inequalities
inequalities unsolved
algebra
inequalities proposed

Let’s consider the real numbers $a, b, c$ satisfying  the condition

\[21 \cdot a \cdot b + 2 \cdot b \cdot c + 8 \cdot c \cdot a \leq 12.\]

Find the minimal value of the expression

\[P(a, b, c) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.\]

Posted: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4208]www.mathlinks.ro/Forum/viewtopic.php?t=4208[/url]

And again, this is an inequalities problem.

It seems with Lagrange if we solve $21b+8c=(21+2c)b^2=(8+2b)c^2$ and then a similar AM-GM solution, of the question that is a bit other with the different topnumbers.

[quote="orl"]Let’s consider the real numbers $a, b, c$ satisfying  the condition $21 \cdot a \cdot b + 2 \cdot b \cdot c + 8 \cdot c \cdot a \leq 12.$ Find the minimal value of the expression $P(a, b, c) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$[/quote]
 This problem should be
Let’s consider the real numbers $a, b, c$ satisfying  the condition $21ab + 2bc + 8ca \leq 12.$ Find the minimal value of the expression $P(a, b, c) = \frac{1}{a} + \frac{2}{b} + \frac{3}{c}.$

Let $a,b,c$ be positive real number and $24a+44b+33c\leq 1.$ Prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 30$$
[size=50](Math and Youth Mag 2017)[/size]
Let $a,b,c$ be positive real number and $2a+4b+7c\leq 2abc.$ Prove that
$$a+b+c \geq \frac{15}{2}$$
[size=50](China  Jilin 2008)[/size]
Let $a,b,c$ be positive real number and $a+2b+3c\leq abc.$ Prove that
$$5a+22b+c \geq 48$$
[size=50](SXTB,(8)2012)[/size]
[url=https://artofproblemsolving.com/community/c6h1269669p6675845]h[/url] [url=https://artofproblemsolving.com/community/c6h1844195p12406132]h[/url]

[quote=sqing]Let $a,b,c$ be positive real number and $2x+4y+7z\leq 2xyz.$ Prove that
$$x+y+z \geq \frac{15}{2}$$[/quote]
$$x+y+z\geq x+\frac{2x+7z}{2xz-4}+z$$

This inequality can be found in Pham Kim Hung's secrets in inequalities


function
trigonometry
integration
logarithms
calculus
calculus computations

http://integrals.wolfram.com gives a polylogarithm term.  Where did you get this problem?

Maybe you wanted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136954]this[/url] problem, which has shown up a million times at least. Otherwise, it seems that the antiderivative is not elementary.


function
real analysis
real analysis unsolved

Let $f$ be continuous odd increasing function, defined on
$\mathbb{R}$ and for all $x_{0}, A,B$ the equation $y''+f(y)=0$ has
unique solution defined on $\mathbb{R}$ such that $y'(x_{0})=A$ and
$y(x_{0})=B$. Prove that any it's solution is periodic.

I asked a professor from my University about this problem and he said that in book Arnold "lectures on classical mechanics" this problem is fully solved using phisics.


algebra
polynomial
ratio
Rational Root Theorem

It is [i]much[/i] easier to use the Rational Root Theorem (if there are rational roots) but here are links for the cubic and quartic formulas:

[url]http://mathworld.wolfram.com/CubicFormula.html[/url]

[url]http://mathworld.wolfram.com/QuarticEquation.html[/url]

[quote="Ceva Theorem"]Does anyone know the formulas for finding the real zeroes for cubic and quartic polynomials? or would it be easier just to use the Ration Root Theorem    :?:[/quote]

Definitely the rational root theorem.

If there are no rational roots, use the newtons approximation thing.  But I doubt you would have to find the irrational solutions to an equation.


conics
ellipse
geometry
geometry proposed

Given an elipse with O is its center. A point M lies on the elipse. The line through O parallel to the tangent at M (of elipse) cuts this elipse at two points N,P. Prove that the area of triangle MNP is equal to 1/2 the area of the elipse.

Sorry, it should be
Given an elipse with O is its center. A point M moves on the elipse. The line through O parallel to the tangent at M (of elipse) cuts this elipse at two points N,P. Prove that the area of triangle MNP is fixed when M moves on the ellipse.

If the tangent point is $ (x_0, y_0)$ and the line passing through the center is $ y\equal{}kx$. The equation of the ellipse is $ b^2x^2\plus{}a^2y^2\equal{}a^2b^2$. Substituting the equation of the chord into the ellipse equation, we get:

$ (b^2\plus{}a^2k^2)x^2\minus{}a^2b^2\equal{}0$

Then we have that the area of the triangle is:

$ S\equal{}\frac{1}{2} \cdot \frac{|kx_0\minus{}y_0|}{\sqrt{k^2\plus{}1}} \cdot \frac{\sqrt{k^2\plus{}1} \cdot \sqrt{\Delta}}{b^2\plus{}a^2k^2}$

where $ \Delta$ is the discriminant of the above equation. Simplifying we get $ S\equal{}ab$.


[quote="Lovasz"]Let n positive real numbers $x_i,i=1,2......n$ in $(a,b)$ and n positive real numbers $y_i,i=1,2,....,n$ not in $(a,b)$ satisfy:
           $x_1+x_2+...+x_n=y_1+y_2+...+y_n$
Prove that: $x_1.x_2.......x_n < y_1.y_2......y_n$[/quote]

i think that the correct is $x_1.x_2.......x_n > y_1.y_2......y_n$


[color=darkblue]Prove that $ \frac{a^2\plus{}b^2}{m_c}\plus{}\frac{b^2\plus{}c^2}{m_a}\plus{}\frac{c^2\plus{}a^2}{m_b} \leq 6D$ where $ D$ is the diameter of the circumscribe circle of the triangle $ ABC$[/color]

http://www.mathlinks.ro/Forum/viewtopic.php?t=105708
http://www.mathlinks.ro/Forum/viewtopic.php?t=76684

  darij


can someone tell me question 11 and 18, i forgot them, and the solution to 18

11. The numbers from 1 to 8 are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same.  What is this common sum?

18. Consider the 12-sided polygon abcdefghijkl, as shown, each of its sides has length 4, and each two consecutive sides form a right angle.  Suppose that ag and ch meet at M.  What is the area of quadrilateral abcm?
solution: C.) 88/5


function
calculus
derivative
real analysis
real analysis unsolved

[quote="bobbydylan"]Given a sequence {an} in R, is there an infinitely differentiable function f, such that the nth derivative of f evaluated at zero is equal to an.[/quote]

Borel's theorem. This has appeared on this forum uncountably many times, I think ...


conics
analytic geometry
advanced fields
advanced fields unsolved

1) Show that any conic in $A^{2}$ is isomorphic to either $A^{1}$ or $A^{1}\setminus \{ 0 \}$.

2) Let $Y \subset P^{n}$ be a non-empty algebraic set and $\theta: A^{n+1}\setminus \{0\}\to P^{n}$ be the map which sends the point with affine coordinates $(a_{0},...,a_{n})$ to the point with homogeneous coordinates $(a_{0},...,a_{n})$. We define the affine curve over $Y$ to be: ${C(Y)= \theta^{-1}(Y}) \cup \{ 0 \}$. I now want to show that $C(Y)$ is an algebraic set in $A^{n}$.

(1) counter-examples: $V(XY)$, $V(X^{2}+Y^{2}+1)$ over $\mathbb Q$

(2) take some homogeneous equations and convince yourself that $C(Y)$ is defined by the same equations

owk


Find all 3-digit positive integers $\overline{abc}$ such that \[ \overline{abc} = abc(a+b+c) , \] where $\overline{abc}$ is the decimal representation of the number.

[quote=silouan]The condition is equivalent to:
$100a+10b+c=abc(a+b+c)$ <=> $9(11a+b)=(a+b+c)(abc-1)$ Now the cases:

i) if $a+b+c=9k$ with $k\leq 3$ .If $a+b+c=9$ the solutions are $a=1,b=3,c=5$ and

another solution $a=1,b=c=4$. If $a+b+c=18$ there are no solutions. If $a+b+c=27$

then again there exist no solutions.

ii) if $abc-1=9m$, then again there are no solutions

iii) if $a+b+c=3n$, $n\leq 8$ and the same time $abc-1=3l$. After all the tests and

tries we have one more solution for $a=1,b=c=4$.

So there exist two numbers. The numbers $135$ and $144$.

This is a problem very boring why you always try and test cases[/quote]

anyone have a better solution without trial and error casework tons of times?


geometry
AMC
AIME
LaTeX
power of a point

Median $AM$ of triangle $ABC$ intersects incircle with radius $r$ at points $P$ and $Q$. Find area of triangle knowing what chord $PQ$ is two times larger than $AP$ and $AP = MQ$.

you need at least one side, but this is very similar to aime 2005I #15...basically use power of a point and equal tangents then stewart to finish

All necessary information to get answer is given (side is not needed) :)

i did not read that carefully, but do you mean in terms of $r$?

My answer is: 13(r square)(sqroot 39)/12

[quote="computer_nuke"]My answer is: $\frac{13r^{2}\sqrt{39}}{12}$[/quote]
Could you post your solutuon?

Hmm, is it correct? It took me really long to do it, but I'll try to figure the Latex codes for everything first, then I'll try my best to explain it.

This answer is correct. Problem is not very easy, I'd say :)

[hide]1.If the median $AM$ of triangle $ABC$ with $AC>AB$ intersects the incircle at points $P$ and $Q$ and  $AP = MQ$ then $2AB=BC$.
Let the incircle is tangent to $AB$ at $L$ and to $BC$ at $K$. By power of a point we have $AL^{2}=AP.AQ=MQ.MP=MK^{2}$. So $AL=MK$, but $AL=p-a$, and $MK=BM-MK=\frac{a}{2}-(p-b)$. From this we obtain $2c=a$ and $AL=\frac{b-c}{2}$.

2. Let the middle part $PQ=nAP$. Then $AL^{2}=AP.AQ=x.(n+1)x$ so $x^{2}=\frac{AL^{2}}{n+1}$ and $AM=(n+2)x$ so $x^{2}=\frac{AM^{2}}{(n+2)^{2}}=\frac{2c^{2}+2b^{2}-a^{2}}{4(n+2)^{2}}$ or $\frac{AL^{2}}{n+1}=\frac{2b^{2}-2c^{2}}{4(n+2)^{2}}$. We obtain $\frac{(b-c)^{2}}{4(n+1)}=\frac{2(b-c)(b+c)}{4(n+2)^{2}}$ or $\frac{(b-c)}{(n+1)}=\frac{2(b+c)}{(n+2)^{2}}$. That is $(n+2)^{2}(b-c)=2(b+c)(n+1)$
In the case $n=2$ $8b-8c=3b+3c$, or $5b=11c$.
In the case $n=1$ this is how I first saw this problem :wink:  $9b-9c=4b+4c$, or $5b=13c$
3. Let $c=5k$, then $a=10k$ and $b=11k$, the semiperimeter is $s=13k$ and by Heron we get that $[ABC]^{2}=13k.3k.2k.8k=s^{2}r^{2}$ or $k=\frac{r\sqrt{13}}{4\sqrt 3}$ and $[ABC]=13kr=\frac{13r^{2}\sqrt{13}}{4\sqrt 3}$[/hide]


Let x, y be real numbers with x>y>0.

You start at the origin, and then at each step, you toss a coin - if it comes up heads, you move x to the right, if its tails, move y to the left. Continue doing this.

What is the probability you will ever be to the left of the origin? (not after each step, but just the probability that will ever happen).

isnt this just a corollary ballot theorem?

hmmm maybe i'm wrong ( i dont know anything about this )
but i'm pretty sure you cant ask that question... or at least the probability would be 1.

Of course you can ask the question, and i'm pretty sure the answers not 1..

The probability is less than 1 (and more than 0).  That follows from the strong law of large numbers.

It's harder to get an exact value for the probability.  To make the problem more manageable, I chose the case x = 2 and y = 1.  Then you get a nice system of difference equations for the problem.  To be more precise, let pi be the probability of ever going to the left of the origin starting at position i.  You get a recurrence relation for pi.  Solving the equations, I found the probability is (:rt5: - 1) / 2.

As long as x/y is rational, you can create a similar system of difference equations, and solve them.  Unfortunately, the equations get more and more complicated to solve.  You have to find the roots of a large-degree polynomial.

You might be able to get somewhere using Catalan numbers and generating functions. It looks messy though.

ok, i'll explain myself...
i can never go left of the origin for the first time after an even number of turns....

there's a 1/2 chance that it goes left after 1 turn (L)
there's a 1/8 chance that it goes left for the first time after 3 turns (RLL)
there's a 2/32 chance that it goes left for the first time after 5 turns (RLRLL or RRLLL)
there's 5 ways to make it go left after 7 turns....
RRRLLLL
RRLRLLL
RRLLRLL
RLRLRLL
RLRRLLL  each of those has a 1/128 chance of happening.... so 5/128?

there's 14 ways to make it go left for the first time after 9 turns.... so 14/512
i wont write them all out....

so... p = (1/2) + (1/8) + (2/32) + (5/128) + (14/512)....
the pattern in the denominators is pretty easy....
so i'll just look at numerators..... 1 , 1, 2, 5, 14......
google that sequence up , lol "1,1,2,5,14"
turns out those are called "catalan numbers"
of the form  2n!/(n+1)!/n!

so, i'm hoping that this series converges to 1.... i haven't thought about how to prove that yet , probably because i dont know how.... but i did put the formula into excel and had it sum up the terms until the numbers go too big for it....

ended up with... after 171 rolls, the chance that its gone left at some point is....
0.93925

( :rt5: -1)/2 is only .618..... so i'm pretty sure that's wrong.. seeing as the first 2 terms of my series are .5 + .125 = .625

but maybe my method is completely faulty and i'm missing a basic law of probability here.

anyways.... .93925 after 171 rolls, think it goes to 1 or not... i'll make a little table

after 11 rolls... p = .7744
after 31 rolls... p = .8601
after 51 rolls....p = .8899
after 71 rolls....p = .9063
after 91 rolls....p = .9170
after 111 rolls..p = .9248
after 131 rolls..p = .9307
after 151 rolls..p = .9354
after 171 rolls..p = .9393

might converge to 1, might not... anyone know any cool irrational numbers between .94 and 1 ?

ok nevermind... i didn't read the problem.... i didn't see the x and y stuff.... heh
thought x=y=1...... yeah, its not p = 1

For x = y, the probability is 1.  I unfortunately do not remember the proof of this -- I would love it if someone would post it, actually.

Actually, Ravi, we can say that the probability is always between 1/2 and 1, since you could always step in the correct direction on the very first try, with probability 1/2.

What happens if you also unbalance the probabilities of stepping to each side?

Yes, Joel, you are right about the lower bound of 1/2.  I was being overly conservative.

For x = y, here is one way to prove that the probability is 1.  After 2n turns, the probability of being at the origin is C(2n, n)/4n, which is on the order of 1/:sqrt:(n).  Summing over all n, we see that the expected number of returns to the origin is infinite.  From there, it is not hard to prove that the probability of eventually being negative is 1.

There are some results known when the probabilities are unbalanced.  Let p be the probability of going right, and q the probability of going left (p + q = 1).  Suppose that p :ge: q (drift to the right).  Suppose that x = y (equal-size steps).  Then  the probability of being eventually negative is q/p.  To show it, you can set up a similar recurrence relation to the one I mentioned before.  In fact, the recurrence is easier here, because the steps are equal size.  (In fact, this approach gives another proof that the probability is 1 when p = q.)

Dealing with unbalanced probabilities and unbalanced steps just makes the problem worse, so I don't have much to say there.

hmm random info...
lower bounds for some cases.....
x/y = 2..... 0.61802 (agrees with that  :rt5: -1 / 2) earlier stated
x/y = 3..... 0.54369
x/y = 4..... 0.51879

for any values of x/y greater than 4... a good approximation (lower bound) is

x/y = m

p  :approx: (1/2) + 1/ (4*2^m) + (m+1) / (8* 2^2m)

these are just the first 3 terms of series that converge to the actual probability... dont know how to figure out a closed form for the whole sequence....

anyways...

if m = 4, that gives .51807
m = 5, .50855
m = 6, .50412
m = 7, .50201

again, these are just lower bounds, but as m gets larger they are very precise.

this just shows that the if m > 4.... p converges very quickly to 1/2 , which makes sense...... if x/y is 10, its pretty unrealistic that you would ever get back to the left if you started with a move 10 spaces to the right.

Yes, if m = x/y is an integer bigger than 1, the probability p is the unique number strictly between 0 and 1 that satisfies pm + 1 = 2p - 1.

I came up with a simple way to prove the formula in my last post.  For convenience, I'll just prove it for the case x = 2 and y = 1, but the proof works whenever x/y is an integer.

Let p be the probability of  going negative (i.e., going to -1).  On the first move, we go randomly to either -1 or +2.  In the first case, we are done.  In the second case, we need to eventually go from +2 to -1.  But that involves three phases: going from +2 to +1, from +1 to 0, and from 0 to -1.  The probability of each is p.  Thus the probability of eventually going from +2 to -1 is p3.  Hence we get the equation p = (1 + p3)/2.

[Note that this gives another proof that when x = y, the probability p = 1.]


suppose [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b05af06a16000a4c5750b131ce0b13fb.gif[/img] are positive integers such that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/c83c1e2f9e91a906cf543fc40f01aaee.gif[/img]
prove or disprove that for every [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f4b1bdeb20ce8d6072430fffd59dc505.gif[/img],we have [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a3338bb887ce19ca926d8cd598d17846.gif[/img]

Hmmm...at least, it is true for $k=2.$

Pierre.

Well,really hard.Then who can give a proof of the case k=2?

For $k=2$, just use the famous Erdös-Selfridge theorem which states that the product of at least two consecutive integers is never a power, and then never a square.
Maybe it is a sledgehammer, but it works  ;)

Pierre.

Excuse me . do you know where can I find the prove of Erdös-Selfridge theorem?   :?:
Thanks.
__

[quote="pbornsztein"]Erdös, Selfridge, 'The product of consecutive integers is never a power', Illinois Journal of Math., 19 (1975), p.292-301.

Pierre.[/quote]

thanks :D  :lol:

[quote="ZetaX"]What was the original problem¿ (there are some possibilities)[/quote]
The original problem, as much as I remember (and I do believe that I remember well, since I noticed the problem as soon as it was posted), was the following:
[quote="suweijie"]suppose $ n_1,n_2,\dots,n_k$ are positive integers such that $ n_1!n_2!\cdots n_k!\equal{}a^k$
prove or disprove that for every $ i,j$,we have $ |n_i\minus{}n_j|\leqslant 1$[/quote]
This statement is [b]not correct[/b]. The counterexample is $ n_1\equal{}n_2\equal{}7$, $ n_3\equal{}8$ and $ n_4\equal{}n_5\equal{}n_6\equal{}9$: \[ 7!\cdot 7!\cdot 8!\cdot 9!\cdot 9!\cdot 9!\equal{}(2^6\cdot 3^3\cdot 5\cdot 7)^6.\]


I have very little experience in math, and I am trying to solve a problem on how to equally spread a variable number of objects along the surface of a sphere . I managed to do this for a circle but a sphere is just to much for me. Can someone tell me in which direction I should be searching for a solution.

Unlike with a circle, it is not possible to spread any number of points out evenly on a sphere, at least in the sense that I am interpreting "evenly."  There are only a few numbers of points for which one can accomplish this.  For a circle, the arrangements are given by the regular polygons.  The 3-D version is the arrangements given by the regular polyhedra, or Platonic solids.  So, that is a good place to start looking.  If you want more information, feel free to ask.

Thank you for your reply.

I already managed to write something that spreads nodes along a circle.
 I have found a website which shows a way do the trick with a sphere. Although it is not presize for every amount. [url=http://www.mpia-hd.mpg.de/homes/stein/]www.mpia-hd.mpg.de/homes/stein/[/url]. But that is not that important to me. I would like to come up with something as presize as possible. I am creating a 3d information visualization enviorment for a school project.

I'm not sure what you mean by "presize" -- are you saying you want to do it for an arbitrary number of nodes?  (Or are you mis-spelling the word precise?)  That's going to be very tough.  One thing you could do is look for semi-regular polyhedra, which will provide semi-regular spreads of nodes that might be useful.  Aside from that, though, I'm not really sure.  I would suggest to a moderator that this get moved to Other Problem Solving topics -- it really requires that someone with a larger over-all knowledge take a look.  It sort-of resembles sphere-packing to me, although I don't see a direct link between the two problems.  Anyhow, best of luck!

Edit: I looked at the link you mentioned and at what is I assume the relevant page therein.  What's wrong with their method that makes it not useful to you?

I meant 'precise', but my English is just not that good. The page I mentioned was an example. I will trie to contact the author but I think his methods are not available for use to everyone. Because he did not publish them on his website.
Thank you for your reactions and I will let you know if and how I found a solution to this problem.


Find the remainder when $ \frac{100!}1\plus{}\frac{100!}2\plus{}\cdots\plus{}\frac{100!}{100}$ is divided by $ 101$.

[hide="dont try this"]
I would say $ 0$ from using some smaller cases.

Try the cases with $ 2! \mod 3$, $ 4! \mod 5$, and $ 6! \mod 7$ where its a mod p, like 101.
Every time, you would get $ 0 \mod p$.

Infact, if you just pair up the terms, $ \frac {100!}{1}$ with $ \frac {100!}{100}$ and any $ \frac {100!}{k}$ and $ \frac {100!}{101 \minus{} k}$ and add them and mod by $ 101$, youll get $ 0$
[/hide]

[quote="pi^2/6"]Find the remainder when $ \frac {100!}1 \plus{} \frac {100!}2 \plus{} \cdots \plus{} \frac {100!}{100}$ is divided by $ 101$.[/quote][hide]Write \[ \frac{1}{1}\plus{}\frac{1}{2}\plus{}\cdots\plus{}\frac{1}{100}\] by combining terms on opposite ends of the summation (so that the denominators would sum to $ 101$): \[ \frac{1}{1}\plus{}\frac{1}{100}\plus{}\frac{1}{2}\plus{}\frac{1}{99}\plus{}\cdots\plus{}\frac{1}{50}\plus{}\frac{1}{51}.\]This equals \[ \frac{101}{1\cdot100}\plus{}\frac{101}{2\cdot99}\plus{}\cdots\plus{}\frac{101}{50\cdot51}.\] Hence, the above expression equals \[ 101!\left(\frac{1}{1\cdot100}\plus{}\frac{1}{2\cdot99}\plus{}\cdots\plus{}\frac{1}{50\cdot51}\right).\] Since all terms in the denominator cancel out with some part of $ 100!$ and $ 101$ does not appear in any denominator, then the remainder is $ 0$.[/hide]

well hmm my assumption got me the right answer but the wrong method.

I think this works when its $ \mod x$ and $ x$ is an odd positive integer greater than or equal to $ 3$.

So

$ \frac{(2k)!}{1}\plus{}\frac{(2k)!}{2}\plus{}...\frac{(2k)!}{2k} \mod (2k\plus{}1)$
$ (2k!)(\frac{1}{1}\plus{}\frac{1}{2}\plus{}...\plus{}\frac{1}{2k}) \mod (2k\plus{}1)$
pairing up the $ \frac{1}{x}$ and $ \frac{1}{2k\plus{}1\minus{}x}$ terms, we can rewrite it as
$ [2k!][2k\plus{}1][\frac{1}{(2k)(1)}\plus{}\frac{1}{(2k\minus{}1)(2)}\plus{}...\plus{}\frac{1}{(k\plus{}1)(k)}] \mod (2k\plus{}1)$

Since everything in that fraction adding thingy will be an integer when multipled by $ (2k!)$, it is $ (2k\plus{}1)(I) \mod (2k\plus{}1)$ which is $ 0 \mod (2k\plus{}1)$

If this were $ \mod 2k$ and it is in the same form, the answer would be $ \frac{(2k\minus{}1)!}{k} \mod 2k$.  For example, if it were a problem like $ \frac{5!}{1}\plus{}\frac{5!}{2}\plus{}...\plus{}\frac{5!}{5} \mod 6$, the answer would simply be $ \frac{(2*3\minus{}1)!}{3} \mod (2)(3)$ which would be $ 4$.

Since 101 is prime, the inverses of 1,2,3,...,100 is just a permutation of those numbers, so in mod 101, it is

100!*( 1+2+...+100)...which has a factor of 101.


[quote="Yosh..."]Find values of $n\in N$ for which the fraction $\frac{3^{n}-2}{2^{n}-3}$ is reducible[/quote]

This problem, is reduced to find $n$ such that $(3^{n}-2)| (2^{n}-3)$
,a clasic number theory problem :P

But I think we must find $n$ such that $gcd(3^{n}-2,2^{n}-3)\not{=}1$ not only $\frac{3^{n}-2}{2^{n}-3}\in N$

Sorry for my english.. (if I'm wrong in understand that problem)
Sorry if wrong place..

The fraction is reducible by $5$ if and only if $n\equiv-1[4]$.


search
number theory unsolved
number theory

Let $ a_1<a_2<a_3<\cdots$ be a not bounded sequence of positive integers, such that the set of all prime divisors of the numbers $ a_1,a_2,a_3,\ldots$ is finite . Prove that for any integer $ k$ (not equal to 0), the set of all prime divisors of the numbers  $ a_1\plus{}k,a_2\plus{}k,a_3\plus{}k,\ldots$ is infinite.

it's kobayashi's theorem and here is a [url=http://www.mathlinks.ro/viewtopic.php?search_id=867701214&t=275813]link[/url] where the reference to kobayashi's paper is mentioned


calculus
derivative
real analysis
vector
geometry
3D geometry
real analysis unsolved

Hello:
I need a hint for showing the following.
Let f :R^n->R be a differentiable function. Let D_k={x:f(x)=k}.
If Df(a) (derivative of f in a) does not vanish for every a in D_k then D_k has 0 measure.

Thanks

Not suitable for Intermediate.
Moved to College Playground.

If $D_{k}$ has positive measure, then by the Lebesgue density theorem there exists $a\in D_{k}$ such that $m(B(a,r)\setminus D_{k})/r^{n}\to 0$ as $r\to0$. Let $v$ be a vector such that $\nabla f(a)\cdot v>0$. Then $\nabla f(a)\cdot u>0$ for all vectors $u$ in some closed cone $C$ with axis $v$. Using the definition of differentiability, one can show that $f(x)>k$ for all $x$ in $(C+a)\cap B(a,r)$ if $r$ is small enough. But $m((C+a)\cap B(a,r))/r^{n}$ is positive and independent of $r$, a contradiction.

(The proof also shows that $D_{k}$ is a $\sigma$-porous set.)

mlok:
Can you think of a proof that does not use the lebesgue density
theorem? Even if its long and messy, I only need a hint in the right
direction.

Thanks a lot

You can prove that a porous set has measure zero w/o Lebesgue density, by considering a near-optimal cover and observing that it can be improved.

mlok

Can you define porous set and near optimal cover?

Thanks


Find all real-valued functions on the positive integers such that f(x + 1019) = f(x) for all x, and f(xy) = f(x) f(y) for all xy.

Any other ways to do besides:

[hide]

f(1) = f(1) 2

So f(1) = 1 or 0.

For any positive a < 1019, a^1018 = 1 (mod 1019) implying

f(a^1018) = f(a)^1018 = f(1)

So f(a) = 1 or 0 for all a s.t. gcd(a, 1019) = 1.

Then all f:N->R for any real m, and a = 1 either or 0 satisfying:

f(x) = a (if gcd(a,1019) = 1)

else f(x) = m

are solutions...is this even correct?

[/hide]

[hide]First off f(1)=0 implies f(n)=0 since f(n)=f(1n)=f(1)*f(n)=0*f(n).

So now let us do f(1)=1

You are correct in stating that f(n)=1 for gcd(x,1019)=1 by the reasoning f(x)^1018=1.

Now we are left to deal with when gcd(x,1019)=1019:

Let f(1019)=y

f(10192)=f(1019)2=f(1019)=y.

Therefore y=0 or 1.

So our three solutions are:

f(n)=0;

f(n)=1;

f(n)=0 for n=1019k, 1 otherwise;[/hide]

No, there are four solutions.
One of them is the Legendre function (related to quadratic residues).

Ah thanks for clearing that up tom, i sorta screwed up in the end. :blush:

Arne, since we systematically solved for the functions, wouldnt the legendre function be equal to one of the 3 offered solutions? At least on the given domain...on another note i dont know what the legendre function is  :(  Thanks again.

[quote="Arne"]One of them is the Legendre function (related to quadratic residues).[/quote]

huh?  Only time I've ever used Legendre was for max/min of problems with a fixed criterion.
i.e. max/min of x2+y2 given x+y=2

[quote="TripleM"]I think you are getting mixed up between Legendre and Lagrange. Two different people :)[/quote]

Yes (Or should it be jacobi ?) .. the other solution is of the form: (Defining only for numbers between 0,1018 - as f(k) = f(k+1019) so one can  just take modulo 1019 )

f(0) = 0
f(k) =  k^509 (mod 1019)    (The answere is either -1 or +1)

(Of course,  one can also have f(k) = 0 as mentioned above)

If I  subsitute  5 instead of 1019 it may look like:
f(0)=0
f(1) = 1 ; f(2) = -1 ; f(3) = -1 ; f(4) = +1

(f(5)= f(0)=0 ; f(6)=f(1)=1 etc...)

[quote]huh?  Only time I've ever used Legendre was for max/min of problems with a fixed criterion.
i.e. max/min of x2+y2 given x+y=2[/quote]

He is refering to Lagrange Multipliers... most definitely.


Okay the rules are simple, I say the name of a category, lets say colors;

[color=black]Okay the category is colors;[/color]

Then the next person to post has to say a color. For example;

[color=black]

Topic: Color

ernie - blue

List;

blue

Scores;

ernie : 1[/color]

Then he gets a point for that round. Then the next person says a color;

[color=black]

Topic: Colors

budi713 - black

List;

blue
black

Scores;

ernie: 1
budi713: 1 [/color]

And then it continues like that, until I change the category. Nobody else is aloud to change the category, and you can't say anything more than once. You must put the information in the same format as above.

The game starts now;

Topic: Units of Time

Wickedestjr: minutes (you can't say minute in singular either)

List;

minutes

Scores;

Wickedestjr

Topic: Units of Time

Wickedestjr - hours

List;

minutes
seconds
hours

Scores;

Wickedestjr: 2
mewto55555: 1

Topic: Units of Time

Wickedestjr: weeks

List;

minutes
seconds
hours
days
weeks

Scores;

Wickedestjr: 3
mewto55555: 2

Topic: Units of Time

james4l: millenniums

List;

minutes
seconds
hours
days
weeks
millenniums

Scores;

Wickedestjr: 3
mewto55555: 2
james4l: 1

Topic: Units of Time

Wickedestjr - months

List;
minutes
seconds
hours
days
weeks
milleniums
months

Scores;
Wickedestjr: 4
mewto55555: 2
james4l: 1

Topic: Units of Time

pacman2812: eons :rotfl:

List;
minutes
seconds
hours
days
weeks
milleniums
months
eons

Scores;
Wickedestjr: 4
mewto55555: 2
james4l: 1
pacman2812: 1

Topic: Units of Time

Wickedestjr - seasons

List;
minutes
seconds
hours
days
weeks
milleniums
months
eons
seasons

Scores;
Wickedestjr: 5
mewto55555: 2
james4l: 1
pacman2812: 1

Topic: Units of Time

pythag011-nanoseconds

List;
minutes
seconds
hours
days
weeks
milleniums
months
eons
seasons
nanoseconds

Scores;
Wickedestjr: 5
mewto55555: 2
james4l: 1
pacman2812: 1
pythag011:  1

Topic: Units of Time

ernie-decades

List;
minutes
seconds
hours
days
weeks
milleniums
months
eons
seasons
nanoseconds
decades

Scores;
Wickedestjr: 5
mewto55555: 2
ernie: 1
james4l: 1
pacman2812: 1
pythag011: 1

Topic: Units of Time

[b]eras[/b]

List;
minutes
seconds
hours
days
weeks
milleniums
months
eons
seasons
nanoseconds
decades
centuries
milliseconds
eras

Scores;
Wickedestjr: 5
mewto55555: 3
ernie: 2
james4l: 2
pacman2812: 1
pythag011: 1

Topic: Units of Time

years

List;
minutes
seconds
hours
days
weeks
milleniums
months
eons
seasons
nanoseconds
decades
centuries
milliseconds
eras
years

Scores;
Wickedestjr: 6
mewto55555: 3
ernie: 2
james4l: 2
pacman2812: 1
pythag011: 1

[hide="Score"]
1. kimmystar94: 14
2. james4l: 13
3. Wickedestjr: 12
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 2
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1 [/hide]
[hide="List"]
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale[/hide]

Score
1. kimmystar94: 14
2. james4l: 13
3. Wickedestjr: 12
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 3
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer

Score
1. kimmystar94: 14
2. james4l: 13
3. Wickedestjr: 12
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 8
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
[b]sheep
pig
lion
wild boar[/b]

Score
1. kimmystar94: 14
2. james4l: 13
3. Wickedestjr: 12
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
[b]tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel[/b]

Score
1. kimmystar94: 14
1. james4l: 14
3. Wickedestjr: 12
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel
dolphin

Score
1. kimmystar94: 14
1. james4l: 14
3. Wickedestjr: 13
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel
dolphin (ooh, I was just about to put that animal)
flamingo

Score
1. kimmystar94: 15
1. james4l: 14
3. Wickedestjr: 13
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel
dolphin
flamingo
ostrich

Score
1. kimmystar94: 15
2. james4l: 14
2. Wickedestjr: 14
4. ernie: 8
4. cf249: 8
6. 1=2: 7
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel
dolphin
flamingo
ostrich
lizard

This has become a pain to read, for I have to scroll down a lot. so would you please hide your f***ing posts?[hide="Score"]1. kimmystar94: 15
2. james4l: 14
2. Wickedestjr: 14
4. ernie: 8
4. cf249: 8
4. 1=2: 8
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1 [/hide][hide="List"]giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel
dolphin
flamingo
ostrich
lizard
buffalo[/hide]

[hide="hide text here"]1. kimmystar94: 15
1. Wickedestjr: 15
3. james4l: 14
4. ernie: 8
4. cf249: 8
4. 1=2: 8
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

List
giraffe
elephant
barn owl
human
rhinoceros
dodo
dogs
snake
Palau owl
cat
rabbit
llama
turkey
horse
cow
bluejay
duckbill
mouse
dinosaurs
platypus
whale
Hammerhead Shark
monkey
bald eagle
hummingbird
killer whale
deer
hen
sheep
pig
lion
wild boar
tiger
frog
octopus
bat
zebra
parrot
wombat
squirrel
dolphin
flamingo
ostrich
lizard
buffalo
falcon [/hide]

[code][hide=hide text here]blah blah smlah[/hide][/code]

[hide="hide text here"]blah blah smlah[/hide]

1. kimmystar94: 15
1. Wickedestjr: 15
3. james4l: 14
4. ernie: 8
4. cf249: 8
4. 1=2: 8
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

New Topic  - Tv shows with the word 'house' in their name.

List;
Little [color=darkred]house[/color] on the prairie.

1. kimmystar94: 16
2. Wickedestjr: 15
3. james4l: 14
4. ernie: 8
4. cf249: 8
4. 1=2: 8
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

New Topic - Tv shows with the word 'house' in their name.

List;
Little house on the prairie.
HOUSE

1. kimmystar94: 16
1. Wickedestjr: 16
3. james4l: 14
4. ernie: 8
4. cf249: 8
4. 1=2: 8
7. mewto55555: 4
7. pacman2812: 4
9. dragon96: 3
10. erabel: 2
10. (^_^): 2
10. cvaneg: 16
10. eggylv999: 2
10. kookamunga: 2
15. pythag011: 1

New Topic - Tv shows with the word 'house' in their name.

List;
Little house on the prairie.
HOUSE
Desperate Housewives


AMC USA(J)MO USAMO Harvard college AIME geometry

[quote="E^(pi*i)=-1"]Unfortunately, the numbers for "Homeschool" are deflated because so many homeschoolers are listed at colleges/math circles/local high schools.  In addition to the 4 you counted (one of whom is me :) ), I see at least 4 more who I know to be homeschooled, and I'm sure there are a few more beyond that.[/quote]

Yeah, for example the two qualifiers from AAMC this year are both homeschooled (I'm one of them).  We also had two other qualifiers that took the AIME at their high school, who aren't on the list.

jmerry, thanks for updating the list.  It looks like you have found some nice additions to my previous list.

As the previous posters have pointed out, I count at least 8 homeschoolers:
[hide="Homeschoolers"]Evan O'Dorney, Jeffrey Manning, John Dorminy, Alex Kandell, Mark Pritt, Thomas Snell, Andrew Ardito, and Schuyler Smith.[/hide]
Last year, Stuyvesant actually had 6 (not 5) qualifiers.  (One qualifier must have been added in after I created last year's chart.)

I will PM you about some more arguable cases.

In addition to the 8 listed by Ravi, David Rolnick is also a homeschooler.

In addition, although it shouldn't be counted in the list above, I recognize at least four more students who homeschooled prior to 9th grade.

[hide="former homeschoolers among 2008 qualifiers"]
David & Nathan Benjamin, Adam Sealfon, and Elizabeth Synge[/hide]

I'll also note that ICAE's number for 08 is probably not accurate.  A significant number of those students listed in other schools in the Detroit/Troy area probably owe much of their qualification to taking classes at ICAE.

Now, can anyone put AoPS on this list :)

Well, if you're going to be that way, the union (with considerable overlap) of the Southern California ARML team, Avid Academy, and Math Zoom claim all 19 of the qualifiers from the greater Los Angeles area, as well as one each from San Diego, North Carolina, and Massachusetts.

The list has been updated with these and several other changes. I knew I was undercounting homeschoolers, but I didn't know exactly what the right count was.

[code]Bellevue     WA 0 0 0 0 0 0 1 3 2 2   8[/code]

Finally! At least we're doing a lot better than in previous years...

EDIT: Why don't we rank by last four years? :P

I believe Lexington's 2007 total should be 5, because I was a qualifier in 8th grade that year, and took all my AMC's there. For some reason I was just listed under my middle school on the list.

[quote="Catalyst of Nostalgia"]I believe Lexington's 2007 total should be 5, because I was a qualifier in 8th grade that year, and took all my AMC's there.[/quote]

You were listed with Lexington last time, but Ravi and I changed our minds. Several other middle schoolers have also been removed from the high schools they took the test at.

Rawr, we're finally on the list!

[code]Hopkins      CT 0 0 1 1 1 1 0 0 2 3   9[/code]

[code]Parkview     GA 0 0 0 0 0 1 1 2 2 2   8
[/code]

Yay, we are also listed! (Although we seem to be the only school in Georgia...)

Aww...we would be:
[code]Lick-Wilmerding       CA 0 0 0 0 0 0 0 0 0 1   1[/code]
I believe.

[quote="timwu"](Although we seem to be the only school in Georgia...)[/quote]
I have several other lines for GA schools: Paideia with 8 total and 6 since 1999, Westminster with 6 and 6, Roswell Central with 5 and 5.

Gogogog Melanie Wood!! [edit: and Chris Mihelich. but note that I gave him a shoutout later on in the post, jmerry =P]

Park Tudor   IN 0 0 2 2 2 1 0 1 2 1 0 0 0 1 2  14

----
Hopefully this other freshmen and I can create another 2-2-2-2 burst like Melanie and Chris did way back when...

[quote="Fanatic"]Gogogog Melanie Wood!![/quote]... and Chris Mihelich, three of the same four years. A younger Mihelich also qualified in 2001 and 2002.

Those were my years, and Melanie was definitely the leader among the IMO team members by virtue of personality.

I'm pretty sure the automatic list is based on where those students took the AIME. Of Gunn's five listed qualifiers this year, four were 10th graders.

[quote="rrusczyk"]I'll also note that ICAE's number for 08 is probably not accurate.  A significant number of those students listed in other schools in the Detroit/Troy area probably owe much of their qualification to taking classes at ICAE.[/quote]
Right.  I contacted Dr. Khetan, head of the ICAE.  He says that for 2008, eight USAMO qualifiers are students of ICAE:
[hide="ICAE 2008 students"]Kevin Wu, Steven Chang, Neil Gurram, Alan Huang, Randy Jia, David Lu, Matthew Vengalil, and Robin He.[/hide]
For 2007, in addition to the five students already on the AMC's list, Dr. Khetan says that six other USAMO qualifiers are students of ICAE:
[hide="ICAE 2007 students"]Robin He, Kevin Wu, Neil Gurram, Alan Huang, Surya Nagaraja, and Rojer Jia.[/hide]
So I recommend bumping up the ICAE numbers for 2007 to 11 and for 2008 to 8.

I've edited ICAE's line in my spreadsheet, but I no longer have the power to edit the original post.

I am leery of allowing overlap- five of those students this year and two last year are listed with Detroit Country Day, another high-ranking school. I prefer to treat the school students attend full time as primary here.

My school just barely missed =\ with a 14-year total of [hide]2[/hide]

Also, Jeremy Hahn is a homeschooler this year.  So we should add 1 to Homeschool and deduct 1 from East Chapel Hill.

[quote="hadasah"]Go CLEMENTS! My friends go there. And clements is really good, the best hs in Texas. And Forgive my stupidity, but what's TJ?[/quote]

[b]Thomas Jefferson High School of Science and Technology, located in Alexandria ^^[/b]

Nobody from the history of my town has ever made it into USAMO, and only 2 ppls made AIME this year, but all that's gonna change when I get to high school!  :D (hopefully)

What town are you from? (I'm a fellow Georgian.)

[quote="serialk11r"]For Gunn that's not including the people who chickened out and went to some random schools to take AMC10B...cough moogra cough.[/quote]

Point being I would have gotten owned at Gunn. I got 1st place where I got it and I got an award

I would like to suggest to someone who knows how to do these things that it might be interesting to update the chart at the beginning of this thread.

If someone sends me the changes they want to make, I'll edit them in.

This year's list isn't final yet, so I'm not posting anything yet. I am, however, updating it behind the scenes, including going back to 1987. If anyone wants to help track down the trickier counts like homeschoolers and ICAE, that would be appreciated.

danggg. my school's actually decent :O lol 20? [well more now but still...] . woot go JMM =]]

dude TJ ftw :lol:

i heard TJ had more people going to usamo this year than all of TX.

of course, TJ is a magnet school which accepts students from like over 70 mile radius, and it impossible to get into...

I count 11 homeschoolers on this year's list.

texas somehow managed to make a bilion carelesses on the aime ii
anyway, i think sartartia ms of stexas did pretty good this year.

Ladue Middle/High School has multiple qualifiers (3 this year,1 last,1 two years ago,3 three years ago)...

There is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272133]new thread[/url] for discussing stuff related to this year's list.


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let $ K\subset L$ be two fields.
Let $ P\in K[X_1,\ldots, X_m]$.
If $ P$ vanishes on $ K^m$, does it vanish on $ L^m$?
If yes, how to prove it when $ K$ is finite (the infinte case is OK I think).
Thank you.

For infinite fields this is not hard to show. It's wrong for finite ones, even if $ m\equal{}1$: take $ x^{q}\minus{}x$ over $ \mathbb F_q$.

If $ m>1$ and $ P$ is supposed to be homogeneous, is the result still wrong for a finite field?

Thanks again ZetaX, I am trying to "force" such a result to be true, so I'll try again to make stronger hypothesis.
Here is another attempt.
$ K$ is of characteristic $ p$. $ P$ is homogeneous of total degree $ \nu$, we suppose $ m>1$, and $ \nu \le p$.
If $ P$ vanishes over $ K^m$, does it vanish over $ L$?

I believe we can say something quite a bit stronger: if $ \rm{deg}\ P \leq q \equal{} |K|$, and $ P$ vanishes on $ K$, then $ P \equal{} \sum_i c_i (X_i^q \minus{} X_i)$ for some $ c_i\in K$.  Hence if $ P$ is homogeneous, it is $ 0$.

Proof: Work in the algebraic closure of $ K$.  The Nullstellensatz tells us that some power of $ P$ lies in $ I \equal{} (X_i^q \minus{} X_i)$, because $ K$ is precisely $ V(I)$.  Why must the $ c_i$ be constants?  Well their terms of degree at least $ q$ are irrelevant, as they would need to cancel, and then we are unable to get large enough powers of $ X_j$ in the ith term for the terms of degree $ > q$ to properly cancel. (this argument seems to work, but I'm having a hard time expressing it cleanly)

EDIT: sorry, there's a flaw in this argument.  It only covers the case where $ P\in I$, but we need to account for $ P^k \in I$ somehow.

More generally, in the polynomial ring $ \mathbb{F}_q\left[X_1,X_2,...,X_n\right]$, those polynomials $ f$ that satisfy $ f\left(x_1,x_2,...,x_n\right)\equal{}0$ for all $ \left(x_1,x_2,...,x_n\right)\in\mathbb{F}_q^n$ form an ideal - and this ideal is generated by $ \left\{X_i^q\minus{}X_i\mid i\in\left\{1,2,...,n\right\}\right\}$. This is a particular case of Theorem 1.1 (the Combinatorial Nullstellensatz) in
[url=http://www.cs.tau.ac.il/~nogaa/PDFS/null2.pdf]Noga Alon, [i]Combinatorial Nullstellensatz[/i][/url].
Moreover, $ \left\{X_i^q\minus{}X_i\mid i\in\left\{1,2,...,n\right\}\right\}$ is a Gröbner basis of this ideal with respect to any monomial ordering, I believe.

Here is something similar (I invented this once as a problem for an algebra course, but it was never used):
The homogeneous ideal of $ \mathbb{F}_q\left[X_1,X_2,...,X_n\right]$ generated by all [i]homogeneous[/i] polynomials $ f$ that satisfy $ f\left(x_1,x_2,...,x_n\right)\equal{}0$ for all $ \left(x_1,x_2,...,x_n\right)\in\mathbb{F}_q^n$ is generated by $ \left\{X_i^qX_j\minus{}X_j^qX_i\mid i,j\in\left\{1,2,...,n\right\}\right\}$.
This makes me think of a projective version of the Combinatorial Nullstellensatz...

  darij


topology
algebra
polynomial
function
linear algebra
linear algebra unsolved

Prove that the set of matrices in $M_{m\times n}$ with determinant equal to zero form a closed set.

Shouldn't that be $M_{n\times n}?$ Only square matrices have determinants.

This is nearly trivial. The determinant is a polynomial function in the entries of the matrix. Polynomials are continuous. The inverse image under a continuous function of a closed set, such as $\{0\},$ is closed.

More generally, the $m\times n$ matrices of rank $\le k$ form a closed subset of $M_{m\times n}$.


[quote="lhplhp"]Let ABC  be a triangle.Choose the pair of points $ A_1;A_2;B_1;B_2;C_1;C_2$   on the sides $ BC;CA;AB$ ,respectively such that $ A_1A_2B_1B_2C_1C_2$ is a convex hexagon with equal opposite sides and the triangle $ AA_1A_2;BB_1B_2;CC_1C_2$ have the same areas.Prove that the lines $ A_1B_2;B_1C_2;C_1A_2$ are concurrency.[/quote]
This problem in Mathematics and youth Magazine.
you shouldn't post  :mad:  :mad:  :mad:


Find all $ f: R\rightarrow R$ such that

$ f(x+yf(x)) = f(x)+xf(y)$


trigonometry
algebra
linear equation
trig identities
Law of Cosines

we have a circle with a diametre of 30cm. we cut a circle of 20cm in diametre and one in 10cm in diametre. we want now to cut another circle. what will its diametre be if we want a circle of maximum diametre?

Okay, I'm really upset because I just wrote up a nice solution to this but when I tried to send it I got "Invalid Session" and then when I hit back, it had disappeared.  So I'll just give my answer, which is [hide]60/7[/hide], and I'll perhaps write the solution again later.

well the answer should be 30/7 or 30/9 or 30/11 or 30/8.
bye

This is a problem very similar to one I posted on the Challenge Board - http://www.artofproblemsolving.com/wforum/viewtopic.php?t=153

I believe the answer JBL gave is correct. But you have to be careful whether you were asking for Diameter or Radius.

well the problem asks for the diametre and it says that the answer is between those numbers. I think that they are asking for the radius, aren't they?

pls help me!

[color=cyan]All right, this is an outline of my solution, not fully explained because I really am annoyed my previous solution was deleted and I don't want to write the whole thing again.
Basically, you use the following theorem: if two circles are tangent, then their centers and the point of tangency all lie on the same line.

The givens give us that the three circles we start with are all mutually tangent, with the two little ones inside the big one.  Call the center of the littlest circle A, of the biggest circle D, of the medium-size circle C, and of the new circle whose diameter you want B.  Also, note that it is intuitively clear that our circle shoudl be tangent to eacho fthe other three.
We can show without a lot of work that
AD = 10
DC = 5
BA = 5 + r
BD = 15 - r
BC = 10 + r.
This is exactly the setup for Stewart's Theorem, a nice theorem that can be derived by applying the law of cosines to the two supplimentary angles at D.  Anyhow, it gives us that
5*(5 + r)^2 + 10*(10 + r)^2 = 15*(15 - r)^2 + 5*10*15
which conveniently converts itself into a linear equation which can be solved to find that r = 30/7 and the diameter (which was the desired length) is 60/7.[/color]


1. Solve in real numbers x,y,z :
  $\{\begin{array}{ccc} x^2=yz+1 \\ y^2=zx+2 \\ z^2=xy+4 \\ \end{array}$

Subtract the third equality from the second one we obtain:
$y^2-z^2=xz-xy-2$ or $(y-z)(y+z)=x(z-y)-2$; $(y-z)(x+y+z)=-2$        (1).
Analogously, by subracting the first one from the third one, we get:
$(z-x)(x+y+z)=3$        (2).
By division (1)/(2) as none of (1) or (2) is equal to 0 we obtain:

$\frac{y-z}{z-x}=-\frac{2}{3}$, $3y-3z=2x-2z$, $z=3y-2x$.

Inserting this expression of z into the first two equalities in the given system, we obtain:
$x^2=y(3y-2x)+1$(3)
$y^2=x(3y-2x)+2, y^2-3xy=2-2x^2, x^2-1=\frac{y^2-3xy}{2}$
Now we insert the expresion of $x^2-1$ into (3):
$-\frac{y^2-3xy}{2}=y(3y-2x), -y^2+3xy-2y(3y-2x)=0,$ $-y^2+3xy-6y^2+4xy=0, 7y(x-y)=0 $, if x-y is not equal to 0, as we get by subracting the second equality from the third one, so y=0. Hence
$x=1, y=0, z=-2,$  or $ x=-1, y=0, z=2$.


algebra
polynomial
search
algebra unsolved

suppose a polynomial P(x)of degree n obeys the relation

[b]P(k)=k/(k+1)[/b] for k=1,2,3.........,n.
find the value of P(n+1) :)


modular arithmetic
algebra unsolved
algebra

It is given that $ n$  is an odd integer greater than 3 , but $ n$ is not a multiple of 3 . prove that $ x^3 \plus{} x^2 \plus{} x$ is a factor of $ ( x \plus{} 1 )^n \minus{} x^n \minus{} 1$

Maybe you meant the divisor to be $ x^2 \plus{} x \plus{} 1$? I ask because the result is not true for n=5.

even  i see that but but if we substitute x = 1  and n=5 it works.

[quote="vijaymenon"]even  i see that but but if we substitute x =  and n=5 it works.[/quote]

I dont understand what you mean.

What hsbhatt said is that for $ n\equal{}5$, we have $ (x\plus{}1)^5\minus{}x^5\minus{}1\equal{}(x^3\plus{}x^2\plus{}1)(5x\plus{}5)\plus{}5x^2\minus{}5$ and so $ x^3\plus{}x^2\plus{}1$ is not a factor of $ (x\plus{}1)^5\minus{}x^5\minus{}1$

[quote="vijaymenon"]What  i meant is x = 1 . it was a typo[/quote]

Oh great !
So your problem is to prove that $ 3|2^n\minus{}2$ for all odd integer greater than $ 3$

Very difficult problem, indeed.  :rotfl:

u seei din mean that . i only told that it is true for x= 1  alone .

We could either

1) Laugh at the poor fellow.
2) Try to figure out what he's actually asking.

In fact, $ x^2 \plus{} x \plus{} 1 | (x\plus{}1)^n \minus{} x^n \minus{} 1$; letting $ \omega$ be a root of $ x^2\plus{}x\plus{}1$,
$ (\omega\plus{}1)^n \minus{} \omega^n \minus{} 1 \equal{} (\minus{} \omega^2)^n \minus{} \omega^n \minus{} 1$, which
in turn is the quantity $ (\minus{}1)^n (\omega)^{2n} \minus{} \omega^n \minus{} 1$. Since $ n$ is odd, this is
$ \minus{} \left( (\omega)^{2n} \plus{} \omega^n \plus{} 1 \right) \equal{} 0$ since $ n$ is not divisible by 3.

So if that's what you meant, there you go. It works a bit too cleanly for it not
to be the original problem.

so if $ x^{2}\plus{}x\plus{}1 | (x\plus{}1)^{n}\minus{}x^{n}\minus{}1$ it also means $ x^{3}\plus{}x^2\plus{} x  | (x\plus{}1)^{n}\minus{}x^{n}\minus{}1$.

Am i right ??

MCQ: Vijay, is your divisor $ x^3\plus{}x^2\plus{}1$ or $ x^2\plus{}x\plus{}1$ or $ x^3\plus{}x^2\plus{}x$?

[quote="Nukular"]We could either

1) Laugh at the poor fellow.
2) Try to figure out what he's actually asking.
...
[/quote]
3) explain him that there are other sections in this forum (different from olympiad section) where he can learn infos for his level.

Ok, so the original problem as intended is:
Prove that $ x^3 \plus{} x^2 \plus{} x \mid (x \plus{} 1)^n \minus{} x^n \minus{} 1$ for all integers $ n \equiv \pm1 \pmod{6}$ that are greater than $ 3$.

[hide]Let $ f(x) \equal{} x^3 \plus{} x^2 \plus{} x$, and $ g(x) \equal{} (x \plus{} 1)^n \minus{} x^n \minus{} 1$.

The roots of $ f(x) \equal{} x(x^2 \plus{} x \plus{} 1)$ are $ x \equal{} 0$, $ x \equal{} e^{\frac {2\pi}{3}}$, and $ x \equal{} e^{ \minus{} \frac {2\pi}{3}}$.

$ g(0) \equal{} (0 \plus{} 1)^n \minus{} (0)^n \minus{} 1 \equal{} 0$.

$ g(e^{\frac {2\pi}{3}}) \equal{} (e^{\frac {2\pi}{3}} \plus{} 1)^n \minus{} (e^{\frac {2\pi}{3}})^n \minus{} 1 \equal{} (e^{\frac {\pi}{3}})^n \minus{} (e^{\frac {2\pi}{3}})^n \minus{} 1$ $ \equal{} e^{\frac {n\pi}{3}} \minus{} e^{\frac {2n\pi}{3}} \minus{} 1 \equal{} \minus{} \left[(e^{\frac {n\pi}{3}})^2 \minus{} (e^{\frac {n\pi}{3}}) \plus{} 1\right]$

If $ r \equal{} e^{\frac {n\pi}{3}}$ is a root of $ \minus{} (r^2 \minus{} r \plus{} 1) \equal{} 0 \Rightarrow r \equal{} e^{\frac {\pm\pi}{3}}$ (which is true for all $ n \equiv \pm 1 \pmod{6}$) then $ g(e^{\frac {2\pi}{3}}) \equal{} 0$.

$ g(e^{ \minus{} \frac {2\pi}{3}}) \equal{} (e^{ \minus{} \frac {2\pi}{3}} \plus{} 1)^n \minus{} (e^{ \minus{} \frac {2\pi}{3}})^n \minus{} 1 \equal{} (e^{ \minus{} \frac {\pi}{3}})^n \minus{} (e^{ \minus{} \frac {2\pi}{3}})^n \minus{} 1$ $ \equal{} e^{ \minus{} \frac {n\pi}{3}} \minus{} e^{ \minus{} \frac {2n\pi}{3}} \minus{} 1 \equal{} \minus{} \left[(e^{ \minus{} \frac {n\pi}{3}})^2 \minus{} (e^{ \minus{} \frac {n\pi}{3}}) \plus{} 1\right]$

If $ r \equal{} e^{ \minus{} \frac {n\pi}{3}}$ is a root of $ \minus{} (r^2 \minus{} r \plus{} 1) \equal{} 0 \Rightarrow r \equal{} e^{\frac {\pm\pi}{3}}$ (which is true for all $ n \equiv \pm 1 \pmod{6}$) then $ g(e^{ \minus{} \frac {2\pi}{3}}) \equal{} 0$.

Since $ g(0) \equal{} g(e^{ \minus{} \frac {2\pi}{3}}) \equal{} g(e^{ \minus{} \frac {2\pi}{3}}) \equal{} 0$, all the roots of $ f(x)$ (which are of multiplicity $ 1$) are roots of $ g(x)$, and thus, $ f(x) \mid g(x)$ as desired. [/hide]

[quote="pco"]
3) explain him that there are other sections in this forum (different from olympiad section) where he can learn infos for his level.
[/quote]

That would be fine, too... but that's not what you did. Anything resembling the following I would consider rude and condescending, and would make me consider not posting here again if that were the way I would expect to be treated.

[quote="pco"]
Oh great!
So your problem is to prove that  $ 3 | 2^n\minus{}2$ for all odd integer greater than $ 3$

Very difficult problem, indeed. :rotfl:
[/quote]

Xantos: You don't need to check both roots of $ x^2\plus{}x\plus{}1$, since they
are conjugates of each other.

[quote="vijaymenon"]
Thank you very much for all ur replies and sorry to everybody to whom i caused any kind of inconvenience including to those who laughed at me .
[/quote]

I hope you don't get the wrong impression of this forum; I think that was a reasonable
problem to post in this section (after working through the typo), and it takes a little
while to acclimate to which forums are appropriate for given types of problems, especially
with varying definitions of "olympiad level."

[quote="Nukular"]...
That would be fine, too... but that's not what you did. Anything resembling the following I would consider rude and condescending, and would make me consider not posting here again if that were the way I would expect to be treated.
..[/quote]
Ok, Ok, I apologize for my rudeness.

Welcome on the olympiad section, vijaymenon.

Dont hesitate to post here any problem you want (and maybe "solve $ z^6\equal{}a^6$", for example, could now be considered as an olympiad problem) and to ask for explanations when you need them. You'll always find kind people to explain any problem on olympiad section.


Let $X$ be a countable metric space with more than one point. Prove that $X$ is completely disconnected.

This has been discussed before, or something very similar to it.

Given a non-empty subset $S$ of the space $X$ and a point $x$ in $S$, the intersection between $S$ and a ball centered at $x$ with radius different from all the distances from $x$ to points of $X$ is both open and closed in $S$.

Thanks grobber! My solution was much more complicated...  ;)


Circles of radius $ 2$ and $ 3$ are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.

[asy]unitsize(3mm);
defaultpen(linewidth(0.7)+fontsize(8));

filldraw(Circle((0,0),5),grey,black);
filldraw(Circle((-2,0),3),white,black);
filldraw(Circle((3,0),2),white,black);
dot((-2,0));
dot((3,0));
draw((-2,0)--(1,0));
draw((3,0)--(5,0));
label("$3$",(-0.5,0),N);
label("$2$",(4,0),N);[/asy]

$ \textbf{(A)}\ 3\pi \qquad
\textbf{(B)}\ 4\pi \qquad
\textbf{(C)}\ 6\pi \qquad
\textbf{(D)}\ 9\pi \qquad
\textbf{(E)}\ 12\pi$

[hide="Solution"]We know from the given that the radius of the third circle is $ 5$ and its area would be $ 25\pi$.
To find the shaded region, we just subtract the areas of the other two smaller circles from the larger circle:
$ 25\pi\minus{}(9\pi\plus{}4\pi)\equal{}\boxed{12\pi}$[/hide]

it is evident that the diameter would we 10 therefore the radius should be 5 now apply area of circle=pi*r^2
you get areas to be 25pi 9pi and 4pi subtract the latter 2 from the first therefore area of shaded part is 12pi. :first:


To prove if $ x_i \in \mathbb{Z}, i\equal{}1,2,...,n,$    $ n\ge 5$, has distinct roots two by two and $ \sum_{i\equal{}1}^{n}x_i\equal{}2n$,then
$ \sum_{1 \le i<j \le n}x_ix_j \neq 2(n\plus{}1)(n\minus{}2)$

From $ \sum_{i<j}x_ix_j\equal{}\frac 12 [(\sum_i x_i )^2\minus{}\sum_i x_i^2], \ \sum_i x_i\equal{}2n$ we get, that $ \sum_{i\equal{}1}^n x_i^2 \not \equal{}4n\plus{}8$.
Because all $ x_i$ distinct for $ n\ge 8$ $ \sum_i x_i^2 \ge \frac{n(n^2\plus{}2)}{12}>4n\plus{}8$ if n is even, $ \sum_i x_i^2 \ge \frac{n(n^2\minus{}1)}{12}>4n\plus{}8$ if n>8 odd.
For n=5,6,7 it is easy to chek, that $ \sum_i x_i^2\equal{}4n\plus{}8,\sum_i x_i\equal{}2n$ and $ x_i$ are distinct had not solution.


In convex quadrilateral $ ABCD$, $ AB\equal{}a$, $ BC\equal{}b$, $ CD\equal{}c$, $ DA\equal{}d$, $ AC\equal{}e$, $ BD\equal{}f$. If $ \max \{a,b,c,d,e,f \}\equal{}1$, then find the maximum value of $ abcd$.

We claim that the answer is $2 - \sqrt3 > \frac14.$ First let us show that this is attainable. Let $\triangle ABC$ be an equilateral triangle of side length $1$. Then, let $D$ be the unique point such that $BD = 1, DA = DC, $ and $ABCD$ is a convex quadrilateral. In this case, $abcd = 1 \cdot 1 \cdot 2 \cos 15 \cdot 2 \cos 15 = 2 - \sqrt 3$.

Let us now show that this is optimal. We will redefine "convex" to permit angles of the quadrilateral to be $180$ degrees. Call a convex quadrilateral satisfying the conditions of the problem a [b] tapir [/b] if it has the maximum possible area. We wish to show that all tapirs have area $ \le 2 - \sqrt 3,$ and we already know that all tapirs have area $\ge 2 - \sqrt 3$. We will call a pair of points $(P_1, P_2)$ [b] tasty [/b] if the distance between them is one.

[b] Lemma 1. [/b] No tasty quadrilateral has three collinear vertices.

[b] Proof. [/b] Suppose that $A, B, C$ were collinear. Then, we have that $AD \cdot DC \cdot CB \cdot BA \le 1 \cdot 1 \cdot \frac14 (AB + BC)^2 \le 1 \cdot 1 \cdot \frac14 \cdot 1 = \frac14 < 2 - \sqrt3,$ contradiction to the fact that $ABCD$ was a tapir.
$\blacksquare$

[b] Lemma 2. [/b] For every tapir $ABCD,$ we have that $(A, C)$ and $(B, D)$ are both tasty.

[b] Proof. [/b] To see this, start with an arbitrary tapir $ABCD$. Suppose that $(A, C)$ was not tasty. If $(A, D)$ is also not tasty, then we can consider the effect of rotating $A$ away from $D$ about $B$. In this way, $\angle ABD, \angle ABC$ are both increasing. Observe that the lengths of $AB, BC, CD$ are preserved by the process, while the length of $AD$ is increased (by the Law of Cosines). Hence, since $ABCD$ was a tapir, this process must somehow break some condition of the problem. Since we assumed $(A, C), (A, D)$ were both not tasty, the only way a condition can break is if $A, B, C$ are collinear. However, this is a direct contradiction to Lemma $1.$ Therefore, we must have that $(A, D)$ is tasty. By similar reasoning, we can establish that $(A, B), (C, B), (C, D)$ are all tasty. However, this implies that $ABCD$ is a rhombus of side length $1$, and so one of $AC, BD$ must be greater than $1$, contradiction to the fact that $ABCD$ is a tapir.
$\blacksquare$

[b] Lemma 3. [/b] All tapirs have at least one side of length $1$.

[b] Proof. [/b] Assume the contrary, for contradiction. Let $\theta_1, \theta_2$ denote $\angle BDA, \angle CDB$ respectively. By Lemma $1$, we know that $\theta_1, \theta_2 >0.$ By Lemma $2$, we know that $BD = 1.$ Consider the effect of rotating $B$ about $C$ in a way that decreases $\theta_2.$ This process clearly preserves $c, d.$ Consider $a^2 b^2 = (d^2 + 1 - 2d \cos \theta_1) (c^2 + 1 - 2c \cos \theta_2)$ as a function of $\theta_1.$ The derivative of this function is $-(d^2 + 1 - 2d \cos \theta_1) \cdot 2c \sin \theta_2+(c^2+1-2c \cos \theta_2) \cdot 2d \sin \theta_1$ by the Product Rule. Since $ABCD$ was a tapir, this derivative must be zero, as otherwise rotating $DB$ about $D$ in some direction increases $abcd.$ Setting this equal to zero, and recalling that $d^2 + 1 - 2d \cos \theta_1 = a^2, c^2 + 1 - 2c \cos \theta_2 = b^2,$ we know:

$$2a^2 c \sin \theta_2 = 2b^2 d \sin \theta_1,$$
which yields

$$\frac{c}{d} \cdot \frac{\sin \theta_2}{\sin \theta_1} = \frac{b^2}{a^2}.$$
By Sine Law in $\triangle CDA$, this implies that $E  =BD \cap AC$ satisfies $\frac{CE}{EA} = \frac{b^2}{a^2}$ and so therefore $BE$ is the $B-$symmedian of $\triangle ABC.$ Analogously, we can obtain that $CE$ is the $C-$symmedian of $\triangle BCD$. The two previous results then yield that $ABCD$ is a cyclic harmonic quadrilateral. Since $AC = BD = 1$, cyclicity yields that $ABCD$ is actually an isosceles trapezoid. Let $EA = EB = x, EC = ED = 1-x$ WLOG, and set $\angle BEC = \theta.$ Then, $abcd$ can be easily computed as:

$$4 \cos^2 (\frac{\theta}{2}) x (1-x) \cdot (x^2 + (1-x)^2 - 2x(1-x) \cos \theta).$$
Noting that $4 \cos^2 (\frac{\theta}{2}) = 2 \cos \theta + 2,$ we can rewrite the above as:

$$[(2 \cos \theta+2) x(1-x)] \cdot [1 - (2 \cos \theta + 2) x (1-x)].$$
Letting $t = (2 \cos \theta + 2)x(1-x)$, the above is just $t(1-t) \le \frac14 < 2 - \sqrt 3.$ This contradicts the fact that $ABCD$ is a tapir.
$\blacksquare$

By Lemmas $1, 2, 3,$ it suffices to show that all tapirs satisfying $CA = AB = BD = 1$ have $abcd \le 2 - \sqrt 3.$ Let $P$ be the point such that $\triangle APB$ is equilateral, and $P, C, D$ are all on the same side of $AB.$ The conditions of the problem imply that $\angle DBA, \angle CAB \le 60.$ Note that this implies that $CD \le 1$ for free.

[b] Case 1. [/b] $P \in \{C, D\}$

Suppose that $P = C$ by symmetry. Let $\angle DBA = 2 \theta$ for some $0 \le \theta \le 30.$ Then, $abcd = 2 \sin \theta \cdot 2 \sin (30 - \theta) = 2(\cos (2 \theta - 30) - \cos 30).$ From here, it is clear that $abcd$ is maximized at $\theta = 15$, which yields $abcd = 2 - \sqrt 3 \le 2 - \sqrt3$. Hence, the case is resolved.

[b] Case 2. [/b] $P \notin \{C, D\}$

Let $\angle CAB =2 \alpha, \angle DBA = 2 \beta,$ where $0 \le \alpha, \beta \le 30.$ Then, we can easily find $AD, BC$ to be $2 \sin \beta, 2 \sin \alpha$ respectively. By the Pythagorean Theorem, it's easy to find that $c = \sqrt{(\cos 2 \alpha + \cos 2 \beta - 1)^2 + (\sin 2 \beta - \sin 2 \alpha)^2}$. Now, consider $bcd$ as a function of $\alpha$. By the same rotating argument used before, its derivative must be zero, and so we have:

$$2 \cos \alpha \cdot c + 2 \sin \alpha \cdot \frac{\partial c}{\partial \alpha} = 0. \qquad (1)$$
Observe that:

$$\frac{\partial c}{\partial \alpha} = \frac{2(\cos 2 \alpha + \cos 2 \beta - 1) (-2 \sin 2 \alpha) + 2(\sin 2 \beta - \sin 2 \alpha)(-2 \cos 2 \alpha)}{2c},$$
and so by $(1)$ we get:

$$4 \cos \alpha \cdot c^2 + 2(\cos 2 \alpha + \cos 2 \beta - 1) (-2 \sin 2 \alpha) + 2(\sin 2 \beta - \sin 2 \alpha)(-2 \cos 2 \alpha) = 0. \qquad (2)$$
Analogously, we obtain that:

$$4 \cos \beta \cdot c^2 + 2(2 \cos 2 \alpha + \cos 2 \beta -1)(-2 \sin 2 \beta)+2(\sin 2 \beta - \sin 2 \alpha)(-2 \cos 2 \beta) = 0. \qquad (3)$$
Now, suppose that $\alpha > \beta.$ Then, as $\cos \alpha < \cos \beta, \cos 2 \alpha + \cos 2 \beta \ge 1, \sin 2 \alpha > \sin 2 \beta, \cos 2 \alpha < \cos 2 \beta,$ it's clear that the LHS of $(2)$ is strictly less than that of $(3)$. However, this is clearly a contradiction since the RHS's of $(2), (3)$ are equal. Therefore, $\alpha \le \beta,$ and analogously we get $\beta \le \alpha.$ In conclusion, we obtain that $\alpha = \beta.$ Now, observe that $abcd = 2 \sin \alpha \cdot 2 \sin \alpha \cdot (2 \cos 2 \alpha - 1),$
which is equal to:

$$4 \sin^2 \alpha \cdot (1 - 4 \sin^2 \alpha).$$
Hence, if we let $\gamma = 4 \sin^2 \alpha,$ we get $abcd = \gamma (1 - \gamma) \le \frac14 < 2 - \sqrt3.$ However, this contradicts the fact that $ABCD$ is a tapir. Hence, Case $2$ is actually absurd.

As we've exhausted all cases, we conclude that all tapirs have $abcd$ equal to $2 - \sqrt 3.$ In fact, it's easy to see that we have actually proven that all tapirs are the same up to rotation and relabeling of vertices.

$\square$


Let $ k;a$ be positive integer numbers such that $ a\plus{}1|a^{4k}\minus{}1$
Prove that $ \frac{a^{4k}\minus{}1}{a\plus{}1}$ isn't a perfect square

I have a solution:

$ \frac{a^{4k}\minus{}1}{a\plus{}1}\equal{}m^2$    $ .........................(1)$
Or, $ \frac{a^{4k}\plus{}a}{a\plus{}1}\equal{}m^2 \plus{}1$
Hence, $ a \; divides \; m^2\plus{}1$
implies, $ a$ is of the form $ 2^t(4r\plus{}1)$
If $ t \ge 2$ then $ m^2 \equiv \minus{}1 \equiv 3 \; mod 4$ which is absurd.
If $ t\equal{}1$then $ a\equal{}2(4r\plus{}1)\equal{}8r\plus{}2\equal{} \; even$ and hence $ m^2$ is odd.
So, $ m^2 \equiv 1 \;mod 8$ which implies
$ a^{4k} \minus{}1 \equiv a\plus{}1 \;mod 8$ or $ a \equiv 6 \;mod 8$ which contradicts the fact that $ a \equiv 2 mod 8$

Hence $ t\equal{}o$and as a result$ a\equal{}4r\plus{}1 \equal{} odd$
By $ (1)$, $ (\frac{m}{2})^2\equal{}[\frac{a^{2k}\plus{}1}{2}] [\frac{a^{2k}\minus{}1}{2(a\plus{}1)}]$

Clearly $ g.c.d (\frac{a^{2k}\plus{}1}{2},\frac{a^{2k}\minus{}1}{2(a\plus{}1)})\equal{}1$  as $ a$ is odd.

Hence $ a^{2k}\plus{}1\equal{}2p^2$ for some natural number $ p$

But there is only one solution to the equation:

$ x^2\plus{}1\equal{}2y^2$ in natural numbers which is $ (1,1)$.

Hence only solutions to $ (1)$ are $ a\equal{}1$ and$ m\equal{}0$


Find the product of x and y if 8^(3x+2) times 3^(2y) = 2^(x) times 27 ^(y+1)

$ 2^{9x\plus{}6}\cdot3^{2y}\equal{}2^x\cdot3^{3y\plus{}3}\implies2^{8x\plus{}6}\equal{}3^{y\plus{}3}$, and the only way for this to be true is if

$ 8x\plus{}6\equal{}0\implies x\equal{}\minus{}\frac34$ and $ y\plus{}3\equal{}0\implies y\equal{}\minus{}3$, and our final answer is

$ \boxed{\frac94}$.


Haber señores....
aqui una encuesta:
Esquema: PAIS {Ptos. en IMO 2005}

PD: Se ha considerado los resultados del ultimo año
para hacer la lista..

Carlos Bravo ;)
Lima -PERU

Como un acto opuesto al nacionalismo que siento por mi patria, voto objetivamente por Peru, razones: llevan chicos jovenes que yatienen suficiente fogueo, son quienes mas participan en este sitio, el año pasado casi nos ganan y bueno, por Colombia no ira por primera vez naie que ya halla ganado medalla previamente...

Aunque eso si, en dos años le apuesto de nuevo a colombia, por el momento hay que ser sinceros: Peru es el mejor

Veo que Peru esta llevando la delantera en esta encuesta..

y lo más curioso es que no solo son peruanos los que votaron
por PERU... :lol:

bueno ... y haber si existen mas opiniones respecto a la performance
de los equipos iberoamericanos en la proxima IMO...

PD: Considerar que debido a la lejania del host de esta edicion,
muchas delegaciones Iberoamericanas no seran de 6 estudiantes...

Saludos,

Carlos Bravo ;)
Lima -PERU

[quote="José"][quote="carlosbr"][quote="José"]Argentina!!!!!!!! (como en el fútbol, ejejej)[/quote]

Alguien de Brasil??

Saludos,

Carlos Bravo ;)
Lima - PERU[/quote]

Jajajaj, qué sarcasmo!!!!!!![/quote]

no te molestes, .. pero sabemos que el equipo brasileño es uno de los más regulares de latinoamerica ;)
y bueno es de suponer q no espemos menos en esta oportunidad ...

Saludos,

Carlos Bravo ;)
Lima - Peru

En efecto la votacion por Peru es alta porque el numero de mathlinkers peruanos en esta comunidad tambien es alto sin embargo cabe resaltar que debido a que la olimpiada se realizara en territorio europeo y en especial slovenia ; los paises europeos tendran algun tipo de ventaja como esa de que los problemas que a los latinos les paracen dificiles a los europeos le parecen faciles debido a al tecnica que usan .De hecho que los participantes latinos la tendran muy dificil en esta olimpiada mundial

Pienso que la competencia entre paises latinos esta reñida no le atino a Peru ya que Peru siempre va con chicos de hasta 16 años ( lo normal ) y en esta ocasion 3 van por primera vez a una IMO .

Terminare diciendole  a los participantes latinos que : Mientras mas dificil vean una prueba mas ganas tengan de resolverla y que los cortes para medalla estaran altos a mi paracer

BRONZE : mas de dos problemas hechos
PLATA: 4 problemas hechos
ORO:5 problemas hechos

asi que no se contenten con hacer dos problemas que  para conseguir medalla Pienso que esta vez habrá que hacer algo mas que eso .

Consejos muy acertados ...DarthTenebrus....

esperamos que los chicos que estaran presentes en el evento .. los tengan presentes .. ;)

Saludos,

Carlos Bravo ;)
Lima -PERU

[quote="carlosbr"]no te molestes, [/quote]

Todo bien

ni lo pensé mucho... Perú será el mejor equipo de Iberoamérica en la IMO

motivos para pensarlo... sobran... los últimos resultados han demostrado una superioridad abismante de los peruanos en las competencias

las personas que votaron también por Perú, aunque sean muchos de ahí, lo hicieron objetivamente...

bueno, sólo eso... saludos

Dos cosas:

La primera:

Sigo apoyando a peru, por resultados del año pasado (segundos despues de colombia), participacion en el foro, resultados ibero y Cono sur. Ademas por la buena preparacion y cobertura de su olimpiada.

La segunda:

No estoy de acuerdo en decir como seran los cortes desde ya. Eso solo se sabe una vez vista la prueba, antes no se puede decir nada, ademas Eslovenia no ha sido un equipo fuerte en IMO, por lo que tendran un jurado muy internacional. No olviden que son los lideres quienes eligen la prueba, no el pais, por eso no hay tanto problema en que sea un  pais europeo, ademas hay que dejar de decir que los europeos tienen alguna ventaja... Creo que hoy en dia con todos los recursos y acceso que tenemos a pruebas y problemas, estamos todos en igualdad de condiciones.

No es de esperarse ningun tipo de cortes, pero eso si les digo algo con toda seguridad: Si sacan 42 tienen oro.

BTW:

hay una cosa que me preocupa de Peru y es que muchos jovenes peruanos me han contado que hay cierfta clase de monopolio de la olimpiada por pocas academias y que las regiones se ven un poco excluidas. Que tan cierto es esto? o realmente el nivel en estas academias es tan alto como para dominar en las pruebas de seleccion? (que es lo que yo creo) Sin embargo de ser asi, creo que esta academia deberia ofrecer cursos abiertamente a todos los estudiantes interesados para que no sea un privilegio de pocos.

IGUAL SOY AJENO A ESTA SITUACION Y SOLO QUERIA SABER QUE TAN CIERTOE S ESTO QUE ME HAN DICHO. :)

[quote="Pascual2005"]Dos cosas:

BTW:

hay una cosa que me preocupa de Peru y es que muchos jovenes peruanos me han contado que hay cierfta clase de monopolio de la olimpiada por pocas academias y que las regiones se ven un poco excluidas. Que tan cierto es esto? o realmente el nivel en estas academias es tan alto como para dominar en las pruebas de seleccion? (que es lo que yo creo) Sin embargo de ser asi, creo que esta academia deberia ofrecer cursos abiertamente a todos los estudiantes interesados para que no sea un privilegio de pocos.

IGUAL SOY AJENO A ESTA SITUACION Y SOLO QUERIA SABER QUE TAN CIERTOE S ESTO QUE ME HAN DICHO. :)[/quote]
Quizás tengas razón al decir que es un privilegio de pocos, pero es un privilegio ganado, la academia Cesar Vallejo busca jóvenes talentos en todo el Perú a través de su Concurso Nacional de Matemática principalmente, y los chicos que están en sus talleres (así se les llama) son chicos que han sobresalido entre otros, se les prepara intensivamente (tienen clases en la mañana y en la tarde) y es por eso que los chicos salen en buenos puestos cuando hay examenes selectivos y no son chicos que solo se hayan formado en la academia, por ejemplo entre mis alumnos hay dos puneños, 1 de Huancayo, 1 de barranca, por eso no es cierto que sea centralizado.

Y que no parezca que es un monopolio lo de las academias, si es que sus alumnos salen bien en los selectivos  es porque son chicos hábiles y se han preparado bastante, no les quiten ese mérito.  :)
Sería interesante que los que te hayan comentado eso, posteen tambien,  así derepente se les puede aclarar más cómo son las cosas... y Pascual gracias por la preocupación.

$Tipe$

Si es asi, entonces me parece muy bien, de igual manera es en colombia, tal vez detras de esos comentarios halla cierta intencion de quitar meritos, o depronto desconocimiento, no lo se... Pero tal cual lo plantean es la misma situacion que en colombia o en EEUU (con la academia EXETER) se vive!

Bueno Si creo que eso de los cortes de las medallas en parte puede variar segun la olimpiada pero no tanto creo yo .
Por lo que yo estaba enterado la shortlist de 30 problemas es seleccionada de un total aproximado de 200 problmas por un comite jurado del pais organizador que no necesariamente deben de ser de eslovenia sino invitados representativos de paises vecinos ( muchos europeos diria yov como bulgaria , rusia , rumania , etc  ) y es por eso que me referia que la shortlist podria tener un estilo europeo de preguntas y aunque son los jefes de delagcaion los que eligen las prueba ellos la elegiran de una shortlist de estilo europeo .

No se hasta que punto tendre razon pero ese era mi comentario

[quote="tipe"]
Quizás tengas razón al decir que es un privilegio de pocos, pero es un privilegio ganado, la academia Cesar Vallejo busca jóvenes talentos en todo el Perú a través de su Concurso Nacional de Matemática principalmente, y los chicos que están en sus talleres (así se les llama) son chicos que han sobresalido entre otros, se les prepara intensivamente (tienen clases en la mañana y en la tarde) y es por eso que los chicos salen en buenos puestos cuando hay examenes selectivos y no son chicos que solo se hayan formado en la academia, por ejemplo entre mis alumnos hay dos puneños, 1 de Huancayo, 1 de barranca, por eso no es cierto que sea centralizado.

Y que no parezca que es un monopolio lo de las academias, si es que sus alumnos salen bien en los selectivos  es porque son chicos hábiles y se han preparado bastante, no les quiten ese mérito.  :)
Sería interesante que los que te hayan comentado eso, posteen tambien,  así derepente se les puede aclarar más cómo son las cosas... y Pascual gracias por la preocupación.

$Tipe$[/quote]

Puedo añadir alguna informacion respecto a este evento:

[url=http://www.conamat.edu.pe/conamat/index.htm]Concurso Nacional de Matematica "CESAR ALLEJO" - CONAMAT[/url]  (la cual cumple este año su 9na Edicion ;) ) organizada por el [url=http://www.colegiobertoltbrecht.edu.pe/colegio/V3/index2.htm]Colegio Bertolt Brecht[/url] que es parte del [url=http://www.ich.edu.pe/principal.htm]Instituto de Ciencias y Humanidades[/url].

Les puedo comentar que este evento cumple una mision de difusion de la cultura de las matematicas a un nivel nacional ya que en la ultima edicion convoco a más de 18 000 estudiantes, definiendo un proceso eliminatorio previo (en setiembre) atraves de  [url=http://www.conamat.edu.pe/conamat/sedes.htm]7 sedes[/url], cubriendo el 100% de las provincias del pais para este año, asegurando asi el buen nivel academico  y competitivo de los estudiantes clasificados en la Fase Final que se da en Octubre.

En mi opnion, esta es una iniciativa muy positiva que ha logrado que más chicos tengan el acceso a la informacion de  las olimpiadas y eventos como CONAMAT, puesto q atraves de estos eventos, se ha difundido gran numero de publicaciones de las ediciones anuales de este evento y otros afines, y claro esta que tambien de pertenecer a los talleres de Bertolt Brecht.

PD:
[quote="tipe"]
por ejemplo entre mis alumnos hay dos puneños, 1 de Huancayo, 1 de barranca, por eso no es cierto que sea centralizado.
$Tipe$[/quote]
Recuerda que mi hermanita (fabycv) y yo somos de cañete ;)
y ella llego a Lima ganando 1er puesto en el 5to CONAMAT a los 10 años ;) ;)

Carlos Bravo ;)
Lima -PERU

[quote="DarthTenebrus"]He visto los comentarios acerca de la academia Cesar vallejo y la evrdad que aqui en el foro hay muchos math linkers peruanos y entre los peruanos hay muchos exalumnos y alumnos del actual Bertolt Brecht o mejor dicho el colegio de la academia Cesar Vallejo

 Aun recuerdo la primera vez que participe en la Olimpiada Nacional de la Cesar Vallejo del año 1996 . Lo gane porque competi en la modalidad de Colegios Nacionales en el tercer año de Secundaria . en aquellos tiempos la unica academia que entrenaba en olimpiadas era la PITAGORAS .

Al año siguiente quize volver a participar pero aunque pase la primera fase no pude dar la segunda porque no habia entregado algunos documentos .  Eso fue lo que me dijeron y fuimos muchos alumnos los que nos quedamos fuera de la segunda fase . ¿ Cual habra sido la verdadera razon ?

YO AFIRMO LO SIGUIENTE :

Ese Concurso llamado CONAMAT tiene vetados a algunos colegios preuniversitarios de LIMA y que por cierto muchos colegios preuniversitarios tiene miedo de enviar a sus alumnos a ese concurso CONAMAT a sus alumnos porque temen que estos puedan ser seducidos a ser cambiados de colegio .

La Cesar Vallejo es muy grande y Es totalmente un MONOPOPLIO que seduce chicos de otras instituciones a formar parate de sus filas . Pienso que si los alumnos escolares tuvieran algo llamado CARTA PASE las cosas no serian tan faciles para la CESAR VALLEJO .

La Academia Trilce por ejemplo realiza muchos concursos interescolares en cada punto del pais y su fin es publicitario y descentralizador la academia Pitagoras hace lo suyo por su parte pero la ACADEMIA CESAR VALLEJO solo hace un concurso que por cierto las pruebas se rinden en LIMA aunque vengan chcios de otras provincias . No solo eso Se cobra la participacion y se vetan a aquellos alumnos de los colegios y obviamente se fortalece su equipo con los alumnso ganadores del concurso .

¿Mi pregunta es porque algunos alumnos son vetados o simplemente son excluidos de dar la fase final ?

Tal vez la respuesta es que aquellos alumnos que no tenga algun interes en la cesar Vallejo mas que de dar la prueba son vetados ya que no le son factibles para los verdaderos interes del concurso y no se proque tanto papeleo como partida de nascimineto y certificado de notas que pdien en ese concurso cuando no es al gran cosa e seguro que ni en la ONEM piden tanto .

En esta ultima calsificacion para la IMO :

hay 3 alumnos del bertolt 1 de saco 1 de pitagoras y 1 de prolog
La ACADEMIA TRILCE ya desaparecio del mapa de als olimpiadas y aunque tiene alumnos muy buenos estos no estan siendo enrtenados . Espero que la Vallejo no entre a seducir a esos chicos porque la SACO OLIVEROS o la PITAGORAS no lo hacen y eso es lo que debe ser .

Algunas Instituciopnes como SACO OLIVEROS y la PITAGORAS son muy fuertes y habran muchso otros colgios emergentes que se daran a conocer poco a poco .

Aunque es ceirto que la CESAR VALLEJO tiene un prestigio ganado eso no le da poder de interferir fuertemente en otros colegios .
Yo apuesto por la libre competencia de las ACADEMIAS pero sin monopolios y eso le dara aun mejores resultados al equipo peruano en las olimpiadas .
Espero no haber sido muy duro pero es asi como yo he visto la situacion desde ya hace muchos años .[/quote]
---------------------------------------------------
[quote="DarthTenebrus"]
He visto los comentarios acerca de la academia Cesar vallejo y la evrdad que aqui en el foro hay muchos math linkers peruanos y entre los peruanos hay muchos exalumnos y alumnos del actual Bertolt Brecht o mejor dicho el colegio de la academia Cesar Vallejo
[/quote]
Para comenzar, yo no he estudiado en Bertolt Brecht, y más aun
he sido delegado de la seleccion del Colegio Nº 20874 de Cañete, y como tal
he supervisado el proceso de las inscripciones de los estudiantes para este  evento,
y puedo dar fe que no existe ninguna preferencia de ningun tipo ni restriccion
para algunos colegios, salvo la prohibicion de los colegios que incumplan las bases
del concurso, las que estan publicadas en la web del evento. (revisar)

[quote="DarthTenebrus"]
Al año siguiente quize volver a participar pero aunque pase la primera fase no pude dar la segunda porque no habia entregado algunos documentos .  Eso fue lo que me dijeron y fuimos muchos alumnos los que nos quedamos fuera de la segunda fase . ¿ Cual habra sido la verdadera razon ?
[/quote]
Los documentos q se exigen son los minimos para verificar que los estudiantes estan
 en el grado academico y tienen la edad q ellos dicen, es solo de verificacion y asi evitar
 suspicacias y suplantaciones,  claro esta cumpliendo los cronogramas establecidos para cada
fase del evento, mientras que en la ONEM (Olipiada Nacional d Estudiantes - Matematica),
 el Ministerio de Educacion ya  cuenta con informacion de los estudiantes que tomaran parte,
puesto que los propios colegios sirven de filtros de control.

[quote="DarthTenebrus"]
... actual Bertolt Brecht o mejor dicho el colegio de la academia Cesar Vallejo ..
[/quote]
Respecto a este comentario, no debemos confundir las cosas, el evento CONAMAT
lo organiza el Colegio Bertolt Brecht y no la Academia Cesar Vallejo, aunque ellos
pertenecen al mismo Instituto de Ciencias y Humanidades no existe una jerarquia
 vertical entre estas instituciones, sino un vinculo de colaboracion horizontal.

[quote="DarthTenebrus"]
La Cesar Vallejo es muy grande y Es totalmente un MONOPOPLIO que seduce chicos de otras instituciones a formar parate de sus filas . Pienso que si los alumnos escolares tuvieran algo llamado CARTA PASE las cosas no serian tan faciles para la CESAR VALLEJO .
[/quote]
Y bueno no comparto este punto de q Bertolt Brecht MONOPOLICE el entrenamiento
de los chicos para las olimpiadas, puesto que existen otros grupos que tambien
realizan tareas de entrenamiento como en los colegios Prolog, Isaac Newton, San Igncio del Cono Norte, etc
la institucion CJEO que organiza eventos de entrenamientos (para mayor informacion
 contactarse con el  prof Eduardo del Carpio).
La preponderancia de los chicos del Colegio Bertotl Brecht se debe a que este cuenta
con un entrenamiento más exigente, y que las personas que dictan alli son bien calificadas.

Debemos tener presente que toda institucion como la Academia TRILCE y su conjunto de concursos cada semana,
la Academia Pitagoras y sus eventos, el Colegio San Ignacio, el colegio Bertolt Brecht con CONAMAT, siempre entrega insentivos como Becas Integras de estudio, libros e inclusive dinero en efectivo, (esto ultimo no se da en CONAMAT), al final la decision de asistir a una institucion u otra, la tiene el estudiante, y naturalmente este optará
por lo mejor para el; pero no podemos decir que un colegio induce a estudiantes de otras instituciones a que se unan
a sus aulas, esto no es correcto, al menos esto no pasa con Bertolt Brecht.

[quote="DarthTenebrus"]
Algunas Instituciopnes como SACO OLIVEROS y la PITAGORAS son muy fuertes y habran muchso otros colgios emergentes que se daran a conocer poco a poco ...

 Yo apuesto por la libre competencia de las ACADEMIAS pero sin monopolios y eso le dara aun mejores resultados al equipo peruano en las olimpiadas .
[/quote]
En este punto si estoy de acuerdo, puesto q existen muchas otras instituciones academicas orientadas a la preparacion pre universitaria, que estan incursionando en entrenamiento en olimpaidas  u otras q dejaron de hacerlo, espero que retomen y/o mejoren sus estrategias para lograr un buen nivel de sus estudiantes y lo mas importante, formar nuevas alternativas a los chicos que buscan incursionar en este mundo de las olimpiadas..

[b]Comentario Final:[/b]
Es normal que la mayoria de estudiantes quieran llegar al Bertolt Brecht, no por presion de este colegio, sino por decision propia, puesto q en este momento, es la institucion que esta haciendo el mejor trabajo a lo q respecta en Olimpiadas de Matematicas, asi justamente como lo hizo la Academia Pitagoras en los '90 y que lograron formar un buen equipo para la IMO '98, donde Leonel Requena y Liz Vivas Inga obtuvieron nuetras 2 primeras y unicas medallas de plata para PERU... ;) ;)
Mi consejo es no crear diferencias entre nosotros, entre colegios y/o academias, sino buscar colaborar como pais, apoyando a los estudiantes interesados en estos temas, sin importar su procedencia, puesto que al final en una IMO no importa en q colegio estudiaste o en q academia estuviste, solo importa tu desempeño como pate del equipo, la seleccion. ;) ;)

PD: Espero haber sido bastante explicito, y evitar extender más esta conversacion.. si alguien necesita alguna informacion o tenga una duda sobre lo q se dice en este post, por favor, eviarmelo en un PM..

Gracias....

Carlos Bravo ;)
Lima - PERU

[quote="DarthTenebrus"]He visto los comentarios acerca de la academia Cesar vallejo y la evrdad que aqui en el foro hay muchos math linkers peruanos y entre los peruanos hay muchos exalumnos y alumnos del actual Bertolt Brecht o mejor dicho el colegio de la academia Cesar Vallejo

 Aun recuerdo la primera vez que participe en la Olimpiada Nacional de la Cesar Vallejo del año 1996 . Lo gane porque competi en la modalidad de Colegios Nacionales en el tercer año de Secundaria . en aquellos tiempos la unica academia que entrenaba en olimpiadas era la PITAGORAS .

Al año siguiente quize volver a participar pero aunque pase la primera fase no pude dar la segunda porque no habia entregado algunos documentos .  Eso fue lo que me dijeron y fuimos muchos alumnos los que nos quedamos fuera de la segunda fase . ¿ Cual habra sido la verdadera razon ?

YO AFIRMO LO SIGUIENTE :

Ese Concurso llamado CONAMAT tiene vetados a algunos colegios preuniversitarios de LIMA y que por cierto muchos colegios preuniversitarios tiene miedo de enviar a sus alumnos a ese concurso CONAMAT a sus alumnos porque temen que estos puedan ser seducidos a ser cambiados de colegio .

La Cesar Vallejo es muy grande y Es totalmente un MONOPOPLIO que seduce chicos de otras instituciones a formar parate de sus filas . Pienso que si los alumnos escolares tuvieran algo llamado CARTA PASE las cosas no serian tan faciles para la CESAR VALLEJO .

La Academia Trilce por ejemplo realiza muchos concursos interescolares en cada punto del pais y su fin es publicitario y descentralizador la academia Pitagoras hace lo suyo por su parte pero la ACADEMIA CESAR VALLEJO solo hace un concurso que por cierto las pruebas se rinden en LIMA aunque vengan chcios de otras provincias . No solo eso Se cobra la participacion y se vetan a aquellos alumnos de los colegios y obviamente se fortalece su equipo con los alumnso ganadores del concurso .

¿Mi pregunta es porque algunos alumnos son vetados o simplemente son excluidos de dar la fase final ?

Tal vez la respuesta es que aquellos alumnos que no tenga algun interes en la cesar Vallejo mas que de dar la prueba son vetados ya que no le son factibles para los verdaderos interes del concurso y no se proque tanto papeleo como partida de nascimineto y certificado de notas que pdien en ese concurso cuando no es al gran cosa e seguro que ni en la ONEM piden tanto .

En esta ultima calsificacion para la IMO :

hay 3 alumnos del bertolt 1 de saco 1 de pitagoras y 1 de prolog
La ACADEMIA TRILCE ya desaparecio del mapa de als olimpiadas y aunque tiene alumnos muy buenos estos no estan siendo enrtenados . Espero que la Vallejo no entre a seducir a esos chicos porque la SACO OLIVEROS o la PITAGORAS no lo hacen y eso es lo que debe ser .

Algunas Instituciopnes como SACO OLIVEROS y la PITAGORAS son muy fuertes y habran muchso otros colgios emergentes que se daran a conocer poco a poco .

Aunque es ceirto que la CESAR VALLEJO tiene un prestigio ganado eso no le da poder de interferir fuertemente en otros colegios .
Yo apuesto por la libre competencia de las ACADEMIAS pero sin monopolios y eso le dara aun mejores resultados al equipo peruano en las olimpiadas .

Espero no haber sido muy duro pero es asi como yo he visto la situacion desde ya hace muchos años .[/quote]

Bueno, veo q tienes una percepcion muy equicovada de la situacion no?

Las academias lo q buscan con lo concursos interescolares es la promocion del estudio de la matematica, y el descubrimiento de nuevos valores para componer las "filas" q mencionas(y me refiero a todas).

He tenido la oportunidad de representar a mi país en algunas ocasiones, y considero ofensivo q digas q mi centro de estudios me "sedujo" para ser uno mas en "sus filas". Hablas de nosotros como si fueramos soldados.

Cuando voy a una olimpiada represento al PERU, en ningun momento siento q voy por parte de tal o cual colegio. Me siento identificado con mi pais, mi ciudad, mis compañeros de equipo(de todos sitios) y mi orgullo propio.

No somos fichas de juego q una academia viene y adquiere, somos estudiantes(TODOS) q se esfuerzan y alcanzan logros para su pais y para si mismos, para nadie mas.

Yo tengo muchas diferencias con mi centro de estudios, muchas cosas en las q estamos en desacuerdo, pero no dejare q hables de mi y de gente q ha entrenado conmigo como si fueramos peones de ajedrez.

Bueno gente tolerancia, vuelvan al tema original de conversación

Definitivamente es cierto lo que dice Jose : Nos hemos salido un poquito del tema despues de lo que dijo Pascual y las réplicas de Tipe respecto a la de pascual .y la mia respecto a la de tipe y la de carlos y Franco con respecto a a la mia .

Esto se que da cierta imagen no se si positiva o negativa del peru a otro paises sin embargo lo unico que hice fue dar mi opinion .

Si este equivocado o no respecto a lo que dije ;  cada peruano sacara sus coclusiones ( algunos mas que otros jeje )  .

No creo que me den replicas porque no creo haber levantado polvo esta vez .

Les pido disculpas a todos los foristas de esta comunidad por mi actuacion .

Pero para mis replicadores sigo pensando lo mismo y si pudiera conversar con ellos personalmente lo haria con gusto .

Está todo perfecto, pero si quieren seguir opinando sobre eso, hagan otro topic, y ahí si van a poder hablar sobre ese tema

Parece que nos equivocamos: Argentina sera primero