Find all natural numbers $ n$ such that $ n, n\plus{}10$ and $ n\plus{}14$ are all prime numbers.

[hide]Take mod 3

$ n \equal{} n (mod 3)$

$ n\plus{}10 \equal{} n\plus{}1 (mod 3)$

$ n\plus{}14 \equal{} n\plus{}2 (mod 3)$

But three consecutive numbers must contain one multiple of $ 3$.

Therefore no such primes exist.[/hide]

[quote="chrischris"][hide]Take mod 3

$ n \equal{} n (mod 3)$

$ n \plus{} 10 \equal{} n \plus{} 1 (mod 3)$

$ n \plus{} 14 \equal{} n \plus{} 2 (mod 3)$

But three consecutive numbers must contain one multiple of $ 3$.

Therefore no such primes exist.[/hide][/quote]

actually there is only one solution $ n\equal{}3$

Yes, chrischris's post shows that the only solution is $ n\equal{}3$, giving the primes $ 3, 13$ and $ 17$.


function
real analysis
real analysis unsolved

Consider a function $f \in C^{(1)}([-1,1])$ such that $f(-1)=f(1)=0.$ Prove that there exists $x\in[-1,1]$ such that $f(x)=({1+x^{2}})f'(x).$

Put $g(x) = exp(-\arctan{x})f(x)$, $g(-1) = g(1) = 0$ thus there exists $c$ such that $g'(c) = 0$ from where the result follows.


analytic geometry
geometry
3D geometry
sphere
geometry unsolved

How to find the cylindrical and spherical coordinate equations for a unit sphere?

[quote="Math^_^"]spherical coordinate equations for a unit sphere?[/quote]

 :rotfl:

  dg


geometry
3D geometry
number theory proposed
number theory

No, because it is a special case of Erdös-Seflridge theorem  :D

Who is Narumi?

Pierre.

What is Erdos-Selflridge theorem , Pbornsztein ?

That theorem states that the product of some consecutive positive integers (more than one, obviously) can never be a perfect square, cube or higher power, if I am not mistaken.

[quote="harazi"]Are there 25 consecutive positive integers whose product is a square?[/quote]
no because the product is a square only if the 25 numbers are all squares but we can't have 25 consecutive square....this is what i think should be right...but i have no enough knowledge to make a proof :? ..is this right? :?:

[quote="Bojan Basic"][quote="EUCLA"]no because the product is a square only if the 25 numbers are all square[/quote]
Like $2\cdot 8=4^2$ or even $2\cdot 2=2^2$.[/quote]
yes i'm wrong...what i said isn't right sorry but i don't know how to say it in english.. :oops:

I know, Pierre, but I was reffering to a little bit easier proof than that of Erdos. By the way, do you happen to have an article with Erdos's proof of the general case?
   About Narumi, I have no idea: the problem was given in a DS and it was called Narumi's theorem.

Yes, I do have the paper of Erdös and Selfridge, but they solved the problem by reducing it to some finite, but many, direct checking by hand....

Pierre.


What is the following number in the sequence,

2, 3, 6, 14, 40, 152, ...

Two days and nothing yet...!
Well, a hint could help!
It is the sum of two very well known sequences!

[quote="JBL"]There is an error!  The third term should be 4 + 2 = 6, shouldn't it?

Those two hints together were too much, by the way.[/quote]
OK, I errased one of the hints. And yes, sorry, there is an error, corrected  :blush:

[quote="bdpiang"]n! +2^n  so next no. in the series is 784,5168[/quote]Good!


integration
function
real analysis
real analysis solved

The base of log is 10 ? That's usually it, but there might be different conventions in different countries...

Thanks. The notation log is also used in Mathematica, but it seems a bit confusing to me. It's usually denoted by ln (natural or neperian logarithm).

i'm a little confused (maybe i'm wrong :D)
1/lnk > 1/lnn
=>
1/n \sum{k=3,n} n1/lnk >1/n \sum{k=3,n}n1/lnn=((n-2)/n)n1/lnn->oo ???

That last thing doesn't go to 00 because n 1/lnn  = [e^(lnn)] 1/lnn  = e and (n-2)/n goes to 1, so it doesn't matter.

[quote="Kuba"]((n-2)/n)n1/lnn[/quote]

You get a good idea,  remember a^b=exp(b.lna)
n1/lnn= exp((1/lnn)lnn)=e

1/n {k=3,n} n^(1/lnk) \geq (n-2)/ne=(1-2/n)e now you know that
lhe limit of the sequence is greater or equal than lim (1-2/n)e=e
Try to find a sequence t(n) such that
 t(n)  \geq  1/n {k=3,n} n^(1/lnk)  \geq  (1-2/n)e  with lim t(n)=e

[quote="Moubinool"]
1/n {k=3,n} n^(1/lnk) \geq (n-2)/ne=(1-2/n)e
Try to find a sequence t(n) such that
 t(n)  \geq  1/n {k=3,n} n^(1/lnk)  \geq  (1-2/n)e  with lim t(n)=e[/quote]

f(x)=n^(1/lnx) is decreasing on [3;n], n^(1/lnk) \leq  \int_[k-1;k]f(x)dx

by summing 1/n {k=3,n} n^(1/lnk) \leq 1/n. \int_[2;n] n^(1/lnx)dx

t(n)=1/n. \int_[2;n] n^(1/lnx)dx , change variable x=ny, dx=ndy

t(n)=  \int_[2/n;1] n^(1/(lnn+lny)) dy = \int C_[2/n;1](y).n^(1/(lnn+lny)) dy where

C_[2/n;1](y)= 1 if y\in [2/n;1] and 0 if y not in [2/n;1] ( caracteristic function )

Fn(y)=C_[2/n;1](y).n^(1/(lnn+lny))=C_[2/n;1](y).exp(lnn/(lnn+lny))

Fn(y)=C_[2/n;1](y).e.exp((-lny)/(lnn+lny))

t(n)= e. \int_[0;1]C_[2/n;1](y).exp((-lny)/(lnn+lny))dy

Please tell me when I have to post my simple solution to this limit.

First of all observe that n^(1/log3)+n^(1/log4)+..+n^(1/logn)>(n-2)e

since n^(1/logn) =e and the terms are decreasing.

Now we give an upper bound for the sum as follows:

Put sum(k=3,..,n)n^(1/logk) = sum1 + sum2 +sum3

where in sum1 : 3 \leq k \leq n^(1/100)

             sum2 : n^(1/100)  <k \leq n/logn

             sum3 : n/logn<k \leq n

We have

 sum1< n^(1/100)*n^(1/log3) \leq n^(1-1/1000)<xn  for every x>0

 sum2<e^(100)*(n/logn)<xn for every x if n>n_0(x)

  sum3< n*n^(1/(logn - loglogn))<ex(1+x)  if n>n_0(x)

thus sum(k=1,..,n) n^(1/logn) < x + 5nx for every x>0 if n>n_0(x)

which proves the result.

I hope it is correct and not too difficult.

[quote="manlio"]Sorry, the last line is:

thus sum(k=1,..,n) n^(1/logn) < e + nx for every x>0 if n>n_0(x)[/quote]

Manlio

sum(k=1,..,n) n^(1/logn)  is n.n^(1/logn)=n.e

Sorry, I am very tired and it is better for me to go to bed.

Naturally it is

sum(k=1,..,n) n^(1/logk) < e + nx for every x>0 if n>n_0(x)


probability MATHCOUNTS AMC AIME \/closed

I'm really only going to have time for one class this spring and I have been debating between Introduction to Counting and Probability or Intermediate Algebra.

I have the Introduction to Counting and Probability book and I've begun to work through it. I haven't encountered any major mishaps up to this point. Would the class be that much more worthwhile over reading the book?

As for Intermediate Algebra, I have a pretty solid command of beginning and early intermed. Algebra and can do the Pre-Test without problems. I'm tending to lean toward this class, especially since the book isn't out yet. Does that appear to be the right decision?

On a random note, I know it's very early, but are there any idea of what classes may be offered in the summer?

If anyone can help and/or give opinions I'd really appreciate it!

If you work through the Counting book closely, then you should probably opt for the algebra class.

This summer, we'll definitely offer Intro #Thy, MATHCOUNTS, and the AIME A Problem Series.  We haven't decided what else, but we'll probably offer the Intermediate Counting class and a couple others.

[quote="funcia"]when will the summer classes end?[/quote]

Most will run June through August.

[quote="horselvr101"]Thanks!!

I really do think I'm going to choose the Interm. Alg. Has anyone around here taken it? What sort of difficulty/time commitment does it usually require? (I'm trying to sort out my very busy shedule  :D ).[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111309


LaTeX
inequalities
inequalities unsolved

Let $a,b,c$ be positive real numbers such that $abc = 1$.

Prove that

\[\frac{a^4}{8a^3b^3 + 1} + \frac{b^4}{8b^3c^3 + 1} + \frac{c^4}{8c^3a^3 + 1} \geq \frac{1}{3}.\]

hahahaha i'm so ashamed of my self... :blush:
Once again i must ask someone to write it in latex...  :)
So someone please kindly write the ineq above in latex please...  :lol:
Thanx!!!  :P

[quote="Detective"]hahahaha i'm so ashamed of my self... :blush:
Once again i must ask someone to write it in latex...  :)
So someone please kindly write the ineq above in latex please...  :lol:
Thanx!!!  :P[/quote]

Done.

Just try to learn, if you want to type inequalities in LaTeX there isn't much you need to learn: the dollar signs, obviously, \sqrt[]{} for roots, and ^ for powers, and \frac{}{} for fractions. That's basically what you need to know.

Many thanx to you arne.

Do you know where i can learn to type in latex?
Or if you don't mind, could you teach me how to type in latex???

hehehe
anyway thanx arne!!! :P

I sent you a PM about it, so just read your private messages.

This looks similar to IMO 2001 Problem 2... I don't have it here with me right now, but I think apply CBS to the denominator separately... then use AM-GM on all 3... of course it sounds easier than actually doing it I'm sure lol

hahaha good thinking hendrata01...

just try to solve it and write down the full solution okay.

Great job hendrata01!!!
Amazing thinking!!!
But unfortunately, you made a fatal mistake.
hahahaha

try to read the problem again okay.

Sorry Soarer you also made a mistake since you followed hendrata01's first step.

The problem is

$\frac{a^4}{8a^3b^3+1}$

NOT

$\frac{a^4}{8b^3c^3+1}$

so you can't say that

$\frac{a^4}{8a^3b^3+1} = \frac{a^6}{a^2+8bc} $ by multiplying the numerator and denominator by $a^2$

[quote="hendrata01"]ok this is my try, although I don't know how to continue from here:

let k = 4/3, then I first claim that $(a^k + b^k + c^k)^2 \geq a^{\frac{k}{2}}(a^2+8bc)$. To prove this, LHS - $a^{2k} = (b^k+c^k)(a^k+a^k+b^k+c^k) \geq 2b^{\frac{k}{2}}c^{\frac{k}{2}}4a^{\frac{k}{2}}b^{\frac{k}{4}}c^{\frac{k}{4}} = 8a^{\frac{2}{3}}bc$, thus LHS $\geq a^{\frac{8}{3}}+8a^{\frac{2}{3}}bc$, then the claim holds.

Now, $\frac{a^4}{8b^3c^3+1} = \frac{a^6}{a^2+8bc} \geq \frac{a^{\frac{20}{3}}}{(a^k+b^k+c^k)^2}$ and similarly for the other terms as well. Then substituting $x = a^k$ and so on, our inequality becomes:

\[
(x^5+y^5+z^5) \geq \frac{1}{3}(x^2+y^2+z^2)^2
\]
with the condition that $xyz = 1$

I feel like the last one should be easier... or at least nicer in forms in my opinion, but I haven't got a chance to do that...[/quote]

LOL ok now I'm inspired by soarer... (although I must admit I copied your technique a little bit  :blush: )

$\frac{\sum x^5}{(x+y+z)^2} (\sum \frac{1}{x}) \geq \frac{\sum(x^2)^2}{(x+y+z)^2}$ (Cauchy)
Then $\sum \frac{1}{x} = xy+yz+zx \leq \sum (\frac{x+y}{2})^2 = \frac{1}{2}(x^2+y^2+z^2+xy+yz+zx) \leq x^2+y^2+z^2  $

So now we only need to show that $x^2+y^2+z^2 \geq \frac{1}{3}(x+y+z)^2$, but this is AM-RMS or cauchy.
Am I right?  :D

oh dang... I posted my reply before you posted yours lol what a waste.... ok the problem IS hard now :huh:

[quote]
LOL ok now I'm inspired by soarer... (although I must admit I copied your technique a little bit  :blush: )

$\frac{\sum x^5}{(x+y+z)^2} (\sum \frac{1}{x}) \geq \frac{\sum(x^2)^2}{(x+y+z)^2}$ (Cauchy)
Then $\sum \frac{1}{x} = xy+yz+zx \leq \sum (\frac{x+y}{2})^2 = \frac{1}{2}(x^2+y^2+z^2+xy+yz+zx) \leq x^2+y^2+z^2  $

So now we only need to show that $x^2+y^2+z^2 \geq \frac{1}{3}(x+y+z)^2$, but this is AM-RMS or cauchy.
Am I right?  :D[/quote]

Hey i told it was a mistake!

so you can't modify the problem into

$\frac{\sum x^5}{(x+y+z)^2}$

c'mon guys
try to solve my problem...
Don't make me disappointed...
it took me about 45 minutes to create this ineq...
hehehe


What positive, two-digit number, when doubled and added to two,
gives a sum that is the original number with its digits reversed?

Two digit number $ xy$ with x and y being the tens and the ones digit, respectively.

$ 2(10x \plus{} y) \plus{} 2 \equal{} 10y \plus{} x$
$ 19x \plus{} 2 \equal{} 8y$
x obviously must be an even number, since 8y is even.  x cannot be 6 or 8 because any number beginning in 6 or 8, when doubled, will give a 3 digit number.
try 4.
$ 19*4 \plus{} 2 \equal{} 78 \equal{} 8y$.  Doesn't divide evenly.

Try 2.

$ 19*2 \plus{} 2 \equal{} 40 \equal{} 8y$
$ y \equal{} 5$
$ xy$ is $ \boxed{25}$.


LaTeX
quadratics
algebra unsolved
algebra

[quote="mavropnevma"]Please
1. Learn a little Latex, or even write plaintext, rather than attach .jpg's;
2. Post to right forum (these are hardly Algebra Unsolved Problems).
Just a personal, friendly advice  :)[/quote]

Well, those problems are actually part of College Algebra as the description of the "Algebra" said. Also, I enlarged the picture. If you are unable to view it, click on it in order to expand it. Of course, I will look forward to the Latex. Right now, those problems are visible, so they should not be problematic. Thanks.

[quote="Anonymous52111"]Please explain each problem with clear steps. Thanks.[/quote]

Are you sure these problems belong to the Olympiad section :?:

Nvm. To close this topic we have:

a)$ x\equal{}8\sqrt{x}\Longleftrightarrow x^2\minus{}64x\equal{}0\Longleftrightarrow x(x\minus{}64)\equal{}0\Longrightarrow x\equal{}0$ or $ x\equal{}64$

b)

$ \sqrt{15\minus{}2x}\equal{}x\Longleftrightarrow x^2\equal{}15\minus{}2x\Longleftrightarrow x^2\plus{}2x\minus{}15\equal{}0\Longleftrightarrow (x\plus{}5)(x\minus{}3)\equal{}0\Longrightarrow x\equal{}3$ or $ x\equal{}5$

c)

$ x^4\minus{}5x^2\plus{}4\equal{}0\Longleftrightarrow (x^2\minus{}1)(x^2\minus{}4)\equal{}0\Longrightarrow x\equal{}\pm 1$ or $ x\equal{}\pm 2$

[quote="Γιώργος"][quote="Anonymous52111"]Please explain each problem with clear steps. Thanks.[/quote]

Are you sure these problems belong to the Olympiad section :?:

Nvm. To close this topic we have:

a)$ x \equal{} 8\sqrt {x}\Longleftrightarrow x^2 \minus{} 64x \equal{} 0\Longleftrightarrow x(x \minus{} 64) \equal{} 0\Longrightarrow x \equal{} 0$ or $ x \equal{} 64$

b)

$ \sqrt {15 \minus{} 2x} \equal{} x\Longleftrightarrow x^2 \equal{} 15 \minus{} 2x\Longleftrightarrow x^2 \plus{} 2x \minus{} 15 \equal{} 0\Longleftrightarrow (x \plus{} 5)(x \minus{} 3) \equal{} 0\Longrightarrow x \equal{} 3$ or $ x \equal{} 5$

c)

$ x^4 \minus{} 5x^2 \plus{} 4 \equal{} 0\Longleftrightarrow (x^2 \minus{} 1)(x^2 \minus{} 4) \equal{} 0\Longrightarrow x \equal{} \pm 1$ or $ x \equal{} \pm 2$[/quote]

For "A," how did you get the part of "x^2 - 64x"?
For "B," how did you get the part of "x^2 = 15 - 2x"?
For "C," explain how you got the part of "(x^2 - 1)(x^2 - 4)."

Thanks for clarifying. I have not been here for so long and therefore am doing this for helping my friend. However, I am trying to understand those problems before being able to explain to my friend who will have a test next week Tuesday. Thanks.

[quote="Anonymous52111"][quote="Γιώργος"][quote="Anonymous52111"]Please explain each problem with clear steps. Thanks.[/quote]

Are you sure these problems belong to the Olympiad section :?:

Nvm. To close this topic we have:

a)$ x \equal{} 8\sqrt {x}\Longleftrightarrow x^2 \minus{} 64x \equal{} 0\Longleftrightarrow x(x \minus{} 64) \equal{} 0\Longrightarrow x \equal{} 0$ or $ x \equal{} 64$

b)

$ \sqrt {15 \minus{} 2x} \equal{} x\Longleftrightarrow x^2 \equal{} 15 \minus{} 2x\Longleftrightarrow x^2 \plus{} 2x \minus{} 15 \equal{} 0\Longleftrightarrow (x \plus{} 5)(x \minus{} 3) \equal{} 0\Longrightarrow x \equal{} 3$ or $ x \equal{} 5$

c)

$ x^4 \minus{} 5x^2 \plus{} 4 \equal{} 0\Longleftrightarrow (x^2 \minus{} 1)(x^2 \minus{} 4) \equal{} 0\Longrightarrow x \equal{} \pm 1$ or $ x \equal{} \pm 2$[/quote]

For "A," how did you get the part of "$ x^2 \minus{} 64x$"?
For "B," how did you get the part of "$ x^2 \equal{} 15 \minus{} 2x$"?
For "C," explain how you got the part of "$ (x^2 \minus{} 1)(x^2 \minus{} 4)$."

Thanks for clarifying. I have not been here for so long and therefore am doing this for helping my friend. However, I am trying to understand those problems before being able to explain to my friend who will have a test next week Tuesday. Thanks.[/quote]

For ''A'' we square both sides and then we get $ x^2\equal{}64x$
Same for ''B''. For ''C'' we can easily see that this is a quadratic equation if we set $ x^2\equal{}t$. Then we have $ t^2\minus{}5t\plus{}4\equal{}0$. Calculating now the discriminant( $ b^2\minus{}4ac$) we have that $ t_1\equal{}1$ and $ t_2\equal{}4$. The rest of it is easy...


Factorize $5^{1985}-1$ into product of three integers; each of them is larger than $5^{100}$.

Hint:   [hide]$x^5-1= (x-1)(x^4+x^3+x^2+x+1)=(x-1)((x^2+3x+1))^2-5x(x+1)^2)$  is easily factorable when $x=5^{397}$  and thus $5x$ is a square of a number.[/hide]

How do you figure that out? It's so tricky!!

Jelly119 -  This one was familiar, (has been posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=x%5E5-1&t=21174]here[/url] before. (just did the search).


geometry
circumcircle
inequalities
inequalities proposed

For the triangle ABC let $ R$ be the circumcircle radius, $ r$ = incircle radius, $ s$ = semiperimeter. Prove that:

1. For any triangle

$ 2\,R \plus{} (3\sqrt{3} \minus{}4)\,r \minus{} s \ge 0$

This inequality has one usual and one unusual propery. Usual is that the inequality becomes equality if, and only if the triangle is equilateral. What is unusual, is that the left part can be $ \it{very\: close}$ to 0 for some triangles that are far from being equilateral (even if we will define $ s\equal{}1$ to avoid very small triangles). So it cannot be proved using the inequality of arithmetic and geometric means or any other inequality which becomes equality when $ a\equal{}b\equal{}c$ only, and is far from equality when $ a \gg b$. This inequality is also very important in proving other inequalities (see below).

2. The following 3 inequalities form a basis for the inequalities of the type $ x\,R\plus{}y\,r\plus{}z\,s \ge 0$ that are true for all triangles:

$ \begin{cases}
2\,R \plus{} (3\sqrt{3} \minus{}4)\,r \minus{} s \ge 0\\
s\minus{}3\sqrt{3}\,r \ge 0  \\
r \ge 0
\end{cases}$

This means that for any true inequality of the type $ x\,R\plus{}y\,r\plus{}z\,s \ge 0$ there exist non-negative numbers A,B,C so that the following identity holds:

$ x\,R\plus{}y\,r\plus{}z\,s \equal{} A\cdot(2\,R \plus{} (3\sqrt{3} \minus{}4)\,r \minus{} s) \plus{} B\cdot(s\minus{}3\sqrt{3}\,r) \plus{} Cr$

Example: $ R\minus{}2\,r \ge 0$ holds. This has been proved using millions of methods, here is one more, by using the basis and the identity:

$ R\minus{}2\,r\equal{}\frac{1}{2} (2\,R \plus{} (3\sqrt{3} \minus{}4)\,r \minus{} s) \plus{} \frac{1}{2}(s\minus{}3\sqrt{3}\,r)$

Note: inequality $ s\minus{}3\sqrt{3}\,r \ge 0$ is well-known, in other words it says that among all triangles that are circumscribed over some circle $ w$, equilateral  has the smallest perimeter.

3. The list of 3 inequalities listed above are the only to form a basis. I mean that they cannot be changes with some other list with saving the same property that any true inequality can be represented as scalar product of the inequalities from the list and non-negative numbers.

4. The following  two statements for the list of 3 numbers $ {x,y,z}$ are equivalent:

Statement 1: Inequality $ x\,R\plus{}y\,r\plus{}z\,s \ge 0$ is true for all triangles.

Stemenent 2: x,y and z do satisfy the following 3 inequalities:

$ \begin{cases}
x \ge 0\\
x\plus{}2\,z \ge 0\\
2\,x\plus{}y\plus{}3\sqrt{3}\,z \ge 0
\end{cases}$

If anyone will find the solution for anything of the above, I'll be happy to read it, please, post it here or e-mail me. I am very interested in short proof of the inequality (1), hope it exists. I will write down my own solution for all of the above and post in this thread at the 1'st of July or earlier.

Thank you!
Sergey Markelov
markelov@mccme.ru

[quote="Markelov"]For the triangle ABC let $ R$ be the circumcircle radius, $ r$ = incircle radius, $ s$ = semiperimeter. Prove that:

1. For any triangle

$ 2\,R \plus{} (3\sqrt {3} \minus{} 4)\,r \minus{} s \ge 0$

[/quote]
If $ a\equal{}y\plus{}z,$ $ b\equal{}x\plus{}z$ and $ c\equal{}x\plus{}y$ then your inequality is equivalent to
$ \sum_{cyc}(x^2\minus{}y^2)^2xy\plus{}(12\sqrt3\minus{}20)\sum_{cyc}xyz^2(x\minus{}y)^2\plus{}\sum_{sym}(x^3y^3\minus{}2x^3y^2z\plus{}x^2y^2z^2)\geq0,$
which obviously true.


By Pythagorian's theorem we have $2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\geq 0.$

[hide="solution"]
I'll get latex sooner or later guys, for now bear with me. I'm using "<=" to mean less than or equal to, in case anyone gets confused.

From a^2 + b^2 = c^2 we have:

(a+b)^2 - 2ab = c^2. Adding (a^2 + b^2) to the left side of the equation and c^2 to the right side, we get:

a^2 - 2ab + b^2 + (a+b)^2 = 2(c^2). Fixing the left hand side we get:

(a-b)^2 + (a+b)^2 = 2(c^2). Therefore, it holds true that:

(a-b)^2 + (a+b)^2 <= 2(c^2) + (a-b)^2. The equality condition is possible if a = b. Continuing on, this inequality implies:

(a+b)^2 <= 2(c^2), which ultimately gives:

a+b <= c * (2 ^ (1/2)), Q.E.D. Again, equality holds if a = b.
[/hide]

You dont need to have latex on your computer.

Just type "$\$$" before and after every latex thing

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1969Problem3

Vo Duc Dien

[quote="chess64"]Let $c$ be the length of the hypotenuse of a right angle triangle whose two other sides have lengths $a$ and $b$. Prove that $a+b\le c\sqrt{2}$. When does the equality hold?[/quote]

Cauchy-Schwarz's inequality gives $(1^2+1^2)(a^2+b^2)\geq (1\cdot a+1\cdot b)^2$

$\Longleftrightarrow 2c^2\geq (a+b)^2\ \text{Pythagorean Theorem}$

$\therefore a+b\leq \sqrt{2}c\ Q.E.D.$

The Alternative

$R.M.S.\geq A.M.$ gives $\sqrt{\frac{a^2+b^2}{2}}\geq \frac{a+b}{2}\Longleftrightarrow a+b\leq \sqrt{2}c$.

let b be base and a the altitude.let angle between c and b be x. b=ccosx and a= csinx. we have to find maximum value of ccosx+csinx which is 2^1/2c.

[hide=sol]
$a+b \le c\sqrt{2}$, so
$(a+b)^2 \le 2c^2$.
Since $a^2+b^2=c^2$,
$a^2+2ab+b^2 \le 2a^2+2b^2$, so
$(a-b)^2 \geq 0$ which is true by trivial inequality.
Going backwards gives proof.
Equality holds when $a=b$. $\square$
[/hide]

[quote=chess64]Let $c$ be the length of the hypotenuse of a right angle triangle whose two other sides have lengths $a$ and $b$. Prove that $a+b\le c\sqrt{2}$. When does the equality hold?[/quote]

We have $a^2+2ab+b^2\leq2c^2$ iff $2ab\leq c^2=a^2+b^2$, which follows immediately from the AM-GM inequality.
Equality holds for a=b, or a 45-56-90 triangle.

[hide=solution]
Note that, by the Pythagorean Theorem and AM-GM, we have $$c^2 = a^2 + b^2 \geq 2\sqrt{a^2b^2} = 2ab.$$ Therefore, we have $$c^2 + 2ab \leq 2c^2,$$ by adding $c^2$ to both sides and substituting $c^2 = a^2 + b^2$ yields $$a^2 + 2ab + b^2 \leq 2c^2.$$ Clearly, both sides are positive because $a,b,c > 0,$ so we can take the square root of both sides to get $$a + b \leq c\sqrt{2},$$ as desired.
Equality holds when $a=b,$ i.e. when the triangle is a right isosceles triangle.
$\square$
[/hide]


geometry
perimeter
probability
geometric transformation
dilation
number theory
prime numbers

24. How many ordered triples of three prime numbers exist for which the sum of the members of the triple is 24?

25. The lenghts of the sides of isosceles triangle ABC are 3x + 62, 7x + 30 and 5x + 50 feet. What is the least possible number of feet in the perimeter of triangle ABC?

26. Janelle averages 40 kilometers per hour biking on level ground. She averages 60% of her level ground speed riding uphill, and she averages 120% of her level-ground speed riding downhill. The course is level for 5 kilometers, uphill for 6 kilometers, and then downhill for 6 kilometers. How many Kilometers per hour is her average speed for the entire course?

27. A bowl contains fewer than 50 marbles, and each is either red, green or blue. The probability of drawing a red marble is 2/5 and the probability of drawing a green marble is 3/7. If two marbles are drawn without replacement, what is the probability that both are blue? Express your anser as a common fraction.

28. Brent likes to dilute his lemonade. he starts with a full cup and drinks 1/3 of its contents. he then fills the cup with water, stirs the contents, and again drinks 1/3 of its contents. He repeats this process until he has consumed one cup of liquid. What part of the original cup of lemonade remains in the cup? Express your answer as a common fraction.

29. The sum of the digits of a three-digit number is 26. The number is multiplied by 7, then by 11, and finally by 13. How many times will the digit 9 occur in the final product?

30. Joe bought a pumpkin that cost 10 cents more per pound than his sister's. Together, the two pumpkins weighed 20 lb, but joe's pumpkin was heavier. Joe paid $3.52, and his sister paid$0.48. How many pounds did Joe's pumpkin weigh?

This is just for fun

PLEASE show ALL your work.

24,27,28,29, and 30 have already been posted a bunch of times...

How long did it take you to type all that? It would be a shame is no one replied because It's a lot of work in calculations.

Ok, I did all of those yesterday night :P , hence I posted two of these :!: [hide]
30. I looked at .48, and started testing factors under 10...i started with 4, and it was correct, so the answer is 16

29. i took 899 and multiplied it by 77 (7 and 11), then 13, to get 899899, which has 4 nines, therefore the answer is 4

28. took 2/3 of 1, 2/3 of 2/3, then 2/3 of 4/9 to get 8/27, so the answer's 8/27

27. i multiplied 6/35 by 5/34 to get 3/117, so the answer's 3/117

26. 40 mph for 1/8 hr, 24 mph for 1/4 hr, and 48 mph for 1/8 hr
i counted the 24 twice, then added all of them to get 136...since i added 4 numbers, i divided 136 by 4 to get 34, so 34's the answer

25. i don't really know how to do this, but the answer's 6

24.we have 3,19,2  we have 5,17,2  and we have 11,11,2
     6 ways                  6 ways           3 ways
added together:15 ways, so the answer's 15
[/hide]
-jorian

[hide]3. $\frac{3}{119}$

for  #25 is x equal to 6 or the answer 6?[/hide]

yes i know, i just mixed up the 117 with the 119 on my paper i used yesterday...sorry,
-jorian

Nobody has solved 25 yet...

[hide="25"]The three sides are $3x+62,$ $7x+30,$ and $5x+50.$

Take each pair of equations and solve for $x.$

$x=\{3,\,6,\,10\}$

We want the smallest value of $x,$ so we choose $3$ (the first and second sides are equal in this case).

$3\cdot3+62+7\cdot3+30+5\cdot3+50=\boxed{187}$[/hide]
@ jorian - It's asking for the smallest possible perimeter, but your answer is wrong anyway... :ninja:

[hide="very easy trick for 29"]It's worth memorizing that $7\cdot11\cdot13=1001$.[/hide]

@ilikepie:whoops, i took the answer i remembered
-jorian

[quote="redcomet46"][hide]25

@jorian do you mean the smallest value of x is 6?

@ilikepie x can be 6,8,10, so the smallest perimeter is 232[/hide][/quote]
Oh... Oops... :ninja:


logarithms
calculus
calculus computations

With yearly inflection of 7% prices are given by $ P(t)\equal{}x(1.07)^t$ where $ x$ is the price in dollars when $ t\equal{}0$ and $ t$ is time in years. When $ x\equal{}1$ how fast are prices rising when $ t\equal{}10$?

Rate of change of prices is given by $ \frac{dP}{dt}\equal{}x(1.07^t\cdot \ln1.07)$.
(We treat $ x$ as a constant).

Therefore, when $ x\equal{}1$ and $ t\equal{}10$, $ \frac{dP}{dt}\equal{}1.07^{10}\cdot \ln 1.07\approx\boxed{.133\ \frac{\text{dollars}}{\text{year}}}$.


calculus
trigonometry
geometry solved
geometry

This is not a quiz, but it's from one of my calculus textbooks. It has something to do with trigonometry first of all, that's why I post it here.

The attachment below shows my problem.

[quote="asjsd"]it is very easy.....  ;)[/quote]

Please don't spam.

As for the problem, here is its geometrical core:

[color=blue][b]Problem.[/b] In space, given a right-angled triangle ABC with the right angle at A, and given a point D such that the line CD is perpendicular to the plane ABC. Denote d = AB, h = CD, $\alpha=\measuredangle DAC$ and $\beta=\measuredangle DBC$. Prove that $h=\frac{d\tan\alpha\tan\beta}{\sqrt{\tan^2\alpha-\tan^2\beta}}$.[/color]

[i]Solution.[/i] Since the line CD is perpendicular to the plane ABC, we have < DCA = 90°, so the triangle DCA is right-angled and thus we get $CD=AC\cdot\tan\measuredangle DAC$. In other words, $h=AC\cdot\tan\alpha$, so that $h^2=AC^2\cdot\tan^2\alpha$, and consequently $h^2\tan^2\beta=AC^2\cdot\tan^2\alpha\tan^2\beta$. Similarly, by considering the right-angled triangle DCB, we get $h^2\tan^2\alpha=BC^2\cdot\tan^2\alpha\tan^2\beta$. But since the triangle ABC is right-angled at A, the Pythagoras theorem yields $BC^2-AC^2=AB^2$, and thus

$h^2\left(\tan^2\alpha-\tan^2\beta\right)=h^2\tan^2\alpha-h^2\tan^2\beta$
$=BC^2\cdot\tan^2\alpha\tan^2\beta-AC^2\cdot\tan^2\alpha\tan^2\beta=\left(BC^2-AC^2\right)\tan^2\alpha\tan^2\beta$
$=AB^2\tan^2\alpha\tan^2\beta=d^2\tan^2\alpha\tan^2\beta$.

Taking the square root, we get

$h\sqrt{\tan^2\alpha-\tan^2\beta}=d\tan\alpha\tan\beta$,

so that $h=\frac{d\tan\alpha\tan\beta}{\sqrt{\tan^2\alpha-\tan^2\beta}}$, and the problem is solved.

  darij

[quote]But since the triangle ABC is right-angled at A, the Pythagoras theorem yields $BC^2-AC^2=AB^2$, and thus

$h^2\left(\tan^2\alpha-\tan^2\beta\right)=h^2\tan^2\alpha-h^2\tan^2\beta$[/quote]

I don't really understand this step.

By the way, how do you enter in all those mathematical symbols like square and alpha and beta, etc.?

[quote="renjer"]I don't really understand this step.[/quote]

???

What exactly don't you understand? $h^2\left(\tan^2\alpha-\tan^2\beta\right)=h^2\tan^2\alpha-h^2\tan^2\beta$ is perhaps the most trivial step in my solution. Maybe you were irritated by the equation $BC^2-AC^2=AB^2$, which is actually used two lines later, namely here:

$\left(BC^2-AC^2\right)\tan^2\alpha\tan^2\beta=AB^2\tan^2\alpha\tan^2\beta$.

[quote="renjer"]By the way, how do you enter in all those mathematical symbols like square and alpha and beta, etc.?[/quote]

That's called LaTeX. See http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 for some documentations.

  Darij


Prove that if $f: I\rightarrow I$ is a continuous function where $I=\left[0,1\right]$, then $f\left(x\right)=x$ for at least one $x\in I$.

Consider the function $g(x) = f(x)-x$. Since $g(0) = f(0) \ge 0$ and $g(1) = f(1)-1 \le 0$, by the intermediate value theorem there exists some $x \in [0, 1]$ for which $g(x) = 0 \Rightarrow f(x) = x$.

The one-dimensional version of the Brouwer fixed point theorem: a continuous function from a closed ball in $\mathbb{R}^{n}$ into itself must have at least one fixed point.

You shouldn't let the fact that it's easy for $n=1$ lull you into complacency about just how difficult, and using what fancy tools, it will be for $n\ge 2.$


I'm pretty happy with my scores this year, but I was wondering if someone could take another look at my solution for round 2 problem 3.  I know it wasn't written very well, but when I looked at the official solution, as the grader told me to do, I didn't see a huge difference, other than that the official solution was written better.  So my question is did I really deserve a 2?  I'm not really sure myself, but I would appreciate if someone would look at it.

My solutions for round 2 are attached.  Problem 3 is on page 4.

Please tell me what you think.

By the way, does this count as an appeal?  I can't seem to find what exactly I am supposed to do for an appeal... If anyone knows, please let me know.

facis, USAMTS protests should be posted in the grading forum, which is at the top of the USAMTS forum. Here's the link: http://www.artofproblemsolving.com/Forum/index.php?f=213. (You must be logged in to access it.)


AMC
AMC 12
USA(J)MO
USAMO
HMMT
Putnam
AMC 10

http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06AMC1012dir-pict.html

Yep.  :D

Do you know when the state stats will be up?

The AMC12 "Grand Statistics" aren't yet linked from the statistics page, but you can see them here:

http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06-stats12.html

wow, average grade 8 and below score is the highest of all :!:

Because 8th graders don't usually take it (at least in public schools I guess), so the ones that do have to be good to want to do it.  Maybe.

Ditto on the state statistics question... been waiting for that a long time.

[quote="adalton"]Wow, some kid got a perfect on both AMC 12 A and B, that's crazy, ever been done before?[/quote]

Yea, Tiankai Liu did it several years.

[quote="adalton"]Wow, some kid got a perfect on both AMC 12 A and B, that's crazy, ever been done before?[/quote]

he is from my state :)
no need to say that he is incredibly smart!(he was a USAMO HM last year~probably a winner this year)

[quote="chess64"]wtf adam was USAMO HM in 9th grade? *considering a possibility of a new role model*... :lol:[/quote]

Well actually in 10th grade but I think he made blue MOP in 9th grade anyway.

:cursing:
darn I didnt get the top Delaware AMC score

I was 2nd though (and first out of AMC 10B!)
shows how pathetic we are that 131.0 is 2nd highest in the state :blush:

[quote="chess64"]Oh yeah he did: http://web.mit.edu/rwbarton/Public/mop.doc

And so did SamE. Hey is Adam in the same grade or is he in 11th grade now?

BTW who is Arnav Tripathy? I think I've heard his name before but I don't remember where... (i think it might be mystic terminator but most likely not :blush:)[/quote]

yup mysticterminator, actually he's from NC I think.

[quote="Elemennop"]He's a guy from Georgia, very smart, wins alot of stuff. Won both of his topics at HMMT, but got 2nd to Nimish overall individual.[/quote]

North Carolinian. I'm really glad he isn't from GA! :) (I know he was on a GA iTest team last year, but that's because he's friends with one of Harrison's friends.)

Oh, my bad. I remembered some FL guy telling me they had a georgia kid on their team, but I guess I must've heard him wrong or whatever.

Him being mystic terminator, I had no clue.

[quote="chess64"]wtf adam was USAMO HM in 9th grade? *considering a possibility of a new role model*... :lol:[/quote]

Well, Reid Barton was a Gold medalist at the IMO or something crazy like that as a freshman.

Reid Barton was the first to win 4 gold medals at the IMO. He is now a 4-time Putnam fellow.

David Benjamin ranks pretty high up there. Qualified for USAMO in 7th and 8th grade. Now he's in 9th.

David Benjamin is awesome. I hear his brother is really good too.

No AMC10 perfects in GA.
That means I must be tied for #1 in the state.
Yay! :D

On the 10B, which I shall presume you took, you are #1 in GA and I am #2 (tied with a 10th grader).\

(The state statistic links work now!!!)


How can we prepare 3-(4-cyclohexen-1-yl)propanoic acid from 4-vinylcyclohex-1-ene?

first perform HYDROBORATION oxidation reaction.you get the substituted -OH at both the double bonds at the less hindered side of the less hindered carbon atom.treat with SiMe3Cl to protect the -OH group in the cyclic structure.Add SOCl2 to get the chloride,treat with KCN and hydrolyse to get the acid and the -OH group in the cyclic structure.Protect the acidic part by forming the ester. then dehydrate to get the final product.
(Is my explanation proper? :wink: )

There are several problems in your proposal:

1) What hydroboration agent will you use? Are you taking into account the possibility of the formation of a bycyclic borane due to the presence of two double bonds? What carbon of the ring double bond will be oxidized?

2) Wouldn't TMSCl protect both -OH groups?

3) In what direction will dehydration occur?

I am using the hydroboration that involves the formation of cyclic intermediate with diborane.

Then what carbon of the ring's double bond will be hydrobotated? Note that I referred to a [i]bycyclic[/i] borane, not a cyclic intermediate.


a convex polylhedron P has 26 vertices, 60 edges, and 36 faces, 24 of wh ich are triangular and 12 are quadrilaterals.  A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face.  How many space diagonals does P have?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=427647307&t=46307]AIME 2004, number 3[/url]


[quote="bpms"]How do you do problems that are like "if there are x circles and y lines how many intersection points are possible?"[/quote][hide] I think we can treat the circles as 2 lines (not sure :maybe: ), so we have $2x+y$ lines. The number of intersection points is $\frac{(2x+y)(2x+y-1)}{2}$. [/hide]

[quote="ragnarok23"]Couldn't you just find the most number of intersections of the amount of lines and add 2 for every line?   :maybe:  :huh:[/quote] Do you mean circles?  :maybe: If you do, I think it's wrong.

EDIT: My above post was incorrect. We can still treat the circles as 2 lines, but these 2 lines cannot intersect. I'm not exactly sure how to account for this.  :maybe:

[quote="lotrgreengrapes7926"][quote="ragnarok23"]Couldn't you just find the most number of intersections of the amount of lines and add 2 for every line?   :maybe:  :huh:[/quote] Do you mean circles?  :maybe: If you do, I think it's wrong.[/quote]
Shoot I read the question wrong...
EDIt:Ok, I'm getting this freaky formula: [hide]$\frac{4xy+(y-1)(y)+2(x-1)(x)}{2}$[/hide]
Which has a $\frac{1}{912.334(e^{\pi i})}$ chance of being correct.  :D

[quote="ragnarok23"][quote="daermon"]what is the formula for only lines?[/quote]
$\frac{(y-1)(y)}{2}$[/quote]
The reason it works is because you need to find how many potential intersections there are. Any two lines can intersect each other. So it is how many ways you can choose two lines. This is $\binom{x}{2}$ which as ragnarok said is also written as $\frac{(y-1)(y)}{2}$.

[quote="bpms"]How do you do problems that are like "if there are x circles and y lines how many intersection points are possible?"[/quote]

None of the lines and circles are coincident, right?

:rotfl: If they were the answer would be $8$ turned sideways.

[quote="ragnarok23"][quote="lotrgreengrapes7926"][hide="Revised Solution"] First, we count the number of intersections with the circles, which equals $x(x-1)$. Then, the number of intersections of the lines. The first line creats $2x$ new intersections, the second line $2x+1$, the third $2x+2$, etc. This totals $2xy+\frac{y(y-1)}{2}$. So, the answer is $\boxed{x(x-1)+2xy+\frac{y(y-1)}{2}}$. [/hide][/quote]
OMFG YOU GOT THE SAME ANSWER AS ME OMG OMG.
 :D[/quote]How did you get this solution. The explanation may be tedious but could you please explain.

i got (correct me if i am wrong) if x=circles y=lines
[hide]
 4xy +  (y-1)(y)(.5) [/hide]

[quote="shybox"]This is how i got it.

[hide] there are 4 circles and everyline intersects the circle 4 times so that is were we get the 4xy. then the llines intersect in the circles. so the number of times the lines intersect in the circle is the equation (y)(y-1).5 because the first line can intersect the second line and the third line can intersect both the first and second. The pattern is adding 1 each time so that is a triangle number. You subtract the 1 from the y because line one doesn't intersect itself. SOrry i am not ver good at explaining things. [/hide][/quote]
Where did you get the there are four circles.

woops :blush:

new explanation and new equation

[hide]             2xy+(y-1)(y)(.5)+(x-1)(x)            We split this equation into 3 parts lines intersecting circles+lines intersecting lines+circles intersecting circles. For lines intersecting circles (2xy) we can treat each circle as 2 lines because each line intersects each circle 2 times. For lines intersecting lines (y-1)(y)(.5) they frm triangle numbers because for example line two intersects line one (1) and line three intersects line 1 and 2 (2) ect.. so you add one each time. However you subtract one because line one does not intersect lne 1. For circles intersecting circlesit is the same idea (x-1)(x). The patern is 2 circles=2 3 circles =6. Sorry one more time cause i am not a good explainer it is easie to explain if you draw it out and find patterns[/hide]

Even better, what's the maximum number of sections those circles and lines can split the plane into? Actually, I've only ever done plane sections with just lines but adding circles shouldn't be that hard.

[quote="bpms"]How do you do problems that are like "if there are x circles and y lines how many intersection points are possible if none are coincident?"[/quote]
the numbers of intersection pints is \[\frac{(2x+y)(2x+y-1)}{2}\]


ratio
geometry
number theory
relatively prime
angle bisector

Please help me with this question:
In triangle ABC, angle bisectors AD and BE intersect at P. If a = 3, b = 5, c = 7, BP = x, and PE = y, compute the ratio x:y, where x and y are relatively prime integers.

This comes from http://155.225.48.46/statemat/stuf2004/Geoprobs.htm

This problem is a basic Geometry of Mass Points Problem:

Spoiler:

[hide]By angle bis. thm, AE:EC=7:3 and CD:DB=5:7. So, we give their lengths these masses and in order to balance each side, we find A has a mass of 3, C has a mass of 7, and B has a mass of 5. So adding 5 and 7 we find E has a mass of 10. Since B has a mass of 5 and E has a mass of 10, BP must have a mass of 10 and PE must have a mass of 5 to balance the line BE. So, x:y=BP:PE=10:5=2:1.

[/hide]

Can you explain the concept of points with masses in more detail?
I found this: http://mathworld.wolfram.com/BarycentricCoordinates.html but it doesn't explain it well.

We'll start in one-dimension:
Imagine a line, or a see-saw, or whatever. It has two endpoints and a center of mass. So, if we put a mass of $a$ on one end and a mass of $b$ on the other, with $a\neq b$, the see-saw will tilt one way or the other, it will be unbalanced. So, in order to balance the see-saw, we make the distance in from the mass $a$ to the center be $b$ and the distance from the mass $b$ to the center be $a$. So, $ab=ba$ and the see-saw is balanced. If we mulitiply one of the endpoints by some $k$, we must multiply the other by $k$ to keep the see-saw balanced, and the same holds true for the distances from the center of mass to the endpoints. The most important part, the mass of the center of mass, is always the sum of the masses of the endpoints.

Now for 2-dimensions:
Start with two line segments $AB$ and $AC$.
$AB$ has a point $F$ that is the center of mass, and masses of 3 on $A$ and 2 on $B$. First, $F$ has mass 5. Second, to balance $AB$, $3*(AF) = 2*(FB)$. So, $AF=2n$ and $FB=3n$ for some $n$, and we get $6n=6n$, so it is balanced. We want to stay as simple as possible, so we say $n=1$ which implies $AF=2$ and $FB=3$. Now, $AC$ has a point $E$ that is the center of mass, and masses of 5 on $A$ and 1 on $C$, so we know $E$ has mass 6. So, to balance $AC$, $5*(AE) = 1*(EB)$, and in simplest form, $AE=1$ and $EB=5$. If we connect the two line segments at $A$, we have a discrepancy: the masses on $A$ differ. So, we need to multiply each by some constant to make them the same. So, multiplying $AB$ by 5 and $AC$ by 3 to get a mass of 15 for $A$ in both. So, now we have $A=15,B=10,C=3,E=18,F=25$. If we connect $B$ and $C$, and give $BC$ a center of mass $D$, we find $D=B+C=10+3=13$. Now we have a triangle $ABC$. If we make the segments $AD$, $BE$, and $CF$ we have a point of intersection $X$ that we wish to balance all segments with. So:
For $AD$, $A=15,D=13$, with a center of mass $X=15+13=28$.
For $BE$, $B=10,E=18$, with a center of mass $X=10+18=28$.
For $CF$, $C=3,F=25$, with a center of mass $X=3+25=28$.
So, the mass of $X$ is equal for all and our triangle is balanced.
In general if two segments meet in a point, analyze each segment individually and if the mass of the point differs from one segment to the other, multiply by constants to make the points have the same mass. Hope that helps.

I'll leave you with this challenge:
For a triangle ABC, with D the midpoint of BC,E the midpoint of AC,F the midpoint of AB, and centroid Z, prove that AZ:ZD=BZ:ZE=CZ:ZF=2:1 using mass points.

Thanks for that explanation.
Solution to problem:
Because they're medians, the verticies of the triangle have weight 1.  Then this means that the midpoints have weight 2, since you combine the weights of the two vertices next to it.  This then means that the centroid cuts the lines AD, BE, and CF into 2:1 ratios, with the 2 part opposite the 2 weight.

It's much easier with mass points.

Medians? I thought the question wanted angle bisectors.

No, my solution to the angle bisector question was in my first post. My second post taught Wumbate Geometry of Mass Points. In that post, I posed a question: Prove that the ratio: (the distance from vertex to the centroid):(the distance from centroid to the opposite side)=2:1. And he proved it.

In the original problem, did a=3 mean that point A had a mass of 3?

No. It meant side a, opposite of vertex A, had length 3.


The correct way to use AA Postulate is to find the length of
something hard to measure such as height of a tree or flag pole.? How else can this postulate be used?

Also, how can I prove that the AA Postulate works or makes sense to students? I know it is a postulate, so I would say it just works?

What real-life object can I use to demonstrate this postulate to students in the most simplest of ways?

This is a school project.  I am not a teacher.  In the project, I must pretend to be a teacher.  Get it?

So basically you're asking us to do your project for you? You've come to the wrong forum.

asking us for help doesnt teach you anything.  it is not hard at all to just wiki it up or search it on mathworld.com   .  there are about 1,810,000 articles on "aa postulate" when you search it on google.

we arent going to do your project for you.  locked

btw...kinda reminds me of interval.  whatever.


modular arithmetic
number theory
divisibility tests
AMC

From Kalva,

n has 4 digits, which are consecutive integers in decreasing order (from left to right). Find the sum of the possible remainders when n is divided by 37.

Before I clicked on the Answers page, my idea was to brute-force the problem.

The solution is n = 9876 = 34 mod 37, or 8765 = 33 mod 37, or 7654 = 32 mod 37, or 6543 = 31 mod 37, or 5432 = 30 mod 37, or 4321 = 29 mod 37, or 3210 = 28 mod 37. So sum = 34 + 33 + 32 + 31 + 30 + 29 + 28 = 217.

How did they know that 9876-34 is divisible by 37? Or 8765-33 is divisible by 37?

To me, it looks like they also brute-forced it.

Kalva's solution doesn't look very good.

[hide="A hint for a better one"] $111 = 3 \cdot 37$ [/hide]
But such calculations are not difficult.  You can figure out the remainder upon division by $37$ just by long division, or by what I like to call "grouping."

For example,

$9876 = 9990-114 = 9990-111-3 = 9990-111-37+34 \equiv 34 \bmod 37$

What's special about $37?$ It's a factor of $111,$ which in turn makes it a factor of $999.$

Now, here's an exercise: what is the remainder when you divide $457,102,321,914,121,032,518,103$ by $37?$ (That's a 24-digit number.)

[hide="A method."]Cast out 999's. Just start adding the 3-digit blocks.
$457+109=566$
$566+321=887=-112$
$-112+914=802$
$802+121=923$
$923+32=955=-44$
$-44+518=474$
$474+103=577$

So that huge number is equivalent to $577\pmod{999}.$

Divide $577$ by $37.$ Take away $370,$ leaving $207.$ That's $15$ less than $222,$ so our number is equivalent to $-15\equiv22\pmod{37}.$

When is a number divisible by $37?$ Precisely when we get zero in this calculation.[/hide]
[hide="Explanation."]Suppose $n=\sum_{k=0}^{n}a_{k}\cdot1000^{k}$ where each $a_{k}$ is an integer between $0$ and $999.$ Then since $1000\equiv1\pmod{999},$ it follows that $1000^{k}\equiv1^{k}\equiv1\pmod{999}.$

Then $n\equiv\sum_{k=0}^{n}a_{k}\pmod{999}.$

This is the same rationale as the more familiar "casting out nines."

Somewhat related: if we use an alternating sum of three-digit blocks instead of a straight sum, we are determining the remainder after dividing by $1001.$ Since $1001=7\cdot11\cdot13,$ this serves as a test for divisibility by $7,11,$ and $13.$[/hide]

Now that the cat's out of the bag,

[hide="A fast solution"] $n$ is going to be of the form

$n = a+10(a+1)+100(a+2)+1000(a+3)$

For $a = 0, ... 6$.  Now, since

$a+10a+100a = 111a \equiv 0 \bmod 37$

And since

$1000 = 999+1 \equiv 1 \bmod 37$

We know that

$n \equiv 10+200+a+3 \equiv 213+a \equiv a+28 \bmod 37$

Which immediately gives us every remainder.  Adding them all,

$7 \cdot 28+\frac{6 \cdot 7}{2}= 196+21 = \boxed{ 217}$ [/hide]

Um, the problem essentially asks for brute force.. n can only be one of seven values, so you just find each mod 37.

[hide="Or, once you calculate the first remainder..."] Notice that the next number is $1111$ less than the previous, and $1111 = 1110+1 \equiv 1 \bmod 37$, so the remainder when dividing by $37$ decreases by $1$. [/hide]

[hide]n is of the form $1000(x+3)+100(x+2)+10(x+1)+x=1111x+3210\equiv x+28\pmod{37}$. x can be any of 0-6 inclusive, for them to be digits. Summing, $\sum_{k=1}^{6}(k+28)=\boxed{217}$.[/hide]


geometry
calculus
integration
calculus computations

Hi, I've got a problem:

[color=blue]Let $R_1$ and $R_2$ be two regions in the plane $\Pi$, let $l$ a line. If a line $l'$ varies parallel to $l$, Prove that if the segments $R_1 \cap l'$ and $R_1 \cap l'$ have the same lenght then $[R_1]=[R_2]$.[/color]

Obviously, [b]it's impossible [/b]to use integrals, it have to be used the suprem theorem.

[quote="Bictor717"]Looks like a 2D analogue of [url=http://mathworld.wolfram.com/CavalierisPrinciple.html]Cavalieri's Principle[/url], though I'm not sure how that is proved.[/quote]

lol, I know that it's the 2D case of the [u]Cavalieri's principle[/u], I've got a proof using the 3D case but I'd like a proof just using supreme theory.


Did anyone else have as bad prizes as ours (Site Award in Division B... lol)? Mine was like a math history textbook for the "math-phobic..."

I LOVE RIPPED USED DOLLAR STORE BOOKS <3

[quote="worthawholebean"]What place did you guys come in?![/quote]
1st in Las Vegas; 3rd in the nation. Sorry to have "rubbed it in," but I just had to do that. :)

Um, Georgia A got the same thing for tying 7th in the nation.  It's top 7 overall in the A Division and top 5 in the B Division.  Unfortunately, I missed my chance of getting my second NSpire cuz I was on the uber-lose B team, which had to compete in the A Division cuz it was too beast for the other B teams.  Our C team competed in the B Division and totally phailed.  But then again, we only had 11 ppl.

I didn't get a prize from ARML, but for coming 2nd in Upstate NY's practice session, I got [url=http://www.amazon.com/Math-Doesnt-Suck-Middle-School-Breaking/dp/1594630399/ref=pd_bbs_2?ie=UTF8&s=books&qid=1212448149&sr=8-2]Math Doesn't Suck[/url] (which I think my coach got at the coach's meeting).  An excellent book :D  :rotfl:

[quote="E^(pi*i)=-1"]Huh, Chesapeake finished 4th in Division B, and all we got was the "What's Happening in the Mathematical Sciences" book - no Mathematica, no Nspire.[/quote]

wait you all got the same books?

we had a pile of 16 different books that were all used (some were even ripped) and were dumb things like one that argued teaching english in schools is obsolete because of word processors but teaching math is still useful.

[quote="junggi"][quote="worthawholebean"][quote="i_like_pie"]Mathematica 6 (student version) + TI-Nspire > math history textbook. Hurray for SFBA team B! :D[/quote]What place did you guys come in?![/quote]
3rd[/quote]

huh, is that in B division?  NC A got 3rd overall and we got TI-Nspires + a mystery gift (almost no one opted for Mathematica).

Does anyone know what the mystery gift is?

Also, GA site winner got Maple.

[quote="b-flat"][/quote][quote="junggi"][quote="worthawholebean"][quote="i_like_pie"]Mathematica 6 (student version) + TI-Nspire > math history textbook. Hurray for SFBA team B! :D[/quote]What place did you guys come in?![/quote]
3rd[/quote][quote="b-flat"]

huh, is that in B division?  NC A got 3rd overall and we got TI-Nspires + a mystery gift (almost no one opted for Mathematica).

Does anyone know what the mystery gift is?

Also, GA site winner got Maple.[/quote]
yeah b division
the a team got 4th a division

[quote="jb05"]Site winner also got "What's Going on in the Mathematical Sciences".[/quote]

You don't need to be a site winner to get it. All you need is a 7+.

[quote="E^(pi*i)=-1"]Oh whoops I forgot that I got that book from scoring 7 on Individuals.  So Chesapeake actually didn't get [i]anything[/i] for 4th place Division B.[/quote]That's pretty stupid...

Oregon was 5th place in Division B, and everyone on the team got Ti Nspire calculators and Mathematica... this feels inconsistent.  Did the different sites have different prizes or something?

As I understand it, if you are a member of one of the top eight teams in the A division or of one of the top five teams in the B division, you will get an Nspire [i]in the mail[/i] if you did not get one on site. I think every such person should have already gotten a letter with instructions on how to receive Mathematica.

If I remember correctly, the SFBA A Team was to receive some other obscure prize as well, and does anyone know when we will actually get our Nspires?

Well, i_like_pie told me he already received it. I have not received mine yet, but I sent in the form to receive Mathematica, though I have it already :D

wow, how did nikita get it that fast lol

[quote="xscapezaer"]Well, i_like_pie told me he already received it.[/quote]
Epic fail. That was a joke. :roll:

According to the NC coaches, the Nspires are sent to the coaches, who then send them to the students, so it will probably be a while before we get them (I was debating about putting the MOP address or my real address).

Someone actually got ACoPS at Penn State. And it was definitely not ripped or anything. That book costs like 60 bucks!

[quote="mathling235"]That book costs like 60 bucks![/quote]

Mathematica costs $ \$$150
TI Nspire costs $ \$$150
Maple costs $ \$$100

ARML saved money buying ACoPS for some people!

I suspect that no purchase was required. All of this was donated. Which, to me, makes the TI-Calcs even more impressive. Donating software is cheap. DOnating actual caluclators can be expensive. And don't forget the 300+ calculators they donated to MATHCOUNTS!

Thanks TI.

I have a feeling that it costs significantly less to create them than the price that they sell them for.

Are NYC Y getting their certificates to Mathematica revoked?

Yeah they don't buy the prizes; I was just responding to the person above me.

[quote="E^(pi*i)=-1"]Oh whoops I forgot that I got that book from scoring 7 on Individuals.  So Chesapeake actually didn't get [i]anything[/i] for 4th place Division B.[/quote]

Everyone on the team will receive a Texas Instruments TI-Nspire calculator and a set of Ultra Tangoes, courtesy of Rex Games.  They didn't have them at the meet, but they gave me a form to send for them.  So they will be coming!

Best,
Dr. Asaro

[quote="gtf"]Agreed, but even that price is significantly more than it takes to produce a software release.[/quote]

What?  The cost of developing software FAR exceeds the cost of hardware in almost every case (think of Windows, video games, Microsoft Office, etc).

In the case of the TI-nSpires, they contain a 90? mhz processor, a nice 16 level grayscale screen, 16mb RAM, 22MB ? flash memory, and a fairly cheap plastic case.  None of that is even close to being expensive.  I would estimate ~$ \$20\minus{}\$25$ to produce it.

Well the software cost only occurs once no matter how many are sold while the hardware cost is per unit. Software is easy to give away since after a long period of time, the price of development has probably already been made up.

That's exactly my point. After 15 years in the software industry including both networking and cell phones, I am well aware of the cost of both software and hardware. At Ericsson, we would do almost anything to shave $ \$$5-10 off of the bill of materials cost, but were perfectly willing to assign several more software engineers to a project if the need was there. When you are looking at donated material, companies that donate hardware (even cheap hardware) are far more impressive to me that software donations. Don't get me wrong, I am VERY appreciative of ALL of the sponsors of ARML. But the ones that donate tangible things are sacrificing just a little bit more.

Tom

[quote="jb05"]Site winner also got "What's Going on in the Mathematical Sciences".[/quote]

Site winner+super lucky dude (points to self)

dude like the top 5 teams got calculators for each kid.  jeff manning got three ti-inspires during the wholle thing and i think almost 2000 dollars in winnings

[quote="worthawholebean"]Has anyone else gotten their calculators yet?[/quote]
Nope... :/
I'm told our coach is looking into it and will get back to us..

Come to think of it I haven't even seen the Mathematica certificates, though that's not a huge deal since it would be my 3rd free copy--didn't even have to win one of them :)

Wolfram gives away too many free copies of Mathematica.  :P

I don't know what happened to the N-Spires either. They're actually coming, right?  :(

Darn, I only have two free copies of Mathematica now. Time to start studying ;)

My coach says we haven't gotten ours either...

So, uh.. any word on the N-spires? I really do want it before I leave for college ^^

[quote="b-flat"][quote="gtf"]Agreed, but even that price is significantly more than it takes to produce a software release.[/quote]

What?  The cost of developing software FAR exceeds the cost of hardware in almost every case (think of Windows, video games, Microsoft Office, etc).

In the case of the TI-nSpires, they contain a 90? mhz processor, a nice 16 level grayscale screen, 16mb RAM, 22MB ? flash memory, and a fairly cheap plastic case.  None of that is even close to being expensive.  I would estimate ~$ \$20 \minus{} \$25$ to produce it.[/quote]
in case you are forgetting the electronic components cost money and they just don't "work" they need to be programmed and that's not very easy to do cause its a low level language like assembly or something.

sorry guys....i'm really new to arml...just got invited to take the pretest today.  Do indivuduals get prizes for being #1 in state or something?

What is this "state" of which you speak?

ARML is both a team competition and an individual competition, on a national scale. Some prizes are given to every team member of the top few teams in Division A and the top few teams in Division B. The calculators that people were talking about in this topic from last year were team prizes. (And no, I don't know the precise definition of "few" in what I just said.) There are also prizes - cash - for the top 10 individuals.

All of this is national - although ARML takes place in several locations, it is a single simultaneous competition. So it's just the top 10 individuals in the U.S. and Canada that get those prizes.

Also: this is a 2008 topic and is the wrong place to continue any conversation. I'm locking the thread.


Consider $ A,B,C \in M_n(\mathbb{C})$ such that $ AB\equal{}BA, AC\equal{}CA$ and $ BC\equal{}CB$. If $ A^3\plus{}B^3\plus{}C^3\equal{}A\plus{}B\plus{}C\equal{}O_n$ then prove that $ ABC\equal{}O_n$.


Relative to a fixed origin O, the points A, B and C have position vectors

$ a \equal{} \minus{}i \plus{} \frac{4}{3}j \plus{} 7k, b\equal{}4i \plus{}\frac{4}{3}j \plus{}2k$ and $ c\equal{} 6i \plus{} \frac{16}{3}j \plus{} 2k$, respectively.

(a) Find the cosine of angle ABC.

The quadrilateral ABCD is a kite K.

(b) Find the area of K.

A circle is drawn inside K so that it touches each of the 4 sides of K.

(c) Find the radius of the circle, giving your answer in the form $ p\sqrt{q} \minus{}q\sqrt{p}$, where $ p$ and $ q$ are positive integers.

(d) Find the position vector of the point D.

I am pretty sure this is the last question of this years maths AEA right?

You can use the dot product for the first part

For B you can use symmetry and your answer to A to find sin(theta) [using $ cos^2(\theta) \plus{} sin^2(\theta) \equal{} 1$ and a certain formula for the area of the triangels, which is ...

For C symmetry is once again your friend - it enables you to find the centre of the incircle. Can you see how B helps you to find the radius of the incircle ? You have a triangle - the area is - and is related to r because ...

For D - you can go on a slog of vector bashing using your current information - however, in this exam I know that you are not required to have any knowledge of the cross product [or should I say - if you do have knowledge of it it is only due to you going beyond the sylabus] which will help alot;

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

because you know the area of the kite and hence ...


Kevin wants to distribute 10 pieces of candy to the 4 teletubbies! In how many ways can he do this if Laa Laa must get at least 3 pieces but Tinky Winky couldn't get more than 6 pieces?

Lets consider a divider argument.  With 10 pieces of candy divided into 4 groups, we have 9 places to place 3 dividers.  But Laa Laa must get three pieces, so lets subtract off his 3 from the total knowing that the recipient of those 3 is already predetermined.  Thus we really only have 7 pieces of candy to distribute.  But now we have to remember to add 4 pieces into the mix, which we will removeat the end by taking back one from each Teletubbie (this is necessary because when we place the dividers, there always has to be at least one object between two dividers).  So there are $ 7 \plus{} 4 \equal{} 11$ items to divide into $ 4$ groups, for $ {}_{10}C_3 \equal{} 120$ ways to distribute the candy.  But in one case Tinky winky will have all $ 7$ pieces that are up for grabs, so there are really only $ 119$ ways to distribute the candy under the given conditions.


algebra
polynomial
inequalities
algebra proposed

Let $P(x)=\sum_{i=0}^na_ix^i$ such that:
1) $\{a_0,a_1,\dots,a_n\}=\{0,1,\dots,n\}$;
2) $P(x)$ has $n$ real roots.

Prove that all roots of $P(x)$ are rational numbers.

[hide="Click to reveal hint"]
Note that first of all, none of the roots can be positive. Thus all roots are non-negative. Now with that in mind factor the polynimial as A(x+r1)(x+r2)...(x+rn) for positive reals r1,r2...,rn. Now use the the fact that {a0,a1..a_n} = {0,1,2,...,n}, and Cauchy Schwarz inequalities applied on the sum of the products of the r's taken k at the time, with the sum of the products taken n - k at a time and  you can do a lot from here.
[/hide]

Hope this helps

Thank you for your hint, but I know the solution, I posted the problem here because I think it is pretty nice and somebody would be glad to solve it.

I certainly was glad to solve it. It was certainly a nice problem.


AMC
AIME
AMC 10
probability
AIME II
AIME I

I took the AIIME II 2006 and scored a 9.  I think the AIME was easier this year because most people in my school's math club could do all of the problem's 1-7 and the last 5 questoons were easier than usual.  Also, I got a 6 on the AIME last year and people said it had gotten way easier that year. Please give your opinions. :P

this year's AIME II is easier than AIME I
AIME I needs a lot of thinking but not much calculations. And if you cannot think out a proper way to solve, you cannot use an easy-complex(stupid) way to solve either.
To AIME II is different, if you cannot think out a proper way(smart) to solve, you can still use a stupid(need calculation) to solve.
like #11, you can sum up(complex way), but also you can use easy way...

I thought the AIME II was significantly easier but since not a lot of people took it (as far as I know, most people took the first one), the cutoff won't change much.

Ehhh I would say this AIME was pretty easy, but the problems (except #13) were like 328094321x more FUN than the AIME I. Well actually maybe it was this one's #15. It was the only problem that actually surprised me when I figured it out.

I disagree with the people saying the aime 2 was much easier.  the first 5/6 questions on both tests were quite approachable to most and easy if you had done any practice aime's.  on aime2, 9, 12 and 13 all were difficult, while 11, 14, and 15 were fairly doable (15 if you saw the trick).  On the AIME1, 10 was much easier than 10 on AIME2 if you just thought about it, 11 was about the same as 11 on aime1, and 12 and 14 were fairly hard.  13 might have looked scary but didn't take long.  I think 15 on the AIME1 was much easier (and much more guessable) than 15 on the AIME2.  Obviously, all of this is just my opinion and I'm sure many people would disagree with me, but I don't think there is a clear winner in terms of difficulty here.

[quote="jen7"][quote="Phelpedo"]I think they are about the same difficulty. AIME I was considerably easy too.[/quote]

I totally agree.  Neither test was as easy as last year's though.  AIME I this year was easier than the AIME II, but I took the second one -- I wish I had taken the first.  :(

Consider #13 on the AIME II--there was no "easy method", and it took significantly longer than that on the first AIME.[/quote]

Hmm, someone else posted that they thought the II was easier, but maybe they were biased because they took the I. Hmm sounds like they were biased, seeing as you thought the I was easier because you took the II. At any rate, it now seems like both contests were about the same difficulty.

Note that I read a statement by Steve Dunbar saying that the AMC tries to make the AIME I and II similar in difficulty level, and only makes the indecies different when absolutely necessary (like last year).

it would have been faster if i had brute forced it :blush:

[quote="noneoftheabove"]I think the comparison of amc 12a to b is the same as AIME1 to AIME2. I think the hard ones were hard and the easy ones were easy for the second one of both, and the others were just fairly average for the first ones. I scored higher on the AIME2 though, by any chance, did you get 822 for number 11?[/quote]

No i got 844. I still don't see where I went wrong. Maybe i should do the problem again. Then again I emember it was such a pain to do... I used the whole front and back of a pice of big scratch paper.

[quote="xxazurewrathxx"][quote="jen7"][quote="Phelpedo"]I think they are about the same difficulty. AIME I was considerably easy too.[/quote]

I totally agree.  Neither test was as easy as last year's though.  AIME I this year was easier than the AIME II, but I took the second one -- I wish I had taken the first.  :(

Consider #13 on the AIME II--there was no "easy method", and it took significantly longer than that on the first AIME.[/quote]

Oh please, #14 took 3 seconds on the AIME II and #15 didn't take 3 seconds but it was doable, unlike AIME I. I might be wrong because I only saw #1, #11, #14, #15 on the AIME II. But everyone at FL agrees that it was a lot easier.[/quote]

wow according to xxazurewrathxx the whole aime II test should've only taken ppl about a minute or so to do!  :rotfl:   I don't think you can demean the AIMe II like that in comparison to the AIME I.  I think ppl are just looking the test they didn't take and solving it in a none stressful environment convincing them that the other test was much easier.  I believe that the AMC board specifically tries to maintain the same level of difficulty for both tests and may even try to make the AIME II harder as students get 2 more wks. to prepare ffor it. O well i'll get off my soapbox now.

No, Chenyu is right, the AIME 2 is definitely noticably easier than the 1 and that should be accounted for. I shouldn't be able to solve the last problem on an AIME by a method I use to solve basic algebra problems on local tournament tests (that is, square and add to get rid of common terms)

[quote="xxazurewrathxx"][quote="jen7"][quote="Phelpedo"]I think they are about the same difficulty. AIME I was considerably easy too.[/quote]

I totally agree.  Neither test was as easy as last year's though.  AIME I this year was easier than the AIME II, but I took the second one -- I wish I had taken the first.  :(

Consider #13 on the AIME II--there was no "easy method", and it took significantly longer than that on the first AIME.[/quote]

Oh please, #14 took 3 seconds on the AIME II and #15 didn't take 3 seconds but it was doable, unlike AIME I. I might be wrong because I only saw #1, #11, #14, #15 on the AIME II. But everyone at FL agrees that it was a lot easier.[/quote]

Hm, and how would you know what everyone in FL thinks? It seems to me that most of the people complaining that the AIME II was easier are AIME I people, and vice versa. I don't think that #15 on the AIME I wasn't "doable." #13 on AIME I was like an AMC10 problem. I must say that it's pretty convenient you only saw 1,11,14,15; why don't you go have a look at the entire test?
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2006

Hmm, I found AIME I much easier (13 unofficially with 2 stupid mistakes). All the problems were like 2 liners if you thought it out.

AIME 2 on the other hand had too much calculation and brute force... I got an 11 with 2 mistakes... But I'm just abnormal.

[quote="MithsApprentice"]I think most people tend to say that the one they took was easier than the other. I cannot, for the love of all that's good, figure out why anyone would do that. The only guess I have which might make a SLIGHT bit of sense is that they're all afraid of having taken the easier one because that would mean it takes a higher score to qualify, and no one wants to be optimistic (remember the "oh no I only got a 12"s?) around here.

Now to be honest, I haven't looked at the AIME II in its entirety; nor have I taken it under testing conditions. But just briefly glancing at the problems gives the impression that it was easier than the first AIME. Besides, if you account for the fact that both TJ and AAST took the second AIME, qualification is supposedly proportional to the number of takers, and both schools regularly churn out a dozen qualifiers a year...where does that put the qualifying index/floor on AIME II?[/quote]

I think they tend to say the one they [i]didn't[/i] take was eaiser...

[quote="MithsApprentice"]I think most people tend to say that the one they took was easier than the other. I cannot, for the love of all that's good, figure out why anyone would do that. The only guess I have which might make a SLIGHT bit of sense is that they're all afraid of having taken the easier one because that would mean it takes a higher score to qualify, and no one wants to be optimistic (remember the "oh no I only got a 12"s?) around here.

Now to be honest, I haven't looked at the AIME II in its entirety; nor have I taken it under testing conditions. But just briefly glancing at the problems gives the impression that it was easier than the first AIME. Besides, if you account for the fact that both TJ and AAST took the second AIME, qualification is supposedly proportional to the number of takers, and both schools regularly churn out a dozen qualifiers a year...where does that put the qualifying index/floor on AIME II?[/quote]
I think you're right about that, MithsApprentice, and I hope AMC does take that into account. Since so few people take the AIME II than the AIME I, it is much more dramatically affected by big schools such as TJ and AAST. If they are basing the indices off of statistics, this might throw the results off.

[quote="amitavadhar"]hmm... it makes you wonder maybe AMC should make two seperate floors and take ppl from each aime based on the proportion of how many ppl took it. Or maybe i'm just overcomplicating things... :oops:[/quote]
I think they might. I hope not because Thomas Jefferson took the II  :(

oh really...? i guess i don't know anything  :P

[quote="Phelpedo"][quote="amitavadhar"]hmm... it makes you wonder maybe AMC should make two seperate floors and take ppl from each aime based on the proportion of how many ppl took it. Or maybe i'm just overcomplicating things... :oops:[/quote]
Isn't that what they did last year?[/quote]
no they had two separate indeces and one floor last year

[quote="nat mc"]no they had two separate indeces and one floor last year[/quote]

That may have been because they only needed one floor. (i.e. the two floors were the same)

[quote="amitavadhar"][quote]like #11, you can sum up(complex way), but also you can use easy way...[/quote]

Yeah, i spent 15 min bruteforcing it but i still missed it. stupid arithmetic mistakes and i was only 12 off too... :([/quote]

I was SO mad when I missed that problem. I spend 20 minutes summing it, and ended up missing the answer by one.  :(

It's called Karma justice for trying to use brute force.

Yeah, but I know some people who bruteforced it and got it right. Oh well, maybe karma will kick back in later and make them fail at USAMO.  :D

[quote="Magixmaniak"][quote="amitavadhar"][quote]like #11, you can sum up(complex way), but also you can use easy way...[/quote]

Yeah, i spent 15 min bruteforcing it but i still missed it. stupid arithmetic mistakes and i was only 12 off too... :([/quote]

hahaha, i had to bruteforce it three whole times to get the correct answer...since i kept messing up over and over again...so that took about 20 minutes[/quote]

[quote="amitavadhar"]Yeah, but I know some people who bruteforced it and got it right. Oh well, maybe karma will kick back in later and make them fail at USAMO.  :D

[/quote][quote="Magixmaniak"][quote="amitavadhar"][quote]like #11, you can sum up(complex way), but also you can use easy way...[/quote]

Yeah, i spent 15 min bruteforcing it but i still missed it. stupid arithmetic mistakes and i was only 12 off too... :([/quote]

hahaha, i had to bruteforce it three whole times to get the correct answer...since i kept messing up over and over again...so that took about 20 minutes[/quote][quote="amitavadhar"][/quote]

well, i'll take you with me when i fail!!  :laugh:

[quote="Phelpedo"]I'm surprised that it only took you 20 minutes to do it, considering that you messed up and everything. That's nearly how long it took me to solve it, and I didn't use brute force.[/quote]

That's the point: brute forcing makes it easier and an almost sure way of solving it if they don't make stupid mistakes.

[quote="Phelpedo"]My point was that it surely took longer than 20 minutes (why don't we have some opinions from other brute-forcers), and in adding up twenty-some numbers up to 6 digits long, there is a high probability for error.[/quote]

Notice the recursive relation: $2S_{n-1}-S_{n-4}$. It's much easier to brute force this way, thinking about only the sums and not the terms.

[quote="Phelpedo"]...You're still adding up very big numbers, and it's still pretty slow. If you're going so far as to find the recursive definition for sums, why not try to actually solve the problem?[/quote]

Think about it: to get to about $S_14$, which is where the numbers become 3-digit, only takes about 2-3 minutes. Then you have 14 more to go, and if you focus, and in particular, organize yourself well (I made a chart for my calculations, each row represented an $S_n$), each recursion should take less than a minute. Naturally, brute-forcing here is much more reliable than attempting to find the elegant solution, so I would pick brute-forcing.

[quote="Phelpedo"]My point was that it surely took longer than 20 minutes (why don't we have some opinions from other brute-forcers), and in adding up twenty-some numbers up to 6 digits long, there is a high probability for error.[/quote]

You only need the last three digits though, so it's really not that bad.


Let $B=\{a_1;a_2;...;a_n\}$,s be a positive integer . Consider functions :
$f_i(x):B ->\{0;1\}$(i=1;2;...;s)
Put $\sum_{i=1}^s\sum_{x\in B}f_i(x) = \alpha n(\alpha\geq1)$
Prove that: $\forall k\leq \alpha, k\in Z^+, \exists i_1;i_2;...;i_k \in \{1;2;...;s\}$such that:
$\sum_{x\in B}\prod_{j=1}^k f_{i_j}(x)\geq \displaystyle\frac{n(\alpha - k+1)}{C_{s}^k}$


[quote="AlexandrosG"]Χαιρετω τους φιλους του forum. Με λενε Αλεξανδρο και μοιραζομαι μαζι σας το ιδιο παθος για μαθηματικα.
Προτεινω το εξης προβλημα Δινονται οι: a,b,c,d>0   με α^4+b^4+c^4+d^4=4   Να αποδειχθει οτι (a+b)(a+c)(a+d)(b+c)(b+d)(c+d)<=64
   (ισως να υπαρχει καποιο καλυτερο φραγμα). :)[/quote]

Καλωσήρθες στην παρέα μας!  :welcomeani:

Εχουμε:

$ LHS\leq (\frac {3\sum{a}}{6})^6 \equal{} (\frac{\sum{a}}{2})^6\leq (\frac {4}{2})^6 \equal{} 64$

Χρησιμοποίησα στην πρώτη ανισότητα την ΑΜ-GM και μετά την:
$ (a \plus{} b \plus{} c \plus{} d)^2\leq 4(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \equal{} 16$

[quote="bilot4"][quote="AlexandrosG"]Χαιρετω τους φιλους του forum. Με λενε Αλεξανδρο και μοιραζομαι μαζι σας το ιδιο παθος για μαθηματικα.
Προτεινω το εξης προβλημα Δινονται οι: a,b,c,d>0   με α^4+b^4+c^4+d^4=4   Να αποδειχθει οτι (a+b)(a+c)(a+d)(b+c)(b+d)(c+d)<=64
   (ισως να υπαρχει καποιο καλυτερο φραγμα). :)[/quote]

Καλωσήρθες στην παρέα μας!  :welcomeani:

Εχουμε:

$ LHS\leq (\frac {3\sum{a}}{6})^6 \equal{} (\frac {\sum{a}}{2})^6\leq (\frac {4}{2})^6 \equal{} 64$

Χρησιμοποίησα στην πρώτη ανισότητα την ΑΜ-GM και μετά την:
$ (a \plus{} b \plus{} c \plus{} d)^2\leq 4(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)\equal{}16$[/quote]

ena typo mono dimitraki-tin idia lysh eixa ki ego.

Sthn teleftaia isotita thelei mikrotero h iso anti gia iso gt to S ton a^4 einai iso me 4 kai oxi to athroisma ton tetragonon.

:blush: Βασικά δεν ήταν typo... apla imoun aprosektos sta dedomena alla eytyxos i lysi dn allazei ki einai eykolo na apodei3oume oti $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\leq \sqrt{4(a^4\plus{}b^4\plus{}c^4\plus{}d^4)}\equal{}4$


Find all function $f: Z \to  R$ such that $f(1)=\frac{5}{2}$ and that $f(x)f(y)=f(x+y)+f(x-y)$, for all integers $x$ and $y$

[quote="Rust"]$y=0$ give $f(0)=2$. From $f(x+1)=\frac{5}{2}f(x)-f(x-1)$ we get $f(n)=2^{n}+2^{-n}$.[/quote]
Why? Can you explain more clearly?


quadratics
number theory proposed
number theory

prove that $3$ is primate-root mudolu every prime number with form of $2^n+1\ \ (n>1)$

Firstly note the well-known fact that if $P = 2^n+1$ is prime then $n = 2^k$ for some $k$.  Now consider the order of 3 in the multiplicative group modulo $P$.  The order of this group is $P-1$, so the order of 3 is a power of 2.  Now $n \ge 2$, so $P \equiv 1 \bmod 4$ and by quadratic reciprocity $\left(\frac{3}{P}\right) = \left(\frac{P}{3}\right) = -1$ since $P \equiv 4^{2^{k-1}}+1 \equiv 2 \bmod 3$.  So the order of 3 modulo $P$ contains as large a power of 2 as the order of the multiplicative group, and hence the order of 3 is as large as possible: that is, 3 is a primitive root.


AMC USA(J)MO USAMO geometry inequalities probability MATHCOUNTS

Hi, all. I'm trying to prepare for the USAMO next next week, but I'm running into some slight difficulties because of the difficulty of the problems. Does anyone have any suggestions as to how to practice more efficiently?

Thanks.

I disagree.  I find that I get almost nothing out of working on a problem for half an hour and then reading a solution.

I prefer to work on a problem for an hour and a half or two hours at a time.  Even if I can't solve it in that amount of time, I continue to work on it another time.  I only look at the solution when I feel completely hopeless, and I think I have no chance of solving it.

t3hggnore, I suggest you work on mostly USAMO #1 and #4, or try other countries' olympiads.  Some olympiads tend to be a little easier or much easier than the USAMO.

hopefully beta is kidding, i suppose that could work if you could apply all the techniques used, but i think that you should not really cram, just do problems for fun, i [b]NEVER[/b] look at answer until i feel that i have a satisfactory solution, just move on, you don't need to feel defeated, after you grow mathematically, you can come back and give it another shot

[quote="pieterminate"]Ahhh, I usually consider a problem a "lost cause" if I can't solve it in .5-1 hour.
But I meant remember the solution as how they reached it anyways.  But that's just me.  :)  :roll:[/quote]

You have 4.5 hour! Unless you had some serious olympiad practice before, every problem on USAMO will be a "lost cause".. You can't really understand how they reached the solution unless you try really hard to solve the problem yourself.

[quote="pieterminate"][quote="beta"][quote="pieterminate"]Ahhh, I usually consider a problem a "lost cause" if I can't solve it in .5-1 hour.
But I meant remember the solution as how they reached it anyways.  But that's just me.  :)  :roll:[/quote]

You have 4.5 hour! Unless you had some serious olympiad practice before, every problem on USAMO will be a "lost cause"..[/quote]

I haven't, but for some odd reason, I can usually solve about 2.5 of them(out of 3) within 2 hours.. :roll:[/quote]

Well congrats! You will definitely be an USAMO winner this year!

I can solve 6.7 out of 6 within 10 minutes

[quote="DPopov"]I'd say take a guess at the topics that you think might show up on this year's USAMO (using past tests as guides) and focus on one or two areas so that you have a better chance of solving that type of problem. Again, it's never a guarantee since last year I tried to focus on inequalities and there wasn't a single one on there. Anyway, that's just my 2 cents...[/quote]

Yeah, that was annoying how there wasn't a single inequality... it was by far my best subject last year. I'm hoping to be a little more balanced this year. At this point, I think focusing on the big picture is more important. Try working on proof-writing and just spending time thinking about problems (unusual, I know :P). The big difference is the immense amount of time you have to solve problems; you don't have 4.5 hours for no reason.

And solving 2.5 out of 3 problems on either day of the USAMO is incredibly difficult... at that level, you're going to the IMO this summer.

[quote="calc rulz"][quote="beta"][quote="pieterminate"]Difficulties as in the problems being to easy  :rotfl:  ;) .
But a good way to practice would be to not do a problem for more than an (half if you are desperate) hour.  Then look at the solutions and remember them.[/quote]

No the best way is definitely to memorize every single problem on kalva and memorize memorize every single solution, you will definitlely do really well on USAMO...

Lol but seriously even the most experienced olympiad problem solvers do not solve USAMO problems that fast, the problems requires experimentation and that takes time.[/quote]

Hmm yeah but if you have 2 weeks to prepare and only want to solve 1 problem, isn't it a good idea to read a lot of solutions just to get an idea what concepts/ideas one tends to need to know to solve these problems?[/quote]

That's what I meant!  :)

[quote="pieterminate"][quote="beta"][quote="pieterminate"]Ahhh, I usually consider a problem a "lost cause" if I can't solve it in .5-1 hour.
But I meant remember the solution as how they reached it anyways.  But that's just me.  :)  :roll:[/quote]

You have 4.5 hour! Unless you had some serious olympiad practice before, every problem on USAMO will be a "lost cause"..[/quote]

I haven't, but for some odd reason, I can usually solve about 2.5 of them(out of 3) within 2 hours.. :roll:[/quote]

I can't tell if you're kidding or if you're Richard McCutchen...

[quote="JSteinhardt"][/quote][quote="pieterminate"][quote="beta"][quote="pieterminate"]Ahhh, I usually consider a problem a "lost cause" if I can't solve it in .5-1 hour.
But I meant remember the solution as how they reached it anyways.  But that's just me.  :)  :roll:[/quote]

You have 4.5 hour! Unless you had some serious olympiad practice before, every problem on USAMO will be a "lost cause"..[/quote]

I haven't, but for some odd reason, I can usually solve about 2.5 of them(out of 3) within 2 hours.. :roll:[/quote][quote="JSteinhardt"]

I can't tell if you're kidding or if you're Richard McCutchen...[/quote]

No he's Brian Lawrence(with a probability of 1.01 because Brian Lawrence had serious olympiad practice). Actuallly probably not, I can solve 2.5 (out of 3 mathcount problems) within 2 hours also.

[quote="beta"][quote="calc rulz"]

Hmm yeah but if you have 2 weeks to prepare and only want to solve 1 problem, isn't it a good idea to read a lot of solutions just to get an idea what concepts/ideas one tends to need to know to solve these problems?[/quote]

Well it's probably better to review all the basic concepts, review what you do know. Blindly reading solutions won't really help that much.[/quote]
More like, you should read the solutions enough so that you are able to explain the answer to someone else.

[quote="probability1.01"][quote="pieterminate"]I haven't, but for some odd reason, I can usually solve about 2.5 of them(out of 3) within 2 hours.. :roll:[/quote]

Ehhh well I can usually read about 2.5 (out of 3) USAMO problems within 2 hours too...[/quote]

Where can I learn to improve my speed reading skills so that they're as good as yours? I can never solve any problems because by the time I read them all, the test is over.

I suggest reading the last sentence of each problem only.

Who would have guessed! Math isn't about problem [i]solving[/i], it's all about problem [i]reading[/i].

AoPR!

[quote="pieterminate"][quote="beta"][quote="pieterminate"]Ahhh, I usually consider a problem a "lost cause" if I can't solve it in .5-1 hour.
But I meant remember the solution as how they reached it anyways.  But that's just me.  :)  :roll:[/quote]

You have 4.5 hour! Unless you had some serious olympiad practice before, every problem on USAMO will be a "lost cause"..[/quote]

I haven't, but for some odd reason, I can usually solve about 2.5 of them(out of 3) within 2 hours.. :roll:[/quote]

Well I suppose that if I looked at the solutions every half hour, I would be able to solve 2.5 too...

Then I can solve $4$ of them looking at the solution every half hour for 2 hours!  :D

No, 30 minutes to read the questions first.

[quote="Fermatprime"]Yeah, I can find 2.5 problems for solutions in 2 hours too.

And you should read the last [i]word[/i], not the last [i]sentence[/i].[/quote]

Err, don't you mean the last [i]letter[/i]? Actually, make that the last [i]period[/i].

Take a ruler and measure the last 42 mm of the problem. Only read that part. That way, you'll be able to spend more time on the problem, and less time wasted reading. ;)

And by the way (no offense) I've only seen pieterminate post solutions to like 3 problem ever, so somehow I doubt he can solve 2.5/3 USAMO problems in actual testing conditions. As in, no solution manual in front of you, no calculator, etc.

I can solve next year's USAMO problems now.

I can solve 2.5 questions easy! Question 1, and then 0.3 of each of the other questions 2-6 :P

[quote="chess64"]
And by the way (no offense) I've only seen pieterminate post solutions to like 3 problem ever, so somehow I doubt he can solve 2.5/3 USAMO problems in actual testing conditions. As in, no solution manual in front of you, no calculator, etc.[/quote]

Well, I've never posted a solution ever, and I can solve 3 USAMO problems in 2 hours in actual testing conditions*...

*if I've seen the solutions before.

But seriously, I don't think the number of solutions posted on AoPS necessarily correlates with skill on USAMO.

And 0.3 of a question = 2 points partial credit. Which is somwhat comforting and somewhat scary. I guess #1 and 2 points PC for 2-6 is 19 points though....

Well he did say that if he can't solve a problem in 1/2 hour, then he goes to the solution. So if we allow 5 minutes to read the problem, 30 minutes of working time, and then 15 minutes to comprehend the solution, we have about 2.5 problems in 2 hours.

[quote="pieterminate"]Hmmm, let's just say that I meant .25 problems and forget this.  :)[/quote]

No, let's not. This is fun.

No, it's annoying. Fractions of a problem? Pssh.

This is a serious post.  At least, the first part is.  Where can I find solutions to the old USAMO problems.  I have the sets from [url=http://www.unl.edu/amc/]www.unl.edu/amc/[/url], but they don't provide solutions.  This is my first time taking it, so by my senior year (I'm a freshman now) I should have three years of USAMO experience, which will hopefully cumulate in IMO participation.  I can solve a few of the old ones, but I really need to check my work and I really want to get a decent score so I can go to the Akamai camp for the top 30 ninth graders.  By the way, does anybody have a reasonable estimate of what my score would need to be to get into the Akamai camp?

I can solve 0.1 out of three problems in nine minutes.  The first tenth is reading the problem correctly, and understanding what it's asking.  So, then, with 540 minutes total, I can solve all six problems.  Or at least understand all of them ten times.

Have a look at [url=http://www.kalva.demon.co.uk/usa.html]the Kalva website.[/url]


linear algebra
matrix
linear algebra unsolved

Let $A$ be a real matrix such that all entries except those from the main diagonal are positive. Prove that all entries of $e^A$ are positive.

Well,it is known that $e^A=e^|trA|$. I don't know if this will help.

You must be kidding. The equality stated is wrong.

one can compute $exp(A+bI)=..$ for a suitable $b>0$. Conclusion follows.


function
algorithm
number theory open
number theory

[quote="dieuhuynh"]Find all $n$ is integer number such that  exist $m$ so $\phi(m)=n$[/quote]

   it is a problem ?

Proposed and Own¿¿¿  :?
Ok, please tell your solution ;)
(the part that it is not odd is trivial, I mean the rest)

oh I have't solution .it is open question

I want to find property for sequence $\phi(n)$

It is difficult problem. If $ord_2(n)=k$, then number of odd prime divisors m not greater than k. If k=0 we have $m=1 \ or \ 2,n=1$. If k=1 $m=n+1$ and n+1 is prime. It give new condition for n: if $ord_2(n)=1\Longrightarrow n+1-prime$.
Because Eiler function is multiplicative it is easy to create algoritm for solution $\phi (m)=n$.


inequalities
inequalities proposed
algebra

The inequality  $9(x+y)(y+z)(z+x)\geq 8(xy+yz+zx)(x+y+z)$  (1)  it seems to be lovely  :D

Using (1)  and after Cauchy
We have $\frac{27}{4}(x+y)(y+z)(z+x)\geq 6(x+y+z)\geq (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2$

Now by using AM-GM  and then  (1)  we have

$(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2\geq 9\sqrt[3]{(x+y)(y+z)(z+x)}\geq 9\sqrt[3]{\frac{8(x+y+z)}{9}}\geq 9 \sqrt[3]{\frac{8 \sqrt{3(xy+yz+zx)}}{9}}\geq 6\sqrt{3}$

and we are done

The inequality is very easy if we substitute $x=tan\frac{A}{2}, y=tan\frac{B}{2},z=tan\frac{c}{2}$ where $A,B,C$ are angles of the triangle.
Then we get easy inequalities(very typical) that are followed by Jensen.

Looks so familiar. I guess I posted my solution too early in the wrong [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85199]thread[/url].

substitute by cotga, cotgb, cotgc
so that a+b+c=180° in a acute triangle
and use jensen

Can someone explain the Jensen solution fully? I tried the cotangent trick but couldn't get it to work.

[quote=xzlbq][quote=Valentin Vornicu]If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that
\[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \][/quote]

But you do this stronger:
If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that

\[\left( 2\,\sqrt {2}+3 \right)  \left( x+y \right)  \left( y+z
 \right)  \left( z+x \right) +8\, \left( {\frac {15}{4}}-2\,\sqrt {2}
 \right) xyz\geq  \left( \sqrt {x+y}+\sqrt {y+z}+\sqrt {z+x} \right) ^{2}\][/quote]

By C-S,need to prove

\[ \left( 2\,\sqrt {2}+3 \right)  \left( x+y \right)  \left( y+z
 \right)  \left( z+x \right) +8\, \left( {\frac {15}{4}}-2\,\sqrt {2}
 \right) xyz\geq 6\,x+6\,y+6\,z\]

<=>
\[x{z}^{2}+y{z}^{2}+{x}^{2}y+x{y}^{2}+{x}^{2}z-6\,xyz+{y}^{2}z\geq 0\]

[quote=xzlbq][quote=Valentin Vornicu]If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that
\[ \frac{27}{4} (x+y)(y+z)(z+x) \geq ( \sqrt{x+y} +\sqrt{ y+z} + \sqrt{z+x} )^2 \geq 6 \sqrt 3. \][/quote]

Stronger is:

If $x,y,z$ are positive real numbers and $xy+yz+zx=1$ prove that

\[ \left( \sqrt {x+y}+\sqrt {y+z}+\sqrt {z+x} \right) ^{2}\geq 6\,\sqrt {
 \left( x+y+z \right) \sqrt {3}}\][/quote]

$W=2z+x+y$
By Holder,need to prove

\[{\frac { \left( {\it \sum} \left(  \left( x+y \right)  \left( 2\,z+x+y
 \right)  \right)  \right) ^{3}}{{\it \sum} \left(  \left( x+y
 \right) ^{2} \left( 2\,z+x+y \right) ^{3} \right) }}\geq 6\,\sqrt {
 \left( x+y+z \right) \sqrt {3}}\]
<=>
${\frac { \left( 8\, \left( x+y+z \right) ^{6}+24\, \left( x+y+z
 \right) ^{4}+24\, \left( x+y+z \right) ^{2}+8 \right) ^{2}}{ \left( 2
\, \left( x+y+z \right) ^{5}+ \left( x+y+z \right) ^{3}+3\, \left( x+y
+z \right) ^{2}xyz+7\,x+7\,y+7\,z-5\,xyz \right) ^{2}}}\geq 36\, \left( x+
y+z \right) \sqrt {3}
$

very hard!

Let's first show the left inequality. After homogenizing, we want to show that:

$$27(x+y)(y+z)(z+x) \ge 4(xy+yz+zx)(\sqrt{x+y} + \sqrt{y+z}+\sqrt{z+x})^2.$$

First of all, notice that $27(x+y)(y+z)(z+x) \ge 24(x+y+z)(xy+yz+zx)$ by AM-GM. Therefore, it suffices to show that

$$24(x+y+z) \ge (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2.$$

However, this is easy, because $(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2 \le 3(x+y+y+z+z+x) = 6(x + y + z).$

Let's now show the second inequality. We want to show that

$$(\sqrt{x+y}+ \sqrt{y+z} + \sqrt{z+x})^2 \ge 6 \sqrt3,$$

whenever $xy+yz+zx = 1$ and $x, y, z$ are nonnegative reals.

If any of $x, y, z$ are bigger than $100$, then we are clearly done. Else, let us restrict $x, y, z$ to the interval $[0, 100].$

Let's prove this with Lagrange Multipliers. The derivative of the LHS with respect to $x$ is

$$2(\sqrt{x+y}+ \sqrt{y+z} + \sqrt{z+x}) (\frac{1}{2 \sqrt{x+y}}+ \frac{1}{2\sqrt{x+z}}).$$

The derivative of the condition w.r.t $x$ is $y+z$. Therefore, we have that:

$$\frac{\frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{x+z}}}{y+z}$$

is cyclically constant. After multiplying by $\sqrt{(x+y)(y+z)(z+x)}$, we get that:

$$\frac{\sqrt{x+y} + \sqrt{x+z}}{\sqrt{y+z}}$$

is cyclically constant. This is easily seen to imply that $x = y =z.$

Hence, the minimum is achieved either when $xyz = 0$, $\max(x, y, z) = 100$, or $x = y = z = \frac{1}{\sqrt3}.$

It's easy to check that all of these cases are wins.

$\square$

Surprisingly simple.
Let $p = x + y + z , r = xyz$. Note that $p^2 \geq 3(xy + yz + zx) = 3 \implies p \geq \sqrt{3}$ and $r \leq \frac{1}{3\sqrt{3}}$ by AM-GM.
First we show the left part. It is equivalent to $$\frac{27}{4}(p - r) \geq 2p + 2\sum_{\text{cyc}} \sqrt{x^2 + 1}$$
By C-S, $\sum_{\text{cyc}} \sqrt{x^2 + 1} \leq \sqrt{3(x^2 + y^2 + z^2 + 3)} = \sqrt{3(p^2 + 1)}$.
So it suffices to show $\frac{19}{4}p - 2\sqrt{3(p^2 + 1)} \geq \frac{27}{4}r$.
Note that the LHS is increasing in $p$ and so $\frac{19}{4}p - 2\sqrt{3(p^2 + 1)} \geq \frac{3\sqrt{3}}{4} \geq \frac{27}{4}r$.
Now let's show the right part. It is equivalent to $$2p + 2\sum_{\text{cyc}} \sqrt{x^2 + 1} \geq 6\sqrt{3}$$.
Since $\sqrt{x^2 + 1}$ is convex, by Jensen we get $2\sum_{\text{cyc}} \sqrt{x^2 + 1} \geq 4\sqrt{3}$. And since $2p \geq 2\sqrt{3}$, we get the result. Done.


Solve the system of equations:
$x-4y=-2\\\\3x-y=5$

[hide]Multiply the first equation by 3.

$3x - 12y = -6$
$3x - y = 5$

Subtract.

$-11y = -11$
$y = 1$
$x = 2$[/hide]


Fie $ ABCD$ un patrulater inscriptibil in care laturile $ [AB]$ si $ [BC]$ sunt mai lungi decat laturile opuse lor. Notam cu $ X$ acel punct al laturii $ [AB]$ si cu $ Y$ acel punct al laturii $ [BC]$ pentru care avem: $ |AX| \equal{} |CD|$ si $ |CY| \equal{} |AD|$, iar cu $ M$ mijlocul segmentului $ [XY]$. Sa se arate ca: $ MA\perp MC$

[b]INDICATIE:[/b]
Fie $ \{E\}\equal{}AB\cap CD, \ \{F\}\equal{}AD \cap BC;$ iar G punctul de intersectie al bisectoarelor interioare ale $ \angle AED$ si $ \angle AFB.$
[b]1).[/b] Avem:  $ ABCD\minus{}inscriptibil \Leftrightarrow EG\perp FG.$ (Proprietate cunoscuta!)
[b]2).[/b] Aratati ca: $ AM\parallel{}EG$ si $ CM\parallel{}FG.$

Notam cu $ P\equal{}MC \cap EB$, iar lungimile laturilor patrulaterului le consideram $ AB\equal{}a, BC\equal{}b, CD\equal{}c, DA\equal{}d$

Dupa cateva calcule obtinem ca $ ED\equal{}\frac{d(dc\plus{}ab)}{b^2\minus{}d^2},EA\equal{}\frac{d^2(dc\plus{}ab)}{b(b^2\minus{}d^2)}\plus{}\frac{cd}{b},PX\equal{}\frac{d(a\minus{}c)}{b\plus{}d}$

$ \Rightarrow ED\equal{}EA\plus{}XP \Rightarrow EP\equal{}EC \Rightarrow \triangle PEC$ este isoscel $ \Rightarrow EG \perp PC$

O alta solutie, mai putin calculatorie se bazeaza pe urmatoarea constructie auxiliara: Notam cu $ E$ simetricul lui $ A$ fata de $ M$.
[b]1).[/b] $ AXEY - paralelogram\Rightarrow YE\parallel{}AB\Rightarrow \angle{BYE}\equiv \angle{ABC}\Rightarrow \angle{ADC}\equiv\angle{CYE}$([i]suplementele unghiurilor anterioare[/i])    (1)
[b]2).[/b] $ |AD| = |YC|; |YE| = |AX| = |CD|$ si (1) $ \Rightarrow \triangle {ADC}\equiv \triangle {CYE}\Rightarrow |AC| = |EC|$  (2)
[b]3).[/b]  $ |MA| = |ME| \ si \ (2)\Rightarrow CM\perp AM.$


If a,b,c are real positive numbers, a,b,c>0, we have the inequalities
$\frac{a^2}{b(a^2+ab+b^2)}+\frac{b^2}{c(b^2+bc+c^2)}+\frac{c^2}{a(c^2+ca+a^2)}$>$\frac{3}{a+b+c}$
$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$>$\sum\frac{a^2+b^2}{2c}$>$\sum a$.

for the 2th
we have $\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} >\sum \frac{a^2+b^2}{2c} \Longleftrightarrow 2(a^4+b^4+c^4)>a^3b+ab^3+a^3c+ac^3+b^3c+bc^3$
and we use the Rearrangement.

for $\sum \frac{a^2+b^2}{2c}>a+b+c$ we have $\sum \frac{a^2+b^2}{2c}>\sum\frac{ab}{c}>a+b+c$

for 1:
$\sum {ab}\sum{a^2+ab+b^2}\sum\frac{a^2}{b(a^2+ab+b^2)}\geq (\sum a)^3$
so$\sum\frac{a^2}{b(a^2+ab+b^2)}\geq \frac{(sum a)^3}{\sum{ab}\sum{a^2+ab+b^2}}$
we just need to prove $\frac{(\sum )^2}{\sum ab}\geq \frac{3\sum{a^2+ab+b^2}}{(\sum a)^2}$
let $\sum a^2=u,\sum ab=v$
then we just need to prove $(u-v)^2\geq 0$
which is easy.