500 milligrams$.


function factorial geometry calculus integration Euler articles

How is (1/x)! defined?

(I'm trying to  understand [url=http://www.inwap.com/pdp10/hbaker/hakmem/geometry.html#item1]HAKMEM item 1)[/url]

Factorial's definition is often extended to include non-integers. (In some math books the function is called gamma function but it is  related to factorial as -
$\Gamma (n+1) = n!$

For example,   $(1/2)! = (\sqrt{\pi }/2)$
(That will give value of all half integers .. eg $(3/2)! = (3/2)((1/2)! )= (3/2)(\sqrt \pi /2)$ etc..

One definition, one can take is the area under the curve $y=x^{n}e^{-x}$ or
the corresponding integral
$n!=\int_{0}^{\infty}x^{n}e^{-x}dx$

Which by the way give the traditional values when $n$ is integer.

Hope this helps.

[quote="13375P34K43V312"]I think that's where $\infty!=\sqrt{2\pi}$ comes from.[/quote]

As  someone pointed out in other thread, "this" actually come from  the computer scientists'  notation.  "$!=$"  means "Not equal to"  and since "infinity" is NOT EQUAL to square-root of  any real number, one can safely say:

$\infty \ \ != \ \ \sqrt{2\pi}$
 :lol:  :)  :lol:

Any way, for those who want  to get some  insight  about non-integer factorial without calculus here is one of my old post about the treatment given by Euler: (Euler was the one who also introduced the integral method mentioned in the above post)

>>>> (Original message follows)>>>
One fun way Euler thought about non-integer factorial (apart from the one where he used integrals - that too was first introduced by Euler): is to use:

 Let $n$ be a positive integer.

$(m+n)! = m! (m+1) (m+2) \dots (m+n)$
which gives:
$m! = \frac{(m+n)! }{ (m+1) (m+2) \dots (m+n) }$
$= \frac{ n ! (n+1)(n+2) \dots (n+m) }{ (m+1) (m+2) \dots (m+n) }$

This obviously is true when $m$  is an integer , but if m is not an integer, one can use this, and consider a VERY large. n ..(so large that you can get as accurate value as you desire -- In mathematical terms you say let n tends to infinity)

$m! \ge \frac{n!(n^{m})}{ (m+1) (m+2) \dots (m+n) }$
and $m! \le \frac{n!(n+m)^{m})}{ (m+1) (m+2) \dots (m+n) }$

when is $n$  very large, the above two values do not differ much, $m!$  is between these two values.

Hope that helps.

This is a practical way to calculate numbers like, say $(1/3)!$ without the use of Calculus.

[quote="Gyan"][quote="13375P34K43V312"]I think that's where $\infty!=\sqrt{2\pi}$ comes from.[/quote]

As  someone pointed out in other thread, "this" actually come from  the computer scientists'  notation.  "$!=$"  means "Not equal to"  and since "infinity" is NOT EQUAL to square-root of  any real number, one can safely say:

$\infty \ \ != \ \ \sqrt{2\pi}$
 :lol:  :)  :lol: [/quote]

Really?  Because I recall finding it on a MathWorld article (where I think they are more precise about their use of the $!$ symbol), although I can't find the exact article at the moment.

http://mathworld.wolfram.com/InfiniteProduct.html
This seems to be the only place I can find it. Equation number 7.

Yes, that's where I found it.  (If anyone could contribute to that "Regularized Product" entry, I'd be very grateful :) )


I'm not sure this problem is elementary. I saw it 8 years ago. It was proposed at a contest, but not for the contest (confusing isn't it?).

Let C be a circle (the pizza). Let P be a point in C. Let d1,d2,d3,d4 be 4 lines concurrent at P such that all the angles between consecutive lines are 45o.

Let S1, S2, S3, S4, S5, S6, S7, S8 be the areas of the regions (slices) determined by the 4 lines inside the circle. Assume they are taken in counterclockwise or clockwise order.

Prove S1+S3+S5+S7=S2+S4+S6+S8.

The result has a more general version, even it is not fully elementary in the sense where it uses a bit of calculus (but it is not difficult at all) :

Let k,n \geq 2, be two integers.
A pizza has the form of a disk. It has been made by several ingredients (tomatos, cheese,...) each of which has the form of a disk covered by the pizza. All of thses disks have a common intersection. choose any point in this common intersection, say P, and from P divides the whole piza into 2kn equal angular sectors (thus each of these is a slice with angle  \pi /kn around P). Then, each person from n takes the slices from n to n, receiving a total of 2k slices.
Then, we may prove that all the persons received an equal part of pizza and of each of the ingredients....

Pierre.

Are you sure it's true, amfulger? I remember you told me about this prb some time ago and you had some doubts because it's obviously not true if we only take 2 lines with angles of  \pi /2 between them, instead of 4 lines with angles of  \pi /4 between them.

pbornsztein's condition n,k \geq 2 avoids the case for 2 lines.
I'm not sure, but pbornsztein seems to be and I think it is true.

[quote="pbornsztein"]I'm sure of what i said. I didn't check the case of two lines, but since i've thought about this problem some time ago, i think i've kept the optimal conditions...
Pierre.[/quote]

For the case of 2 lines, my problem which is different from pbornsztein's, is not true.
Let me tell you first why my problem is different from pbornsztein's:
I still take from 2 to 2 but he takes from 1 to 1 (so there is one person that gets all the pizza  :D ) if he removes the condition n,k \geq 2. If he keeps the condition, the problem has no meaning for him, so you may say he gets no pizza  :( .
I didn't specify this (that I take from 2 to 2 for every number of lines) in the problem text, because I gave the problem for 4 lines, but now I wanted you to understand what grobber was talking about.

Now why is my problem not true for 2 lines?
Take a square inscribed in the circle and take P very close to a vertex. The you will get that the difference between the 2 areas is the area of the square (or very close to it), which is obviously not 0, so the areas are not equal.

Can someone with some geometry software check this out? I'm really not convinced it's true.

this pizza theorem is tru for all n=4,6,8,10.... where n is the number of the chords .in the problem amfulger posted n=4  adn the problem can be solved using calculus and polar coordinates.in case of n=odd ,the difference in the areas simpleifies to
 \pi /n - tan ( \pi /n),which is nonzero (in case my calculations are correct)

cheers!!

One question- that point P is on the cirle or inside the circle?

I think it must be inside the circle, but it makes you wonder what happens if it tends to be on the circle.


linear algebra
matrix
algebra unsolved
algebra

Solve sytem equation:
$ \left\{\begin{matrix}
\left | y \right |\equal{}\left | x\minus{}3 \right | &  & \\
 (2\sqrt{z}\minus{}2\plus{}y)y\equal{}1\plus{}4y&  & \\
 x^{2}\plus{}z\minus{}4x\equal{}0&  &
\end{matrix}\right.$

i think with your problem would be nicer if the second equation is
$ (2\sqrt {z} \plus{} 2 \minus{} y)y \equal{} 1 \plus{} 4y$ with your original problem i must use the general solution to quartic equation

if you are Vietnamese, you can see it here [hide]đề thi chọn đội tuyển Nghệ An năm học 2008-2009[/hide]


vector
linear algebra
linear algebra unsolved

Assume $ detA\equal{}detB\equal{}0$. prove that $ det(E\minus{}AB)\equal{}0$ if only if $ det(E\minus{}BA)\equal{}0$

To prove $ det(E \minus{} AB) \equal{} 0$ iff $ det(E \minus{} BA) \equal{} 0$

only to prove: $ rank(E \minus{} AB) < n$ iff $ rank(E \minus{} BA) < n$

and $ rank(E \minus{} AB) < n$ iff there exists a non-zero vector $ X$ such that: $ (E \minus{} AB)X \equal{} 0$.

so we get: $ rank(E \minus{} AB) < n$ iff there exists a non-zero vector $ X_{o}$ such that: $ ABX_{o} \equal{} X_{o}$

think about non-zero vector $ BX_{o}$.

we have $ BA(BX_{o}) \equal{} BX_{o}$. so we get the conclusion: $ BX_{o}$ is the non-zero vector of the equation $ (E \minus{} BA)X \equal{} 0$.

so we have: $ rank(E \minus{} BA) < n$.

then we prove the problem :huh:


[quote="236factorial"]I think that the sum of the recipricols of two numbers is
$\frac{\text{sum of two integers}}{\text{product of two integers}}$

Is that true, and if it is, why?[/quote]

$\frac1{a}+\frac1{b}=\frac{b}{ab}+\frac{a}{ab}=\frac{a+b}{ab}$


calculus
integration
trigonometry
logarithms
calculus computations

OK, here is another one...

$ \int_0^{\dfrac{\pi}{4}}\;\left(\dfrac{x}{\sin\,x}\right)^n\;\textbf dx$

where $ n$ is an integer $ \ge 1$


trigonometry
geometry
logarithms
probability
geometric series

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=301536]Refer here for the full set of problems[/url]. Note, there are two forms (some problems appear in only one).

- If $ \cos{\theta} \equal{} \frac {1}{x}$ for some angle $ \theta$ and $ \cos(2\theta) \equal{} x \minus{} 1$, then $ x \equal{} \sqrt [a]{b}$ for integers $ a$ and $ b$. What is the minimum value of $ ab$?

- Let $ \displaystyle\sum_{k \equal{} 1}^{2008} {k^2} \equal{} p$ and $ \displaystyle\sum_{k \equal{} 1}^{2009} {k^3} \equal{} q$. How many primes divide both $ p$ and $ q$?

- Examine the figure in the pdf. If $ AC \equal{} 5$, then what is the area of $ EFGH$, given that $ ABCD$ is a square and $ GDC$, $ CFB$, $ BEA$, and $ AHD$ are equilateral?

- If $ 3f\left(\frac {1}x\right) \plus{} f(2x) \equal{} \log_2{2x}$, determine $ f(32)$.

- A target board is made from five concentric circles, forming 4 rings that surround the center bullseye. The width of each ring and the diameter of the bullseye are identical. Millinova keeps throwing darts at the board until she hits the bullseye. Flarien does so as well independently of Millinova. Every dart successfully hits the board, but it is completely random where exactly each hits. What is the probability that Millinova used the same amount of darts as Flarien?

Hint:

- If $ 3f\left(\frac {1}x\right) \plus{} f(2x) \equal{} \log_2{2x}$, determine $ f(32)$. [b]Can you make another equation out of this?[/b]

[hide="answer"]try x=1/32 and x=16 and you come out with a system of equations.  Solve and you're done.[/hide]

[color=green]Moderator: Fixed your hide tag. Hide tag's can't have quotation marks in them.[/color]

For the $ cos \theta \equal{} \frac{1}{x}$ one,

[hide="My_Solution"]$ cos(2\theta) \equal{} 2cos^2(\theta)\minus{}1$

$ x\minus{}1 \equal{} 2(\frac{1}{x})^2\minus{}1$

$ x \equal{} 2(\frac{1}{x})^2$

$ x^3 \equal{} 2$

$ x \equal{} \sqrt[3]{2}$

$ ab \equal{} 6$

Not too sure how to prove this is the minimum, except that $ x$ was in its simplest form.
[/hide]

Solution for the geometry one:
[hide]Let $ a$ equal the side length of $ ABCD$, $ b$ equal the altitude length of the equilateral triangles, and $ c$ equal the side length of $ EFGH$.
Then $ a = \frac 5 {\sqrt 2}$ and $ b = \frac {\sqrt 3} 2 a = \frac {5 \sqrt 3} {2 \sqrt 2}$. Because the diagonal length of $ EFGH$ is $ c \sqrt 2$,
\begin{align*}
c \sqrt 2 &= a + 2b \\
c \sqrt 2 &= \frac {5 + 5 \sqrt 3} {\sqrt 2} \\
c &= \frac {5 + 5 \sqrt 3} 2 \\
c^2 &= \boxed {\frac {25(2+\sqrt 3)} 2}
\end{align*}[/hide]

[quote="sa314"]Solution for the geometry one:
[hide]Let $ a$ equal the side length of $ ABCD$, $ b$ equal the altitude length of the equilateral triangles, and $ c$ equal the side length of $ EFGH$.
Then $ a = \frac 5 {\sqrt 2}$ and $ b = \frac {\sqrt 3} 2 a = \frac {5 \sqrt 3} {2 \sqrt 2}$. Because the diagonal length of $ EFGH$ is $ c \sqrt 2$,
\begin{align*} c \sqrt 2 & = a + 2b \\
c \sqrt 2 & = \frac {5 + 5 \sqrt 3} {\sqrt 2} \\
c & = \frac {5 + 5 \sqrt 3} 2 \\
c^2 & = \boxed {\frac {25(2 + \sqrt 3)} 2} \end{align*}[/hide][/quote]

That's awfully complicated. I suggest you use the triangle area formula or the law of cosines.

Can someone help me out on the last problem here, the one with the target board?

-- apologies for the revive.

Let the radius of the bullseye be $1$. That means the width of each ring is $2$. So the radii of the concentric circles from smallest to largest is $1$, $3$, $5$, $7$, $9$.
The area of the the bullseye is $\pi$ and the area of the target board is $81\pi$. So there's a $\frac{1}{81}$ chance of hitting the bullseye and $\frac{80}{81}$ chance of not hitting it. Let's say Flarien hits the bullseye with the first dart, probability is $\frac{1}{81}$. Then the probability Millinova does it is $\frac{1}{81}$. Total probability $\left(\frac{1}{81}\right)^2$. Let's they hit it with the second. The probability is $\left(\frac{80}{81}\cdot \frac{1}{81}\right)^2$. So the total probability is an inifinite geometric series in the form
$\sum_{k=0} ^{\infty} \left[\frac{1}{81}\cdot\left(\frac{80}{81}\right)^k\right]^{2}=\frac{1}{81^2}\sum_{k=0}^{\infty} \left(\frac{80}{81}\right)^{2k}=\frac{1}{81^2}\cdot \frac{81^2}{161}=\boxed{\frac{1}{161}}$


find all solution of the equation
$\cos^{4}x+\sin^{4}x=\frac{1}{4}$

[hide]
We know that

$\cos^{2}x+\sin^{2}x=1$.

If we square this and substitute in the given equation, we get

$2\cos^{2}x\sin^{2}x=\frac{3}{4}$.

Since

$2\cos x\sin x=\sin2x$,

We can multiply both sides by 2 and factor to get

$\sin^{2}2x=\frac{3}{2}$.

But the LHS is always $\leq 1$ and the RHS is $>1$, so there are  [b]no solutions[/b].
[/hide]

the equoition is equivalente to $\sin^{2}xcos^{2}x=\frac{3}{8}$ but
in other hand we know that from $(cosx-sinx)^{2}\geq0$ and
$(cosx+sinx)^{2}\geq0$ we get $|\sin^{2}xcos^{2}x|\leq\frac{1}{8}$
so the equiotion has no solution.
sorry about latex.

Use ^ for exponents, so x^y gives $x^{y}$ and x^{yz} gives $x^{yz}$ (but x^yz gives $x^{y}z$), and use \cos for $\cos$.  < and >  give $<$ and $>$, while \geq and \leq give $\geq$ and $\leq$.  You can use the edit button to change your post and make it legible, if you want.

[quote="me@home"][quote="E^(pi*i)=-1"][hide]
We know that

$\cos^{2}x+\sin^{2}x=1$.

If we square this and substitute in the given equation, we get

$2\cos^{2}x\sin^{2}x=\frac{3}{4}$.

Since

$2\cosh\sinh=\sin2x$,

We can multiply both sides by 2 and factor to get

$\sin^{2}2x=\frac{3}{2}$.

But the LHS is always $\leq 1$ and the RHS is $>1$, so there are  [b]no solutions[/b].

EDIT: I don't know why the third equation above is not showing up properly, but I can't fix it.  It should read 2*cos(x)*sin(x)=sin(2x).
[/hide][/quote]

It's because you wrote: \cosx \sinx but latex can't recognize the tag "cosx" or "sinx". Just put a space between the x and the trig function :)[/quote]

Thanks!  I should have known that.

Ya Latex is annoying like that... I wish there was a WYSIWYG latex editor or something...

k ya now (E^(pi*i)=-1)'s solution makes sense before it said

$2=\sin{2x}$ and that made no sense...

Would we forbid complex plane then should agree with military people. They speak that in a wartime the values of a sine can exceed unit considerably.   :lol:


algebra
polynomial
linear algebra
matrix
vector
superior algebra
superior algebra solved

Let $M \in M_3(R)$ satisfy $M^3=I_3, M \neq I_3$.
Prove that $\exists P \in GL_3(R)$ such that:
$M=P^{-1}\left(\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right)P$

Step 1: Show that the minimal and characteristic polynomials of $M$ are both $x^3-1$. Proof: The irreducible factors of $x^3-1$ are $x-1$ and $x^2+x+1$. The only products of these with degree 3 are $(x-1)^3$ and $x^3-1$. In the first case, the minimal polynomial of $M$ must be $x-1$; the only such matrix is the identity, which was specifically excluded. In the second case, the characteristic polynomial is square-free, so it is also the minimal polynomial.

Step 2:
Theorem (special case of Rational Canonical Form): If $M$ is a $3\times 3$ matrix over any field $F$ with minimal polynomial $x^3+ax^2+bx+c$, $M$ is similar to a matrix of the form $\begin{bmatrix}0&0&-c\\1&0&-b\\0&1&-a\end{bmatrix}$.
Proof: Find a vector $v$ such that $v,Mv,M^2v$ are linearly independent*. Express $M$ with respect to the basis $v,Mv,M^2v$.
This solves the problem.

*I don't remember exactly how to do this- but I know it works, and you can always find a $v$ such that the minimal polynomial of $M$ is the minimal polynomial for $M$ on the subspace generated by $M^kv$.


Is it better to do AMC 12 tests for preparation for AMC 10?  Or is it  better to do the last few questions of AMC 10 tests?

Definitely do some AMC 10 problems and make sure you can solve the first 15-20 problems. The last few problems on the AMC 10 appear as mid-AMC 12 problems, so you can also work on AMC 12 tests if you want.

Also, I would do entire tests, not just the hardest problems of a competition. Never assume you can do easy problems just because you can do hard ones. Doing full tests is also good preparation from avoiding making stupid mistakes.

I'd say do AMC 12's, just because there are more problems on the AMC 12's, than the last few problems from AMC 10's.

EDIT: I would disagree with darkdieuguerre. You should start doing AMC 12's a little bit before you think you're ready for them. Also, you should start doing AIME's a little bit before you think you're ready for them. And for most people, being able to do the last problems probably means you can do the earlier problems. I do agree with him though, that you should do full tests, just to get used to the time pressure.


Diophantine equation
number theory proposed
number theory

Solve this equation:($ y$ $ \in$ $ Z$)

$ 1\plus{}2007 y^2$ $ \equal{}$ $ k^2$

[hide="my solution"]$ 2007y^2\equal{}(k\minus{}1)(k\plus{}1)$
$ 2007$ can be expressed as ±$ 2007*($±$ 1)\equal{}$±$ 223*($±$ 9)\equal{}$±$ 669*($±$ 3)$
we find out the solutions in these [hide="cases"]
I.  ± $ 2007\equal{}k\plus{}1$ and ±$ y^2\equal{}k\minus{}1$
II.  ±$ 2007\equal{}k\minus{}1$ and ±$ y^2\equal{}k\plus{}1$
III.±$ 223\equal{}k\plus{}1$ and ±$ 9y^2\equal{}k\minus{}1$
IV.±$ 223\equal{}k\minus{}1$ and ±$ 223\equal{}k\plus{}1$
V. ±$ 669\equal{}k\minus{}1$ and ±$ 3y^2\equal{}k\plus{}1$
VI. ±$ 699\equal{}k\plus{}1$ and ±$ 3y^2\equal{}k\minus{}1$[/hide]
the only solutions are:
$ (k,y)\equal{}(224,\plus{}5),(224,\minus{}5)$[/hide]

that's a pell equation so it has infinitely many answers that can be find easily(just find the fundamental answer).
of course we must add $ (y,k)\equal{}(0,\pm 1)$ to the answers...am i missing something?

[quote="shoki"]that's a pell equation so it has infinitely many answers that can be find easily(just find the fundamental answer).
of course we must add $ (y,k) \equal{} (0,\pm 1)$ to the answers...am i missing something?[/quote]
I forgot the case when $ (y,k) \equal{} (0,\pm 1)$.Of course there are infinitely solutions,but not in set of $ Z$(at least i couldn't find it).If so,Can you show some easily solutions as you said above.

[quote="Obel1x"] Of course there are infinitely solutions,but not in set of $ Z$(at least i couldn't find it).If so,Can you show some easily solutions as you said above.[/quote]

If a solution $ (k,y)$ exists, so exist $ (224k\plus{}10035y,5k\plus{}224y)$ and $ (224k\minus{}10035y,\minus{}5k\plus{}224y)$

So for example : $ (1,0)\to(224,5)\to(100351,2240)\to(44957024,1003515) ...$


The corrected problem is :-
[b]Problem:-[/b]

Show that for any real nos. $ A_3 , A_4 , ..... A_{85}, $ the roots of the eqn.
$ A_{85} X^{85} +\ A_{84} X^{84} +\  .....+\  A_3 X^3 +\ 3 X^2 +\ 2X +\ 1 =\ 0  $

are NOT all real

I think the question is not prove that the roots are not real, but :
Prove that not all the roots are real.

Let $x_i$ be the roots, where $i=1, \cdots, 85.$
Let $P = \prod x_i$ and $a = a_{85}.$
Then using the relations between roots and coefficients, we have :
$P = \frac {-1} a$ and $\sum_ i \frac P {x_i} = \frac 2 a$ and $\sum_{i<j} \frac P {x_ix_j} = - \frac 3 a.$
Thus :
$\frac 4 {a^2} = (\sum_i \frac P {x_i} )^2 = \sum_i \frac {P^2} {x_i^2} + 2 \sum_{i<j} \frac {P^2} {x_ix_j} = \sum_i \frac {P^2} {x_i^2} + \frac 6 {a^2}.$
It follows that $\sum_i \frac {P^2} {x_i^2}  = - \frac 2 {a^2} < 0$ which proves that the $x_i$ cannot be all real.

Pierre.


inequalities
LaTeX
inequalities proposed

a)Prove that$\forall a,b,c >0$ we have :
                  $\sum \frac{a^2-bc}{\sqrt{b^2+c^2+3a^2}} \ge 0$    $(1)$ :)  :)
b)Prove that $\forall a,b,c>0$ are the sides of a triangle ,we have:
                   $\sum \frac{a^2-bc}{b^2+c^2+3a^2} \le 0$               $(2)$  :)  :)

[quote="nthd"]a)Prove that$\forall a,b,c >0$ we have :
                  $\sum \frac{a^2-bc}{\sqrt{b^2+c^2+3a^2}} \ge 0$    $(1)$ :)  :)
[/quote]
Anybody can prove that with $k>3$?I think $k\le 4$ :)

[quote="nthd"][quote="nthd"]a)Prove that$ \forall a,b,c > 0$ we have :
                  $ \sum \frac {a^2 \minus{} bc}{\sqrt {b^2 \plus{} c^2 \plus{} 3a^2}} \ge 0$    $ (1)$ :)  :)
[/quote]
Anybody can prove that with $ k > 3$?I think $ k\le 4$ :)[/quote]
For $ k \equal{} 4$ it's wrong.
Try $ a \equal{} 4$ and $ b \equal{} c \equal{} 5.$
I think, for all $ 0 \leq k\leq2\sqrt2 \plus{} 1$ and for all positive $ a,$ $ b$ and $ c$  the following inequality holds:
\[ \sum_{cyc} \frac {a^2 \minus{} bc}{\sqrt {b^2 \plus{} c^2 \plus{} ka^2}} \ge 0
\]

I have a ideal for two problem.Since$ \sum (a^2\minus{}bc)(b\plus{}c)\equal{}0$*
Used it and Chebysev ineq we can prove $ \sum \frac{a^2\minus{}bc}{b^2\plus{}c^2\plus{}ka^2} \geq 0$ for all $ k\leq 2$
By the way,i think * is useful in proving a) b)

[quote="Math pro"]It is from Pham Kim Hung 's book! :lol:[/quote]
Stop it here now,math pro,Why do you often post some sentences and it is not valuables.You should learn how to post by $ Latex$ and post your proof fully,Don't only post some comments


induction
Ring Theory
function
linear algebra
linear algebra unsolved

Here is a marvelous problem: prove using linear algebra that a finite unitary boolean ring is isomorphic to $(Z/2Z)^n$ for a certain $n$.

There are different definitions of Boolean algebra.  The most standard one makes the problem trivially false.  The other one (every element satisfies $x^2 = x$) may be what you mean.   I don't see the relevance of linear algebra, it is pure commutative algebra (ring theory) unless I am missing something.

The definition is $x^2=x$ for all $x$. Now, indeed, it may seem strange what could linear algebra do here, but in fact it allows a very rapid and nice solution. The classical solutions I knew for this problem were quite long and boring, but this one is very quick and nice.

I think I have a solution, however, without a linear algebra but
it is short, if I didn't make a mistake.
 Suppose that $ B $ is a finite unitary boolean ring. Let's prove
by induction that $ B $ is isomorphic to some $  ( \mathbb{Z} / 2
\mathbb{Z} )^{n} $. First of all, if $ x \in B $ then $ (2x)^{2} =
2x $ so $ 4x = 2x $ so $ 2x = 0 $. So if we have $ x,y \in B $,
then $ (x+y)^{2} = x+y $ so $ x^{2}+xy+yx+y^{2} = x+y+xy+yx = x+y
$ so $ xy = -yx =yx $, therefore $ B $ is commutative.
 If every element is $ 0 $ or $ 1 $, then it is
$ \mathbb{Z} / 2 \mathbb{Z} $. Now suppose that we have some $ 0,1
\neq a \in B $. Define $ B',B'' \subset B $, $ B' = \{ x \in B |
ax = x \} $ and $ B'' = \{ x \in B | ax = 0 \} $. It is easy to
see that $ B',B'' $ are proper subsets of $ B $, because $ 1
\notin B' $ and $ a \notin B'' $. Also it is easy to see that $
B', B'' $ are boolean subrings of $ B $ , however the unit in $ B'
$ is $ a $ and in $ B'' $ is $ 1-a $, so they are unitary. Also $
B' \cap B'' = 0 $ and $ B = B' + B'' $ because for every $ x \in B
$ we have $ x = ax + (1-a)x $. Finally, if we have $ x \in B' $
and $ y \in B'' $ then $ xy = axy = ayx = 0 $. Therefore we can
write $ B \cong B' \oplus B'' $ as rings. Now, by induction on
size of $ B $, we know that $ B' \cong ( \mathbb{Z} / 2 \mathbb{Z}
)^{k} $ and $ B'' \cong ( \mathbb{Z} / 2 \mathbb{Z} )^{m} $ for
some $ k,m $ therefore $ B \cong B' \oplus B'' \cong ( \mathbb{Z}
/ 2 \mathbb{Z} )^{k} \oplus ( \mathbb{Z} / 2 \mathbb{Z} )^{m}
\cong ( \mathbb{Z} / 2 \mathbb{Z} )^{k+m} $.

I hope it is correct.

I like your solution, Leva, it is short and nice. Here is the solution with linear algebra: first of all, it is simple to see that $2=0$ and that our ring $A$ can be seen as a  vectorial space $E$ over $Z/2Z$. Next, consider $e_1,...,e_n$ a base and the endomorphismes $ g_i(x)=e_ix$. They are idempotent and since our ring is commutative (easy to prove), they are simultaneously diagonalisable. Thus we can find a base $f_1,...,f_n$ in which these maps are diagonal. It follows that $xf_i$ is collinear with $f_i$ for all $i$ and since $f_if_j=f_jf_i$ and $f_i, f_j$ are not collinear, we must have $f_i f_j=0$. Now, define a linear function $t(f_i)=(0,0,...,1,0,..,0)$ and observe that in fact it is an isomporhism of rings since $f_i f_j=0$. Isn't it nice?

Sorry for reviving it like this, especially when two beautiful solutions have been posted, but I had also posted one which was incomplete, I thought it was wrong, and now I wanted to put it back up, although it doesn't even come close to matching the other two  :blush:. I think it's a bit different though, and that's why I'm posting it.

If $|A|=2^n$, we'll prove by induction on $n$ that there is a unique $n$-uple of non-zero elements $a_1,\ldots,a_n\in A$ s.t. $a_ia_j=0,\ \forall i\ne j$.

Suppose we have shown it for $n-1$, and consider a maximal ideal $M$ of $A$. For every $x\in A\setminus M$, $M$ must contain $1+x$, so $|M|=2^{n-1}$. $M$ is a unital boolean ring, by taking the unit to be the union of all its elements, where the union of two elements $x,y\in A$ is defined as $x+y+xy$ (we make use of the fact that boolean rings are, in fact, the same as boolean algebras, so we take the unit in $M$ to be the supremum of all the elements in $M$), so, by induction, there is a unique set of non-zero elements $a_1,\ldots,a_{n-1}\in M$ s.t. $a_ia_j=0,\ \forall i\ne j$.

Take any $x\in A\setminus M$. Assume WLOG the $a$'s which when multiplied with $x$ do not yield zero are $a_1,\ldots,a_t,\ t\le n-1$. The set $\{xa_1,\ldots,xa_t,a_{t+1},\ldots,a_{n-1}\}$ is a set of non-zero elements in $M$ with pairwise null products, so it must be the same as $\{a_1,\ldots,a_{n-1}\}$. It's easy to see from here that $xa_i=a_i,\ \forall i\in\overline{1,t}$, and if we take $a_n=x+a_1+\ldots+a_t$, the set $\{a_1,\ldots,a_n\}$ will have pairwise null products (and non-null elements).

The uniqueness of the set also follows easily: first of all, we cannot have $xy=0,\ x,y\in A\setminus M$, since $M$ is prime, and we can't have more than $n-1$ non-null elements with null pairwise products, because they are independent over $\mathbb Z_2$, and they yield a group with more than $2^{n-1}=|M|$ elements, so any such set must contain $a_1,\ldots,a_{n-1}$. The $n$'th element is a linear combination with coefficients in $\mathbb Z_2$ of $a_1,\ldots,a_n$, and it follows immediately from here that it must be $a_n$.

We have thus proven the existence and uniqueness of a set $\{a_i\}$ of $n$ non-null elements with pairwise null products in $A$ (the uniqueness was used in the induction step; what's important here is the existence), and we can now map $a_i$ to the element of $\mathbb Z_2^n$ having $1$ on the $i$'th position and zero everywhere else, thus getting a ring isomorphism from $A$ to $\mathbb Z_2^n$, as desired.

Sorry for the double post and for reviving such an old topic, but I was thinking about this subject earlier today, and even though the problem is simple, I think it's a nice mathematical fact :); here's another proof (or a sketch of a proof):

We introduce an order relation on $A$ (our ring) in an obvious manner: $x\le y$ iff $xy=x$.

Now let $a_1,\ldots,a_n$ be the (distinct) minimal non-zero elements of $A$ with respect to this order. For every $x\in A$ and every $i\in\overline{1,n}$, we have either $xa_i=a_i$ or $xa_i=0$, and if $xa_i=0,\ \forall i$, then $x=0$ (because every non-zero element of $A$ dominates at least one $a_i$, since the $a_i$'s are all the non-zero minimal elements in $A$). It's easy to see now that given $x\in A$, if $I=\{i\in\overline{1,n}\ |\ xa_i=a_i\}$, then $x=\sum_{i\in I}a_i$, and that the map sending $a_i$ to the element of $(\mathbb Z/2\mathbb Z)^n$ having $1$ on its $i$'th position and $0$ everywhere else can be extended to an isomorphism.


AMC
AIME
USA(J)MO
USAMO
algebra
binomial theorem

How many combinations are possible if you can choose from any of 8 toppings?

[hide="what I think"]A teacher first thought it was $ 8!$, but  I think its $ \frac {8!} {8!}$ plus  $ \frac {8!} {7!}$ plus $ \frac {8!} {6!}$ and on...[/hide]

[quote="Shortgeek"]You're both wrong; it's $ \binom{8}{8} \plus{} \binom{8}{7} \plus{} \binom{8}{6} \plus{} \binom{8}{5} \plus{} \binom{8}{4} \plus{} \binom{8}{3} \plus{} \binom{8}{2} \plus{} \binom{8}{1} \plus{} \binom{8}{0}$, I think. I can't evaluate that right now, but a combination doesn't matter about order.[/quote]

Ahem... the binomial theorem means that the answer is $ 2^8$. However, there is a much simpler approach.

Given each topping, it could be either on the (pizza?) or not. Since there are two choices (on or not) for each topping and there are 8 toppings, the answer is simply $ 2^8 \equal{} \boxed{256}$ combinations.

Just to cheer you up, I will list my recent failures.

- Failed AMC 10A - 112.5
- Failed AMC 12B - 93
- Failed AIME (unofficial, as I didn't qualify) - 10
- Failed Chapter - 43
- Failed State - 39
- Failed USAMO - 0

Now my mistakes:

- Couldn't count to 6, thereby getting the first question on the AMC 12B wrong.
- Thought that $ \sqrt{50} \equal{} 2\sqrt 5$.

WHAT? YOU CONSIDER A 10 ON THE AIME FAILING?  :rotfl:

...I feel worse now. Don't try to cheer me up, it'll be worse than that.

Well... OK. I technically got a $ 67\%$ on the AIME, which translates to a $ D\plus{}$, so I barely passed (60 and under is failing). Whatever.

[quote="Yongyi781
- Failed AMC 10A - 112.5
- Failed AMC 12B - 93
- Failed AIME (unofficial, as I didn't qualify) - 10
- Failed Chapter - 43
- Failed State - 39[/quote]
These are hardly failures.  These are great scores.

:D Hey, I was trying to make Shortgeek feel better!

[size=59]Maybe I didn't do that great a job...[/size]


MATHCOUNTS
geometry
geometric transformation
reflection

dingzhou participate in The Race For Flying People in hope of winnign 2 dollars. The track is between two walls. One wall is 300 m high, the other wall is 500 m high. The distance between the two wall is 1200 m. All flyers have to start from top of the 300 m wall, and touch the ground at any point, and fly back up the top of the 500 m high wall. dingzhou want to calculate the shortest possible track length, but due to his poor math abilities, he cannot, so he consult the strange world of artofproblemsolving at the mathcounts forum. Please help dingzhou to find the shortest possible path, if you do, he might even give you 1% of the prize money...[b]might[/b]...

dingzhou wishes you luck

[hide="hinto"]Reflect the top of the 300 tall thing over the ground, and connect it to the top of the other one. That's the distance you want.[/hide]

Heh, forgot to end the code. Twice.

why thank you xpmath, dingzhou will give you the 2 cents if he ever see you and recognize you.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190195

Read all of the posts and you should get a good idea of how to find the shortest distance on any path.


Hello everyone,

I do not understand the following exercises from the Checkup section of my precalculus book, and I would greatly appreciate any help and explanation. Thank you in advance.

---

1. Let $ f(x)$ be a third-degree polynomial function with real coefficients. When $ y \equal{} f(x)$ is graphed, it is found that $ f(2) \equal{}\minus{}3$ and $ f(3) \equal{} 1$. Which statement MUST be true?

A) $ f(x)$ has exactly one real zero in the interval $ [2,3]$.
B) $ f(x)$ has exactly one real zero in the interval $ [\minus{}3,1]$.
C) $ f(x)$  has exactly one or three real zeros in the interval $ [2,3]$.
D) $ f(x)$ has exactly one or two real zeros in the interval $ [\minus{}3,1]$.

I chose A) as the answer because the function only crosses the x-axis once in the interval [2,3]. The book's answer was [hide]C[/hide].

2. $ f(x) \equal{} 4x^{5}\minus{}x^{3}\minus{}32x^{2}\plus{}8$

How would I factor this polynomial equation to find the zeros?

3. Find a polynomial function $ f$ for which -3 is a zero of multiplicity 2 and $ 1\minus{}\sqrt{2}$ is also a zero.

My answer is $ (x\plus{}3)^{2}(x\minus{}1\plus{}\sqrt{2})$. Would this be correct?

1)If $ f(2) \equal{}\minus{}3 < 0,f(3) \equal{} 1 > 0$ that means that $ f$ has at least a root in $ (2,3)$.
So, $ f$ can have one or three roots in $ (2,3)$
$ f$ can't have two real roots, because the third must be complex, and the number of complex roots of a polynomial with real coefficients is even.

2) $ f(x) \equal{} x^{3}(4x^{2}\minus{}1)\minus{}8(4x^{2}\minus{}1) \equal{}$
$ \equal{} (x^{3}\minus{}8)(4x^{2}\minus{}1) \equal{} (x\minus{}2)(x^{2}\plus{}2x\plus{}4)(2x\minus{}1)(2x\plus{}1)$

3) If $ f\in\mathbf{R[X]}$ then is correct.
If $ f\in\mathbf{Q[X]}$ then $ f$ has also the root $ 1\plus{}\sqrt{2}$

1) Graph a function like $ x^{3}\minus{}x \equal{} 0$ and look at what happens in the interval $ [\minus{}2,2]$. What's stopping the same thing from happening for $ f(x)$ in $ [2,3]$ in your problem?

2) The post above has the correct solution. Here is a way to figure it out on your own.

First, separate the first two terms from the last two. So consider these two separately

$ 4x^{5}\minus{}x^{3}$, $ \minus{}32x^{2}\plus{}8$.

The first two terms are divisible by $ x^{3}$, and the last two are divisible by $ \minus{}8$. So factor them as such.

$ x^{3}(4x^{2}\minus{}1)$, $ \minus{}8(4x^{2}\minus{}1)$

Notice the $ 4x^{2}\minus{}1$ are identical. So if you add these back up (getting what you had before), by the distributive property you would get

$ (x^{3}\minus{}8)(4x^{2}\minus{}1)$

It should be easier now. Binomials are not too difficult to work with.

Basically, the way to recognize when this trick will work is to recognize the relationship between groups of terms. Here, if you examine $ 4x^{5}$ and $ \minus{}x^{3}$, you find that one is $ \minus{}4x^{2}$ times the other. Looking at the other two terms, you have the same relationship ($ \minus{}32x^{2}\equal{}\minus{}4x^{2}\cdot 8$). This is a sign that you definitely can factor something.

This is a fundamental, common pattern, though it comes in disguised forms sometime. For instance, the factorization $ a^{2}\plus{}2ab\plus{}b^{2}\equal{} (a\plus{}b)^{2}$ is very obvious (besides being extremely well known) if you write $ a^{2}\plus{}2ab\plus{}b^{2}\equal{} a^{2}\plus{}ab\plus{}ab\plus{}b^{2}$, when it's easy to see the last equation can be written as $ a(a\plus{}b)\plus{}b(a\plus{}b)$ and then $ (a\plus{}b)^{2}$.

With respect to all of the things you're concerned with, 3) is completely correct.


function
limit
calculus
calculus computations

Hi, I'm having trouble solving this problem, can anyone help me please ?

Suppose we wish to model harvesting a fishing ground where we can control the number of boats. Assume that the fish population dynamics are logistic. Let x(t) be fish population
units as a function time of time units t. Let the fishing boat numbers be F. Then we model this with the equation

dx/dt = rx(1 − x/K) − F C(x),

where r,K > 0 are constants and C(x) is the catch rate per boat as a function of fish population. We assume that the C(x) is an increasing function of fish population, that it is approximately proportional to the fish population when the population is small, and approaches a limiting upper bound as the fish population grows large.

(i) Show that

C(x) =x/(x0 + x), x0 > 0

satisfies the assumptions above.

Thank You.

So you want to show that: (i) $C(x)=\frac{x}{x_{0}+x}$ is increasing; (ii) $\lim_{x\to 0}\frac{C(x)}{x}$ exists and is positive; (iii) $\lim_{x\to \infty}C(x)$ exists.

(i) $C(x)=1-\frac{x_{0}}{x_{0}+x}$. The larger values of $x$ decrease the fraction that is being subtracted.

(ii) $\lim_{x\to 0}\frac{C(x)}{x}=\lim_{x\to 0}\frac{1}{x_{0}+x}=\frac{1}{x_{0}}$.

(iii) $\lim_{x\to \infty}C(x)=\lim_{x\to \infty}\frac{1}{1+x_{0}/x}=\frac{1}{1}=1$.


geometry
circumcircle
rectangle
parallelogram
geometry unsolved

Let $ ABC$ be a triangle and $ (O)$ be its circumcircle, with center $ O$. Let $ D$ be the intersection point of the tangent lines of $ (O)$ at vertices $ B$, $ C$. Denote by $ E$ an arbitrary point lies on $ (O)$, $ AE$ intersects $ DB$, $ DC$ at $ M$, $ N$, respectively. Let $ F$ be the intersection of $ BN$ and $ CM$, $ T$ be the intersection of $ AD$ with $ BC$. Prove that $ E$, $ F$ and $ T$ are collinear.

Let ED meet BC at S and let AE meet BC at P.

Case 1: E is on the circumcircle arc BC containing A, line DP does not intersect (O). Project (O) to a circle $ \mathcal O'$ and the line DP to infinity. Then B'C' is a diameter of (O') $ B'N' \parallel C'N' \parallel S'E' \parallel T'A' \perp B'C'$ and $ A'E' \parallel B'C',$ B'C'N'M' and S'T'A'E' are rectangles of equal heights B'M' = S'E' = T'A' = C'N' symmetrical WRT the berpendicular bisector of B'C', having common diagonal intersection F'.

Case 2: E is on the circumcircle arc BC opposite to A, line DP intersects (O). Project (O) to a circle  $ \mathcal O'$ with center $ P \equiv P'$. Then B'C' and A'E' are diameters of  $ \mathcal O',$ $ P \equiv P'$ their common midpoint, $ B'N' \parallel C'N' \parallel A'T' \parallel E'S',$ $ B'M'C'N' \sim S'E'T'A$ are centrally similar parallelogram with similarity center at their common diagonal intersection $ P \equiv P',$ and $ B'N' \parallel C'M' \parallel S'A '\parallel T'E'$ have the common point at infinity F'.

Thank you very much dear [b]Yetti[/b], for your interest. I'm sorry that I haven't read your solution yet, but I think I have another solution, and let me show it.

Denote points $ S \equal{} MN\cap BC$. We have $ E$, $ F$ and $ T$ are collinear if and only if $ (BCTS) \equal{} (NMES)$, i.e. $ (MNAS) \equal{} (NMES)$, and it's equivalent to $ \frac {AM}{AN}: \frac {SM}{SN} \equal{} \frac {EN}{EM}: \frac {SN}{SM}$ $ \iff\frac {AM\cdot EM}{AN\cdot EN} \equal{} \frac {SM^2}{SN^2}$ $ \iff\frac {MB^2}{NC^2} \equal{} \frac {SM^2}{SN^2}$ $ \iff\frac {BM}{BS}\cdot\frac {CS}{CN}\cdot\frac {SN}{SM} \equal{} 1$, what is truly.

$ \bullet$ This proof is a my present for [b]Virgil Nicula[/b].

[b]P.S.[/b] A trigonometric solution can be found.


Find the maximum of $ u \equal{} Re(z^3)$ on the unit square [0,1] x [0,1]

[hide="Compute"]
$ z^3\equal{}(x^2\minus{}y^2\plus{}2ixy)(x\plus{}iy)\equal{}x^3\minus{}3y^2x\plus{}3ix^2y\minus{}iy^3$
$ Re(z^3)\equal{}x(x^2\minus{}3y^2)$, the maximum value will be taken if $ x\equal{}1$,$ y\equal{}0$, the value is $ 1$.
[/hide]

note you can (very) slightly simplify the computation by first applying the maximum principle to claim the point in question must lie on the boundary, and then drawing a sketch to find it must be on the right boundary


advanced fields
advanced fields unsolved

Show that any map $S^{n}\to S^{n}$ of odd deegree must carry some pair of antipodal points into a pair of antipodal points.

If n is even the problem is not so hard. Since, if $-f\neq f(-Id)$ then $f$ and $f(-Id)$ is homotopic to each other. This fact shows that $deg f=deg f(-Id)=(-1)^{n+1}deg f$. Hence deg f must equal to 0.

My book only has quite informal discussion on the degree of maps (it is very introductory), but I think the following works:

Suppose $f: S^{n}\rightarrow S^{n}$ is a map of odd degree that does not carry any pair of antipodal points onto a pair of antipodal points. Then, if $g$ is the antipodal map, $f(x)$ and $f(g(x))$ is not a pair of antipodal points. Let $h(x)$ be the midpoint of the great circle joining $f(x)$ and $f(g(x))$ (the midpoint chosen on the shortest piece connecting the points). $h$ has even degree, since $h(x) = h(g(x))$. Hence also $f(x)$ has even degree, since $f(x)$ and $h(x)$ are homotopic.


Can someone tell me some inequality book available on the net for free :D

Try these:

http://my.netian.com/~ideahitme/eng.html
http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html


modular arithmetic
number theory
relatively prime
number theory unsolved

Let $ m$ be a positive integer, let $ a_1, a_2, ..., a_{\phi(m)}$ be the integers between $ 1$ and $ m$ that are relatively prime to $ m$, and write the quantity

$ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\frac{1}{a_3}\plus{}\cdots\plus{}\frac{1}{a_{\phi(m)}}$

as a fraction $ \frac{A_m}{B_m}$ in lowest terms.

Find the value of $ A_m {\pmod m}$ and prove that it is true.

$ a_m\equiv e 0\mod m$.
For any $ a_i$, s.t. $ (a_i,m) = 1$ exist $ b_i$, s.t. $ b_ia_i\equiv 1\mod m$ and $ a_i\to b_i$ is bijective.
Therefore $ \sum_i \frac {1}{a_i}\equiv \sum_i b_i \equiv 0\mod m$.


Solve this system of equations
a+b+c+d=12
abcd=27+ab+ac+ad+bc+bd+cd  if a,b,c,d are positive real numbers. :D

It's awfully symmetric, so I tried $ a\equal{}b\equal{}c\equal{}d\equal{}3$, and it worked.

That system of equations actually has an infinite number of positive real solutions

By $ AM\minus{}GM$, $ 12\equal{}a\plus{}b\plus{}c\plus{}d\ge 4\sqrt[4]{abcd} \rightarrow \sqrt{abcd}\le 9$ (1)

Then $ abcd\minus{}27\equal{}ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\ge 6\sqrt[6]{a^3b^3c^3d^3}\equal{}6\sqrt{abcd}$
$ abcd\minus{}6\sqrt{abcd}\minus{}27\ge 0$
$ \minus{}3\le \sqrt{abcd}\le 9$, $ \sqrt{abcd}\le 9$ (2)

From (1) and (2), equality holds, so $ a\equal{}b\equal{}c\equal{}d\equal{}3$

In fact, (2) states that $ \sqrt{abcd} \geq 9$, to obey that inequality. Now it works.


analytic geometry
graphing lines
slope
calculus
calculus computations

At each point (x,y) on a certain curve, the slope of the curve is 4xy. If the curve contains the point (0,4), then its equation is:

A) y = e^(2x^2) + 4
B) y = e^(2x^2) + 3
C) y = 4e^(2x^2)
D) y^2 = 2x^2 + 4
E) y = 2x^2 + 4

$ \frac{dy}{dx}\equal{}4xy$

$ \frac{1}{y}dy\equal{}4xdx$
Differentiate both sides:
$ \ln(y)\equal{}2x^2\plus{}C$
Our initial condition is (0,4):
$ \ln(4)\equal{}2(0)^2\plus{}C$
$ C\equal{}\ln(4)$
$ \ln(y)\equal{}2x^2\plus{}\ln(4)$
$ y\equal{}e^(2x^2\plus{}\ln(4))$
$ \boxed{y\equal{}4\cdot e^{2x^2}}$


Find all positive integers $n$ such that $\frac{3^n+1}{2^n+1}$ is an integer.

Does the prposer has a solutio or  is it an open problem? ;)  It seems hard until now!....

ah... sorry  :lol:
 well we don't have a solution, but i would like to know if someone from this forum knows this problem, at least to know if it is an open one or not...


logarithms inequalities function calculus

Solve the inquation: $\log^{2}_{9}x\ge\log^{2}_{3}\sqrt{x-\frac{1}{4}}$

[hide="I must be mistaken"]
Just $(\ln_{3}^{9})^{4}\geq x-.25, 16.25 \geq x$
[/hide]

I don't get it.
What's the argument of the first log?
or does this $\log^{x}_{y}$ notation mean something else?

I assumed it was log base 10 and then the underscored number was the argument

Sorry every body. It's my mistake :blush:  And I've just edited it. :D

$\log_{3}q=a\iff 3^{a}=q\iff 9^{a}=q^{2}\iff \log_{9}q^{2}=a\iff \boxed{\log_{9}q^{2}=\log_{3}q}$

$\log_{9}x>\log_{9}x-\frac14$ is always true, thus I find that:
\[\log^{2}_{9}x>\log^{2}_{3}\sqrt{x-\frac{1}{4}}\]
for all $x$.

Yup, I'm still confused...

$\log_{9}x>\log_{9}x-\frac14$   is not always true, so your answer is incorrect  :)
here is mine  [hide]  $\log^{2}_{9}x\ge\log^{2}_{3}\sqrt{x-\frac{1}{4}}\iff (\log{x}-2\log\sqrt{x-\frac{1}{4}})(\log{x}+2\log\sqrt{x-\frac{1}{4}})\geq{0}$
and by solving this inequality, we find  ${x}\in[\frac{1+\sqrt{65}}{8},+\propto[$ [/hide]

But log is an increasing function, are you considering log |x| as the function? (This might be standard Im not sure... it's used in my calculus textbook)

[quote="yfehehey"]$\log_{9}x>\log_{9}x-\frac14$   is not always true, so your answer is incorrect  :) [/quote]

Of course it's always true (so long as the relevent expressions are defined, that is $x > \frac{1}{4}$).  The answer (s)he gave is incorrect for entirely different reasons, namely that $x > y$ does not imply $x^{2}> y^{2}$.

i got no solution... but i think i did something wrong...

[quote="me@home"]But log is an increasing function[/quote]
yeah, but you need to mention that ${x}$ is greater than $\frac{1}{4}$ then, you have to study the cases in which
$\log_{9}x>\log_{9}x-\frac14\Longrightarrow\log^{2}_{9}x\ge\log^{2}_{3}\sqrt{x-\frac{1}{4}}$ is correct.  try it and tell me if you got the same result


Sorry but there aren't really too many.  Probability is difficult and heavily intuitive.  I'm sure wikipedia has something but if not, check the AoPS books because they definitely do.

um...yes i agree with JRav i used to be aweful at probablity but i bought the intro to it by David Patric and im not saying im a master at it now but im decent!  On thing that is not directly related to probablity but could be used in a probablity question are combination and permutation.  Youll see them everywhere even if your not aware of them.

Combination -- order of things (eg order that gumballs are picked from a hat doesnt matter)  we have nCk said "n in combinations of k" (eg how many ways can 3 people be picked randomly from a group of twelve.  we have 12 in combinations of 12) the formula for this is $ \frac {n!}{(k!)\times((n \minus{} k)!)}$

permutations -- order matters (eg how many ways can we arrange 5 books on a shelf) so for this the fomula is just $ n!$

Practice probablity problem involving combination and permutation.

What is the probablity that when 15 students (8 boys and 7 girls) are arranged in a line no two girls and no two boys are standing next to each other?

I have a friend taking a class in probability and even he complains how hard it is.

[quote="MathIsGreat"]Can you guys give me probability formulas?[/quote]

There are not many formulas for probability.  If you know what you're doing, then you really shouldn't need a formula. :)

[quote="stevenmeow"]The most basic thing about probability is that it is defined as Good outcomes over Total outcomes.
   DO NOT confues this with odds, which is defined as Good outcomes over Bad outcomes[/quote]

The odds against something is bad:good, which is pretty important as well.

Now, probability isn't that hard if you get the hang of it.
Get the total correct outcomes over the total possible outcomes.
Usually combinations is used for total possible outcomes and counting as well.

Yes, agreed. Just learn the concept of permutations and combinations and make sure you understand them, and then the rest is pretty much intuitive.

Although, one useful tool for counting probility is counting the opposite of something and subtracting from 1.
Make sure you also understand casework.

Normally for probability problems, you want to think, okay, how many total possibilities (denominator)? Then you just figure out, how many of them are "success (numerator)?"

I agree with the above posts, but it does take a while to get that intuition.  It comes with doing many types of problems.

if you really want to get good i strongly recomend intro to counting and probability by DPatrick,  like JRav said it is important to try many different types of problems which this book provides.  I bought it and it really is great!!!

combination formula
   n!
  ---
 k!*(n-k)!

[quote="MathIsGreat"]Can you guys give me probability formulas?[/quote]

Formula = plug and chug = bad.

Instead of blindly applying formulas, you really should know what you are doing.

Besides, probability doesn't have many formulas,

I haven't come up against any.


In Paul Zeitz's ACoPS book, he solves the problem:

Prove that \[ (\frac{1}{2})(\frac{3}{4})\cdots(\frac{2n-1}{2n})\leq\frac{1}{\sqrt{3n}}.\]

His first attempt is to show that if $ P(n)$ is true, then $ P(n+1)$ is true: \[ {(\frac{1}{2})(\frac{3}{4})\cdots(\frac{2n-1}{2n})}(\frac{2(n+1)-1}{2(n+1)})\leq\frac{1}{\sqrt{3n+3}}.\]

But he fails, saying that this inequality is untrue for $ n = 1$ and is caused by a "very weak" assertion.
One question is, how can you decide if an induction hypothesis is too weak or too strong? What are the signs?

He next replaces $ 3n$ with $ 3n+1$, and shows this to be an equality. How can he use this for induction? Shouldn't it be $ 3(n+1)$ since $ n$ is an integer, and he must prove it for each $ n+1$ in order to establish induction, and not just add $ 1$ to $ 3n$?

Thanks for any help!

[quote="darkprince"]One question is, how can you decide if an induction hypothesis is too weak or too strong? What are the signs?[/quote]

It's too weak if it doesn't solve the problem and it's too strong if it's false, I suppose.  :P

[quote="darkprince"]He next replaces $ 3n$ with $ 3n\plus{}1$, and shows this to be an equality. [/quote]

Could you give more details?  I don't remember this proof well, but from what I remember this is how he strengthens the inductive hypothesis, not how he inducts.

Sometimes when we prove inequalities we can instead of proving the inequality we are trying to prove, prove a "stronger" inequality. Say I want to show $ f(x,y,z)\leq g(x,y,z)$ but the expression $ g(x,y,z)$ isnt conventint to work with. Instead I prove $ f(x,y,z)\leq h(x,y,z)$ where its obvious that $ h(x,y,z)$ is smaller than $ g(x,y,z)$ which will immedialte prove that $ f(x,y,z)\leq g(x,y,z)$. The inequality $ f(x,y,z)\leq h(x,y,z)$ is referred to as a "stronger inequality". But the point in doing that is because it might be messy to work with the original inequality. So we shift to a stronger but nicer looking inequality.

When we want to prove,
$ \frac{1}{2}\cdot ...\cdot\frac{2n\minus{}1}{2n}\leq\frac{1}{\sqrt{3n}}$
We can do it with induction by defining $ P(n)$ equal to the entire product on the LHS. Now we want to show that $ P(n\plus{}1)$ is true if we do that we get,
$ P(n)\cdot\frac{2n\plus{}1}{2n\plus{}2}\leq\frac{1}{\sqrt{3n\plus{}3}}$.
But its more convienet to work with the weaker inequality $ P(n)\leq\frac{1}{\sqrt{3n}}$ so,
$ \frac{1}{\sqrt{3n}}\cdot\frac{2n\plus{}1}{2n\plus{}2}\leq\frac{1}{\sqrt{3n\plus{}3}}$.
But this isnt true (take $ n\equal{}1$ for instance).

So instead we hope,
$ \frac{1}{2}\cdot ...\cdot\frac{2n\minus{}1}{2n}\leq\frac{1}{\sqrt{3n\plus{}1}}$.
Note this is a stronger inequality. It might not be true but we try it.
We accept $ P(n)\leq\frac{1}{\sqrt{3n\plus{}1}}$ and try to show $ P(n\plus{}1)\leq\frac{1}{\sqrt{3n\plus{}4}}$.
Its more convient to try to show,
$ \frac{1}{\sqrt{3n\plus{}1}}\cdot\frac{2n\plus{}1}{2n\plus{}2}\leq\frac{1}{\sqrt{3n\plus{}4}}$.
Now multiple stuff out and it works out (easier said, but messier done).

sylow_theory, I understand now!
You try to find an inequality that is much less messier than the original expression; in this case, $ \frac{1}{\sqrt{3n}}$ is not convenient, so you replace it with something that is smaller than $ \frac{1}{\sqrt{3n}}$, such as $ \frac{1}{\sqrt{3n\plus{}1}}$, and Zeitz finds this to be an equality. Making it smaller will not refute the original inequality since we are actually tightening the boundaries (I think that's where they got the name 'stronger inductive hypothesis' from). Therefore, \[ P(n) \equal{}\frac{1}{\sqrt{3n\plus{}1}}\leq\frac{1}{\sqrt{3n}}.\]

I understand perfectly now. Thanks a lot for your help!

@t0rajir0u: Yes, you're right; he does strengthen the inductive hypothesis by adding $ 1$ to $ 3n$, not prove the induction itself. I wasn't being clear when I said prove the induction - I meant that at a later step, he uses the stronger hypothesis, $ \frac{1}{\sqrt{3n\plus{}1}}$, and then inducts, proving the original inequality. Sorry! :blush: Referring to your first response, I'm guessing that making it too strong make the inequality untrue, and making it too weak will not achieve anything at all. I'm new to induction, so I wouldn't know for sure, but intuitively, it seems right...
Thanks for your help!

[quote="darkprince"]sylow_theory, I understand now!
You try to find an inequality that is much less messier than the original expression; [/quote]
Correct, just to tell you the bad news this doesn't always work. But its a good way to procede.


geometry
3D geometry
factorial
modular arithmetic
AMC
AIME
number theory

How many positive integer cubes divide $ 3!\cdot 5!\cdot 7!$?

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 5\qquad
\textbf{(E)}\ 6$

[hide="Solution"]We have $ 3!5!7! \equal{} 2^83^45^27$
We can form cubes with $ 2^3, 2^6, 3^3$.
So there are 5 possible solutions from those cubes, and $ 1^3$ must be counted as well, to get $ \boxed{\textbf{(E)}\ 6}$.[/hide]

grn_trtle

wouldn't it be $ 3\plus{}2\plus{}1\equal{}5$

$ 1^3, 2^3, 4^3, 8^3, 3^3, 6^3$.
I didn't make that very clear though.

A bit more explanation:

[hide="Click for solution"]
We write our given expression as a product of primes:

$ 3! \cdot 5! \cdot 7! \equal{} 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$.

A cube that is a factor of this must have a prime factorization of the form $ 2^p \cdot 3^q \cdot 5^r \cdot 7^s$, where $ p \equiv q \equiv r \equiv s \equiv 0 \pmod{3}$, that is, $ p$, $ q$, $ r$, and $ s$ are all multiples of $ 3$.  There are $ 3$ possible values for $ p$ ($ 0,3,6$).  There are $ 2$ possible values for $ q$ ($ 0,3$).  The only value for $ r$ and for $ s$ is $ 0$.  Thus, there are $ 3 \cdot 2 \cdot 1 \cdot 1 \equal{} 6$ distinct cubes that divide $ 3! \cdot 5! \cdot 7!$, or answer choice $ \boxed{\textbf{(E)}}$.
[/hide]

Thanls AIME 15

Ignore my previous response

how does 8^3 divide this number? When we factor we get 2^8, and not 2^9(this would prove that 8^3 divides this number).

Prime factorize: $2^8 \cdot 3^4 \cdot 5^2 \cdot 7$.
Then, the answer is just $\left( \lfloor \frac{8}{3} \rfloor + 1 \right) \left( \lfloor \frac{4}{3} \rfloor + 1 \right) \left( \lfloor \frac{2}{3} \rfloor + 1 \right) \left( \lfloor \frac{1}{3} \rfloor + 1 \right) = 3 \cdot 2 \cdot 1 \cdot 1 = 6$.


[quote="behdad.math.math"]what about when $ a_i$ and $ b_j$ are not equal to 0.[/quote]
Then just let $ b_j$ be really close to $ 0$. Or fx $ a_1 \equal{} a_2 \equal{} .... \equal{} a_{10} \equal{} 100$ $ b_1\equal{}b_2\equal{}...\equal{}b_{20}\equal{}1$ $ b_{21} \equal{} 980$


probability
integration
algebra
function
domain
calculus
probability and stats

Let (B_t) à brownien motion, what is the value of  P( B_1>=0 and B_2 <= 0 ) , in my calculation i get 1/16 but i'm not sure !! can any one help ?

you could compute (since $ B_1$ and $ B_2\minus{}B_1$ are independent):
\[ P[ (B_1 \geq 0) \cap (B_2 \leq 0 ) ] \equal{} \int_{x\equal{}0}^{\infty} P[B_1 \in [x; x\plus{}dx] ] P[B_2\minus{}B_1 \leq \minus{}x]\]
which is equal to:
\[ \int_{x\equal{}0}^{\infty} \frac{e^{\minus{}\frac{x^2}{2}}}{\sqrt{2 \pi}} \int_{y\equal{}\minus{}\infty}^{\minus{}x}\frac{e^{\minus{}\frac{y^2}{2}}}{\sqrt{2 \pi}} dy dx \equal{} \frac{1}{2 \pi}\int_{x \in \mathbb{R}} \int_{y \in \mathbb{R}} e^{\minus{}\frac{x^2\plus{}y^2}{2}} 1_{x>0}1_{y<\minus{}x} dx dy \]
which is also equal to (draw the domain $ \{ (x>0) \cap (y < \minus{}x)\}$)
\[ \frac{1}{8} (\frac{1}{2 \pi}\int_{x \in \mathbb{R}} \int_{y \in \mathbb{R}} e^{\minus{}\frac{x^2\plus{}y^2}{2}}  dx dy) \equal{} \frac{1}{8} \]

thank you alekk, i made a big mistak , P(B1>=0,B2<= 0 )=P(B1>=0,B2-B1<=-B1 )=P(B2-B1<=-B1 | B1>=0)*P(B1>=0) , P(B1>=0)=1/2 ,and the integral that u calculated was my P(B2-B1<=-B1 | B1>=0)  !! so i got 1/16 !


the equation is given by this in 3D dimension
$ \frac{x^2}{9}\plus{}\frac{y^2}{10}\plus{}\frac{z^2}{64}\equal{}\minus{}1$ and $ z\equal{}16$

The most useful thing for you to do is to create a picture (through Mathematica or some site that will graph functions).  Then you'll be able to see where your surface lies with respect to $ z\equal{}16$ and hence you can set up your integral.

[quote="Javkhaa"]$ \frac {x^2}{9} \plus{} \frac {y^2}{10} \plus{} \frac {z^2}{64} \equal{} \minus{} 1$[/quote]  What was this really supposed to mean?

maybe i wrote it wrong but that's an equation

It's an equation with no real solutions -- one doesn't usually ask for the volume of an empty set.


ARML
linear algebra
matrix
geometry
3D geometry
tetrahedron

heres a thing that will really tickle your minds

A Roman Boy was given this problem:

Take the number one hundred and one, affix to it half a dozen or if you please six, lastly put fifty. Then you will see what every good boy should be.

I know the answer. I just want see if anyone still have a room for a good common sense :D

This is just for fun dont worry :lol:

For one thing the answer is not 101 + 6 + 50 = 157

This should belong in the fun and games forum. A mod please move this.

oh really?

prove it by answering it! :P

wow thats great!!! How did you know thats the answer???

Because I know roman numerals :P.  [hide="Other reasons I figured the problem out"]1) the roman numeral for 157=CLVII doesn't make a word (that I know of).
2) The word "affix" hardly has a double meaning.  It means "append" a lot more often than it means "add," and since I write computer programs that use that command a lot, I guess I didn't think of it that way.[/hide]

This problem is a similar situation to one in Michael Crichton's "Timeline."  Can anyone find it?

You also did say it was a Roman boy :P.  Sed Romani non linguam Anglicam dixerunt!  I don't think civil is a word in Latin (although civis is).

ooohhhh ur so funny :rotfl:  :rotfl:  :rotfl:

maybe im the one who has no room for common sense :blush:

[quote="drunner2007"]Draw exactly 4 lines that pass through all 9 pts without lifting your pencil

.    .    .
.    .    .
.    .    .[/quote]

I love this one!

Hint (don't look until you've tried, really):
[hide]
a line doesnt need to start and end with one of the given points
[/hide]

[quote="drunner2007"]Draw exactly 4 lines that pass through all 9 pts without lifting your pencil

.    .    .
.    .    .
.    .    .[/quote]

[hide="hint"]
Or, you can start at a given point but extend your line past the grid![/hide]

[quote="ankur87"][quote="drunner2007"]Here is a similar problem. Don't look too hard at strict math. Let your creativity guide you.

Make the equation true

102 = 100[/quote]

[hide="my guess"]
Would you put it in Roman Numeral and move one I from the 102 side to the 100 side?

[/hide][/quote]

How about: $10^2 = 100$?  :lol:

Or better yet. $102 \neq 100$.

[quote="drunner2007"]Draw exactly 4 lines that pass through all 9 pts without lifting your pencil

.    .    .
.    .    .
.    .    .[/quote]

basically, one phrase: "Think Outside the Box!!!"

sometimes you need to use ur common sense ;)  that is...........if you still have a common sense :rotfl:  :rotfl:  :rotfl:

[quote="probability1.01"]One equals one.  Two equals two.  One hundred equals one hundred, too.

100 = 102 :D[/quote]

That's brilliant!

That is pretty brilliant. I have Cartoon Network, but I choose not to watch it....

[quote="probability1.01"]Sometimes you need to spare those who don't have cartoon network or cable for that matter.  Just tell us what you're talking about please.  :?[/quote]

[b]Agreed[/b]

im just asking which came first the egg(chicken's egg to be specific) or the chicken?

and to find a clue you will need to watch cartoon network

if you dont have cable............you could just guess it.

Lol... You remind me of me on my first forum  :P

And by the way, cartoon network is my favorite channel, and I have not seen a single clue. Unless you're talking about cow & chicken which was cancelled a long time ago.

I also vote to move this to the games and fun factory...

Sry for not moving this sooner  :blush: .  Off to fun and games this thread goes.

can anyone pls answer my questions..............pls...pls...pls..pls...pls...

[quote="cindy28"]i think i also have some trivias here in my sleeve.

find a number that when multiplied by itself the answer is 12345678987654321.

its patterned u know..... ;)[/quote]

I have a question--probably stupid:

why
ABC/XYZ=ABCABC/XYZXYZ  ?
or even repeat the pattern more times?
while A,B,C,X,Y,Z are digits   (i.e. A=1,B=3,C=6-->ABC=136)

what a problem!!!!!
 :o
[hide]is there an answer to that problem???

if there is pls tell me[/hide]

[quote="cindy28"]heres another!

in algebra, which can first the $chicken$ or the $egg$? how? why? and whom did you learn it?

[hide]if your watching cartoons youll know which came first[/hide][/quote]

[b]Tell us what ur talking about.  I don't see any clue on Cartoon Network!![/b]

[quote="zeb"]Probably everyone knows this one, but:

Use 6 matches to make four triangles.[/quote]

[hide]
build a tetrahedron :D [/hide]


integration
trigonometry
real analysis
real analysis unsolved

Compute

$\int_{0}^{\pi}\frac{xsinx}{1+cos^{2}x}$

 :D

Let $y=\pi -x$ and $I=\int_0^{\pi}\frac{x\sin x}{1+cos^2x}dx$. Then \[ I=\int_0^{\pi}\frac{(\pi -y)\sin y}{1+cos^2y}dy=\pi\cdot\int_0^{\pi}\frac{\sin y}{1+cos^2y}dy-I  \]
It results that \[ I=\frac{\pi^2}{4}.  \]

I don't think that is correct.

I agree that:

$2I = \pi \int_0^{\pi} \frac{\sin y}{1+\cos^2 y} \, dy$

Let $u = \cos y$ then $\int_0^{\pi} \frac{\sin y}{1+\cos^2 y} \, dy = \int_{-1}^1 \frac1{1+u^2} = \frac{\pi}{2}$

Then $2I = \frac{\pi^2}{2} \rightarrow I = \frac{\pi^2}{4}$


We have prime numbers $p,q$ such that:

$q \mid p^2+1$ and $p \mid q^2-1$.

Prove that the number $p+q+1$ is composite.

$p|q^2-1$ is the same as $p|(q-1)(q+1)$

Because $p$ is prime it has to divide $q-1$ or $q+1$. If p divides $q+1$ the statement is obviously true, because $p+q+1$ is divisible by $p$.
So, suppose that $p|q-1$. It means that there is integer $k$ such that $pk = q-1$ so $p=\frac{q-1}{k}$.
It's obvious that the equation $a^2 + 1 \equiv 0 \mod{q}$ has at most two solutions because $q$ is prime. Now, we write: $p^2 + 1 \equiv 0$ and we transform it into: $\frac{(q-1)^2}{k} + 1 \equiv 0$ (we can do this because $k$ is not $0 \mod{q}$) which is equivalent to:
$k^2 + 1 \equiv 0$. From this we deduce that $k+p \equiv 0 \mod{q}$ or $p \equiv k \mod{q}$ and because $p, k$ are smaller than $q$ $p+k=q$ or $p=k$. So:
$p(q-p) = q-1$ or $q = p+1$, which means that $p=2$ and $q=3$ which doesn't satisfy the first condition (but $p+q+1$ is composite anyway ; )
or $p^2 = q-1$, when therefore $p=2$ and so $q=5$; then  $p+ q+ 1 = 8$ is obviously composite.

Hmm, it looks a little strange to me...

Another way:

$p \, | \, q-1$, so $q=kp+1$ for $k \in \mathbb{N}$.

$q \, | \, p^2+1$, so $q \, | \, p^2+1-(kp+1)$, so $q \, | \, p(p-k)$, so $q \, | \, p-k$.

If $p>k$, then $p-k\ge kp+1$, which is equivalent to $(1-k)(p+1) \ge 2$, contradiction.

If $p<k$, then $k-p \ge kp+1$, which is equivalent to $(k+1)(1-p) \ge 2$, contradiction.

So $p=k$, so $q=p^2+1$. Only solution of this equation is $(p,q)=(2,5)$ then $p+q+1=8$.