THE WEBSITE IS DIFFERENT

gasp

I feel so alone in this world

what can I count on anymore?

even the AMC website changes!

I like the new change. It had been getting a little... messy.

Glad you noticed the new format.

We are experimenting with a new format that I expect will make it easier to maintain the wesite.  I hope it will also make it easier to navigate.

Please post suggestions and ideas for improvements, we'd be glad to hear them.

We are aware that a number of links are currently broken and everything may not be working perfectly.  Please be patient as we get those fixed, we are aware of them.

Steve Dunbar
AMC Director

Nothing seems different to me. :huh:  :huh:  :huh:

After I tried to access the problem dir. page it appears a 404 error message with a remarkable [quote]Absence Makes the Heart Grow Fonder[/quote]  Hmm..
I love the new appearence of the website.  :)

BTW, It is always fun and makes me happy while reading many of MysticTerminator's posts, they are just so funny. :D

Haha yea, it is quite different, but I like it. It's kinda funny to see the CSS and XHTML compatible signs  :D

For those of you who said nothing is different, are you sure you're on this site: http://www.unl.edu/amc/index.shtml ?? :D

Ok I see the difference now  :wink:  All the links lead to the old page though.

Oh I see. I thought you guye were talking about AOPS. :lol:

[quote]The American Mathmatics Competitions:[/quote]
That's on the front page - you might want to check the spelling.

It's fixed now.  There are probably more, RuthAnn is working fast to get the pages all converted.

Thanks for pointing it out.

Steve Dunbar
AMC Director

I can't seem to find many of the pages that were available on the old website, such as the USAMO Archives.  :maybe:  Will these still be available? If so, where are they?

I don't like the new one. Too bright for a math site  :wink:

I can't find the pages as easily as before since the links are so small or I am losing my sight.

[quote="Karth"][quote="bpms"]Notice how the topic is OH MY GOD ! [b]s[/b] :D[/quote]

I think Arnav wished to call forth his plethora of god[b]s[/b] due to his polytheistic religion  :P .[/quote]
:rotfl:
Wait, my religion is polytheistic too...

[quote="nebula42"]I can't seem to find many of the pages that were available on the old website, such as the USAMO Archives.  Will these still be available? If so, where are they?[/quote]

is this it?: http://www.unl.edu/amc/e-exams/e8-usamo/archiveusamo.shtml

overall the site looks a lot sleeker and more professional (not that the old one wasn't professional or anything  :P )

the picture on the homepage is amazingly awesome. i like how its so...pyramidy  :P

[quote="MysticTerminator"][quote="Karth"][quote="bpms"]Notice how the topic is OH MY GOD ! [b]s[/b] :D[/quote]

I think Arnav wished to call forth his plethora of god[b]s[/b] due to his polytheistic religion  :P .[/quote]

:what?:[/quote]

DUDE YOU KNOW IT'S TRUE. It's okay... I have a polytheistic religion too. I find it particularly useful, as I can add multiple gods to my pantheon. Some of my recent gods are:

Karamata
rrusczyk
AMC

Wow, the site is amazingly different, it looks wonderful!!!  The AMC 8 info is much easier to find now.

Smiles,
lyra

[quote="bpms"]Some of my recent gods include:
Brahma,
Vishnu,
Shiv,
Ram,
Hanuman,
Krishna
the list goes on and on. :D
Also, to prevent this from being spam, 1996 and 1997 have broken links for the list of USAMO qualifiers, so I can't see if Reid Barton qualified in sixth like I was wondering about...[/quote]

Hah. Nicee same here. Just to make this a good post, have the USAMO, AMC archives been updated yet? If not, when will it be?  :(


Nat wrote this addition puzzle for his parents. The sum is the amount of money he needs in order to purchase a computer game. Each distinct letter represents a different digit. The letter E represents an even digit. How many dollars does Nat need to purchase his game? Express your answer as a decimal to the nearest cent.\\
$ \$$T.EN\\
$ \plus{}$ .EN\\
$ \overline{\$ N.AT}$\\

The only way I can think is to do some guess and check.

$ E$ has to be even, so let's try:

$ 8.69 \plus{} 0.69 \equal{} \boxed{9. 38}$

EDIT: no guessing is needed

TEN+EN=NAT, so T is also even

E is greater than 5 because otherwise, T=N; and N=T+1, so N is odd, and 2N=T mod 10, so N=-1 mod 10, so N=9 and T=8. E is even, and its greater than 5, so E=6; and the price of teh game is 9.38


Αν $ x$, $ y$ θετικοί πραγματικοί να δείξετε ότι: $ x^{y}$ $ \plus{}$ $ y^{x}$ $ >$ $ 1$

Την άσκηση την έλυσα μόλις τώρα και θα έλεγα πως πιο πολύ ανισότητα ειναι παρά Θ.Α. Είναι αρκετά δύσκολη, για αυτό δίνω ένα απο τα κούμπια της.
[hide]Άνισοτητα Bernouli (παίξτε λίγο με τη μορφή της)[/hide]

Εχει συζητηθει πολλες φορες.Λυση υπαρχει με bernulli και αναλυση απο οσο ξερω εγω.
H ιδεα στην Bernulli ειναι να γραψεις το x σαν (1-(1-x)) και στην αναλυση να χρησιμοποιησεις fermat...  :)

Βάζω την λύση μου με Bernoulli, γιατί μου φαίνεται πως είναι διαφορετική από αυτή που πρότεινες.

1) Για $ x$ μεγαλύτερο  ή ίσο του άσου, το ζητούμενο είναι προφανές.
2) Για $ x$ μικρότερο του άσου, έχουμε $ (1 \plus{} x/y)^{1/x} > 1 \plus{} 1/y$, που ισοδυναμεί με $ y/(x \plus{} y) < y^{x}$. Συμμετρικά: $ x/(y \plus{} x) < x^{y}$. Προσθέτοντας κατά μέλοι $ x^{y} \plus{} y^{x} > 1$.


$\displaystyle ab^2cosA = bc^2cosB$
From this we get (do calculations...)
$a^4+c^4=b^2(a^2+c^2)$

similarly we get $b^4+c^4=a^2(b^2+c^2)$
              and    $a^4+b^4=c^2(a^2+b^2)$

however the following ineq holds $2(a^4+c^4)\geq(a^2+c^2)^2$
Thus  $2b^2\geq(a^2+c^2)$

In the same way $2c^2\geq(a^2+b^2)$
              and      $2a^2\geq(b^2+c^2)$

Add these three ineqs to get $2a^2+2b^2+2c^2\geq b^2+c^2+b^2+a^2+a^2+c^2$

so in every ineq the equality holds that is

$b^2+c^2=2a^2$
$a^2+c^2=2b^2$
$b^2+a^2=2c^2$

Now it is easy to see that $a^2=b^2=c^2$

[size=150]([/size]just think of the system
2x=y+z
2y=x+z
2z=x+y  [size=150])[/size]

Finally the triangle is equilateral....
Quite tiring my solution
Maybe there is some more elegant.... :)  :P

[quote="indybar"]In triangle $ABC$, we know that $ab^2cosA = bc^2cosB = ca^2cosC$. What can we tell about it?[/quote]

We have $ab^2cosA = bc^2cosB = ca^2cosC$

$\Rightarrow \frac{{bCosA}}{c} = \frac{{cCosB}}{a} = \frac{{aCosC}}{b}$

$\Rightarrow \frac{c}{{bCosA}} = \frac{a}{{cCosB}} = \frac{b}{{aCosC}}$

Using Sine Rule,  we can write this as

$ \Rightarrow \frac{{SinC}}{{SinBCosA}} = \frac{{SinA}}{{SinCCosB}} = \frac{{SinB}}{{SinACosC}}$

$\Rightarrow 1 + \frac{{TanA}}{{TanB}} = 1 + \frac{{TanB}}{{TanC}} = 1 + \frac{{TanC}}{{TanA}}$

which gives us $A=B=C$

Hence the traingle is equilateral.

[quote="shobber"][quote="vidyamanohar"]
$\Rightarrow 1 + \frac{{TanA}}{{TanB}} = 1 + \frac{{TanB}}{{TanC}} = 1 + \frac{{TanC}}{{TanA}}$

which gives us $A=B=C$
[/quote]
Why?[/quote]

$Let\,\,\frac{p}{q} = \frac{q}{r} = \frac{r}{p} = x$

$\Rightarrow p = qx = rx^2  = px^3  \Rightarrow x = 1$

Hence $p = q = r$

Hope this makes the things clear.

Actually, the solution given by vidyamanohar
is much shorter and of course much nicer than mine.
However can you take a look at my solution
and tell me whether it is true, just to know if it is valid......

Thanks! :)  :)

how about a triangle where:

$a^2 = ca(cos B) + ab(cos C) - bc(cos A)$

thought of this when I read the original question..


topology
real analysis
real analysis unsolved

Let $\sigma$ be the set of all infnite sequences of $0$'s and $1$'s. Define the metric $d(a,b)=\sum \frac{|a_{n}-b_{n}|}{2^{n}}$ where $a=(a_{n})$ and $b=(b_{n})$ are points in $\sigma$.
[b]i)[/b]Prove that $\sigma$ is compact. [b]ii)[/b]Prove that $\sigma$ is homeomorphic to the Cantor set.

Ok, since you didn't give any ideas, I give my idea:D Since we have a metric here, it might be helpful to check whether the space is sequentialy compact or not instead of covering compact.

Sketch of proof: [b]i[/b])(This is due to my friend Hossein Mohades, and the last part is from me.)
First we show that every sequence in $\sigma$ has a Cauchy subsequence and then we prove that the space is complete, then according to my comment above, which I couldn't edit anymore!, we will be done.
Suppose we are given an infinite sequenece.It is clear that there exists a subsequence whose first-place elements are the same (zero or one). Pick a subsequence of this subsequence such that their second-place elements agree, continuing this way we have found a Cauchy subsequence(choose $\epsilon$ in this form: $\frac{1}{2^{N}}$.Using this trick again one can show that the space is complete ).

[b]ii[/b])Every member of the Cantor set is of the form $\sum_{k=1}^{\infty}\frac{a_{n}}{3^{k}}$ where $a_{n}\in \{0,1\}$. Define $f: \sigma \to C$ which doubles the elements of each sequence. Since $\sigma$ is first countable, assuming $x_{n}\to x$ easily yields $f(x_{n})\to f(x)$. QED


Find all solutions to \[ 2^{a}+3^{b}=5^{c}, \]

    for positive integers $a, b, c$.

See : http://www.mathlinks.ro/Forum/viewtopic.php?t=48431 . :)

The solutions are : (1, 1, 1) and (4, 2, 2).


limit
number theory proposed
number theory

Find all the pairs of natural numbers $(a,b)$ such that for any natural $n$ there exists natural $c$ such that $a^n+b^n=c^{n+1}$

Suppose $|a|\neq |b|$ then we choose $|b|>|a|$.
$a^{2n}+b^{2n}=c^{2n+1} \Rightarrow \frac{a^{2n}}{b^{2n}} + 1 =\frac{c^{2n}}{b^{2n}}\times c \Rightarrow \lim_{n\rightarrow +\infty} \Big(\frac{c}{b}\Big)^{2n} = \frac{1}{c}$, it's possible only if $c=1$ and $b=1$ or $-1$ so $(a,b)$ is $(0,1)$ or $(0,-1)$ (and of course $(1,0)$ and $(-1,0)$)
If $|a|=|b|$ then there is at most one positive since a+b=c^2, we set $a$ positive and $b=\delta a$ with $\delta=1$ or $-1$
Then two cases for $\delta$ works, if $\delta=-1$ then for $n$ odd we take $c=0$ and $n$ even we have $2a^n=c^{n+1}$ then $a=-b=c=2$. if $delta=1$ it's the same thing $a=b=c=2$.

Consequently we have $(0,1),(1,0),(0,-1),(-1,0),(2,-2),(-2,2),(2,2)$


Hi everyone!

is there anybody here who gave the astronomy olympiad this year or at any time in the past.i was in the astronomy olypiad team last year.i have joined this group only recently.i just want to inform you all that an olympiad such as astronomy exists because many ppl claim that they never receive any information ,or worse still, never get to know abt it.

Udbhav

Hello,
    My brother in Standard IX has given his INAO - Junior this year.What are the cut offs usually?
Sathej

I did the INAO Junior. It wasn't too bad but I didn't do too wonderfully either. Anyway there is always next year.
   Ashwath Rabindranath

Please post the result here when it comes out ---- it seems HBCSE hasn't put it online yet!!!

Hmmm.....I wrote the astronomy junior as well....it had nothing much to do with astronomy though -- most of it were basic physics { Even then , I really didn't get all of them right but then that's a different story...}......I dont seem to find any place on the web where we get more than one question paper........(at least for the junior level.....I haven't really searched for anything in the senior level...}....

Ok, I basically wanted two things --

1) Does anyone know a good site where you can find previous astro olympiad papers of India and also some papers from other countries and/or papers of the International Astronomy Olympiad ?

2) If someone remembers the questions, please start a thread for the discussion of those questions......it might not only help in building up a question bank kind of thing, but will also help in everyone understanding the various ways in which each person looks at the sum.

[quote="udi_s37"]hi!
    last year i was in juniors.the highest was 73 and the cut off point was abt 45-50.for seniors the highest was 84(by abhishek dang  :lol: )the cut off point was abt 50-55.it changes year by year but a score of abt 55-65 can surely get u in.(the scores were'nt supposed to be shown but we did some mischief with the comps at the olympiad :lol: )

Udbhav[/quote]
This cutoff is out of 150 or out of 100 ?


induction
Mandelbrot Competition
Pascal\u0027s Triangle

I was going over a Mandelbrot Competition from the early 1990s earlier today, and I came across this problem. It was a #7, so I don't expect it to be trivial, but it looks very interesting, so I would like to know how to solve it.

Evaluate the sum

19/93+19*18/93*92+19*18*17/93*92*91+...+19!/93*92*91*...*75.

I tried writing it as 19C1/93C1+19C2/93C2+19C3/93C2+...+19C19/93C19, but I didn't get anywhere with that. Would someone mind helping please? Thanks.

From first impressions it looks as if the sum will telescope in some way. However I haven't been able to find out how so heres a slightly messier proof by induction:
Let f(a,b) for b>a be a/b + a(a-1)/b(b-1) + ... + a!/b(b-1)..(b-a+1).
We now prove that f(a,b) = a/(b-a+1).
By multiplying f(a,b) by b/a we get b/a f(a,b) = 1 + f(a-1,b-1).
f(1,x) = 1/x = 1/(x-1+1) so its true for a=1.
Suppose its true for a=k, now f(k,x) = k/(x-k+1) for all x>k
f(k+1,x+1) = x/k f(k,x) - 1 = ... = (k+1)/(x-k+1) for all x+1>k+1.

So by induction the formula is true and thus the sum we want is equal to 19/75.

Anyone see a simpler way to do it (I'm pretty sure there will be.)

Nice induction; don't think I've seen it done that way before.

Your instinct is right about there being a non-inductive way to do it.  Play around with some simple cases (start from 3/4, 3/5, 3/6) & see if you see a pattern...

(Disclaimer: I was a co-creator of the Mandelbrot (story for another day - how we got the thing started) and was one of the writers of the test this appeared on, though I don't think I wrote this question...)

Any more hints?

This is basically what I thought as soon as I saw it..
hmm small cases, answer seems to be 19/75. hmm everything else must cancel out in some way, maybe theres some sort of telescoping thing going on, maybe we could pair up each bit and add them in some other way.. nothing seems to be working. Ah fine I'll use induction :).

3/93+(3*2)/(93*92)+(3*2*1)/(93*92*91)
=
((3*92*91)+(3*2*91)+(3*2*1))/(93*92*91)

sub 19 for 3 and keep going.there probably is a much easier way but that is the pattern I found....OH WAIT...since there are 19 consecutive factors in the denominator...and we can factor out a 19 out of the numerator...those will cancel leaving a sum...which I can't find...well crap...a little help pls?

Take a look at those numbers in the expressions I came up with:

3/4 + 3*2/4*3 + 3*2*1/4*3*2 = ([color=red]3[/color]+[color=red]2[/color]+[color=red]1[/color])/[color=red]4 [/color]

3/5 + 3*2/5*4 + 3*2*1/5*4*3 = ([color=red]6[/color]+[color=red]3[/color]+[color=red]1[/color])/[color=red]10[/color]

3/6 + 3*2/6*5 + 3*2*1/6*5*4 = ([color=red]10[/color]+[color=red]4[/color]+[color=red]1[/color])/[color=red]20[/color]

Where have you seen those numbers before?

I think I get it...the differences between the numbers in each pattern are the numbers in the pattern b4 it...

P1:1,2,3,4

P2:1,3,6,10=1,1+2,1+2+3,1+2+3+4

P3:1,4,10,20=1,1+3,1+3+6,1+3+6+10

I haven't seen these numbers before but that is the pattern I came up with...is this what you are looking for?

I'm not sure if this helps, but I think that the numbers are the diagonals of Pascal's triangle.

Ok, after numerous hints, I think I got it. If your original fraction is n/r, then the answer will be (r choose r-n+1)/(r choose r-n). Let me just check my math...

Ok after Mr. Rusczyk's hint and Bookworm's call to reality, I was able to observe the positions of these numbers on Pascal's triangle. I am not sure exactly why they are so neatly arranged on Pascal's triangle, which is what I am working on now, but in any case, I was able to use block-walking (outlined in AoPS) to calculate that final answer.

I have no idea how that works but it does...can anyone explain triplem's method for me?

I believe this is correct; all you have to do is prove it now. It's pretty easy to prove with induction.


easy!

see the numbers ${a_{1}}^{2},{a_{1}}^{2}+{a_{2}}^{2}, ... ,{a_{1}}^{2}+{a_{2}}^{2}+...+{a_{n}}^{2}$

if no one of them are a multiple of n, two of them has the same rest when divided by n. so, your difference are a multiple of n!


Determine the pairs of integer $ (x,\ y)$ such that $ 55x^2\plus{}2xy\plus{}y^2\equal{}2007$.

[hide="sigh"]$ 54x^2\plus{}(x\plus{}y)^2\equal{}2007\equal{}9*223$

We can try various values of $ x^2$ up to 36 and we see that only $ x^2\equal{}9$ and $ (x\plus{}y)^2\equal{}1521$ works. $ (3,26)$, $ (3, \minus{}32)$, $ (\minus{}3, \minus{}26)$, $ (\minus{}3, 32)$[/hide]


function
analytic geometry
graphing lines
slope
calculus
calculus computations

The only one implicit problem I stuck with is:

Find the equations of the tangent lines to the following curve at the indicated points:

21. $ \ln(xy) \equal{} 2x$ at $ (1,e^2)$.

Consider $ y$ as a function of $ x$ and differentiate both sides of your equation with respect to $ x$ using the chain rule when needed. Now (try to) answer the following questions in this order:
a) What do you get?
b) Can you solve the resulting equation for $ y'$?
c) What value of $ y'$ do you get for $ x\equal{}1$, $ y\equal{}e^2$?
d) How is $ y'$ related to the slope of the tangent line?
e) What is the equation of the tangent line?


In $ \triangle ABC$, let the incircle touches side $ BC,CA,AB$ at points $ D,E,F$ respectively. Let $ M,N,P$ be the midpoints of sides $ BC,CA,AB$. Let $ DF$ cut $ MN$ at $ Y$, $ DE$ cut $ MP$ at $ X$, $ XY$ cut $ BC$ at $ A'$.

Prove that $ \frac{AX}{XA'}\equal{}\frac{AY}{YA'}$.

It is easy to prove that $ A,X,Y$ lie on a line. ( Let $ AI\cap FD\equal{}Y'$ . It is well-known that $ CY'\perp AY'$. Therefore $ ZA\equal{}ZY'$, $ \angle ZY'A\equal{}\angle BAY'$ -> $ Y\equal{}Y'$)
Let $ DF\cup AC\equal{}Z$.
$ (AA',XY)\equal{}(AC,EZ)\equal{}1$....

[quote="easternlatincup"][quote="Little Gauss"]
$ (AA',XY) \equal{} (AC,EZ) \equal{} 1$....[/quote]

Can you explain this more?[/quote]

he meant \[ (AA',XY) \equal{} (AC,EZ) \equal{} \minus{}1 \ and \ (AB,CD)\equal{}k \\ means \  \frac{\frac{AC}{CB}}{\frac{AD}{DB}}\]


group theory
abstract algebra
calculus
integration
number theory
superior algebra
superior algebra unsolved

This may be rather subjective, but does anyone have interesting examples of normal subgroups N of some G and the corresponding quotient groups G/N?

One interesting fact: $G/Z(G) \cong \textrm{Inn}\ G$, the group of inner automorphisms of $G$.

Try to prove the isomorphism yourself.

Also, prove that $\textrm{Inn}\ G \lhd \textrm{Aut}\ G$. The group $\textrm{Aut}\ G/\textrm{Inn}\ G$ is the group of "exterior automorphisms" of $G$.

[quote]What is "interesting"¿[/quote]

As long as you find it interesting.
For example, such groups that occur in other branches of mathematics outside algebra may be interesting (don't know of any myself)

An important example from number theory:
The invertible ideals $a$ (meaing that there is an ideal $b$ with $ab = (1)$) of a (number theoretic, especially integral) ring $R$ build an abelian group $P(R)$ by ideal multiplication. In those, we can take the principle fractional ideals to be those of type $rR$, $r \in Quot(R)$; this ones build a subgroup $J(R)$.
Then the Picard group is defined to be $Pic(R) : = P(R)/J(R)$.
For the special case of a dedekind ring, this is the same as the class group.


geometry
geometric transformation
rotation
symmetry
counting
distinguishability
Alcumus

A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular to each other. Each sector is to be painted with one of four colors. How many distinct ways can the table be painted? (A color may be used on more than one sector, but paintings that are the same after a rotation are not considered distinct.)

Is there a good way to do this and refraining from excessive listing?

Listing isn't excessive.  It's a team round, so $ 4$ people, $ 20$ minutes, $ 10$ questions, so one person gets $ 8$ minutes to solve this problem.  That's a lot of time, for this simple listing.

Yes but this is [i]me[/i] we're talking about, the idiot who can't count for his life.

[quote="mathemagician1729"]
Is there a good way to do this and refraining from excessive listing?[/quote]

When I tried this problem, to me it was not that much of excessive listing, but imagining a case and putting colors in. There are a few subcases but it eventually works out to the best.

Case I: All the same color
[hide]Because there are only 4 colors, there are only 4 choices for this one.[/hide]

Case II: Half is the same color, the other half is the same but different color
[hide]
There are two possible ways to paint: One color on top of the other, or same color diagonal to each other. This means there are 2 ways to paint with the same color.

How can I pick 2 colors out of 4. 4C2= 6

So that is 6*2= 12 ways
[/hide]

Case III: Three are the same color, and one is different
[hide]Because of rotation symmetry, all we need to think about is how many ways I can pick that one color for the three same regions, and the other color that is different.

That would 4*3= 12 ways[/hide]

Case IV: Two of the same color, two that are different.
[hide]How many ways can I paint this: same colors over the different colors, same colors diagonal to each other. That make two ways.

Let's first do Method 1:
[hide]That's just (4*3*2)/(2*2)= 6, for each color to be used twice. Since there are four colors that makes 6*4= 24 arrangements for this way.[/hide]

Method 2:
[hide]Again you will reach 24 arrangements as last. BUT because this is diagonal, you have to account for rotational symmetry. This cuts thing is in half to make only 12 possible ways.[/hide][/hide]

Case V: All are different
[hide]Set a color at the top, say red. From there are only 3*2= 6 ways to arrange the other colors. Because of rotational symmetry, it does not matter which color you set at the top and there are only 6 ways here[/hide].

4+12+12+24+12+6= [b]70 [/b]arrangements in total.

This may seem confusing without a diagram (to find a solution with a diagram click on the following link and read solution #10 [url]http://mathcounts.org/Document.Doc?id=223[/url])

It may not be MUCH of an easier way, but approaching things systematically avoids any mistakes in the solution.

[quote="jhangil"]Is there an easier way for this? I was thinking that since there you could choose from four different colors on each section it would be 4^4 but you have to count for rotational symmetry so you divide by 4 to get 64. I am not sure about this rotational symmetry (should do more Alcumus). Someone help![/quote]

The thing is, what if you have $ 4$ of the same color?  Then it is only counted once when you do $ 4^4$, but you still divided $ 4$ out.  Same thing with $ 3$ and $ 2$ of the same color.

Ha, I knew that it was around 64 at state, so I just guessed like 90 or something. I'm surprised I actually can get this now. Distinguishability problems are the ones I phail most at.

I thought this would be easier....

If you pick a random sector... Any sector... You have 4 choices for it. So you put down any random color. Then, you pick another random, unselected sector. You still have 4 colors because in the problem, it says you can repeat the colors. You continue this process until you have colors on every sector.

In Conclusion, it turns out to be 4^4 because you have 4 for every slot.

(I think...  :wink: )

Then you overcount for rotations..

For example
let Y be yellow B be blue R be red and G be green.

YB
RG

is the same as

RY
GB

yet you count them as the same.

Oh I should really read first.

IGNORE POST

I've seen a really good solution in Alcumus invoking rotational symmetry (but it's not there anymore?).

Basically they divided the $ 4^4$ cases counting rotations in to the $ 4\cdot1\equal{}4$ that are counted once (all one color), the $ 4\cdot3\equal{}12$ that are counted twice (each sector is the same color as the opposite one), and the $ 4^4\minus{}4\minus{}12\equal{}240$ that are counted four times (all other ones), and then divided by the number of rotations for each to get $ \frac41\plus{}\frac{12}2\plus{}\frac{240}4\equal{}\boxed{70}$.

My solution that I found like right after states was to first partition in to the sets $ \{4,0,0,0\},\{3,1,0,0\},\{2,2,0,0\},\{2,1,1,0\},\{1,1,1,1\}$, which represent the number of times each color is used (so the set $ \{2,1,1,0\}$ means that one color is used twice, two others are each used once, and the last is not used), which actually was pretty easy to do. This probably doesn't work too well for higher numbers of sectors, though (but the aforementioned method still works well :) ).

[quote="Mewto55555"]But the stupid kid on our team who can never get #10 but still insists on doing it put 64. Although we wouldn't have had enough time anyway, we barely got the obtuse triangle one down in time.[/quote]

 :P

Anyhow, we beasted that obtuse triangle one...and we're definitely changing our strategy this year. :wink:

Edit: What are you talking about, AIME15? :huh:

Definition of [b]beasted[/b]: Utilized the Law of Cosines and abused the TI-84.

Spam belongs in the Missouri Forum, not here.


The product of two of the four zeros of the quartic equation

$x^{4}-18x^{3}+kx^{2}+200x-1984=0$ is $-32$

Find $k$.

[hide="Solution"]Let the roots be $a,b,c,d$. Then by Vieta we have the system \begin{eqnarray}a+b+c+d &=& 18\\ ab+ac+ad+bc+bd+cd &=& k\\ abc+abd+acd+bcd &=&-200\\ abcd &=&-1984\end{eqnarray} and it's given that $ab=-32$. Plug that into $(4)$ to get \[cd=62\] Then $(3)$ becomes \[-32c-32d+62a+62b=-200\] Rewrite it as follows and couple it with $(1)$: \begin{eqnarray*}62(a+b)-32(c+d) &=&-200\\ (a+b)+(c+d) &=& 18 \end{eqnarray*} Substitute $u=a+b, v=c+d$ to get \begin{eqnarray*}62u-32v &=&-200\\ u+v &=& 18 \end{eqnarray*} The solution to this system is $u=4, v=14$

Now $(2)$ becomes \begin{eqnarray*}k &=& ab+ac+ad+bc+bd+cd\\ &=&-32+a(c+d)+b(c+d)+62\\ &=& 30+(a+b)(c+d)\\ &=& 30+uv\\ &=& \boxed{86}\end{eqnarray*} [u]NOTE[/u]: From $ab=-32,a+b=4$ and $cd=62, c+d=14$ it can be easily found that the roots are $x_{1}=8, x_{2}=-4,x_{3,4}=7\pm i\sqrt{13}$
[/hide]

[quote="JBL"]Could a moderator please delete this post and the three which precede it as spam?  Thank you.[/quote]Weiyan, please do not post the answer to the problem without justification. Thank you.

Posts deleted. :)


algebra
polynomial
complex analysis
linear algebra
linear algebra unsolved

Let A and B  be two nxn matrices  with elements in a field F.  Suppose that  B has more eigenvalues  than A.
 Prove that there is no  polynomial h such  that h(A)=B.

It is easy. Let $ K$ be the algebraic closure of $ F$. If $ B\equal{}h(A)$ and $ spectrum(A)\equal{}(\lambda_i)\subset{K}$ then $ spectrum(B)\equal{}(h(\lambda_i))$. Thus if $ A$ has $ r$ distinct eigenvalues then $ B$ has at most $ r$ distinct eigenvalues.
You may replace $ h$ with an holomorphic function.


If a cyclic quadrangle has sidesw $ a,b,c,d$ and semiperimeter s,its area K prove that
 $ K^{2}\equal{}(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)$

Check out the [url=http://en.wikipedia.org/wiki/Brahmagupta%27s_formula//url]Brahmagupta's formula[/url]


$x+y+4 \geq 6\sqrt[6]{xy}$

For positive $x,y$

Nice solution. I had a slightly different one using AMGM twice...

$x+y+1\geq 3 \sqrt[3]{xy}$

$x+y+4 \geq 3(\sqrt[3]{xy}+1) \geq 6\sqrt[6]{xy}$


limit
real analysis
real analysis unsolved

prove that, given $ f(x),\ g(x)$ defined in a neighbourhood of c,
if $ f,g \nrightarrow 1$ (I mean "do not tend to one for $ x\rightarrow c$), and $ f,g>0$, then $ f\thicksim_{c} g\Longrightarrow log(f)\thicksim_{c} log(g)$.

You should replace "do not tend to $ 1$ as $ x\to c$" (which is the negation of $ \lim _{x\to c}f(x) \equal{} 1$) by a stronger hypothesis: "bounded away from $ 1$ near $ c$" (which means $ \exists \epsilon,\delta>0$ such that $ |f(x)\minus{}1|>\epsilon$ whenever $ |x\minus{}c|<\delta$).


Source:Dr. Zeitz from mathpath, Slightly paraphrased

1)There is a woman, named Albatross, who has 5 weights. She can make any weight up to 31 lbs. What are the weights of the rocks.

2)There is a woman, named Bobert, who has 5 weights. She can find the weight of any rock up to 121 lbs. What are are the weights of  the rocks.

[i]This is a good example of using Albatross and Bobert in a word problem thank you.[/i]

A+Math inserted that comment above. Moderators, BAN, JK.

1. [hide]Use the base two number system[/hide]

The answer to question number 1:

[hide]1,2,4,8,16[/hide]

[quote="IntrepidMath"]1)There is a woman, named Albatross, who has 5 weights. She can make any weight up to 31 lbs. What are the weights of the rocks.

2)There is a woman, named Bobert, who has 5 weights. She can find the weight of any rock up to 121 lbs. What are are the weights of  the rocks.

[i]This is a good example of using Albatross and Bobert in a word problem thank you.[/i]

A+Math inserted that comment above. Moderators, BAN, JK.[/quote]

 Problem 1 and Problem 2 are worded differently.  What does "she can find the weight of any rock..." have to do with things.  Fisrt your talking about weights, then rocks, it seems very vague.........

ATTENTION MATHPATH ATTENDEES:

I do not mean to pick on this particular posting, as I have seen MathPath questions given in several AoPS Forums.

This is a gentle suggestion that it would be good etiquette to seek permission from your MathPath teachers before posting their unpublished problems on this Web Site.  My educated hunch is that the staff will be happy to share their problems, and that you will get broad approval.  But, [b]you should ask[/b].  [If you are already doing this, my commendations!]

Whether you seek permission or not, it is also a good practice to list the source of your problems.  I have been trying to encourage this for all posts in the Middle School Forums.  For this particular problem a notation, such as "paraphrased from a MathPath problem by Dr. Paul Zeitz" might be appropriate.

[quote="rcv"]ATTENTION MATHPATH ATTENDEES:
I do not mean to pick on this particular posting...[/quote]
Oh, I did not know that it was wrong. I am sorry. Thanks for the information.  And yes, this was one of the problems Dr. Zeitz gave us, but I copied it EXACTLY. anyway, I am sorry and I will not do this again.

EDIT: Oh, I see what you mean. I'll ask him...

[quote="IntrepidMath"]Oh, I did not know that it was wrong. I am sorry. Thanks for the information.  And yes, this was one of the problems Dr. Zeitz gave us, but I copied it EXACTLY. anyway, I am sorry and I will not do this again.[/quote]
Not necessarily *wrong*.  Just that you might ask Dr. Z's permission.  If permission is granted, please do it again and again.  He has some *great* problems.

I think I have the answer to # 2

[hide]
The weight of the rocks are 24.0, 24.1, 24.2, 24.3, and 24.4.

does it matter if they are consecutive?
[/hide]


[quote="usaha"]is it 1260?[/quote]
Yep.  :)

[quote="usaha"]is it 1260?[/quote]
Solution.......

[quote="rnwang2"]If $a$ and $b$ are natural numbers and $5a^{3}=6b^{4}$, what is the smallest possible value of $a+b$?[/quote]
[hide]
The RHS's got a factor of 6, so the LHS does too. The LHS's got a factor of 5, so the RHS does too. Then $a=2^{a}\cdot 3^{b}\cdot 5^{c}$ and label the exponents of $b$ similarly $d,e,f$. Then you've got $3a=1+4d$, $3b=1+4e$ and $3c+1=4f$ so the minimums are $a=3, d=2, b=3, e=2, c=1, f=1$ so $a=2^{3}\cdot 3^{3}\cdot 5=1080$, $b=2^{2}\cdot 3^{2}\cdot 5=180$, $a+b=1080+180=\boxed{1260}$.[/hide]

Well, here is my solution:
[hide] We know that $5a^{3}=6b^{4}$.

Thus, b must have a factor of 5 and a must have a factor of 6.

Say that a=6n and b=5m.

Plugging everything back in, we have that $5(6^{3})(n^{3})=(5^{4})(6)(m^{4})\Rightarrow (6^{2})n^{3}=(5^{3})m^{4}$

Now, note that n will need a factor of 5 and m will need a factor of 6.

Say that n=5c and m=6d.

Plugging our numbers back in, we get that $(6^{2})(5^{3})(c^{3})=(5^{3})(6^{4})(d^{4})\Rightarrow c^{3}=(6^{2})d^{4}$

Now, note again that c will need a factor of 6.  Thus, say that c=6x.

Plugging the value of c in, we get that $(6^{3})(x^{3})=(6^{2})d^{4}\Rightarrow 6x^{3}=d^{4}$

Now, again, d will need a factor of 6.  Say that d=6y.
Now, we have that $6x^{3}=(6^{4})y^{4}\Rightarrow x^{3}=(6^{3})(y^{4})$

Now, we can say that the lowest value of x is 6 and that of y is 1.  Thus, we have that d=6 and c=36.  Therefore, n=180 and m=36.  Finally, we have that a=1080 and b=180.  Hence, $a+b=1260$.  [/hide]

this might be helpful

if p|n^2, (a prime divides an integer squared), then p^2|n^2

Or generally, if $p | n^{k}$ then $p | n$.  Or even more generally, if $p | ab$ then $p | a$ or $p | b$.

[hide="So, a somewhat faster solution"] We are given $2, 3 | a$ and $5 | b$, so the problem only involves these three primes.  Let $a = 2^{x}3^{y}5^{z}$ and let $b = 2^{p}3^{q}5^{r}$.  Then

$2^{3x}3^{3y}5^{3z+1}= 2^{4p+1}3^{4q+1}5^{4r}$

Then

$3x = 4p+1$
$3y = 4q+1$
$3z+1 = 4r$

It is not hard to find the smallest natural solutions, which are

$x = y = 3, p = q = 2$
$z = 1, r = 1$

Giving

$a = 2^{3}3^{3}5 = 1080$
$b = 2^{2}3^{2}5 = 180$
$a+b = \boxed{ 1260 }$. [/hide]


Can anybody explain the proof for the quadratic formula?  I read it in AoPS vol. 1 and I don't really understand it.

Well it's just a rearrangement of the general form of a quadratic, $ax^2+bx+c=0$.
Divide by $a$ and subtract $\frac{c}{a}$ from both sides: $x^2+\frac{bx}{a}=-\frac{c}{a}$
Complete the square: $x^2+\frac{bx}{a}+\displaystyle\left(\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}$
$\displaystyle\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\frac{b^2}{4a^2}$
$x=-\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}$
$x=-\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Well we begin by completing the square.

$ax^2+bx+c = 0$
$a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} = 0$
Rearranging, we see that
$(x+\frac{b}{2a})^2 = \frac{c-\frac{b^2}{4a}}{a}$
Now, we take the square root of both sides and isolate $x$ to get
$x = \sqrt{\frac{c-\frac{b^2}{4a}}{a} }\pm \frac{b}{2a}$ which can be rearranged to
$x = -\frac{b}{2a}\pm\sqrt{\frac{b^2}{4a^2} - \frac{4ac}{4a^2}}$
Thus,
$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$Q.E.D.$

For those who don't like fractions, my old Algebra I textbook did this proof:

$ax^2+bx+c=0$
$4a^2x^2+4abx+4ac=0$
$4a^2x^2+4abx=-4ac$
$4a^2x^2+4abx+b^2=b^2-4ac$
$(2ax+b)^2=b^2-4ac$
$2ax+b=\pm\sqrt{b^2-4ac}$
$2ax=-b\pm\sqrt{b^2-4ac}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
QED

[quote="nutz_for2.718281828"]How about backwards?[/quote]

You're really not supposed to do that. I can prove that 1=2 backwards:

$1=2$
$1 \cdot 0=2 \cdot 0$
$0=0$
Q.E.D.

[quote="daermon"]Thanks guys i think i have it now  :lol:

by the way Mathnerd314, I was reading ur signature.  What is 0! equal to?[/quote]

I know I am not Mathnerd314, but 0!=1.

That is why punctuation is important.

[quote="nutz_for2.718281828"]How about backwards?

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$2ax=-b\pm\sqrt{b^2-4ac}$
$2ax+b=\pm\sqrt{b^2-4ac}$
$(2ax+b)^2=b^2-4ac$
$4a^2x^2+4abx+b^2=b^2-4ac$
$4a^2x^2+4abx=-4ac$
$4a^2x^2+4abc+4ac=0$

$\frac{4a^2x^2+4abc+4ac}{4a}=\frac{0}{4a}$

$ax^2+bx+c=0$

QED[/quote]

You are proving the converse of the statement.  We haven't figured out anything about the statement whether or not the converse is true or not.  So, we must prove the direct statement and not the converse.  I know this doesn't make any sense, but yeah.

In chapter 2 of intro to geometry, there is a whole page about converses. This is one of the excercises:

Write the converse of each of the following statements, then identify whether or not that converse is true

If two teams are playing in the World Cup Finals, then the teams must be playing soccer.
If two of the angles of a triangle add to 80 degrees, then one of the angles must be 100 degrees.
If a river is the longest river in the world, then it must be the Nile.
If an animal is a duck, then it must be a bird.

in nutzfor2.718281828's proof, it is like saying if all squares are rectangles, all rectangles are squares. Obviously, that's garbage.

Thanks for agreeing with what I said. :lol:


calculus
integration
calculus computations

hello, your integral leads to an elliptic function.
Sonnhard.


Put $abc=r$, $a+b+c=p$, and $ab+bc+ca=q$, the given inequality reduces to
\[\frac{q^{2}+2p^{2}-3q+pq}{p^{2}+pq+q^{2}}+\frac8{2+p+q}\leq2.\]
Now, it is solvable using Schur $p^{3}\geq4pq-9$, and maybe $q^{2}\geq 3p$.

I also let $a+b+c,ab+bc+ca$ but I feel your last stepp needs to be clarified.


[quote="NIE26"]If a,b,c are positive real numbers, such as a+b+c=1.
minimal: P= $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{13}{a^{2}+b^{2}+c^{2}}$[/quote]

[hide]By AM-HM,

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9$

$a^{c}+b^{2}+c^{2}-2(ab+bc+ac)=1-2(ab+bc+ac)$

whose maximum is obviously 1

The answer is then $9+13=\boxed{22}$[/hide]

[quote="SplashD"][quote="pvthuan"]That's wrong![/quote]

How?  :huh:

is it becaue i hsve to conside the expression as a whole?

[hide]so is the snswer 48?[/hide][/quote]
In the first step, the inequality holds when $a=b=c$, but the rest, we have $a=b=0$ for equality.

when the equality holds for your answer?

@ me@home : Yor Cauchy is wrong .
Check it again .

The answer is not $48$  :(

Try $a=\frac{1}{2}$, $b=\frac{1}{3}$, $c=\frac{1}{6}$.

if the problem is $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a^{2}+b^{2}+c^{2}}$, maybe it is more solveable for me  :D

[quote="me@home"] \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{13}{a^{2}+b^{2}+c^{2}}\]  \[=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{13}{a^{2}+b^{2}+c^{2}}\right) (a+b+c)^{2}\[ =\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{13}{a^{2}+b^{2}+c^{2}}\right) (a^{2}+b^{2}+c^{2}+2ab+2bc+2ca)\geq (2(a+b+c)+\sqrt{13})^{2}=(2+\sqrt{13})^{2}\] By Cauchy Schwartz
so $P=17+4\sqrt{13}$
is that right? it is awefully messy... why the 13?[/quote]
oh no I am so sorry, I made such a stupid mistake... it's \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{13}{a^{2}+b^{2}+c^{2}}\geq (2(\sqrt{a}+\sqrt{b}+\sqrt{c})+\sqrt{13})^{2}\] if i'm lucky doesn't rms-am fix this up (i hope there is the same equality condition) \[4\sqrt{a+b+c+13/4}\geq 2\sqrt{a}+2\sqrt{b}+2\sqrt{c}+\sqrt{13}\]  \[16*17/4\geq RHS\] if equalities help me at all, we have minimum when the above is equal...
so that's my answer (eek i don't want to check the equalities yet...)

me@home, you are wrong becase the sign is opposite.

i know the sign is opposite, i was hoping for equality conditions to line up... but i'm pretty sure they don't.

Why they don't?, i had solution for it, but must prove up one theorem.
You can help me one the shorter way, thanks.


$ x$ ,$ y$ ,and $ z$ are reals strictly positives such that $ x\plus{}y\plus{}z\equal{}1$

prove that
$ \sqrt{xy\plus{}z}\plus{}\sqrt{yz\plus{}x}\plus{}\sqrt{zx\plus{}y}>\equal{} 1 \plus{}\sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx}$

[quote="greatestmaths"]$ x$ ,$ y$ ,and $ z$ are reals strictly positives such that $ x \plus{} y \plus{} z \equal{} 1$

prove that
$ \sqrt {xy \plus{} z} \plus{} \sqrt {yz \plus{} x} \plus{} \sqrt {zx \plus{} y} > \equal{} 1 \plus{} \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}$[/quote]
Use $ \sqrt{(x\plus{}y)(x\plus{}z)}\geq x\plus{}\sqrt{yz}.$ :wink:

Substitute $ x \equal{} \frac{1}{x}, y \equal{}\frac{1}{y} , z\equal{}\frac{1}{z}$
It's equivalent to prove     $ \sqrt{x\plus{}yz}\plus{}\sqrt{y\plus{}zx}\plus{}\sqrt{z\plus{}xy}\ge\sqrt{xyz}\plus{}\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}$
with condition $ {1\over x}\plus{}{1\over y}\plus{}{1\over z}\equal{}1.$
which is APMO 2002
See here : http://www.mathlinks.ro/viewtopic.php?p=474885#474885


calculus
integration
logarithms
calculus computations

Evaluate $ \int_0^1 \frac{(1\minus{}2x)e^{x}\plus{}(1\plus{}2x)e^{\minus{}x}}{(e^x\plus{}e^{\minus{}x})^3}\ dx.$

Set $ u\equal{}1\plus{}e^{2x}$

$ \int_{0}^{1}\frac{(1\minus{}2x)e^{x}\plus{}(1\plus{}2x)e^{\minus{}x}}{\left( e^{x}\plus{}e^{\minus{}x}\right) ^{3}}dx\allowbreak$

$ \equal{}\allowbreak \int_{2}^{1\plus{}e^{2}}\left( \frac{1}{2u^{2}}\plus{}\left( \frac{1}{u^{3}}\minus{}\frac{1}{2u^{2}}\right) \ln \left( u\minus{}1\right) \right) du\allowbreak$

$ \equal{}\allowbreak \frac{e^{2}}{\left( 1\plus{}e^{2}\right) ^{2}}$


In triangle $ ABC, AB \neq AC$. Let $ D$ be the midpoint of side $ BC$, and let $ E$ be a point on median $ AD$. Let $ F$ be the foot of perpendicular from $ E$ to side $ BC$, and let $ P$ be a point on segment $ EF$. Let $ M$ and $ N$ be the feet of perpendiculars from $ P$ to sides $ AB$ and $ AC$, respectively. Prove that $ M,E,$ and $ N$ are collinear if and only if $ \hat{BAP} \equal{} \hat{PAC}$. :roll:  :P

We have:
$ [ADB] \equal{} [ADC]\Rightarrow AB\sin{\widehat{EAB}} \equal{} AC\sin{\widehat{EAC}} \Rightarrow \frac {\sin{\widehat{EAB}}}{\sin{\widehat{EAC}}} \equal{} \frac {AC}{AB}$.

$ \Rightarrow \frac {[AEM]}{[AEN]} \equal{} \frac {AM}{AN}.\frac {\sin{\widehat{EAB}}}{\sin{\widehat{EAC}}} \equal{} \frac {AM}{AN}.\frac {AC}{AB}$

On the other hand:

$ \frac {[PEM]}{[PEN]} \equal{} \frac {PM}{PN}.\frac {\sin{\widehat{EPM}}}{\sin{\widehat{EPN}}} \equal{} \frac {PM}{PN}.\frac {\sin{\widehat{ABC}}}{\sin{\widehat{ACB}}} \equal{} \frac {PM}{PN}.\frac {AC}{AB}$.

Thus: $ E,M,N$ collinear iff $ \frac {[PEM]}{[PEN]} \equal{} \frac {[AEM]}{[AEN]} \Leftrightarrow \frac {AM}{PM} \equal{} \frac {AN}{PN}\Leftrightarrow \cot{\widehat{PAB}} \equal{} \cot{\widehat{PAC}}$

And the proof is complete!


What is the sum of all two digit numbers with exactly 12 factors?

I've seen a bunch of problems on some practice rounds with number-of-factor questions. Is there some sort of quick method to find number of factors?

[hide]the number of factors can be found by taking the exponents of all the numbers in the prime factorization adding one, and multiplying it all...

Example
12=2^2*3
(2+1)(1+1)=6
or
36=2^2*3^3
(2+1)(2+1)=9
35=5*7
(1+1)(1+1)=4

It works because, for example, if 2^2 was part of the prime factorization, a factor can be divided by 2^0, 2^1, or 2^2, so that gives three possibilites.  I thought I figured out something cool when I found this out, but then I found it in my math textbook...phoey

2^5*3=96
2^3*3^2=72

is it 168?[/hide]

Sorry, my formula is right, but I forgot to consider something like 2^2*3*5
or 2*2 instead of 4, so listen to that guy below

[quote="mathgeniuse^ln(x)"][quote="ch1n353ch3s54a1l"][hide]I got 392.
The twelve factor number can be written as $a\times b\times c^2$, $a^{11}$, $a^3\times b^2$, or $a^5 \times b$. $a^{11}$ never produces a two digit number, and then i just went through each case and got $96, 72, 60, 84,$ and $90$ which adds up to $392$. [/hide][/quote]

You added those wrong.

It is supposed to be 402.

Right method, wrong answer.[/quote]
318+84=402  :D


Solve in $\ N$ the equation $\ 2^a=1+3^b5^c$.

It give $a=2d$ (by residue mod 4). Therefore $2^d-1=3^b,2^d+1=5^c$ (because $2^d-1\not =5^c$). It give $a=4,b=1=c.$


AMC
AIME
trigonometry
geometry
function
MATHCOUNTS
AMC 10

Lots of people have told me that the amc 12 is a good indication of whats on the aime. Anyone mind giving some advice?

AIME questions are a bit more tricky. Also, the fact that the answer must be an integer 000-999 doesn't help at all, since the questions end up being "find the remainder when n is divided by 1000" or "If, fully simplified, T is written in the form $\frac{\sqrt{a}+b}{c^d}$,  find a+b+c+d. It's horribly easy to make a stupid mistake along the way, and with only 15 questions, each one hurts.

Know trig identities and complex number properties inside out - if you're a TI-89 "cheater" on the AMC 12's, better watch out.

The topic distribution, on average, is similar to that of the AMC 12, though the AIME usually won't have that many "freebie" questions.

Bring the protractor, ruler, and graph paper; if you get good with them, you can make some very accurate speed estimates for geometry or modeling problems.

Actually, I find that the answer being an integer helps out in many cases. I remember this one circle and chords problem (where the last step you're squarerooting a product...if that helps anyone find out what I'm talking about) back in the 80's that could only be solved if you limited yourself to integers.

Well...
For every AMC problem, you could do it brute force, but there's an elegant and easy way, right?  Similarly, for every AIME problem, there's an elegant way: just harder to find.  I'd suggest trying some to get some idea.  Don't get stumped too long on one problem: just go on and come back later.  But remember that the problem progressively get harder.  Lastly, remember there's no penalty for guessing.  So when you are almost done, just fill any boxes which you haven't done yet with random numbers or estimated guesses.

You should know your identities and be able to recognize things such as \[ 1+x+x^2+x^3+...+x^{17}=\frac{x^{18}-1}{x-1} \]
Remember that you don't have a calculator, so check your work and arithmetic THOROUGHLY: I've made large numbers of careless mistakes in the past for failing to do so.  If you're relatively new to the AIME, I would suggest focusing on the first 10 problems or so, and once done with whichever of those and fully checked (you don't want to use up all of your brainpower doing other problems before you check these), you can go on to the last 5, the tough ones.

But in general, you have more time than the AMC, so you can experiment a little bit more with problems.  Lastly, [b]don't forget to stop every now and then and breath[/b]...it's a long one

[quote="amirhtlusa"]AIME: harder than last year[/quote]

Can you give a past year that you think it will be similar to? :D

Well, [i]if[/i] the AMC is a good indicator of what's on the AIME then I would say the AIME will be similar to last year's with an emphasis toward geometry and combinatorics.

The format of AIME answers is advantageous to people solving the questions.  Anyone who's taken a lot of practice AIMEs off of Kalva would probably tell you that :D  Even though they can basically put the answer in any form, they have to specify the form for you to be able to give an answer.  Sometimes you can look at the form to see if you're going in the right direction.  They're better than nothing, at the least.

I don't think every AMC problem can be brute forced.  To offer one example, take the function problem from this year $f(x)+f(1/x)=x$.  How would you brute force that one?  Geometry problems are often times not susceptible to brute force also.

[quote="chess64"][quote="joml88"]...with an emphasis toward geometry and combinatorics....[/quote]

NOOOO!!!! :( :mad: :wallbash_red:

Oh well. Time to start studying geometry. I was hoping it would be an algebra/NT-emphasized test so that I'd get >8.[/quote]

It's all speculation....but looking at last year's AIME, there was 6 geometry, 4 NT, 3 combinatorics, and 2 algebra on the A.  But really, that's just a cursory glance at them.  Usually, NT and combinatorics problems overlap a little bit.  Some of the geometry problems required a good knowledge of algebra, etc.

[quote="jackb1115"][quote="amirhtlusa"]my guess
AIME: harder than last year
Floor: 7[/quote]

Are you counting the expansion in your analysis?

JB[/quote]
yeah, that's why i siad the floor is 7, other wise i guess it would have been 8 or 9...

to Chess64: i think it would like the Mid-90's test...

[quote="beta"]I don't know but it should be

around 4-4-4-4 for geometry-algebra-combinatorics-number theory

because most likely there will be a problem that's a combination of two subjects.[/quote]

Yep, and the 16th question is extra credit! Yes!! That means I have a chance to get the 16 that is my goal...

But yeah... I think it will be fairly balanced (like most AIMEs). Even last year's wasn't hugely unbalanced, especially if you consider the fact that #1 was a trivial geometry problem.

[quote="amirhtlusa"]to Chess64: i think it would like the Mid-90's test...[/quote]

Ooh, you mean like '94 maybe? That AIME was fun...I can answer like 9 questions on it. :yup:

No there's no 16th question, it's simply 4 geometry, 4 algebra, 4 number theory, 4 combinatorics questions, but the set of number theory problems overlaps the set of combinatorics problems by 1, so from Principle of Incluison/Excustion there are 4+4+4+4-1=15 problems WOOT, actually algebra and number theory overlaps too oh well.

Anyone have input with what I should do?

Scenario: Ive been studying for MATHCOUNTS, I got a 137 on the AMC10a, I've only studied book one of AoPS.  Should I start piling through the AIME problems I have or just go there and bomb it so I can study for it next year?

Um, at this point, take a set. If you are borderline, I'd focus on AIME sets. But considering that you're in middle school, it's best to learn the AOPS II Volume because once you master that, I'd say you can safely get above 10 on the AIME's... but that's just my opinion.

kyyuanmathcount: I had the same scenario last year (well, almost) :). I would suggest doing a couple of old AIME exams (preferably 2000-2005). The first 4 or 5 AIME questions should be in reach of a MATHCOUNTS competitor.

[quote="1/(ln x)"]The first 4 or 5 AIME questions should be in reach of a MATHCOUNTS competitor.[/quote]

Err, I'm a little skeptical on the validity of this statement, since the abilities of "a MATHCOUNTS competitor" range from USAMO qualifier to barely knows math. A 137 on the AMC10A would probably translate to like 5-7 on an average AIME. Of course that's assuming you didn't get super lucky or bomb the AMC and you don't get super lucky or bomb the AIME.

[quote="kyyuanmathcount"]I did a practice(the alternate of 2003, #1-9 on the AMC website).  The first 3 were quite easy, I failed miserable on numbers 4 and 5, I was able to do 6 and 8 but I hadn't learned the material in 7 and 9.[/quote]

In this case, you seem to have a reasonable chance at qualifying for USAMO. I think practice AIMEs are worthwhile for you, but you don't really know exactly how much you want to practice this until you actually do a few full tests.


geometry
3D geometry
AwesomeMath
summer program
AMC
USA(J)MO
USAMO

Sooo...what summer camps you all going to?
I have this problem, I'm gonna take living skills (and my mom really wants me to so uh...) which is 6/25-7/20 :O which means I can't go to any math camps :( I am so jealous of you AMP people...
Suggestions? Apparently CTY is full or something, or the deadline passed. EPGY is eh...

Apparently EPGY also has a good "Cube program" or so they said at the EPGY competition. Can't you do awesomemath?

[quote="Sly Si"]I'm doing EPGY!  :)  Just like last year!

EPGY is not "eh", it's lots of fun![/quote]
I know it was fun and all, but I sorta want to learn math/physics somewhere...and EPGY sorta isn't the greatest place for that. And its a bit expensive.
AMP overlaps 1 week with summer school :(

[quote="Ubemaya"]I didn't make USAMO.

I quit math.[/quote]

if you didnt make the usamo
then that reduces the amount of ppl taking it this year to about 50 or so
and no one form teh bay area

so i highly doubt that

[quote="bowei"]im going to cosmos this summer
its a lot less mathy than amp
but more project-oriented

gives some material other than just problem solving[/quote]
Crap application period closed...the picture on their website is sorta discouraging though, cuz it seems to me that whenever any program tries to put stuff with circuitry to make it looks cool, the program sucks.

[quote="davidyko"]Don't do a summer camp, stay at home, be lazy, do some math, and pwn next year.  :P[/quote]
That what I WANT to do, but my mom says I have to go to a camp :(

Go to debate camp so you can do policy seriously next year :P

I can't think of camps in an acceptable time range off the top of my head. Have you looked at the AoPSWiki list?

[url=http://www.awesomemath.org]AMP[/url] was so fun...

Both of them are during summer school time, so I can't go to either :(
I actually got pretty desperate, and looked at the MathZoom website, since that would probably be better than going to the places my mom wants me to go to. It coincides with AMP (timewise).
I DID go to a debate camp, just haven't done debate...don't think I want to go to one, I need a more passive camp where I can do math all day instead of pay attention.

[quote="junggi"][quote="bookaholic"]CANADA/USA MATHCAMP IS BETTER![/quote]
wrong

i second amp[/quote]

_________________
Procrastination is the FASTEST way to get things done.
Talent=Apparent Skill/Work Ethic
If you like chess, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193031 and post if interested in the 5 minute BLITZ TOURNAMENT!

opeth break.

[youtube]AY5LRReFYus[/youtube]

serialk11r gets dragonforce punishment.

[youtube]9vE5LP0keEE[/youtube]

why

_________________
Procrastination is the FASTEST way to get things done.
Talent=Apparent Skill/Work Ethic
If you like chess, go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193031 and post if interested in the 5 minute BLITZ TOURNAMENT!

because junggi gasp is trying to discourage serialk11er from coming to amp?

my mom said she would sign me up next year :P
argh mom found a (probably) crappy camp so yay 3 weeks to waste at yale.

[quote="serialk11r"][quote="moogra"]Why do you even want to go to a camp if all the good ones are gone?[/quote]
Dunno, might get bored at home? Today she tells me she's not forcing me to go, when yesterday she was...[/quote]
idiot...
you need to learn to take advantage of opportunities
when she says she's not forcing you to go
you need to just come right out and say
OK
IM NOT GOING

I could tell she wanted me to go though, so I just said "Uh...I don't know I'll tell you tomorrow or something." (we have a window of like 2 days to sign up so waiting a day will kill the chance) My mom is hella into the "respect authority" thing, and she takes stuff like that as an insult for some reason so I have to be really careful.

Better Quote:

"Be a man, use your hand!"

Wow CTY is officially a piece of s***. They send me another email saying "Yay you're enrolled in our class now!"...

well I went to AMP... it was so awesome! (as suggested in its name) :lol: You should consider doing it next year.

Wow. By the looks of your goals, you're either very ambitious or very beast. Either way, I'm sure you'll make it to AMP... it's not that hard...


calculus
integration
logarithms
trigonometry
calculus computations

oh !   :|  i just now found out i already asked $\textbf{(i)}$ before,
and that the answer is already [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149116]here[/url] ! sorry.....  :(

i suppose now you only have to solve $\textbf{(ii)}$

Wait, let me just get something straight. In the link you posted, the expression was $(\ln ({\cos{x}}))^{2}$ but in this post you had $\ln^{2}\cos^{2}{x}$? Are they the same?

[quote="BanishedTraitor"]Wait, let me just get something straight. In the like you posted, the expression was $(\ln ({\cos{x}}))^{2}$ but in this post you had $\ln^{2}\cos^{2}{x}$? Are they the same?[/quote]

(no, the answers will [u]not [/u] be identical)  :|

note:
$\ln^{2}\cos^{2}{x}\;=\;\boxed{4}\;(\ln \cos{x})^{2}$

so, the [b]latter [/b]answer will just be 4 times as big ! the technique/method for solving the 2 integrals will remain identical.


Here's a not too hard problem.

Prove that the sum of any two consecutive triangular numbers is always a perfect square.

Yeah, that's right. It wasn't hard at all, was it?  :P

nope, just use a couple of variables, and vuala, we got it

[quote="nat mc"][quote="G-UNIT"]You can do it nat mc's way or the hard way...............

[hide="Click here to see text pop out of nowhere"]

$\frac{n(n+1)}{2} + \frac{n+1(n+2)}{2} = \frac{2n^2 + 4n + 2}{2} = \frac{4n^2 + 8n + 4}{4} = \frac{2n + 2}{2} * \frac{2n + 2}{2}= (n+1)^2 $  :rotfl:  :rotfl:  :rotfl:  :rotfl: [/hide][/quote]
:rotfl: Why would someone do that?[/quote]

it makes the problem more challenging :rotfl:  :rotfl:  :rotfl:


Why does $ (x^x)^4$ equal $ 5^x$?  Thanks in advance!

$ (x^x)^4 \equiv 5^x$? :rotfl:

If you mean to solve for $ x$ such that $ (x^x)^4 \equal{}5^x$, then $ x^4\equal{}5 \Leftrightarrow$ $ x\equal{}\pm ^4 \sqrt{5}$ and $ x\equal{}\pm ^4 \sqrt{5} i$ for $ x \neq 0$

If $ 0^0\equal{}1$ (I am not sure for this), then $ 0$ will be a solution also.

The line $ y\equal{}5x$ intersects the graph of $ x^y\equal{}y^x$ at one point in the first quadrant.  Compute the exact $ (x.y)$ coordinates of the point.

Then the question you meant to ask was "why do these conditions imply this identity?"

Plug it in.  $ x^{5x} \equal{} (5x)^x$.  What does that tell you?

[quote="stephencheng"]If $ 0^0 \equal{} 1$ (I am not sure for this), then $ 0$ will be a solution also.[/quote]
Just for clarification, $ 0^0$ is undefined.

In some contexts, yes.  In others, it makes perfect sense to define it to be one.  For example, here is the following very basic property of sets.

[b]Property:[/b]  If $ A$ and $ B$ are two sets and $ A^B$ is the set of functions from $ B$ to $ A$, then $ |A^B| \equal{} |A|^{|B|}$.

How many functions are there from the empty set to itself?  One!  Every set has an identity function.

[quote="t0rajir0u"]Then the question you meant to ask was "why do these conditions imply this identity?"

Plug it in.  $ x^{5x} = (5x)^x$.  What does that tell you?[/quote]

\begin{align*}
x^{5x} &= 5^xx^x\\
x^{4x}&=5
\end{align*}

Oh, I get it now.  Thanks a lot!


[quote="kiet"]Graph the following inverses
y = arcsec x
y = arccsc x
Identify asymptote, intercepts, and symnetry[/quote]

to find the inverses, switch the x and the y's

ill do the first one:

x=arcsec y

so y=sec x

go here to see what it looks like:
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2000/Umberger/EMAT6690smu/Day6/Day6.html

there is a symmetry about the y-axis and asymptotes at $\pm \frac{k\cdot \pi}{2}$ where k is an odd number


We have $ \frac{a^4}{2b^2\minus{}ab}\ge\frac{5a^2\minus{}3b^2}2$, since it's equivalent to $ \frac{(a\minus{}b)^2(2a^2\plus{}9ab\plus{}6b^2)}{2(2b\minus{}a)b}\ge0$, which is clearly true. Adding this cyclically we get the desired inequality. Equality holds if and only if $ a\equal{}b\equal{}c$.

[quote="bimbo"]$ \frac{(a\minus{}b)^2(2a^2\plus{}9ab\plus{}6b^2)}{2(2b\minus{}a)b}\ge0$, which is clearly true. [/quote]
What if $2b-a<0$?

[quote="dgreenb801"]Show that for positive a,b,c, where none is double another,
$ |\frac {a^4}{2b^2 \minus{} ab} \plus{} \frac {b^4}{2c^2 \minus{} bc} \plus{} \frac {c^4}{2a^2 \minus{} ac}| \ge a^2 \plus{} b^2 \plus{} c^2$[/quote]
It's wrong! Try $(a,b,c)=(0.00466, 1.976, 1.02)$.


LaTeX
combinatorics unsolved
combinatorics

Adjust this expression to the simplest form:

(0^2)*C(n,0) + (3^2)*C(n,3) + (6^2)*C(n,6) + ... + (|_n/3_|^2)*C(n,|_n/3_|)

where C(n,k)= n!/[k!(n-k)!] and |_x_| denote the down integer part of number x.
Examples:|_1,2_|=1, but also |_1,8_|=1.

According to my opinion, the sequence suggested above doesn't make enough sense or this expression is more logical for me:
(0^2)*C(n,0) + (3^2)*C(n,3) + (6^2)*C(n,6) + ... + ((3*|_n/3_|)^2)*C(n,3*|_n/3_|)
And I also know, how to solve it:
Let's take it very quickly, the key is to utilize the qualities of roots of x^3 = 1 (I think it's known) and De Moivre's and Binomial theorem.
{SUM...(3k)^2*C(n,3k)...for k=0 to k=|_n/3_|} = 1/3*n(n-1)*[2^(n-2)-2 cos[(n-1)pi/3] + 1/3*n*[2^(n-1)-2 cos[(n-3)pi/3]=
= 1/3*n(n+1)*2^(n-2) - 2/3*n^2*cos[(n-1)PI/3] - 2/3*n*sin[(n-2)PI/3].
And we may adjust it more... but what about the demanded problem? Is it the previous task really wrong?
Thanks.

Can you rewrite your post with LaTeX? Please!


Given $n\geq k$ find the best constant $r_(n,k)$ such that for all positive $ x_1,x_2,...,x_n$ we have the inequality $\sum_{i_1,i_2,...,i_k=1,n} {\frac{1}{x_{i_1}+x_{i_2}+...+x_{i_k}}}\geq r(n,k)\cdot\sum_{i=1,n} {\frac {1}{x_i+x_{i+1}+...+x_{i+k-1}} }$.
    Here the indices are taken modulo n.

A stupid question: does $i_k=1,n$ mean $i_k=1,...,n$?
(ok I know it is very obvious...)

Yes, it means that. Sorry for the stupid notation.

I know that notation $i=\overline{1,n}$ is in use (in Russia at least).


Gnorizei kaneis me sigouria , kai isws kai to pou mporw na to vrw diatipomeno kai apodedigmeno , to an to prosegkistiko theorima tou Wiestrass isxiei kai stin klasi twn oloklirwsimwn sinartisewn ?Eimai sxedon vevaios pws isxiei apla den mporw na vrw tin apodiksi tou an kai kati exw skeftei kai monos :P

Λοιπόν, η πιο γενική διατύπωση του Stone-Weierstrass που ξέρω λέει τα εξής (στις παρενθέσεις θα βάζω ό,τι αφορά το "κλασικό" θεώρημα Weierstrass).

Έστω ένας συμπαγής και Hausdorff τοπολογικός χώρος $ X$ (π.χ. το $ [0,1]$).

Έστω $ C(X,\mathbb{R})$ το σύνολο των πραγματικών συνεχών συναρτήσεων του $ X$, εφοδιασμένο με τη μετρική ομοιόμορφης σύγκλισης, δηλ. $ \|f\| \equal{} \sup_{x\in X}\{f(x)\}$.

Ως άλγεβρα του $ C(X,\mathbb{R})$ ορίζουμε έναν υπόχωρο $ A$ του $ C(X,\mathbb{R})$ που είναι κλειστός ως προς τον πολ/σμό, δηλ. $ f,g\in A \implies f\cdot g\in A$ (π.χ. πολυώνυμα στο $ [0,1]$).

Θα λέμε ότι η $ A$ "διαχωρίζει σημεία" όταν για κάθε $ x,y\in X$ υπάρχει $ f\in A$ ώστε $ f(x)\neq f(y)$ (αυτό ισχύει στα πολυώνυμα; π.χ. πάρε το $ (t\minus{}x)(t\minus{}y)$).

ΟΥΦ!!!

[b][color=red]Θεώρημα[/color][/b]: Η μόνη κλειστή (στην τοπολογία ομοιόμορφη σύγκλισης) άλγεβρα του $ C(X,\mathbb{R})$ που διαχωρίζει σημεία και που περίεχει τις σταθερές συναρτήσεις είναι ο ίδιος ο χώρος των συνεχών συναρτήσεων $ C(X,\mathbb{R})$. Ισοδύναμα, η κλειστή θήκη κάθε άλγεβρας που περίεχει τις σταθερές συναρτήσεις και που διαχωρίζει σημεία είναι ο $ C(X,\mathbb{R})$.

Κάτι πιο γενικό από αυτό δεν ξέρω :(

Το κλασικό θεώρημα Weierstrass προκύπτει θεωρώντας την άλγεβρα των πολυωνύμων πάνω στο $ [0,1]$: για κάθε $ \epsilon>0$ και για κάθε συνεχή συνάρτηση στο $ [0,1]$, υπάρχει πολυώνυμο $ p$ ώστε $ \|p\minus{}f\|\leq\epsilon$.

Tώρα, σε ό,τι αφορά ολοκληρώσιμες συναρτήσεις... για γράψε τη διατύπωση που έχεις στο μυαλό σου, και βλέπουμε ;) Ωστόσο, αν είναι να προσπαθήσεις να προσεγγίσεις ολοκληρώσιμες συναρτήσεις στην ομοιόμορφη μετρική με συνεχείς συναρτήσεις... θα έχεις πρόβλημα. Πράγματι, έστω ολοκληρώσιμη $ f$ και έστω ότι μπορείς να βρείς ακολουθία πολυωνύμων (ή κάποιας άλλης άλγεβρας του $ C(X,\mathbb{R})$) $ f_n$ που να συγκλίνει ομοιόμορφα στην $ f$, a la Weierstrass (δηλ. διάλεξε $ p_n$ ώστε $ \|f\minus{}p_n\|\leq \frac{1}{n}$). Τότε, επειδή οι $ f$ είναι συνεχείς στο συμπαγή χώρο $ X$ και συγκλίνουν ομοιόμορφα, η $ f$ θα πρέπει οπωσδήποτε να είναι συνεχής. Άρα, αν η $ f$ είναι ολοκληρώσιμη και όχι συνεχής... πρόβλημα στην αρχική υπόθεση :(

Εσύ τί έχεις στο μυαλό σου? Γενικά πάντως, οι ολοκληρώσιμες συναρτήσεις δεν συμπεριφέρονται καθόλου καλά με την ομοιόμορφη μετρική. Ιδίως αν μιλάμε για Lebesgue ολοκληρώσιμες... τα πράγματα μπορούν να γίνουν πολύ, πολύ άσχημα.

Hope this helps! Cheers,

Durandal 1707

PS: Σου χρωστάω μία απάντηση σε pm... αλλά είναι τόσο ευαίσθητο το θέμα, που δεν ξέρω από που ν'αρχίσω ;)

[quote="Durandal"]Λοιπόν, η πιο γενική διατύπωση του Stone-Weierstrass που ξέρω λέει τα εξής (στις παρενθέσεις θα βάζω ό,τι αφορά το "κλασικό" θεώρημα Weierstrass).

Έστω ένας συμπαγής και Hausdorff τοπολογικός χώρος $ X$ (π.χ. το $ [0,1]$).

Έστω $ C(X,\mathbb{R})$ το σύνολο των πραγματικών συνεχών συναρτήσεων του $ X$, εφοδιασμένο με τη μετρική ομοιόμορφης σύγκλισης, δηλ. $ \|f\| \equal{} \sup_{x\in X}\{f(x)\}$.

Ως άλγεβρα του $ C(X,\mathbb{R})$ ορίζουμε έναν υπόχωρο $ A$ του $ C(X,\mathbb{R})$ που είναι κλειστός ως προς τον πολ/σμό, δηλ. $ f,g\in A \implies f\cdot g\in A$ (π.χ. πολυώνυμα στο $ [0,1]$).

Θα λέμε ότι η $ A$ "διαχωρίζει σημεία" όταν για κάθε $ x,y\in X$ υπάρχει $ f\in A$ ώστε $ f(x)\neq f(y)$ (αυτό ισχύει στα πολυώνυμα; π.χ. πάρε το $ (t \minus{} x)(t \minus{} y)$).

ΟΥΦ!!!

[b][color=red]Θεώρημα[/color][/b]: Η μόνη κλειστή (στην τοπολογία ομοιόμορφη σύγκλισης) άλγεβρα του $ C(X,\mathbb{R})$ που διαχωρίζει σημεία και που περίεχει τις σταθερές συναρτήσεις είναι ο ίδιος ο χώρος των συνεχών συναρτήσεων $ C(X,\mathbb{R})$. Ισοδύναμα, η κλειστή θήκη κάθε άλγεβρας που περίεχει τις σταθερές συναρτήσεις και που διαχωρίζει σημεία είναι ο $ C(X,\mathbb{R})$.

Κάτι πιο γενικό από αυτό δεν ξέρω

Το κλασικό θεώρημα Weierstrass προκύπτει θεωρώντας την άλγεβρα των πολυωνύμων πάνω στο $ [0,1]$: για κάθε $ \epsilon > 0$ και για κάθε συνεχή συνάρτηση στο $ [0,1]$, υπάρχει πολυώνυμο $ p$ ώστε $ \|p \minus{} f\|\leq\epsilon$.

Tώρα, σε ό,τι αφορά ολοκληρώσιμες συναρτήσεις... για γράψε τη διατύπωση που έχεις στο μυαλό σου, και βλέπουμε   Ωστόσο, αν είναι να προσπαθήσεις να προσεγγίσεις ολοκληρώσιμες συναρτήσεις στην ομοιόμορφη μετρική με συνεχείς συναρτήσεις... θα έχεις πρόβλημα. Πράγματι, έστω ολοκληρώσιμη $ f$ και έστω ότι μπορείς να βρείς ακολουθία πολυωνύμων (ή κάποιας άλλης άλγεβρας του $ C(X,\mathbb{R})$) $ f_n$ που να συγκλίνει ομοιόμορφα στην $ f$, a la Weierstrass (δηλ. διάλεξε $ p_n$ ώστε $ \|f \minus{} p_n\|\leq \frac {1}{n}$). Τότε, επειδή οι $ f$ είναι συνεχείς στο συμπαγή χώρο $ X$ και συγκλίνουν ομοιόμορφα, η $ f$ θα πρέπει οπωσδήποτε να είναι συνεχής. Άρα, αν η $ f$ είναι ολοκληρώσιμη και όχι συνεχής... πρόβλημα στην αρχική υπόθεση

Εσύ τί έχεις στο μυαλό σου? Γενικά πάντως, οι ολοκληρώσιμες συναρτήσεις δεν συμπεριφέρονται καθόλου καλά με την ομοιόμορφη μετρική. Ιδίως αν μιλάμε για Lebesgue ολοκληρώσιμες... τα πράγματα μπορούν να γίνουν πολύ, πολύ άσχημα.

Hope this helps! Cheers,

Durandal 1707

PS: Σου χρωστάω μία απάντηση σε pm... αλλά είναι τόσο ευαίσθητο το θέμα, που δεν ξέρω από που ν'αρχίσω [/quote]

Kita egw exw kata nou to klasiko theorima Wiestrass pou leei oti kathe sinexis sinartisi se ena klisto diastima mporei na prosegkistei apo ena polionimo , dld an Pn(x) ena polionimo kai F(x) sinexis sto [a,b] ,tote gia kathe e>0 |F-Pn|<e 'h allios oti iparxei akolouthia polionimwn Pn---->F
Tha mporousa na antikatastisw afto me to eksis ,

 Estw mia sinartisi F oloklirosimi kata Riemann sto [a,b] (dld F sxedon pantou sinexis) tote iparxei akolouthia polionimwn Pn tetoia wste Pn---->F (|F-Pn|<e , gia kathe e>0) omiomorfa ....Dld i i pn na sigklinei omiomorfa sxedon pantou....(pou nomizw oti isxiei...alla den eimai apolita sigouros)

Gia na ginw ligo pio safis safto tha pw pou skeftika kai pws mou vgike oil afti i aporia...Eida tin askisi tou mathimtikou diagonismou gia panepistimio , kai eixe ena erotima pou elege : estw f monotoni sto [1,00) (afto sinepagete oti i f einai sxedon pantou sinexis kai fusika kai oloklirosimi) , kai oti to olokliroma apo 1 ews apiro tou (f-x)/x^2 sigklinei , na dixthei oti f/x--->1 otan x--->00  kai skeftika to eksis : Na parw mia prosegkisi tis f apo ena polionimo , an afto to polionimo twra einai vathmou  megalitero 'h isw tou 2 tote to olokliroma den tha mporouse na sigklinei , ara tha prepei na einai vathmou eite midenikou eite vathmou 1 , an einai midenikou tote pali den tha sinekline to olokliroma , ara thelw ena polionimo tis morfis ax+b , a<>0  , an to a einai diaforo tou 1 tote
x(a-1)+b/x^2 pali den tha sinekline to olokliroma ara thelw a=1 , to  b/x^2 kserw oti sigklinei ara thelw  tin prosegkisi me to polionimo  x+b....An isxiei to Wiestrass opws rwtaw olo afto einai sosto , an den isxiei , tote fusika einai kai lathos i oli silogistiki.....

P.S. I plaka einai oti vrika kai ena akoma theorima pou mou dixnei oti an isxioun oooola ta proigoumena tote i f den prosegkizete aplos apo to x+b alla gia tin akrivia vgenei f(x)=x+b ............

 :roll:  :oops:  :maybe:  :blush:  :idea:  :?:  :|  :huh:

Λοιπόοον,

[quote]Estw mia sinartisi F oloklirosimi kata Riemann sto [a,b] (dld F sxedon pantou sinexis) tote iparxei akolouthia polionimwn Pn tetoia wste Pn---->F (|F-Pn|<e , gia kathe e>0) omiomorfa ....Dld i i pn na sigklinei omiomorfa sxedon pantou....(pou nomizw oti isxiei...alla den eimai apolita sigouros)[/quote]

Ναι, αυτό σου έλεγα στο post μου ότι δεν ισχύει και γι'αυτήν την περίπτωση ανέφερα το πώς να κατασκευάσεις αντιπαράδειγμα. Να το κάνω πιο συγκεκριμένο: πάρε $ f(x) \equal{} 0$ στο $ [0,1)$, $ f(1) \equal{} 1$. Έστω ότι υπάρχει ακολουθία πολυωνύμων που συγκλίνει ομοιόμορφα στην $ f$. Αφού τα πολυώνυμα είναι συνεχείς συναρτήσεις και συγκλίνουν [i]ομοιόμορφα[/i] στην $ f$, θα πρέπει και η $ f$ να είναι συνεχής, όπερ άτοπο.

Sorry... :(

Επίσης, πρόσεχε λίγο πιο πολύ τις έννοιες σύγκλισης που χρησιμοποιείς... να συγκλίνει μία ακολουθία συναρτήσεων κατά σημείο σε μία συνάρτηση σχεδόν παντού, ok. Να συγκλίνει μία ακολουθία συναρτήσεων ομοιόμορφα σε μία συνάρτηση, επίσης ok.

Να συγκλίνει μία ακολουθία συναρτήσεων ομοιόμορφα σε μία συνάρτηση σχεδόν παντού... αν μαντεύω σωστά αυτό που εννοείς, δεν είναι τόσο ok. Η ομοιόμορφη σύγκλιση δεν ξεχνάει εύκολα τα σύνολα μέτρου $ 0$.

Για να καταλάβεις ποιό είναι το πρόβλημα, θεώρησε την $ f(x) \equal{} \frac{1}{\sqrt{x}}$ όταν $ 0<x\leq 1$, και $ f(0) \equal{} 0$. Αυτή είναι μια χαρά ολοκληρώσιμη στο $ [0,1]$ αλλά δεν μπορείς να την προσεγγίσεις ομοιόμορφα με πολυώνυμα σε κανένα υποδιάστημα του $ [0,1]$ που να έχει το $ 0$ ως σημείο συσσώρευσης... δηλ. ξέχνα και την "ομοιόμορφη σχεδόν παντού" σύγκλιση :(

[quote]estw f monotoni sto [1,00) [...] Na parw mia prosegkisi tis f apo ena polionimo , an afto to polionimo twra einai vathmou megalitero 'h isw tou 2 tote to olokliroma den tha mporouse na sigklinei[/quote]

Εεεεεεπ, βιάζεσαι πάρα πολύ εδώ πέρα :P

Πρώτο πρόβλημα: το πεδίο ορισμού σου εδώ πέρα είναι το $ [1, \plus{} \infty)$ που δεν είναι συμπαγές... άρα ξέχνα το θ/μα Weierstrass εξ'ολοκλήρου (ακόμα και την ομαλότατη $ e^x$ δεν μπορείς να την προσεγγίσεις ομοιόμορφα με πολυώνυμα στο $ [1, \plus{} \infty)$).

Δεύτερο πρόβλημα: αν $ \|f \minus{} g\| < \epsilon$ και η $ f$ είναι $ L^1$, αυτό δε σημαίνει ότι και η $ g$ θα είναι $ L^1$ (π.χ. πάρε $ f \equal{} 0$ και $ g \equal{} \epsilon/2$). Οπότε και να βρεις κάτι που να είναι ομοιόμορφα κοντά στη συνάρτησή σου, αυτό δε σημαίνει κάτι για τα ολοκληρώματά τους μιας και δε μιλάμε για [i]συμπαγή[/i] σύνολα.

[quote]P.S. I plaka einai oti vrika kai ena akoma theorima pou mou dixnei oti an isxioun oooola ta proigoumena tote i f den prosegkizete aplos apo to x+b alla gia tin akrivia vgenei f(x)=x+b ............ [/quote]

Η $ f(x) \equal{} \sqrt {x} \plus{} x$ ικανοποιεί τις συνθήκες της εκφώνησης, αλλά δεν είναι της μορφής $ f(x) \equal{} x \plus{} b$.   :|

Βασικά, αν μπορείς, καθαρόγραψε την άσκηση και βάλτη στο topic για ασκήσεις ΑΕΙ, είναι νοστιμούλα :)

Cheerio,

Durandal 1707

[quote="Durandal"]Λοιπόοον,

[quote]Estw mia sinartisi F oloklirosimi kata Riemann sto [a,b] (dld F sxedon pantou sinexis) tote iparxei akolouthia polionimwn Pn tetoia wste Pn---->F (|F-Pn|<e , gia kathe e>0) omiomorfa ....Dld i i pn na sigklinei omiomorfa sxedon pantou....(pou nomizw oti isxiei...alla den eimai apolita sigouros)[/quote]

Ναι, αυτό σου έλεγα στο post μου ότι δεν ισχύει και γι'αυτήν την περίπτωση ανέφερα το πώς να κατασκευάσεις αντιπαράδειγμα. Να το κάνω πιο συγκεκριμένο: πάρε $ f(x) \equal{} 0$ στο $ [0,1)$, $ f(1) \equal{} 1$. Έστω ότι υπάρχει ακολουθία πολυωνύμων που συγκλίνει ομοιόμορφα στην $ f$. Αφού τα πολυώνυμα είναι συνεχείς συναρτήσεις και συγκλίνουν [i]ομοιόμορφα[/i] στην $ f$, θα πρέπει και η $ f$ να είναι συνεχής, όπερ άτοπο.

Sorry... :(

Επίσης, πρόσεχε λίγο πιο πολύ τις έννοιες σύγκλισης που χρησιμοποιείς... να συγκλίνει μία ακολουθία συναρτήσεων κατά σημείο σε μία συνάρτηση σχεδόν παντού, ok. Να συγκλίνει μία ακολουθία συναρτήσεων ομοιόμορφα σε μία συνάρτηση, επίσης ok.

Να συγκλίνει μία ακολουθία συναρτήσεων ομοιόμορφα σε μία συνάρτηση σχεδόν παντού... αν μαντεύω σωστά αυτό που εννοείς, δεν είναι τόσο ok. Η ομοιόμορφη σύγκλιση δεν ξεχνάει εύκολα τα σύνολα μέτρου $ 0$.

Για να καταλάβεις ποιό είναι το πρόβλημα, θεώρησε την $ f(x) \equal{} \frac {1}{\sqrt {x}}$ όταν $ 0 < x\leq 1$, και $ f(0) \equal{} 0$. Αυτή είναι μια χαρά ολοκληρώσιμη στο $ [0,1]$ αλλά δεν μπορείς να την προσεγγίσεις ομοιόμορφα με πολυώνυμα σε κανένα υποδιάστημα του $ [0,1]$ που να έχει το $ 0$ ως σημείο συσσώρευσης... δηλ. ξέχνα και την "ομοιόμορφη σχεδόν παντού" σύγκλιση :(

[quote]estw f monotoni sto [1,00) [...] Na parw mia prosegkisi tis f apo ena polionimo , an afto to polionimo twra einai vathmou megalitero 'h isw tou 2 tote to olokliroma den tha mporouse na sigklinei[/quote]

Εεεεεεπ, βιάζεσαι πάρα πολύ εδώ πέρα :P

Πρώτο πρόβλημα: το πεδίο ορισμού σου εδώ πέρα είναι το $ [1, \plus{} \infty)$ που δεν είναι συμπαγές... άρα ξέχνα το θ/μα Weierstrass εξ'ολοκλήρου (ακόμα και την ομαλότατη $ e^x$ δεν μπορείς να την προσεγγίσεις ομοιόμορφα με πολυώνυμα στο $ [1, \plus{} \infty)$).

Δεύτερο πρόβλημα: αν $ \|f \minus{} g\| < \epsilon$ και η $ f$ είναι $ L^1$, αυτό δε σημαίνει ότι και η $ g$ θα είναι $ L^1$ (π.χ. πάρε $ f \equal{} 0$ και $ g \equal{} \epsilon/2$). Οπότε και να βρεις κάτι που να είναι ομοιόμορφα κοντά στη συνάρτησή σου, αυτό δε σημαίνει κάτι για τα ολοκληρώματά τους μιας και δε μιλάμε για [i]συμπαγή[/i] σύνολα.

[quote]P.S. I plaka einai oti vrika kai ena akoma theorima pou mou dixnei oti an isxioun oooola ta proigoumena tote i f den prosegkizete aplos apo to x+b alla gia tin akrivia vgenei f(x)=x+b ............ [/quote]

Η $ f(x) \equal{} \sqrt {x} \plus{} x$ ικανοποιεί τις συνθήκες της εκφώνησης, αλλά δεν είναι της μορφής $ f(x) \equal{} x \plus{} b$.   :|

Βασικά, αν μπορείς, καθαρόγραψε την άσκηση και βάλτη στο topic για ασκήσεις ΑΕΙ, είναι νοστιμούλα :)

Cheerio,

Durandal 1707[/quote]'

Thnx gia tis episimanseis...Tha to psaksw kai pereterw !Nai den perase apo to mialo mou to [0,00) gia tin simpagotita :P giati poli apla den mou kolousan ta alla opote itan to telefteo pou tha pigena na eksetasw:P.Aftos itan o enalakitkos tropos gia na to kanw , alli itan i vasiki mou lisi :)

Oso gia tin katharografi diskolo na to kanw giati den kserw Latex ...

Me ton simbolismo tou Durandal sto post #2 :
A) Isxuei to idio theorima gia ton $ C(X,\mathbb{C})$ ?,  breite ena antiparadeigma otan X=[0,1] !
B) Mporeite na breite ena antiparadeigma otan to X den einai metrikos xoros ( alla einai simpagis kai Hausdorff !) ?
C) Mporeite na breite ena antiparadeigma otan to X einai to Cantor set ?
D)(Diskolo) Mporeite na breite ena antiparadeigma oste to X na min periexei kapoio antigrafo tou Cantor set ?

Basika i apodeiksi pou exo gia to A, xreisimopoiei to C, alla eimai sigouros oti prepei na iparxei kai kapoio pio efkolo paradeigma. Episeis i apodiksi pou exo gia to B xrisimopoiei to D, alla pali prepei na iparxei kati pio efkolo. Sto D to diskolo kommati einai na dixeis oti den mporeis na embithiseis to cantor set sto xoro sou.

To problima exei sxedon lithei apo mia seira apo ergasies (W.Rudin, e.t.c) gia poious xorous isxuei to theorima over C (xoris na baloume tin extra sinthiki tis autoepisinapsis). I apantisi einai i exis :
1) Ean o X periexei kapoio antigrafo tou Cantor tote den isxuei.
2) Ean to theorima isxuei gia ton X, tote o X einai scattered space. Auto simainei oti kanena kleisto iposinolo tou den einai perfect.
3) Kathe compact metric scattered space periexei kapoio antigrafo tou Cantor set.

Stergios


Here when it says consecutive digits, it means like 654321, in a consecutive decreasing order.  its pretty bad wording though.  I was confused at first too.

hahah yeah, i read it wrong on the test :(

i thought things like 27262524 were valid....

and so i answered 666 (0+1+2+3..... 36)

oh well

[quote="billzhao"]inbase 10, a digit is commonly accepted as a number between 0 and 9[/quote]

Yes, but an integer can be composed of more than one digit.

Another question with bad wording was #4.  Looking back I cant see any reason to prefer the official answer over mine.  Maybe I misunderstood what was meant by the question...

4. A square has sides of length 2. Set S is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set S enclose a region whose area to the nearest hundredth is k. Find 100k.

My answer: 314 (area of the circle created by the midpoints)
Official answer: 86 (area of the region between the midpoints and the square)

To me it seems like this question is purposely deceptive if not wrong...

Ian:

I think the wording for #4 was fine, because it asked for the region enclosed by the locus, not the region between the locus and the square.

Ooooh, heh -- yeah (im feeling kinda dumb right now>.<)

[quote="Fierytycoon"]I didn't take the AIME 1 (I take the AIME 2), but I just took a look at the AIME 1 problems for this year and found the wording for #1 very confusing:

[quote]1. The digits of a positive integer n are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 37? [/quote]

The "four consecutive integers" mentioned are supposed to be digits, but with this wording, wouldn't a value of 47464544 be valid as well, since 47, 46, 45, 44 are indeed "four consecutive integers in decreasing order"?[/quote]

I think that the wording on this question is good.  It says that "the digits" are "positive integers".  The digits of your 8 digit number are not "four consecutive integers in decreasing order".

This year's AIME did seem to have more questions that were hard to read and parse than usual and I think that contributed to more careless errors and might have taken 1 off the AIME floor.  I think that young math students are simply going to have to focus on their verbal interpretation skills.  I think this perfectly reasonable since problem solving is not simply a matter of solving equations, but a matter of framing and interpretting problems as well.

[quote]1. The digits of a positive integer n are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 37?[/quote]

While I think the conventional interpretation of this problem is "best", I completely understand fierytycoon's objection.

Suppose the problem has stated:

[quote]1.  The eight digits of a positive integer n are four consecutive integers in decreasing order...[/quote]

Then there would be little question that 99989796 through 13121110 were the positive integers called for.

My sentence structure is identical to the AIME's wording, except I added an adjective to modify their use of the word "digits".

If my modified sentence is gramatically correct, then I think fierytycoon has identified an alternate interpretation of the original sentence that is also correct.

[quote="rcv"]
Suppose the problem has stated:

[quote]1.  The eight digits of a positive integer n are four consecutive integers in decreasing order...[/quote]

Then there would be little question that 99989796 through 13121110 were the positive integers called for.

My sentence structure is identical to the AIME's wording, except I added an adjective to modify their use of the word "digits".

If my modified sentence is gramatically correct, then I think fierytycoon has identified an alternate interpretation of the original sentence that is also correct.[/quote]

It's not a matter of grammar.  It's a matter of definition.  Your problem makes no sense because, by definition, 8 digits can't [i]be[/i] four consecutive integers.  They can [i]construct[/i] four consecutive integers, but that's different.

While I do have sympathy for the large number of good students who seem to have missed number 1, I think that the problem was well-stated and the errors people made were not simply a matter of misinterpretation, but a matter of not reading closely enough.


floor function
number theory proposed
number theory

Find the greatest positive integer    $x$   such that   $23^{6+x}$  divides the number

$2000!$  , (where $2000!=1 \cdot 2 \cdots 2000$)


probability
expected value
probability and stats

Chapter:Probability; Examples 32C, Problem 17

A bag contains a coin of value M and a number of other coins whose aggregate value is m. A person draws one at a time till he draws the coin M. Find the value of his expectation.

[hide="answer"] $ \frac{m}{2}\plus{}M$ if we include the desired coin [/hide]
[hide="solution"] Every other coin has 0.5 chance of being selected. So every other coin adds to the expected value half of its value. [/hide]


trigonometry
MATHCOUNTS
geometry
trig identities
Law of Sines

[quote="Ignite168"]Given a circle with radius 3.5, find the chord length of an 86 degree arc.

You can probably approximate.[/quote]

[hide]After spproximating, you get $\boxed{4.77398852}$[/hide]

[quote="Ignite168"]Given a circle with radius 3.5, find the chord length of an 86 degree arc.

You can probably approximate.[/quote]

[hide]If we didn't approximate then we would have to use trig and that is not allowed in this forum.

We can approximate that that is a 90 degree angle and hence the chord length would be $\frac{7\sqrt{2}}{2}$.[/hide]

Well, here is the solution:

Is this mathcounts level, or am I solving it the hard way?

[hide="A SOLUTION!!!"]Draw 2 radii and connect them, with the angle between them $86^\circ$.  Now, there is a triangle with sides of length 3.5, 3.5, and $x$.  The angles are 86, 47, and 47.  Use the law of sines, and solve.

$\frac{3.5}{\sin47^\circ}=\frac{x}{\sin86^\circ}$

$x=\frac{3.5\sin86^\circ}{sin47^\circ}$

$\boxed{x\approx4.77398852}$[/hide]

Umm, treething, your answer isn't very close.

Your answer would be the correct answer, but I don't think you are allowed to use Trig on MathCounts Forum.  And Ignite, what do you mean by approximate?

[quote="b-flat"]Umm, treething, your answer isn't very close.[/quote]
He did the arc, not the chord. Probably.

[quote="b-flat"]
Umm, treething, your answer isn't very close.[/quote]

Knowing, Treething, he probably did the arc.

I don't think trig is even allowed in getting started even unless its really basic (including to the admins). Then again, some of the stuff in getting started is harder than the stuff in intermediate, but nothing ever gets moved.

[quote="b-flat"]Then we need better moderators  :D

To find the arc length (probably what he meant to ask because that is mathcounts level):

[hide]Because the length of the arc is $86^\circ$, and the total number of degrees in a circle is $360^\circ$, the length of the arc is $\frac{86}{360}=\frac{43}{180}$ of the entire circumference.  The entire circumference is $2\pi3.5=7\pi$.  Now, multiply to get:

$\boxed{\frac{301\pi}{180}\approx5.253441049}$[/hide]

Yes, this is close to treethings answer.[/quote]

He asked for the chord length because if he did ask for the arc length, we wouldn't need to approximate. The questions would just say 'Express your answer as a common fraction in terms of $\pi$'

[quote="lotrgreengrapes7926"][quote="b-flat"]Well, if it involves trig, it is not mathcounts level, so that is why I said he might have meant arc length instead.[/quote]
Well he said you could approximate, which is what mathgenius did to get $\frac{7\sqrt2}{2}$.[/quote]

that's not approximating, that's estimating.

[quote="SplashD"]
that's not approximating, that's estimating.[/quote]

Actually that is approximating since estimating usually involves whole numbers.  Right?

But who cares, they both practically mean the same thing.

[quote="b-flat"][quote="mathgeniuse^ln(x)"][quote="SplashD"]
that's not approximating, that's estimating.[/quote]

Actually that is approximating since estimating usually involves whole numbers.  Right?

But who cares, they both [b]practically[/b] mean the same thing.[/quote]

Practically does not mean that they are synonyms.  They mean different things, like SplashD said.[/quote]

Yeah, what he said.

[quote="Ignite168"]You were supposed to draw the chord and divide it up into two right triangles and use trig. I didn't think it was off-difficulty, I'm sorry. I thought it was tons easier than a lot of MathCounts problems.[/quote]

it is

[quote="Ignite168"]You were supposed to draw the chord and divide it up into two right triangles and use trig. I didn't think it was off-difficulty, I'm sorry. I thought it was tons easier than a lot of MathCounts problems.[/quote]

It is easier than a lot of MathCounts problems, but we are not allowed to use trig in the forum.  So that is what the dissent is coming up for.

[quote="Ignite168"]You were supposed to draw the chord and divide it up into two right triangles and use trig. I didn't think it was off-difficulty, I'm sorry. I thought it was tons easier than a lot of MathCounts problems.[/quote]

While the problem was considerably easier than a lot of mathcounts problems, it involved knowing trigonometry, which is not a topic in MathCounts.  It is not necessarily the difficulty of the topic so much as the type of problem and knowledge needed to solve it that makes mathcounts problems mathcounts problems.


LaTeX
logarithms
MATHCOUNTS
geometric series

Evaluate

2/2/2/2/2/2/2/2/2/2/2/2/.......
(unfortunately this latex is unparseable or potentially dangerous latex formula.Error: 7)

this is meaning 2 divided by (2 divided by (2 divided (...)))

So for this I got multiple answers which I don't understand.

Solution #1
When experimenting, what 2/2 is and what 2/2/2 is, you get 1 or 2. If you continue, it turns out to be if there is an even number of 2s then the answer is 1 if there are an odd number of 2s then the answer is 2. Since infinity isn't neither odd nor even then the answer must be $ 1 and/or 2$.

Solution #2
x=2 divided by (2 divided by (2 divided (...)))
$ x\equal{}\frac{2}{x}$
$ x^2\equal{}2$
$ x\equal{}\sqrt2$

Solution #3
Same as #2 but putting x in the third fraction like this.
$ x\equal{}\frac{2}{\frac{2}{x}}$
$ x\equal{}2*\frac{x}{2}$
the 2's crosses out so
$ x\equal{}x$
therefore the infinite fraction equals anything?

Can you tell me what the correct answer is and what is wrong with my wrong answers?

Thanks

my guess is that the fraction is undefined =p

that doesnt help at all right? :P

i'll get back to you when i get a better answer....

It is undefined, sort of like how formulas show that $ 1\minus{}1\plus{}1\minus{}1...\equal{}1/2$

I would say that it is 0, considering that it approaches it but never makes it.

Using the infinite geometric series formula, with $ r \equal{} \minus{} 1$ , $ 1 \minus{} 1 \plus{} 1 \minus{} 1 \plus{} 1 \minus{} 1\cdots \equal{} \frac {1}{2}$

Which is obviously false because of the imposed requirement that $ |r| < 1$.

Oh, I forgot about that requirement :blush:

Well then, what's the series sum to? Not $ \infty$ nor $ 1$ nor $ 0$ Half the time it equals 1, half it equals 0. The average value of the series is $ \frac {1}{2}$ but whether or not the is the value I do not know.

[quote="mewto55555"]Well then, what's the series sum to? Not $ \infty$ nor $ 1$ nor $ 0$ Half the time it equals 1, half it equals 0. The average value of the series is $ 1/2$ but whether or not the is the value I do not know.[/quote]

it doenst converge to anything

youll learn it in calculus.

in order for it to converge to a certian number, the nth term test at infinity must equal 0, clearly this one does not

The series $ 1 \minus{} 1 \plus{} 1 \minus{} 1 \plus{} 1 \minus{} 1 \pm ...$ doesn't sum to [url=http://en.wikipedia.org/wiki/Summation_of_Grandi%27s_series]anything[/url] (in the normal sense).  Assigning it a formal sum is occasionally done for physical reasons.

Similarly, the fraction you're trying to decipher doesn't equal anything.  It's not a well-defined expression.  Taking the base-$ 2$ logarithm of it gives the above series, so the two questions are related.

I'm not saying it converges. The $ 1 \minus{} 1 \plus{} 1...$ diverges, but what does it diverge to? Not $ \infty$ or a number. So is it undefined? And has this gone beyond Mathcounts level? So is the original question $ 2^{1\minus{}1\plus{}1...}$?

[quote="mewto55555"]I'm not saying it converges. The $ 1 \minus{} 1 \plus{} 1...$ diverges, but what does it diverge to? [/quote]

The (partial sums of the) series oscillates between $ 1$ and $ 0$.  It doesn't diverge "to" anything.

[quote="mewto55555"]And has this gone beyond Mathcounts level?[/quote]

In all likelihood, yes.  Series summation is usually considered a high school topic.

[quote="mewto55555"] So is the original question $ 2^{1 \minus{} 1 \plus{} 1...}$?[/quote]

More or less.


Here's a cool one!  I'll post the solutions later, once there are a few answers.  :)

The value of [x] is the greatest integer less than or equal to x.
What is the arithmetic mean of the 10 members of the set
{[-$ \pi$],[$ -\frac{1}{2}$],[0],[$ \frac{1}{2}$],[0.689],[$ \frac{\pi}{4}$],[$ \frac{\pi}{3}$],[2],[$ \sqrt{5}$],[$ \pi$]
Express your answer as a common fraction.

[hide="Answer"]The value of [x] is the greatest  integer less than or equal to x.  We  are asked to find the arithmetic mean  of the 10 members in the set. [-pi] = -4      [-1/2 ] = -1 [0] = 0         [ 1/2 ] = 0 [0.689] = 0  [ pi/4 ] = 0 [ π/3 ] = 1       [2] = 2 [ 5] = 2     [pi] = 3 -4 + -1 + 1 + 2 + 2 + 3 = -5 + 3 + 5 = 3 The mean is:  3/10[/hide]

Did you get this from the contest this year? I remember this problem.

This was on the 2007 chapter sprint round.

oops! :blush: I'll post an older one next time!

That's ok, just remember - If you know the source, please put it in the "source" bar of the thread.

[quote="archimedes1"]Is that equivalent to the "Subject Description" bar?[/quote]

Whoops, yep, sorry, they must have changed it on me :(.

[quote="kimmystar94"]hey!
I remember doing this!!!
(sorry..)
..
it's good to remember that[] is the greatest integer function..
but i can never get it to show on my TI-89...
is it even possible, tho??
I somehow got it on the TI-84, but not on mine... :huh:[/quote]

Use the catalog... its the roof/floor functions

This question was one of the 5 I got wrong

 :(

but my chapter round was stupid. 41/46 was first

[quote="xpmath"]Well, that isn't too bad. At mine, the highest was 45 (Allen Yuan).

By one of the only 5 you got wrong, do you mean 5 wrong on sprint, or overall[/quote]
Umm cant you see hes purposely spamming.


When observing a die, one can see at most 3 faces at a time. Of the natural numbers 5, 6, 7, ..., 15, which cannot be the sum of the observed faces?

HINT: There are multiple values.

are we to assume that the die is regular in that opposite faces add up to 7?

[quote="Ubemaya"]When observing a die, one can see at most 3 faces at a time. Of the natural numbers 5, 6, 7, ..., 15, which cannot be the sum of the observed faces?[/quote]
[hide]
13 is not possible.
if there are only 2 faces, it would require at least a 7
it has to be 3 faces.
when there is 3 faces, it would need a 6 as one of them, since with 5, the highest is 3+4+5, which is 12.
with the other two faces, the possibilities are 3+4 and 2+5
however, the 3 and 4 are on opposite sides, and 2 and 5 are on opposite sides, so there is no way to see a thirteen
[/hide]

There are more.

I see why you would give that answer, but this is poorly worded.
It is possible that you only see one face, or two faces.

with one or two faces it is still not possible to see a 13...i said it in my solution. with two faces it would require a 7

and the problem only asks for one number, and as long as 13 can't be seen, its the answer

The problem is messed up.

1) There are 3 answers.
2) It is assuming that you MUST see 3 faces at a time.


Prove that in each triangle $ABC$ with heights $h_a, h_b\ \text{and}\ h_c$ and semiperimeter $p$ we have

$ h_a^{2}+h_b^{2}+h_c^{2}\leq p^2$

Let K be the area of the triangle.

We have $h_a = \frac{2K}{a}$

It remains to show $4K^2(\sum \frac{1}{a^2}) \leq (a+b+c)^2$

Substitute $a = y+z$ etc.  It becomes $4(x+y+z)(xyz)(\sum \frac{1}{(y+z)^2}) \leq 4(x+y+z)^2$

It remains to show $x+y+z \ge (xyz)(\sum \frac{1}{(y+z)^2})$.  This is equivalent to $\sum \frac{1}{xy} - \frac{1}{(x+y)^2} \ge 0$.

But $(x+y)^2 \ge xy$, so it is true.

Socrates  , you can  use  the ineq [tex]h_a\leq\sqrt{s(s-a)}[/tex] . To prove  this one , just  square  it and  use  the  formula  [tex]h_a=2\frac{\sqrt{s(s-a)(s-b)(s-c)}}{a}[/tex]

[quote="cezar lupu"]Come on, you could have at least posted $l_a +l_b+l_c\leq p^2$ where $l_a ,l_b ,l_c$ are the lenghts, of bisectors of a triangle. ;)[/quote]

Are you sure that that is true? (The left hand side and right hand side have different "degrees", you see, the inequality is not homogenous)

Actually it is

$\displaystyle l_a^{2} +l_b^{2}+l_c^{2}\leq p^2$

and the proof for this is quite easy

$l\leq\sqrt{p(p-a)}$

It'd be nice to have a collection of these geometric inequalities somewhere on the forum. Does one already exist? I could not find it in Inequalities Theorems section in the Olympiad forum.

Thanks.


function
geometry
3D geometry
sphere
Support
calculus
calculus computations

Hey, i'm doing some population growth type stuff right now in calc, and i'm having some difficulties with this problem:

----------------------------------------------
The world population increases about 2% per year. In 1990 it was about 5,000,000,000 people. We will represent the earth as a sphere with a radius of 3960 miles.

Assume that planet earth has just enough resources to support a maximum of 40 billion people. propose a specific logistic function or a model of the earth's population.

Given Info: $f(t)= L/(1+ce^{-kL(t-t_{0})})+d$
$L$, $c$, $k$, $d$, and $t_{0}$ are constants

does anyone know this? i really don't know how i'm supposed to approach this. does it have something to do with finding the equation for the maximum? or am i just reading that wrong?  :huh:


Prove that: $\forall n\in\mathbb{Z},n> 0,$ we have:
$\frac{1}{n+1}\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)\ge \frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)$

Adding $\frac{1}{n+1}\left( \frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2n}\right)$ to both sides, and letting $H_{k}$ be the $k$th harmonic sum or w/e, it would suffice to prove that

$\frac{H_{2n}}{n+1}\ge \frac{(2n+1)H_{n}}{2n(n+1)}$

or

$2nH_{2n}\ge (2n+1)H_{n}$

This is true since for $1 \le k \le n$,

$2n(n+2k) = 2n^{2}+4nk \ge 2n^{2}+2nk+n+k = (n+k)(2n+1)$

$2n \left( \frac{1}{k}+\frac{1}{n+k}\right) = \frac{2n(n+2k)}{k(n+k)}\ge \frac{2n+1}{k}$

Probably there's an easier way but this was the first thing I thought of and it seems to have worked.


Using the Chicken McNugget Theorem(AoPS Wiki it if you don't know what it is), our answer is $ (3 \times 7) \minus{} (3 \plus{} 7) \equal{} 11$.

[quote="GameBot"]Tim and Kurt are playing a game in which players are awarded either 3 points of 7 points for a correct answer. What is the greatest score that cannot be attained?[/quote]


probability
geometry
MATHCOUNTS
number theory
AMC 10
\/closed

I'm in 6th grade right now. I'm currently reading the Intermediate Algebra book, Volume 2: And Beyond, and I finished Introduction to Algebra, Introduction to Counting and Probability, Introduction to Number Theory, Introduction to Geometry, and Volume 1: The Basics.

I purchased Intermediate Counting and Probability, but I haven't started reading it yet.

Based on this information, does anyone have any suggestions on what class to take?

Thanks :lol:

well you are in a good shape

have you taken the amc, mathcounts, etc

how good are you at the intro series

I suggest looking at the diagnostic tests.  First, go through the diagnostics for the Intro classes and Algebra 2, to make sure you have them down.   If you have all of that covered, then I would recommend Algebra 3.  But if you have a lot of trouble with the PostTest on any of the Intro classes or Algebra 2, I would recommend the corresponding class.

I am in the same situation as you....
Do you think Advanced Mathcounts is a good class for me?
I made it to state in mathcounts.(Score of 37)
What class do you reccommend?

If you scored 37 at state mathcounts I think you'll be fine if you just go through past tests.

If you're going to take a MATHCOUNTS class, it should be the advanced one.  You might also look into one of the subject classes or the AMC 10 Problem Series.


The sum of the squares two numbers is 34. The square of their sum is 64. Find the product of the two numbers.

[hide="Solution"]

Let $ a$ and $ b$ be the two numbers, such that,

$ a^2\plus{}b^2\equal{}34$ and $ (a\plus{}b)^2\equal{}64$,

Note that $ (a\plus{}b)^2\minus{}(a^2\plus{}b^2)\equal{}2ab$ and by substituting we see that $ 2ab\equal{}64\minus{}34\implies ab\equal{}\boxed{15}$
[/hide]

Hello, we have

$ x^2 \plus{}y^2\equal{}34$ and $ (x\plus{}y)^2\equal{}64$

$ x^2 \plus{}2xy\plus{}y^2\equal{}64$, so $ xy\equal{}15$


vector
function
linear algebra
linear algebra unsolved

Let $ n$ be a positive integer and $ F$ a field. Let $ W$ be the set of all vectors $ (x_1,...,x_n)$ in $ F^n$ such that $ x_1\plus{}...\plus{}x_n\equal{}0$
 $ (a)$ Prove that $ W^0$ consists of all linear functionals $ f$ of the form
                            $ f(x_1,...,x_n)\equal{}c\displaystyle\sum_{k\equal{}1}^{n}x_k$
 $ (b)$ Show that the dual space $ W^*$ of $ W$ can be 'naturally' identified with the linear functionals
$ f(x_1,...,x_n)\equal{}c_1x_1\plus{}...\plus{}c_nx_n$
of $ F^n$ which satisfy $ c_1\plus{}...\plus{}c_n\equal{}0$

Let $ e_{i}$ be the vector with its $ i$-th component $ 1$ and zeros elsewhere. As for part a, one side of inclusion is obvious. For the other side, note that $ f(e_{i})\equal{}f(e_{i}\minus{}e_{j}\plus{}e_{j})\equal{}f(e_{i}\minus{}e_{j})\plus{}f(e_{j})\equal{}f(e_{j})$, for all $ i, j$ and $ f$ in $ W^{0}$.
For part b, I am not sure what you mean by "naturally identified", but the following works for what I have in mind:
Consider these $ (n\minus{}1)$ linearly independent vectors over $ F$ as linear functionals: $ e_{1}\minus{}e_{i}$.

About part $ (b)$: this is problem from the book so i just wrote what it was written there without any change. I also wondered what does this mean but I proved that all functions in $ W^*$ can be written in that form and probably this is what author means by 'natural identified'.


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that $GL_{2}(\mathbb{C})$ has no subgroup isomorphic to $A_{5}$.

Hi.

If you know something about presentations you can prove it by finding the charactertable of $A_{5}$. There is no irreducible character of dimension 2, so there can't be an irreducible presentation of dimension 2 (=special kind of homomorphism $A_{5}\to GL_{2}(\mathbb{C})$). One can easily find that there can't be reducible presentations either.


[quote="ccy"]Tyler rolls four standard 6-sided dice and finds that the product of the numbers rolled is 450. What is the sum of the numbers that were rolled?[/quote]
[hide]$450=5\times2\times3\times5\times3$. since this is five numbers and we need four, we multiply the smallest. $2\times3=6, 6+5+5+3=\boxed{19}$[/hide]

[hide]only one possibility so it is 19[/hide]


How do you solve this..

Six people are chosen to participate in a contest to win a car. There are six keys placed in a box. The six contestants will in random order take a key from the box. If the key starts the car, they win. If the key does not, it is discarded and the next person tries. Which position (1st-6th) gives you the largest probabilty of winning

does only 1 key start the car? if so, it doesn't matter the order.

[hide]

first person has probability $\frac{1}{6}$

second person has probability $\left(\frac{5}{6}\right)\left(\frac{1}{5}\right)$

third person has $\left(\frac{5}{6}\right)\left(\frac{4}{5}\right)\left(\frac{1}{4}\right)$

and so on. all these products telescope down to $\frac{1}{6}$ so each person has equal probalility.

[/hide]