[quote="Math pro"]chebysev or jensen is nice chosee![/quote]
Can U detail your proof ?

[quote="onlylove_math"]Let $ a;b;c \ge\ 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {(1 \minus{} a)^2}{1 \plus{} (1 \minus{} 2a)^2} \plus{} \frac {(1 \minus{} b)^2}{1 \plus{} (1 \minus{} 2b)^2} \plus{} \frac {(1 \minus{} c)^2}{1 \plus{} (1 \minus{} 2c)^2} \le\ \frac {6}{5}$[/quote]
My friend,it quite hard.Here is my proof|
With$ a \plus{} b \plus{} c \equal{} 1$.Your ineq <=>
$ \sum \frac{a^2}{a^2 \plus{} (b \plus{} c)^2}\geq \frac{3}{5}$
And it is true since C-S ineq
Is your proof,my dear

[quote="onlylove_math"][quote="Allnames"][quote="onlylove_math"]Let $ a;b;c \ge\ 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {(1 \minus{} a)^2}{1 \plus{} (1 \minus{} 2a)^2} \plus{} \frac {(1 \minus{} b)^2}{1 \plus{} (1 \minus{} 2b)^2} \plus{} \frac {(1 \minus{} c)^2}{1 \plus{} (1 \minus{} 2c)^2} \le\ \frac {6}{5}$[/quote]
My friend,it quite hard.Here is my proof
With$ a \plus{} b \plus{} c \equal{} 1$.Your ineq <=>
$ \sum \frac {a^2}{a^2 \plus{} (b \plus{} c)^2}\geq \frac {3}{5}$
And it is true since C-S ineq
Is your proof,my dear[/quote]
Are U sure your proof is right? My solution is different from your proof.
Can U detail ?  :)[/quote]
My proof is true (i think so and if i have mistakes,iam sorry)
$ LHS \equal{} \sum \frac {(b \plus{} c)^2}{(a \plus{} b \plus{} c)^2 \plus{} (b \plus{} c \minus{} a)^2} \leq \frac {6}{5}$
<=> $ \sum \frac {(b \plus{} c)^2}{2(a^2 \plus{} (b \plus{} c)^2)}\leq \frac {6}{5}$
and it <=> the ineq that i have just said
Post your proof,please

[quote="Allnames"][/quote][quote="onlylove_math"][quote="Allnames"][quote="onlylove_math"]Let $ a;b;c \ge\ 0$ and $ a + b + c = 1$. Prove that:
$ \frac {(1 - a)^2}{1 + (1 - 2a)^2} + \frac {(1 - b)^2}{1 + (1 - 2b)^2} + \frac {(1 - c)^2}{1 + (1 - 2c)^2} \le\ \frac {6}{5}$[/quote]
My friend,it quite hard.Here is my proof
With$ a + b + c = 1$.Your ineq <=>
$ \sum \frac {a^2}{a^2 + (b + c)^2}\geq \frac {3}{5}$
And it is true since C-S ineq
Is your proof,my dear[/quote]
Are U sure your proof is right? My solution is different from your proof.
Can U detail ?  :)[/quote][quote="Allnames"]
My proof is true (i think so and if i have mistakes,iam sorry)
$ LHS = \sum \frac {(b + c)^2}{(a + b + c)^2 + (b + c - a)^2} \leq \frac {6}{5}$
<=> $ \sum \frac {(b + c)^2}{2(a^2 + (b + c)^2)}\leq \frac {6}{5}$
and it <=> the ineq that i have just said
Post your proof,please[/quote]
Thank Allnames. My proof use ineq Jensen (now i have not enough time )

[quote="onlylove_math"][/quote][quote="Allnames"][/quote][quote="onlylove_math"][quote="Allnames"][quote="onlylove_math"]Let $ a;b;c \ge\ 0$ and $ a + b + c = 1$. Prove that:
$ \frac {(1 - a)^2}{1 + (1 - 2a)^2} + \frac {(1 - b)^2}{1 + (1 - 2b)^2} + \frac {(1 - c)^2}{1 + (1 - 2c)^2} \le\ \frac {6}{5}$[/quote]
My friend,it quite hard.Here is my proof
With$ a + b + c = 1$.Your ineq <=>
$ \sum \frac {a^2}{a^2 + (b + c)^2}\geq \frac {3}{5}$
And it is true since C-S ineq
Is your proof,my dear[/quote]
Are U sure your proof is right? My solution is different from your proof.
Can U detail ?  :)[/quote][quote="Allnames"]
My proof is true (i think so and if i have mistakes,iam sorry)
$ LHS = \sum \frac {(b + c)^2}{(a + b + c)^2 + (b + c - a)^2} \leq \frac {6}{5}$
<=> $ \sum \frac {(b + c)^2}{2(a^2 + (b + c)^2)}\leq \frac {6}{5}$
and it <=> the ineq that i have just said
Post your proof,please[/quote][quote="onlylove_math"]
Thank Allnames. My proof use ineq Jensen (now i have not enough time )[/quote]
If my proof is true ,i think it is nicer than yours
At all,your problem is very nice


induction
superior algebra
superior algebra solved

I know that at some point Acchileas was talking about a problem concerning rings but I don't seem to find it anymore - can anyone remember which one it was?

It's been over a month ago, so let's see if I remember it. It was a question of grobber:

Let A be a commutative ring of order 2n+1, where n is different from 4. Assume that for any nonzero nonunit element x of A, we have either x^2=x or x^2=-x. Then A is a field.

Yeah, I think that's the one. It was lost when there was that loss of information. I'm amazed. How did you remember it? I had forgotten it myself and I posted it! :D

:)  I tried to solve it for a couple of days, so that's why I remember it. Unfortunately, I did not save anything of my (incomplete) work.

I have a proof for this, but it is huge. I'll give you the ideas.

First, I know that it is in most deffinitions of the ring that it has a unit element. I assumed that A has 1.

-the characteristic of A is an odd prime number; (remember the characteristic of A is the smallest n such that n*1=1+...+1 (n times) =0.)

-if G is a commutative group whose elements different from e have order p (a prime number), then |G| is a power of p. (Use Cauchy or induction)

Assume A has noninvertible elements different from 0. Then

-the characteristic of A is 3;
-if x is an invertible element of A, then x 2 =1.
-if M={x in A| x invertible and x <> \pm 1} and N={a in A| a not invertible and a<>0} then 2|M|=|N|
- there is t positive integer such that |M|=2 t-2.

We have A={ \pm 1,0}  \cup M \cup N and the 3 sets are disjoint, so |A|=3+|M|+|N|.
(A,+) is a group where all the elements have order 1 or 3, so there exists r integer such that |A|=3 r=2n+1.
We have 3 r-1=2 t-1.
After solving this equation we obtain no contradictory solution.
So A has no noninvertible and not 0 element, so A is a field.

another solution (maybe easier, or at least full :D)

suppose A is not a field and let S be the set of idempotent elments in A. It must have at least 3 elements, since 0,1 are in S and if a is a non-invertible element then either a or -a is in S.  Consider that a is in S. Also let b be an invertible element.

Obviously ab is not invertible so (ab)2 = \pm ab => a2b2 = \pm ab => a2b = \pm a => ab = \pm a. But b is a randomly chosen invertible element, and because 2 cannot be the characteristic of the ring, we have that 2 is a unit, therefore 2a = \pm a => 3a=0 for all non-invertible a-s, which means that 3=0.

Returning we have two possible cases: if ab=a => (b-a)2=a2-2ab+b2=a+ab+b2=2a+b2=b2-a=b2-a2 = (b-a)(b+a). If b-a is invertible, then b-a = b+a => 2a = 0 => a=0 contradiction. So b-a is not invertible, which means that b2-a  = (b-a)2 = b-a or a-b. But, if b<>1, we have b2<>b => b2 = 2a - b = -(a+b), that is (a+b) is invertible.  But (a+b)2 = a2 +2ab + b2= a+2a +b2 = b2 = (b2)2 => b2 = 1 => -a-b = 1 => b=-a-1 or b=1.

In the second case ab=-a and in the same way we find that b=-1 or b=a+1, thus all the units of A are \pm 1, \pm (a+1), and then A can only be { 0,1,-1, -a,a,-a-1,a+1,a-1,-a+1}, which would mean that n=4 a contradiction.

Thus A is indeed a field. :D

amfulger I am not sure that your solution is good. Please detail it (especially the equation 3r-1 = 2t-1!)

The equation 3r-1=2k-1, gives solutions only for r=1 and r=2.
If r=2, we get 2n+1=3 r, so n=4 (impossible)
If r=1, we get 2n+1=3, so A is izomorphic to Z_3, so it is a field (remember I assumed for a contradiction that A is not a field).

you still haven't solved the equation. maybe I am too tired but I don't see it now :)

Let n=r-1 and m=k.
Let's solve for our god's will the equation 3 n=2 m-1.

If n=0 then m=1.
If n>1 then 3|2 m-1, so m is even. Substitute
m by 2m, and get 3 n=4 m-1
We have 3 n-1=1+4+...+4 m-1.

If n=1 we get m=1 and the older m=2.
If n>1 then 3|1+4+...+4 m-1, so 3|m. Substitute m by 3m. And get 3 n=64 m-1. But the last is divisible by 7, so we get a contadiction.

So the only solutions to the initial equation are (n=0 and m=1) and (n=1 and m=2).
See my previous proof to understand how this affects the problem.
Hope you understand it Valentin, if you are not too tired.
You must have studied too much for tomorow's algebra test.

yeah now that part of your proof is ok. there are more parts to explore. indeed your solution is huge :D

[quote="amfulger"]
-the characteristic of A is an odd prime number;
-if G is a commutative group whose elements different from e have order p (a prime number), then |G| is a power of p. (Use Cauchy or induction)
Assume A has noninvertible elements different from 0. Then
-the characteristic of A is 3;
-if x is an invertible element of A, then x 2 =1.
[/quote]
My part of the proof above is covered in Valentin's proof.
What remanines unsolved is this

-if M={x in A| x invertible and x <> \pm 1} and N={a in A| a not invertible and a<>0} then 2|M|=|N|.
To prove this, prove that if a is in N then exactly one of a+1 and a-1 is invertible - I think Valentin did smth similar in his proof.
Then prove that if x is in M then x+1 and x-1 are in N.
Write A\{-1,0,1} as a disjoint union of sets of the form {a-1,a,a+1}. This is possible. For example take A/Z_3 as additive groups and pick one a from each factorization class.
It will be clear that in each subset of the form {a-1,a,a+1} we will have one element from M and 2 from M. It should be easy to deduce that |M|=2|N|.

- there is t positive integer such that |M|=2 t-2. (for this notice that W=M \cup{ \pm 1} is a multiplicative group in which all elements have order 2 or 1, and use that Group statemnt above to find t such that |W|=2 t)

We have A={ \pm 1,0}  \cup M \cup N and the 3 sets are disjoint, so |A|=3+|M|+|N|.
As an additive group A has characteristic 3, so all elements have order 1 or 3, so there is a r such that |A|=3 r.


The TeXer seems not to like some stuff and gives out "You have an empty LaTeX formula in your post. Please correct it before submitting the post.".
At least the following causes this: [code]$0$[/code]
(I would understand it for $\emptyset$ instead :D )

It works it the TeXer, but it doesn't work in posts.

That's the TeXer for me ;)

But why are you quoting my post for your answer, fear that one may not know what you are answering to ;)

I don't know... :ninja:

I guess I shouldn't. :)

It doesn't render in posts, I think, for me.  :(  Maybe its because you have Mod powers.

[hide][img]http://i96.photobucket.com/albums/l167/adasarathy/emptylatex.png[/img]
is the screen I see.  :( [/hide]

Same here. I don't think anyone would want to post a 0, but it's still supposed to work.

I wanted today :P  ( http://www.mathlinks.ro/Forum/viewtopic.php?t=116639 )

@anirudh: I see the same $\mod$ AoPS <-> MathLinks.

I've rewritten today some of the $\LaTeX$ parsing on the Forum (in order to be able to quote dollars). As it turns out I forgot that the function empty() in php returns the argument string '0' true :)

This should be fixed now: $0$


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ G$ be a finite group. If $ H$ is a subgroup of $ G$ and $ N$ is a normal subgroup of $ G$. If $ |H|$ and $ |G: N|$ are relatively prime, then $ H$ is a subgroup of $ N$. Thanks in advance.

take integers $ u,v$ such that $ u|H| \plus{} v |G: N|\equal{}1$. then applying twice lagrange, we get $ h \equal{} h^{|H|u} h^{v|G: N|} \equiv 1$ mod $ N$, i.e. $ h \in N$ for every $ h \in H$.


A small puck of mass $m$ is released from rest a height $h$ above the ground and slides
along a frictionless track. The track contains a circular "loop-the-loop" of radius $r$ as
in Figure 1. Determine the angle $\theta_{0}$ relative to the vertical at which the puck loses
contact with the track. Express the answer as a function of the height $h$, the radius $r$,
the mass m and the gravitational field $g$. Assume $h > r$ so that the puck has a chance
to fall off.
[color=green]
Please provide good solutions and not only loose [hide="answers"]
$\cos{\theta_{0}}=\frac{2}{3}\left(\frac{h}{r}-1\right)$
[/hide].[/color]

The system is conservative. It begin to lose contact when the normal is equal to centripetal force

${m g h= m g (r+\cos(\theta ) r)+\frac{m v^{2}}{2}}$
${{\frac{m v^{2}}{r}=m g \cos}(\theta ) \Rightarrow v^{2}=r g \cos(\theta )}$

${m g h=m g (r+\cos(\theta ) r)+\frac{m }{2}r g \cos(\theta ) }$
${2 h=r (2+3 \cos(\theta ))}$
${\cos(\theta )= \frac{2 (h-r)}{3 r}=\text{ }\frac{2}{3}\left(\frac{h}{r}-1\right)}$

err... This is not a Advanced Physics problem, dude...

If you considere friction, so, it will be a advanced physics problem.

c ya

[quote="yagaron"]The system is conservative. It begin to lose contact when the normal is equal to centripetal force
[/quote]
no that's not the right condition for loosing contact.the right condition is as the ball or whatever tries to fly off the surface it has to do gradually so the normal force acting on it $\rightarrow$ $0$ so finally when it flies off the surface the normal force becomes $0$ .

[b]@yagaron[/b] your condition gives $mg \cos \theta_{0}$ $-$ $N$ $=$ $N$
which obviously is not the right one. :rotfl:

Please. read what I said

actually this a resolutions of a problem of ITA in the site http://www.cursoanglo.com.br/webstander/index.asp

dont say idiot things dude...

at the moment the ball begin to lose contact the normal is equal the centripetal force. but after does not have centripetal force or normal force

what i said is $N = m v^{2}/r$

you could try change your glass, because you are reading things wrongly

[quote="yagaron"]Please. read what I said

actually this a resolutions of a problem of ITA in the site http://www.cursoanglo.com.br/webstander/index.asp

[/quote]

sorry i don't understand the language there :rotfl:

[quote="yagaron"]
dont say idiot things dude...

you could try change your glass, because you are reading things wrongly[/quote]

don't act smart .Be within your limits .this is supposed to be a scientific discussion and remember u r not the only one with a dictionary of slangs.
 :starwars:  :bomb:  :bruce:  :diablo: .
you are again wrong :D

u say

[quote="yagagron"]at the moment the ball begin to lose contact the normal is equal the centripetal force. but after does not have centripetal force or normal force

what i said is $N = m v^{2}/r$

[/quote]
first of all there is no force called centripetal force.There is simply centripetal accelaration.Centripetal force is just a concept which arises due to other forces.

secondly when the particle moves $\frac{mv^{2}}{r}$ $=$ $mg \cos \theta-N$
so your statement becomes equal what i wrote before and hence you are wrong. :lol:
i don't understand after getting the basic equation wrong how did you get the answer . :P
surely ....

nope yagaron, pardesi is right!
Think about it:
newtons law at an arbitrary point of the loop(before falling):
$mg\cos \theta+N =m\frac{v^{2}}{r}$
So when does it lose contact with the loop?
of course when the loop doesnt exert any force on the mass, that is when $N=0$ :wink:

[quote="Thaakisfox"][quote]first of all there is no force called centripetal force.There is simply centripetal accelaration.Centripetal force is just a concept which arises due to other forces. [/quote]

No pardesi, thats wrong. When a particle is moving on a circular path, then the direction of the particles velocity changes, hence it has acceleration. So since it has acceleration there must be some kind of force exerted on it. Force is the cause, and acceleration is what it causes. So if there is no force how can there be acceleration? So centripetal force is not just a concept, its a real force. If there was no centripetal force, then there would be no circular motion.[/quote]

Thaakisfox centripetal force is no real force like gravitational,electromagnetic...etc
it arises due to these .when a particle moves in a circle there is accelaration and thus there must be some forces that must be acting on the particle these forces might have gravitational,EM, nuclear...or whatever origin.These forces when put together must give an accelaration of $\frac{v^{2}}{r}$ towards the centre .their With proper algaebric signs in between them they constitue what is called the '[i]centripetal force[/i]' because they 'produce' centripetal accelaration.
[quote="Thaakisfox"]
$mg \cos \theta$ $+$ $N$ $=$ $\frac{mv^{2}}{r}$
[/quote]
there should be negative sign before $N$ .

[quote]Thaakisfox centripetal force is no real force like gravitational,electromagnetic...etc
it arises due to these .when a particle moves in a circle there is accelaration and thus there must be some forces that must be acting on the particle these forces might have gravitational,EM, nuclear...or whatever origin.These forces when put together must give an accelaration of  towards the centre .their With proper algaebric signs in between them they constitue what is called the 'centripetal force' because they 'produce' centripetal accelaration. [/quote]

First of all, what do you mean by real force?
Centripetal force is not a real force? If there is no centripetal force then there is no circular motion. I think you are mixing centripetal with centrifugal force. centrifugal is force is not a real force it just exists in accelerating frame.
So if what you say is correct than in outer space far from every planet where there is no gravity, EM-force, nuclear force etc. a particle cannot move on circular path?
or if on a horizontal plane, gravity acts only downwards, and there is no EM, then there cannot be circular motion?
I know you are trying to say that there must somekind of force acting on a particle, to make it go on a circular path, but you havent said it correctly :wink:

[quote="Thaakisfox"][quote]there should be negative sign before  .
[/quote]
It depends where you are writing it. in the upper half of the loop, where the puck will fall off, the direction of the normal force is the same as the direction of the radial component of gravity force. so there is $+$ sign before $N$. :wink:[/quote]
no in the upper half the radial component is [i]not i[/i]n the same direction as that of the normal force. Referring to the original diagram by color till
$\theta$  $\leq$  $\frac{\pi}{2}$

normal is opposite to the radial component of gravity and that is what we are referring here to   :)

[quote="Thaakisfox"][quote]Thaakisfox centripetal force is no real force like gravitational,electromagnetic...etc
it arises due to these .when a particle moves in a circle there is accelaration and thus there must be some forces that must be acting on the particle these forces might have gravitational,EM, nuclear...or whatever origin.These forces when put together must give an accelaration of  towards the centre .their With proper algaebric signs in between them they constitue what is called the 'centripetal force' because they 'produce' centripetal accelaration. [/quote]

First of all, what do you mean by real force?
[/quote]
by that i a mean  a force that has a so called 'source' like garvitational force has mass bodies as source , EM has charged particles as source.
[quote="Thaakisfox"]I think you are mixing centripetal with centrifugal force. centrifugal is force is not a real force it just exists in accelerating frame. [/quote]
No i am not.
[quote="Thaakisfox"]So if what you say is correct than in outer space far from every planet where there is no gravity, EM-force, nuclear force etc. a particle cannot move on circular path? [/quote]
Yes of course if no force is acting on the particle  how can it accelarate.
[quote="Thaakisfox"] if on a horizontal plane, gravity acts only downwards, and there is no EM, then there cannot be circular motion? [/quote]
i don't get that . but see gravity alone can produce centripetal accelaration.Our satellites are the best example

Finally there must be some kind of force that must be acting on a particle to produce centripetal aceelaration centripetal force is a dependent term  it has to have it's  origin in graviatational , EM, nuclear strong ,weak whatever , independently it's meaningless.

[quote="pardesi"]first of all there is no force called centripetal force.There is simply centripetal accelaration.Centripetal force is just a concept which arises due to other forces. [/quote]

[b]pardesi[/b] you are wrong as [b]Thaakisfox[/b] said, please read [url=http://en.wikipedia.org/wiki/Centripetal_force]this[/url] and you will see, perhaps you mean centrifugal force?

[color=red]Edit: This a misunderstanding see below[/color]

[quote="color"][quote="pardesi"]first of all there is no force called centripetal force.There is simply centripetal accelaration.Centripetal force is just a concept which arises due to other forces. [/quote]

[b]pardesi[/b] you are wrong as [b]Thaakisfox[/b] said, please read [url=http://en.wikipedia.org/wiki/Centripetal_force]this[/url] and you will see, perhaps you mean centrifugal force?[/quote]
yes i read that and please read this  an extract from the above passage

 [quote="Wikipedia"]Hence it is a[u] force requirement[/u], [u]not a particular kind of force. Any force (gravitational, electromagnetic, etc.) can act as a centripetal force[/u]. The term centripetal force comes from the Latin words centrum ("center") and petere ("tend towards").[/quote]
so this is exactly as i had explained to Thaakisfox taht centripetal force arises due to other force

[b]Color[/b]
see [b]thaakisfox[/b] had written this

[quote="Thaakisfox"]So if what you say is correct than in outer space far from every planet where there is no gravity, EM-force, nuclear force etc. a particle cannot move on circular path?[/quote]
 so i think  it is clear to you by now who has the misconception.

also u can read my posts above. :wink:

[quote="Thaakisfox"]I dont understand why its not in the same direction[/quote]
the normal force acts outward to the surface and here it is radially outward.
also since $mg$ is in the direction shown by you the component is along radially inward direction

Ok I see, you stressed that it is a label for other forces
like string tension or gravity, but in each of these cases the particular force
which is being labeled centripetal force exists. But centrifugal force is in principle fictional,
that difference is important to remember and I see you know this.
Now the misunderstanding is hopefully gone, and we can focus on the problem  :lol: .

Yes Pardesi is right I had the misconception. :oops:

Got to think about basics too, looks like I got too much carried away to more complicated physics :D

[quote="Thaakisfox"][quote]the normal force acts outward to the surface and here it is radially outward.
also since  is in the direction shown by you the component is along radially inward direction
[/quote]

No the normal force acting on the puck acts inwards, and the normal force acting on the loop acts outwards. :)[/quote]
 no it's the other way round .
now say you have a block on a table .the normal force acts on it outwards to the table and opposite to the direction of gravity otherwise if we go by what you say then there will be  a net force downwards .here by newton's third law strong form we get the normal force avts on the table inwards as seen from block. :wink:
here we are in exactly same situation. normal force has it's origin in EM so here it is a kind of repulsive force due to the close contact between the molecules of the surface

no its the other way around than you say:
how can the normal force acting on the puck be outwards?
Think about it, the loop "wants" to push the puck to stay on the circular path in the upper half too, so if the force exerted by it on the puck was pointing outwards, it would want to do just the opposite, so the force exerted by the loop on the puck throughout the movement must be pointing inwards :wink:

No it is not a question of wanting or not wanting it is just that the normal force s outward normal
[url="click"]http://en.wikipedia.org/wiki/Normal_force[/url]

In order for the mass to make all the circular path, the minimum velocity at the highest point of the circular path (point P) must be such that $N = 0 \Rightarrow F_{c}= P \Rightarrow v_{min}^{2}= gR$. When the mass is abandoned from the height $h$ (point A), by applying conservation of energy (because there is no friction) we get that the velocity with which the mass arrives at P is given by $v^{2}= 2g(h-2r)$. So, in  order for the mass to make the whole circular path we must have

$2(h-2r) \ge r \Rightarrow 2h-4r \ge r \Rightarrow h \ge \frac{5}{2}r$.

As color has made the condition $h > r$, so I will consider $r < h < \frac{5}{2}r$. So the mass will leave contact with the circular surface at a point Q before P, and in that situation we must have $N = 0 \Rightarrow F_{c}= P\cos \theta \Rightarrow v_{Q}^{2}= gr\cos \theta.$ Now let's apply the conservation of mechanical energy between points A and Q:

$mgh = mgr(1+\cos \theta)+\frac{1}{2}mgr\cos \theta \Rightarrow h = r\left(1+\frac{3\cos \theta}{2}\right)$
$\Rightarrow  \cos \theta = \frac{2}{3}\left(\frac{h}{r}-1\right)$.

[quote="yagaron"]for the one that said that there is not centripetal force

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

let's delete this article of wikipedia. and all websites on google.

thing. if there is a acceleratation. using newtons law, there is a force.

[quote]Thaakisfox centripetal force is no real force like gravitational,electromagnetic...etc [/quote]

man, you should learn Einstein., because there is no gravitational force, actually the gravitational force is a inertial force.  :rotfl:
because the geometry of the universe.

dude.. this guy is a spammer? oh god...

my "wrong" solution is equal to the solution of the best school in brazil. wooow... prepotent...

just pretend that we dont see his posts.

and another thing  $N = m g \cos(\theta)$

is easy to see that. people are having trouble with that[/quote]   :spam:  :spam:  :spam:  :diablo:    :bruce:  :starwars:   :furious:(only [b]10[/b] allowed at a time so....)
Well yagaron [b]most[/b] people in Mathlinks-Aops are sensible enough to know which is a spam and which isn't :P
I am really sorry that the[b] best [/b] school in Brazil teaches you such(which i am sure isn't true)

Pardesi, come on man. dont quote from wikipedia just think. If the normal force on the puck would be outwards then the loop wouldnt push the puck on a circular path. The normal force by the puck on the loop looks outwards. :wink:

[b]@Thaakisfox[/b]
Now Say that you need $x$ amount of force to produce a necessary accelaration and if you are given something more tahn that then u surely need to reduce it here the normal force plays that role of 'reducing' the net force.
The normal force is supposed to or 'has' to act away from the surface no matter what the body is undergoing or how it is moving .Normal force is again a self adjusting force it changes it's value with the 'need of time'.

[b]P.S.[/b]: How is that [b]color[/b] makes such beautiful diagram.do you need any software to do so

[quote="Thaakisfox"]Yes exactly
[quote]The normal force is supposed to or 'has' to act away from the surface no matter what the body is undergoing or how it is moving .[/quote]
If its direction is outwards as you say then it will be acting towards the surface, so now you are encountering your own words.
 :wink:[/quote]
ofcourse i won't do that.
 ok u answer this question of mine in what direction does a table exert normal force on the block kept on it and you will get the answer here ahen i said outward i meant i  was sitting on the block and looking above.

But this is not a horizontal block. its a loop :D
So then look at my diagram a few posts ago.  that what you said just before is just what i drew on the diagram.  :wink:

who denies that this is not an loop.but no matter that a force can't change its' 'nature' when acting between two different surfaces. :wink:

So then you agree that the normal acting on the puck points inwards?

well i assumed the block moves above the track and hence i got the normal force upward and if you assumed it downward then youi will get it inward.
this is because the [b]normal force always points away from surfaces[/b] as i have been stressing .

Who said that normal force doesnt point away from surfaces? I didnt.
So we can end the conversation then. It probably was a misunderstanding, since you were referring to the block, and I was to the given problem. But of course both of us know it correctly, since we discussed it now. :D

[hide="solution"]Conservation of energy yields: $mgh=\frac12 mv^{2}+mgr(1+\cos{\theta})$
thus speed of puck determined by $v^{2}=2g(h-r(1+\cos{\theta}))$.
Track exerts a normal force on puck $N$. Total force in radial direction is given by
$-N-mg\cos{\theta}$. RAdial acceleration is $-\frac{v^{2}}{r}$ and we have
$-m\frac{v^{2}}{r}=-N-mg\cos{\theta}$ or $N=m\frac{v^{2}}{r}-mg\cos{\theta}$

we have $v^{2}$ above, so we get
$N=mg\left( \frac{2h}{r}-2-3\cos{\theta}\right)$

Note that $N\geq 0$ and one loses contact when $N$ vanishes.
This occurs at $\theta_{0}=\frac23\left( \frac{h}{r}-1 \right)$
For the puck to reach the top the loop we must have $\boxed{ h\geq \frac{5r}{2}}$[/hide]


function
inequalities
inequalities proposed

Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\plus{}2abc\geq\frac {247}{54}$

Now,I use $ p,q,r$ method
We have $ p\equal{}1,q\in (0,\frac{1}{3}],r\in [\frac{4q\minus{}1}{9},\frac{1}{27}]$
$ f(q,r)\equal{}54\minus{}193q\plus{}108qr\plus{}247r\minus{}108r^2$
Easy to see that function is increasing function
$ \Rightarrow f(q,r)\ge f(q,\frac{4q\minus{}1}{9})\ge 0$
Equality holds for $ a\equal{}0,b\equal{}c\equal{}\frac{1}{2}$ and their permutations

We can prove that
          $ f(a,b,c)\geq f(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2})$

[quote="arqady"][quote="vuthanhtu_hd"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} 2abc\geq\frac {247}{54}$[/quote]
[quote="vo thanh van"]
Equality holds for $ a \equal{} 0,b \equal{} c \equal{} \frac {1}{2}$ and their permutations[/quote]
Are you sure?

The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0$ and $ a \plus{} b \plus{} c \equal{} 1.$
Prove that
\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {27}{2}abc\geq5
\]
[/quote]
Nice, arqady.  :)
The inequality $ r \ge \max \left\{ {0,\frac{{\left( {1 \plus{} q} \right)^2 \left( {1 \minus{} 2q} \right)}}{{27}}} \right\}$ is strong enough for this equality but I don't like it. Can you post your proof?

[quote="arqady"]Honey_S, the Mixing Variables method.  :wink:[/quote]

Denote $ f\left( {a,b,c} \right) \equal{} \sum {\frac {1}{{a \plus{} b}}} \plus{} \frac {{27}}{2}abc \minus{} 5$ then we will prove that
\[ f\left( {a,b,c} \right) \ge f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c} \right)
\]
where $ c \equal{} \min \left\{ {a,b,c} \right\} \Rightarrow c \le \frac {1}{3}$. But we have
\[ f\left( {a,b,c} \right) \minus{} f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c} \right) \equal{} \left( {a \minus{} b} \right)^2 \left[ {\frac {1}{{\left( {1 \plus{} c} \right)\left( {c \plus{} ab} \right)}} \minus{} \frac {{27}}{8}c} \right] \ge 0
\]
since $ 27c\left( {1 \plus{} c} \right)\left( {c \plus{} ab} \right) \le 27.\frac {1}{3}.\left( {1 \plus{} \frac {1}{3}} \right)\left( {\frac {1}{3} \plus{} \frac {1}{4}} \right) \equal{} 7 < 8$
Therefor $ f\left( {a,b,c} \right) \ge f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c} \right)$ and it suffices to prove
\[ f\left( {t,t,c} \right) \ge 0
\]
with $ 2t \plus{} c \equal{} 1$. And it's true since
\[ f\left( {t,t,c} \right) \equal{} \frac {{\left( {1 \minus{} 2t} \right)\left( {1 \minus{} 3t} \right)^2 \left( { \minus{} 3t^2 \plus{} t \plus{} 1} \right)}}{{2t\left( {1 \minus{} t} \right)}} \ge 0
\]
We have done.  :)
And $ {k \equal{} \frac{{27}}{2}}$ is also the best contanst for this inequality.

[quote="Honey_S"][quote="arqady"]Honey_S, the Mixing Variables method.  :wink:[/quote]

Denote $ f\left( {a,b,c} \right) \equal{} \sum {\frac {1}{{a \plus{} b}}} \plus{} \frac {{27}}{2}abc \minus{} 5$ then we will prove that
\[ f\left( {a,b,c} \right) \ge f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c} \right)
\]
where $ c \equal{} \min \left\{ {a,b,c} \right\} \Rightarrow c \le \frac {1}{3}$. But we have
\[ f\left( {a,b,c} \right) \minus{} f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c} \right) \equal{} \left( {a \minus{} b} \right)^2 \left[ {\frac {1}{{\left( {1 \plus{} c} \right)\left( {c \plus{} ab} \right)}} \minus{} \frac {{27}}{8}c} \right] \ge 0
\]
since $ 27c\left( {1 \plus{} c} \right)\left( {c \plus{} ab} \right) \le 27.\frac {1}{3}.\left( {1 \plus{} \frac {1}{3}} \right)\left( {\frac {1}{3} \plus{} \frac {1}{4}} \right) \equal{} 7 < 8$
Therefor $ f\left( {a,b,c} \right) \ge f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c} \right)$ and it suffices to prove
\[ f\left( {t,t,c} \right) \ge 0
\]
with $ 2t \plus{} c \equal{} 1$. And it's true since
\[ f\left( {t,t,c} \right) \equal{} \frac {{\left( {1 \minus{} 2t} \right)\left( {1 \minus{} 3t} \right)^2 \left( { \minus{} 3t^2 \plus{} t \plus{} 1} \right)}}{{2t\left( {1 \minus{} t} \right)}} \ge 0
\]
We have done.  :)
And $ {k \equal{} \frac {{27}}{2}}$ is also the best contanst for this inequality.[/quote]

Thank you !

[quote="arqady"]
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0$ and $ a \plus{} b \plus{} c \equal{} 1.$
Prove that
\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {27}{2}abc\geq5
\][/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $(a+b)(a+c)(b+c)=9uv^2-w^3$, we see that our inequality is equivalent to $f(w^3)\leq0$,
where $f(w^3)=w^6+A(u,v^2)w^3+B(u,v^2)$.
Hence, $f$ gets a maximal value, when $w^3$ gets an extremal value, which happens, when $w^3=0$ or
when two numbers from $\{a,b,c\}$ are equal.
These two cases very easy to check. :wink:


Let $ X_1,X_2,...,X_n$ be (not necessary independent) discrete random variables. Prove that there exist at least $ n^2/2$ pairs
$ (i,j)$ such that \[ H(X_i\plus{}X_j) \geq \frac 13 \min_{1 \leq k \leq n} \{ H(X_k) \},\] where $ H(X)$ denotes the Shannon entropy of $ X$.

[i]GY. Katona[/i]

Consider the graph $G$ with $ n$ vertices, in which $ i$ and $j$ are connected iff we have
$$H(X_i+X_j)< \frac{1}{3} \min_{1\leqslant k \leqslant n}\{H(X_k)\}.$$
We will show that $G$ is a triangle-free graph.
Assume for the sake of contradiction that we can find $3$ vertices $m$, $ n$ and $p$ that form a triangle in $G$.

Notice that
$$H(X_m)=H\big( (X_m+X_n)+(X_m+X_p)-(X_n+X_p)\big) \leqslant H(X_m+X_n)+H(X_n+X_p)+H(X_p+X_m).$$
Therefore
$$\min_{1\leqslant k \leqslant n}\{ X_k \}\leqslant H(X_m)\leqslant 3\max \{H(X_m+X_n),H(X_n+X_p),H(X_p+X_m)\}.$$
which is a contradiction since $m$, $n$, and $p$ are connected.

So $G$ is triangle free and $G$ has no self-loops.
Therefore, by Turan, we know that $G$ has at most $n^2/4$ edges.

Hence the number of pairs we are searching for is at least $n^2-2\cdot(n^2/4)=n^2/2.$


Here are the problems of the Lithuanian TST for IMO 2005:

Problem 1
http://www.mathlinks.ro/Forum/viewtopic.php?p=209614#209614

Find the smallest integer $n$ such that an $n\times n$ square can be partitioned into $40\times 40$ and $49\times 49$ squares, with both types of squares present in the partition, if
a) $40|n$; b) $49|n$; c) $n\in \mathbb N$.

Problem 2
http://www.mathlinks.ro/Forum/viewtopic.php?p=209613#209613

Let $ABCD$ be a convex quadrilateral, and write $\alpha=\angle DAB$; $\beta=\angle ADB$; $\gamma=\angle ACB$; $\delta= \angle DBC$; and $\epsilon=\angle DBA$. Assuming that $\alpha<\pi/2$, $\beta+\gamma=\pi /2$, and $\delta+2\epsilon=\pi$, prove that
\[(DB+BC)^2=AD^2+AC^2\]

Problem 3
http://www.mathlinks.ro/Forum/viewtopic.php?p=209612#209612

The sequence $a_1, a_2,..., a_{2000}$ of real numbers satisfies the condition
\[a_1^3+a_2^3+...+a_n^3=(a_1+a_2+...+a_n)^2\]
for all $n$, $1\leq n \leq 2000$. Prove that every element of the sequence is an integer.

only 3 problems ?
If not, then could you please post the rest of them also.

Thanks

Alas, the problems are only 3. And only 1 TST...


inequalities
geometry
geometric transformation
reflection
algebra
polynomial
inequalities proposed

In Old and new inequalities book, VasC proposed the following nice statement
If $ a,b,c$ are the side lengths of a triangle, then
\[ 3\left( \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{}3\right) \ge 2\left( \frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}\minus{}3\right)\]
Then in Mathematical reflection, Hungkhtn proposed the general
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{}3 \ge k\left( \frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}\minus{}3\right)\]
Now, here is my inspiration from these problems
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2} \minus{}3 \ge k\left( \frac{b^2}{a^2}\plus{}\frac{c^2}{b^2}\plus{}\frac{a^2}{c^2}\minus{}3\right)\]
Regards
:)

[quote="can_hang2007"]In Old and new inequalities book, VasC proposed the following nice statement
If $ a,b,c$ are the side lengths of a triangle, then
\[ 3\left( \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\right) \ge 2\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \minus{} 3\right)
\]
Then in Mathematical reflection, Hungkhtn proposed the general
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3 \ge k\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \minus{} 3\right)
\]
Now, here is my inspiration from these problems
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \minus{} 3 \ge k\left( \frac {b^2}{a^2} \plus{} \frac {c^2}{b^2} \plus{} \frac {a^2}{c^2} \minus{} 3\right)
\]
Regards
:)[/quote]
No one like this? :( Arqady, where are you?

The largest value of $ k$ is $ \frac 1{2}$. This is a known inequality (inequality of Walker, 1970), which can be obtained from

$ (ya^2\plus{}zb^2\plus{}xc^2)(za^2\plus{}xb^2\plus{}yc^2) \ge (xy\plus{}yz\plus{}zx)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$,

for $ x\equal{}\frac 1{c^2}$, $ y\equal{}\frac 1{a^2}$, $ z\equal{}\frac 1{b^2}$ (see Algebraic Inequalities, page 109).

[quote="Markelov"][quote="can_hang2007"]
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3 \ge k\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \minus{} 3\right)
\]
[/quote]

Hello [b]Can Hang[/b], hello [b]Vasc[/b], hello [b]forum[/b]!

I want to advance a hypothesis. I expect that
\[ k_{max} \equal{} \frac {1}{46}\left(43 \plus{} 4\sqrt {2} \minus{} \sqrt {344\sqrt {2} \minus{} 235}\,\,\right)
\]
This $ k$ is the root of the polynomial $ 23k^4 \minus{} 86k^3 \plus{} 125k^2 \minus{} 86k \plus{} 23$ and is approximately $ 0.713$. I expect there is an identity for this $ k$ simular to the identity for the original Vasc problem ([url=http://www.mathlinks.ro/viewtopic.php?p=1167938#1167938]link[/url]). But this is a hypothesis only, no identity (at least now). Maybe you can prove or disprove this using some other method?

Thank you!
Sergey Markelov
markelov@mccme.ru[/quote]

You can try with $ a\equal{}1,b\equal{}3$ and $ c \to 5$.

[quote="Honey_S"][quote="Markelov"][quote="can_hang2007"]
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3 \ge k\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \minus{} 3\right)
\]
[/quote]

Hello [b]Can Hang[/b], hello [b]Vasc[/b], hello [b]forum[/b]!

I want to advance a hypothesis. I expect that
\[ k_{max} \equal{} \frac {1}{46}\left(43 \plus{} 4\sqrt {2} \minus{} \sqrt {344\sqrt {2} \minus{} 235}\,\,\right)
\]
This $ k$ is the root of the polynomial $ 23k^4 \minus{} 86k^3 \plus{} 125k^2 \minus{} 86k \plus{} 23$ and is approximately $ 0.713$. I expect there is an identity for this $ k$ simular to the identity for the original Vasc problem ([url=http://www.mathlinks.ro/viewtopic.php?p=1167938#1167938]link[/url]). But this is a hypothesis only, no identity (at least now). Maybe you can prove or disprove this using some other method?

Thank you!
Sergey Markelov
markelov@mccme.ru[/quote]

You can try with $ a \equal{} 1,b \equal{} 3$ and $ c \to 5$.[/quote]
But $ a,b,c$ are lengths of a triangle.

[quote="y-a-s-s-i-n-e"]hi !!

where is that book can you give me athe link of  Algebraic Inequalities  please?[/quote]

[img]http://books.gil.ro/covers/AIONM.jpg[/img]

Vasile Cîrtoaje, "Algebraic Inequalities - Old and New Methods", Gil publishing house: Zalau 2006.

Here are 3 links for details:

http://www.mathlinks.ro/viewtopic.php?t=100399
http://www.librarie.net/carte.php?nia=376&id=57823
http://www.cip.ifi.lmu.de/~grinberg/faq.html#ineqsbooks

Hope some day this book will be available online... I see no reasonable way to buy from Romania to Russia (they ship using DHL, cost of shipping to Russia is 6 times more then the book price), so I do not have this book in my library :-(. Probably this is a good book - a lot of good guys recommend it.

Sergey Markelov
markelov@mccme.ru

I'm so sorry Markelov.
\[ k_{m{\rm ax}}  \equal{} \frac{1}{{46}}\left( {43 \plus{} 4\sqrt 2  \minus{} \sqrt {344\sqrt 2  \minus{} 235} } \right)\]

You are right, Sergey.
$ k_{max} \equal{} f\left(\frac {1 \plus{} \sqrt2 \minus{} \sqrt {2\sqrt2 \minus{} 1}}{2}\right),$ where $ f(x) \equal{} \frac {x^3 \minus{} x \plus{} 1}{x^3 \minus{} x^2 \plus{} 1}.$

[quote="Vasc"]The largest value of $ k$ is $ \frac 1{2}$. This is a known inequality (inequality of Walker, 1970), which can be obtained from

$ (ya^2 \plus{} zb^2 \plus{} xc^2)(za^2 \plus{} xb^2 \plus{} yc^2) \ge (xy \plus{} yz \plus{} zx)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$,

for $ x \equal{} \frac 1{c^2}$, $ y \equal{} \frac 1{a^2}$, $ z \equal{} \frac 1{b^2}$ (see Algebraic Inequalities, page 109).[/quote]
Your solution in the book is very nice, my idol. But I would like to see another solution, since my solution is much different... :)
[quote="Markelov"][quote="can_hang2007"]
Determine the largest constant such that the following inequality is holds for all $ a,b,c$ be the side lengths of a triangle
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3 \ge k\left( \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \minus{} 3\right)
\]
[/quote]

Hello [b]Can Hang[/b], hello [b]Vasc[/b], hello [b]forum[/b]!

I want to advance a hypothesis. I expect that
\[ k_{max} \equal{} \frac {1}{46}\left(43 \plus{} 4\sqrt {2} \minus{} \sqrt {344\sqrt {2} \minus{} 235}\,\,\right)
\]
This $ k$ is the root of the polynomial $ 23k^4 \minus{} 86k^3 \plus{} 125k^2 \minus{} 86k \plus{} 23$ and is approximately $ 0.713$. I expect there is an identity for this $ k$ simular to the identity for the original Vasc problem ([url=http://www.mathlinks.ro/viewtopic.php?p=1167938#1167938]link[/url]). But this is a hypothesis only, no identity (at least now). Maybe you can prove or disprove this using some other method?

Thank you!
Sergey Markelov
markelov@mccme.ru[/quote]
Your result is correct, Markelove. And what is your solution? :)
Thanks, I used the same techniqus for both two problems. :)


geometry
incenter
circumcircle
vector
geometry proposed

Let $ ABCD$ be quadrilateral such that $ DC \equal{} DB \equal{} CA$.$ AC$ and $ BD$ meet at $ Q$. $ I$ be incenter and $ O$ be circumcenter of triangle $ DQC$

Show that $ OI\perp AB.$

[hide="hint"]
we can use vectors to solve it,
or we can use similarity between $ OIL$ and $ ABD$ where $ L$ is the midpoint of arc $ CQD$
and finally (for the ugly method) we can try to prove $ IA^2\minus{}IB^2\equal{}OA^2\minus{}OB^2$. :)
[/hide]

[quote="shoki"][hide="hint"]
we can use vectors to solve it,
or we can use similarity between $ OIL$ and $ ABD$ where $ L$ is the midpoint of arc $ CQD$
and finally (for the ugly method) we can try to prove $ IA^2 \minus{} IB^2 \equal{} OA^2 \minus{} OB^2$. :)
[/hide][/quote]

Could you please give me a link because i don't know the source

[quote="shoki"]me,too , 'don't know the source;(and if i had known the link i would have told u!) :)[/quote]
this is result of USA TST 2006, Problem 6

[hide]http://www.mathlinks.ro/viewtopic.php?p=841255#841255[/hide]


function
real analysis
real analysis unsolved

¿Do there exist functions f and g such that f(x)= O(g(x)) and g(x)=O(f(x))?

Maybe: $ f(x) \equal{} g(x) \equal{} x^2$ because $ x^2$ belongs to the complexity class $ x^2$. I think this is true for any power of $ x$ and maybe for many functions more. I have only an intuitive definition of big $ O$ so I may be saying nonsense.

The formal definition of $ f(x)\in O(g(x))$ is:

$ f(x)\mbox{ is }O(g(x))\mbox{ as }x\to\infty$ if and only if $ \exists \;x_0,\exists \;M > 0\mbox{ such that } |f(x)| \le \; M |g(x)|\mbox{ for }x > x_0.$

So I think that if you take $ f(x) \equal{} x^2 \minus{} x$ and $ g(x) \equal{} x^2$, is clear that for $ x_0 \equal{} 0$, $ M \equal{} 1$ $ \Rightarrow$ $ f(x)\in O(g(x))$ on the other hand, for $ x_0 \equal{} 2$, $ M \equal{} 2$ ($ 2x^2 \minus{} 2x\geq x^2$ from $ x \equal{} 2$ $ x\geq 0$) we get $ g(x)\in O(f(x))$.

Well i think the notation depends on the book youre reading. many authors use "equalities"  with the O'S.  Nevertheless the meaning is still the same which is the definition in your last reply. EA!


[quote="4everwise"]Is there a fast way to factor polynomial equations such as $x^4+4x^3-x^2-20$?[/quote]

Try syntetic division or you could replace $x^2$ with $x$, thus replacing the others too....and then just square root your answer.

[quote="chess64"][quote="4everwise"]Is there a fast way to factor polynomial equations such as $x^4+4x^3-x^2-20$?[/quote]

No  :D

Especially since 2 of its roots aren't real, one is $2$, and one is $-3.923791294...$[/quote]

I think he meant any equations with higher degrees than $2$ though  ;) He just made a random example on the top of his head, anyway, like I said, synthetic division or substitution for a lower degree of $x$

wasn't it proven that there is no efficient method of factoring all polynomials but only those especially designed polynomials? i mean the only tools we have are synthetic division and that root principle to find rational roots...

Yeah, it depends on the polynomial, for example,

\[f(x)=x^4+x^3+x^2+x\]

Can be factored easily:

$f(x)=x^3(x+1)+x(x+1)=(x^3+x)(x+1)$
$f(x)=(x^3+x)(x+1)=x(x^2+1)(x+1)$

Cant you do it by remainder factor theorems? which is the easiest thing

there isnt really an 'efficient' way. the best ways are the theorems learned in algebra 2: synthetic division, decartes rule of signs, multiplicity properties, etc. i happen to favor my TI89  :D but uh... i become too reliant on it sometimes

[quote="Palytoxin"]are you sure? because if i expand the polynomial back, i get $f(x)=x^4+x^3+x+1\neq x^4+x^3+x^2+x$[/quote]

Fixed, I always get messed up when I type math on the computer without doing it on paper first  :blush:

There are methods for finding rational roots, but there is a general way to solve the quartic. If you look on http://mathworld.wolfram.com/QuarticEquation.html the four solutions $z_1,z_2,z_3,z_4$ to the equation $z^4+a_3z^3+a_2z^2+a_1z+a_0=0$ are given by equations (36) through (39), listed on the right, in terms of the coefficients and the numbers $R,D,E$ which are defined by (40) through (42). If you have some time on your hands you could apparently read how the general quartic is solved start to finish, which would be very interesting. It looks pretty intimidating to me though. :? Once you go past degree $4$ there's no general solution by extracting roots and such. This was first proven (incompletely) by a possible ancestor of mine, Paolo Ruffini.  :D

:lol:  i prefer to use rational root principle instead of memorizing that quartic formula, which is less practical than rational root and synthetic division. ok im going to sleep  :sleeping:

Two more common ways are Descartes' Rule of Signs and the Rational Root Theorem.

The first states that the maximum number of positive real roots in $f(x)$ is the number of sign changes between successive coefficients in $f(x)$ and the maximum number of negative real roots is the number of sign changes in $f(-x)$.
For example, say we want to chase after the roots of $p(x)=5x^5-4x^4+2x^3+6x^2+3x-1$.
The number of sign changes in $p(x)$ is three ($5\rightarrow-4, -4\rightarrow2, 3\rightarrow-1$), so there is a maximum of three positive real roots.
The number of sign changes in $p(-x)$ is two ($-2\rightarrow6, 6\rightarrow-3$), so there is a maximum of two negative real roots.
More information: http://mathworld.wolfram.com/DescartesSignRule.html

The rational roots theorem says that for a polynomial in the form $a_nx^n+a_{n-1}x^{n-1}+\ldots+a_2x^2+a_1x+a_0$, all the rational roots can be expressed as a divisor of $a_0$ divided by a divisor of $a_n$.
You can find out more about it here: http://mathworld.wolfram.com/RationalZeroTheorem.html

I think AoPS volume 2 mentions some other guidelines (such as finding upper and lower bounds) in Chapter 6, but although useful, I don't use them as frequently as the two mentioned above, so I can't explain them very well.  :blush:

Hope that helped though.


For a positive integer n, let $w(n)$ denote the number of distinct prime
divisors of n. Determine the least positive integer k such that
$2^{w(n)} \leq k \sqrt[4]{n}$
for all positive integers n.

Let $(p_n)$ the sequence of prime numbers.
- If $w(n) > 6$ then $n \geq 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17^{w(n)-6} \geq  2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 16^{w(n)-6}$
- If $w(n)=m\in[1,6]$ then $n \geq \prod_{1 \leq j \leq m}p_j \geq \prod_{1 \leq j \leq m}p_j \times \prod_{m< j \leq 6}\frac{p_j}{16}=2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 16^{w(n)-6}$
So in both cases, $16^{w(n)} \leq \frac{16^6}{2 \times 3 \times 5 \times 7 \times 11 \times 13}n$ $\Rightarrow 2^{w(n)} \leq 5 \sqrt[4]{n}$.
If $n=2 \times 3 \times 5 \times 7 \times 11 \times 13$, then $\frac{2^{w(n)}}{\sqrt[4]{n}}>4$ so $k=5$.

Oh...we used this one for our presTST (qualifying round) this year. :)

Pierre.


geometry
3D geometry
search
algorithm
videos

so i got my cube...

i was trying to solve the cube with the solutions floating in google, but i couldnt...

does someone know the easiest way - so that a complete beginner could do it - to solve the rubik's cube?

Thanks.  :)

that was exactly what i was doing to solve the cube...

but i didnt think that method is easy...

could you explain specifically how to do it?

Here is a basic method for cube solving:

http://www.alchemistmatt.com/cube/Denny3x3/index.htm

How far are you and what do you know already? I recommend first learning how to solve it by Dan Brown's method (search Youtube).

I learned how to solve it through the free solution hints booklet. :)

There are very few algorithms, and the longest is only $ 8$ turns...

If you don't have the booklet I can post it here for you. :)

THE beastliest way to learn to solve the cube is to do it yourself without any kind of help because that is the way i got faster. I first used to do it by the basic layer-by-layer method (if you want to learn this, go to Thrawst on Youtube, he has a lot of videos about this thing) then I realized that it was extremely slow so, i started making my own method. If I had done this before, it would have been less of a hassle cause i kept relating the layer-by-layer technique to my new faster technique and kept skrewing up.

hm dont have time to make my own way :P

oh and ur right, i take like 2 mins by layer

wth 2 mins by layer??

omg it took me like 5 minutes to solve it by layers...

ok now i will look up dan brown's method

u get better with time

but i bet 2 mins is a lot on aops, btw

Yes it is ernie :wink:

When i learned, I could get 38 w/ the booklet method.

The [url=http://www.rubiks.com/solving-center/3x3_guide/]solution[/url] right off of rubiks.com


vector
inequalities
induction
number theory open
number theory

Find all integer solutions $ (a_1,a_2,...,a_n)$.

[quote]$ a_1^2 - 1 = a_2$
$ a_2^2 - 2 = a_3$
:whistling:..so on..:whistling:
$ a_n^2 - n = a_1$ [/quote]

My original problem(Not unsolved).

Obviosly, there are not integer solutions if $ n\equiv 1,2\mod 4$.
If $ n\equiv 0,3\mod 4$ may be solutions. For example $ n\equal{}4,a_1\equal{}0,a_2\equal{}\minus{}1,a_3\equal{}\minus{}1,a_4\equal{}\minus{}2$.

[quote="Rust"]$ n \equal{} 4,a_1 \equal{} 0,a_2 \equal{} \minus{} 1,a_3 \equal{} \minus{} 1,a_4 \equal{} \minus{} 2$.[/quote]

There are more solutions.

Case $ n\equal{}2$ yields by substitution and check of divisors $ a_1\in\{\pm 1\}$, which does not provide a solution

Case $ n\equal{}3$ yields by substitution and check fo divisors $ a_1\in\{\pm 1,\pm 2\}$ and $ a_1\equal{}1$ determines the only solution, i.e. the solution vector is $ (1,0,\minus{}2)$

Case $ n>3$

First we will prove the following lemma (actually, we need only parts of it  :blush: )

Lemma: If for some $ i\geq 3$ the inequality $ |a_i|\geq i$ or $ i\equal{}2, |a_2|>2$ holds, then

(i) $ \forall k\geq i: a_k>k$ and $ a_1>n\plus{}1$ and

(ii) $ a_i<a_{i\plus{}1}<\ldots<a_n<a_1$

Proof of (i): Note, that $ a_{i\plus{}1}\minus{}(i\plus{}1)\equal{}a_i^2\minus{}(2i\plus{}1)\geq (i\minus{}1)^2\minus{}2>0$ holds and now apply induction (making the identification $ a_{n\plus{}1}\equal{}a_1$ at the end). The case $ i\equal{}2,|a_2|>2$ yields $ a_3>3$.

Proof of (ii): Direct calculation yields $ a_{i\plus{}1}\minus{}a_i\equal{}\frac{1}{4}\left[(2a_i\minus{}1)^2\minus{}(1\plus{}4i)\right]\geq\frac{1}{4}\left[(2i\minus{}1)^2\minus{}(1\plus{}4i)\right]\equal{}i(i\minus{}2)$. Now apply (i) and note that for $ i\equal{}2,|a_2|>2$ we have $ a_3>3$.

We can assume that the conditions of the lemma are not fulfilled, since otherwise $ a_1>n\plus{}1$ and therefore $ a_2>\max\{2,a_1\}$ will hold, yielding the contradiction $ a_1<a_2<\ldots <a_n<a_1$. In particular, we can restrict to $ |a_2|\leq 2$.

Case $ a_2\equal{}0$ direct calculation yields $ a_1\equal{}\pm 1$ and for the following coefficients we have $ a_3\equal{}\minus{}2,a_4\equal{}1,a_5\equal{}\minus{}3,a_6\equal{}4,a_7\equal{}10>7$. This gives $ n\equal{}3$, which has already been considered before.

Case $ |a_2|\equal{}1$ is only possible for $ a_1\equal{}0,a_2\equal{}\minus{}1$ and the following coefficients are $ a_3\equal{}\minus{}1,a_4\equal{}\minus{}2,a_5\equal{}0,a_6\equal{}\minus{}5,a_7\equal{}19>7$. This gives $ n\equal{}4$ with the unique solution vector $ (0,\minus{}1,\minus{}1,\minus{}2)$.

Case $ |a_2|\equal{}2$ does not allow any integer solution for $ a_1$

[quote="hjbrasch"]the solution vector is $ (1,0, \minus{} 2)$

This gives $ n \equal{} 4$ with the unique solution vector $ (0, \minus{} 1, \minus{} 1, \minus{} 2)$.
[/quote]

Great! :P


Tickets to a concert cost $ \$16$ for adults and $ \$13$ for children. A group of adults and children spent $ \$154$ for its tickets. How many adults were in the group?

We have $ 16x\plus{}13y\equal{}154$, where $ x$ is the number of adults and $ y$ is the amount of children.  If we substitute values for $ y$ into it, we find that $ y\equal{}2$, because $ 154\minus{}26\equal{}128\equal{}8 \times 16$.  Thus, $ x\equal{}\boxed{8}$.

Couldnt one take this mod13?

EDIT: Another occurence of 154!


Hope u don't count INMO selected students as olympians.  :lol:

this is of the form $ a^4 \plus{} 4b^4$ . Use the Sophie German identity( or whatever it's called!!) So they are not prime.

obviously anyone who knew the identity n wrote inmo are olympians  :D
Yes that's right the Sophie Germain identity.

[u][b]The identity:[/b][/u]
$ a^4 \plus{} 4 \cdot b^4$  can be factorised as : $ a^4 \plus{} 4 \cdot b^4 \plus{} 4 \cdot a^2 \cdot b^2 \minus{} 4 \cdot a^2 \cdot b^2 \implies (a^2 \plus{} 2 \cdot b^2)^2 \minus{} (2 \cdot a \cdot b)^2 \implies$ it can be factorised as its of the form $ a^2 \minus{} b^2$   :)

INMO awardees are those who qualified the INMO and reached the camp.  I didn't.  :D

ok but still that doesnt make you a non olympian  :D


Are we allowed to talk about the first round mandelbrot questions for this year?

i got a 7 score total ouf of like 14 on the last mandelbrot which is horrible. I am taking geometry only though


#11.

How would you go about graphing $ |x| \plus{} |y| \le \sqrt {2(x^2 \plus{} y^2)} \le 2 \times Max(|x|,|y|)$?

#21

How do you find the real solutions of the inequality $ |x \minus{} 1| \plus{} |x \plus{} 2| < 5$?
The authors of the contest problem book solved it with some sort of line graph but I don't understand how.

11: I think that's always true. The first inequality is RMS-AM (or just square and simplify) and the second is like the triangle inequality or something.

aznlord1337, could you explain the cases to me?

SamE, the test says that the inequality I typed above can represented by a circle with a circumscribed and inscribed square centered at the origin. I'm guessing that the graphic representation of the inequality as they have it written is when equality holds (?) It still doesn't make sense to me...

[quote="aufha"]How do you find the real solutions of the inequality $ |x \minus{} 1| \plus{} |x \plus{} 2| < 5$?
[/quote]

There are three cases:

CASE 1:
$ x < \minus{} 2\implies x \minus{} 1 < 0,\ x \plus{} 2 < 0$
$ \implies |x \minus{} 1| \plus{} |x \plus{} 2| \equal{} (1 \minus{} x) \plus{} ( \minus{} x \minus{} 2) \equal{} \minus{} 2x \minus{} 1 < 5$
$ \implies x > \minus{} 3$.

CASE 2:
$ \minus{} 2\leq x < 1\implies x \minus{} 1 < 0,\ x \plus{} 2\geq0$
$ \implies |x \minus{} 1| \plus{} |x \plus{} 2| \equal{} (1 \minus{} x) \plus{} (x \plus{} 2) \equal{} 3 < 5$, which is always true, so all such $ x$ in CASE 2 satisfy the inequality.

CASE 3:
$ x\geq 1\implies x \minus{} 1\geq 0,\ x \plus{} 2 > 0$
$ \implies |x \minus{} 1| \plus{} |x \plus{} 2| \equal{} (x \minus{} 1) \plus{} (x \plus{} 2) \equal{} 2x \plus{} 1 < 5$
$ \implies x < 2$.

Putting this together, we find the solution set to be $ \boxed{( \minus{} 3,2)}$.

I think what they want you to do is graph these: $ c\equal{}|x|\plus{}|y|$, $ c\equal{}\sqrt{x^2\plus{}y^2}$ and $ c\equal{}2\max(|x|,|y|)$, where $ c$ is a constant. At least, that would give you what they want (the equality case is where they all touch).

The inequalities hold everywhere because the first square is inscribed in the circle, which is inscribed in the second square.


inequalities
geometry
3D geometry
sphere
triangle inequality
real analysis
real analysis solved

The p-norm is defined as
$\| x \|_{p} = \Big ( \sum_{i} |x_{i}|^{p} \Big )^{1/p}$

How do I show that $p$ has to be larger or equal to one? Apperently I have to show that the triangle inequlity won't hold for $p < 1$. I might be able to do it using the convexity property together with a closer look at $|t|^p$. Can you help me?

Thanks!

Mike

You're trying to disprove the triangle inequality, so all you need is a single counterexample. Try this:

Let $x=(1,0,0,\dots)$ and $y=(0,1,0,\dots).$ Then $||x||_p=||y||_p=1$ but $||x+y||_p=2^{1/p}.$ Then a necessary condition for the triangle inequality to be true is that $2^{1/p}\le2$, which means that $p\ge 1.$

If $0<p<1,\,d(x,y)=||x-y||_p^p$ is a metric, but since it doesn't scale correctly under scalar multiplication, it's not a norm.

there are many ways to prove that,that is one method(graphically):
we know that every 1-sphere(sphere in p-normed space with radius=1) in normed space is a convex set.
So if we'll see a 1-sphere in normed space with p-norm we'll  have :
for p >= 1 the sphere will be really a convex set
but when p<1 then the sphere will not be a convex
you can try it with any mathprogram , I watch this in Derive


The most number of bags is [hide]nineteen.  We can get this by using triangular numbers.  The formula for the nth triangular number is n(n+1)/2.  The nth triangular number is the sum of all numbers 1 to n.  Since each bag has a different number, to maxamize the number of bags, we use the triangle numbers.  The 19th triangular number is 190, therefore there are 19 bags.[/hide]

While I agree with smiley's answer and it's validity, I would argue that the most number of bags could also be ...

[hide]

slightly larger than 19.  The problem does not state that bags contain only marbles.  Therefore, it is possible that bags of marbles can be placed in other bags, so long as the rule, "no two bags may contain the same number of marbles", is preserved.

One solution I found was 34.  I have not yet proven that 34 is the largest number of bags, but it should provide someone a launching point for a proof.

Start with 19 bags, with the nth bag containing n marbles.  Then, place the bags inside other bags in such a way that the sum of the marbles in the exterior bag is unique.  I came up with 8 more bags (marble counts of 20-26 & 29) by combining 1 + 19, 3 + 18, 5 + 17, etc.  Then repeat by adding those bags (marble counts of 41,45,49,55,86,104,190) until there is only one bag containing all the marbles.

Perhaps a less symmetrical approach would yield a greater "maximum".

This alternative thinking may provide a different answer than the expected one.  While the answer may not be any MORE correct, with a well-formed proof, it should not be considered any LESS correct, and could possibly cause everyone else's answer to be less correct for lack of completeness (if they didn't state their assumption that bags only contained marbles).[/hide]

I welcome and encourage comments on this line of thinking.

bob123 wrote:The answer is [hide]190 because you can put bags in each other. Put a marble in the first bag and put it in the second bag, put a marble in the second bag and put it in the third bag... and so on, meaning you could use 190 bags.[/hide]

Nice work!

Wasn't this a MathCounts question a while back?

I remember this from when I was training for States...

I was trying to figure out

[hide]

how I could do what you did, MCrawford, but I got stuck in thinking I would put 1 marble in each bag (which violates the rule) and then combining those.  Great work![/hide]

I can't imagine how 189 bags and 190 marbles can fit in one bag... but in the world of mathematics, I guess that's OK.

If I were to dispose of 190 marbles and 189 ziplog bags (for some reason), I could probably do so in a kitchen-sized garbage bag.

The wonderful thing about the world of mathematics is that if you can't prove it is false, assume it is true unless proven otherwise!  If you don't believe this, provide a counterexample...

[quote="MCrawford"]What is the greatest number of bags that can be used to hold 190 marbles if each bag must contain at least one marble, but no two bags may contain the same number of marbles?[/quote]

Is this actually a MATHCOUNTS problem?  If so, can somebody please provide a reference.  I don't recall seeing any "trick" questions from MATHCOUNTS, since I've been coaching.

I just pull the questions out of old tests and don't document which ones I use.  It actually didn't occur to me this was a trick question until I saw the responses.  It's possible that it wasn't even meant that way.

Thanks Mathew.  Perhaps someone else will run across the problem, and post a reference.

I certainly hand it to JGouldie and bob123 for thinking outside the box.  If I were their coach, and they put 190 as the answer on a competition, I would forcefully argue that 190 is a correct answer, given the wording of the problem.  But, as I inferred in my previous post, I don't think MATHCOUNTS would intentionally use a trick question in their competitions.

Yeah, out of the hundreds and hundreds of MATHCOUNTS problems I worked in middle school, I don't recall a trick question either.

I heard that the original answer to this problem was 19, but when one mathlete refuted the problem, both the answers of 19 and 190 were accepted.

That's a very rare mistake on the part of MATHCOUNTS.  They are typically better at weeding out problems like that than most competitions.

That problem was at a National competition one year and a student filed a protest that 19 was the correct answer and he was given credit.  If you have the "Greatest MATHCOUNTS problems" bok that MATHCOUNTS sells, i think they explain the situation there.  At least that's where I think I read about it.


Okay, I found this problem that I've tried REALLY hard on. So it's Sir Gallant was to cross a moat 15 loogas wide. No people are to become wet, no tools are allowed, and so on and so forth, only the boards are allowed and every person must get over. There are two boards, both which are 14.8 loogas.
I think that you would somehow make a triangle, but I'm not very sure...  :blush:
Thanks!

Hmm...I don't think you're supposed to make a triangle though because that would make the board the hypotenuse which would require it to be even longer.

Who says you can't attach two boards together? :P

Also, the triangle methods don't work, as, if one board can't reach across, then a triangle surely won't (unless, of course, you extend the boards).

So you mean something like this? Or?...
Why doesn't the triangle method work? Can you prove it or?...
You're not allowed to attach them with glue or any thing of that sort. It's just a restriction for the problem.

Howabout he vaults over? :D   He then throws it back, and then everyone repeats the process!  :D  :D

Also, you can't cross the river by jumping, or any thing like that... you have to be walking/running/jogging ect. on the boards... You have to use all of the boards if I didn't mention it... >.<

As eli140 said, if one of the boards is too short to bridge the gap, then a triangle certainly is too short. I don't think I see a good way if you're not allowed to attach boards.

Wait, why would it be too small?
Wouldn't it just be an obtuse triangle instead of a right triangle? And couldn't you have the boards not exactly touching on the tips but like what I drew? I mean, it's still a triangle...

Place both boards parallel to each other.  Walk over it until you reach the end of the board.  Since it is a moat, there will be crocodiles, and you wait for one to make its head above water, and then you jump on it across. :P

Their in ruins... I don't think there will be any crocodiles... >.>

Uh.. I'm pretty sure this isn't possible! Is there an answer?  :o

Yeah, there should be an answer... The answer is going to be given on Friday... Not that I know the answer....

Ah, I saw a similar problem before. The problem I saw had a circle around a square. The solution was to go to the corner of the square and just one of the boards will reach across.

Well the width is 15 loogas, using the 45, 45, 90 triangle, that means that the diagonal is 15 square root of 2. Which is longer than 14 loogas so that won't work...

Well, is there someone on the other side so they can place on board from the other side of the moat?

There is no one on the other side. NO ONE. Also, both boards must be used.
Wait, could you use the diagonal? I know in the last one I mentioned you couldn't but what if you used the two boards as a triangle, would it be possible?
Wait, how does that work? You can't glue them together...

Wait, how does that work? You can't use anything to stick them together...

Who said they were sticking?  Just keep them as close as possible...Wait, you should be in school right now, right?

EDIT: How wide is each plank?

Yes, but they would fall into the water, and you can't have the boards touch the water.
No, I get home at 3:10.

Ok. Now the only way I can think of it is 007MATH's way. Cause it doesn't say how steep it is. And I don't think it's possible. Maybe the no solution is the answer.  :maybe:

Ugh I just know the person who posed this problem is from my math class ;) (because my teacher gave the same problem). Okay. So here is the answer (I'm pretty sure about it but I might be wrong). (assume the planks are line segments ;))

Put one of the boards across a corner such that it forms a $ 45 \minus{} 45 \minus{} 90$ right triangle. Since the hypotenuse (the board) has length $ \sqrt{2}$ times the leg, we can do $ \frac{14.8}{\sqrt{2}}$ which is approximately $ 10.5$. (This might be a little hard to visualize so I will supply a picture).

 [geogebra]4150051b8751b0e6d17e6df37d1f8cba8fb0e212[/geogebra]

(Sorry that I'm so bad at geogebra and asymptote  :oops: and you need java to view it. Assume the lines I've drawn are parallel).

Okay so (hopefully) you can see that the right triangle that looks like a $ 45 \minus{} 45 \minus{} 90$ triangle has leg length of about $ 10.5$ (all you need is an estimate for this problem).

Now we are left with proving that the 2nd plank can reach the corner of the island from the midpoint of the 1st plank. First we need to find the length of the "hypotenuse" of that right triangle.

The longest leg on the very right of the "busy" part of the diagram is $ 15$. Since the leg of the $ 45 \minus{} 45 \minus{} 90$ triangle is $ 10.5$, what is left in between is $ 4.5$. So the "hypotenuse" of the triangle involving the midpoint of the 1st plank and the corner, is $ \sqrt{{4.5}^2 \plus{} {15}^2} \equal{} \sqrt{245.25}$.

Phew! We are almost done now :).

So what is left to prove is that $ {7.4}^2 \plus{} {14.8}^2 \ge 245.25$. Simplifying this, we get $ 54.76 \plus{} 219.04 \ge 245.25$ which is true, AND WE ARE DONE!!

So the two logs are: One of the legs of the 45-45-90, and the other stick that holds the other leg of the 45-45-90?

uh yes... the 2nd log is the one on the midpoint of the first log

Wait. So there's a corner? Hm. Well it didn't say there wasn't! :D Smart kid there! And don't worry. I don't even know how to work Geogebra...

OH HAHA I JUST FOUND A REALLY SIMPLE PROOF  :D  :D  which really fails because I didn't find it when I posted the solution :rotfl:

 [geogebra]b22170ccb1e0df46bc5f4f80dd7efb58f346be06[/geogebra]

The 45-45-90 triangle has hypotenuse of $ 14.8 \text{"loogas"}$ so median from the hypotenuse is $ \frac{1}{2} \times 14.8 \equal{} 7.4$. That means the combined spanning length of the two boards is $ 7.4 \plus{} 14.8 \equal{} 22.2$. The diagonal from the outside corner to the inside corner is $ 15\sqrt{2}$. We just need to make sure that $ 22.2 \ge 15\sqrt{2}$. Since $ \sqrt{2} \approx 1.414$, we get $ 22.2 \ge 15 \times 1.414 \equal{} 21.21$ which is true. :D

You'll agree that this is much simpler than my first solution :)

Perhaps you could do the same thing just on the curve of the moat (depending on how big/small the moat is)

Well, that would depend on the circumfrence of each circle. The inner and outer "land" I mean...

sigh, it would be [b]impossible[/b] if the moat was circular. we are given the moat is rectangular

There is nothing in the statement of the problem as to what the shape of the moat is. So it is possible that it could be circular as well as rectangular. joyofpi
is correct, though. It would depend on the circumference of the circles. If the circumferences are too large the boards may not be adequate.


[color=olive]Show that every integer greater than 11 is the sum of two composite integers.[/color]

Do I need to use induction for this question?  I've tried, but it's not working.  Here's my work so far:

Let P(n): n=a+b for n>11, where a and b are composite integers.

Initial step: P(12) says that 12 is the sum of two composite integers because 12=4+8=6+6 where 4, 6 and 8 are all composite integers.

Induction step: Let k>11 and suppose P(k) is true and k=a+b.
We want to show that P(k+1) is true.
k+1=a+b+1

This is what i've tried, but actually I'm not sure if I'm on the right track.  Just give me some suggestions.  Thanks very much !!!:)

I don't think that induction will work here. Try considering the seperate cases for when n is even and when it is odd, remembering that all even numbers greater than 2 are composite.

[hide="Solution"]If n is even then we have 4 and n-4 are composite and sum to n.
If n is odd we have 9 and n-9 are composite and sum to n[/hide]


AMC USA(J)MO USAMO geometry inradius inequalities

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

We use the standard technique for dealing with problems involving incircles. We let the incircle be tangent to the triangle at points which split the sides into segments of length $x$ and $y$, $x$ and $z$, and $y$ and $z$, such that the sides of the triangle have lengths $x+y$, $x+z$, $y+z$.

Now we have right triangles with angles of $A/2$, $B/2$, $C/2$, and with legs $x$ and $r$, $y$ and $r$, and $z$ and $r$.

Also note that $s = (2x+2x+2z)/2 = x+y+z$.

We rewrite the equation

$\left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2$

as

$\dfrac{z}{r}^2 + 4\dfrac{y}{r}^2 + 9\dfrac{x}{r}^2 = \left(\dfrac{6(x+y+z)}{7r}\right)^2$

We can multiply out by $r^2$ and simplify to

$49(z^2+4y^2+9x^2) = 36(x+y+z)^2$

The most obvious approach is to expand, which we take:

$49z^2 + 196y^2 + 441x^2 = 36x^2 + 36y^2 + 36z^2 + 72(xy+xz+yz)$

or

$13z^2 + 160y^2 + 405x^2 = 72xy + 72xz + 72yz$

If we try to make squares, we'll need to break up the numbers $13,160,405$ into sums of squares. We can easily see that $13 = 2^2+3^3$ and after some computations that $160=12^2+4^2$ and $405 = 9^2+18^2$. Thus we can factor:

$(2z-18x)^2 + (3z-12y)^2 + (9x-4y)^2 = 0$

And we conclude that

$z=9x, z=4y, x=4/9y$

Putting $k=y/9$ we get $x=4k, z=36k, y=9k$ and the sides of the triangle must be $13k, 40k, 45k$. Thus the triangle is similar to the triangle with side lenghts $13,40,45$, and we have completed the problem.

Just an interesting side note to solve the

$49(x^2+4y^2+9z^2) = 36(x+y+z)^2$ condition.

We can apply Cauchy to get

$49(x^2+4y^2+9z^2)\left(1+\frac{1}{4}+\frac{1}{9}\right) \ge 49(x+y+z)^2$
$49(x^2+4y^2+9z^2) \ge 36(x+y+z)^2$

So by the equality condition on Cauchy, we know

$x = \frac{2y}{\frac{1}{2}} = \frac{3z}{\frac{1}{3}}$
$x = 4y = 9z$.
$\frac{x}{36} = \frac{y}{9} = \frac{z}{4} = k$.

So $x = 36k, y = 9k, z = 4k$, giving sides of $13k, 40k, 45k$.

Solution: We first prove a lemma.
Lemma: $\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{s}{r}$.
Proof: Note that $\cot\frac{A}{2}=\frac{s-a}{r}$. We have similar relations for $\cot\frac{B}{2}$ and $\cot\frac{C}{2}$. Hence our sum is simply \[\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\frac{(s-a)+(s-b)+(s-c)}{r}=\frac{3s-2s}{r}=\frac{s}{r}. \]
$\Box$
Now we employ Cauchy and our Lemma to get \[\left(\left(\cot\frac{A}{2}\right)^2+\left( 2\cot\frac{B}{2}\right)^2+\left( 3\cot\frac{C}{2}\right)^2\right) \left(1+\frac{1}{4}+\frac{1}{9}\right)\]
\[\ge\left(\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}\right)^2=\frac{s^2}{r^2}.\]
The left side of the inequality simplifies to \[\frac{49}{36}\left(\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2\right),\] so our inequality is actually \[\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left( 3\cot\frac{C}{2}\right)^2\ge\frac{(6s)^2}{(7r)^2}. \] Since $\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2=\frac{(6s)^2}{(7r)^2}$, the inequality is an equality case. Therefore, \[ \frac{\cot\frac{A}{2}}{\cot\frac{B}{2}}=4,\] \[\frac{\cot\frac{B}{2}}{\cot\frac{C}{2}}=\frac{9}{4}, \]and \[\frac{\cot\frac{C}{2}}{\cot\frac{A}{2}}=\frac{1}{9}. \] Since we proved that $\cot\frac{A}{2}=\frac{s-a}{r}$, our system simplifies to \[\frac{-a+b+c}{a-b+c}=4, \]\[\frac{a-b+c}{a+b-c}=\frac{9}{4}\]\[\frac{a+b-c}{-a+b+c}=\frac{1}{9}.\] This is equivalent to the three equations \[5a-5b+3c=0, \] \[5a+13b-13c=0, \]and \[5a+4b-5c=0. \]Subtracting the third from the second, we find that $\frac{9}{8}b=c$. Using this fact, we can reduce the first equation to \[5a-5b+\frac{27}{8}b=5a-\frac{13}{8}\implies \frac{40}{13}a=b. \]Therefore, our triple $(a,b,c)$ is \[(a,b,c)=\left(a,\frac{40}{13}a,\frac{45}{13}a\right). \] Clearly the minimal $a$ such that all sides are integers is $13$, so our triangle $\Delta ABC$ is similar to the triangle with side lengths $13, 40, 45$, which also has the smallest integer side lengths. We are done. $\blacksquare$

Wow. I was laughing at how un-geometric this problem looked, as it was the only geometry problem on the USAMO 2002. Then I actually tried it and realized it was not geometry.

Let $x = s-a$, $y = s-b$, $z = s-c$ in the usual fashion, then the equation reads \[ x^2 + 4y^2 + 9z^2 = \left( \frac67(x+y+z) \right)^2. \]
However, by Cauchy-Schwarz, we have
\[ \left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2  \]
with equality if and only if $1 : \tfrac12 : \tfrac13 = x : 2y : 3z$, id est $x : y : z = 1 : \tfrac14 : \tfrac19 = 36 : 9 : 4$. This is equivalent to $y+z : z+x : x+y = 13 : 40 : 45$. $\blacksquare$

You can tell this is not a geometry problem because eliminate the cotangents right away to get an algebra problem... and then you realize the problem claims that one equation can determine three variables up to scaling, at which point you realize it has to be an inequality (otherwise degrees of freedom don't work). So of course, Cauchy-Schwarz...

Denote $x, y, z$ as the intouch lengths from $A, B, C$ respectively.
The given equation is \[ x^2+4y^2+9z^2 = 36(x+y+z)^2 = (6x+6y+6z)^2 \]
Additionally, Cauchy yields \[ (x^2+(2y)^2+(3z)^2) \ge (6x+6y+6z)^2 \]
So equality holds in Cauchy, thus $x=6k, y=\frac{3k}{2}, z=\frac{2k}{3}$. Therefore the answer is $13:40:45$.

[quote=MithsApprentice]Let $ABC$ be a triangle such that
\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2,  \]
where $s$ and $r$ denote its semiperimeter and its inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers.[/quote]

Let \[ \begin{cases} x=s-a, \\ y=s-b, \\ z=s-c. \end{cases} \] Then we have $$\left(x^2+(2y)^2+(3z)^2\right)\left(1+\frac{1}{4}+\frac{1}{9}\right) \ge (x+y+z)^2$$ proving that \[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 \ge \left( \dfrac{6s}{7r} \right)^2,  \] hence $x:y:z=36:9:4$. Consequently, our desired triangle has side lengths $13, 40, 45$.

We know $\cot(A/2) = \tfrac{s-a}{r}$, and letting $x=s-a,y=s-b,z=s-c$, we get
\[ x^2+4y^2+9z^2 = \frac{36}{49}(x+y+z)^2. \]
We can rewrite this as
\[ (x^2+4y^2+9z^2)(k_1^2+k_2^2+k_3^2) = (k_1x+2k_2y+3k_3z)^2 \]
where $k_1=6/7,k_2=3/7,k_3=2/7$, making use of the fact that $k_1^2+k_2^2+k_3^2=1$. However, this is the equality case of Cauchy, hence $\frac{x}{k_1}=\frac{2y}{k_2} = \frac{3z}{k_3}$. This tells us that the ratio of the sides of the triangles are rational numbers, implying the desired conclusion.

Let $x=s-a, y=s-b, z= s-c.$ Then the equation is
\[x^2+4y^2+9z^2 = \frac{36}{49}(x+y+z)^2\]
But by Cauchy-Schwarz, we know that
\[\left( 1 + \frac{1}{4} + \frac{1}{9} \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge (x+y+z)^2 .\]
The equality case occurs when $x:y:z = 1 : \tfrac{1}{4} : \tfrac{1}{9}.$ If we scale this up by $36,$ we have $x:y:z = 36:9:4.$ Finally, using the definitions of $x,y,z$ tells us our triangle is one with side lengths $13,40,45.$ $\blacksquare$

[quote=baldeagle123]Sorry if this is a stupid question but is the proof of similarity in the fact that we used a general form of the side lengths and the relations between $\cos \frac{x}{2}$ to the parts of a triangle to derive the similar triangle T?[/quote]

I'm not sure If I understand your question, but what I did was use Cauchy to find triangle similar to $T$. We got that the ratio of the sides are $1:\tfrac14:\tfrac19.$ But:
[quote=USAMO 2002/2]Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers.[/quote]
So we scale up $36$ so that this holds. Since all sides were scaled up by an equal amount, it remains similar.

Observe that
\[s=r[\cot A/2+\cot B/2+\cot C/2].\]
Let $\cot A/2=x,\cot B/2=y,\cot C/2=z$ so we have
\[x^2+4y^2+9z^2=36(x+y+z)^2/49.\]
Rearranging, we have
\[13x^2+160y^2+405z^2=72[xy+yz+zx].\]
Rewrite this equation as follows:
\[(2x-18z)^2+(3x-12y)^2+(9z-4y)^2=0.\]
Hence we have $y=x/4,z=x/9$. This implies $x:y:z=36:9:4$. Since
\[\cot A/2+\cot B/2:\cot B/2+\cot C/2:\cot C/2+\cot A/2=c:a:b,\]
we get $a:b:c=13:40:45$, so we are done.

We set $x = s - a$, $y = s - b$, and $z = s - c$ where $s$ is the semiperimeter. Recall that $\tan \left( \frac x 2 \right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}$. Using these facts, we plug in and the equation reduces to basically \begin{align*}
    \frac{x}{yz} + \frac{4y}{xz} + \frac{9z}{xy} = \frac{36}{49} \cdot \frac{(x + y + z)^2}{xyz},
\end{align*} which simplifies to \begin{align*}
    x^2 + 4y^2 + 9z^2 = \frac{36}{49} (x + y + z)^2.
\end{align*} However, observe the following application of Cauchy-Schwarz: \begin{align*}
    (x^2 + 4y^2 + 9z^2)(6^2 + 3^2 + 2^2) \geq 36(x + y + z)^2,
\end{align*} so we know that for equality to hold we need $\frac x 6 = \frac{2y}{3} = \frac{3z}{2}$, or $x = 4y = 9z$. It's easy from here to check that the smallest integer solution for $(a, b, c)$ is $(13, 40, 45)$.

This is rather amusing - hence this post. Solved with [b]L567[/b].

Let $x,y,z$ be $s-a, s-b,s-c$, respectively, and notice by Sine Rule in the triangles involving vertices, incenter and intouch points we have that $$49(x^2+4y^2+9z^2) = 36(x+y+z)^2$$ which by the CS equality case means that $(x,y,z) = (4,9,36)$ and consequently we have that $\triangle ABC$ is similar to the triangle with side-lengths $13,40,45$. $\blacksquare$ \\

Recall that $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ if $A+B+C=\pi$. Let $[ABC]$ be the area of triangle $ABC$.
Notice that $$\frac{s}{r}=\frac{[ABC]^2}{sr^3}=\frac{s-a}{r}\cdot \frac{s-b}{r}\cdot \frac{s-c}{r}=\tan\left(90-\frac{A}{2}\right)\tan\left(90-\frac{B}{2}\right)\tan\left(90-\frac{C}{2}\right)=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2},$$ and also that $$\left(\frac{s}{r}\right)^2 = \left(\left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 \right)\left(1+\frac{1}{4}+\frac{1}{9}\right) \geq \left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right)^2.$$ Clearly equality is achieved, thus $\cot \frac{A}{2}=36n,\cot \frac{B}{2}=9n,\cot \frac{C}{2}=4n$ for some number $n$. But we also know that $36n+9n+4n=36n\cdot 9n\cdot 4n$, so $n=\frac{7}{36}$. The rest is easy, resulting in triangle $ABC$ being similar to a triangle with side lengths of $13$, $40$, and $45$.

[hide=Spoiler]Totally wasn't too lazy to write the final part of the proof

[quote=USAMO 2002 P2]Let $ABC$ be a triangle such that
\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2,  \]
where $s$ and $r$ denote its semiperimeter and its inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers.[/quote]
Let $x=s-a$, $y=s-b$, and $z=s-c$, we have that $$\cot \frac{A}{2}=\frac{s-a}{r}=\frac{x}{r},\quad \cot \frac{B}{2}=\frac{s-b}{r} =\frac{y}{r}, \quad\cot \frac{C}{2}=\frac{s-c}{r}=\frac{z}{r}\implies$$
$$x^2+4y^2+9z^2=\frac{36}{49}(x+y+z)^2$$ By Cauchy-Schwarz Inequality we get that $$(x^2+4y^2+9z^2)(1+\frac{1}{2}+\frac{1}{3})\geq (x+y+z)^2\implies x^2+4y^2+9z^2\geq \frac{36}{49}(x+y+z)^2$$
Since we know that these two expressions are equal we can consider the equality case: $$x:2y:3z=1:\frac{1}{2}:\frac{1}{3}\implies x:y:z=1:\frac{1}{4}:\frac{1}{9}=36:9:4\implies x+y:y+z:z+x=\boxed{45:11:40}.\blacksquare$$

Write $x=s-a, y=s-b, z=s-c$. Then, the expression becomes
\[(\frac{x}{r})^2 + (2\cdot \frac{y}{r})^2 + (3\cdot \frac{z}{r})^2 = (\frac67 \cdot \frac{x+y+z}{r})^2\]
But by Cauchy,
\[x^2 + (2y)^2 + (3z)^2\cdot (1 + \frac12^2 + \frac13^2) =\lVert \begin{bmatrix} x \\ 2y \\ 3z\end{bmatrix} \rVert^2\cdot \lVert \begin{bmatrix} 1\\ \frac12 \\ \frac13\end{bmatrix} \rVert^2 \geq (x+y+z)^2\]
Note that $1+\frac14+\frac19 = \frac{49}{36}$, so $\frac{x}{r})^2 + (2\cdot \frac{y}{r})^2 + (3\cdot \frac{z}{r})^2\geq \frac{36}{49} \frac{x+y+z}{r}^2$. Thus, the equality case of Cauchy must hold true, so $x:2y:3z = 1: \frac12: \frac13$, so $x:y:z = 1:\frac14 : \frac19 = 36:9:4$, and thus $BC:AC:AB = y+z:x+z:x+y = 13:40:45$ and we're done. $\blacksquare$.

Let $D,E,F$ be the contact points of the incircle on $BC,AC,AB.$ Let $AF=AE=u,BF=BD=v,CD=CE=w.$
We have:
\begin{align*}
\cot\frac{A}{2} &=\frac{u}{r} \\
2\cot\frac{B}{2} &= \frac{2v}{r} \\
3\cot\frac{C}{2} &= \frac{3w}{r}
\end{align*}
The equation becomes $\tfrac{u^2+4v^2+9w^2}{r^2}=\tfrac{36(u+v+w)^2}{49r^2}$ which rearranges to $13u^2+160v^2+405w^2-72wu-72uv-72vw=0.$ In fact, we can complete the square to get \[(2u-18w)^2+(3u-12v)^2+(9w-4v)^2=0\] Therefore, we conclude $u:v:w=36:9:4.$ Our triangle is $BC=13k,AC=40k,AB=45k.$

Using $\cot \frac A2 = \frac{s-a}r$, the equation is equivalent to $$(s-a)^2+4(s-b)^2+9(s-c)^2 = \frac{36s^2}{49}.$$ But $$((s-a)^2+4(s-b)^2+9(s-c)^2)(6^2+3^2+2^2) \geq (18s-6(a+b+c))^2 = 36s^2,$$ so equality must hold everywhere. Thus the side lengths must be in the ratio $13:40:45$.

[hide]
By Cauchy
\begin{align*}
&[(6r)^2+(3r)^2+(2r)^2]\left[\left(\cot\frac{A}{2}\right)^2+\left(2\cot\frac{B}{2}\right)^2+\left(3\cot\frac{C}{2}\right)^2\right]\\
\ge&\left(6r\cot\frac{A}{2}+6r\cot\frac{B}{2}+6r\cot\frac{C}{2}\right)^2\\
=&~(6s)^2
\end{align*}
with equality at $\cot\tfrac{A}{2}=4\cot\tfrac{B}{2}=9\cot\tfrac{C}{2}$, i.e., $(s-a):(s-b):(s-c)=36:9:4$.
Pairwise addition gives $\boxed{(13,40,45)}$.

Note that $\cot{\tfrac{A}2}=\tfrac{s-a}r$, $\cot{\tfrac{B}2}=\tfrac{s-b}r$, $\cot{\tfrac{C}2}=\tfrac{s-c}r$.
Then,
\begin{align*}
\left(\frac{s-a}{r}\right)^2+\left(2 \cdot \frac{s-b}{r}\right)^2+\left(3 \cdot \frac{s-c}{r}\right)^2&=\left(\frac{6s}{7r}\right)^2\\
\implies (s-a)^2+4(s-b)^2+9(s-c)^2&=\frac{36}{49} s^2.
\end{align*}
But by Cauchy,
\begin{align*}
((s-a)^2+4(s-b)^2+9(s-c)^2)\left(\left(\frac11\right)^2+\left(\frac12\right)^2+\left(\frac13\right)^2\right) &\ge s^2\\
\implies (s-a)^2+4(s-b)^2+9(s-c)^2 &\ge \frac{36}{49} s^2,
\end{align*}
so we have an equality. Therefore, $(s-a) : 2(s-b) : 3(s-c)=1 : \tfrac12 : \tfrac13$, so $(s-a) : (s-b) : (s-c)=36 : 9 : 4$. Hence, the answer is $(13, 40, 45)$.

$\color{magenta} \boxed {\textbf{SOLUTION P2}}$
$\color{red} \textbf{Geo Marabot Solve 6}$
The equation turns into,
$$x^2 + 4y^2 + 9z^2 = \frac{36}{49} \left(x+y+z \right)^2$$
By [b]Cauchy-Schwarz Inequality,[/b]
$$\left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2 $$
Equality holds if and only if $$1 : \tfrac12 : \tfrac13 = x : 2y : 3z \implies x : y : z = 1 : \tfrac14 : \tfrac19 = 36 : 9 : 4 \implies a : b : c= 13 : 40 : 45\blacksquare$$

Denote the side opposite $A$ by $a$, and define $b$ and $c$ similarly.
Refer to the diagram, where $I$ is the incentre of triangle $ABC$, and $D$ the foot of the altitude from $I$ onto $AB$.
Observe that $$\cot \frac{A}{2} = \frac{AD}{AI} = \frac{s-a}{r}.$$
Similarly, $$\cot \frac{B}{2} = \frac{s-b}{r}, \cot \frac{C}{2} = \frac{s-c}{r}.$$
Substituting these into the original equation gives us $$\left(\frac{s-a}{r} \right)^2 + \left(2 \cdot \frac{s-b}{r} \right) + \left( 3 \cdot \frac{s-c}{r} \right) = \left( \frac{6s}{7r} \right)^2$$
$$\implies (s-a)^2 + 4(s-b)^2 + 9(s-c)^2 = \frac{36}{49} s^2.$$
But observe that $$(s-a)^2 + 4(s-b)^2 + 9(s-c)^2  = \frac{(s-a)^2}{1} + \frac{(s-b)^2}{\frac{1}{4}} + \frac{(s-c)^2}{\frac{1}{9}}$$ $$\ge \frac{(s-a+s-b+s-c)^2}{1+\frac{1}{4} + \frac{1}{9}}$$ $$= \frac{s^2}{\frac{49}{36}}$$ $$=\frac{36}{49}s^2,$$
so equality must hold in the step with Titu's lemma, i. e. $s-a = 4(s-b) = 9(s-c)$.
Simplifying, we get $b+c-a = 4(a+c-b) = 9(a+b-c)$.
WLOG let $c = 1$.
Then solving the resultant system of equations gives us $a = \frac{13}{45}, b  = \frac{8}{9}$.
Therefore, $a:b:c = \frac{13}{45}:\frac{40}{45}:1 = 13:40:45$, and so triangle $ABC$ is similar to one with sides $13, 40, 45$. $\square$

Let $I$ be the incenter of $ABC$, and $D, E, F$ be its $A$-, $B$- and $C$-intouch points respectively. Now let \[x = AE = AF, y = BF = BD, z = CD = CE\text{.}\] Therefore we have
\begin{align*}
    \left(\frac{x}{r}\right)^2 + \left(\frac{2y}{r}\right)^2 + \left(\frac{3z}{r}\right)^2 &= \left(\frac{6(x + y + z)}{7r}\right)^2 \\
    (x^2 + (2y)^2 + (3z)^2)\left(\frac{1}{1} + \frac{1}{4} + \frac{1}{9}\right) &= (x + y + z)^2
\end{align*}
which is the equality case of Cauchy-Schwarz. Therefore
\begin{align*}
    x : 2y : 3z &= 1 : \frac{1}{2} : \frac{1}{3} \\
    x : y : z &= 1 : \frac{1}{4} : \frac{1}{9} \\
    x : y : z &= 36 : 9 : 4\text{.}
\end{align*}
Now we have
\begin{align*}
    AB : BC : CA &= (x + y) : (y + z) : (z + x) \\
    &= 45 : 13 : 40\text{.}
\end{align*}
Therefore $ABC$ is similar to a triangle with side lengths $45$, $13$ and $40$.


trigonometry
inequalities proposed
inequalities

If $ x,y,z$ are real numbers and $ A,B,C$ are three angles of a triangle, then
$ x\sin A\plus{}y\sin B\plus{}z\sin C \le \sqrt{(1\plus{}x^2)(1\plus{}y^2)(1\plus{}z^2)}.$


Hi
I am getting stuck with the following sequence :
Prove that $ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n}i^{n(n \minus{} 1)}}{n}$  converges

Any help is highly appreciated

$ n(n\minus{}1)$ is always even.  That tells you...

[quote="t0rajir0u"]$ n(n - 1)$ is always even.  That tells you...[/quote]

Thanks for you reply BUT Blease note that $ i^{n(n - 1)}$   [b]not[/b] $ {( - 1)^{n(n - 1)}}$ where $ i^{ 2} = - 1$

$ S\equal{}\sum_{n\equal{}1}^\infty \frac {( \minus{} 1)^{n}i^{n(n \minus{} 1)}}{n}$
[hide="parities"]
$ n\equal{}4k \to n(n\minus{}1)\equal{}4k(4k\minus{}1)\equal{}4(4k^2\minus{}k)$
$ n\equal{}4k\minus{}1 \to n(n\minus{}1)\equal{}(4k\minus{}1)(4k\minus{}2)\equal{}2(8k^2\minus{}6k\plus{}1)$
$ n\equal{}4k\minus{}2\to n(n\minus{}1)\equal{}(4k\minus{}3)(4k\minus{}2)\equal{}2(8k^2\minus{}10k\plus{}3)$
$ n\equal{}4k\minus{}3 \to n(n\minus{}1)\equal{}(4k\minus{}4)(4k\minus{}3)\equal{}4(4k^2\minus{}7k\plus{}3)$
[/hide]
 [hide="then maybe some algebra"]
$ S\equal{}\sum_{k\equal{}1}^\infty\left[\frac {( \minus{} 1)^{4k\minus{}3}}{4k\minus{}3}\plus{}\frac {( \minus{} 1)^{4k\minus{}2}i^{2(8k^2\minus{}10k\plus{}3)}}{4k\minus{}2}\plus{}\frac {( \minus{} 1)^{4k\minus{}1}i^{2(8k^2\minus{}6k\plus{}1)}}{4k\minus{}1}\plus{}\frac {( \minus{} 1)^{4k}}{4k} \right]$

$ \equal{}\sum_{k\equal{}1}^\infty\left[\frac {\minus{}1}{4k\minus{}3}\plus{}\frac {\minus{}1}{4k\minus{}2}\plus{}\frac {1}{4k\minus{}1}\plus{}\frac {1}{4k} \right]$

$ \equal{}\sum_{k\equal{}1}^\infty \left( \minus{}\frac{8k\minus{}5}{(4k\minus{}3)(4k\minus{}2)} \plus{} \frac{8k\minus{}1}{4k(4k\minus{}1)}\right)$

$ \equal{}\sum_{k\equal{}1}^\infty \left( \minus{}\frac{4(2k)\minus{}5}{(2(2k)\minus{}3)(2(2k)\minus{}2)} \plus{} \frac{4(2k\plus{}1)\minus{}5}{(2(2k\plus{}1)\minus{}3)(2(2k\plus{}1)\minus{}2)}\right)$

$ \equal{}\sum_{m\equal{}2}^\infty \frac{(\minus{}1)^{m\minus{}1}(4m\minus{}5)}{(2m\minus{}3)(2m\minus{}2)}\equal{}\sum_{m\equal{}2}^\infty \frac{(\minus{}1)^{m\minus{}1}(m\minus{}\frac{5}{4})}{(m\minus{}\frac{3}{2})(m\minus{}1)}$

this is a decreasing alternating sequence so it must converge, and should be equal to S[/hide]

Or, more generally, if $ a_k$ is a periodic sequence, then $ \sum_{k}\frac{a_k}k$ converges if and only if the sum of $ a_k$ over the period is $ 0$. (Abel-Dirichlet criterion). In our particular case, the period is $ 4$ and all we need is to add up $ 4$ numbers $ 1\plus{}(\minus{}1)\plus{}(\minus{}1)\plus{}1$ to figure out the answer :).

See also Kent Merryfield's theorem and the generalization [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231623]here[/url].


geometry
parallelogram
incenter
trapezoid
inradius
trigonometry
geometry unsolved

Dear Mathlinkers,
let ABCD a square, M a point on segment CD, I, X, Y the incenters wrt MAB, MAD, MBC,
P, Q the meetpoints of the parallel to AD passing through X with BI, AC,
S, R the meetpoints of the parallel to AD passing through Y with AI, BD.
Prove that PQRS is a parallelogram.
Sincerely
Jean-Louis

wow,it seems nice
could you give a solution?

$ a \equal{} AB \equal{} ...$ is the square side and $ r_x, r_y$ radii of the incircles $ (X), (Y).$ Triangles $ \triangle ASY \sim \triangle AXD$ are similar, having equal angles at the common vertex $ A$ and at the vertices $ Y, D.$ Likewise, triangles $ \triangle BPX \sim \triangle BYC$ are also similar $ \Longrightarrow$

$ XP \equal{} \frac {CY \cdot BX}{BC} \equal{} \frac {2r_y(a \minus{} r_x)}{a},\ YS \equal{} \frac {DX \cdot AY}{AD} \equal{} \frac {2r_x(a \minus{} r_y)}{a}.$

As $ AQXD, BRYC$ are isosceles trapezoids,

$ a \minus{} PQ \equal{} 2r_x \plus{} XP \equal{} 2 (r_x \plus{} r_y) \minus{} \frac {2r_xr_y}{a} \equal{} 2r_y \plus{} YS \equal{} a \minus{} SR$

$ \Longrightarrow$ $ PQ \equal{} SR.$

Dear Yetti and Mathlinkers,
thank you for your nice and short proof.
The shortness of your proof can it give a one without calculation?
What do you think ?
Sincerely
Jean-Louis

@yetti

Actually, seeing that a-PQ is symmetrical in r_x and r_y, it shows that PQ = SR.

Best regards,
sunken rock

I do not think this qualifies as "without calculation". I can similarly say $ XYRQ$ is isosceles trapezoid, $ \frac {YS \minus{} XP}{2} \equal{} r_x \minus{} r_y$ $ \Longrightarrow$ $ XYSP$ is also isosceles trapezoid $ \Longrightarrow$ $ PS \parallel QR,$ but that does not qualify either, still have to calculate $ XP, YS.$ It is also possible to calculate $ PQ \equal{} RS$ based on $ AS \parallel MX, BP \parallel MY,$ but this calculation is about three times as long.

[size=117][color=darkred][b]Very nice [u]Jayme[/u]'s problem and [u]Yetti[/u]'s proof ![/color] [color=darkblue]Thank you for these. [/b][/color][/size]

[color=darkblue]Suppose $ AB \equal{} 1$ and denote $ m(\angle MAD) \equal{} 2x$ , $ m(\angle MBC) \equal{} 2y$ .

Denote the lengths $ r_1$ , $ r_2$ , $ r$ of the inradius for $ AMD$ , $ BMC$ , $ AMB$ respectively.

Prove easily that  $ \left\|\begin{array}{ccc} r_1 \equal{} \frac {\tan x}{1 \plus{} \tan x} & \implies & \tan x \equal{} \frac {r_1}{1 \minus{} r_1} \\
 \\
r_2 \equal{} \frac {\tan y}{1 \plus{} \tan y} & \implies & \tan y \equal{} \frac {r_2}{1 \minus{} r_2} \\
 \\
r \equal{} \frac {(1 \minus{} \tan x)(1 \minus{} \tan y)}{2(1 \minus{} \tan x\tan y)} & \implies\ & \boxed {\frac 1r \equal{} \frac {1}{1 \minus{} 2r_1} \plus{} \frac {1}{1 \minus{} 2r_2}}\ (1)\end{array}\right\|$ .

Since $ \tan 2x \plus{} \tan 2y \equal{} 1$ obtain second independent relation w.r.t. the position of $ M\in [CD]$ :

$ 4(r_1 \plus{} r_2)(1 \plus{} r_1r_2) \equal{} 12r_1r_2 \plus{} 2\left(r_1^2 \plus{} r_2^2\right) \plus{} 1$ $ \implies$ $ \boxed {1 \plus{} 2(1 \minus{} r_1)^2(1 \minus{} r_2)^2 \equal{} 2(1 \minus{} r_1r_2)^2}\ (2)$ .

Also show easily that $ \boxed {PQ \equal{} SR \equal{} \tan\left(\frac {\pi}{4} \minus{} x\right)\tan\left(\frac {\pi}{4} \minus{} y\right) \equal{} (1 \minus{} 2r_1)(1 \minus{} 2r_2)}\ (3)$ ,

i.e. $ \boxed {PQ \equal{} RS \equal{} \frac {AU\cdot BV}{AB}}$ , where $ U\in BI\cap AD$ and $ V\in BI\cap BC$ . [/color]

[quote="jayme"]...The shortness of your proof can it give a one without calculation?
What do you think ? ...[/quote]

$ MY \parallel BP$ $ \Longrightarrow$ $ \angle XMI \equal{} 90^\circ \minus{} \angle YMC \equal{} \angle QPB \equal{} \angle 180^\circ \minus{} \angle IPX$ $ \Longrightarrow$ $ XPIM$ is cyclic and similarly, $ YSIM$ is cyclic. According to the 1st paragraph of [url]http://www.mathlinks.ro/viewtopic.php?p=1484793#1484793[/url], $ XIYM$ is also cyclic $ \Longrightarrow$ $ XPISYM$ is cyclic, $ XPSY$ is an isosceles trapezoid. As $ XQRY$ is also an isosceles trapezoid, $ PQRS$ is a parallelogram.

But to prove the underlying theorem about the pairwise common tangents of 4 directed circles, these tangents touching a single directed circle, requires adding/subtracting a bunch of line segments, not to mention adding/subtracting angles to show that $ XPIM, YSIM$ are cyclic. So, is it without calculation ? :P

Dear Mathlinkers,
for a proof see:  http://perso.orange.fr/jl.ayme  vol.5,  le résultat de Larrosa Canestro, p.25
Sincerely
Jean-Louis