am i right to say that the net force on a particle moving along a circular path is its centripetal force?

It's the net force perpendicular to the particle's velocity, or the net force pointing towards the center of the circular path. Yes, it is the centripetal force. It's like summing forces.

No.
It can be stated as..
"The net force on a particle moving along a circular path [b][i]uniformly[/i][/b] is its centripetal force."

In non-uniform cases, along with the radial (centripetal force) there will be tangential force which causes angular acceleration

yea that makes sense because if it were moving uniformly along a circle its tangential velocity would be constant so its only acceleration and thus force would be the one changing its direction, i.e its centripetal force.


[quote="dogseatcheese"][quote="ch1n353ch3s54a1l"][quote="dogseatcheese"]In this 5 by 5 grid of dots, one sqare is shown in the diagram. Including this square, how many different sizes of squares can be formed using 4 dots of this array as vertices.

.  .  .  .  .
.  .  .  .  .
.  .  .  .  .
.  .  .  .  .
.  .  .  .  .    :arrow:  this is supposed to be a 5 by 5 grid of dots :!:

Please show all of your work. Is there a way to figure out this problem other than drawing and counting?

Thanks[/quote]
[hide]
You have 16 1x1 squares ($4^2$) , 9 2x2 squares ($3^2$), 4 3x3 squares ($2^2$), and 1 4x4 square. $16+9+4+1=30$ [/hide][/quote]

Actually, the problem asks for [b]different[/b] squares. And remember, the lengths of the sides of the squares don't have to be integer lengths. :P[/quote]

huh? "different" squares meaning...  :P sorry

the sides of the squares don't have to be integer lengths, it all depends on the length of a unit side. but that doesnt' really matter because you have to choose four points that are gridded in the square. So if the unit were for instance 1.5, you could have a side length of 4.5 which isn't an integer, if that's what you mean  :) i might be wrong though

I'm pretty sure the only way is to check different possibilities.

As for sides not being an integer length, it means that you can, for example, have diagonal points connected to give a side length of $\sqrt{2}$ or $\sqrt{3}$, or something like that.

I assume the dots are evenly spaced both up-and-down and sideways? and it asks for how many different sizes? then I think I've got it

[hide]
all these indicate side lengths
4*4, 3*3, 2*2, 1*1,
which is 4
and also the squares formed kind of sideways with sidelenths of
sqrt 2, sqrt 5 and sqrt 10
which is 3, so the answer is 7
[/hide]

[quote="k8reindeer"][hide]
all these indicate side lengths
4*4, 3*3, 2*2, 1*1,
which is 4
and also the squares formed kind of sideways with sidelenths of
sqrt 2, sqrt 5 and sqrt 10
which is 3, so the answer is 7
[/hide][/quote]

Actually,

[hide]You're missing $2\sqrt{2}$.  [/hide]

[hide]
going vertically and horizontally let the distance between the dots be one unit.

the total number of squares that are $1 \times 1$ is $20$

the total number of squares that are $2 \times 2$ is $9$

the total number of squares that are $3 \times 3$ is $4$

the total number of squares that are $5 \times 5$ is $1$

going at $45^{\circ}$ one side of a square is $\sqrt{2}$ units

the total number of squares that are $\sqrt{2} \times \sqrt{2}$ is $9$

the total number of squares that are $2\sqrt{2} \times 2\sqrt{2}$ is $1$

there are $44$ squares with $6$ different size squares.
[/hide]
This is relatively simple if you have a grid with a small number of points, but how would you determine the number of squares and the different size squares with $n \times n$ points with $n$ being a large number?

Actually, the answer to this problem is 8.

So in short, the only real way to figure out these types of problems is to count it all out? :(

I guess it is easier with a small amount of dots.....  :|

I think that would be the only way. And yes, it is complicated, but that's why they never give you large numbers of dots to work with. ;)

You can kind of do it more quickly, but there's no formula to just plug in numbers and find the answer for a certain size grid. Basically, take every combination of two positive integers up to $N$ in an $N \text{x} N$ grid, then use Pythagorean on those, and put down all of the distinct answers that you get.

You are correct JF.

Obviously, I didn't give this enough thought.

By counting the number of squares that are $\sqrt{5} \times \sqrt{5}$, I came up with a count of $8$.

By counting the number of squares that are $\sqrt{10} \times \sqrt{10}$, I came up with a count of $4$.

So there are, indeed, $8$ different sizes of squares, bringing the total number of squares generated to $56$.

Good eye JF.


LaTeX
geometry
geometric transformation
dilation
ratio
analytic geometry
percent

ok, since there is a Marathon Problem Solving in Getting Started, and the Intermediate math forum, i thought it would be nice to have a Chemistry Marathon as well.
so the idea is the same, just solve a problem and post new problem,
i think it would be better to post Olympiad problems, but i'll start with a very easy one.
[size=150][b]Problem 1[/b][/size]
in the Equilibrum:
$ 2A_{(aq)} \plus{} 3B_{(aq)}\Leftrightarrow C_{(aq)} \plus{} 2D_{(aq)}$
what happens to the equilibrium constant, $ K$ if the volume of the container is doubled?

The K value does not change, for only temperature can change that. However, the equilibrium will shift to the side with more moles, in this case the reactant side.

I'll post a problem shortly

[size=150][b]Problem 2[/b][/size]
The reaction for the formation of Nitrosyl Chloride is:

$2NO(g) + Cl_2(g) \Leftrightarrow 2NOCl(g)$

The pressures of each gas at $25^{\circ}C$ equilibrium were found to be:
$P_{NOCl} = 1.2 \text{ atm}$
$P_{NO} = 5.0 \times 10^{-2} \text{ atm}$
$P_{Cl_2} = 3.0 \times 10^{-1} \text{ atm}$

Calculate the value of $K$ for this reaction.

[hide="2"]$k_p=1920 atm$
since $k_p=k_c(RT)^{\Delta n}$
$k_c=\frac{k_p}{RT}(\Delta n=-1)=46917.12$
[/hide]
problem 3 coming up in couple of minutes

[size=150][b]Problem 3 [/b] [/size]
assume for the reaction:
$Fe{}_3O{}_4 _{s}+4H{}_2{}_{g}\Leftrightarrow 3Fe_{s}+4H{}_2O{}_{g}$ the $k_c=16$ what is the ratio of $[H_2]$ and $[H_2O]$, in the equilibrium?

yeah....i'll edit the problem
otherwise it would be impossible to get it... :)

haha yeah, i was confused, but then i realized it just might have been a typo or something.

In this case, the equilibrium may be expressed as $\displaystyle\frac{[H_2O_2]^4}{[H_2]^4} = 16$

Take the fourth root of each side and then take the reciprocal to get a ratio of [b]1/2[/b].

[size=150][b]Problem 4 [/b] [/size]
The remnants of a fire showed a Carbon-14 decay rate of 3.1 counts per minute per gram of Carbon.  Assuming the decay rate of Carbon-14 in wood today is 13.6 counts per minute per gram of Carbon, calculate the age of the remants.  The half-life of Carbon-14 is 5730 years.

Giving it a try.. probably wrong, though.

[hide]$\frac {3.1}{13.6} = (\frac 12) ^ t$
Solve for t: $t \approx 2.133$ so the answer would be $2.133 \times 5730= 1.2 \times 10^5$[/hide]

that looks right to me  :) could you post problem 5?

[quote="sky9073"]Cool. okay..
[size=150][b]Problem 5[/b][/size]
When 0.55g of pure benzoic acid, C7H6O2, is dissolved in 32.0g of benzene, the freezing point of the solution is measured to be 0.36'C lower than the freezing point value of 5.50'C for the pure solvent. Calculate the molecular weight of benzoic acid in benzene.[/quote]
i don't exactly understand what you're asking for,
are you asking for the mass of the the Benzoic acid dissolved? ;)

I think perhaps the value of the freezing point depression constant for benzene $(K_f = {\rm 4.9\ K\ kg\ mol}^{-1})$ should have been included in the problem?  Also, knowing its benzoic acid makes the problem confusing if the goal is obtaining the molecular weight - most would prefer using the periodic table for that calculation!

what...?
if you only want the molecular weight of $M{}_{C{}_7H{}_6{}_O2}=122 g/mol$ ;)
tell me if that's what you wanted...then i'll post the new problem

Enh. Just randomly picked one from somewhere but didn't realize it was so lame...
Go ahead.

[hide]B because 0.47=0.34-(-0.13)[/hide]

Someone else post another problem

Problem: How many moles of hydrochloric acid are needed to completely neutralize 5 moles of sodium hypochlorite?

I got [hide]3.85[/hide] moles. Someone check my work please. Sorry about not being fluent in LaTeX.

[hide]
I thought neutralize meant final pH=7, so I knew that 5 moles would be too much.
HCl + NaClO --> HClO + NaCl
I set [HClO]=[HCl]=x, and [NaClO]=5-x
ClO(-) + H20 <--> HClO + OH(-)
I plugged the concentrations (assuming 1L of solution) into the K_b expression.
10^-14/(3.0*10^-8)=10^-7*x/(5-x)
x=50/13=3.85

I also happen to have an exam on this tomorrow so hopefully I'm doing this right and don't fail my exam...
[/hide]

Both of these answers are incorrect. In this case, neutralize means to react with all moles of base present.

[quote=skywalker1]has anybody tried mixing acid w/ bleach? Renember that this is never a good idea, because there is a particular gas released which you may want to take int  account[/quote]

RIP i just might cause of ded NSB results

$\ce{2 HCl(aq) + NaClO(aq) -> NaCl(aq) + Cl2(g) + H2O(l)}$
Thus 2 moles of hydrochloric acid are needed to neutralize each mole of sodium hypochlorite, and thus 10 moles of hydrochloric acid are needed to neutralize 5 moles of sodium hypochlorite.

This is correct. Does anyone else want to post another problem?

[quote=skywalker1]A student has a 1 L solution of 2 molar potassium permanganate and a 2 ml Solution of 1.5 molar sulfuric acid. He adds and equal volume of each to a beaker with slightly acidic medium. Balance the reaction that occurs.[/quote]

No reaction: Neither sulfate nor permanganate can be further oxidized. Obviously neither can potassium. The only option would be to oxidize the [b]H$_2$O[/b] to [b]O$_2$[/b] but I don't believe this occurs. Hence no reaction.

Rafay what happened to your latex? It looks pretty scary :P

oh I combine latex with aops bold to type compounds :P

easier and looks much nicer too[quote=rafayaashary1]
No reaction: Neither sulfate nor permanganate can be further oxidized. Obviously neither can potassium. The only option would be to oxidize the $\ce{H2O}$ to $\ce{O2}$ but I don't believe this occurs. Hence no reaction.[/quote]


I was wondering if there was a rule for going between bases for divisibility.

For example, 11 is 1 mod 10. But if we use the fact that all of those numbers are 1 and 0 and try to change it from base 2 (though its not in base 2 right now) to base 10, we get 3 is 1 mod 2, which is still true. is this a coincidence?

Also, 1101 is 1 mod 11. If we do the same thing we get 13 is 1 mod 3. again, true.

i've been searching for a generalization, but i cant quite hit it. any help would be appreciated!

I don't quite understand what you are trying to generalize

Are you trying to take a number like

$ 1101_b$ and change b from 10 to 2 and look for patterns in divisibility?

Yeah its hard to explain but I think my examples are easier to understand.

I only looked into this in base 2:

We know that 1101 = 1 mod 11 in base 10.

Since we are using 1s and 0s, we can "pretend" that this is in base 2.
1101 base 2 = 13 base 10
1 base 2 = 1 base 10
11 base 2 = 3 base 10.

Then if we substitute we get 13 = 1 mod 3. This, again, is a true congruence.

The reason both of the examples you gave work is that the first is equivalent to $ 10 | 10$ and the second is equivalent to $ 11 | 1100$, both of which are true in any base.  Can you see why now?  Can you generalize?

[hide="Hint"] Think polynomials in $ b$. ;) [/hide]

I think the reason is
[hide]$ \sum_{i\equal{}0}^{n}{a_ib^i}\equiv a_0 \pmod b$[/hide]

right?

Not quite.

[hide="Consider..."] that $ 10_b \equal{} b, 11_b \equal{} b \plus{} 1, 1100 \equal{} b^2 \plus{} b^3$.  Hence the given divisibilities, interpreted in base $ b$, mean

$ b | b$
$ b \plus{} 1 | b^3 \plus{} b^2$

both of which are clearly true for any $ b$. :) [/hide]

Thank you for your in-depth replies.  :)

I think you mean that my examples worked out by chance, because I presented such a case to begin with.

A few more, more complex examples have proven to me that my conjecture does not always hold true. But I still wonder if there is a relatively simple rule that can tell me when this works. t0rajir0u, I think you presented that to me, but frankly, I don't understand completely. How can we tell when it will "work"?

Another example I found, in case it helps anyone:

$ 1000100010001 \equiv 4 \mod{101}$

Thanks again!

[quote="t0rajir0u"][quote="buzzer11"]t0rajir0u, I think you presented that to me, but frankly, I don't understand completely. How can we tell when it will "work"?[/quote]

Here's the deal:  a list of digits interpreted in an arbitrary base $ b$ is a polynomial in $ b$.  If you want a divisibility relation to work for more than one base (more than one value of $ b$), a [i]sufficient[/i] (but not necessary) condition is that one polynomial [b]divides the other[/b].  For example, because

$ b \plus{} 1 | b^3 \plus{} 1$

we know that

$ 11_b | 1001_b$

in any base $ b$.  If we restrict our digits to $ 1, 0$, we can use any $ b \ge 2$.  :)[/quote]

Oh! I understand now. Thanks again!  :)


geometry
vector
trigonometry
geometry unsolved

Dear Mathlinkers ,can anyone solve it  ? :)

[quote="CCMath1"]Let $ {O}$ be a fixed point in $ {\triangle ABC}$ , moving point $ {P}$ satisfies: $ {\vec{OP} \equal{} \vec{OA} \plus{} \lambda (\frac {\vec{AB}}{|\vec{AB}|sinB} \plus{} \frac {\vec{AC}}{|\vec{AC}|sinC})}$,$ {\lambda > 0}$, then the locus of point P must pass through which "center" of the triangle ? :)[/quote]
Denote $ AB \equal{} c,BC \equal{} a,CA \equal{} b$
See that $ |\vec{AB}| \sin B \equal{} |\vec{AC}| \sin C \equal{} \frac {bc}{2R}$. Let $ t \equal{} \lambda \frac {2R}{bc} > 0$ then:
$ \vec{OP} \equal{} \vec{OA} \plus{} t( \vec{AB} \plus{} \vec{AC} )$.
$ \vec{AB} \plus{} \vec{AC} \equal{} 2\vec{AD}$, where $ D$ is the midpoint of the side $ BC$
So $ P$ moves along the median from $ A$ to $ BC$, and it therefore passes through the centroid.

Well done! Mathias_DK! That is right , you solution is nice :wink:


[quote="mr. math"]It takes 130 Joules of work to compress a certain string 0.10 m. What is the force constant of this spring? To compress the spring an additional 0.10 meters, does it take 130 joules, more than 130 joules, or less than 130 joules? Verify your answer with a calculation.

I got the force constant but am having trouble solving the second part of the problem.[/quote]

1) .5kx^2 = 130

Solve for k substituting x=.1.

2) Ans is more

Logically speaking the resistance offered to your force is more, so you need to do more work.

Mathematically

.5k(.01) = 130

.5k(.04) = 4x130

so 3x130 is the energy requried. Which is more.

Hope you understand.


inequalities
Support
inequalities proposed

Let $a,b,c>0$. Prove that \[\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\]

[quote="nayel"]Let $a,b,c>0$. Prove that
\[\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\ge (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \]
[/quote]
It's also Iran 2005. :wink:

Does this support a generalization? something with $\sum_{sym}$ maybe...


Heres a good question I came across the other day:

Using the numbers $ 1, 3, 4, 6$ obtain the number $ 24$ by using each number only once and by only using the mathematical operations +-/* (and brackets).

[hide="solution"]$ 6/(1\minus{}3/4) \equal{} \dfrac{6}{1\minus{}\dfrac{3}{4}} \equal{} 24$[/hide]

Very nice :) How long did it take you?

Heres another one:

A mother gave birth to her son when she was 21 years old. In 6 years (from today), she will be 5 times older then her son. Where is the son's father?

How is that even possible?
If $ x$ is the mother's age and $ y$ is the son's, we have
$ x\plus{}6\equal{}5(y\plus{}6) \Leftrightarrow  x \equal{} 5y\plus{}24$
and
$ x\equal{}y\plus{}21 \Leftrightarrow y\plus{}21\equal{}5y\plus{}24 \Leftrightarrow \minus{}3\equal{}4y \Leftrightarrow$ the daughter is $ \minus{}\frac{3}{4}$ years old.

Maybe it's one of those stupid trick questions, seeing as the question asks "Where is the son's father?"  :P

This problem has been posted before.

Here is a link to it:
[url]http://artofproblemsolving.com/Forum/viewtopic.php?t=37312&start=500&sid=42d719b067cfa45254b6e4249f97f1d8[/url]

Rather disturbing answer...

I see; $ \minus{}\frac 34$ year $ \equal{} 9$ months before birth.


induction
probability
combinatorics unsolved
combinatorics

How to prove this:
$%Error. "display" is a bad command.
\sum_{i=0}^n \binom{m}{i} \binom{n}{i} = \binom{n+m}{m}$
in two ways:
induction and combinatorically...

Now combinatorcally, on the left we choose the same number of elements from two groups with size n and m... but how to explain the answer on the right?

and the induction doesn't work for me here... I am trying to hold m, and do it on n, but still I get no answer...

thanx for you help.

[quote="grobber"]The sum on the left can be written as $\sum\binom mi\binom n{n-i}$, and we can interpret it as the number of ways of choosing $m$ objects from $m+n$ objects: for each $i\le m$, we choose $i$ objects from a group of $m$ objects, and the other $n-i$ from another group of $n$ objects (assume these two groups are fixed from the beginning).[/quote]

Thanx!!
Now if I am trying to do this with induction I get this:
since m and n have the same meaning we can hold one of them and iterate on the other. So we decide to use n for that purpose:

First we show for n = 0:
$  \binom{m}{0} \binom{0}{0} = 1 = \binom{m}{m}$ And it's ok.

So we assume it is true for n = n;
$\sum_{i=0}^n \binom{n}{i} \binom{m}{i} = \binom {m+n}{m}$ and we prove for:
$\sum_{i=0}^{n+1} \binom{n+1}{i} \binom{m}{i} = \binom{m+n+1}{m}$
From here we want to reach the assumption somehow... so I do this:
$\sum_{i=0}^n \binom {n+1}{i} \binom {m}{i}  - \binom {n+1}{n+1} \binom {m}{n+1} = \binom {m+n}{m} + \binom {m+n}{m-1}$
$\sum_{i=0}^n \binom {n}{i} \binom {m}{i} + \sum_{i=0}^n \binom {n}{i-1} \binom {m}{i}   - \binom {n+1}{n+1} \binom {m}{n+1} = \binom {m+n}{m} + \binom {m+n}{m-1}$
Now we can at least remove some elements
$ \sum_{i=0}^n \binom {n}{i-1} \binom {m}{i}   -  \binom {m}{n+1} = \binom {m+n}{m-1}$

And I don't know how to continue .... and in all probability it's silly these things I am doing here...


[quote="filomen"]If $ a,b,c > 0$, then

$ \sum\frac {3a^3 \plus{} 7abc}{(b \plus{} c)^2}\geq\frac {3}{2}[\frac {a(a^2 \plus{} 2b) \plus{} b(b^2 \plus{} 2c) \plus{} c(c^2 \plus{} 2a)}{a^2 \plus{} b^2 \plus{} c^2}]$[/quote]
Try $ a\equal{}b\equal{}c\equal{}0.1.$ :wink:

[quote="arqady"][quote="filomen"]If $ a,b,c > 0$, then

$ \sum\frac {3a^3 \plus{} 7abc}{(b \plus{} c)^2}\geq\frac {3}{2}[\frac {a(a^2 \plus{} 2b) \plus{} b(b^2 \plus{} 2c) \plus{} c(c^2 \plus{} 2a)}{a^2 \plus{} b^2 \plus{} c^2}]$[/quote]
Try $ a \equal{} b \equal{} c \equal{} 0.1.$ :wink:[/quote]

i am sorry my friends  :blush:


AMC
AIME
function
algebra
polynomial
AoPS Books

Functions

I've noticed that I consistently can't figure out how to do alot of the functions and polynomial questions.  How does one go about trying to learn alot about functions?

Sets

Same question. You know, find the alternating sum of the subsets or something.  There have been quite a few questions on it.

Use the aops books, they have some stuff about that on functions and sets. Also the intermediate algebra course has a lesson on functions, that helped me a lot.

The only other advice I could give you is to keep working on the Aime functions/sets problems. This seems to be one of those things where you get better at once you do lots of problems.


Can you program a factorial computing program?
For example input 7 and compute the value of 7!

There is already a function on the calculator for factorials, under the Math menu, I think.

Well, for a start, the factorial is already built in.

Other than that, if you really really really wanted to, you could do recursion (with a FOR loop) or other methods.

I have it memorized:
[MATH][< (left)][4]

Edited to Klebian.

actually, that gives you the logic operator [quote]not([/quote]

To get to factorial, you don't hit the [2nd] key

Sniff.
On mine, you do.
Maybe I had it wrong?
I think mine gets factorial if you go to math, then hit left, then hit 4.


From point $ A$, Leigh walked 40 yards south, 60 yards west, 10
yards north, and 20 yards east to point $ B$. What is the length, in
yards, of $ \overline{AB}$?

Draw it and get a 30-40-[b]50[/b] triangle.
Note that for any Pythagorean triple (i.e. 3-4-5) or a-b-c,
na-nb-nc is also a triple.


calculus
integration
trigonometry
limit
calculus computations

hello folks.  :D

try this one:

$ \int_0^1\;\dfrac{\sin^{\minus{}1}\left(\sqrt{\,1\minus{}x^2\,}\right)}{(2\minus{}x^2)\,\sqrt{\,1\minus{}x^2\,}}\;\textbf dx$

it's enough to calculate $ \int_{\minus{}1}^{1}$ by complex variable :
countor is $ 0\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}0$ where $ 0$ is circle of $ \minus{}1$ and next $ 0$ is circle of $ 1$  and $ \sin^{\minus{}1}$ rewrite as $ Ln...$

My substitution is x=cost and we have arcsin(sint)=t.
We have $ \int_{0}^{pi/2}\frac{t}{2\minus{}(cost)^2}dt$
My second substitution is t=u+pi/4 .
My final answer is $ \frac{\sqrt{2}pi^2}{8}$

[quote="the_wonder"]Where is my mistake?[/quote]

$ \int_{0}^{1} \frac {arcsin{\sqrt []{1 \minus{} x^2}}}{\left( 2 \minus{} x^2\right)\sqrt []{1 \minus{} x^2}}dx \equal{} \lim_{\epsilon \rightarrow 0} \int_{0}^{1 \minus{} \epsilon} \frac {arcsin \sqrt []{1 \minus{} x^2}}{\left( 2 \minus{} x^2\right)\sqrt []{1 \minus{} x^2}}dx$

note : $ 0< \epsilon < 1$


Prove that if $ a,b,c,d\geq0$, then
\[ \frac{99\minus{}21\sqrt{21}}2\plus{}abcd\plus{}a^3\plus{}b^3\plus{}c^3\plus{}d^3\geq ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd.\]
When does equality occur?

[quote="pvthuan"]Prove that if $ a,b,c,d\geq0$, then
\[ \frac {99 \minus{} 21\sqrt {21}}2 \plus{} abcd \plus{} a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3\geq ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd.
\]
When does equality occur?[/quote]
To prove this inequality, it suffices to consider the following cases (no EV theorem ;) )
1. $ a\equal{}b,c\equal{}d$
2. $ a\equal{}b,d\equal{}0$
3. $ b\equal{}d\equal{}0$
:)


AMC
USA(J)MO
USAMO
quadratics
algebra
complex numbers

USAMO 19??

Let r be one of the roots of the quadratic equation x(1-x) = 1; the other root is 1-r.  Show that for n = 1, 2, 3, ....

if 3|n, r^n + (1-r)^n = 2*(-1)^n
if not, r^n + (1-r)^n = (-1)^(n-1)

[quote="mathfanatic"]Can you rigorously prove that this pattern will hold?  I don't think that observing the pattern and converting it into an equation will get full credit...[/quote]

By the recurrence it'll hold, becoz
a_(6k+1)=a_1
a_(6k+2)=a_2
...

Okay.  Looks good to me (not that I'm a very good judge...)

[quote="ral"]Or you could of course just use complex numbers and it will be over in a second. :D[/quote]

nah, you'd have to go back to non-complex afterwards, I think it would take a lot more time than what we got now...

Well, Peter, I think that in this case using complex numbers is really the shortest way to do it ... and the most straightforward :) As ral says, it's over in one second :)

oh well if you guys first think of that method that's fine, I just thought of siuhochung's method right when I saw the problem, so for me there was no shorter way possible :D

perhaps it's "nicer" to get it with a line less or so, but I would only have used complex numbers when this method failed.


[hide="fractions"]

[code]$\frac{a}{b}$[/code]

makes $ \frac{a}{b}$[/hide]

[quote="hsiljak"][hide="fractions"]

[code]$\frac{a}{b}$[/code]

makes $ \frac {a}{b}$[/hide][/quote]
Is there really a reason to hide that? :P
[quote="mathwizarddude"](1) $ a \plus{} x^2 \equal{} 2006, b \plus{} x^2 \equal{} 2007, c \plus{} x^2 \equal{} 2008$ and $ abc \equal{} 3$. Find
the value of: $ a/bc \plus{} b/ac \plus{} c/ab \minus{} 1/a \minus{} 1/b \minus{} 1/c$.

(2) If $ x \plus{} \sqrt {xy} \plus{} y \equal{} 9, x^2 \plus{} xy \plus{} y^2 \equal{} 27,$ find the value of:
$ x \minus{} \sqrt {xy} \plus{} y$.

P.S. How do you do fractions using the code?[/quote]
[hide="Overkill on 2"]We know that $ \sqrt {\frac {x^2 \plus{} xy \plus{} y^2}{3}} \equal{} 3$. By the Power Mean Inequality, this is greater than or equal to $ \frac {x \plus{} \sqrt {xy} \plus{} y}{3}$, implying that $ 9\geq x \plus{} \sqrt {xy} \plus{} y$. Since we know equality is met, we have $ x^2 \equal{} xy \equal{} y^2 \Rightarrow x \equal{} y \Rightarrow x \equal{} y \equal{} 3$. From here, we can find $ x \minus{} \sqrt {xy} \plus{} y \equal{} 3 \minus{} 3 \plus{} 3 \equal{} 3$.[/hide]
EDIT: Eww at my ugly solution, just look at Altheman's nice solution instead...

(2)  $ (x \plus{} y \plus{} \sqrt {xy})(x \plus{} y \minus{} \sqrt {xy}) \equal{} (x \plus{} y)^2 \minus{} xy \equal{} x^2 \plus{} y^2 \plus{} xy$

then the answer is $ 3$...of course assuming that there is a solution for $ x,y$

$ x \plus{} y \equal{} 6$, and $ \sqrt {xy} \equal{} 3$... $ xy \equal{} 9$,

[quote="Brut3Forc3"][quote="hsiljak"][hide="fractions"]

[code]$\frac{a}{b}$[/code]

makes $ \frac {a}{b}$[/hide][/quote]
Is there really a reason to hide that? :P
[/quote]

Not really got to do anything with the problem, so I hid it...  :lol:

$ |a \minus{} b| \equal{} 1$, $ |b \minus{} c| \equal{} 1$, and $ |c \minus{} a| \equal{} 2$ by subtracting equations.

The fraction when simplified is $ \frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{abc} \equal{} \frac {\sum (b \minus{} c)^2}{2abc} \equal{} \frac {1 \plus{} 1 \plus{} 4}{2} \equal{} 3$.

and bruteforce, that doesn't work since you don't know that x,y are positive reals or whatever

[quote="Altheman"]and bruteforce, that doesn't work since you don't know that x,y are positive reals or whatever[/quote]
Well, obviously $ x,y$ must both be positive, or both be negative. If both $ x$ and $ y$ are negative, then having $ x'\equal{}\minus{}x$ and $ y'\equal{}\minus{}y, \frac{x'\plus{}y'}{2}\geq\sqrt{x'y'} \Rightarrow \sqrt{x'y'}\minus{}(x'\plus{}y')<0$...

what if $ xy\equal{}5$ and $ x\plus{}y\equal{}1$?


advanced fields
advanced fields unsolved

prove that the subset of countable set is also countable.

That's classic.
See for example : http://plato.stanford.edu/entries/set-theory/primer.html#7


Suppose that $x,y,z$ are positive reals such that $xyz=1$, prove that if

\[x^k+y^k+z^k\geq \frac{1}{x^k}+\frac{1}{y^k}+\frac{1}{z^k}\]

holds for $k=1$, then it holds for any real $k>0$.

Just the small substitution $ z=\frac{1}{xy}$, a factorization and we are done...

Let $x=\frac a b$, $y=\frac b c$, $z=\frac c a$. The inequality is
\[ a^{2k}c^k+b^{2k}a^k+c^{2k}b^k\geq c^{2k}a^k+a^{2k}b^k+b^{2k}c^k\]
equivalent to $(a^k-b^k)(b^k-c^k)(c^k-a^k)\geq0$ and we are done.

You posted this solution just when I was about to post it myself :)

Its  the same  problem  from  4 th  round Russian  mo 1999 (Titu Andreesu 1999-2000  page 142)


geometry
circumcircle
Euler
trigonometry
ratio
parallelogram
Pythagorean Theorem

Let $G,H,K$ be the centroid, the orthocenter, the Lemoine point of $\triangle ABC$ respectively and denote $a=BC, b=CA, c=AB$.
If $b^2 + c^2 = 2a^2$ show that:

(1) $OK$ is perpendicular to $AK$

(2) $GK$ is parallel to $BC$

(3) $\ \ AK^2  = BK \cdot CK$

(4) $\ \ BH^2  + CH^2  = 2AH^2 $

(5) the pedal triangle of $G$  is isosceles

(6) the locus of $A$, when $B,C$ are fixed, is a circle

_____________________________________________

[color=blue]Edit:[/color] Typo error in (1) fixed.

Leon

[quote="Leon"]Let $G,H,K$ be the centroid, the orthocenter, the Lemoine point of $\triangle ABC$ respectively and denote $a=BC, b=CA, c=AB$.
If $b^2 + c^2 = 2a^2$ show that:

(1) $OH$ is perpendicular to $AK$
...
[/quote]
Let $O$ be the circumcenter and $R$ the circumradius of the triangle $\triangle ABC$. The distance $OH$ between the circumcenter and the orthocenter of any triangle is given by

$OH^2 = 9R^2 - (a^2 + b^2 + c^2)$

Since the points $O, G, H$ lie on the Euler line and $GH : OH = 2 : 3$, the distance $GH$ between the centroid and the orthocenter is given by

$GH^2 = \frac 4 9 OH^2 = 4R^2 - \frac 4 9  (a^2 + b^2 + c^2)$

The distance $AH$ from the vertex $A$ to the orthocenter is

$AH = 2R \cos \alpha$

$AH^2 = 4R^2 (1 - \sin^2{\alpha}) = 4R^2 - a^2$

The distance $AG$ from the vertex $A$ to the centroid is $\frac 2 3$ of the median $AA'$ from this vertex (where $A'$ is the midpoint of the side $a = BC$) and this median length is given by

$AA'^2 = \frac 1 4 (2b^2 + 2c^2 - a^2)$

$AG^2 = \frac 4 9 AA'^2 = \frac 1 9 (2b^2 + 2c^2 - a^2)$

If the triangles sides $a, b, c$ satisfy the condition $b^2 + c^2 = 2a^2$, then

$GH^2 = 4R^2 - \frac 4 9  (a^2 + b^2 + c^2) = 4R^2 - \frac 4 3  a^2$

$AG^2 = \frac 1 9 (2b^2 + 2c^2 - a^2) = \frac 1 3 a^2$

$AH^2 = 4R^2 - a^2$ [color=white].[/color] (as before)

$AG^2 + GH^2 =  \frac 1 3 a^2 + 4R^2 - \frac 4 3  a^2 = 4R^2 - a^2 = AH^2$

By Pythagorean theorem, the angle $\measuredangle AGH = 90^o$ is then right, i.e., the median line $AG$ (and not the symmedian line $AK$) from the vertex $A$ to the side $a = BC$ is perpendicular to the Euler line $OH \equiv GH$.

[quote="Leon"]Let $G,H,K$ be the centroid, the orthocenter, the Lemoine point of $\triangle ABC$ respectively and denote $a=BC, b=CA, c=AB$.
If $b^2 + c^2 = 2a^2$ show that:

(1) $OK$ is perpendicular to $AK$
...[/quote]
Then the proof of (1) continues by demonstrating that the triangles $\triangle AGH \sim \triangle AKO$ are similar by SAS. The angles $\measuredangle OAK = \measuredangle HAG$, because the orthocenter $H$ is the isogonal conjugate of the circumcenter $O$ and the symmedian (Lemoine) point $K$ is the isogonal conjugate of the centroid $G$. Squares of the sides $AO, AH, AG$ are equal to

$AO^2 = R^2$

$AH^2 = 4R^2 - a^2$

$AG^2 = \frac{a^2}{3}$ [color=white].[/color] (when $b^2 + c^2 = 2a^2$)

We have to calculate the side $AK$  using the condition $b^2 + c^2 = 2a^2$. Let $P$ be the intersection of the symmedian $AK$ with the side $a = BC$ of the triangle $\triangle ABC$. The symmedian $AK \equiv AP$ divides the side $a = BC$ in the square of the ratio of the adjacent sides $b = CA$ and $c = AB$ (see Honsberger's Episodes, Chapter 7):

$\frac{PC}{PB} = \frac{b^2}{c^2}$

$PC = \frac{b^2}{b^2 + c^2}\ a = \frac{b^2}{2a}$

Using the cosine theorem for the triangles $\triangle ABC, \triangle APC$,

$c^2 = a^2 + b^2 - 2ab \cos{\widehat C}$

$AP^2 = b^2 + PC^2 - 2b \cdot PC \cos{\widehat C} = b^2 + \frac{b^4}{4a^2} - \frac{b^3}{a} \cos{\widehat C}$

Calculating $\cos{\widehat C}$ from the 1st equation and substituting to the 2nd one,

$\cos{\widehat C} = \frac{a^2 + b^2 - c^2}{2ab}$

$AP^2 = b^2 + \frac{b^4}{4a^2} - \frac{b^3}{a} \cdot \frac{a^2 + b^2 - c^2}{2ab} = b^2 + \frac{b^4}{4a^2} - \frac{b^2}{2a^2}\ (2b^2 - a^2) =$

$= \frac 3 4 b^2 \left( 2 - \frac{b^2}{a^2} \right) = \frac 3 4 \cdot \frac{b^2c^2}{a^2}$

The symmedian point $K$ divides the symmedian $AP$ in the ratio (see Honsberger's Episodes, Chapter 7)

$\frac{AK}{KP} = \frac{b^2 + c^2}{a^2} = 2$

$AK = \frac 2 3 AP$

$AK^2 = \frac 4 9 AP^2 =  \frac 1 3 \cdot \frac{b^2c^2}{a^2}$

$\left( \frac{AK}{AG} \right)^2 = \frac{b^2c^2}{a^4}$

Now we have to show that square of the ratio $\frac{AO}{AH}$ is the same. Combining the usual formula and Heron's formula for the area $\triangle$ of the triangle $\triangle ABC$, the circumradius $R$ can be calculated in terms of the sides $A, b, c$ (see the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28146]Four points in xyz space[/url]).

$\triangle^2 = s(s - a)(s - b)(s - c) = \left( \frac{abc}{4R} \right)^2$

$[(2bc)^2 - (b^2 + c^2 - a^2)^2] R^2 = a^2b^2c^2$

$(4b^2c^2 - a^4) R^2 = a^2b^2c^2$

$R^2 = \frac{a^2b^2c^2}{4b^2c^2 - a^4}$

$\left( \frac{AO}{AH} \right)^2 = \frac{R^2}{4R^2 - a^2} = \frac{a^2b^2c^2}{4a^2b^2c^2 - (4b^2c^2 - a^4) a^2}= \frac{b^2c^2}{a^4}$

Consequently, $\frac{AO}{AH} = \frac{AK}{AG}$. Since the angles $\measuredangle OAK = \measuredangle HAG$ are equal, the triangles $\triangle AGH \sim \triangle AKO$ are similar by SAS. As a result, the angles $\measuredangle AGH = \measuredangle AKO$ are also equal and according to the previous reply, the angle  $\measuredangle AGH = 90^o$ is right.

[quote="Leon"]Let $G,H,K$ be the centroid, the orthocenter, the Lemoine point of $\triangle ABC$ respectively and denote $a=BC, b=CA, c=AB$.
If $b^2 + c^2 = 2a^2$ show that:
...

(2) $GK$ is parallel to $BC$
...[/quote]
The centroid $G$ divides the median $AA'$ (where $A'$ is the midpoint of the side $a = BC$) in the ratio $\frac{AG}{AA'} = \frac 2 3$. The symmedian point $K$ divides the symmedian $AP$ (where $P$ is the intersection of the symmedian from the vertex $A$ with the side $A = BC$) in the ratio (see Honsberger's Episodes and also the previous reply)

$\frac{AK}{AP} = \frac{b^2 + c^2}{a^2 + b^2 + c^2}$.

Under the condition $b^2 + c^2 = 2a^2$, this ratio becomes $\frac{AK}{AP} = \frac 2 3$. Hence, the triangles $\triangle AGK \sim \triangle AA'P$ with the common angle $\measuredangle GAK \equiv \measuredangle A'AP$ are similar, which implies that the lines $GK \parallel A'P \equiv BC$ are parallel.

[quote="Leon"]...
(3) $\ \ AK^2  = BK \cdot CK$
...[/quote]

Let $P, Q, R$ be the intersections of the symmedians $AK, BK, CK$ with the triangle sides $a = BC, b = CA, c = AB$. The segment lengths $PC, QA, RB$ can be calculated in terms of the triangle sides $a, b, c$ from the fact that the points $P, Q, R$ divide the sides $a, b, c$ in the ratios $\frac{PC}{PB} = \frac{b^2}{c^2},\ \frac{QA}{QC} = \frac{c^2}{a^2},\ \frac{RB}{RA} = \frac{a^2}{b^2}$:

${PC} = \frac{b^2}{b^2 + c^2}\ a$

${QA} = \frac{c^2}{c^2 + a^2}\ b$

${RB} = \frac{a^2}{a^2 + b^2}\ c$

The segment lengths $AP, BQ, CR$ of the symmedians from the vertices $A, B, C$ can be calculated in terms of the triangle sides $a, b, c$ using the cosine theorem for the triangles $\triangle ABC, \triangle APC, \triangle BQA, \triangle CRB$, just like the segment length $AP$ was calculated in the 2nd reply:

$AP^2 = b^2 + PC^2 - 2b \cdot PC \cos{\widehat C} = b^2 + \frac{a^2b^4}{(b^2 + c^2)^2} - 2ab\ \frac{b^2}{b^2 + c^2} \cdot \frac{a^2 + b^2 - c^2}{2ab} =$

$= b^2 + \frac{a^2b^4}{(b^2 + c^2)^2} - \frac{b^2(a^2 + b^2 - c^2)}{b^2 + c^2} = \frac{b^2c^2(2b^2 + 2c^2 - a^2)}{(b^2 + c^2)^2}$

$BQ^2 = \frac{c^2a^2(2c^2 + 2a^2 - b^2)}{(c^2 + a^2)^2}$

$CR^2 = \frac{a^2b^2(2a^2 + 2b^2 - c^2)}{(a^2 + b^2)^2}$

The segment lengths $AK, BK, CK$ of the symmedians from the vertices $A, B, C$ to the symmedian point $K$ can be calculated in terms of the triangle sides $a, b, c$ from the fact that the symmedian point $K$ divides the symmedians $AP, BQ, CR$ in the ratios $\frac{AK}{KP} = \frac{b^2 + c^2}{a^2},\ \frac{BK}{KQ} = \frac{c^2 + a^2}{b^2},\ \frac{CK}{KR} = \frac{a^2 + b^2}{c^2}$:

$AK^2 = \left(\frac{b^2 + c^2}{a^2 + b^2 + c^2}\right)^2 AP^2 = \frac{b^2c^2(2b^2 + 2c^2 - a^2)}{(a + b^2 + c^2)^2}$

$BK^2 = \left(\frac{c^2 + a^2}{a^2 + b^2 + c^2}\right)^2 BQ^2 = \frac{c^2a^2(2c^2 + 2a^2 - b^2)}{(a + b^2 + c^2)^2}$

$CK^2 = \left(\frac{a^2 + b^2}{a^2 + b^2 + c^2}\right)^2 CR^2 = \frac{a^2b^2(2a^2 + 2b^2 - c^2)}{(a + b^2 + c^2)^2}$

$\frac{AK^4}{BK^2 \cdot CK^2} = \frac{b^2c^2(2b^2 + 2c^2 - a^2)^2}{a^4(2c^2 + 2a^2 - b^2)(2a^2 + 2b^2 - c^2)}$

Under the condition $b^2 + c^2 = 2a^2$, this simplifies to

$\frac{AK^4}{BK^2 \cdot CK^2} = \frac{b^2c^2 \cdot (3a^2)^2}{a^4 \cdot 3c^2 \cdot 3b^2} = 1$

Taking the square root of the last equation,

$AK^2 = BK \cdot CK$

[quote="Leon"]...
(4) $\ \ BH^2  + CH^2  = 2AH^2 $
...[/quote]

The segment lengths $AH, BH, CH$ of the altitudes from the vertices $A, B, C$ to the orthocenter $H$ are equal to $AH = R \cos{\widehat A}$, $BH = R \cos{\widehat B}$, $CH = R \cos{\widehat C}$.

For example, let $D$ be the intersection of the line $BO$ with the circumcircle $(O)$ Since the angles $\measuredangle BAD = \measuredangle BCD = 90^o$ are both right, the lines $AH \parallel CD$, both perpendicular to $BC$, are parallel and the lines $CH \parallel DA$, both perpendicular to $AB$, are also parallel. Hence, the quadrilateral $AHCD$ is a parellogram and $AH = CD, CH = DA$. Since the angles $\measuredangle BDA = \measuredangle C$, spanning the same arc $AB$ of the circumcircle $(O)$ are equal, $CH = DA = 2R \cos{\widehat C}$. Since the angles $\measuredangle CDB = \measuredangle A$, spanning the same arc $BC$ of the circumcircle $(O)$, are also equal, $AH = CD = 2R \cos{\widehat A}$

$BH^2 + CH^2 = 4R^2 (\cos^2{\widehat B} + \cos^2{\widehat C}) =  4R^2 \left[2 - (\sin^2{\widehat B} + \sin^2{\widehat C})\right] =$

$= 8R^2 - (b^2 + c^2) = 8R^2 - 2a^2 = 8R^2 (1 - \sin^2{\widehat A}) = 8R^2 \cos^2{\widehat A} = 2 AH^2$

[quote="pestich"]If Leon put 6)  at the top ...
[/quote]

This proposition is patently false. Proof:

[quote="Leon"]Let $G,H,K$ be the centroid, the orthocenter, the Lemoine point of $\triangle ABC$ respectively and denote $a=BC, b=CA, c=AB$.
If $b^2 + c^2 = 2a^2$ show that:

(1) $OK$ is perpendicular to $AK$

(2) $GK$ is parallel to $BC$

(3) $\ \ AK^2  = BK \cdot CK$

(4) $\ \ BH^2  + CH^2  = 2AH^2 $

(5) the pedal triangle of $G$  is isosceles

(6) the locus of $A$, when $B,C$ are fixed, is a circle[/quote]

I am not sure if you are aware of this, lieutenant, but a false proposition implies anything.

[quote="Leon"]...
(5) the pedal triangle of $G$  is isosceles
...[/quote]
Let $A', B', C'$ be the midpoints of the sides $a = BC, b = CA, c = AB$. Using the parallelogram law and the given relation $b^2 + c^2 = 2a^2$, the median lengths are:

$m_a = AA' =\frac 1 2 \sqrt{2b^2 + 2c^2 - a^2} = a \frac{\sqrt 3}{2}$

$m_b = BB' =\frac 1 2 \sqrt{2c^2 + 2a^2 - b^2} = c \frac{\sqrt 3}{2}$

$m_c = CC' =\frac 1 2 \sqrt{2a^2 + 2b^2 - c^2} = b \frac{\sqrt 3}{2}$

Let $\alpha_b = \measuredangle A'AB$, $\alpha_c = \measuredangle A'AC$, $\beta_c = \measuredangle B'BC$, $\beta_a = \measuredangle B'BA$,  $\gamma_a = \measuredangle C'CA$, $\gamma_b = \measuredangle C'CB$ be the angles into which the medians $m_a = AA', m_b = BB', m_c = CC'$ split the triangle angles $\alpha =  \measuredangle CAB$, $\beta = \measuredangle ABC$, $\gamma = \measuredangle BCA$. Using the cosine theorem for the triangles $\triangle A'AB, \triangle A'AC, \triangle B'BC, \triangle B'BA, \triangle C'CA, \triangle C'CB$, cosines of these these angles are given by

$\cos \alpha_b = \frac{c^2 + m_a^2 - \frac 1 4 a^2}{2cm_a} = \frac{2c^2 + a^2}{2ac \sqrt 3}$

$\cos \alpha_c = \frac{b^2 + m_a^2 - \frac 1 4 a^2}{2bm_a} = \frac{2b^2 + a^2}{2ab \sqrt 3}$

$\cos \beta_c = \frac{a^2 + m_b^2 - \frac 1 4 b^2}{2am_b} = \frac{2c^2 + a^2}{2ac \sqrt 3}$

$\cos \beta_a = \frac{c^2 + m_b^2 - \frac 1 4 b^2}{2cm_b} = \frac{4c^2 - a^2}{2c^2 \sqrt 3}$

$\cos \gamma_a = \frac{b^2 + m_c^2 - \frac 1 4 c^2}{2bm_c} = \frac{4b^2 - a^2}{2b^2 \sqrt 3}$

$\cos \gamma_b = \frac{a^2 + m_c^2 - \frac 1 4 c^2}{2am_c} = \frac{2b^2 + a^2}{2ab \sqrt 3}$

From these formulas, it is clear that $\alpha_b = \beta_c$ and $\alpha_c = \gamma_b$, but it does not work for $\beta_a \neq \gamma_a$ (unless $b = c$).

Let $P, Q, R$ be the vertices of the pedal triangle with respect to the centroid $G$ on the sides $a = BC, b = CA, c = AB$. The quadrilateral $ARGQ$ is cyclic, because its opposite angles at the vertices $R, Q$ are both right. The angles $\alpha_b = \measuredangle RAG = \measuredangle RQG$ are equal, because they span the same arc $RG$ of the quadrilateral circumcircle. The angles $\alpha_c = \measuredangle QAG = \measuredangle QRG$ are also equal, because they span the same arc $QG$ of the quadrilateral circumcircle. Similarly, from the cyclic quadrilaterals $BPGR$ and $CQGP$, we find that $\beta_c = \measuredangle PBG = \measuredangle PRG$ $\beta_a = \measuredangle RBG = \measuredangle RPG$ and $\gamma_a = \measuredangle QCG = \measuredangle QPG$ $\gamma_b = \measuredangle PCG = \measuredangle PQG$. The internal angles of the pedal triangle $\triangle PQR$ are equal to

$\measuredangle RPQ = \measuredangle RPG + \measuredangle QPG = \beta_a + \gamma_a$

$\measuredangle PQR = \measuredangle PQG + \measuredangle RQG = \gamma_b + \alpha_b$

$\measuredangle QRP = \measuredangle QRG + \measuredangle PRG = \alpha_c + \beta_c$

As a result,

$\measuredangle PQR = \gamma_b + \alpha_b = \alpha_c + \beta_c = \measuredangle QRP$

and the pedal triangle $\triangle PQR$ is isosceles with $PQ = PR$.

[quote="Leon"]...
(6) the locus of $A$, when $B,C$ are fixed, is a circle
...[/quote]
The last proposition is trivial. Under the condition $b^2 + c^2 = 2a^2$, the median from the vertex $A$ to the side $a = BC$ is given by

$m_a = AA' = \frac 1 2 \sqrt{2b^2 + 2c^2 - a^2} = a \frac{\sqrt 3}{2}$

If the points $B, C$ are fixed, the length and the midpoint $A'$ of the side $a = BC$ are also fixed, which gives the length of the median $m_a = A'A$. Hence, the locus of possible vertices $A$ is a circle $(A')$ centered at the midpoint $A'$ of the side $a = BC$ and with radius $m_a = a \frac{\sqrt 3}{2}$.


invariant
algorithm
graph theory
combinatorics unsolved
combinatorics

1.In the Parliament each member has at most three enemies. Prove that the house can be separated into two houses, so that each member has at most one enemy in his own house.

2. $ 2n$ ambassadors are invited to a banquet. Every ambassador has at most $ n\minus{}1$ enemies. Prove that the ambassadors can be seated around a round table, so that nobody sits next to enemy.

Can someone solve it using 'invariant theorey' - not graphs, please? Unfortunately I am not good at graph theory and I think that I will not understand such a solution....

Thanks...

[quote="Invariant"]1.In the Parliament each member has at most three enemies. Prove that the house can be separated into two houses, so that each member has at most one enemy in his own house[/quote]

this one is a classic problem,I will post a solution which appeared in a book:

[b]solution.[/b] initially, we separate the members in any way into the two houses. Let $ H$ be the total sum of all the enemies each member has in his own house. Now suppose $ A$ has at leat two enemies in his own house. Then he has at most one enemy in the other house. If $ A$ switches houses, the number $ H$ will decrease. This decrease cannot go on forever. At some time, $ H$ reaches its absolute minimum. Then we have reached the required distribution.

[quote="Invariant"]
2. $ 2n$ ambassadors are invited to a banquet. Every ambassador has at most $ n \minus{} 1$ enemies. Prove that the ambassadors can be seated around a round table, so that nobody sits next to enemy.
[/quote]
[b]solution.[/b]First, we seat the ambassadors in any way. Let $ H$ be the number of neighboring hostile couples. We must find an algorithm which reduces this number whenever $ H>0$.let $ (A,B)$ be a hostile couple with $ B$ sitting to the right of $ A$.now if there were couples $ A',B'$ with $ B'$ sitting to the right of $ A'$ and $ (A,A')$ and $ (B,B')$ were friendly couples,then if we reverse the arc $ BA'$ then $ H$ will be reduced...
hus it remains to be shown that such a couple always exists.We start in $ A$ and go around the table counterclockwise.We will encounter at leat $ n$ friends of $ A$.To their right,there are at least $ n$ seats.They cannot all be occupied by enemies of $ B$ since $ B$ has at most $ n\minus{}1$ enemies.Thus,there is a friend $ A'$ of $ A$ with right neighbor $ B'$,a friend of $ B$.

this one is a famous theory in graph:
let $ G$ be a linear graph with $ n$ vertices.Then $ G$ has a Hamiltonian path if the sum of the degrees of any two vertices is equal to or larger than $ n$...
here is a link for this theorem:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159829[/url]


calculus
integration
trigonometry
calculus computations

Evaluate $\int_{0}^{\pi/2}\int_{0}^{\pi/2-\theta}\cos^{4}(\theta+\phi) \sec^{2}\phi\, d\theta\, d\phi$.

I was surprised to see no attempts!!!. Here is the hint. Change the order of integration and the rest follows trivially.


This thread is to play more than one word game. So for the first week we play one game, the next week a different game, etc.

This is a free for all topic so you won't have to meet a deadline to join.

Now the first game is a rhyming game. You have to think of a word that rhymes with the current word. It's not that easy, because when you run out of easy words, you'll have to think of harder words. So if you're on the at rhyming position, you may say "democrat" "aristocrat" or even "buerocrat" you can only post one word per post(you can post comments) and your word must rhyme with the last word. You also can't use the same word twice.

here's a few very important rules: NO PHRASES! NO made up words.

Now, to make this game even harder than it seems, I'll extend this game for 2 weeks. I'll check back on the 18th.

Here is the first word:  cat

P.S. this is not pointless.

If this is to go on, then the first rule is that there will [u]not[/u] be 1 word per post. If this game is going to be like that, where people post 1 word and then something like "Just making the message bigger", I'll lock the topic.

$1$ cat
$2$ doormat

@ Klebian:
Stop being a gruff-and-grum and join in. We should encourage mbabbitt2003 in creative issues.

bat

let's get to the tricky words!!!

you: like, what tricky words? :P

1.at
2.matt
3.hat
4.pat
5.fat
6.rat
7.sat
Now for words 2+ syllables?


algebra
polynomial
quadratics
algebra unsolved

Let $ a,b\in R$ be such that the quadratic polynomial $ f(x)\equal{}x^2\plus{}ax\plus{}b$ has no nonnegative real roots. Prove that there exist two polynomials $ g,h$ whose coefficients are nonnegative real numbers such that $ f(x)\equal{}\frac{g(x)}{h(x)}$ for all real numbers $ x$.

$ f(x)\equal{}x^2\plus{}ax\plus{}b\equal{}(x\minus{}c)(x\minus{}d)$
then $ b\equal{}cd$ and $ a\equal{}\minus{}(c\plus{}d)$
with $ c,d <0$ then $ b>0$ and $ a>0$
$ g(x)\equal{}f(x)$ with nonnegative coefficients
$ h(x)\equal{}1$ works

Is it ok?

[quote="mszew"]$ f(x) \equal{} x^2 \plus{} ax \plus{} b \equal{} (x \minus{} c)(x \minus{} d)$

[/quote]

What??

I said $ f(x)$ has nonnegative real roots!

Posted before.
Chose $ h(n)\equal{}(x\plus{}1)^n$
For $ n$ large enough we will have all coefficient of $ (x^2\plus{}ax\plus{}b) (x\plus{}1)^n$ are nongative.


$ABC$ is a triangle whose angles are all smaller than $120$.
Equilateral triangles $AFB, AEC, BDC$ on the sides of and exterior to triangle $ABC$. It can be easily verified that $AD, EB$ and $FC$ are concurrent at a point $S$.
Prove that $SD+SE+SF=2(SA+SB+SC)$

[hide]Very transparent problem, but quite nice.

$AFBS$ is cyclic. Ptolemy: $\left|AS\right|+\left|BS\right| = \left|SF\right|$
Sum up, and done.[/hide]


In the last couple of years we haven't had much participation from SC in the USAMTS. I would like to see all the students on this forum and all the students from the All-State Math team participate this year. I've said that the USAMTS was the most influential math contest that I participated in during my years in high school in that it really taught me how to apply all my tools towards a problem, and to be persistent and try new ideas. It's the contest that in my opinion is the closest to what "real" math is like. Even if you can only solve 2 or 3 of the problems it's still worth sending your solutions in.

http://www.usamts.org/

Best of luck,
Marcus

I'll probably submit solutions this year.  Usually I forget or get distracted, but I think I'll do it this time.


I have this rubiks cube where i accidentally did a move that i was not trying to do and ended up with this. I am wondernig if there is an easier way to solve this. Anyway. I have finnished the first two layers (yellow was the starting color). and on the top layer, all of the mini cubes are roted the right direction: all of the white spaces facing up. But, three of the pieces with two colors are in the wrong positions. I need to rotate these clockwise without messing up any other colors. I think this can be done. Please tell me if there is a trick to this or i'll have to solve it  the natural way.

L2 U B F' L2 B' F U L2

I think that should do it.

are u sure that that is how it is oriented? because it seems as if there is a missing blue piece and one extra red

Thats like impossible because there are two Red-Whites.  I think your cube is misprinted.
:)

oops... the red one on the green side is supposed to be blue.

I'm not sure about tjhance's algorithm, but I'm sure Hamster1800's works (although it is faster if you hold the cube differently).

I tried tjhance's and it works. It's a nice 2-generator. Mine is how I learned it from the Petrus method. How do you hold your cube when you do it?

I hold the cube with the 3 out of place edges on top, the middle one facing me. Maybe I just don't know the finger tricks for your way that would make it faster.

I wanted to test tjhance's algorithm, but my cube's stickers all fell off, and I can't seem to find my old one, which is very slow anyway.

I usually hold the cube with the three out-of-place edges in the back, left, and right positions, so that I am moving the up and right faces

With white on top, try this:
R U Ri U R U U Ri followed by a U
Thats step six in the process a Seven-Step guide
R=Right Face
L=Left Face
B=Back Face
D=Down Face
F=Front Face
U=Upper Face
?i=Inverse ( Counter Clockwise )
Everything w/o i is turned clockwise
that might solve it.
:?: :?:

minicon's algorithm rotates corners. If you look at the diagram you'll see all the corners are already oriented correctly.

[quote="moogra"][quote="tjhance"]Also B' U B' U' B' U' B' U B U B2 should work[/quote]
so does
(y)
R' U R' U' R' U' R' U R U R2
I find that way easier to do than Back moves[/quote]

Yes, I agree; I only used back moves when posting the algorithm because of how the cube was positioned in ashpotter's diagram.

yeah that's why  the (y) was there meaning rotate it clockwise

just hold the cube with the orange away from you and the white on top.
Then perform:
F2
U
L
R'
F2
L'
R
U
F2

Then it should be solved!!! :)


linear algebra
matrix
linear algebra unsolved

Find all matrices $A,B\in M_n(C)$ such that for any nilpotent matrix $X$, the matrix $AX+XB$ is also nilpotent.

For all $i\ne j$, if $X$ is the matrix with $1$ on $(i,j)$ and $0$ everywhere else, then $AX+XB$ has $A$'s $i$'th column as its $j$'th column and $B$'s $j$'th line as its $i$'th line. All its powers are of the same form, eventually having non-zero entries only on the $j$'th column and the $i$'th line, and the $(i,i),(j,j)$ entries of $(AX+XB)^k$ are the $k$'th powers of the $(i,i),(j,j)$ entries in $AX+XB$, which means that these must be zero. These two entries, however, are the $(j,i)$ entry of $B$ and the $(i,j)$ entry of $A$ respectively, so this shows that $A,B$ must be diagonal.

In order to prove that $A$ must actually be scalar (and the same will be true for $B$), we can observe that for every invertible $T,\ TAT^{-1}$ and $TBT^{-1}$ have the same property as $A,B$, so $TAT^{-1}$ must be diagonal for all invertible $T$, from which it easily follows that $A$ is scalar.


Let S = {1, 2, 3, ..., 1982}.
Determine the maximum number of elements of a set A such that :
(1) A is a subset of S ;
(2) There do not exist numbers x, y, z in A such that xy = z.

Good problem!
If you know answer then you know solution.

Arne, where did you read problems from sovjet olympiads?

Do you mean you know the answer Myth ?

You can find the problems at http://www.kalva.demon.co.uk .

Of cause I know answer!
I have great red book "Problems of sovjet mathematical olympiads" with solutions. I learned this book very well when I was preparing to IMO and to national olympiads.

So what's the solution myth ? I've no idea ...

To Valentin. Yes, I took part in IMO97

To Arne. You must exlude 43 numbers: 2,3,...,44. Now you will find solution, I think.

Ignore this message, I didn't see answer of Myth.

Now I realize ... I didn't expect it to be so simple ! Thanks !


Does there exist a convex pentagon, all of whose vertices are lattice points in the plane, with no lattice point in the interior?

No,there is not.suppose $A_1..A_5$is the smallest one($A_i(x_iy_i)$),then$x_i\equiv x_j,y_i\equiv y_j(mod 2)$,then the midpoint X is also lattic points.then X be in the interior or there is a smaller pentagon
( I will rewrite this solution tommorow :oops: )

[quote="Hawk Tiger"]( I will rewrite this solution tommorow :oops: )[/quote]reminder... ;)

If "MM" is what I think it is ([i]Mathematics Magazine[/i]) then Peter's reference (MM, Problem 1409, Gerald A. Heur) is incorrect. Moreover, I think the real problem here is the following:

[color=darkblue][b]Theorem 1[/b][/color] (Russian Mathematical Olympiad 2000). [color=blue]If $ ABCDE$ is a convex pentagon having the vertices on lattice points, then the "smaller" pentagon determined by the diagonals of $ ABCDE$ contains (in its interior or on its boundary) a lattice point. [/color]

Although Kalva doesn't give a solution, I suspect the official solution is by combining a minimality argument with Pick's theorem several times (as in Titu Andreescu's 2000-2001 book).

However, viewing Hawk Tiger's nice observation I can continue using his idea. Let me first rewrite in a more understandable way the things he said:

[i]Proof of Theorem 1[/i]. Since a lattice point can have only one of the four formes (odd, odd), (even, even), (odd, even), (even odd), two of the vertices of $ ABCDE$ will have the same form, and therefore the midpoint $ M$ of the segment determined by these two points is a lattice point as well. Therefore we have two cases:

For sake of clarity, in the following I will denote by $ S(XYZUV)$ the "small" pentagon determined by the diagonals of $ XYZUV$:

1. Suppose $ M$ is in the interior of a diagonal of the pentagon $ ABCDE$; assume WLOG that $ M$ is the midpoint of $ AC$. If it lies inside (on the boundary) of $ S(ABCDE)$ we are done. If it doesn't, one of the two pentagons $ ABMDE$ and $ MBCDE$ is convex. Let this one be $ MBCDE$. This convex pentagon is a lattice pentagon contained in the initial $ ABCDE$. We observe that $ S(MBCDE)$ is included in $ S(ABCDE)$. Thus, by inducting on the number of lattice points contained in the initial convex pentagon, $ S(MBCDE)$ will contain a lattice point, and then so will $ S(ABCDE)$. We are done.

2. Suppose $ M$ is in the interior of an edge of the pentagon $ ABCDE$; assume WLOG that $ M$ is the midpoint of $ AB$. The pentagon $ MBCDE$ is convex. Now, we do not have anymore that $ S(MBCDE)$ is included in $ S(ABCDE)$, but we notice that $ S(MBCDE)$ is included in the reunion of $ S(ABCDE)$ with the triangle, say $ PQR$, bounded by the three lines $ BD$, $ BE$, $ AC$. Thus, by induction, $ S(MBCDE)$ contains a lattice point $ M'$. If $ M'$ is in $ S(ABCDE)$ we are done. Suppose it isn't. Thus, it is in $ PQR$, and the pentagon $ AM'CDE$ is convex, with $ S(AM'CDE)$ incldued in $ S(ABCDE)$. By induction, once again, we conclude that $ S(AM'CDE)$ contains a lattice point, and therefore (according to the inclusion argument) $ S(ABCDE)$ does as well. This proves Theorem 1.

EDIT: Here is an older topic about Theorem 1 (which Darij pointed out to me), but there's no complete solution: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5743[/url] (iura's remark is a part of the official solution).


probability
function
calculus
calculus computations

[quote="Beginner"]Verify the identity
\[\sum_{x=0}^{\infty}\sum_{y=1}^{\infty}\sum_{m+n=2}^{x+y+1}\text{Pr}(X=m) \cdot \text{Pr}(Y=n)= \mu_{x}+\mu_{y}-1 \]
where $\mu_{x}= \sum_{i=1}^{\infty}\text{Pr}(X \geq i)$

 :lol:  :lol:[/quote]

this forum include people different nationality and in other country is other defination  some mathematical formul.
what is it $Pr(X=m)$. is it true: x,y,n,m is natural number?

The original poster presumably meant to include the following pieces of information:
$X$ and $Y$ are random variables on the nonnegative integers.  $\textrm{Pr}(A)$ is the probability of event $A$.

And there's some really bad notation going on there, since $\mu_{x}$ is a function of the random variable $X$, not the dummy variable $x$.

Sorry guys...
I've realize that a few minutes after posting, but having trouble to get here...

Yes $\mu_{x}$ is a Mean of random variable $X$... and $\text{Pr}$ is a probabilistic function.


calculus
integration
function
real analysis
real analysis solved

does the integral $\int_{0}^{\infty} x.sin(x^3-bx) dx$ converge where b is an arbitrary real ?

Yes, it converges for all $b.$ First, choose a positive $a$ such that $a^2>b.$ The integral from $0$ to $a$ is the integral of a continuous function on a closed bounded interval, hence convergent. We need only consider the integral from $a$ to $\infty.$ We write that as

$\int_a^{\infty}\frac{x}{3x^2-b}(3x^2-b)\sin(x^3-bx)dx = $

$=-\frac{x}{3x^2-b}\cos(x^3-bx)|_a^{\infty}+ \int_a^{\infty}\frac{3x^2-6x-b}{(3x^2-b)^2}\cos(x^3-bx)dx$

The integral then converges by comparison to the integral of $\frac1{x^2}.$


[quote="davidli"]What is the maximum number of points of intersection of two congruent squares that do not share a common line segment?[/quote]
[hide="unsure"]2 per a side makes 8[/hide]

[hide="idk..."]i think it's 8 but somthing tells me that it's wrong...[/hide][/list]

i personally think it's 4. draw a picture with 8 then.

jorian

[quote="Vaele"]Ummm...

If you rotated the square, wouldn't it touch to corners?
Basic Geometry:

There is only one line through 2 points...you guys are saying theres 2...

Or am I reading it wrong?[/quote]
Yes, you are reading it wrong.  They're not saying that there's more than one line through two points, they say there's more than one point on any line.  The squares are congruent, so it would look like an 8-pointed star.

[hide="Answer"]take a square, then rotate it. You get 8 intersection points[/hide]

[quote="davidli"]What is the maximum number of points of intersection of two congruent squares that do not share a common line segment?[/quote]
where does the problem say anything about rotating the square?

[quote="#H34N1"][hide="Solution"][img]http://i96.photobucket.com/albums/l167/adasarathy/SquareIntersection.gif[/img]
Count the number of intersections (in the black).[/hide][/quote]

[hide="mistake in picture"]There are only 7 black ones.... and 8 intersection points.[/hide]

[quote="davidli"]Yes, it's 8. Draw a square. Then rotate that square. You can get 2 intersections each side.[/quote]

wow. I had 8 in the picture in my head, but it just seemed like 4 for some reason. thanks.

jorian

[quote="ckck"][quote="InsanelyBored"][quote="davidli"]What is the maximum number of points of intersection of two congruent squares that do not share a common line segment?[/quote]

[hide="not sure seems too easy"] 2 per side, 2*4=8[/hide][/quote]
That solution looks a lot familiar.[/quote]
:huh:


Let W be vector subspace of 2*2 matrices over R such that W is spanned by {AB-BA such that A,B are 2*2 matrices over R},then:
1)prove that dim(W)>=3 by verifyind E12,E21,E11-E22 all belong to W.
2)Considering the trace function tr:M2(R)-->R ,prove that dim(W)=3.
3)Let f:M2(R)-->R be a linear map such that f(AB)=f(BA),then prove that f=c.tr,for some c in R.Hence prove that f=tr if and only if f(Id)=2.

any help or suggestions rare appreciated..thank you

its a nice problem and (1),(2) are really helpfull..

as a hint: show that $ E_{ij}E_{kl}\equal{}E_{il}d_{jk}$, where $ d$ is the delta of Kronecker.

1) We have

$ \begin{bmatrix} 1 & 1 \\
0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\
0 & 0 \end{bmatrix} \minus{} \begin{bmatrix} 0 & 1 \\
0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\
0 & 0 \end{bmatrix} \equal{} \begin{bmatrix} 0 & 1 \\
0 & 0 \end{bmatrix}$,

$ \begin{bmatrix} 0 & 0 \\
1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 \\
1 & 0 \end{bmatrix} \minus{} \begin{bmatrix} 0 & 0 \\
1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\
1 & 1 \end{bmatrix} \equal{} \begin{bmatrix} 0 & 0 \\
1 & 0 \end{bmatrix}$,

$ \begin{bmatrix} 0 & 1 \\
0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\
1 & 0 \end{bmatrix} \minus{} \begin{bmatrix} 0 & 1 \\
1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\
0 & 0 \end{bmatrix} \equal{} \begin{bmatrix} 1 & 0 \\
0 & \minus{} 1 \end{bmatrix}$,

and since these matrices are linearly independent, $ \dim W \geq 3$.

2) If $ \dim W \equal{} 4$, then we would have $ I_2 \in W$: but then there would have to be matrices A,B such that $ AB \minus{} BA \equal{} I_2$, which is impossible, since $ tr(AB\minus{}BA) \equal{} 0$ and $ tr(I_2) \equal{} 2$. Therefore, $ \dim W \equal{} 3$.

3) Start by observing that $ Ker(f) \equal{} W \equal{} \left \{\begin{bmatrix} a & b \\
c & \minus{} a \end{bmatrix}: a,b,c \in \mathbb{R} \right \}$. Therefore,

$ f \left(\begin{bmatrix} a & b \\
c & d \end{bmatrix} \right) \equal{} f \left(\begin{bmatrix} a & 0 \\
0 & d \end{bmatrix}\right) \plus{} f \left(\begin{bmatrix} 0 & b \\
c & 0 \end{bmatrix}\right) \equal{} f \left(\begin{bmatrix} a & 0 \\
0 & d \end{bmatrix}\right) \plus{} 0 \equal{} f \left(\begin{bmatrix} a & 0 \\
0 & d \end{bmatrix}\right)$ .

Now, the third matrix equality in my previous post shows that $ f \left(\begin{bmatrix} 1 & 0 \\
0 & 0 \end{bmatrix}\right) \equal{} f \left(\begin{bmatrix} 0 & 0 \\
0 & 1 \end{bmatrix}\right) \equal{} k \in \mathbb{R}$, which means that

$ f \left(\begin{bmatrix} a & b \\
c & d \end{bmatrix} \right) \equal{} a f \left(\begin{bmatrix} 1 & 0 \\
0 & 0 \end{bmatrix} \right) \plus{} d f \left(\begin{bmatrix} 0 & 0 \\
0 & 1 \end{bmatrix} \right) \equal{} (a \plus{} d) k \equal{} k tr$.

The last part of the problem is now immediate.


inequalities
trigonometry
function
inequalities proposed

[quote="toan hoc"]Prove that in any triangle ABC, the following inequality holds:
$ \frac {{\cot A.\cot B.\cot C}}{{\sin A.\sin B.\sin C}} \le \left( {\frac {2}{3}} \right)^3$[/quote]
Take a look at the function $ f(x) \equal{} ln \left ( \frac{\cos x }{\sin^2 x} \right )$.
Then $ f''(x) \equal{} \frac{3 \cos^2 x \\ \minus{} 1}{\sin^2 x \cdot \cos^2 x}$. From here we can easily prove with Karamata that in order to get the maximal value of the $ LHS$ two of $ A,B,C$ has to be the same. Let $ A\equal{}B$. Then $ \frac {{\cot A.\cot B.\cot C}}{{\sin A.\sin B.\sin C}} \equal{} \frac {{ \minus{} \cos^2 A \cdot \cos 2A}}{{\sin^4 A \cdot \sin^2 2A}}$. Use $ \sin^2 2A \equal{} 4 \sin^2 A \cdot \cos^2 A$, $ \cos 2A \equal{} 1 \minus{} 2 \sin^2 A$, $ \cos^2 A \equal{} 1 \minus{} \sin^2 A$ and substitue $ \sin A \equal{} x$. Then we have:
$ \frac {{ \minus{} \cos^2 A \cdot \cos 2A}}{{\sin^4 A \cdot \sin^2 2A}} \equal{} \frac{2x^2\minus{}1}{4x^6} \le \frac{8}{27} \iff$
$ f(x) \equal{} 32x^6 \minus{} 54x^2 \plus{} 27 \ge 0$. Let $ y \equal{} x^2$:
$ 32y^3 \minus{} 54y \plus{} 27 \equal{} (4y\minus{}3)^2(2y\plus{}3) \ge 0$
So equality iff $ y \equal{} \frac{3}{4} \iff x \equal{} \frac{\sqrt{3}}{2}$ (Since $ x \ge 0$). So equality iff $ A\equal{}B\equal{}C\equal{}\frac{\pi}{3}$. (Since $ A,B,C \le \pi$)


$ \sqrt[3]{x}\plus{}\sqrt[3]{2x\minus{}3}\equal{}\sqrt[3]{12(x\minus{}1)}$

How many real x values are there satisfing the above equality

[quote]elevamos al cubo
\[ \root 3 \of x  \plus{} \root 3 \of {2x \minus{} 3}  \equal{} \root 3 \of {12x \minus{} 12} /(\;)^3\]
\[ x \plus{} 3\left( {\root 3 \of x } \right)^2 \root 3 \of {2x \minus{} 3}  \plus{} 3\root 3 \of x \left( {\root 3 \of {2x \minus{} 3} } \right)^2  \plus{} 2x \minus{} 3 \equal{} 12x \minus{} 12\]
factorizamos
\[ \root 3 \of x \root 3 \of {2x \minus{} 3} \left( {\root 3 \of x  \plus{} \root 3 \of {2x \minus{} 3} } \right) \equal{} 3x \minus{} 3\]
pero como [quote]$ \root 3 \of x  \plus{} \root 3 \of {2x \minus{} 3}  \equal{} \root 3 \of {12x \minus{} 12}$[/quote] entonces:
\[ \root 3 \of x \root 3 \of {2x \minus{} 3} \root 3 \of {12x \minus{} 12}  \equal{} 3x \minus{} 3/(\;)^3\]
\[ 12x\left( {2x \minus{} 3} \right)\left( {x \minus{} 1} \right) \equal{} 27\left( {x \minus{} 1} \right)^3\]

\[ \left( {x \minus{} 1} \right)\left( {x^2  \minus{} 6x \plus{} 9} \right) \equal{} \left( {x \minus{} 1} \right)\left( {x \minus{} 3} \right)\left( {x \minus{} 3} \right) \equal{} 0 \Rightarrow x \equal{} 1 \vee x \equal{} 3\]
[/quote]

Substitute $ \sqrt[3]{x}\equal{}a$, $ \sqrt[3]{2x\minus{}3}\equal{}b$. Then it is equivalent with:

\[ a\plus{}b\equal{}\sqrt[3]{4(a^3\plus{}b^3)}\]

\[ \Leftrightarrow (a\minus{}b)^2(a\plus{}b)\equal{}0\]

$ \Rightarrow x\in \{1,3\}$


[quote="mathboy_cntHDVN"]sorry
$\sum\frac{a^{m}}{b^{n}}\geq\sum\frac{a^{m-k}}{b^{n-k}}$[/quote]

What is the property of $a,b,m,n,k$ ?

$a,b,c>0 and k<min(n;m)$
        $\frac{a^{m}}{b^{n}}+\frac{b^{m}}{c^{n}}+\frac{c^{m}}{a^{n}}\geq\sum\frac{a^{m-k}}{b^{n-k}}$
                 true or fales

thank , but can you post realy solution ?

Wait, isn't that clear enough? :) (By weighted AM-GM :wink:)

[quote="mathboy_cntHDVN"]we cannot use AM_GM  :lol:  :D[/quote]

Hmm... I didn't bother to check the solution by BoesFX, but the problem is definitely solvable by Weighted AM-GM, so what's wrong with using it? (There are three terms with real weights)

[quote="mathboy_cntHDVN"]can you prove this inequaliti ??????????????? :lol:[/quote]

It is already proven above

[quote="mathboy_cntHDVN"]k ;m;n is not negative  :maybe:[/quote]

Of course! If k is negative then the inequality flips. If m,n are negative then the inequality may or may not flip. If you want either of these then obviously the inequality is false.

oh ; thanks weiquan  :roll:
if for n number this problem is true or false

my friend call me that he used cauchy_svac  :)
but he don't call how to used ?  :lol:
what do you think ?


There are 2008 congruent circles on a plane such that no two are tangent to each other and each circle intersects at least three other circles. Let $ N$ be the total number of intersection points of these circles. Determine the smallest possible values of $ N$.

well i believe that the minimum should be 2008, it holds when we put four in a group and the minimum number of intersection points among 4 circles is 4.

But... it's quite a hard problem to tackle :(
(well i suppose it would not really be that hard coz CHKMO traditionally consist of moderate questions~~ probably trivial to most experienced mathlinkers~)

First by enumeration,it is easy to show that on each circl,there is
 at least three points.Now,if a point belongs to n circles ,then each
 of these circle get score$\frac{1}{n}$ from this point.Now, we will
 prove that the sum of score on each circle is no less than 1.
let $A_{1}$,$A_{2}$,.......$A_{t}$ are all on the same circle,and
 belong to at least two circles so the circle passes $A_{i}$ is no more
than t,so the score of this circle $\ge\frac{1}{t}*t$=1,
Hence the score of all the circle is at least 2008,so there are at least 2008 points


inequalities
function
ratio
inequalities proposed

Give $ a,b,c  \geq 0 $and [tex] min \lbrace {a,b,c} \rbrace \geq \frac14 max \lbrace {a,b,c} \rbrace [/tex].Prove that:
[tex]  \sum ab . \sum\frac {1}{(a+b)^2}  \geq \frac94 +  \sum \frac{(a-b)^2}{16(a+b)^2}  [/tex] :D
([tex] \sum [/tex] here is cycle sum)

No idea.I really want to have another solution .
Everybody,please try it. :)

Really no ideas.So I will have Master post his solution.
Shoot your goal,master... :D

what do you mean by $\sum$ cycle sum ? l don't see the necessity of these symbols. l thought that this symbol sigma was used only like that :
$\displaystyle \sum_n^{i=0}$ or $\displaystyle \sum_{i<j}$

l am really tired today : it's $\displaystyle \sum_{i=0}^n$

Well,If you want to use$ \displaystyle\sum_{i=1}^n$ then it can be writeln:
[tex]\displaystyle\sum_{i=1}^3 a_ia_{i+1}.\displaystyle\sum_{i=1}^3\frac{1}{(a_i+a_{i+1})^2}\geq \frac{9}{4} + \displaystyle\sum_{i=1}^3\frac{(a_i-a_{i+1})^2}{16(a_i+a_{i+1})^2}[/tex] where $a_4=a_1$
However,it is really tired for me to write down that one.So I use sum cycle:
[tex] \sum ab=ab+ac+bc=f(a,b)+f(b,c)+f(c,a) [/tex]where [tex]f(a,b)=ab;(a,b)->(b,c)->(c,a)[/tex] is a cycle. :)
Okay,I will write the full form again:
[tex](ab+ac+bc)(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2})\geq[/tex]
[tex]\frac94+\frac{(a-b)^2}{16(a+b)^2}+\frac{(b-c)^2}{16(b+c)^2}+\frac{(c-a)^2}{16(c+a)^2}[/tex]
Now,you can try it,right? :lol:

huh, that notation is completely new for me, and l had some problems with that.
l think l have a solution but before write it l will check it out

now i have the solution
but can you tell me why can you find 1/16
it is very nice and enough

Hi,I am very eager to see your both solutions. :D
About the number [tex] \frac{1}{16}[/tex],maybe you can find it too when you solve,it depends one the number[tex] \frac14[/tex] :)

I have waited for 3 months since I posted this one,but I still haven't no new solution yet :(

Dear arthur,I have been waiting for you day by day.
And I am so tired now...zz...zz... :cool:

No one want to wake me up.Help me,I don't want to sleep anymore time. :(
Where are you,arthur?
Trust me,it doesn't exist only a "buffalo" solution.

I think Arthur just dreamt he had a solution to this problem.

So, what about a solution?
The iranian one is not simple, this is definetly not so :D

We can write the inequality as follows
$\sum{\frac{a}{b+c}}\geq \frac{3}{2}+\frac{5}{16}\sum{\frac{(b-c)^2}{(b+c)^2}}$.
Since
$\sum{\frac{a}{b+c}}- \frac{3}{2}=\sum{\frac{(b-c)^2}{2(a+b)(a+c)}}$,
the inequality becomes
${\sum{A(b-c)^2}} \geq 0$,
where $A=8(b+c)-5a-5\frac{a^2+bc}{b+c}$,
Suppose $a\geq b\geq c$.  We get $B\geq 0$,  $C\geq 0$  and  $A+2B \geq 0$. This completes the proof.
To prove that $A+2B \geq 0$, we use the inequalities
$\frac{b^2+ac}{a+c}\leq b$  and   $24c\geq 6a$. :)

Yes,it is similar to my solution.Thank you,I have thought that I must sleep forever.
This proof can be done for this problem too:
[tex] min\lbrace{a,b,c}\rbrace \geq n max\lbrace{a,b,c}\rbrace [/tex] where [tex] n \in[0,1] [/tex].Then:
[tex] (ab+ac+bc)(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(a+c)^2}) \geq \frac94+\frac{n(a-b)^2}{4(a+b)^2}+ \frac{n(b-c)^2}{4(b+c)^2}+\frac{n(c-a)^2}{4(c+a)^2}[/tex]
The equality occurs when $a=b=c$ or $na=nb=c$
And of course,when n=0 it is Iran problem :)

yes my solution was completely wrong but l couldn't say it because my modem had been broken.

you're not  very kind  with me, Arne...

Sorry :) It's rather a coincidence, I didn't mean to be unkind to you Arthur.

Dear Vasc,could you please post again your full solution ?
I am really a careless person.You know,I thought that you wrote $ 2A+B \geq 0 $. :blush:
Everyone,enjoy the general one too. :D

Dear Vasc,

I also, like Anh Cuong, don't understand why $2A + B \geq0 $ is the necessary condition for the case $ A \leq0 $

I have also tried to verify $ 2A +B \geq0 $ writig

$16(b+c) -10a -10\frac{a^2+bc}{b+c} \geq 5b +5\frac{b^2+ac}{a+c}-8(c+a) $

$16(b+c) -10a -5b+8(c+a) \geq 10\frac{a^2+bc}{b+c} +5\frac{b^2+ac}{a+c}$

but $ \frac{b^2+ac}{a+c} \leq b $

so it is sufficient to prove

$ 11b +24c -2a \geq 5b +10\frac{a^2+bc}{b+c} $

but $ 24c \geq 6a $

so it is sufficient to prove

$4a + 6b \geq 10\frac{a^2+bc}{b+c} $

after some calculation becomes

$ 4ab +4ac +6b^2 \geq 10a^2 +4bc $

which is not true but probably I am in error.

So I am very grateful to you if you post your detailed  solution about  this LAST STEP

Thank you very much, Vasc.

Dear Anh Cuog,

I am very interested also in your solution that you said it is very similar to Vasc solution.

This is really a BEAUTIFUL ineq and I am very curious in all variants of it.

Thank you very much.

Well,in fact my solution is almost different from Vasc's.It is just the same a little.Okey I will post it after Vasc writes down his full solution :) .

Anh Cuong and Manlio,

I dont write that $2A+B\geq 0$, but $A+2B\geq 0$. This one is true, but it helps us nothing.
I was wrong in the final of the proof, which is really much more difficult.
I propose now the following way to prove that ${\sum{A(b-c)^2}} \geq 0$, where
$A=8(b+c)-5a+5\frac{a^2+bc}{b+c}$, ..
We may write
$A=8(b+c)-10a-5\frac{(a-b)(a-c)}{b+c}$, ..
and the inequality becomes
$\sum{[8(b+c)-10a](b-c)^2 \geq \frac{5(a-b)^2(b-c)^2(c-a)^2}{(a+b)(b+c)(c+a)}}$.
Now we consider $a\geq b\geq c$  and use the substitutions $b=c+y$ and $a=b+x=c+x+y$,
where $x\geq 0$  and $y\geq0$. Also, we have $3c\geq x+y$. We may replace everywhere $c$ with $(x+y)/3$. After some calculation we obtain (I hope this) a trivial inequality in $x$ and $y$.

why may you replace $c$ with $\frac{x+y}{3}$? e.g., when trying to prove the inequality for $a=b=c$, you would replace $c$ by 0, which desharpens the inequality...

Peter

Peter,
After the substitutions $b=c+y$ and $a=c+x+y$, we obtain the inequality
$f(c,x,y)\geq0$.
But $f$ is incresing relating to $c$. Thus
$f(c,x,y)\geq f(\frac{x+y}{3},x,y)$
and remains to show that $f(\frac{x+y}{3},x,y)\geq0$.
The condition $c\geq\frac{x+y}{3}$ follows from $4c\geq a$, according to
the problem statement.  :)

thanks,

that's the argument i thought of, but i don't immediately see why this function is increasing in c...could you please show?
btw if it is actually increasing in c, then we would have equality if a=b(since we have equality if a=b=c). but as anh cuong said that we have only equality if a=b=c or na=nb=c. so your function is not increasing in my mind...

Peter

i just recognize that it's complete nonsense i'm talking, sorry. just ignore me  ;)

Peter

Well,what about a complete proof.Everybody,try it :)

I have tried to follows Vasc topic but it becomes very complicated to finish so I

must again invoke Vasc help for a detailed solution.

Thank you very much.

Manlio,
We have
$f(c,x,y)=(3c+4x+8y)x^2+(3c-5x-y)y^2+(3c+4x-y)(x+y)^2-\frac{5x^2y^2(x+y)^2}{2(2c+x+2y)(2c+y)(2c+x+y)}$

It suffices to replace $3c$ with $x+y$ in the first three terms, and $2c$ with $0$ in the last term (at denominator). It is sufficient to show that
$10x^3+18x^2y+xy^2\geq\frac{5x^2y^2(x+y)^2}{2(x+y)y(x+y)}$.
But this is trivial.  ;)

Anh Cuong, now when I done all calculation for Manlio, I saw that the best (maximum possible) ratio
$max{a,b,c}/min{a,b,c}
is really equal to 4. A beautiful inequality! :)

Thank you very much Vasc.

A BEAUTIFUL solution for a BEAUTIFUL inequality.

Now I would like to see your solution, Anh Cuong.

Thank you very much.

Actually - in my personal opinion this inequality is not so beautiful :?

I partially agree with Arne.
Beautiful inequality must have beatiful idea or have perfect form. What beautiful idea lies in this inequality?
Undoubtedly, this inequality is far from simple, and we must thank Anh Cuong for invention and thank Vasc for solution.

Yes,I can't say this is a simple inequality but the solution is not too hard.Nice idea doesn't lie in the inequality,it lies in the solution.
That is what we feel interesting in Maths.
No problem is exactly it is,it allways holds some mysterys that we found and have never found yet.
Sorry if I write too much :)
And here is my solution.In fact the inequality is equal to:
[tex] \sum (ac-bc)^2(\frac{2}{c^2(a+b)(b+c)}-\frac{5}{4(ac+bc)^2}) \geq 0 [/tex]
Assume that $ a \geq b \geq c$ and [tex] A=\frac{2}{a^2(a+b)(a+c)}-\frac{5}{4(ab+ac)^2}... [/tex]
We can easily see that $A \leq B \leq C$.So all we have to do is prove that $A+B \geq 0$.
Since[tex]  \frac{1}{(ab+ac)^2}+\frac{1}{4(ab+ac)^2} \leq \frac{1}{(ab+ac)^2}+\frac{c}{b(ab+ac)^2} = \frac{1}{a^2b(b+c)} [/tex]
Hence we have:
[tex] A+B \geq \frac{2}{a^2(a+b)(a+c)}+\frac{2}{b^2(b+a)(b+c)}-\frac{1}{a^2b(b+c)}-\frac{1}{b^2a(a+c)} = \frac{2a+2b+c}{(a+c)(b+c)}(a-b)^2}\geq0[/tex]
And this stops my solution too :)

Sorry, but I cannot easily see that $ A \le B \leq C$

Please, help!!! :(

Thank you very much.

I don't agree with you Myth.
Maybe the problem looks not beautiful but the idea that leads to solution is rather nice!
And its hardness increases its beauty.

Thank you very much,jura :)
Dear Manilo,sorry but I have made mistake we don't need to prove that $ C \geq B \geq A $.We just need $ B +C \geq 0 $ and $ B \geq 0 $,and this is really trivial.
[tex] B=\frac{2}{b^2(a+b)(c+b)}-\frac{5}{4(ab+bc)^2} \geq \frac{2}{b^2(a+b)(c+b)} - \frac{1}{b^2a(b+c)} [/tex]
[tex]  =\frac{a-b}{b^2a(a+b)(b+c)} \geq 0[/tex]

Sorry, Anh Cuong but I don't see the proof :

1) how do you  prove $ B+C \geq 0 $  ?

2) your proof of $ A+B \geq 0 $ is not correct.

3) why $B+C \geq 0$  and $ B \geq0 $ implies that

$c^2(b-c)^2C+b^2(c-a)^2B+a^2(b-c)^2A \geq 0 $

For me these facts are not obvious.

I am obstinate as I don't understand this beautiful proof and this get me depressed :( .

Probably I am very stupid  :blush:

Uh,sorry.
[tex] A+B \geq \frac{(a-b)^2(2a+2b+c)}{a^2b^2(a+b)(b+c)(c+a)} \geq 0 [/tex]
For the the proof [tex] B+C \geq 0 [/tex] ,it is similar to [tex] A+B \geq 0 [/tex].
I am sure that that solution will work,but now I think I can make it easier.We can prove that $ C \geq 0 $.
We have:
[tex] C=\frac{2}{c^2(a+c)(b+c)}-\frac{5}{4(ac+bc)^2} \geq \frac{2}{c^2(a+c)(b+c)}- \frac{1}{c^2a(a+b)} \geq \frac{2}{c^2(a+c)(b+c)}-\frac{1}{c^2a(b+c)} =\frac{a-c}{c^2a(a+c)(b+c)} \geq 0 [/tex].
Since [tex] ab-bc \geq ab -ac \geq 0 [/tex]  then [tex] f(a,b,c) \geq (A+B)(ab-ac)^2+C(ac-bc)^2 \geq 0 [/tex]  :)
I hope it is ok now :maybe:

You discover the extraordinary beauty of this ineq when you kill it.

Thanks to Anh Cuong I have understood it.

You can prove $B \geq0 $ and $C\geq0$  so $ B+C \geq 0$ but this

doesn't help you as $A$ can be negative.

But fortunately you can prove that

$ A+C \geq0 $ and $A+B \geq 0 $  are true.

So you can use $B \geq 0$ and $A+C \geq 0$

or alternatively

$C \geq0 $ and $A+B \geq0 $ to kill the ineq.

Is it right , Anh Cuong?

[quote="Anh Cuong"]If $ n > 1$ and $ x,y,z\in[1,n],$ then

$ (yz \plus{} zx \plus{} xy)\left[\frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2} \plus{} \frac {1}{(x \plus{} y)^2}\right]$

$ \geq \frac {9}{4} \plus{} \frac {(y \minus{} z)^2}{4n(y \plus{} z)^2} \plus{} \frac {(z \minus{} x)^2}{4n(z \plus{} x)^2} \plus{} \frac {(x \minus{} y)^2}{4n(x \plus{} y)^2},$

with equality if $ x \equal{} 1, y \equal{} z \equal{} n.$[/quote]$ \sum{\frac {yz \plus{} zx \plus{} xy}{(y \plus{} z)^2}} \minus{} \frac {9}{4} \minus{} \sum{\frac {(y \minus{} z)^2}{4n(y \plus{} z)^2}}$

$ \equal{} \sum{\frac {(y \minus{} z)^2\left[n(y \plus{} z) \minus{} (n \plus{} 1)x\right]^2\left[(21n \plus{} 32)\left(x^2 \plus{} yz\right) \plus{} (27n \plus{} 76)x(y \plus{} z)\right]}{12n(n \minus{} 1)(4n \plus{} 9)(y \plus{} z)^2(z \plus{} x)^2(x \plus{} y)^2}}$

$ \plus{} \sum{\frac {(y \minus{} z)^2(ny \minus{} z)(nz \minus{} y)\left[49\left(x^2 \plus{} yz\right) \plus{} (48n \plus{} 59)x(y \plus{} z)\right]}{12n(n \minus{} 1)(4n \plus{} 9)(y \plus{} z)^2(z \plus{} x)^2(x \plus{} y)^2}}\geq0.$


calculus
quadratics
complex numbers
calculus computations

overview: when using newton's method, you can solve for real roots, and there is a geometric interpretation, but when you have complex roots, you can also solve for them using his method, but why does this work
sorry about the wall of text, you can skip the first part if you know what i am talking about

consider newton's method for finding intersections
you have a number x_0 that is close to the intersection to a root of an equation
then you draw a tangent line to the curve at that point, and it produces a point closer to the original point
repeating this process will give one good approximations for roots of an equation,
(this is the geometric interpretation)
considering a sequence of numbers generated by this idea recursively is
$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$

of course this works well for the real roots of an equation, but i have found that this works with equations with complex solutions as well
consder $f(x) = x^2 +1$ we know that they are $x = i, -i$
if you have a ti-83 calculator, you can test this for your self
input
$X - \frac{x^2 + 1}{2x} \rightarrow X$ (the right arrow is the sto buttton) it basically takes some already defined x and makes it another x
if you have x as a real number, then the sequence just oscilates back and forth (makes sense geometrically)
but if you start with x being complex
input
$4i + 1 \rightarrow X$
and then run it through
$X - \frac{x^2 + 1}{2x} \rightarrow X$
then the series does in fact converge towards $i$
but my question is: How can you interpret this geometrically? if you have the complex numbers, you cannot graph it and interpret the way that one does with the real solutions
i have found that this is invaluable for my math class because if there are difficult equations with all complex roots, i can easily solve for them and the teacher does not know what i am doing
but i would like to know how this works, sorry if this is an easy question, the only calculus that i know, i have learned on my own
and my calculus book offers no explanation whatsoever to this

Newton's method is fundamentally the same in $\mathbb{C}$ as it is in $\mathbb{R}$, and it works for the same reasons, with the same quadratic convergence. There's four dimensions in play, and I wouldn't try to visualize that geometrically.

Suppose $f(x)=ax+bx^2+O(x^3)$. We iterate $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{ax_n+bx_n^2+O(x_n^3)}{a+2bx_n+O(x_n^2)}= x_n-x_n\left(1+\frac bax_n-2\frac bax_n+O(x_n^2)\right)$ $=-\frac bax_n^2+O(x_n^3)$.
If you work this a little more carefully, you get a proof of convergence, if you started close enough to the root. Multiple roots slow down convergence, but they are easy to deal with.

At long range, it can be hard to predict which root you converge too, and you can draw some very pretty fractal pictures depicting where these iterations go (as long as you've got something more complicated than a quadratic).

ok, i see, there is an algebraic way to show the convergence, but not necessarily a geometric way since there would be 4 dimentions,
knowing this brings me to my second question, when doing the convergence, how would you know which root it will approach
using $f(x) = x^2 +1$
i notice the convergence tends towards the root with the sign i input
i.e. if i start with
$a + bi \rightarrow X$
then this converges to the positive root $i$
but if i start with
$a - bi \rightarrow X$
then it converges to $-i$
would one say that there is a general rule to this, i.e. if it is closer on the compled plane to a certain root, then it will converge there
is there a general way to show this? this is a pretty easy example, but what about like $x^8 -1$, does the input actually converge towards the root that is closest on a complex plane?

[quote="Altheman"]is there a general way to show this? this is a pretty easy example, but what about like $x^8 -1$, does the input actually converge towards the root that is closest on a complex plane?[/quote]
No. If you graphed which root the iteration converged to based on initial location, the picture would look something like the picture you put under your handle.

In the quadratic case, it does converge to the closer root, but that is only for quadratic polynomials.

My argument shows that you converge to a root if you start close enough- there's a lot of the plane that's not "close enough" to any root. It will almost always converge to some root, but it can take a very long time to get close.

thats cool, i never had any idea how the fractals worked


The perimeter of a rectangle is 32 cm. Its length is three times its width. How many centimeters are in the length of the rectangle?

Let $ l$ be the length, then the width is $ \frac{l}{3}$ and we have \[ 2\left(l\plus{}\frac{l}{3}\right) \equal{}32 \Rightarrow l\equal{}\boxed{12}\]


algebra
polynomial
inequalities
inequalities proposed

Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (a)\ \ \ \ \ \  \sum x(x\minus{}y)(x\minus{}z)(x\minus{}6y)(x\minus{}6z)\plus{}\frac {16(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2}{x\plus{}y\plus{}z} \ge 0$;

$ (b)\ \ \ \ \ \  \sum x(x\minus{}y)(x\minus{}z)(x\minus{}7y)(x\minus{}7z)\plus{}\frac {81(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2}{4(x\plus{}y\plus{}z)} \ge 0$.

Equality holds in (b) for
$ x\equal{}y\equal{}z$;
$ x\equal{}0$, $ y\equal{}z$;
$ x\equal{}0$, $ y\equal{}2z$;
$ \frac x{7}\equal{}y\equal{}z$
(or any permutation).

[quote="Vasc"]Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (b)\ \ \ \ \ \ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} 7y)(x \minus{} 7z) \plus{} \frac {81(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{4(x \plus{} y \plus{} z)} \ge 0$.

Equality holds in (b) for
$ x \equal{} y \equal{} z$;
$ x \equal{} 0$, $ y \equal{} z$;
$ x \equal{} 0$, $ y \equal{} 2z$;
$ \frac x{7} \equal{} y \equal{} z$
(or any permutation).[/quote]
Let $ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^2$, $ xyz \equal{} w^3.$
Hence, $ \sum_{cyc} x(x \minus{} y)(x \minus{} z)(x \minus{} 7y)(x \minus{} 7z) \plus{} \frac {81(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{4(x \plus{} y \plus{} z)} \equal{}$
$ \equal{} Aw^6 \plus{} Bw^3 \plus{} C,$ where $ A \equal{} \minus{} \frac {729}{4u} < 0$, $ B \equal{} B(u,v^2)$ and $ C \equal{} C(u,v^2)$.
Thus, $ Aw^6 \plus{} Bw^3 \plus{} C$ gets the minimal value on bounders of $ w^3.$
But $ w^3$ is a free term of degree three polynomial with three real roots $ x,$ $ y$ and $ z$.
Id est, we obtain the bounders of $ w^3$, when two of $ x$, $ y$ and $ z$ are equal or when $ w^3 \equal{} 0.$
Hence, we need check your inequality for $ y \equal{} z \equal{} 1$ and for $ z \equal{} 0$,
which quits true.

[quote="arqady"][quote="Vasc"]Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (b)\ \ \ \ \ \ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} 7y)(x \minus{} 7z) \plus{} \frac {81(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{4(x \plus{} y \plus{} z)} \ge 0$.

Equality holds in (b) for
$ x \equal{} y \equal{} z$;
$ x \equal{} 0$, $ y \equal{} z$;
$ x \equal{} 0$, $ y \equal{} 2z$;
$ \frac x{7} \equal{} y \equal{} z$.[/quote]
Hence, we need check your inequality for $ y \equal{} z \equal{} 1$ and for $ z \equal{} 0$,
which quits true.[/quote]
Yes, Arqady.  :lol:  Using this method, we can prove that

$ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} ky)(x \minus{} kz) \plus{} \frac {(k \plus{} 2)^2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{4(x \plus{} y \plus{} z)} \ge 0$

for $ k\ge 6$, and

$ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} ky)(x \minus{} kz) \plus{} \frac {4(k \minus{} 2)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{x \plus{} y \plus{} z} \ge 0$

for $ 2\le k\le 6$, but I think we can't prove either the last inequality for $ k < 2$, or

$ \sum (y \plus{} z)(x \minus{} y)(x \minus{} z)(x \minus{} ky)(x \minus{} kz)\ge \frac {(k \plus{} 2)^2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{x \plus{} y \plus{} z}$

for $ k\ge 0$.

[quote="Vasc"][quote="arqady"][quote="Vasc"]Let $ x,y,z$ be nonnegative real numbers. Prove that

$ (b)\ \ \ \ \ \ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} 7y)(x \minus{} 7z) \plus{} \frac {81(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{4(x \plus{} y \plus{} z)} \ge 0$.

Equality holds in (b) for
$ x \equal{} y \equal{} z$;
$ x \equal{} 0$, $ y \equal{} z$;
$ x \equal{} 0$, $ y \equal{} 2z$;
$ \frac x{7} \equal{} y \equal{} z$.[/quote]
Hence, we need check your inequality for $ y \equal{} z \equal{} 1$ and for $ z \equal{} 0$,
which quits true.[/quote]
Yes, Arqady.  :lol:  Using this method, we can prove that

$ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} ky)(x \minus{} kz) \plus{} \frac {(k \plus{} 2)^2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{4(x \plus{} y \plus{} z)} \ge 0$ (1)

for $ k\ge 6$, and

$ \sum x(x \minus{} y)(x \minus{} z)(x \minus{} ky)(x \minus{} kz) \plus{} \frac {4(k \minus{} 2)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{x \plus{} y \plus{} z} \ge 0$ (2)

for $ 2\le k\le 6$.[/quote]
We see that $ \sum x(x\minus{}y)(x\minus{}z)(x\minus{}ky)(x\minus{}kz)$ is in degree 5th, so if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, and after multiplying both sides for $ p\equal{}x\plus{}y\plus{}z>0,$ the degree of $ r$ in this expression is still $ 1$. Moreover, the degree of $ r$ in the epxression $ (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2$ after transforming into $ pqr$ form is $ 2$ and its coefficient is negative, therefore in (1) and (2), after transforming into $ pqr$ form, we can rewrite both inequality in the form $ f(r) \equal{}Ar^2\plus{}Br\plus{}C \ge 0,$ where $ A \le 0.$ It shows that $ f(r)$ is concave and from a known theorem (I have also posted on the forum already), we only need to consider both inequality in the cases $ xyz\equal{}0$ and $ (x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}0.$

I am very sorry as my English is actually bad.

The following two inequalities are nice and no easy.

If $ x,y,z$ are nonnegative real numbers, then

$ (a1)\ \ \ \ \ \  \sum x(x\minus{}y)(x\minus{}z)(2x-y)(2x-z) \ge \frac {24(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2}{x\plus{}y\plus{}z}$;

$ (b1)\ \ \ \ \ \  \sum (y+z)(x\minus{}y)(x\minus{}z)(x\minus{}2y)(x\minus{}2z) \ge \frac {16(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2}{x\plus{}y\plus{}z} $.

[quote="Vasc"]

If $ x,y,z$ are nonnegative real numbers, then

$ (a1)\ \ \ \ \ \  \sum x(x-y)(x-z)(2x-y)(2x-z) \ge \frac {24(x-y)^2(y-z)^2(z-x)^2}{x+y+z}$;
[/quote]
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ ($v\geq0$), $xyz=w^3$ and $u^2=tv^2$.
Hence, our inequality is equivalent to $24w^6+(127u^2-169v^2)uw^3+6(3u^2-4v^2)^2(2u^2+v^2)\geq0$.
Thus, for $t\geq\frac{169}{127}$ our inequality is true.
Since $\Delta=(7u^2-9v^2)(823u^4-1865u^2v^2+1024v^4)$, it remains to prove our inequality for $1\leq t\leq\frac{9}{7}$.
Hence, it's enough to prove that $w^3\geq\frac{169uv^2-127u^3+\sqrt{\Delta}}{48}$.
Since $(x-y)^2(x-z)^2(y-z)^2\geq0$ gives $w^3\geq3uv^2-2u^3-2\sqrt{(u^2-v^2)^2}$, it remains to prove that
$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\geq\frac{169uv^2-127u^3+\sqrt{\Delta}}{48}$
or $31u^3-25uv^2-96\sqrt{(u^2-v^2)^3}\geq\sqrt{\Delta}$ or
$31t-25-96\sqrt{\frac{(t-1)^3}{t}}\geq\sqrt{\frac{(7t-9)(823t^2-1865t+1024)}{t}}$.
Easy to show that $31t-25-96\sqrt{\frac{(t-1)^3}{t}}>0$ for all $t\in\left[1,\frac{9}{7}\right]$.
Hence, after squaring of the both sides we need to prove that
$46t^2-91t+45\geq2(31t-25)\sqrt{\frac{(t-1)^3}{t}}$
or ${46t-45\geq2(31t-25)\sqrt\frac{t-1}{t}}$ or $(24t-25)^2(4-3t)\geq0$, which is obvious.

We can prove $(b1)$ by the same way.

[quote="Vasc"]
The following two inequalities are nice and no easy.

If $ x,y,z$ are nonnegative real numbers, then

$ (a1)\ \ \ \ \ \  \sum x(x\minus{}y)(x\minus{}z)(2x-y)(2x-z) \ge \frac {24(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2}{x\plus{}y\plus{}z}$;
[/quote]
By using the method you mentioned before, we could get the following result:
If $ x,y,z $ are non-negative real numbers and $ k \le 2 $, then
\[ \sum_{cyc}{x(x-y)(x-z)(x-ky)(x-kz)} \ge \frac{4(2-k)(x-y)^2(y-z)^2(z-x)^2}{x+y+z} \]
with equality for $ x=y=z, $ or $ x=0,y=z $ or $ x=ky=kz $
((include any cyclic permutation

[quote="BSJL"] By using the method you mentioned before, we could get the following result:
If $ x,y,z $ are non-negative real numbers and $ k \le 2 $, then
\[ \sum_{cyc}{x(x-y)(x-z)(x-ky)(x-kz)} \ge \frac{4(2-k)(x-y)^2(y-z)^2(z-x)^2}{x+y+z} \]
with equality for $ x=y=z, $ or $ x=0,y=z $ or $ x=ky=kz $
((include any cyclic permutation[/quote]
If $x,y,z$ are nonnegative real numbers and
 \[\alpha_k&=\left\{
\begin{array}{ll}
4(k-2), & k\le 6
 \\[10pt]
\dfrac{(k+2)^2}{4}, & k\ge 6
 \\[10pt]
\end{array},
\right.\] then \[\sum x(x-y)(x-z)(x-ky)(x-kz)+\alpha_k\frac{(x-y)^2(y-z)^2(z-x)^2}{x+y+z}\ge 0.\]
See Proposition 2.4 in
http://www.sciencedomain.org/abstract.php?iid=360&id=6&aid=2653


modular arithmetic
trigonometry
geometric series
calculus
calculus computations

We know that series $\sum_{i=1}^\infty a_n$ is convergent and for all positive integers k and l  we have $|\sum_{i=0}^\infty a_{nl+k}|<\infty$. Can we conlude from these facts that $\sum_{i=1}^\infty a_n$ is absolutely convergent?

In my opinion (intuition) it's true, i.e. this series is absolutely convergent but I'm not sure how to show it (maybe applying any integrals).

Thanks for help
Nobody1111

Just cover the whole set of natural numbers with a few arithmetic progressions, i.e. choose some $x_i,y_i,\ i=\overline{1,k}$ s.t. every natural is $\equiv x_i\pmod {y_i}$ for at least one $i$. After that, observe that $\sum_n |a_n|\le \sum_i\sum_n|a_{ny_i+x_i}|$. Am I missing something? :?

grobber- you're missing that we don't know that any of the series are absolutely convergent.

Counterexample: $\frac1ne^{ni}$ (I could take real or imaginary parts if I wanted to). Summation by parts tells us that $\sum_{n=0}^N \frac1{nk+l}e^{(nk+l)i}=\sum_{n=0}^N\left[\left(\frac1{nk+l}- \frac1{nk+k+l}\right)S_{k,l}(n)\right]+\frac{S_{k,l}(N)}{(N+1)k+l}$
$=\sum_{n=0}^N\frac{k\cdot S_{k,l}(n)}{(nk+l)(nk+k+l)}+ \frac{S_{k,l}(N)}{(N+1)k+l}$,
where $S_{k,l}(n)=\sum_{m=0}^ne^{(mk+l)i}= \frac{e^{li}-e^{(nk+k+l)i}}{1-e^{ki}}$. Since the series $\sum\frac{k}{(nk+l)(nk+k+l)}$ is absolutely convergent and $|S_{k,l}(n)|\le \frac2{|1-e^{ki}|}$, the transformed series is absolutely convergent and $\sum_{n=0}^N \frac1{nk+l}e^{(nk+l)i}$ converges.

Of course, the original series $\sum\frac1ne^{ni}$ is not absolutely convergent- its absolute values are the harmonic series.

I read $\sum|a_{nk+l}|$, that's why I wrote that.

Thanks

But what if we assume that this series must be series of Real numbers (not complex)?

As I said, take the real parts in my example. I only used the complex exponential because sum formulas for geometric series are easier to remember than trigonometric sum formulas. The same bound $|S_{k,l}(n)|\le\frac1{|\sin \frac k2|}$ would still apply.


Why does carbonyl mainly form complexs with central atom in 0 oxidation state :?:

Yes $ \ce{Ni(CO)4}$ remember Mond's process? :)

but the low oxidation state is actually a prerequisite os Synergy effect ...so why is the converse it's answer :)

The answer for this may perhaps be found on the nature of the bond between the carbon of CO and the metal. This is not the regular dative covalent bond found in most complex ions, in which both electrons are provided by the ligand. In the case of metal carbonyls, the bond M-CO arises as follows: first, there is a sigma bond that forms from the overlap of a nonbonding electron pair from carbon with some sort of dsp hybridized orbital from the metal; and second, there are two pi bonds that arise from overlap of filled d-orbitals from the metal with two pi antibonding orbitals from the carbon of CO. These two last bonds require that the metal have d-electrons, and that the metal is in a relatively low oxidation state (less that +2). Note that the M-CO bond is a triple bond.

yes, CO has a pi electron acceptor, and has empty pi orbitals, with the correct symmetry to overlap with metal t2g orbitals, forming pi bonds(back bonding).


group theory
abstract algebra
number theory unsolved
number theory

Let $m$ be an odd composite number, and let $G$ denote the set of those reduced residue classes $a (mod$ $m)$ such that $a^{(m-1)/2} \equiv \left( {a \over m} \right)$ $(mod$ $m)$, where $\left( {a \over m} \right)$ is Jacobi's symbol. Show that $G$ is a proper subset of the collection of reduced residue classes $(mod$ $m)$.

Let $Z_m$ be the collection of reduced residue classes mod. $m=p_1^{i_1}p_2^{i_2}...p_k^{i_k}$ with pairwise distinct primes $p_j$ and positive integers $i_j$.
That $G$ is non-empty is trivial.
1. case: $m$ is not a perfect square, so one of the $i_j$ is odd, let it be $i_1$. Now there is an $a$ with $\left( a \over p_1 \right) =-1$.
Because of chinise remainder theorem, there is an $b$ that satisfies $a \equiv b(mod(p_1))$ and $b \equiv 1(mod(p_j)) \Rightarrow b^{(m-1)/2} \equiv 1(mod(p_j))$ for all other $j$. You have $\left( b \over m \right) =-1$, but $b^{(m-1)/2} \not\equiv -1 (mod(m))$, so you get that $b$ lies not in $G$

2. case: $m$ is a perfect square:
You get $\left( a \over m \right) =1$ for all $a$ coprime to $m$, but because of the fact that $Z_{p_1^2}$ is cyclic of order $p_1(p_1-1)$ with a primitive element $a$, $a^{(m-1)/2} \equiv 1 (mod (p_1^2))$ would need that $p_1(p_1-1)|(m-1)/2$, but $p_1$ cannot divide both $m$ and $(m-1)/2$, so if you choose (chinise remainder theorem) an $b$ that satisfies $a \equiv b(mod(p_1))$ and $b \equiv 1(mod(p_j))$ for all other $j$, you get $b^{(m-1)/2} \not\equiv 1 (mod(m))$, so $b$ lies not in $G$

If all elements of the group $(\mathbb F_m^*,\cdot)$ would be in $G$ then results that $a^{m-1}\equiv 1(\mod m)$ for all $a\in \mathbb F_m^*$. But we know by Cauchy that if $p$ is prime and $p$ divides the order of a group then there is an element which has the order $p$.

If $m=\prod p_k^{\alpha_k}$ with at least one $\alpha_k>1$ then for $p=p_k$ there is an element of order $p$ and we will have that $p|\varphi(m), p|m$ and $p|m-1$ contradiction.

If $m=p_1\cdot p_2\cdots p_r$ with all $p_i$ distinct. There exists $x$ s.t. $\left(\frac x{p_1}\right)=-1$ and we can choose(by the Chinese Remainder Theorem) an element $a$ s.t. $a\equiv x(\mod p_1)$ and $a\equiv 1(\mod p_i), i>1$. We will have that $-1\equiv\left(\frac am\right)=a^{(m-1)/2}\neq -1$ contradiction.

So $G$ is a proper subgroup of $(\mathbb F_m^*,\cdot)$

Kuba: Where from do you get $-1\equiv\left( \frac am\right)$

sorryy.. that was a little bit wrong...  I edited my post


who watches it?
yeah i know its from like the 80s

It's been your avatar for like years.  Other than that, what is it?

never heard of it, so i searched it on youtube and found this theme song

[youtube]lP_2YffJhmg[/youtube]

Are you serious? I thought MacGyver was like a TV show that everyone knows but doesn't watch anymore. Like Hogans Heroes or MASH.