induction
geometry
calculus
integration
geometric series

I thought these questions were substantially harder than last year's problems. ([url]http://www.math.umd.edu/highschool/mathcomp/PartII2008.pdf[/url])

1) I originally skipped this one and jumped straight to the second problem, but finally went back and got it after about one minute of drawing out potential divisions.

2) This seemed a lot easier than #1 only because I knew how to deal with such squares.

3) I made the false assertion that all numbers of the form $ k^2 \plus{} 1$ are winning positions for Player II; I ironically only checked up to $ 4^2 \plus{} 1$ and it turns out $ 5^2 \plus{} 1$ fails after brute forcing the problem on my computer.

4) Considered it, but was unsure of how to prove either part.

5) Stupidly said that the power exists as base 2 with a flawed argument about multiplying n-digit numbers in any base. The argument is actually valid for base 5 and not base 2 upon further inspection, although the argument doesn't exactly hold up.

I'm looking at two correct problems, so I can kiss my chances of an Honorable Mention goodbye. (I might still be County Winner though...)

yeah, problems this year are somewhat difficult in my taste and I wrote really messy and maybe incomplete solutions to the last two problems. :wink:

Yeah it was harder.

1. I proved it for 1 mod 3, 0 mod 3, and 2 mod 3. I think I got this one.
2. I actually thought this one was hard to approach. I assumed it wasn't a square and then found a contradiction.
3. I used induction starting at 2 and then proving it for all 2+5n. I think there are infinitely.
4. I didn't know how to do this one. I wrote down some faulty stuff.  :(
5. Same as 4!

AHHH

[quote="greentreeroad"][quote="buzzer11"]
3. I used induction starting at 2 and then proving it for all 2+5n. I think there are infinitely.
[/quote]

Interesting, could you please show the solution.

I made too many careless mistakes when writing the proofs. :(  Hope graders will still understand what I mean[/quote]

I don't think it's actually valid. It fails because Player I can subtract multiples of 25 (such as 25, 225, 625, etc.) For example, n=27 results in a win for Player I, not Player II. This is because if Player I subtracts 25, then Player II will be left with n=2 and will lose.

The basis of buzzer11's induction (I assume; this is how I was going to do it until I caught my error) is that all squares are congruent to either 1 or 4 when considered mod 5. So, Player I has to subtract a square, which will either be 1 or 4 (mod 5). (EXCEPT he can actually subtract something that's 0 (mod 5), like 25). Then Player II subtracts the number 1 or the number 4 (as needed) to maintain n=2 (mod 5) when it's Player I's turn. This will eventually reach n=2 and a loss for Player I (win for Player II). Unfortunately, if Player I subtracts something that's 0 mod 5, then it reverses the situation and Player II loses, unless he subtracts something that's 0 mod 5 as well. However, he can't always subtract something that's 0 mod 5 (like 25), if the number has dropped below 25.

[quote="buzzer11"]Oops.. that is what I did. But are there an infinite number of starting points?[/quote]

I don't have a conclusive proof, but I brute forced the problem up to 20 million and it looks like there are an infinite number of valid positions.

I think a really big part of the proof is that Player I must be forced to choose only 1 when 2 are available. Proving that Player I can be forced to 2 from infinitely many starting points should do it.

1) Very simple.

2) Also easy, although there are elegant and ugly ways of doing it.  I was lucky to find one of the former.  If you guess the answer from small cases and then write things as sums of geometric series it works out nicely.

3) I've studied Combinatorial Game Theory previously, so this didn't take me long.  There are an infinite number.  Assume that there are only finitely many and let $ n$ be the largest.  Consider $ n^2\plus{}2n$.  Subtracting squares, the lowest the first player can get is $ 2n$.  Thus we will end up above $ n$, and thus in a first-player win.  But now the "first player" is actually the second player, so the second player wins for $ n^2\plus{}2n$, contradiction.

4) Blah, I'm bad at geometry, and I didn't really have time for it because of . . .

5) This problem took me an hour to solve, but I finally did get it right.  Use induction, and say that $ k$ digits works.  Divide into three cases: $ k$ digits exists in base $ 2$, $ k$ digits exists in base $ 5$ and $ 5^k$ is less than twice the power of ten, and exists in base $ 5$ with $ 5^k$ greater than the power of ten.  The first case is not hard.  The other two work if you see that $ 2^k\equal{}a10^h$, $ 5^k\equal{}b10^j$, for $ 1<a,b<10$, so $ 10^k\equal{}ab10^{h\plus{}j}$, so $ ab\equal{}10$.

So 4 problems.  I had a pretty low Part 1 score (80) due to stupid mistakes, but hopefully I can at least get into the top 10 . . .

[quote="E^(pi*i)=-1"]2) Also easy, although there are elegant and ugly ways of doing it.  I was lucky to find one of the former.  If you guess the answer from small cases and then write things as sums of geometric series it works out nicely.[/quote]

It's messy. I didn't get the immediate 333...3 * 333...34 until right after the competition.

Do you mean the geometric sequences?  It is pretty clean:

$ n$ can be written as

$ (10^{4017}\plus{}\cdots\plus{}10^0)\plus{}(10^{2009}\plus{}\cdots\plus{}10^0)\plus{}3$

or

$ \frac{10^{4018}\minus{}1}{10\minus{}1}\plus{}\frac{10^{2010}\minus{}1}{10\minus{}1}\plus{}3\equal{}\frac{10^{4018}\plus{}10^{2010}\plus{}25}{9}$.

The answer is

$ 3(10^{2008}\plus{}\cdots\plus{}10^0)\plus{}2\equal{}\frac{10^{2009}\plus{}5}{3}$

Square this, and you're done.  :)

Can anyone put up solutions for 2 and 3? I'm curious to see the geometric series approach, and I messed up #3, so...

I have posted solutions for 2 and 3 in my last two posts . . .

[quote="E^(pi*i)=-1"]
3) I've studied Combinatorial Game Theory previously, so this didn't take me long.  There are an infinite number.  Assume that there are only finitely many and let $ n$ be the largest.  Consider $ n^2 \plus{} 2n$.  Subtracting squares, the lowest the first player can get is $ 2n$.  Thus we will end up above $ n$, and thus in a first-player win.  But now the "first player" is actually the second player, so the second player wins for $ n^2 \plus{} 2n$, contradiction.
 [/quote]

I solve number 3 using same idea. I use number $ n^2\plus{}n\plus{}1$ However, due to time limit, in my solution I stupidly stated that after player 1 moves first, the game will become a game with player 2 as "first player" and n+1 stones(should be at least n+1). I expect to get around 0-10 in this problem now. :(

Problem 4 can also be generalized into this.
http://web.archive.org/web/20040101111657/www.kalva.demon.co.uk/towns/tn80s.html

[quote="E^(pi*i)=-1"]That shouldn't be a big deal - I'm sure they'll know what you meant.[/quote]

Hope so. :wink:

[quote="E^(pi*i)=-1"]Do you mean the geometric sequences?  It is pretty clean:

$ n$ can be written as

$ (10^{4017} \plus{} \cdots \plus{} 10^0) \plus{} (10^{2009} \plus{} \cdots \plus{} 10^0) \plus{} 3$

or

$ \frac {10^{4018} \minus{} 1}{10 \minus{} 1} \plus{} \frac {10^{2010} \minus{} 1}{10 \minus{} 1} \plus{} 3 \equal{} \frac {10^{4018} \plus{} 10^{2010} \plus{} 25}{9}$.

The answer is

$ 3(10^{2008} \plus{} \cdots \plus{} 10^0) \plus{} 2 \equal{} \frac {10^{2009} \plus{} 5}{3}$

Square this, and you're done.  :)[/quote]

What if the proof stopped at $ \frac {10^{4018} \plus{} 10^{2010} \plus{} 25}{9}$? We can just say that the bottom is $ 3^2$ and the top is $ (10^{2009} \plus{} 5)^2$, right?

Yes, that's true.  The actual square root is not needed.

EDIT: Although you should mention either the lemma that the square root of an integer is either integral or irrational, or note that $ 10^{2009}\plus{}5$ is divisible by $ 3$ to get full credit.

Hey, I just discovered that E^(pi*i)=-1 had posted UMD part 2 problem 1 here before: http://www.mathlinks.ro/Forum/viewtopic.php?t=148847 :D

I wonder how many people already knew it

1: Saw this one the day before, easy anyway

2: Same as most of the ones above, except I didn't use geometric series, I said 333....3 = 999....9/3 = (10^n - 1)/3

3: Let n be the largest, consider n^2+n+1

4: Got part a by using 'midpoints of a quadrilateral from a parallelogram' over and over, missed part b

5: Puzzled over this one for a long time, the solution turned out to be only a couple of lines. #5 is always frustrating like that.


We say that a positive integer is happy if can expressed in the form $ (a^{2}b)/(a \minus{} b)$ where $ a > b > 0$ are integers. We also say that a positive integer $ m$ is evil if it doesn't a happy integer $ n$ such that $ d(n) \equal{} m$. Prove that all integers happy and evil are a power of $ 4$.

hey, could you explain your question a bit more? and what is $ d(n)$ i.e. number of divisors or sum of divisors?

$ d(n)$ is the number of divisors of $ n$.

[hide=Solution]We have the two following observations:
[b]Property 1.[/b] If $m$ is a happy number then so is $k^2m$ for all $k \in \mathbb{Z}^+$.
[u][i]Proof.[/i][/u] There exists integers $a>b>0$ such that $m=\frac{a^2b}{a-b}$ so $k^2m= \frac{(ka)^2(kb)}{ka-kb}$, which is a happy number.   $\blacksquare$

[b]Property 2.[/b] There always exists a prime divisor $p$ of any happy number $m$ such that $2 \mid v_p(m)$.
[u][i]Proof.[/i][/u] Write $m$ as $\frac{a^2b}{a-b}$ and let $\gcd (a,b)=d, a=da_1,b=db_1$ with $\gcd (a_1,b_1)=1$. We have $m= \frac{d^2a_1^2b_1^2}{a_1-b_1}$. We can see that $a_1 >1$ (otherwise if $a_1=1$ then $a \le b$, a contradiction) so there exists prime $p \mid a_1$. Since $p \nmid b_1(a_1-b_1)$ we follows $2 \mid v_p(m)$.   $\blacksquare$

Back to our problem, we denote $m$ as an happy and evil number. If there exists odd divisors of $m$ or $m=2^k(2c+1)$ with $c \ge 1, c \in \mathbb{Z}$. We can easily find a happy $n$ such that $d(n)=m$. Indeed, we starts with $l=\frac{3^2 \cdot 2}{3-2}=3^2 \cdot 2$ and by applying Property 1, we times this number by $3^{2(c-1)}$ and $2^{2^k-2}$ to get a new happy number $n= 3^{2c-2} \cdot 2^{2^k-2} \cdot l = 3^{2c} \cdot 2^{2^k-1}.$ Hence, $d(n)=2^k(2c+1)=m$, a contradiction. Thus, $m$ can only have even prime divisor, or $m=2^k$. Applying Property 2 we follows that $m=4^k$.

We check and see that $4^k$ is a happy number, indeed, $4^k = \frac{(2^k)^2 \cdot 2^{k-1}}{2^k-2^{k-1}}$. From property 2, we follows that for any happy number $n$ then there exists odd prime divisor of $d(n)$. This gives $4^k$ as an evil number.

Thus, all the happy and evil integers are power of $4$.
[/hide]


calculus
integration
trigonometry
symmetry
calculus computations

For real numbers $ a,\ b\ (ab\neq 0)$, evaluate $ \int_0^{\pi} (a\sin x\plus{}b\cos x)^6 dx$.

Since $ \cos x$ has odd symmetry around $ \frac {\pi}{2}$, and $ \sin x$ has even symmetry around $ \frac {\pi}{2}$, we conclude that all terms having an odd exponent for $ \cos x$ will integrate to 0.  We then have the equivalent integrand
\[ a^6\sin^6x \plus{} 15a^4b^2\sin^4x\cos^2x \plus{} 15a^2b^4\sin^2x\cos^4x \plus{} b^6\cos^6x \equal{} \frac {a^6(1 \minus{} \cos(2x))^3 \plus{} 15a^4b^2(1 \minus{} \cos(2x))^2(1 \plus{} \cos(2x)) \plus{} 15a^2b^4(1 \minus{} \cos(2x))(1 \plus{} \cos(2x))^2 \plus{} b^6(1 \plus{} \cos(2x))^3}{8}.
\]
Notice now that $ \cos(2x)$ and $ \cos^3(2x)$ integrate to 0 since $ 2x$ makes a complete revolution from 0 to $ 2\pi$, and that $ \cos^2(2x) \equal{} \frac {1 \plus{} \cos(4x)}{2}$ integrates to $ \frac {\pi}{2}$ since $ 4x$ makes two complete revolutions from 0 to $ 2\pi$, so that the integral is
\[ \frac {1}{8}\left(a^6(\pi \plus{} \frac {3\pi}{2}) \plus{} 15a^4b^2(\pi \minus{} \frac {\pi}{2}) \plus{} 15a^2b^4(\pi \minus{} \frac {\pi}{2}) \plus{} b^6(\pi \plus{} \frac {3\pi}{2})\right) \equal{} \frac {5\pi}{16}(a^6 \plus{} 3a^4b^2 \plus{} 3a^2b^4 \plus{} b^6) \equal{} \frac {5\pi}{16}(a^2 \plus{} b^2)^3.
\]
Edited by daniel73: 2nd stupid calculation mistake in the day...  :blush:

Thanks to kunny for pointing out that, in the original post, a calculation mistake had been made...


geometry
3D geometry
Ring Theory
superior algebra
superior algebra unsolved

Is $(XV-YU,\ XW-ZU,\ YW-ZV)$ a prime ideal of $\mathbb C[X,Y,Z,U,V,W]$?

It is the ideal generated by the $2\times 2$ minors of $\left(\begin{array}{ccc}X&Y&Z\\U&V&W\end{array}\right)$.

I have a geometric proof for the problem which could be exotically summarized as:
The vanishing locus of the ideal $(XV-YU,\ XW-ZU,\ YW-ZV)$ is the cone over the blow-up of $\mathbb C^3$ in $(0,0,0)$. The blow-up of $\mathbb C^3$ at $(0,0,0)$ is an irreducible variety, so the cone over it is also irreducible.

I am also curious about the existence of an algebraic proof.


function
invariant
calculus
integration
real analysis
real analysis unsolved

What are the positive measurable functions $f: [0;1] \to \mathbb{R}^{+}$ satisfying for all $x \in [0;1]$:
$f(x) = \sum_{1}^{\infty}\frac{1}{(n+x)^{2}}f(\frac{1}{n+x})$

Seems you are after the invariant measures for the continuous fraction transformation on $[0,1)$ :).  It is well-known and easy to check that $\frac 1{1+x}$ is such a function. There are no really different $L^{1}$ positive functions just because the fact that the mapping is ergodic does not depend on the choice of the function and, for any bounded measurable function $\phi$, the integral $\int_{0}^{1}\phi f$ can be found as the limit of the ergodic averages (though, of course, you can multiply your function by a constant). As to the non-integrable functions, there are none: each such function would give you a measure space with infinite measure $f\,dx$ and an ergodic transformation on that measure space. So, the ergodic averages should tend to $0$ a.e. for any integrable with respect to the measure $f\,dx$ function $\phi$. But we can take $\phi=\chi_{\{f<a\}}$ for sufficiently big $a$ to ensure that $\int_{0}^{1}\phi>0$, which will immediately show that the ergodic averages must tend to some positive number a.e. and we'll get a contradiction. Of course, this solution is highly non-elementary and if anyone finds a simpler approach avoiding the use of the ergodic theorem, I'll be happy to see it.

yep, I did that too .. but I find that the proof of the fact that the transformation is ergodic is non trivial. Any easy proof for that ?

I admit that the standard proof is not completely trivial but it is not too hard either. Basically, it boils down to assuming that $E$ is an invariant set of strictly positive measure and using the invariance for constructing a partition of $[0,1]$ into arbitrarily small intervals such that the portion of points of $E$ in each interval is bounded away from $0$ by some fixed constant. By the Lebesgue theorem, this is enough to conclude that almost every point belongs to $E$. Is this the proof you know? I spent a couple of hours of my free time during the last two days trying to see whether I could come up with some clever trick but I didn't succeed and, anyway, the approaches I tried relied on something more advanced than just the Lebesgue theorem.


topology
advanced fields
advanced fields unsolved

Let X be a compact hausdorff space. Prove that if two distinct points are in diferent conected components, then we can separate them by a separation of X.

It suffices to prove that the quasicomponent of an arbitrary point $x$ is connected.

Assume the contrary. Since the quasicomponent is closed and not connected, it can be written as a disjoint union of closed sets $A,B$, with $x\in A$, say. Since $A,B$ are disjoint closed subsets of a compact Hausdorff and hence normal space, we can find disjoint open subsets $U_{A}\supset A,\ U_{B}\supset B$, and another open subset $W_{A}\supset A$ such that $\overline{W_{A}}\subset U_{A}$.

Now I'll use the following

[b]Lemma[/b] (see the proof below)

In this setting, if $x$ can be separated from every point of a closed set $K$, then $x$ can be separated from $K$ (i.e. there is a clopen set containing $x$ and disjoint from $K$).

Since $x$ can be separated from every point of the closed set $X\setminus(W_{A}\cup U_{B})$, the lemma tells us that there is a clopen set $x\in S\subset W_{A}\cup U_{B}$.  The set $S\cap\overline{W_{A}}$ is, on the one hand, closed, and on the other hand open, because $S\cap\overline{W_{A}}=S\cap U_{A}$. This means that it's a clopen set which contains $x$ and is included in $U_{A}$, thus being disjoint from $B$. $B$, however, is a subset of the quasicomponent of $x$, so $x$ cannot be separated from any point of $B$. We have a contradiction.

[b]Proof of the lemma[/b]:

For every $y\in K$, let $x\in U_{y},\ y\in V_{y}$ be a separation of $x,y$. The compact set $K$ is covered by $V_{y},\ y\in K$, so it's also covered by finitely many of them, say $V_{y_{i}},\ i=\overline{1,n}$. Now $x\in\bigcap_{i}U_{y_{i}},\ K\subset\bigcup_{i}V_{y_{i}}$ is a separation of $x,K$.


Prove that in all triangle $ABC$ the following inequality holds:

      $\sum\frac{b^{2}m_{b}+c^{2}m_{c}}{m_{a}^{2}+m_{b}m_{c}}\geq\frac{16}{9}\cdot(m_{a}+m_{b}+m_{c})$ ;)

Sign reversed I suppose.
Apply something i call duality (The substitution $a -> m_a$, $m_a -> \frac{3}{4}a$)
then it remains to prove $a+b+c \ge \sum \frac{b^3+c^3}{a^2+bc}$
which has the sign reversed.
In fact when you reverse it it can be easily solved by cauchy.


induction
trigonometry
real analysis
real analysis solved

For 1)

use the fact that lim(x-->0) ((1+x)^k-1)/x=k
and rewrite the limit like so:
lim(x-->0) [((1+ax)^(1/m)-1)/x-[((1+bx)^(1/n)-1)/x]=a/m-b/n

cheers! :D :D

I have something for 2)

well you denote L_n=lim_{x->0}(1-cosx.cos2x...cos(nx))/x^2  and then compute the limit using induction!
L_1=1/2
L_(k+1)=L_k++lim(x-->0) [1-cos(k+1)x]/x^2=L_k+(k+1)^2/2
so you sum all these from k=1 to n and get that the limit is:
n(n+1)(2n+1)/12

cheers! :D :D

The idea get  the limit by induction is new for me
(1-coxx...cos((k+1)x)/x^2 =
(1-cosx...cos(kx))/x^2 + cosx.cos2x...cos(kx)(1-cos((k+1)x)/x^2 =
and tend x->0, I see.

there is GM a problem which looks like moubi's :D
maybe it has been posted b4... anyway it's kind of easy
lim{x->0}(a_1*a_2*...a_n*x^n-trig(xa_1)*...*trig(xa_n))/x^(n+2)
where trig \in {sin,tg,sh,th}, a_i \in  R, i=1,..,n

It seems there was a lot of problems like 2) posted by Lagrangia

Yes, I have posted a lot of problems like that, but now I am a big fan of solving them with Landau, induction sometimes ... stinks! :D :D

cheers!


induction
floor function
combinatorics unsolved
combinatorics

Let N be the number of people standing on a circle shooting at each other clockwise.
(ie a kills b. c kills d,...)The question is who will survive in the end ?

Let $N = 2^k + r$, where $0 \leq r < 2^k$ and label the people $1, 2, 3, \ldots$. We can easily prove by induction that the $(2r+1)$th person will survive in the end. The induction hypothesis is assuming it's true for all $N \leq 2^k$ and then consider all $N \in (2^k, 2^{k+1})$. After the first round, w/o going back to the 1st person, we would have $\lfloor \frac{N}{2} \rfloor \leq 2^k$ people left still alive, then renaming the people and using the induction hypothesis, we get the desired result.


Which compound contains both ionic and covalent bonds, and why?

1 - CaCO3
2 - PCl3
3 - MgF2
4 - CH2O

--

Given the reaction NH3 + HCl --> NH4Cl, ammonia molecules (NH3) act as a base because they

1 - accept hydrogen ions
2 - accept hydroxide ions
3 - donate hydrogen ions
4 - donate hydroxide ions

The answer key says that the answer is 1. Why couldn't it be 4?

Thanks in advance!

The first answer is $ \mbox{CaCO_{3}}$.  The ionic bond is between the calcium and the carbonate, and the covalent bonds are between the carbon and oxygen.

In the second question, it can't be 4 because there isn't any hydroxide in the reaction.  Look at the reaction: ammonia receives a hydrogen ion in order to become ammonium.  It couldn't be anything else.

If they give me the reaction as written, I'll assume both reagents are gases. You may be used to seeing their aqueous solutions (easier to handle), called ammonium hydroxide (with hydroxide ions) and hydrochloric acid, but both compounds are gases, and organic chemists frequently use them as gases (no water, no ions).
The question relates to the different views/definitions of acids and bases by Arrhenius, Bronsted, and Lewis. Without water, Arrhenius is out of luck.


calculus
integration
analytic geometry
combinatorics unsolved
combinatorics

A square is divided into $ n^2$ equal squares and the diagonals of each little square are drawn. Determine, in terms of $ n$, the total number of isosceles right-angled triangles of all sizes.

Count by which vertex we choose as the right angle: for a vertex an integral distance from all sides and a triangle whose legs are parallel to the sides of the square we get
\[ \sum_{a \equal{} 0}^n \sum_{b \equal{} 0}^n \min(a, b) \plus{} \min(a, n \minus{} b) \plus{} \min (n \minus{} a, b) \plus{} \min(n \minus{} a, n \minus{} b)
\]
(one term for each of the four orientations; $ a, b$ are the distances of our selected vertex from two sides (i.e. its coordinates).)

For the same vertices but with legs at $ 45^\circ$ angles with the sides (hypotenuse parallel to the sides) we get the similar sum
\[ \sum_{a \equal{} 0}^n \sum_{b \equal{} 0}^n \min(a, b, n \minus{} a) \plus{} \min(a, b, n \minus{} b) \plus{} \min (a, n \minus{} a, n \minus{} b) \plus{} \min(b, n \minus{} a, n \minus{} b).
\]
Finally, for vertices at half-integer distances from the sides we have only triangles of the second sort, and we get the sum
\[ \sum_{a \equal{} 1}^n \sum_{b \equal{} 1}^n \min(a, b, n \plus{} 1 \minus{} a) \plus{} \min(a, b, n \plus{} 1 \minus{} b) \plus{} \min (a, n \plus{} 1 \minus{} a, n \plus{} 1 \minus{} b) \plus{} \min(b, n \plus{} 1 \minus{} a, n \plus{} 1 \minus{} b).
\]
(Here the coordinates of the point are $ (a \minus{} \frac {1}{2}, b \minus{} \frac {1}{2})$.)

Now we just have to sum these things.  $ \sum_{a \equal{} 0}^n \min(a, b) \equal{} \frac {b(b \plus{} 1)}{2} \plus{} b(n \minus{} b)$, and summing this for $ b \equal{} 0..n$ gives $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$, and the total contribution from the first case is just four times this.  The second and third cases are slightly uglier but still perfectly doable -- one must split into cases for the parity of $ n$.  The final result is [url=http://www.research.att.com/~njas/sequences/A100583]A100583[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


Just thought some of you might be interested in a stupid game:

http://www.kingsofchaos.com/recruit.php?uniqid=6u9jxp4d

Join the war! Whee!

[quote="joml88"]Hate to revive old threads, but this one's only one post long so it's not like I'm reviving a 45 page long post that requires going through half of it to figure out what's being discussed.

Anyway, just wanted to say that I know the original creator of the game.  We go to the same school and run cross country together.  He's a really cool guy!

I actually haven't checked the game out myself yet, but I'll get around to that sometime :P[/quote]

Pretty sure the 4 creators went to TJ.

Yep.  The original creator went to MOP as well.


@Sinajackson: be careful, x, y, z need not be nonnegative.

We had this inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=101919 as well, but nothing useful was posted... Some references would be nice (who are Walker and Tooren?).

[b]EDIT:[/b] I have found the references: It's problem E 2245 of the American Mathematical Monthly. The only complete solution I can find in AMM writes everything in terms of $ a\plus{}b\plus{}c$, $ bc\plus{}ca\plus{}ab$ and $ abc$. Not the peak of beauty, if you ask me...

  Darij

[quote="sinajackson"]Maybe this Hint is not so much useful, because i think it makes the problem more complicated, but maybe it can help too!

[hide="Hint"]Let $ b\plus{}c\minus{}a \equal{} x,c\plus{}a\minus{}b \equal{} y,a\plus{}b\minus{}c \equal{} z$ then we have $ c \equal{}\frac{x\plus{}y}{2},b \equal{}\frac{x\plus{}z}{2},a \equal{}\frac{y\plus{}z}{2}$ and the inequality will be in the form:

$ (x\plus{}y)^{2}(y\plus{}z)^{2}(z\plus{}x)^{2}(x\plus{}y\plus{}z)^{3}xyz\geq$

$ \left(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}xy\plus{}yz\plus{}zx\right)^{3}\left(x^{2}\plus{}xz\plus{}xy\minus{}yz\right)\left(y^{2}\plus{}yz\plus{}yx\minus{}xz\right)\left(z^{2}\plus{}zx\plus{}zy\minus{}yx\right)$

But we have:

$ x^{2}\plus{}xz\plus{}xy\minus{}yz \equal{} x^{2}\plus{}x(z\plus{}y)\minus{}yz\leq x^{2}\plus{}\frac{x^{2}\plus{}(z\plus{}y)^{2}}{2}\minus{}yz \equal{} x^{2}\plus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}$

similarly:

$ y^{2}\plus{}yz\plus{}yx\minus{}xz\leq y^{2}\plus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}$

$ z^{2}\plus{}zx\plus{}zy\minus{}yx\leq z^{2}\plus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}$

so now we only need to prove that:

$ (x\plus{}y)^{2}(y\plus{}z)^{2}(z\plus{}x)^{2}(x\plus{}y\plus{}z)^{3}xyz\geq$

$ \left(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}xy\plus{}yz\plus{}zx\right)^{3}\left(x^{2}\plus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}\right)\left(y^{2}\plus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}\right)\left(z^{2}\plus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}\right)$

Now can [color=green]S.M.V Theorem[/color] Help, Maybe! :maybe:
[/hide][/quote]

Very nice, and S.M.V theorem you can find on hungkhtn'inequality book

[quote="darij grinberg"]
I have found the references: It's problem [b]E 2245 [/b]of the American Mathematical Monthly. The only complete solution I can find in AMM writes everything in terms of $ a\plus{}b\plus{}c$, $ bc\plus{}ca\plus{}ab$ and $ abc$. Not the peak of beauty, if you ask me...

  Darij[/quote]
Darij,
What is the Volume where contain solution of this problem?

[quote="N.T.TUAN"]Darij,
What is the Volume where contain solution of this problem?[/quote]

Unfortunately, I couldn't find part I of the solution.
Part II: Vol. 79, No. 9. (Nov., 1972), pp. 1034-1035.
Part III: Vol. 80, No. 7. (Aug. - Sep., 1973), pp. 809-810.

  Darij


integration
function
probability
LaTeX
calculus
calculus computations

Let $f(x)$ be a continuous positive function. For $0\le x \le 1$
we have the following:
$\displaystyle \int^{1}_0 f(x)\,dx=1$
$\displaystyle \int^{1}_0 xf(x)\,dx=a$
$\displaystyle \int^{1}_0 x^2f(x)\,dx=a^2$
where $a$ is a given real number. Find all $f(x)$ functions

Just use Cauchy-Bunyakowski for integrals. You will imediately obtain that $x\sqrt{f(x)}$ and $\sqrt{f(x)}$ are collinear functions. Absurd  :?

If you interpret $f$ as a probability density, the variable has mean $a$ and variance zero, which makes it a point mass.
A point mass does not have a probability density.

By the way- use \le in $\LaTeX$ to get $\le$.


If $ \left|ax^3 \plus{} bx^2 \plus{} cx \plus{} d\right|\leq1$ on $ \left[ \minus{} 1,1\right]$, then

$ |a|\leq 4,$

with equality only if $ 3a\plus{}4c\equal{}b \equal{} d \equal{} 0.$
[hide="Hint."]Let $ f(x) \equal{} \left(ax^3 \plus{} bx^2 \plus{} cx \plus{} d\right)^2$, then

$ f^2(1) \plus{} f^2( \minus{} 1) \plus{} 2f^2\left(\frac {1}{2}\right) \plus{} 2f^2\left( \minus{} \frac {1}{2}\right)\leq6\Longrightarrow |a|\leq 4.$[/hide]


geometry
geometric transformation
homothety
projective geometry
power of a point
radical axis
geometry theorems

Consider three circles in the plane, none of which completely lies inside another. For any two of these three circles, consider the two external common tangents. These two tangents intersect at some point.

Prove that the three points obtained in this way are collinear. [Monge's theorem.]

[i]Alternative formulation.[/i] Let $S_{1}$, $S_{2}$, $S_{3}$ be three circles in the plane. Let the two external common tangents to the circles $S_{2}$ and $S_{3}$ (we assume that these tangents exist) intersect each other at a point $A_{1}$. Similarly define two points $A_{2}$ and $A_{3}$. Prove that the points $A_{1}$, $A_{2}$, $A_{3}$ are collinear.

[quote="romano"]Consider three circles in the plane, none of which completely lies inside another. For any two of these three circles, consider the two external common tangents. These two tangents intersect at some point.

Prove that the three points obtained in this way are collinear. [Monge's theorem.][/quote]
I think we can prove it by Menelaus and by Desargues.

[Moderator edit: See also
http://www.cut-the-knot.org/proofs/threecircles.shtml
http://www.cut-the-knot.org/Curriculum/Geometry/MongeTheorem.shtml
http://www.cut-the-knot.org/Curriculum/Geometry/MongeDesargues.shtml
]

I thought the problem little bit more and found the following proof of Monge's theorem. If some of the circle has equal radii, then at least one of the point will be in infinity, and thinking in the view of projective geometry, all lines will pass though the infinity. Now suppose all of the circles have different radii. Then there are three points $A_1$, $A_2$, $A_3$ in plane which are homothetic centers of two of circles. Now every homothetic which maps circles to circles has coefficient different from 1. Therefore product of two coefficients is also different from 1. But this means that the composition of homotheties is a homothety, so the three points are indeed on the same line.

[quote="puuhikki"]I have just found a Finnish paper which says we can form a triangle $A,B,C$ where $A,B$ and $C$ are centers of the circles and then theorem follows from Menelaus' theorem. So I believe it is not necessary to use Desargues' theorem. The paper does not give any more hints. I have to go and think more.[/quote]
Menelaus is easy.
And if we look at the diagram carefully, we can find out that it is very similar to the diagram of Desargues' theorem. So we can connect the centres and construct some angle bisectors, then use Desargues. But that is more like a trick, too difficult to come up with...

$\text{1000's POST!}$

[url=http://mathworld.wolfram.com/MongesCircleTheorem.html]http://mathworld.wolfram.com/MongesCircleTheorem.html[/url]
It's easy to show the triangle formed by the centers is perspective with the triangle in the "middle". The result follows.
It is really tricky...I bet I cannot figure it out if shobber did not mention centers in his last post... :(

A very easy solution follows by realizing that the points you want to show that are collinear, are actually the centers of similarity of the circles taken 2 by 2 respectively. If the circles are $S_1, S_2, S_3$, consider the homothety that transforms $S_1$ into $S_2$, and then the homothety that transforms $S_2$ to $S_3$. Finally consider the homethety that transforms $S_1$ into $S_3$. After that, the problem is trivial

Actually, with a little more care, we could prove that the line these points make is actually the radical axis of the circles which are tangent to the all the 3 circles internally and externally. Use poles and polars :D.

oh .. i found another thing.mangoes theorem is the special form of another theorem.(i forgot the name)

instead of circle ,we will prove it for each three sphere. ;)

[url]http://www.geom.uiuc.edu/~banchoff/mongepappus/MP.html[/url]

The above url has something on the 3-D extension.
    Ashwath

What a nice problem!

Since now, I had been solved it by homothecy , ceva and Menelaus but I forgot

my proof by pole and polar or something.again i thought and find a solution, im pretty sure

that this one is different from the other one.

I post this solution for my dear friend $Number 1$ and all of the friends who love

Pole and polar.

$Theorem 1$:

Suppose $A, B$ are two points on circles $(C), (C')$, respectively. And consider

$M$ as the intersection of the tangents to circles $(C), (C')$, at points $A, B$

Respectively.

$M$ is on radical axis of two circles, if and only if $A, B$ are ant homologous.

$Theorem 2$:

Suppose $(C), (C'), (C\")$, are three circles. The radical axises of:

$(C),(C')$  and $(C),(C'')$  and $(C'),(C'')$  are concurrent , and their

Intersection, $S$ is named radical center.

(Note: Pascal theorem is proved by pole and polar, then each work which we can do

By Pascal, can be done by pole and polar.)

Ok let's come back to our problem.

Suppose $(C), (C'), (C\")$, are three circles. And $S$ is their radical center.

$M$ is the outer homothechy center of circles $(C), (C')$.

$N$ is the outer homotechy center of circles $(C'), (C\")$.

$O$ is the outer homotechy center of circles $(C), (C\")$.

We should prove that $M, N, O$ are collinear.

As we know, power of $S$ to each of the circles $(C), (C'), (C\")$, is equal.

Then the length of tangents from $S$ to circles are equal.

And we can draw a $(S)$ circle with center $S$, which is orthogonal to the

Three circles.

Suppose circle $S$, intersects $(C), (C'), (C\")$ at points $A, B$ and $C, D$ and

$E, F$, respectively. $SB, SC$ are tangent to $(C), (C')$ and $S$ is on their

Radical axis .then according to $Theorem 1$, $B, C$ are ant homologous,

And are collinear with $M$, because $M$ is their homotechy center.

By using similar way we will conclude these facts:

$A, D, M$ are collinear.

$C, F, N$ are collinear.

$D, E, N$ are collinear.

$A, F, O$ are collinear.

$B, E, O$ are collinear.

Now the best thing is to use Pascal theorem, we can also use pole and polar

But as I said at first, every fact which can prove by Pascal theorem, can also

 Prove By pole and polar.

In hexagon $B, C, F, A, D, E,$ we have:

The intersection of $BC, AD$ which is $M$,

The intersection of $CF, AD$ which is $N$,

And the intersection of $AF, BE$ which is $O$,

Are collinear.

And the problem is solved.

In fact, the problem is proved by pole and polar but we made it shorter

By using , Pascal theorem.

Dear $Number 1$ if u are disagree with it, I can post the proof of Pascal

Theorem by pole and polar.

After that everything will be OK.

And life will be more enjoyable.

 Each time we solve a math problem, we will become more and more nearer to

God and will know him better and better…

[quote="Ashegh"]Suppose  are two points on circles , respectively. And consider

 as the intersection of the tangents to circles , at points

Respectively.

 is on radical axis of two circles, if and only if  are ant homologous.

[/quote]

I'dont know what ant homologus means.

Can you pleas explain?

if there where a diagram it would be better...

hmmmm...

suppose $S$ is the outer homotechy center of circles $(C)$ and $(C')$.from $S$,draw an arbitrary line which

intersects$C$ at $A,B$ .  and $(C')$ at $C,D$ respectively.

points$A,D$ are anthomologous

and points $C,B$ are also anthomologous.

sorry dear number 1,  i cant explain it better... :blush:

Ashegh, will you please post the proof of pascal's theorem using poles and polars
P.S  dont use Brianchon's theorem

[quote="Ashegh"]
$Theorem 1$:

Suppose $A, B$ are two points on circles $(C), (C')$, respectively. And consider

$M$ as the intersection of the tangents to circles $(C), (C')$, at points $A, B$

Respectively.

$M$ is on radical axis of two circles, if and only if $A, B$ are ant homologous.
[/quote]

Dear Ashegh

Salam. I am from Tabriz and glad to see a patriot here.

I am trying to solve a problem. You can see it in this link.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=464495

I think above-mentioned theorem by you can help me.
What is the name of theorem you mentioned and where can I find more info in this regard?

[quote="Debdut"][quote="puuhikki"]I have just found a Finnish paper which says we can form a triangle $A,B,C$ where $A,B$ and $C$ are centers of the circles and then theorem follows from Menelaus' theorem. So I believe it is not necessary to use Desargues' theorem. The paper does not give any more hints. I have to go and think more.[/quote]

Post your paper here.[/quote]
I don't have that paper anymore.

Dear Mathlinkers,
who is the author of the collinearity mentioned above :  Monge or d'Alembert?
Any ideas?
Sincerely
Jean-Louis

I solved this problem with only Menelaus theorem


MIT college AMC USA(J)MO USAMO Princeton

hey, guys, i was really surprised to get rejected from MIT today.  I guess a big reason is because I applied as international.  I wonder how good you have to be in order to get into MIT as an international.
Here are my stats

SAT: Math 800
Verbal 750
Writing 640

GPA: 3.71/4.00

Major Awards:
Intel Finalist
2 times USAMO
Mandelbrot National 10th place

Major activities:
Varsity tennis for 4 years, first single last year
I also started two clubs at school last year

P.S. Sorry Geehoon, it seems like I won't be able to go to school with yo after all.

Don't worry man, it is very hard. I have a good friend who is an IMO gold medalist who got rejected from mit last year... and he had many more things other than math stuff..so yeah it is hard..

BTW, im also into tennis thats cool.. im gonna play in the mit team..!
the guy that i talked you about got into princeton, so im sure you will get into a great univ!

thanks for the encouragement, man.  yeah, i also applied to a bunch other schools, hopefully i can get into one of them.


geometry
3D geometry
sphere
rectangle
combinatorics proposed
combinatorics

I was just thinking about this problem, don't know if it is true, false, or open ended.

1. Color the circumference of a certain circle with two colors. Is it true that there always be exists a rectangle, such that all 4 vertices of the rectangle are on the circumference, and all the vertices are colored with the same color?

2. Color the surface of a certain sphere with two colors. Is it true that there always be exists a rectangular cuboid, such that all 8 vertices of the rectangular cuboid are on the surface, and all the vertices are colored with the same color?

Clearly not.
Divide the points on the circumference of the disk (or of the sphere) into pairs of diametrally opposite points and, for each pair, color one of the point red and the other blue. :wink:

Pierre.


find a closed form for
$\sum_{i=1}^{n}(i^{2}+1)i!$
in terms of n.


There's a board of m*n squares.
[u]Question1[/u] Find the number of ways to color black the board so that no 2 colored square are next to the other.
[u]Question2[/u] Find the number of ways to color black k unit squares of the board so that no 2 colored square are next to the other.

 With n=1 the problem is not very difficult. I think that all of you know the result. :)  :) (if you don't know you should try because it's an interesting one)
 However with n>1 it becomes really complicated.
 For the question1 I've solved for n=2,3 but I have no idea for the question2 :D  :D  :D

What do you mean by 'k squares'? You have to choose k unit squares among the m*n and color them so as to no two which are adjacent have the same color? Or is it kxk squares?

Moreover what is the number of colors???

Pierre.

Thanks for having mentioned me. Because English is my second language so my expression may be not exact. :)  :)  :)


trigonometry
inequalities
inequalities proposed

Prove that in any triangle we have:

$ \frac{3}{\sum \sin^{2}(\frac{A}{2})} \leq \sum \frac{a(b+c)}{r_{a}(r_{b}+r_{c})}$. When do we have equality? (standard notations)

I hope this is original and even more: correct!

cheers! :cool:

That's strange. The more or less easy problems posted on MathLinks get solved, with some exceptions, the hard problems also become solved after some time, even conjectures are sometimes being proven on this board, but on the other hand, some trivial problems are still waiting for their solution. An example is this problem here. At first we remember the well-known identities $r_br_c=s\left(s-a\right)$, $r_cr_a=s\left(s-b\right)$ and $r_ar_b=s\left(s-c\right)$, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Using these identities,

$r_a\left(r_b+r_c\right) = r_ar_b+r_cr_a = s\left(s-c\right)+s\left(s-b\right) = s\left(2s-c-b\right)=sa$,

and thus,

$\frac{a\left(b+c\right)}{r_a\left(r_b+r_c\right)} = \frac{a\left(b+c\right)}{sa} = \frac{b+c}{s}$.

Hence,

$\sum \frac{a\left(b+c\right)}{r_a\left(r_b+r_c\right)} = \sum \frac{b+c}{s} = \frac{2a+2b+2c}{s} = 4$

(remember that $s=\frac{a+b+c}{2}$). Hence, it remains to prove the inequality

$\frac{3}{\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}} \leq 4$.

This is clearly equivalent to

$\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2} \geq \frac34$.

Using the formula $\sin^2\frac{x}{2} = \frac12 \left(1-\cos x\right)$, we can rewrite this in the form

$\frac12 \left(1-\cos A\right) + \frac12 \left(1-\cos B\right) + \frac12 \left(1-\cos C\right) \geq \frac34$.

This simplifies to $\cos A + \cos B + \cos C \leq \frac32$, what is well-known.

  Darij

Maybe it is really simple. But you need to find great aesthete, which will want to solve this monster (at first sight, of course) ;)


function
algebra
difference of squares
special factorizations

If $ f(x) \equal{} ax \plus{} b$ and $ f^{ \minus{} 1}(x) \equal{} bx \plus{} a$ with $ a$ and $ b$ real, what is the value of $ a \plus{} b$?

$ \textbf{(A)} \minus{} \!2 \qquad \textbf{(B)} \minus{} \!1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

[quote="crazypianist1116"]$ f^{ \minus{} 1}(x)$ implies $ x \equal{} ay \plus{} b \rightarrow \frac {x}{a} \minus{} \frac {b}{a} \equal{} bx \plus{} a$ which means $ \frac {1}{a} \equal{} b$ and $ \frac { \minus{} b}{a} \equal{} a$. The only values that satisfy $ (a,b)$ are $ ( \minus{} 1, \minus{} 1)$. Thus we have $ a \plus{} b \equal{} \minus{} 2 \boxed A$.[/quote]

I don't think that's right. For one, consider the function $ f(x)\equal{}x$. :)
[hide="My Solution"]We have $ f(f^{\minus{}1}(x))\equal{}f^{\minus{}1}(f(x))\equal{}1$. The first gives us $ f(bx\plus{}a)\equal{}a(bx\plus{}a)\plus{}b\equal{}abx\plus{}a^2\plus{}b\equal{}1$. The second gives us $ f^{\minus{}1}(ax\plus{}b)\equal{}b(ax\plus{}b)\plus{}a\equal{}abx\plus{}b^2\plus{}a\equal{}1$.
Equating the 2, we get $ a^2\plus{}b\equal{}b^2\plus{}a \iff a^2\minus{}b^2\equal{}a\minus{}b$. Using difference of squares, we get $ a\plus{}b\equal{}\boxed{\textbf{(D)}\ 1}$.[/hide]

Well...$ f(x) \equal{} x$ doesn't work because then $ f^{ \minus{} 1}(x) \equal{} 1$ according to the problem

Oh...and it's $ f^{ \minus{} 1}(f(x)) \equal{} x$ ...I think...

And crazypianist's answer is correct

[hide="Official Solution"]
Since $ f(f^{\minus{}1}(x))\equal{}x$, it follows that $ a(bx\plus{}a)\plus{}b\equal{}x$, so $ ab\equal{}1$ and $ a^2\plus{}b\equal{}0$.  Hence $ a\equal{}b\equal{}\minus{}1$, so $ a\plus{}b\equal{}\minus{}2$, or $ \boxed{\textbf{(A)}}$.
[/hide]


Let $ \sum\limits_{i \equal{} 1}^n {\frac{1}{i}}  \equal{} \frac{k}{m}$ and $ \left( {k,m} \right) \equal{} 1$,
Try to prove k is odd and m is even...

It is true because the power of $ 2$,which divides $ \sum_{i\equal{}1}^{n}\frac{n!}{i}$,is less than the power of $ 2$,which divides $ n!$.

Because $ m\equal{}LCM (1,2,...,n)$ then m is even!

If $ \sum_{i \equal{} 1}^1\frac {1}{i} \equal{} \frac{k}{m}$ with $ (k,m) \equal{} 1$, then $ k \equal{} m \equal{} 1$. But now $ m$ is odd.

[quote="tdl"]Because $ m \equal{} LCM (1,2,...,n)$ then m is even![/quote]

That's not enough to prove the problem. Otherwise, it could prove $ \sum_{k\equal{}1}^n \frac{1}{k}$ has denominator divisible by 3 for $ n\geq 3$, but $ \sum_{k\equal{}1}^6 \frac{1}{k}\equal{}\frac{49}{20}$ does not.

[quote="zkw"]Let $ \sum\limits_{i \equal{} 1}^n {\frac {1}{i}} \equal{} \frac {k}{m}$ and $ \left( {k,m} \right) \equal{} 1$,
Try to prove k is odd and m is even...[/quote]
[hide="Solution"]
We proceed through induction.  Clearly, when $ i \equal{} 2$, we seet that $ \frac {1}{2} \plus{} 1 \equal{} \frac {3}{2}$, so $ k \equal{} 3$ is odd and $ m \equal{} 2$ is even.  Now, assume that $ \frac {1}{1} \plus{} \frac {1}{2} \plus{} \frac {1}{3}\cdots \plus{} \frac {1}{n} \equal{} \frac {2k \plus{} 1}{2j}$, where $ k$ and $ j$ are integers.  Thus, $ \frac {1}{n \plus{} 1} \plus{} \frac {2k \plus{} 1}{2j} \equal{} \frac {2j \plus{} (2k \plus{} 1)(n \plus{} 1)}{(n \plus{} 1)(2j)}$.  Let $ 2j \equal{} 2^x\cdot (2y \plus{} 1)$ and let $ n \plus{} 1 \equal{} 2^z\cdot (2w \plus{} 1)$ and $ z$ can be $ 0$.

[b]Lemma: $ x\ne z$[/b]
[i]Proof: [/i] Assume that $ x \equal{} z\implies 2^x|(n \plus{} 1)$.  Notice that $ 2j \equal{} \text{lcm} [1, 2, 3\cdots n]$, so $ x$ the exponent of the largest power of $ 2$ below $ n$.  Now, it is clear that if $ n \plus{} 1 > 2^{x \plus{} 1}$, then $ n\ge 2^{x \plus{} 1}$, so $ 2^{x \plus{} 1}$ is among $ 1, 2 \cdots n$, which means that the exponent of the largest power of $ 2$ below $ n$ is $ x \plus{} 1$, when it is actually defined to be $ x$.  Also, $ n \plus{} 1\ne 2^{x \plus{} 1}$, because then $ x \equal{} z \equal{} x \plus{} 1\ne x$.  Thus, $ 2^{x}\le n < n \plus{} 1 < 2^{x \plus{} 1}$.  However, there is no multiple of $ 2^x$ strictly between $ 2^x$ and $ 2^{x \plus{} 1}$, so we have a contradiction.  Thus, $ x\ne z$.

Without loss of generality, let $ x > z$.  With this in mind, the new fraction is $ \frac {2j \plus{} (2k \plus{} 1)(n \plus{} 1)}{(n \plus{} 1)(2j)} \equal{} \frac {2^z[2^{x \minus{} z}\cdot (2y \plus{} 1) \plus{} (2k \plus{} 1)(2w \plus{} 1)]}{2^{x \plus{} z}(2y \plus{} 1)(2w \plus{} 1)}$, which is
\[ \frac {2^{x \minus{} z}\cdot (2y \plus{} 1) \plus{} (2k \plus{} 1)(2w \plus{} 1)}{2^x\cdot (2y \plus{} 1)\cdot (2w \plus{} 1)}
\]
Since $ x\ne z$, we have that $ 2|2^{x \minus{} z}\implies 2^{x \minus{} z}\cdot (2y \plus{} 1) \plus{} (2k \plus{} 1)(2w \plus{} 1) \equal{} \text{Odd \plus{} Even} \equal{} \text{Odd}$ and obviously, $ 2^x\cdot (2y \plus{} 1)\cdot (2w \plus{} 1)$ is even.  Thus, the denominator of the new fraction is even and the numerator is odd, as desired. [/hide]

[quote="t0rajir0u"]Directly:  write

$ \sum_{i \equal{} 1}^{n} \frac {1}{i} \equal{} \frac {\frac {n!}{1} \plus{} \frac {n!}{2} \plus{} \frac {n!}{3} \plus{} ... \plus{} \frac {n!}{n} }{n!}$.

Let $ k \equal{} 2^a$ be the largest power of $ 2$ less than or equal to $ n$.  Let $ A$ be the largest integer such that $ 2^A | n!$.  Then $ 2^{A \minus{} a} | \frac {n!}{k}$ but a greater power of $ 2$ divides every other term in the numerator.

It follows that once we divide numerator and denominator by $ 2^{A \minus{} a}$ the denominator will be even but the numerator will be odd.  :)[/quote]
See my post above.

The general version reads:
Let some integer $ k$ be given.
Let $ m$ be the smallest integer $ >1$ that is coprime to $ k$. Show that for $ n \geq m$, the denominator of $ \sum_{\substack{s\equal{}1 \\ \gcd(s,k)\equal{}1}}^n \frac{1}{s}$ is divisible by $ m$.

Now you can sharpen that property further.


How long?

Me? Till I rot and die probably.

i'll quit when there's no more challenge, a looooooooooong time from now . . .  :weightlift:

[quote="mustafa"]Forever and ever and ever and ever :rotfl:\

edit: Who said quitting soon?[/quote]
treething, probably

me, the last option, for obvious reasons

3 people are quitting soon? stop kidding around lol.

I'm staying for a long long long long long long time!!!

Till I am out of high school  :)  Too much free time.

And it's good for asking and improving in math, and possibly when I am taking higher level math do other stuff.

[quote="asdf2345"][quote="junggi"][quote="asdf2345"]till the end of time. star ocean anyone?[/quote]
lol cool game
I borrowed from friend for about 3 months, got stuck really early, didnt touch it for about 2+1/2 months and then tried again and got addicted but friend wanted it back lol[/quote]

20 bucks at gamestop, pretty cheap if you ask me and worth it too. Tri-Ace rules!!!, not meant to be advertising of any sort.[/quote]
hey, not every parents let their kids buy games(in fact they don't let me play them anymore =().

Or you can mod your GameCube, or get a swap magic disc if it's PS2, and XBoxes are crap so no one cares about them and then download the game and burn it onto a mini DVDr or a CDr.

This is getting off topic (like everything usually does...).

[quote="junggi"][/quote][quote="asdf2345"][quote="junggi"][quote="asdf2345"]till the end of time. star ocean anyone?[/quote]
lol cool game
I borrowed from friend for about 3 months, got stuck really early, didnt touch it for about 2+1/2 months and then tried again and got addicted but friend wanted it back lol[/quote]

20 bucks at gamestop, pretty cheap if you ask me and worth it too. Tri-Ace rules!!!, not meant to be advertising of any sort.[/quote][quote="junggi"]
hey, not every parents let their kids buy games(in fact they don't let me play them anymore =().[/quote]

My parents don't let me buy video games or play them, but that never stopped me

I have money and I have means of transportation; simple, if you think about it

I will continue buying video games as long as I stay on AOPS, which is like I said till the end of time (hopefully that got me on topic)

Are you kidding me? i only joined this year, but i really like AoPS. It's really interesting and challenging.  :omighty:
If there is no more challenge, i might retire, but i really think this is a good way to learn math, and i would much rather have it than those teachers' lectures that numb your mind. :D

I think it's something you just know...like an instinct sort of thing. I guess it comes from liking math a lot  :lol:

[quote="Klebian"][quote="fredbel6"]All you people (except one) have only been members for a few months so how can you tell? :roll:

I joined in July 2003 and made a few posts.  I didn't post anything between September 2003 and April 2004.

It's only in 2005 that I became really active.[/quote]
3rd old member of this topic!

I'll probably stop visiting as often as I do during/soon after high school. But who knows.[/quote]
I'm sorry, what do you mean by "3rd old member of this topic"? :maybe:

all thru high school for me since I like responding andgetting help in math related type specifics.

I dunno, I just put through high school.

[quote="now a ranger"]yeah, in college it's highly unlikely people have a sufficient amount of time to study and also stay updated on AoPS. Especially because of the supposedly BORING hw assignments I hear about.[/quote]
I have time for AoPS. And I always will. :D
And my homework isn't boring.

AoPS is amazing. I'm going to use it until I die (shudders at thought of my death)

When I become old and crippled, I will make sure I have a labtop to do Aops while death becains.....

Probably when I get married and have children. :D

A lot of people think they'll be on AoPS as long as they can, but they get busy and leave.

Yeah this is kind of funny, because it's from 2006 and there's users saying they will be here till they die. Umm they probably left already.


function
calculus
derivative
induction
limit
algebra proposed
algebra

Find all the functions $f: \mathbb R \rightarrow \mathbb R$ satisfying :
$(x+y)(f(x)-f(y))=(x-y)f(x+y)$ for all $x,y \in \mathbb R$

the second equation Mr harazi is:
$f(x+2)=\frac{x+2}{x-2}(f(x)-f(2))$

This is just a typo mistake.
And the important is that he killed it in two lines.
He proved that its Easy problem. :mad:

yeah it's a quit simple prob tell me AYMANE did you participe int the TST's this YEAR? :?:

i'm ok with it if the function is continous! :blush:

[quote="AYMANE"]Here is an other solution:
1-tend $y$ to $-x$ then$f(x)-f(-x)=2xf'(0)$ take $x=1$then$f'(0)=\frac{f(1)-f(-1)}{2}$
2-take$x+y=1$ then $f(x)-f(1-x)=(2x-1)f(1)=f(x)-f(-(1-x))=f(x)-(f(x-1)-2a(x-1))$
==>$f(x)-f(x-1)=(2x-1)f(1)-2a(x-1)$
3-take $y=-1$ then$f(x-1)=\frac{x-1}{x+1}(f(x)-f(-1))$
==>$f(x)-\frac{x-1}{x+1}(f(x)-f(-1))=(2x-1)f(1)-2a(x-1)$
==>$f(x)=(f(1)-a)x^2+ax$
Hope did not make mistake on computing :?[/quote]

but there's NO condition about continous and differentiable   :oops_sign:  :!:

This is a tricky problem !! it seems easy , but under conditions exam , nobody knows how to deal with boring computations !! this problem needs just dare and^patience and time then it's really easy , however , only 2 students out of 11 could solve it . (i got caught with it) and i say to Ayman that he can't use derivatives , and it remains impossible proove continuity!!

Hi every body,Sorry for the delate i were on training
Math-sipo,Chang hwan and Wisemaniac are right to tell me i am not allowed to use continuity, but some time like in this problem we can use it indirectly and this was what i want to transmite to you:
i wrote:$f'(0)=\frac{f(x)-f(-x)}{2x}=a$ lets prove it:
Put $g(x)=\frac{f(x)-f(-x)}{2x}$ for $x<>0$ (E)
then $a=g(1)$ we have $g$  even.
1-$g(x+y)=\frac{x}{x-y}g(x)-\frac{y}{x-y}g(y)$ for $x-y<>0$
Take $y=-2x$ in (E)==>$g(-x)=g(x)=\frac{x}{3x}g(x)+\frac{2x}{3x}g(-2x)$==>$g(x)=g(2x)$
By induction easy to prove:$g(nx)=g(x)$
Take $x=1$==>$g(n)=g(1)=g(-n)$
Take $x=\frac{1}{n}$==>$g(\frac{1}{n})=g(1)$==>$g(\frac{m}{n})=g(1)$
So $g(x)=g(1)$ for all rationnals.
2-Take in (E) $x$ real non rational and$y=\frac{-1}{2^n}$==>$g(x-\frac{1}{2^n})=\frac{x}{x+\frac{1}{2^n}}g(x)+\frac{\frac{1}{2^n}}{x+\frac{1}{2^n}}g(\frac{-1}{2^n})=\frac{x}{x+\frac{1}{2^n}}g(x)+\frac{\frac{1}{2^n}}{x+\frac{1}{2^n}}g(1)$
Then $\lim_{n\to\infty}g(x-\frac{1}{2^n})=g(x)$
Do the same then $\lim_{n\to\infty}g(x+\frac{1}{2^n})=g(x)$ So $g$ is continious
And thus $g(x)=g(1)$ for all reals.
If you wont use continuity you can take$x=z-\frac{1}{2^n}$and$y=-z+\frac{2}{2^n}$ where $z$ is real non rational, use them in (E) and tend $n$ to infinity then $g(z)=g(1)$
And we done :roll:

Hi math-sipo. this exercice is quite easy I solved it in the exam in 2 mn. I'll be in Mexico this year:MAR2.

ok! mon ami so i will say good luck for u all! moroccan team!
u can do it!
allez Maroc!!!! :lol:

[quote="harazi"]You have $ f(x+1)=\frac{x+1}{x-1} (f(x)-f(1))$. Thus $ f(x+2)=\frac{x+2}{x} (\frac{x+1}{x-1}(f(x)-f(1))-f(1))$. But $  f(x+2)=\frac{x+2}{x} (f(x)-f(2))$. Eliminate $ f(x+2)$ among these relations and you will find a first degree equation in $f(x)$. From here it is easy.[/quote]

Is the answer $f(n) = \frac{n}{2}f(2)-\frac{n(n-2)}{n-1}f(1)$?

Probably, I did not make the computations since the two equations show our function is linear and then it's easy.

[quote="Bojan Basic"]I believe that the answer is $f(x)=2ax-ax^2$ for any $a$ (derived from harazi's solution).[/quote]

I don't understand. Is my result wrong? Can you show your work?

Sorry, I made a mistake, but here are the calculations.

From \[\frac{x+2}x\left(\frac{x+1}{x-1}\bigl(f(x)-f(1)\bigr)-f(1)\right)=\frac{x+2}{x-2}\bigl(f(x)-f(y)\bigr)\] you get \[(x^2-x-2)\bigl(f(x)-f(1)\bigr)-(x^2-3x+2)f(1)=(x^2-x)\bigl(f(x)-f(2)\bigr)\] from where is easily to find \[f(x)=2xf(1)-x^2f(1)+\frac{x^2f(2)-xf(2)}2\]Now put $x=1$ and $x=2$ to obtain $f(1)=a$ and $f(2)=b$ for any $a$ and $b$, and hence follows \[f(x)=\left(\frac b 2-a\right)x^2+\left(2a-\frac b 2\right)x\] or, if $\frac b 2-a=k$, \[f(x)=kx^2+(a-k)x\]


Anyone?
I got this one from another problem:
Let $ P(x) \equal{} (...((x \minus{} a_1)^2 \minus{} a_2)^2 \minus{} ... \minus{} a_{n \minus{} 1})^2 \minus{} a_n$.
Prove that there always exists n numbers $ a_1,a_2, \ldots ,a_n \in R$ to satisfy these equations:  $ P(1)\equal{}0, P(2)\equal{}0, \ldots ,P(n)\equal{}0.$
In fact, we don't have to show the formulas of $ a_i$ (you can use induction).


Prove that there are infinitely many [u]even[/u] positive integers $n$ such that  $p^2+ n$ is not prime number for each prime number $p$

[quote="paladin8"]Take $n = 2p, 4p, 6p, \ldots$?[/quote]

No, you misunderstood the condition.It means that for all prime number $p$ the number $p^2+n$ is not prime

just let $n=6k+2$ and for all $p\neq3$,$3|p^2+n$,and we only need to let $n+3^2=6k+11$ not be a prime, it's easy to know there is infinite $k$ satisfied it.


Da schon nächstes Wochenende das nächste Seminar auf uns wartet, wollte ich einfach mal fragen, wie ihr so nach Bad Homburg kommt, wann ihr ankommt und wer denn das erste Spiel am Rennautomaten bestreiten wird :D
.
Was mich betrifft, kann ich das momentan leider nicht sagen, da ich meine Fahrkarte zwar schon vor Wochen gebucht habe, sie zu diesem Zeitpunkt aber nicht finden kann :D. Werde wohl so ca. 15:15 in der Jugendherberge sein.

hmm also ich werd wahrscheinlich wieder klassischerweise so ziemlich als erster in bad homburg sein, war ich zumindest bis jetzt immer(manchmal war michael müller nur schon von mir da) - aber ich werd wohl erst n spiel bestreiten, wenn irgendwer gegen mich antritt :D - freu mich schon *g*
aber meine fahrkarte hab ich grad keinen bock zu suchen, deswegen weiß ich nich so genau, wann cih da sein werde

hey, beim 2. oder 3. war ich als erstes dar und hab mich ne 1/2h gelangweilt ;)

Also planmäßíg komm ich um 13:15 in Bad Homburg an. Wie lange ich dann brauche um die JH zu finden ist ne andere Sache ;). Auf jeden Fall bin ich ganz gespannt darauf endlich mal die Kotelettentiere Kennen zu lernen  :D .

bevor ichs selber vergesse:

[quote="mein anfahrtsplan"]Karlsruhe Hbf  18.03.05  ab  10:54  Gleis 4   IC 2372 InterCity
Frankfurt(Main)Hbf  18.03.05  an  12:18  Gleis 13

Frankfurt(Main)Hbf  18.03.05            Fußweg 10 Min.
Frankfurt Hbf (tief)  18.03.05

Frankfurt Hbf (tief)  18.03.05  ab  12:39  Gleis 104   S 5 S-Bahn Richtung: Bad Homburg
Bad Homburg  18.03.05  an  13:02

Dauer: 2:08[/quote]

komme dann um 13:02 an... erster? keine ahnung, eher nicht...

  dg

ich komme erst um 14:45 an. geht diesesmal leider nicht früher  :(

muss meine angaben leider ändern, da ich soeben meine fahrkarte gefunden habe :D

Komme erst um 14:15 an.

Bis dahin, eine gute Anreise


counting
distinguishability
number theory

How many ways are there to choose m objects of n different types, given that you can take as many of each type as you want (order doesn't matter, or it would just be $ n^m$)? I've seen this appear often, but have never seen an answer.

What if the n types are all indistinguishable?

It's still balls and urns. I am truly embarrassed that I posted this topic, I have done balls and urns a million times before, but for some strange reason couldn't recognize it after thinking for a long time this time.

Isn't balls and urns for distributing indistinguishable objects to distinguishable objects?
In this case my question is regarding distributing indistinguishable to indistinguishable (with duplicates)...

Also how about distinguishable to indistinguishable?

n distinguishable objects into m distinguishable bins - $ n^m$ (multiplication principle)

n indistinguishable objects into m distinguishable bins - $ \binom{n \plus{} m \minus{} 1}{n}$ (balls-in-boxes)

n distinguishable objects into m indistinguishable bins - computationally difficult (related to the theory of [url=http://en.wikipedia.org/wiki/Partition_of_a_set]partition of sets[/url])

n indistinguishable objects into m indistinguishable bins - computationally difficult (related to the theory of [url=http://en.wikipedia.org/wiki/Partition_(number_theory)]partition of numbers[/url])

Maybe an example will help:

You have an infinite amount of donuts (they're in your back yard, go check if you don't believe me!). You want to create a box of 10 of them. There are two types, jelly-filled and creme-filled. How many ways can you make the box?

Well, let's look at the box as 10 slots: _ _ _ _ _ _ _ _ _ _

What if we placed a divider in there?

_ _ _ _ | _ _ _ _ _ _

Let's say everything before the divider is jelly and everything after it is creme. Now it's easy; there are obviously $ \binom{11}{1}\equal{}\boxed{11}$ places to put the divider, so this is your answer.


What is the coefficient of x^99 in the polynomial expansion of

   (x - 1)(x - 2)(x -3) ... (x - 100). Good luck... :)

The degree of polynomial is 100

hence coefficient of $x^{99}, b =-(sum of roots)= \frac{-100(101)}{2}=-5050$

Ok, here's a little bit harder question :) (or maybe not?)

What are the coefficients of $x^{98}$ and $x^{97}$?

You use the viete formulas (I think that's their name).  I can't explain them well, especially in this context.  They are supposed to be used to find the sum of products of roots.

But anyway... I think coefficient of $x^{98}$ would be $1 \cdot 2 + 1 \cdot 3 + ... + 2 \cdot 3 + 2 \cdot 4 + ... + 99 \cdot 100$
And I think the coefficient of $x^{97}$ would be $1 \cdot 2 \cdot 3 + 1 \cdot 2 \cdot 4 + ... + 1 \cdot 3 \cdot 4 + ... + 98 \cdot 99 \cdot 100$.  I think it would be negative though...
Both of which would take a while to compute without a calculator.

Hopefully you notice a pattern, as this is the viete formulas.  $\frac a a$, where $a$ is the lead coefficient is obviously $1$.  $\frac b a$ where $b$ is the coefficient of the term with degree one less than the polynomial's degree, is equal to the sum of all the possible products of roots with one root in each term (ie. sum of all the roots).  $\frac c a$, where $c$ is the coefficient of the term with degree two less than the polynomial's degree, is equal to the sum of all possible products of roots with two roots in each term.  As shown above, I multiplied all the roots together, two in each term, and add them up.  The pattern keeps going until you get to $\frac x a $ where $x$ is the constant term, which is equal to the product of all roots (a well known identity).

I don't know how the parity works with viete's formulas though.  I think it alternates based on degree of the polynomial, whether it is even or odd.  Like I said above, I think the coefficient of $x^{97}$ would be negative, but I'm not sure.

[quote="deej21"]
I think coefficient of $x^{98}$ would be $1 \cdot 2 + 1 \cdot 3 + ... + 2 \cdot 3 + 2 \cdot 4 + ... + 99 \cdot 100$
And I think the coefficient of $x^{97}$ would be $1 \cdot 2 \cdot 3 + 1 \cdot 2 \cdot 4 + ... + 1 \cdot 3 \cdot 4 + ... + 98 \cdot 99 \cdot 100$.  I think it would be negative though...
Both of which would take a while to compute without a calculator.[/quote]
You're on the right track, deej21. Here's a hint for $x^{98}$.
$(1+2+3+...+99+100)^2=(1^2+2^2+3^2+...+99^2+100^2)+S$
Do you see the relationship between $S$ and the coefficient of $x^{98}$? :)

Yeah, I think $S=2c$ where $c$ is coefficient of $x^{98}$.  Because $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, $S$ in that case is all the possible products of two of the numbers * 2... yeah.

Also
$S= \sum _{n=1} ^{100} (n^3 - n^2)$
Right?

[quote="deej21"]Yeah, I think $S=2c$ where $c$ is coefficient of $x^{98}$.  Because $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$, $S$ in that case is all the possible products of two of the numbers * 2... yeah.[/quote]
Yup. The coefficient of $x^{98}$ is $\frac{1}{2}S$.

[quote="deej21"]Also
$S= \sum _{n=1} ^{100} (n^3 - n^2)$
Right?[/quote]
I think so.
$\sum_{n=1}^{100} n^3=\left(\frac{n(n+1)}{2}\right)^2=\left(\sum_{n=1}^{100} n \right)^2$

So $S=\left(\sum_{n=1}^{100} n \right)^2-\sum_{n=1}^{100} n^2=\sum _{n=1} ^{100} (n^3 - n^2)$


$V_{n}$ be the Vandermonde determinant $V(1,2,...,n)$

prove that

$\ln(V_{n})= \frac{n^{2}.\ln(n)}{2}-\frac{3n^{2}}{4}+(\frac{2\pi}{2}-\frac{1}{4})n-\frac{\ln(n)}{12}+k+O(\frac{1}{n})$

where $k$ is a constant


inequalities
algebra
polynomial
inequalities proposed
3-variable inequality
133

It's just a coincidence that Nam and I proposed the same inequalities. We like it a lot since it looks simple, has two equality cases but it's not that simple to solve.

Here it is:

Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that \[\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\geq2.\] You can find a nice and short solution in new Vasc's inequality book.

I saw this solution in Vasc's book but I can't post it because it belong to Vasc's book but I have a idea to make it stronger:If you saw this solution can you do it?
Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\ge 3-\frac{2}{\sqrt{x+z^{2}}+\sqrt{y+x^{2}}+\sqrt{z+y^{2}}}\ge 2$ :)

I am able to prove it in the case when one of the numbers is zero. :blush:
Take WLOG $z=0$.And then $x+y=1$.Now we have to prove \[\sqrt{x+y^{2}}+\sqrt{y}+x\geq 2\] .But$x+y=1$ therefore we set $x=1-y$ and the ineq is equivalent to \[\sqrt{y^{2}-y+1}+\sqrt{y}\geq y+1 \iff\]  \[y^{2}-y+1+y+2\sqrt{y(y^{2}-y+1)}\geq y^{2}+2y+1\] and after canceling we have to prove \[\sqrt{y(y^{2}-y+1)}\geq y\iff\]  \[y(y-1)^{2}\geq 0\] .This proves the equality case $(x;y;z)=(0;1;0)$ and the permutations.

What about $a=b=1={1 \over 3}$.
Can someone post a complete solution?

[quote="gemath"]
Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\ge 3-\frac{2}{\sqrt{x+z^{2}}+\sqrt{y+x^{2}}+\sqrt{z+y^{2}}}\ge 2$ :)[/quote]
Can hungkntn or Vasc or anybody do it?  :D  :D

Zhao Bin and I have checked the solution in your book and noticed that there is something wrong. Can you post it here, Vasc?

Vasc's  solution is the following
Let z the min of {x,y,z} then $\sqrt{x}+\sqrt{y}\geq \sqrt{z}+\sqrt{x+y-z}$

By the above inequality
$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}\geq \sqrt{c+b^{2}}+1-c$
so it suffices to peove that
$\sqrt{c+a^{2}}+\sqrt{c+b^{2}}\geq 1+c$  or

by Squaring and AM-GM
$\sqrt{c+a^{2}}\sqrt{c+b^{2}}\geq c+ab$

which is true by Cauchy.

I haven't really seen the proof in VASC's book,I just see the proof that Thuan sent to me.
After some thinking,I think the proof is right.
[quote="silouan"]Vasc's  solution is the following
Let z the min of {x,y,z} then $\sqrt{x}+\sqrt{y}\geq \sqrt{z}+\sqrt{x+y-z}$

By the above inequality
$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}\geq \sqrt{c+b^{2}}+1-c$
.[/quote]

I think we should assume $a \ge b \ge c$ or $c \ge b \ge a$ when we prove the inequality by the above one.It is OK because the inequality is cyclic.But I think it is necessary to say it out. :D

[quote="Vasc"]The left inequality is not true for x=0, y=1/3, z=2/3. :P[/quote]
oh that's right Vasc sorry for the mistake  :) >but two problems following I believe it is true:
Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\ge 3-\frac{2}{\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}}\ge 2$ and
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\ge 1+\frac{\sqrt{x+z^{2}}+\sqrt{y+x^{2}}+\sqrt{z+y^{2}}}{2}\ge 2$
hope you are interested in  :D

[quote="pvthuan"]It's just a coincidence that Nam and I proposed the same inequalities. We like it a lot since it looks simple, has two equality cases but it's not that simple to solve.

Here it is:

Let $x,y,z$ be nonegative real numbers such that $x+y+z=1$, prove that
\[\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\geq2. \]
You can find a nice and short solution in new Vasc's inequality book.[/quote]
We need a lemma : If $a,b,c,d \ge 0$ and $a+b=c+d;|a-b| \le |c-d|$ then we have $\sqrt{a}+\sqrt{b}\ge \sqrt{c}+\sqrt{d}$.
Applying the lemma we have
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\ge (x+y)+\sqrt{z+y^{2}}+\sqrt{z+x^{2}}\ge x+y+\sqrt{(\sqrt{z}+\sqrt{z})^{2}+(x+y)^{2}}=1-z+\sqrt{4z+(1-z)^{2}}=2$

[quote="pvthuan"]Let $ x,y,z$ be nonegative real numbers such that $ x \plus{} y \plus{} z \equal{} 1$, prove that

$ \sqrt {y \plus{} z^{2}} \plus{} \sqrt {z \plus{} x^{2}} \plus{} \sqrt {x \plus{} y^{2}}\geq2.$[/quote]$ \prod\left[T\pm\sqrt {y(x \plus{} y \plus{} z) \plus{} z^2}\pm\sqrt {z(x \plus{} y \plus{} z) \plus{} x^2}\pm\sqrt {x(x \plus{} y \plus{} z) \plus{} y^2}\right]$

$ \equiv F(T),$

$ F[2(x \plus{} y \plus{} z)]$

$ \equal{} \minus{} \frac {16}{3}(x \plus{} y \plus{} z)^2\sum_{cyc}zx(y \minus{} z)^2\left(24x^2 \plus{} 61y^2 \plus{} 48z^2 \plus{} 107yz \plus{} 69zx \plus{} 75xy\right)$

$ \leq0\Longrightarrow$

$ 2(x \plus{} y \plus{} z)\leq\sqrt {y(x \plus{} y \plus{} z) \plus{} z^2} \plus{} \sqrt {z(x \plus{} y \plus{} z) \plus{} x^2} \plus{} \sqrt {x(x \plus{} y \plus{} z) \plus{} y^2},$

which is the great root of the eighth degree polynomial $ F(T).$

See also here : http://www.mathlinks.ro/viewtopic.php?t=216119

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=539304

Let $x,y,z$ be nonegative real numbers such that $x+y+z=1.$ [url=https://artofproblemsolving.com/community/c6h1996040p13928225]Prove that[/url]
 $$2\leq \sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}<\frac{11}{5}$$

Let $a,b,c$ be a permutation of $x,y,z$ and $a\geq b\geq c$.

Then

 $(a+b)^2\geq \max\{x+y^2,y+z^2,z+x^2\}$

$(b+c)^2\leq \min\{x+y^2,y+z^2,z+x^2\}$

$(a+b)^2+(b+c)^2+(c+a)^2=x+y^2+y+z^2+z+x^2$

which implies that

$\{(x+y)^2,(y+z)^2,(z+x)^2\}$ majorizes $\{x+y^2,y+z^2,z+x^2\}$

Since $f(x)=\sqrt{x}$ is a concave function on $[0,\infty)$, then by Karamata inequality, we have

$$\sqrt{x+y^2}+\sqrt{y+z^2}+\sqrt{z+x^2}\geq\sqrt{(x+y)^2}+\sqrt{(y+z)^2}+\sqrt{(z+x)^2}=2$$

[quote=sqing][hide=*][quote=yleo]Let $x,y,z$ be a permutation of $a,b,c$ and $x\geq y\geq z$.

Then

 $(x+y)^2\geq \max\{a+b^2,b+c^2,c+a^2\}$

$(y+z)^2\leq \min\{a+b^2,b+c^2,c+a^2\}$

$(x+y)^2+(y+z)^2+(z+x)^2=a+b+c+a^2+b^2+c^2$

which implies that

$\{(a+b)^2,(b+c)^2,(c+a)^2\}$ majorizes $\{a+b^2,b+c^2,c+a^2\}$

Since $f(x)=\sqrt{x}$ is a concave function on $[0,\infty)$, then by Karamata inequality, we have

$$\sqrt{a+b^2}+\sqrt{b+c^2}+\sqrt{c+a^2}\geq\sqrt{(a+b)^2}+\sqrt{(b+c)^2}+\sqrt{(c+a)^2}=2$$[/quote][/hide]
$x\geq y\geq z$
[b]No.[/b][/quote]

Thanks, I have edited.

I have a different idea to cope with this problem.
To put it in a brief way, that's
$\sqrt{a+b^2}  = \sqrt{\frac{a}{a+1}(a+c)^2+\frac1{a+1}(a+b)^2}$
I tried Jensen and it came out as a Rearrangement Inequality which appeared a wrong order.
I tried the Minkowski  Inequality too and it failed to fit in the (1,0,0) equality case

Anyone to help? I think I'm around the corner with this equation


Ring Theory
Support
videos
articles
blogs

http://cosmos.bcst.yahoo.com/scp_v3/viewer/index.php?pid=16598&rn=49750&cl=1113838&ch=68276&src=news

I remember hearing that when he wanted someone dead, he had that person's best friend do it, or the best friend's family will be killed.  I don't think you can get more cruel than that.  He deserved it.

What do you think?

I disagree with his sentencing and hope that he receives a different punishment through the appeals process.

But something that I think is much more to note would be this: http://biz.yahoo.com/prnews/061030/nym224.html?.v=44 It claims that they were willing to accept the US' demands yet the US did not response in the hours before they invaded. If this is true, we could've avoided the entire war. Then again, it could just be BS spouts by him although he claims to have proof.

I find it hard to disagree that, if anyone deserves the death penalty, then he must be one of the most deserving. But killing him benefits the US and their allies very little - they could easily afford to keep him locked up - but it will surely trigger some increased violence, which is something the occupying forces should be trying to avoid. Then again, his sentencing comes from the Iraqi law, not by American request, I presume.

And as for the story you linked to, Ignite, can it really be believed? Why has no-one heard about this before? What motive would the US have for rejecting it? Why would Saddam yield to this ultimatum at the last minute?

[quote]Then again, his sentencing comes from the Iraqi law, not by American request, I presume. [/quote]
Cough, cough

[quote]Why would Saddam yield to this ultimatum at the last minute?[/quote]
Certain death?  :wink:

When you put it that way... yeah, life in prison is better.

[quote="13375P34K43V312"]I see death sentences as unethical. It's murder in itself. Anyway, if you kill him, he's not really techincally feeling the punishment. Life in prison please.

I bet in 400 years, people will be looking at the way we killed people for their crimes the way Americans today look at hanging people.[/quote]

He deserves to go to h*ll as quickly as possible.

[quote]Quote:
Then again, his sentencing comes from the Iraqi law, not by American request, I presume.

Cough, cough [/quote]

I'm not exactly an expert on Iraqi law, but what sentence do you think that it reserves for the murder of 140+ people? I'm not saying the Americans didn't play a part in bringing him to trial- but they didn't affect the sentence.

How exactly do americans look at hanging by the way? I find it much less cruel than, for example, the electric chair.

And here is some news: US was always behind Saddam in his killings and without their help he could never rise to power. But they collared him when he grew disobedient. They never thought much of disloyal agents ...

[quote="Ignite168"]He deserves to go to h*ll as quickly as possible.[/quote]
If he's going to burn in h*ll for all eternity, what difference does another 30 years sitting a prison cell make?

[b]tim[/b]: [b]pieterminate[/b] might have meant to imply that Iraqi law has been heavily influenced by the US (overtly or not), so it's a bit questionable to make a big deal over the distinction between Iraqi and US law. It's what US people wanted to happen to him (except those who are opposed to the death penalty).

[b]bubka[/b]: I agree, there's not that much difference, morally speaking, between hanging and our current methods. Maybe people think that hanging was bad because people didn't always die right away and sometimes a botched hanging would result in a really unpleasant experience for a criminal? These people should read the following account of Ethel Rosenberg's execution (by electric chair):
[hide="hidden because its grisly"] [quote]She died a lot harder. When it appeared that she had received enough electricity to kill an ordinary person and had received the exact amount that had killed her husband, the doctors went over and pulled down the cheap prison dress, a little dark green printed job, and placed the steto ... stetho ... can't say it ... placed the stethoscopes to her and then looked around and looked at each other rather dumbfounded and seemed surprised that she was not dead.

And she was given more electricity which started again and a kind of a ghastly plume of smoke rose from her head and went up against the skylight overhead. After two more of those jolts, Ethel Rosenberg had met her maker. [/quote][/hide]
But I think it might be more appropriate to say that people here in the future might look back on our executions from this time period and think of it in the same way that we think of the [i]lynchings[/i] from our past now....

When in 400 hundred years people look back they will probably be astonished for mankind still had death penalty. Most unjustifiedly in developed countries since I don't want to judge poor countries and taking into account the present world order, with the same ruler. Death penalty is the acknowledgement of the impossibility of rehabilitation aswell as the absolute absence of errors in the judicial process. In general it should be abolished. This doesn't mean I believe in a possible Sadam rehab, in his extreme case DP is useless anyway. As has been remarked life in prison would serve equally to prove the magnitude of his crimes, and would also serve to put an example of what a civilised society is.
I must say I was surprised when Sadam got captured alive. After so many crimes on his head it seems pretty coward of him not to have a bullet saved for himself knowing he was going to end this way. But again this wouldn't be the issue if there wasn't war on Iraq in the first place.
Islam needs help, non-agressive help from within, not the alienation and oil-paid 'help' is receiving.

Ursula

You look at the controversy surrounding the death penalty today, and it's all about DNA evidence and yadda-yadda-yadda. He performed these murders in front of thousands of people! That's the only unethical thing here. Also on his resume, starting a full-fledged war. The only other modern person on that list: Adolf Hitler. What happened to him? He died. Not by force of law, sure, but sure shootin' he's dead. And by putting him in jail, we always run the risk that he could somehow escape. He's not a good person, none of these radical Muslims are. I think that every TV station in America, around the world for that matter, should televise some guards kicking the stand out from underneath him and watch his filthy little neck get strangled ruthlessly by a piece of rope. We can't run the risk of keeping him alive.

[quote="Ignite168"]He deserves to go to h*ll as quickly as possible.[/quote]

*gasp* You shouldn't say that! He was cruel, but the question is "Why?". No excuse like "My mother didn't give me twinkies." would satisfy me, but some rational excuse, please.

Gee...maybe killing innocent victim after innocent victim after innocent victim after innocent victim after innocent victim after...

I think he should have lived in misery instead of just being put in eternal rest. After all the pain he has caused some Iraqi people....

severe punishment should be inevitable

[quote="Ignite168"]I disagree with his sentencing and hope that he receives a different punishment through the appeals process.
[/quote]
What would your sentence be, and why is it a better one?

I think Saddam Hussein deserved it. He was a cruel killing dictator.

It is not possible to "deserve" death. If he is being sentenced for "revenge", then it is for the wrong reasons.

If there is plausible cause to believe that him being alive will cause increased violence, then that would be a legitimate reason to kill him.

Actually I really don't understand the whole "he deserved it" talk. How can you profess to know that someone "deserves" to die? What is your rationale for this?

[quote="kstan013"]You look at the controversy surrounding the death penalty today, and it's all about DNA evidence and yadda-yadda-yadda. He performed these murders in front of thousands of people! That's the only unethical thing here. Also on his resume, starting a full-fledged war. The only other modern person on that list: Adolf Hitler.[/quote]

So are you saying that George W. Bush is in no way responsible for beginning the war? Or his father in part, for putting Saddam Hussein in the position of power immeaditally?  I would argue otherwise.

[quote="kstan013"]He's not a good person, none of these radical Muslims are.[/quote]

Are all radicals bad people?  Or is it just radical Muslims?  I am guessing (correct me if I'm wrong) that you have not met every radical Muslim, and cannot genralize like that.  I am sure that you would say I was wrong if I said that everyone with the name "Tom" was stupid and should be killed because they would never amount to anything or contribute to society.

[quote="kstan013"]I think that every TV station in America, around the world for that matter, should televise some guards kicking the stand out from underneath him and watch his filthy little neck get strangled ruthlessly by a piece of rope. We can't run the risk of keeping him alive.[/quote]

Killing someone in the United States, at least, is much costlier than keeping them imprisoned for the rest of their life - the legal system is tied up with their appeals, other cases are displaced, etc.  What risk would we be running by keeping him in a maximum security cell for the remainder of his life?  Killing him is sure to spring new violence - killing more people.

[quote="kstan013"]You look at the controversy surrounding the death penalty today, and it's all about DNA evidence and yadda-yadda-yadda...[/quote]

There are these posts coming from "America" that you know, I begin to be sincerely worried. The most discouraging thing is that before I start to write, I'm almost convinced it's useless. If what you want is to televise

[quote="kstan013"]..some guards kicking the stand out from underneath him and watch his filthy little neck get strangled ruthlessly by a piece of rope. We can't run the risk of keeping him alive.[/quote]

don't know why I have this feeling that you are not to be convinced by me, i mean, where do you get those words from? filthy little neck? strangled ruthlessly? Sounds like sixteenth century pirates slang to me. Have you thought about the issue carefully? We have come to a point where there's a big problem about Iraq, do you think it is all Hussein's fault? Have you read about Hussein's coming to power? Have you read about the other many dictators around the world with many deaths on their heads, that your beloved America has supported fully, and still hang around(Pinochet the clearest example)? So how does America decide which dictator will be taken down? I wonder

[quote="kstan013"]He's not a good person, none of these radical Muslims are[/quote]

.. and this you know because? Are you suggesting DP for all radical muslims? Have you asked yourself the question, why is it that Islam is a religion where fundamentalism is relatively common? Do you think one should ask oneself that question? In case you were born in Iraq or Yemen or whatever, are you so sure you wouldn't be a fundamentalist? Oh, but since you didn't what we should do is to erase this plague muslims from the face of the earth?

boy, hopefully I won't see no filthy little necks strangled ruthlessly this time. I don't think they will televise it in many countries in Europe where by the way DP doesn't exist for some time now. You're also wrong about the controversy about DP being based only in DNA evidence.

And I assume when you refer to the "full-fledged war" you are refering to the Gulf War when Iraq invaded Kuwait starting it because I don't see many responsibilities in Iraq starting this last one, specially after the mass destruction weapons smoke clouds and after Iraq's intention to cooperate with UN right before the US decided they have had enough yadda-yadda

Ursula

[quote="JSteinhardt"]Um also I think it should be pointed out that killing Saddam does not improve the state of anybody's life, and so seems a bit pointless?[/quote]

How about backing up your statements.

Do you really think that there isn't anyone in Iraq, who may have lost loved ones because of Saddam, who wants to see retribution given? Do you think that they aren't any people who are so angry at Saddam for ruling the way he did and want to see him die?

Plus, many Republicans lives will be benefitted from Saddam dying. It creates good publicity.

[quote="pieterminate"]How about new conflicts that would be caused by saddam's death?
[url=http://www.chicagotribune.com/news/nationworld/chi-0611060218nov06,1,16059.story?coll=chi-newsnationworld-hed9]......[/url][/quote]

Shi'ites and Kurds will be mad if Saddam Hussein isn't sentenced to death.

[quote="Ignite168"]How about backing up your statements.

Do you really think that there isn't anyone in Iraq, who may have lost loved ones because of Saddam, who wants to see retribution given? Do you think that they aren't any people who are so angry at Saddam for ruling the way he did and want to see him die?

Plus, many Republicans lives will be benefitted from Saddam dying. It creates good publicity.[/quote]

How about backing up yours? Someone achieving "retribution" is not the same as the state of their life being improved. I would be very interested in seeing you show to me that the state of someone's life was improved because Saddam Hussein was dead rather than in costudy.

Now if you prefer to argue that they are happier (which is not what I said, but which I also agree with), then you would have to show that retribution leads to happiness.

I would argue that the exact opposite occurs. The attainment of retribution only supports negative emotions, such as anger, vengeance, etc., and perhaps even reminds them of the pain incurred by Saddam. I could launch into a very long dissertation about negative emotions and what actually constitutes happiness but I think I could do just as well by having you Google any of "Socrates", "Plato", "Aristotle", "Augustine", "Aquinas", "stoicism", "cynicism", or "epicureanism" and have the benefit of 7 major philosophies that all seem to say the same thing.

Okay, so it's obvious everyone is swallowing the story like they should :

mean dictator terrorizes own people and neighbours, ousted by brave allied forces and now tried for his crime.

 :|

Here we go again .....:

[img]http://www.gwu.edu/~nsarchiv/NSAEBB/NSAEBB82/handshake300.jpg[/img]
December 1983 :
Donals Rumsfeld is head of a USA-delegation in Iraq, during the Iran-Iraq war, and meets Saddam Hussein.

Watch the video here : [url]http://www.gwu.edu/~nsarchiv/NSAEBB/NSAEBB82/[/url]

[img]http://upload.wikimedia.org/wikipedia/en/a/a7/Halabja1.jpg[/img]
March 1988 :
the poison attack on the Kurdish village Halabja in Iraq.

As we can read here ... :
[url=http://news.bbc.co.uk/onthisday/hi/dates/stories/march/16/newsid_4304000/4304853.stm]BBC article[/url]
US Defence Intelligence Agency blamed Ira[u]N[/u], not IraQ for the attack.

In the months before the Iraqi invasion, this Halabja attack was a very popular argument to make a case against Iraq.

And last but not least :

Sunday, November 5, 2006 : Saddam Hussein is sentenced to death
Tuesday,November 7,2006 : Midterm elections in the USA

[quote="funcia"][color=red][size=200]Hey he is an enemy combatant!!!! NOW THAT BUSH HAS MORE POWER HE SHOULDN'T OF GIVEN SADDAM A TRIAL!!![/size][/color]
Funcia is really mad.[/quote]
Uhm, since when does Bush have "more power"?And why would that imply that Saddam shouldn't have been given a trial? :huh:

in my opinion Death Sentence is against Democracy.
So what kind of Democracy is US bring to Iraq.

The same kind of slightly twisted democracy they have themselves, where the death penalty is enforced.  :(

But anyway, what do you think the majority of Iraqis (the Sh'ites at least) want?

If anyone finds a poll of Iraqis over whether Saddam should be killed, I would be interested to see it. Also, are there any Iraqi Mathlinkers who could voice an opinion on the topic?

edit: I did in the end find [url=http://www.csmonitor.com/2006/1108/dailyUpdate.html]this selection of extracts from Iraqi blogs[/url], and the opinion there was quite divided.

[quote="tim1234133"]The same kind of slightly twisted democracy they have themselves, where the death penalty is enforced.  :(

But anyway, what do you think the majority of Iraqis (the Sh'ites at least) want?

If anyone finds a poll of Iraqis over whether Saddam should be killed, I would be interested to see it. Also, are there any Iraqi Mathlinkers who could voice an opinion on the topic?[/quote]
Well, the idea is that the tribunal was completely indepedent and had nothing to do with the USA.
And the fact that he is sentenced to death two days prior to the midterm elections is of course coincidence.

[quote="JSteinhardt"]Um also I think it should be pointed out that killing Saddam does not improve the state of anybody's life, and so seems a bit pointless?[/quote]Yes it will, everyone will be happy that yet another evil force in the world has vanished from the face of the planet (fighting evil is good)

[quote="JSteinhardt"]His existence does not cause evil.

His seat of power does cause evil.

He has already been removed from his seat of power. Killing him will not prevent any evil that has not already been prevented by his detainment.[/quote]
That is what i mean.

[quote="tim1234133"]But anyway, what do you think the majority of Iraqis (the Sh'ites at least) want? .[/quote]

What about Sunnis,who are supporting Saddam.
How would be their reaction, if Saddam is death?

[quote="JSteinhardt"]He has already been removed from his seat of power. Killing him will not prevent any evil that has not already been prevented by his detainment.[/quote]

You're thinking in quite utilitarian (and simplistic) terms.  The world is more complicated than that.  Consider these possible justifications for his death:

- In the interest of justice, as an execution of a law against criminals,
- In the interest of justice, as retribution for the death and misery he has caused,
- To free up time, energy, space, and money that would otherwise be wasted in confining him and keeping him alive,
- To reassure the populace and boost morale

http://en.wikipedia.org/wiki/Military_Commissions_Act_of_2006

http://www.msnbc.msn.com/id/15321167/
[size=167] FOR THOSE WHO DIDN'T UNDERSTAND ME.[/size]
Also-the US had a lot to do with Saddam having power.

[quote="fredbel6"]
And last but not least :

Sunday, November 5, 2006 : Saddam Hussein is sentenced to death
Tuesday,November 7,2006 : Midterm elections in the USA
[/quote]

reading this topic from the beginning i wondered would anyone notice this "strange "coincidence""

Well it sure didn't help the Republicans...if that's what you people are implying...

Yea. The Dems won on the platform of anti-Iraq war.

SO? People saw through the disguise!! They knew that we ourselves helped get Saddam in power.


[quote="Lazarus"]The sum of the first six terms of a geometric series is 8. If the sum of the next six terms is 18, what is the sum of terms 1, 2, 3, 10, 11, and 12?[/quote]
[hide="solution"]
Let first term be $ q$, ratio between terms $ r$
Sum of first 6:
$ q(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1) \equal{} 8$
Sum of next 6:
$ q(r^{11} \plus{} r^{10} \plus{} r^{9} \plus{} r^8 \plus{} r^7 \plus{} r^6) \equal{} qr^6(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1) \equal{} 18$
Dividing this equation by the one for the sum of first 6 gives $ r^6 \equal{} \frac {9}{4}$
sum of 4th to 9th:
$ q(r^8 \plus{} r^7 \plus{} r^6 \plus{} r^5 \plus{} r^4 \plus{} r^3) \equal{} r^3q(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1)$
Using Substitution, $ r^3q(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1) \equal{} 8r^3$
Since $ r^6 \equal{} \frac {9}{4}$, by taking the square root of both sides, $ |r^3| \equal{} \frac {3}{2}$
The sum of the 4th to 9th terms is $ 12$.

Subtracting that from the total of the first 12 terms, $ 18\plus{}8\equal{}26$, the answer is $ 14$.
[/hide]

[quote="ProtestanT"]
[hide="solution"]
Let first term be $ q$, ratio between terms $ r$
Sum of first 6:
$ q(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1) \equal{} 8$
Sum of next 6:
$ q(r^{11} \plus{} r^{10} \plus{} r^{9} \plus{} r^8 \plus{} r^7 \plus{} r^6) \equal{} qr^6(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1) \equal{} 18$
Dividing this equation by the one for the sum of first 6 gives $ r^6 \equal{} \frac {9}{4}$
sum of 4th to 9th:
$ q(r^8 \plus{} r^7 \plus{} r^6 \plus{} r^5 \plus{} r^4 \plus{} r^3) \equal{} r^3q(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1)$
Using Substitution, $ r^3q(r^5 \plus{} r^4 \plus{} r^3 \plus{} r^2 \plus{} r^1 \plus{} 1) \equal{} 8r^3$
Since $ r^6 \equal{} \frac {9}{4}$, by taking the square root of both sides, $ |r^3| \equal{} \frac {3}{2}$
The answer is $ 12$
[/hide][/quote]

not quite =)

read carefully

[quote="Lazarus"]

not quite =)

read carefully[/quote]
fixed.

[hide]We can take the geometric mean of the sums of the first 6 and last 6 terms and subtract that from the sum of the first 12 terms to get $ 14$.[/hide]

Note that in a geometric series, the geometric mean of terms $ x$ and $ y$ is the term $ \frac{x\plus{}y}{2}$. Therefore, term 4 is the geometric mean of terms 1 and 7; similarly, term 5 is the geometric mean of terms 2 and 8, and we can extend this pattern (term 9 is the geometric mean of terms 6 and 12). As a result, the sum of terms 4 to 9 is the geometric mean of the sum of terms 1 to 6 and the sum of terms 7 to 12.

[quote="Lazarus"]Note that in a geometric series, the geometric mean of terms $ x$ and $ y$ is the term $ \frac {x \plus{} y}{2}$. Therefore, term 4 is the geometric mean of terms 1 and 7; similarly, term 5 is the geometric mean of terms 2 and 8, and we can extend this pattern (term 9 is the geometric mean of terms 6 and 12). As a result, the sum of terms 4 to 9 is the geometric mean of the sum of terms 1 to 6 and the sum of terms 7 to 12.[/quote]

Are you sure?  I thought that the geometric mean of a and b is defined to be the square of a times b.  In that case we take the geometric mean of 18 and 8 and we get the square of 144 or 12.  Then we subtract this from 26 to get 14.
However, is there a proof for the reason this works?


analytic geometry
geometry
parallelogram
vector

Three points on a cartesian plane are on [b]A[/b](0,4), [b]B[/b](-1,1) and [b]C[/b](2,2). List all possible coordinates of [b]D[/b] so that ABCD is a parallelogram. One solution is by graphing. However, here is a more challenging one. Solve algebraically.

Source: Me :D

Heres a hint, not very quick but its simple algebra and it works...
[hide]In parallelogram ABCD, AB || CD, BC|| AD and AB = CD, BC = BD [/hide]

[quote="Farenhajt"]Very easy if you

[hide]use vectors[/hide][/quote]

I am not really good at vectors. I would appreciate if you can show me the solution using it.  I had a long time doing it using manual algebra.

First, $\overrightarrow{AB}=\langle -1,-3\rangle, \overrightarrow{BC}=\langle 3,1\rangle, \overrightarrow{CA}=\langle -2,2\rangle$

Vector addition by the rule of the parallelogram gives us three possibilites:

1) $\overrightarrow{AD_1}=\overrightarrow{AB}+\overrightarrow{AC}=\langle -1,-3\rangle+\langle 2,-2\rangle=\langle 1,-5\rangle$, and now $\overrightarrow{OD_1}=\overrightarrow{OA}+\overrightarrow{AD_1}=\langle 0,4\rangle +\langle 1,-5\rangle=\langle 1,-1\rangle$. Therefore $D_1(1,-1)$

The other two cases are similar and equally easy:

2) $\overrightarrow{BD_2}=\overrightarrow{BA}+\overrightarrow{BC}$

3) $\overrightarrow{CD_3}=\overrightarrow{CA}+\overrightarrow{CB}$

[quote="frt"]The problem restricts the order of the verticies...[/quote]

As far as I know, there's no RULE about ordering the vertices of any polygon. There exists only the usual PRACTICE (whose only merit is simplification of the formulas, since it is "neat" to have $A_n$ between $A_{n-1}$ and $A_{n+1}$ for ANY $n$), but I've never seen a geometry book that states something like: "Vertices are ALWAYS named either clockwise or counterclockwise, and no other order is allowed."

Japanese rules might differ, though.

everywhere I've seen does say that order matters.

[hide]
Going from B-->A we have (x+1, y+3)
Going from C-->D we have the same, so plug in $(2+1, 2+3) = (3, 5)$
[/hide]

[hide]
Going from A-->B we have (x-1, y-3).
From C-->D, we plug in, $(2-1, 2-3) = (1, -1)$.

Going from C-->A we have (x-2, y+2).
From B-->D we have $(-1-2, 1+2) = (-3, 3)$.

Now that I'm thinking about it, I'm not sure how to solve this algebraically if there is no restriction on order. Would you have to do 3 different solvings? (waste of time)
[/hide]

[quote="guile"][quote="frt"]Technically, since the problem says "List all possible coordinates of D so that [b]ABCD[/b] is a parallelogram", there's only one possible location for $D$... (The problem restricts the order of the verticies, so we can't have ADBC or ABDC) :ninja:

Anyway, even if you don't know vectors, you can still use the fact that the two diagonals of a parallelogram bisect each other, which gives you pretty much the same calcultion as Farenhajt's solution.[/quote]

We shouldn't consider the order.  We are only looking for the possible coordinates of the points. I just used ABCD to label the coordinates.[/quote]
The question should then be worded for what points D can the points A, B, C, and D form a parllelogram.  ABCD is defined to be a quadrilateral with B next to A, C next to B, D next to C, and A next to D.


[quote][b]Dr Pepper[/b] is making good on its promise of free soda now that the release of Guns N' Roses' "Chinese Democracy" is a reality. The soft-drink maker said in March that it would give a [b]free soda[/b] to everyone in America if the album dropped in 2008. "Chinese Democracy," infamously delayed since recording began in 1994, goes on sale Sunday.

"We never thought this day would come," Tony Jacobs, Dr Pepper's vice president of marketing, said in a statement. "But now that it's here, all we can say is: The Dr Pepper's on us."

[b]Beginning Sunday at 12:01 a.m., coupons for a free 20-ounce soda will be available for 24 hours on Dr Pepper's Web site. They'll be honored until Feb. 28.[/b]
[/quote]

[url]http://news.yahoo.com/s/ap/20081120/ap_on_en_mu/guns_n_roses_dr_pepper[/url]

so pretty much free soda, where can you go wrong with that

oh and you can discuss your enjoyment of the soda, gunsnroses, etc. all here :)

[url]http://www.drpepper.com/freeDrPepper/?icamp=hp_dpfree_coupon[/url]

there's the link...less than 10 hours left to get YOUR free dr. pepper


geometry
circumcircle
power of a point
radical axis
geometry solved

Let $ M$ and $ N$ be two distinct points in the plane of a triangle $ ABC$ such that $ AM: BM: CM \equal{} AN: BN: CN$. Prove that the line $ MN$ contains the circumcenter of triangle $ \triangle ABC$.

Amazing result!
My solution uses many geometric facts, I believe there is a much more elementary solution.
Spilt $AM:BM:CM=AN:BN:CN$ into $\frac{AM}{BM}=\frac{AN}{BN}=\alpha$ and $\frac{BM}{CM}=\frac{BN}{CN}=\beta$. Let $X\neq X'$ be the points on AB such that $\frac{AX}{BX}=\frac{AX'}{BX'}=\alpha$. Y and Y' are defined similarly. Then M and N are the intersection points of the circles with diameters XX' and YY'(the Apollonian circles). Note that MN is the radical axis of the two circles, thus it suffices to show O has the same powers to them. Let D and D' be the midpoints of AB and XX' respectively. The power of O to the circle with diameter XX' is $OD'^2-D'X^2=OD^2+DX^2-D'X^2=OD^2+DX\cdot DX'$. Note that $\frac{AX}{BX}=\frac{AX'}{BX'}=\alpha$ implies A,X,B,X' are harmonic conjugate. Since D is the midpoint of BC, $DX\cdot DX'=CD^2$ and therefore $OD'^2-D'X^2=OD^2+DX\cdot DX'=OD^2+CD^2=R^2$. Similarly, we can also prove that O has the power $R^2$ to the circle with diameter YY'.

pretty much the same, I guess

Lemma:

Suppose $\omega$ is a circle of Apollonius with $P \in \omega \Rightarrow \frac{\overline{AP}}{\overline{BP}} = \alpha$. Then if $P'$ is the inverse of $P$ in the circumcircle, we know $\frac{\overline{AP'}}{\overline{OA}} = \frac{\overline{AP}}{\overline{OP}} \Rightarrow \overline{AP'} = \frac{\overline{AP} \cdot \overline{OA}}{\overline{OP}}$, so as a similar result holds for $\overline{BP'}$, we have $\frac{\overline{AP'}}{\overline{BP'}} =$ $\frac{\displaystyle \frac{\overline{AP} \cdot \overline{OA}}{\overline{OP}}}{\displaystyle \frac{\overline{BP} \cdot \overline{OB}}{\overline{OP}}} =$ $\frac{\overline{AP}}{\overline{BP}} = \alpha \Rightarrow P' \in \omega$, so $\omega$ maps to itself and is thus orthogonal to the circumcircle.

Proof: Applying this result to the Apollonius circles that define $M$ and $N$, we see that they are inverses in the circumcircle and thus are collinear with the circumcenter.

Is there something wrong with the following argument?

From the given,
\[\frac{AM}{AN} = \frac{BM}{BN} = \frac{CM}{CN}.\]
So, A, B, and C lie on the circle of Apollonius with respect to M and N and this common ratio.  But this circle is  symmetric with respect to MN, so its centre lies on MN.

[quote="ehsan2004"]Let $M$ and $N$ be two distinct points in the plane of a triangle $ABC$ such that $AM: BM: CM=AN: BN: CN$. Prove that the line $MN$ contains the circumcenter of $\triangle ABC$.[/quote]

There's an stronger fact.

Given $\triangle ABC$ and two points $P$ and $P'$ such that $AP: BP: CP=AP': BP': CP'$. Then $P$ and $P'$ are inverses wrt the circumcircle of $\triangle ABC$.

So it follows that $P$ and $P'$ are collinear with the circumcenter of $\triangle ABC$.