You use Chebyshev inequality in a wrong way. If $a+b+c+d=4$ then certainly
\[a^{2}+b^{2}+c^{2}+d^{2}\ge 4.\]

[quote="manlio"]But $(\frac{1}{a}-a)$ can be $<0$ !!! :wink:[/quote]
If $(\frac{1}{a}-a) <0$,then?

I believe that this way is hardly true. You should try another one, chienthan.

For example if $(\frac{1}{a}-a)<0$ you cannot write $(\frac{1}{a}+a)(\frac{1}{a}-a) \geq 2(\frac{1}{a}-a)$

if numbers are in $(0,3)$ i can prove it by convexiti of the function .but i will find an other solution.
hi [b]hungkhtn[/b] can u post ur theorem ( i mean smvt and E?) because i do not now anything about them $thinks$. :D

About SMV, you can see

http://www.mathlinks.ro/Forum/viewtopic.php?t=113248

Yes, Vasc, it was also on my note few minutes ago. The solution is based on an idea of Vardach.

I would like to see your solution, Thuan and VASC, if it is not mixing variable.

i remember mr. Schwarz told us that Vasc's inequality holds even for $5$ variables and for the degree $3,4$ also.

I proved already that the inequality
\[\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}\ge a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\]
holds for all $n\in\{1,2,...,9\}$.

However, I would like to see a simple solution for $n=4$, that the reason why I post this problem. I am also very curious of how Cauchy-Schwarz can be applied to prove this case $n=4$ and even $n=5$ as Pohoatza's post. Could you show it? Pohoatza?

I also wait a solution of VASC and Thuan about their improvement. Hope that I have not to wait long then.

The following inequality is true and stronger

\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}-4 \ge \sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4).\]

[quote="Vasc"][quote="hungkhtn"] The following inequality is true and stronger
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}-4 \ge \sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4). \]
[/quote]
I think the inequality below is true and stronger
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}+4 \ge 2(a^{2}+b^{2}+c^{2}+d^{2}). \]
[/quote]

Nice one, VASC,

I build this inequality from the general case of $n$ so I don't check the case $n=4$ much. I just use the general result to replace this particular case. For all, we only need to check the inequality when $n-1$ numbers equal, so your problem can be done similarly as mine, VASC, just a bit different in the last step. I prove it by elementary method, not mixing variable, EV or SMV, too.

Dear VASC and others,

Do you think that it is time now to show a solution to the initial problem? We did discuss a lot, with strong problem, but a nice solution to the first is still hidden,

[quote="hungkhtn"]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}. \]
[/quote]

Yes, toanhocmuonmau.  :lol: It is also my solution.

A general problem:
 Let $a_{1},a_{2},...,a_{n}\geq 0 such that: \sum a_{i}=n(n\geq 4)$.
Prove that:
  $\sum\frac{1}{a_{i}^{n-2}}\geq \sum a_{i}^{2}$

[quote="hungkhtn"][quote="Vasc"][quote="hungkhtn"] The following inequality is true and stronger
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}-4 \ge \sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4). \]
[/quote]
I think the inequality below is true and stronger
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}+4 \ge 2(a^{2}+b^{2}+c^{2}+d^{2}). \]
[/quote]

Nice one, VASC,

I build this inequality from the general case of $n$ so I don't check the case $n=4$ much. I just use the general result to replace this particular case. [/quote]

My general case of $n$ is the following.

[i]If $a_{1},a_{2},...,a_{n}$ ($n\ge 4$) are positive numbers such that $a_{1}+a_{2}+...+a_{n}=n$, then
\[\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}-n \ge \frac 8{n}(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n). \]
[/i]
The cases $n=4$ and $n=8$ are very interesting.

In the topic about how to construct an inequality here

http://www.mathlinks.ro/Forum/viewtopic.php?t=143631

I did post the following stronger inequality

$\bigstar$ (Hungkhtn's, SI Inequalities A story of Maths, problem 5.3.4)
Let $a_{1},a_{2},...,a_{n}$ be non-negative real numbers such that $a_{1}+a_{2}+...+a_{n}=n$. Prove that
\[n^{2}\left(\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}-n\right) \ge 10(n-1)\left(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n\right).\]

My solution is very elementary.

The following is more sharper than the first
\[\sum\frac{1}{a}\ge 3+\frac{3}{4}\sum a^{2}\]
and it has 2 equalities :)

Maybe

$\sum \frac 1{a}\ge 1+\frac 3{4}\sum{a^{2}}$ for $\sum a=4$.

This was posted in 2004 on this forum for $n$ numbers.

[quote="toanhocmuonmau"][quote="Vasc"][quote="hungkhtn"]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}. \]
[/quote]

I think the following stronger inequality holds for  $a+b+c+d=4$:

${\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}+\frac{1}{da}}\ge a^{2}+b^{2}+c^{2}+d^{2}$.[/quote]
Using the known $x^{2}+y^{2}\le \frac{(x+y)^{4}}{8xy}\quad \forall x,y \ge 0$ $(*)$, we have
\[a^{2}+c^{2}\le \frac{(a+c)^{4}}{8ac}=\frac{(a+c)^{4}bd}{8abcd}\le \frac{(a+c)^{4}(b+d)^{2}}{32abcd}, \qquad b^{2}+d^{2}\le \frac{(b+d)^{4}(a+c)^{2}}{32abcd}\]
We need to prove
\[{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}+\frac{1}{da}}\ge \frac{(a+c)^{4}(b+d)^{2}}{32abcd}+\frac{(b+d)^{4}(a+c)^{2}}{32abcd}\]
Or
\[32 \ge (a+c)(b+d)((a+c)^{2}+(b+d)^{2}) \]
It is true by $(*)$. :) Equality holds if and only if $a=b=c=d=1$. :)[/quote]

It is just now I can read your proof. Very nice one, toanhocmuonmau. Congras!

Do you find a nice proof for $n=5$?

[quote="hungkhtn"]I proved already that the inequality
\[\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}\ge a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\]
holds for all $n\in\{1,2,...,9\}$.
[/quote]
It's true for $n=10$ and it's wrong for all $n\geq11$. :wink:

[b]Ji Chen:[/b]
Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that$$ a^{2}+b^{2}+c^{2}+d^{2}\leq 2\left(\frac{1}{ac}+\frac{1}{bd}\right).$$

[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}.\][/quote]
Let $a,b,c,d$ be real numbers such that $abcd=1$. Prove that$$ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\geq ab+bc+cd+da$$
(Proposed by Mircea Becheanu, University of Bucharest, România)

Let $a,b,c$ be real numbers such that $abc=1$. Prove that:$$a^2+b^2+c^2 \geq a+b+c .$$

[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}.\][/quote]
[url=https://artofproblemsolving.com/community/c6h181229p996138]the stronger one[/url]
[url=https://artofproblemsolving.com/community/c6h1370494p10534020]here
[/url]

Let $a,b,c,d$ be positive real numbers such that $a <b <c <d $ and  $a+b+c+d=4$. Prove that$$4\leqslant \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leqslant \frac{a}{d}+\frac{d}{a}+2$$
(M.S.Vovos)

3-var version:Let $a,b,c$ be positive real numbers such that $a+b+c=3$,then
\[\sum {\frac{1}{{{a^2}}} \ge \sum {{a^2}} }  + 0.548\sum {{{\left( {a - b} \right)}^2}} \]
strong in this form.

[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}.\][/quote]
Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. For the integer $n\geq2$, prove that
$$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n} +\frac{1}{d^n}\ge a^n+b^n+c^n+d^n.$$
[url=https://artofproblemsolving.com/community/c6h1370494p10534020]here[/url]

[quote=hungkhtn]I proved already that the inequality
\[\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}\ge a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\]
holds for all $n\in\{1,2,...,9\}$.

[hide=*]However, I would like to see a simple solution for $n=4$, that the reason why I post this problem. I am also very curious of how Cauchy-Schwarz can be applied to prove this case $n=4$ and even $n=5$ as Pohoatza's post. Could you show it? Pohoatza?

I also wait a solution of VASC and Thuan about their improvement. Hope that I have not to wait long then.

The following inequality is true and stronger

\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}-4 \ge \sqrt{3}(a^{2}+b^{2}+c^{2}+d^{2}-4).\][/hide][/quote]
[quote=arqady]
It's true for $n=10$ and it's wrong for all $n\geq11$. :wink:[/quote]

...

[quote=arqady]See here:
https://artofproblemsolving.com/community/c6h144103p3588112[/quote]
Thanks.

[quote=hungkhtn]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}.\][/quote]
Let $ a,b,c,d$ be positive numbers sach that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. [url=https://artofproblemsolving.com/community/c6h181229p996138]Prove that[/url]
$$ \frac 1{ab}\plus{}\frac 1{bc}\plus{}\frac 1{cd}\plus{}\frac 1{da}\ge a^2\plus{}b^2\plus{}c^2\plus{}d^2$$
[url=https://artofproblemsolving.com/community/c6h1856322p12541678]h[/url]

[quote=sqing]Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1a_2\cdots  a_n=1.$ Prove that $$\frac{1}{a^n_1}+\frac{1}{a^n_2}+\cdots+\frac{1}{a^n_n}\geq a_1+a_2+\cdots+a_n.$$[/quote]

Nice but trivial by Am-Gm inequality multiple times


Prove that if $x,y,z>0$ then we have:

$ \frac{3}{2} ( \sum \frac{x^{2}}{y^{2}} + 1)  \geq \frac{x}{y} +2 \sqrt{ \frac{y}{z} }+3 \sqrt[3]{ \frac{z}{x} } \geq 6$.

made by Ioan Dancila.

cheers! :D :D

3+3sum(x^2/y^2) = sum(x^2/y^2) + 3 + 2sum(x^2/y^2)
>= sum(x^2/y^2) + 3 + 2*3 (xyz/xyz)^2 by AM-GM
= (x^2/y^2 + 1) + (y^2/z^2 + 3) + (z^2/x^2 + 5)
>=2 x/y + 4 (y/z)^1/2 + 6 (z/x)^1/3 by AM-GM on each bracket, proving the left inequality (obviously, equality occurs iff x = y = z = 1).

To prove the right inequality, note that

1/6 x/y + 1/3 (y/z)^1/2 + 1/2 (z/x)^1/3
<=(xyz/xyz)^1/6 = 1, by the weighted AM-GM inequality.  Again, equality occurs iff x = y = z = 1.


It's \frac, not frac.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=25780 .

  darij


the     following    problem  maybe   a  Olympiad   one:

http://www.mathlinks.ro/viewtopic.php?p=1685878#1685878


Balanced chemical equation:
C3H5(NO3)3(s) → 3CO2(g) + 2.5H2O(g) + 1.5N2(g) + 0.25O2(g)

I've been thinking for a while, and can't quite answer why it's so explosive.  Any help?

Hrmm.. your question sparked my interest.. taken from wikipedia:

"Nitroglycerin and any or all of the diluents mentioned above can certainly deflagrate, or burn. However, the explosive power of nitroglycerin is derived from detonation: a shock propagates through the fuel-rich medium at a supersonic speed. In other words, the initial decomposition sets up a pressure gradient that induces decomposition in contiguous material, creating a fast-moving transition zone, which (due to the nature of the material) can detonate any unstable or explosive material it encounters. This generates a self-sustained cascade of near instantaneous pressure-induced decomposition into gas of the explosive material which grows upon itself exponentially. This is quite unlike deflagration, which depends solely upon available fuel, regardless of pressure or shock."

A couple of reasons...

First, obviously a very negative $\Delta H$.

Second, a very positive $\Delta S$

Third, (very likely) a very low-energy transition state, in turn leading to a high reaction rate.

Fourth, (likely) a low specific heat (Kopp's rule gives $265\ \text{J deg}^{-1}\text{ mol}^{-1}$; the actual value is unknown to me).

One must be very specific in one's questions if one is to obtain a precise answer.  Unfortunately, in most cases, one must be more familiar to the answer than one is to know enough to ask a percise answer.

There are two dimensions to the question about nitroglycerin (the reactive chemical in dynamite that made blasting much safer and fueled the Nobel prizes, and the start of the Nobel chemical company).  The first dimension is the amount of energy released, which is driven by the high bond strength of the formed nitrogen-oxygen bonds formed in the reaction of nitroglycerin with oxygen in burning or exploding.  (This has been addressed in the other responses to your question.)

The second dimension is the speed of the reaction in other words how fast the energy is released.  This is an independent physical property.  One traditional way to measure this is to determine how fast a flame front will progress in an open trench.  This is why nitroglycerin is not used as a fuel for internal combustion engines -- it simply delivers the energy too fast.  It is also not used as a gunpower substitute for the same reason -- it delivers the energy too fast.  The mechanics of converting the released chemical energy is insufficient for mechanical or projectile use.  If one were to put kerosene in a open trench and lit one end of the vapor, one could walk faster than the flame front progressed.  This is a relatively slow release of energy, useful in jet engines and the like.  One would have to run to keep up with an methyl ether fire (a member of my lab did so and got his beaker into the hood before the entire beaker and the flame front coincided, avoiding a more memorable incident.)  If one were to do this with nitroglycerin, the flame front would progress at a speed greater than mach 1- instantaneous for most real world considerations.  In this speed of progression lies the explosive force (and great danger) of nitroglycerin.  The explosion is due to the generation of great quantities of gas that puts great pressure on whatever is about to be impacted, be it a beaker, hand or face.

This is too dangerous for casual experiementation.  If one is not working with folks who have experience with these chemicals, one may well end up earning a Darwin Award, perhaps even demonstrating the theory that led to the naming of this award.

Explosives should not be experiemented with casually.

Explosives should not be experiemented with casually.

Explosives should not be experiemented with casually.

I hope I have addressed your question in a way that will lead to you and the readers of this to not conduct experiements that you and all who love you will regret.


A larger square table top divided into 100 squares (on a 10 square by 10 square set-up) of 12 cm. on a side of each smaller square is marked out on the larger sheet of the table top.

In order to win a prize, a player has to roll one coin down a chute so that the coin rolls onto the large sheet and lands completely within a smaller square and not touching a line.

If a player rolls a 5 cm. diameter coin down the chute, are the chances less than even or greater than even, that the coin will come to rest not touching a line?

Assuming that the coin stays center on the table:
[hide]
The locus of the points where the coin's center can be winning is a lot of 7 by 7 squares in the middle of the grid squares. Area: 49. So, the odds are about $ \frac{49}{144} \approx 34.0 percent < \frac{1}{2}$ for one square. Since this stays the same for all of the interior of the board, and is only slightly different at the edges (that is, if the coin stays completely on the table), my answer would be less than even.
[/hide]
Is that right? Sorry, can't figure out how to put the percent symbol.

[quote="math_explorer"]Sorry, can't figure out how to put the percent symbol.[/quote]

The code for $ \%$ is \%.  :)

Wait, math_explorer,

assuming the coin stays on the table,

just condense this problem down to one 5 cm. diameter coin

being dropped onto one 12 cm. by 12 cm. square.  It must

stay inside this one square and not touch any of the four

lines of the square.  Please reconsider with this clarification/hint.

Your post looked right, except you would want "probability" instead of "odds."

Good work.

Oops... I think my subconscious did that.  :blush:


Prove that

$ i \tan(3\pi/11) \equal{} (x^3\minus{}1)/(x^3\plus{}1) \equal{} x^{10}\plus{}x^9\minus{}x^8\minus{}x^7\minus{}x^6\plus{}x^5\plus{}x^4\plus{}x^3\minus{}x^2\minus{}x$

where $ x\equal{}e^{2\pi i/11}$.


A friend of mine spoke to me about someone he knew a few years ago who could track down sports games with mathlab. For instance he could track down where shots were taken from and so on but I have no clue how to do this and neither did my friend, just wondering if any of you have experience doing this.

I think I've found it

http://www.cs.ubc.ca/nest/lci/thesis/fhli/videoclips/


factorial
algebra
polynomial
probability and stats

Say $ X$ is distributed Poisson($ \lambda$).

What is $ E(X^r)$, $ r$ a nonnegative integer?

My idea is to express $ E(X^r)$ in terms of $ E(X),E((X)(X \minus{} 1)),E((X)(X \minus{} 1)(X \minus{} 2))$,etc, but I can't find a simple formula for it.

you're on the right track.

Let $ (x)_n$ denote the falling factorial, that is

\[ (x)_n \triangleq x(x-1)(x-2)\ldots(x-n+1)
\]

1)  it's straightforward to verify that if $ X\sim\text{Poisson}(\lambda)$, then

\[ E((X)_n) = \lambda^n
\]

2)  if you know some combinatorics (i actually had to look this up), then, as polynomials,

\[ t^n = \sum^n_{k=0} S(n,k)(t)_k
\]

where $ S(n,k)$ are the Stirling numbers of the second kind.

3)  it follows from linearity of expectations that

\begin{eqnarray*}
E(X^n) &=& E\left(\sum^n_{k=0} S(n,k)(X)_k\right)  \\
&=&\sum^n_{k=0} S(n,k)E((X)_k) \\
&=&\sum^n_{k=0} S(n,k)\lambda^k
\end{eqnarray*}

I don't think you're going to get a much more satisfying answer than this.


To play this game, watch the video in the previous post and wait until the related videos comes up at the end.  Or just press the menu button.  Pick one of those videos and post it, preferrably [i]not[/i] the same video.  :P

[youtube]pcYbnxA-qDY[/youtube].

[quote="miyomiyo"][youtube]Tr1qee-bTZI[/youtube]

Please, take 15 minutes out of your day to watch this.[/quote]

Actually, for arithmetic, use MCGrawHill, and for Algebra, use AoPS text.

[youtube]I-b5D5941AM[/youtube]

[youtube]VjgidAICoQI[/youtube]

What in the world...?


If $ a,b,c$ are positive reals such that $ abc \equal{} a \plus{} b \plus{} c$ prove that
\[ (2ab \minus{} 1)(2bc \minus{} 1)(2ca \minus{} 1)\ \geqslant\ \frac1{2(a \plus{} b \plus{} c)^2}\]

Is equivalent with :

\[ 2(a\plus{}b\plus{}c)(2a\plus{}2b\plus{}c)(2a\plus{}b\plus{}2c)(a\plus{}2b\plus{}2c)\ge 1\]

[quote="dgreenb801"]It seems too weak.
$ 2ab \minus{} 1 \equal{} \frac {2abc}{c} \minus{} 1 \equal{} \frac {2(a \plus{} b \plus{} c)}{c} \minus{} 1 \equal{} \frac {2(a \plus{} b)}{c} \plus{} 1$
So the product on the left side must be greater then 1.
So we have to show $ 2(a \plus{} b \plus{} c)^2 \ge 1$, or $ 2(a \plus{} b \plus{} c)^3 \ge abc$, which is obvious.[/quote]

Indeed , the stronger :
\[ (2ab \minus{} 1)(2bc \minus{} 1)(2ca \minus{} 1)> \frac {2916}{(a \plus{} b \plus{} c)^2}\]
is also true for $ a,b,c > 0$ with $ abc \equal{} a \plus{} b \plus{} c$ , but very easy again :( .

I think
$ (2ab \minus{} 1)(2bc \minus{} 1)(2ca \minus{} 1) \ge \frac {3375}{(a \plus{} b \plus{} c)^2}$ is true.
We have
$ 2ab \minus{} 1 \equal{} \frac {2abc}{c} \minus{} 1 \equal{} \frac {2(a \plus{} b \plus{} c)}{c} \minus{} 1 \equal{} \frac {2(a \plus{} b)}{c} \plus{} 1$
So we have to show
$ (\frac {2(a \plus{} b)}{c} \plus{} 1)(\frac {2(b \plus{} c)}{a} \plus{} 1)(\frac {2(c \plus{} a)}{b} \plus{} 1) \ge \frac {6750}{(a \plus{} b \plus{} c)^2}$
By Holder's inequality,
$ (\frac {2(a \plus{} b)}{c} \plus{} 1)(\frac {2(b \plus{} c)}{a} \plus{} 1)(\frac {2(c \plus{} a)}{b} \plus{} 1) \ge (2 \sqrt [3] {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}} \plus{} 1)^3$
But $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8 abc$, so this is $ \ge 125$. So we have to show
$ 125 \ge \frac {3375}{(a \plus{} b \plus{} c)^2}$, or $ (a \plus{} b \plus{} c)^3 \ge 27abc$, which is obvious.

[quote="dgreenb801"]I think
$ (2ab \minus{} 1)(2bc \minus{} 1)(2ca \minus{} 1) \ge \frac {6750}{(a \plus{} b \plus{} c)^2}$ is true.[/quote]

It fails for $ a\equal{}b\equal{}c\equal{}\sqrt{3}$ ;)


Let's start with geometry.

[size=150]Problem 1[/size]

Let $ BB', CC'$ be altitudes of triangle $ ABC,$ and assume $ AB\not\equal{}AC$. Let $ M$ be the midpoint of $ BC, H$ the orthocenter of $ ABC$, and $ D$ the intersection of $ BC$ and $ B'C'$. Show that $ DH$ is perpendicular to $ AM$.

Preparation for USAMO  :wink:.

OK...

[hide="Minor Hint"]Radical axes either all concur or are parallel.[/hide]
[hide="Hint"]
First note what happens if we consider the point $ E$ on $ AM$ such that $ EH$ is perpendicular to $ AM$. And then go from there.[/hide]
[hide="Bigger hint"]Extend $ AM$ to form a parallelogram. Cyclic quads![/hide]

sweet marathon you have here.

and no i can't do it.

I guess I need to start practicing now, since its only 3 weeks away.

Ill try to post as many proof problems as I can.

^unimpossible, I feel for you and youkow.  You two are probably more deserving than me of making it, seriously.

Btw, did you guys see David Liang (7th grade) on the list?  That guys my hero, even though Ive never met him.


LaTeX
quadratics
arithmetic series
algebra
quadratic formula
arithmetic sequence

I think you can solve this with an arithmetic sequence. I am not good at using latex so I will use regular typing. ar[hide]ith sequence formula.
an= a1 + (n-1)d
5000=10+10n-10
n=50
I hope I am correct.[/hide]

[quote]Lance Armstrong wants to ride to his friend's house 5000 miles away.
He rides his bike 10 miles on September 1.
He then rides another 20 miles on September 2.
Each subsequent day, he rides 10 miles more than he rode the day before.
On what day will he show up at his friend's house?[/quote][hide]
We have the [b]sum[/b] of an arithmetic series: $S_n = \frac{n}{2}[2a + (n-d)d]$

We have: $a = 10,\;d = 10,\;S_n = 5000$

Hence: $\frac{n}{2}[20 + 10(n-1)] = 5000\;\;\;\Rightarrow\;\;\;n^2 + n - 1000 = 0$

which has the positive root: $n \;= \;\frac{-1 + \sqrt{4001}}{2} \;= \;31.126...$

Therefore, Lance arrives on the 32nd day: $\text{October 2.}$[/hide]

[hide]well first you can divide everything by 10, the numbers will then be a bit easier to work with so you have:
n(n+1)/2=500
n^2+n-1000=0
Using the quadratic formula, I find x is 31.126...so he will arrive after 32 days. There are 32 days in September, so he will arrive [b]October 2nd[/b][/hide]

[color=darkred]
Brute force =D !

[hide="yummy solution o_o"]
10
30
60
100
150
210
280
360
450
550
660
780
...
too... tired...
[/hide]

[hide="screw that and try a different method o_o"]
Formula for triangular numbers:

n(n+1)/2

You have to find the smallest value of n where that expression > 5000/10.

n(n+1)/2 > 500
n(n+1) > 1000
I think n = 32 (please don't eat me if I'm wrong, I don't taste good)

After I wonder how the heIl Lance Armstrong managed to bike 310 miles in one day (he doesn't bike a full 320 miles on October 2nd), I got...
Thirty days hath April, September, June and November.
October 2nd.
[/hide]
[/color]

I don't understand what is wrong with my solution though.

Hello hello!  What you noticed was that the distances he travels are in arithmetic sequence, which is correct.  Thus, you did arrive at the (correct) result that on the 50th day, Lance will travel 5000 miles.  However, you neglected to take into account that at that time he will already have ridden for 49 days, covering a considerable distance.  Thus, the question you needed to solve was not, "What value of n makes 10n greater than or equal to 5000?" but rather, "Which value of n makes 10 + 20 + 30 + ... + 10n greater than or equal to 5000?"

[quote]Thus, you did arrive at the (correct) result that on the 50th day, Lance will travel 5000 miles.[/quote]hmm...There is a small typo... missing an extra zero between 500 and 5000?
Hope this helps ..

Haha, I realized that just as I sat down to dinner.  That was another mistake, which made your first seem more reasonable -- you divided 5000 by 10 and got 50.

Thanks for the corrections. Sorry for showing my mathematical weakness.

Hey, don't apologise.. making mistakes is the best way to learn!


I'm going to do some self-independent studying and that I'm going to use my textbooks to take notes on important thing and maybe like create a problem and solve by myself. I'm going to upload it on website so other people can see and point out if I'm doing wrong or not.

I won't post on AoPS. I would post on my own website and I guess it's both good for me and others because it would be like reading notes to other people and getting ideas on in case going wrong during my study.

I'll start today and if I make some progress, I'll post on here.

Wish me luck :)

Good luck.

I'd look at it if you put a link.

There are two lessons (or notes :) for myself) that I made.

The link to there is:

http://www.dhost.info/jwec05/My%20Websites/Number%20one/math%20lesson.htm

The first one is polygon and the second one is on complex number. Please read through and find if there is anything wrong.

[quote="Silverfalcon"]I won't post on AoPS. I would post on my own website and I guess it's both good for me and others because it would be like reading notes to other people and getting ideas on in case going wrong during my study.

I'll start today and if I make some progress, I'll post on here.

Wish me luck :)[/quote]First of all good luck :)
Second of all I don't understand how you will not post on AoPS-MathLinks, but still "I'll post on here".
Finally, maybe it's hard for people to tell you their opinions when the subject at hand is somewhere else (it would be easier to quote) ;)

[quote="Silverfalcon"]Um.. For the second one, Valentine Vornicu, I posted on my website nothing but in case it will get messed up or it might take a lot of space on this public forum.. You know, this note is my stuff and well, if this takes too much space on the aops forum, it would be like taking spaces off from public forum, ya know..

That's all... But thanks! :)[/quote]I didn't understood what you said. You first said you ain't gonna write it here, then you said you will :P :)

I think that if you keep everything into a single topic nobody will mind ;)

[quote="LynnelleYe"]What's weird is that some textbooks (like mine) call b the 'imaginary part' of a+bi, and other textbooks (like whatever you're using, apparently) call bi the imaginary part. Maybe you should mention that...[/quote]

The imaginary part of $a+bi$ is $b$. That is quite universal.

I agree that the imaginary part of $a + bi$ should be $b$, which is how I've seen it in almost all books.  Unfortunately, the AoPS book says that the imaginary part is $bi$.  On the other hand, it says that $\mathrm{Im}(a + bi) = b$.  That distinction is confusing to me, because I pronounce "Im" as "the imaginary part of".

Perhaps the authors felt that, from the point of view of English, it is weird for the imaginary part to be a real number.  I agree it is weird from that point of view, but it is much more convenient mathematically.

[quote="Silverfalcon"]360 = sum of ANY exterior angles in polygon[/quote]

I don't suppose it's wrong mathematically, but it sounds funny. Although, if you're using for yourself, it doesn't matter.


Hi, I was wondering why there are no solid geometry problems in the IMO.

I was looking thruogh the past years' IMO questions , and there were some really good questions on geometry .... But somehow , the weightage for geometry seems very little .... { I could get only about 1 or 2 geometry questions in one paper. }
 :?

They chose not to include solid geometry because people didn't like them?

[quote="G-UNIT"]how does that answer my question?[/quote]

I laffed.

my guess is that some of the problems can be greatly simplified with multivariable calculus. and these olympiad stuff promote non-calculus algebra

[quote]I was wondering why there are no solid geometry problems in the IMO.[/quote]

Solid geometry is not taught in most countries.  The IMO jury and problem committees are dominated by people from the majority of countries that do not teach this subject, and many of them consider such problems unfair to "their" students.

It's a bit harder to illustrate and put clear proofs for solid geometry.  People would run into complications while grading.

Solid geometry not being taught in most countries, was the reason stated to me by a couple of IMO team leaders.  I don't know if that is the full story, but I doubt that difficulty of writing up the solutions rigorously has much to do with it (there are several other subjects with the same difficulty, that make it onto the IMO).

i have noticed that there is at least one solid geometry problem at early IMO.

[quote="fleeting_guest"]Solid geometry was taught in the early-IMO countries at the time of the early IMO.[/quote]
Solid geometry is an important part in Chinese high school course. What about other countries?


geometry
circumcircle
incenter
angle bisector
geometry proposed

Let $ ABC$ be equilateral triangle inscribed in circumcircle $ (O)$. Let $ D$ be some point on the smaller arc of $ BC$. Let $ I_1,I_2,I_3$ be the incenters of triangle $ DBC,DCA,DAB$, respectively. Prove that the line through $ A,B,C$ perpendicular to $ I_2I_3,I_3I_1,I_1I_2$ concurrent. :D

[b]hint[/b]: Why don't you use the [b]Carnot theorem[/b] to prove it?? :D

I shall try solving, before receiving full proof from its author...
First, let's re-name I_1, I_2 and I_3 as 1, 2 and 3 respectively.

It's known that the perpendiculars from the vertices of a triangle onto the sides of another triangle are concurrent iff the perpendiculars from the vertices of the 2nd triangle onto the sides of the first one are concurrent (is this Carnot's, or just a consequence?).

I shall show that the perpendiculars from 1 to BC, 2 to AC and 3 to AB concur at the Fermat's point of the triangle 123..

I name M and N the midpoints of the small arcs BD and CD, M' the point on the circle O such that B is the midpoint of the arc MM', and N' such as C is the midpoint of the arc NN'.
The points 1 and 3 lie on the circle C(M, BM) and, of course on the angle bisectors of <BAD and <BCD, while 1 and 2 lie on the circle C(N, NC), that is, 13M and 12N are equilateral triangles. The lines M3 and N2 pass through A and the angle <MAN = 30 degs (by construction), this making 13 and 12 perpendicular, or <M12 = <N13 = 150 degs, that is, AM12 and A31N are cyclic and <1M2 = <1A2 ( 1 ) and <1N3 = <1A3 ( 2 ), therefore M, 2 and N' are collinear, and so are N, 3 and M'.
Now it's easy to see that <MAC + <N'MA = 90 degs, that is, 2M and AC are perpendicular, and same are 3N and AB.
But 2M and 3N pass through the Fermat point of the triangle 123, hence the line joining 1 with Fermat point will be perpendicular to BC.

Best regards,
sunken rock

Very nice solution :D [b]sunken rock[/b]. I cannot tell anything, just want to remind you all a very important outcome of [b]Carnot theorem[/b]:
[b]"The perpendiculars from the vertices of a triangle onto the sides of another triangle are concurrent iff the perpendiculars from the vertices of the 2nd triangle onto the sides of the first one are concurrent"[/b] :lol:

If $I$ is incenter of $ \triangle ABC,$ then according to [url=http://en.wikipedia.org/wiki/Japanese_theorem_for_concyclic_quadrilaterals]Mikami and Kobayashi theorem[/url], $II_1I_2I_3$ is a rectangle. Thus, it remains to prove that $ \triangle II_2I_3$ and $\triangle ABC$ are orthologic. Let $ X,Y$ be the projections of $ I_2$ and $ I_3$ onto $ AB$ and $ AC$ and $ Q \equiv{} I_2X \cap I_3Y.$ Easy angle chasing gives that complete quadrangle $DI_2QI_3$ has excenter $A$ $\Longrightarrow$ $AX \equal{} AY$ $ \Longrightarrow$ Perpendiculars $ I_2X$ and $ I_3Y$ meet on the angle bisector of $ \angle BAC.$ Hence, $ \triangle ABC$ and $ \triangle II_2I_3$ are orthologic $\Longrightarrow$ Perpendiculars from $ A,B,C$ to the sidelines of $\triangle II_2I_3$ concur.


function
Putnam
algebra proposed
algebra

Find all functions $ f : Z\rightarrow Z$ for which we have $ f (0) \equal{} 1$ and $ f ( f (n)) \equal{} f ( f (n\plus{}2)\plus{}2) \equal{} n$, for every natural number $ n$.

[quote="tdl"]Find all functions $ f : Z\rightarrow Z$ for which we have $ f (0) \equal{} 1$ and $ f ( f (n)) \equal{} f ( f (n \plus{} 2) \plus{} 2) \equal{} n$, for every natural number $ n$.[/quote]
So we don't have any info on negative integers?

[quote="aviateurpilot"]$ f(f(n)) \equal{} f(f(n \plus{} 2) \plus{} 2) \equal{} n$
$ \Rightarrow f(f(f(n))) \equal{} f(f(f(n \plus{} 2) \plus{} 2))$
$ \Rightarrow f(n) \equal{} f(n \plus{} 2) \plus{} 2$
so $ f(n \plus{} 2) \equal{} f(n) \minus{} 2$
gives $ f(2k) \equal{} f(0) \minus{} 2k$ and $ f(2k \plus{} 1) \equal{} f(1) \minus{} 2k$
where $ f(0) \equal{} 1\ and \ f(1) \equal{} f(f(0)) \equal{} 0$
so $ f(2k) \equal{} 1 \minus{} 2k\ and\ f(2k \plus{} 1) \equal{} \minus{} 2k$
so $ \forall n\in\mathbb{Z}: \ f(n) \equal{} 1 \minus{} n$[/quote]
You are wrong .
Solution is :
$ f(0)\equal{}a,f(1)\equal{}b$ where $ a\equiv 0 (\mod 2) ,b\equiv 1 (\mod 2)$ and
$ f(2n)\equal{}a\minus{}2n$ and $ f(2n\plus{}1)\equal{}b\minus{}(2n\plus{}1)$
Easy to check this function satisfy condition.

[quote="TTsphn"][quote="aviateurpilot"]$ f(f(n)) \equal{} f(f(n \plus{} 2) \plus{} 2) \equal{} n$
$ \Rightarrow f(f(f(n))) \equal{} f(f(f(n \plus{} 2) \plus{} 2))$
$ \Rightarrow f(n) \equal{} f(n \plus{} 2) \plus{} 2$
so $ f(n \plus{} 2) \equal{} f(n) \minus{} 2$
gives $ f(2k) \equal{} f(0) \minus{} 2k$ and $ f(2k \plus{} 1) \equal{} f(1) \minus{} 2k$
where $ f(0) \equal{} 1\ and \ f(1) \equal{} f(f(0)) \equal{} 0$
so $ f(2k) \equal{} 1 \minus{} 2k\ and\ f(2k \plus{} 1) \equal{} \minus{} 2k$
so $ \forall n\in\mathbb{Z}: \ f(n) \equal{} 1 \minus{} n$[/quote]
You are wrong .
Solution is :
$ f(0) \equal{} a,f(1) \equal{} b$ where $ a\equiv 0 (\mod 2) ,b\equiv 1 (\mod 2)$ and
$ f(2n) \equal{} a \minus{} 2n$ and $ f(2n \plus{} 1) \equal{} b \minus{} (2n \plus{} 1)$
Easy to check this function satisfy condition.[/quote]
but we have $ f(0)\equal{}1$ in the execice
and we can find $ f(1)\equal{}fof(0)\equal{}0$
so your $ a\equal{}1$ and your $ b\equal{}0$

this is putnam 1992 A1
http://www.mathlinks.ro/Forum/viewtopic.php?t=185988

Simultaneous subbing goes brrr...

$n\to 0 \implies f(0)=1.$
$n\to f(n) \implies f(n)=f(n+2)+2$
$ \implies f(n+2)=f(n)-2.$
Now induction yields the solution: $\boxed{f(n)=1-n},$ which works.


The radius of a circle is 6 meters. A square is inscribed in
the circle. What is the number of square meters in the area
of the square?

[asy]size(70); draw(Circle((0,0),1)); draw((1,0)--(0,1)--(-1,0)--(0,-1)--cycle); draw("6 m",(0,0)--(1,0),N,fontsize(8pt));[/asy]

The square's diagonal is 12, so its side length is 12/(sqrt2), or 6sqrt2. Square this for the area and you get 72 square meters.

Alternately, the area of a square is half the square of the length of its diagonal.  Since the length of its diagonal is $ 6\cdot2 \equal{} 12$, the area is
\[ \frac {12^2}2 \equal{} \frac {144}{2} \equal{} \boxed{72}\]


On side AB of square ABCD, right traingle ABF with hypotenuse AB is drawn externally to the square. E is the point of intersection of hte diagonals of the square. If AF=6, BF=8, find EF.

I used analytic geometry to get umm...7:rt2:.  I'm too tired for explanation.

Ptolemy's theorem is probably the nicest way to do this . . . but it's certainly doable with analytic geometry, or with sythetic (non-analytic) techniques.

With Ptolemy's, its like a 1-step thing, right?

You get the side length is 10, and then it's immediate by Ptolemy.


[quote="math92"]Three people share a car for a period of one year and the mean number of kilometers travelled by each person is 152 per month. How many kilometers will be travelled in one year?[/quote]

[hide="answer"]

152*3=456
456*12=$\framebox{5472}$[/hide]

[quote="math92"]Three people share a car for a period of one year and the mean number of kilometers travelled by each person is 152 per month. How many kilometers will be travelled in one year?[/quote]
[hide]
What do you mean by sharing the car? Do they each have if for themselves for a part of the year, or do they just use the car the entire year?... im sorta confused..

152x12=1824

or

152x3x12=5472

[/hide]


floor function
number theory
prime numbers

To determine whether a number $ \textit{N}$ is prime, we must test for divisibility by every prime less than or equal to the square root of $ \textit{N}$. How many primes must we test to determine whether $ 2003$ is prime?


In the next North American tournament there are 12 more American than Canadian teams (only US and Canada are in). If there are c Canadian teams enrolled, what’s the percentage of Canadian teams?

Is the answer supposed to be in terms of c?

[hide="Solution?"]Canadian: $ c$
American: $ c \plus{} 12$
Total: $ 2c \plus{} 12$
Canadian/Total=$ \boxed{\frac {c}{200c \plus{} 1200}}$[/hide]

I multiplied the denominator by 100 to achieve a percentage...

You're a bit off (You multiply the numerator by $ 100$ to make it a percentage), so:

\[ \boxed{\frac{100c}{2c\plus{}12}}\]

Oh, ya....forget my solution, sri340. No idea what I was thinking there. go with LaTeX's.


Bob had parked his car under a tree and was heading toward his friend's house when a vicious dog lunged toward him. Luckily for Bob, he was just beyond the dogs reach. Since his friend wasn't home, Bob decided to leave, but he had a slight problem. The dog, straining at the end of the chain, was following Bob's every move. With the chain attached to the tree, the dog had access to both doors. How would Bob be able to get to the door without the dog being able to reach him?

How did Bob get from the car to the door in the first place?

I'll tell ya how....
[hide]The friend's door and the car were on the same side of the dog[/hide]

[quote="moogra"][quote="besttate"]1) Who is Shadow??
2) [hide]Just have Bob walk in big loops so the dog's chain gets wrapped around the tree, shortening it enough so Bob can walk to his car.
[/hide][/quote]

1. Shadow? Where does it say shadow? I'm guessing the dog anyways[/quote]

ihatepie edditted shadow out


The squares of a 9x9 chessboard are fills with numbers $ {1,2,3...,81}$. Prove that there exist two squares with a common side, such that the difference between the numbers in them is at least $ 6$.

[hide="not really pigeonhole but..."]
Suppose the contrary: all adjacent squares have difference at most 5.  Notice that if the bottom-left most square is $ x$, then the square above it is at most $ x \plus{} 5$, the square above that is at most $ x \plus{} 10$ and so on.  Then the top-left square is at most $ x \plus{} 35$, and the top-rightmost square is at most $ x \plus{} 70$.  So any two squares have a difference of at most 70, but there's a square labeled 1 and a square labeled 81 with difference 80, so we have a contradiction.

[/hide]


[img]http://i27.photobucket.com/albums/c187/toughtop/Severity.jpg[/img]

[img]http://i27.photobucket.com/albums/c187/toughtop/Severity1.jpg[/img]

Please help guys. Thanks in advance...

:D Good luck to you all.

Sketches of the solutions...
[hide="19"]
The first one factors to $(x^{2}-x+1)(2x+a)$, which means the roots are $-\frac{a}{2}, \frac{1 \pm i\sqrt{3}}{2}$, so the difference of the two quadratic roots is nonreal.  Then, the sum of two roots can be equal to the thrid one if and only if $a=-2$
[/hide]
[hide="20"]
I'm assuming it's talking about the $f$ given.
We have to solve the equation $x^{3}-3x^{2}+3=x \iff (x-3)(x-1)(x+1)=0$
So the fixed points of that $f$ sum to $-3$.
[/hide]
[hide="16"]
Well, expanding it out, we get the leading coefficient only affected by the $x$ terms, and the constant term only affected by the constant terms.  Then, we get the leading coefficient to be $(-1)^{n+1}3^{(n+1)(2n+1)}$, and the constant term to be $\frac{(-1)^{n}}{3^{(n+1)(2n+1)}}$.
The degree is, of course, $2n+2$, because we have a product of $2n+2$ monomial terms...[/hide]

I think this is posted in the wrong forum  :wink:


A circle is inscribed in a larger regular hexagon, and a smaller regular hexagon is inscribed in the circle. What is the ratio of the area of the smaller hexagon to the area of the larger hexagon? Express your answer as a common fraction.

[asy]draw((5,8.7)--(10,0)--(5,-8.7)--(-5,-8.7)--(-10,0)--(-5,8.7)--cycle);
draw((5.8,10)--(11.6,0)--(5.8,-10)--(-5.8,-10)--(-11.6,0)--(-5.8,10)--cycle);
draw(Circle((0,0),10));[/asy]

Sorry, I don't understand what you did at all :blush:


Well...I have read the linked post, but I still have a question:

[b]When are USAMO qualifiers from international papers announced?[/b]

I mean...how many days before the USAMO.

Like, today is only 16 days before the test -__-;;;

This is the same exact question as the one in your previous post.

I am an American citizen in Taiwan, and I've qualified for the USAMO three years in a row now. I've always gotten it arranged. Have some patience!


Let U be a subspace of R^n and define V to be the set of vectors w such that w belongs to R^n and w is orthogonal to U.

Prove that V is a subspace of R^n and dim(U)+dim(V)=n.

Proving "subspace" requires proving closed under addition and scalar multiple, right?
So is V closed under addition?  Under scalar multiple?

Basically it is to prove that V is closed under addition and scalar multiple operations, and in addition to that prove that dim(U)+dim(V)=n.

Basically I could prove that it is closed under scalar multiplication, but not addition (at least it is not so obvious to me)

And there is [b]no need[/b] for you to make such a comment, transseries. I [b]WOULD NOT[/b] be posting this problem in this forum if I had even the slightest idea about solving this problem.[/b]

[quote="simonlin_00"]And there is [b]no need[/b] for you to make such a comment, transseries. I [b]WOULD NOT[/b] be posting this problem in this forum if I had even the slightest idea about solving this problem.[/quote]
It was a rude comment, but it had a point. We don't like dealing with people who post routine exercises like this and then act completely helpless. It's much more productive to give a few hints so you can solve it yourself- but we need to know how much you know to do that.

Saying that a vector is orthogonal to some fixed vector is a linear equation. Saying that it's orthogonal to some fixed collection of vectors is a system of linear equations. Write that down; what does it look like here?


Each cell of a 100 x 100 board is painted with one of four colors, so that each row and each column contains exactly 25 cells of each color.

Show that there are two rows and two columns whose four intersections are all different colors.

More generally, let us color a $4n x 4n$ square with each row and column having $n$ of each color.  Each column has $(4n)*(3n)/2=6n^2$ pairs of rows in it which are of different color.  Thus there are $24n^3$ pairs of squares which are in the same column, but different colors.  Since there are fewer than $8n^2$ pairs of rows, some pair of rows contains strictly more than $3 n$ columns in which they differ.

I claim those two rows can be used to make our rectangle.  To see this, let the colors be labelled 1,2,3,4 and let $f(i,j)$ be the number of columns which are color $i$ in the first row, $j$ in the second.

If both $f(1,2)$ and $f(3,4)$ are nonzero, we are done.  Assume WLOG $f(1,2)=0$.  If both $f(1,3)$ and $f(2,4)$ are nonzero, we are done.  Assume WLOG $f(1,3)=0$.  $f(1,4)$ cannot be 0 now, since there are strictly fewer than $n$ columns in our pair of rows which match.  Therefore we are done unless $f(2,3)=0$, which we shall now assume.

This gives us the relationships:

$f(1,1)+f(1,4)=n$ (as there are $n$ squares of color 1 in the first row)
$f(2,2)+f(2,4)=n$ (as there are $n$ squares of color 2 in the first row)
$f(3,3)+f(3,4)=n$ (as there are $n$ squares of color 3 in the first row)
$f(1,4)+f(2,4)+f(3,4) \leq n$ (there are $n$ squares of color 4 in the second row)
$f(1,1)+f(2,2)+f(3,3)<n$ (the two rows match in fewer than $n$ places.

The sum of the three equations contradicts the sum of the two inequalities.


[quote="peregrinefalcon88"]i need to know how to find the date of my mathcounts regional competition so i no it wont interfere with hmmt[/quote]

I suggest you Google it.  But it's always in February.

Most places haven't finalized their times for regionals yet.

You may want to ask a coach or veteran Mathlete, in my chapter, it's always the same weekend in February, this year I have feeling it will be on Valentine's day :P


function
search
real analysis
real analysis unsolved

$|f(x+y)-f(x-y)|\leq \frac{1}{3},\forall x,y$ so we are allowed to define $a_n=a_n(x,y)=f(x+ny)-f(x-ny)$.
We have that $|a_1|\leq \frac{1}{3}$(and also $|a_n|\leq \frac{1}{3},\forall n$).Define $c_1=\frac{1}{3}$.We will search a better upper bound ,named $c_n$,of $|a_1$|(in fact it will be an upper bound for all $|a_k|,\forall k\in [1..n]$).We will show that $c_n$ goes to 0. From this it follows that $f$ is constant.

From $\left|\sum^{n+1}_{k=1}3^ka_k\right|\leq 1$ we get $3^n |a_{n+1}|\leq c_n(1+3+...+3^{n-1}) +\frac{1}{3}$.
So $|a_{n+1}|\leq \frac{c_{n}}{2}+\frac{1}{3^{n+1}}$.So $|a_{1}|\leq \frac{c_{n}}{2}+\frac{1}{3^{n+1}}$.We get a sequence  that  verifies $c_{n+1}\leq \frac{c_{n}}{2}+\frac{1}{3^{n+1}}$.It is easy to check that $c_n$ goes to 0.
I hope it's ok.

[quote="Peter"]Suppose that $f: \mathbb{R}\rightarrow\mathbb{R}$ fulfils $\left|\sum^n_{k=1}3^k\left(f(x+ky)-f(x-ky)\right)\right|\le1$ for all $n\in\mathbb{N},x,y\in\mathbb{R}$. Prove that $f$ is a constant function.[/quote]
Let $x,y \in \mathbb{R}$ and
\[ T_n := \sum^n_{k=1}3^k\left(f(x+ky)-f(x-ky)\right) \]
Then $\left|T_n-T_{n-1}\right| \le \left|T_n\right| + \left|T_{n-1} \right| \le 2$, so for all $x,y \in \mathbb{R}, n \in \mathbb{N}$:
\[ \left |3^nf(x+ny)-f(x-ny)\right| \le 2 \]
Let $z \in \mathbb{R}$ and $x = \frac{z}{2}, y = \frac{z}{2n}$ gives for all $n \in \mathbb{N}$:
\[ \left | f(z) - f(0) \right | \le \frac{2}{3^n} \]
And hence $f(z) = f(0)$ for all $z \in \mathbb{R}$ and $f$ is constant.


recurrence relation
quadratics
algebra
functional equation
IMO Shortlist

Let $ \mathbb{R}^\plus{}$ be the set of all non-negative real numbers. Given two positive real numbers $ a$ and $ b,$ suppose that a mapping $ f: \mathbb{R}^\plus{} \mapsto \mathbb{R}^\plus{}$ satisfies the functional equation:

\[ f(f(x)) \plus{} af(x) \equal{} b(a \plus{} b)x.\]

Prove that there exists a unique solution of this equation.

[hide="answer"]Answer

$ f(x) = bx$[/hide]
[hide="solution"]Solution

Put $ u_0 = k, u_n+1 = f(u_n)$. Then $ u_n+2 - b(a+b)u_n+1 + au_n = 0$. That is a standard linear recurrence relation, so we can immediately solve it. The associated quadratic factorises as $ (x - b)(x + a + b) = 0$, so the general solution is $ u_n = Abn + B(-a-b)n$. Hence $ u_n/(a+b)n = A(b/(a+b))n + (-1)nB$. But $ (b/(a+b))n → 0$ and $ lhs$ is always non-negative. Hence $ B = 0$. So $ A = k$. Hence $ f(k) = u_1 = bk$. It is easy to check that this does indeed satisfy the relation given. [/hide]

I've got a question what are the neccesary conditions to apply this method since if I use this to solve $f(f(x))=x$ which is equivalent to $a_n+2=a_n$ with $a_0=x$ I get a unique solution $f(x)=x$, but the function $f(x)=1/x$ is also a valid solution.

[quote=Ramanujan1057894736842]I've got a question what are the necessary conditions to apply this method since if I use this to solve $f(f(x))=x$ which is equivalent to $a_n+2=a_n$ with $a_0=x$ I get a unique solution $f(x)=x$, but the function $f(x)=1/x$ is also a valid solution.[/quote]
You won't get $f(x)=x$. You'll get $f^{2n}(x)=x$ with no information about $f(x)$ (here $f^k(x)$ denotes $k$ times application of $f$)

Fix $x$, and let $a_n = f^n(x)$ (the $n$-th iterate of $f$). Then,
\[a_{n+2} +a.a_{n+1} - b(a+b)a_n = 0,\]
which has characteristic polynomial
\[x^2+ax-b(a+b),\]
which has roots $-a-b$ and $b$. Hence,
\[a_n = u.(-a-b)^n+v.b^n,\]
with $u,v$ constant. Therefore,
\[a_n = \frac{xb-f(x)}{a+2b}(-a-b)^n+\frac{(a+b)x+f(x)}{2b+a}b^n,\]
which, unless $f(x) = bx$, is negative for some large $n$. Therefore, $f(x) = bx, \ \forall x$, which is a solution.

Define $f(x_{k-1})=x_k.$ Then the original equation becomes \begin{align*}x_{k+2}=b(a+b)x_k-ax_{k+1} \\ \implies x_k=c_1b^k+c_2(-a-b)^k,\end{align*} where $c_1$ and $c_2$ are non-negative constants. We also have $x_0=c_1+c_2,$ and $x_1=c_1b-c_2(a+b).$ Notice that $c_2=0$ for $x_k$ to be non-negative. Thus $$x_0=c_1\implies f(x_0)=x_1=bc_1=bx_0.$$ Hence the solution is $f(x)=bx ~~\forall x,$ which fits.


Please tell me is this right, if not, where did I made the mistake?

m^2 + 6m + 9 - n^2
(m^2 + 6m) + (9 - n^2)
m(m + 6) + (9 - n^2)

Thank you all[/code]

[quote="towersfreak2006"]Is this what you're looking for?
$m^2+6n+9-n^2$
$(m^2+6n+9)-n^2$
$(m+3)^2-(n)^2$
$(m+n+3)(m-n+3)$[/quote]

Everything is right but it should be $6m$ not $6n$

how about this one? I don't think it's posible, but see if someone can solve this for me
14m^2 + 4mn - 15n

It can't be factored.
Notice that the expression is a quadratic in $m$, and the discriminant, $\sqrt{16+4(14)(15)} = 2\sqrt{214}$, since this is not an integer, the expression can't be factored.


trigonometry geometry calculus AMC AIME USA(J)MO USAMO

Many schools finish trig, geometry, and everything else that is tested on the AMC competitions by Grade 10. As printed right on the contest, all problems can be solved using precalculus methods. After that, much of the content we learn cannot be applied to the contests.

Hence, many of us also do not improve significantly in the 11th and 12th grades vs. 10th unless we have practiced since then. I've noticed that at my school, many seniors end up getting approximately the same scores as the 10th and 11th graders.

Since everyone Grade 10 and above pretty much has an equal chance at the tests (except the fact that you will be more proficient with more practice) why is it that they should get an easier route?

Most of them do very poorly on the Olympiad anyway. Last year, didn't half of them get 0's?

Also, it should be pointed out that many people who qualify for Olympiad-level tests have skipped grades in math, and not every school allows that. My school, for example, has a no-skip policy unless you skipped in middle school and it's beyond their control. When contestants skip grades, they have obvious advantages. I think the cutoffs, should they exist, should be determined by the COURSE contestants are taking, and not their respective ages or grade levels.

JBL and Complex Zeta have both amazed me with descriptions of the middle school educations they received in geometry, in private schools, but I think the fact remains that most top-level math competitors learn most of their math by self-education, almost more despite school than because of school. The mandatory wastage of time in most school programs is a drag, but everyone has opportunity to use their free time to learn more math, and yet more math, and still more math, so I think it is possible to make it into the top echelon if that is what you really desire.

On grade-skipping and math acceleration, I agree with you that current math standards in the United States are too easy (for EVERYONE), but I also point out that you may not want to fall into [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]the calculus trap[/url]. Knowing mathematics well has never been about only knowing the meager topics of the standard curriculum; putting the topics together and finding their unity and understanding their details takes extra time and effort beyond school homework, and can't be rushed.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=33140&highlight=#33140]my very first post to this site[/url] for my advice on how a young learner should prepare for deep understanding of mathematics. Some of that advice to parents and to young learners can also be applied by high school students.

Well, pragmatically, it encourages younger students to study more for AIME / USAMO, investing the the future, and leading to a stronger crop of older USAMO qualifiers and MOPpers two years later.  As far as I'm concerned, anything that gets students interested in proofs at a young age is good. :-)

It also rewards students who have natural problem-solving ability but have not had the opportunity to be overly accelerated in math, and students who did not get the head start in problem solving practice provided by MathCounts.

[quote="tAsDoOm"]
Since everyone Grade 10 and above pretty much has an equal chance at the tests (except the fact that you will be more proficient with more practice) why is it that they should get an easier route?
[/quote]

[quote="tAsDoOm"]
Also, it should be pointed out that many people who qualify for Olympiad-level tests have skipped grades in math, and not every school allows that. My school, for example, has a no-skip policy unless you skipped in middle school and it's beyond their control. When contestants skip grades, they have obvious advantages. I think the cutoffs, should they exist, should be determined by the COURSE contestants are taking, and not their respective ages or grade levels.[/quote]

That's really rather flawed logic. I (for the most part) agree that cutoffs should be determined, to some extent, by course level; however, claiming that people who have skipped grades in math have obvious advantages. I skipped twice (once in grade 2 and once earlier this year) and it's not helped me much. Second, skipping a full GRADE in school (rather than just a year of maths) is very, very [i]dis[/i]advantageous. Not only does it mean one less year of possibly taking the test for the student in question, but they are already being challenged more in math, so they may not have time to study the extracurricular material necessary to do well on tests like the AIME.

And, contrary to what you seem to be saying, there are advantages to age-based cutoffs. First, all 7th, 8th, 9th, 10th graders get in easier without having to elbow their way through the seniors. Second (and this is much more important) IT GIVES THE YOUNGER STUDENTS AN OPPORTUNITY TO TAKE THE TEST [b]BEFORE[/b] they reach their last year of eligibility. If you think this is a bad thing, OK. Just take the USAMO without practicing at all in a test environment and tell me how you do.

A student who is exceptional only at mathematics may take math classes beyond their grade level, but probably will never officially "skip a grade".

A student who is very good at all subject areas may officially "skip a grade".

In MATHCOUNTS, for example, we have 8th grade students who are as young as 12 and other 8th grade students who are as old as 15.

AMC-10 allows students as old as 17.5 years old to compete.  The AMC-12 cutoff is 19.5 years.

Is it "fair" to the mature 12 year old 8th grader, who is good at math and language arts and science, and is taking 8th grade courses to be competing against a 15-year old 8th grader, who is taking advanced high-school math classes.

Research shows a good deal of development occurs in the brain between the ages of 12 and 15, especially in the area of reasoning and critical thinking.  [Not new brain cells, but new connections, and significantly faster nerve impulse activity].

In a very real sense, MATHCOUNTS (and AMC-8) penalize the 12 year old because they were also exceptionally good at language arts.

I can also argue against basing eligibility on what grade of math course the student is taking.  Our school district used to offer Geometry to 8th graders.  Should an 8th grader in our district be disallowed from competing in MATHCOUNTS or AMC-8 because they take Geometry at the high school?  Other districts in our state still offer Geometry to 8th graders.  Should they also be disallowed from competing in MATHCOUNTS or AMC-8?

Note that I am not meaning to pick on MATHCOUNTS or AMC.  MATHCOUNTS is the premier program for middle-school students.  AMC has the premier competitions for high-school students.

Different programs have slightly different rules.  Most nationwide programs have standardized on using the official "grade" as the measure to compare students.  I'm sure they understand the disparities.  I think AMC goes out of their way to recognize the younger students who choose to take higher-level tests.  And their testing and advancement structure encourages students to take the AMC-12 over the AMC-10, once they have the coursework covered by the AMC-12.

You just have to know and live by the rules the various competitions have established.  The year MATHCOUNTS was changing from a limit of twice per lifetime to three times per lifetime, I joked with the State MATHCOUNTS champion if he wanted to repeat the 8th grade.  (I don't think he was amused by my suggestion.)

Since it is the grade that determines eligibility for anything, that is what should be considered.  If being younger bars one from entering a program and being in a certain grade bars the same person from participating in a contest it would be terribly unfair to the child who had little choice in the decisions that were made by his teachers and parents.  Go either by grade or by age.

In sports those who practice with older kids because of the training they received still get to compete with their peers in age or grade depending on the type of sport and level of competition.

If the math course determined the eligibility, there would be several 10 and 11 year olds who cannot be moppers.  You have to be a teenager to be a mopper and there are students who will finish all school math offerings by the time they are that age.


trigonometry
limit
real analysis
real analysis unsolved

Hello peoples,

I think this is a trick question... well sort of :P

[img]http://img133.imageshack.us/img133/472/picture8ox1.png[/img]

for part (a) i get that the cosine Fourier Series for f(x) = cos(x) to be:

[img]http://img138.imageshack.us/img138/6114/picture9sq2.png[/img]

i hope that is ok, but its part (b) that is troubling me...

is all that happens as [img]http://img138.imageshack.us/img138/7436/picture10gf7.png[/img] is that the cosine Fourier series of cos(x) goes to 0?

i am guessing i am missing some trick to this question?

Cheers! :D

Sarah

As $\alpha\to\pi$, most of the terms go to zero by inspection.
The $n=1$ term in the sum deserves closer inspection: what is $\lim_{\alpha\to \pi^{-}}\frac{-2\alpha\sin\alpha}{\alpha^{2}-\pi^{2}}$?
That's a $\frac{0}{0}$ indeterminate limit, and it should be 1.

ah yep i see now.  i need to be more careful with the n = 1 term when alpha = pi

thanks for the help! :D

Sarah


[color=blue] [b] Problem. [/b]
Let $a, b\in\mathbb{R}$ such that $0< a< b$. Find all functions $f: [a, b]\to [a, b]$ such that

                                                      \[ f(x)\cdot f(f(x))=x^{2}, \]
 for any $x\in [a, b]$.

 [/color]

Nice problem, here is my solution:
[hide]
Let's fix $x$ and we put $a_n=f^{(n)}(x)$ ($f^{(n)}(x)=f(f(f(...(f(x))...)$ n times $f^{(0)}(x)=x$)
We get $a_{n+2}a_{n+1}=(a_n)^2$
Put $b_n=ln(a_n)$ then we get
$b_{n+2}=-b_{n+1}+2b_n$
so b is linear recurentional sequence.
$b_n=c_1*1^n+c_2*(-2)^n$
but $a_i \in[a;b]$, so $b_i \in[ln(a);ln(b)]$
but if $c_2\neq 0$ we can take such $n$ for which $b_n<ln(a)$, so $c_2=0$ and $b_n=k$, $k=const$, so $a_n=const$, but $a_0=x$, so $f(x)=a_1=a_0=x$. So $f(x)=x$. Check the answer.
[/hide]


Yay, useless statistics! Number of 2006 USAMO qualifiers by the first letter of first name.

First, see if you can predict which letter is the most common, and which letter is the rarest.

Here go:

[hide][code]j -- 61
a -- 43
d -- 38
s -- 35
m -- 34
c -- 23
r -- 20
t -- 19
k -- 19
p -- 17
y -- 17
b -- 14
g -- 12
h -- 12
w -- 12
e -- 11
n -- 11
z -- 8
v -- 6
i -- 6
f -- 5
x -- 3
q -- 3
l -- 3
u -- 1
o -- 0[/code]
[/hide]

lol i was close
i guessed A cause it was a vowel :)

I was surprised by the most common letter too.

BTW, it seems a lot of Alexes are qualifying. Just something I noticed.

*runs and changes name to Alex*

now im ready to qualify for the usamo next year. :)

Two other $q$'s qualified?   :P  We should form a club.

i guessed D with no apparent reason.
you should do a stat with last names. i think C is most common. and "chen" might be the most appearing last name.

score! i guessed J for most common, and U for least common... i was close =D

oh, and btw, my name is alex too =D

watch the Ks get into jusamo by the dozens
heh ehe

One of these days the C's are going to beat everyone...
Yea.  I guessed C.  :)

Anyone want to look at statistics over multiple years?

You asked for it: USAMO Qualifiers by First letter of Last name.

See if you can guess the most and least common letters again! This is a trick question this time: Two letters tied for the most common!

[hide][code]l -- 47
c -- 47
s -- 39
h -- 30
b -- 25
k -- 24
w -- 23
m -- 20
z -- 20
g -- 18
f -- 17
p -- 15
t -- 14
y -- 13
j -- 12
r -- 11
d -- 11
a -- 8
n -- 8
v -- 7
e -- 5
x -- 5
o -- 5
i -- 3
u -- 3
q -- 1[/code]

There seem to be a lot of Chens taking the USAMO.[/hide]

First names starting with O:
Year Oleg Others
2005: 1 0
2004: 2 0
2003: 1 1

Clearly, Oleg is the best "O" name to have, with three individuals qualifying a total of four times since 2003.

[quote="paladin8"]Yay. I'm the top statistic.[/quote]

Me too. Apparently J is for Just Too Good.

[quote="joml88"][quote="jb05"]Yay! I got a total of 86 (frequency of first initial+frequency of last initial). If only I had a fun name like Oliver Queen...[/quote]

108 - beat that! :D

Eh, if only I actually made the USAMO...then it would be 110 :P[/quote]

Wow; that's over twice my total (47.)

[quote="joml88"][quote="jb05"]Yay! I got a total of 86 (frequency of first initial+frequency of last initial). If only I had a fun name like Oliver Queen...[/quote]

108 - beat that! :D

Eh, if only I actually made the USAMO...then it would be 110 :P[/quote]

You didn't make it?

Wow..I deserved to make it...but you definitely did much more...

[quote="rem"]Yay for all Alexes making Usamo!
Imagine all the winners being Alexes :rotfl:  :rotfl:[/quote]

Yes, that's a good idea. Maybe they should make tiebreakers depend on alphabetical ordering of first name.

yay! arnav > brian > matt > sherry > stephen > yi > zach

hehe
J definitely wins.
i wonder why y and z are so popular

Maybe it's because it's 3:45 AM and I've been writing code for the last four hours but I found this topic exceedingly amusing.

[quote="Philip_Leszczynski"]Who is Matt?

By the way, Stephen would actually beat you.  :P[/quote]

turns out he did :lol:

good thing I didn't put hesterberg though (could anybody beat him?)

Sadly, yes--looking through the list, there are two Aarons, an Abraham, and possibly another Adam who would beat me. At least I'd make the team :)

I, too, was a J who made USAMO  :) .

My total is 91.... fairly high but not a 108... JC or JL, i can think of a couple (jeffrey chen)

[quote="Go Beyond"][quote="randomdragoon"]
There seem to be a lot of Chens taking the USAMO.[/quote]

No surprise. Chen is one of the largest family names in China. Also Li, Zhang, Liu... are the largest. :D[/quote]

Wang is still the most common in my opinion, my mom knows at least  20 Wangs (counting each individual family member though)

Ha, my initials are JC.

Well my guess J was kinda based on the fact that my name, my brother's name, my other brother's name, my mom's name, one of my uncles' names, one of my cousins' names, and a couple other family names all start with J.


A=2^29

A has nine digits
Which digit doesn't involve in A

[quote="JRav"][hide]4[/hide][/quote]

How did you solve?

The sum of all $ 10$ digits is $ 45\equiv 0\pmod{9}$, and $ 2^{29} \equal{} 2^2\cdot(2^3)^9\equiv 4\cdot( \minus{} 1)^9\equiv 5\pmod{9}$, hence what's lacking from the lot is $ 4$.

The problem statement should make clearer that $ A$ has nine digits, all of which are distinct.

If you look closer into the wording, it's clear - a digiT (not digiTS) missing when there's nine of them can only mean all of the nine digits involved are distinct (otherwise there would be digitS missing). Though I agree it could've been made more explicit.

Similar problem.

Prove that $ 2^{99}$  involves at least one digit which occurs at least 4 times.
This is also true for $ 2^{98}$ and $ 2^{97}$.


logarithms
ratio
integration
calculus
probability and stats

Let $ X_0$ have a uniform distribution on $ [0,1]$, and for $ n \ge 1$, $ X_{n \plus{} 1}$ has a uniform distribution on $ [0,X_n]$.
Find the almost sure limit of $ \frac {\ln{X_n}}{n}$.

Doesn't the law of large numbers usually involve some kind of sum? :maybe:  Can you give more hints how to solve this?

If $ Y_n$ are iid random variables that are uniformly distributed on $ [0,1]$, then $ X_n\equal{}\prod_0^n Y_j$.

:blush:  I don't understand how I missed that... That's it from me today, back to learnin PDE  :lol: .

[quote="fedja"]If $ Y_n$ are iid random variables that are uniformly distributed on $ [0,1]$, then $ X_n \equal{} \prod_0^n Y_j$.[/quote]
I m not finding it quite obvious. Can anyone explain how is the above true?

To say $ X_{n \plus{} 1}$ is uniformly distributed on $ [0, X_n]$ is the same as saying $ X_{n \plus{} 1} \equal{} Y_{n \plus{} 1} \cdot X_n$ where $ Y_{n \plus{} 1}$ is uniformly distributed on $ [0, 1]$ (with appropriate independence).

JBL, can you explain how is $ X_{n\plus{}1}\equal{} Y_{n\plus{}1}\cdot X_{n}$?

Define $ Y_n$ to be the ratio $ \frac {X_{n \plus{} 1}}{X_n}$.

That integral doesn't appear to converge.  We are given that $ X_{n \plus{} 1}$ is uniform in $ [0, X_n]$, which is equivalent to saying that $ \frac {X_{n \plus{} 1}}{X_n}$ is uniform in $ [0, 1]$.

[Minor edit for grammar.]


Columbia calculus geometry AMC USA(J)MO USAMO counting

This is not a post about the positives and negatives of graduating early.  I am fully aware of what it would take to graduate early and the millions of reasons not to do it.  What I'd like to know is how greatly it would affect my chances of getting into a top school such as Columbia which is where I'd like to go if I can get in after junior year.  I am currently in tenth grade, and am greatly considering the fact of graduating after eleventh.

I took AP Calculus BC last year and got a score of 5 (with an AB subscore of 5).  I also took AP Physics C Mechanics and got a score of 5.  This year I am taking AP Physics C Electricity and Magnetism, AP World History, AP Chemistry, AP Economics Micro and Macro, and AP Java AB.  Next year I will be taking AP US Government, AP Statistics, AP Biology, AP English Language, AP Latin Literature, and AP Psychology.

I am taking the PSAT in October and would take the SAT I in the Spring.  I am going to take the SAT II Math IIc and the SAT II Physics as well within the next few months.  At the end of the year, I'd like to take the SAT II World History and SAT II Chemistry.  My GPA is 97.5/100, unweighted (my school does not weight or use class rank).

Last summer, I did research in geophysics at MIT.  This coming summer, I plan on continuing research except in more of a theoretical physics area hopefully regarding something like Quantum Mechanics or String Theory.

My goal is to make the USAMO this year.  I am assistant captain of my school's math team, captain of my school's chess team, a member of my school's mock trial club, a member of my school's Quizbowl club, and technical manager of my school's newspaper.  I have some math and chess awards, and I expect to have some science research awards by the end of this year.

How realistic are my chances?  Thanks.

You're doing many of the right things, but it's really hard to evaluate chances at a school like Columbia with a low base rate of acceptance. A lot of the applicants who don't make it in are hardly distinguishable from those who do.

I definitely wish you well in trying to wrap up high school early--I would gladly have done the same at your age if only I had prepared as well as you have. I can't predict the chances for acceptance at the most selective schools, but it does seem you are on the right track.

Do you have a plan B in the event of plan A not working as you desire?

I haven't read anything above, so I'm not sure if this is the slightest relevant, nontheless its quite interesting.... :lol:  My local university runs an accelarated, high-school programme called "transition".  Its a two-year programme that covers 5-years of highschool. ( In Canada 8th Grade is the beginning of high-school). Meaning transition kids end up graduating at the end of 9th grade, and then proceed onto university.

  Its a pretty intense programme with very little free-time, my math teacher used to teach there and now considers it child-abuse...

my plan B would be to stick high school out my senior year and apply again to the highly selective schools where i would have even a greater chance of being accepted (since i would be of the regular age).  of course there are 2nd tier schools such as NYU I could apply to as well.  i'm really hoping that i can get into columbia a year early though, because my senior year in high school would be very pointless.  anything i could do to improve my chances?  unfortunately, i don't do sports or anything like that.  i play cello but not at any great level, so i don't think that would be of much help either.

FWIW -  There are many  schools (including some Ivy )  who will admit  a promising student  who is in Junior year.
 Apart from Columbia and other Ivy leagues, you may like to check out other schools, say USC  (Resident Honors Program) which will    admit  students  who are in their  Junior year (and may  award  a full tution scholarship - if  they are really impressed with you :))  and work with your high shcool so that you get the highschool diploma  too (( to  work-out any remaining requirtemnts etc)... I knew this was there a few years ago, so check out their site, if the program still exists.
(Of course, other places may have similar programs so check them out too)

Added later:   Point to note about  these special programs (eg Resident honors program of USC)  is that there are additional resources (eg special housing, (with less freedom than average freshman :D)  , advisers etc) which  are available for those who are coming in early)

Stanford sent this to me when I emailed then concerning applying early (I am currently a junior)

[b]

Thank you for contacting the Office of Undergraduate Admission at
Stanford
University. We have admitted younger applicants who were academically
competitive in the past. We do not discriminate against a student based
on
age, but we will consider both the academic achievement and the
student's
overall maturity level and ability to adjust to college life regardless
of
their age.

However, we would strongly recommend taking advantage of the full four
years to complete high school while making the most of your time both
academically and personally. If you like, you might even consider
taking
courses at a local community college during your final year.

As we do not schedule appointments, you are welcomed to drop by the
office
or give us a call. Please bear in mind, however, that we will not be
able
to review any materials you may bring to the office.

If you have any additional questions, please feel free to contact our
office at (650) 725-5294.

Sincerely,

The Office of Undergraduate Admission
Stanford University

[/b]

I decided to just have a really slow and easy senior year instead. [/b]

It's a bit like the folklore college essay listing activities such as cancer research, international peace negotiations, felling charging elephants with a single shot between the eyes, and so on.

The set of AP exams above is, obviously, overkill for a school at the level of Columbia, especially in tenth grade and with 1 or 2 years left to prolong the list.   Barring a criminal record, or major deficiencies in your grades or bank account, admission is more or less a given.

Bear in mind, though, that  "research in geophysics" will be decoded by anyone with a clue as "interned at a lab", and that the chances of doing research in string theory at this level (i.e. anything beyond working menial calculations for the actually qualified) are very, very low.  It is generally advisable to discuss such things in realistic, neutral, objective and factual terms rather than exaggerations or speculations about the future, and that is especially true in the context of accelerated college admission, where maturity is a specific consideration.

You sound way more prepared than I ever was for college.
I'd like to add that in the case that you do not get in after your junior year, you might still be able to take classes at Columbia during your senior year. I took Introduction to Number Theory during the fall term of my senior year at Columbia, with the help of my school's math department assistant principal. Just check with someone in your school.
The experience was worthwhile, even though I was not officially enrolled in the class and do not plan on claiming credit. You get a feel for the school, know what you might be getting into. Doing something like this can show the colleges that you have the maturity and time management necessary to succeed. In fact, with your credentials, I'd say you might be able to do it next year, and further improve your chances of getting in early.


geometry
geometric transformation
reflection

Pretend that you are camping.  While you are searching for natural berries, you see that your tent is on fire! :)
You are 200 feet from the river, you're tent is 240 feet downstream from you're current location, and the tent is 120 ft. away from the river bank.  Thankfully, you have a bucket with you.  You must run to the river and to your tent.

1) What is the shortest distance?
2) How far did you run with the pail full and pail empty?

[hide]

Let us reflect the image off the river bank. Let A be your current location, B the tent, and B' the reflection. So, we need to find AB'. We have a right triangle with legs 320 and 240 (the distance from the riverbank to A and distance from riverbank to B'; distance tent is downstream). So, the distance is $ \sqrt{[8(40)]^2\plus{}[6(40)]^2}\equal{}\sqrt{40^2(8^2\plus{}6^2)}\equal{}400$

Let the river be CD, where C is closer to A and D is closer to B. Also, let the intersection of the river and AB' be X. We have $ \triangle{ACX}\sim\triangle{B'DX}$. So, $ \frac{200}{120}\equal{}\frac{5}{3}\equal{}\frac{AX}{400\minus{}AX}$. So, AX=distance without water=250. This distance with water is 400-250=150.

[/hide]


Point P lies on side AB of a convex quadrilateral ABCD. Let ! be the incircle of triangle
CPD, and let I be its incenter. Suppose that ! is tangent to the incircles of triangles APD
and BPC at points K and L, respectively. Let lines AC and BD meet at E, and let lines
AK and BL meet at F. Prove that points E, I, and F are collinear.


So like I'm going with the grissom team and so is dynamo. hey I CAN MEET DYNAMO (and ash and evil ed i think). Cool. Hopefully I shall get more than 50% of the questions in individual right this time. hmmm I get the feeling I have a room to myself, since there are 17 people. GO TEAM GRISSOM!

This shall be fun. :) I am very excited.

Oh snap if I had made usamo grissom could not be in division B, or I could not go. or we would be in division A, according to the email.

Haha. I found out I was on ARML this week. I did one practice and got a 3/8  :dry: So if this post contains ARML stupidity it's because idk anything about ARML.
Just wondering, but why are you going with grissom? and do you know, since there are 17 of us, which two are not on the team?
hey I CAN MEET XINKE! The all-famous Xinke...  :o  You won't be impressed by me, I'm not smart, and... I'm white :(
yes, you will meet ash and sid. They are not like each other except for being brown, short, and geniuses so don't expect them to be the same.
haha yeah go Grissom. weird, 3 of our people dont even go to Grissom.
im excited about the competiton cuz its experience, but i'll weigh us down so i'm nervous.
how close were you to usamo?

11 points by amc 10. Horrendous amount of points by amc 12. I need to up my average practice aime score to 12 (I should stop dreaming) since like my real aime score both years have been 2 or 3 less than average practice aime scores....

hmmmm so like i wanted to go to arml and I told Mr. Crawford, and he said Grissom might send a team, and he thought that maybe I could go with grissom, and it worked out well from there.

Hmmmm, you have practiced more for arml than I have this year..... :o

hmmm you shall be less impressed by me. I like to do problems very slowly, if at all possible, to triple check everything, and the only reason i am ok at math is cuz I kind of work hard at it. I consider you and tony and botong smarter than me.

by the way they are changing arml rules this year to make it 10 problems instead of 8.

I would have asked you if you wanted to go if the grissom teacher had said that she had more room on the team, but right now, there's already a full team and 2 alternates. Also, registration ended 2 days ago. Sorry.


algebra
polynomial
algorithm
real analysis
real analysis unsolved

Given an even degree polynomial in $ \mathbb{R}[x]$, is there any deterministic way of finding out if it has a real root? ( like we use the discriminant for degree 2 polynomials)

please see

[url]http://en.wikipedia.org/wiki/Root-finding_algorithm[/url]

[url=http://en.wikipedia.org/wiki/Sturm%27s_theorem]Sturm's theorem[/url] is even more relevant here.

Thanks.

But more than an algorithm I was looking for a condition involving the coefficients which would give a conclusion. Rather than evaluating the polynomial at some points.

I would hazard a guess that Sturm's theorem is the best you're going to get. Be satisfied with that.

WLOG, assume that the leading coefficient of your $ 2n$ degree polynomial is positive. You're asking whether $ p(x)$ is ever negative. You could phrase that as asking about the minimum value of $ p(x).$ Where is that minimum located? At one of the roots of $ p'(x),$ which is a $ 2n\minus{}1$ degree polynomial. Can you find the roots of $ p'(x)$ by "a condition involving the coefficients which would give a conclusion"? Let's put it this way: you do know about the Abel-Galois impossibility results for the solvability of polynomials of degree 5 and above?

Now if you want to se if you can find something specifically for the fourth degree case, go look at the Tartaglia-Cardano-Ferrari formulas and see what you can do.


search
LaTeX
trigonometry
Olimpiada de matematicas

me disculpo de entrada por que pues no me he puesto a mirar como funciona y por ende no se como usar el latex.el caso es que
 si x pertenece al intervalo cerrado cero, pi medios donde x es un angulo dado en radianes demostrar que

     cos(cosx)>=sex

nuevamente les pido disculpas por no usar el latex.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=317043477&t=279103]contestación[/url]

En latex:

si $ x$ pertenece al intervalo $ [0,\frac {\pi}{2}]$ donde $ x$ es un angulo dado en radianes demostrar que:

$ \cos(\cos(x)) \geq \sin(x)$.


Was the 2003 AMC 10B just really easy? Over winter break, I have averaged about 130 on the AMC 10. This evening, on the AMC 10 test I just did, I got a 145.5, not answering one of them from misreading. Was this to be expected from the difficulty or did I suddenly get smarter?

Try comparing the AMC 10 statistics.  [url=http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2003-10a/03stats10.html]Here is the 2003 AMC 10/12 statistics.[/url]  The AIME cutoff was 121, which is higher than the usual 120, so I would assume it was slightly easier.

Same here, I just took it and got a 144, but I usually average a 130-135.

I just took the 2003 12A and 12B and scored the same as my AMC 10 average, i just think 2003 was a lot easier than all the other years for some odd reason

I also remember hearing that the 2003 AIME was ridiculously easy compared to other years, so maybe the MAA decided to just make everything easier that year.

I think that year was slightly easier because my test scores were higher that year.

Ya, i think it was easier :P
Compare to other people s score

hmmm... Try the 2004 AMC10 B... It was quite a bit tougher than the 2003 one.

[quote="batteredbutnotdefeated"][quote="AwesomeToad"]Um, now, is the AIMe cutoff 120 or higher (100 or higher for AMC 12) or is it 1% (5% for AMC 12)[/quote]

It's either one, I believe.[/quote]

Whichever is more less.

[quote="AwesomeToad"]What do you mean "more less"? which one gets most people?[/quote]

Whichever has a lesser score.  If the top 1% was people with scores of 140+; then 120+ would be cutoff.  If top 1% was 100+, then I think 100+ would be cutoff.


I was looking for roots for this fraction to split it into as many factors as possible.

$f(x)=\frac{2x^4+x^3-5x^2+8x-3}{x^3+x^2-2x+3}$

First of i found out that $0.5$ was a root to the nominator.

$f(x)=\frac{(x-0.5)(2x^3+2x^2-4x+6)}{x^3+x^2-2x+3}$

...but from here i couldn't split it further. Maybe some of you can help me one step further.

Thx in advance for your help.

[hide="very big hint"]try factoring that 2 out of the second polynomial in the numerator[/hide]

lol, this was really great. Just two seconds after i posted the thread i found the solution to my problem (well, quite simple way when you take a little more time instead of following the book's guidelines). I clicked the delete button but i couldn't deleted my own thread since it is only allowed by the moderators. Then i wrote a reply to answer the result to my own problem and found that in the same time i wrote the reply my internet connection was gone :lol: Didn't help with a restart of my comp. Here is what i wrote in the reply (which i saved):

------------------------------------------------------------------------------------

OK, i can't delete it, but now i think i know how to handle this. Plz tell me if this is right, thx.

$f(x)=\frac{(x-0.5)(2x^3+2x^2-4x+6)}{x^3+x^2-2x+3}$

$f(x)=\frac{2(x-0.5)(2x^3+2x^2-4x+6)}{2(x^3+x^2-2x+3)}$

$f(x)=2(x-0.5)$

$f(x)=2x-1$

------------------------------------------------------------------------------------

With the exception that i first timed it up by 2 and then removed the equal fractions instead of just dividing with 2 i seem to have done everything right. Thx for your reply though i didn't need it, but thx anyway (how would you know) ;)


for some who is interested in math, would you suggest acturial science or being an analyst or trader at wall street

also which is more high paying?

[quote="JRav"]If you want to make ungodly sums of money, become a successful investment banker with a big company.  Of course, this requires an MBA.[/quote]
Pretty much no Wall Street type jobs [i]require[/i] MBAs anymore. Also, the only way this strategy works is if you're willing to do that for your whole life. Those people at Merrill Lynch or whatever whose ratings of particular stocks can cause huge swings in those stocks once announced have spent decades moving up in those companies. If you want to get rich quick (not super quick but faster than at a big i-bank), work for a hedge fund.

If you like programming, being a Wall Street quant type person is probably about the best you can do and stay in the "math-analytical" sort of realm. Being an i-banker or a prop trader is not nearly as analytical a profession.

Wel.....i wanna be an actuary co mathematician is that possible?

[quote="123456789"]How much do actuaries, quants, etc. get paid?[/quote]
It varies quite a bit, but generally a lot. This should not be your sole motivation for working such a job, though. (Probably, if it is your sole motivation, you won't get one of the best jobs.)

Just my opinion- having worked for a large (top 10) P&C insurance company.  It was not always the most interesting, but it had it's moments.  What I will say is that the pay was good (10 years ago I started at $ \$$44K with no exams passed) and the hours were very good - the only extra time I put in was studying for exams. When I decided not to pursue exams, which was fairly accepted in my department, it definitely became a job worth having with a young family. As for making money - they paid$ \$$2k automatic raise for each exam passed.  If you passed the exam on the first try, there was an added bonus.  There were a few young gung-ho types (under 26) who passed all the exams first time and I know their salary was at least close to $ \$$100k. I have had a few friends work as investment brokers and they said the money was insane (one claimed over$ \$$1 mil in a year before bonuses) but the hours could be trying as well.  He basically said there was only one relationship in his life.  But he is in his late 30's and essentially retired.

The previous poster's discussion of salary is totally consistent with my own personal knowledge of actuary salaries.  Actuaries make a respectable amount of money, but to put it bluntly, it's a pittance compared to what people with comparable math skills are able make in other jobs such as investment banking.  (Career actuaries have to be fairly smart people since they have to pass a series of grueling exams.)  It's also less than what you could make in computer programming or engineering.  On the other hand, actuaries don't have to work particularly hard and have good job security.

Here is some salary data on actuaries:

[url]http://www.actuaryjobs.com/salary.html[/url]

[quote="mr.affable"]It seems that actuaries stay within probability and statistical application.  Does this field offer much in terms of expanding one's knowledge mathematically?  In other words, is it more of a "business" career, or a mathematic career?[/quote]

I think it is more of a business career. Probably you know most of the math but rather have to get used to certain models, assumptions and other domain knowledge.

I heard actuaries have better job satisfaction? If you're an investment banker or analyst you might only get to sleep from 2am to 7am, and bankers have to work very long hours

i interned with quants this summer. I mean they know math but not like we do.....(problem solving)

I think that job satisfaction depends on the person.. everyone likes different things! You cannot conclude that every banker hates his job and has less job satisfaction than the actuaries ;)


LaTeX
number theory unsolved
number theory

Determine all integer solutions of the equation $ x^n\plus{}y^n\equal{}1994$,where $ n \geq 2$

Obviously, $ x,y$ are both odd. It's easy to conclude that $ n\le3$ because $ 5^5 > 1994$ and $ 2.3^5 < 1994$ and the case $ n \equal{} 4$ follows easily from Fermat's little theorem.Now we need to check 2cases, which are easy to do.
[hide="Some hints"](the main idea in the case $ n \equal{} 3$ is to use modulo 7 and conlude that one of the numbers is divisible by $ 7$, say $ x$, and since $ 14^3 > 1994$, so $ x \equal{} 7$ but then $ y^3 \equal{} 1651$-impossible)[/hide]

But x and y need not be positive, right?

in that case we shall have to solve two equations
$ x^n\minus{}y^n\equal{}1994$ and $ x^n\plus{}y^n\equal{}1994$

the first one is only possible if n is odd cuz else it ll lead to second..
expand it...ll post the solution as soon as poss..
in a hurry..!

I don't understand. Why n needs to be smaller than 3 or equal ?

If I write : 3^5 < 1994 and 5^5 > 1994 ?

I missed one case because I thought it's obvious. It's true because we know that x,y are both odd and since $ 5^5>1994$ then they should be 1 or 3. Now, I forgot to write the fact that $ 3^6>1994$ which implies that there aren't any solution for $ n\ge 5$. But, as I mentioned, the case $ n\equal{}4$ is impossible because of the Fermat Little Theorem and so $ n\le3$.

[size=150]x^2=4a+1=3b+1, thus a=3k, x^2=12h, for some natural h, similarly for b, but 12 does not divide 1994, hence no solutions.....[/size]

Can someone give me the link for using latex here?[/url]

for latex, simply put \$ sign in between text

if $ n$ even, so simply
if $ n$ odd, then we can assume $ n$ be a prime odd.
So $ x \plus{} y|1994$ and $ x \plus{} y$ be a positive integer even.
$ \rightarrow x \plus{} y \equal{} 2,1994$
1)Case $ x \plus{} y \equal{} 1994$
We have $ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \plus{} x^{n \minus{} 2}y \plus{} . . . \plus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1}) > x \plus{} y \equal{} 1994$.
2)Case $ x \plus{} y \equal{} 2$, assume $ x > 0$.
$ \rightarrow n|1992$.
$ \rightarrow n \equal{} 3,83$.
$ x^n \minus{} (x \minus{} 2)^n \equal{} 1994$.
If $ x \minus{} 2\leq 0$, impossible.
If $ z \equal{} x \minus{} 2 > 0$, we have:
$ 1994 \equal{} (z \plus{} 2)^n \minus{} z^n\geq 2^n$, so $ n \equal{} 3$ and we have equation:
$ (z \plus{} 2)^3 \minus{} z^3 \equal{} 1994$ has no root on $ Z$.

[quote="Thjch Ph4 Trjnh"]We have $ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \plus{} x^{n \minus{} 2}y \plus{} . . . \plus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1}) > x \plus{} y \equal{} 1994$.[/quote]

Actually, for odd $ n$: $ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \minus{} x^{n \minus{} 2}y \plus{} . . . \minus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1})$ (alternating signs). Although, it doesn't really make a difference.

Anyway, here is a complete solution:
[hide]WLOG, let $ x \ge y$. If $ x \equal{} y$, then $ 2x^n \equal{} 1994 \Rightarrow x^n \equal{} 997$, which is impossible. So, $ x > y$.

If $ x$ and $ y$ are even, then $ 2^n | x^n \plus{} y^n \equal{} 1994$, which is a contradiction since $ n \ge 2$ and $ 4 \nmid 1994$.

If one of $ x,y$ is even, and the other is odd, then $ x^n \plus{} y^n \equal{} 1994$ is odd, a contradiction.

Thus, $ x,y$ are both odd.

Case I: $ n$ is even.
Then, $ x^n , y^n > 0$. Also, $ (x,y)$ is a solution iff $ (\pm x, \pm y)$ are all solutions. So WLOG, let $ x > y > 0$.

If $ n \ge 8$, and $ x \ge 3$, then $ x^n \plus{} y^n \ge 3^8 \equal{} 6561$, a contradiction.
Thus, $ 1 \le y < x \le 1$, a contradiction.
Therefore, there are no solutions for even integers $ n \ge 8$

If $ n \equal{} 6$ and $ x \ge 5$, then $ x^6 \plus{} y^6 \ge 5^6 \equal{} 15625$, a contradiction.
Thus, $ 1 \le y \le x \le 3 \Rightarrow x^6 \plus{} y^6 \le 2 \cdot 3^6 \equal{} 1458$, a contradiction.
Therefore, there are no solutions for $ n \equal{} 6$.

If $ n \equal{} 4$ and $ x \ge 7$, then $ x^4 \plus{} y^4 \ge 7^4 \equal{} 2401$, a contradiction.
Thus, $ 1 \le y < x \le 5 \Rightarrow x^4 \plus{} y^4 \le 2 \cdot 5^4 \equal{} 1250$, a contradiction.
Therefore, there are no solutions for $ n \equal{} 6$.

If $ n \equal{} 2$ and $ x \ge 45$, then $ x^2 \plus{} y^2 \ge 45^2 \equal{} 2025$, a contradiction.
Thus, $ 1 \le y < x \le 43$.
Note that $ 2y^2 \le x^2 \plus{} y^2 \equal{} 1994 \le 2x^2$
Therefore, $ 33 \le x \le 43$ AND $ 1 \le y \le 31$.
Testing all $ 11$ possible values of $ x$ yields $ (x,y) \equal{} (37,25)$ as the only possible solution with $ x > y > 0$.
Thus, the only solutions for $ n \equal{} 2$ are $ (x,y) \equal{} (\pm 37 , \pm 25) ; (\pm 25 , \pm 37)$.

Case II: $ n$ is odd.
Obviously, either $ x > y > 0$ or $ x > 0 > y$, and $ |x| > |y|$.

Since $ |x|,|y|$ are both odd integers, $ |x| \ge |y| \plus{} 2$.
Since $ x^n \plus{} y^n \equal{} 1994$, we must have $ |x|^n \minus{} |y|^n \le 1994$.
So, we have $ (|y| \plus{} 2)^n \minus{} |y|^n \le |x|^n \minus{} |y|^n \le 1994$.

If $ n \ge 7$, then we have: $ (|y| \plus{} 2)^n \minus{} |y|^n \ge 3^7 \minus{} 1^7 \equal{} 2186$, a contradiction.
Thus, there are no solutions for odd $ n \ge 7$.

If $ n \equal{} 5$, and $ |y| \ge 3$, then we have: $ (|y| \plus{} 2)^5 \minus{} |y|^5 \ge 5^5 \minus{} 3^5 \equal{} 2882$, a contradiction.
Therefore, we must have $ |y| \equal{} 1$, and thus, $ x^5 \pm 1 \equal{} 1994$ which has no integer solution for $ x$.
Thus, there are no solutions for $ n \equal{} 5$.

If $ n \equal{} 3$, and $ |y| \ge 19$, then we have: $ (|y| \plus{} 2)^3 \minus{} |y|^3 \ge 21^3 \minus{} 19^3 \equal{} 2402$, a contradiction.
Therefore, we must have $ |y| \le 17$.
Since $ y < x$, we have $ 2y^3 < x^3 \plus{} y^3 \equal{} 1994 \Rightarrow y \le 9$.
Therefore, $ \minus{} 17 \le y \le 9$. Testing all $ 13$ possible values of $ y$ yield no integer solutions for $ x$.
Thus, there are no solutions for $ n \equal{} 3$.

Therefore, the only integer solutions to $ x^n \plus{} y^n \equal{} 1994$ with $ n \ge 2$ are:
$ (x,y,n) \equal{} (\pm 37 , \pm 25 , 2); (\pm 25 , \pm 37 , 2)$. [/hide]

[quote]
We have $ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \plus{} x^{n \minus{} 2}y \plus{} . . . \plus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1}) > x \plus{} y \equal{} 1994$.
[/quote]
Sorry.:D
$ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \minus{} x^{n \minus{} 2}y \plus{} . . . \minus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1})$.
$ 2(x^{n \minus{} 1} \minus{} x^{n \minus{} 2}y \plus{} . . . \minus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1}) \equal{} x^{n \minus{} 1} \plus{}x^{n \minus{} 3}(x \minus{} y)^2 \plus{} . . . \plus{}y^{n \minus{} 3}(x \minus{} y)^2 \plus{} y^{n \minus{} 1}\geq x^{n \minus{} 1} \plus{} y^{n \minus{} 1} \geq |x| \plus{} |y|\geq 1994$

[quote="soumik"][size=150]x^2=4a+1=3b+1, thus a=3k, x^2=12h, for some natural h, similarly for b, but 12 does not divide 1994, hence no solutions.....[/size]
[/quote]

this is wrong soumik


Λοιπόν , θα πέσετε από τα σύννεφα !

  Οι τούρκοι , μόλις έμαθαν ότι στην Ελλάδα κυκλοφορεί με  25 ευρώ το IMO COMPENDIUM, παρήγγειλαν - πόσα λέτε ; -  [color=red]20000 [/color]αντίτυπα !!! Οι Έλληνες - μαθηματικοί και μαθητές - αγόρασαν λιγότερα από 10 αντίτυπα.
                Είπατε τίποτα ;;;

    Μπάμπης

ΤΙ ΝΑ ΠΕΙΙΣ ? ΕΔΩ ΔΕΝ ΧΩΡΟΥΝ ΛΟΓΙΑ ΠΑΡΑ ΜΟΝΟ ΤΟ ΜΥΝΗΜΑ ΝΑ ΠΡΟΤΡΕΨΟΥΜΕ ΟΛΟΥΣ ΤΟΥΣ ΜΑΘΗΤΕΣ ΚΑΙ ΤΟΥΣ ΜΑΘΗΜΑΤΙΚΟΥΣ ΝΑ ΑΓΟΡΑΣΟΥΝ ΤΟ ΒΙΒΛΙΟ  ΑΛΛΑ ΤΙ ΝΑ ΓΙΝΕΙ ? ΔΕΝ ΑΦΟΡΑ Η ΠΡΟΟΔΟΣ ΤΗΣ ΕΛΛΑΔΑΣ ΣΤΑ ΜΑΘΗΜΑΤΙΚΑ ΩΣ ΟΜΑΔΑ  :mad:  ΠΡΟΗΓΕΙΤΑΙ ΒΛΕΠΤΕ Η ΓΙΟΥΡΟΒΙΖΙΟΝ  :mad:  :mad:
ΧΑΙΡΕΤΑΙ


In $ \triangle ABC$

$ AB=t^2, AC=t, BC=1,t>1$

$ \angle A=x^o,\angle C=x^o+60^o,
\angle B=120^o-2x^o$

Prove that  $ t <\sqrt{2} .$

[hide]Here's my approach which pops up from my head:
Applying laws of cosines we have: AB^2=1+t^2-2t.cosACB.Therefore, we have:t^4-1-t^2+2t.cosACB=0(3)
Now, let's calculate cosACB. We see that cos(ACB)=sin(90-ACB)=sin(30-x)(2). In addition we also have:sinx/1=sin(x+60)/t, therefore,t=sin(x+60)/sinx.=cos(30-x)/sinx(1)
Plugging (1) and (2) into (3) and bashing with algebra, we can find the value of x. From that, we can calculate the value of t, which will be<sqrt(2).
Or another way to do with it is solving the inequality: cos(30-x)/sinx>1 to get:
sqrt(3)/2.cosx>1/2.sinx, which means cotg(x)>1/sqrt(3).Using this bound to prove that: [sqrt(2)-1/2].sinx>sqrt(3)/2.cosx, which means: sqrt(3)/(sqrt(2)-1/2)<tg(x) with 0<x<120 is not hard.[/hide]

hello, the theorem of cosine gives $ \cos(x)=\frac{t^2+t^4-1}{2t^3}$ (1), the theorem of sine gives $ \frac{\sin(x)}{\sin(x+60^{\circ})}=\frac{1}{t^2}$ (2).
From (2) we get $ x=\arctan(\frac{\sqrt{3}}{2t{}^2{}^-1})$, substituting in (1) we have $ \frac{1}{\sqrt{1+\frac{3}{(2t^2-1)^2}}}=\frac{1}{2}\frac{t^2+t^4-1}{t^3}$. Solving by $ t$ we have ${ t=\sqrt{\frac{1}{4}+\frac{\sqrt{13}}{4}+\frac{\sqrt{-2+2\sqrt{13}}}{4}}}<\sqrt{2}$ as desired.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the theorem of cosine gives $ \cos(x) = \frac {t^2 + t^4 - 1}{2t^3}$ (1), the theorem of sine gives $ \frac {\sin(x)}{\sin(x + 60^{\circ})} = \frac {1}{t^2}$ (2).
From (2) we get $ x = \arctan(\frac {\sqrt {3}}{2t^2^ - 1})$, substituting in (1) we have $ \frac {1}{\sqrt {1 + \frac {3}{(2t^2 - 1)^2}}} = \frac {1}{2}\frac {t^2 + t^4 - 1}{t^3}$. Solving by $ t$ we have ${ t = \sqrt {\frac {1}{4} + \frac {\sqrt {13}}{4} + \frac {\sqrt { - 2 + 2\sqrt {13}}}{4}}} < \sqrt {2}$ as desired.
Sonnhard.[/quote]
Sorry Dr. Sonnhard Graubner! I think your solution was exactly the same as what I did in my post before. You should give credit to me, Dr. Graubner. :roll:

[quote="Dr Sonnhard Graubner"]hello, the theorem of cosine gives $ \cos(x) = \frac {t^2 + t^4 - 1}{2t^3}$ (1), the theorem of sine gives $ \frac {\sin(x)}{\sin(x + 60^{\circ})} = \frac {1}{t^2}$ (2).
From (2) we get $ x = \arctan(\frac {\sqrt {3}}{2t^2^ - 1})$, substituting in (1) we have $ \frac {1}{\sqrt {1 + \frac {3}{(2t^2 - 1)^2}}} = \frac {1}{2}\frac {t^2 + t^4 - 1}{t^3}$. Solving by $ t$ we have ${ t = \sqrt {\frac {1}{4} + \frac {\sqrt {13}}{4} + \frac {\sqrt { - 2 + 2\sqrt {13}}}{4}}} < \sqrt {2}$ as desired.
Sonnhard.[/quote]

Have any simple idea to prove it? :D

[hide]Let $ D$ be a point on $ \overline{AB}$ such that $ DA \equal{} DC$ and $ u \equal{} CD$.

Clearly $ 2u\cos x \equal{} t$(*),then apply cosine low to $ \triangle ABC$ and $ \triangle DBC$,thus we have

$ t^4 \plus{} t^2 \minus{} 2t^3\cos x \equal{} 1$(**) and $ u^2 \plus{} 1 \minus{} u \equal{} (t^2 \minus{} u)^2$(***).

by(*) and (**)  $ u \equal{} \frac {t^4}{t^4 \plus{} t^2 \minus{} 1} < 1$ because $ t > 1$,

thus, by (***) $ t^2 \equal{} u \plus{} \sqrt {u^2 \minus{} u \plus{} 1} \equal{} u \plus{} \sqrt {1 \minus{} u(1 \minus{} u)} < 1 \plus{} \sqrt {1 \minus{} 0} \equal{} 2$ as desired.[/hide]

In Chinese statements version:

where $ r\equal{}t ,$


[quote="Silverfalcon"][quote="Zootieroaz"]Can anyone help me how to claculate 2137 to the 753th power?[/quote]

Exact value?

That will be by computer (I don't think calculator can handle that). If you mean last *** digits (*** is number, like two digits, etc), then you can use Fermat's or Binomial Theorem.[/quote]

how would u use the binomial theorem to find the last..say 3 digits of that number?

Well, it doesn't "always" work. If you have nice numbers, it's pretty nice.

If you want to find out the last three digits of the expansion of $9^{9999}$, we can write this as $(10-1)^{9999}$ and use binomial theorem to consider only few terms that will matter in calculation.

the answer is 42 :-)

(think hitchhiker's guide to galaxy)

it would take a super computer to calculate that, then u'd probly spend a year reading the answer, so just be satisfied that it will be really really big.

you could always deal with it in terms of logs to make its calculation simpler, or get a quick estimate

[quote="sen"]what exacltly is this binomial theorem ?  :blush:[/quote]

http://www.google.com/search?client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial_s&hl=en&q=binomial+theorem&btnG=Google+Search

It takes TWENTY-EIGHT letters and TWO clicks to find the answer.