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summer program

pretty much like the old tug of war games, except with 3 teams

Team Philadelphia Cream Cheese, Team Bertolli, Team Fruit Loops

you have to join the team with the least members, and if the teams have the same number of members, you have to join the team with the least point, and if that still doesn't do it, then its your choice.

oh and you get 3 points each turn, use them anyway you want. (kinda like hurt and heal)

A team loses when they reach -75

Same general rules, no posting twice in an hour, no double-posting.

and copy and paste the stuff below and update the scores and stuff.

[b]+3 to Team Philadelphia Cream Cheese[/b]

[b]Team Philadelphia Cream Cheese: 3[/b]
dingzhou

[b]Team Bertolli: 0[/b]

[b]Team Fruit Loops : 0[/b]

Huh?  :huh: I thought you either +3 to yourself, +2 to yourself and -1 to somebody, +1 to yourself and -2 to somebody, +1 to yourself and -1 to somebody, and -1 to somebody else, -3 to somebody, or -2 to somebody and -1 to somebody else. (i.e., total = 3)

But... oh well:

+3 to Team Fruit Loops

[b]Team Philadelphia Cream Cheese: -3[/b]
dingzhou

[b]Team Bertolli: -3[/b]
1=2

[b]Team Fruit Loops : -3[/b]
leoxnlin

plus 3 to team FL

Team Philadelphia Cream Cheese: -6
dingzhou

Team Bertolli: -6
1=2

Team Fruit Loops : 0
leoxnlin
dreambold22

[quote="1=2"]You forgot to subtract 3 from each team.

[b]+3 to Team Bertolli [/b]

[b]Team Philadelphia Cream Cheese: 0 [/b]
dingzhou

[b]Team Bertolli: 0 [/b]
1=2

[b]Team Fruit Loops : -6[/b][/quote]

you only get 3 points total...so leoxnlins right...

but who cares...

+3 to PCC

Team Philadelphia Cream Cheese: -3
dingzhou

Team Bertolli: -6
1=2

Team Fruit Loops : 0
leoxnlin
dreambold22

+3 T B

Team Philadelphia Cream Cheese: -3
dingzhou

Team Bertolli: -3
1=2

Team Fruit Loops : -3
leoxnlin
dreambold22

+3 Team Philadelphia Cream Cheese

Team Philadelphia Cream Cheese: 0
dingzhou
math154

Team Bertolli: -3
1=2

Team Fruit Loops : -3
leoxnlin
dreambold22

wait what? Oh I get it now. Sorry. The scores should be the ones below me.

+3 to Bertolli

Team Philadelphia Cream Cheese: 9
dingzhou
math154

Team Bertolli: 9
1=2

Team Fruit Loops : 6
leoxnlin
dreambold22

-3 to Bertolli

Team Philadelphia Cream Cheese: 9
dingzhou
math154

Team Bertolli: 6
1=2

Team Fruit Loops : 6
leoxnlin
dreambold22

-3 to Philadelphia Cream Cheese

Team Philadelphia Cream Cheese: 6
dingzhou
math154

Team Bertolli: 6
1=2
Brut3Forc3

Team Fruit Loops : 6
leoxnlin
dreambold22

Hi everyone!

+3 to Philadelphia Cream Cheese

Team Philadelphia Cream Cheese: 9
dingzhou
math154
budi713

Team Bertolli: 6
1=2
Brut3Forc3

Team Fruit Loops : 6
leoxnlin
dreambold22

+3 to Bertolli

Team Philadelphia Cream Cheese: 9
dingzhou
math154
budi713

Team Bertolli: 9
1=2
Brut3Forc3

Team Fruit Loops : 6
leoxnlin
dreambold22

+2 to Team Fruit Loops, -1 to Bertolli
Team Philadelphia Cream Cheese: 9
dingzhou
math154
budi713

Team Bertolli: 8
1=2
Brut3Forc3

Team Fruit Loops : 8
leoxnlin
dreambold22
ProtestanT

+2 to Team Fruit Loops, -1 to PCC
[b]Team Philadelphia Cream Cheese: 8[/b]
dingzhou
math154
budi713

[b]Team Bertolli: 8[/b]
1=2
Brut3Forc3

[b]Team Fruit Loops : 10[/b]
leoxnlin
dreambold22
ProtestanT

+3 Bertolli
[b]Team Philadelphia Cream Cheese: 8[/b]
dingzhou
math154
budi713

[b]Team Bertolli: 11[/b]
1=2
Brut3Forc3

[b]Team Fruit Loops : 10[/b]
leoxnlin
dreambold22
ProtestanT

Hey! You can only post once per hour!

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -57
ra5249
PI-Dimension
1=2
alexhhmun
ernie

Note: My post at 8:57 was NOT a post. It was merely correcting the scores. Therefore, I am not posting twice in a hour.

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -60
ra5249
PI-Dimension
1=2
alexhhmun
ernie

We won
a long time ago

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -63
ra5249
PI-Dimension
1=2
alexhhmun
ernie

Meh.  It's like 3-on-1.

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -60
ra5249
PI-Dimension
1=2
alexhhmun
ernie

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -66
ra5249
PI-Dimension
1=2
alexhhmun
ernie

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -69
ra5249
PI-Dimension
1=2
alexhhmun
ernie

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -72
ra5249
PI-Dimension
1=2
alexhhmun
ernie

Team Ninja: 87
chenhsi
splatyango
budi713
evilhamster
james4l

Team seti: -75 = PWN!
ra5249
PI-Dimension
1=2
alexhhmun
ernie

I don't think we should start another one, since there is already one in the other thread.

meh.  that was 4-on-1.

oh well

it took like 5 pages

xD


superior algebra
superior algebra unsolved

Let $ A$ be a commutative ring and $ S$ a multiplicative subset of $ A$. Show that if $ A \neq A_s$ then $ A_s$ is flat over $ A$ but not faithfully flat.

Obviously I've no problem showing that $ A_s$ is flat over $ A$. It's the other part of the problem that is causing problems. :)

$ A \to A_S$ induces an isomorphism $ A \otimes A_S \to A_S \otimes A_S$.


geometry
calculus
integration
3D geometry
perimeter

Solve the following problem [u]without calculus[/u].

A goat is tethered to a circular hay silo of radius 10 ft by means of a rope 10 :pi: feet long. Find the area of the region the goat can reach.

i don't see how this problem is possible without calculus.  It's basically about integrating a polar function.

Zennyk, do you mean that the rope slides along the silo?
If so, the goat can reach a circle of $(10\pi+10)^2.\pi - 10^2.\pi$ right?
If not, I neither got any clue how to do this without calculus... :?

I think the rope is fixed to a point of the sylo. The curve forming the boundary of the area which the goat can cover resembles a cardioid, but I'm not sure that's what it is.

Isn't this deceptively simple?  The radius of the silo is 10ft, so the circumference is 20 :pi:, and half of this is 10 :pi: .  That means that the rope would reach from where it is fixed to exactly the opposite point on the other side of the silo.  This means the silo is tangent to the circle that the rope makes (center where it is fixed to the silo) at the point oposite where he rope is fixed.  This means that the grazable area is the area of a circle radius 10 :pi: ft minus the area of a circle radius 10 ft which is
:pi: x (10 x :pi:) 2 - :pi: 10 2
= 100(:pi:3-:pi:) ft

Maybe I've got something really wrong, this is only the first time I've posted on this board and it's 12:35am...

I think I sort of cheated here. Grobber has the right idea. When I said "without calculus" I was thinking "no integrals." I used a limit for my solution though, which is technically part of calculus. So no integrals are allowed, but limits are fair game.

I am pretty sure that the curve in question is a cardioid as well.  If you look at http://mathworld.wolfram.com/Cardioid.html, they give a way to generate the cardioid which seems like it would fit the requirements of this problem.

Given that, we have that the area is:

3/2pi(10pi)^2 - Kcircle= 150*pi^3 - 100pi

Though it resembles one, the curve is actually not a cardioid. Consider that the curve in question can be constructed of a semicircle and part of an Archimedean spiral.

Full_ 360 etc... (The silo interferes)

Statiing the obvious (Or not so obvious). No it is not that simple.   For simplicity let the center be at (0,0),  and the goat is tiead at (10,0).
(let us call the direction  from  (0,0) to (0,10) as North etc)

Goat can freely Graze Semi-circle (Radius 10 pii, center at 10,0)  East of the center.

Goat can reach the point (-10,0)  at other points west of (10,0) part of the rope would 'wrap' around the silo, and then  go as a tangent to the silo.
 . This portion (Or tather two  parts - northern and sourthern  - goat has to approach from one side) would  be involute of the circle - well that is what it is called in math language) (Ir resembles, but is not equivalent to, a cardioid.)

Total area comes out to be [b](5/6)pi^3*R^2[/b]  (Here R=10)
(Assuming I did not make any mistake)

It is really easy  using calculus, .. If I have to I can do it without calculus by dividing in small triangles but that would be artiificial like  soiving Rubiks Cube with my toes - (which BTW I did long ago  to impress my then future wife)...

Added later:  One can also take a polygon (say 2n-gon with 2p perimeter with goat attached to a corner with  p lenght of  rope ).  ... It is not so difficult to add all  the areas, and let the limit n tends to infinity... For example for a square silo there is one large (3/4) part of the circle with radius p and two small (radius p/2) quarter circles makeing  the area (7/8)*pi*p^2... do this for 2n-gon .. not hard at all. (something like  (pi*p^2)(1/2+ 1/(2n) +  (1/n)(((1-1/2n)^2+(1-2/2n)^2+(1-3/2n)^2+...)

The challenging (if you want to call it that) part of that problem is to account for the fact that the side lengths of the 2n-gon must be adjusted as n increases so that the radius of the polygon remains constant.


function
real analysis
real analysis unsolved

#1.
Prove that if $ f$ is continuous at $ c$ then there exists $ \delta > 0$ such that $ f$ is bounded at $ (c\minus{}\delta,c\plus{}\delta)$

#2
Prove that if $ f$ and $ g$ are continuous at c, then $ fg$ is continuous at $ c$.

These follow almost immediately from the definition. We'll use the $ \epsilon \minus{} \delta$ definition here, but we could also do it using sequences (an equivalent definition).

If $ f$ is continuous at $ c$ then $ (\forall \epsilon > 0)(\exists \delta > 0)(x\in\langle c \minus{} \delta,c \plus{} \delta\rangle\Rightarrow f(x)\in\langle f(c) \minus{} \epsilon,f(c) \plus{} \epsilon\rangle)$. Take, say, $ \epsilon \equal{} 1$. Thus, there exists $ \delta > 0$ such that, on $ \langle c \minus{} \delta,c \plus{} \delta\rangle$, $ f(c)\minus{}1<f(x) < f(c) \plus{} 1$. As $ f(c)$ is constant, $ f$ is bounded on that interval.

If $ f$ and $ g$ are continuous at $ c$, then $ (\forall \epsilon > 0)(\exists \delta_1,0)(x\in\langle c \minus{} \delta_1,c \plus{} \delta_1\rangle\Rightarrow f(x)\in\langle f(c) \minus{} \epsilon,f(c) \plus{} \epsilon\rangle)$ and $ (\forall \epsilon > 0)(\exists \delta_2 > 0)(x\in\langle c \minus{} \delta_2,c \plus{} \delta_2\rangle\Rightarrow g(x)\in\langle g(c) \minus{} \epsilon,g(c) \plus{} \epsilon\rangle)$.

We now want to prove a similar statement about $ fg$. Obviously, $ |f(x)g(x) \minus{} f(c)g(c)| \equal{} |f(x)g(x) \plus{} g(x)f(c) \minus{} g(x)f(c) \minus{} f(c)g(c)| \equal{}$ $ \equal{} |g(x)(f(x) \minus{} f(c)) \plus{} f(c)(g(x) \minus{} g(c))|\leq |g(x)|\cdot |f(x) \minus{} f(c)| \plus{} |f(c)|\cdot |g(x) \minus{} g(c)|$. Take any $ \epsilon > 0$.

We know (from #1) there exists $ \delta_1 > 0$ such that $ |x \minus{} c| < \delta_1\Rightarrow |g(x)| < M$ for some $ M\in\mathbb{R}_ \plus{}$. Also, we know (as $ \frac {\epsilon}{3M},\frac {\epsilon}{3(|f(c)| \plus{} 1)} > 0$) there exist $ \delta_2 > 0$ such that $ |x \minus{} c| < \delta_2\Rightarrow |f(x) \minus{} f(c)| < \frac {\epsilon}{3M}$ and $ \delta_3 > 0$ such that $ |x \minus{} c| < \delta_3\Rightarrow |g(x) \minus{} g(c)| < \frac {\epsilon}{3(|f(c)| \plus{} 1)}$.

Take $ \delta \equal{} \min\{\delta_1,\delta_2,\delta_3\}$ and we have
\[ |x \minus{} c| < \delta \Rightarrow |g(x)|\cdot |f(x) \minus{} f(c)| \plus{} |f(c)|\cdot |g(x) \minus{} g(c)| < \frac {M\epsilon}{3M} \plus{} \frac {|f(c)|\epsilon}{3(|f(c)| \plus{} 1)} < \frac {\epsilon}{3} \plus{} \frac {\epsilon}{3} < \epsilon\textrm{.}\]
Thus, for each $ \epsilon > 0$ we have $ \delta > 0$ such that $ |f(x)g(x) \minus{} f(c)g(c)| < \epsilon$, which is exactly what we wanted.

(I wanted to be completely rigorous in this proof. We could easily have finished #2 more naturally, not involving $ \frac {\epsilon}{3(|f(c)| \plus{} 1)}$ and such, but this really gets us directly to $ \epsilon$ in the final sum.)


[hide]Add the first and second to get 2X + 5Z = -18. Add the second and third to get 4X + 5Z = -26. Subtract those two do get 2X = -8, so X = -4. Plug that back into one of the previous equations to get Z = -2. Plug both of those into the first original equation to get Y = 1.

The answer is (-4, 1, -2).[/hide]

[quote="jenny221"]Solve the following system:
X + y + z = -5
X – y + 4z = -13
3x + y + z = -13[/quote]
[hide]$x+y+z=-5;\,x-y+4z=-13;\,3x+y+z=-13$

Add the first two equations to get $2x+5z=-18.$ Add the second two equations to get $4x+5z=-26.$ We find $x=-4$ and $z=-2.$ Then, we plug these values into one of the original equations to find $y=1.$

$(-4,1,-2)$[/hide]

[quote="jenny221"]Solve the following system:
X + y + z = -5
X – y + 4z = -13
3x + y + z = -13[/quote]

[hide="solution"]We see, from the first and third equations, that x=-4. We plug that into the first and second euqations:

y+z=-1
-y+4z=-9

5z=-10

z=-2

-6+y=-5,

y=1

(-4,1,-2)[/hide]


Let $ a,b,c$ be positive real numbers and let $ (x,y,z)$ be a permutation of $ (a,b,c).$ Denote
$ A\equal{} \frac{a}{b\plus{}y}\plus{}\frac{b}{c\plus{}z}\plus{}\frac{c}{a\plus{}x},$
and
$ B\equal{}\frac{a}{b\plus{}z}\plus{}\frac{b}{c\plus{}x}\plus{}\frac{c}{a\plus{}y}.$
Prove that $ \max\{A,B\} \ge \frac{3}{2}.$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers and let $ (x,y,z)$ be a permutation of $ (a,b,c).$ Denote
$ A \equal{} \frac {a}{b \plus{} y} \plus{} \frac {b}{c \plus{} z} \plus{} \frac {c}{a \plus{} x},$
and
$ B \equal{} \frac {a}{b \plus{} z} \plus{} \frac {b}{c \plus{} x} \plus{} \frac {c}{a \plus{} y}.$
Prove that $ \max\{A,B\} \ge \frac {3}{2}.$[/quote]
Thank you very much for your nice gift, can_hang  :lol:
We only need to prove: $ A \plus{} B \ge 3$
On the other hand, by Cauchy Schwarts, we have: $ A \plus{} B \ge 4\sum\ \frac {a}{2b \plus{} y \plus{} z} \ge \frac {4(a \plus{} b \plus{} c)^2}{\sum\ a(2b \plus{} y \plus{} z)} \ge 3$
Because acorrding the hypothesis of the problem, $ (x,y,z)$ be a permutation of $ (a,b,c).$
We have done  :)


A  problem which looks simple ....

if   a=  3^(1/3)

Then  what is      (a^(a^(a^(a ......)

The power tower continues (till infinity...)

[hide="a start?"]let the power tower be x

x=a^x

lnx=xlna

lnx=(x/3)ln3

(3/x)lnx=ln3

x^(3/x)=3

3 is an obvious solution, so my hunch is x=3...[/hide]

My guess is 2.4780526802 :) Well, at least to some number of dp. Don't know if that has an exact expression.

Should it? It doesn't, it converges to the number I had above.

Triple M is, of course,  correct.  [hide]( 3^(1/3) is also happens to be   (2.478...)^(1/2.478) .....  :D

P.S   Notice that I did not choose  fourth-root of  4, in which case the "obvious answer"  4, (perhaps a little more obviously :)  would not be correct, (correct answer would be  2)

[hide]Because it is also square root of 2.   (4^(1/4) = 2^(1/2)
The problem  came around here becuase   See:
 [url=http://artofproblemsolving.com/Forum/viewtopic.php?p=149733#p149733]artofproblemsolving  form[/url]

[quote] In fact a *VERY HARD* problem can be made by replacing sqrt(2) with cube_root(3).. where the obvious answer (3) would be wrong.. .. May be I will ask the problem in the Pre-Ollympiad section.. Smile[/quote]

So if   x =  (y)^(1/y)  Then the infinite power tower of y would  be  x. The trick is, there are two values of y for a given x,  and you have to   choose the value of y which is less than  e.)

And yes x has to be between (e)^(-e) and e^(1/e) for the power tower to converge... [/hide][/hide]

I think I have it:

[hide]Let x=a^(a^(a^(a^.... where a=3^(1/3)
then this can be rewritten as x=a^x since the power tower goes on infinitely.
cubing both sides we get x^3=3^x where x=3 is clearly a solution (the only positive one at  least).  In general, if a=n^(1/n) for positive integers n then using the same process we can deduce that a^(a^(a^(a^.... =n[/hide]

Jeffrey & AMD2357:

[hide]As you may have seen from  the above post(s), the obvious answer (3) is [b]not[/b] correct.  Sorry  it is a little tricky..

[quote]we get x^3=3^x where x=3 is clearly a solution (the only positive one at least).[/quote]

x=3 is cleary a solution, but so is,( may be not very clear but positive and real)   the one given by TripleM
 x= 2.47805268 ...
 try it  using a calculator .. and it is the  this value which  happens to be correct. (See the post above for more information..)

Following TripleM's reasonign , You can prove  if x < 2.5 ,   a^x < (3)^((2.5/3) = 2.5
  a^(5/2)  = 3^(5/6)  < (5/2)    (You can check  3^5*2^6 < 5^6 )

so the tower can never be greater than 2.5  and  the answer can not be greater than 2.5.

Hope this has been fun and helpful.[/hide]


For $ a,b,c$ positive reals

$ \sum (\frac{a}{b\plus{}c})^3 \plus{}9\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq \sum\frac{a}{b\plus{}c}$

I tried to solve it following can_hang2007 method, but I failed  :blush:

This is what I did:

Setting $ x\equal{}\frac{2a}{b\plus{}c}$ and cyclic, we have

$ \sum xy\plus{}xyz\equal{}4$ and $ \sum x \geq 3$

Setting $ p\equal{}\sum x$ , $ q\equal{}\sum xy$ and $ r\equal{}xyz$ we have that our inequality is equivalent with

$ p^3\minus{}16p\plus{}3r(4\plus{}p) \geq 0$.

Now, using the well known  $ r \geq \frac{p(4q\minus{}p^2)}{9}$  we have

$ 4\equal{}q\plus{}r \geq q \plus{} \frac{p(4q\minus{}p^2)}{9}$

that is

$ r \geq \frac{p(16\minus{}p^2)}{4p\plus{}9}$

and finally our inequality is reduced to

$ p^3\minus{}3p^2\minus{}16p\plus{}48 \geq 0$

or

$ (p\minus{}3)(p\minus{}4)(p\plus{}4) \geq 0$

but IT IS FALSE as $ p \geq 3$ only :(

Have, please, anyone any nice idea to solve this nice inequality.

Thank you very much. :)

[quote="nguoivn"]It's also my hint (I found it before can_hang). I think you see it here  :)
http://www.mathlinks.ro/viewtopic.php?t=249791
Dear manliko, you had some mistakes in calculus. Check it again  :)[/quote]

Thank you very much for your answer, nguoivn.

I will check it again. :blush:

This one can be solved by Mixing Variable method. It is not very hard to prove I think :wink: Before I post my solution, let manlio solve it again  :)

[quote="nguoivn"]It's also my hint (I found it before can_hang). I think you see it here  :)
http://www.mathlinks.ro/viewtopic.php?t=249791
Dear manliko, you had some mistakes in calculus. Check it again  :)[/quote]

Dear nguoivn, sorry if I am so stupid :blush:  :blush: , but I cannot see my mistake. Can you please help me to find it?

Thank you very much fr your help. :)

[quote="zaizai-hoang"]This one is really nice and special. The following is stronger result :
\[ \sum \left(\frac {a}{b \plus{} c}\right)^3 \plus{} \frac {9abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq \sum\frac {a}{b \plus{} c} \plus{} \frac {4H\times \left[abc \plus{} \sum ab(a \plus{} b)\right]}{(a \plus{} b)^3(b \plus{} c)^3(c \plus{} a)^3}
\]
which $ H \equal{} \sum a^6 \minus{} \sum a^2b^2(a^2 \plus{} b^2) \plus{} 3a^2b^2c^2$[/quote]

Nice this proof by expansion, but I would like to see a proof by using can_hang2007 and nguoivn method. :wink:

Thank you very much.

[quote="zaizai-hoang"]what is "can_hang2007 and nguoivn method" ?! pqr technique ?![/quote]

Yes, it is pqr method? :wink:

I cannot solve this problem using this method.  :oops:

[quote="nguoivn"][quote="zaizai-hoang"]what is "can_hang2007 and nguoivn method" ?! pqr technique ?![/quote]
No, our hint can't be called the method.
In this problem, the nice hint which can_hang and me found is:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
By this lenma, we can change this ineq to x, y, z and pqr complete easily the proof  :)

@manlio: You can see my proof here. From it, I think you'll find your mistake  :)[/quote]
I checked your proof and I think it is not correct, nguoivn. You can check it again ;)
[quote="zaizai-hoang"]The interesting thing is the lemma. Why they found out the lemma? Is it well-known?! Anyway, pqr technique solved it easily or there are a lot of method can help us prove it by some ways. How about the stronger which was posted above ? Notice that: $ H\ge 0$ which is Schur's inequality  :)[/quote]
Both inequalities can be proved easily with Vornicu Schur. Can you try it? ;) :)

For the moment, I don't like using Vornicu Schur's ineq to prove inequalities (and also SOS,SS :D ). Both of them can be solved only Schur's ineq  :)
Here is my solution: (I like this one than the second solution  :D )
Problem is equivalent to:
Let $ a,b,c\ge 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Prove that:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc\ge 4(a \plus{} b \plus{} c)
\]
Without loss of generality we assume that: $ b \equal{} max\{a,b,c\}$
We call:
\[ f(a,b,c) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc \minus{} 4(a \plus{} b \plus{} c)
\]
Therefore, we only need prove:
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge 0
\]

\[ \iff (b \minus{} c)^2\left(\frac{2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0
\]
And we also have:
\[ f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right) \equal{} \frac {a^2(a^2 \plus{} 5a \plus{} 8)(a \minus{} 1)^2}{(a \plus{} 1)^3}\ge 0
\]
We are done  :wink:

[quote="zaizai-hoang"]Both of them can be solved only Schur's ineq  :)[/quote]
I am curious to see it ;). I hope that you dont use computer to have the direct expansion ;)

[quote="nguoivn"][quote="zaizai-hoang"]what is "can_hang2007 and nguoivn method" ?! pqr technique ?![/quote]
No, our hint can't be called the method.
In this problem, the nice hint which can_hang and me found is:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
By this lenma, we can change this ineq to x, y, z and pqr complete easily the proof  :)

@manlio: You can see my proof here. From it, I think you'll find your mistake  :)[/quote]

Thank you very much, nguoivn for your interest in this very nice inequality.

Sorry, but I also checked your solution and I think it is wrong. It is very strange that a so nice inequality is hard to prove by pqr method. However thank you very much zaizai-hoang for your nice solution. I tried to solve it by Vornicu-Shur, but for me it is hard to work on the term $ \sum (\frac{a}{b\plus{}c})^3$. A hint please?

Thank you very much, can_hang2007. :)

[quote="zaizai-hoang"]The interesting thing is the lemma. Why they found out the lemma? Is it well-known?! Anyway, pqr technique solved it easily or there are a lot of method can help us prove it by some ways. How about the stronger which was posted above ? Notice that: $ H\ge 0$ which is Schur's inequality  :)[/quote]
It isn't well-known. When I am solving this problem, I want to change it to pqr.
So, I had used Cauchy Schwarts:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}$
But this estimate isn't stronger enough with this ineq. And naturally, I want to find a better lenma with the form:
$ \sum\ \frac {a}{b \plus{} c} \geq\ \frac {k(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} \frac {(12 \minus{} 8k)abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
And it's easily to see that $ k \equal{} 1$
--------------------------
But I see my silly mistakes: with $ p \leq\ 2$, LHS $ \geq\ 0$, not $ \leq\ 0$
Verry sorry. I'll try it again   :oops:

Following zaizai-hoang proof I tried to compute

\[ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge 0
\]
and I obtained

$ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\equal{} \frac{(b^3\plus{}c^3\minus{}4b\minus{}4c)(b\plus{}c\plus{}bc)\plus{}36bc\minus{}9b^2c^2}{b\plus{}c\plus{}bc} \minus{}\frac{52(4\minus{}bc)^2(b\plus{}c\plus{}bc)\minus{}68(4\minus{}bc)(b\plus{}c\plus{}bc)^2\minus{}32(b\plus{}c\plus{}bc)^3}{(b\plus{}c\plus{}4)^3}$

but how t prove this is equal to
\[ \iff (b \minus{} c)^2\left(\frac {2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0
\]

Thank you very much, zaizai-hoang.

[quote="manlio"][quote="Honey_S"]Check and tell me if there is any mistake. Thank you. :)[/quote]

Thank you very much, Honey_S.  :)

I will check it and I will tell you. :)[/quote]

I checked it very carefully and it is Ok.  :)

You have only to change your proof that $ X,Y,Z$ are positive this way:

$ X = \frac{2a^2}{b+c}+a-\frac{abc}{b+c}(\frac{1}{a+b}+\frac{1}{a+c}) \geq \frac{2a^2}{b+c}+a-\frac{abc}{b+c}(\frac{1}{b}+\frac{1}{c}) =  \frac{2a^2}{b+c}$

and it works also for $ Y,Z$  :wink:

Also you made a typo in last line; it should be rewritten as

$ X-Y= (\frac{2a^2}{b+c}-\frac{2b^2}{c+a})+(a-b)-\frac{abc}{a+b}(\frac{1}{b+c}-\frac{1}{a+c}) =
\frac{2(a^4-b^4)+2ab(a^2-b^2)+2c(a^3-b^3)+abc(a-b)}{(a+b)(b+c)(c+a)} \geq 0$ as $ a\geq b$

I hope to be right, however thank you very much again for your proof. :)

Thank you for checking my proof, manlio. :)

Thanks manlio and thanks Honey_S because of your nice proof.
I very like your proofs by Voni Schur  :lol:

[quote="zaizai-hoang"]

Here is my solution: (I like this one than the second solution  :D )
Problem is equivalent to:
Let $ a,b,c\ge 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Prove that:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc\ge 4(a \plus{} b \plus{} c)
\]
Without loss of generality we assume that: $ b \equal{} max\{a,b,c\}$
We call:
\[ f(a,b,c) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc \minus{} 4(a \plus{} b \plus{} c)
\]
Therefore, we only need prove:
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge 0
\]

\[ \iff (b \minus{} c)^2\left(\frac {2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0
\]
[/quote]

I am struggling with this proof. :(  I cannot prove that

\[ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge (b \minus{} c)^2\left(\frac {2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0
\]

Can anyone, please, help me?  :oops: :oops:

Thank you very very very much. :)

[quote="zaizai-hoang"]

Here is my solution: (I like this one than the second solution  :D )
Problem is equivalent to:
Let $ a,b,c\ge 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Prove that:
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc\ge 4(a \plus{} b \plus{} c)
\]
Without loss of generality we assume that: $ b \equal{} max\{a,b,c\}$
We call:
\[ f(a,b,c) \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc \minus{} 4(a \plus{} b \plus{} c)
\]
Therefore, we only need prove:
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge 0
\]

\[ \iff (b \minus{} c)^2\left(\frac {2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0
\]
[/quote]

Dear zaizai-hoang, can you please give me a hint to obtain

$ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge (b \minus{} c)^2\left(\frac {2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0$

Thank you very much. :)

[quote="manlio"]Following zaizai-hoang proof I tried to compute
\[ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)\ge 0
\]
and I obtained

$ f(a,b,c) \minus{} f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right) \equal{} \frac {(b^3 \plus{} c^3 \minus{} 4b \minus{} 4c)(b \plus{} c \plus{} bc) \plus{} 36bc \minus{} 9b^2c^2}{b \plus{} c \plus{} bc} \minus{} \frac {52(4 \minus{} bc)^2(b \plus{} c \plus{} bc) \minus{} 68(4 \minus{} bc)(b \plus{} c \plus{} bc)^2 \minus{} 32(b \plus{} c \plus{} bc)^3}{(b \plus{} c \plus{} 4)^3}$

but how t prove this is equal to
\[ \iff (b \minus{} c)^2\left(\frac {2(b \plus{} 2)^2(21b^2 \minus{} 6b \plus{} 2)}{(b \plus{} c \plus{} 4)^3}\right)\ge 0
\]
Thank you very much, zaizai-hoang.[/quote]

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum (\frac{a}{b\plus{}c})^3 \plus{}9\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq \sum\frac{a}{b\plus{}c}$

[/quote]
Let $\frac{2a}{b+c}=x$, $\frac{2b}{a+c}=y$ and $\frac{2c}{a+b}=z$.
Hence, $xy+xz+yz+xyz=4$ and we need to prove that $\sum_{cyc}(x^3+3xyz-4x)\geq0$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ ($v>0$) and $xyz=w^3$.
Hence, the condition does not depend on $u$ and $4=xy+xz+yz+xyz=3v^2+w^3\leq3u^2+u^3$, which gives $u\geq1$.
We need to prove that $f(u)\geq0$, where $f(u)=27u^3-27uv^2+12w^3-12u$.
But $f'(u)=81u^2-27v^2-12>0$.
Hence, by $uvw$ it remains to prove our inequality in the case $y=x$.
The condition gives $z=\frac{2-x}{x}$ and $f(u)\geq0\Leftrightarrow (x-1)^2(x-2)^2(2x^2+3x+2)\geq0$. Done!

[quote=manlio]For $ a,b,c$ positive reals

$ \sum (\frac{a}{b\plus{}c})^3 \plus{}9\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq \sum\frac{a}{b\plus{}c}$

[/quote]

We have
\[LHS-RHS=\frac{1}{4}\sum{\frac{(2a^2+3ab+7ac+2b^2+7bc+7c^2)(a-b)^2(a+b-c)^2}{(c+a)^3(b+c)^3}}\ge{0}\]

[quote=zaizai-hoang]This one is really nice and special. The following is stronger result :
\[ \sum \left(\frac {a}{b \plus{} c}\right)^3 \plus{} \frac {9abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq \sum\frac {a}{b \plus{} c} \plus{} \frac {4H\times \left[abc \plus{} \sum ab(a \plus{} b)\right]}{(a \plus{} b)^3(b \plus{} c)^3(c \plus{} a)^3}
\]
which $ H \equal{} \sum a^6 \minus{} \sum a^2b^2(a^2 \plus{} b^2) \plus{} 3a^2b^2c^2$[/quote]

We have
\[LHS-RHS=\frac{\sum{a(a-b)(a-c)}\cdot \sum{a^2(a^2-b^2)(a^2-c^2)}}{(a+b)^3(b+c)^3(c+a)^3}\ge{0}\]

[quote=Grotex]Let $a,b,c$ be non-negative.
Prove that
$$19\sum (\frac{a}{a+b})^3+75\frac{\prod a}{\prod (b+c)}\leq 11\sum \frac{a}{b+c}.$$[/quote]
What happens for $c=0$ and $a=b=1$? :maybe:

[quote=unrulykid3]$\sum \left ( \frac{a}{a+b} \right )^2+\frac{4abc}{(a+b)(b+c)(c+a)}\geq \sum \frac{a}{a+b}+\frac{1}{4}$[/quote]

What about this problem?

[quote=unrulykid3]$\sum \left ( \frac{a}{a+b} \right )^2+\frac{4abc}{(a+b)(b+c)(c+a)}\geq \sum \frac{a}{a+b}+\frac{1}{4}$[/quote]
It's wrong.

It means
$$\sum \left ( \frac{a}{a+b} \right )^2+\frac{4abc}{(a+b)(b+c)(c+a)}\geq \sum \frac{a}{a+b}-\frac{1}{4}$$ and it's just
$$\left(\sum_{cyc}\frac{2a}{a+b}-3\right)^2\geq0.$$

[quote=Nguyenhuyen_AG][quote=arqady]It means
$$\sum \left ( \frac{a}{a+b} \right )^2+\frac{4abc}{(a+b)(b+c)(c+a)}\geq \sum \frac{a}{a+b}-\frac{1}{4}$$ and it's very nice![/quote]
Yes. It's Gabriel Dospinescu old result.[/quote]

Can you show me how to prove this, sir?

[quote=unrulykid3]Can you show me how to prove this, sir?[/quote]
Note that
\[\frac{a^2}{(a+b)^2} = \frac{a}{a+b}-\frac{ab}{(a+b)^2},\]
inequality become
\[\frac14+\frac{4abc}{(a+b)(b+c)(c+a)} \geqslant \sum \frac{ab}{(a+b)^2},\]
equivalent to
\[\sum \left(\frac{a-b}{a+b}\right)^2  \geqslant 2 - \frac{16abc}{(a+b)(b+c)(c+a)},\]
or
\[\sum \left(\frac{a-b}{a+b}\right)^2  \geqslant 2 \sum \frac{(a-b)(a-c)}{(a+b)(a+c)},\]
\[ \left(\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right)^2 \geqslant 0.\]
Done.


geometry
circumcircle
angle bisector
geometry proposed

In triangle $ ABC$, let $ E,F$ be points on sides $ AB,AC$ such that $ BE\equal{}CF$. Let $ D$ be the other intersection of the circumcircles of triangles $ ABF$ and $ ACE$. Prove that $ AD$ is the angle bisector of $ \angle BAC$.

[color=darkblue]Connect $ D$ with points $ A,E,F,C$. Then $ \angle{BAD}\equal{}180^\circ\minus{}\angle{BFD}\equal{}\angle{DFC}$. By analogy $ \angle{DFC}\equal{}\angle{AED}$. In triangles $ AED$ and $ FCD$ we have : $ AE\equal{}CF, \angle{BAD}\equal{}\angle{DFC}, \angle{DFC}\equal{}\angle{AED} \Rightarrow \triangle{AED} \equiv \triangle{FCD}$, so $ ED\equal{}CD$. We obtained that chords $ ED$ and $ CD$ are equal, so $ \angle{EBD}\equal{}\angle{DBC}$ and $ BD$ is bisector of angle $ BAC$.[/color]

Your solution is good, but you totally messed up the points!  :D

$ < BED \equal{} < DCB, < EBD \equal{} < DFC \equal{} >\Rightarrow \triangle{AED} \equiv \triangle{FCD} \equal{} > BD \equal{} DF \equal{} > AD$bisector


Do you guys thinks that MAA will publish a book containing AIME dropped questions (I'm sure there are hundreds) so we can use more authentic problems to practice?

In the book, [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883856409/qid=1111894404/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/102-8707774-9105711?v=glance&s=books&n=507846]Contest Problem Book V[/url], there is a chapter with dropped AIME problems.

Yes, I know of those 14 or so problems, but there must be many more every year....


calculus
integration
function
calculus computations

First remove the factors with even powers. If nothing remains, the answer is $1$. Suppose that $f_{i_{1}}\dots f_{i_{s}}$ ($i_{1}<\dots i_{s}$) are raised to an odd power. Note that $f_{i_{1}}(x+2^{-i_{1}})=-f_{i_{1}}(x)$ and $f_{i_{m}}(x+2^{-i_{1}})=f_{i_{m}}(x)$ for all $1<m\le s$. This should imply that the integral equals $0$.


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

well if it is finite, that's easy and we don't even need isomorphism at all.

If it is infinite, then all I can conclude is 0B and 1B (addition and multiplication inverse) belongs to A. (unfortnately that is very obvious...)

after spending half an hour on this seemingly easy and obvious question, I give up. So how to show B C A?

oh btw, A and B are intended to be field. Although it might be better off to start them as group, then work your way up to ring, integral domain, field I think. Also I have no idea if A does equal to B or not. I think it should be true, but no guarantee at the moment

For practically all algebraic structures, the answer is no. A proper subX can be isomorphic the the entire X if both are large enough.

For groups: Let $G_0$ be the free abelian group on $x_0,x_1,\dots,x_n,\dots$ and $G_1$ be the free abelian group on $x_1,x_2,\dots,x_n,\dots$. $G_1\subset G_0$, $G_1\neq G_0$, and $G_1$ is isomorphic to $G_1$.

It's not hard to translate the spirit of this example into something like a field


Is there anyway we can combine some intro classes and have them at the same classroom? Cuz I like want to do a mock AMC (if someone can do it) and take it and there isn't any better place than on the classroom. Like we can have bunch of people from different intro and we can get some problems from different area and make something like mock AMC (I'd love to have one in AMC-10 level) and for the problems, we can get someone to LaTeX it and post on the classroom. So, you can like scroll up and down the page and take the test and since it's classroom, you don't have to refresh your page or for asking question.

Only thing I'm saying that combining more classes, like intro geometry + intro counting on same classroom would be that there would be more people and more people means better thing. And after all, this whole thing that Admins are trying to do is how this classroom working regularly can determine the use of charoom in public. I mean, if the purpose is to get people interested in doing and attending it, why not make a like competitional. I mean, I don't know for sure but I think, at least what I think is a great idea. I don't go to the classroom often because I know not many people are on my classroom so why go? But if it's like competitional thing, most people would come to at least to see what it looks like.

We can get like some people to stack up some problems or something.

That would be nice.

Is this possible? Or is this really bad idea or just nothing?

First off, when making a rather "formal" post like this, please like don't like say like so like much. It's extremely difficult to tell what you're trying to say.

Now, as for the issue with different classes combining...It completely defeats the point of having different classes in the first place. I'm not too sure all the details of what all the details of your suggestion are, but it just seems like more chaos than anyone would care to pay money to go through.

In my opinion, it's best to keep the atmosphere of the classes the way it is. If we wanted to take mock tests, those can be ordered from AMC. The purpose of going to class is not to just do problems but to learn how to do them. Certainly not everyone in a class is at the same level of problem solving (even if they are all "intro" students) so why not just let everyone learn at their own pace without so much competition? There's plenty of it out there already.

I think the admin is doing a good job of conducting the classes so that they're accessible and friendly to a wide range of students.

What's so special about the classroom?  I think that pdf's are just fine.  Also, the admins have asked not to dump large posts of TeX into the the classroom.  I don't know about you but I think a mock AMC test in LaTeX would violate that request ten times over.

If someone has a test they should just administer it as usual.  Then students who are in Intro (or Intermediate/Olympiad) classes can discuss it in the classroom during and Open time.

[quote="Silverfalcon"]Is there anyway we can combine some intro classes and have them at the same classroom?[/quote]

This is how the open times work for students in the school - all the Intros in one room, the Interms in another, etc.


This is a question from my Real Analysis text.

Prove that if two norms on V have the same unit ball, then the norms are equal.

Thanks in advance..

Look at the intersection of the ball with a line $ \{\lambda v\}$. If $ \{\lambda: \|\lambda v\|_1<1\}\equal{}\{\lambda: \|\lambda v\|_2<1\}$, what can you say about $ \|v\|_1$ and $ \|v\|_2$?


symmetry
geometry
geometric transformation
reflection
calculus
derivative
function

[quote="silouan"]Find the minimum value of
$\sqrt{(x-1)^{2}+4}+\sqrt{(x-3)^{2}+16}$[/quote]
[hide="Hint"]Try regarding the two terms as distances between two points.
[hide="More_specifically"]The first term is the distance between $(x,0)$ and $(1,2)$, and the second term is the distance between $(x,0)$ and $(3,4)$.[/hide][/hide]

[hide]Let $A(1,2)$, $B(3,4)$ and $C(x,0)$. We want to minimize $|AC|+|BC|$. Let $A'(1,-2)$. Clearly, $|AC|+|BC| = |A'C|+|BC| \geq |A'B| = 2\sqrt{10}$, and equality is achieved when $C$ lies on $A'B\leftrightarrow y+5 = 3x$, i.e. $x=\frac{5}{3}$.[/hide]

[quote="silouan"]Find the minimum value of
$\sqrt{(x-1)^{2}+4}+\sqrt{(x-3)^{2}+16}$[/quote]

[hide]The question is essentially asking us to find the minimum value of the sum of the distances between teh points $(x,0)$, $(1,2)$, and $(x,0)$, $(3,4)$

Notice that by symmetry the minimum value is the distance between the reflection of $(1,2)$ across the x-axis and $(3,4)$.

This is just then $\sqrt{2^{2}+6^{2}}=2\sqrt{10}$[/hide]

[hide="Calculus Method"]
Or find the derivative of the function, which is:
${\frac{x-1}{\sqrt{((x-1)^{2}+4)}}+\frac{x-3}{\sqrt{((x-3)^{2}+16)}}}$

And solve for X, where the derivative is zero.

This works out to be $\boxed{x = \frac{5}{3}}$
[/hide]

Er, that's a really horrible way to do this problem when Minkowski's Inequality solves it instantly.

Edit:  Your derivative is also incorrect.  As is, your function is always positive.

Can you explain Minkowski's to me?
I looked at wiki and mathworld, by those didn't make any sense.
If you could apply it to this problem, that would be nice.


pigeonhole principle
number theory proposed
number theory

A number is called homogeneous if its decimal representation contains only 1 number, eg: 2222 or 333.
Prove there are infinitely many homogeneous numbers which are divisible by 2008.

If we could show that there is a homogeneous number of form $ \overbrace{1111...1}^{n}$, which is divisible by $ 251(251\equal{}2008: 8)$.
It can be proved by Dirichlet principle.
We have that the numbers
$ 1$
$ 11$
$ 111$
$ 1111$
$ .$
$ .$
$ .$
$ \overbrace{11...11}^{251}$
So one of the numbers is divisible by 251 or there are two numbers, whose difference is divisibe by 251.
As $ 1111...11\minus{}111...111\equal{}111...1100...00\vdots 251 \Rightarrow 111...11\vdots 251$ as ($ (10^{\alpha},251)\equal{}1$)
So there is a number that $ \overbrace{111...11}^{n}\vdots 251 \Rightarrow \overbrace{888...88}^{n}\vdots 2008$, so $ \overbrace{888...88}^{kn}\vdots 2008$.

can u kindly describe how u used the dirichlet principle???as i know it deals with finding infinitely many primes.is it another version of dirichlets theorem???please help to understand.thanks

Pigeonhole. Dirchlets box principle. Same difference. taking the results mod 251 gives the pigeons, the values 0,1,2,3,...,250 the boxes. Either 2 lie in the same box, or 1 lies in the box 0.


What is the positive difference between the median and the mode of the data given in the stem and leaf plot below? In this plot $ 5|8$ represents 58.

\begin{tabular}{|c|c|}\hline
  \textbf{Tens} & \textbf{Units} \\ \hline
  1 & $ 2 \hspace{2mm} 3 \hspace{2mm} 4 \hspace{2mm} 5 \hspace{2mm} 5$ \\ \hline
  2 & $ 2 \hspace{2mm} 2 \hspace{2mm} 2 \hspace{2mm} \hspace{2mm} \hspace{2mm} \hspace{1.5mm}$ \\ \hline
  3 & $ 1 \hspace{2mm} 1 \hspace{2mm} 8 \hspace{2mm} 9 \hspace{2mm} \hspace{1.9mm}$ \\ \hline
  4 & $ 0 \hspace{2mm} 1 \hspace{2mm} 2 \hspace{2mm} 3 \hspace{2mm} \hspace{1.9mm}$ \\ \hline
  5 & $ 2 \hspace{2mm} 8 \hspace{2mm} 9 \hspace{2mm} \hspace{2mm} \hspace{2mm} \hspace{1.5mm}$ \\\hline
   \end{tabular}

Median=middle # when arranged in numerical order
Mode=number that occurs most often
31-22=9.

$ \begin{tabular}{|c|c|}\hline
\textbf{Tens} & \textbf{Units} \\ \hline
1 & 2 3 4 5 5 \\ \hline
2 & 2 2 2 \\ \hline
3 & 1 1 8 9 \\ \hline
4 & 0 1 2 3 \\ \hline
5 & 2 8 9 \\\hline
\end{tabular}$


None the less, I can not figure this out.  How do you find the value of: $ \sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)$.

hello , what about using the formulas of  double angle sinus to get $ 3/16$

okay, yea, so I thought that it was that easy, but, I did not get the intended answer and felt dumb.  Care to explain?

by the way: sinus? ahahah.

:rotfl:  :blush:  :rotfl:  Oh sorry I mean double angle sine.

I have seen many problems like this one ,but honestly I'm not good at trig Im just giving you some thoughts

One quick way to do it is to use the sine triple angle formula $ \sin 3x \equal{} 3 \sin x \minus{} 4\sin^3 x$, by setting $ \sin 3x \equal{} \sin 60 \equal{} \frac{\sqrt{3}}{2}$ we know that the distinct solutions for $ \sin x$ are $ \sin 20,\ \sin 40,\ \minus{}\sin 80$. By letting $ y \equal{} \sin x$ we get from the triple angle formula that $ \frac{\sqrt{3}}{2} \equal{} 3y \minus{} 4y^3$ and from Viete's formulas the product of the roots of $ y$ is equal to $ \\minus{}\frac{\sqrt{3}}{8} \equal{} \minus{}\sin 20 \sin 40 \sin 80$. Multiplying both sides by $ \sin 60 \equal{} \frac{\sqrt{3}}{2}$ gives you the identity.

There are also various other ways to do it for example by constructing the polynomial in $ \sin x$ whose roots are the solutions to $ \sin 9x \equal{} 0$.  :)


calculus
integration
logarithms
Putnam
calculus computations

Could any one recommend me an advanced calculus textbook that focuses on solving challenging problems as opposed to focusing on theory.  A book that would cover advanced topics in evaluation of integrals and sums, such as converting between the two etc.  The type of problems I'd like to see would be:

$ \displaystyle\int_0^1\ln{x}\ln{(1 \minus{} x)}dx$ (Whose solution I am not requesting as I already know it)

Just get a problem book and a theory book. Alas most problem books I know of do not have any solutions.

I already know calculus and wrote a manuscript for my own textbook that teaches calculus the problem solving way with advanced problems (like that one), but I wanted to see what methods/techniques/tricks/problem types I may have left out to create an exhaustive compendium of the types of problems.

I don't know of such a book. Like I said, most of the challenging problem books I know of don't include (worked) solutions. There are some books geared towards Putnam etc, but they are of course not calculus specific.

Hey,
[quote="t0rajir0u"]
The Red and Green Book(s) of Mathematical Problems contain many challenging Putnam-level problems, many of which use calculus, and they do include solutions.
[/quote]
Are these the books you were referring to?
[url]http://www.amazon.com/Red-Book-Mathematical-Problems/dp/0486694151[/url]
[url]http://www.amazon.com/Green-Book-Mathematical-Problems/dp/0486695735[/url]

In addition, with respect to preparing for the Putnam exam besides the following,
[i]The William Lowell Putnam Mathematical Competition Problems & Solutions - Three Volume Set[/i] (Gleason/Larson/Kedlaya)
[i]Putnam and Beyond[/i] (Gelca/Andreescu)
[i]The Art and Craft of Problem Solving[/i] (Zeitz)
[i]Problem-Solving Through Problems[/i] (Larson)
[i]Problem-Solving Strategies[/i] (Engel)

Any others you would suggest t0rajir0u?

Thanks,

-APSStudent


Divide the number $ 1301$ into four three digit parts $ abc$, $ def$, $ ghi$ and $ jkl$ such that:

$ abc\plus{}def\plus{}ghi\plus{}jkl \equal{} 1301$

and

$ abc \equal{} a^{3}\plus{}b^{3}\plus{}c^{3}$
$ def \equal{} d^{3}\plus{}e^{3}\plus{}f^{3}$
$ ghi \equal{} g^{3}\plus{}h^{3}\plus{}i^{3}$
$ jkl \equal{} j^{3}\plus{}k^{3}\plus{}l^{3}$

[quote="i_like_pie"]In the beginning, $ abc$ is a three digit number. What about the second part? Do you mean the product of the digits or the number?[/quote]

I mean the number only...

[hide="With some somewhat systematical guessing"]
I figured if a number started with 1, it would have a 5 (125 is the greatest cube less than 200), so I found 153
I figured if a number started with 3, it would have a 7 (343 is the greatest cube less than 400), so I found 371 and 370
I figured if a number started with 4, it would have a 7 (343 is the greatest cube less than 500), so I found 407

$ 153\plus{}370\plus{}371\plus{}407\equal{}1\plus{}125\plus{}27\plus{}27\plus{}343\plus{}0\plus{}27\plus{}343\plus{}1\plus{}64\plus{}0\plus{}343\equal{}1301$[/hide]

Yeah, it is well known (sort of) that there are only four three digit numbers with the property that the sum of the cubes is the number itself.


Harvard college MIT Princeton Yale Stanford percent

Would anyone happen to know exact figures for the average SAT scores of admitted freshmen at the top-notch schools? I'm specifically looking for:

Harvard
MIT
Princeton
Yale
Stanford
UPenn
Rice
Caltech
Berkeley

but please include any other schools that would fall in the same category of prestige.

It's also be of help if someone would happen to know how to adjust the scores to be accurate for the new 2400 system. (And I don't mean simply multiplying by 150%)

Google "common data set" and [name of school] and you'll get some good figures. (Some schools don't put their Common Data Set data on their Web sites, but many do.) Having score breakdowns based on a common methodology helps compare schools better.

http://registrar1.princeton.edu/data/common/cds2003.pdf

Percent submitting SAT scores 100%

25th Percentile 75th Percentile
SAT I Verbal 680 770
SAT I Math 690 790

Percent of first-time, first-year (freshman) students with scores in each range:
SAT I Verbal SAT I Math
700-800 71 73
600-699 25 25

http://web.mit.edu/ir/cds/2003/cds2003.html

Percent submitting SAT scores: 92 Percent submitting ACT scores: 19

25th Percentile 75th Percentile
SAT I Verbal 	680   760
SAT I Math 	740   800

Percent of first-time, first-year (freshman) students with scores in each range:

         SAT I Verbal           SAT I Math
700-800 	64.4  91.8
600-699 	30.7  8.1

That'll get you started.

The SAT's are being changed, and as I recall also watered down,
over time so this particular statistic is probably less important for the top schools.

If you are a visibly math-directed student applying to the
top 6 or so schools, their admissions process will largely consist
of ranking you within a pool of math-types.
Near-perfect math SAT, 5 on calculus AP, and similar for
related exams (math, physics, computer subject tests)
will be at or near par for the course within this pool, and
would be typical of most of the applicant pool at the top engineering
schools.     So it's not very informative, and they will be
more interested in other distinctions such as olympiads or
evidence of non-mathematical competence.  A high verbal
score would be helpful if you're a math type.

Do any of the schools provide bivariate distributions,
i.e. percentage of student with both math and verbal
scores in specified ranges?

It is plain that the U.S. News and World Report assumption that most applicants have the SUM of the verbal and math averages of the average applicant to their schools is wrong, but I've found remarkably few sources of information on combined scores (what one might call total SAT I scores) of college applicants. That may be partly because most colleges, by policy, count an applicant as having a combined score made up of the applicant's sum of the highest verbal score submitted and highest math score submitted, so that many people have a better combined score (for college admission purposes) than they ever obtained in a single sitting. The College Board does have a chart of single-sitting combined scores, for which there is no convenient search on its Web site, but which I found by a Google search for the document's title, [url=http://www.collegeboard.com/prod_downloads/about/news_info/cbsenior/yr2003/pdf/2003CBSVM.pdf]SAT Percentile Ranks for Males, Females, and Total Group[/url].

Thanks for the interesting discussion.

This link tabulates 25th and 75th percentile SAT scores for a lot of the major schools (for some reason they leave out small "liberal arts" places like Williams, Amherst and Swarthmore):

http://www2.acs.ncsu.edu/UPA/peers/current/research_intensive/sat.htm

[quote="pseudon"]This link tabulates 25th and 75th percentile SAT scores for a lot of the major schools (for some reason they leave out small "liberal arts" places like Williams, Amherst and Swarthmore):[/quote]

Thanks for sharing the link. Alas, that link has the usual methodological error of assuming that the 25th percentile level of COMBINED scores is simply the sum of the 25th percentile level verbal score and 25th percentile math score, which is surely an incorrect assumption.


[quote="N.T.TUAN"]Let $ x,y,z>0$ such that $ xy+yz+zx=2(x+y+z)$. Prove that $ xyz\leq x+y+z+2$.[/quote]
$ xyz\leq x+y+z+2\Leftrightarrow$
$ \Leftrightarrow4xyz\left(\frac{x+y+z}{xy+xz+yz}\right)^{2}\leq x+y+z+\frac{xy+xz+yz}{x+y+z}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^{4}y^{2}+x^{4}z^{2}+3x^{3}y^{3}-2x^{4}yz-x^{3}y^{2}z-x^{2}z^{2}y-x^{2}y^{2}z^{2})\geq0,$
which true by Muirhead.

[quote="N.T.TUAN"]Let $ x,y,z>0$ such that $ xy+yz+zx=2(x+y+z)$. Prove that $ xyz\leq x+y+z+2$.[/quote]
Let $ a=x-1,b=y-1,c=z-1$
$ \Rightarrow ab+bc+ca=3$ and $ a,b,c \ge-1$
We need to prove that: $ abc \le 1$

Assume that $ a \le b \le c$
If $ b \le 0$ then $ x \le y \le 1 \Rightarrow xyz \le z \le x+y+z+2$
If $ a \ge 0$  then  $ abc \le 1$ by AM-GM inequality.
If $ a \le 0 \le b \le c$ then $ abc \le 0 <1$
Done.

We know that '' if $ x,y,z>0$ and $ xyz=x+y+z+2$ then $ xy+yz+zx\geq 2(x+y+z)$'' (*).

Now assume that $ xyz>x+y+z+2$ then $ z(xy-1)>x+y+2$ therefore $ xy>1$ and $ z>\frac{x+y+2}{xy-1}: =z'>0$. By (*) we have $ xy+yz'+z'x\geq 2(x+y+z')$ therefore $ x+y>2$ and $ z'\geq \frac{2(x+y)-xy}{x+y-2}=z$. Contradiction!


[quote="rustam"][quote="AlvaroRecoba"]Calculate $u'_{xx}$, $u'_{xy}$, $u'_{yy}$ in the same way to get $d^2u$[/quote]

$u'_{x^2}=f''_{11}+2xf''_{12}+2f'_2+2x(f''_{11}+2xf''_{12})$
and so on...
but what should i do next?
How i'll find $d^2u$?
 :(

$d^2u=(\frac{\partial }{\partial x}dx+\frac{\partial }{\partial y}dy)^2u$?[/quote]

Your calculation of $u''_{xx}$ is correct.
My teacher once told me that the use of the term $d^2u$ is somewhat inaccurate and sometimes misleading...
I think we can interpret $d^2u$ as $u''_{xx}dx^2+u''_{xy}dxdy+u''_{yx}dydx+u''_{yy}dy^2$, though I'm not sure...

[quote="AlvaroRecoba
My teacher once told me that the use of the term $d^2u$ is somewhat inaccurate and sometimes misleading...
I think we can interpret $d^2u$ as $u''_{xx}dx^2+u''_{xy}dxdy+u''_{yx}dydx+u''_{yy}dy^2$, though I'm not sure...[/quote]

Now I know how to solve this kind of problems  :)
$u''_{xx}dx^2+u''_{xy}dxdy+u''_{yx}dydx+u''_{yy}dy^2$ is correct and since $u''_{xy}=u''_{yx}$ we can write the differential in the operater form:
$d^2u=(\frac{\partial }{\partial x}dx+\frac{\partial }{\partial y}dy)^2u$

Thanx.


$\triangle ABC$ has $\angle C=90$ and the height from $C$ to $AB$ is $12$. If the perimeter is $P$, express $AB=h$n terms of $P$.

Remember, try and simplify the algebra as much as possible.

[hide="a possible solution"]let the legs be $x$ and $y$, then we have

$x^{2}+y^{2}=h^{2}$, $xy=12h$, and $x+y+h=P$

$(x+y)^{2}=(P-h)^{2}\implies h^{2}+2(12h)=(P-h)^{2}\implies$
$24h=P^{2}-2Ph\implies h=\frac{P^{2}}{24+2P}$[/hide]


1)ab+2ac+3b^2+6bc-5a-13b+4c-10 (follow power of a and power of b)
2)a^2+3b^2+4ab+2ac+6bc-4b+4c-4

I don't understand.
please write all method

thank you for every answer

The quickest way is to assume that a factorization exists, and then set one of the variables to zero, and then factoring the result. [Think about the process backwards; suppose we are given a factorization and set one of the variables to zero] For the first one, let's try setting $ b \equal{} 0$ (because $ b$ appears the most): $ 2ac \minus{} 5a \plus{} 4c \minus{} 10$, which happens to factor to $ (a \plus{} 2)(2c \minus{} 5)$. Now a bit of trial-and-error allows us to re-insert $ b$:
\[ ab \plus{} 2ac \plus{} 3b^2 \plus{} 6bc \minus{} 5a \minus{} 13b \plus{} 4c \minus{} 10 \equal{} (a \plus{} 3b \plus{} 2)(b \plus{} 2c \minus{} 5)
\]
[hide="Now try the second, (answer)"]
$ a^2 \plus{} 3b^2 \plus{} 4ab \plus{} 2ac \plus{} 6bc \minus{} 4b \plus{} 4c \minus{} 4 \equal{} (a \plus{} 3b \plus{} 2)(a \plus{} b \plus{} 2c \minus{} 2)$
[/hide]

you could also do some intense grouping

a(b+2c)+3b(b+2c)-5a-13b+4c-10
(a+3b)(b+2c)-5a-15b+2b+4c-10
(a+3b)(b+2c)-5(a+3b)+2b+4c-10
(a+3b)(b+2c-5)+2(b+2c-5)
(a+3b+2)(b+2c-5)

and then do problem 2 in the same fashion


function
integration
irrational number
real analysis
real analysis unsolved

Prove that:

$ \sum_{k \equal{} 1}^{\infty} \frac {1}{k^4} \equal{} \frac {\pi^4}{90}$

It should be $ \pi^4.$ Anyway I'd like to see a proof of this nice result. I've heard that one can find an expression for $ \zeta(2k).$

A related question: can $ \zeta(3)$ be expressed as $ \frac{\pi^3}{\text{number}}$?

[quote="Black"]It should be $ \pi^4.$ Anyway I'd like to see a proof of this nice result. I've heard that one can find an expression for $ \zeta(2k).$

A related question: can $ \zeta(3)$ be expressed as $ \frac {\pi^3}{\text{number}}$?[/quote]No. I don't think we have any closed way of expressing it.

http://www.mathlinks.ro/viewtopic.php?t=180928
also about $ \zeta(3)$ we know only that it's irrational number (Apery constant) and irrational measure is equal to $ \approx 5,5$ (if I remember right)
we don't know nothing about $ \zeta(2k \plus{} 1)$ for $ k \geq 2$
smoe resutl you can find in google: by Zudilin.
irrationality of $ \zeta (3)$ is same as consider :
$ \int_{0}^{1}\int_{0}^{1}\frac {P_{n}(x)P_{n}(y)}{1 \minus{} xy}dxdy$ as $ n \to \infty$ where $ P_{n}(x)$ is Legendre polynomial.
more information try to find in google: note on the irrationality on $ \zeta(2)$ and $ \zeta(3)$ Frist Bakers.


algebra
system of equations
algebra unsolved

[i]  Solve the following system of equations in $ \mathbb{R}$ :

   $ \{\begin{array}{l}x^2 + (x - y)^2 \le 4 \\
y^2 + (x - y)^2 \ge 4\end{array}$[/i]

[quote="nguyenvuthanhha"][i]  Solve the following system of equations in $ \mathbb{R}$ :

   $ \{\begin{array}{l}x^2 + (x - y)^2 \le 4  :
y^2 + (x - y)^2 \ge 4\end{array}$[/i][/quote]

$ y=2\sqrt{2}$.....$ -4\geq x \geq 4$ i am wrong sure :blush:


[quote="mostel"][i]Συζητούσα με τον Ράλλη για πολλή ( :P ) ώρα αν θα τη βάλλω τελικά στο φόρουμ, αλλά πιστεύω πως καλό είναι να υπάρχει. [/i]

Έστω η ακολουθία:

$a_{n}=\sqrt{1+\left(1+\frac{1}{n}\right)^{2}}+\sqrt{1+\left(1-\frac{1}{n}\right)^{2}}$ με $n\geq1$

α) Να αποδειχθεί ότι ο $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{20}}$ είναι ακέραιος.

β) Ποια η τιμή της παράστασης;

[i]Το β ερώτημα είναι αφιερωμένο στον Ράλλη  :rotfl:  [/i][/quote]

[color=red]LYSH[/color]

Einai eukolo na doume (pollaplasiazontas arithmiti paronomasti me ti sizigi parastasi kai vgazontas apo ton arithmiti to $\frac{1}{n^{2}}$ koino paragonta mesa apo tis rizes) oti:

$\frac{1}{a_{n}}=\frac{\sqrt{2n^{2}+2n+1}-\sqrt{2n^{2}-2n+1}}{4}$

Ara $\frac{4}{a_{n}}=\sqrt{2n^{2}+2n+1}-\sqrt{2n^{2}-2n+1}$

kai etsi

\begin{eqnarray}\frac{4}{a_{n}}+\frac{4}{a_{n+1}}&=&\sqrt{2n^{2}+2n+1}-\sqrt{2n^{2}-2n+1}+\sqrt{2(n+1)^{2}+2(n+1)+1}-\sqrt{2(n+1)^{2}-2(n+1)+1}\nonumber \\ &=&\sqrt{2n^{2}+2n+1}-\sqrt{2n^{2}-2n+1}+\sqrt{2n^{2}+6n+5}-\sqrt{2n^{2}+2n+1}\nonumber \\ &=& \sqrt{2n^{2}+6n+5}-\sqrt{2n^{2}-2n+1}\nonumber \\ &=& \sqrt{2(n+1)^{2}+2(n+1)+1}-\sqrt{2n^{2}-2n+1}\nonumber \end{eqnarray}

Sinexizontas loipon me ton idio tropo diagrafontas tous antistoixous orous exoume oti

\begin{eqnarray}\frac{4}{a_{1}}+\frac{4}{a_{2}}+\frac{4}{a_{3}}+\cdots+\frac{4}{a_{19}}+\frac{4}{a_{20}}&=&-\sqrt{2\cdot 1^{2}-2\cdot 1+1}+\sqrt{2\cdot 20^{2}+2\cdot 20+1}\nonumber \\ &=&-1+29 \nonumber \\ &=& 28 \nonumber \end{eqnarray}

diladi $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \frac{1}{a_{19}}+\frac{1}{a_{20}}=7$

Alexandros

Y.G. Elpizw na min exw xasei kamia praksi kapou. Epivevaiwse mostel ean to apotelesma einai ok.

Τέλεια Αλέξανδρε !!!  :lol:

[quote]β) Ποια η τιμή της παράστασης;

Το β ερώτημα είναι αφιερωμένο στον Ράλλη  :rotfl: [/quote]
Thanks gia tin afierosi... :P
Ayti akrivos tin anadromiki sxesi xrisimopiisa +ego..
a..+episis i monadikes times tu dikti $i<1000$ gia ton opio to a8risma ton antistrofon ine akereos ine gia $i=2,20,695$..+genikotera i ari8mi gia tus opius $(n+1)^{2}+(n+2)^{2}=k^{2}$ opote duleuontas me pi8agories triades,(an dn exo kani kanena la8os,stis praksis)..prokiptun ayta ta i...

 :D


Clicking on a Blog link produced the following interesting message:

"[username's] blog has been marked for deletion and will be deleted once a Board Administrator purges all similar weblogs that are marked for deletion."

What is the shared similar attribute of certain weblogs that marks them for deletion? What proportion of weblogs are deleted?

It's basically the users wish to delete a blog. An admin can finish up. Since some of the blogs are public (ie contributors and other people have replied, etc.) it is ensured that someone (for example someone finds out his password) doesn't delete the blog.


A partition of n is an increasing sequence of integers with sum n. For example, the partitions of 5 are: 1, 1, 1, 1, 1; 1, 1, 1, 2; 1, 1, 3; 1, 4; 5; 1, 2, 2; and 2, 3. If p is a partition, f(p) = the number of 1s in p, and g(p) = the number of distinct integers in the partition. Show that Sum f(p) = Sum g(p), where the sum is taken over all partitions of n.

Hmm, isn't this USAMO problem just Stanley's Theorem (from searching mathworld)? Was this an actual theorem before the 1986 USAMO? Just curious.

there was an IMO question to prove OI = sqrt(R(R-2r)), which I thought was used in the proof of euler's inequality..

perhaps a common proof for stanley's theorem wasn't known? (btw, it would also help if someone posted the solution, please. I don't know the proof)

[quote="ComplexZeta"]Yeah, this is well-known now, but in the 1986 USAMO, I was less than a year old, and I didn't know what a partition was, so I'm not sure how well-known it was back then.[/quote]

Pffft, what were you, like a late bloomer or something?

yea, even I knew what a partition was in 1986.

I was just barely unable to take the 1986 USAMO because I was born 8 days after it took place, so I'll make up for not having known about partitions, let alone Stanley's Theorem, at the time by proving it now:

[hide]

We'll show that the number of partitions of n with at least k 1s is the same as the number of partitions of n with a part of size k.

Indeed, there is a bijection: if p is a partition of n with at least k 1s,  remove k 1s from p and replace them with one k to produce a partition of k with a part of size k.  This is clearly invertible, hence a bijection.

Now all we have to do is sum over all k, 1 <= k <= n.  Our left hand side is

# of partitions w/ >= 1 1s + # of partitions w/ >= 2 1s + ... # of partition w/ >= n 1s.

A partition with k 1s is counted k times in this sum, so this is the same as the total number of 1s in all the partitions.

Summing the right hand side gives

# of partitions w/ a part of size 1 + # of partitions with a part of size 2 +... # of partitions w/ a part of size n.

A partition with k distinct parts is counted k times in this sum, so this is the same as the total number of distinct parts over all the partitions.

Setting those two equal yields the desired result.

[/hide]

The problem can also be solved with generating functions: I did that first, because that requires less cleverness, but it's too messy to write up and not elegant enough for it to be worth putting up on the forum.

Also, from http://www-math.mit.edu/~rstan/ec/newerr.pdf, the history of the theorem according to Richard Stanley himself:

<i>The following historical remarks concerning this exercise may be of interest. I discovered the result in 1972 and submitted it to the Problems and Solutions section of the Amer. Math. Monthly. It was rejected with the comment "A bit on the easy side, and using only a standard argument." My guess is that the editors did not understand the actual statement and solution of the problem. I had mentioned the result to Daniel I. A. Cohen, who included the case k = 1 as Problem 75 of Chapter 3 in his book Basic Techniques of Combinatorial Theory, Wiley, New York, 1978. For this reason the case k = 1 is sometimes called "Stanley's theorem." An independent proof of the general case was given by Kirdar and Skyrme, as mentioned in the text (page 59). The generalization from k = 1 to arbitrary k was independently found by Paul Elder in 1984, as reported by R. Honsberger, Mathematical Gems III, Mathematical Association of America, 1985 (page 8). For this reason the general case is sometimes called "Elder's theorem." A further proof was given by A. H. M. Hoare, Amer. Math. Monthly 93 (1986), 474-476.</i>

So it appears the result was fairly new in 1986.


Let $ a,b,c,d$ be [b]REAL [/b]numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$.Prove that:
         $ a^3\plus{}b^3\plus{}c^3\plus{}d^3 \leq 8$.
Any hints(or even solutions) would be appreciated.If it has been posted before let me know the link please.I found this one in one of Vasc's books
but it is an introductory problem so it shouldn't be that hard but I am struggling finding a solution :(

We have $ x^3 \plus{} y^3 \le (x^2 \plus{} y^2)^{\frac {3}{2}} \Leftrightarrow 2x^4y^2 \plus{} 2x^2y^4 \plus{} x^2y^2(x \minus{} y)^2\ge 0$ for real numbers $ x$, $ y$.
Hence $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \le (a^2 \plus{} b^2)^{\frac {3}{2}} \plus{} (c^2 \plus{} d^2)^{\frac {3}{2}} \le (\sqrt {a^2 \plus{} b^2} \plus{} \sqrt {c^2 \plus{} d^2})^{\frac {3}{2}}$ and then square and work with AmGm.

or you can do it like pain rinnegan^^ that's much more elegant :P

[quote="new_member"]I don't understand.Can you become more specific?
The inequality you wrote is wrong,i think.[/quote]

I'm pretty sure the inequality i wrote is correct ( with the condition $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$ )

You said you want a hint . From that i think you should able to develop it by yourself :P

[quote]I'm pretty sure the inequality i wrote is correct ( with the condition a^2+b^2+c^2+d^2=4 )

You said you want a hint . From that i think you should able to develop it by yourself Razz[/quote]

Pain,what drugs are you taking?Obviously you can say that [b]AT LEAST[/b] one of $ a,b,c,d$ should be $ \leq 2$ otherwise the
sum will be $ > 4$ so yea your inequality is true.Your hint helped me so much towards the solution of the problem.LOL!!!
i

[quote="new_member"]Let $ a,b,c,d$ be [b]REAL [/b]numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
         $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \leq 8$.
Any hints(or even solutions) would be appreciated.If it has been posted before let me know the link please.[/quote]
[color=green]inequalities marathon[/color] post #21 ,(problem 10).
http://www.mathlinks.ro/viewtopic.php?t=299899&start=20

I don't understand why $ a,b,c,d \leq 2$.I give up about thinking more about this problem so
if anyone can explain,thanks.

$ a^2\le 4$ . Is it clear now ? :huh:

No offense , but you're the one taking drugs :D

Guys, let's put some order here  :)

From $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$ follows not just $ |a| \leq 2$, but all of them, since for example $ 4 \geq b^2$.

Now $ c^3 \leq 2c^2$, the more if $ c<0$.

So $ d^3 \plus{} c^3 \plus{} b^3 \plus{} a^3 \leq 8$, with equality if $ p\equal{}2$ and $ q\equal{}r\equal{}s\equal{}0$.

How's that.

[quote="mavropnevma"]Guys, let's put some order here  :)

From $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$ follows not just $ |a| \leq 2$, but all of them, since for example $ 4 \geq b^2$.

Now $ c^3 \leq 2c^2$, the more if $ c < 0$.

So $ d^3 \plus{} c^3 \plus{} b^3 \plus{} a^3 \leq 8$, with equality if $ p \equal{} 2$ and $ q \equal{} r \equal{} s \equal{} 0$.

How's that.[/quote]

Is the same as mine because $ |a|\le 2\Rightarrow a\le 2$ ;)

Finally i understood this solution!
Thanks mavropnevma for explaining step by step how you got the solution of the problem.
I am sorry pain but i didn't get your hint at all.It was more confusing actually to me than enlightening.LOL
I spent so much time working with $ AM\minus{}GM$ and $ Cauchy\minus{}Schwarz$ trying to find a solution and i just couldn't believe
such an "easy" solution was possible.Anyways,everything is fine now.

[hide]Lesson of the day:I am never asking for hints again.Only solutions  :rotfl: [/hide]


limit logarithms invariant function inequalities Gauss algebra

Let $a$ and $b$ be reals greater than or equal to 1.

Let $a_{1}$ be the real number that satisfies $a^{b}=a_{1}^{a_{1}}$
and let $b_{1}$ be the real number that satisfies $b^{a}=b_{1}^{b_{1}}$

Also, let $a_{n+1}$ be the real number that satisfies ${a_{n}}^{b_{n}}={a_{n+1}}^{a_{n+1}}$
and let $b_{n+1}$ be the real number that satisfies ${b_{n}}^{a_{n}}={b_{n+1}}^{b_{n+1}}$.

[b]a.[/b] Does $\lim_{n\to\infty}[a_{n}]=\lim_{n\to\infty}[b_{n}]$ for all positive inputs $a$ and $b$ greater than 1? Prove or disprove.

[b]b. [/b]If the above is true, can one find a general formula for $\lim_{n\to\infty}[a_{n}]$?

[b]c.[/b] In the cases of $(a,b)=(2,3)$ and $(a,b)=(5,6)$, it appears that the (apparent) limit is between the geometric and harmonic means. Is this always the case for inputs greater than 1?

[quote="The Zuton Force"][b]b. [/b]If the above is true, can one find a general formula for $\lim_{n\to\infty}[a_{n}]$?[/quote][hide="b"]We can start from
${a_{n}}^{b_{n}}={a_{n+1}}^{a_{n+1}}$,
${b_{n}}^{a_{n}}={b_{n+1}}^{b_{n+1}}$.

Taking logarithms,
$b_{n}\ln(a_{n})=a_{n+1}\ln(a_{n+1})$,
$a_{n}\ln(b_{n})=b_{n+1}\ln(b_{n+1})$,

and then taking the product of the two previous equalities,
$a_{n}b_{n}\ln(a_{n})\ln(b_{n})= a_{n+1}b_{n+1}\ln(a_{n+1})\ln(b_{n+1})$,
or equivalently,
$\ln({a_{n}}^{a_{n}})\ln({b_{n}}^{b_{n}})= \ln({a_{n+1}}^{a_{n+1}})\ln({b_{n+1}}^{b_{n+1}})$,
we've reached an invariant!

So, if we call
$A=\lim_{n\to\infty}a_{n}$, $B=\lim_{n\to\infty}a_{n}$,
then
$\ln(a^{a})\ln(b^{b})=\ln(A^{A})\ln(B^{B})$.

If we assume that [b]a[/b] is true, that is,
$A=B=M$,
then
$\ln(a^{a})\ln(b^{b})=(\ln(M^{M}))^{2}$.

Finally,
$M=f^{-1}(G(f(a),f(b)))$,
where
$G(x,y)$ is the geometric mean of $x,y>0$, and
$f(x)=\ln(x^{x})$,
$f^{-1}(x)=\frac{x}{W(x)}$,
related to the $W$ Lambert function.

The final expression is
$\boxed{M=\frac{\sqrt{\ln(a^{a})\ln(b^{b})}}{W(\sqrt{\ln(a^{a})\ln(b^{b})})}}$
[/hide]

Excellent. Btw, part c should say arithmetic, not geometric

I've been looking for information about means.
[hide="a"]This problem is just a special case of the so-called "Gauss composition of strict means".

A function $f(a,b)$ is called a "strict mean" if
- $f$ is continuous  and
- if $a\neq b$, $\min\{a,b\}<f(a,b)<\max\{a,b\}$
(From the two condirions, it follows that $f(a,a)=a$).
Notice that $f$ is not necessarily symmetric like the usual means.

Given two strict means, $f_{1},f_{2}$, one can construct two sequences with the following recurrence
$a_{0}=a$, $a_{n+1}=f_{1}(a_{n},b_{n})$
$b_{0}=b$, $b_{n+1}=f_{2}(a_{n},b_{n})$

From the properties of strict means, it is known that $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=f(a,b)$,
and $f$ is also a strict mean, called the Gauss composition of $f_{1}$ and $f_{2}$.

The arithmetic-geometric mean is very well known, and the arithmetic-harmonic mean turns out to be the geometric mean.

The case of this problem is the Gauss composition of
$f_{1}(a,b)=c$, $c^{c}=a^{b}$,
$f_{2}(a,b)=c$, $c^{c}=b^{a}$,
which are asymmetric but strict means, because if $1\leq a<b$,
$a^{a}<a^{b}<b^{b}$,
$a^{a}<b^{a}<b^{b}$.[/hide]

[quote="lordWings"]I'm having a nice time with this problem.
I hope that our top-level AOPSers will take a look at it and give some insight.

[hide="c"]If $f$ is an invertible function, then I'll call
$M_{f}(a,b)=f^{-1}(\frac{f(a)+f(b)}{2})$
the $f$-mean of $a$ and $b$.

The following means are $f$-means!
$f(x)=x$ $\to$ arithmetic mean $A$.
$f(x)=\frac{1}{x}$ $\to$ harmonic mean $H$.
$f(x)=\ln x$ $\to$ geometric mean $G$.
$f(x)=x^{n}$ $\to$ $n$-th power mean $A_{n}$.

With a slight transformation of my latest post, the mean in your problem is also a $f$-mean, where
$\boxed{f(x)=\ln(\ln(x^{x}))}$.

A visual way to understand the definition:
Plot the function $f$, and mark the points $(a,f(a))$ and $(b,f(b))$.
If we find the point of the plot whose ordinate is the arithmetic mean of the ordinates,
$(m,\frac{f(a)+f(b)}{2})$,
then the abscissa, $m$, is the $f$-mean.

Looking at the plot, it looks like:
If $f$ is increasing and convex, then $M_{f}>A$.
If $f$ is increasing and concave, then $M_{f}<A$.
If $f$ is decreasing and convex, then $M_{f}<A$.
If $f$ is decreasing and concave, then $M_{f}>A$.

Following this, I think it shouldn't be difficult to establish an inequality between $f$-means simply "comparing" which $f$ is "more convex".

Am I wrong?[/hide][/quote]
[hide="Final answer to c"]Let $M$ be the mean defined in the problem, and $H$ the harmonic mean.
We have that $M(1.5,2)<H(1.5,2)$ but $M(3,2)>H(3,2)$.
So, we've found a counter-example and the answer to [b]c[/b] is "no".[/hide]
And more! [hide="c"]How can we compare two $f$-means?

We'll start from
- two bijective (invertible) functions $f_{1},f_{2}$ in a common domain $D$,
- the two related $f$-means $M: D^{2}\to D$:
$M_{1}(a,b)={f_{1}}^{-1}(\frac{f_{1}(a)+f_{1}(b)}{2})$,
$M_{2}(a,b)={f_{2}}^{-1}(\frac{f_{2}(a)+f_{2}(b)}{2})$,
- the auxiliar function $f_{12}(x)=f_{1}({f_{2}}^{-1}(x))$ (be careful, the domain $D'$ of this function is the range of $f_{2}$).

We can say that
$M_{1}(a,b)<M_{2}(a,b)$ for all $a,b\in D$ iff
- $f_{1}$ is increasing in $D$ and $f_{12}$ is concave in $D'$, or
- $f_{1}$ is decreasing in $D$ and $f_{12}$ is convex in  $D'$.
And vice versa.
[hide="Proof"]$M_{1}(a,b)<M_{2}(a,b)$ for all $a,b\in D$ iff
${f_{1}}^{-1}(\frac{f_{1}(a)+f_{1}(b)}{2})<{f_{2}}^{-1}(\frac{f_{2}(a)+f_{2}(b)}{2})$ for all $a,b\in D$,
which is equivalent to
${f_{1}}^{-1}(\frac{f_{1}({f_{2}}^{-1}(c))+f_{1}({f_{2}}^{-1}(d))}{2})<{f_{2}}^{-1}(\frac{c+d}{2})$ for all $c,d\in D'$.
If $f_{1}$ is increasing in $D$, that's
$\frac{f_{1}({f_{2}}^{-1}(c))+f_{1}({f_{2}}^{-1}(d))}{2}<f_{1}({f_{2}}^{-1}(\frac{c+d}{2}))$ for all $c,d\in D'$,
which is exactly the definition of concavity.
If $f_{1}$ is decreasing, the inequality is reversed (convexity).
[/hide]
If $f_{1},f_{2}$ are twice differentiable in $D$, this translates to
$M_{1}(a,b)<M_{2}(a,b)$ for all $a,b\in D$ iff
the sign of ${f_{1}}'(x)$ in $D$ is different to the sign of ${f_{12}}''(x)$ in $D'$.
[hide="More"]We could even prove that this is equivalent to the symmetric
$\frac{{f_{1}}''(x)}{{f_{1}}'(x)}<\frac{{f_{2}}''(x)}{{f_{2}}'(x)}$,
(and not the intuitive "which function is more convex" ${f_{1}}''(x)<{f_{2}}''(x)$ that I thought),
but I won't use it in this problem.[/hide]
From here, taking the $f$-mean in our problem, with
$f_{1}(x)=\ln(\ln(x^{x}))=\ln(x\ln(x))=\ln(x)+\ln(\ln(x))$,
with domain $x>1$ and
${f_{1}}'(x)=\frac{\ln(x)+1}{x\ln(x)}>0$,
we can compare it to other means:
[hide="Arithmetic mean"]$f_{2}(x)=x$,
$f_{12}(x)=\ln(x)+\ln(\ln(x))$ with domain $x>1$,
${f_{12}}'(x)=\frac{\ln(x)+1}{x\ln(x)}$,
${f_{12}}''(x)=-\frac{(\ln(x))^{2}+\ln(x)+1}{(x\ln(x))^{2}}<0$ in $x>1$,
so $M_{1}(a,b)<M_{2}(a,b)=A(a,b)$.[/hide]
[hide="Geometric mean"]$f_{2}(x)=\ln(x)$,
$f_{12}(x)=x+\ln(x)$ with domain $x>1$,
${f_{12}}'(x)=1+\frac{1}{x}$,
${f_{12}}''(x)=-\frac{1}{x^{2}}<0$ in $x>1$,
so $M_{1}(a,b)<M_{2}(a,b)=G(a,b)$.[/hide]
[hide="Harmonic mean"]$f_{2}(x)=1/x$,
$f_{12}(x)=-\ln(x)+\ln(-\ln(x))$ with domain $0<x<1$,
${f_{12}}'(x)=\frac{-\ln(x)+1}{x\ln(x)}$,
${f_{12}}''(x)=\frac{(\ln(x))^{2}-\ln(x)-1}{(x\ln(x))^{2}}$, which hasn't always the same sign in $0<x<1$
(the sign of $z^{2}-z-1$ in $1<z<e$ changes at $z=\phi$, golden constant),
so sometimes $M_{1}(a,b)<M_{2}(a,b)=H(a,b)$ and sometimes $M_{1}(a,b)>M_{2}(a,b)=H(a,b)$
if we have values of $a,b$ at both sides of $e^{1/\phi}=1.855\dots$.[/hide][/hide]


In a triangle $ABC$ with $AB < BC < AC$, points $A',B',C'$ are
such that $AA' \perp BC$ and $AA' = BC$ (A' and A are from different sides of BC), $BB' \perp CA$ and $BB' = CA$(B' and B are from different sides of AC),
and $CC' \perp AB$ and $CC' = AB$(C' and C are from different sides of AB), as
shown on the picture http://www.imo.org.yu/othercomp/Mex/MexMO96.pdf Problem 6. Suppose that
$\angle AC'B$ is a right angle. Prove that
the points $A',B',C'$ are collinear.

The triangles $\bigtriangleup ABA'$ and $\bigtriangleup CC'B,$ are congruent because they have $AA' = BC,$ $AB = CC'$ and $\angle BAA' = \angle C'CB$ $($ from $AA'\perp BC$ and $CC'\perp AB$ $)$ and so, we have that $BA' = BC'$ $,(1)$ and $\angle ABA' = \angle BC'C$ $,(2)$

We denote as $H,$ the orthocenter of $\bigtriangleup ABC$ $($ concurrency of $AA',$ $BB',$ $CC'$ $)$ and it is easy to prove that $\bigtriangleup BA'C',$ is a right-angled isosceles triangle, at $B.$

$\angle A'BC' = \angle ABA'-\angle ABC' = \angle BC'C-\angle ABC' = 180^{o}-\angle BC'H-\angle ABC'$ $\Longrightarrow$ $\angle A'BC' = 90^{o}$

$\Longrightarrow$ $\angle BA'C' = \angle BC'A' = 45^{o}.$

By the same way we can prove that  $\angle AB'C' = \angle AC'B' = 45^{o}.$ Hence the points $A',$ $B',$ $C',$ are collinear and the proof is completed.

Kostas Vittas.


Construct an algebraic equation [tex]f(x)=0[/tex] ([tex]f\in \mathbb {Z}[x][/tex]) such that [tex]f(\sqrt{3}-\sqrt{2})=0[/tex] and [tex]deg f [/tex] would be minimal. Justify your answer.

$X=\sqrt{3}-\sqrt{2}\Longrightarrow x^2=5-2\sqrt{6}\Longrightarrow x^2-5=-2\sqrt{6}$

$\Longrightarrow (x^2-5)^2=(-2\sqrt{6})^2\Longrightarrow x^4-10x^2+1=0$

Kunny didn't prove that $\deg f=4$ is minimal.
Indeed, if $\sqrt{3}-\sqrt{2}$ is a root, then all $\pm\sqrt{3}\pm\sqrt{2}$ are roots (well known trick). So $\deg f\geq 4$.

[quote]if [tex]\sqrt{3}-\sqrt{2}[/tex] is a root, then all [tex]\pm\sqrt{3}\pm\sqrt{2}[/tex] are roots (well known trick).[/quote]
How to prove this? Sorry, if I am missing something trivial.

After substitution to the equation you will obtain something like $a+b\sqrt 3+c\sqrt 2+d\sqrt 6=0$, where $a,b,c,d$ are integers, which depends on coeeficients of the equation and numbers $(\sqrt 3)^2$ and $(\sqrt{2})^2$. So if you will change sign of some summands in $\sqrt 3-\sqrt 2$ you will obtain $a\pm b\sqrt 3\pm c\sqrt 2\pm d\sqrt 6=0$, since, clearly, $a=b=c=d=0$.


I had this problem:

We are given some numbers, say:
$ 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3$
($ 4$ appears $ 4$ times & $ 3$ appears $ 16$ times)

Their $ \text{sum}$ is $ 64$.

The problem asks to determine the smallest numbers $ k$ such that the multi-set of numbers can be partitioned in $ \frac {\text{sum}}{k}$ subsets, with the numbers in each subset having sum equal to $ k$. Also, provide one way to do this for the computed $ k$.

In the example, $ k \equal{} 16$, with the partition:
$ 4,3,3,3,3$
$ 4,3,3,3,3$
$ 4,3,3,3,3$
$ 4,3,3,3,3$

What I sent: I took the divisors $ k$ of $ \text{sum}$ one by one, computed all the ways in which the sum $ k$ can be obtained from some of the given numbers (DP), and then backtracked on those solutions. I tried various optimizations to the bkt, but they all kind of failed miserably.

Any other ideas? It seems that one can do a little better.
I know it doesn't seem interesting, but I spent a lot of time writing the code for the above solution, so I don't see why I shouldn't waste more time posting it here.

(This is really closer to the partition problem than subset-sum, but these are already pretty close [it took me about 5 seconds to come up with both directions of the mapping reduction] that it doesn't matter.)

Anyway, how exactly are you backtracking?

What I came up with:

Assign each element of the multiset an index; you can then express each subset as an n-tuple like (16,4) (i.e.: in the multiset given, assigning $ 3$ an index of $ 0$ and $ 4$ and index of $ 1$). Once you have found all subsets summing to $ k$, you must determine whether there exists some combination of them whose union is the entire multiset. Since these subsets can be expressed as n-tuples like (4,1), this is simply a multidimensional knapsack problem, which can easily converted to a single-dimensional knapsack problem by expressing each n-tuple as a base-whatever integer.

Wow, I totally missed this latter interpretation.  :) The only problem with this approach would be that I'd need a LOT of space. One question though: some solutions can appear several times... how can you handle them elegantly?

About my backtracking, I just backtracked... I just tried all the solutions, maintaining the sum of indices under the index of the sum.

I'm sorry, I don't understand either of your issues. What do you mean by "some solutions can appear several times?" How would this take a lot of space?

If you mean a set can be repeated in a solution, the knapsack formulation already takes care of that. As for space, you don't need much more space that whatever you need to find the various subsets of a given sum.

[quote="perfect_radio"]
I know it doesn't seem interesting, but I spent a lot of time writing the code for the above solution, so I don't see why I shouldn't waste more time posting it here.[/quote]

as a 2 year-old sidenote, in order to waste less time on an implementation maybe you'd want to use a language to prototype your idea in Haskell for example.


Given a $\Delta ABC$ construct point(s) $X$ on $AB$ such that $AX*BX=CX^{2}$. Prove that there are at most 2 such points.
 If there is only one point prove $CX$ is angle bisector.
If there are 2 points $X_{1}$ and $X_{2}$ ($X_{1}$ between $A$ and $X_{2}$, prove that $\angle ACX_{1}=\angle BCX_{2}$

The two possible locations for X are the intersection of the circle with diameter CO and the line AB. I shall justify this later today, but it uses the intersecting chords theorem on CX produced to the circumcircle.


algebra function domain calculus derivative vector Gauss

I cannot solve the following problem:

Let $ U \subset {R^n}$ be an open and bounded domain with smooth boundary $ \partial(U)$
and denote by $ 1_U$ its characteristic function.  Compute $ \nabla_{1_U}$ in the sense of distributions
and determine its support.

The derivative is $ \nu(x)\delta_{\partial U}(x)$, where $ \nu(x)$ is the outer normal vector to $ \partial U$ at $ x$. This is exactly the Ostrogradsky-Gauss theorem.


inequalities
geometry
function
area of a triangle
Heron\u0027s formula

Prove that:

$ \frac {\sum^n_{i \equal{} 1}\sum^i_{k \equal{} 0} \frac {(p \plus{} 2ki)^{k \plus{} 2}}{k \plus{} 2}}{4n} > A$

where A is the area of triangle ABC and p is the perimeter.

i actually derived this and i don't see how you can make an application of this......

[hide="Solution that I have a lot of doubts about being correct."]

If n is greater, the left side will be larger.

So let's say n = 1.

Then we get that we want to prove $ 3p^2 \plus{} 2(p\plus{}2)^3 > 24 A$

Using AM-GM on s-a,s-b, and s-c, where s is the semiperimeter, and Heron's, we get that $ p^2 \ge 4A\sqrt{27}$

And the thing we want to prove is obvious...

I think I made a mistake.
[/hide]

you made lots of them......

first, you forgot about s in the heron's formula.

Unless you prove it, you can't say that the function is increasing in terms of n (what if the numerator is similar to the series $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\dots$?? (and denominator is obviously linear)

also, you didnt prove that it is $ >$, not $ \geq$

[hide="Heron's"]I got an upper bound for (s-a)(s-b)(s-c) in terms of s, using am-gm, then square rooted that, and multiplied by s, to get the thing.[/hide]

[hide="Is increasing"]Hmm...for the second sums, you can ignore everything except the first two terms, and they will be larger than the other terms, and because if $ \frac{a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n}{n} \ge \frac{a_n}{n} If$a_1 \ge a_2,$ etc.$[/hide]

[hide="Greater than or equal to or no equal"]The 12\sqrt{3} > 24 thing is true...[/hide]

Hmm... did you make a typo with the k in the denomiantor actually being i?

Because then the left side migth not be strictly increasing.

i is the variable for the sum, not the imaginary i.

your proof is correct, but it is strict.

Also, there's a real nice one using some variance of AM-GM, and the heron's formula which you caught on to.


probability AMC USA(J)MO USAMO AIME counting distinguishability

Which states are you all from?

Are your states any good?

Which state/s do you think have a good chance of winning at Nationals?

I think the Virgin Islands have a good shot at it.

I'm from California, and I think California's gonna win  :D

Texas, man, Definitely!
WE HAVE MARK ZHANG and DENNIS MOU!!!

(Screaming battle-cry at the top of lungs) OOOOOOOOOOOOOHHHHHHH-IIIIOOOOOOOOOOOOOOOOOOOOOO!!!!!

From what I heard Indianna had <snip>.

So I think they are going to win.
[color=darkblue]
[size=75][Edited by moderator rcv.  Do not discuss scores until after April 2.][/size][/color]

[quote="elementsofmathkid"]Go Nebraska!!!!! :)  :)  :)  :lol:[/quote]

I agree with her. We're gonna win!

Our best score was <snip>. I doubt Nebraska is going to win.

[color=darkblue][size=75][Edited by moderator rcv.  Do not discuss scores until after April 2.][/size][/color]

Oh, come on WM. Be optimistic. We'll never win with an attitude like that.

And btw- I TOLD YOU [u][b]NOT[/b][/u] TO POST SCORES, BUT NO, YOU WOULDN'T LISTEN TO ME!!!! :)  :)  :)  :lol:  :P

Washington is gonna win soooooooooooo bad. We are gonna SCHOOL you!!!

[quote="Klebian"]Texas, man, Definitely!
WE HAVE MARK ZHENG and DENNIS MOW!!![/quote]

WHO IS KLEBIAN!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?

I AM DENNIS AND MY LAST NAME IS SPELLED MOU!!

MARK'S LAST NAME IS SPELLED ZHANG!!

hehe...

Yea, we're gonna win, as long as we can beat California.

Calm down dude, the guy was praising you... :lol:

nat mc why are you still bitter over the whole indiana thing no offense but i mean who still cares they won and now get over it even if indiana is better

btw tarquin is going to 9th

fcms is first colony middle school with kevin chen

well IN did do really bad...

and tex*s is now a cus word....AoPS needs to have filters

Rofl, you middle schoolers are ... um, peculiar in ways.

You'd think that you're talking about rival colleges or high schools, heh.

I think that the top contenders will be Calafornia, Texas, Illinois, and Florida.  I was suprised :o  that only one Texan made it into the nationals countdown round, after looking at there amazing scores at state and district! I am  :o  that some people voted for Delaware. i didn't make it past district, and i beat all of delawares's scores! :ninja:

Well actually, MA had the second best state scores(Indiana had the best), but we ended up in fourth place overall, so you can never tell.

[quote="Iamnotleet"]Yep, I'm surprised at how well Teaxs did...
Neal Wu was...? I knew him in Kindergarten xD[/quote]

 :ninja: you serious? You live in Louisiana? Or did he move there later? Wow, you're lucky...wait a second... how do you still remember him? :huuh:

[quote="G-UNIT"][quote="Iamnotleet"]Yep, I'm surprised at how well Teaxs did...
Neal Wu was...? I knew him in Kindergarten xD[/quote]

really? was he a child prodigy or something?  :P[/quote]I moved to Texas later on...
He wasn't really a child prodigy... I was in a school with the only GT class thingie in Baton Rouge at the time... so was he (Figures...)
My mom is a friend of a friend of Neal Wu's mom...
And we used to be babysitted by the same person o.O

no offense to anyone but i heard that louisana's academics were kind of bad in general but neal wu's academics are good :D

K, I think natmc has EVERY RIGHT to be very bitter -- I'm actually kind of shocked at how everyone cares so much about team performances -- I'm just concerned about individuals! And nat_mc was up there with the benjamins (from what i recall from trips over to the Benjamins' house)

It'd suck to be David Benjamin, actually, since he and nathan are basically the same age, they probably take the same lessons (homeschool), 'cept david's older -- so poor him, his brother has a better chance for fame.

Indiana's REALLY tough this year -- I remember in 2004 when I heard i was moving to Indiana i thought it'd be a walk in the park to make it to nats compared to Illinois (who had then had a few consecutive AWESOME teams) -- boy, things changed in a year pretty quickly! There's only ONE spot left on the Indiana team (Sichen, John Ryan, Nathan B. all have secured slots) -- and personally, if I rob someone out of it -- I could care less -- unless its that one person I know really well.

One thing I've learned from Mathcounts, is that there are always "spots" open, so don't assume anything. If you're good, you'll get a spot.

I hope so! My brother nabbed a spot when he DEFINITELY wasn't supposed to take it (5th in school, 3rd in state the same year, and the 5th in school thing wasn't a fluke in the least bit!)


Let $ A\in C^{m\times n}$, show
$ rank (I_m \minus{} AA^ \minus{} ) \equal{} m \minus{} rank(A)$, where $ A^ \minus{} \in A\{1\}$, i.e., $ AA^ \minus{} A \equal{} A$.
If you have got a solution, please let me know
 miwalin@163.com

[hide]I use $ \forall X \in M_{pxq}{(\mathbb{C})},Y \in M_{qxp}{(\mathbb{C})}$
We have :$ Rank(XY) \leq min(Rank(X),Rank(Y))$ and if $ p \equal{} q$ ,$ Rank(X \plus{} Y)\leq Rank(X) \plus{} Rank(Y) \leq Rank(XY) \plus{} q$
We have :
$ Rank(AA^{ \minus{} })\leq Rank(A) \equal{} Rank(AA^{ \minus{} }A) \leq Rank(AA^{ \minus{} })$ then $ Rank(A) \equal{} Rank(AA^{ \minus{} })$
                               $ m \equal{} Rank(I_m) \leq Rank(I_m \minus{} AA^{ \minus{} }) \plus{} Rank(AA^{ \minus{} }) \equal{} Rank(I_m \minus{} AA^{ \minus{} }) \plus{} Rank(A) \leq Rank((I_m \minus{} AA^{ \minus{} })AA^{ \minus{} }) \plus{} m \equal{} Rank(AA^{ \minus{} } \minus{} AA^{ \minus{} }AA^{ \minus{} }) \plus{} m \equal{} Rank(AA^{ \minus{} } \minus{} AA^{ \minus{} }) \plus{} m \equal{} m$
Hence : $ Rank(I_m \minus{} AA^{ \minus{} }) \plus{} Rank(A) \equal{} m$[/hide]


This one is proposed by Mr.Janous:
n>=2,Determine the minimum $C(n)$ such that ${\sum_{i=1}^{n}{\sqrt{x_{i}}}\leq \sqrt{\prod_{i=1}^{n}(x_{i}+C(n))}}$ holds for all $x_{i}\geq 0$
Quite hard for me.............

Coulde you post you solution,Vasc? Or Could you give me some hint?
Thanks.

[quote="Vasc"]A nice hint.
See problem 6-(b) in Algebraic Inequalities (chapter 8, page 388).  :lol:[/quote]

can you tell me where i can find this book?

For Iurion:
See
http://www.mathlinks.ro/Forum/viewtopic.php?t=100399
or
http://www.gil.ro/en/

I am too upset that I can't get the book.....................................................I don't think it is a nice hint,Vasc :(
Who can help me?

[quote="Hawk Tiger"]I am too upset that I can't get the book.....................................................I don't think it is a nice hint,Vasc :(
[/quote] You are right. It is not a hint. It is just the required inequality.

Thank you very very much!!!
I am so grateful...
And it is an unblieveble solution..... :lol:

are there some digital edition of this book???

There is only printed edition distributed by Gil Publishing House.

You can deduce the problem to consider the case: $x_{1}=x_{2}=...=x_{n-1}$  :wink:


logarithms
inequalities
geometry
ceiling function
triangle inequality

Recently someone gave me triangle question with the sides of the triangle all being in log form. I know that the question involves the triangle inequality. I just don't know how to figure out the logs. The problem goes something like this: [i]if a triangle has side lengths log 12 and log 75(base 10 is implied) Then what is the smallest possible side length of the third side?[/i] I think this is the problem... If it isn't then I will post the correct problem later today.

If you were to put the sides directly on top of one another, the side length would be $ \log{75}\minus{}\log{12}\equal{}\log\frac{75}{12}$.  Since this is a degenerate triangle, you can move the sides infinitesimally away, but either way, the infimum is still $ \log\frac{75}{12}$.

If you wanted to use the triangle inequality, that would be largest possible side length.  Did you mean that?

Here is the actual problem...

The lengths of the sides of a triangle with positive area are log 12, log 75, and log n. Find the number of possible values of n. Remember base 10 is implied.

Is $ n$ an integer?

[hide="Assuming that`s true,"]
The minimum side length is $ \log75\minus{}\log12\equal{}\log\frac{75}{12}$, so the minimum value of $ n\equal{}\left\lceil\frac{75}{12}\right\rceil\equal{}7$.

The maximum side length is $ \log75\plus{}\log12\equal{}\log(75\cdot12)\equal{}\log900$, but that would result in a degenerate triangle. Thus, the largest possible $ n\equal{}899$, so we have $ 899\minus{}7\plus{}1\equal{}\boxed{907}$ possible values for $ n$.
[/hide]

In a triangle the sum of two sides is strictly bigger then the other, so if the sum of two sides equal the other then it is not a triangle, but instead three collinear points.

oh right. Didn't see you used equality instead of inequality.

[quote="batteredbutnotdefeated"]should it be 893... :blush:[/quote]
Oops, yeah that should be $ 893$, not $ 907$. I fail arithmetic?


Congratulations to:

[b]Acton-Boxborough Regional High School[/b] for claiming the Large School championship with 134 points! The second-place team was [b]Lexington High School[/b].

[b]Choate-Rosemary High School[/b] for claiming the Medium School championship with 88 points! The fourth-place team was [b]Bedford High School[/b] with 73 points.

[b]Hopkins School[/b] for claiming the Small School championship with 95 points! The second-place team was Worcester Academy with 90 points.

Good job everyone, at least Bedford beat Sharon! :lol:

The top individual score was 18, followed by 17, 16, 15, ..., 1, and 0. (...)

Apparently, based on general consensus, Rounds 3, 4, and 5 were the hardest rounds. Worthawholebean mentioned protesting a problem on Round 5. Which problem did you protest, and why?

Some of the results are up! http://www.mathmeets.com/2009_new_england_meet/index.php

5/2 missed a special case; all $ k>\minus{}25$ EXCEPT $ k\equal{}0$ work, as $ k\equal{}0$ only intersects once.

More data:

The average scores for rounds 1-6 are:

1: 3.578
2: 4.000
3: 2.028
4: 2.978
5: 1.318
6: 2.190

Using this data, the rounds in order of difficulty, decreasing, are: 5, 3, 6, 4, 1, 2. However, it should be noted that 2 free points were given to everybody on Round 4, so it should be harder than is listed here (but not by a full 2 points).

The number of people who scored a 6 on rounds 1-6 are:
1: 54
2: 58
3: 27
4: 20
5: 10
6: 21

Using this data, the rounds in order of difficulty, decreasing, are: 5, 4, 6, 3, 1, 2.

The average score overall was 7.983 out of 18.

10 people received a score of 18 on this meet.

Which rounds did you do? I did rounds 1, 5, 6.

gahhh we bombed (3rd small schools). I only got a 14, because I forgot shoelace formula totally (5/3) and made an arithmetic error (1/1 :oops: ). This kinda sucked.

We bombed even worse. We sent our best members on rounds 1, 5, 6, and our worst members on rounds 2, 3, 4 (even though I repeatedly told my coach not to). As a result, even though we were winning with a perfect on round 1, we bombed rounds 2, 3, 4, and that was what killed us. This also happened during the state meet, but we did well enough on rounds 5 and 6, as well as the team round, to be able to pick ourselves back up. This did not happen this time.

Here's Bedford's data:

Round 1: 18 (Tied with Westborough and Worcester Academy for the only two schools in the ENTIRE New England with a perfect on round 1)
Round 2: 9 (This was very bad - we had the third lowest score in medium schools.)
Round 3: 6 (OK, but not that great either.)
Round 4: 6 <- (This means nobody from Bedford got a single question right on this round...)
Round 5: 8 (Best score in medium schools along with Choate)
Round 6: 14 (Beat all other schools in medium schools here)

In fact, for each of 1, 5, 6 we either beat or tied EVERYBODY from medium schools. However, our scores for rounds 2, 3, 4 were just inexcusable.

In addition, we got a 12 instead of 15 on the team round becuase somebody misread #6 and thought the angles had to be integers, and as a result put $ (31^\circ,44^\circ)$...

Conclusion: If NEAML required you to only send 3 members instead of 6, we would've won.

[quote="Yongyi781"]Lol what is with this. Team round problems are 3 points each in large schools?[/quote]
That correction has been made to the official team sheets, but has not been updated.  My bad.

Also, there are quite a few spelling mistakes such as:

Chen, Y[color=red]u[/color]ngyi $ \implies$ Chen, Y[color=green]o[/color]ngyi
Ra[color=red]u[/color]i, Shreyas $ \implies$ Ra[color=green]v[/color]i, Shreyas

Also, on the 2009 MAML data, my name is reversed (lol?), so my name on the site is Yongyi, Chen (where Yongyi is my last name, Chen is my first name) whereas it should be Chen, Yongyi (where Chen is my last name, Yongyi is my first name).

Average score for each round:

R1: 3.58
R2: 4.00
R3: 2.03
R4: 2.98 (0.98/4 if you discount 2)
R5: 1.32
R6: 2.19

So the hardest round combination was 3/5/6, followed by 3/4/5 and 4/5/6.

edit: whoops, didn't see Yongyi's post above.

what is shoelace formula again?
yea 8th in small schools but I could have so easily had an 18 :(
there were some beast people there (sergei bernstein for example)

I did rounds 4, 5, and 6 and got a 13.

Worthawholebean, did they consider $ k>\minus{}25$ to be the answer? If so, that is wrong.

[quote="calc rulz"]I did rounds 4, 5, and 6 and got a 13.

Worthawholebean, did they consider $ k > \minus{} 25$ to be the answer? If so, that is wrong.[/quote]They accepted $ k\neq0$ as optional, as they didn't want to have to regrade papers I guess.

Their line of thinking was probably two-fold:

(1) "If we couldn't get it, we won't mark off other people who didn't get it."
(2) "Round 5 was already a killer as it is; we don't want more 0's."

I don't think they've ever marked people down due to an appeal, it would be too much regrading I guess.

Connecticut states did that this year on 6/2.

[quote="Yongyi781"]Also, there are quite a few spelling mistakes such as:[/quote]
I grabbed all of the individual data from mathmeets.com and transferred it to my "archive" on wocomal.org.  If you notice spelling errors anywhere in the statistics (NEAML, MAML, any year) please send a PM to let me know.  Unfortunately I don't have access to the original scoresheets, which makes it much more difficult to get the names right.

My version of the NEAML information is [url=http://www.wocomal.org/statistics/neaml/2009/2009n_individual.html]here[/url].  I'll keep updating it until I get no more PMs.

Thanks a lot.

[Edited -- I had linked to the MAML rankings rather than the NEAML rankings.]

[quote="Yongyi781"]Congratulations to:

[b]Acton-Boxborough Regional High School[/b] for claiming the Large School championship with 134 points! The second-place team was [b]Lexington High School[/b].

[b]Choate-Rosemary High School[/b] for claiming the Medium School championship with 88 points! The fourth-place team was [b]Bedford High School[/b] with 73 points.

[b]Hopkins School[/b] for claiming the Small School championship with 95 points! The second-place team was Worcester Academy with 90 points.[/quote]

this presentation of results is rather suspicious

Indeed. I forgot to mention that Lexington High School had 128 points, and that second and third place team for medium schools were Winchester and Westborough, with 76 and 75 points, respectively.


[quote="chien than"]Let $x;y;z>0$.Prove that:
${\sqrt{\frac{y^{2}z^{2}}{x^{2}(y+z)^{2}}+\frac{x^{2}y^{2}}{z^{2}(x+y)^{2}}}\ge \sqrt{\frac{y^{2}}{(x+y)^{2}}+\frac{y^{2}}{(y+z)^{2}}}}$[/quote]

right? if so, just use rearrangement. for that, note that the inequality is symmetric in z, x. :wink:

When, $x = z = 2, y = 1$, we have $LHS = \frac{4}{9}< 1$..
So it's false..


Prove whether the function f(x) is injective, surjective or bijective.  Also prove whether it is an involution.  When:
f(xf(x) + f(y)) = x^3 + y

An involution is bijective (both injective and surjective). Plug $ x \equal{} 0$ and get $ f(f(y)) \equal{} y$ - the definition of involution.


function
algebra proposed
algebra
Functional Equations

Find all function $f: \mathbb{Z}\to \mathbb{Z}$ such that :
 \[f(x^3+y^3+z^3)=f^3(x)+f^3(y)+f^3(z)  \quad \forall x,y,z \in \mathbb{Z}\]

You will find a solution (in french) at :
http://www.animath.fr/tutorat/dossier_03044sol.pdf

Pierre.