A [i]walk[/i] of length $ n$ on the integer lattice $ \mathbb{Z}^2$ is simply a sequence of ordered pairs $ a_1 \equal{} (0,0),a_2,\ldots,a_n$, where $ \|a_{j \plus{} 1} \minus{} a_j\| \equal{} 1$ for $ 1\leq j\leq n \minus{} 1$ (that is, the walk can only move to one of the four adjacent lattice points at each step).  A walk is [i]self-avoiding[/i] if $ a_i\not \equal{} a_j$ for $ 1\leq a_i\not \equal{} a_j\leq n$ (that is, you can't return to lattice points you've already visited).

prove that:

i) there are at least $ C_1\cdot 2^n$ self avoiding walks of length $ n$
ii) there are at most $ C_2\cdot 3^n$ self avoiding walks of length $ n$
iii) there are at least $ C_3\cdot C_4^n$ self avoiding walks of length $ n$ for some $ C_4\in (2,3)$
iv) there are at most $ C_5\cdot C_6^n$ self avoiding walks of length $ n$ for some $ C_6\in (2,3)$

$ C_1,\ldots,C_6$ are constants which you can choose yourself.

as an aside, solving iii) and iv) shows that the [url=http://mathworld.wolfram.com/Self-AvoidingWalkConnectiveConstant.html]connective constant[/url] of the lattice $ \mathbb{Z}^2$ is strictly between $ 2$ and $ 3$.  it's an open problem as to what its exact value is (there's no reason to expect that it's any sort of known constant though)

no interest?

for example, consider the set of walks of length $ n$ that only go up or left.  there are clearly $ 2^n$ of these, all of which are self avoiding, so this proves i) with $ C_1\equal{}1$.

i'm going to bump this thread one more time in case there is any interest  :)

Part (ii):  The first move has 4 choices.  After that, since you can't go back to the place you just came from, each move has at most 3 choices.  So the number of walks is at most $ 4 \cdot 3^{n \minus{} 1}$.

[b]Part (iv): [/b]  Let $ T(n)$ be the number of self-avoiding walks of length $ n$.  My previous argument showed that $ T(n \plus{} 1) \le 3 T(n)$ for all $ n \ge 1$.  Iterating, we get $ T(n \plus{} 3) \le 27 T(n)$.  But of those 27 suffix walks of length 3, two will crash us into the pre-suffix.  (It's the same way that 3 left turns or 3 right turns will bring you back when driving.)  So we get $ T(n \plus{} 3) \le 25 T(n)$ for $ n \ge 1$.   That gives us $ T(n) \le c \left(\sqrt [3]{25}\right)^n$.

[b]Part (iii):[/b]  As blahblahblah pointed out, we can restrict to Up (U) and Right (R) moves and get $ 2^n$ walks.  For example, when $ n \equal{} 5$, we get 32 walks.  In addition, consider the walk RULUR (where L means left).  That goes from say the origin to (1, 2).  Furthermore, it stays within the rectangle bounded by those two points.  So for 5 moves, we get at least 33 walks that are effectively up-right.  So the number of self-avoiding walks is at least $ 33^{n/5}$ when $ n$ is a multiple of 5.  Even if $ n$ is not multiple of 5, we can finish with say a handful of moves to the Right.

ravi's solutions to ii) and iv) are identical to mine, and my solution to iii) is similar.

consider the collection of self avoiding random walks of length $ n$ which only move up, left, or right.  let $ x_n$ be the proportion of them which end with an up step.  then $ T(n\plus{}1) \equal{} 3x_nT(n)\plus{}2(1\minus{}x_n)T_n \equal{} x_nT_n\plus{}2T_n$

it's clear that $ x_n\geq 1/3$ uniformly in $ n$ (you can always move up to finish a walk, but you can't always move left or right -- if this isn't clear, you can get it by a simple induction) and hence for all $ n$, $ T(n\plus{}1)\geq 7/3 T_n$, which shows that $ T(n)\geq c(7/3)^N$.

That's a nice solution.  You are basically counting the number of words of length $ n$ from the alphabet ULR in which no L and R are adjacent.  We can get an exact recurrence for the number of such words, which leads to a closed-form expression.  If my calculation is correct, the number of such words is on the order of $ (\sqrt {2} \plus{} 1)^n$.


Calculate the mean A and the median B of this set of fifteen automobile speeds collected by a police officer:	72;  80;  62;  71;  71;  64;  65;  73;  66;  78;  81;  66;  64;  77;  66  .
What does  A – B  equal?

A.-2.6	B.-0.6	 C.-4.4	D.1.4	E.4.4

[hide]A.
Speeds collected: 72,80,62,71,71,64,65,73,66,78,81,66,64,77,66.
Add them together, then divide by 15, the number of readings.
 1056/15= 70.4mph A = 70.4

B.
To find the median, we look at the middle of the road....
Actually we order them in numerical order.
72,80,62,71,71,64,65,73,66,78,81,66,64,77,66

62,64,64,65,66,66,66,71,71,72,73,77,78,80,81
and the find the one in the middle.  Therefore B=66.
A(70.4)-B(66)=4.4

So our answer is E) 4.4.[/hide]


function
floor function
superior algebra
superior algebra unsolved

For a group $\left( G, \star \right)$ and $A, B$ two non-void subsets of $G$, we define $A \star B = \left\{ a \star b : a \in A \ \text{and}\ b \in B \right\}$.

(a) Prove that if $n \in \mathbb N, \, n \geq 3$, then the group $\left( \mathbb Z \slash n \mathbb Z,+\right)$ can be writen as $\mathbb Z \slash n \mathbb Z = A+B$, where $A, B$ are two non-void subsets of $\mathbb Z \slash n \mathbb Z$ and $A \neq \mathbb Z \slash n \mathbb Z, \, B \neq \mathbb Z \slash n \mathbb Z, \, \left| A \cap B \right| = 1$.

(b) If $\left( G, \star \right)$ is a finite group, $A, B$ are two subsets of $G$ and $a \in G \setminus \left( A \star B \right)$, then prove that function $f : A \to G \setminus B$ given by $f(x) = x^{-1}\star a$ is well-defined and injective. Deduce that if $|A|+|B| > |G|$, then $G = A \star B$.

[hide="Question."]Does the last result have a name?[/hide]

(a) $A = \{z+n\mathbb{Z}: z \text{~even}\}$, $B = \{z+n\mathbb{Z}: z \text{~odd}\}$.
(b) $x^{-1}a \in B$ would imply $a \in A \star B$, so that $f$ is well-defined. it's obvius that $f$ is injective; especially $|A| \leq |G \backslash B|$, i.e. $|A|+|B| \leq |G|$, and the claim follows.

@-oo-
If n=3, the set A is the whole group... (and B too)

@perfect_radio:
$A: =\{z+n\mathbb{Z}| gcd(z,n)=1\}\cup\{n\mathbb{Z}\}$ and
$B: =\{z+n\mathbb{Z}| gcd(z,n)\neq 1\}$

ok, I meant the following:

$A = \{\overline{0},\overline{2},..., \overline{2 \left\lfloor n/2 \right\rfloor}\}$
$B = \{\overline{1},\overline{3},...,\overline{1+2 \left\lfloor (n-1)/2 \right\rfloor}\}$


We have a trifluoro-ceatic acid solution of molar concentration $0,01 \frac{mol}{dm^{3}}$ and a ceatic acid solution of molar concentration $0,01 \frac{mol}{dm^{3}}$.
a, give the pH of the solutions.
b, In what ratio should we mix the two solutions if we want the pH    of the mixed solution to be the arithmetical middle of the original pH of the solutions?

I assume that's meant to be triflouroacetic acid and acetic acid.

We need to know the pKa for each of these acids - do you have that available?

I do suspect that triflouroacetic acid is a strong acid or a "nearly-strong" acid with a very low pKa. In that case, the pH of the 0.01M solution should be close to 2.

The pKa for acetic acid : 4,75
yes triflouroacetic acid can be considered a strong acid, so for this concentration pH=2


A cruise boat travels 84 miles downstream in 3 hours and returns to its starting point upstream in 12 hours. Find the speed of the stream

[hide]Let the speed of the boat be $r$ and the speed of the stream be $s.$

We know $3(r+s)=84,$ or $r+s=28,$ and $12(r-s)=84,$ or $r-s=7.$ Solving, we find $r=\frac{35}2$ and $s=\boxed{\frac{21}2}.$[/hide]


[quote="diane"]we can use the identity:

 a^3+b^3+c^3-3abc=1/2(a+b+c)[(a-b)²+(b-c)²+(c-a)²][/quote]
Can you solve it?  :oops:

[quote="hunter_cvp_ic"]Let a,b,c be real number such :$a^{3}+b^{3}+c^{3}=0$.Prove that $[ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}](a^{4}+b^{4}+c^{4}) \geq (a^{2}+b^{2}+c^{2})^{3}.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99973

sorry i dont solve it ,it is just an idea


Euler
number theory unsolved
number theory

find all rational x,y such that:
$3x^{2}+1=4y^{3}$

@ L. Euler : How can you deduce that if $(1+a)^{3}+(1-a)^{3}= (2c)^{3}$, then $|a| = 1$ ?

@ Rust : Yup, because $a,b,c,d$ are integers.. $x$ and $y$ are rationals..

[quote="RDeepMath91"]@ L. Euler : How can you deduce that if $(1+a)^{3}+(1-a)^{3}= (2c)^{3}$, then $|a| = 1$ ?

[/quote]

Fermat's Last Theorem  :D
it is useful in olympiad problems :D

Yup, I know... But as far as I know, the Fermat's Last Theorem holds only for positive integers, not for all nonzero integers...  :maybe:

Hm.. Do u mean like this : $(p$ is an odd prime)
If $a,b,c > 0$, then $a^{p}+b^{p}=c^{p}$ doesn't have integer solutions..
If $a > 0 , b < 0 , c > 0$, then $(-b)^{p}+c^{p}=a^{p}$ doesn't have integer solutions..
If $a > 0, b < 0, c < 0$, then $a^{p}+(-c)^{p}= (-b)^{p}$ does'nt have integer solutions..
If $a,b,c < 0$, then $(-a)^{p}+(-b)^{p}= (-c)^{p}$ doesn't have integer solutions..

So, we can extend the famous theorem to nonnegative integers..  :)
Thx a lot, L. Euler..
What about the second and third case ?  :|

second case there is mistake
$lbl=3$:
-)$d=1$ gives $a^{2}+3=12c^{3}$

I prove all case by letting $\frac{q}{p}$.
and we can extend there is no rational solution $x^{n}+y^{n}=z^{n}$
except 0

Thx for the correction..  :)
And thus, we can use the same trick to do the case $a^{3}+3 = 12c^{3}\Rightarrow (3+a)^{3}+(3-a)^{3}= (6c)^{3}\Rightarrow |a| = 3, c = 1$..
I think we can do the rest by the same way.. Anway, thx a lot for giving me a new trick..  :wink:


modular arithmetic
number theory proposed
number theory

Let $ p$ be a prime $ k$ a positive integer and $ r$ an integer. Prove that for infinitely many $ n$ we have $ \binom{2n}{n}=r\pmod (p^{k})$.

[quote="harazi"]Let $ p$ be a prime $ k$ a [i]positive[/i] integer and $ r$ an integer. Prove that for infinitely many $ n$ we have $ \binom{2n}{n}= r\bmod p^{k}$.[/quote]
Something must be wrong. Indeed, if $ n \in \mathbb{N}$ and $ v_{2}(\cdot)$ is a dyadic valuation, then
\[ v_{2}(C_{2n}^{n}) = v_{2}(2^{n}\cdot n!)-2v_{2}(n!) = n-v_{2}(n!) > n-n\cdot \sum_{k=1}^\infty \frac{1}{2^{k}}= 0. \]
Hence $ v_{2}(C_{2n}^{n}) \ge 1$, for any $ n \in \mathbb{N}$. Consequently the modular equation $ C_{2n}^{n}\equiv 0 \bmod 2^{k}$ is impossible, whenever $ n, k \in \mathbb{N}$. Here as usual, $ \mathbb{N}= \{1, 2, \ldots\}$ and $ C_{2n}^{n}= \binom{2n}{n}$, for any $ n \in \mathbb{N}$. Maybe do we need to suppose $ p \ge 3$? :maybe:

Oh, sorry, I forgot to add that $ p$ is odd.  :D

Well!It looks like a problem posted by me on this forum for a long time ago,which states that there are no a natural number $ p$ satisfying that our series is periodic in modulor $ p$ directly except $ 1$ and $ 2$.
  Obviously yours is different and harder a lot!

Let $ p^l>n$. Consider $ (p^l+n)!=(p^l+n)(p^l+n-1)...(p^l+1)(p^l)!$, therefore
${ \binom{2p^ll+2n}{p^l+n}=\frac{(2p^l+2n)...(2p^l+1)}{(p^l+n)....(p^l+1)}\binom{2p^l}{p^l}=2\binom{2n}{n}(1+O(p^l))}.$ Generally
$ \binom{2mp^l+2n}{mp^l+n}=\binom{2m}{m}\binom{2n}{n}(1+O(p^l)).$
Multiplicative group $ G\mod p^k$ is cyclic for odd prime. Therefore sufficiently if we find n for $ g$, suth that $ g=\binom{2n}{n}\mod p^k$.
Let g is odd ( else we can take ${ p^k-g}$). Let $ g=2n_0+1$,
$ \binom{2n_0+2}{n_0+1}=2(2n_0+1)\binom{2n_0}{n_0}.$
Let $ l\ge k, p^l>2n_0$, then work \[ n=n_0+1+n_0\sum_{i=1}^{p^{k-1}(p-1)-1}p^{il} +p^{p^k(p-1)}\sum_{i=1}^{p^{k-1}(p-1)p^l,}\]
i.e. $ g=\binom{2n}{n} \mod p^k$.


geometry
incenter
circumcircle
trigonometry
perpendicular bisector
geometry proposed

Given a triangle $ ABC$. Let $ D,E$ be the intersection point of bisector $ \hat{ABC}$ with $ AC$ and the intersection point of bisector $ \hat{ACB}$ with $ AB$ respectively. Construct an equilateral triangle $ DEF$ another side to $ \hat{A}$. Prove that $ F$ lie on $ BC$ if and only if $ \hat{A}\equal{}60^o$

[color=darkred]Actually $ F$ lies on $ BC$ in 2 cases:
1)$ \angle{A}=60^\circ$
2)$ \angle{B}=120^\circ$ or $ \angle{C}=120^\circ$[/color]

[size=150][b]I.[/b][/size]

Proof:
[color=darkblue]
1) $ \angle{A}=60^\circ$. Let $ I$ be incenter, $ M$ midpoint of $ ED$ and let perpendicular bisector of the segment $ DE$ intersect $ BC$ at $ F$. We'll prove that $ \triangle{EDF}$ is equilateral.
Quadrilateral $ ADIE$ is cyclic, so $ \angle{EDI}=\angle{DEI}=30^\circ$ $ \Rightarrow$ $ EI=ID$ and $ MF$ passes through $ I$.
$ \angle{MIE}=60^\circ$, $ \angle{EIB}=60^\circ$ $ \Rightarrow$ $ \angle{BIE}=60^\circ=\angle{EIB}$, so quadrilateral $ BEIF$ is kite-shaped and $ EI=IF=ID$ $ \Rightarrow$ $ I$ is circumcenter of the triangle{EDF}, then $ \angle{EDF}=\frac{1}{2}\angle{EIF}=60^\circ$ $ \Rightarrow$ $ \triangle{EDF}$ equilateral.

2) $ \angle{B}=120^\circ$. Let circumcircle of the triangle BED intersect $ BC$ at $ F$. Then $ \angle{EFD}=\angle{BED}=60^circ$ and $ \angle{FED}=\angle{FBD}=60^\circ$ $ \Rightarrow$ $ \triangle{EDF}$ equilateral.[/color]

[size=150][b]II.[/b][/size] [color=green]$ F$ lies on $ BC$. We have to prove that $ \angle{A}=60^\circ$ or $ \angle{B}=120^\circ$.

 By sine law in $ \triangle{EFC}$ and $ \triangle{EDC}$ we have:

$ \|\begin{array}{cc} \frac{EF}{\sin{\angle{ECF}}}=\frac{EC}{\sin{\angle{EFC}}} \\ \\ \frac{ED}{\sin{\angle{ECD}}}=\frac{EC}{\sin{\angle{EDC}}}\end{array} \Rightarrow \sin{\angle{EDC}}=\sin{\angle{EFC}}$.

By analogy $ \sin{\angle{BED}}=\sin{\angle{BFD}}$.

So we have 3 cases:

1)  $ \angle{BED}=\angle{BFD}$ and $ \angle{EDC}=\angle{EFC}$
Then $ \angle{BEF}=\angle{BFE}=90^\circ-\frac{1}{2}\angle{B}$ and $ \angle{CFD}=\angle{CDF}=90^\circ-\frac{1}{2}\angle{C}$. Because $ \angle{BFE}+\angle{CFD}=120^\circ$ we have $ 180^\circ-\frac{1}{2}(\angle{C}+\angle{B})=120^\circ \Leftrightarrow \angle{A}=60^\circ$.

2) $ \angle{BED}+\angle{BFD}=\pi$ and $ \angle{EDC}+\angle{EFC}=\pi$.
Then quadrilaterals $ BEDF$ and $ CDEF$ are cyclic, so $ \angle{B}=180^\circ-\angle{EDF}=120^\circ$ and $ \angle{C}=120^\circ$, impossible.

3) $ \angle{BED}+\angle{BFD}=\pi$ and $ \angle{EDC}=\angle{EFC}$ (or $ \angle{EDC}+\angle{EFC}=\pi$ and $ \angle{BED}=\angle{BFD}$)
Then triangle{CFD} is isosceles and quadrilateral $ BEDF$ is cyclic $ \Rightarrow$ $ \angle{B}=120^\circ$.[/color]


Let $ (x_n)$ be a bounded sequence for each $ n \in N$ let $ s_n: \equal{}$sup{$ x_k: k$[u]>[/u]$ n$} and $ S\equal{}inf${$ s_n$}.  Show that there exists a sub sequence of $ (x_n)$ that converges to S.

Ok so clearly $ s_1$[u]>[/u]$ s_2$[u]>[/u]....[u]>[/u]$ s_n$.

Since it is decreasing and bounded then $ s_n$ converges to $ S$.  Furthermore there exists a subsequence of $ s_n$ that is convergent that much also converge to S.  My only problem is I don't know how to relate this back to $ (x_n)$

$ S\equal{} \inf s_n \to \forall k, \exists n_k, s_{n_k} \minus{}S< \frac{1}{2k}$

$ s_{n_k} \equal{} \sup_{j \ge n_k} x_j$

so for a fixed n, k,

$ \exists m_k, s_{n_k}\minus{}x_{m_k}< \frac{1}{2k}$

$ \to |x_{m_k}\minus{}S|<\frac{1}{k}$

then $ x_{m_k} \to S$ as $ k \to \infty$


integration
logarithms
real analysis
real analysis unsolved

Note that $\arctan(ax)-\arctan(bx)=\int_{bx}^{ax}\frac{dt}{1+t^{2}}$

Make the substitution $t=xu$ and this becomes

$\int_{b}^{a}\frac{x\,du}{1+x^{2}u^{2}}$

So our original problem becomes

$\int_{0}^{\infty}\int_{b}^{a}\frac{1}{1+x^{2}u^{2}}\,du\,dx$

$=\int_{b}^{a}\int_{0}^{\infty}\frac{1}{1+x^{2}u^{2}}\,dx\,du$

$=\int_{b}^{a}\frac{\pi}{2u}\,du= \frac{\pi}2\ln\left(\frac{a}{b}\right).$

This can be, and has been, generalized. There are related topics to be found.

[quote="Kent Merryfield"]This can be, and has been, generalized. There are related topics to be found.[/quote]

Yep, look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103776&start=20]here[/url].

I'm looking for an interpretation of the method of this problem in terms of nonnegative random variables. Haven't yet found what I'm looking for.


How about a chess tournament?

post your moves in this thread. NO EDITING!!!!!!! NO FRITZ, RBYKA, CHESSMASTER, etc. AS A HELPER!! YOU MAY CONSULT CHESS BOOKS, BUT NOT CHESS SOFTWARE (this is on the honor system).

signups for chess tourney starting now.

anyone can join.

first, if ur going to join, tell me ur USCF rating, or estimated rating (plz be honest), and tell me if you want a quad paring, or swiss (my style, first seed plays middle seed, etc.)

signups are here, PLZ JOIN!!!!!!! (limit 12) (use this format- (name)|(rating)|(quad or swiss) when signing up plz!)

o, i might not keep up with this, so if someone wants to help, that would be greatly appreciated. TIme control= 3 days per move.
1. Joe10112|1327|quad

[quote]How about a chess tournament?
[/quote]

Awesome!

[quote]post your moves in this thread. NO EDITING!!!!!!! NO FRITZ, RBYKA, CHESSMASTER, etc. AS A HELPER!! YOU MAY CONSULT CHESS BOOKS, BUT NOT CHESS SOFTWARE (this is on the honor system).[/quote]

Good rule.

[quote]first, if ur going to join, tell me ur USCF rating, or estimated rating (plz be honest)[/quote]

My USCF rating is currently 1945 (although I'm pwning a tounrament I'm in currently)

[quote]and tell me if you want a quad paring, or swiss (my style, first seed plays middle seed, etc.)
[/quote]

idc.

[quote]1. Joe10112|1327|quad[/quote]

1. Joe10112|1327|quad
2. BOGTRO|1945|idc

WOW Bogtro, u are gonna pwn my face at chess =).

how old are you?

im 11.

my rating sucks. (my friends say i shood be 1400+) :)

I'm 12. When I was 11 I was around 1850.

Eh i might as well register. USCF rating is 2100, give or take 50 points.

I don't care about the pairings. I think that the time control should be 3 days/move, but allow only 30 minute's thought at most. But whatever.

I might forget about this, so.....

hmmmmm

sorry guys, but there are too less people.

i guess i will abandone this.  :(

if you still want to play, say so now.......

I'll join...but correspondence on AoPS seems too hard and time consuming.

We can have a 5 0 tourn or some other tourn on some chess website, preferably on ICC (i like icc).

It's so weird.  sometimes i preform 2000+, sometimes i preform like 1300 :(  (I'm 1650 uscf)

How about just start a clean game without like a tourney or something. Like when someone post a new chess board, anyone can move the next move by getting the picture and just photoshop it.

I use chess.com, which is great for hosting correspondence tourneys btw... make those games unrated and we'll have a go.

Unrated USCF, but skill is around 1200.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $ A$ be an integral domain with ideals $ I_i,J_i$ for $ 1 \leq 1 \leq r$. Then $ \bigoplus_{i \equal{} 1}^r I_i \cong \bigoplus_{i \equal{} 1}^r J_i$ implies that $ I_1 \ldots I_r \cong J_1 \ldots J_r$.

[hide="Solution:"]Eisenbud, Commutative Algegra, Exc. 19.4 (incl. solution in Appendix).[/hide]


In a circle of radius 42, 2 chords of length 78 intersect at a point whose distance from the center is 18.  the two chords divide the interior of the circle into four regions.  two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form m:pi:-n:sqrt: d, where m,n, and d are positive integers and d is not divisible by the square of any prime number.  find m+n+d.

aime 1995 #14

the problem is entirely doable, i just couldn't figure out what they were looking for.

i thought it sounded like they were looking for the area of DPC or APB, but it turns out they were looking for the area APD or BPC

can someone tell me how to interpret this problem?

Let the center of the circle be O. Pythagorean theorem gives us the distance between midpt. of BD M and O, which is [tex]9\sqrt3[/tex].And PM=9.
ALso DP+PM=39. DP=30. Note that angleAPD=angleDOA=[tex]\theta[/tex].  So DPOA is cyclic quad. And OPM=60. So angleOAD=angleADO=60=angleDOA=[tex]\theta[/tex]=60.DP=30,AP=48.  Everything else should be easy.

yea, the problem wasn't too difficult.  i didn't have much trouble doing once i figured out what they were looking for.  problem is, i didn't understand what they were looking for until i searched up the problem on kalva, which gave a much clearer interpretation.

alright maybe i should bold this:

[b] [u]how does the wording of the problem signify that they are asking me to find the area of APB or BPC?[/u]  it sounded like they were asking for the areas of one of the two sectors which were not equal in area.[/b][/b]

[quote="Nosepuppy"]... the two chords divide the interior of the circle into four regions.  two of these regions are bordered by segments of [b]unequal[/b] lengths, ...[/quote]

Since the two chords AC and BD have the same lengths, it can be shown that  :tri: APD :cong:  :tri: BPC. So PD=PC and PA=PB. Both regions DPC and APB are bordered by segments of [b]equal[/b] lengths. So these are not the regions you are looking for.

ah.   i see where i misinterpreted it.  thanks


Sorry, but I don't remember if it has been already posted.

For $ a,b,c$ positive reals such that $ \sum ab \equal{}1$

$ 64(\sum a)^{4}\geq 243(a\plus{}b)^{2}(b\plus{}c)^{2}(c\plus{}a)^{2}$

Let $ x\equal{}a\plus{}b,y\equal{}a\plus{}c,z\equal{}b\plus{}c$ then your inequality is $ \Longleftrightarrow$:
$ (\sum x)^{4}\geq 243(xyz)^{2}$.Now using 4th consequence in   [url=http://www.mathlinks.ro/viewtopic.php?t=165132]ABC theorem[/url]
We must consider only one cas $ x\equal{}y\Longleftrightarrow b\equal{}c$.But then inequality will be just a polynomial(use that $ 2ab\plus{}b^{2}\equal{}1$).Now i hope you can finish it.


trigonometry quadratics function algebra

1. If cos2x-2cosx=2, find cos2x+2cosx.

2. If 3sinA+4cosB=(sqrt5), what is the max value of 4sinA+3cosB?

3. For 0<x<180, find the sum of the solutions to tan^2 x - 2009tanx +1 =0.

hello, to 3) solving the quadratic equation we get $ \tan(x)\equal{}\frac{2009}{2}\minus{}\frac{3}{2}\sqrt{448453}$ or $ \tan(x)\equal{}\frac{2009}{2}\plus{}\frac{3}{2}\sqrt{448453}$ so we get
$ x_1\equal{}\arctan\left(\frac{2}{2009\plus{}3\sqrt{448453}}\right)$
$ x_2\equal{}\arctan\left(\frac{1}{2}(2009\plus{}3\sqrt{448453}\right)$
Sonnhard.

For #1, we have $ 2\cos^2 x \minus{} 2\cos x \minus{} 3 \equal{} 0 \implies \cos x \equal{} (1 \pm \sqrt7)/2$.
So $ \cos 2x \plus{} 2\cos x \equal{} 2\cos^2 x \plus{} 2\cos x \minus{} 1 \equal{} 4\pm \sqrt7 \plus{} 1\pm \sqrt7 \minus{}1\equal{} 4 \pm 2\sqrt7$.
I hope there's a faster solution.

hello, let $ f(A,B)\equal{}4\sin(A)\plus{}3\cos(B)$, then we have from the condition $ \cos(B)\equal{}\frac{\sqrt{5}}{4}\minus{}\frac{3}{4}\sin(A)$, plugging this in our function we get
$ f(A,\frac{\sqrt{5}}{4}\minus{}\frac{3}{4}\sin(A))\equal{}\frac{7}{4}\sin(A)\plus{}\frac{3}{4}\sqrt{5}$ and because $ |\sin(A)|\le1$ we have $ |f(A,\frac{\sqrt{5}}{4}\minus{}\frac{3}{4}\sin(A))|\le \frac{7}{4}\plus{}\frac{3}{4}\sqrt{5}$ . This will be obtained for example for $ A\equal{}\frac{\pi}{2}$.
Sonnhard.


LaTeX
inequalities unsolved
inequalities

Please use LaTeX!! (for some LaTeX help, you can visit [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]this site[/url])

You problem is probably this:

[color=blue](1)Given arbitrary $x,y,z>0$. Find minimum value of the expression:[/color]
$\frac{x}{y^2}+z^2 +\frac{y}{z^2}+x^2+\frac{z}{y^2}+x^2$

maybe you meant this one:

[color=blue](2)Given arbitrary $x,y,z>0$. Find minimum value of the expression:[/color]
$\frac{x}{y^2+z^2} +\frac{y}{z^2+x^2}+\frac{z}{y^2+x^2}$

My problem is 2 and i think it has a minimum

[quote="jin"]I think the second one doesn't have a minimum.[/quote]
Yes, I agree.

hi

well i dont think so

put \[ x\to 0, y=z=t\to  infinity \]

then value of expression becomes 0

there must be some other conditions like i think many of u might be knowing the q

\[ x^2 +y^2 + z^2 = 1 \]
then prove that minimum of the 2nd is \[ \frac{3\sqrt 3}{2} \]

if \[ x^2 + y^2 +z^2 = k \]
then similarly the min is \[ \frac{3\sqrt 3}{2\sqrt k} \] :)


A number which when divided by $ 10$ leaves a remainder of $ 9$, when divided by $ 9$ leaves a remainder of $ 8$, by $ 8$ leaves a remainder of $ 7$, etc., down to where, when divided by $ 2$, it leaves a remainder of $ 1$, is:

$ \textbf{(A)}\ 59 \qquad\textbf{(B)}\ 419 \qquad\textbf{(C)}\ 1259 \qquad\textbf{(D)}\ 2519 \qquad\textbf{(E)}\ \text{none of these answers}$

[quote="Brut3Forc3"]A number which when divided by $ 10$ leaves a remainder of $ 9$, when divided by $ 9$ leaves a remainder of $ 8$, by $ 8$ leaves a remainder of $ 7$, etc., down to where, when divided by $ 2$, it leaves a remainder of $ 1$, is:
$ \text{(A)}\, 59 \qquad\text{(B)}\, 419 \qquad\text{(C)}\, 1259 \qquad\text{(D)}\, 2519 \qquad\text{(E)}\, \text{none of these answers}$[/quote]That's D

So all the numbers from 1 to 10 divide n+1. So n+1=lcm(1,2,3,4,5,6,7,8,9,10). So n+1=2520. So n=2519.
EDIT: Yes, lcm

n is just 1 less than the least common multiple of the numbers from 1 to 10.
So we have $ n \equal{} 2^3 \cdot 3^2 \cdot 5 \cdot 7 \minus{} 1 \equal{} 2519$.

dgreenb801, right idea but I think you mean lcm not gcd.  anyway all numbers of the form $ 2520k\minus{}1$ fit the problem.  In any case, you could also just try each answer to see if it has the given properties.


does this series converges or diverges?
$\sum_{n=1}^{\infty}\ln(\frac{n^{2}+1}{n^{2}})$

courtigrad, you must be more specific than that.
$1/1+1/2+1/3+...=\infty$ even though the limit of $1/n$ where the limit of n taken to infinty is 0.

[quote="courtrigrad"]its divergent by the comparison test, because $\ln (\frac{n^{2}+1}{n})> \ln(n)$ which is a divergent series[/quote]It's over $n^{2}$, not just n.


6.  You will do two experiments with a rubber band to analyze the laws of thermodynamics.
Experiment 1.  You will need a wide rubber band and a stiff upper lip.  (Your lip is a good detector of small temperature changes.)  Hold the rubber band against your lip and quickly stretch it to its limit.  Wear your safety glasses just in case the rubber band breaks.
a.  What did you feel?  Was the process of stretching the rubber band exothermic or endothermic?  Give reasons.  What is the sign of q for this process?

is heat produced or supplied, i will know the sign for q. i just need an explanation why it would be exothermic or endothermic

When you stretch a rubber band, heat will be produced (exothermic). This is because you rearrange the molecules of the rubber band when you stretch it; the rearrangement causes molecules to collide, releasing energy in the form of heat. The rubber band will feel cool when you release it because it needs to absorb energy to return the molecules to their original, tighter formation.

We can also explain this in terms of entropy. When you stretch the rubber band, you put its molecules in a more orderly formation; there are less possibilities for molecule conformations in a stretched band. As entropy decreases, heat is released, which is why the rubber band heats up. When you release the band, the molecules will go back to their less ordered state (higher entropy) and absorb heat, making the band feel cooler.

I hope this helps answer your question.

KrazyNerd, are you sure that is true?  I haven't looked online or tried the experiment myself, but I see an exothermic contributor in both direction: friction.  Is this accounted for in the contraction step?  The energy required for squashing the molecules into their original spots should be supplied by elastic potential energy stored in the band, no?

Also, I am not sure that your entropic explanation is completely rigorous--entropy is not always directly connected to heat.  I suppose though that if the stretching is done reversibly, things should work out well...

I actually did this experiment way back in freshman physical science, and the band does get cooler when you let it contract again. I got much of my entropic explanation from a site that explains the experiment in terms of entropy. I guess I did make it sound like entropy is always connected to heat...sorry. :blush: I need to go study that some more...

that definitely explains it thanks man, but that's weird how if you stretch it heat is produced,


Five socks, colored blue, brown, black, red, and purple are in a drawer. In how many different ways can we choose three socks from the drawer if the order of the socks does not matter?

there are 5.
you choose 3.
5 choose 3 is 10.

you got 60, but each of your ways was repeated 6 times in total. each combination is repeated 3*2*1 times, if you think about it. you do
$ \frac{5\cdot 4\cdot 3}{3\cdot 2\cdot 1}$

[quote=chipmunk88]I thought it was 60 because 5 choose 3 is 5*4*3, but the answer is 10 :maybe:  :?:[/quote]

Same but now i get it


Hi everybody,

Umm when i take practice amc 12's, i can finish 1-20 in about 50 minutes, but struggle with the last 5. Would one recommend just to work through previous 21-25's or AIME 1-5's?

All help is appreciated

how much time is allotted on the amc12? is it 90 or 75?

Definitive answer:
(1) On the contest cover, item 7, bold face, 75 Minutes, see pages 201-21 of the Teachers' Manual
http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-AMC_1012A,2009.pdf
(2) Teachers' Manual, page 4, second sentence on the page
(3) Teachers' Manual, page 5 in Section II, item 7
(4) On the AMC 12 webpage, first sentence
http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

Assuming that you get 1-20 right, I'd suggest working through AIME 1-5, as if you can solve those, you should be able to make headway on 21-25.

Really it depends on your goals. If you don't care about USAMO or anything, work on 21-25's on AMC 12.

If you want to get further, focus on AIME.

on this point, i would differ with the rest.

I can safely crank out the first 7-8 AIME problems, but 21-25 on the AMC still trip me up. So I'd say do actually 20-25 to be safe for like 6-7 sets of problems and then move to AIME. If you commit, this won't take long at all. And it's much more efficient than doing problems you know you can get right. The only purpose for taking the whole AMC12 sessions at this point would be to train your speed :\

Solving all of 1-7 on the aime is easier than solving all of 21-25 on amc12.

I have to agree;  I solved 1-5 and 8 on AIME this year, but got like 1/5 on last five of AMC 12.


$ABCD$ is a trapezoid with $AB$ parallel to$CD$.$M$is the midpoint of $AD$, $\angle MCB = 150^{\circ}$, $BC = x$ and $MC = y$. Find area $ABCD$ in terms of $x$ and$y$.

Let $CM$ intersect the extension of $BA$ at $P$. then because $CD\|PA$ and $DM=MA$, we can say that triangles $DMC$ and $MPA$ are equal and $PM=CM=y$. Also $S_{ABCD}=S_{DMC}+S_{CMAB}=S_{MPA}+S_{CMAB}= S_{CPB}= \frac{1}{2}CP.CB.sin150^{\circ}=\frac{1}{2}2yxsin150^{\circ}=xysin150^{\circ}$.
Thus the answer is $S_{ABCD}=xysin150^{\circ}$ :)


advanced fields
advanced fields unsolved

Hello,
I would like to prove that the following are equivalent:
(1)  (Compactness Theorem; Gödel-Mal'cev) Let $ T$ be any set of first-order axioms. If for every finite subset $ T_0$ of $ T$ there is a model of all the axioms in $ T_0$, then there is a single model of all the axioms in $ T$.

(2) If $ T \cup {\psi}$ is a set of first order sentences and $ T \models \psi$, then there is a finite $ T_0 \subseteq T$ such that $ T_0 \models \psi$.

For now, I was able to prove that (2) follows from (1), but don't know how to prove that (2)=>(1).
In the book it is taken from (J. Barwise, Handbook of Mathematical Logic) it says: "let $ \psi$ be some absurd $ \psi$ like $ \exists x (x \neq x)$", but I don't see how we can use it.

Any ideas are welcomed!

[size=75]edit: minor mistypes.[/size]

let $ \psi$ be some false sentence. then $ T$ has a model if and only if $ T \models \psi$. from this (2) => (1) is clear.


Points M and N are given on sides AD and BC of a rhombus ABCD. Line MC meets the segment BD at T and line MN meets the segment BD at U. Line CU intersects the side AB at Q and the line QT intersects the side CD at P.

Show that $\triangle QCP$ and $\triangle MCN$ have equal area.

This is a nice review of the Area Ratios.

$\frac{\triangle{POC}}{\triangle{TQC}}=\frac{PQ}{TQ}\Longrightarrow \triangle{PQC}=\frac{PQ}{TQ}\triangle{TQC}$

Since AD is parallel to BC, so $\frac{PQ}{TQ}=\frac{BD}{BT}$, thus we must have:$\triangle{PQC}=\frac{BD}{BT}\triangle{TQC}$.

Similarly we have:

$\frac{\triangle{TQC}}{\triangle{TUC}}=\frac{QC}{UC}\Longrightarrow \triangle{TQC}=\frac{BD}{UD}\triangle{TUC}$

$\frac{\triangle{TUC}}{\triangle{MUC}}=\frac{TC}{MC}\Longrightarrow \triangle{TUC}= \frac{BT}{BD}\triangle{MUC}$

$\frac{\triangle{MUC}}{\triangle{MNC}}=\frac{MU}{MN}\Longrightarrow \triangle{MUC}= \frac{DU}{DB}\triangle{MNC}$

Pluge all the thing into our original equality, we shall have:

$\triangle{PQC}=\frac{BD}{BT}\cdot \frac{BD}{DU}\cdot \frac{BT}{BD}\cdot \frac{DU}{BD}\cdot \triangle{MNC}=\triangle{MNC}$


Hello people

Please can someone tell me if LaTeX can be used to convert the troff-formatted paper
into a latex-formatted paper (file with latex commands).

thanks

jack

Check this out: [url]http://www.tex.ac.uk/tex-archive/support/troff-to-latex/[/url]

I have no idea if it works or not, as I wouldn't touch troff with a 10-foot pole.  :)

I assume you want troff->LaTeX (as you said in your message) and not LaTeX->troff (as you said in your subject title)

Hey
 thankx for the site, but i have no clue how to convert it..
and i dont think the zipped version of the file works..
i have a pretty small file to convert....can u plzz look into it and lemme know how to convert it?

thankz again

I know nothing about troff, so I can't help you any more than I did.  :(

If it's small, perhaps just rewriting the paper from scratch in LaTeX is the simplest way to go.


Ok, it's 2:14 AM now, and I am going to stay up for another hour.  This is one of the lastest times that I have ever gone to sleep on a school night.  What about you guys?  When do you go to bed?  [b]Please post the time at which you go to sleep and your grade[/b]

Yeah uh bubs I added an option. -.- After 3AM is much too...uh...broad.

i usually go to sleep at around 11:30.. and im a *generally* A student

9th grade, A student, I go to bed 9-10. :D However, if I could get by without that much sleep, I would, but I need it.

11th grade: 12-1:30 AM. The reason I sleep late is because I have a policy of doing math first, homework second. Though doing math didn't help me in recent competitions, I got a really bad score in the most recent math competitio I took.

Any time from 10:00 to 3:30, depending on workload. I'd say I average about 12:30 to 1:30. I can go through a day of school and then sports pretty well on less than 3 hours of sleep. Of course, that usually means I'm dependant of Sweettarts the next day.

i go to bed at 9, but go to sleep at 10 lol :rotfl:

12th grade.

Usually I go to sleep somewhere between 1 and 2 on school nights. On weekends I pretty much just stay up until I fall asleep involuntarily, at which point I go to bed. :lol:

[quote="WindSlicer"]From 11 to 12, it's the latest I can get to bed without being dead the next morning at 5:30 when I wake up. It works out fairly nicely.

EDIT: And I'm an A student and 11th grade[/quote]
Holy cow. You get up at WHAT time?! :what?:

I went to bed at 6:20 (AM) last night/this morning. That's probably later than most of you get up for school (high schoolers).

I sleep at 10 PM, being a cross country athlete, I need the sleep, except I don't fall asleep for a while. So, I'm quite tired during the day.

12th grade, usually 12:30-1:00. If I don't have much homework and other stuffs to do, I go to bed earlier.

Many of my Japanese friends go to bed much later though(around 2:00-3:00 a.m.) :roll:

11th grade.  I average 12-1 am usually, occasionally an extreme like after 4 am or if I'm really tired, I'll sleep early around 9ish.  I think it's mostly due to my inefficiency and tendency to get distracted.  Procrastination is like a disease, but it keeps things interesting.  :P

[quote="mwahahahaha"][quote="236factorial"][quote="Derek"]At 11. Any unfinished work = screw this, I will do it tomorrow.

I'm a junior.[/quote]

Procrastinator  :P[/quote]

there is absolutely NOTHING wrong with procrastination...
it is an art of the most difficult kind - a skill that can only be learned through intense training and grueling practice (a.k.a. doing homework due in 3rd period during 2nd, or doing homework due in 5th period in the 5 minutes between 4th ending and teacher collecting)
it's a really useful thing to know how to do...esp if you forget you had hw, which happens a lot to me since i don't write my hws down.

and you all sleep wayyyyyyyyyy too much (except Miths, who sleeps the correct amt of time)
2-3 hours a night is plenty enough for the body...you just gotta supplement some sugar, caffeine, and perhaps some adrenaline if you have a test or something
i go to sleep at times ranging from 2am-5:30am and wake up at 6am.  Here's a little secret: getting two hours of sleep feels 10x more re-energizing than getting anything in the 4-6 range...i dunno why...but it does

oh and i'm an A student in senior year of high school  :P[/quote]

haha, I so agree.  My favorite is when I decide to just leave off writing a paper for english, and then having to fake sick in the class prior to "go to the bathroom" and type it up in the library.

In my school in certain subjects(Like math ),you get beaten up if you don't do your homework,so out of fear i do it the previous day,i hate doing homework at the lat moment bcos my handwriting is really bad and then it becomes worse,i have to improve upon it.

The time I fall asleep is quite varied (usually from 9:00 to about 2 AM.)  I usually wake up at 5:45, but sometimes, If I need to do homework, I will wake up at atound 3.

Junior, A Student

I'm an 8th grader, and I try to go to bed at 10, but I can't help but waste time until 10:30.

But I get up at 7:15  :o and school starts at 8  :o

I'm a 7th grader and I go to sleep from 10-11 because my parents gave me a lights out time of 10. I wake up 6:30-7:00 and actually get moving around 7:15. The spare time there is usually spent reading.

This is an OLD topic..

Anyway, I'm in 8th grade and I usually go to bed 11PM-2AM.

[quote="diamondcastledad"]children should go to sleep at a decent time, otherwise they become irritable and irresponsible and perform poorly in school[/quote]

Well most children here on AoPS do not just want to perform good in school,they do much more so that's why if they don't sleep less and do their other important extra math stuff,they become even more tensed and irritable :)

I'm in 6th grade and I sleep from 9-11 PM on a school night, wake up at 6:15 5 days a week. On weekends, I'll go to bed 11-12 PM. I'm not the kind of person who likes to sleep too late (past midnight).

11 to 12 though I can do without sleep for a Day or two.

9-11 pm, 6th grade (10yrs)  I once stayed up to 1:47 AM when I was 7.

I go to bed 7-12PM and wake up around 1-5AM.

I am in *** and I go to sleep at 10 PM...it works out fine

did y ou know, when you go to sleep at differnt times every day, your body doesn't know when to go to sleep and gets messed up and tired etc


topology
real analysis
real analysis unsolved

assume that $f: U\subset \mathbb{R}^{n}\rightarrow V\subset \mathbb{R}^{n}$ ($U,V$are open sets)such that $f$ is homorphism and $V$ is bounded .does it follow that $I\rightarrow m^{*}(f(I))$ is absolute continuous?

A homeomorphism can map sets of zero measure into sets of positive measure. For example, $(x_{1},x_{2},\dots,x_{n})\mapsto (x_{1}+h(x_{1}),x_{2},\dots,x_{n})$, where $h\colon\mathbb R\to\mathbb R$ is the Cantor staircase.


arithmetic sequence
algebra unsolved
algebra

Do there exist distinct positive integers $a$, $b$, $c$ such that $a$, $b$, $c$, $-a+b+c$, $a-b+c$, $a+b-c$, $a+b+c$ form an arithmetic progression (in some order).

Do these numbers have to be consecutive elements of this arithmetic sequence? Or it's not important?

I spent so much time on this problem  :(  :blush:

$x = -a+b+c$, $y = a-b+c$, $z=a+b-c$, $t = a+b+c$ $(1)$
Then $\prod^{cyclic} (x-y-z-t) = 0$ as you can check.

If $x,y,z,t$ is an arithmetic progression, say, $A, A+r, A+2r,A+3R$ in some order then
expanding the cyclic $\prod^{cyclic} (x-y-z-t)$ with these values gives :

$16A(A+r)(A+2r)(A+3r) = 0$ i.e. one (and only one) of the term is null.

Looking back at $(1)$, we can get convinced that there are no solutions.

Though it's too much old thread but..... Here is my solution...

          We use the lemma if $x, y, z$ are three distinct reals in A.P. then x±a, y±a, z±a
 are in also A.P. Where $\a in {R}$
          If $a, b, c, b+c-a, c+a-b, a+b-c, a+b+c$ are three reals in A.P. then $a, b, c$ are in A.P. & $b+c-a, c+a-b, a+b-c, a+b+c$ are in also A.P.

In the $2nd$ A.P. series we add $a+b+c$ to each term.
$\implies that\ 2(b+c), 2(c+a), 2(a+b), 2(a+b+c)$ then
Again we add $-2(a+b+c)$ to each term to the last A.P. terms.
Now we get $-2a, -2b, -2c, 0$ are in A.P. Now, simplifying $3b=2a=6c=k$ by using the formula $\frac{a+c}{2}=b$ if $a, b, c$ in A.P.
Putting the value $a=3k, b=2k, c=k$, We have the A.P. series $3k, 2k, k, 0, 2k, 4k, 6k$. Which contradiction as the $k, 0, 2k$ are not in A.P.

VOILÀ.....

Let $a<b<c$ then
 $a+b-c = min (a, b, c, -a+b+c, a-b+c, a+b-c, a+b+c)$ and $a+b+c=max (a, b, c, -a+b+c, a-b+c, a+b-c, a+b+c)$ so
AP has difference $d=\frac{(a+b+c)-(a+b-c)}{6}=\frac{c}{3} \to c=3d$

So $x_1=a+b-c,x_4=a+b$
$a+b-c<a<b<a+b \to a=x_2=a+b-c+d \to b=2d, b=x_3=a+b-c+2d \to a=d$
But then $b=2d=a-b+c \to $ contradiction

I think the correct problem is:\\\\
Do there exist three distinct positive real numbers $a$, $b$, $c$ such that $a$, $b$, $c$, $-a+b+c$, $a-b+c$, $a+b-c$, $a+b+c$ form a $7$-term arithmetic progression in some order.\\\\
Source: HBCSE

there are 7 terms so it doesnt really matter


Outline the steps to synthesize anthracene from benzyl chloride in good yield.

(this is not my own question. I read this in a book and thought i would share with you people. Please hide your answers. To give a hint, dont think too big. Its actually very simple)

[hide="Possible answer"]1) Benzyl chloride + DMSO -----> benzaldeyde

2) Benzyl chloride + Mg/ether, then benzaldeyde -----> phenyl-(o-tolyl)methanol

3) phenyl-(o-tolyl)methanol + 300ºC -----> anthracene.[/hide]

First step can also be achieved by using hexamine(urotropine).

Actually the solution says, prolonged heating with AlCl3 gives you 9,10 dihydro anthracene, oxidation of which gives you anthracene.

Very interesting, but are you sure that gives a good yield?


conics
parabola
algebra unsolved
algebra

find all solutions:

$x^{4}+y^{4}+8xy+4=0$

no, all solutions(including complex ones)

think to get an equation of ( parabola or heperbola ) :wink:


Will the MAA be releasing the general statistical data, perfect scorers' list, etc for the AMC this year?

Is there anything that would lead you to believe otherwise?

If I recall correctly, the statistics were up before AIME I last year, however, they aren't up yet.  Also I don't think perfect scorers on the AMCs have been asked for photographs or the like.

If you have a direct question, you should PM AMCDirector or send an email.


A rectangular field is half as wide as it is long and is completely enclosed by $ x$ yards of fencing. The area in terms of $ x$ is:

$ \textbf{(A)}\ \frac {x^2}{2} \qquad\textbf{(B)}\ 2x^2 \qquad\textbf{(C)}\ \frac {2x^2}{9} \qquad\textbf{(D)}\ \frac {x^2}{18} \qquad\textbf{(E)}\ \frac {x^2}{72}$

[hide]
Let the length be denoted $ l$. We know that the width is $ \frac{l}{2}$, and that the perimeter, which is $ l\plus{}l\plus{}\frac{l}{2}\plus{}\frac{l}{2}$, is equal to $ x$. This simplifies to $ x\equal{}3l$.  The area of the field is $ l \times \frac{l}{2}\equal{}\frac{l^2}{2}$. Because $ l$ is equivalent to $ \frac{x}{3}$, we have: $ \frac{(\frac{x}{3})^2}{2}\equal{}\frac{x^2}{18}\equal{}\boxed{D}$.
[/hide]


When I tried to start a new blog, since I deleted my old one days ago, it wouldn't work. It gave me this message of error:

Select blog: no value available for this field.

Select blog: the value choosen is not available in the list.

Click Here to set additional options for your blog

Click Here to return to the Index

If any of you have suggestions, then gracias!

A subject of this has already been posted. (But in the wrong section)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158633


Euler
logarithms
inequalities
integration
limit
calculus
calculus computations

We know:
$\sum_{1}^{n}\frac{1}{n}=lnn+\gamma+o(1)$ where $\gamma$ called Euler constant and $\gamma=0,577...$ Tell me how determin $\gamma$ ?

[quote="mathmanman"]Let $H_{n}= \sum_{k=1}^{n}\frac{1}{k},$ $u_{n}= H_{n}-\ln n$ and $v_{n}= H_{n}-\ln(n+1).$
Let $\epsilon > 0$ be given. For any positive integer $n,$ we have
$0 \leq \gamma-v_{n}\leq u_{n}-v_{n}= \ln(n+1)-\ln n = \ln(1+\frac{1}{n}).$
So,
$0 \leq \gamma-v_{n}\leq \epsilon \Leftarrow \ln(1+\frac{1}{n}) \leq \epsilon \iff n \geq \frac 1{e^{\epsilon}-1}\Leftarrow n = \left[\frac 1{e^{\epsilon}-1}\right]+1.$
If $n_{0}= \left[\frac 1{e^{\epsilon}-1}\right]+1,$ $v_{n_{0}}$ is an approximate value of $\gamma$ up to $\epsilon.$[/quote]
Yes I Thank you, but
tell me: how do you know that there exist $\gamma$? how do you know this :$0 \leq \gamma-v_{n}$ ? If I don't know this: there exist gamma. therefore I can take for gamma this value $\gamma=0,1$ and your inequality is wrong.

[quote="Extremal"]how do you know that there exist $\gamma$?[/quote]

If $A_{n}= \sum_{k=1}^{n}\frac{1}{k}$ then
$A_{n}-1 \le \int^{n}_{1}\frac{1}{x}\,dx = ln n \le A_{n}-\frac{1}{n}$

Now, substracting $A_{n}$ to each side and then multiplying by -1 we get
$\frac{1}{n}\le A_{n}-ln n \le 1$

Now taking limit when $n\longrightarrow\infty$,

$0 \le \lim_{n\to\infty}(A_{n}-ln n) = \gamma \le 1.$

Carcul

Yep, this uses only my $(u_{n})$ and allows you to conclude that $\gamma \in [0, 1].$

But now, if you use also $(v_{n})$ then you can show a bit more, namely $\gamma \in [0.4, 0.9].$
Indeed, let $n$ be a positive integer. Then you can easily show that $(v_{n})$ is monotone increasing. Besides,
$v_{n}= H_{n}-\ln(n+1) = u_{n}-\ln(1+\frac{1}{n}) = u_{n}+o(1) = \gamma+o(1).$
So $(v_{n})$ is increasing and converges towards $\gamma.$ We deduce that :
\[\forall n \in \mathbb{N}, v_{n}\leq \gamma \leq u_{n}.\]
Especially,
$0.4 = 1.5-1.1 \le \frac{3}{2}-\ln 3 = v_{2}\leq \gamma \leq u_{2}= \frac{3}{2}-\ln 2 \leq 1.5-0.6 = 0.9.$

And now, I think my first post is fully justified. :)


Find all functions: $ f: R \to R$ such that $ f(x\plus{}y)\equal{}x^2.f(\frac{1}{x})\plus{}y^2f(\frac{1}{y})$for all $ x;y \in R^*$

It is an not hard problem from math and young magazine .
After 2 month I will posted my solution .


geometry
trigonometry
real analysis
real analysis theorems

How to find the equation of the curve which is tangent to a family of curves $F(x,y;\phi)=0$? For example
\[\{\begin{array}{l}x=v_{0}t\cos\phi \\ y=v_{0}t\sin\phi-\frac12gt^{2}\end{array},\; o\leq\phi\leq\pi \]

With $\phi$ as the third variable, the equation $F(x,y,\phi)=0$ determines a surface $S$ in $\mathbb R^{3}$. The curve you are looking for is the projection of the part of $S$ where the tangent plane is vertical. Its equation is $\frac{\partial F}{\partial \phi}=0$.

I notice that in your example the curves are not described by an equation $F(x,y,\phi)=0$, but you can eliminate $t$.

[quote="mlok"]The curve you are looking for is the projection of the part of $S$ where the tangent plane is vertical. Its equation is $\frac{\partial F}{\partial \phi}=0$.
[/quote]
May you (or anyone) explain it particularly? :blush:


Three flower beds overlap as shown. Bed A has 500 plants, bed B has
450 plants, and bed C has 350 plants. Beds A and B share 50 plants,
while beds A and C share 100. What is the total number of plants?

[asy]/* AMC8 1999 #9 Problem */
draw((0,0)--(3,0)--(3,1)--(0,1)--cycle);
draw(circle((.15,0), .5));
draw(circle((3,-0.25), .65));
label("C", (.2,-0.25));
label("B", (3.25, -0.25));
label("A", (1.5, 0.45));[/asy]

Use PIE [Principal of Inclusion-Exclusion]: $ 500 \plus{} 450 \plus{} 350 \minus{} 50 \minus{} 100 \implies 1300 \minus{} 150 \implies \boxed{1150}$


I just got this error:

[hide="Error"]
SQL requests not achieved

DEBUG MODE

SQL Error: 1194 Table 'pt' is marked as crashed and should be repaired

SQL Request: SELECT p.*, pt.post_subject, pt.post_sub_title, pt.post_text, pt.bbcode_uid, r.rating FROM phpbb_posts p, phpbb_posts_text pt LEFT JOIN phpbb_post_ratings r ON ( r.post_id = pt.post_id AND r.user_id = 44541) WHERE p.topic_id = 289192 AND pt.post_id = p.post_id ORDER BY p.post_time DESC, p.post_id DESC LIMIT 0, 20

Line : 419
File : class_posts.php
[/hide]

Someone please explain!!

The site is doing some sort of maintance. I myself keep up a site, and when they are working with code, it usually returns some sort of SQL error or something. Looks like they are doing something with the AoPS classes.

Site crashed for a bit. I think we are back to normal now.


Hello everyone.

[b] Question [/b]

Which is much better -- Being a generalist or being a specialist?

A specialist is one who has a specialized (and often profound) knowledge in his/her area of interest (i.e. specialization).

A generalist is one who mirrors the idiom - [b] Jack of all trades and master of none [/b], a person who doesn't have specialized knowledge in any field, but has broad knowledge across many disciplines.

So, I would like to know which is much better -- being a generalist or being a specialist with profound knowledge (in his/her area of specialization)?

Share your views/suggestions/opinions on this particular question.

That highly depends, but modern mathematics requires a lot of specialisation, whereas a good amount of vast knowledge is helpfull, too.

[quote="ZetaX"]That highly depends[/quote]  I think this is the only right answer :).  Some tasks require very specialized knowledge of certain subjects, while other tasks require the ability to link ideas or skills that are not closely related.  And, of course, perspective matters: there are mathematicians who are relative jacks-of-all-trades in mathematics who nonetheless could be viewed as a specialist by anyone who isn't himself a mathematician.

Why are they mutually exclusive?  Haven't you heard of the idiom [b]know something about everything and everything about something[/b]?  :)

Aside from mathematics, same "that highly depends" works for medicine as well. There is no such thing as "better" between osteopathic and allopathic although allopathic, which are "specialists" in the sense that they do not do family practice, is more lucrative and has higher number of years in residency.

It depends. Is your job highly specialized? Then it's better to be a specialist. Is it not? Then it's better to be a generalist.


[b]Source: 2001 State Sprint Round #24[/b]

This is one of the questions that bugged me on the 2001 State Sprints. It's a good problem if you're systematic but a nightmare if you aren't. I'm sure there is a pretty short way to do this one... as with all sprint problems.

[i]The number 4 can be written as a sum of one or more natural numbers in exactly 5 ways: 4, 3+1, 2+1+1, 2+2, and 1+1+1+1; and so it is said to have 5 partitions. What is the number of partitions for the number 7?[/i]

Good Luck...

hmmm.... I got 15 but I don't know that I'm right...... i just listed the combinations and crossed out any that were the same by doing their sums starting at 7 and doing the differnt 6s and 5s etc.. try it and let me know if you get the same result

[hide]AGeneson, I just gave my kids this test last week, and I'm sure that's(15) right.  I did it myself just to be sure :) [/hide]

Is there a faster method of doing that? I did that year and I was only one off...

Ya.. If I were doing this sprint in any state but California :lol:, I would definately use the time spent on this problem to check... I guess I'm just not systematic enough.. anyways the rest of the sprint was pretty easy, so if you did check, then you'd have plenty of time to work on the problem....

AGenson's method would be the preferred method in this case. That's the easiest way to do it, and it has the least chance of skipping a possible combination.

[quote="dengmi"][hide]The answer is 15.
There's no formula I can see for it, but...
1/1 partition
2/2 partition (One more than before)
3/3 partition (One more than before)
4/5 partition (Two more than before)
5/7 partition (Two more than before)
6/11 partitions (Four more than before)
7/15 partitions (Four more than before)
Do I see a pattern?[/hide][/quote]

I tried that and thought I was on to something, untill.....

[hide]
8/22 partitions (7 more than before)
9/30 partitions (8 more than before)

Assuming I didn't miss any
[/hide]

[quote="filletwho"]The partitions of seven come up alot: I would memorize this one.[/quote]
In that case, can someone list the partitions of seven? It would be a good reference for us all... again I'm not all that systematic so I'd have quite a bit of trouble trying to come up with this list... otherwise I'd list it...  ;)

[quote="cj"][quote="Scrambled"]so far, humans have no formula for the amount of partitions a number has[/quote]

There is a formula obtained by Hardy in the early 20th century.
Here are some more: 10/42; 11/56; 12/77; 13/101; 26/2436. ....[/quote]

Will the formula ever be used in Mathcounts? Or are we better off memorizing the number of partitions for basic numbers and the specific partitions of certain numbers such as 7?

--Adi

I just thought over this problem and found a pretty easy way to do this one...
The most important thing to do is to list the numbers in numeric order from greatest to least, or else the potential for careless errors can rise and you will have to waste time canceling.
Anyways, I went from bottom up, starting with the 7 ones.
I then listed 3 twos and a 1, 2 twos and 3 ones, and finally a two and five ones.
The hardest part was the next one, but if you kept to what I was doing, it becomes pretty easy. I kept listing the maximum number of times a number can occur, and then filled in the remainder if possible with ONLY LESSER NUMBERs.I then listed the ways it would work with lesser amounts of each number in the same fashion. So by doing this, I got 331, 322, 3211, 31111 for three. Then I went to four, and did 43, 421, and 4111. Then, I moved on to 5, which gave me 52 and 511. Finally I did the two last ones, 61 and 7 to complete my list. By summing up the combinations starting with each number, I got 1+3+4+3+2+1+1 (starting with 1,2,3,4,5,6, and 7, respectively) to get 15.
I've tried this method with 6 and 8 and its the easiest way once you are familiar with it that I can think of to finish off this problem. I know this seems confusing because it's a really hard method to explain but it is really easy to do....


geometric inequality
trigonometry
geometry
function
area of a triangle
IMO Shortlist

Let $ ABC$ be an acute-angled triangle. Let $ L$ be any line in the plane of the triangle $ ABC$. Denote by $ u$, $ v$, $ w$ the lengths of the perpendiculars to $ L$ from $ A$, $ B$, $ C$ respectively. Prove the inequality $ u^2\cdot\tan A \plus{} v^2\cdot\tan B \plus{} w^2\cdot\tan C\geq 2\cdot S$, where $ S$ is the area of the triangle $ ABC$. Determine the lines $ L$ for which equality holds.

Note that if the line $ L$ lies entirely outside the triangle, we can decrease each of $ u,v,w$ by sliding $ L$, keeping it parallel to its initial position, until it passes through one of the vertices. [b]Therefore we need only consider lines that pass through the interior (including border) of the triangle.[/b]

Of those lines, we can further slide in a parallel fashion and decrease the sum of $ u,v,w$ until the setup is equivalent to:

$ A$ and $ B$ lie on opposite sides of $ L$, $ L$ passes through $ C$.

So relabel diagram if necessary to this.
This is true because $ u\plus{}v$ remains constant as we slide around [i]iff[/i] $ B$ and $ C$ are on opposite sides of $ L$. Meanwhile $ w$ decreases until it is zero, the minimum, which is just when $ L$ is through $ A$.

Back to the inequality. We can show it is equivalent to \[ \frac{u^2}{b\cos A}\plus{}\frac{v^2}{b\cos A}\geq c\]
If $ u\equal{}b\sin( A\plus{}X)$ then $ v\equal{}a\sin (B\minus{}X)$
So we need show
\[ b\frac{\sin^2(A\plus{}X)}{\cos A}\plus{}a\frac{\sin^2(B\minus{}X)}{\cos B}\geq c\]
We will verify by checking boundary cases as well as zeros of deriv. (as function of $ X$)
The deriv. is zero when:
\[ b\frac{\sin 2(A\plus{}X)}{\cos A} \equal{} a\frac{\sin 2(B\minus{}X)}{\cos B}\]
\[ \iff\frac{\sin 2(A\plus{}X)}{\sin 2A} \equal{} \frac{\sin 2(B\minus{}X)}{\sin 2B}\]
\[ \iff \cos 2(A \plus{} X \minus{} B) \minus{} \cos 2(A\plus{}X\plus{}B) \equal{} \cos 2(B\minus{}X\minus{}A) \minus{} \cos 2(A\plus{}B\minus{}X) \]
\[ \iff \cos 2(A\plus{}B\plus{}X) \equal{} \cos 2(A\plus{}B\minus{}X) \iff \sin 2(A\plus{}B) \sin X \equal{} 0\]
Thus we need only check boundaries; i.e., $ X\equal{}0,\frac{\pi}{2}$; but at both we clearly see equality attained ($ a\cos B \plus{} b\cos A \equal{} c$)

So we have proven inequality with equality when $ L$ is one of the sides of the triangle.

[quote="me@home"]This is true because $ u \plus{} v$ remains constant as we slide around [i]iff[/i] $ B$ and $ C$ are on opposite sides of $ L$. Meanwhile $ w$ decreases until it is zero, the minimum, which is just when $ L$ is through $ A$.[/quote]

Sorry to revive such an old topic, but I don't quite see how $ u\plus{}v$ is constant implies the minimum of the expression occurs when $ w$ is minimum.  Can someone elaborate on me@home's solution?


calculus
algebra
polynomial
search
geometry
integration
number theory

who are preparing for isi entrance 2010
can we discuss a pre-exam discussion like the topics we are giving stress and any other thing related to this. :)

Very good idea,I shall appear for the test in 2010.But I think the preparation will be more about checking one's ability from easy to challenging problems.The book 10+2 has ISI  problems.But there are only a few subjective problems.I am looking for some new problems of this level.It will be very helpful if someone tells me a source.There are some well known books for olympiad problems but I think olympiad problems are not particularly like those problems that appear in isi entrance.

i think it is not,isi problems has a great similarity with olympiad problems.Also i watch that in the B.Stat entrance paper of 2004 has a repeat problem no 9 of INMO 1996 problem no 3.ISI has a extra part of calculus which olympiad has not.But I think calculus problems of ISI is very easy.For good problems you can purchase Polynomials by E.J.Barbeau and Mathematical Circles by Fomin,Genkin,Itenberg.
you can see also an old but nice book in download section of this site' USSR Olympiad book.'

abt dat UUSR Olympiad book site.....cud u provide the specific link plz?

Can anyone give me a link to download "MATHEMATICAL CIRCLES - RUSSIAN EXPERIENCE" by FOMIN? :D

$ 1$.try it from rapidlibrary-rapid search engine or amazon.com.
$ 2$.mathlinks-olympiad section-high school contest-download-USSR Olympiad book.

Well,I know that one inmo problem and several rmo problems came in isi entrance which is very unfair.I think the difficulty of ISI problems is like that of RMO though there are  few routine problems.When I said that about olympiad problems i didn't mean RMO problems.I referred those problems that appears in olympiad problem books like mathematical olympiad challenges,treasures etc. which teaches you lot special topics in number theory,geometry and combinatorics which are interesting but not similar to isi problems. There one can find lot of food for the brain.But In ISI entrance you only have to know basic principles.There almost all problems have very short and nice solutions.
I have the books you mentioned.Actually I was looking for books that will contain some challenging problems from calculus.

I have never heard about that book,but russian books are generally good.

ISI ADMISSION NOTICE is released .I watched it from my mobile internet.But I do not find it in Computer Internet.I am puzzled . Can anyone reply me. What is the fact :(

check ISI admission  notification on http://www.isical.ac.in/~deanweb/isient_announcement.html

[quote="tamal banerjee"]calculus-Did you read the book Differential and Integral Calculus Vol.1 by N.Piskunov.I found it great help full for theoritical knowledge required to solve ISI problems :D but it's problem collection is poor :([/quote]

For good collection you can try the collection of books from AMS

I had cleared the ISI B.Math entrance last year with rank 56, the written test was not that tough, but the Interviews are a bit harder. I failed there. :(  :mad:

Can anyone suggest books containing good collection of calculus problems reuired at the Isi level

Someone please post the questions tomorrow after the exam.

Shall we discuss here -[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=362&t=348532[/url]


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

[quote="sigma"]Let $2n\ (n\geq 1)$ points on a sphere, prove that there is a plane cut the sphere into two pieces such that each piece contain $n$ points.[/quote]

Let's take a plane nonparallel to every line passing through some 2 points of our set.
(And additional condition is that the plane doesn't intersect the sphere)

Lets move this plane continuously and parallel to itself (in the direction of perpendicular from the sphere's centre to the plane)
By the selection of the plane
There wouldn't be such position of plane that at least two points of the set belongs to the plane.
So there will be position of the plane with equal quantity of the set points in the two half-spaces during this move.

V:)


calculus
integration
function
real analysis
real analysis unsolved

$ \int_0^{\infty}\frac{x^{n\minus{}1}}{(x^2\plus{}a^2)^{\frac{n}{2}}}dx$, $ a \in R$, $ n>\equal{}2$ is natural number.

It's comparable to $ \frac{x^{n\minus{}1}}{x^n}$ for large $ x$, hence divergent.

how to calculate: $ \int_0^{\infty}\frac {x^{n \minus{} 2}}{(x^2 \plus{} a^2)^{\frac {n}{2}}}dx$, $ a \in R$, $ n \geqslant 2$ is natural number.

Reduce it to Beta function, I got $ \frac {a^{n \minus{} 1}}{2}B(n \minus{} 1,\frac {5}{2})$, although I maybe way off (I was calculating it on post-it blocks :blush: ). You should go like this; exchange variables so you get $ a$ out of the integral, then exchange $ x^2 \plus{} 1 \equal{} : u$, then $ u \equal{} : \frac {1}{t}$.

Edit:  ...And it is wrong, should be $ \frac{a^{n\minus{}1}}{2}B(\frac{n\minus{}1}{2},\frac{1}{2})$. That's it for me today  :roll: .


vector
calculus
derivative
function
linear algebra
matrix
linear algebra unsolved

Since Vectors are most nearly related to Matrices, I am posting this question here! Hope to get your opinion fast! Need help immediately!

IS Derivative wrt to a vector defined in ANY way in any field - no matter if its higher maths! ??? By this I mean that the derivative of $\vec{a}$ wrt $\vec{b}$ is say $\vec{c}$ means that when $\vec{b}$ is changed slightly, $\vec{a}$ also changes depending on $\vec{c}$ .... is such a definition used in any field.... is a special operator defined for this...... or is something else used... or is this all meaningless.... please help me on this guys.... fast!!!  :D

If you're differentiating with respect to a vector, how are you supposed to divide?

The answer is that you can differentiate a vector valued function with respect to a variable, but not with respect to a vector.

But you can certainly define it ina  special way like hundreds of things in Maths!!  :lol: SO a re you sure there is NO such topic anywhere in Math???

Well, there is something- but you have to use the linear operator definition. The derivative of a function from $\mathbb{R}^n$ to $\mathbb{R}^m$ is an $m\times n$ matrix, and the entries are the partial derivatives. The second derivative is a bilinear operator ($m\times n\times n$), and it only gets more complicated from there.


If we have an empty room, and every week there is a 0.6 probability for a person to join, 0.2 probability for a person to leave, and 0.2 probability for no change (only one action can be performed per week), what is the probability that there are at least 40 people in the room after 104 weeks, given that only one person may enter or leave the room each week?

All I could come up with was that the solutions are $ (40,0,64),(41,1,62)...(40\plus{}n,n,64\minus{}2n)$ for $ n\equal{}\{0,1,2,...,32\}$.  So the probability should be $ \sum_{n\equal{}0}^{32} {104 \choose 40\plus{}n}{64\minus{}n \choose n}(.6)^{40\plus{}n}(.2)^n(.2)^{64\minus{}2n} \equal{}$ $ \sum_{n\equal{}0}^{32} {104 \choose 40\plus{}n}{64\minus{}n \choose n}(.6)^{40\plus{}n}(.2)^{64\minus{}n}$. This seems extremely crude, and pretty useless without a calculator/computer.   Is there a more elegant approach to this problem?

What happens when a person is supposed to leave, but there are no people? Do we say nothing happens, or that there are $ \minus{} 1$ people?

Perhaps there is a .6 probability that a person enters and .4 that nothing happens until there is at least one person.


geometry
trigonometry
geometric transformation
reflection

Consider the configuration of six squares as shown on the picture. Prove that the sum of the area of the three outer squares ($ I,II$ and $ III$) equals three times the sum of the areas of the three inner squares ($ IV,V$ and $ VI$).

use the law of cosines repeatedly, and the fact that cos (180-x) = - cos (x)

[b]Problem:[/b] On the sides of $ \triangle ABC,$ we construct outwardly the squares $ABED,$ $CHFB,$ $ALKC.$ Let $ S_1,S_2,S_3$  be the areas of the squares with side length $ LD,EF,HK.$ Prove that
\[[ABED] + [CHFB] + [ALKC]= \frac{_1}{^3}(S_1 + S_2 + S_3) \]
[b]Solution:[/b] Let $ B'$ be the reflection of $ B$ about $ A.$ Then it is easy to see that $ \triangle ADL$ and  $\triangle AB'C$ are congruent. Thus if $ M$ is the midpoint of $ BC,$ we have $ DL = B'C = 2AM.$ This means that $ DL,FE,HK$ are twice the corresponding medians $m_a,m_b,m_c$ of $ \triangle ABC.$ From ${m_a}^2 + {m_b}^2 + {m_c}^2 = \frac {_3}{^4}(a^2 + b^2 + c^2),$ we deduce then the desired relation.

Dear Mathlinkers,
you can see
http://jl.ayme.pagesperso-orange.fr/   vol. 5 la figure de Vecten p. 104-106

Sincerely
Jean-Louis


A guy just bought the iPod Movie, it costed 317 with a tax of 10 percent.
The guy broke his iPod the next day, because of warranty, he got a new one.
The next day, he broke it again, once again, he got a new one.
How much did Apple lose, if they don't have to pay tax, and they sell their products at 3 times the amount it costs to make them?

[hide]Apple has to give him 3 ipods.  They spend $\frac13$ of the price on each one.  That means they spent the equivolent of $\frac13\cdot3=1$ ipod.  He payed them the equivolent of 1 ipod (the tax goes to the government), so they didn't lose anything $(1-1=0)$.  (The government gained $\$37.10$)

He must be clumsy to break 2 ipods in 2 days...[/hide]


I tried switching my avatar, but it says debug mode and won't let me upload it. When I try to upload from my comp, it doesn't work, why?!?!
I use AoPS skin and Safari. I have Mac OS X Leopard as my software.

[quote="solafidefarms"]How's he going to use IE7 on his mac? ;)[/quote]There are ways :D

[quote="7h3.D3m0n.117"]Have you tried uploading it with IE7 or Mozilla?[/quote]

Unless it's a cookie problem, that won't matter in the least.

I'm pretty sure avatar settings don't work with cookies....

Could it be the site I'm trying to upload from? Also, how do I delete cookies again?
When I try from my computer, it says no avatar selected, what must I do after I select it on the browse option?

 Here is the error report:

Unable to upload file

DEBUG MODE
Line : 259
File : usercp_avatar.php

any administrators know what that one is?

I will take a look at this error, but unfortunately I can't do that until later next week.

thanks anyway, I'll try to figure it out

[quote="solafidefarms"]How's he going to use IE7 on his mac? ;)[/quote]

I believe my old school computers had IE6 (before 7 came out) on their Macs.

That might have been IE5.5 for Mac Edition.

Why would internet explorer work where firefox didn't?


If $P(x)$ is an $n$ degree polynomial with $P(k)=2^{k}$ for $k=1,2,3,\dots,n$.
Find the value of $P(n+1)$.

[hide=" Maybe a useful hint?"]
You might be able to use the $n^{\text{th}}$-differences of the function, since they will be constant. [/hide]

(edit: Is $P(x)$ uniquely defined?)

Where did you get the original problem?  I think it specified $P(x)$ is an integer polynomial, or it specified $k = 0, 1, 2, ... n$.


Determine the extreme values of

$ \sin A \plus{} \sin B \minus{} \sin C$

$sinA+sinB-sinC=2cos\frac{C}{2}cos\frac{A-B}{2}-sinC\le 2cos\frac{C}{2}(1-sin\frac{C}{2}) < \frac{3\sqrt{3}}{2},$
but $sinA+sinB-sinC<sinA+sinB<2<\frac{3\sqrt{3}}{2},$
Why?


So about what date will the problems be posted and the registration updated to this year?

Should I be worried that last year, USAMTS problems were first posted in the one week before summer ended?

Sorry if this an inconsiderate or in any way annoying question to the USAMTS organizers.  I'm just worried that there might be a new website with the actual problems posted elsewhere... :maybe:

It's the same old website. :)

The problems may be posted as soon as tomorrow; they will definitely be posted within the next couple of weeks.


There are 24 volumes of an encyclopedia on a bookshelf. In how many ways may 5 volumes be selected, if no two consecutive volumes may be chosen?

[hide]
$\frac{24!}{(24-5)! \cdot 2! \cdot 5!}\rightarrow \frac{12 \cdot 23 \cdot 22 \cdot 21 \cdot 20}{5!}\rightarrow \frac{2550240}{120}\rightarrow 21252$
[/hide]

[quote="eavqg.whbeg"]I got a different answer   :huh:
[hide]
If we consider a possible way of choosing the five books, and then remove one book from between each pair, we are left with the same problem without restrictions (restriction=no two consecutive volumes may be chosen). So the answer is
$\binom{20}{5}=15504$.
[/hide][/code][/quote]
Thats whta I got.

[quote="AstroPhys"]My solution is invalid, I misread the problem.
$\left( \stackrel{20}{5}\right)$ seems to be correct.[/quote]
Latex: Use $\binom{p}{q}$ for combinations.

I realized that after I posted I had used the wrong command  :oops:  but was too lazy to fix it.  :wink:

[hide]
... B_B_B_B_B ...
(B is a chosen book, _ is any other book).
Now we need to place 15 books between the B's and between the B's and ...

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=15 \implies \binom{20}{5}$
[/hide]


combinatorics
college contests
inequalities
real analysis
algebra

For a real number $ x$, let $ \|x \|$ denote the distance between $ x$ and the closest integer. Let $ 0 \leq x_n <1 \; (n\equal{}1,2,\ldots)\ ,$ and let $ \varepsilon >0$. Show that there exist infinitely many pairs $ (n,m)$ of indices such that $ n \not\equal{}
m$ and \[ \|x_n\minus{}x_m \|< \min \left( \varepsilon , \frac{1}{2|n\minus{}m|} \right).\]

[i]V. T. Sos[/i]

It's enough to prove there exists at least one pair $(n,m)$ and then use the same arguments for $(x_i)\,, j>\max(n,m). $
Fix some $n$ and consider $x_j, j=1,2,\dots, n$. Let's order them in increasing order $0\leq x_{j_1}<x_{j_2}<\dots <x_{j_n}<1 $.

Now, suppose on the contrary the statement is not true. In particular it implies:
\[\|x_{j_s}-x_{j_{s+1}}\| \geq \min \left( \varepsilon , \frac{1}{2|j_s-j_{s+1}|} \right)\,,\,s=1,2,\dots,n \,\,\,\,\,\,\,(1) \]
where by $x_{j_{n+1}}$ we mean $x_{j_1}$.

Clearly $\sum_{s=1}^n \|x_{j_s}-x_{j_{s+1}}\| \leq 1$ and summing up (1) we get:
\[ 1 \geq \sum_{s=1}^n \min \left( \varepsilon , \frac{1}{2|j_s-j_{s+1}|} \right) \,\,\,\,\,\,\, (2)\]

The idea is to find an upper bound of $\sum_{s=1}^n |j_s-j_{s+1}|$, then use $AM-HM$ inequality and get a contradiction.
Fix some interval $ [k,k+1]\,,\, k \in \{1,2,\dots,n \} $ and let's see how many intervals $[j_s, j_{s+1}]$ can contain $[k, k+1]$ as a subinterval. Since every point among $1,2,\dots,k$ and $k+1,\dots,n$ can be an end of at most two intervals $[j_s, j_{s+1}]\,,\,s=1,\dots,n$, it implies $[k,k+1]$ is contained in at most $2\cdot \min(k,n-k)$ such intervals. Therefore:
\[\sum_{s=1}^n |j_s-j_{s+1}|\leq 2 \sum_{k=1}^n \min(k,n-k)\leq \frac{n^2}{2} \,\,\,\,\,\,\, (3) \]
Consider at first the case when $\frac{1}{2|j_s-j_{s+1}|} \leq \varepsilon\,,\, s=1,2,\dots,n$. Using (2), (3) and AM-HM yields:
\[ 1 \geq \frac{1}{2}\sum_{s=1}^n \frac{1}{|j_s-j_{s+1}|} \geq 1\]
Note that the equality above is only attained when all $|j_s-j_{s+1}|$ are equal. Since it's not possible, we get a contradiction.
Suppose now, $\frac{1}{2 |j_s-j_{s+1}|} > \varepsilon $ for $s\in I$, where $I$ is some not empty set of indices. Denote $k=|I|$. Clearly $1\leq k\leq 1/\varepsilon$. This time we have:
\[ 1 \geq \varepsilon + \frac{1}{2}\cdot \frac{(n-k)^2} {(n^2/2 - k)} \]
But $k$ is bounded from above with $1/\varepsilon$, not depending on $n$, thus choosing big enough $n$ leads to a contradiction.

dgrozev, you say $[k,k+1]\subseteq [j_s,j_{s+1}]$, which implies $j_{s+1}\ge k+1$. Then, how can it be $[k,k+1]\subseteq [j_{s+1},j_{s+2}]$ ? Because

$[j_s,j_{s+1}]\cap[j_{s+1},j_{s+2}]=j_{s+1}$, not $[k,k+1]$. If $[k,k+1]\notin\left([j_s,j_{s+1}]\cap[j_{s+1},j_{s+2}]\right),$ how do those intervals contain it?

Sorry, maybe I should have made it clearer. By $[j_s, j_{s+1}]$ I meant the interval which end points are $j_s$ and $j_{s+1}$. Since $j_1,j_2,\dots,j_n$ is a permutation of $1,2,\dots,n$ of course it may happen $j_{s+1}<j_s.$ So, we want to know how many intervals with endpoints $j_s,j_{s+1}$ contain $[k,k+1]$ as a subinterval.

dgrozev, thank you very much for the feedback! May I ask how you came up with the exact expression $1\ge\varepsilon+\dfrac12\cdot\dfrac{(n-k)^2}{\left(\frac{n^2}2-k\right)}$?

I see that:

$\lim\limits_{n\to\infty}\dfrac{(n-k)^2}{n^2-2k}=\lim\limits_{n\to\infty}\dfrac{1-\frac{2k}n+\frac{k^2}{n^2}}{1-\frac{2k}{n^2}}=1$ and

$\begin{aligned}\dfrac{(n-k)^2}{n^2-2k}&<0&\forall n<\sqrt{2k}\\0\le\dfrac{(n-k)^2}{n^2-2k}&<1&\forall n\ge k\end{aligned}$

and that we should get $1-\varepsilon<\dfrac{(n-k)^2}{n^2-2k}$ for sufficiently large $n$, but I couldn't figure out how you came up with that expression.

Using $(2)$ we get
$$1 \geq \sum_{s=1}^n \min \left( \varepsilon , \frac{1}{2|j_s-j_{s+1}|} \right) = \sum_{s\in I} \varepsilon+\frac{1}{2}\sum_{s\notin I} \frac{1}{|j_s-j_{s+1}|}$$
	The first term of the RHS is at least $\varepsilon.$ The number of the summands of the second sum is $n-k.$ Further
	$$\sum_{s\notin I} |j_s-j_{s+1}|\le \frac{n^2}{2}-k$$
	since $\sum_{s\in I} |j_s-j_{s+1}|\ge \sum_{s\in I} 1 =k.$ Thus applying AM-HM, we get
	$$\sum_{s\notin I} \frac{1}{|j_s-j_{s+1}|} \ge \frac{(n-k)^2} {(n^2/2 - k)}$$
	which yields
	$$\frac{1}{2}\sum_{s=1}^n \frac{1}{|j_s-j_{s+1}|} \ge \frac{1}{2}\cdot \frac{(n-k)^2} {(n^2/2 - k)}.$$

Did you mean 'applying AM-HM'  instead of 'GM-HM'?

Thank you for the explanation!


Prove that

$ \frac{\sin A \sin B \sin C}{\sin A \plus{} \sin B \plus{}\sin C}\geq3 \frac{\cos A \cos B \cos C}{\cos A \plus{} \cos B \plus{} \cos C}$

[quote="Ligouras"]Prove that in any triangle $ ABC$ there is the inequality $ \frac {\sin A \sin B \sin C}{\sin A \plus{} \sin B \plus{} \sin C}\geq3 \cdot\frac {\cos A \cos B \cos C}{\cos A \plus{} \cos B \plus{} \cos C}$ .[/quote]

[b]Proof.[/b] Suppose w.l.o.g. that $ \triangle ABC$ is acute. Since $ \sum\sin A \equal{} \frac sR$ and $ \sum \cos A \equal{} \frac {R \plus{} r}{R}$ obtain

$ \frac {\sin A \sin B \sin C}{\sin A \plus{} \sin B \plus{} \sin C}\geq 3\cdot \frac {\cos A \cos B \cos C}{\cos A \plus{} \cos B \plus{} \cos C}$ $ \Longleftrightarrow$ $ \prod\tan A\ge \frac {3s}{R \plus{} r}$ . Since $ \sum\sin 2A \equal{} \frac {2S}{R^2}$

obtain $ \prod\tan A \equal{} \sum\tan A \equal{} 2\cdot \sum\frac {\sin^2A}{\sin 2A}\ \stackrel{C.B.S.}{\ge}\ 2\cdot\frac {\left(\sum\sin A\right)^2}{\sum \sin 2A} \equal{} \frac sr\ge\frac {3s}{R \plus{} r}$ because $ R\ge 2r$ .


Find all $ n \in \mathbb{N}$ such that $ 121 \mid n^{10}\minus{}1.$

if we want to find all solutions we obviously need to find all numbers from $ 1$ to $ 60$ for which we have this divisibilty and after getting all the numbers satisfying them , if $ 1 \le l \le 60$ and $ 121 | l^{10}\minus{}1$ then the answers will be $ 121k \pm l$
do you have another method :?:


function
quadratics
algebra
polynomial
calculus
quadratic formula
real analysis

I’ve just finished my textbook, “Forgotten Calculus”. The last chapter is on multivariable calculus. Disappointingly, while there were some “pure math” problems on finding and classifying relative extrema, there were no real-world applications of this technique, i.e optimization problems.

So I decided to make one up, or rather base it on a marginal profit problem in the text. So what I’ve done below is not an “assigned” question per se. So the book has no answer for me to check myself by.

I would therefore ask any of you kindly, mentoring types out there to tell me if I am going about this right or wrong (I suspect the latter) and if need be correct me. To make it  easier I show pretty much all of my work, although it does make this post pretty long.

I also need some guidance on interpretation.

Lastly, if anyone can recommend any sources which have well-explained, worked examples in multivariate optimization, I’d be grateful. I don't think I need (or have time for) a whole book; even 5-10 examples could help me feel better about my ability in this area.

[b][u]The premise[/u][/b]

Total annual profit z from a small “summer hobby” farm is determined as

z = f(x,y) = 10x^3 + 20y^2 – 10xy

when x = weekly ad spending = 10 and
y = acres farmed = 5.

This much came from the text.

[u][b]The question and strategy summary[/b][/u]

I’d like to find out which combination(s), if any, of ad spending and acres maximizes (or minimizes) profit.

I will attempt to do so by trying to locate and evaluate the critical points, and hope they make some sense.

By the way: as a reference point, I evaluate the original function at values given, i.e

If

f(x,y) = 10x^3 + 20y^2 – 10xy

then

f(10, 5) = 10 (10)^3 + 20(5)^2 – 10 (10) (5)
= 10,000 + 500 - 500
=10,000

[u][b]Find critical points[/b][/u]

Start by taking the partials f[subscript]x and f[subscript]y

f[subscript]x = 30x^2 – 10y
f[subscript]y = 40y – 10x

Next, solve for x and y by setting each partial equal to 0, and solving as a “system of equations”, if this is the right term.

0 = 30x^2 – 10y
0 = 40y – 10x

Rearrange the bottom equation…
0= 30x^2  - 10y
0= -10x  + 40 y

Put them into a common format…
0 = 30x^2 – 0x – 10y
0 =   0x^2 – 10x + 40y

Multiply the top equation by 4 to get the “y”s to cancel out when I then “add” the two  equations together….

0 = 120x^2 – 0x – 40y
0 =   0x^2 – 10x + 40y
___________________
0 =  120x^2 – 10x +0

Use the quadratic formula (done on a calculator, work not shown) to solve for x

[b]Point 1: x = .08333[/b]  (displayed in this explanation showing only 3 threes, but in calculations, using 6 threes)

[b]Point 2: x = 0[/b]

Solve for y at Point 1 by plugging the Point 1 x value into one of the first partials set equal to 0

f[subscript]x = 30x^2 – 10y

0 = 30x^2 – 10y
0 = 30(.08333)^2 – 10y
0 = .208333 – 10 y
.208333 = 10y
[b]Point 1:  y = 48
So Point 1 = (.08333; 48; z)[/b]

Solve for y at point 2 using the same method, only using the Point 2 x value

f[subscript]x = 30x^2 – 10y
0 = 30x^2 – 10y
0 = 30 (0) – 10y
0 = 10y
[b]Point 2: y = 0
So Point 2 = (0; 0; z)[/b]

Find z for Point 1

Plug the Point 1 x and y values into the original function

z = 10x^3 + 20y^2 – 10xy
= 10 (.08333)^3 + 20(48)^2  - 10 (.08333)(48)
= .005787 + 46,080 – 39.999
[b]Point 1: z = 46,040
So Point 1 = (.08333; 48; 46,040)[/b]

Find z for Point 2 using the same method, only plugging in Point 2 values

z = 10x^3 + 20y^2 – 10xy
= 0 + 0 -0
[b]Point 2: z = 0
So Point 2 = (0; 0; 0)[/b]

[u][b]An aside [/b][/u]

At this point, before properly classifying the critical points, I was thinking ahead: “Ok, this makes sense – Point 1 is going to be a maximum, at which the maximum profit is 46,060, which is within the same order of magnitude as the profit under the premise conditions, and therefore seems plausible; while Point 2, showing 0 profit when there is no ad spending and no acres to far, makes perfect sense as a minimum”

As you’ll see below, I was wrong…

[b][u]Classifying the critical points[/u][/b]

Let D (x,y) = [ ( f[subscript]xx ) ( f[subscript]yy ) ]  - (f[subscript]xy )^2

From above, for reference:

f[subscript]x = 30x^2 – 10y
f[subscript]y = 40y – 10x

So

 f[subscript]xx = 60x
f[subscript]yy = 40
f[subscript]xy = 10

So

D= [ (60x) (40) ]  - (10)^2
[b]D= 240x – 100[/b]

For Point 1:

D = 240(.08333)  - 100
D =  20 – 100
[b]D = - 80
This is negative; thus Point 1 is a saddle point[/b]

For Point 2:
D = 240(0)  - 100
[b]D =   – 100
This is negative; thus Point 2 is also a saddle point[/b]

Concluding questions

First: Have I conceived and / or executed the solution wrong?

Second: I must say this is somewhat unsatisfying. Assuming I’ve done this right,  I’m not sure what to make of the answer. I know what a drawing of a saddle looks like but frankly don’t understand what it means in this case – a profitability problem with no minima or maxima, only two saddle points. How would you plan an optimal business strategy based on this?

If you’ve read this far, thanks for your attention.

Regards,

lingyai

The problem is that you picked an equation that isn't well-suited to this kind of problem.

In real life, if you realized that your revenues followed that function, you would shout "OMFG" and run around finding investors. Why? Look at the second partials. If you keep increasing $x$ and $y$, your profit will keep increasing at an increasing rate! This never happens in real life.

If you want to make this kind of problem for yourself, and your profit equation is going to be a polynomial, you want the highest-degree terms to have negative coefficients. Then you'll get answers that make sense.

It looks like you did all the math right, though. :)

(I'm a little horrified you used a calculator to solve $120x^2 - 10x + 0 = 0$, though... Just factor it, $10x(12x-1) = 0$. $x = 0$ or $x = 1/12$. :P)

Thanks for that feedback. The only other one I've received so far has been one advising me how to format subscripts.

Well, at least I have the technique right.  But I'd really like some more appropriate examples. Know of any good resource, off the top of your head? This text, Forgotten Calculus, is great in some respects but is pretty stingy with application problems.  Are you a math educator by the way?

Yes, in retrospect I didn't have to use the quadratic equation, I was just in a certain mode where I was concentrating on getting this posted as quickly,  and bullet-proofing my work as much, as possible.

 Take care,

Ken (lingyai)

[quote="lingyai"]Know of any good resource, off the top of your head?[/quote]
No, I'm afraid not. But there are many calculus books out there. Try your local library, and just flip through whatever they have, see if there's anything good. I took a multivariable class at Texas A&M, and we used Stewart's book. But I didn't do the homework (it wasn't graded, and I was lazy back then), so I don't know what the problems were like.

[quote="lingyai"]Are you a math educator by the way?[/quote]
No, I'm still a student. :D I'm quite flattered! I'd like to teach math some day, but I'll need to get a degree or two first. :P


calculus
vector
real analysis
real analysis unsolved

This problem is taken from [color=red]Rudin's Principles of Mathematical Analysis[/color]

chapter 9 problem 5

Assume $ A \in L(\mathbb{R}^{n}, \mathbb{R})$, show that there exists only one vector

$ y \in \mathbb{R}^{n}$, such that $ Ax \equal{} x\cdot y$

and $ \parallel{}A\parallel{} \equal{} |y|$, where : $ \parallel{}A\parallel{} \equal{} \sup_{|x|\leq1}|Ax|$

[color=red]Edit[/color]: if $ y\equal{}(y_{1},y_{2},...,y_{n})$, then $ |y|\equal{}\sqrt{y_{1}^{2}\plus{}y_{2}^{2}\plus{}...\plus{}y_{n}^{2}}$ :D

Let $ e_1,e_2,..,e_n$ --orthogonal base in $ R^n$.
we consider $ y\equal{}(Ae_1,Ae_2,...,Ae_n)$. Let $ x\equal{}(x_1,..,x_n)$.
Then $ Ax\equal{} A(\sum x_i e_i ) \equal{} \sum A x_i e_i \equal{} \sum x_i A e_i \equal{} x\cdot y$.
Now, suppose than there exists $ y_1,y_2$ such that $ Ax\equal{}xy_1\equal{}xy_2$ for all $ x\in R^n$.
Then $ x(y_1\minus{}y_2)\equal{}0$ for all $ x\in R^n$, so $ y_1\minus{}y_2\equal{}0$.

$ \parallel{}A\parallel{}\equal{}\sup_{\parallel{}x\parallel{}\le 1} |Ax| \equal{} \sup_{\parallel{}x\parallel{}\le 1} {\sum x_i y_i} \le \sup_{\parallel{}x\parallel{}\le 1} \left(\sum x_i^2 \sum y_i^2\right)^{\frac12}$. it follows from Helder inequality.
Then
$ \parallel{}A\parallel{}\le \sup_{\parallel{}x\parallel{}\le 1} \left(\sum x_i^2 \sum y_i^2\right)^{\frac12} \equal{} \left( \sum y_i^2\right)^{\frac12}   \sup_{\parallel{}x\parallel{}\le 1}  \left(\sum x_i^2\right)^{\frac12} \equal{} \left( \sum y_i^2\right)^{\frac12} \equal{} |y|.$
And
$ |A\frac{y}{|y|}|\equal{}\sum \frac{y_i}{|y|}y_i \equal{} \frac{\sum y_i^2}{|y|}\equal{}|y|$, so $ \parallel{}A\parallel{}\equal{}\sup_{|y|\le 1}\ge |y|.$


Assume that x(1),x(2)...x(n) are non negative numbers with
(1) x(i)<=x(i+1) i=1,2,3...n-1
(2) x(1)+x(2)+x(3)+...+x(n)=m
then how many pairs of (x(1),x(2),x(3),...x(n)) are there altogether?
Try to find the answer and give a proof.
Thanks.[/hide]

This is not very clear. Could you please rewrite it?

I think he wants to say,
$ x_1,x_2,...,x_n$ are non negative numbers such that $ x_i \le x_{i \plus{} 1}$ for $ i\equal{}1,2,...,n\minus{}1$ and $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} m$.
Find the number of solutions


Prove that the neccessary and sufficient condition for $p$ and $p+2$ to be prime is $4[(p-1)!+1]=0$ mod $p^2+2p$.

Hi, in your statement a "$+p$" is missing on the LHS.
Anyway, it was posted (and solved) before : http://www.mathlinks.ro/Forum/viewtopic.php?t=49074 .
:)


function
trigonometry
inequalities
complex analysis
absolute value
complex analysis unsolved

Determine all entire functions $f$ such that for every $z$, $|f(z)| \le |\sin z|$.

Is it a correct argument to say that since $f$ has zeros where $\sin$ does $\frac{f(z)}{\sin z}$ is also an entire function and it has absolute value less than 1 (perhaps by continuity in the zeros of $\sin$). In particular it is bounded and therefore constant. Hence, $f(z) = c\sin z$ for some $c$ with $|c| \le 1$.

This seems to generalize so that if you take any entire function $g$, the only entire functions $f$ satisfying $|f(z)| \le |g(z)|$ are $f(z) = c\cdot g(z)$ for some $c$ with $|c| \le 1$.

This seems to be such an incredible result that I don't really believe it. Is it right?

you have to check the multiplicity of the zeroes, otherwise there could be some problems of continuity..

Why would that be a problem? If $g$ has a zero of mult. $n$, won't $f$ have a zero of at least multiplicity $n$ ? Or are you just saying I should state that to complete the argument?

A good thing to notice is that $sin(z)$ doesn't have multiple zeroes?, so it should be ok (and it seems the statement you have is also true, as long as we're talking about entire without multiple zeroes...)

Why would the statement be untrue because of zeros of higher order than $1$? If $g(z)$ has a zero at $z = p$ of multiplicity $n$, then $f$ will have a zero of at least multiplicity $n$ at $p$. This is because $\frac{|g(z)|}{|z-p|^{n-1}}\to 0$ as $z \to p$, and by the given inequality the same is true for $\frac{|f(z)|}{|z-p|^{n-1}}$. It's clear that it follows that $f$ has a zero of at least multiplicity $n$ at $p$.

I do not see what is wrong with my more general statement  :?:

There is nothing wrong with your more general statement. Being smaller than an entire function everywhere is a very strong condition.
Multiple zeros aren't a problem at all- the quotient is uniformly bounded, so any isolated singularities are automatically removable.


$a^2+b^2+c^2=1$, show $\sum_{cyc}ab(a^2+ab+b^2)\le 1$

Who has the easiest proof for that?

[quote="Together"]We have $(a^2+b^2+c^2)^2 \geq 3(a^3b +b^3c+c^3a)$
[/quote]

But how would you prove this? :D

Arne, the inequality you are reffering to is Vasc's. ;)

I know, I know, I was being a bit ironic/sarcastic. ;)

[quote="TomciO"]Homogenize first:
$\sum ab(a^2 + ab + b^2) \leq 1$ is equivalent to: $\sum ab(a^2 + ab + b^2) \leq (a^2+b^2+c^2)^2$
Expand and reduce to get:
$a^4 + b^4 + c^4 + a^2b^2 + b^2c^2 + c^2a^2 \geq a^3b + ab^3 + b^3c + bc^3 + c^3a + ca^3$
But this can be rewritten as:
$\frac{1}{2}((a-b)^2(a^2+b^2) + (b-c)^2(b^2+c^2) + (c-a)^2(c^2+a^2)) \geq 0$
which is obviously true.[/quote]

Its also  Schur  for  $r=2$


inequalities
calculus
ratio
inequalities solved

Prove that if $ a$, $ b$, $ c$ are nonnegative reals, then

$ a^5 \plus{} b^5 \plus{} c^5\geq 5abc\left(b^2 \minus{} ac\right)$.

We denote : x=a/b ; y=c/b
The inequality becomes :
x 5 +y 5 +1+5x 2 y 2 -5 xy \geq 0
If xy \geq 1 , we have
x 5  +y 5  \geq 2 \sqrt (x 5 y 5 ) \geq 2xy \geq xy
4x 2 y 2 -4xy+1=(2xy-1) 2  \geq 0
=>q.e.d
If xy<1 , we have:
x 5 +y 5  +1/36+1/36+1/36 \geq (xy)^(5/6) \geq xy
5x 2 y 2 -4xy+11/12=(xy \sqrt 5-2/ \sqrt 5) 2 +7/60>0
=>q.e.d

Note: If  xy \geq 1  then  ac  \geq b 2   so the  original  inequality is  trivial.

note  3: This  one  is  not  a  UK  94  problem...at  least  round  1  and  2  mo  that  year

It's a problem from FIST 1994 (Final IMO Selection Test) in the UK.

I found a very short solution. WLOG $b = 1$. Take $k = \sqrt{ac}$. Then \[2k^5 + 5k^4 - 5k^2 + 1 \geq 0\] since \[(2k + 1)(k^2 + k - 1)^2 \geq 0\] and that's basically it.

[b]First solution :[/b] This ineq is homogeneous , so we can suppose $ b =1 $
 We have : $ a^5+c^5+1 = S^5-5S^3P+5SP^2+1 $ , where $ S =a+c , P=ac $
  The ineq is equivalent to $ (S^2-4P)(S^3-SP) +SP^2+5P^2-5P+1 \geq 0 $
  We have $ S \geq 2\sqrt{P} $ , so it's enough to show :
   $ 2k^5+5k^4-5k^2+1 \geq 0 $ , where $ k=\sqrt{P} $
  This's proved by Arne  ;)
 [b] Second solution :[/b] Use calculus
  Put $ f(b)=a^5+b^5+c^5-5abc(b^2-ac) $
  $ f'(b) = 5(b^4-3b^2ac+a^2c^2) $
  $ f'(b) =0 \iff b= \frac{\sqrt5 +- 1}{2}\sqrt{ac} $
  Put $ \frac{\sqrt5 + 1}{2} = g $ , which is golden ratio  :lol:
  We have : $ g^2=g+1 \to g^5=5g+3 $
  $ f(b) $ reach minimum at $ b=g\sqrt{ac} $
  $ f(g\sqrt{ac})= a^5+(5g+3)(ac)^{\frac{5}{2}}+c^5-5(g+1)(ac)^{\frac{5}{2}} = (a^{\frac{5}{2}}-c^{\frac{5}{2}})^2 \geq 0 $  ;)

[quote=blahblahblah]Put $ac=1$ by homogeneity.  Then $a^5+c^5\geq 2$.  So we want to show that $a^5+2\geq 5b(b^2-1)$

But this is equivalent to $(x+2)(x^2-x-1)^2\geq 0$, which is obvious.

Equality holds iff $a=c=t$ and $b=\left(\frac {1+\sqrt 5}{2}\right)t$, $t\geq 0$.[/quote]

What is the main reason you put $ac+1$? like could you put like that at every problem?

[quote=David_Kim_0202][quote=blahblahblah]Put $ac=1$ by homogeneity.  Then $a^5+c^5\geq 2$.  So we want to show that $a^5+2\geq 5b(b^2-1)$

But this is equivalent to $(x+2)(x^2-x-1)^2\geq 0$, which is obvious.

Equality holds iff $a=c=t$ and $b=\left(\frac {1+\sqrt 5}{2}\right)t$, $t\geq 0$.[/quote]

What is the main reason you put $ac=1$? like could you put like that at every problem?[/quote]

[quote=djmathman]You can do this in this problem because the inequality is [b]homogeneous[/b] (that is, it still remains true if $(a,b,c)$ is replaced by $(2a,2b,2c)$ or $(\pi a,\pi b,\pi c)$ or whatever).  Thus, we can scale our triple of nonnegative numbers appropriately to assume $ac=1$.[/quote]

Oh thanks so like if the degrees of the two sides are the same, you can do for all $k \in \mathbb{R}\ (a, b, c) \rightarrow (ak,bk,ck)$


[quote="Forte"]Let triangle $ ABC$ with right angle at $ A$ and $ BA$ having a length of $ 6$, and likewise $ AC$ has a length of $ 8$, if $ 2$ points on the hypotenuse $ P$ and $ Q$ are chosen such that $ APQ$ is an equilateral triangle, compute $ [APQ]$. {Area of triangle APQ}

I would like a Trigonometric solution for this, although Geometric ones are welcome too.[/quote]

Hi
the area will be : ( 192 sqr3 / 25 ) = 13.302 cm^2
the complet solution sooner
mrmohamed

[quote="Forte"]Let triangle $ ABC$ with right angle at $ A$ and $ BA$ having a length of $ 6$, and likewise $ AC$ has a length of $ 8$, if $ 2$ points on the hypotenuse $ P$ and $ Q$ are chosen such that $ APQ$ is an equilateral triangle, compute $ [APQ]$. {Area of triangle APQ}

I would like a Trigonometric solution for this, although Geometric ones are welcome too.[/quote]
Hi all

[url=http://www.servimg.com/image_preview.php?i=458&u=12899977][img]http://i19.servimg.com/u/f19/12/89/99/77/almonk12.gif[/img][/url]

with my best wishes
mrmohamed

[quote="mrmohamed"]
 the area will be : ( 192 sqr3 / 25 ) = 13.302 cm^2
[/quote]
mrmohamed,

the answer does not have centimeters (or any other specifically named units
raised to any positive integer) in it, because there are no specific units
mentioned in the problem.

If your value is correct, then, for example, the answer could be expressed as
$ 13.302$ units squared, or
$ 13.302$ $ {units}^2$.

***EDIT***


$a,b,c$ are positive reals. Prove that:
$(\frac{a+b+c}{3})^{a+b+c}\geq a^bb^cc^a$

Shouldn't source be changed to "Kiran Kedlaya's Notes on Inequalities".  :)

[quote="paladin8"][quote="Mathx"]$3(\frac{ab}{a+b+c}+\frac{ac}{a+b+c}+\frac{bc}{a+b+c})$
$\geq 3(a^{\frac{b}{a+b+c}}b^{\frac{c}{a+b+c}}c^{\frac{a}{a+b+c}})$
[/quote]

How do you get that?[/quote]
the general form of AM >= GM


search
number theory unsolved
number theory

For positive integers $ n$, $ f_n \equal{} \lfloor2^n\sqrt {2008}\rfloor \plus{} \lfloor2^n\sqrt {2009}\rfloor$. Prove there are infinitely many odd numbers and infinitely many even numbers in the sequence $ f_1,f_2,\ldots$.

[quote="cosinator"]Similar to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2010018844&t=154073[/quote]

But harder...  :wink:

It seems like that according to the link provided. ><

[quote="cosinator"]It boils down to simply proving $ \sqrt {2009}\pm\sqrt {2008}$ is irrational, no?[/quote]
yup...
It is rather strange that they still propose problems which can be solved by these standard tricks....I've seen this method when I was in 9th grade and since that time I have solved so much problems using this trick,but I must admit that this method sometimes does not work...if this trick doesn't work I strongly suggest to use dirichlet theorem about irrational numbers...

I think there is a very nice solution by considering this base 2. If the sequence was all odd after a while, then that would mean the nth digit of (2008)^.5 is always different from the nth digit of (2009)^.5 when working base 2. therefore (2008)^.5+(2009)^.5 must be rational as can be seen from it having all 1s after a certain point in the base 2 expansion, but its not difficult to show this is false. If the sequence was even after a certain points, the nth digits would be the same and we would only have to show (2009)^.5-(2008)^.5 is not rational which is also not difficult.

For each $n \in \mathbb{N}$, let $a_n = 1$ if $\{2^n \sqrt{2008}\} > \frac12$ and $0$ else. Similarly define $b_n$ for $\sqrt{2009}.$ Assume, for contradiction, that $f_1, f_2, \cdots$ had the same parity after a while. Then, $f_{n+1} - 2f_n = a_n + b_n$ also has invariant parity after a while. Therefore, we have that for big enough $n \ge N$, either $a_n = b_n$ or $a_n = 1-b_n$. This implies that $2^N\sqrt{2008}$ and $2^N \sqrt{2009}$ either have the same fractional part or add up to an integer. Hence, it suffices to show that $\sqrt{2008} \pm \sqrt{2009}$ is not a rational number. Since they are both algebraic integers, we just need to show that they are not integers. However this is easy to see since $\sqrt{2008} + \sqrt{2009} \in  (89, 90)$ and $\sqrt{2009} - \sqrt{2008} \in (0, 1).$
$\square$

Let the binary representation of $\sqrt{2008}$ be $\overline{\ldots a_0.a_1a_2\ldots}_2$ and the binary representation of $\sqrt{2009}$ be $\overline{\ldots b_0.b_1b_2\ldots}_2$. It is easy to see that $V_i$ is even if and only if $a_i=b_i$.
Suppose that there are finitely many odd $V_i$, so every sufficiently large $i$ satisfies $a_i=b_i$. Then $\sqrt{2009}-\sqrt{2008}$ has in binary form eventually terminates, so it's rational. But this isn't true by squaring, as $2009 \cdot 2008$ isn't a square.
Likewise, if there are finitely many even $V_i$, every sufficiently large $i$ satisfies $a_i\neq b_i \implies a_i+b_i=1$ since $a_i,b_i \in \{0,1\}$. Then $\sqrt{2009}+\sqrt{2008}$ has a binary expansion that's eventually all $1$'s, which can be "re-expressed" as terminating (since $1_2=0.\overline{1}_2$), so it's rational. But this isn't true by squaring either. $\blacksquare$

Let $\{\sqrt{2008}\} = 0.\overline{a_1a_2a_3\cdots a_k\cdots}$ and $\{\sqrt{2009}\} = 0.\overline{b_1b_2b_3\cdots b_k\cdots}$ in their binary representations.

If there are only finitely many odd $f_i$, for all $k>N$ we have $b_k + a_k = 1$, so $$\sqrt{2008} + \sqrt{2009} = \frac{n+0.\overline{111\cdots}}{2^N} = \frac{n+1}{2^N}$$ for some positive integer $n$ is rational, contradiction.

If there are only finitely many even $f_i$, for all $k>N$ we have $b_k = a_k$, so $$\sqrt{2009} - \sqrt{2008} = \frac n{2^N}$$ for some positive integer $n$ as the repeating parts cancel out. This is rational, which is again a contradiction.


Find all function $ f: Q^ \plus{} \rightarrow Q^ \plus{}$ such that:
1)$ f(x) \plus{} f(\frac {1}{x}) \equal{} 1$.
2)$ f(2x \plus{} 1) \equal{} \frac {1}{2}f(x)$.


limit
linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Construct an example of matrices $ A, B\in M_n(\mathbb{R})$ such that  the limits $ \displaystyle\lim_{n\to\infty}A^n$ and $ \displaystyle\lim_{n\to\infty}B^n$ exist but the limit $ \displaystyle\lim_{n\to\infty}(AB)^n$ does not exist.

Let $ A\equal{}\begin{bmatrix}1&1\\0&\frac12\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}\frac12&0\\1&1\end{bmatrix}.$

Then $ \lim_{n\to\infty}A^n\equal{}\begin{bmatrix}1&2\\0&0\end{bmatrix}$ and $ \lim_{n\to\infty}B^n\equal{}\begin{bmatrix}0&0\\2&1\end{bmatrix}.$

$ AB\equal{}\begin{bmatrix}\frac32&1\\\frac12&\frac12\end{bmatrix}.$

The characteristic polynomial of $ AB$ is $ x^2\minus{}2x\plus{}\frac14$ and the eigenvalues are $ 1\pm\frac{\sqrt{3}}2.$

Since the spectral radius of $ AB$ is $ 1\plus{}\frac{\sqrt{3}}2>1,$ we must have that $ (AB)^n$ contains arbitrarily large entries for large $ n$ and cannot tend to a limit.


Hi everyone,
I am doing the assignment, got a few problems and not sure about my answer. Could anyone please help me to solve some of these questions?
Thank you in advance!
Have a nice day!
FlyWheel

For #1, check out the Lambert W-Function on math world.  http://mathworld.wolfram.com/LambertW-Function.html

"Please ensure that you attach a signed statement to your assignment that it is your own work?"

I'm not sure that I can, in good faith, help you with such an assignment (and I note that you have already posted some of the questions before)

Edit: I deleted this post after reading what blahblahblah wrote, I missed the notice at the top of the page.


In ABC triangle D is the midpoint of CB and P is the midpoint of AD. If the line that passes through BP intersects AC at Q then AQ / AC = ?

$ 1: 3$. I solved it here: http://www.mathlinks.ro/viewtopic.php?t=222934