Let $ x,\ y$ be non zero real numbers such that $ x^2 \plus{} y^2 \equal{} a^2$ for a positive constant number $ a$.

Find the maximum and minimum values of $ \frac {1}{ax \plus{} y \plus{} 1}$.

Assume x=a cos(p) ; y=a sin(p)

Max. and Min. values may easily be found for

ax+y+1 = (a^2)cos(p)+ asin(p)+1

by using calculus,which will be Min. and Max. values of  1/(ax+y+1)


Hello Everyone,

I need some help with this problem:

 If $\cos{10x} + \cos{5x} = 0$, $0< x < 2.2 \pi$, find the exact average of all solutions for $x$.

[b]
Let $5x$ be $\varphi$, then the equation becomes $\cos{2\varphi}+\cos\varphi=0$
That is, $2\cos^2{\varphi}+\cos\varphi-1=o$
Then it becomes much easy to solve.[/b]

If you are lucky,  one of the answer will jump out as   $12$ degrees (or $\pi / 15)$ .  If you call this angle $\alpha = \pi/15$  all    values would be $ \alpha, 3\alpha , 5\alpha,  .,   ... 29\alpha$ ...... etc.. or 15 such values.

One of the ways is to use $\cos (n\theta) = (1/2)( z^n+z^{-n}) $ where  $z = e^{i\theta} $  and then the above reduces to $z^{20}+z^{15}+z^5+1 = 0 $
or $(z^{15}+1)(z^5+1) = 0 $
and the roots are given by $z^{15 }= -1  = e^{i \pi} $  etc.. rest is simple.

BTW Another way to 'see' the same is to rewrite the original eq: as

$\cos (10x) = - \cos (5x) = \cos ( \pi -5x)$
which gives  $10 x = \pi - 5x  + 2n\pi $  (last term is added to get all the solutions
which again gives the above solutions..

[quote="fanaticsm"]So what would be the answer  :?[/quote]

Are you sure you can not  find it yourself[hide] by taking the averages of  these 15 values? (Assuming that you don't have a typo and meant  $0<x<2\pi$  the answer would be $\pi $ [/hide]

Gyan, I think he means what are the values that he needs to take the average of (all of them)...

That was in Gyans first post:

[quote="Gyan"] If you call this angle $\alpha = \pi/15$  all    values would be $ \alpha, 3\alpha , 5\alpha,  .,   ... 29\alpha$ ...... etc.. or 15 such values.
[/quote]


What is the remainder when 1,493,824 is divided by 4?

A number $ a_na_{n\minus{}1}...a_2a_1 \equiv a_2a_1 \pmod{4}$.  So we have $ 1493824 \equiv 24 \equiv \boxed{0}$.

hello, since $ 24\equiv 0 \mod 4$ we have $ 1493824\equiv 0 \mod 4$ and the remainder is zero.
Sonnhard.

In order for a number to be divisible by 4 the number formed by the tens and units digit is also divisible by 4. Since 14938[u]24[/u] ends in 24 and 24 is divisible by 4 that means the remander is 0.


[quote="nr1337"]1.  A number n is perfect if the sum of the divisors of n is 2n.  Prove that no perfect number is 2 times another perfect number.
2.  Let  $x_1, y_1,$ and $z_1$ be the lengths of the sides of a triangle, and $x_2, y_2,$ and $z_2$ be the lengths of the sides of a triangle similar to the first one.  Also let $a_1$ be an altitude of the first triangle and $a_2$ be the corresponding altitude on the second triangle.

\[
\frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{z_1}{z_2} = c
\]

for some c.  Does

\[
\frac{a_1}{a_2} = c?
\]

3.  Show that there are infinitely many polynomials P(x) of degree >=1 such that P(x) is composite for every integer x>=1.[/quote]

3. [color=white] Consider the polynomial p(x) = (x+1)(x+2) to start with, now for evert integer x>= 1 one of the terms will be even, so  2 | p(x) for all x >= 1) This can be expanded with (x+1)(x+2)(x+3), etc. [/color]

Is this okay?

#1: Proof by contradiction.  let p,q be the perfect numbers, with p=2q. p is even, and it is well known that all even perfect numbers are of the form 2^(k-1) * (2^k - 1), where 2^k - 1 is prime. if q is even, q is of the form 2^(l-1)*(2^l - 1), and so
2^(k-1) * (2^k - 1) = 2^(l) * (2^l - 1). then k-1=l since the highest power of 2 dividing LHS = highest power of 2 dividing RHS. then 2^k-1=2^l-1, 2^k=2^l, k=l which contradicts k-1=l. Now, if q is odd, 4 does not divide p, i.e. 4 does not divide 2^(k-1), so k=2 or 1. when k=1, 2^k-1 = 1 is not prime :). when k=2, 2^k-1=3 giving the perfect number 6, which implies p/2=3 is another perfect number, which is false. together we have the desired contradiction.
#2: yes  :) let area of first triangle = A1, area of second triangle = A2. then A1/A2= c^2, but a1 = 2*A1/x1, a2= 2*A2/x2, so a1/a2=(2*A1/x1) / (2*A2/x2)
= (A1/A2) * (x2/x1) = c^2* (1/c) = c.

[quote="nr1337"][quote="lemonparade"]3. [color=white] Consider the polynomial p(x) = (x+1)(x+2) to start with, now for evert integer x>= 1 one of the terms will be even, so  2 | p(x) for all x >= 1) This can be expanded with (x+1)(x+2)(x+3), etc. [/color]

Is this okay?[/quote]

Almost but you forgot the case where it is of degree 1.

vinoth_90_2004:  #2 is right.  I will check #1 tomorrow but I have to go now.[/quote]

You can just have 2x+2, 2(2x+2), etc... :)

Another solution to #1:

[hide]If n is a perfect number, then the sum of divisors of n is 2n.

If 2n is a perfect number, then the sum of divisors must be 4n.

But sum of divisors of 2n=(2+1)(sum of divisors of n)=6n, Contradiction.[/hide]

Vinoth and lemonparade, your solutions are correct.  Beta, you said that if n is a perfect number, then the sum of the divisors of 2n is 6n.  6 is a perfect number.  The sum of the divisors of 12 is 1+2+3+4+6+12 = 28, which is not equal to 6*6=36.

[quote="beta"]oops, that works if and only if n is odd.[/quote]

And so far we don't even know if there are any odd perfect numbers.

My point is:
If n is odd and sum of divisors of n is k, then sum of divisors of 2n is 3k.
Although I forgot that it doesn't work when n is even.


algebra
difference of squares
special factorizations

[quote="nonie"]The difference of squares of two consecutive odd numbers is 200. What are the two numbers?[/quote]
[hide]
$(n+2)^{2}-n^{2}=4n+4=200$
$n=49$

$\boxed{49,51}$[/hide]

[quote="nonie"]The difference of squares of two consecutive odd numbers is 200. What are the two numbers?[/quote]

[hide]$n^{2}-(n-2)^{2}=200$
$n^{2}-(n^{2}-4n+4)=200$
$n^{2}-n^{2}+4n-4=200$
$4n-4=4(n-1)=200$
$n=\frac{200}{4}+1$
$n=51$.

Then, the solution is $49;51$[/hide]

[hide]plugging $y=x-2$ into $x^{2}-y^{2}=200$ gives [color=red]$51,49$[/color][/hide]

51 and 49 is what I got too. I love this forum, I can do like everything on it =)

[quote="leela"] \[(x+2)^{2}-x^{2}=200\]  \[4x+4=200\]  \[x=49\] The  numbers  are  49 and  50[/quote]

See that the numbers are not consecutive, they're odd consecutive numbers.

[quote="moogra"][quote="nonie"]The difference of squares of two consecutive odd numbers is 200. What are the two numbers?[/quote]

[hide]x+x+2=200
2x=198
x=99
x+2=101[/hide][/quote]

difference of squares, not sum  :wink:

[hide="solution"]
The first number is $2k+3$ and the second is $2k+1$
$(2k+3)^{2}=4k^{2}+12k+9\\ (2k+1)^{2}=4k^{2}+4k+1\\$
Their difference is $8k+8=200\Rightarrow k+1=25\Rightarrow k=24$
So our numbers are $49$ and $51$[/hide]


unless i am looking over something obvious, [hide]1[/hide]

You are overlooking something not so obvious, but still there...
Suppose you label the people 1 2 3 4 5 6 7 8 9 0, and they're sitting in that order.  Then after the switch, no two consecutive increasing numbers can be next to each other (increasing left to right).  There are several ways in which this can be done.  They could all switch around so that they go 0 9 8 7 6 5 4 3 2 1 if they wanted to, or they could switch to 1 0 3 2 5 4 7 6 9 8
But is the thing we're trying to count the number of ways to change seats?  Or the number of possible permutations?  Though I don't really know how to count every permutation...I hope this gives an idea...
Suppose we put 0 in one seat.  Then 1 cannot be to 0's right.  So there are 8 possibilities for that seat.  The problem is now that the whole game will change if we put 9 in that seat...i'm not good with permutation, someone else will probably have a better soltuion than I can find...if nobody else posts, i'll try to find a solution.

ugh... this is strange, how do you know where to start because what happens if they rotate like $ABCDEFGHIJ$ and $BCDEFGHIJA$? is that considered different because they are sitting different seats...

Consider those two the same.  If two situations like those two are to be considered different, you can just multiply by 10 at the end.

How would you do this with combinations? Would it just be $10C10$?

No, because $10C10$ is 1, and we already decided that is not true.

[hide]Assume person $A$ sits in seat $1$.  Assigning all of the seats without being concerned with the special rule gives $9!$  Now subtract the situations where the special rule is broken.  I believe this is the only way to efficiently solve the problem.  I have tried assigning seat $1$ to a person, then seat $2$ to a person, and so on, I have tried assigning person $A$ to a seat, person $B$ to a seat, and so on, and I have tried assigning all of the odd-numbered seats and then all of the even-numbered seats.  All of these would take a long time.[/hide]

Matt276eagle's method sounds good, actually.  You could count those by taking the number of situations where B is to the right of A, then the number where C is to the right of B, then subtract the number of situations whereyou have ABC because those are counted twice.  So on.  I have to go now though, so I can't find a solution...

Unfortunately, it wouldn't be that easy, because of the situations where person $A$ sits next to person $B$ and person $B$ sits next to person $C$.  You would end up double-counting those situations and getting an answer too low.

[hide]I suggest counting the situations where all ten people sit to the right of the person they were sitting next to before ($1$), then the situations where exactly nine people sit to the right of the person they were sitting next to before ($0$), and so on.[/hide]

Pretend that instead of in a circle, it is in a line with ten seats.
_ _ _ _ _ _ _ _ _ _

Seat 1 has 10 possible people to sit in it.
Seat 2 has 9 possible people to sit in it.
Seat 3 has 8 possible people to sit in it.
Et cetera, etc.

10*9*8*7*6*5*4*3*2*1 would be the answer if it was in a line.

Since it is in a circle with ten seats, we must divide by ten.

So the answer is:

9*8*7*6*5*4*3*2*1=
362880  :huh: the answer seems a little bit too big though :(

i got the same answer as orangejuice's, which is $9!$. in fact, for $n$ people sitting in a table, we get $(n-1)!$ different ways for $n$ people to sit in a round table. as K81o7 said in post #5 on answering my question in #4, we divide the total permutations of 10 elements in a line by 10, i.e. $\frac{10!}{10}=9!$ and in general $\frac{n!}{n}=(n-1)!$.

That's the total number of seating arrangements...doesn't discount the ones where people have the same people on the right...

yeah, i think so. because in a circle it doesn't matter.

[quote="liangchene"]Consider 10 people sitting around a circular table. How many differnt wayz can they change seats so that each person has a different neighbor to the right?
canbody can any idea to do this question?[/quote]

orangejuice and Palytoxin, both of you are ignoring the fact that each person must have a different neighbor to the right.

I know.  I was trying to back you up. :lol:

It is an old post but thought it might complete the thread if I posted the answer I got based on my solution which, coincidentally, is same as the one in this link [url]http://homepage.math.uiowa.edu/~sokratov/2008m150/derangements.pdf[/url].

It is a nonconsecutive circular permutation expression (look for expression for $C_n$ in Theorem 7, page 2 of the above pdf file). For n=10, you get 120288 as the total number of such nonconsecutive circular permutations.

You can find the same expression in the following pdf file as well (look at example 6.3).
[url]http://www.math.ust.hk/~mabfchen/Math391I/Inclusion-Exclusion.pdf[/url]


Show that $\frac{x_1}{\sqrt {1-x_1}} + \frac{x_2}{\sqrt {1-x_2}} + . . . + \frac{x_n}{\sqrt {1-x_n}}\geq \frac{\sqrt{x_1}+\sqrt{x_2} + . . . + \sqrt{x_n}}{\sqrt{n-1}}$ for any real positive numbers such that $x_1+x_2+...+x_n=1$ and $n\geq 2$

[i]Edited by Myth[/i]

It was given in China before and in any case it was already discussed. Use Jensen to minorize the left hand side and Cauchy for the right hand side. Or directly Titu's lemma for the left hand side.

Maybe RaduSorici will be so kind to correct his post?
I see only something like
\frac{\sqrt{x_1}+\sqrt{x_2} + . . . + \sqrt{x_n}}{\sqrt{n-1}}" alt="\frac{x_1}{\sqrt {1-x_1}} + \frac{x_2}{\sqrt {1-x_2}} + . . . + \frac{x_n}{\sqrt {1-x_n}}\frac{\sqrt{x_1}+\sqrt{x_2} + . . . + \sqrt{x_n}}{\sqrt{n-1}}" style="border: 0; vertical-align: middle;" onclick="newWindow=window.open('http://www.artofproblemsolving.com/Forum/latexrender/latexcode.php?code=%5Cdisplaystyle+%5Cfrac%7Bx_1%7D%7B%5Csqrt+%7B1-x_1%7D%7D+%2B+%5Cfrac%7Bx_2%7D%7B%5Csqrt+%7B1-x_2%7D%7D+%2B+.+.+.+%2B+%5Cfrac%7Bx_n%7D%7B%5Csqrt+%7B1-x_n%7D%7D%26gt%3B%3D%5Cfrac%7B%5Csqrt%7Bx_1%7D%2B%5Csqrt%7Bx_2%7D+%2B+.+.+.+%2B+%5Csqrt%7Bx_n%7D%7D%7B%5Csqrt%7Bn-1%7D%7D','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=375,height=250,left=200,top=100');" />

Indeed, using "< =" may cause [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27508]problems[/url]. Use "\geq" or "\leq" instead. And please preview before posting.


Two ping pong players, $A$ and $B$, agree to play several games. The players are evenly matched; suppose, however, that whoever serves first has probability $P$ of winning that game (this may be player $A$ in one game, or player $B$ in another). Suppose $A$ serves first in the first game, but thereafter the loser serves first. Let $P_{n}$ denote the probability that $A$ wins the $n$th game. Show that $P_{n+1}=P_{n}(1-P)+(1-P_{n})P$.

[hide]
Before winning game $n+1$, Person A can either have won or lost game $n$. $P_{n}(1-P)$ represents the probability that Person A wins game $n$, multiplied by the probability of Person A winning game $n+1$, which is equal to $1-P$ since Person B will serve first in game $n+1$. $(1-P_{n})P$ is the probability that Person A will lose game $n$, and then win game $n+1$. The sum of the two is then the probability that Person A wins game $n+1$.
[/hide]

Thanks, Adunakhor. That's exactly what I did in my solution, though I wasn't very confident about it, because it seemed to be too simple, which isn't typical to the most of other problems from the same book. ;-)


If $ f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $ f(x) \plus{} f(y) \minus{} f(x\plus{}y) \in \mathbb{Z}$ for all $ x,y \in \mathbb{R}$, prove that $ f(x) \equal{} ax \plus{} b$ for some $ a \in \mathbb{R}, b \in \mathbb{Z}$

[quote="bakerbakura"]If $ f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $ f(x) \plus{} f(y) \minus{} f(x \plus{} y) \in \mathbb{Z}$ for all $ x,y \in \mathbb{R}$, prove that $ f(x) \equal{} ax \plus{} b$ for some $ a \in \mathbb{R}, b \in \mathbb{Z}$[/quote]
$ f(x) \plus{} f(y) \minus{} f(x\plus{}y) \equal{} n \in \mathbb{Z}$
Consider the function $ g(y) \equal{} f(x) \plus{} f(y) \minus{} f(x\plus{}y)$. Since $ g(y) \in \mathbb{Z}$ and $ g$ is continuous $ g(y)$ must be constant. So $ g(y) \equal{} g(0) \equal{} f(0)$. Hence $ f(x) \plus{} f(y) \minus{} f(x\plus{}y) \equal{} f(0) \in \mathbb{Z} \forall x,y \in \mathbb{R}$.
Let $ g(x) \equal{} f(x) \minus{} f(0)$. Then $ g(x) \plus{} g(y) \equal{} g(x\plus{}y)$. And since $ g(x)$ is continous, we see that $ g(x) \equal{} ax, a \in \mathbb{R}$. Let $ b \equal{} f(0)$, then $ f(x) \equal{} ax \plus{} b$. And since $ b \equal{} f(x) \plus{} f(y) \minus{} f(x\plus{}y) \in \mathbb{Z}$, we are done.


chess- just make sure it doesn't trick u

i have reversi and backgammon  :lol:

[quote="nutz_for2.718281828"]Do you have some? I have checkers, chess, backgammon, mancala, 4-in-a-row, tiic-tac-toe, and reversi. My favorite is mancala.

P.S. Do you have some strategies on each level?[/quote]

I own in mancala I haven't lost in a couple years.


Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\angle BAC$ is acute.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

Actually, the hypothesis that $PA,PB,PC$ are sides of a triangle is not really needed. We can prove the following: if $\angle BAC\ge\frac\pi 2$, then $PA^2\le PB^2+PC^2\ (*)$ for all points $P$ in the plane.

Suppose the contrary. Divide the plane into four regions by drawing the perpendicular bisectors of $AB,AC$. If $(*)$ doesn't hold, then, since $PA>PB,PC$, and we're assuming that $\angle BAC\ge\frac\pi 2,\ P$ must lie in the region completely contained in the semiplane bounded by $BC$ which does not contain $A$. Now keep $P,B,C$ fixed, and push $A$ along $(PA$ until $\angle BAC=\frac\pi 2$. $(*)$ still doesn't hold, so the problem is reduced to the case when $\angle BAC=\frac\pi 2$.

Let the parallel through $P$ to $AB$ and the parallel through $A$ to $PB$ intersect in $D$, and let $T=AC\cap PD$. We have $PA^2=AT^2+PT^2$ and $PB^2+PC^2=AD^2+PC^2=(AT^2+DT^2)+(PT^2+TC^2)$, which is $\ge AT^2+PT^2$, meaning that $(*)$ does hold, and contradicting our assumption.

Place $ABC$ on the coordinate plane.

WLOG, assume $A$ be the point $(0,0)$ and $B$ be $(1,0)$.

Let $C$ be $(x,y)$.

We want to show that $x > 0$.

Let $P$ be $(a,b)$.

By the given condition, we have $PA^2 > PB^2+PC^2$, giving

$a^2+b^2 > (a-1)^2+b^2+(a-x)^2+(b-y)^2$.
$0 > (a-1)^2+x(x-2a)+(b-y)^2$.

Now suppose $x \le 0$. The $(b-y)^2$ term can be arbitrarily small, so we have

$(a-1)^2+x(x-2a)+(b-y)^2 \ge (a-1)^2+x(x-2a) = a^2-(2+2x)a+(1+x^2)$.

The discriminant of this (as a quadratic in $a$) is $(2+2x)^2-4(1+x^2) = 8x \le 0$, so

$0 > (a-1)^2+x(x-2a)+(b-y)^2 \ge a^2-(2+2x)a+(1+x^2) \ge 0$

since it can have at most one root for $a$.

But that gives a contradiction, so $x > 0 \Rightarrow \angle BAC$ is acute.

this is one of my first solutions using vectors, and i think it's nice  :D

Let $A$ be the origin. then, we have that $PA^{2}>PB^{2}+PC^{2}$ is equivalent to
$\overrightarrow{P}^{2}>(\overrightarrow{P}-\overrightarrow{B})^{2}+(\overrightarrow{P}-\overrightarrow{C})^{2}$

which is equivalent to

$2\overrightarrow{B}\cdot\overrightarrow{C}>\overrightarrow{P}^{2}+\overrightarrow{B}^{2}+\overrightarrow{C}^{2}-2\overrightarrow{P}\cdot\overrightarrow{B}-2\overrightarrow{P}\cdot\overrightarrow{C}+2\overrightarrow{B}\cdot\overrightarrow{C}=(\overrightarrow{P}-\overrightarrow{B}-\overrightarrow{C})^{2}\geq 0$

which implies that angle $BAC$ is acute.

Let $ PA,PB,PC$ be $ a,b,c$ respectively, and $ AB,BC,CA$ be $ z,x,y$ respectively.

Assume for contradiction that $ \angle BAC$ is not acute, that is, that $ x^2 \ge y^2 \plus{} z^2$. By the condition in the problem, $ a^2 > b^2 \plus{} c^2$. Therefore, by Cauchy-Schwarz, $ a^2x^2 > (b^2 \plus{} c^2)(y^2 \plus{} z^2) \ge (by \plus{} cz)^2 \implies ax > by \plus{} cz$.

But by Ptolemy's Inequality on quadrilateral $ ABPC$, $ ax \le by \plus{} cz$; contradiction.

We will show that $ AP^{2}\plus{}BC^{2}\le AB^{2}\plus{}AC^{2}\plus{}BP^{2}\plus{}CP^{2}$. Note that this will immediately imply that $ \angle{BAC}$ is acute, since we have $ PA^{2}>PB^{2}\plus{}PC^{2}$, which means that $ BC^{2}<AB^{2}\plus{}AC^{2}$, or $ \angle{A}$ is acute.

The claim is simple with coordinates  :wink: . Apply the appropriate transformations preserving the sign of $ AB^{2}\plus{}AC^{2}\plus{}BP^{2}\plus{}CP^{2}\minus{}AP^{2}\minus{}BC^{2}$ remains the same, so that $ A\equal{}(0,0),B\equal{}(0,1),C\equal{}(a,b),P\equal{}(c,d)$. We compute this to be

$ 1\plus{}(a^{2}\plus{}b^{2})\plus{}(c^{2}\plus{}(d\minus{}1)^{2})\plus{}((a\minus{}c)^{2}\plus{}(b\minus{}d)^{2})\minus{}(c^{2}\plus{}d^{2})\minus{}(a^{2}\plus{}(b\minus{}1)^{2})$
$ \equal{}(a\minus{}c)^{2}\plus{}(b\minus{}d\minus{}1)^{2}\ge0$,

so we're done.

Much like paladin8's solution, let $A$ be $(0,0)$ and $B$ $(1,0)$. Let $C$ be $(m,n)$ with $m,n$ constants, WLOG $n>0$. It suffices to show $m>0$.

The locus of possible points $P$ is the locus of all points such that $PA^2 > PB^2+PC^2$ or

$x^2+y^2>(x-1)^2+y^2+(x-m)^2+(y-n)^2$

Rearranging,

$0>-2x+1+x^2-2mx+m^2+(y-n)^2=m^2-2mx+[(y-n)^2+(x-1)^2]$. Note that this inequality necessarily has solutions, as $P$ is one of them.

By the quadratic formula,
$\frac{2x-\sqrt{4x^2-4[(y-n)^2+(x-1)^2]}}{2}<m<\frac{2x+\sqrt{4x^2-4[(y-n)^2+(x-1)^2]}}{2}$
Clearly, the upper bound is positive (and real, as our inequality has solutions). The lower bound is also real, but we must prove it's positive. The lower bound simplifies to:

$x-\sqrt{x^2-[(y-n)^2+(x-1)^2]}$ which we must prove is greater than 0.

$x-\sqrt{x^2-[(y-n)^2+(x-1)^2]} > 0 \Longleftrightarrow  x \ge \sqrt{x^2-[(y-n)^2+(x-1)^2]} \Longleftrightarrow  (y-n)^2+(x-1)^2>0$
which is always true except when $y=n$ and $x=1$. However, if that is the case, we have $m=0$, so $C$ is located at $(0,n)$. That means that $PB^2+PC^2=n^2+1=PA^2$, meaning the triangle with side lengths $PA,PB,PC$ is not obtuse, contradiction. Thus, $m>0$ and $ABC$ is acute.

WLOG let $B$ be at $(-1,0)$ and $C$ be at $(1,0)$.  Let $P$ be at $(x,y)$ and let $A$ be at $(a,b)$.  Now the condition that $PA$ is equal to the side opposite the obtuse angle is equivalent to $PA^2>PB^2+PC^2$ so
\begin{align*}(x-a)^2+(y-b)^2&> (x+1)^2+y^2+(x-1)^2+y^2 \\
a^2+b^2&>2xa+2yb+x^2+y^2+2 \\
2(a^2+b^2)&>(x+a)^2+(y+b)^2+2\geq2 \\
a^2+b^2&>1 \\
\end{align*}
Thus $(a,b)$'s distance from the origin is greater than $1$, so $A$ is outside the circle with diameter $BC$ therefore $\angle BAC$ is acute.

[hide="Non-coord bash solution"]
Let $A_1, B_1, C_1$ be the feet of the perpendiculars from P to $BC, AC, AB$. Note that $AC_1PB_1$ is a cyclic quadrilateral with diameter $AP$, so by the extended law of sines,
$\frac{a}{\sin A}=2R$
$\frac{B_1C_1}{\sin A}=AP$
$B_1C_1=a\frac{AP}{2R}$
And similarly for the other sides.
Now we have
$AP^2>BP^2+CP^2$
$\frac{B_1C_1^2}{a^2}>\frac{A_1C_1^2}{b^2}+\frac{B_1A_1^2}{c^2}$
Suppose FTSoC that BAC is not acute. Then we have
$a^2\ge b^2+c^2$
$B_1C_1^2>(\frac{A_1C_1^2}{b^2}+\frac{B_1A_1^2}{c^2})(b^2+c^2)$
By cauchy-shwarz, this in turn implies
$B_1C_1^2>(A_1C_1+A_1B_1)^2$
A violation of the triangle inuequality on $A_1B_1C_1$.
[/hide]

There are many fairly simple solutions to this problem, however, I would like to point out that it can be trivialized if one knows a certain ([b]quite useful[/b], in my opinion) [b]lemma[/b].
[hide="Solution"][i][u]Lemma.[/u][/i]
If $x, y, z$ are real numbers, and $A, B, C$ are the angles of a triangle, then \[x^2 + y^2 + z^2 \ge 2xy\cos A + 2yz \cos B + 2zx \cos C.\]
[i][u]Proof.[/u][/i] (Note, this can also be proved using Jensen's Inequality, and I believe there is also a (much nicer) vector argument)
We rewrite the inequality as a quadratic in terms of $x$: \[x^2 - 2x(y\cos A + z\cos C) + y^2 + z^2 - 2yz \cos B \ge 0.\] By completing the square, we rewrite again, as \[(x - (y\cos A + z\cos C))^2 + y^2 + z^2 - 2yx \cos B \ge (y \cos A + z \cos C)^2.\] But clearly our first square term is nonnegative. Therefore it suffices to show that \[y^2 + z^2 - 2yx \cos B \ge (y \cos A + z \cos C)^2\]\[\iff y^2 + z^2 \ge y^2 \cos^2 A + z^2 \cos^2 C + 2yz \cos A \cos C + 2yz \cos B\]\[\iff y^2(1 - \cos^2 A) + z^2(1 - \cos^2 C) \ge\]\[\ge 2yz \cos A \cos C + 2yz \cos (180^{\circ} - A - C)\]\[\iff y^2 \sin^2 A + z^2 \sin^2 C \ge 2yz \cos A \cos C - 2yz \cos (A + C)\]\[\iff y^2 \sin^2 A + z^2 \sin^2 C \ge 2yz \sin A \sin C\] which is true, by AM-GM. This concludes the proof of lemma.

Now, we solve the problem, using our lemma.
Suppose, for the sake of contradiction, that $\angle BAC = A$ is obtuse.
Since $PA, PB, PC$ make up an obtuse triangle, and the obtuse angle is opposite the side congruent to $\overline{PA}$, we see that $\overline{PA}$ has the greatest side length in the obtuse triangle. Hence $PA^2 > PB^2 + PC^2.$
By the Law of Cosines on $\triangle ABP, \triangle ACP$, we rewrite this as \[PA^2 > (PA^2 + c^2 - 2PA \cdot c \cdot \cos \angle PAB) + (PA^2 + b^2 - 2PA \cdot b \cdot \cos \angle PAC)\]\[\implies 2(PA \cdot b) \cos \angle PAC + 2(PA \cdot c) \cos \angle PAC > PA^2 + b^2 + c^2. \; \text{(1)}\]
But this is an immediate contradiction by our lemma. We let our three real numbers be $PA, b, c$ and our three angles with sum $180^{\circ}$ be $\angle PAB, \angle PAC, (180^{\circ} - A).$ Then the lemma implies that \[{PA^2 + b^2 + c^2 \ge 2(PA \cdot b) \cos \angle PAC + 2(PA \cdot c) \cos \angle PAC + 2bc\cos (180^{\circ} - A}). \; \text{(2)}\]
But $90^{\circ} < A < 180^{\circ} \implies \cos (180^{\circ} - A)  > 0.$ Therefore $2bc\cos (180^{\circ} - A) > 0 \implies$ \[{2(PA \cdot b) \cos \angle PAC + 2(PA \cdot c) \cos \angle PAC + 2bc\cos (180^{\circ} - A}) >\]\[> 2(PA \cdot b) \cos \angle PAC + 2(PA \cdot c) \cos \angle PAC. \; \text{(3)}\]
But combining $(3)$ with $(2)$, we see that \[PA^2 + b^2 + c^2 > 2(PA \cdot b) \cos \angle PAC + 2(PA \cdot c) \cos \angle PAC\] which contradicts $(1).$
Since we have arrived at a contradiction, we conclude that our hypothesis was flawed, and hence $\angle BAC = A < 90^{\circ}.$
The proof is complete.[/hide]

We show the contrapositive.
Assume that $ABC$ is obtuse at $A$, we will show that, for any point $P$, $PA^2\le PB^2+PC^2$. Note that
\[
PC^2=AC^2+PA^2-2(PA)(AC)\cos{PAC}\\
PB^2=AB^2+PA^2-2(PA)(AB)\cos{PAB}
\]
Then,
\[
PC^2+PB^2=AB^2+AC^2+2PA^2-2(PA)(AC)\cos{PAC}-2(PA)(AB)\cos{PAB}.
\]
We wish, then, to show that
\[
AB^2+AC^2+PA^2\ge2(PA)(AC)\cos{PAC}+2(PA)(AB)\cos{PAB}.
\]
Note that, because $\angle PAB+\angle PAC=\angle BAC$ is obtuse,
\begin{align*}
&& \cos{(PAB+PAC)} &< 0\\
\Rightarrow\;\; && \cos{PAB}\cos{PAC} &< \sin{PAB}\cos{PAC}\\
\Rightarrow\;\; && \cos^2{PAB}\cos^2{PAC} &< \sin^2{PAB}\cos^2{PAC}\\
&& &=(1-\cos^2{PAB})(1-\cos^2{PAC})\\
&& &=1-(\cos^2{PAB}+\cos^2{PAC})+\cos^2{PAB}\cos^2{PAC}\\
\Rightarrow\;\; && \cos^2{PAB}+\cos^2{PAC} &<1.
\end{align*}
Thus, we can find an $x$ such that
\[
\cos^2{PAB}\le x\\
\cos^2{PAC}\le 1-x
\]
whence we find that

\[
\cos{PAB}\le \sqrt{x}\\
\cos{PAC}\le \sqrt{1-x}
\]
Finally, we use AM-GM to get
\begin{align*}
AB^2+xPA^2+AC^2+(1-x)PA^2 &\ge 2(AB)(AP)\sqrt{x}+2(AC)(AP)\sqrt{1-x}\\
&\ge 2(AB)(AP)\cos{PAB}+2(AC)(AP)\cos{PAC}
\end{align*}
and we conclude.

One more solution (I think this works?):

Let $x, y, z$ be the lengths of $PA, PB, PC$ respectively.  Fix $P, x, y, z$.  Then note that $A, B, C$ lie on the three circles centered at $P$ with radii $x, y, z$ respectively.  In order to maximize $\angle BAC$, we choose A arbitrarily and then B and C to maximize $\angle BAP$ and $\angle CAP$ by making $AB$ and $AC$ tangent to the circles of radius $y$ and $z$ respectively.  Then it suffices to show that $\angle BAP + \angle CAP < 90$, which is equivalent to $\sin^-1(y/x)+\sin^-1(z/x) < 90$.  This can be seen by considering the triangle $XYZ$ with sidelengths $x, y, z$ (opposite the corresponding vertices per the standard notation); because $\angle YXZ$ is obtuse, we have $\sin^-1(y/x)+\sin^-1(z/x) < \angle XYZ + \angle XZY < 90$, and we are done. $\blacksquare$

[hide=Solution]
Assume that $\angle{A}$ is not acute. Then $a^2\geq b^2+c^2$. Thus, $$\frac{\left(a^2-b^2\right)PB^2+\left(a^2-c^2\right)PC^2}{2}\geq PB\cdot PC\sqrt{\left(a^2-b^2\right)\left(a^2-c^2\right)}\geq bc\cdot PB\cdot PC,$$ so rearranging, we get $$PB^2+PC^2\geq\left(\frac{PB\cdot b+PC\cdot c}{a}\right)^2.$$ But the RHS is $\geq PA^2$ by Ptolemy's Inequality, so $$PB^2+PC^2\geq PA^2.$$ Thus, the triangle formed by $PA$, $PB$, and $PC$ is not obtuse, contradiction. Thus, $\angle{BAC}$ is acute.

Q.E.D.
[/hide]

[hide=Solution]Note the condition is equivalent to $PA^2>PB^2+PC^2$. For the sake of contradiction, suppose that $\triangle ABC$ is not acute and that $BC^2\ge AB^2 + AC^2$. Multiplying the two inequalities, it follows that $$(PA\cdot BC)^2>(PB^2+PC^2)(AB^2+AC^2)\ge (PB\cdot AC + PC\cdot AB)^2.$$ However, by Ptolemy's Inequality in quadrilateral $ACPB$ we have that $AB\cdot PC +  AC\cdot PB \ge AP\cdot BC$, a contradiction.[/hide]

someone told me a solution that seems too easy,
If $\angle{BAC}>90$, and $PA,PB,PC$ forms a obtuse triangle with $PA$ being the opposite side of
the obtuse angle, $PA>PB,PA>PC$ so $$\angle{PCA}>\angle{PAC},\angle{PBA}>\angle{PAB}$$
$$\angle{PCA}+\angle{PBA}>\angle{PAC}+\angle{PBC}=\angle{BAC}$$,
=> $\angle{BCA}+\angle{CBA}>\angle{BAC}$ which is a contradiction.

Using Ptolemy's inequality and Cauchy-Schwarz, \begin{align*} 	PA \cdot BC 	&\le PB \cdot AC + PC \cdot AB \\ 	&\le \sqrt{(PB^2+PC^2)(AB^2+AC^2)} \\ 	&< \sqrt{PA^2 \cdot (AB^2+AC)^2} = PA \cdot \sqrt{AB^2+AC^2} \end{align*} meaning $BC^2 < AB^2+AC^2$, so $\angle BAC$ is acute.

Is this solution correct?
Because of the obtuse angle we know that $PA>PB$ and $PA>PC$, thus we know that $P$ lies on the side of the perpendicular bisectors of $AB$ and $AC$ which don't contain $A$ and at the same time it lies in the plane of the triangle $ABC$, which obviously is impossible when $\angle BAC\ge 90^{\circ}$(by seeing where is the circumcenter) .

Let $\angle BPC=\alpha,\angle CPA=\beta,\angle APB=\gamma$. Then we need to show
\[PB^2+PC^2-2PB\cdot PC\cos\alpha=BC^2<AB^2+AC^2=2PA^2+PB^2+PC^2-2PB\cdot PA\cos\gamma-2PC\cdot PA\cos\beta\iff\]
\[0<4PA^2-4PC\cdot PA\cos\beta-4PB\cdot PA\cos\gamma+4PB\cdot PC\cos\alpha = \]
\[(2PA-PC\cos\beta-PB\cos\gamma)^2+4PB\cdot PC\cos\alpha-(PC\cos\beta+PB\cos\gamma)^2.\]
It is clear that $PA>(PB+PC)/2$, so we can instead show an inequality with $PA=\sqrt{PB^2+PC^2}$. By homogenizing we can assume $PA=1$ and let $PB=y,PC=z$ with $y^2+z^2=1$. Then the desired inequality is
\[1\ge z\cos\beta+y\cos\gamma-yz\cos\alpha=z\cos\beta+y\cos\gamma-yz\cos(\beta+\gamma).\]
With respect to $\beta$, the derivative of this expression is $-z\sin\beta+yz\sin(\beta+\gamma)$. Similarly, with respect to $\gamma$, the derivative of this expression is $-y\sin\gamma+yz\sin(\beta+\gamma)$. If one of $y,z$ is equal to zero the inequality is obvious, so suppose neither is. Then we must have $\sin \beta=y\sin(\beta+\gamma),\sin\gamma=z\sin(\beta+\gamma)$. Let $p=e^{i\beta},q=e^{i\gamma}$ so we have the equations \[p-1/p=y(pq-1/pq)\implies p^2q-q=z(p^2q^2-1),q-1/q=z(pq-1/pq)\implies pq^2-p=z(p^2q^2-1).\]
Since this is a system of two quadratic equations, it has at most four solutions: as $\sin$ is cyclic mod $\pi$, we can restrict our search to choices of $\beta,\gamma\in [-\pi/2,\pi/2]$. Moreover, if $\beta=0$ or $\gamma=0$ the equation is not quadratic (its only solution is actually $\gamma=0$) so we only expect three solutions. If $y=\sin\theta,z=\cos\theta$ with $\theta\in (0,\pi/2)$ we expect the solutions to be $(\beta,\gamma)=(0,0),(\theta,\pi/2-\theta),(-\theta,\theta-\pi/2)$. Since the last one is just the second with the sign flipped and $\cos$ does not care about signs, we have two cases. We need to show $1\ge y+z-yz$ and $1\ge y^2+z^2$, the latter of which is trivially true. To see the first, write it suffices to show
\[1\ge y+z-yz\iff 0\ge -(1-y)(1-z),\]
which is trivial.

Assume for the sake of contradiction that $\angle BAC$ is not acute, so $BC^2 \ge AB^2+AC^2$. Multiplying this with $PA^2 > PB^2+PC^2$, we have
\[(PA \cdot BC)^2 > (PB \cdot AB)^2+(PB \cdot AC)^2+(PC \cdot AB)^2+(PC \cdot AC)^2. \qquad \qquad (1)\]
By Ptolemy's Inequality, we have $PA \cdot BC \le PC\cdot AB+PB \cdot AC$, so
\[(PA\cdot BC)^2 \le (PC\cdot AB)^2+(PB\cdot AC)^2 +2(PC\cdot PC \cdot AB \cdot AC).\]
Combined with $(1)$,
\[(PC\cdot AB)^2+(PB\cdot AC)^2 +2(PC\cdot PC \cdot AB \cdot AC) >(PB\cdot AB)^2+(PB\cdot AC)^2+(PC \cdot AB)^2+(PC \cdot AC)^2,\] which means that
\[2 PC \cdot PC \cdot AB \cdot AC > (PB \cdot AB)^2+(PC \cdot AC)^2.\]

However this inequality is false by AM-GM so we have a contradiction. Hence $\angle BAC$ must be acute.

QM-AM and Median Length Formula when $AP$ passes through the midpoint $M$ of $BC$, $2(AM^2+PM^2)>PB^2+PC^2$ implies $AB^2+AC^2>BC^2$

Suppose that $\angle BAC$ is obtuse. Let $A=(0,0), B=(x_b,y_b), C=(x_c,y_c)$ and $P=(x_p,y_p)$, with $x_b,y_b>0$ and $x_c,y_c<0$. We have that:
$$BC^2>AB^2+AC^2\implies x_a^2+y_a^2>(x_p-x_b-x_c)^2+(y_p-y_b-y_c)^2-2(x_bx_c+y_by_c) (i).$$
The statement gives us that $PA^2>PB^2+PC^2\implies-(x_bx_c+y_by_c)>x_a^2+y_a^2 (ii)$. With $(i)$ and $(ii)$:
$$x_bx_c+y_by_c>(x_p-x_b-x_c)^2+(y_p-y_b-y_c)^2\geq0$$
which is a contradiction. $\blacksquare$

Let $M, N$ be midpoints of $AP$ and $BC$, respectively. For the points $A, B, P, C$; let's apply [color=#f00][b]Euler's quadrilateral formula[/b][/color],
$$ AB^2 + BP^2 + PC^2 + CA^2 = AP^2 + BC^2 + 4MN^2 \geq AP^2 + BC^2 .$$
Given that $AP^2 > BP^2 + PC^2$. Thus,
$$ AB^2 + AC^2 > BC^2 .$$
and we get $\angle BAC$ is acute.

(Lokman GÖKÇE)

[quote=Mogmog8]Notice $AP^2>BP^2+CP^2$ and $AB\cdot CP+AC\cdot BP\ge AP\cdot BC.$ Suppose FTSOC that $\angle BAC\ge 90.$ Then, $BC^2\ge AB^2+AC^2.$ Hence, $$(AB\cdot CP+AC\cdot BP)^2\ge (AP\cdot BC)^2>(BP^2+CP^2)(AB^2+AC^2),$$ a contradiction by Cauchy-Schwarz. $\square$[/quote]

Regarding this: Based on my diagram, it seems like it should be quadrilateral APCB, so $AB \cdot CP + AP \cdot BC \geq AC \cdot BP$. Am I being dumb? (I have P to the right of triangle ABC, and it does seem to satisfy PA > PB > PC).

By the condition, $AP^2>BP^2+CP^2.$ By Ptolemy, $AP\cdot BC \le AC\cdot PB + AB\cdot PC.$ By C-S inequality, \[(AC^2+AB^2)(PB^2+PC^2)\ge (AC\cdot PB+AB\cdot PC)^2 \ge AP^2 \cdot BC^2.\] Since $BP^2+CP^2< AP^2$ we must have $BP^2+CP^2> BC^2$ which is what we wanted to prove.

Assume by way of contradiction that $PA^2 > PB^2 + PC^2$ and simultaneously $A$ is obtuse, i.e., $BC^2 > AB^2 + AC^2$. Multiplying these two inequalities provides us with \[PA^2 \cdot BC^2 > AC^2(PB^2 + PC^2) + AB^2(PB^2 + PC^2).\] However by Ptolemy's theorem on any four arbitrary points, we have that
\begin{align*}
PA \cdot BC &\leq AC \cdot BP + AB \cdot CP \\
\implies PA^2 \cdot BC^2 &\leq (AC \cdot BP + AB \cdot CP)^2 \\
\end{align*}
However combining our two results gives us that \[AC^2(PB^2 + PC^2) + AB^2(PB^2 + PC^2) < (AC \cdot BP + AB \cdot CP)^2 \iff AC^2 \cdot PC^2 + AB^2 \cdot PB^2 < 2AC \cdot BP \cdot AB \cdot CP.\] However by AM-GM on the LHS of the inequality, we obtain that $2AC \cdot PC \cdot AB \cdot PB < 2AC \cdot PC \cdot AB \cdot PB$, which is a contradiction. Therefore we cannot both have $PA^2 > PB^2 + PC^2$ and $BC^2 > AB^2 + AC^2$. $\blacksquare$

Suppose otherwise. Then $AB^2+AC^2 > BC^2$ and $PA^2 > PB^2+PC^2$. But by Ptolemy's inequality, $$(PB^2+PC^2)(AB^2+AC^2) < (PA)^2 (PC)^2  \leq (AB \cdot PC + AC \cdot PB)^2.$$ This is a contradiction by C-S.

Let $a$, $b$ and $c$ be positive reals such that $a^2 > b^2 + c^2$ – then, we assign $PA = a$, $PB = b$ and $PC = c$. Imagine fixing $A$ and $P$ while varying $B$ and $C$. In the worst case situation, both $\angle PAC$ and $\angle PAB$ are maximized. This happens when $\angle PBA = \angle PCA = 90^{\circ}$ (consider drawing circles of radius $b$ and radius $c$ centered at $P$.) Even in this worst case situation, we have
\[ AB^2 + AC^2 = (a^2 - b^2) + (a^2 - c^2) > b^2 + c^2 = PB^2 + PC^2,\]
and since $ABPC$ is cyclic, this implies that $\angle BAC$ is acute.


i want to get all problems of AMC,AIME,USAMO from http://www.kalva.demon.co.uk/,but i can't get on it!if you can go,please put them on this forum.thank you!(all all problems of AMC,AIME,USAMO )
or you can send them to my e-mail( tiangaorenwofei08@163.com ).

Yeah, Kalva seems to be down.  But by the way, kalva only has AIME's and USAMO's.  There does not seem to be any comprehensive free source of AMC's and AHSME's.

I understand that it was urgent to you, but please do not double post.

(I saw this post in "Other problem solving topics" forum)

The attached file is all the files I could possibly get off of that site...

They're AIME..

I think I posted this before though.


This is our Team

1. George Arabidze
2. George Nadiradze (HM IMO 2006, 14 pts, and 529!=23^2  :blush: )
3. Rati Gelashvili (Bronze Medal IMO 2006, 18 pts)
4. Beqa Ergemlidze
5. Levan Varamashvili
6. Nika Machavariani

GEO is ready to fight for 3 medals better than bronze this year

But clearly we need to know perfect cubes by heart  :D

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

Gold Or Nothing

Congratulations to Levani and Beqa !!!   :lol:
gelash, do you know , is Levani and Beqa in Georgian IOI team, too ?

Good luck !!!

Levan is in IOI team

This is another Beqa I guess :D

We took gold, silver, bronze and 3 honourable mentions,

good result I guess

Congratulations guys,

Thats a fantastic result,


Find the smallest positive integer $n$ such that \[2^{1989}\; \vert \; m^{n}-1\] for all odd positive integers $m>1$.

Or we can use LTE:
If $n$ odd then $v_2(m^n-1)=v_2(m-1) \ge 1989$, a contradiction. For example, $m=3$. Thus, $n$ is even.
We get $v_2(m^n-1)=v_2(m-1)+v_2(m+1)+v_2(n)-1 \ge 1989$.
Since for all odd positive integers number $m>1$ then $2^{1989}|m^n-1$ then we must have $v_2(n) \ge 1989$. It follows that $n=2^{1989} \cdot k$ with $k \in \mathbb{N}^*$.
Thus, $n=2^{1989}$ is the smallest of $n$.

More general: If $2^k|m^n-1$ then $2^k|n$ with $m$ is odd.

Min ord(m－1)＋ord(m＋1)－1 is $2$.So min n is $2^{1987}$.:)


$ x\ln x \minus{} x \plus{} \frac{1}{2} \equal{} 0$

I'm not completely sure but your equation might not be "solvable" in the simple sense of the word. I don't see any clever manipulations or strategies that might help. I got the equation down to $ x^x \equal{} e^x/(e^(1/2))$ and didn't down how to proceed further since there was a variable in both the exponents and the base of the equation

im absolutely terrible at logs, i only know the basic properties but i have no idea how to proceed from that point.

@bane -- are you asking for help or are you trying to challenge us.  If you are trying to challenge us the   answer would be very helpfull!

I don't see how to get an exact answer. My TI-89 gave approximate solutions:
[hide]$ x\equal{}0.1866823089, 2.155535204$[/hide]

[quote="alanchou"][quote="bane"]$ x\ln x \minus{} x \plus{} \frac {1}{2} \equal{} 0$[/quote]

All I see is an equation.  Are we supposed to solve for $ x$, $ x^2$, $ \ln x$, or what?[/quote]
I'm assuming for $ x$?

Most transcendental equations don't have closed-form solutions.  I'm not sure what you're trying to accomplish here.

No, it's not a challenge for you:) I was just curious if there is any algebraic way to solve it, but as I see you share my doubt.

@alanchou: yes, for $ x$, but you can solve it for whatever you've mentioned :wink:

[quote="bane"]$ x\ln x \minus{} x \plus{} \frac {1}{2} \equal{} 0$[/quote]
[hide="Solution"]
Applying properties of exponents, isolating the variable,  and putting the "base e" there, we have $ \ln_e x^x\equal{}\minus{}\frac{1}{2}$
Using more properties of logs, we get $ x^x\equal{}e^\minus{}\frac{1}{2}$
You can use a calculator to get what 123456789 said. [/hide]

MrRubix, for your exponent, in Latex, put braces around it so it looks like an exponent.

[quote="MrRubix"][quote="bane"]$ x\ln x \minus{} x \plus{} \frac {1}{2} \equal{} 0$[/quote]
[hide="Solution"]
Applying properties of exponents, isolating the variable,  and putting the "base e" there, we have $ \ln_e x^x \equal{} \minus{} \frac {1}{2}$
Using more properties of logs, we get $ x^x \equal{} e^ \minus{} \frac {1}{2}$
You can use a calculator to get what 123456789 said. [/hide][/quote]

Hey, wait.  Where did that middle term go?


$\text{Let}\; x,y,z\; \text{be positive real numbers such that}\; x^2+y^2+z^2=3,\; \text{ prove that}$
$x^3y^2+y^3z^2+z^3x^2+xyz\leq4,\;\;\text{and}\;\;\; x^4y+y^4z+z^4x+xyz\leq4.$

[quote="t0rajir0u"]We should prove a stronger inequality, $x^5 + y^5 + z^5 + xyz \le 4$, which implies both by Rearrangement, but I haven't the time to do it  :([/quote]
are you sure?what about $x^2=3,y=z=0$???

Come on, come on. It is not hard to "invent" such kind of inequalities but it is certainly not easy to prove them at all. Someone may say that Murhead or Schur can work for these. Oh, I do not like using Murhead or Schur very much.

[quote="ghjk"]ta co:xyz=<1
suy ra:4-xyz>=3
bay gio ta chung minh:3>=x^3*y^2+y^3*z^2+z^3*x^2
x^2+y^2+z^2>=VP(DUNG vi x,y,z=<1)[/quote]

No Vietnamese here. Your proof is NOT correct.

[quote="neworder"][quote]are you sure?what about x^2=3, y=z=0[/quote]

x,y,z are positive reals, so y, z can't be 0.[/quote]
But we can take $y=z\to 0$ and $x\to \sqrt3$ :)

Sorry, sorry, that inequality doesn't work  :blush:


[quote="mathking123"][quote="mathking123"]The only rule is that you must quote the person above you and add a line to it.[/quote] waiting for someone to join...[/quote]c'mon people...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242339

That was pretty similar to this one, except it was description instead of adding a line. Same thing.

[quote="evilhamster"]These kinds of things usually get locked by mods.[/quote]

Good call. [b]Locked[/b].


integration
algebra
polynomial
real analysis
real analysis unsolved

Let $ f \in L^1[0,1]$. Suppose that

$ \int_{[0,1]}f(y)\sin(xy)dy\equal{}0$ for all $ x \in[0,1]$. Show that $ f \equal{} 0$ a.e.

have you noticed that $ \int_{[0;1]} P(y^2)f(y) \sin(xy) dy \equal{} 0$ for any polynomial $ P \in \mathbb{R}[X]$ ?

why is that? Can you be a bit more elaborate?

Just differentiate with respect to $ x$ a few times. ;)


While that video looks funny, please note that downloading it will result in a new toolbar appearing on your browser and getting lots and lots of popup ads.


Another problem:
Prove that:
$a^2 + b^2 + c^2 \ge 4\sqrt{3}S + (a - b)^2 + (b - c)^2 + (c - a)^2$

The second one is a theorem somewhere.... i wont ruin it by posting a solution

[quote="bénabar"][quote="Huynh Anh Hao"]Let a triangle. Prove that:
$\frac{1}{cosA + cosB + cosC} \ge \sum \frac{1}{cosA + 1}$[/quote] No condition on $A$? :?[/quote]
Yes!!! ABC is an arbitraty triangle.


Do you think that Runescape is dum or kewl?

well, i've only played runescape for about a week a year ago, but, i found it kind of boring.

its kindof boring, because it takes so long to do everything. :mad:

i played runescape before, but it got really boring after a while :P

it takes like 10 minutes to get from one corner of the map to the other.

Yeah...Runescape needs more...eh...stuff. *nods* Tell them to build more. Otherwise, it's not a bad game really. Quite addictive after a while.

well, the game has to be at least somewhat good if so many people are willing to pay money to play it.

they should make superspeed cheatcodes in runescape, or make you be able to buy transports with gold. :(

[quote]they should make superspeed cheatcodes in runescape, or make you be able to buy transports with gold. [/quote]

if they were to make cheat codes for transportation, they would have to make cheat codes for everything else too, wouldn't they? ;)

its only fun when u play with your friends at the same time

[quote="h78reg"][quote]it would be OK if they didn't make it take so long to do everything. I think that they may try to make it that way so that people play more.[/quote]

well, they already have millions of people playing it. i don't think they're THAT interested in getting more users.[/quote]

I didn't mean more people, I meant that people who already play would play more.


if $s(n)$  is the sum of digits of $n$ prove for any $n>2$ $s(9^n)>9$. :blush:

$s(n)=0(mod \ 9).$ If s(n)=9 then n is even. Let $s_0(n)$ is sum digits for $10^0,10^2,10^4, \dots$ and $s_1(n)$ is sum digits for $10^1,10^3,10^4,\dots$. We have $s(n)=s_0(n)+s_1(n), s_0(n)-s_1(n)=(-2)^n(mod \ 11)$. If s(n)=9, then 2|n and it give contradition.


Find the values of the constants a and b that make the function continuous everywhere

x+6 if x less than or equal to -1
ax^2+bx if x is greater than -1, less than 2
2 if x is greater than or equal to 2

Just find the endpoints of the other two pieces of the function, and then find an $a$ and $b$ so that the quadratic connects with those.

[hide]
When $x=-1$, we find that the value of $y$ is $5$. We need $ax^2+bx$ to equal $5$ at $x=-1$. We also need $ax^2+bx$ to equal $2$ at $x=2$

$a(-1)^2+b(-1)=5$ and $a(2)^2+b(2)=2$
$a-b=-1$ and $4a+2b=2$

$a=b-1$
Substitute and,
$a=1,~b=0$[/hide]


trigonometry
calculus
derivative
complex numbers

Find the maximum value of sinx^cosx + cosx^sinx (That's "sinx raised to cosx..")

[hide]How about sin(x) = 2mn/(m^2+n^2) and cos(x) = (n^2-m^2)/(n^2+m^2).  Now lets fix n=1 and let m be an even number and increase to infinity.  Now sinx goes to 0 and cosx goes to -1, and both have odd denominators.  So sinx^cosx becomes arbitrarily large, and cosx^sinx goes to 1, making the sum grow without bound.[/hide]

I tried by choosing sinx = 10^-n and cosx = -sqrt(1-sin^2x)

You get the expression becomes 10^n+1 for large n and hence does not have an upper bound.

By choosing, sinx = -10^-n and a similar choice of x, you get -10^n+1. So it does not have a lower bound.

Well the problem with your way is that cosx^sinx is a negative number raised to a fractional power with even denomionator, and is thus not defined.

How did you choose those other representations of sine and cosine?

well (m^2-n^2 , 2mn, m^2+n^2) is a well known formula for pythagorean triples, so modifying it gives rational values for sine and cosine.

Okay guys.  I think it's pretty obvious that hsbhatt wants a maximum on a certain interval, so we'll let that interval be $ \left[0,\frac{\pi}{2}\right]$ because that's really the only part we can deal with.  Then we could look at the convexity of $ f(x) \equal{} \sin{x}^{\cos{x}}$ and use Jensen's, hopefully.  The important part to realize is that by setting $ y \equal{} \frac{\pi}{2} \minus{} x$, we have $ f(y) \equal{} \cos{x}^{\sin{x}}$.  Thus we have to find a bound on $ f(x) \plus{} f(y)$, which was why we would want to use Jensen's.  (Six "w"s in a row! Yay for alliteration!)

[hide="hrmm..."]
Logarithmic differentiation for the win!
[/hide]

I tried differentiation and, to me, it looks quite messy.  Though, maybe defining the interval will help?

Two doubts: 1.I am told that the problem with using \10^{-n} is that I would have a negative number raised to a power with an even denominator. But in complex numbers wouldnt the limit still go to 1.
2. Could i get around this difficulty by using say 11^-n


algebra
polynomial
function
complex analysis
domain
algebra proposed

$P,Q,R$ are non-zero polynomials that for each $z\in\mathbb C$, $P(z)Q(\bar z)=R(z)$.
a) If $P,Q,R\in\mathbb R[x]$, prove that $Q$ is constant polynomial.
b) Is the above statement correct for $P,Q,R\in\mathbb C[x]$?

Probably I'm blind now, but what is wrong with the following argument: for any $z$ on the unit circle we have $P(z)Q(\frac{1}{z})=R(z)$. This is an identity between rational functions, because it holds for infinitely many $z$. Similarly, by taking $z$ on the circle of radius $2$ we find that $P(z)Q(\frac{4}{z})=R(z)$ for all $z$. Thus because $P$ is nonzero, we deduce that $Q(z)=Q(4z)$ and thus $Q$ is constant. What is wrong with this?

The statement is true in the general case, when $P,Q,R$ are complex polynomials.

In the plane, consider the differential operators $\frac\partial{\partial z}=\frac{1}{2}\left(\frac\partial{\partial x}-i\frac\partial{\partial y}\right)$ and $\frac\partial{\partial \bar z}=\frac{1}{2}\left(\frac\partial{\partial x}+i\frac\partial{\partial y}\right)$. The condition for a function $f$ to be holomorphic is $\frac{\partial f}{\partial\bar z}=0$.

Since $Q(\bar z)=\frac{R(z)}{P(z)},\ z\mapsto f(z)=Q(\bar z)$ is holomorphic in any domain noit containing the roots of $P$. The map $z\mapsto\bar f(z)$ is also holomorphic, because it's a polynomial in $z$ having as coefficients the conjugates of the coefficients of $Q$. This means that on the one hand we have $\frac{\partial f}{\partial\bar z}=0$, and on the other hand we have $\frac{\partial\bar f}{\partial\bar z}=0\Rightarrow\frac{\partial f}{\partial z}=0$. We thus get $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial z}+\frac{\partial f}{\partial\bar z}=0$, and similarly, $\frac{\partial f}{\partial y}=0$, which implies that $f$ is constant.

Edit:

I was a bit too late I guess :).

Take $j=0,1,2,...$ successively in the statement $\forall r \in \mathbb{R}, Q(i^{\frac{1}{2^j}}r)=Q((-i)^{\frac{1}{2^j}}r)$ to prove the result for $Q\in \mathbb{C}[x]$ as this means that the coefficients of $x^k$ in $Q$ where $\nu_2(k) = 0,1,2,...$ are 0.

We claim that $\overline{Q}\in\mathbb C[z]$. Suppose (in the interest of contradiction) that $P\nmid R$, and note by the Fundamental Theorem of Algebra that there exists some $\zeta\in\mathbb C$ for which $P(\zeta)=0\neq R(\zeta)$. It follows that $P(\zeta)Q(\overline{\zeta})=0\neq R(\zeta)$, impossible.

Now observe that $Q\cdot \overline{Q}$ takes values only in $\mathbb R$. It is well-known (and easily shown) that any such polynomial is constant, and thus so too must $Q$ be $\blacksquare$


For $a,b,c \geq 1$, prove that:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} + 2\frac{ab+bc+ca}{a^2+b^2+c^2} \geq 2 + a+b+c$
Have fun with it!

This ineqality is equivalent to
    \[ \sum_{cyc}\frac{(a-b)^2}{b} \geq \sum_{cyc}\frac{(a-b)^2}{a^2+b^2+c^2} \]
Which is obvious since $\frac1b \geq \frac{1}{b^2} \geq \frac{1}{a^2+b^2+c^2}$.

[quote="Vegetius"]For $a,b,c \geq 1$, prove that:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} + 2\frac{ab+bc+ca}{a^2+b^2+c^2} \geq 2 + a+b+c$
[/quote]
The inequality still holds if we replace the condition $a,b,c\geq 1$ by $a,b,c>0$ with $a+b+c\ge 3$.

[quote]The inequality still holds if we replace the condition $a,b,c\geq 1$   by $a,b,c>0$  with $a+b+c \geq 3$ .[/quote]
It's ok. And my proof still works.
$(a^2+b^2+c^2) \geq \frac{(a+b+c)^2}{3} \geq (a+b+c) \geq b$
Hence $\frac1b \geq \frac{1}{a^2+b^2+c^2}$ leads to the answer.

[quote="Vegetius"]For $ a,b,c \geq 1$, prove that:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} 2\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \geq 2 \plus{} a \plus{} b \plus{} c$
Have fun with it![/quote]
Suppose that $ c\equal{}min\{a;b;c\}$ then this inequality becomes:
$ (a\minus{}b)^2[\frac{a\plus{}b}{ab}\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2}]\plus{}(a\minus{}c)(b\minus{}c)[\frac{b\plus{}c}{ac}\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2}] \geq 0$
From $ c\equal{}min\{a;b;c\}$ we have $ \frac{a\plus{}b}{ab}\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2} \geq 0$ and $ \frac{b\plus{}c}{ac}\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2} \geq 0$
Done!


i think it stands for  limitation.  But i don't know its mathematical relevance.

$\lim$ means limit. This should be in the Calculus Forum...

$\lim_{x \rightarrow 0} f(x)$ is the limit of $f(x)$ as $x$ approaches $0$.


Do any of you know of a good basic-to-intermediate computer programming book that you would recommend?  Thanks in advance.

Problem solving & program design in C, and another one in C++

http://www.amazon.com/exec/obidos/ASIN/0321198034/qid=1120920970/sr=2-1/ref=pd_bbs_b_2_1/103-1730386-0784607

This is used in our school. It's powerful and teaches you the basics.

Moreover, I have mentioned before in this forum that Dr. Deitel's books are also useful for beginners.
Like "C How to program " , "C++ How to program", "Java How to program" etc...

What programming language will you use?

I will probably begin with C, and then see if I like others better.

Well to start with, C is perfect. You may also start with C++ since it is an improvement of C

I'd say start with C, since C++ is a lot more complicated than C.

But C++ opens the door for OOP and I think that C++ makes life easier than C.
Suppose you want to pass a variable by reference, theoretically, there is nothing in C that does this for you! Instead you have to directly use pointers and this brings up a lot of difficulties and problems. Whereas C++ has a pass by reference operator in language definition and easy to use. This is just an example..
 :)

I'd start with Python. C++ is hard to learn in my opinion, and C....is outdated.

As for books, Learning Python from O'Reilly is what I've used to learn Python. Anything by Wrox Press should be good too.

Billy

Well, I was going to say Java, but I thought the original post was about C and C++.  But I'd say Java is the best one to learn, since it's getting more and more widely used, uses OOP, and is also the AP CS language, which you can use to get college credit.

C family are the most commonly used in teaching. And regarding Java, it's also a nice choice. But, someone if wants to learn things closer to the machine level, C/C++ is better choice.

Don't get stuck in the mindframe that only these "mainstream" languages are worth learning. There are many great languages out there that are interesting, efficient, and powerful but don't recieve much press because they're not targeted for the enterprise. Many of them take a completely different approach to programming that goes beyond the syntatical differeces of, say, Java and C#.

Haskell ( http://www.haskell.org/ ) -- A functional programming language that is specially suited for higher-level mathematics and scientific programming. Give it a look.

Ruby ( http://www.ruby-lang.org/en/ ) -- An OO scripting language that has a much more interesting syntax than Java. Much more flexible, and 'code blocks' are neat.

Caml ( http://caml.inria.fr/ ) -- Supports imperative, functional, and OOP programming. Also one of the most efficient languages -- even outpreforms natively-compiled C in many cases.

So, don't fool yourself that Java, C, and VB are the 'only' languages out there. Explore. You might learn something.

You're right. Why should someone always recommend C, C++, Java etc...

But, it depends on what kind of programming style he would like to follow?
Of course, for mathematicians, functional programming languages are better than imperatives. ML is also nice and strongly typed.

Ruby is VERY easy to learn, expecially with that nice ebook about it. And OCaml is on my list of languages to learn sometime soon

I downloaded Ruby and decided to learn it. It looks cool...


Ok, this is wierd. The question asks if the probabiity that an unfair coin will come up heads is 2/3, what is the probability that the number of heads after 50 tosses is even. If you just do the problem purely mathematically you get the answer is $\frac{1+(\frac{1}{3})^{50}}{2}$ which is VERY close to 1/2 but if you think about it you get 1/2 because no matter how many times you get a heads the number of them will occilate between odd and even. So 1/2 makes sense but is definately wrong, so how is is wrong?

I get the part about 1/2 making sense, but why is it definitely wrong? I would expect the probability to tend toward 1/2, so everything seems to work to me...

Wait...first of all, how would you figure out the probability?

well if you work it out with no logic (use binomial theorem) you get (1+(1/3)^50)/2 which tends towards 1/2, with logic you get 1/2 and the two are not equal.

Could you explain a bit more how you get the probability to be 1/2 with logic?

[quote="bigbadwolf"]if you get a heads then the number of heads occilates between odd and even, so regardless of the probability of getting a heads the probability of getting an even number of them is alwas 50:50 because it occilates between exactly one other item, getting a head does not change the probability of getting an even number of them. (Somewhere this is wrong)[/quote]

I think it's a "logical" step with a mathematically incorrect assumption. Basically you assume the probability of getting a head SHOULDN'T change the probability of an even number of heads showing up; the only way to prove that assumption is through calculating the actual probability. And since the calculations show that this is not true, it's just a false assumption.

For instance, apply your same logic to a single flip. The probability SHOULD, by the same logic, be $\frac{1}{2}$, but it's actually $\frac{1}{3}$.


function
trigonometry
real analysis
real analysis unsolved

the function f(x) = a* !sinx! + (a^2) *!cosx! + g(a) has a period equal to (pie/2) ,then the least a>0 is

a) 2
b)1/3
c)1/4
d)none of these

please note ! sinx! means modulus of sinx

please help me to find the ans

$ f(x) \equal{} a |\sin x| \plus{} a^2 |\cos x| \plus{} g(a)$.

What is this $ g$?

'a' can be any number from the interval $ [-\cot(\frac{\pi e}{2}), -\tan(\frac{\pi e}{2})]$, so I suppose that $ a_{min} = -\cot\frac{\pi e}{2} \approx 0.632728$.

"pie" was meant to be $ \pi$, not $ \pi e$ ;)

I started to write my post asking if what you say is true, but gave up thinking it's not possible.. But, it seems I was wrong! So, if that is the case, I suppose the only two answers are $ a \equal{} 1$ and $ a \equal{} 0$, so this gives us the answer $ a \equal{} 1$.


I found the following generalisation of IMO 2001/2 (the original problem being the case n = 3):

Let $n \geq 2$ be a positive integer and let $x_1$, $x_2$, ..., $x_n$ be positive reals. Prove:

\[\sum \frac{x_i}{\sqrt[n - 1]{x_i^{n - 1} + (n^{n - 1} - 1)\cdot \prod_{k \neq i} x_k}} \geq 1.\]

where the summation ranges over all i.

Did anyone see this before? I hope it is original.

I just see a atex error !. By the way, I don't think anything related to that problem can be more general than the following generalization from Komal:
  If $ a_1,...,a_n $ have product 1 and k is real smaller tran 1, then $\frac{1}{(1+(n^k-1)a_1)^{1/k}}+...\geq 1$. This one is extraordinary difficult!

I read what you wrote and it's clear that your generalization is weaker than that from Komal!

Hm yes, that one is even more general :(

I think that $k$ is greater than 1 in the Komal Inequality, isn't it ?

Yes, of course, I'm sorry. Indeed, k is at least 1.

Who (What) is Komal ?
I found this general inequality solved in Crux and in GM.

Komal is a hungarian magazine. If I'm not wrong, it's the eldest mathematical magazine in the world, but I'm not sure. Anyway, this problem appeared as an advanced (and it really is!) problem in November 2001. I know you also gave two beautiful solutions for this problem, as far as I remember if k is natural.


geometry
circumcircle
ratio
geometric transformation
reflection
parallelogram
geometry proposed

Let triangle ABC be the acute triangle inscribed in circumcircle (O). Let H be the orthocenter of triangle ABC. Through H, we draw line d, d' respectively parrallel to AB, AC, meets AB, AC at M, N. The circle that takes AH be its diameter intersects (O) at Q. Prove that A, M, N, Q are concyclic. :D

That's too easy
MY/AY=YH/YC  NX/AX=HX/BX
MY/NX=YH/YC*AY/AX*BX/HX  notice BX/YC=AX/AY
MY/NX=YH/XH=AHsin∠BAH/AHsin∠CAH= sin∠BAH/sin∠CAH= BCsin∠BAH/BCsin∠CAH =BY/CX
⊿QBY∽⊿QCX   =>  BY/CX=QY/QX  => MY/NX=QY/QX ∠QYA=∠QXA => ⊿QYM∽⊿QXN
=>  ⊿YQX∽⊿MQN  => ∠MQN=∠YQX=∠YAX   =>  M,Q,A,N are concyclic

$ E, F$ are feet of the B-, C-altitudes. $ (K), (P)$ are circles with diameters $ BC, AH,$ intersecting at $ E, F.$ $ \angle ENH \equal{} \angle FMH \equal{} \angle A$ $ \Longrightarrow$ $ \frac {\overline {MF}}{\overline{NE}} \equal{} \frac {\overline {FH}}{\overline{EH}} \equal{} \frac {\overline {FB}}{\overline{EC}}.$ Ratio of powers of $ M$ to $ (O), (P)$ is equal to ratio of powers of $ N$ to $ (O), (P)$:

$ \frac {p(M, (O))}{p(M, (P))} \equal{} \frac {\overline{MA} \cdot \overline{MB}}{\overline{MA} \cdot \overline{MF}} \equal{} \frac {\overline{MB}}{\overline{MF}} \equal{} 1 \plus{} \frac {\overline{FB}}{\overline{MF}} \equal{}$ $ 1 \plus{} \frac {\overline{EC}}{\overline{NE}} \equal{} \frac {\overline{NC}}{\overline{NE}} \equal{} \frac {\overline{NA} \cdot \overline{NC}}{\overline{NA} \cdot \overline{NE}} \equal{} \frac {p(N, (O))}{p(N, (P))}.$

Therefore, $ M, N$ are on the same circle coaxal with $ (O), (P),$ i.e., passing through $ A, Q.$ :lol:  :D  :maybe:  :blush:  :(  :)  :mad:  :o  :oops:  :P  :wink:  :roll:  :?:  :!:  :arrow:  :idea:  :rotfl:  :|  :ninja:  :huh:

Let $ D, E,F$ the feet of the altitudes issuing from $A,B,C.$ Negative inversion through pole $H$ taking 9-point circle $\odot(DEF)$ into circumcircle $(O)$ takes $\odot(AEHF)$ into line $BC$ $\Longrightarrow$ Inverse of $ Q$ is the second intersection of $BC$ with $\odot(DEF),$ i.e. the midpoint $ K$ of $BC.$ The inverse of $ M$ and $ N$ are the reflections $S$ and $ R$ of $C$ and $ B$ about $ HM$ and $ HN.$ Thus, we need to show that $K,D,R,S$ are concyclic. Indeed,

$ \angle RDS \equal{}  \angle RHS   + \angle HBD  +  \angle HCD \equal{} 2\angle BAC$

If lines $BR$ and $CS$ meet at point $P,$ then $ HCPB$ is a parallelogram and $HSPR$ is cyclic with circumcenter $ K$ $\Longrightarrow$

$ \angle RKS \equal{}2\angle RHS \equal{}2\angle BAC \Longrightarrow \angle RKS \equal{} \angle RDS$  $\Longrightarrow$ $ DKSR$ is cyclic $\Longrightarrow$ $APMN$ is also cyclic. $\square$

[quote="yetti"]Ratio of powers of $ M$ to $ (O), (P)$ is equal to ratio of powers of $ N$ to $ (O), (P)$:
...........Therefore, $ M, N$ are on the same circle coaxal with $ (O), (P),$ i.e., passing through $ A, Q.$ [/quote]
I don't know why this is true. (I mean I haven't ever seen this theorem) Could you give me a reference where I can get a proof of this fact?

[quote="yetti"] :maybe:  :blush:  :(  :)  :mad:  :o  :oops:  :P  :wink:  :roll:  :?:  :!:  :arrow:  :idea:  :rotfl:  :|  :ninja:  :huh:[/quote]
But I don't understand why all these emoticons are for.... :maybe: If it is a reply for two emoticons of mathVNpro, then I'd like to remind you that there about 100 more emot's in the "more emoticons" :wink:  :spider:

(a) Read a more general proof then the posted problem at [url]http://www.mathlinks.ro/viewtopic.php?t=121062[/url]. :bruce:

(b) These are mixed emotions. I know how to get the spider.

Cool! :lol: (Both, your proof  and the last emoticon)


[quote="Hatucdao"]Let $x,y,z \in [-1,1]$. Prove that:
$\sqrt{1-xy+y^2}+\sqrt{1-yz+z^2}+\sqrt{1-zx+x^2} \ge 3$[/quote]

I just think, this ineq, or the same kind ones of it, maybe useful.
$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}\ge\sqrt{c^2+ca+a^2}$

[quote="Lovasz"]I just think, this ineq, or the same kind ones of it, maybe useful.
$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}\ge\sqrt{c^2+ca+a^2}$[/quote]

I don't agree with you :D Have you strict thought about the problem?  :(

This problem is strange but not strict. Indeed, we have the stronger result:

Let $x_1,x_2,...,x_n$ are real numbers satisfy $x_ix_j \le 1$ for i<j . Prove that:
$(1-x_1x_2+x_2^2)(1-x_2x_3+x_3^2)...(1-x_nx_1+x_1^2) \ge 1$


function
geometry
3D geometry
real analysis
real analysis unsolved

Prove that if $0<an<1$ then one cannot find function $f$ defined on $[0,1]$ with values in $R^n$ such that for all $x,y$ we have $||f(x)-f(y)||\geq |x-y|^a$.

I think we could do something like this (but it's not fully rigorous):

Consider any set $A\subseteq[0,1]$ s.t. $f(A)$ is bounded. Since $\|f(x)-f(y)\|\le\varepsilon\Rightarrow|x-y|\le\varepsilon^t$, where $t=\frac 1a>n$, it means that any covering of $f(A)$ with $N$ cubes of side $\varepsilon$ yields a covering of $A$ with $\le N$ intervals of length $\varepsilon^t$, for any $\varepsilon>0$ (well, not exactly; give or take multiplication with a constant which depends only on $n$, so let's not complicate things :)).

Now, if $\varepsilon>0$ is small enough, then we can approximate $f(A)$ very well, in the sense that we can cover it with fewer than $m(f(A))\cdot\frac 1{\varepsilon^{t'}}$ cubes of side $\varepsilon$, where $n<t'<t$, and $m$ is the outer Lebesgue measure in $\mathbb R^n$. This yields a covering of $A$ with fewer than $m(f(A))\cdot\frac 1{\varepsilon^{t'}}$ intervals of length $\varepsilon^t$, giving it an outer measure smaller than $m(f(A))\cdot\varepsilon^{t-t'}$, which goes to $0$ as $\varepsilon\to 0$.

That means that for all bounded $S\subseteq\mathbb R^n,\ f^{-1}(S)$ has null measure, and this is clearly impossible.

I believe the basic intuition behind the problem is that trying to find such a function is just like trying to cram a set of [url=http://mathworld.wolfram.com/CapacityDimension.html]capacity dimension[/url] greater than $n$ into $\mathbb R^n$, which is impossible.


Harvard college articles geometry email Support probability

Politics always seem to stir up bitter, harsh feelings and is the talk of so many people. Why is this? There could be several options:

1. The people themselves want power themselves,
2. The people want a someone supportive,
3. The people want to have their own way without having to sacrifice anything,
4. Other.

Also, what is up with all those people who go to the field of politcs, when they know that it will most likely ruin their reputation as a person. I mean look at all those paraodies of Sarah Palin on SNL. How can that person keep their head up when such mockingness about them?

Disscuss what your opinion is.

Personally, I think McCain and Palin need to go home. They really aren't suited to be the next leaders of the country. Seriously, who wants an old geriatric and an inexperienced hockey mom who believes she can wave to Russia and expect something back besides a nuke on Alaska. I mean, and then she tries to go to random countries just to serve the impression that she's a worthy vice president candidate, and it clearly isn't working. And then there are McCain's policies that are completely flawed. Is anyone actually gonna spend 5000 dollars on "healthcare" if they're given it? Heck no! They'll probably go on vacation and then complain about it later. Though I found it hilarious how Joe the Plumber was mentioned so often, that was interesting...I think Obama is definitely the better candidate: he has promise and more hope for our country. He makes sense in his policies, he knows what to say, it's clearly evident that he's intelligent and up to the job. After all he did graduate from Harvard Law School, let's see some of that in action!

The Joe plumber thing was so dishonest. Plumber has a low working class connotation to it, while in reality plumbers have quite high pay (at least in Sweden). So McCain makes it sound like the tax raises will affect "average people". What a coincidence his name was Joe too. Standard McCain shenanigans where specific instances of something work as counterexamples to overall statistics.

And on-topic (see I started ranting OT without even realizing it), I think it's probably 3. Most people feel that they are right about every issue and nothing else is acceptable. That's why these debates are so pointless, out of a perspective of intellectual debate at least.

See my point has been made.
Not to point you out ajai, but you can see that politics has a strong grip on many people, of us intelectual youths bieng one.

I would have to also agree that it is 3. That may be the reason people have become the 'most powerful' animal on Earth. Our own greed yet our unwillingness (is that a word?) to sacrifice anything of our own. It's why our species is still on this planet, 'rising' above all. We don't have claws, fur, special adaptations, we have a brain.

[quote="Kalle"]The Joe plumber thing was so dishonest.[/quote][url=http://www.msnbc.msn.com/id/27207215/]Literally[/url]. It points to the Republican economic message: vote for lower taxes on the rich, because you might become rich some day. This doesn't appeal to me, but I'm a strange person who doesn't really care about money.

USA politics right now is particularly nasty, because of the very sharp difference between the two parties. Also, many of the issues are fundamentally moral points, and those opinions are held very strongly.

Somehow politics and weather are two things that pop up in almost every casual conversation. I am always amused by the fact that people can realize they don't understand how the weather changes, have no real influence on atmospheric processes, are tolerant to other people preferences as far as having or not having cold winters or hot summers is concerned, etc., but when it comes to political issues where (with rare exceptions) they have about as much understanding and influence as in the weather case, the discussions suddenly become much more heated and opinionated.

The usual outcome of the political arguments is that everybody remains at his original point of view but gets the strong impression that his opponent (and other people thinking like him) are to blame for all the problems in the world, while the real things to blame are general ignorance of the population, dishonesty and selfishness of the politicians (from all parties), outward stupidity of decision makers (which is not about making mistakes: everybody makes mistakes all the time, but about being extremely stubborn in sticking to the wrong decisions even after it becomes clear to everyone that they lead nowhere) and other things like that. If everybody had asked himself "What did I do to make this world better?", he would almost certainly realize that a) he has not tried much and b) most of what he tried failed. This might make a person a bit less eager to try to enforce his personal opinions and preferences on others and to find whom to blame and a bit more inclined to spend some time thinking of what can be done given the current circumstances instead.

We are all too much used to the idea that there is an outside enemy guilty of all our problems (be it an opposite party, the muslim terrorists, the greedy rich, the criminal poor, the delusional believers in God, the amoral atheists, or whoever else), while everybody's worst and most real enemy is himself. What I hate most about McCain is not his tax or foreign policies (I'm not an expert in either field, so I would let the reality to pass its judgement on both) but his idea that one can win the election by just showing how bad his opponent is. It makes me think that he will stick to the same idea of finding an external enemy and trying to eliminate it in all other cases as well. Nazi once tried the simplified version of this approach with Jews as the main enemy to blame for everything, and, even if we disregard all moral issues, we can still pass the verdict that the approach turned out to be plainly inefficient for solving any of the society problems and left Germany lying in ruins in less than $ 8\equal{}2\cdot 4$ years. I just do not want something similar for the USA (or any other country).

Well, it seems that, like Kalle I "went off topic without realizing it", so I'd better stop here.

[quote="jmerry"]
USA politics right now is particularly nasty, because of the very sharp difference between the two parties. Also, many of the issues are fundamentally moral points, and those opinions are held very strongly.[/quote]

There are more than a few people who think things are so nasty now because there is not a sharp difference between the two parties.  There's something to that, I think.  On a scale of libertarian to communist, both parties are very, very close to each other.  Just look at the argument over tax policies.  If you were to start from absolute scratch (not from where taxes are right now, but be zapped into a new world and mind wiped clean of the status quo), the two would seem to be proposing pretty similar tax policies.  Obama doesn't want to put us back to where we were under Carter, and McCain isn't looking to push us back to an age before income taxes.  Neither is advocating a shift from income taxes to a VAT tax, etc.  Their policies on Iraq are converging.  Their arguments over Iran are whether or not to meet with preconditions (which gets my vote for most moronic heated dispute in the debates).  They basically agree on climate change, but have small differences on how to go about marginal change.  I could go on and on.  Same was true with Bush v. Kerry, same with Bush v. Gore, Clinton v. Dole, Clinton v. Bush, etc.  I think you have to go back to Carter v. Reagan to get very large, clear differences between the candidates (OK, maybe Mondale v. Reagan, but does Mondale really count?)

And fedja - Hitler and the Nazis?  But didn't you hear -- Hitler was a community organizer, too!  (Yes, I'm being facetious, and not calling Obama Hitler, I'm merely suggesting that references to the Nazis with respect to McCain or Hitler are rarely instructive or constructive, or even in the ballpark of accurate.  Comparing McCain's camp sliming Obama to the Nazis sounds just about as on-target as McCain supporters saying Obama is a terrorist.  But I suppose in this election, [url=http://en.wikipedia.org/wiki/Godwin's_law]Godwin's law[/url] is in full force...)

FWIW, fedja, I should add that I couldn't more strongly agree with nearly everything you wrote before you wrote the word "Nazi", and I do find McCain's apparent instinct to define people who disagree with him as not only wrong, but evil, as among his ugliest and most dangerous personality traits.  Obama seems to believe that another could differ in opinion from him and not be a bad person.  It's not clear to me if he believes that another could differ from him [i]and still be right[/i], but I think it's very unlikely McCain feels that way about the issues he cares about (which I think are a small subset of the issues he'd face as President).

[quote="rrusczyk"]
On a scale of libertarian to communist, both parties are very, very close to each other. [/quote]

But the scale of libertarian to communist is (albeit likely intentionally) quite crude. I happen to like the 2D political spectrum of [url]idealog.org[/url]/ Their two axes are the so-called traditional dilemma of politics, namely "order versus freedom" and the so-called modern dilemma of politics, namely "freedom versus equality".

This is still quite crude, but I believe more accurate than the original 1-dimensional scale. They then divide the plane into four quadrants--the libertarian quadrant prefers "freedom" on both accounts. The Communitarian (or communist) quadrant, in the words of the test, believes the purpose of government is to "correct wrongs" so is willing to sacrifice freedom on both accounts. Your scale is like a diagonal line passing through these two quadrants.

But the traditional "liberal versus conservative" scale is a diagonal line that is perpendicular to yours. And in fact, I believe this scale actually better represents the two parties than does the libertarian versus communist scale. While they aren't [i]all that[/i] far apart on this scale either, it at least makes a bit more sense to look at them on this scale, and even more sense to look at them in the plane in general.

Certainly, the linear scale is a poor one, as we hashed out in an earlier thread arguing over what "conservative" really means.

Mainly, I think if you go down the list, issue by issue, and look at them in *many* dimensions, these two are not that terribly far apart -- the two party system naturally produces candidates that gravitate to the middle, and then fight for little slices of special interest groups that might put one or the other over the top.  Every once in a while there's a big shift when we reach a very unattractive equilibrium (like we did in the 70s, and like we may now if the financial crisis isn't contained).

I'm not saying it doesn't matter who's President -- some marginal changes can have very big long-term effects (card check, for example, might be one of them), but I am saying that I don't think the bitterness of the battle comes from the candidates being so different.  I also don't think it's even all that new.  It's just that there's much more of it out there for consumption if you want to look at it.

[quote="fedja"]Somehow politics and weather are two things that pop up in almost every casual conversation. I am always amused by the fact that people can realize they don't understand how the weather changes, have no real influence on atmospheric processes, are tolerant to other people preferences as far as having or not having cold winters or hot summers is concerned, etc., but when it comes to political issues where (with rare exceptions) they have about as much understanding and influence as in the weather case, the discussions suddenly become much more heated and opinionated.
[/quote]

Actually fedja, people do have the power to vote. One person may not be a big difference but even just one city can be partially influential.

[quote="rrusczyk"]
And fedja - Hitler and the Nazis?  But didn't you hear -- Hitler was a community organizer, too!  (Yes, I'm being facetious, and not calling Obama Hitler, I'm merely suggesting that references to the Nazis with respect to McCain or Hitler are rarely instructive or constructive, or even in the ballpark of accurate.  [/quote]
Ah, no, I didn't really mean what you seem to be finding in my post. I just wanted some extreme example of this kind of thinking when bringing the Nazi in, neither less, nor more. Perhaps, I just got carried away a bit, but, as I said, "everyone makes mistakes all the time" :).

As to gravitating towards the middle, I'm inclined to agree to the extent that you do not have much choice in what to say/promise if you want to get elected to the office. The real difference, if there is any, will manifest itself only [i]after[/i] the election is over. We do not demand from presidents what we would demand from an architect whom we hire to build our house, i.e., to present clear blueprints of the proposed construction and then to follow them as closely as the circumstances allow. Basically, the whole election process is much more about whom we are more inclined to trust to make the crucial decisions for the next few years than about what those decisions will look like. Of course, the candidates are loosely bound by their pre-election promises, but I do not think these bounds will stop any of them from doing exactly the opposite to (or just something completely different that may be of much greater importance than) all that has been promised.

I also find, say, Al Gore and Bush quite different as far as their way of thinking is concerned. I don't want to go into any details now, but, really, do they look like identical twins in different garments to anyone? To me they look rather as two creatures from different planets in the same uniform :). Of course, up to a certain point, the uniform you wear matters more than the character you have, but when it comes to something beyond the ordinary (and one good thing about the USA system is that we, probably, can have a mannequin for the president and the ordinary things will still be taken care of in the same way they are now), the uniform stops playing the leading role and the character shows up in full. I surmise that Al Gore's handling of 9/11 would be very different from Bush's (though, of course, I cannot prove anything here). On the other hand, perhaps, I am just too naiive and all politicians that make it to the top are really more or less alike... I only know that mathematicians that make it to the top are as different from each other as day and night.

As to the bitterness of the battle, I agree that it comes not as much from the differences in candidates as from the simple fact that there is only one chair and two persons with one of the ultimate goals in life to take the seat. Different or alike, they will fight like cats and dogs under such circumstances.

@Dojo Well, if the usual political conversations had started with analyzing the current situation, proceeded to discussing what might be the best course of action (without necessarily agreeing on some panacea, but with pointing out the positive and the negative sides of each solution) and ended by looking at how each candidate proposes to handle the issues and figuring out which of them is closer to the ways each participant in the conversation finds best, I would agree with you (though with this style of conversations, I do not see where the heat or gruelness would come from), but I'm yet to see anything like that. Also, do not overestimate the "right to vote": it is just the right to combine all your personal opinions into a binary choice and then have this choice diluted in a huge pool. Doesn't seem like a great deal to me. Getting people's signatures under a petition to the mayor or the governor concerning some particular local issue, writing newspaper articles about the things you are concerned about, and other similar approaches may certainly bring more changes and are much closer to the idea of being able to tell the others and the government what to do than just exercising your right to vote, but, again, I'm yet to see a typical political conversation to concentrate around something like that.

[quote="fedja"]
I also find, say, Al Gore and Bush quite different as far as their way of thinking is concerned. I don't want to go into any details now, but, really, do they look like identical twins in different garments to anyone? To me they look rather as two creatures from different planets in the same uniform :). Of course, up to a certain point, the uniform you wear matters more than the character you have, but when it comes to something beyond the ordinary (and one good thing about the USA system is that we, probably, can have a mannequin for the president and the ordinary things will still be taken care of in the same way they are now), the uniform stops playing the leading role and the character shows up in full. I surmise that Al Gore's handling of 9/11 would be very different from Bush's (though, of course, I cannot prove anything here). On the other hand, perhaps, I am just too naiive and all politicians that make it to the top are really more or less alike... I only know that mathematicians that make it to the top are as different from each other as day and night.
[/quote]

I think we'll have to agree to disagree on this one.  Sure, their way of thinking may be different (one appears to think, the other doesn't -- who knows if either appearance is really true; keep in mind, if Gore were such a brilliant man, how did he come up with such an awful 2000 campaign).  But I think their conclusions probably would not be that different.  There *might* have been more action on climate issues under Gore, though this would have been tough with a Republican Congress.  Moreover, he didn't really touch these issues as Vice-President, or mention them much.  He probably wouldn't have touched them as President until his second term for re-electability reasons, and by then he would have had his hands full, because I believe we still would have been in Iraq, judging from Gore's pre-2000 record (he was among the most hawkish Democrats, no matter what political grandstanding he did about Iraq when the worm turned and it was OK to bash Bush again).  The main difference is that Gore probably would have handled our allies better than Bush (hard to imagine things would have been worse).  And he wouldn't have made his first call for sacrifice be "go shopping".  The current financial crisis would have been exactly the same under either man, and our tax code wouldn't look that different from now (the difference between W and Clinton's tax code is not that large when you consider the historical, or even international, range of taxes).

All that said, I'm not complaining that the candidates are pretty close--large changes have large unintended consequences.  Small changes to complex systems are probably much safer, with the large ones left as a last resort for emergencies.  Are we in a financial emergency now?  Iceland probably is, but it's not so clear if the rest of the world is.

Ah, I did think of one big difference that would have happened if Gore defeated Bush: right now, McCain would be coasting to victory, and Obama would not have run (but he would have been first in line in 2012).

[quote="fedja"]On the other hand, perhaps, I am just too naiive and all politicians that make it to the top are really more or less alike... I only know that mathematicians that make it to the top are as different from each other as day and night.
[/quote]

I would guess that the mechanism for selecting the top mathematicians is considerably different than that choosing top politicians.  (One would hope!  Electing your best mathematicians -- that's a funny idea.)

I think their personalities and temperaments can be very different, but their actual policies very similar.  Perhaps this isn't a result of what the two candidates actually would do, but rather what they will be able to do, considering that (fortunately) we are not electing them dictator.  What Bush wanted to do was drastically different from what Gore wanted in areas like Social Security, but it was clear that it wasn't going to happen.  Comparing Obama and McCain, I'd say one of the biggest differences is simply temperament, with a substantial advantage to Obama.  I only hope he has the courage to stand up to the extremes of his party the way McCain once stood up to the extremes of his (where did that guy go?).

In retrospect, I don't really agree with my previous part. The really nasty stuff isn't about policy.
See [url=http://www.rightwingwatch.org/content/hate-you-can-believe-acorn-deluged-threatening-and-racist-voicemails-and-emails]here[/url] for a few examples (Warning: offensive language and ideas). Stirring up hate like that is one of the few things I will unequivocally call evil.

On the Bush/Gore hypothetical, I really don't think we'd be in Iraq. That war was a pet project of Bush and a few of his advisers. Without the personal motivation, Gore would most likely have been content to leave Iraq as a nasty little dictatorship that could only harm itself.

[quote="rrusczyk"]The current financial crisis would have been exactly the same under either man, and our tax code wouldn't look that different from now (the difference between W and Clinton's tax code is not that large when you consider the historical, or even international, range of taxes).[/quote]What is significant is the difference in the government's finances. A Gore administration would have needed to borrow far less money, and that would have given them more room to act on the current mess.

[quote="jmerry"]On the Bush/Gore hypothetical, I really don't think we'd be in Iraq. That war was a pet project of Bush and a few of his advisers. Without the personal motivation, Gore would most likely have been content to leave Iraq as a nasty little dictatorship that could only harm itself.[/quote]

Gore's history suggests otherwise, I think.  He was one of the 10 Democrats who supported the first Gulf War, for example.  He's definitely not from the peace wing of the Democratic Party.  He was one of the more vocal within the Clinton administration about pressing Iraq, and encouraged Clinton in the us of extraordinary rendition (something also associated with Bush's post-9/11 conduct).  I don't think Gore would have antagonized Iran as much as Bush has, which would definitely have been a step in the right direction.

Regardless, we'll still never know whether or not Iraq has been a good idea.  My guess is that looking back 30 years from now, it will look like the Korean War does now.  The Korean War was certainly a huge boon to the South Koreans, but it's not so clear it was worth the great many deaths suffered by troops from all over the world.  In the long run, the Iraq War will likely be a huge moral good for the Iraqis.  Its value to the US and Britain will be much more questionable, if not obviously negative.

[quote="jmerry"]What is significant is the difference in the government's finances. A Gore administration would have needed to borrow far less money, and that would have given them more room to act on the current mess.[/quote]

This is a [i]very[/i] good point.  Gore would have fought much more with Congress than Bush did over budgets, and would have resisted such a large tax cut.  We'd still be in bailout land, and the incumbents would be toast in this election, but the long-run damage wouldn't be as bad.

(Your other point about the nastiness not being about policy is spot-on as well.  It's us-against-them in their cultures.  I'm a weird hybrid of both cultures, so I just look at the fight as bringing out the worst in both.)

What's the argument here -- Gore isn't a peacenik so he would have fabricated a case for war and aggressively sold it to the American public with basically no sustained media scrutiny?  This seems rather far-fetched, don't you think?  (Perhaps especially the lack of media scrutiny.)  The idea that we must attack Iraq, with or without valid pretense, was deeply ingrained in the neocons Bush and Cheney brought along with them -- what evidence is there that Gore would have surrounded himself with a group of similar nutcases?  I have no doubt that had the attacks of September 11 been carried out under a Gore presidency (not something that's entirely clear), we would have ended up in Afghanistan -- there's a non-crazy connection between the two and, as you note, Gore is no pacifist.  What's the non-crazy connection that would have him running to Iraq?

Gore was pretty well-connected with the neo-cons, too, and was not only not a peacenick, but was actively hawkish.  Again, I point to the fact that he was among the small minority of Democrats who voted for the first Gulf War.  He was also a strong proponent of the use of extraordinary rendition in the Clinton administration (stronger than Clinton, at the very least).  I think support for this policy certainly correlates positively with pre-emption and invading Iraq.  I'm not saying this is conclusive proof; I'm just saying the stand "Gore would have kept us out of Iraq" has very little to commend itself, either.  Just the words of Gore himself well after it was clearly popular among his base to say so.

I think it's basically unknown what would have happened, but I think it's at least on the order of 50/50 that we would have been in Iraq anyway.  (I think greater than 50% likely.  And if less than 50%, it's not anywhere near 0%, and if we could go to the replay in a different world, I wouldn't give 2-1 odds in either direction.)  Again, look at his record prior to 2000, not his "darling of the liberals" image post 2002.  (Speaking of which, kind of amazing how the media came to embrace him after helping sink him in 2000.)  He was a very different person in 2003 than he was as a professional politician.  And he might have been even more different still as President.  We just don't know.

Thinking about this more, I think our discussion of Gore & whether or not he would support the war ties in a bit with jmerry and I eventually agreeing that the parties are not that far apart -- that is, it's not issues or general principles that guide the ugliness, but rather an us versus them mentality.

Witness our last four major interventions of choice (I don't count Afghanistan, which clearly would have happened with anyone as President):

Iraq I: a Republican war (Bush I): Republicans all supported, very few Democrats supported
Bosnia: a Democratic war (Clinton): Democrats all supported, virtually no Republicans supported (and take a look at some of the awful stuff folks like DeLay had to say about Clinton regarding it)
Iraq II: a Republican war (Bush II): Republicans support, Democrats for it before they are against it

(I'm speaking here of the politicians of the parties.)

What's consistent here is that the guiding principle that can be said to determine whether or not a given politician supports it is "Is it our side's war?"  I believe very few politicians made principled decisions on the issues at hand (though they could all rationalize flopping this way or that), and might even call it an act of political courage that Gore voted for Iraq War I.  (To be fair, he might have been making the same calculation Hillary and Biden made in voting for Iraq War II -- that is, you could only be President if you voted yes.  Didn't work out for Hillary, and Obama was fortunate enough to be out of the line of fire on that one.)  Anyway, what I'm saying is that I think most politicians' views on most topics are determined by the direction of the wind.  [i]And that's exactly the result that a republic form of government is designed to produce[/i].  When someone goes against the wind of his party, pay attention.  There may be longer-range political games being played, or they may really mean it.

That's why I think Gore's support of Iraq War I cannot be easily dismissed, and part of the reason I'm convinced the bitterness in politics has very little to do with issues.

It's also why conservatives still are not comfortable with McCain due to his illegal immigrant stand (and McCain-Feingold before that), no matter what he says since then.  He took a stand that all but killed his chance of being President, and no matter how much he backed off since, in the very low-probability universe in which McCain wins this election, there would very likely be significant progress along the lines of his original proposal.  It means something when you cross your team.  It doesn't always mean something good, but it means more than when you simply agree with your team.  And this is part of the reason I think we have very little to measure Obama by.  It's clear where McCain crosses his team; it's not nearly as clear with Obama, though he's given more than a few hints that education is one area where he is.  And that's a good thing.

I thought about this one day, and here is the answer I came up with:

At first, the democrats and republicans are a tiny bit to the left and right. Then one group feels the need to make the average opinion move their way, so they have to move even further to the left or right. Then the other has to do it even more, and so on. For example, it goes
45-55 to 45-65 to 25-65 to 25-85 to 5-85, then it approaches a limit and both become fully the other way.

If you want to see the American epicenter of pathetic, corruption-filled politics, come to Illinois. Our last governor is in prison, our current governor is about to be impeached, and there are four major [url=http://en.wikipedia.org/wiki/Richard_M._Daley#Controversies]controversies[/url] surrounding Chicago major Richard M. Daley. And if you start looking into our history you will find even more outstanding characters. It is really a fantastic show here.

gotta luv chicago.

Politics is grueling because the only two parties(the independents don't stand a chance :roll: ) usually attack each other like mad dogs when it comes to deciding leadership. They find every little thing(AN F IN SECOND GRADE GYM CLASS! DO YOU WANT A WIMP FOR PRESIDENT?!) they can on their opponent to get people to vote for them instead of their opponent. It's all about comparisons, and mud-slinging gets painful to watch after a while. Somebody should call a halt to it. Lots of people have mud in their eyes.


For x,y,z real positive such that xy+yz+zx=1 prove that

x(1-y 2 )(1-z 2 )+y(1-z 2 )(1-x 2 )+z(1-x 2 )(1-y 2 ) \leq
(4 \sqrt 3)/9

We will see that this one reduces to a trivial one via some simple computations. Indeed sum x(1-y^2)(1-z^2)=sum x-sum x(y^2+z^2)+xyzsum xy=(sum x)(sum xy)-sum x(y^2+z^2)+xyz=4xyz. So we must prove that xyz<=sqrt3/9, which is trivial.


Our math club has 20 members and 3 officers: President, Vice President, and Treasurer.  However, one member, Ali, hates another member, Brenda.  How many ways can we fill the offices if Ali refuses to serve as an officer if Brenda is also an officer?

I'm asking (and reviving) this because I thought the answer was $ 20\cdot19\cdot18\equal{}6840\minus{}6\equal{}6834$ ways.

[quote="elevate"]Wait I'm confused. Why didn't we divide 6840 by 3! first?[/quote]

The reason you don't divide by $3!$ is that the three positions are different.  You're not just choosing three officers...you're choosing a president, a VP, and a secretary.

[hide="Solution"]
So, with that said, there are $20 \cdot 19 \cdot 18 = 6840$ permutations with no restrictions.  If Ali and Brenda are both officers, there is $1$ other slot available. for the other $18$ members.  Thus, there are $18$ ways to pick the three people, and $3! = 6$ ways to pick offices for them, for a total of $18\cdot 6 = 108$ invalid cases.  This leaves $6840 - 108 = \boxed{6732}$ valid groupings.
[/hide]

[quote=YT2019][quote=math154]Use complementary counting. There are $ 20\cdot19\cdot18\equal{}6840$ ways to do this without restrictions, and $ 3\cdot2\cdot18\equal{}108$ invalid cases in which both Ali and Brenda are officers. Thus, there are $ 6840\minus{}108\equal{}\boxed{6732}$ ways.[/quote]

How is it $3\cdot2\cdot18=108$ which both Ali and Brenda are officers? I got $18$ ways.[/quote]

It is not 18 ways because the order in which Ali, Brenda, and the other officer is chosen doesn't matter.
In other words, after picking the 3 people, there are (3 * 2 * 1) = [b]6[/b] ways to assign them different positions. That is why it is 18 * 6 = [b]108[/b] and not 18.

[quote=YT2019][quote=math154]Use complementary counting. There are $ 20\cdot19\cdot18\equal{}6840$ ways to do this without restrictions, and $ 3\cdot2\cdot18\equal{}108$ invalid cases in which both Ali and Brenda are officers. Thus, there are $ 6840\minus{}108\equal{}\boxed{6732}$ ways.[/quote]

How is it $3\cdot2\cdot18=108$ which both Ali and Brenda are officers? I got $18$ ways.[/quote]

Because there are 3*2 ways to assign what positions Ali and Brenda are.

[quote=LeonidasTheConquerer][quote=jmiao]Hate is a strong word[/quote]

ayo fresh comments??[/quote]

freshh


At the Adladskjhabon High School, there are 1000 students, and 1000 lockers in a row. The first student walks down the row and opens every locker. The second student walks down and closes every other locker. The third student walks down and opens every third locker that is closed and closes every third locker that is open. This continues until everyone has gone. How many lockers are open at the end?

Each locker is opened and closed n times if it has n factors, meaning those that are open have been opened and closed an odd number of times. Odd number of factors means square number. There are $\boxed{31}$ square numbers between 1 and 1000.

[hide]the only numbers with an odd number of factors are perfect squares and there are 31 squares between 1 and 1000[/hide]

@Treething:
are you sure its 33?

[list]odd factors work so just take the square root of 1000 and take the floor value.[/list]

Generalizing:
If there are $N$ lockers, then after changing the states,
there will be $\lfloor\sqrt{N}\rfloor$ open lockers.


How might I enter a double sum into a text editor?  For instance, when I do a double integral I just use \iint.  However, \ssum doesn't work and \sum\sum won't allow me to put a subscript under the entire double sum.  Could someone tell me how to do this?

Thank you.

OK I was able to do something similar, but I can't get the sums large enough:

example:  $\substack{\sum\sum \\ \textrm{something here}}$

How can I make those sigmas larger?  Thank you.

How about this: $\sum\sum\atop{\textrm{Leibniz}}$ ?? ;)

Exactly $1$ month$-47\textrm{min}$ to get the answer :rotfl: