Hey guys, a really really general question about USAMTS. You know the fourth question on the current round. Well, it says "describe the strategy and prove it works." How exactly do you prove a strategy? Is it sufficient just to explain how you arrived at the strategy? Are there sources where I can read about game theory and stuff on the internet?
   And btw, does anyone know how high you have to score on USAMTS to get into AIME? It's my first year doing it see...

i think you just have to show that no matter what, the opponent cannot overcome the strategy. its hard to explain it without giving the answer away...

a perfect on USAMTS garners only a 100 on the AMC12, so i wouldnt suggest doing USAMTS only.

You should know some terminology for game theory. Suppose we are dealing with a symmetric game, which means that the two players can each make the same moves. (For example, Tic Tac Toe is not a symmetric game, since one marks Xs and one marks Os, but Dots and Boxes is a symmetric game.)

A position is said to be [b]winning[/b] if it is possible to get from that position to a losing position.

A position is said to be [b]losing[/b] if it is not possible to get from that position to a losing position.

You can usually figure out whether some given position is either winning or losing. For example, often in games such as Nim, the person who takes the last stick loses. (The other variant is the person who takes the last stick loses, but we won't discuss that now.) So suppose we have a simplified version of Nim, where there is one pile of 2003 sticks. Each person may take 1, 2, or 3 sticks. The person who takes the last stick(s) wins. So who wins? Here is a strategy for the first person to win: on the first turn, the first player should take 3 sticks. Then 2000 sticks are left. On every turn from then on, the second player may take 1, 2, or 3 sticks. If the second player takes one stick, the first player should take three. If the second player takes two, the first player should also take two. If the second player takes three, then the first player should take one. Then after the first person has moved, there will always be a multiple of four sticks left. There is no way the second person can change that. Thus the first person wins.

I like Simon's answer to your first few questions.

Anyway, the cutoff to qualify for the AIME has been 90% after 3 rounds for the past couple years.  So what is that...  68/75?  Yeah.  That percentage is then taken to be the number of points that go toward your USAMO index, so the maximum the USAMTS can contribute is 100 points.  (So take the AMC even if you've crushed the USAMTS.)


Find the maximum or minimum  value of
$ x^3 \plus{} y^3 \minus{} 2(x^2 \plus{} y^2)$ for all $ x,y \in R.$

Sorry, i make some mistake. i make the correction now.

[quote="4865550150"]Find the maximum value of
$ x^3 \plus{} y^3 \minus{} 2(x^2 \plus{} y^2)$ for all $ x,y \in R.$[/quote]

You must have made some mistakes. When $ x \to \infty$ ...


Sanki ikinci soruyu daha önce bir erden hatırlıyorum

hatırlamanız doğal.. sorular kitapta yazdığına göre eski 2. aşama sorularındanmış. bu arada 1999-2000 yılları 2. aşamaları elinde olan var mı? varsa erencankizildag@gmail.com adresine yollarsa sevinirim.


Have someone write down a 13-digit number and then reverse the digits.  Subtract the smaller number from the other( if the original number has less than 13 digits then the reverse number is less than 13-digits) then have them draw a circle around anyone of the digits. Asks them to say what the uncircled digits are,  after hearing the numbers you will be able to tell them the circled digit.  How does the trick work? Why does it work?

I'm not sure I understand what you're asking -- what numbers can I choose from to circle?  What digits do I read off?

I have edited the problem so that you will understand it better.  You asks the person to circle anyone of the digits from the
subtracted sum and then read all the digits that are uncircled so that with this information you should be able to find the circled digit.

N = a1...a13
N'= a13...a1

Then |N-N'|=b1..b13 is divisible by 9 and 11 (just look mod 9 and mod 11).

So here is a short algorithm which, I think, works :

Let c1 .. c12 the given (=uncircled) digits.
Let a = sum ci

. if (a mod 9) != 0 then there is a unique digit d which makes a+d divisible by 9.
. if (a mod 9) = 0 then there is ambiguity mod 9 between 0 and 9 but there is a unique digit d (in {0,9}) which makes sum(-1)^i*ci +/- d divisible by 11.

In each case d is the wanted digit.


[hide="One fourth of the way there"]$\frac{1}{\frac{1}{x}+\frac{1}{y}}=6^{20}$ simplify the fraction, $\frac{xy}{y+x}=6^{20}$, multiply over the denominator, $xy=6^{20}x+6^{20}y$ move the RHS to the LHS, $-6^{20}x+xy-6^{20}y=0$, factor, $(x-6^{20})(y-6^{20})=6^{40}$. Then since the two parts on the left have to be factors of the right, let $x$ and $y$ be $6^{20}+a$ and $6^{20}+b$. This doesn't change the number of answers, or the magnitude of either number so it doesn't affect the $x<y$ rule. So now we have $ab=6^{40}$ with $a<b$. Now the only thing left to do is to take all combinations of $2^m3^n$ where $0 \le m, n \le 40$ for $a$ and $b$ and get rid of all combinations where $a>b$. Then take the combinations of $2^m3^n$ and add $6^{20}$ and it's done. I'm not too sure how to find whether or not, for example, $3^5 \cdot 2^{40} ? 3^{35}$  is less than/greater than.[/hide]

[quote="joml88"]The harmonic mean of two positive numbers is the reciprocal of the arithmetic mean of their reciprocals.  For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}.$[/quote]

$\frac{2}{\frac{1}{x}+\frac{1}{y}} = 6^20 \Longleftrightarrow \frac{xy}{x+y} = 2^{19}3^{20} \Longleftrightarrow (x-2^{19}3^{20})(y-2^{19}3^{20}) = 2^{38}3^{40}.$

$2^{38}3^{40}$ has $39(41)$ factors.  Every factorization of it has one term greater than the other except its square root, when $x=y=2^{19}3^{20}$.  However, each factorization can be switched; in other words $x$ can be, say $2^{9}3^{10}$ and $y= 2^{29}3^{30}$ OR $y$ can be $2^{9}3^{10}$ and $x= 2^{29}3^{30}$.  Therefore, one must take half of the remaining factorizations after subracting 1 for when $x=y$.

Your final answer is now $\frac{1}{2}[39(41)-1] = \fbox{799}.$

All of my answer will be wrong  :(
Started out with $\frac{1}{\frac{1}{x}+\frac{1}{y}}=6^{20}$ instead of $\frac{2}{\frac{1}{x}+\frac{1}{y}}=6^{20}$. I like your solution pkothari13.

[quote="pkothari13"]
$ 2^{19}3^{20} \Longleftrightarrow (x \minus{} 2^{19}3^{20})(y \minus{} 2^{19}3^{20}) \equal{} 2^{38}3^{40}.$

[/quote]
Can someone please explain how they got this?


Find the smallest positive integer a for which:

1001x + 770y= 1000000 + a

is possible, and show that it has then 100 solutions in positive integers

P.S.: This is my first post and I must use it to greet all people of this wonderful site!

Yes, I do this but I don't know a formula to find pairs (x,y) satisfying the equation...

it's not very hard using Euclide's algorithm and the theorem of Bezout. which enables to find solutions of equations like $ax+by=c$

[quote="dante"]Sorry, what is Bezout's theorem?[/quote]
Integers a and b are coprime iff there exist integers x and y such that ax+by=1.

[quote]Integers a and b are coprime iff there exist integers x and y such that ax+by=1.[/quote]

Thanks! I knew of that as Euclid's Lemma.


I think the result is too short, because you can use AM-GM and Schwartz.
but we can prove it to>=9/168.

[quote="Ly Tieu Long"]but we can prove it to>=9/168.[/quote]
sorry , i dont get you


Four couples are at a party. Four people of the eight are
randomly selected to win a prize. No person can win more
than one prize. What is the probability that both members
of at least one couple win a prize? Express your answer as a
common fraction.

The probability that both members of at least one couple will win a prize is the same as one minus the probability of each couple winning one prize. So...

$ 1\minus{}1*\frac67*\frac46*\frac25\equal{}\frac{27}{35}$

How'd you get those fractions though?

Or mayb I'm just being really dumb
>.<

I'm sorry.  The probability of each couple winning one prize is:

Probability of first prize going to a couple without a prize: 1
Probability of second prize going to a couple without a prize (3 couples out of 7 total people--since person who won first prize is ineligible): $ \frac67$
Probability of third prize going to a couple without a prize (2 couples out of 6 total people--since people who won first two prizes are ineligible): $ \frac46$
Probability of fourth prize going to a couple without a prize (1 couple out of 5 total people--since people who won first three prizes are ineligible): $ \frac25$

So probability of each couple winning one prize is:  $ 1*\frac67*\frac46*\frac25 \equal{} \frac8{35}$

$ 1 \minus{} \frac8{35} \equal{} \frac {27}{35}$

Hope this is more helpful.

The last poster's reasoning is probably the best.  The important idea to remember is sometimes it's easier to compute the probability that something WON'T happen, and then you know the probability it WILL happen is 1 minus your computation.

If the fractions of the last problem are intimidating, you can learn how to do it with the "n choose k" function, and the idea that "probability=(# of chances of success)/(total # of events)".***

So, let's count successes:  I need to choose one person from each couple.  There are 2 ways to do that for the first couple, 2 ways for the second, 2 ways for the third and 2 ways for the fourth.  So in total, 2x2x2x2=16 ways to succeed.  How many total events are there?  That's where the "n choose k" function comes in.  It'll tell you how many ways there are total of picking four people out of eight.  Grab any basic probability reference, it's in there.  In our case:  8 choose 4=8x7x6x5/4x3x2x1.  This is quickly reduced to 70 by cancelling.  So, using *** we have 16/70=8/35....Then 1-8/35 is 27/35 as above.
Both methods are probably about the same speed, once you're used to them.  So, just pick the one that you remember the fastest :)  Definitely learn how to use the "n choose k" function though!

A note:

$ n$ choose $ k$ can be written as $ nCk$ or $ \binom{n}{k}$. These are called combinations. If you do not understand combinations, there's an entry on them in my blog, link's in my signature.

What iin did is called complementary counting. This means we count what we don't want, and subtract this from 1. More on this is my blog, too.


ARML
geometry
trigonometry
trapezoid
analytic geometry
ratio
trig identities

Regular polygon $ ABCDEFGHIJKLMNOPQR$ has side length $ 1$. The area of polygon $ ARML$ can be expressed as $ \frac {3(\sqrt {3} \plus{} \sqrt {2 \plus{} 2x})}{4}$. What is the value of $ 16x^3 \minus{} 12x \plus{} 11$?

18.

If calculators are allowed, we could simply bash this with the $ \frac {s^2 n}{4 \tan{\pi / n}}$ formula, find the actual area, solve for $ x$, and then evaluate. If not, then I'm not sure...

Ummm, doesn't that formula only apply to regular n-gons so how would it help for finding the area of $ ARML$?

[hide="Another possibility"]
Use complex roots of unity.  Let $ \theta\equal{}\frac{2\pi}{18}$ Let the points A through R be the 18th roots of unity $ e^{ik\theta}, k\equal{}0$ to $ 17$ and fix B at $ (1,0)$ with the alphabet continuing in counterclockwise order.

$ ARML$ is thus an isosceles trapezoid where the vertices are $ A\equal{}e^{i17\theta}$, $ R\equal{}e^{i16\theta}$, $ M\equal{}e^{i11\theta}$, and $ L\equal{}e^{i10\theta}$.  The bases of this trapezoid are then $ AL\equal{}2cos(17\theta)$ and $ RM\equal{}2cos(16\theta)$.  The height of the trapezoid is $ sin(16\theta)\minus{}sin(17\theta)$.  The area comes straight from the average of the bases times the height.

All that this would change is the side length of the 18-gon, but that only multiplies all the areas by a scalar anyway.  This method might not give the area in the form that we would require to solve for $ x$ though.  Maybe a sort of backwards triple angle formula or something would make it in the right form, but my guess is that some fact about the cubic will make this alot easier.
[/hide]

Ooh, trigtastic!

[hide="This is as short as I could make it while still showing a fair amount of work."]
The figure is a trapezoid whose diagonals are both $ 120^{\circ}$ cords, so their lengths are $ r\sqrt {3}$. One can use the LoS to show $ r \equal{} 1/2\sin 10^{\circ}$. The angle between the chords is $ 20^{\circ}$, so the area of the trapezoid is $ d_1d_2\sin\theta/2 \equal{} \frac {3}{4\tan 10^{\circ}} \equal{} \frac {3\tan 80^{\circ}}{4}$.

So it looks like we have $ \tan 80^{\circ} \minus{} \tan 60^{\circ} \equal{} \sqrt {2 \plus{} 2x}$. The difference-of-tangents formula can be re-derived by combining fractions:
\[ \tan A \minus{} \tan B \equal{} \frac {\sin (A \minus{} B)}{\cos A\cos B},
\]
which in our case gives $ \tan 80^{\circ} \minus{} \tan 60^{\circ} \equal{} \frac {\sin 20^{\circ}}{\sin 10^{\circ}\cos 60^{\circ}} \equal{} 4\cos 10^{\circ}$, so now we have the relation $ x \equal{} 8\cos^2 10^{\circ} \minus{} 1 \equal{} 4\cos 20^{\circ} \plus{} 3$. Are you sick of trig yet? Well too bad! We press on...

Triple angle for cosine looks like $ \cos 3A \equal{} 4\cos^3 A \minus{} 3\cos A$, which means the final answer is $ 4\cos 3A \plus{} 11$, where $ x \equal{} \cos A$. Here I get stuck, since our value for $ x$ is clearly larger than $ 5$. My gut instinct is that the problem-writer multiplied by powers of $ 2$ rather than divided whilst contriving this kooky problem. I may be wrong, of course.

For example, if $ \sqrt{2\plus{}2x}$ had been replaced by $ \sqrt{8\plus{}8x}$ in the original statement, we would have
\[ 4\cos 10^{\circ} \equal{} \sqrt{8\plus{}8x} \quad \Rightarrow \quad \cos 10^{\circ}\equal{}\sqrt{\frac{1\plus{}x}{2}} \quad \Rightarrow \quad x \equal{} \cos 20^{\circ},
\]
from which $ 16x^3\minus{}12x\plus{}11 \equal{} 4(\cos 60^{\circ})\plus{}11 \equal{} \boxed{13}$, which should have been the guess (based on the poser's subject description).

My initial reaction while working through this problem was I would never wish it on any ARML participant. I've had problems half this hard shot down as "way too complicated / contrived / obscure / etc.", and I would argue that in a short time-limit environment there are fairer ways of testing everyone -- not just the smarty-pants AoPS types -- than a problem that relies in part on triple-angle formulas. I would totally use this sorta problem on a team selection test, though  :wink:
[/hide]

I dont think its a trapeziod...im probably being retarded though...

If you really wanted to bash....
[hide]
Consider the 18gon formed by the 18 roots of unity on the complex plane(let the verticies be A', B'...). Find the coordinates of A'R'M'L' and use shoelace. However, this is bigger than the original 18gon. to find the ratio between the areas, let the center of the original be Z.(new one is Z') using the similarity between ABZ and ABZ' and using law of sines, we find that(i skipped a TON of algebra, and i hope its right) the new quadrilaterals is $ 2\sin\frac{\pi} {18}$ times bigger than the old one. hence, the area is $ 4\sin^2\frac{\pi} {18}$ times more. So multiply $ BASH*4\sin^2\frac{\pi} {18}$ and youre done.
[/hide]

I dare someone to do that.


algebra
polynomial
search
number theory solved
number theory

Prove that $p(x)=(x^2+x)^{2^n}+1$ is irreducible in $\mathbb Z[x]$

This was posted at least five times before... (I recommend search button)
http://www.mathlinks.ro/Forum/viewtopic.php?t=76611
http://www.mathlinks.ro/Forum/viewtopic.php?t=70003
http://www.mathlinks.ro/Forum/viewtopic.php?t=49799
http://www.mathlinks.ro/Forum/viewtopic.php?t=14780
http://www.mathlinks.ro/Forum/viewtopic.php?t=311

in all the links given i don't see a simple solution, what about this one : take X=x+1 => P(x)=(X²-X+1^)^(2^n)+1
and the result immediately follows from eisenstein criterion for p=2, any mistake?

[quote="maskman"]in all the links given i don't see a simple solution, what about this one : take $X=x+1 \, \Rightarrow \. P(x) = (X^2-X+1)^{2^n}+1$
and the result immediately follows from eisenstein criterion for p=2, any mistake?[/quote]

You have $(X^2-X)^{2n}+1$... which brings you nowhere, I think.


calculus
derivative
function
inequalities
Putnam
real analysis
real analysis solved

Let $ f: R-->R $ be a function has $3_{th} $ derivative  . Prove that : There exists $ a $ such that :
 $ f(a)f'(a)f''(a)f'''(a) \geq 0 $

Assume that each $f(x),f'(x),f''(x),f'''(x)$ is positive or negative for all $x$. (If one of them is zero, the ineq holds)
WLOG, assume that $f''(x)>0$ and $f'''(x)>0$. (replace $-f$ by $f$ and replace $f(x)$ by $f(-x)$ if necessary)
$f''(x)>0$ implies that $f'$ is strictly increasing
$f'''(x)>0$ implies that $f'$ is strictly convex : $f'(y) > f'(x) + f''(x)(y-x)$. Fix $x$ and let $y$ tend to $+\infty$ we know that $f'(y) \to +\infty$ hence $f'(x)>0$ for all $x$.

Now $f'(x)>0$ implies that $f$ is strictly increasing
and $f''(x)>0$ implies that $f$ is strictly convex
A similar argument gives $f(x)>0$ for all $x$

Hence $f(x)f'(x)f\"(x)f'''(x)>0$, the inequality holds.

It is a Putnam problem but I can't remember the year.

[quote="liyi"]It is a Putnam problem but I can't remember the year.[/quote]
Source: 59th Putnam 1998 Problem A3


We are given equilateral triangle with side $n$ which is divided into $n^{2}$ congruent equilateral triangles with side $1$. Let's call these small triangles plates. Each plate is either black or white. We perform the following operation:
we choose a plate which has common sides with at least two plates of diiferent colour from chosen plate and then we change the colour of our plate. For every $n\geq 2$ decide whether exists a configuration of colours that enables to perform infinite sequence of operations

Here is a solution in Polish. Can someone translate it into English?

Nazwijmy odcinkiem granicznym wspólny bok dwóch płytek, których widoczne
strony maja rózne kolory. Oczywiscie liczba odcinków granicznych jest
nieujemna i nie przekracza liczby m wszystkich odcinków bedacych wspólnym
bokiem dwóch płytek. Zbadamy, jak zmienia sie liczba odcinków granicznych
w wyniku wykonania dozwolonego ruchu.
Kazda z n2 płytek ma wspólne boki z co najwyzej trzema innymi płytkami.
Odwrócenie płytki jest dopuszczalne, jezeli co najmniej dwa z takich
boków sa odcinkami granicznymi. Przypuscmy, ze odwracamy płytke P.
Wówczas odcinki graniczne moga pojawic sie albo zniknac jedynie na bokach
płytki P. Ponadto bok płytki P jest po odwróceniu odcinkiem granicznym
wtedy i tylko wtedy, gdy przed odwróceniem nie był on odcinkiem granicznym.
Wynika stad, ze w wyniku wykonania dozwolonego ruchu liczba odcinków
granicznych zmniejsza sie.
Udowodnilismy w ten sposób, ze z dowolnego poczatkowego ułozenia
płytek mozna wykonac nie wiecej niz m ruchów.
Odpowiedz: Dla zadnego n2 nie istnieje ułozenie pozwalajace wykonac
nieskonczony ciag ruchów.

Here is my solution:

Let G be a graph with $n^{2}$ vertices. (each vertice assigned to one of the plates in the triangle)
two vertices are 'linked' by one edge if they have one side in common and your color is the same.

now, see that, after one operation, our number of edges will diminute, because if from one vertice out 2 edges, will out 1, and if out 3, will out zero. (we can do one operation if, from one vertice, 2 edges are outing)

so, our problem is solved, because the number of edges always will decrease! (and the operations are limited!)


given atriangle ABC , a=4,b=5 , c = 6
show that  C=2A

What do you mean by C=2A?

<C = 2<A?

Use law of cosines or sines.

[quote="ashrafmod"]Given a triangle $ABC\ ,\ a=4\ ,\ b=5\ ,\ c = 6$ show that  $C=2A\ .$[/quote]
Generally, $C=2A\Longleftrightarrow$ $c^{2}=a(a+b)\ .$


function
factorial
ratio
algebra
linear equation

[quote="pewthian"]Hi, I am new to this forum. I hope someone can help me with this problem which I have tried to solve for a long time and still can't obtain the solution. Given a discrete function:

f(x) = binom(a + x, x) * binom(b + N-x, N-x) where x = 0,1,...N

where a,b>0 are constants and N a positive integer, what is the value of x that gives the maximum value for f(x)? Note that binom(n,r) is a choose function giving the binomial coefficients.

Thanks.[/quote]

$f(x)=\binom{a+x}{x}\cdot \binom{b+n-x}{n-x}$

Binomials have factorials, and factorials are messy, so if possible we'd like to get rid of them.  To find potential maxima, we only need to compare adjacent terms; because of the factorials, comparing differences is messy.  But comparing ratios of consecutive terms is quite straight-forward -- you just need to know the value of $x$ at which the ratio switches from being $\geq 1$ (your sequence is increasing) to being $\leq1$ (your sequence is decreasing), and you end up only having to solve a linear equation for $x$.


function
floor function
LaTeX
ceiling function

for which numbers $k\lfloor x \rfloor=\lfloor kx \rfloor$?

if you're talking about an arbitrary $k$ you're only guaranteed that its right if $x$ is an integer. It will work for different values of $x$ based on what $k$ is.

It works whenever $x - \lfloor x \rfloor < \frac{1}{k}$.

Wait, how do you write the floor and ceiling functions in LaTeX?

\lfloor x\rfloor

\lceil x\rceil

You'll want to use \left\lfloor and \right\rfloor if the argument is a fraction or something.


At a particular school, all 43 students take chemistry, biology, or both. The chemistry class is three times larger than biology class, and 5 students are taking both classes. How many people are in the chemistry class?

Well as you see I will show another solution, it has the same basic idea but with some differences..
The solution ::
Say that the number of students taking Chemistry is [i]x[/i] and the number of students taking biology is [i]y[/i] ,, then we know that the number of chemistry students = 3 times the number of biology students,, so we get that [i]y[/i] = [i]x[/i]/3 . Now we make our equation 43 = x + x/3 -5 (43 is the number of all the students and we subtract 5 since we counted the students who studied both subjects twice )
We solve the equation for [i]x[/i] and we get that the number of chemistry students =[i]x[/i]=36
This is how i solved it .. I hope it helps :) ..


algebra
system of equations
advanced fields
advanced fields unsolved

How to prove that the system of equations
$\begin{cases}x \text{ cos }y=x^{2}+y^{2}-1\\ y\text{ cos }x=\text{tan }2\pi(x^{3}+y^{3}) \end{cases}$
has a solution $(x_{0},y_{0})$ s.t. $x_{0}^{2}+y_{0}^{2}\leq 1$. According to the book, one should use the fact that a continuos odd map $f: \mathbb{D}^{2}\to \mathbb{R}^{2}$ on the boundary, (i.e. $x\in \mathbb{S}^{1}\implies f(-x)=-f(x))$ has a point $x_{1}\in\mathbb{D}^{2}$ s.t. $f(x_{0})=0$.

I made a stupid calculation mistake. If I define $f(x,y)=\left (x\text{ cos }y-x^{2}-y^{2}+1,y\text{ cos }x-\text{ tan }2\pi (x^{3}+y^{3})\right )$ then $f(x,y)=f(-x,-y)$ for all $(x,y)$ inside the unit circle.


A set $A \subset [0,1]$ is called [i]negligible[/i] if it is Lebesgue measurable with Lebesgue measure zero. Let $\mathcal{F}$ is a totally ordered family of negliglible subsets of $[0,1]$ (totally ordered means that, for arbitrary $A,B$ in the family, we either have $A \subseteq B$ or $B \subseteq A$). Must the union $\bigcup_{A \in \mathcal{F}} A$ be negliglble?

 :roll:

Start with a well-ordering $x_{\alpha}$ of $[0,1]$, indexed by an initial segment of the ordinals. For each ordinal $t$, let $S_t=\cup_{\alpha<t}\{x_\alpha\}$. Let $M$ be the least index such that $S_M$ is not negligible, and let our family consist of $S_\alpha$ for $\alpha<M$.

Since $S_M=\cup_{\alpha<M}S_\alpha\cup x_M$ and $S_M$ is not negligible, $\cup_{\alpha<M}S_\alpha$ is not negligible.

By the way, I've met Noam Elkies- he gave a rather incomprehensible lecture at MOP, and dominated at chess.

:)

The following is exactly the same,  but how misterious it seems stated this way:

[i]
Suppose that every such union were negligible. This would mean that the collection of all negligible sets satisfies the condition of Zorn's lemma, and hence it has a maximal element. This is clearly absurd: if $A \subset [0,1]$ is negligible and $x \in [0,1] \setminus A$, then $A \cup \{x\}$ is also negligible.
[/i]

 :)


algebra
polynomial
linear algebra
matrix
induction

Let $ A$ be an $ n\times n$ matrix with characteristic polynomial $ f(t)\equal{}(\minus{}1)^nt^n\plus{}a_{n\minus{}1}t^{n\minus{}1}\plus{}\cdots \plus{}a_1t\plus{}a_0$. Prove that $ f(0)\equal{}a_0\equal{}\det(A)$, and also that $ f(t)\equal{}(A_{11}\minus{}t)(A_{22}\minus{}t)\cdots (A_{nn}\minus{}t)\plus{}q(t)$, where $ q(t)$ is a polynomial of degree at most $ n\minus{}2$.

The first one follows directly from definition of characteristic polynomial.

The second one follows from the fact that the $ x^{n\minus{}1}$'s coefficient of the characteristic polynomial is $ \minus{}tr(A)$.

How can you prove the second one from scratch (just using induction on $ n$)?

That is the way I did it. It isn't all that difficult. If you would like I can outline the proof I did. (right now I'm about to go to bed so it will have to be sometime tomorrow)


If point P lies on the line 3x+5y=15, and is equidistant from the axes, what are the coordinates for P?

[hide](15/8,15/8)[/hide]

thats what i got im so smart!!!!!

[hide]x=y because it is the same distance from both of the axes, so now we have
3x + 5x = 15
8x = 15
x = 15/8
y = 15/8[/hide]

I am not completely satisfied with the answer to this question.

i posted the other answer in a separate post but someone deleted it!
 :bruce:
[hide](-15/2,15/2) is the other answer for y=-x[/hide]


geometry
geometric transformation
homothety
incenter
projective geometry
geometry proposed

Given a circle $ T$, a triangle $ ABC$ lying in the interior of $ T$. A circle internally tangent to circle $ T$ at point $ A_1$ is tangent to the rays $ BA, CA$ beyond the vertex $ A$. Similarly one defines points $ B_1, C_1$ lying on $ T$. Prove that, lines $ AA_1, BB_1, CC_1$ are collinear.

Let $ I$ be the incenter of triangle $ ABC. AA_1\cap OI \equal{} \{L\}. IM, O_aN\perp AC$.
Applying Menelaus's theorem for triangle $ IOO_a$ we have:
$ \frac {LI}{LO}.\frac {A_1O}{A_1O_a}.\frac{AO_a}{AI} \equal{} 1$
$ \Rightarrow \frac {LI}{LO} \equal{} \frac {R_a}{R}.\frac {IM}{O_aN} \equal{} \frac {r}{R}$
Similarly $ BB_1, CC_1$ pass through $ L$. We are done.

Name $ H$ the center of the inverse homothethy of the incircle $ (I)$ of the triangle $ ABC$ and circle $ T$ . $ A_1$ is center of the direct homothethy of $ T$ and the circle $ T'$ that is internally tangent , $ A$ is center of the inverse homothety of $ (I)$ and $ T'$. Apply Monge's theorem, $ A,A_1,H$ are collinear. Similarly $ B,B_1,H$ and $ C,C_1,H$ are collinear.


$ f(x) \equal{} \dfrac{abx \minus{} (a^{2} \plus{} ac \plus{} c^{2})}{b^{2}x \plus{} bx}$

$ f^{n}(x) \equal{} x$

$ n \equal{} ?$

[hide="answer"]n=3[/hide]

good luck!

[hide="Easy"]
Check until 3, where really 3 is solution. Now, every 3k is a solution, and any other isn't.

Proof: Suppose the contrary: $ n\equal{}3k\plus{}a,a<3,a\neq 0$. Then $ f^n(x)\equal{}f^{3k}(x)\circ f^a(x)\equal{}f^a(x)\neq x$ [/hide]

[quote]Check until 3, where really 3 is solution. Now, every 3k is a solution, and any other isn't. [/quote]

It is enough to say this, or you can say the period is 3. the second part isnt completely necissary.


I was obviously wrong, check the third post.

No. Can you find a Taylor series of $ f(x) \equal{} e^x$? Has it any roots? Also, if your statement would be true then for a bounded function $ f$ you could find the Taylor expansion of $ f(x) \plus{} a$ only for some values of $ a$ - when a root exists. In general, the Taylor series does not depend in any truly meaningful way with the roots of a function. On the other hand, what needs to be true is that the function $ f$ must be well defined in a neighborhood of $ a$ for it to have a Taylor expansion around $ a$.

When you say it has to be well defined in the neighborhood of a, do you mean the limit has to exist at a? I.e. you can't have a taylor series for ln(x) centered at x=0 because the limit does not exist.

Or by neighborhood well defined you need f to be a mapping $ f: \mathbb{R} \rightarrow \mathbb{R}$?


find all n in N satisying the condition: n!=sqr(x)+sqr(y)
here x and y are integer numbers.

What is given?  :huh:
I mean : what about $x = (n!)^2$ and $y=0$?

Pierre.

Perhaps he means $x^{2}+y^{2} = n!$...?  If that's the case, I have no idea, but I'd surmise that n= 8 is the biggest.

This reminds me... someone once told me that there is an easy proof of the fact that $n!$ is never a square, without resorting to Bertrand's lemma.  Does anyone know if such a proof exists?


1) Let $ a,b,c,d$ be real numbers. Prove that
\[ a(a \plus{} b)^3 \plus{} b(b \plus{} c)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3\geq0.
\]
2) Let $ a,b,c,d$ be real numbers. Prove that
\[ a(a \plus{} b)^3 \plus{} b(b \plus{} c)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3\geq\frac18(a \plus{} b \plus{} c \plus{} d)^4.
\]

hi !!!!!!!!
I think by answering the 2 we automaticly answer to 1
so : put a>b>c>d  => a+b>b+c>c+d>d+a
by using Chebychev innequality we have :
S>= 1/4(a+b+c+d)((a+b)^3 + (b+c)^3 + (c+d)^3 +(a+d)^3)
>= 1/4(a+b+c+d)(1/16*(2*(a+b+c+d))^3)
= 1/8(a+b+c+d)^4

[quote="iverson_h3"]a>b>c>d  => a+b>b+c>c+d>d+a
[/quote]
wrong,take $ a\equal{}100,b\equal{}3,c\equal{}2,d\equal{}1$ :wink:

[quote="iverson_h3"]put a>b>c>d  => a+b>b+c>c+d>d+a[/quote]
why c+d>d+a :?:

[quote="pvthuan"]1) Let $ a,b,c,d$ be real numbers. Prove that
\[ a(a \plus{} b)^3 \plus{} b(b \plus{} d)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3\geq0.
\]
[/quote]

Did you mean $ a(a \plus{} b)^3 \plus{} b(b \plus{} c)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3\geq0.$ ?

[quote="ringos"]
Did you mean $ a(a \plus{} b)^3 \plus{} b(b \plus{} c)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3\geq0.$ ?[/quote]

Yes, ringos. I have edited it.

If you think they are too easy, the following is hard enough.

Prove that for all real numbers $ a,b,c,d$, we have the inequality
\[ a(a\plus{}b)^5\plus{}b(b\plus{}c)^5\plus{}c(c\plus{}d)^5\plus{}d(d\plus{}a)^5\geq\frac1{32}(a\plus{}b\plus{}c\plus{}d)^6.\]

Unfortunately, take $ a \equal{} \minus{} 1,b \equal{} 3,c \equal{} \minus{} 2,d \equal{} 1$, we have

$ a(a \plus{} b)^3 \plus{} b(b \plus{} c)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3<0$
$ a(a \plus{} b)^5 \plus{} b(b \plus{} c)^5 \plus{} c(c \plus{} d)^5 \plus{} d(d \plus{} a)^5<0$

 :wink:

Thank you, shalex !
Regret that two above nice inequality are wrong.I have mistaken in calculating :oops:
But we have:
Let $ a$, $ b$, $ c$,$ d$ be positive real numbers.Prove that

1) $ a(a \plus{} b)^3 \plus{} b(b \plus{} c)^3 \plus{} c(c \plus{} d)^3 \plus{} d(d \plus{} a)^3 \geq \frac {1}{8}(a \plus{} b \plus{} c \plus{} d)^4$;
2) $ a(a \plus{} b)^5 \plus{} b(b \plus{} c)^5 \plus{} c(c \plus{} d)^5 \plus{} d(d \plus{} a)^5 \geq \frac {1}{32}(a \plus{} b \plus{} c \plus{} d)^6$

Generally,we also have:
Let $ a$, $ b$, $ c$,$ d$  be positive real numbers.Then for all odd natural number k the inequality
$ a(a\plus{}b)^k\plus{}b(b\plus{}c)^k\plus{}c(c\plus{}d)^k\plus{}d(d\plus{}a)^k \geq \frac{1}{2^k}(a\plus{}b\plus{}c\plus{}d)^{k\plus{}1}$
is true.

That 's a great pro! I'l think of it in some day

Assume that $ a\plus{}b\plus{}c\plus{}d\equal{}1$. Use Jensen Inequality we have

$ a(a\plus{}b)^k\plus{}b(b\plus{}c)^k\plus{}c(c\plus{}d)^k\plus{}d(d\plus{}a)^k \ge [a(a\plus{}b)\plus{}b(b\plus{}c)\plus{}c(c\plus{}d)\plus{}d(d\plus{}a)]^k$

It suffices to show

$ a(a\plus{}b)\plus{}b(b\plus{}c)\plus{}c(c\plus{}d)\plus{}d(d\plus{}a) \ge \frac{1}{2}$

Which is equivalent to

$ (a\minus{}c)^2\plus{}(b\minus{}d)^2 \ge 0$

It's done, and the equality holds for $ a\equal{}c, b\equal{}d$  :wink:

A solution is nice and easy. It is amazing! Congratulation, shalex! :D


When the expression $ 8^{10} \times 5^{22}$ is multiplied out (written in decimal notation), how many
digits does the number have?

they made the $2^{30}\times 5^{22}$ turn into $2^8\times 10^{22}$


LaTeX
number theory unsolved
number theory

No.
I'll prove inductively the following statement:
For $n\in \mathbb{N}$ and $k_1,k_2,...,k_n, a\in \mathbb{N}$ there exist arbitrarily large $m\in \mathbb{N}$ such that all numbers $k_ia^m+1$, $i=1,...,n$ are composite.

For $n=1$ let $p$ be a prime divisor of $k_1a^t+1$ for some big positive integer $t$. Then $k_1a^{pt}+1\equiv k_1a^t+1\equiv 0 (p)$ and $k_1a^{pt}+1>p$, so it is composite.
Now assume, that this is true for $n<l$. Let $p|k_la^t+1$ for some $p$ prime and some positive integer $t$. Then $k_la^{t+r(p-1)}+1\equiv k_la^t+1\equiv 0 (p)$, so $k_la^n+1$ is composite for $n=t+r(p-1)$, $r=1,2,...$. Now it's enough to consider numbers $(k_i*a^t)\cdot (a^{p-1})^r+1$, $i=1,2,...,l-1$ and find $r$ large enough for which they are all composite (inductive hypothesis).

I have non-inductive solution. Sorry for not typing in LATEX

Remark that

k.2^m + 1 = k.2. [2^(m-1) -1]  + (2.k +1)

For each i = 1,n, we could find an odd prime number p_i divides 2.k_i + 1 ( p_1, p_2,..., p_n are not necessarily pairwise distinct)

Let m take the value

M. (p_1 -1 ). (p_2 -1). (p_3 -1 )...(p_n -1)  + 1

then for each i=1,n we have

2^(p_i - 1) - 1 divides 2^(m-1) - 1. Using Fermat theorem we get  2^(m-1) -1 is divisible by p_i, that yields  p_i divides
k_i.2^m + 1

So all the number k_i. 2^m + 1 are composite for large enough M

[quote=shobber]Does there exist $ 2002$ distinct positive integers $ k_1, k_2, \cdots k_{2002}$ such that for any positive integer $ n \geq 2001$, one of $ k_12^n \plus{} 1, k_22^n \plus{} 1, \cdots, k_{2002}2^n \plus{} 1$ is prime?[/quote]

[hide=Solution]We'll generalize this for $F>2002$
Choose $X = {F^F \prod_{i=1}^{F^F} ({p_i-1})}$ to be extremely large where $p_i|k_x+1$
By Fermat's Little Theorem, $2^X k_i +1 \equiv k_x+1 \equiv 0 \pmod {p_i}$,and by the size of $2^x k_i+1>p_i$,so by conclusion no such integers exist.
[/hide]

The answer is no. The problem is very easy once you realize $2002$ doesn't matter and you can deal with each $k_i$ separately. Simply pick some prime divisor $p_i \mid 2k_i+1$ for each $k_i$, and set $$n=N(p_1-1)(p_2-1)\cdots(p_{2002} - 1) + 1.$$ By Fermat's little theorem this implies that $$k_i \cdot 2^n + 1 \equiv 2k_i + 1 \equiv 0 \pmod p,$$ so all of the terms can be composite.


7/x+3 +  3/6x-2=  4/x+3x-2   solve for x...a problem i have been trying and havent gotten...please show work


Please also post anything else interesting (where you are, how you think you'll do, any pre-comp preparations or foods to eat, etc...)

Dream - dadadadlalalalala - in dreamland far, far away. Not really nervous except for Shaunak killing me in chapters lalalalalalala. Other than that nothing ... lalalalalalala.

yes...different states take chapters on different days. Chapter competitions are usually held in February where as State competitions are held in March. This year my chapter competition is on V-Day. And state is on March 6.

I am in Indiana...most of you know that...my team is most likely going to place first in our chapter...we have the past several years. None of the schools in our conference or whatever you call it are a very big threat.

why is held on different days? (doesn't that make it possible for cheating ) like you have a cousin in another a state who has taken it and tells yhou the solutions. (i remember that in utah last year several questions were tkaen out because they wer mentioned in teh newspaper in earlier compeitions of other states)

I'm in Ohio!  On the 28th!  This is my first and last year!


a,b and c are positive real numbers such that for each positive integer n there exists a triangle $T_{n}$ with sides $a^{n},b^{n},c^{n}$

Prove that all the $T_{n}$ are equilateral

[hide="Is it true?"] Let $b = c = 1$ and let $0 < a < 1$, then we have a series of isosceles triangles with a smallest side approaching $0$. [/hide]


calculus
geometry
3D geometry
function
similar triangles
calculus computations

Hi guys. I guess this problem is pretty easy to you. I solved it also. Just need a confirmation. It belongs to an assignment that will pretty much decide my grade.

Water drains from a conical tank with height 12 ft and diameter 8ft into a cylindral tank that has base 400π sqaure feet. The depth, h, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

(Volume of cone is V=1/3 * π * r^2 * h)

a. Write an expression for the volume of water in the conical tank as a function of h
b. At what rate is the volume of water in the conical tank changing at h=3
c. At what rate is the depth of the water in the cylindral tank changing when h=3

Here is my answer
a. [hide]16/3πh[/hide]
b. [hide]-144π[/hide]
c. [hide]9/25[/hide]

Thanks.

Anhnguyen, you have failed to take account of the fact that $r$ is not constant, but rather a function of $h$. Draw a diagram showing a cross-section of the cone and use similar triangles to find out how $r$ is related to $h$. Your expression for $V$, when multiplied out, should involve $h^3.$


If we count by 3s starting with 1, the following sequence is obtained: 1, 4, 7, 10, … What is the 100th number in the sequence?

The $ n$th number is $ 3n\minus{}2$, so the $ 100$th number is $ 3\times100 \minus{} 2\equal{} 298$.


We had like 10 posts in a month, so we are sucking now

It might be because of:

Winter break- christmas eve etc.

Vacations-

Many forums are not posted in often, so it's not uncommon.

yeah I guess, I just went to some random ones and saw that there were almost no posts

Yup. It is dying. The topics that are only about 6 from the top are like a month old.

Many of the state forums are dying since there aren't that many people and there isn't much to talk about; at least from my observations.

HUZZZZZZAHHH!!  :P

There would be more to talk about if everyone posted in their own state forums, not the ones for the competitions.
 :P


I have this question which I've been struck on for a long time:

A spotlight on the ground shines on the outside wall of a parking garage 12m away. If a 2-m tall man walks toward the garage at a speed of 0.75m/s, how fast is the height of the man's shadow on the garage wall decreasing when he is 4m from the building?

I have no clue how to solve it. :oops:

All I got was:

dx/dt = 0.75m/s
It should be a triangle but y is constant (2)

But the triangle size always move because y = 2 always.

Help please?

There's more than one triangle in the picture- I don't believe you've said anything about the actual shadow yet.

Actually, I got it. 12/y = d/2 ratio b/w the two triangles. Thanks anyways


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14444[/img]

 :(

[url]http://www.polarprof.org/geometriagon/wesercizio.asp?c=545437&n=867[/url]

could you give me an example of a circumscribed square  i cant understand

It is a square whose sides are each tangent to one circle or the other

  :(


geometry
trapezoid
ratio
geometry proposed

Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ be on a line through $D$
such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from
$D$. Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$, respectively. Prove that $AN$
is perpendicular to $NM$.

Let $B_1,C_1,M_1$ be the projections of $B,C,M$ respectively on the line through $A$ parallel to $BC$. We have to prove that $N$ lies on the circle $(ADM)$, which coincides with the circle $(ADM_1)$.

The four points $A,D,B_1,E$ are concyclic, so the lines $AD,B_1E$ are antiparallel in the angle formed by the lines $B_1C_1,EF$. The in the same manner we deduce the fact that $AD,C_1F$ are antiparallel in the same angle. From these two observations we get $B_1E\|C_1F$, which means that $M_1N\|B_1E\|C_1F$, because $M_1,N$ are the midpoints of the sides $B_1C_1,EF$ of the trapezoid $B_1EFC_1$. This, in turn, means that the lines $AD,M_1N$ are antiparallel in the angle formed by the lines $AM_1,DN$, i.e. $A,D,M_1,N$ are concyclic, Q.E.D.

I actually got this problem from the MOC-Book by Titu Andreescu. After I solved it I looked at the "official" solution in the book and the first thing I thought after reading it was: "Hey, I've just found the ultimate IMO Nr 6 Problem: Find a more complicated solution than that one" :lol: - A problem that I thought to be almost impossible. How wrong I was. When I saw the solution on Kalva, I almost fell off the chair (I did not bother to read it).

So here's my (hopefully right) solution:

[hide]

$AEBD$ and $AFDC$ are concyclic so $\angle AED = \angle ABD$ and $\angle AFE = \angle ACD$.
Thus $\triangle ABC\thicksim \triangle AEF$. Because $M$ and $N$ divide $BC$ and $EF$ in the same ratio, the triangles $BAM$ and $EAN$ are similar, thus $\frac{AN}{AM} = \frac{AE}{AB}$ and $\angle BAM = \angle EAN \Rightarrow \angle NAM = \angle EAB$. Thus the triangles $AMN$ and $ABE$ are similar and hence $\angle ANM = \angle AEB = 90^\circ$.

We also see that $M$ and $N$ do not necessarily have to bisect $BC$ and $EF$, they only have to divide them in the same ratio.
[/hide]

Here is my solution:

  Median  AM divides segment between feet of AB and AC on side BC in equal parts,

  so it is quite  ordinary and natural for a circle built on AM as diameter cut in half a

  segment passing  through one of the intersection points (D)  of circles built

  on AB and AC as diameters.   Angle ANM looks at the diameter AM.

  Thank you.

  M.T.

Here is a equivalent enunciation (a wellknown problem):

"Let $w_1,w_2$ be two fixed secant circles and the (fixed) points $\{A,D\}\subset w_1\cap w_2$, $B\in w_1$, $C\in w_2$ so that $D\in (BC)$ and $AD\perp BC$. For the mobile points $E\in w_1$, $F\in w_2$ such that $D\in EF$ we note the middlepoint $N$ of the segment $[EF]$. Ascertain the geometrical locus of the point $N$."

[u]Answer:[/u] The in demand geometrical locus of the point $N$ is the circle with the diameter $[AM]$, where the point $M$ is the middlepoint of the segment $[BC]$.

[u]A solution.[/u] $\triangle AEF\equiv \triangle ABC\Longrightarrow \widehat {ANF}\equiv \widehat {AMC}\Longrightarrow$
the quadrilateral $ANDM$ is cyclic $\Longrightarrow NA\perp NM.$


geometry
circumcircle
ratio
inradius
angle bisector
similar triangles
geometry proposed

1. Let $ABC$ be a triangle, $T$ its centroid and $S$ its incenter.
Prove that the following conditions are equivalent:

(1) line $TS$ is parallel to one side of triangle $ABC$,

(2) one of the sides of triangle $ABC$ is equal to the half-sum of the other two sides.

2. Let $ABC$ be an isosceles triangle with $\angle A=\angle B=80^\circ$. A straight line passes through $B$
and through the circumcenter of the triangle and intersects the side $AC$ at $D$. Prove that $AB=CD$.

1. Suppose 2a=b+c. Let AS cut BC at D. By angle bisector theorem, $\frac{AI}{ID}=\frac{AB}{BD}=\frac{AC}{CD}=\frac{AB+AC}{BC}=\frac{b+c}{a}=2$. The result follows since the centroid divides a median into 2 segments of the ratio 2:1.

2. I think it's pretty well known, just construct a point E s.t. triangle EAB is equilateral.

btw, these 2 problems should belong to the Intermediate Forum.

[hide="Solution to Problem 1"]
The first condition is equivalent to the distance from $T$ to one side $c$ equaling the distance from $S$ to that same side. The distance from $S$ is obviously the inradius, which is $\frac{A}{s}$ ($A$ is the area of the triangle). The distance from $T$ is 1/3 of the altitude to that side because of similar triangles, which is also equal to $\frac{2A}{3c}$. Setting our two quantities equal, we have $2s = 3c$, or $2c = a + b$. Incidentally, this is equivalent to the second condition, and we are done.
[/hide]


search
combinatorics unsolved
combinatorics

There are nine people attending a meeting.Some of them are friends.
It is known that there exists no four people such that any of two are friends
Prove or disprove,nine people can be partioned into four groups such that any two people in the same group are not friends.

I think this follows from [url=http://en.wikipedia.org/wiki/Tur%C3%A1n%27s_theorem]Turan[/url].

Consider a graph $ G$ with $ k$ vertices and containing no $ \ell$ - clique. By Turan, in an maximal connected graph with no $ \ell$ - clique, any two partitions are bipartite. If there exists more than $ \ell \minus{}1$ partitions, we can find a clique.

[quote="Talegari"]I think this follows from [url=http://en.wikipedia.org/wiki/Tur%C3%A1n%27s_theorem]Turan[/url].

Consider a graph $ G$ with $ k$ vertices and containing no $ \ell$ - clique. By Turan, in an maximal connected graph with no $ \ell$ - clique, any two partitions are bipartite. If there exists more than $ \ell \minus{} 1$ partitions, we can find a clique.[/quote]
Did you mean if a graph contains no  $ K_4$ then it must be a tripartite graph?
But it is not true...
Very sorry if I misunderstand you :roll:

It is known that Ramsey numbers $ R(3,4) \equal{} R(4,3) \equal{} 9$, so there must be three persons $ x,y,z$ not knowing each other. For the remaining six, first add edges (edge = people are friends) until any new edge creates a $ K_4$ (thus creating an edge-extremal graph; at the end we will remove those edges, which are only to our disadvantage). Denote the remaining persons $ a,b,c,d,e,f$. There must exist some pair not friends, wlog $ a\perp b$ (notation for lack of edge). Hence adding this edge must create some $ K_4$, wlog with $ c, d$, so there exist edges $ ac, ad, bc, bd, cd$.
If also $ ce$ and $ de$, then $ a\perp e$ and $ b\perp e$. If now also $ cf$ and $ df$, then $ a\perp f$ and $ b\perp f$, and also $ e\perp f$, so we can take $ \{ x,y,z \}, \{ a,b,e,f \}, \{ c \}$ and $ \{ d \}$. Thus we must have wlog $ c\perp f$ (or $ d\perp f$), so we can take $ \{ x,y,z \}, \{ a,b,e \}, \{ c,f \}$ and $ \{ d\}$.
Similarly if we assume first $ cf$ and $ df$.
It remain the cases when $ c\perp e$ and $ d\perp f$, when we can take $ \{ x,y,z \}, \{ a,b \}, \{ c,e \}$ and $ \{ d,f \}$, and similarly when $ c\perp f$ and $ d\perp e$.
Finally, when $ c\perp e$ and $ c\perp f$, but $ de$ and $ df$, then the only bad continuation is $ ef, af, be$ and $ a\perp e, b\perp f$ (or similar).
There remain triangles $ dac, dbc, daf, dbe$ and $ def$. If $ x\perp d, y\perp d$ and $ z\perp d$, we are done, since we can take $ \{ x,y,z,d \}, \{ a,e \}, \{ b,f \}$ and $ \{ c \}$. So $ dx$ and/or $ dy$ and/or $ dz$. Now it's becoming a little messy, but at least the problem has been reduced to checking some not so large number of cases!
(in all, the triplet $ \{ x,y,z \}$ may be now split).

I finally found a reference for this problem.
It is from[url=http://www.mathlinks.ro/viewtopic.php?search_id=281014044&t=170287]Bulgaria TST 2005[/url]


I haven't seen Madagascar, and i liked Shrek 2 better than Shrek.  By the way  :welcome: to AoPS.  Nice avatar.

[quote="chess64"]What's madagascar? A movie?[/quote]

It's highly popular right now. In fact, it's been on the top 10 list for movie profits for like months. :-) Bad movie. Don't go see it.

Although that line "Dang you, dang you all to heck." That was hilarious.

Madagascar was a bad movie....it had some REALLY funny parts in it, but overall it was a bad movie.

[quote="math92"]Wow, it seems you guys don't like madagascar.  Maybe it was for younger children.[/quote]

No. It is family appropriate, but it's not exactly a Lion King or a Barney type film. The movie is just stupid, despite the good animation.

Why on earth isnt The Incredibles on that poll?

I thought that the Incredibles and Finding Nemo were great movies!  THey should be on the poll

has anyone seen national treasure GREAT MOVIE and pacifier is ok too i heard the madagascar stinks except people were obsessed with i like to move it move it?(no idea what it is) shrek was ok but shrek 2 was definitely better. finding nemo was good but shark tale was a rip off of nemo. the incredibles was a good movie

I liked National Treasure but this thread seems to be about animated movies. :P In my opinion, The Incredibles is clearly one of the best out. Shrek is good too, although I must say the lil Fairy Godmother thingy in Shrek II kinda killed some of it.

[quote="juicybooty911"]only 2 choices?[/quote]

It's just to see which one is better.

I think that Shrek was an awesome movie... Saw it like 20 times in a row though... It's still a great (but boring after 20 times) movie.

As for madagascar, my sister liked it but she's 12.

[quote="mathbomberII"]i haven't seen Madagascar yet but from teh votes it seems like Shrek is way better[/quote]

Your'e right  :P

i've never seen madagascar, but I've heard from my friends it was a good one, but from you, it sounds awful!


conics
ellipse
linear algebra
matrix
calculus
derivative
function

Suppose I multiply all points on the unit disc with the matrix $ A$ as defined below:
$ A \equal{} \left( \begin{array}{cc}
2 & 3 \\
0 & 4  \end{array} \right)$
This will result in an ellipse. Now I'd like to find out how far the farthest point on this ellipse is from the origin. This problem is supposed to be solvable without using derivatives and without ever multiplying $ A$ with a unit vector.

Calculate the greatest eigenvalue of the (positive) matrix $ A^tA$.

Counter question: Where do you know from that the result is an ellipse? How is it parametrized? ;)

Sorry, I dont follow... I know that it's an ellipse since the problem states that :)

What you are looking for is the maximum of the function $ f: S^1\to\mathbb R$ with $ f(x)\equal{}x^t(A^tA)x$. Since $ B\equal{}A^tA$ is symmetric, we can orthogonally diagonalize $ B$. This implies (?) that we can assume WLOG that $ B$ is diagonal where the diagonal elements are the the eigen values of $ A^tA$.

Does this help you?


hey is anybody here doing engineering who has applied for kvpy in EA or EB stream. please post here or pm me.
 i'd like to know what kind of  projects have been made and what's the allowed scope and syllabus and so on.

This link might be of some use to you:

http://www.orkut.com/CommMsgs.aspx?cmm=453797&tid=2463773829331124275&kw=KVPY+award+winning+projects


analytic geometry
trigonometry
ratio
absolute value
real analysis
real analysis solved

I generally don't enjoy this type of problem, but my college friend needed help with it and it ended up being pretty interesting. I got an answer, but he said the computer was telling him it was wrong. Let me know what you get.

A clock has a $10\text{cm}$ long minute hand and a $5\text{cm}$ hour hand. How fast is the distance between the end of the minute hand and the end of the hour hand changing (in $\frac{\text{cm}}{\text{hr}}$) at $3\text{ o'clock}$?

My answer: $-\frac{5900\pi}{\sqrt{125}}\frac{\text{cm}}{\text{hr}}$

Put the coordinate origin at the clock center, x-axis through 3 o'clock and y-axis through 12 o'clock. Let $A = 5\ \text{cm}$, $B = 10\ \text{cm}$ be the lengths of the hour and minute hands, $\omega = 2 \pi\ \text{rad/hr}$ the angular velocity of the minute hand and let the time be $t = 0$ at midnight (or at noon). The endpoints $H, M$ of the hour and minute hands have coordinates

$H = [x_H, y_H] = [A \sin{\frac{\omega}{12} t}, A \cos{\frac{\omega}{12} t}]$

$M = [x_M, y_M] = [B \sin{\omega t}, B \cos{\omega t}]$

The distance $HM$ between the hand endpoints is

$HM = \sqrt{(x_H - x_M)^2 + (y_H - y_M)^2} = \sqrt{\left( A \sin{\frac{\omega}{12} t} -  B \sin{\omega t} \right)^2 + \left(A \cos{\frac{\omega}{12} t} - B \cos{\omega t} \right)^2}$

$\frac{dHM}{dt} = \frac{\left( A \sin{\frac{\omega}{12} t} -  B \sin{\omega t} \right) \left( \frac{\omega}{12} A \cos{\frac{\omega}{12} t} - \omega B \cos{\omega t} \right) - \left( A \cos{\frac{\omega}{12} t} -  B \cos{\omega t} \right) \left( \frac{\omega}{12} A \sin{\frac{\omega}{12} t} - \omega B \sin{\omega t} \right)}{\sqrt{\left( A \sin{\frac{\omega}{12} t} -  B \sin{\omega t} \right)^2 + \left(A \cos{\frac{\omega}{12} t} - B \cos{\omega t} \right)^2}}$

Obviously, at the time $t = 3\ \text{hr}$,

$sin{3 \omega} = \sin{6 \pi} = \sin 0 = 0$
$\cos{3 \omega} = \cos{6 \pi} = \cos 0 = 1$

$\sin{\frac{\omega}{4}} = \sin{\frac{\pi}{2}} = 1$

$\cos{\frac{\omega}{4}} = \cos{\frac{\pi}{2}} = 0$

$\frac{dHM}{dt} = \frac{-\omega AB \left( 1 - \frac{1}{12} \right)}{\sqrt{A^2 + B^2}} = -\frac{2 \pi \cdot 5 \cdot 10 \cdot \frac{11}{12}}{\sqrt{5^2 + 10^2}} = \frac{11 \pi \sqrt 5}{3}\ \text{cm/hr}$

That makes 3 different results, so here is a sanity check: Let the time be $t = 0$ at 3 o'clock. Consider the endpoint $H$ of the hour hand moving vertically down from the 3 o'clock position at the tangent velocity $u = -\frac{\omega}{12} A$ and the enpoint $M$ of the minute hand move horizontally right from the 12 o'clock position at the tangent velocity $v = \omega B$. Then

$H = [A, -\frac{\omega}{12}At]$

$M = [\omega B t, B]$

$HM =  \sqrt{(x_H - x_M)^2 + (y_H - y_M)^2} = \sqrt{(A - \omega B t)^2 + \left( B + \frac{\omega}{12}At \right)^2}$

$\frac{dHM}{dt} = \frac{-\omega B (A - \omega Bt) + \frac{\omega}{12} A \left( B + \frac{\omega}{12} At \right)}{\sqrt{(A - \omega B t)^2 + \left( B + \frac{\omega}{12} At \right)^2}}$

At the time $t = 0$

$\frac{dHM}{dt} = \frac{-\omega AB \left( 1 - \frac{1}{12}\right)}{\sqrt{A^2 + B^2}} = -\frac{2 \pi \cdot 5 \cdot 10 \cdot \frac{11}{12}}{\sqrt{5^2 + 10^2}} = \frac{11 \pi \sqrt 5}{3}\ \text{cm/hr}$

This is what I did.

edit: i just realized the ratio of the minute hand's angular velocity to that of the hour hand is 12, not 60  :blush:

\[D=\sqrt{(x_m-x_h)^2+(y_m-y_h)^2}\]
Differentiate with respect to time.
\[D'=\frac{(x_m-x_h)(x_m'-x_h')+(y_m-y_h)(y_m'-y_h')}{\sqrt{(x_m-x_h)^2+(y_m-y_h)^2}}\]
Now we need to find equations for $x_m,x_h,y_m,y_h$: ($t$ is in hours)

$x_m=10\sin12\cdot2\pi{t}$
$x_m'=240\pi\cos24\pi{t}$.

$x_h=5\cos2\pi{t}$
$x_h'=-10\pi\sin2\pi{t}$.

$y_m=10\cos12\cdot2\pi{t}$
$y_m'=-240\pi\sin24\pi{t}$.

$y_h=-5\sin2\pi{t}$
$y_h'=-10\pi\cos2\pi{t}$.

These give the position of the hour and minute hands at time $t$ (hours), where $t=0$ corresponds to $3\text{ o'clock}$. Thus

\[D'|_{t=0}=\frac{(0-5)(240\pi-0)+(10-0)(0-(-10\pi))}{\sqrt{(0-5)^2+(10-0)^2}}=-\frac{1100\pi}{\sqrt{125}}\frac{\text{cm}}{\text{hr}}=-\frac{220\pi}{\sqrt{5}}\frac{\text{cm}}{\text{hr}}.\]

First, I noticed that your answers were both positive, but I'm pretty sure the answer should be negative. They're moving closer together because the minute hand is moving to the right and the hour hand is barely moving with respect to it.

(a few edits in the following few sentences..)

Also, I set up a simpler problem, where the hour hand is stationary and the tip of the minute hand is moving to the right (linearly) at a constant speed: $\left(\frac{12\text{ revolutions}}{hr}\right)\left(\frac{2\cdot10\text{cm}\cdot\pi}{\text{revolution}}\right)=240\pi\frac{\text{cm}}{\text{hr}}$. I graphed $D(t)$ and found $\frac{dD}{dt}$ with my calculator and got a number that was slightly less (but greater in absolute value) than my answer for the original problem, which is what would be expected.

[quote="yetti"][quote="AntonioMainenti"] Thus

\[D'|_{t=0}=\frac{(0-5)(240\pi-0)+(10-0)(0-(-10\pi))}{\sqrt{(0-5)^2+(10-0)^2}}=-\frac{1100\pi}{\sqrt{125}}\frac{\text{cm}}{\text{hr}}=-\frac{220\pi}{\sqrt{5}}\frac{\text{cm}}{\text{hr}}.\]

First, I noticed that your answers were both positive, but I'm pretty sure the answer should be negative. They're moving closer together because the minute hand is moving to the right and the hour hand is barely moving with respect to it.[/quote]

My answers are both negative (both expressions have the minus signs even after substituting numbers). Just the final fractions (one a copy of the other) have the minus sign missing, otherwise, the value is correct.

Another sanity check: the angular velocity of the [b]minute[/b] hand is $2 \pi\ \text{rad/hr}$ (1 full revolution per hour). The orbital velocity of the minute hand enpoint is $10\ \text{cm} \cdot 2 \pi\ \text{rad/hr} = 20 \pi\ \text{cm/hr}$. Surely, the absolute value of the result has to be less than that. You considered the angular velocity of the [b]hour[/b] hand to be $2 \pi\ \text{rad/hr}$, but it is $\frac{2 \pi}{12}\ \text{rad/hr}$ ( 1 full revolution per 12 hours).[/quote]I see now. Thanks for clearing that up. This is why I hate doing related rates problems..


Q:Prove that a trapezium with non-parallel sides equal is cyclic.

An isosceles trapezoid has congruent pairs of base angles and the 2 angles on one leg are supplementary, so the opposite angles are supplementary and the trapezoid is cyclic.


Yup yet another mathleague.org offering.

Post how you felt about the contest: I felt the sprint round was disappointingly easy, the target round was good difficulty, and the countdown? Let's just say I got lucky - twice.

Sprint - I don't know, maybe Chapterish difficulty
Target - I don't know, maybe slightly less than Chapterish difficulty

I guess the difficulty might have been about right for middle schoolers...
[hide]
but dang, that was boring
[/hide]

Also, what happened to Miller Team #1?

@Ube, sure, you are too old  :D

[quote="james4l"]Also, what happened to Miller Team #1?
[/quote]

Hmm... I wonder where Kevin Lei and Nikhil Buduma were?

nikhil, klei were just gone for some reason

mieh i failed written AND cd today.  *cries*

wow i added wrong about 5 times on cd today

lol, 6th-7th-8th scores were pretty funny

6th:

1st: 29?
2nd: 28?
3rd: 27

or something like that

7th

1st: 45 (james4l)
2nd: 39 (me) :wallbash_red:
3rd: 38

8th

1st: 45 (miller4math)
2nd: 44 (ecstaticpotter)
3rd: 39 (bowser)

mieh.

notice difference between 6th and 7th/8th scores :D

hm, ubey saw me fail countdown?  not good.

[quote="bowei"]oh so that explains why you weren't at Scioly.
who else was there? i know ccy + bookaholic were at scioly w/ me.[/quote]
Albert doesn't do scioly though, doesn't he? I had to explain to him what scioly was just a month ago.
Also, scioly=silly.

AAAAAAA NOW EVERYONES GOING TO KNOW WHO I AM >:[ o well its just ANYONE WHO REMEMBERS ehh just kidding...? me::winner_third:
actually no
victor and johnny:  :winner_first:  :winner_second: me:  :dry: hahaha

>.< i failed....hmm.....
3 stupid mistakes on sprint :wallbash: + bombed target round  :wallbash_red:  :wallbash_red: = 3rd place?? i am amazed
but i was just lucky that kevin lei or nikhil buduma didnt go
and U B E Y M A Y A M Y M O M S A W Y O U R M O M @_____@ i hope you can read that...theres like 10 O's and 20 A's and 30 M's...HI I THINK THAT YOUR REALLY SMART

yesterday there was a 1 point gap between 1st and 2nd
AND THEN A 5 POINT GAP BETWEEN 2ND AND 3RD BWAHAHAHA
i just got lucky XD yay now i have a pretty plaque or however you spell it  :D

i hope i dont fail aime!!  i practiced a lot and i have one more practice tomorrow....before the real test.....GASP  :wow: I WILL NOT FAIL!!! if i tell myself i wont fail then maybe i wont :jump:

what?? how come theres a smiley limit?? >____<

yo please limit # of smileys to like
1

its annoying to read a post spammed with smileys
so i didnt read your post
ok
?

Well that wasn't really too many smileys... it only becomes spammy when you add 5+ in a row.

Also darn I got a 41 on that test... misread 3 on sprint and 1 on target (for example on #7 on target I thought that the vertices of the square were on (-10, 0) and (0, 4), because when I read it my brain automatically assumed "it couldn't be on (4,0) because they wouldn't make the problem THAT trivial").

What contest are you guys talking about?

lol edgar

that is you right?

what grade are you in?

what was your score?

O_O

i want to know akshay's score

time to email him

and for countdown

you should have seen my scratch paper

i looked at it afterwards and found 7 adding erros and 2 multiplication errors

fail?

mathleague.org has a better description than what i can give.

[quote="weird888"][quote="junggi"]STOP QUOTING 80 LINE POSTS AND REPLYING WITH A ONE LINE POST[/quote]

junggi, plz stop with the remarks and stuff, it wasnt an 80 line post, and people have there reasons...

anyway... what tournament is this???[/quote]
what
its like basic forum etiquette
you dont reply to a page long post with one line
especially if that one line doesnt concern the whole post
like cut out the parts you arent responding to
i dont want to scroll down a whole page while rereading a huge post and then read a one line post which is irrelevent to most of the post

Okay, I agree. No more pointlessly long quotations with pointlessly short responses.

[quote="james4l"]Also, what happened to Miller Team #1?
[/quote]
well u see the dumb rules divided our scores by 2...
and we got 41.5 or something...
lol and the third place team was 47.5
 :dry:


function
limit
real analysis
real analysis unsolved

Let $ f_n: N\rightarrow C$ a sequence of complex functions on the positive integers such that $ \sum_{m\equal{}1}^{\infty}|f_n(m)|<\infty$ for all $ n$.

Suppose that $ \lim_{n\rightarrow\infty} \sum_{m\equal{}1}^{\infty}f_n(m)y_m\equal{}0$ for every bounded complex sequence $ \{y_m\}$.

Prove that $ \lim_{n\rightarrow\infty} \sum_{m\equal{}1}^{\infty}|f_n(m)|\equal{} 0$.

I just guess, maybe I am wrong  :lol:

write $ f_{n} = f^{1}_{n} + if^{2}_{n}$ so we can assume $ f_{n}$ is a real -valued sequence

chose bounded sequences  and $ ({y_{m})}$, st : $ 0\leq \sum_{m = 1}^{k}|f_{n}(m)|\leq \sum_{m = 1}^{k}f_{n}(m)y_{m}$

since $ \sum_{m = 1}^{k}f_{n}(m)y_{m}\rightarrow 0$ when $ k,n\rightarrow \infty$ so $ \sum_{m = 1}^{k}|f_{n}(m)|\rightarrow 0$

I think this is in exact consequance of  the uniform boundedness principle or Banach-Steinhaus theorem.


Find the product of the roots of the equation $ 18t^2 \plus{} 45t \minus{}500 \equal{}0$.

hello, solving the given quadratic equation we get $ t_1\equal{}\frac{25}{6}$, $ t_2\equal{}\minus{}\frac{20}{3}$ and their product is $ \minus{}\frac{250}{9}$.
Sonnhard.

To make it faster (since it's Countdown), we use Vieta's.  In a quadratic equation $ ax^2\plus{}bx\plus{}c$, the product of the roots is $ \frac{c}{a}$.  So in this case, we do $ \minus{}\frac{500}{18}\equal{}\boxed{\minus{}\frac{250}{9}}$, as stated by Dr Sonnhard Graubner.


Prove that the sides a,b,c of any triangle suck that $ a \plus{} b \plus{} c \equal{} 3$   satisfy the inequality
i/$ \sum\frac {a}{b\plus{}c^2}>\equal{}\frac{3}{2}$

you have posted the same problem here,an hour ago:
[url]http://www.mathlinks.ro/viewtopic.php?t=199993[/url]
  :huh:


[u][i]Problem[/i][/u]

 [i]Let two sequences $ \{a_{n}\}$ , $ \{b_{n}\}$ defined by :

         $ a_{0} \equal{} b_{0} \equal{} 1$

          $ a_{n \plus{} 1} \equal{} a_{n} \plus{} b_{n}$

          and     $ b_{n \plus{} 1} \equal{} a_{n} \plus{} 3 b_{n}$   with all    $ n \in N , n \geq 1$

          Find  $ \lim_{n\to\infty}\frac {a_{n}}{b_{n}}$[/i]

                  [i]Nice problem[/i]
 :lol:

  [i]  Love An Forever

 Hùynh Võ Phương An[/i]

Dividing the two equations we get:

$ \frac{b_{n\plus{}1}}{a_{n\plus{}1}}\equal{}1\plus{}\frac{2 b_n}{a_n\plus{}b_n}\equal{}1\plus{}\frac{2}{\frac{a_n}{b_n}\plus{}1}$

Now, we know that:

$ \lim_{n \rightarrow \infty} \frac{a_n}{b_n}\equal{}\lim_{n \rightarrow \infty} \frac{a_{n\plus{}1}}{b_{n\plus{}1}}$

and suppose that the limit actually exists and equals L. Now, let $ n \rightarrow \infty$ and get:

$ \frac{1}{L}\equal{}1\plus{}\frac{2}{L\plus{}1}$

$ L\plus{}1\equal{}(L\plus{}1)L\plus{}2 L$

Now, we have:

$ L_{1,2} \equal{} \pm \sqrt{2}\minus{}1$

but since $ a_0,\, b_0 \geq 0$ and $ a_n < a_{n\plus{}1}$ and $ b_n < b_{n\plus{}1}$ we conclude that $ a_n,\, b_n \geq 0$ so $ L$ must be positive and hence we have:

$ L \equal{} \sqrt{2}\minus{}1$


When looking at a book at a hypothetical online book store, you can click "surprise me" to see a random page from the book. On average, how many clicks will it take for you to see the entire book?

[hide="Solution"]
Suppose there are $p$ pages. Consider $p$ steps; in each step, the total number of distinct pages viewed increases by 1.

Suppose $n$ distinct pages of $p$ have been viewed. The probability that the next page will be a new one is $\frac{p-n}{p}$. Hence, the expected number of pages that must be viewed to reach $n+1$ distinct pages viewed is given by:
$\sum_{j=1}^{\infty}j \left( \frac{p-n}{p}\right) \left( \frac{n}{p}\right)^{j-1}= \frac{p}{p-n}$

Now, we just have to sum this over all of the $p$ steps:
$\sum_{n=0}^{p-1}\frac{p}{p-n}$
which cannot be further simplified.
[/hide]

Wow, that was fast!

My friend The Purple Cow came up with a somewhat easier-to-read version of your final formula:  [hide]Let n be the number of pages. The expected number of clicks is then

$n \sum_{k=1}^{n}\frac{1}{k}$[/hide]


When I look at Resources, I can't download PDF version.
What's the matter with that?

kunny

i try to download the jbtst and i failed too. please answer kunny's question.

Problem was detected and (finally) fixed :)

Yay Valentin's back at last...

I think this has been an issue for a while, right?
Well thanks for fixing it!

I can't really post anything but does the PDF file database get automatically updated when a new problem is added, or does the physical PDF have to get updated?

The PDF file gets updated when someone edits the Resources page for that particular contest (e. g. changes the links to the posts with the problems, or does as if he would change them but changes nothing), [i]not[/i] when the posts with the problems are edited. Hence, the PDF file may be slightly obsolete compared with the HTML resources page.

@Valentin: thanks a lot on my part!

  darij

[quote="darij grinberg"]The PDF file gets updated when someone edits the Resources page for that particular contest (e. g. changes the links to the posts with the problems, or does as if he would change them but changes nothing), [i]not[/i] when the posts with the problems are edited. Hence, the PDF file may be slightly obsolete compared with the HTML resources page
  darij[/quote]Unfortunately you are wrong Darij. The pdf's are updated as you change stuff in the posts themselves ;)

I noticed this because there is some trouble with images etc. that should be looked into, did you yourself write the code that generates the pdf?

[quote="darij grinberg"][quote="Valentin Vornicu"]Unfortunately you are wrong Darij. The pdf's are updated as you change stuff in the posts themselves ;)[/quote]

Nice update ;) .

  Darij[/quote]It's not update, it's the way I designed it from the first time ;)

me@home: there's a new version coming up soon, so I'm not really working on this one anymore. Have a bit of patience, Resources 2.0 will be here :P


In any triangle $ABC$. Prove that:
\[bc(b^{2}-c^{2})\cos A+ca(c^{2}-a^{2})\cos B+ab(a^{2}-b^{2})\cos C=0\]

[hide]Plug $\cos A={b^{2}+c^{2}-a^{2}\over 2bc}$ etc. and expand to obtain mass cancelation.[/hide]


calculus
integration
limit
real analysis
real analysis unsolved

how to integrate

1/(1+x+x^(2n))?
i know that usually it should be factorized but how...isn't there an easier way
this is par of

problem: a in(0,1). find lim (n->+infinity) of integral(from -a to a) of dx/[(1+x^2+x^(2n))(1+e^x)].

notes:the integral in the problem simplifies to integral(from 0 to a)dx/[(1+x^2+x^(2n)) after representing it like integral(-a to 0)+integral(0 to a) and changing x=-t in the first.

If I understand your problem corectly you need to calculate the following limit

$\lim_{n\rightarrow \infty}\int_{0}^{a}\frac{dx}{1+x^{2}+x^{2n}}=\int_{0}^{a}\frac{dx}{1+x^{2}}=\arctan(a)$.

The limit follows based  on the Bounded convergence Theorem, since

$0<f_{n}(x)=\frac{1}{1+x^{2}+x^{2n}}\leq 1$ for all $n\in N$ and for all $x\in (0,a)$ where $0<a<1$.

Note that $f_{n}(x)\rightarrow \frac{1}{1+x^{2}}$ for $x\neq 0$ and hence the result.


geometry
perimeter
calculus
3D geometry
sphere
conics
ellipse

$Question 1$
Prove that of all right triangles with given perimeter, the isosceles right triangle has maximum area. One can easily do this using techniques from calculus; can you find a proof which does not use calculus?

$Question 2$
Let $P$, $Q$, $R$, and $S$ be four distinct points on a circle in clockwise order, and let $A$, $B$, $C$, and $D$ be the midpoints of the arcs $PQ$, $QR$, $RS$, and $SP$, respectively. Prove that the line segments $AC$ and $BD$ are perpendicular.

$Question 3$
Imagine that you have a calculator which can only do the following operations on any real number:
$i$. add
$ii$. subtract
$iii$. divide  by $24$;
$iv$. raise to the $4$th power.
Show how this calculator may be used to compute the product of any two real numbers.

$Question 4$
Ms X has stored $100$ kg of gold dust in her house and has invited Ms Y to take as much of it as she likes. On the first day Ms Y takes $1\%$ of the gold, and leaves the remaining $99\%$. The next day she returns and takes $2\%$ of the available gold, and leave the rest. The next day she takes $3\%$ of the available gold, and leave the rest. This continues until the $100$th day, when she takes $100\%$ of the gold, leaving none. On which day did she take the most gold?

$Question 5$
A follicularly challenged person is out in a walk in a park when a bee spots him. The bee is amused by the sight of his shining pate. If the person's head can be approximated to a sphere of diameter $d$ and the bee is at a distance $h$ above the person's head, what fraction of the area of the head can the bee see?

$Question 6$
If two triangles have the same area and also the same perimeter, does it follow that the triangles are congruent? If so, give a proof; or else give a counterexample.

[quote="fizsle"]I don't know the actual theorem but i think by number theory u can prove it... product of two equal numbers is greater than that of products of two other numbers having the same sum... :maybe:[/quote]

We need to prove that $a^{2}> (a-b)(a+b)$ (as $a+a=a-b+a+b$). Multiplying out we get $a^{2}> a^{2}-b^{2}$, which is true (as we know $b>0$).

[hide="Hint"]
Use this together with Heron's Formula.
[/hide]

For #1, you just use the same method you would if you were to find the maximum area for a given amount of fence you have.

[hide="q4"]
$\text{nth}=\frac{(100)!n}{(100-n)!100^{n}}$
Then find the maximum...
I have no clue
[/hide]
The diagram is for question 2

[quote="fizsle"]I don't know the actual theorem but i think by number theory u can prove it... product of two equal numbers is greater than that of products of two other numbers having the same sum... :maybe:[/quote]
but it's only true if you prove that $a+b$ is constant-$c$ is constant' but I'm not sure it's true. how can we use heron?
for question 3, we can work in base 24 and then its esay but I think you didn't meant it.
$Q4.$ in the 99th day?
for $Q6$ I know it doesn't correct but how do we find a counterwxample in a nice way?

6. false
fix a base, then the locus of constant area for the last side is a parallel line, the locus for perimeter is an ellipse (with the endpoints of the base as the foci), their intersection is the third vertex

however if we sufficiently extend the length of the base so it is greater than any side of the first triangle, then we have a different triangle (intersection is guaranteed because the ellipse gets bigger while the line gets closer

Does anyone have a solution for the other problems i.e. q4, q3?
Thanks


linear algebra
matrix
vector
linear algebra unsolved

Hi all,

let A a n*n matrix of range r. Consider the following (n-1)*(n-1) matrix B such that B(i,j)=det( A(1   ,1)   A(1  ,i+1)
                       A(j+1,1)  A(i+1,j+1) )
Find the range of B ...

bye.

I'm sure by "range" you mean "rank", right?

yeah that's .. and to conclude, think about the LU decomposition ...

It can;t always be $r-1$. If the first column is $0$, for example, it can be $0$.

If all element of the first column of $A$ is zero $a_{i1}=0$ then $B=0$, and
$rank(B)=0$ :D

If $a_{11} \neq 0$ and if there exist $a_{i1}\neq0, a_{1j} \neq 0$
the rank of $B$ is $rank(A)-1$


Find the minimum natural number $n$ with the following property: between any collection of $n$ distinct natural numbers in the set $\{1,2, \dots,999\}$ it is possible to choose four different $a,\ b,\ c,\ d$ such that: $a + 2b + 3c = d$.

Denote by M a set of numbers such that for any four different numbers we have $a+2b+3c\neq d$.Denote by m the number of its elements which are $a_1<a_2<....<a_m$. We will prove that the maximum value of $m$ is 833.

We have that the sets $A=\{3a_1+2a_2+a_x: x\ge 3,a_x\le a_m-3a_1-2a_2\}$ and $B=\{a_x: a_x\ge a_m-3a_1-2a_2\}$ are disjoint, and their union is a subset of $(a_m-3a_1-2a_2,999] \cap \mathbb{N}$.

We used the fact that $\min B>a_m-3a_1-2a_2$

mmm...

so wich is the minimun value of $ n$??

:huh:

@ElChapin:
according to xirti the minimum value of $ n$ is $ 999\minus{}833 \equal{} 166$

mmm... I don't have idea of wath he tried to do then...

166 non even in dreams...

833 the maximun without the property, but even 834 doesn't work...

835 seems to be the answer, but I can't prove that you can alwas find $ a$, $ b$, $ c$, $ d$.

elchapin 835 is the answer  :lol:
see [url=http://www.kalva.demon.co.uk/ibero/isoln/isol983.html]here[/url] :wink:

Initially, will select the last $k$ integers so that when equality occurs above. Since the smallest element of the set, we have:

$3a +2(a +1) +a+2 < 999 \iff a<166$

Thus, the set $\{165.166, ..., 999\}$, $k = 835$, satisfies the conditions of the utterance. Thus, $n \ge 835$. A good start would be to see if $835$ is the solution or if $n> 835$ ...

divide the initial set into two subsets $A = \{1,2, ..., 166\}$ and $B = \{167,168, ..., 999\}$. Clearly, they are selected at least two elements of $A$, so suppose we have selected $n +2$ elements in $A$, which form a subset $A'= \{x_1, x_2, ..., x_ {n +2}\}$, and $833-n$ elements in $B$, forming the set $B'= \{y_1, y_2, ..., y_ {833-n}\}$, $n \ge 0$. Are $x_1, y_1$ the smallest elements in $A, B$, respectively. Note that $x_1 \le 167 - (n +2)$ and $167 + y_1 \le n$. Thus, $x_p$ is any element in $A$ and different from $x_1$, which follows:

\[167 \le y_1+2x_p+3x_1 \le 167+n+2 \cdot 166+3(167-(n+2))\]
\[167 \le y_1+2x_p+3x_1 \le 994-2n\]

Thus, the expression $ y_1+2x_p+3x_1$ is always an element in $B$.Mas note that $x_p$ can take $n +1$ values. And $B'$ has $833-n$ elements, at least one of these $n + 1$ is in $B'$! Therefore, $835$ is the smallest value of n with this property.


[color=green]1)[/color]Show that a convex polygon is a pentagon if and only if the number of sides is equal to the number of diagonals.

[color=green]2)[/color] Denote $ a$ and $ d$ the lenght of the side a the diagonal of a regular pentagon, show the relation:

$ \frac{d}{a} \minus{} \frac {a}{d} \equal{} 1$

Thanks in advance

[hide]1) $ n\equal{}\frac{n(n\minus{}3)}{2}$, so $ n\equal{}5$.

2) Sine theorem: $ d\equal{}2a\cos36^\circ$

$ d^2\minus{}a^2\equal{}ad$ is equivalent to $ 4\cos^236^\circ\minus{}1\equal{}2\cos36^\circ$. This can easily be proven.[/hide]

I dont think that  is necessary to kill this nice problem with trigonometry. The proportion d/a-a/d=1 follows from being the side of a  regular pentagon the golden section of its diagonal.


geometry
3D geometry
tetrahedron
counting
distinguishability
rotation
geometric transformation

Each face of a regular tetrahedron is painted either red, white, or blue.  Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical.  How many distinguishable colorings are possible?

A.) 15   B.) 18   C.) 27  D.) 54   E.)81

Whats the fastest/easiest way to solve this problem.  The solution in the AMC 12 packet is kind of long.

The fastest way to solve this problem is probably with [b]Burnside's Lemma[/b] (see http://en.wikipedia.org/wiki/Burnside's_lemma ), although that isn't exactly elementary.  The "longer" method is much more elementary and, as long as you are carefully organizing your work, just as good (for this problem).

Hopefully this is right

[hide="Solution"]
One color: 3 ways to color all four faces one color.

Two colors: WLOG let the colors be red and blue.  We can have

1 red     3 blue              2 red     2 blue           3 red     1 blue

so only 3 possible ways.  There are $ \binom{3}{2}$ ways to choose two colors, so $ 3\times 3\equal{}9$ ways for two colors.

All Three colors: Clearly we must use the color pattern 1   1   2, so two colors are used once, and one color is used twice.  There are 3 ways to pick the color that is used twice, so there are only 3 ways.  The final total is
$ 3\plus{}9\plus{}3\equal{}15$, so A.[/hide]

t0rajir0u, that link doesn't work, it should be: http://en.wikipedia.org/wiki/Burnside%27s_lemma

What does wiki mean by "one identity element fixing all 36 elements of X"?  Can someone explain burnside's lemma to me?

Burnside's Lemma deals with [url=http://en.wikipedia.org/wiki/Group_action]group actions[/url].  In this case, we are dealing with the group of rotations of a tetrahedron.  The [url=http://en.wikipedia.org/wiki/Identity_element]identity element[/url] of this group is the rotation of $ 0^{\circ}$ in every direction - in other words, it fixes every face.

The group-theoretic approach is a little heavy for this problem, and is largely unnecessary.  I mention it only as a reasonably fast general technique for problems of this nature.  It would be much easier if we wanted to discuss, say, colorings of an icosahedron.

but why dosen't this work :
3 choices of paint for each side (including bottom side) so: 3*3*3*3.
divide this by 3, because you can rotate the tetrahedron three times and the colorings would still be in the same order.  so the answer is 27.

There aren't exactly $ 3$ rotations corresponding to colorings that are the same.  Count more carefully.


[quote]One real number is written in each box of a rectangle table . It is allowed to change the sign of all the numbers in any row or column.Prove using these operations one can make the sum of the numbers in the table non-negative.[/quote]

If the sum is negative, can't we just flip the sign of every number in the table?


algebra
polynomial
calculus
derivative
real analysis
real analysis unsolved

A polynomial of deg n (n>3) has all its roots real and distinct. Assume them ordered. The AM of the first 2 and the AM of the last 2 cannot be roots of the derivative of the polynomial.

PS: All of these were proposed for the 11'th grade.

Hey.. isn't this one, a problem of Titu Andreescu? Cause I remember it from somewhere... and I think he was the author! :D

thank you grobber for posting them! :D

cheers!

Yeah, it was proposed by Titu Andreescu.

Let $r_1, \dots, r_n$ be the roots of the polynomial $f$.  Then it is known that
\[\frac{f'(x)}{f(x)} = \sum_{k=1}^n \frac{1}{x - r_k} .\]
From there, I think it is pretty quick to finish off the problem.


Putnam
algebra
polynomial
college contests

Find all real polynomials $ p(x)$ of degree $ n \ge 2$ for which there exist real numbers $ r_1 < r_2 <  ... < r_n$ such that

(i) $ p(r_i) \equal{} 0, 1 \le i \le n$, and

(ii) $ p' \left( \frac {r_i \plus{} r_{i \plus{} 1}}{2} \right) \equal{} 0, 1 \le i \le n \minus{} 1$.

[b]Follow-up:[/b]  In terms of $ n$, what is the maximum value of $ k$ for which $ k$ consecutive real roots of a polynomial $ p(x)$ of degree $ n$ can have this property?  (By "consecutive" I mean we order the real roots of $ p(x)$ and ignore the complex roots.)  In particular, is $ k \equal{} n \minus{} 1$ possible for $ n \ge 3$?

in your follow-up you require the roots to be distinct?
because even n could be reached with $ x^n$ :)

Yes; I forgot to add the condition that the roots be sorted.

n-1 can not be reached! $ \frac{r_i\plus{}r_{i\plus{}1}}{2}$ can not be a root of $ p'$ if $ r_i,r_{i\plus{}1}$ are the two smallest (biggest) roots of $ p$.

Find all real polynomials $ p(x)$ of degree $ n \ge 2$ for which there exist real numbers $ r_1 < r_2 <  ... < r_n$ such that

(i) $ p(r_i) = 0, 1 \le i \le n$, and

(ii) $ p' \left( \frac {r_i + r_{i + 1}}{2} \right) = 0, 1 \le i \le n - 1$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

a) Prove that a group of order $ 30$ cannot be simple.
b) Prove that a group of order $ 42$ cannot be simple.
c) Prove that a group of order $ 45$ cannot be simple.

b) Show that any group of order $ 42$ has exactly one $ 7$-Sylow subgroup.
c) How many $ 3$-Sylow subgroups does a group of order $ 45$ have? - In fact, it can be shown that there are no simple groups of order $ p^{2} \cdot q$.

 The idea underlying both proofs above is [b]simple[/b] to express: if G is a finite subgroup and H is the only subgroup of G of order |H|>1, then H is necessarily a normal subgroup of G. Do you know why?

[quote="coquitao"]Do you why?[/quote] Why?  :oops:

[quote="mr.starlight"]a) Prove that a group of order $ 30$ cannot be simple.[/quote]

a) We wish to show that at leat one of the prime divisors of $ 30$ is such that $ n_{p} \equal{} 1$. If it were not the case, we'd have $ n_{2} > 1, n_{3} > 1,$ and $ n_{5} > 1$... Can you derive a contradiction from the previous assumption now?

[quote="mr.starlight"][quote="coquitao"]Do you know why?[/quote] Why?  :oops:[/quote]

For any $ x \in G$ it is known that $ xHx^{ \minus{} 1}$ is a subgroup of G. Is it clear to you that $ |H| \equal{} |xHx^{ \minus{} 1}|$? SInce H is the only subgroup of G that has order |H|, it follows that $ H \equal{} xHx^{ \minus{} 1}$ for every $ x \in G$ and whence the result.

[quote="coquitao"]

a) We wish to show that at leat one of the prime divisors of $ 30$ is such that $ n_{p} \equal{} 1$. If it were not the case, we'd have $ n_{2} > 1, n_{3} > 1,$ and $ n_{5} > 1$... Can you derive a contradiction from the previous assumption now?

[/quote]

In that scenario we'd have $ 9$ three-Sylow subgroups and $ 6$ five-Sylow subgroups. Since the intersection of any two three-Sylow subgroups is trivial (being in a group or order 30), we have that there are $ 24$ elements of order $ 3$ in our group. By the same token, we can show that the group has at leat $ 8$ elements of order $ 5$, and since $ 24 \plus{} 8 > 30$, we're done.


In convex hexagon $ABCDEF$, triangles $ABC, CDE, EFA$ are similar.
$\angle BAC = \angle DCE = \angle FEA$
$\angle BCA = \angle DEC = \angle FAE$.
For what condition of these three triangle,
that triangle $ACE$ is equilateral if and only if triangle $BDF$ is equilateral?

Can anybody find the condition?  :maybe:


[quote="Yankzgodzilla55"]Hey. I need some help with two problems:

1. (square root of 5x) + (square root of x+4)=8

-i really got stuck after squaring a zillion times and then foiling, a huge pain

2.

square root of 4x+ square root of 16x+ square root of 64x+3/2 is equal to 2 (square root x) plus 1 (outside the square root)

if anyone wants to post this using latex and i will reply if you are correct, feel free to. thanks a ton!  :D  :D[/quote]
1. $ \sqrt{5x}\plus{}\sqrt{x\plus{}4}\equal{}8$
2. $ \sqrt{4x}\plus{}\sqrt{16x}\plus{}\sqrt{64x\plus{}\frac{3}{2}}\equal{}2\sqrt{x}\plus{}1$

If anyone answers this correctly, I MIGHT give them a free cookie. :]

[hide="#1"]$ \sqrt {5x} \equal{} 8 \minus{} \sqrt {x \plus{} 4}$

$ 5x \equal{} 64 \minus{} 16 \sqrt {x \plus{} 4} \plus{} x \plus{} 4$

$ 68 \minus{} 4x \equal{} 16 \sqrt {x \plus{} 4}$

$ 17 \minus{} x \equal{} 4 \sqrt {x \plus{} 4}$

$ 289 \minus{} 34x \plus{} x^2 \equal{} 16(x \plus{} 4)$

$ x^2 \minus{} 50x \plus{} 225 \equal{} 0$

$ (x \minus{} 45)(x \minus{} 5) \equal{} 0$

$ x \equal{} 45 (extraneous)$ or $ \boxed{x \equal{} 5}$[/hide]

[hide="#2"]$ \sqrt {4x} \plus{} \sqrt {16x} \plus{} \sqrt {64x} \plus{} \frac {3}{2} \equal{} 2 \sqrt {x} \plus{} 1$

$ 2\sqrt {x} \plus{} 4 \sqrt {x} \plus{} 8 \sqrt {x} \plus{} \frac {3}{2} \equal{} 2 \sqrt {x} \plus{} 1$

$ 12 \sqrt {x} \equal{} \minus{} \frac {1}{2}$

$ \sqrt {x} \equal{} \minus{} \frac {1}{24}$

$ \boxed{x \equal{} \frac {1}{576}}$[/hide]

[quote="modularmarc101"][hide="#1"]$ \sqrt {5x} \equal{} 8 \minus{} \sqrt {x \plus{} 4}$

$ 5x \equal{} 64 \minus{} 16 \sqrt {x \plus{} 4} \plus{} x \plus{} 4$

$ 68 \minus{} 4x \equal{} 16 \sqrt {x \plus{} 4}$

$ 17 \minus{} x \equal{} 4 \sqrt {x \plus{} 4}$

$ 289 \minus{} 34x \plus{} x^2 \equal{} 16(x \plus{} 4)$

$ x^2 \minus{} 50x \plus{} 225 \equal{} 0$

$ (x \minus{} 45)(x \minus{} 5) \equal{} 0$

$ \boxed{x \equal{} 45}$ or $ \boxed{x \equal{} 5}$[/hide]

[hide="#2"]$ \sqrt {4x} \plus{} \sqrt {16x} \plus{} \sqrt {64x} \plus{} \frac {3}{2} \equal{} 2 \sqrt {x} \plus{} 1$

$ 2\sqrt {x} \plus{} 4 \sqrt {x} \plus{} 8 \sqrt {x} \plus{} \frac {3}{2} \equal{} 2 \sqrt {x} \plus{} 1$

$ 12 \sqrt {x} \equal{} \minus{} \frac {1}{2}$

$ \sqrt {x} \equal{} \minus{} \frac {1}{24}$

$ \boxed{x \equal{} \frac {1}{576}}$[/hide][/quote]
For #1 don't forget about extraneous solutions  :wink:

Errr... Why is the one I had different from yours?

For one, I had: *4x + *16x + *64x+3/2 = 2*x + 1

For two, I had: *4x + *16x + *64x + *... + *4^n+3/2 = 2*x + 1

Let * equal the radical symbol. And the first radical symbol of number one extends throughout the first half of the equation. All the radical symbols of number two extends throughout the first half of the equations. Sorry if this sounds extremely confusing.

And btw, the cookie is still up for grabs. :3

[quote="Yankzgodzilla55"]number 1 right

but number 2

square root of 4x and everything goes together in one big problem

like the square root of 4x+ square root of 16x

this is all under one big square root thanks![/quote]
So for #2 it should be
$ \sqrt{4x\plus{}\sqrt{16x\plus{}\sqrt{64x\plus{}\frac{3}{2}}}}\equal{}2\sqrt{x}\plus{}1$
Right or no?

In that case, let $ y \equal{} 4x$, then

$ \sqrt {y \plus{} \sqrt {4y \plus{} \sqrt {16y \plus{} \frac {3}{2}}}} \equal{} \sqrt {y} \plus{} 1$

$ y \plus{} \sqrt {4y \plus{} \sqrt {16y \plus{} \frac {3}{2}}} \equal{} y \plus{} 2 \sqrt {y} \plus{} 1$

$ \sqrt {4y \plus{} \sqrt {16y \plus{} \frac {3}{2}}} \equal{} 2\sqrt {y} \plus{} 1$

$ 4y \plus{} \sqrt {16y \plus{} \frac {3}{2}} \equal{} 4y \plus{} 4 \sqrt {y} \plus{} 1$

$ \sqrt {16y \plus{} \frac {3}{2}} \equal{} 4 \sqrt {y} \plus{} 1$

$ 16y \plus{} \frac {3}{2} \equal{} 16y \plus{} 8 \sqrt {y} \plus{} 1$

$ \frac {1}{2} \equal{} 8 \sqrt {y}$

$ y \equal{} \frac {1}{256}$

$ 4x \equal{} \frac {1}{256}$

$ \boxed{x \equal{} \frac {1}{1024}}$

now chang just added another question

2.

can anyone solve that please. thanks. number 1 is already completed. thanks!

And trust me, you're REALLY going to want this cookie. :)

[quote="gaussintraining"]@modularmarc101: I got $ \frac {1}{1024}$. I think your error was at the point where you got to "$ \frac {1}{2} \equal{} 8\sqrt {y}$". Rather than squaring the expression, you square rooted it.[/quote]

I know, I fixed it like 5 minutes after finishing the solution, cuz I hadn't noticed. Look at it. Thanks anyways  :)

[quote="iEatEmoKids"]Here's what I got. For the question. Nikhil stop being a butthead. :]

[img]http://i263.photobucket.com/albums/ii133/Jcwo137/Mathproblem.jpg[/img]

And it's okay to stare in awe at my epically awesome artistic math diagram.[/quote]

Okay, the thread starter, who is my friend, has #1 as a completely separate question. His #2 is the same as my #1, and my #2 is also a completely separate question, but we need to know it. Sorry for the confusion, and thanks in advance.

PS: Cookie.  :D

1. $ \sqrt {5x} \plus{} \sqrt {x \plus{} 4} \equal{} 8$
$ \sqrt {5x} \equal{} 8 \minus{} \sqrt {x \plus{} 4}$

Square both sides;

$ 5x \equal{} 64 \minus{} 16\sqrt {x \plus{} 4} \plus{} x \plus{} 4$
$ 4x \minus{} 68 \equal{} \minus{} 16\sqrt {x \plus{} 4}$
$ 68 \minus{} 4x \equal{} 16\sqrt {x \plus{} 4}$
$ 17 \minus{} x \equal{} 4\sqrt {x \plus{} 4}$

Square both sides again;

$ 289 \minus{} 34x \plus{} x^2 \equal{} 16x \plus{} 64$
$ x^2 \minus{} 50x \plus{} 225 \equal{} 0$
$ (x \minus{} 5)(x \minus{} 45) \equal{} 0$
$ x \equal{} 5$
or
$ x \equal{} 45$

2. $ \sqrt {4x \plus{} \sqrt {16x \plus{} \sqrt {64x \plus{} \frac {3}{2}}}} \equal{} 2\sqrt {x} \plus{} 1$
Square both sides;

$ 4x \plus{} \sqrt {16x \plus{} \sqrt {64x \plus{} \frac {3}{2}}} \equal{} 4x \plus{} 4\sqrt{x} \plus{} 1$
$ \sqrt {16x \plus{} \sqrt {64x \plus{} \frac {3}{2}}} \equal{} 4\sqrt{x} \plus{} 1$

Square both sides;

$ 16x \plus{} \sqrt {64x \plus{} \frac {3}{2}} \equal{} 16x \plus{} 8\sqrt{x} \plus{} 1$
$ \sqrt {64x \plus{} \frac {3}{2}} \equal{} 8\sqrt{x} \plus{} 1$

Square both sides;

$ 64x \plus{} \frac{3}{2} \equal{} 64x \plus{} 16\sqrt{x} \plus{} 1$
$ \frac{3}{2} \equal{} 16\sqrt{x} \plus{} 1$
$ \frac{1}{2} \equal{} 16\sqrt{x}$
$ \frac{1}{32} \equal{} \sqrt{x}$

Square both sides;

$ \frac{1}{1024} \equal{} x$

Yes, that's correct, but on #1, x=45 is an extraneous solution.

[quote="game.guy"]

Square both sides;

$ 64x \plus{} \frac {3}{2} \equal{} 64x \plus{} 16\sqrt {x} \plus{} 1$
$ \frac {3}{2} \equal{} 16\sqrt {x} \plus{} 1$
$ \frac {1}{2} \equal{} 16\sqrt {x}$
$ \frac {1}{32} \equal{} \sqrt {x}$

Square both sides;

$ \frac {1}{1024} \equal{} x$[/quote]
$ 64x \plus{} \frac {3}{2} \equal{} 64x \plus{} 16\sqrt {x} \plus{} 1$

$ \frac {3}{2} \equal{} 16\sqrt {x} \plus{} 1$

$ \frac {1}{2} \equal{} 16\sqrt {x}$

$ \frac {1}{32} \equal{} \sqrt {x}$

Square both sides

=====================================================================

When your fractions are running into each other (as here one of your denominators is touching
one of your numerators, and one of your numerators is touching a denominator), then at least
insert a horizontal space between affected lines, as I have suggested, for readability.


Έστω $ p>=1$ και $ f,g: [0,1]\longrightarrow R$ ολοκληρώσιμες συναρτήσεις . Να αποδείξετε ότι $ (\int^{1}_{0}\left|f+g\right|^{p})^{\frac{1}{p}}\leq(\int^{1}_{0}\left|f\right|^{p})^{\frac{1}{p}}+(\int^{1}_{0}\left|g\right|^{p})^{\frac{1}{p}}$

Ummm, αυτή είναι απλώς η ανισότητα Minkowski για το διάστημα $ [0,1]$ και προκύπτει απο μία εφαρμογή της τριγωνικής ανισότητας και μετά Holder. Aν απλώς ψάχνεις για την απόδειξή της, πιστέυω ότι μέχρι και η Wikipedia θα την έχει :) Αλλιώς, πες μου να γράψω κανονικά τη λύση...

Πάντως, αυτή είναι απίστευτα σημαντική και χρήσιμη ανισότητα για όσους κάνουν ανάλυση αφού είναι απλώς η τριγωνική ανισότητα για την $ L^p$ νόρμα... καλά έκανες και την ανέβασες!

Cheerio,

Durandal 1707

[quote="Γιώργος"]
Εαν γίνεται γράψε κανονικά τη λύση της....
[/quote]

Dum dum dum, sure, η λύση σε spoiler, για να τη δοκιμάσουν και όσοι δεν την ξέρουν:
[hide="Proof"]
Κατ'αρχάς, ας αγνοήσουμε τις προφανείς περιπτώσεις $ p \equal{} 1$ ή $ \int |f \plus{} g|^p \equal{} 0$. Επίσης, για να μη γράφουμε πάρα πολλά, ας θέσουμε $ \|f\|_p \equal{} \left(\int|f|^p\right)^\frac {1}{p}$, όπου $ f$ μία ολοκληρώσιμη συνάρτηση. Ισοδύναμα λοιπόν, θέλουμε να δείξουμε ότι η $ \|\cdot\|_p$ ικανοποιεί την τριγωνική ανισότητα για $ p\geq 1$: $ \|f \plus{} g\|_p \leq \|f\|_p \plus{} \|g\|_p$.

Από την κλασσική τριγωνική ανισότητα έχουμε: $ |f \plus{} g|^p \leq \left(|f| \plus{} |g|\right)|f \plus{} g|^{p \minus{} 1}$.
Έτσι, σε συνδυασμό με την ολοκληρωτική μορφή της Hölder για $ q \equal{} \frac {p}{p \minus{} 1}\implies q(p \minus{} 1) \equal{} p$:
\[ \int|f \plus{} g|^p \leq \int(|f| \plus{} |g|)|f \plus{} g|^{p \minus{} 1}\implies \\
\int|f \plus{} g|^p \leq \|f\|_p \cdot \| |f \plus{} g|^{p \minus{} 1}\|_q \plus{} \|g\|_p \cdot \| |f \plus{} g|^{p \minus{} 1} \|_q \equal{} \\
\equal{} \left(\|f\|_p \plus{} \|g\|_p\right) \left(\int|f \plus{} g|^p\right)^\frac {1}{q} \equal{} \left(\|f\|_p \plus{} \|g\|_p\right) \cdot\frac {\int |f \plus{} g|^p}{\|f \plus{} g\|_p}
\]
απ'όπου και προκύπτει το ζητούμενο.
[/hide]

Cheerio,

Durandal 1707


The first sum telescopes:

$ \sum_{n\equal{}2}^{\infty} \frac{1}{F_{n\minus{}1} F_{n\plus{}1}} \equal{} \sum_{n\equal{}2}^{\infty} \frac{1}{F_{n\plus{}1} \minus{} F_{n\minus{}1}} \left( \frac{1}{F_{n\minus{}1}} \minus{} \frac{1}{F_{n\plus{}1}} \right) \equal{} \sum_{n\equal{}2}^{\infty} \frac{1}{F_n F_{n\minus{}1}} \minus{} \frac{1}{F_{n\plus{}1} F_n} \equal{} \frac{1}{F_1 F_2} \equal{} 1$

as does the second: since each term is $ F_n$ times the terms of the above, that sum is

$ \sum_{n\equal{}2}^{\infty} \frac{1}{F_{n\minus{}1}} \minus{} \frac{1}{F_{n\plus{}1}} \equal{} \frac{1}{F_1} \plus{} \frac{1}{F_2} \equal{} 2$

[quote="tdl"]Prove that: With $ F_n$ is Fibonacci number
$ \sum_{n \equal{} 2}^{\infty}\frac {1}{F_{n \minus{} 1}.F_{n \plus{} 1}} \equal{} 1,\sum_{n \equal{} 2}^{\infty}\frac {F_n}{F_{n \minus{} 1}.F_{n \plus{} 1}} \equal{} 2$[/quote]

[hide]
Note that $ F_{n\plus{}1} \equal{} F_n \plus{} F_{n\minus{}1}$ implies

$ 1 \equal{} \frac{F_{n\plus{}1}}{F_n} \minus{} \frac{F_{n\minus{}1}}{F_n}$,

where upon dividing both sides by $ F_{n\minus{}1}F_{n\plus{}1}$ yields

$ \frac{1}{F_{n\minus{}1}F_{n\plus{}1}} \equal{} \frac{1}{F_n F_{n\minus{}1}} \minus{} \frac{1}{F_n F_{n\plus{}1}}$.

Therefore for any positive integer $ N \ge 2$,

$ \sum_{n\equal{}2}^N \frac{1}{F_{n\minus{}1} F_{n\plus{}1}} \equal{} \frac{1}{F_2 F_1} \minus{} \frac{1}{F_N F_{N\plus{}1}}$,

and taking the limit as $ N \rightarrow \infty$ gives 1, as claimed.

Similarly, we find

$ \frac{F_n}{F_{n\minus{}1} F_{n\plus{}1}} \equal{} \frac{1}{F_{n\minus{}1}} \minus{} \frac{1}{F_{n\plus{}1}}$,

from which we obtain

$ \sum_{n\equal{}2}^N \frac{F_n}{F_{n\minus{}1} F_{n\plus{}1}} \equal{} \frac{1}{F_1} \plus{} \frac{1}{F_2} \minus{} \frac{1}{F_N} \minus{} \frac{1}{F_{N\plus{}1}}$.

Again, taking the limit as $ N \rightarrow \infty$, we find the second sum evaluates to 2.
[/hide]


All competitors please IM me on AIM. My screen name is Mr Chen 484.

Meeting will be at 3 PM Eastern Time

You're not there :maybe: I am waiting from 2:45-3:00 :maybe:

not enough people. I think I might postpone it. Anyone have any ideas?

[quote="anirudh"]Maybe next weekend?

Seriously...two of the questions out of 4 were like waaay past mc and I will say that they're minimum target by all menas[/quote]
Have you ever seen a national target? I mean an official one, not just the first 25 that are distributed, I mean all of the questions. Some of those final questions could be in the last ten of a state sprint easily.

Now that I think about it, and because of requests, it probably should be a Saturday 3-hour session. Anyone have any suggestions? Afternoon, morning, etc?

AFternoon please before 4:00. Maybe 12-3?

that's even worse from my perspective :P

-jorian

We should have a 3 hr session as Now a ranger said, when we do round robin.

Wait, can I be a competitor? I think I IMed you.

[quote="now a ranger"]Okay Here are the competitors so far:

Bpms
Anirudh
SplashD
Jhredsox
Ragnarok23
Eliross2
redcomet
david li[/quote]

jw, but when did i sign up? :|
-jorian

I'd be up for it -- timing is a bit whack as of right now though.

[quote="kyyuanmathcount"]To bpms.

You mean countdown right :P?

Yeah, or moreso, have you tried to compete with the people on TV at nationals (or been there for that matter).  Unless you can easily beat them, you will not have a chance in the increased pressure of nationals.[/quote]
Yes I meant countdown :rlx:
Also, I agree with what you said unless you are like me and do better under a ton of pressure.

Anyone who wants to CD today, I'm online. Please IM me. Within an hour of this post or less.

Is it possible to do something like this by email? Because I have an aim, but... long story short, its not working and its not going to any time soon, and I have no way of getting a new one (really long story :roll: ). But I would really like to do this. :)

[quote="anirudh"]Use AIM Express :lol: It works fine that way. When is this going to be?[/quote]

Don't use AIM Express. I had to use it on the last countdown, and it went so slow. The message gets to you about a second later, and it takes about a second for your messages to show up.

No, see, I had that, but then I got linux wich has gaim. Then my dad disconnected it , but the normal aim doesn't work because it's not supported by my browser. :(


geometry
topology
advanced fields
advanced fields unsolved

Let $ X$ be a topological space and $ Y$ a subset of $ X$. Then $ \dim Y \leq \dim X$ (we use the Krull dimension).
I saw a proof of this statement:
Let $ Y_1 \subsetneq Y_2$ be two irreducible closed subsets of $ Y$. Let $ X_1,X_2$ be their respective closures in $ X$. Then $ X_i$ are irreducible and $ X_1 \subsetneq X_2$. It follows that $ \dim Y \leq \dim X$.

See (Algebraic Geometry and Arithmetic Curves By Qing Liu, Reinie Erné, p 68)

Now I have questions about the proof. My main question is:Why should the closures of $ X_i$ be irreducible?
Why is it enough to prove just for a chain with 2 irreducible closed sets? The dimension can be infinitely large!

To give a proof I started with writing $ Y_i\equal{}Y \cap W_i$ with $ W_i$ a closed set in $ X$ for $ i\equal{}1,2$. You can see that $ W_1 \subsetneq W_2$ but I can not see why $ W_i$ is irreducible!

Claim 1: If $ Y_1 \subset X$ is irreducible, then $ \bar{Y_1}$ is irreducible.
Proof: Suppose not, we can write $ \bar{Y_1} \equal{} Z_1 \cup Z_2$, where $ Z_1,Z_2$ are closed in $ \bar{Y_1}$, and they are not the whole $ \bar{Y_1}$. Then $ Y_1 \equal{} (Y_1 \cap Z_1) \cup (Y_1 \cap Z_2)$. Irreducibility of $ Y_1$ implies that $ Y_1 \cap Z_1 \equal{} Y_1$ or $ Y_1 \cap Z_2 \equal{} Y_1$. Suppose the first case is true (the other is similar), then $ Y_1 \subset Z_1 \Rightarrow \bar{Y_1} \equal{} Z_1$ (because $ Z_1$ is closed in $ \bar{Y_1}$). This shows that $ Z_1 \equal{} \bar{Y_1}$, contradiction.

For your other question, note that whatever chain of irreducible closed sets $ Y_1 \subset Y_2 \subset ...\subset Y_n$ you have in $ Y$, taking closure gives you a corresponding chain $ X_1 \subset X_2 \subset ... \subset X_n$ in $ X$, thus the supremum of length of chains in $ Y$ must be smaller or equal to that of $ X$.

if $ X_i$ is the closure of $ Y_i$ in $ X$, then $ X_i \cap Y$ is the closure of $ Y_i$ in $ Y$, that is $ Y_i$. thus $ Y_i \equal{} Y \cap X_i$ and from this you see that $ X_1 \neq X_2$.


Solve for n:

$ \frac{3}{2}\binom{n}{3} \equal{} \binom{n\plus{}1}{2}$

I expanded into the equation and simplified up to $ \frac{n^2\minus{}3n\plus{}2}{n\plus{}1} \equal{} 0$, but how do I solve for n?

Thanks.

Hmm didn't realize you could subtract, I cancelled out too much.

Thanks.

Anymore combination problems..that are a bit harder?

Thanks a lot

Actually, I guess you could argue that 0 works, too. There are 0 ways to choose 3 objects when there are not objects, and no ways to choose 2 objects out of only one object.

True, so most likely $ 0$ could also be an answer.

Edit: for fryeggs

Here are two problems:

1. Find the value of the sum $ \binom{n}{0}\plus{}2\binom{n}{1}\plus{}2^2\binom{n}{2}\plus{}\cdots \plus{}2^n\binom{n}{n}$

2. Find the value of the sum $ \binom{n}{0}\minus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\cdots \plus{}(\minus{}1)^n\binom{n}{n}$

[hide="Answers"]
The first one is just $ (1\plus{}2)^n\equal{}\binom{n}{0}\plus{}2\binom{n}{1}\plus{}2^2\binom{n}{2}\plus{}\cdots \plus{}2^n\binom{n}{n}\equal{}3^n$

The second one is $ (1\minus{}1)^n\equal{}\binom{n}{0}\minus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\cdots \plus{}(\minus{}1)^n\binom{n}{n}\equal{}0$
[/hide]

Problem:
Find the number of zeros at the end of $ \binom{54}{13}$