Do you guys know when the official AMC 10/12 B results are coming out? I know its after the AMC people grade it, but whats a rough estimate of the release day? A month or so?

It depends on how quickly whoever ran it at your school sends back the answer sheets. They are graded on a rolling basis and come back whenever the AMC is done with them. For answers sent immediately after the AMC A, results have been coming back throughout the past week, so roughly 12 days.

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

Steve Dunbar
MAA Director, American Mathematics Competition


The answer is -16 because there's no parenthesis, so you take the sign after you do the square (if there were ()'s around the -4, it would've been 16).

[quote="math92"]THis problem was in my math packet
$-4^2$
i don't know whether the answer is
[hide]16 or -16[/hide][/quote]

It's -16 because you can look that it this way:

$-1*4^2$

You have to to the exponent before multiplication so it's $-16$

[quote="math92"][quote="ifai"]$-4^2=-16$ and $(-4)^2=16$[/quote]
is it 16 or -16?[/quote]

-16, that was the problem, right?

it was -16 but ifai gave me two answers.

[hide]ok when you have $-4^2$ the square sign only applies to the $4$, so you pretend the negative sign is not there, square it, and then add the negative sign in front of whatever you get $-(4)(4) = -16$

but if it's $(-4)^2$, then the square applies to the negative sign too, so it's $(-4)(-4)$, two negative gives you a positive, and $4*4 = 16$, so the answer is $16$[/hide]


where x,y,z >=0
          prove that  :
 x/y + y/z+z/x >=  (x+y)/(y+z) +(y+z)/(x+y) +1 .

[quote="tranthanhnam"]where x,y,z >=0
          prove that  :
 $\frac xy + \frac yz+\frac zx \geq  \frac{x+y}{(y+z} +\frac{y+z}{x+y} $+1 .[/quote]

[quote="tranthanhnam"][quote="tranthanhnam"]where x,y,z >=0
          prove that  :
 $\frac{x}{y} + \frac{y}{z}+\frac{z}{x} \geq  \frac{x+y}{(y+z} +\frac{y+z}{x+y} $+1 .[/quote][/quote]


quadratics
modular arithmetic
number theory proposed
number theory

Find all nonnegative integers $a<2007$ so that the congruence $x^{2}+a\equiv 0\pmod{2007}$ has exactly two nonnegative integer solutions $x<2007$.

[quote="Rust"]... or $(\frac{-a}{223})=1, \ 9|a$.[/quote] Well, that's not true since $x^{2}\equiv 0\pmod 9$ has three solutions modulo $9$ (ie $0,3,6$) so $x^{2}+a\equiv 0 \pmod{2007}$ would have a number of solutions divisible by $3$ if $9\mid a$...and we want the congruence to have exactly two solutions :P


geometry
incenter
angle bisector
perpendicular bisector
geometry unsolved

Triangle $ ABC$ has incentre $ I$ and the incircle touches $ BC, CA$ at $ D, E$ respectively. Let $ BI$ meet $ DE$ at $ G$. Show that $ AG$ is perpendicular to $ BG$.

I was also pretty sure that the innitial question was wrong....
Alternative.
Let $ ED$ intersect $ AB$ at $ K$.
The $ (A,L,B,K)$ (note that $ AD,CL,BE$ are concurent at the georgonne point) is a harmonic 4-tumple.($ L$ is the intersection of the incircle with the $ AB$).
By well known lemma in order to prove that $ AG$ is perpendicular to $ BG$ it suffices to prove that $ GB$ is bisector of $ \angle LGK$,which can be proved by an easy angle chasing :wink:

$ AG$ perpendicular to $ BG$.(not $ BC$)

It is $ AG$ perpendicular to $ BG$
sorry for the error

is there any alternate solution possible?

Let AG intersect BC at H.
Then $ \frac {AB}{BH} \equal{} \frac {AG}{GH}$ by the angle bisector theorem.
By Menelaus, $ \frac {DC}{DH} \cdot \frac {HG}{GA} \cdot \frac {AE}{EC} \equal{} 1$. But $ DC \equal{} EC$, so $ \frac {AG}{GH} \equal{} \frac {AE}{HD}$.
So $ \frac {AB}{BH} \equal{} \frac {AE}{HD} \equal{} \frac {AB \minus{} AE}{BH \minus{} HD} \equal{} \frac {AB \minus{} AE}{BD}\equal{}1$, since $ AB\equal{}AF\plus{}FB\equal{}AE\plus{}BD \implies AB\minus{}AE\equal{}BD$. So $ AB \equal{} BH$, so $ BG \perp AH$.

[quote="saif"]we need to prove that $ IGEA$ is cyclic
it's easy because
$ \angle EGI\equal{}\angle B/2 \plus{} \angle EDB\equal{}\angle B/2 \plus{}\angle B/2 \plus{}\angle A/2\plus{} \angle C\equal{}180^{o}\minus{}\angle A/2$
so it's done :D[/quote]

also,
$\angle EGI\equal{}\angle B/2 \plus{} \angle EDB\equal{}\angle IDB\plus{}\angle IDE $
but $\angle IDB\equal{}90^{o} $ and $\angle IDE\equal{}\angle C/2 $
thus $ \angle EGI\equal{}\angle B/2 \plus{}90^{o}\plus{}\angle C/2 $
which is equal to  $180^{o}\minus{}\angle A/2.$
Done.
 :wink:

Dear Mathlinkers,
old problem but always fruitfull...
See,
http://perso.orange.fr/jl.ayme  An unlikely concurrence, revisited and generalized  vol. 4 p. 5
Sincerely
Jean-Louis

Let incircle touch $AB$ at $F$.just prove $FIGE$ is cyclic.$BG$ is perpendicular bisector of $FD$,so $FG=GD$,$FI=ID$ and $IG$ is common.so $\angle IFG=\angle IDG$,also $\angle IDG=\angle IEG$,hence $FIGE$ is cyclic.

we need to prove AGEI IS cyclic
IAB+ABI=AIG
A/2+B/2=(180-C)/2=DEC=AEG
THEREFORE AIG=AEG
AGEI IS CYCLIC
SO WE ARE DONE

[hide=goal is to prove AIGE cyclic]so we join $AI$  and mark $BI \cap AC=X$, we notice that $\angle{IED}=\frac{C}{2}$ and since $\triangle{IEX}$ is right angled we have $\angle{EIX}=90^{\circ}-(\frac{B}{2}+C)$\\
in $\triangle{EGI}\\
\angle{IGE}=180-\frac{A}{2}$\\
which proves $AIGE$ to be cyclic. now since $IE \perp AC$ we have $\angle{AGI}=90^{\circ}  \implies AG \perp BG$ $\blacksquare$[/hide]


It is well known that normed space $X$ is banach iff every absolutely convergent series in $X$ is
also convergent (in norm). But is the following true?

$X$ is banach iff
every series $\sum f_{j}$ converges (in norm), where $\parallel f_{j}\parallel \leq 2^{-j}$.

True.

Let $\sum a_{j}$ be a positive series (series with positive terms) converging to one. If every series $\sum f_{j}$ (from $X$) converges in norm whenever $\parallel f_{j}\parallel \leq a_{j}$ for all $j$, we will call the series $\sum a_{j}$ a [b]good series[/b] for $X$. Otherwise, call it a bad series.

Let $0=n_{0}< n_{1}< n_{2}< \cdots$ be a sequence of natural numbers and define $A_{k}= \sum\{a_{j}| n_{k-1}< j \le n_{k}\}$ for each $k=1,2,3,\cdots$. We say $\sum A_{j}$ is a [b]group series[/b] of $\sum a_{j}$. We can prove that the series $\sum A_{j}$ is a good series if and only if $\sum a_{j}$ is a good series.

Finally, given two positive series $\sum a_{j}$ and $\sum b_{j}$ both converging to one, there exists a positive series $\sum c_{j}$ converging to one such that $\sum a_{j}$ and $\sum b_{j}$ are both group series of $\sum c_{j}$. Therefore $\sum a_{j}$ is a good series if and only if $\sum b_{j}$ is a good series.

[quote="redkimchi"]
Finally, given two positive series $\sum a_{j}$ and $\sum b_{j}$ both converging to one, there exists a positive series $\sum c_{j}$ converging to one such that $\sum a_{j}$ and $\sum b_{j}$ are both group series of $\sum c_{j}$. [/quote]

How to find $\sum c_{j}$ such that $\sum a_{j}$ and $\sum b_{j}$ are both group series of $\sum c_{j}$?

[quote="kalazealot"]
How to find $\sum c_{j}$ such that $\sum a_{j}$ and $\sum b_{j}$ are both group series of $\sum c_{j}$?[/quote]

Let $(A_{j})$ be the sequence of partial sums of $\sum a_{j}$, i.e. $A_{n}= a_{1}+a_{2}+\cdots+a_{n}$. Let $(B_{j})$ be the sequence of partial sums of $\sum b_{j}$. Then $(A_{j})$ and $(B_{j})$ are strictly increasing sequences of positive numbers converging to one. There is strictly increasing sequence $(C_{j})$ of positive numbers converging to one such that $(A_{j})$ and $(B_{j})$ are both subsequences of $(C_{j})$. Define $c_{j}= C_{j}-C_{j-1}$ for each $j$.


blogs
number theory unsolved
number theory

Let $ n \equiv 1 \ mod \ 4$ be a non-square positive integer such that $ n \equal{} a^2 \plus{} 4b^2$ for some integers $ a$ and $ b$.
Assume that $ x^2 \minus{} ny^2 \equal{} \minus{}1$ has integer solution $ (x, \ y) \equal{} (t_1, \ t_2)$. Show that $ n \equal{} rs$ for some $ r, \ s$ where $ rx^2 \minus{} sy^2 \equal{} a$ has an integer solution. What do you deduce when a prime greater than 2 is expressed as a sum of two squares?

I have cited an article in my blog which gives you the answer.:

[url=http://blog.csdn.net/northwolves/archive/2008/11/24/3364010.aspx]Which Numbers are the Sum of Two Squares?[/url][/url]


myFTP.exe crashes when browsing ML/AoPS sporadically (Chrome). While it doesn't seem to affect the computer's performance, it gets annoying to have popups pop up every 15-20 minutes.

Do you really want myFTP.exe installed? It's generally considered malware/virus.

http://vic.zonelabs.com/tmpl/body/CA/virusDetails.jsp?VId=46504


Hey, Moubi, why have put it in the unsolved problems section? You know that it is solved in 'Supermath'....

Pierre.

Supermath
Hypermath
Megamath
are good books in french for preparing IMO.

http://www.amazon.fr/exec/obidos/search-handle-form/171-2165527-1494653

[quote="Moubinool"]Supermath
Hypermath
Megamath
are good books in french for preparing IMO.

[/quote]
wrote by Pierre Bornsztein
on mathlinks as  pbornsztein


If $a_i>0$ for $i=1,2, ... ,n$ then prove that

$\prod^{n}_{k=1}(1+\frac{1}{a_k})(1+\frac{a_k}{k^2})\geq(n+1)^2$

Enjoy it!! :)  :)

Besides I'm thinking of creating a topic (in pre-olympiad section)
which will put up inequalities , not very easy but also not too hard,
for people who are interested in ineqs. Everyone will be able to propose
one, then it will be discussed and opinions will be exchanged.
What do you think? :)

My solution.
$(1+\frac{1}{a_k})(1+\frac{a_k}{k^2})\geq (1+\frac{1}{k})^2$ by Cauchy. Next multiply these and we are done

That's it silouan!
Well done!! :)

What about the question????


group theory
abstract algebra
superior algebra
superior algebra unsolved

1)Given a group $ G$ with $ H<G$ can there be an isomophism:
$ \phi: G\mapsto H$.

If #1 is false then the following will be false but if not what about #2?

2)Given a field $ E$ with $ F<E$ can there be an isomorphism:
$ \phi: E\mapsto F$

If #2 is false then the following will be false but if not what about #3?

3)Given a non-closed field $ F$ with closure $ \bar F$ can there be an isomorphism:
$ \phi: F\mapsto \bar F$

3) of course not, the property "algebraically closed" is preserved by field isomorphisms.
1) yes, consider infinite copies of something, e.g. $ \bigoplus_{\mathbb{N}}\mathbb{Q}\cong \bigoplus_{2 \mathbb{N}}\mathbb{Q}\subset \bigoplus_{\mathbb{N}}\mathbb{Q}$.
2) yes, I quote an example due to [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=19822]Lima[/url].

consider the field of rational numbers $ \mathbb{Q}$, an infinite countable set $ X$ of variables and another variable $ y$. then we have

$ \overline{\mathbb{Q}(X)}\subseteq \overline{\mathbb{Q}(X)}(y) \subset \overline{\overline{\mathbb{Q}(X)}(y)}\cong \overline{\mathbb{Q}(X)(y)}\cong \overline{\mathbb{Q}(X)}$

[quote="-oo-"]3) of course not, the property "algebraically closed" is preserved by field isomorphisms.[/quote]
That what I thought. Can you provide the details.

[quote]
1) yes, consider infinite copies of something, e.g. $ \bigoplus_{\mathbb{N}}\mathbb{Q}\cong \bigoplus_{2 \mathbb{N}}\mathbb{Q}\subset \bigoplus_{\mathbb{N}}\mathbb{Q}$.
[/quote]
After I posted this and left I realized an easier example. Consider $ \left< \mathbb{Z},+\right>$ and choose $ 2\mathbb{Z}= \{2n|n\in \mathbb{Z}\}$.

EDIT: Ignore my request to post a proof to #3, its easy. I actually was thinking of something else. Maybe I post it.


Sasha and Trudy are playing tic-tac-toe. Sasha agrees to give
Trudy two pieces of candy for each game that Sasha loses. Trudy agrees to give Sasha three pieces of candy for each game that Sasha wins. After playing thirty games, Sasha has the same number of pieces of candy with which she started. If there were no ties, how many games did Sasha win?

[hide="Don't know if it's the fastest way, but..."]$ sasha \equal{} o \plus{} 3w \minus{}2l$

and

$ w\plus{}l \equal{} 30$

w = wins

l = losses

o = original # of candy

So we have that

$ o \equal{} o \plus{} 3w \minus{}2l$

$ 3w\equal{}2l$

$ l \equal{} 3w/2$

Plug back into $ w\plus{}l \equal{} 30$

$ w \plus{} 3w/2 \equal{} 30$

$ 5w/2 \equal{} 30$

$ \boxed{w\equal{}12}$[/hide]

Slightly shorter:[hide] Since Sasha gains 3 pieces when she wins, and she loses 2 pieces when she loses, Sasha must win 2/3 as many times as she loses; in other words, every time she wins 2 games, she loses 3. (She'll gain 6 candies and lose 6 candies, giving her no extra candy.) Then, she wins 2/5 of her total games, and for 30 games, we easily compute the answer to be 12 games.[/hide]


...went out today by email, and the postal mailed copy with all information will go out in tomorrow's mail, meaning it should arrive no later than Monday, May 18 depending on speed of mail delivery.

Invitation levels are exactly as detailed in my other post.

Steve Dunbar
MAA Director, American Mathematics Competitions

Is there any way for a Canadian to go to the MOP?? I just want to be there. I understand that MOSP is the preparation for Americans.. but if i scored past the cutoff , can i be a standby student?? or any other ways to BE THERE?


[quote="Round 1, Problem 3"]Find all monotonic functions $ f: \mathbb{R}\to\mathbb{R}$ such that $ f(x) \plus{} 2x \equal{} f(f(x))$.[/quote]

[quote="Differ"]$ g(x) \equal{} \frac {f(x) \minus{} x}{2}$
The domain of $ g(x)$ is $ \mathbb{R}$.
$ g(f(x)) \equal{} x$
So $ g(x)$ is the inverse of $ f(x)$. Thus, $ f(x)$ is surjective and as it is monotonic, it is continuous.[/quote]

I don't see how $ g(f(x)) \equal{} x$ implies that $ g(x)$ is the inverse, or that $ f(x)$ is surjective (don't you need to show that $ f(g(x)) \equal{} x$?), unless I'm being stupid...and if it doesn't, how would one prove that $ f(x)$ is surjective?

I was hoping someone would answer this, because I am confused as well

How could $ f(g(x)) \equal{} x$ imply that the function is surjective?
For example. $ f(x)\equal{}e^x$ has an inverse and is continuous and monotonic, but not surjective..right? unless you specify the codomain as $ (0,\infty)$

I wasn't even sure why the proof required continuity, I thought it required it to be surjective because if $ \max \left(\frac{f(x)}{x}\right) \equal{} \frac{f(a)}{a}$ for $ a\in \mathbb{R}$
for $ \max \left(\frac{f(x)}{x}\right) \equal{} \max \left(\frac{f(f(x))}{f(x)}\right)$ we would only require $ f(x)\equal{}a$ to have a solution
(this is assuming the maximum even exists..)


number theory
greatest common divisor
modular arithmetic
number theory unsolved

Given $ n>1$ an odd number. Let
\[ S\equal{}\{x\in\mathbb{Z}\mid\text{gcd}(x,n)\equal{}\text{gcd}(x\plus{}1,n)\equal{}1,1\leq x\leq n\}\]
Show that if $ P$ is the product of all elements in $ S$, then $ n\mid P\minus{}1$.

At least for $ n\equal{}p$ prime you can do it, since then $ S\equal{}\{1,2,\ldots,p\minus{}2\}$ and so $ P\equal{}(p\minus{}2)!$.
According to Wilson's theorem $ (p\minus{}1)! \equiv \minus{}1 \pmod{p}$. But $ (p\minus{}1)! \equal{} (p\minus{}1)\cdot ((p\minus{}2)!) \equiv \minus{}(p\minus{}2)! \pmod{p}$ and you are done. Try to extend this.

just show that $ P \equiv [(p \minus{} 2)!]^a \mod n$ and then use wilson like in mavropnevma's post

[quote="wauwau"]just show that $ P \equiv [(p \minus{} 1)!]^a \mod n$ and then use wilson like in mavropnevma's post[/quote]what is $ a$? Can you write full solution please?


percent
geometry
3D geometry
sphere
ratio

[quote="MCrawford"]The radius of a circle is increased by 100%.  By what percent is the area of the circle increased?[/quote]

[hide="Solution"]This means that the radius of the circle is doubled, so the area is quadrupled. 400% of the area minus 100% of the area is 300% of the area.[/hide]

[color=indigo][hide]
The area of the circle gets 4 times as big, so it is increased 300%.[/color][/hide]

[hide]
cirlce gets 4x bigger. 300% increase.[/hide]

[hide="answer"]2*2 = 4
please try to understand my stupidty of showing the answer, 4

(if that makes sense to you)

so, it's 400%
but hold on,
you have to subtract the original 100%

which is 300%[/hide]

[quote="MCrawford"]The radius of a circle is increased by 100%.  By what percent is the area of the circle increased?[/quote]

[hide="Answer"]
300%
[/hide]

[hide="click here if you wish to do so..."]
300%[/hide]

[hide]300 percent--1=:pi:,2=4:pi:, so it is 4 times the original, increasing 300 percent.[/hide]

yeah, when you double the circle, it increases 4 times the amount

for a sphere, when you double that, it increases 8 times. remember that

[hide]Because the radius doubled, you square the ratio to find that the area increased by 4 times. Subtrace 400% by 100% and you get 300%[/hide]


trigonometry
function
limit
algebra unsolved
algebra

[hide="hint for 2"]We can rewrite the cosine as the real function of an imaginary power of $e$.[/hide] This might lead to the correct solution.

Are you sure this is in the correct forum?

[quote="Rust"]a) $a_{n}$ increase[/quote]
why? how do you have it?


geometry
circumcircle
analytic geometry
parameterization
power of a point
radical axis
geometry unsolved

Assume that a triangle $ABC$ is not equilateral, and let $0<t<\frac12$ be a real number. Let points $A_1$ and $A_2$ be given on the side $AB$, points $B_1$ and $B_2$ on the side $BC$, and points $C_1$ and $C_2$ on the side $CA$ such that
$\frac{AA_1}{AB}=\frac{A_2B}{AB}=\frac{BB_1}{BC}=\frac{B_2C}{BC}=\frac{CC_1}{CA}=\frac{C_2A}{CA}=t$.
Prove that the radical axis of the circumcircles of the triangles $A_1B_1C_1$ and $A_2B_2C_2$ depends only of the triangle $ABC$, but does not depend on the value of $t$.

For convenience redefine the points such that $ A_1,A_2$ lie on $ BC,$ $ B_1,B_2$ lie on $ CA$ and $ C_1,C_2$ lie on $ AB$ respectively. Then $ \omega_1 \equiv \odot(A_1B_1C_1)$ and $ \omega_2 \equiv \odot (A_2B_2C_2).$ Using  barycentric coordinates with respect to $ \triangle ABC,$ we get

$ A_1 \ (0: 1 \minus{} t: t) \ , \ B_1 \ (t: 0: 1 \minus{} t) \ , \ C_1 \ (1 \minus{} t: t: 0)$

$A_2 \ (0: t: 1 \minus{} t) \ , \ B_2 \ (1 \minus{} t: 0: t) \ , \ C_2 \ (t: 1 \minus{} t: 0)$

Plugging the coordinates of $ A_1,B_1,C_1$ and $ A_2,B_2,C_2,$ respectively into the general equation of the circle $ a^2yz \plus{} b^2zx \plus{} c^2xy \minus{} (x \plus{} y \plus{} z)(\alpha_ix \plus{} \beta_iy \plus{} \gamma_iz) \equal{} 0$ yields the parameters of $ \omega_1,\omega_2$ as

$\alpha_1 \equal{} \frac {b^2t^3(1 \minus{} t) \plus{} c^2t(1 \minus{} t)^3 \minus{} a^2t^2(1 \minus{} t)^2}{t^3 \plus{} (1 \minus{} t)^3}$

$\alpha_2 \equal{} \frac {b^2t(1 \minus{} t)^3 \plus{} c^2t^3(1 \minus{} t) \minus{} a^2t^2(1 \minus{} t)^2}{t^3 \plus{} (1 \minus{} t)^3}$

$\beta_1 \equal{} \frac {c^2t^3(1 \minus{} t) \plus{} a^2t(1 \minus{} t)^3 \minus{} b^2t^2(1 \minus{} t)^2}{t^3 \plus{} (1 \minus{} t)^3}$

$\beta_2 \equal{} \frac {c^2t(1 \minus{} t)^3 \plus{} a^2t^3(1 \minus{} t) \minus{} b^2t^2(1 \minus{} t)^2}{t^3 \plus{} (1 \minus{} t)^3}$

$ \gamma_1 \equal{} \frac {a^2t^3(1 \minus{} t) \plus{} b^2t(1 \minus{} t)^3 \minus{} c^2t^2(1 \minus{} t)^2}{t^3 \plus{} (1 \minus{} t)^3}$

$ \gamma_2 \equal{} \frac {a^2t(1 \minus{} t)^3 \plus{} b^2t^3(1 \minus{} t) \minus{} c^2t^2(1 \minus{} t)^2}{t^3 \plus{} (1 \minus{} t)^3}$

$ \alpha_1 \minus{} \alpha_2 \ , \beta_1 \minus{} \beta_2 ,  \gamma_1 \minus{} \gamma_2$ have common factor $ \frac {t^3(1 \minus{} t) \minus{} t(1 \minus{} t)^3}{t^3 \plus{} (1 \minus{} t)^3}$

Thus, equation of the radical axis $ \tau$ of $ \omega_1,\omega_2$ is indeed independent of $t.$ Namely,

$ \tau \equiv (b^2 \minus{} c^2)x \plus{} (c^2 \minus{} a^2)y \plus{} (a^2 \minus{} b^2)z \equal{} 0$

Which is the trilinear polar of the Steiner's point $ X_{99}$ of $ \triangle ABC.$ Therefore, $ \tau$ passes through the centroid $ G(1: 1: 1)$ and symmedian point $ K(a^2: b^2: c^2)$ of $ \triangle ABC.$

I have just discovered the source of this problem:

[url=http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=3133376&fulltextType=RA&fileId=S0013091500031175]R. Tucker, [i]Geometrical Note[/i], Proceedings of the Edinburgh Mathematical Society / Volume 11 / pp 57 - 60[/url]. (With public-readable PDF.)

According to some of my notes from 4 years ago, the midpoint between the symmedian points of triangles $A_1B_1C_1$ and $A_2B_2C_2$ also seems to lie on the line joining the centroid and the symmedian point of triangle $ABC$. I have never checked or proved this, but maybe somebody else can come up with an idea (or a counterexample) now. (He/she would have my blessing to publish it then; there is nothing I have done for this conjecture except of conjecturing it.)

[quote="darij grinberg"]According to some of my notes from 4 years ago, the midpoint between the symmedian points of triangles $A_1B_1C_1$ and $A_2B_2C_2$ also seems to lie on the line joining the centroid and the symmedian point of triangle $ABC.$[/quote]
This conjecture is true, unfortunately I don't have a synthetic proof. Again, using the barycentric coordinates of $A_1,B_1,C_1$ WRT $\triangle ABC,$ we get that

$|B_1C_1|^2=S_A(1-2t)^2+S_Bt^2+S_C(t-1)^2$

$|C_1A_1|^2=S_A(t-1)^2+S_B(1-2t)^2+S_Ct^2$

$|A_1B_1|^2=S_At^2+S_B(t-1)^2+S_C(1-2t)^2$

Thus, barycentric coordinates $(x_1:y_1:z_1)$ of the symmedian point $K_1$ of $\triangle A_1B_1C_1$ WRT $\triangle ABC$ are given by

$K_1 \equiv |B_1C_1|^2 \cdot (0:1-t:t)+|C_1A_1|^2 \cdot (t:0:1-t)+|A_1B_1|^2 \cdot (1-t:t:0)$

$x_1=S_A(1-t)t+S_B(3t^3-t^2-2t+1)+S_C(-3t^3+8t^2-5t+1)$

$y_1=S_A(-3t^3+8t^2-5t+1)+S_B(1-t)t+S_C(3t^3-t^2-2t+1)$

$z_1=S_A(3t^3-t^2-2t+1)+S_B(-3t^3+8t^2-5t+1)+S_C(1-t)t$

Similarly, barycentric coordinates $(x_2:y_2:z_2)$ of the symmedian point $K_2$ of $\triangle A_2B_2C_2$ WRT $\triangle ABC$ are given by

$x_2=S_A(1-t)t+S_B(-3t^3+8t^2-5t+1)+S_C(3t^3-t^2-2t+1)$

$y_2=S_A(3t^3-t^2-2t+1)+S_B(1-t)t+S_C(-3t^3+8t^2-5t+1)$

$z_2=S_A(-3t^3+8t^2-5t+1)+S_B(3t^3-t^2-2t+1)+S_C(1-t)t$

Clearly, $x_1+y_1+z_1=x_2+y_2+z_2,$ thus coordinates $(\overline{x}:\overline{y}:\overline{z})$ of midpoint $\overline{K}$ of $K_1K_2$ are found by just adding the homogeneous coordinates of $K_1,K_2.$

$\overline{x}=2S_A(1-t)t+a^2(7t^2-7t+2)$

$\overline{y}=2S_B(1-t)t+b^2(7t^2-7t+2)$

$\overline{z}=2S_C(1-t)t+c^2(7t^2-7t+2)$

Eliminating $t$ from the coordinates of $\overline{K},$ we get that the locus of $\overline{K}$ is precisely the line through $X_2$ and $X_6,$ namely $(b^2-c^2)x+(c^2-a^2)y+(a^2-b^2)z=0.$


Show that any set of $ n$ integers has some subset whose sum is divisible by $ n$.

consider Let a_i be the set, and let $ b_1 \equal{} a_1$ , $ b_2 \equal{} a_1 \plus{} a_2$ , $ b_3\equal{}a_1\plus{}a_2\plus{}a_3$ etc. all the way up to $ b_n \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n$

Now, if b_i is congruent to b_j modulo n with i>j then b_i-b_j is divisible by n. Therefore, assume all b_i are different modulo n. By pigeonhole, if no b_i is 0, then 2 are congruent modulo n, so one must be congruent to 0 modulo n, so that b_i is divisible by n.

Q.E.D.

[quote="rofler"]Let a_i be the set, and let $ b_1 \equal{} a_1 b_2 \equal{} a_1 \plus{} a_2$ etc. all the way up to $ b_n \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n$
Q.E.D.[/quote]

How are $ b_i$'s defined?

Sry, latex and spaces do not go well together. I made it clearer now.


What is the geometric mean of $6\frac{1}{4}$ and 100?

well.. i forgot what the difference was...

the arithmetic mean of 2 numbers is (a+b)/2

is the geometric mean axb/2?

$6 \frac{1}{4}$ times a number r equals the geometric mean, and this geometic mean times r again = 100.

[quote="236factorial"][quote="B4k4ka"]so its 6.25 x r^2 = 100?[/quote]

YES!!  :D[/quote]
yay! now ON with the problem!!!

[hide="could it be?"]
$100=6xr^2
100/6=r^2
r= root 100/6
r = root 600/36
r= 1/6 root 600
r = 10/6 root 6$
[/hide]

Uhhhh, it was 6.25 and 100, not 6 and 100  :roll:

You found the multiplier, but not the geometric mean (multiply 6 1/4 by r).

[quote="buzzer11"][hide]6.25 X 100 = 625. sqrt 625 = 25[/hide][/quote]

That's the way I like  :D

Arithmetic mean for n positive real numbers $a_1, a_2, ..., a_n$
\[\frac{a_1+a_2+...+a_n}{n}\]

Geometric mean for n positive real numbers $a_1, a_2, ..., a_n$
\[\sqrt[n]{a_1a_2...a_n}\]


algebra
polynomial
Rational Root Theorem

Prove $\sqrt{2}$ is irrational

i know this is easy, but i think i have another way that i have not seen before

[hide]

we all know euclid's proof by contradiction, but here is another one

consider $f(x)=x^2-2$, by the rational root theorem, if $r$ is a root, then it is in $r \in \{\pm1,\pm2\}$

but we know that none of those are roots, so none of the roots are rational, thus $\sqrt{2}$ is irrational

what do you think? is this correct, or is there problem with this? it seems rigirous to me...

[/hide]

The rational root theorem should be perfectly rigorous, but if you're not sure, you could prove it first. (Not that much work, just multiply by the denominator enough times.)

This probably wouldn't be very efficient for $\sqrt{2}$, but there are some cases where there might not be another possibility. Like this...

$\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}$


1.Line HO cuts the circle at A and B with A near H
2.Bissector AH cuts the circle at C and D
3.Triangle BCD is the answer


prove the sum of the squares of two consecutive fibbonaccis is a nother one and find out which one.
can u generalize even furhter?

There are a LOT of various identities with Fibonacci numbers.

And a journal (The Fibonacci Quaterly) is entirely dedicated to this sequence and relatives.

Pierre.

Is it true?  :? There is such journal (?!)

Have a look here :
http://www.engineering.sdstate.edu/~fib/

It deals only with the Fibonacci sequence or, by extansion, to Lucas' sequences. There are articles and theorems, some quite deep. And a problem corner.
I don't know how many relations do there exist involving Fibonacci numbers or Lucas' numbers or both or...

Pierre.


In a factory, there are 10,15, and 20 jobs of types I, II, and III, respectively, waiting to be processed on a machine (e.g. these may be physically different parts, or orders for one type of product for different customers). The machine operator randomly selects parts one at a time for processing (i.e. each part is equally likely to be selected at any time). What is the probability that the first four jobs processed are of types II, II, III, II and in this exact order?

Further problem:

The probability from above question can be written as $\frac {j}{k}$ where $(j,k) = 1$. Find $j+k$

Is the order  important in this question ? :?

um, i got 65/4257...here's how i did it:

for the first part, you have 15/45 chance of picking a part II, for the 2nd, you have 14/44 (since it's without replacement), etc. You get therefore:

(15/45)(14/44)(20/43)(13/42)=65/4257

thus, j+k=4322.

Cheers!

How is "Easy" a source?  And why are most of the questions you post, silverfalcon, classified as easy by you?

Oops. :blush:

That was typo, joml88.

Maybe "easy" standard is hard to do. I guessed I just said it on my level or general level.

Not sure how level it is but I'm trying to not post anything too difficult for sure.

i think most high schoolers shouldve learned about this kinda probability. But, Silverfalcon, should probably say easily solvable instead of easy next time.  ;)

Can't say there's much of a difference between something that's easy and something easily solvable in my book :D


[quote="marko avila"]hola a todos, oigan ¿será que ustedes tenga este libro: elementary number theory? me dicen que es muy bueno. y no lo he podido conseguir. ¿alguien de casualidad tiene un pdf de este libro? bueno si alguien lo tiene le agradecería mucho si me ayuda a conseguirlo. muchas gracias[/quote]

Bueno Marco, un autor o algun dato para acotar la busqueda seria bueno. GIL me devuelve mas de 350 libros con ese titulo. Pedir el libro "Elementary Number Theory" es tanto como pedir el libro titulado "Linear Algebra", es un topico, por ende podras encontrar cualquier cantidad de libros titulados asi.

Ahora, pues no se cual sera el nivel al que te encontres, sin embargo a mi me han recomendado el libro de Teoria de Numeros por Carlos Ivorra Castillo. No lo he ojeado en lo absoluto, pero quien me lo recomendo dice que pueden extraersele cosas del tipo de olimpiadas. Se puede encontrar en la pagina del autor, solo googlea el nombre. Eso si, puede ser un poco teorico.

Tambien, por regla de la universidad, no puedo descargar nada que este bajo derechos de autor, y como ha habido un gran problema con las redes P2P, quien las use aqui se expone a una auditoria del departamento de computo, y como a nadie le gustan las auditorias, prefiero no usarlas, pero si el archivo que queres se puede consegir, se consegira mediante eMule o Gnutella, podes probar.

Finalmente, vivis en Mexico, no deberia haber demaciado problema para comprarlo por Internet en Amazon (una vez que sepas bien cual es el que queres) y que te lo lleven a la puerta de la casa por un precio bastante razonable!

Autores:Gareth A. Jones y J. Mary Jones. Gracias por tu ayuda.

djimenez: Encontré el libro que me sugeriste. Está muy bueno¡¡ El único problema es que se mete con un tema que he intentado conocer mejor; incluso puse un post en la sección de number theory buscando información sobre este tema. El tema es 'gaussian integers'.  De  casualidad tu o alguien tiene alguna referencia sobre este tema? Estaríamos hablando de algo que tenga desde lo más basico hasta un nivel 'intermedio' obueno un nivel en el que tenga la informacion suficiente que podría necesitar para la olimpíada. Lo que puedan darme será muy bien recibido. muchas gracias.

marko encontraste el libro online o lo tienes en formato .pdf?

Bueno,

El libro, por las criticas que tiene online, parece ser un buen libro, pero no esta en la biblioteca del Tech como para darle un vistazo. En Amazon cuesta 35 dolares, se que el shipping es gratis en los Estados, pero no se cuanto sera hasta tu ciudad en Mexico. Lo que podrias hacer es que, si alguien va a venir a los Estados, pues que lo compre aqui y pida el envio rapido. Usualmente con ese lo tenes al dia siguiente o a los dos dias en tu casa!

Los engeros gausianos son basicamente ceros de polinomios monicos en $\mathbb Z[x]$. No se que querras saber de ellos, pero hay toneladas de informacion sobre ellos colgada en la web.

manuel: lo encontré en pdf. solo sigue las instrucciones de djimenez y lo encientras muy facil.  :D


#8

There are 12 distinct empty boxes labelled with the numbers 1, 2, 3, ..., 12 respectively. Find the number of ways to distribute 8 identical balls into the 12 distinct boxes so that the total number of balls in boxes 1, 2, and 3 is even and the total number of balls in boxes 4, 5, and 6 is odd.

The official solution to the above question is as follows:

One easily sees that the number of ways is the coefficient of $x^{8}$ in the expansion (even part of $(1-x)^{-3}$) $\times$ (odd part of $(1-x)^{-3}$) $\times (1-x)^{-6}$

\[=[\frac{(1-x)^{-3}+(1+x)^{-3}}{2}] \times [\frac{(1-x)^{-3}-(1+x)^{-3}}{2}] \times (1-x)^{-6}\]

\[=\frac{1}{4}[(1-x)^{-12}-(1-x^{2})^{-6}]\]

Therefore, number of ways $=\frac{1}{4}[{{8+11}\choose{8}}-{{4+5}\choose{4}}] = \frac{1}{4}[{{19}\choose{8}}-{{9}\choose{4}}] = 18864.$

The problem, however, is that i don't "see it as easily as one sees". So could anyone explain how the solution begins by connecting the problem with the coefficients of some negative power of a binomial?

You can put no balls into the first box ($x^{0}$), one ball into the first box ($x^{1}$), two balls into the first box ($x^{2}$), etc. This is represented by the generating function

$1+x+x^{2}+...$.

Notice what happens if you square that:
$1+2x+3x^{2}+...$
Due to the fact that $x^{a}x^{b}=x^{a+b}$, this squared generating function represents two boxes. The coefficient of $x^{0}$ is the number of ways to put a total of 0 balls into the two boxes. The coefficient of $x^{1}$ is the number of ways to put a total of 1 ball into the two boxes, etc. Likewise, cubing the generating function $1+x+x^{2}+...$ gives us a generating function representing three boxes.

$(1+x+x^{2}+...)^{3}=\left(\frac{1}{1-x}\right)^{3}=(1-x)^{-3}$

Since the number of balls in the first three boxes is even, we only take the even part to represent that. For boxes 4,5,6, we only want the terms of odd powers.

For the remaining six boxes, they are simply represented by $(1-x)^{-6}$

Multiply it all together to get our final generating function:

$(\text{Even part of }(1-x)^{-3})(\text{Odd part of }(1-x)^{3})(1-x)^{6}$ Since there are eight balls, you want the coefficient of $x^{8}$.

Hope that helps.


Numb3rs-20

The Simpsons-20

Family Guy-20

Grey's Anatomy-20

Mythbusters-20

The Mole-20

American Idol-20

CSI's-20

Numb3rs-20
The Simpsons-20
Family Guy-20
Grey's Anatomy-20
Mythbusters-20
The Mole-21
American Idol-20
CSI's-18

Is it -1 or -2? I used the latter.

Numb3rs-20
The Simpsons-21
Family Guy-20
Grey's Anatomy-20
Mythbusters-20
The Mole-21
American Idol-18
CSI's-18

Numb3rs-20
The Simpsons-21
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-21
American Idol-18
CSI's-18

Numb3rs-20
The Simpsons-22
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-21
American Idol-18
CSI's-16

Numb3rs-20
The Simpsons-22
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-18
CSI's-14

Numb3rs-20
The Simpsons-23
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-16
CSI's-14

Numb3rs-20
The Simpsons-23
Family Guy-18
Grey's Anatomy-20
Mythbusters-23
The Mole-22
American Idol-16
CSI's-14

+2 Mythbusters.

crazy:

1: who did you attack?
2: giving 2 is prohibited.

undoing
1 to d'oh simpsons
Numb3rs-20
The Simpsons-24
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-14
CSI's-14

The simpsons-23
Numb3rs-20
The Simpsons-24
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-13
CSI's-14

-1 for American Idol

stop making ilegal moves, please.

Numb3rs-20
The Simpsons-24
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-14
CSI's-14

Numb3rs-20
The Simpsons-25
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-14
CSI's-14

+1 The Simpsons
and how am I making illegal moves?

you must add 1 to someone and delete 2 from another.

+1 the simpsons, -2 AI

Numb3rs-20
The Simpsons-25
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-22
American Idol-12
CSI's-14

[quote="cf249"]+1 The Mole, -2 simpsons

Numb3rs-20
The Simpsons-23
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-23
American Idol-12
CSI's-14[/quote]

I propose you an alliance. Please don't attack The Simpsons, and the final 2 will be Mole-D'oh

Numb3rs-20
The Simpsons-24
Family Guy-18
Grey's Anatomy-20
Mythbusters-21
The Mole-23
American Idiot (sorry idol)-10
CSI's-14

1 The Simpsons - 34
2 The Mole - 26
3 Mythbusters - 22
4 Numb3rs - 20
5 Family Guy - 16
6 CSI's - 3

I don't even know what The Mole is but Family Guy's better.

1 The Simpsons - 34
2 The Mole - 24
3 Mythbusters - 22
4 Numb3rs - 20
5 Family Guy - 17
6 CSI's - 3

1 The Simpsons - 34
2 The Mole - 25
3 Mythbusters - 22
4 Numb3rs - 20
5 Family Guy - 15
6 CSI's - 3

1 The Simpsons - 35
2 The Mole - 25
3 Mythbusters - 22
4 Numb3rs - 20
5 Family Guy - 15
6 CSI's - 1

lol no one is willing to hammer csi so -2 csi and +1 simps

Mole-lovers: Kill Mythbusters. They are mole-enemies.

+1 to d'oh, -2 molebusters

[color=green]1 The Simpsons - 36 [/color]
[color=olive]2 The Mole - 25 [/color]
[color=yellow]3 Mythbusters - 20
3 Numb3rs - 20 [/color]
[color=orange]5 Family Guy - 15 [/color]
[color=red]6 CSI's - 1 [/color]

1 The Simpsons - 36
2 The Mole - 26
3 Mythbusters - 20
3 Numb3rs - 20
5 Family Guy - 15

CSI kill and +1 Mole

1 The Simpsons - 34
2 The Mole - 27
3 Mythbusters - 20
3 Numb3rs - 20
5 Family Guy - 15

1 The Simpsons - 35
2 The Mole - 27
3 Numb3rs - 20
3 Mythbusters - 18
5 Family Guy - 15

The Simpsons - 33
The Mole - 25
Numb3rs - 20
Mythbusters - 19
Family Guy - 16

alingaling's didn't count, so I changed it back to the way chinesemaniac had it to get;

The Simpsons - 33
The Mole - 27
Numb3rs - 20
Mythbusters - 19
Family Guy - 15

And I'll change it to;

The Simpsons - 34
The Mole - 25
Numb3rs - 20
Mythbusters - 19
Family Guy - 15

Illegal move

The Simpsons - 34
The Mole - 27
Numb3rs - 20
Mythbusters - 17
Family Guy - 15

The Simpsons - 32
The Mole - 27
Numb3rs - 20
Mythbusters - 18
Family Guy - 15

The Simpsons - 35
The Mole - 25
Numb3rs - 20
Mythbusters - 17
Family Guy - 15

Well, is anybody going to help me finish this post?


calculus
integration
logarithms
calculus computations

A cup of coffee is warmed in $90^\circ C.$ It left in the room of $10^\circ C$ for three minutes, it reached $70^\circ C.$ How much time does it takes to go down to $55^\circ C$ from the first ? Note that  room temperature is fixed, descending speed of temperature is proportional to the difference of temperature to the that of surrouding.

And in 20 minutes the cup is down to less than $22^{\circ}C$ and it's time to dump it out because it's not doing a very good job of warming me up. Why is the room this cold anyway? Someone should complain to the management.

[hide="Some numbers and letters one-click away :P "]
By what you said I get:
($T$ = temperature, $t$ = time, $T_{a}$ = ambient temperature)

$\frac{dT}{dt}$~$(T-T_{a})$
$\frac{dT}{dt}=k(T-T_{a})$, where $k$ is constant of proportionality
$T'=kT-kT_{a}$, $T'=\frac{dT}{dt}$ just to visualise it better (note that this is a differential equation)

$T=T_{H}+T_{P}$, where $T_{H}$ and $T_{P}$ are solutions to homogenous and particular differential equations

$T_{H}'-kT_{H}=0$, substitute $T_{H}=Ae^{at}$
$aAe^{at}-kAe^{at}=0$
$a=k$, $T_{H}=Ae^{kt}$

$T_{P}=b$, where $b$ is some constant
$0=kb-kT_{a}\Rightarrow  T_{P}=b=T_{a}$

$T(t)=T_{H}+T_{P}=Ae^{kt}+T_{a}$
Now I got the formula, just need to calculate coefficients and plug-in.

$T(0)=A+10C=90C \Rightarrow  A=80C$
$T(180)=80C*e^{180k}+10C=70C$
$e^{180k}=\frac{3}{4}, k=\frac{\ln\frac{3}{4}}{180}$
$T(t)=80C*e^{kt}+10C=55C, e^{kt}=\frac{9}{16}, t=\frac{\ln\frac{9}{16}}{k}=360$

So, it takes the coffee 6 minutes to cool down to 55C.
[/hide]
Excellent problem kunny, I learned something new. :)

Thanks :) Your answer is correct.
This problem is related to Newton's cooling law.

(My $<22^{\circ}$ after 20 minutes comment was based on the same solution to the same differential equation as in ozolexil's post.)

An alternate solution, with less visible calculus:

The cup started at $80^{\circ}$ above ambient and was down to $60^{\circ}$ above ambient in 3 minutes. That means it has a "3/4 life" of 3 minutes. In another 3 minutes, we'll be down by another factor of 3/4. In other words, after 6 minutes we'll be at $\left(\frac34\right)^{2}\cdot 80^{\circ}=45^{\circ}$ above ambient - which is $55^{\circ}C.$ A lucky guess, of course, but we're done.

Yes, my solution is the same as yours, Kent Merryfield. :)


Let $x,$ $y$ and $z$ are non-negative real numbers. Prove that
$x^{4}y+x^{4}z+y^{4}x+y^{4}z+z^{4}x+z^{4}y+6xyz(xy+xz+yz)\geq8xyz(x^{2}+y^{2}+z^{2}).$


function
calculus
derivative
calculus computations

Looking for some help on parts of this problem:

Suppose that the function f has a continuous second derivative for all x, and that f(0) = 2, f'(0) = -3, and f"(0) = 0. Let g be a function whose derivative is given by g'(x) = e^-2x(3f(x) + 2f'(x)) for all x.

(c) Given that g(0) = 4, write an equation of the line tangent to the graph of g at the point where x = 0.

(d) Show that g"(x) = e^-2x(-6f(x) - f'(x) + 2f"(x)). Does g have a local maximum at x = 0? Justify your answer.

Thanks in advance.


Any books out there to improve speed?
Basically in pre-calculus levels and below, especially probability.

To improve your speed, reading a book won't help nearly as much as doing problems.  Just find some old AMC 10 or MathCounts problems, and get to work.

If you are really interested in arithmetic speed tricks, however, Google "Vedic math tricks" and look for stuff like [url=http://www.vedicmaths.org/Introduction/Tutorial/Tutorial.asp]this[/url].


geometry
3D geometry
tetrahedron
AMC
AIME
ratio
similar triangles

Here is a problem that I liked from my first AIME:

3. $T_{1}$ is a regular tetrahedron. Tetrahedron $T_{2}$ is formed by connecting the centers of the faces of $T_{1}$. Generally, a new tetrahedron $T_{n+1}$ is formed by connecting the centers of the faces of $T_{n}$. $V_{n}$ is the volume of tetrahedron $T_{n}$. $\frac{V_{2006}}{V_{1}}=\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find the remainder when $m+n$ is divided by $1000$.

man....the 4th time I saw this one.

[hide]
although it looks like pure geo., it's actually number theory.  draw the altitude of one of the side and the altitude of the base of the bigger tetrahedron, then you get similar triangles and see that the side ratio is $1 : 3$, so the volumne ratio is $1 : 27$.  The thing becomes $\frac{1}{(3^{3})^{2005}}$.  So we have $3*9^{3007}+1$ mod by 1000, or $3*(10-1)^{3007}+1$.  Binomial expand, we see only $-\binom{3007}{2}10^{2}+\binom{3007}{1}10-1$ left over (the rest are all divisible by 1000).  calculate that, multiply by 3, and add 1 to get the answer.
[/hide]

[quote="drunner2007"]i'm assuming that you are correct but I don't see how you know that the side ratio of $V_{1}$ to $V_{2}$ is $\frac{1}{3}$. I see that the height is in $\frac{1}{3}$ ratio, but not the base being in $\frac{1}{9}$ ratio. also, how can you show that the entire tetrahedron is similar?[/quote]

basically you can show that all the sides are equal in both so they are similar

oh yeah. just think about rotating it. it works by symmetry. makes sense  :lol:

To find that the side ratio was $\frac{1}{3}$ I used the fact that the vertices of the new tetrahedron was at the centroid of the faces so the centroid is $\frac{1}{3}$ of the altitude of the face.


Split the fraction up and do two geo sums.

This is kinda easy for interm.

[quote="bpms"]Evaluate $\sum_{n=0}^\infty (\frac{3^{n}+5^{n}}{8^{n}})$[/quote]
[hide]
Try to find a way of splitting it into two fractions.  Then it is pretty much downhill from there.[/hide]

I can't believe I didn't think of splitting it. :stupidity:


Let T be a triangle of area 6 square inches and perimeter 12 inches. Show
that the triangle T can be cut into 2006 triangles each of which has perimeter strictly
larger than 6 inches.

This is nice problem. You can done it by aplly Dirichlet.

Even better.
Triangle T can be cut into $ N$(as great as you want) triangles each of which has perimeter strictly
larger than $ a$ inches, where $ a < 9.3054072$.


What is the maximum number of non-overlapping regions created by four distinct lines in a plane?

[quote="Jongao"]Test:
0 lines create 1 region
1 line create 2 regions
2 lines create 4 regions
3 lines create 7 regions
4 lines create 11 regions
For some reason, I feel like I'm missing something...[/quote]

No you're correct.  But isn't there some sort of relationship?

Uh all of you are correct, but why is the answer 7???? Were the solution makers thinking 3 lines?

Hmmmmmmm....

$ \frac{n^2\plus{}n\plus{}2}{2} \equal{} \frac{n^2 \plus{} n}{2} \plus{} \frac22 \equal{} \frac{n(n\plus{}1)}{2} \plus{} 1$

basically the same thing...

When you draw the nth line, it can, at most, split n parts into 2n parts, an increment of n.
So when you draw 0 lines, you have 1 region.
Then, solution follows.


Find all functions $f,g: \mathbb R^{+} \to \mathbb R^{+}$ such that
\[ f(x)\equal{}\frac 1x \plus{} \frac 1{f(g(x))},\ g(x)\equal{}\frac 1x \plus{} \frac 1{g(f(x))},\]
for all positive reals $x$ and $y$.

[quote="Diogene"]Find the functions $ f,g: R^ + \rightarrow R^ +$( positive real numbers) such that :
$ f(x) = \frac 1x + \frac 1{f(g(x))},\ g(x) = \frac 1x + \frac 1{g(f(x))}$
 :cool:[/quote]

I think this problem is very difficult.
Beside the trivial solution $ f(x) = g(x) = \frac 2x$, there exist infinitely many solutions.

==========================  example of solutions =====================
For example, considering the cases where $ f(x) = g(x)$ and $ f(x) = \frac 1x + \frac 1{f(f(x))}$, we have solutions like :

Let $ p > 1\in\mathbb N$
Let $ A = \{x = 2^u(p + 1)^{\frac v2}p^{\frac w2}$, $ \forall u,v,w\in\mathbb Z\}$
Let $ B = \mathbb R^ + \backslash A$

Consider the relation $ \sim$ defined in $ B\times B$ by $ x\sim y$ $ \iff$ either $ xy = (p + 1)p^n$, either $ \frac xy = p^n$ for some $ n\in\mathbb Z$
This clearly is an equivalence relation.
Let then $ h(x): B\to B$ any choice function associating to $ x\in B$ a representant for it's equivalence class.

Let $ x\in B$. Since $ x\sim h(x)$ :
either $ \exists n\in\mathbb Z$ such that $ xh(x) = (p + 1)p^n$
either $ \exists n\in\mathbb Z$ such that $ \frac {h(x)}x = p^n$
and , for a given $ x$, it's impossible to have two different integers $ n_1$ and $ n_2$ verifying such equalities :
$ xh(x) = (p + 1)p^{n_1} = (p + 1)p^{n_2}$ would imply $ n_1 = n_2$

$ \frac {h(x)}x = p^{n_1} = p^{n_2}$ would imply $ n_1 = n_2$

$ xh(x) = (p + 1)p^{n_1}$ and $ \frac {h(x)}x = p^{n_2}$ would imply $ x^2 = (p + 1)p^{n_1 - n_2}$ and so $ x\in A$ and so $ x\notin B$

So there is a well defined function $ n(x): B\to\mathbb Z$ such that either $ xh(x) = (p + 1)p^{n(x)}$, either $ \frac {h(x)}x = p^{n(x)}$

Then we can define $ f(x) = g(x)$ as :

$ \forall x\in A$ : $ f(x) = \frac 2x$

$ \forall x\in B$ such that $ xh(x) = (p + 1)p^{n(x)}$ : $ f(x) = h(x)p^{ - n(x) - 1}$ $ =\frac{p+1}{px}$

$ \forall x\in B$ such that $ \frac {h(x)}x = p^{n(x)}$ : $ f(x) = \frac {(p + 1)p^{n(x)}}{h(x)}$ $ =\frac{p+1}x$

[hide="verification that this function fits the requirements"]
1) $ x\in A$
Then $ f(x) = \frac 2x$ and $ f(x)\in A$. So $ f(f(x)) = \frac 2{f(x)} = x$ and $ \frac 1x + \frac 1{f(f(x))} =$ $ \frac 1x + \frac 1x = f(x)$
And so $ f(x) = \frac 1x + \frac 1{f(f(x))}$

2) $ x\in B$ such that $ xh(x) = (p + 1)p^{n(x)}$
Then $ f(x)=h(x)p^{ - n(x) - 1}$ and so $ f(x)\in B$ and $ f(x)\sim h(x)$ and so $ h(f(x)) = h(x)$

So $ f(x) = h(f(x))p^{ - n(x) - 1}$ and $ \frac {h(f(x))}{f(x)} = p^{n(x) + 1}$ and so $ n(f(x)) = n(x) + 1$

So $ \frac {h(f(x))}{f(x)} = p^{n(f(x))}$ and $ f(f(x)) = \frac {(p + 1)p^{n(f(x))}}{h(f(x))}$ ${ = \frac {(p + 1)p^{n(x) + 1}}h(x)}$

Then $ \frac 1x + \frac 1{f(f(x))}$ $ = \frac {h(x)}{(p + 1)p^{n(x)}} + \frac {h(x)}{(p + 1)p^{n(x) + 1}}$ $ = \frac {(p + 1)h(x)}{(p + 1)p^{n(x) + 1}}$ $ = \frac {h(x)}{p^{n(x) + 1}}$ $ = f(x)$

And so $ f(x) = \frac 1x + \frac 1{f(f(x))}$

3) $ x\in B$ such that $ \frac {h(x)}x = p^{n(x)}$
Then $ f(x) = \frac {(p + 1)p^{n(x)}}{h(x)}$ and so $ f(x)\in B$ and $ f(x)h(x) = (p + 1)p^{n(x)}$ so $ f(x)\sim h(x)$ and so $ h(f(x)) = h(x)$

So $ f(x)h(f(x)) = (p + 1)p^{n(x)}$ and $ n(f(x)) = n(x)$

Then $ f(x)h(f(x)) = (p + 1)p^{n(f(x))}$ and so $ f(f(x)) = h(f(x))p^{ - n(f(x)) - 1}$ $ = h(x)p^{ - n(x) - 1}$

So $ \frac 1x + \frac 1{f(f(x))}$ $ = \frac {p^{n(x)}}{h(x)} + \frac {p^{n(x) + 1}}{h(x)}$ $ = \frac {(p + 1)p^{n(x)}}{h(x)}$ $ = f(x)$

And so $ f(x) = \frac 1x + \frac 1{f(f(x))}$

Q.E.D

[/hide]

========================== end of example ================================

I would be very interested, Diogene, on informations about the origin of the problem (it's a crazy problem for olympiad, or olympiad preparation, in my opinion).

And I would be very interested in your own general solution.

Thanks in advance.

Hello pco,
 I was inspired  by http://www.mathlinks.ro/viewtopic.php?t=307928
The functional system $ f(x) \equal{} \frac ax \plus{} \frac 1{f(g(x))},\ g(x) \equal{} \frac ax \plus{} \frac 1{g(f(x))}$ has only the solution $ f(x) \equal{} g(x) \equal{} \frac {a \plus{} 1}x$ if $ a > 1$
If $ a \leq 1$ the probleme is very hard, and I am not much more advanced than you .

 :cool:


Find all the functions $f: Z\to Z$ satisfy:
           $f(x+f(f(x)))=f(f(x+y)-y$

What is meant by $Z-$¿ (Is that $-$ just a mistake¿)
And there is an unclosed bracket...


abstract algebra
superior algebra
superior algebra unsolved

1. Let $ G$ be a finite abelian group. Prove that there exist $ x_0,x_1,x_2,...,x_k$ such that $ x_0^{a_0}x_1^{a_1}x_2^{a_2}...x_k^{a_k}$ uniquely generates the elements of $ G$ for $ 0\le a_m\le |x_m|$ for $ 0\le m\le k$.

2. Let $ a\in G$. Prove that $ (a^{\minus{}1})^{\minus{}1}\equal{}a$.

The first problem requires a fair amount of machinery and is called the Fundamental Theorem of Finite Abelian Groups. You should probably consider homomorphisms of the form $ f(g) \equal{} g^k$ for various values of $ k$.

In the second one, you somewhat need to specify what axioms we are assuming for a group, since this is sometimes itself taken as an axiom for groups. However, I think we can at the very least prove this assuming that the left and right identities are the same. Also, your question is a bit vague. I'm assuming you want to show that the left and right inverses of $ a$ are the same.

Let $ f(a)$ be the right-inverse of $ a$. Then clearly $ af(a) \equal{} e$, where $ e$ is the identity. Then $ af(a)f(f(a)) \equal{} f(f(a))$. But $ af(a)f(f(a))f(a) \equal{} f(f(a))f(a)$, so $ af(a) \equal{} f(f(a))f(a)$. Multiply by $ f(f(a))$, then we get $ ae \equal{} f(f(a))e$, so $ a \equal{} f(f(a))$. Thus the right-inverse of the right-inverse of $ a$ is $ a$. So if $ af(a) \equal{} e$, then $ f(a)a \equal{} e$ (since $ f(a)f(f(a)) \equal{} e$ and $ a \equal{} f(f(a))$). This is what we wanted to show.


real analysis
calculus
derivative
integration
function
rudin
calculus computations

I have finished Calculus 1 and 2, both single variable calculus courses, going through limits, derivatives, integration, and sequences and series including taylor series. I have a very good grasp of the material, but am looking for a rigorous treatment. I have also worked through How to Prove It, so I have a good understanding of proofs.

I have the feeling spivak calculus is more towards learning calculus the rigorous way the first time, and I will see a lot of the things I have already seen. Should I just do baby Rudin and skip spivak?

Spivak says in the intro that he feels a better title would be "An Introduction to Analysis".  While it is true that you will see many things that you have already seen, you are almost certain to see a ton of things you haven't.  Do you know the rigorous $ \epsilon\minus{}\delta$ definition of limits and how to use them?  Do you know the integral is rigorously constructed and how to prove the fundamental theorems of calculus?  Are you comfortable with all of the tests for convergence?  If you know all of these things and are comfortable, then you are probably ready to go to analysis.  Even if you do know these things, though, Spivak goes through complex functions and power series, a construction of the real numbers, and he even proves the irrationality of $ \pi$ and the fact that $ e$ is transcendental, so there is still something for everyone.

Thanks,

I know how to do epsilon delta proofs, although past simple limits my abilities fall through. I do know how the integral is constructed, and the fundamental theorem of calculus proof I have only seen once back in Calc 1, so I guess ill need a refresher.

The tests for convergence is probably my strongest area.


calculus
integration
real analysis
real analysis solved

For 0 < s < 1 evaluate the integral,

\int_0^{\infty} x^{s-1}/(x+1) dx.

y= x^s=> x= y^q with q=1/s>1
dx=qy^(q-1)dy
the integral becomes with this change variable
q\int_{0 to +oo}dy/(1+y^q) = q. \pi /q.sin(\pi /q)
\pi sin(\pi s)


calculus
integration
quadratics
Vieta
algebra
algebra unsolved

Hi, here are two problems that i am not able to solve. Solutions are unknown, so plz help.

 :arrow:  Find all INTEGRAL values of a for which the quadratic equation [b](x-a)(x-10)+1=0[/b] has INTEGRAL roots.

 :arrow: Let a and b be the roots of the equation [b]x^2-10cx-11d=0[/b] and those of [b]x^2-10ax-11b=0[/b] are c,d then what is the value of a+b+c+d, when none of the roots are equal ?

ANSWERS
[hide]1::Dunno
2::1210[/hide]

[hide="1)"]$(x-a)(x-10)+1 = 0$.

So then we have $(x-a)(x-10) =-1$.  But, since $x, a, 10$ are integers, both factors must be integers.  Thus, one of them is $1$, and the other is $-1$.

Case 1: $x-a = 1$, $x-10 =-1$.
So, $x = 9$, and therefore $a = x-1 = 8$.

Case 2: $x-a =-1$, $x-10 = 1$.
So, $x = 11$, and therefore $a = x+1 = 12$.[/hide]

thx tynan
any1 knows the solution for the second?

[quote="Ilthigore"]$x^{2}-10cx-11d=(x-a)(x-b)$
$x^{2}-10ax-11b=(x-c)(x-d)$

We obtain (by Vieta's theorem) the system:
a+b=10c
c+d=10a
ab = -11d
cd = -11b

Which can probably solved if you want to.[/quote]

Yes, it can be solved.  But after I got everything in terms of a, I got a nasty cubic, which I didn't feel like solving.


How many distinct sums can you get from adding three numbers from set { 1,2,3,11,12,13,21,22,23}

Help would be appreciated...

can they be repeated, for example 11+11+11?

Well, the maximum you can have is 23+22+21=66
minimum is 1+2+3=6
which is 61 numbers. If that helps. Now you have to rule out the ones you cana't make.

[hide="hint"]casework of 10s digits[/hide]


factorize  $ x^4 \plus{}16x \minus{} 12$

and $ x^4 \minus{}16x \plus{}12$

hello, by the ansatz
$ x^4\plus{}16x\minus{}12\equal{}(ax^2\plus{}bx\plus{}c)(dx^2\plus{}ex\plus{}f)$ we get the system
$ ad\equal{}1,bd\plus{}ae\equal{}0,cd\plus{}be\plus{}af\equal{}0,ce\plus{}bf\equal{}16,cf\equal{}\minus{}12$.
Solving this we have $ a\equal{}\frac{2}{e},b\equal{}\minus{}\frac{4}{e},c\equal{}\frac{12}{e},d\equal{}\frac{1}{2}e,e\equal{}e,f\equal{}\minus{}e$. Let $ e\equal{}2$ so we have
$ x^4\plus{}16x\minus{}12\equal{}(x^2\minus{}2x\plus{}6)(x^2\plus{}2x\minus{}2)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, by the ansatz
$ x^4 \plus{} 16x \minus{} 12 \equal{} (ax^2 \plus{} bx \plus{} c)(dx^2 \plus{} ex \plus{} f)$ we get the system
$ ad \equal{} 1,bd \plus{} ae \equal{} 0,cd \plus{} be \plus{} af \equal{} 0,ce \plus{} bf \equal{} 16,cf \equal{} \minus{} 12$.
Solving this we have $ a \equal{} \frac {2}{e},b \equal{} \minus{} \frac {4}{e},c \equal{} \frac {12}{e},d \equal{} \frac {1}{2}e,e \equal{} e,f \equal{} \minus{} e$. Let $ e \equal{} 2$ so we have
$ x^4 \plus{} 16x \minus{} 12 \equal{} (x^2 \minus{} 2x \plus{} 6)(x^2 \plus{} 2x \minus{} 2)$.
Sonnhard.[/quote]
It is necessary to confirm the following.
$ x^4 \plus{} 16x \minus{} 12 \neq (ax \plus{} b)(cx^3 \plus{} dx^2 \plus{} ex \plus{} f)$
$ (a, b, c, d, e, f \in \mathbb{Z})$

I found how to solve others in Mathlinks.
http://www.mathlinks.ro/viewtopic.php?t=222785


ok, my teacher gave us a handout to do for homework, but i cant find an answer to one of the questions (that isnt some wierd decimal)...

Solve the equation(for x)...

$ \sqrt{x\plus{}16}\equal{} x\minus{}\sqrt{x\plus{}7}$

make sure that there arent any extraneous solutions

$ \sqrt{x\plus{}16}\equal{}x\minus{}\sqrt{x\plus{}7}$

lol i guess and checked and its 9  :D

Or:

Square to get

$ x\plus{}16\equal{}x^2\plus{}x\plus{}7\minus{}2x\sqrt{x\plus{}7}$
$ x^2\minus{}9\equal{}2x\sqrt{x\plus{}7}$
$ x^4\minus{}18x^2\plus{}81\equal{}4x^2(x\plus{}7)$
$ x^4\minus{}4x^3\minus{}46x^2\plus{}81\equal{}0$

9 works

$ (x\minus{}9)(x^3\plus{}5x^2\minus{}x\minus{}9)\equal{}0$
There aren't any more nice solutions. However, I claim those are all extraneous:

Proof:

$ x\minus{}\sqrt{x\plus{}7} > 0 \implies x > \sqrt{x\plus{}7}$. Since both sides are positive, we can square to get
$ x^2 > x\plus{}7$
$ x^2\minus{}x\minus{}7 > 0$
$ x > \frac{1 \pm \sqrt{29}}{2}$

Since x is positive, $ x > 3$.

Consider the polynomial $ x^3\plus{}5x^2\minus{}x\minus{}9$. Note that from 3 up, this polynomial is a increasing function. plugging in 3 gives $ 27\plus{}45\minus{}3\minus{}9\equal{}60$. All numbers above this are larger than 0. Thus, all other solutions are extraneous.

ok, ty!

i got to $ x^4\minus{}4x^3\minus{}46x^2\plus{}81$ and thought that it was prime.... (lame)

Starting with BOGTRO's factorization, $ (x \minus{} 9)(x^3 \plus{} 5x^2 \minus{} x \minus{}9) \equal{} 0$, any rational roots of $ x^3 \plus{} 5x^2 \minus{}x \minus{}9$ have to be rational factors of $ \frac{\minus{}9}{1}$, i.e. $ x \in \{\pm 1, \pm 3, \pm 9\}$. None of which work, so $ 9$ is the only rational root of the original polynomial.


What is the smallest value that

√(49+a^2-7a．√2)
+
√(a^2+b^2-ab．√2)
+
√(50+b^2-10b)

can have for positive real numbers a and b?

please help!

hello, do you mean
$ \sqrt{49\plus{}a^2\minus{}7a\sqrt{2}}\plus{}\sqrt{a^2\plus{}b^2\minus{}ab\sqrt{2}}\plus{}\sqrt{50\plus{}b^2\minus{}10b}$?
Sonnhard.

Not sure for it:
$ \sqrt {(a - \frac {7}{\sqrt {2}})^2 + \frac {49}{2}} + \sqrt {(\frac {a}{\sqrt {2}} - b})^2 + \frac {a^2}{2}} + \sqrt {(b - 5)^2 + 25}\ge \sqrt {(a + \frac {a}{\sqrt {2}} - \frac {7}{\sqrt {2}} - 5)^2 + (\frac {7}{\sqrt {2}} + \frac {a}{\sqrt {2}} + 5)^2$
This expression is on the form:
$ ka^2 + la + z$ with $ k > 0$ so it has minimum on $ - \frac {D}{4a}$ where $ D$ is the discriminant of the trinomial.

do you want the answer in terms of a and b?

[hide="hint 1"]
Try a geometric interpretation
[/hide]

[hide="hint 2 (give away)"]
Use Cosine Rule
[/hide]


logarithms geometry trigonometry AMC AIME probability blogs

Well, it's back to school. And what better way to start it with a Mock AMC 12?  :)

My friend started on a Mock AMC 12 a couple of weeks ago and I helped him write a few problems. It still seems a bit too easy for AMC 12 though.  :o

The AMC 12 should be ready by mid-September. Who wants to try it out?

I'll help write problems if you need any.

That would be nice. We have plenty of intermediate problems, but not enough really difficult problems. Another logarithm problem would be nice too.

If you're going to be writing problems, then you can't take it of course. However, you can help vet the test.

I'm in, although I'm a little afraid of your problems.  I've seen your other threads, and you make some tough problems.

I'm in (to participate, not to write problems)

Moi aussi :D

btw, i'm assuming we're supposed to time ourselves, right?

Je veux aussi.  :lol: Well that's probably wrong. Anyways I'll take it! I have to focus on GEOMETRY!! Specially relating trigonomatry to it.

Hello. Is it alright if I take it on Friday at the regular time (6:00 PM ET). Also, where can we download the PDF, also if you are going to give instructions tomorrow, where should we go to get the instructions? Thanks.

Hello. I would like to take the Friday one if it is still possible. Is there a solid time for it?

Can i take Friday too, i have volunteering on the weekend. Thx

I think I might be able to do it, but I'm not sure...

[quote="PowerOfPi"]What time is it at tomorrow (Friday)? If it is in the morning/afternoon, I will take it then. If not, then I will take it on Saturday.[/quote]

You will have to take it on Saturday then. It's 6:00 P.M. today.

Okay, where and what time on Saturday?

EDIT: Didn't see other thread, sorry. I have a suggestion: Why don't you just make it so the survey doesn't have to be completed in the competition time, and instead, you have to submit the survey to see your results?

Also, the time zone for all the discussed times would be useful.

Please look at past posts in this thread or the administration date. Specifically stated there. We've got about 2 hours to testing time.

Hello. I am going to take it in two hours, and I was wondering how I will get the contest (will you post it or PM it?). Also, if you are going to post it, will it be here or in your personal blog? Also, can I PM answers to you? Thanks.

Sign up and take it tomorrow. Check the Administration thread too.

Everyone go here and take it!!!!

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1634205#1634205


Points A,B,C, and D form a square with the side length of x.

E is a point which externally divides AB into 3:1, F is a point which externally divides BC into 3:1, G is a point which externally divides CD into 3:1, and H is a point which externally divides DA into 3:1, Find the area of EFGH.

Challenge: Can you generalize the results?

Hint for the Challenge: [hide]Try changing the ratio (3:1). What happens? what if the ratio was x:y?[/hide]

Spoiler:

[hide]When you connect EFGH, it forms a square. the side length of the square is sqrt((3/4x):^2:+(1/4x):^2:)= sqrt(5/8x:^2:), so the area is sqrt(5/8(x:^2:)):^2:= 5/8(x:^2:).

Generalize: so if it were y:z, and the side length of the orignal square is x, the side length of the smaller square would be sqrt((xy/(y+z)):^2: + (xy/(y+z)):^2:)= sqrt(((y:^2: + z:^2:)x:^2:)/(y:^2: + 2yz + z:^2:)), and the area would be (y:^2:+z:^2:)x:^2:/(y:^2:+2yz+z:^2:)[/hide]

I hope that's right...

I'm sorry, but that is incorrect.
The question states [quote="question"]E is a point which [b]externally[/b] divides AB into 3:1, F is a point which [b]externally[/b] divides BC into 3:1, G is a point which [b]externally[/b] divides CD into 3:1, and H is a point which [b]externally[/b] divides DA into 3:1, Find the area of EFGH.[/quote]
I think you've mistake [b]exterior[/b] division for [b]interior[/b] division.

If the question stated [quote="if the question stated..."]E is a point which internally divides AB into 3:1, F is a point which internally divides BC into 3:1, G is a point which internally divides CD into 3:1, and H is a point which internally divides DA into 3:1, Find the area of EFGH.[/quote]
, you would be right, though.

oh... okay... what's the difference between interior division and exterior division??

Interior Division: A point (X) [b]on line AB[/b] which divides line AB into AX:BX=m;n
Formula for finding X:( (mB1+nA1)/m+n, (mB2+nA2)/m+n )
(A is at (A1,A2) and B is at (B1, B2))

While Exterior Division is
A point (X) on the extention of line AB in which AX:BX=m:n
Formula for finding X:( (mB1-nA1)/m-n, (mB2-nA2)/m-n )

I thought Exterior Division is a nice subject to focus on because usually you don't learn about this in school. All you learn is the midpoint formula, [b]which was originally derived from the above mentioned Interior Division Formula (or whatever it is called in English)[/b]

Ok, I did this kind of thing in the Chinese books, but never really knew the names, but here goes:

[hide]Ok, E externally divides AB so AE:BE = 1:3 (just checking), then E would have to be on the "B" side... like closer to B. So then BE would equal 1/2 AB since

BE = 1/3x + 1/3 BE

2/3 BE = 1/3 x

2 BE = x

so BE = 1/2 x

Now then... we know that. EFGH is a square, because triangle AHE, BEF, GCF, and HDG (they're triangles right? well I think they are.) are congruent due to LL congruency since the legs are all 1/2 x and 3/2 x... so then

angle EBF is right since it is supplement to ABC which is right since it's a vertex of the square... (not sure if I'm using correct term but oh well...) , so then

angle BEF + BFE = 90

and angle BFE = angle AEH, then BEF + AEH = 90.... so HEF is right ,and so are the others... so then we find the length of the sides of the square....

AH = 1/2 x and AE is 3/2 x so then

HE^2(which is the area of EFGH) = 1/4 x^2 + 9/4 x^2 = 5/2 x^2

so is it 5/2 x^2?[/hide]

generalize? Like find like if the ratio is m:n or some unknown?

[quote="Tare"][quote="Chinaboy"]generalize???? Like find like if the ratio is m:n or some unknown???[/quote]
Right.
And please note that using ? 3 or 4 times is unnecessary and/or considered spamming.
(I'm sorry I had to say that, but it's my job...forgive me martin :( )[/quote]

LoL, it's okay. I'll cut down on it.

And when generalizing for m:n, can I assumre that m=>n?

Ok... here goes

[hide]ratio is m:n (I think it would be the same result either way when one is bigger than the other)

So now, assumming m is larger.

BE = (n/m)(x+BE)

BE = (n/m)x + (n/m)BE

((m-n)/m)BE = (n/m) x

(m-n)BE = nx

BE = (n/(m-n)) x

AE = x + (n/(m-n))x

AE = (m/(m-n))x

HE^2(the area) = HA^2(which is equal to BE^2) + AE^2

HE^2 = x^2(m^2/(m-n)^2) + n^2/(m-n)^2)

HE^2 = (m^2 + n^2)/(m-n)^2 i hope..[/hide]

Wait, how is doing this ????????? spamming, especially since its the end of a sentence?

I'm just curious.


[tex] \begin{gathered}
  {\text{If }}x,y,z > 0{\text{ and }}x^4  + y^4  + z^4 = 3 {\text{ then }} \hfill  x + y + z \leqslant 3 \hfill{\text{, }}  xy + yz + zx \leqslant 3{\text{ and }}xyz \leqslant 1 \hfill  \end{gathered}

 [/tex].

made by Szoeroes Al.

cheers! :D :D :cool:

x^4+y^4+z^4+1=4 using AMGM we find xyz=<1
Now using  \sqrt (((x)+(y)+(z))/4) >= (x+y+z+1)/4 using this twice we find: x+y+z=<3
(x+y+z)=<9 out of this we get= xy+yz+zx=<3

Actually, you don't even need the "+1" ! Just proceed as follows:

We have $\frac{x+y+z}{3}\leq \sqrt[4]{\frac{x^{4}+y^{4}+z^{4}}{3}}$ and thus x+y+z <= 3. Further, $\sqrt[3]{xyz}\leq \sqrt[4]{\frac{x^{3}+y^{3}+z^{3}}{3}}$, so that xyz <= 1. At last, we know that $3\left( xy+yz+zx\right) \leq \left( x+y+z\right) ^{2}$, and hence xy + yz + zx <= 3.

  Darij


R$ <<$wL

what does the double arrow mean?

it means  that the value of r IS MUCH MUCH LESS THAN VALUE OF wL

No, I think he's saying the right thing.

[url]http://en.wikipedia.org/wiki/Table_of_mathematical_symbols[/url]

[quote="Ihatepie"]are you joking? because much much less isn't really a mathematical term.[/quote]

Usually one will understand how much "much much" is given the context.

In the context of electronic/electric circuits (I assume the O.P. refers to a filter or integrating circuit of some kind), terms $ A$ in $ A<<B$ are in most cases more than $ 10$ times smaller than $ B$, but taking it even smaller won't hurt - if the problem is about synthesis of a circuit. Usually, $ A<<B$ is used when approximating a sum $ A\plus{}B$ with $ B$.

It's often less than the other value by several orders of magnitude. Sometimes can be considered negligible, I guess.


Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . Locked!


calculus
integration
vector
calculus computations

Problem was to verify Divergence theorem by evaluating surface integral and the triple integral using:
F(x,y,z) = 2xi - yzj + z^2k and the surface sigma is the paraboloid z = x^2 + y^2 capped by the disk x^2 + y^2 <=1 in the plane z=1.

 Using divergence theorem, I get the outward flux to be 4pi/3

To use surface integral, I first find flux over the disk and add to that flux over paraboloid.
For the disk, I find that the unit normal vector giving outward orientation is k and find the flux over disk is pi.
For the paraboloid, I find unit normal vector giving outward orientation is -2xi - 2yj + z^2k
using this, I get flux over paraboloid to be -pi/3

Adding the two, I get 2pi/3, which is not the value I got from divergence theorem (the textbook lists 4pi/3 as the correct answer)

Why the discrepancy? *i am tempted to believe that I messed up the orientation of the parabolid, but am unsure*

Of course you did. The outward normal to the paraboloid must have negative vertical component.


Let$A,B,C,D$ be different points in a plane , find the intersection of the lines
$AB$ and $CD$ just with a compass.

P.S My teacher told me that if any thing if constructable by compass and ruler then it is
constuctable by just a ruler or a compass. any idea how we can prove this?

google Mascheroni constructions, I think, for the compass-only part

for the ruler-only part, it's not true. Anything you construct with a ruler (lines through two points, intersections of lines) is invariant under a projection, and this certainly isn't true of many standard constructions, i.e. the midpoint of a segment.


A man spent two-thirds of his money and misplaced one-third of the remainder, leaving him with $ \$22$. How much money did he originally have?

if he has $ M$ dollars, then he spent $ \frac{2}{3}M$ dollars and has $ \frac{1}{3}M$ dollars left.  Of this, he phailed to keep track of his money and lost 1/3 of that, or $ \frac{1}{9}M$ dollars, so he has $ \frac{1}{3}M\minus{}\frac{1}{9}M\equal{}\frac{2}{9}M$ dollars left, which is $ \$22$.  Hence, \[ \frac{2}{9}M\equal{}22\Rightarrow M\equal{}\boxed{99}\]


If I want start study about fermat's last theorem...which book are good?? help me please..thnx..

You're going need considerable background if you actually want to understand the proof.  Start by learning abstract algebra, then you'll need to learn some elliptic curve and representation theory, which is advanced stuff.

If you want a nice expository book on it, read "Fermat's Enigma" by Simon Singh.

You don't know if he made this thread for showing off (why shouldn't he just want to know how to prove FLT¿).
Stop being so unfriendly, stop calling people "nooby" or "jerks", stop using chatspeak, stop necromancing for no appearant reason, stop this needless flood of smileys, and generally stop acting so childish, stupid and silly. What's the point of _your_ post¿
And the next post of this type and language will just get deleted.

[quote="Yoni2b"]I can't actually help with the FLT itself, but an amazing book about it, which I hope will inspire you is http://www.simonsingh.net/Fermats_Last-Theorem_The_Book.html[/quote]

Meh I tried reading it last year, but stopped a third of the way in because the style of writing was not.. stimulating. The content was ok; his brief mention of Galois' duel was the most most exciting part =\


Find all functions $ f: \mathbb Z \to \mathbb Z$ such that
\[f(m + n) + f(mn - 1) = f(m)f(n) + 2 ,\]
for all $m,n \in \mathbb Z$.

[quote="Xaenir"]Find all functions $ f: Z \to Z$ such that
$ %Error. "eqalign" is a bad command.
{ \& f(m + n) + f(mn - 1) = f(m)f(n) + 2 \cr$;$ \forall m,n \in Z \cr}$[/quote]

$ n=0$ $ \implies$ $ f(m)+f(-1)=f(m)f(0)+2$

If $ f(0)\neq 1$, this implies $ f(x)=c$ and so $ 2c=c^2+2$, which is impossible. So $ f(0)=1$ and $ f(-1)=2$

Let then $ f(1)=a$

$ n=1$ $ \implies$ $ f(m+1)=af(m)-f(m-1)+2$ and this allow us to compute all values of $ f(x)$. Let's go :

$ f(-1)=2$
$ f(0)=1$
$ f(1)=a$
$ f(2)=a^2+1$
$ f(3)=a^3+2$
$ f(4)=a^4-a^2+2a+1$
$ f(5)=a^5-2a^3+2a^2+a$

$ m=n=2$ $ \implies$ $ f(4)+f(3)=f(2)^2+2$ $ \iff$ $ (a^4-a^2+2a+1)+(a^3+2)=(a^2+1)^2+2$ $ \iff$ $ a(a-1)(a-2)=0$ and so $ a\in\{0,1,2\}$

$ m=2,n=3$ $ \implies$ $ f(5)+f(5)=f(3)f(2)+2$ $ \iff$ $ 2(a^5-2a^3+2a^2+a)=(a^3+2)(a^2+1)+2$
Checking this equality with $ a=0$, $ a=1$ and $ a=2$ implies $ a=2$

So $ f(m+1)=2f(m)-f(m-1)+2$ which is easily solved as $ \boxed{f(m)=m^2+1}$

And it's easy to check that this solution fits the initial equation.


I would like to buy some of the books which are only available at this site.

On this site is the information given that you can also buy the books by Highperception, because that saves a lot of money (the shipping from the USA is more than 40 dollars and that is quite a lot for a high school guy...). But highperception doesn't exist any longer, so I would like to know if there is a possibility to buy the books at another place with less shipping costs?

(Sorry for my bad English)

As far as I know, Highperception is still open and still sells our books.

http://www.highperception.com


Let ABC be a triangle and D the midpoint of BC .Let P be a point on AD such that angle BAC+angle BPC=180 degree. Prove that AC.BP=AB.CP


1) Make "Always Render PNG" the standard (for people who are not logged in). Although it may cause some trouble for the unlucky 56k chaps (who can log in and change the preference back), it makes the whole Wiki look a lot better.

2) This may be hard/time-consuming to write, but linked wiki/forum log in? Instead of having to log-in in both places... once your logged into one, your logged into the other (and vice-versa)?.

The AoPSWiki is looking great! :D

[quote="surge"]1) Make "Always Render PNG" the standard (for people who are not logged in). Although it may cause some trouble for the unlucky 56k chaps (who can log in and change the preference back), it makes the whole Wiki look a lot better.[/quote]

Yes, I agree.  (I made this suggestion a little ways back but forgot about it or just moved on to other things.)

For reasons that would take a long time to explain, we made a deliberate decision previously not to change those aspects of the AoPSWiki, though that's not to say we won't change them in the future.


Oops...
Sorry.... Move it please to ineq...

[quote="Ahiles"][color=darkblue] Find all $ (x,y,z) \in R^3$, for which
\[ 2x\sqrt {y \minus{} 1} \plus{} 2y\sqrt {z \minus{} 1} \plus{} 2z\sqrt {x \minus{} 1}\ge xy \plus{} xz \plus{} yz
\]
[/color][/quote]
We set $ \sqrt{x\minus{}1}\equal{}a,\sqrt{y\minus{}1}\equal{}b,\sqrt{z\minus{}1}\equal{}c$
So we have to find the numbers that satisfy
$ 2(a^2\plus{}1)b\plus{}2(b^2\plus{}1)c\plus{}2(c^2\plus{}1)a\geq(a^2\plus{}1)(b^2\plus{}1)\plus{}(b^2\plus{}1)(c^2\plus{}1)\plus{}(c^2\plus{}1)(a^2\plus{}1)$
But this is true only as equality,because
$ (a^2\plus{}1)(b^2\plus{}1)\geq(a^2\plus{}1)(2b)$ and cyclic...
So we have that $ (a\minus{}1)^2\equal{}0,(b\minus{}1)^2\equal{}0,(c\minus{}1)^2\equal{}0$, so
$ a\equal{}b\equal{}c\equal{}1$ and $ x\equal{}y\equal{}z\equal{}2$

[quote="tuandokim"][quote="Ahiles"][color=darkblue] Find all $ (x,y,z) \in R^3$, for which
\[ 2x\sqrt {y \minus{} 1} \plus{} 2y\sqrt {z \minus{} 1} \plus{} 2z\sqrt {x \minus{} 1}\ge xy \plus{} xz \plus{} yz
\]
[/color][/quote]
I think it must be liked this:
$ 2x\sqrt {y \minus{} 1} \plus{} 2y\sqrt {z \minus{} 1} \plus{} 2z\sqrt {x \minus{} 1}\le xy \plus{} xz \plus{} yz$
and
[quote="Dimitris X"]
$ 2(a^2 \plus{} 1)b \plus{} 2(b^2 \plus{} 1)c \plus{} 2(c^2 \plus{} 1)a\geq(a^2 \plus{} 1)(b^2 \plus{} 1) \plus{} (b^2 \plus{} 1)(c^2 \plus{} 1) \plus{} (c^2 \plus{} 1)(a^2 \plus{} 1)$
[/quote]
must be liked this
$ 2(a^2 \plus{} 1)b \plus{} 2(b^2 \plus{} 1)c \plus{} 2(c^2 \plus{} 1)a\leq(a^2 \plus{} 1)(b^2 \plus{} 1) \plus{} (b^2 \plus{} 1)(c^2 \plus{} 1) \plus{} (c^2 \plus{} 1)(a^2 \plus{} 1)$[/quote]
Are you sure????We want to FIND $ x,y,z$ not to prove it :wink:


A bus has 12 stops, numbered 1-12. The bus has place for 20 passengers. Each day the bus arrives empty at stop 1, and the last passenger leaves at stop 12 (or before that). If passengers A and B get on the bus at the same stop, they are not allowed to both leave at the same stop. What is the maximal number of passengers that can use this bus each day?

OK, I'll give you the answer, you just need to prove it:

[hide]The maximal number of passengers is 50.[/hide]

A simple one. Without the restrictions on the size of the bus, we can have at most 66 passengers, with maximum 36 passengers at the same time. So we should throw away at least 16. And good candidates for throwing away are those who travel too much while paying the same amount. So throw away those who travel more than 6 stops and one of those six who travel exactly 6 stops (an unlucky guy).

Another question is the following.

Let us, in contrast to what I've written, give the preference to those travelling far, because others can walk or use other routes. What is the longest possible accumulative distance (in stops) that can be achieved?


A sequence $ (u_n)$ is defined by $ u_0\in[0;2)$, $ u_n^2 \minus{} (n \plus{} 1)u_{n \plus{} 1} \equal{} 1$ for every $ n\in \mathbb N$ (n = 0,1,2,...).
Find the limit of the sequence $ u_n$

I'm waiting a solution to this nice problem. Thanks.

1) If $ |u_k|\leq 1$ for some $ k\geq0$, $ |u_n|\leq 1$ for all $ n\geq k$.
This is easy to prove by induction.
2) $ u_n\equal{}n\plus{}2$ if $ u_0\equal{}2$.
It is easy to prove by induction.
3) $ u_k\in [0, 1]$ for some $ k\geq0$ .
It is obvious when $ u_0\leq 1$. Now we suppose this is not true for some $ u_0\in (1,2)$. We notice that the series $ u_n$ can not jump from a greater-than-1 number to a negative number. Then $ 1<u_n<n\plus{}2$ for $ n\geq0$. Hence, we have
$ \frac{n\plus{}3\minus{}u_{n\plus{}1}}{n\plus{}2\minus{}u_n}\equal{}\frac{n\plus{}2\plus{}u_n}{n\plus{}1}>\frac{n\plus{}3}{n\plus{}1}$.
So
$ \frac{n\plus{}3\minus{}u_{n\plus{}1}}{2\minus{}u_0}>\frac{(n\plus{}2)(n\plus{}3)}2$, or
$ u_{n\plus{}1}<n\plus{}3\minus{}(2\minus{}u_0) \frac{(n\plus{}2)(n\plus{}3)}2$.
Obviously the inequality cannot hold for any $ n>0$ because the RHS will be negative for a big enough $ n$. A contradiction.

Now we actually show $ |u_n|\leq 1$ whenever $ n>k$ for some $ k$.
Hence $ |u_{n\plus{}1}|\leq\frac1{n\plus{}1}$ implies the limit is $ 0$.


limit
abstract algebra
function
superior algebra
superior algebra unsolved

Let $g ( n )$ be the number of groups of order $\leq n$.
Calculate (if exist)
\[ \lim_{n \rightarrow \infty} \frac{g ( n )}{n}  \]
In particular how many groups of order $n$ is there ''in the average'' ?
Also, what are the most "frequent" groups: simple groups, solvable groups, abelian groups, cyclic groups ?

just a guess, but what makes you think the limit with just n in the lower part is gonna converge?
my guess is that is grows much more quickly

for abelian groups there is a function [url]http://mathworld.wolfram.com/AbelianGroup.html[/url]

I found a nice paper of Ram Murty, somewhat related to my question:
http://www.mast.queensu.ca/~murty/density.dvi
It's from 1983, the question was raised before I was born ;)


that's interesting( not so much) but I don't know if i've chosen the right section to put this problem in or not.
We have several numbers and we decide to add another number to our group.this new number must be really new( not seen before) and the arithmatic mean of two or more old numbers. prove that starting with 0, 1 we can produce all the rational numbers in [0,1]. :)

Well,easily we can obtain every ratinal number a/b that b is power of 2.
Now considering no f these numbers with sum 1 e can obtain 1/n.
After that problem is trivial.


floor function
AMC
AIME
search
ratio
algebra
polynomial

Consider sequences that consist entirely of $ A$'s and $ B$'s and that have the property that every run of consecutive $ A$'s has even length, and every run of consecutive $ B$'s has odd length. Examples of such sequences are $ AA$, $ B$, and $ AABAA$, while $ BBAB$ is not such a sequence. How many such sequences have length 14?

[hide]
Let $ A_n$ be the number of such sequences of length $ n$ that end in $ A$, and $ B_n$ be the number of such sequences that end in $ B$.  We can form a sequence in $ A_n$ by appending two $ A$'s to any sequence that is two characters shorter, so $ A_n\equal{}A_{n\minus{}2}\plus{}B_{n\minus{}2}$.  We can form a sequence in $ B_n$ by appending one $ B$ to a sequence that ended in $ A$, or two $ B$'s to a sequence that ended in $ B$, so $ B_n\equal{}A_{n\minus{}1}\plus{}B_{n\minus{}2}$.

It is easy to work out that $ B_1\equal{}1$, $ A_1\equal{}0$, $ B_2\equal{}0$, $ A_2\equal{}1$.  Using the recursion repeatedly gives $ A_{14}\plus{}B_{14}\equal{}172$.
  [/hide]

[hide] We use recursion, with casework on the character a string ends in. Call such a string good. If a string of length n ends in A, it must have been formed by adding two A's to a string of length n-2 (for n>2), ending in either A or B. If it ends in B, it must have been formed by either adding two B's to a string of length n-2 (for n>2) ending in B or a string of length n-1 ending in A.

Let $ A_{n}$ and $ B_{n}$ denote the number of strings of length $ n$ ending in A and B, respectively. By the above, $ A_{n}\equal{}A_{n\minus{}2}\plus{}B_{n\minus{}2}$, and $ B_{n}\equal{}B_{n\minus{}2}\plus{}A_{n\minus{}1}$, for all $ n>2$.

Now, we use the fact that $ A_{1}\equal{}0,A_{2}\equal{}1,B_{1}\equal{}1,B_{2}\equal{}0$, and crank it out.

$ A_{3}\equal{}1,B_{3}\equal{}2$;
$ A_{4}\equal{}1,B_{4}\equal{}1$;
$ A_{5}\equal{}3,B_{5}\equal{}3$;
$ A_{6}\equal{}2,B_{6}\equal{}4$;
$ A_{7}\equal{}6,B_{7}\equal{}5$;
$ A_{8}\equal{}6,B_{8}\equal{}10$;
$ A_{9}\equal{}11,B_{9}\equal{}11$;
$ A_{10}\equal{}16,B_{10}\equal{}21$;
$ A_{11}\equal{}22,B_{11}\equal{}27$;
$ A_{12}\equal{}37,B_{12}\equal{}43$;
$ A_{13}\equal{}49,B_{13}\equal{}64$;
$ A_{14}\equal{}80,B_{14}\equal{}92$.

Thus, our answer is $ A_{14}\plus{}B_{14}\equal{}80\plus{}92\equal{}\boxed{172}$.[/hide]

(edit) dangit, e^pi*i beat me. well, there is another solution using casework on the number of B strings (0,2,4)

Casework is superfluous.  Here is a bijective solution.

[hide]
We replace "14" with "$ 2n$". We first note that we must have an even number of chunks of $ B$'s, because of parity issues.  We then note that every chunk of $ B$'s except the last one must end in the sequence $ BAA$, since we need at least two $ A$'s to separate it from the next chunk of $ B$'s. The last chunk of $ B$'s must, of course, end with a $ B$.  Thus our sequence must look like this :
\[{{{{{{{ \boxed{\quad A\text{'s} \quad}} \boxed{\quad B\text{'s} \quad}} BAA \boxed{\quad A\text{'s} \quad}} \cdots \boxed{\quad B\text{'s} \quad}}BAA \boxed{\quad A\text{'s} \quad}} \boxed{\quad B\text{'s} \quad}} B \boxed{\quad A\text{'s} \quad}} , \]
where each box holds an even number of letters (possibly zero).

If we want a sequence with $ 2k$ chunks of $ B$'s, then we have $ (2n - 6k +2)$ letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have $ (n - 3k + 1)$ pairs of letters to put into $ 4k + 1$ boxes. By a classic balls-and-bins argument, the number of ways to do this is
\[ \binom{n+k+1}{4k+1} . \]
It follows that the total number of desirable sequences is
\[ \sum_k \binom{n+k+1}{4k+1} . \]
For $ n=7$, this evaluates as $ \binom{8}{0} + \binom{9}{4} + \binom{10}{8} = 1+ 126 + 45 = 172$.  $ \blacksquare$ [/hide]

Boy Soprano: judging by the form of your answer (similarity to Fibonacci and the like), there should be a direct recursion for even-length strings...I'm a little lazy to find it however.

This is not the correct problem statement. See the topic Ignite started about this one.

I got enormously lucky semi-brute forcing this one... I found that one can only have an even number of b-strings, and that the number of a-strings has to be within one of the number of b-strings, and so generated the number of possibilities of A-string, B-string, A-string; A-string, B-string, A-string, B-string, etc.  I was very surprised that it all added up in the end.

Dammit, my fault everyone. Could a mod edit the problem statement, as I can't do it anymore?

[quote="mihail911"]I have a question regarding why my solution doesn't work.
i did A LOT of casework. The number of A's in the sequence has to be 0,2,4,6,8,10,12,14. For 0 As the number of sequences is obviously 1. for 2, 13. For every number of As>2, i consider each pair of As as one Big [b]A[/b]. thus the number for 4 A's is $ \binom{12}{2}$ i did similarly for the rest to get the outrageously large number of 610.  :([/quote]

OMG, that's what i got as my guess.  I just added up 14 choose 0, 13 choose 1, etc., up to 7 choose 7, and I got 610. :D

[quote="MellowMelon"]You do realize $ 610$ is not the correct answer? ($ 172$)[/quote]

 :| Yes...  I'm not that stupid. :wink: I'm just saying...

[quote="caffeineboy"][quote="mihail911"]I have a question regarding why my solution doesn't work.
i did A LOT of casework. The number of A's in the sequence has to be 0,2,4,6,8,10,12,14. For 0 As the number of sequences is obviously 1. for 2, 13. For every number of As>2, i consider each pair of As as one Big [b]A[/b]. thus the number for 4 A's is $ \binom{12}{2}$ i did similarly for the rest to get the outrageously large number of 610.  :([/quote]

That solution only works if the requirement on the sequence is the misstatement above. In the problem, there is the additional requirement that " every run of consecutive B's has odd length." Your answer is correct for the problem statement above, though.[/quote]

dangit carl.  :D

I like the "recursive" and "balls and urns" solutions that have already been presented.  But here's a straightforward counting method I came up with last week, requiring some [not excessive] casework.

We organize by the number of runs of $ B\text{'s}$.  The number of runs of $ B\text{'s}$ must be even, so the total number of letters is even.  There are 3 cases:  zero runs of $ B\text{'s}$; two runs of $ B\text{'s}$; or four runs of $ B\text{'s}$.  [Six runs of $ B\text{'s}$ would yield a string longer than 14 characters.]

If there are [b]zero runs[/b] of $ B\text{'s}$.  The string consists of 14 $ A\text{'s}$.  [$ 1$ solution]

If there are [b]four runs[/b] of $ B\text{'s},$ we start with the minimum possible sequence:  $ B\thinspace AA\thinspace B\thinspace AA\thinspace B\thinspace AA\thinspace B.$  We add markers to the ends:  $ O\thinspace B\thinspace AA\thinspace B\thinspace AA\thinspace B\thinspace AA\thinspace B\thinspace O.$  There are 9 groups of letters, which correspond to 9 places where we can add letters.  We have to add 4 letters to get to a string of length 14.  We always have to add letters two at a time.  If we add letters to an existing $ AA$ group, we must add more $ A\text{'s}$.  If we add letters to an existing $ B$ group, we must add more $ B\text{'s}$.  If we add letters to the endpoints, we must add $ A\text{'s}$.

Now, in this case, we can add 4 letters in any of the 9 places.  [$ 9$ solutions].
Or, we can add 2 letters to each of two places chosen from the 9.  9C2.  [$ 36$ solutions.]

If there are [b]two runs[/b] of $ B\text{'s},$ we start with the minimum possible sequence:  $ B\thinspace AA\thinspace B.$  We add markers to the ends:  $ O\thinspace B\thinspace AA\thinspace B\thinspace O.$  There are 5 groups of letters, which correspond to 5 places where we can add letters.  We have to add 10 letters to get to a string of length 14.  As before, we always have to add letters two at a time.

- Add 10 letters to one position.  [$ 5$ solutions]
- Add 8 letters to one position, 2 letters to another.  5x4.  [$ 20$ solutions]
- Add 6 letters to one position, 4 letters to another.  5x4.  [$ 20$ solutions]
- Add 6 letters to one position, 2 letters to two others.  5 x 4C2.  [$ 30$ solutions]
- Add 4 letters to two positions, 2 letters to another.  5C2 x 3.  [$ 30$ solutions]
- Add 4 letters to one position, 2 letters to three others.  5 x 4C3.  [$ 20$ solutions]
- Add 2 letters to each of five places.  5C5.  [$ 1$ solution]

[Note that we have to add five pairs of letters.  There are 7 partitions of 5.  Each subcase corresponds to one of the partitions of 5, so I have high confidence I have counted all of the subcases.]

Total:  $ 1\plus{}45\plus{}126\equal{}172$

[Also note that Boy Soprano's "balls and urns" method is a natural extension to this method.]

[hide="recursion solving start"]Expanding $ a_n$ as defined in the math jam, we get
\begin{eqnarray*}a_n&=&a_{n-2}+b_{n-2}= a_{n-2}+a_{n-3}+b_{n-4}\\
&=&a_{n-2}+a_{n-3}+a_{n-5}+a_{n-7}+a_{n-9}\ldots\\
&=&a_{n-2}+\sum_{k=1}^{\left\lfloor\dfrac{n}{2}\right\rfloor-1} a_{n-(2k+1)}\end{eqnarray*}
replacing $ a_n$ with $ a_{n-2}$ you get the same terms except for 3:
\begin{eqnarray*}a_n-a_{n-2}&=&a_{n-2}-a_{n-4}+a_{n-3}+(a_{n-5}-a_{n-5}+a_{n-7}-a_{n-7}\ldots)\\
&=&a_{n-2}-a_{n-4}+a_{n-3}\\
a_n&=&2a_{n-2}-a_{n-4}+a_{n-3}\end{eqnarray*}[/hide]
If anyone can do anything to solve this, that would be good.  I tried setting $ a_n=c^n$ and got nowhere.

Assuming your recursion is correct, the explicit solution to $ a_n \equal{} 2a_{n\minus{}2} \minus{} a_{n\minus{}4} \plus{} a_{n\minus{}3}$ depends on the roots to the quartic equation $ x^4 \minus{} 2x^2 \minus{} x \plus{} 1 \equal{} 0$.  In a contest situation, this would not be the best way of doing it (especially considering the above has no rational solutions).

Yeah, I know, I tried that already.  And I solved this one right on the AIME using the above methods. :)

For those searching for a closed form answer.  Following answer (without solution) shows how hopeless this would be without a calculator.

[hide]Adapted from [url]http://algo.inria.fr/bin/encyclopedia?Search=ECSnb&argsearch=484[/url]

Let $ \alpha_1, \alpha_2, \alpha_3, \alpha_4$ be the roots of $ \alpha^4\plus{}\alpha^3\minus{}1\equal{}0$

$ \alpha_1\approx \minus{}1.380278$
$ \alpha_2\approx \plus{}0.819173$
$ \alpha_3\approx \minus{}0.219447 \minus{} 0.914474 i$
$ \alpha_4\approx \minus{}0.219447 \plus{} 0.914474 i$

$ S_n\equal{}\sum_{j\equal{}1}^4{\frac{16 \plus{} 73\alpha_j \minus{} 3\alpha_j^2 \minus{} 12\alpha_j^3}{283\alpha_j^{2n\plus{}5}}}$

$ S_{14}\equal{}172$[/hide]

Interesting References:
  [url]http://www.research.att.com/~njas/sequences/A062200[/url]
  [url]http://www.anaphoria.com/meruone.PDF[/url].  [See sequence $ E_n$]

[quote="123456789"][quote="rcv"]For those searching for a closed form answer.  Following answer (without solution) shows how hopeless this would be without a calculator.[/quote]That's really interesting. How do people think of these explicit formulas? Is it more about patterns, or is there any generic way to find explicit formulas?[/quote]For linear recurrence relations - those in the form $ a_n : \sum_{i\equal{}0}^j k_ia_{n\plus{}j\minus{}i}\equal{}0$, so that there is nothing but multiples of previous terms in the relation, a closed form can be found that is the sum of $ j\plus{}1$ geometric sequences, where the common ratios are the roots of the polynomial $ \sum_{i\equal{}0}^j k_ix^{j\minus{}i}$. For example, with the Fibonacci sequence, which has the relation $ F_{n\plus{}2}\minus{}F_{n\plus{}1}\minus{}F_n\equal{}0$, the common ratios of the two geometric sequences in the closed form are the roots of $ x^2\minus{}x\minus{}1$. There's a section in AoPS vol. 2 that explains it better than I can.

There's a very general method we can apply here:  the answer is related to counting walks of length $ n$ on the graph with adjacency matrix

$ \left[ \begin{array}{cccc}
 0 & 1 & 0 & 0 \\
 1 & 0 & 1 & 0 \\
 1 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 \\ \end{array} \right].$

The first two vertices are $ A$s and the second two vertices are $ B$s.  The graph is constructed so that whenever you step on the first $ A$ vertex you have to do it an even number of times to get to a $ B$ vertex, and similarly for the $ B$ vertices.  This is conceptually neater than coming up with multiple sequences (which just count paths between multiple pairs of vertices), and I have hope that this leads in a natural way to Boy Soprano II's solution.  (Consider the similar combinatorial formula for the Fibonacci numbers.)

I just did this by doing some case work. A group of 2A's can always be combined with another group. So as long as you make sure to get the different possibilities for B lengths you can do some simple ball and urn (I think this is in that category, I've never seen it used outside of AoPS).

For example:
If you have 8 A's then you can think of it as |AA|AA|AA|AA| with the |'s representing where you can add B's. There are three different cases for lengths.
1) B,B,B,BBB
2) B,BBBBB
3) BBB,BBB

There are 17 cases in total, but they are simple to calculate and an alternative to the recursive method.

[quote="Octahedr0n"][quote="E^(pi*i)=-1"]...so $ A_n \equal{} A_{n \minus{} 2} \plus{} B_{n \minus{} 2}$.  We can form a sequence in $ B_n$ by appending one $ B$ to a sequence that ended in $ A$, or two $ B$'s to a sequence that ended in $ B$, so $ B_n \equal{} A_{n \minus{} 1} \plus{} B_{n \minus{} 2}$...[/quote]

I don't get this part of the solution.  Could someone please explain it to me?  Is there anything that I need to know about recursion?

Thanks![/quote]

So, we can divide the sequences in $ B_n$ into two groups: those which end in exactly one $ B$, and those which end in multiple $ B$'s.  The first group is formed by taking a sequence in $ A_{n\minus{}1}$ and appending a $ B$.  For the second group, we must end in at least three $ B$'s due to parity constraints, so these sequences take a sequence in $ B_{n\minus{}2}$ and append two $ B$'s to it.  Thus $ B_n\equal{}A_{n\minus{}1}\plus{}B_{n\minus{}2}$.

Did that make sense?

If I'm understanding E^(pi*i)=-1's solution correctly, using a recursion means building small sequences and adding on more As and Bs to it.  However, since I haven't done much recursion work, could you please explain how you got
[quote="E^(pi*i)=-1"]It is easy to work out that $ B_1 \equal{} 1$, $ A_1 \equal{} 0$, $ B_2 \equal{} 0$, $ A_2 \equal{} 1$.  Using the recursion repeatedly gives $ A_{14} \plus{} B_{14} \equal{} 172$.[/quote]
I get the concept of a recursion, but I don't know how to compute it.  Thanks!

This problem is a recursion problem but here is another way without it.

Call a B sequence a sequence of consecutive B's. Notice that there has to be an even number of B sequences and the maximum number of B sequences is 4.

Now we have an extreme amount of cases:

B B B B
BBB BBB B B
BBB B B B
BBBBB B B B
no B's
B B
BBB B
BBBBB B
BBBBBBB B
BBBBBBBBB B
BBBBBBBBBBB B
BBB BBB
BBBBB BBB
BBBBBBB BBB
BBBBBBBBB BBB
BBBBB BBBBB
BBBBBBB BBBBB

Now essentially we do sticks and stones on each of these cases with an A between each B sequence acting as a divider and multiply by the number of ways to arrange the B sequences

If we do all that work out we get our answer to be $\boxed{172}$.

Can anybody explain why this solution is wrong?

Sol: Let $X_n$ denote the number of such sequences of length $n.$ It is easy to see that $X_0=X_1=X_2=1$. Now, we see that if we have a sequence  of length $k$ that starts with A, it needs two As, and then some other valid sequence of length $k-2,$ starting with $A$ or $B$. If a sequence starts with $B,$ then we just have to choose when to "insert" the B into the string(or not). If we insert the A as the second letter, then we need two As followed by a string of length $k-3,$ and if the second and third letters are B and the fourth letter is an A, then we need a string of length $k-5,$ and so on. Using this recursion, we get the following values:

$X_0=X_1=X_2=1$
$X_3=X_1+X_0=2$
$X_4=X_2+X_1=2$
$X_5=X_3+X_2+X_0=2+1+1=4$
$X_6=X_4+X_3+X_1=2+2+1=5$
$X_7=X_5+X_4+X_2+X_0=8$
$X_8=X_6+X_5+X_3+X_1=5+4+2+1=12$
$X_9=X_7+X_6+X_4+X_2+X_0=17$
$X_{10}=X_8+X_7+X_5+X_3+X_1=27$
$X_{11}=38$
$X_{12}=59$
$X_{13}=86$
$X_{14}=129$.

Where is the error in my solution?

Well, I think it is a little bit difficult to think of the dynamic programming solution. Here is my solution for this problem:
As someone mentions above, we can take "AA" as a new element so that no matter how many "AA" gets together, we can make sure that it always produce an even number of "A". Now we just put some sticks to the gap between different "AA".(We should take notice that "B" should always be even number)
(1) "B“ = 2 and "A" = 12 :  As we see, the only possible way is to put two "B" into _AA_AA_AA_AA_AA_AA_ , so $ \binom{7}{2}$ = 21
(2) "B" = 4 and "A" = 10 : As we see, similarly with (1), we have 6 sticks and two types (B BBB and B B B B) all together, so $ \binom{6}{4}$+ 6*5 = 45
(3) "B" = 6 and "A" = 8 : As we see, five sticks ok ? and three types (BBB BBB and BBBBB B and also BBB B B B, we should notice that B B B B B B can not be put here coz 6>5),  so $ \binom{5}{2}$+ 4*5 +4*5 = 50
(4) "B" = 8 and "A" = 6 : As we see, 4*3+4*3+4+6=34 (BBB BBBBB and BBB BBB B B and B B B BBBBB)
(4) "B" = 10 and "A" = 4 : As wee see, 3*2*2+3=15 (B BBBBBBBBB and BBB BBBBBBB and BBBBB BBBBB)
(5) "B" = 12 and "A" = 2 : As we see, 2*3 = 6 (B BBBBBBBBBBB and BBB BBBBBBBBB and  BBBBB BBBBBBB)
(6) "B" = 0 and "A" = 14 " As wee see, only one case, "AAAAAAAAAAAAAA"
 In total, as it mentioned above, we have the final result 21+45+50+34+15+6+1=172


[quote="yaa"]Let a,b,c>0 and abc=1. prove that:
$ \frac {a}{{{b^3} \plus{} 2}} \plus{} \frac {b}{{{c^3} \plus{} 2}} \plus{} \frac {c}{{{a^3} \plus{} 2}} \ge 1$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=304320


inequalities
algebra
polynomial
inequalities proposed

Find the condition for $\frac{a}{b}\ (b\neq 0)$ such that the following inequality satisfies for all real numbers $x$.

\[\frac{1}{3}\leqq \frac{x^2-ax+a^2}{x^2+bx+b^2}\leqq 3\]

Multiplying out on each side, I get $2x^2+(3b+a)x+(3b^2-a^2)\ge0$ and $2x^-(3a+b)x+(3a^2-b^2)\ge0$ for all $x$.  Thus the discriminants of these two polynomials must be negative, i.e. $(3b-a)^2-8(3b^2-a^2)<0$ and $(3a+b)^2-8(3a^2-b^2)<0$.  Rearranging, and letting $k=\frac{a}{b}$, we get $3k^2+2k-5<0$ and $5k^2-2k-3>0$.

The first inequality requires $k\in\left(-\frac{5}{3},1\right)$, and the second requires $k\not\in\left(-1,\frac{5}{3}\right)$.  Combining the two gives $k\in\left(-\frac{5}{3},-1\right)$.

(I've been a bit hasty with the actual number crunching here, but I think the approach is right...)

Thank you for your reply, Diarmuid.
Regrettably your answer is wrong a little bit, but your approach is right.

kunny


can someone crack these and show all work, so i can figure them out. thanx. please do them step by step!!!
-3/y+6(4)=6

5-(2y)6+7(6)=15(3)

thanx guys..

Didn't you post homework problems before? I'm sorry, but I'm only giving hints.

[hide="hint"]
Use the associative, commutative and distributive property of multiplication. Also remember that division is the inverse operation of multiplication.
[/hide]

this has already been posted in classroom basics so ill let them answer it

doesnt belong here anyways

locked


There isn't really much "solution" on this problem.

[hide="View"]

For $k = 1$, we get 0.

For $k = 2$, we get $\frac{1}{2}$.

For $k = 3$, we get $\frac{1}{3}$.

For $k = 4$, we get $\frac{1}{8}$.

For $k = 5$, we get $\frac{1}{30}$.

But since $0 + \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{30} > 0.99$, the maximum value of $n = 4$. So, C.[/hide]

SnowStorm, right idea but wrong letter.

Ohh... Great! i'm so stupid :D. It was obviously :P


topology
advanced fields
advanced fields unsolved

A surjective local homeomorphism is not always a covering, even if it has finite fibers (for example,
$ p: (0,2)\to S^{1}, x\mapsto (cos2\pi x,sin2\pi x)$).
What "reasonable" condition should we assume to make such a local homeomorphism a covering map? For example, is it true that a surjective local homeomorphism with finite CONSTANT fibers is a covering?

The answer to the second question is NO: take the same map on the interval $ [0,2)$.

I was wrong: with $ [0,2)$ it is not a local homeomorphism.

Consider a Hausdorff space $ X$ and a surjective local homeomorphism $ f\colon X\to Y$ satisfying $ \#f^{\minus{}1}(y)\equal{}n$ for all $ y\in Y$. Now take any $ y\in Y$ and set $ f^{\minus{}1}(y)\equal{}\{x_1,\ldots,x_n\}$. For each $ x_i$, choose an open neighbourhood $ U_i$ such that the $ U_i$ are mutually disjoint and $ f|_{U_i}$ is a homeomorphism onto an open subset of $ Y$. Set $ V\equal{}\bigcap_{i\equal{}1}^nf(U_i)$. It is clear that $ f^{\minus{}1}(V)$ contains $ n$ homeomorphic copies of $ V$, one contained in each $ U_i$. By the assumption that all fibres have precisely $ n$ elements, $ f^{\minus{}1}(V)$ cannot contain any additional points.

There are counterexamples if $ X$ is allowed not to be Hausdorff: let $ X_1$ be the real line "with doubled origin" (i.e., $ \mathbb R\times\{0,1\}$ modulo the equivalence relation $ (x,0)\sim(x,1)$ if $ x\ne0$), $ X_2\equal{}\mathbb R\setminus\{0\}$, $ X\equal{}X_1\sqcup X_2$ with the obvious map to $ Y\equal{}\mathbb R$. owk


The number 210 is the product of two consecutive positive integers and is also the product of three consecutive integers. What is the sum of those five integers?

\[ 210\equal{}20\cdot 21\equal{}5\cdot 6\cdot 7\rightarrow 20\plus{}21\plus{}5\plus{}6\plus{}7\equal{}\boxed{59}\]