[quote="mdk"]Let p be a prime number.
(a)  Show that p divides 2^p-2 .
(b)  Show that  P^2 divides 2^(p^2)-2^p[/quote]

Uh...should these be in Pre-Olympiad?
[hide="hint on 1"]
Order of the group of reduced residue class mod p under multiplication is $ p-1$, so for each $ k$ not divisible by $ p$, $ k^{p-1}\equiv 1\mod{p}$.[/hide]

[hide="hint on #2"]
Factor out $ 2^{p}$, and then instead use the group of reduced residue classes modulo $ p^{2}$ under multiplication.[/hide]

Or, alternatively, (though I assume the whole purpose is not to use this)
[hide]Fermat's Little Theorem[/hide]

I understand that these approaches use the same, group-theoretic/pigeonhole idea, so maybe you don't want people to use that, but that seems like a very short, insightful, generalizable way to do it...


quadratics
floor function
calculus
integration
algebra
number theory proposed
number theory

Hey guys.. I just came up with this problem which - once you know the method - isn't that difficult :)

Show that:
$ k| \left[\left( \frac {2k \plus{} 1 \plus{} \sqrt {4k^2 \plus{} 1}}{2} \right)^n \right], k \in \mathbb{N}$

Any solutions and/or suggestions on how to make this problem harder is appreciated:)

[hide="hint"] Let $ r_1 \equal{} \frac {2k \plus{} 1 \plus{} \sqrt {4k^2 \plus{} 1}}{2}$ and $ r_2 \equal{} \frac {2k \plus{} 1 \minus{} \sqrt {4k^2 \plus{} 1}}{2}$.  $ r_1,r_2$ are roots of the quadratic equation
\[ x^2 \minus{} (2k \plus{} 1)x \plus{} k \equal{} 0
\]
and therefore for all $ c_1,c_2\in\mathbb{R}$, we have that $ a_n \equal{} c_1 r_1^n \plus{} c_2 r_2^n$ is a solution of the recurrence equation
\[ a_n \equal{} (2k \plus{} 1)a_{n \minus{} 1} \minus{} ka_{n \minus{} 2}
\]
Notice also that for each $ k$, $ 0 < r_2 < 1$ and so for sufficiently large $ n$ we have that $ a_n \equal{} \left \lfloor c_1r_1^n\right \rfloor \pm 1$.  Thus to prove the original statement it suffices to show $ k \mid a_k \pm 1$ (which sign depends on the choice of $ c_2$).

Though it remains to choose convenient integral initial conditions $ a_0,a_1$ so that $ c_1 \equal{} 1$, and $ c_2$ is small enough so that you can bound $ 0 < c_2r_2^k < 1$.  I think $ a_0 \equal{} 0$ and $ a_1 \equal{} 2k \plus{} 1$ would work.
[/hide]

Maybe it was a little more than just a hint? :):):)

Cool you solved it that way - I was actually using induction. Do you have a suggestion on how to make this harder?

Edit: Just have $ c_1 \equal{} c_2 \equal{} 1$

$ r_1r_2 \equal{} k$, because $ r_2 > k \plus{} 0.5$, $ r_2 < 1$. Therefore
$ a_n \equal{} r_1^n \plus{} r_2^n \minus{} 1 \equal{} (r_1 \plus{} r_2)(r_1^{n \minus{} 1} \plus{} r_2^{n \minus{} 1}) \minus{} r_1r_2(r_1^{n \minus{} 2} \plus{} r_2^{n \minus{} 2}) \minus{} 1 \equal{} (2k \plus{} 1)(a_{n \minus{} 1} \plus{} 1) \minus{} k(a_{n \minus{} 2} \plus{} 1) \minus{} 1 \equal{} (2k \plus{} 1)a_{n \minus{} 1} \minus{} ka_{n \minus{} 2} \plus{} k$.
Because $ a_1 \equal{} k,a_2 \equal{} 4k^2 \plus{} 2k \plus{} 1$, we get $ a_n \equal{} a_{n \minus{} 1}\mod k \equal{} 1\mod k$ for $ n > 2$.


find x so that 0<=x<=90 and sin^3x+1/sin^4x=cos^3x+1/cos^4x

This seems simple enough:

$0=sin^3 x-cos^3 x+\frac 1{sin^4 x} - \frac 1{cos^4 x} = (sinx - cosx)(sin^2 x + sinx cosx + cos^2 x) + (\frac 1{sin^2 x} - \frac 1{cos^2 x})(\frac 1{sin^2 x} + \frac 1{cos^2 x}) = (sinx - cosx)(1+sinxcosx)+\frac{sin^2 x - cos^2 x}{sin^4 x cos^4 x} = (sinx - cosx)(1+sinxcosx+\frac{sinx+cosx}{sin^4 x cos^4x})$.  However, the equation is undefined for $x=0, \frac{\pi}{2}$ and for $x \in (0,\frac{\pi}{2})$, $sinx,cosx > 0$.  Therefore, $sinx = cosx$ and the only solution in the given interval is $x=\frac{\pi}{4}$.

Well,I have the general one.This comes from an inequality.
Give [tex] x\in (0, \frac{\pi}{2}) [/tex] and $ m,n $are integers.
Solve this equation:
    [tex] sin^nx+\frac{1}{sin^mx}=cos^nx+\frac{1}{cos^mx} [/tex] :)

anhcuong ! your problem is very nice .can you solve my problem ?

[quote="Anh Cuong"]Well,I have the general one.This comes from an inequality.
Give $x\in (0, \frac{\pi}{2})$ and $m,n$are integers.
Solve this equation:
    $sin^{n}x+\frac{1}{sin^{m}x}=cos^{n}x+\frac{1}{cos^{m}x}$ :)[/quote]
It is a time to show your solution.


modular arithmetic
group theory
abstract algebra
number theory
greatest common divisor
induction
geometry

Find all integers $n>1$ such that $n^{2}\, |\, (2^{n}+1)$.

This is IMO Problem 3, 1990.

I would like to know a basic solution. :lol:

Not to pass premature judgment or anything, but I don't think IMO problems belong in the Intermediate Forum. Moved...

[hide="doesnt work  so dont point out its flaws"]

Lemme work out the kinks, or you can for me if you want.

I believe $n=1, 3$.

Suppose $n$ is even. Then $2|n^{2}|2^{n}+1$ which is a contradiction since the third one is odd. From now on $n$ is odd.

Lemma: $2^{n}\equiv2\pmod{n}$ for odd $n$.
Proof: Consider the group of non-zero residues modulo $n$. Call this group $G$. Obviously, $G$ has order $n-1$. Now consider the subgroup, $H$, obtained by taking $2^{k}$ modulo $n$, for $k>0$. By Lagrange's Theorem, $ord(H) | ord(G) = n-1$. We also know that $2^{ord(H)}\equiv 1 \pmod{n}$ since $n$ is odd. Therefore, $2^{n-1}\equiv1\pmod{n}$ and $2^{n}\equiv 2 \pmod{n}$.

Obviously, if $n^{2}|2^{n}+1$, $n|2^{n}+1$, so $2^{n}+1 \equiv 0 \pmod{n}$ and by the lemma, this is equivalent to $3 \equiv 0 \pmod{n}$, which is true for $n=1,3$.
[/hide]
Hopefully I didn't do anything wrong...

[quote="tjhance"][quote="Hamster1800"][hide="doesnt work"]

Lemme work out the kinks, or you can for me if you want.

I believe $n=1, 3$.

Suppose $n$ is even. Then $2|n^{2}|2^{n}+1$ which is a contradiction since the third one is odd. From now on $n$ is odd.

Lemma: $2^{n}\equiv2\pmod{n}$ for odd $n$.
Proof: Consider the group of non-zero residues modulo $n$. Call this group $G$. Obviously, $G$ has order $n-1$. Now consider the subgroup, $H$, obtained by taking $2^{k}$ modulo $n$, for $k>0$. By Lagrange's Theorem, $ord(H) | ord(G) = n-1$. We also know that $2^{ord(H)}\equiv 1 \pmod{n}$ since $n$ is odd. Therefore, $2^{n-1}\equiv1\pmod{n}$ and $2^{n}\equiv 2 \pmod{n}$.

Obviously, if $n^{2}|2^{n}+1$, $n|2^{n}+1$, so $2^{n}+1 \equiv 0 \pmod{n}$ and by the lemma, this is equivalent to $3 \equiv 0 \pmod{n}$, which is true for $n=1,3$.
[/hide]
Hopefully I didn't do anything wrong...[/quote]

$2^{9-1}=256\equiv 4 \mod 9$[/quote]

Hence why I edited it to say it doesn't work.

sorry i posted that before i saw the edit

How about this?

Take the expression modulo $n+1$. Since $n\equiv-1 \pmod{n+1}$, $n^{2}\equiv 1 \pmod{n+1}$. Thus, we need $2^{n}\equiv 0 \pmod{n+1}$. This means that $n+1$ is a power of 2. (only powers of 2 divide powers of 2).

Now how do I get from here to 1,3?

[quote="t0rajir0u"][quote="Hamster1800"]Take the expression modulo $n+1$.
[/quote]

Clearly not valid.[/quote]

right...that was stupid.

So now that I've been stupid twice in this thread, anyone wanna show me how to not be stupid?

So here's what I know I can do:

$2^{n}+1 \equiv 0 \pmod{n^{2}}$
$2^{n}+1 \equiv 0 \pmod{n}$
$2^{n}\equiv-1 \pmod{n}$
$2\cdot 2^{n-1}\equiv-1 \pmod{n}$.
Therefore, if $n$ is prime,
$2 \equiv-1 \pmod{n}$ and $n|3$ so $n=3$. Now what if $n$ isn't prime?

Suppose $n = kp$ and it works and $p$ is prime.

$p^{2}k^{2}| 2^{p}2^{k}+1$
$p | 2^{p}2^{k}+1$
$p | 2 * 2^{k}+1$
$2^{k+1}+1 \equiv 0 \pmod{p}$
$2^{k+1}\equiv-1 \pmod{p}$

I have no idea if any of this helps, but it should all be valid.

I'm still gonna post this because I didn't find a post in kunny's thread that helped me.

[hide="how about this..."]
Clearly it works for $n=1$. Otherwise, let $p$ be the smallest prime divisor of $n$, and $p>2$ because it is clearly false for even $n$. Then we must have
\[2^{n}\equiv-1\pmod{p^{2}}\Rightarrow 2^{2n}\equiv 1\pmod{p^{2}}\]
Let $d$ be the order of 2 modulo $p^{2}$. Then $d$ divides both $2n$ and $p(p-1)$. Now observe that $\frac{n}{p}$ and $\frac{p-1}{2}$ are relatively prime (since any prime divisor of $\frac{p-1}{2}$ is less than $p$, so could not divide $n$.) Therefore the gcd of $2n$ and $p(p-1)$ is $2p$, so we get
\[2^{2p}\equiv 1\pmod{p^{2}}\]

\[(2^{p}-1)(2^{p}+1)\equiv 0\pmod{p^{2}}\]
Then at least one of $2^{p}-1$ or $2^{p}+1$ must be divisible by $p$. But $2^{p}\equiv 2\pmod{p}$ so $1\equiv 0$ or $3\equiv 0$. It follows that $p=3$.

Now we can write $n=3^{k}\cdot m$ where $m$ is not divisible by 2 or 3. Then since $2^{n}+1$ must be divisible by $3^{2k}$, consider the greatest power of 3 dividing $2^{3^{x}}+1$ for each integer $x$. It turns out that there are $x+1$ factors of 3. This is easily seen by induction: it is true for the base case $x=1$; for the inductive step, use sum of cubes:
\[2^{3^{x+1}}+1=(2^{3^{x}}+1)(2^{2\cdot 3^{x}}-2^{3^{x}}+1) \]
The second factor is congruent to 3 modulo $3^{x+1}$; since $x\geq 1$, there is exactly one factor of 3 in the second factor. This completes the inductive step.

Now we can factor $2^{n}+1$ into
\[(2^{3^{k}}+1)(2^{3^{k}(m-1)}-2^{3^{k}(m-2)}+\cdots+1) \]
The second factor contains no factors of 3 since it is congruent to $m$ modulo 3. Therefore, all factors of 3 come from $2^{3^{k}}+1$, and there are $k+1$ of them. But we need $2k$ of them, so $k+1\leq 2k\Rightarrow k\leq 1$. So we must have $k=1$ as well.

Now $n=3m$ so $2^{3m}\equiv-1\pmod{m^{2}}$. If $m=1$, then $n=3$. Otherwise, let $q$ be the smallest prime factor of $m$. (oh no, not this again.) Let $e$ be the order of 2 modulo $q^{2}$, so $e$ divides $6m$ (therefore $m$ since $q$ is relatively prime to 6) and $q(q-1)$. But the gcd of $m$ and $q(q-1)$ is $q$, so $2^{q}\equiv 1\pmod{q^{2}}$. In particular, $2\equiv 1\pmod{q}$ which is BAD!

So our only solutions (if you could actually follow this proof) are $\boxed{n=1,3}$.
[/hide]


geometry
3D geometry
sphere
calculus
integration
rotation
perimeter

Find the volume that remains after a hole of radius 3 is bored through the center of a solid sphere of radius 5.

So what I'm thinking is the equation x^2 + y^2 = 5. Perhaps
Would it be the  Pi [integral] (x-[Root]5)^2 - (x-[Root]3)^2)^2?
No that's not it. I'm not sure where to go from here.

[/code]

Do this integration with shells of revolution about the x-axis.

Draw the circle x² + y² = 5².

Draw a line y = 3 through the circle.
Draw an elemental strip (horizontal) in the circle segment between y = 3 and y = 5, of thickness dy, and length 2x.

Now rotate this strip about the x-axis.

Elemental volume is dV = 2piy.2x.dy (perimeter*cylinder height*thickness)

Total volume is V = int[y=3 to y=5] 4pi.xy dy

Now substitute for x from the eqn of the circle and evaluate.

I got V = 256pi/3


trigonometry
percent
function
algebra proposed
algebra

Find all real solutions $x$ of the equation

$\cos\cos\cos\cos x=\sin\sin\sin\sin x$.

(Angles are measured in radians.)

[quote="José"][hide="zero percent sure"] Well, it doesn't care of which number I put in the calculator, the cos cos cos cos part is always equal to the same number, aprox. $0.999847741$, but then, when I put INV sen INV sen INV sen INV sen, it gave me error. Since, no solution...[/hide][/quote]

Zero percent is good... do you really think that the function $\cos\cos\cos\cos x$ is constant??

By the way, a hint to everybody: Don't try to find a constant C such that $\cos\cos\cos\cos x\geq C\geq\sin\sin\sin\sin x$ for every x; such a constant doesn't exist.

  darij

[quote="darij grinberg"]Zero percent is good... do you really think that the function $\cos\cos\cos\cos x$ is constant??[/quote]

That was lol-worthy.

I wrote a solution in an old thread called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7344]Trig Problem[/url].

Sometimes if you iterate many times it will seem that the function is constant because your calculator is not accurate enough ...

[quote="José"][quote="darij grinberg"]Find all real solutions $x$ of the equation

$\cos\cos\cos\cos x=\sin\sin\sin\sin x$.[/quote]

[hide="zero percent sure"] Well, it doesn't care of which number I put in the calculator, the cos cos cos cos part is always equal to the same number, aprox. $0.999847741$, but then, when I put INV sen INV sen INV sen INV sen, it gave me error. Since, no solution...[/hide][/quote]

0.999847741  is the solution for cos x = x in degree

[quote="José"][quote="darij grinberg"]
Zero percent is good... do you really think that the function $\cos\cos\cos\cos x$ is constant??

By the way, a hint to everybody: Don't try to find a constant C such that $\cos\cos\cos\cos x\geq C\geq\sin\sin\sin\sin x$ for every x; such a constant doesn't exist.

  darij[/quote]

Yes, every number I tried the calculator gave me the same result... for what value for $x$ the left side (the cos one) does not give me the result I put in my post before?[/quote]

Can anyone answer me???????

At first, the problem deals with radians and not with degrees.
At second, when doing it with degrees, you need to compute it up to more digits, e.g. $x=0$ gives $0.999847741531084$, $x=90$ gives $0.999847741545212$ (everytimes used degree wich in reality would make no sense).

But I second the question of darij: do you really think that it is constant¿

[quote="ZetaX"]But I second the question of darij: do you really think that it is constant¿[/quote]

I really don't know, if I knew I'd not ask... why do you ask me in that way, as if I were crazy?

[quote="blahblahblah"]Without being condescending, if you even have to ask that question, you should not be posting in the olympiad forums in the first place.

There are lower level forums for a reason, and if you cannot contribute anything constructive to higher ones, you should post primarily in the lower level ones.[/quote]

Calm down man, I took my calc and tried cos cos cos cos x for some x's and got the same result, that's why I asked... what's wrong with that? I only thought that the function cos cos cos cos x was constant........ because I tried 10 different numbers and saw that the result was the same!!!!!!! I'm not wrong to post? Or that spamming too?

Put $f(x)=x+x^5$.  Calculate $(f\circ f\circ f\circ f\circ f)(.0001)$.  Are you going to tell me that this function is also constant?

The point is just this - if problems are obviously too difficult for you, why clutter the discussion?  On one hand you go into middle school topics and ridicule their questions as being easy, and then you go into the olympiad section and post things that indicate that the problems are obviously too hard for you.  There is a middle ground somewhere, and I suggest you find it.

What you say blahblahblah, I don't want to argue again.... continue solving the problem then

Well, there is a more elegant way than to look at several intervals and then show that the images of the functions do not intersect in those intervals. This was my first idea in the TST though, but as we were not allowed to use calculators, this could hardly be done by hand as the estimates are quite sharp.

First, it is easy to see that if $x$ is a solution, then $\sin(x)>0$ and $0<\sin(\sin(x))<1$ and $0<\cos(\cos(x))<1$. It is well-known that $|\sin(x)|\leq |x|$ with equality only for $x=0$.

By $\sin^{(4)}(x)$ etc. i mean $\sin(\sin(\sin(\sin(x))))$ in the following.

$\sin^{(4)}(x)=\cos^{(4)}(x)$
$\Rightarrow \sin^{2}(\sin^{(3)}(x))=\cos^{2}(\cos^{(3)}(x))=1-\sin^{2}(\cos^{(3)}(x))$
$\Rightarrow \sin^{2}(\sin^{(2)}(x))+\cos^{2}(\cos^{(2)}(x))\geq \sin^{2}(\sin^{(3)}(x))+\sin^{2}(\cos^{(3)}(x))=1$
$\Rightarrow \sin^{2}(\sin^{(2)}(x)) \geq \sin^{2}(\cos^{(2)}(x))$
$\Rightarrow \sin^{(2)}(x) \geq \cos^{(2)}(x)$ (because $0<\sin(\sin(x))<1$ and $0<\cos(\cos(x))<1$)
$\Rightarrow \sin^{2}(\sin(x)) \geq \cos^{2}(\cos(x))=1-\sin^{2}(\cos(x))$
$\Rightarrow 1=\sin^{2}(x)+\cos^{2}(x) > \sin^{2}(\sin(x))+\sin^{2}(\cos(x))\geq1$ (because $\sin(x)=0$ and $\cos(x)=0$ ist not possible simultaneously),

contradiction.

Peter


if the pediatic dosage is 200mg/kg/day
how much medication should be administrated per day to an infant who weighs 13.2 lb ? and if the medication is to be administrated three times per day,how much medication should be administered per dose?
what do you think guys??

Once you convert pounds into kg, the problem is a simple exercise in dimesional analysis.  And I think you mean "pediatric", not "pediatic".


function
number theory
divisibility tests
algebra proposed
algebra

Find all functions $ f: \mathbb{N} \to \mathbb{N}$ such that for any $ n \in mathbb{N}$:
1) $ S(f(n))\equal{}S(n)$ ($ S(x)$ is the sum of the digits of number $ x$)
2) $ f(f(n)) | f(n)\plus{}1$

[quote="pmvald"]Find all functions $ f: \mathbb{N} \to \mathbb{N}$ such that for any $ n \in mathbb{N}$:
1) $ S(f(n)) \equal{} S(n)$ ($ S(x)$ is the sum of the digits of number $ x$)
2) $ f(f(n)) | f(n) \plus{} 1$[/quote]

$ 3|n$ $ \implies$ $ 3|S(n)$ $ \implies$ $ 3|S(f(n))$ $ \implies$ $ 3|f(n)$ $ \implies$ $ 3|f(f(n))$ $ \implies$ $ 3|f(n)\plus{}1$ which is impossible since $ 3|f(n)$

So no solution.

Take n=3, consider the divisibility by 3, contradiction. So there's no such function.

Edit: sorry, i didn't see pco's solution.


ratio
ceiling function
logarithms
floor function
real analysis
real analysis solved

"Let S be the sequence of positive integers divisible only by 2 or 3 arranged in ascending order. [The sequence starts: 1, 2, 3, 4, 6, 8, 9, 12, 16,... ]
Prove that the ratio of successive terms tends to 1."

It suffices to prove that whenever $a+b$ is large enough, there is a pair $a',b'$ s.t. $\frac{2^{a'}3^{b'}}{2^a3^b}\in[1,1+\epsilon)$. This is, however, pretty clear: Take $a'=\lceil\log_2{(2^a3^b)}\rceil,\ b'=0$. The ratio is $\ge 1$, and it's $\le 1+\frac 2{2^a3^b}\to 1$ as $a+b\to\infty$.

[quote="grobber"]It suffices to prove that whenever $a+b$ is large enough, there is a pair $a',b'$ s.t. $\frac{2^{a'}3^{b'}}{2^a3^b}\in[1,1+\epsilon)$. This is, however, pretty clear: Take $a'=\lceil\log[2]{(2^a3^b)}\rceil,\ b'=0$. The ratio is $\ge 1$, and it's $\le 1+\frac 2{2^a3^b}\to 1$ as $a+b\to\infty$.[/quote]

I believed something similar at the first sight.
Consider $a\"$ as the fractional part of that $log_2 {(2^a3^b)}$.
Why the fractional part of the $log_2 {3^b}$ should tends to $0$ for large $b$ - the behaviour of $3^b$ in terms of modulo $ 2^a$ is an open problem related to Collatz conjecture.
Then i think it follows that $\frac{2^{a'}3^{b'}}{2^a3^b} = \frac{1}{2^{a\"}}$ so it is not finished, or am i wrong?

Consider $2^a3^b = C$ where $C$ - constant.
Then $a+log_2 {3} * b = log_2 {C}$.
In the carthesian system, eqaution $2^a3^b = C$ is represented as a line. It is suffice to show that for large $C$, between infintesimal difference of lines corresponding to constants $log_2 {C}$ and $log_2 {C} + e$ there is always some point with integer coordinates.

Yes, I now realize what I did wrong. Not only was it wrong, it was pretty stupid as well :). I'll take another whack at it :).

I have some results related about - like distribution of the terms of sequence among some interval, but i am not still able to prove it. I will try tommorow again.

Anyway, I think it suffices to show that there are sequences of positive integers $a_n,b_n,c_n,d_n$ s.t. $\frac{2^{a_n}}{3^{b_n}}>1,\to 1,\ \frac{3^{c^n}}{2^{d_n}}>1,\to 1$.

It should also suffice to prove the existence of $a_n,b_n$, because the existence of $c_n,d_n$ is proven analogously.

This problem is, however, equivalent to showing that $a_n\ln 2-b_n\ln 3>0,\to 0$, or, in other words, $a_n\log_3 2-b_n>0,\to 0$, and this really is clear because the set of values of $\{m\log_3 2\}$ is dense in $(0,1)$, so we can find values of $\{a_n\log_3 2\}$ arbitrarily close to $0$, and then we can take $b_n=\lfloor\{m\log_3 2\}\rfloor$.


If someone wants the solutions i am here

Personally, I don't need your solution, since I created the first one.  :P

Silouan I don't know whatis your solution,but about the first inequality i proved that

       $S>=sum(sqrt(x+y)(x+z))-(x+y+z)>=sqrt(3(xy+yz+zx))$
 If you want to post your solution that's ok with me.I'm corious if yours is different than mine.

For the second one:
WLOG assume $\sum x^n=1$
Then by cauchy,
$L.H.S. \ge \frac{1}{\sum x^n(y+z)^m}$
By power mean, $(y+z)^m \le 2^{m-1}(y^m+z^m)$, then
$\frac{1}{\sum x^n(y+z)^m} \ge \frac{1}{2^{m-1}(\sum x^n(y^m+z^m))}$
Since ${x^n}$ and ${y^m+z^m}$ are oppositely sorted, by chebyshev
$\frac{1}{2^{m-1}(\sum x^n(y^m+z^m))} \ge \frac{1}{2^{m-1}(\sum x^n)(2 \sum x^m)/3}=
\frac{3}{2^m(\sum x^n)(\sum x^m)}=\frac{3}{2^m(\sum x^m)}$
Again by chebyshev, $1=\sum x^n \ge \frac{1}{3}(\sum x^{n-m})(\sum x^m)$
So,
$\frac{3}{2^m(\sum x^m)} \ge \frac{3}{2^m\frac{3}{\sum x^{n-m}}}=R.H.S.$

I think m have to be s.t:[tex]m\geq 1[/tex]

[quote="silouan"]My friend Cesar Lupu a have another solution. I suppose that x+y+z=1. I use B.C.S , transfomations and the inequality Nesbitt etc. My solution is a bit big.See you[/quote]
can you post your solution?

siuhochung,I'll post my solution as soon as I learn  some more $Latex$.If I'll post it now I'm sure you don't understand
 my bad writing. ;)

harazi posted this on the thread "intriguing ineq" and said to assume $x+y+z=1$. but i can't continue from there...

anyway, cezar, you could post at least the main ideas of your soln        :D  (i'm really interested to see it)

Here is my proof:Chebysev's ineq says me that:
[tex]L.H.S\sum x^{n-m}.(\frac{x}{y+z})^m\geq \frac13(\sum x^{n-m})(\sum(\frac{x}{y+x})^m)\geq R.H.S[/tex](since Nesbitt's inequal)

cezar lupu, i know one solution of this ineq already. I just want to know siloan's solution..

I think Cezar refers to the second proof given in Old and New Inequalities, initially found by Vasc. The idea is that using Cauchy we can prove the LHS is at least $\sum{\sqrt{(x+y)(x+z)}}-(x+y+z)$. To prove this just write the LHS in the form $ (a+b+c)(\frac{y+z}{b+c}+\frac{z+x}{c+a}+\frac{x+y}{a+b})-2(x+y+z)$ and apply Cauchy. Next, try to prove using substitution $ x+y=w^2$ and so on the nice inequality ${\sum{\sqrt{(x+y)(x+z)}}\geq x+y+z+\sqrt{3(xy+yz+zx}}$. Sooner or later, you'll find it's just Schur inequality.

[quote="harazi"]I think Cezar refers to the second proof given in Old and New Inequalities, initially found by Vasc.[/quote]

can you tell me where can i find this book, 'cos i saw it mentioned many many times on this forum :?:  (i looked for it, and it isn't available at any library)


geometry
MIT
college
Stanford
number theory

Based on the explanation on the AMSP homepage,

I get that we can pick 2 courses to take during the 3-week camp.

The fact that we can take [b]2 [/b]courses, and that there are [b]4[/b] areas of mathematics in which I want to improve

seems sad  :(

Is there any way I can take more than 2 courses....? I know it's "being greedy" but all the courses seem

so interesting and fun...  :D [/b]

I remember that this issue came up when I was a student (two summers ago). Essentially, you already have 6 hours of classes and problem sessions a day (in addition to whatever you want to do in your free time). I don't think that you could really do 4 areas and take anything meaningful from any of them because it would be too much. So I cannot speak officially, but I think that the purpose is that you'll learn the most by concentrating in two areas. Another thing that you should have in mind is that while you are there, you can get a sense of how to learn a certain subject. So once the camp is over, you'll be able to study effectively on your own.

During AwesomeMath's very first year (2006), students were organized into color-coded groups: white, yellow, red, blue, and black. Each professor gave his or her lectures to most of the groups. I remember we covered topics in algebra, geometry, number theory, and combinatorics, but 3 weeks was not enough time to learn (in other words "get better at") all these subjects. So I came home with mathematical tidbits instead of a workable understanding of a major subject.

So it may not be the best idea to cover so many subjects.

I was faced with the same problem.
I had to beg my parents (and I mean beg....) and convince them that I wouldn't goto MIT, Cal-Tech, or Stanford if I didn't get a complete understanding of problem solving by going to both locations.
Really, I wouldn't have been convinced that you can learn 2 subjects over 3 weeks if it wasn't for the glowing reviews of this camp. It seems that it would be hard to go in depth on four areas in a 3 week span.


function
complex analysis
logarithms
algebra
domain
topology
complex analysis unsolved

Does there exist a function $f(z)$, holomorphic in the disk $|z| \le 1$, such that $f(1/n) = 1/2^{n}$ for all positive integers $n$?

Intuitively the answer is no since $e^{-\ln 2 \frac{1}{z}}$ has a singularity at $z=0$. I would like to somehow apply some function to this to remove the singularity, so that we can use the uniqueness principle... but I can't think of how to do it.

I dont believe it... The previous problem was to determine whether there was a function analytic in the open unit disk $D$ which satisfies $f(1/n) = 1/\sqrt{n}$, $n=2,3,\ldots$, and unless I am totally mistaken this would imply an analytic square root existing in $D$.

Suppose that $f$ exists. Write it as
$f(z)=z^{k}g(z)$ where $g(0) \neq 0$ is also analytic for $|z|<1$, so continuous.
$g(\frac{1}{n})=\frac{n^{k}}{2^{n}}$, and letting $n$ goes to infinity, you see that $g(0)=0$, contradiction. Such a function does not exist.

Thank you alekk.

julien_santini, is it possible that you could expand on what you were saying?

[quote="julien_santini"]There's a theorem saying that $f$ can take any values on a given discrete set.[/quote]
Which misses the point that $\{1/n\},$ as a subset of the unit disk, isn't a discrete set because it has a limit point in the domain of interest.

[quote="Kent Merryfield"]Which misses the point that $\{1/n\},$ as a subset of the unit disk, isn't a discrete set because it has a limit point in the domain of interest.[/quote]

Well, I believe I misquoted the theorem (which must apply only to entire functions, or something like that). But what about the set $\{\frac{1}{n}\}$ ? Endowed with the induced topology, it really seems like discrete, no matter if it has an accumulation point in the unit disk, since this accumulation point does not belong to the set ... (?)

but the accumulation point belongs to the unit disk ! and when one imposes the value on the set $\{\frac{1}{n}\}$, it also imposes the value at $z=0$, by continuity.


Let $ a_{1},a_{2},\cdots,a_{n}$ be positive real numbers satisfying $ a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{n} \equal{} 1$. Prove that
\[\left(a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} \cdots \plus{} a_{n}a_{1}\right)\left(\frac {a_{1}}{a_{2}^2 \plus{} a_{2}} \plus{} \frac {a_{2}}{a_{3}^2 \plus{} a_{3}} \plus{} \cdots \plus{} \frac {a_{n}}{a_{1}^2 \plus{} a_{1}}\right)\ge\frac {n}{n \plus{} 1}\]

I check two case
If $ a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} \cdots \plus{} a_{n}a_{1}\ge \frac {1}{n}$
we have $ \frac {a_{1}}{a_{2}^2 \plus{} a_{2}} \plus{} \frac {a_{2}}{a_{3}^2 \plus{} a_{3}} \plus{} \cdots \plus{} \frac {a_{n}}{a_{1}^2 \plus{} a_{1}}\ge \frac {n^2}{n \plus{} 1}$ since AMGM inequality
and If $ a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} \cdots \plus{} a_{n}a_{1}\le \frac {1}{n}$
Use Cauchy Schwarz inequality
  $ RHS \ge \frac {a_1}{\sqrt {a_2 \plus{} 1}} \plus{} \cdots \plus{} \frac {a_n}{\sqrt {a_1 \plus{} 1}}\ge \frac {(a_1 \plus{} \cdots \plus{} a_n)^2}{a_1\sqrt {a_2 \plus{} 1} \plus{} \cdots \plus{} a_n\sqrt {a_1 \plus{} 1}}$
And now
$ (a_1\sqrt {a_2 \plus{} 1} \plus{} \cdots \plus{} a_n\sqrt {a_1 \plus{} 1})^2\le (a_1 \plus{} \cdots \plus{} a_n)(a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} \cdots \plus{} a_{n}a_{1} \plus{} a_1 \plus{} \cdots \plus{} a_n) \le  \frac {n \plus{} 1}{n}$
we have done

This is another method

First denote $ T\equal{} \sum_{i\equal{}1}^{n} \frac {a_{i}}{a_{i\plus{}1}},  (a_{n\plus{}1}\equal{}a_{1})$, then applying AM-GM Inequality we get $ T \ge n$.
Next applying Caughy Inequality we get $ \sum_{i\equal{}1}^{n} \frac {a_{i}}{a_{i\plus{}1}^2\plus{}a_{i\plus{}1}}\ge \frac {T^2}{\sum_{i\equal{}1}^n \frac {a_{i}(a_{i\plus{}1}\plus{}1)}{a_{i\plus{}1}}}\equal{} \frac {T^2}{T\plus{}1} \ge \frac {nT}{n\plus{}1} (T \ge n)$

Finally applying Caughy again we have $ (\sum_{i\equal{}1}^n a_{i}a_{i\plus{}1})(\sum_{i\equal{}1}^{n} \frac {a_{i}}{a_{i\plus{}1}^2\plus{}a_{i\plus{}1}}) \ge (\sum_{i\equal{}1}^n a_{i}a_{i\plus{}1})(\frac {nT}{n\plus{}1}) \ge \frac {n}{n\plus{}1}(\sum_{i\equal{}1}^n a_{i})^2)\equal{}\frac {n}{n\plus{}1}$

Q.E.D.


Can someone offer explanations of how to solve the following two handbook geometry problems?
(Sorry, I don't think I can include diagrams...)

Warmup 9, problem 8:
In the regular octagon ABCDEFGH, what is the number of degrees in the measure of angle DAF?

[hide="this is what I have so far"]I know that each interior angle of a regular octagon is 135 degrees.  I can sort of believe that angle DAF is one third of angle HAB (which is 135 degrees) but I don't see how to easily prove that.[/hide]

Workout 5, number 9:
Segment AB measures 4 cm and is a diameter of circle P. In triangle ABC, point C is on
circle P and BC = 2 cm. What is the area of the shaded region? Express your answer
as a decimal to the nearest tenth.  (The shaded region is the entire region inside the circle and outside the triangle.)

[hide]Angle ACB kind of looks like a right angle, but is it actually guaranteed to be one?  If not, how would we determine the area of the triangle?[/hide]

[quote="mathmom"]
Warmup 9, problem 8:
In the regular octagon ABCDEFGH, what is the number of degrees in the measure of angle DAF?[/quote]

Call $\angle DAF$ $x$. Notice that, after drawing line segments $AF$ and $AD$, the octagon is divided into an irregular quadrilateral and two congruent trapezoids, $AFGH$ and $ABCD$. Then the measure of $\angle HAF$ is equal to the measure of $\angle DAB$. Call these angles both $y$. Notice that $z = 135 - 2y$. Remember that the sum of the angles in a trapezoid is 360. $\angle B$ and $\angle C$ are both 135 degrees. $\angle D = y$, so $360 = 135 + 135 + 2y$. Solving for $y$, we see that $y = 45$. Now $z = 135 - 2(45) = \fbox {45}$. Maybe there is a way to prove that $\angle A$ is trisected by $\angle DAF$, but I don't know.

[quote="mathmom"]
Workout 5, number 9:
Segment AB measures 4 cm and is a diameter of circle P. In triangle ABC, point C is on
circle P and BC = 2 cm. What is the area of the shaded region? Express your answer
as a decimal to the nearest tenth.  (The shaded region is the entire region inside the circle and outside the triangle.)
[/quote]

First, you have to notice that $\angle C$ is a right angle. [b]Any angle inscribed in a semicircle is a right angle.[/b] Here's a proof: http://mathforum.org/library/drmath/view/54662.html. Next, recognize that $\triangle ABC$ is a 30-60-90 triangle, so $AC = 2\sqrt{3}$. This is the height of the triangle. Then the area is $A = \frac{bh}{2} = \frac{(2\sqrt{3})(2)}{2} = 2\sqrt{3}$.

edit: sorry, i didn't finish the question yet  :blush:

The area of the circle is $A = \pi r^2 = 4\pi$. The final answer is $4\pi - 2\sqrt{3}$.

For the first problem:

If you draw Angle DAF, you'll get quadrilateral FADE, Two of the angles, angles AFE and ADE, are easily 90 degrees (because AB and FE are parallel), and the remaining one, angle DEF, is 135 degrees. Thus, angle DAF has to be 45 degrees.

1st problem:
[hide]
After my first drawing, I stared at the inaccurate thing for a few minutes, then drew a more accurate one.  I then realized that Trapezoid AFGH had 2 angles measuring 135, the other two had to equal 90.  Because it is a regular octagon, the two angles must be equal.  So each of those is 45 :^o:.  Line AD is perpendicular to Line AH, making angle BAD 45 degrees.  The two angles add up to 90 degrees. 135-90=45[/hide]

[quote="chess64"]
First, you have to notice that $\angle C$ is a right angle. [b]Any angle inscribed in a semicircle is a right angle.[/b] Here's a proof: http://mathforum.org/library/drmath/view/54662.html. [/quote]

Thanks.  This was the piece we were missing on this one.  It [b]looked [/b]like a right angle, but we weren't sure that it was guaranteed to be one.

Thanks again  :)

Thanks for all the replies.  I am still a bit stuck on the first problem.

When you draw AF, how do you know (prove) that it is parallel to GH, or (equivalently) perpendicular to EF.  It all kind of seems obvious when you look at it, but I still don't see quite how to prove it.

[hide="I can prove that AB is parallel to EF and perpendicular to GH "]several ways -- easiest seems to be extending AB to meet GH at P, then triangle HAP has two 45 degree angles, so the other must be 90 degrees.  Do it again for GH perpendicular to EF and you end up with AB parallel to EF.[/hide]

And I can see that you must use the fact that all the sides are equal length to get to AF being parallel to GH, but I can't quite see how to get there.

Thanks again for everyone's help :)

[quote="mathmom"]Thanks for all the replies.  I am still a bit stuck on the first problem.
When you draw AF, how do you know (prove) that it is parallel to GH, or (equivalently) perpendicular to EF.  It all kind of seems obvious when you look at it, but I still don't see quite how to prove it.
[/quote]

Trapezoid $AFGH$ has angles $y$, $y$, $135$, and $135$, where $y=\angle HAF = \angle AFG$. Using the sum of the angles of a quadrilateral, $135+135+2y=360 \Rightarrow y=45$. Then we know that $\angle AFE = 135-y = 90$. Does this help?

[quote="mathmom"][quote="chess64"]
Trapezoid $AFGH$ has angles $y$, $y$, $135$, and $135$, where $y=\angle HAF = \angle AFG$. Using the sum of the angles of a quadrilateral, $135+135+2y=360 \Rightarrow y=45$. Then we know that $\angle AFE = 135-y = 90$. Does this help?[/quote]

No ;)

When you call $AFGH$ a "trapezoid", you assume $AF \parallel GH$, but I wanted to see a proof of that.   (From $AF \parallel GH$ I can get to $\angle AFE = 90$, or vice versa, but how do we get to one of these without assuming the other.)

Can we prove $\angle HAF = \angle AFG$ due to the inherent symmetry of the regular octagon, without assuming $AFGH$ is a trapezoid?  It seems "obvious" but that is the step I am missing.  If we can prove that, then we prove that $AFGH$ really is a trapezoid, and the rest follows from there.[/quote]

Sorry. :( I'll try again. :)

Well, $AF$ has to be parallel to $GH$, because $\angle H=\angle G$. If $GH$ isn't parallel to $AF$, then $\angle H \neq \angle G$, but we know that $\angle H=\angle G$, so this can't be true.

Does that work?

[quote="chess64"]
Well, $AF$ has to be parallel to $GH$, because $\angle H=\angle G$. [/quote]

$\angle H=\angle G$ alone doesn't prove that $AF \parallel GH$; you also need to use the fact that the lengths of the sides $FG$ and $AH$ are the same.

I can "see" that with the angles and side lengths being the same, the top and bottom must be parallel, but I'm still not convinced that we've [i]proven [/i]it.

I think to prove it what we need to do is connect $AG$ and $FH$ to create congruent isosceles triangles $\triangle AHG$ and $\triangle FGH$.  (They are congruent because they have equal legs and vertex angles.)  Since they are congruent, $AG = FH$.  Also because they are congruent, we have $\angle HGA = \angle GHF$.  Using this and the fact that $\angle FGH = \angle AHG$ we know that $\angle FGA = \angle AHF$.  Then we have that $\triangle FGA$ is congruent to $\triangle AHF$ (using side-angle-side) which means that $\angle HFA = $\angle GAF$.  Combine that with the fact that we know $\angle GFH = \angle HAG$ and we have the fact finally, that $\angle GFA = \angle HAF$
And from there we know that $FGHA$ is a trapezoid, and we can proceed with either of the original proofs given in this thread.  :D

Now that I've convinced myself that it is indeed provable, I'd sure like to be convinced that my method was overkill. ;)

Thanks.

For those Mathletes who have taken Geometry...

Circumscribe a circle around the regular hexagon.  We have eight equally spaced points, so the central angle between adjacent vertices is $360/8=45$ degrees.  The central angle DOF is 90 degrees.

Any inscribed angle is half of the corresponding central angle.  Thus, $\angle DAF=45^o$

[quote="mathmom"]I'd sure like to be convinced that my method was overkill. ;)[/quote]
Can we make a less formal symmetry argument?

Rotate the figure 22.5 degrees counterclockwise from the illustration shown above.  The figure is symmetric about a vertical line through point O.  The perpendicular bisectors of GH, FA, EB, and DC are all vertical and they all pass through point O.  Thus $GH || FA || EB || DC$.

[quote="rcv"]Any inscribed angle is half of the corresponding central angle.  Thus, $\angle DAF=45^o$[/quote]

That's a little gem that seems worth remembering.

Circumscribing a circle definitely makes the problem easier, if you know either that theorem, or the one about the angle inscribed by a diameter being a right angle, which is of course just a special case of that one.

Using the latter, connect $A$ to $E$ and you have $\angle AFE$ is inscribed by a semicircle, and is thus a right angle.  Then you have two right angles and a 135 degree angle in quadrilateral $AFED$, leaving 45 degrees for $\angle FAD$ $QED$ :)

[quote="rcv"]
Can we make a less formal symmetry argument? [/quote]

I don't know, can we?  :P

[quote]
Rotate the figure 22.5 degrees counterclockwise from the illustration shown above.  The figure is symmetric about a vertical line through point O.  The perpendicular bisectors of GH, FA, EB, and DC are all vertical and they all pass through point O.  Thus $GH || FA || EB || DC$.[/quote]

Hmm, I think I find that convincing. ;)

I think I like the "inscribe in a circle" method best, though.  I will add the relevant theorem to the list of "geometry theorems useful for mathletes" that I am trying to assemble ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33791]see my other thread[/url])

Thanks rcv and chess64.


Two circles $C_{1}$ and $C_{2}$ intersect at points $P$ and $Q$. Their common tangent, closer to $P$ than to $Q$, touches $C_{1}$ at $A$ and $C_{2}$ at $B$. The tangents to $C_{1}$ and $C_{2}$ at $P$ meet the other circle at points $E \not = P$ and $F \not = P$ , respectively. Let $H$ and $K$ be the points on the rays $AF$ and $BE$ respectively such that $AH = AP$ and $BK = BP$ . Prove that $A,H,Q,K,B$ lie on a circle.

let $L$ be the intersection between $BP$ and $AF$. i claim that $L\equiv H$, and $L$ lies on the circumcircle of $\triangle ABQ$.
firstly, $\angle FPQ=\angle LBQ$, but $\angle FPQ=\angle LAQ$, thus, $\angle LBQ=\angle LAQ$, so $ABQL$ is cyclic.
now, note that $\angle BAP=\angle AQP$, thus, $\angle APL=\angle PAB+\angle PBA=\angle BQP+\angle PQA=\angle AQB$, but $\angle AQB=\angle ALP$, so $\angle APL=\angle ALP$, thus $AL=AP$, so $H\equiv P$.
thus $H$ lies on the circumcircle of $\triangle ABQ$. in the same way $K$ will lie on that circumcircle, so the problem is proved.

This is just essentially [url=https://artofproblemsolving.com/community/c6h79788p456609]APMO 1999 P3[/url]

[hide=Solution]Essentially we want to prove $\odot(AHBQ)$ cyclic.Since $\angle AQB=\angle AQP+\angle BQP=\angle PAB+\angle PBA=\angle APL=\angle ALB$ so we are done.$\blacksquare$[/hide]

Here is my solution for this problem
[b]Solution[/b]
We have: $(PB; PH) \equiv (PB; PA) + (PA; PH) \equiv (AB; AP) + (BP; BA) + \dfrac{\pi}{2} - \dfrac{(AF; AP)}{2} \equiv (QA; QP) + (QP; QB)$ $+ \dfrac{\pi}{2} - \dfrac{(PF; PE)}{2} \equiv (QA; QB) + \dfrac{\pi}{2} - \dfrac{(PF; PQ) + (PQ; PE)}{2} \equiv (QA; QB)+ \dfrac{\pi}{2} - \dfrac{(BP; BQ) + (AQ; AP)}{2}$ $\equiv (QA; QB) + \dfrac{(QB; QA) + (AP; AB) + (BA; BP)}{2} \equiv (QA; QB) + \dfrac{(QB; QA) + (PA; PB)}{2}$ $\equiv (QA; QB) + (QB; QA) \equiv 0$ (mod $\pi$)
Then: $B$, $H$, $P$ are collinear
Similarly: $A$, $K$, $P$ are collinear
So: $(HB; HA) \equiv (PA; PH) \equiv (PK; PA) \equiv (KB; KA) \equiv (QB; QA)$ (mod $\pi$) or $A$, $H$, $P$, $K$, $B$ lie on a circle

[hide = Proof] Straightforward angle chasing.

Let $H' = BP \cap AF$. It suffices to show $H = H'$.

$\bold{Claim:}$ $BQH'A$ is cylic.

$\bold{Proof:}$ We have the following angles:

\begin{align*}
\angle FPQ &= \angle PBQ \\
&= \angle H'BQ \\
&= \angle FAQ \\
&= \angle H'AQ.
\end{align*}

Thus we have $H'BAQ$ is a cyclic quadrilateral. Now we claim that $\angle APH' = \angle AH'P \implies H = H'$.
Now we also have:
\begin{align*}
\angle AH'P &= \angle AQB \\
&= \angle AQP + \angle PQB \\
&= \angle BAP + \angle PBA \\
&= \angle APH'.
\end{align*}

Thus we have $H = H'$. Now we have that $B, P, H$ and $K, P, A$ are separately collinear. Thus we have that as $\angle PKB = \angle AKB = \angle APH = \angle AHB$, we have $A, H, K, B$ are concyclic. Thus we have $A, H, Q, K, B$ are concyclic, as desired. $\blacksquare$ [/hide]


you can set up 4 equations based on the given inequality and then graph them and shade in the area they bind in

[quote="lotrgreengrapes7926"]There was a problem similar to this in our Math League.

The vertices of a square are at $(1,0),\ (2,1),\ (3,0),\ (2,-1)$, with its interior shaded. Find [b]one[/b] inequality to describe the graph.[/quote]

Just wondering, but was the answer:
[hide] $|x-2|+|y| \le 1$ if the line connecting the vertices is solid; if the line is dashed, then $|x-2|+|y| < 1$  ?[/hide]

[quote="deimos"][quote="guile"]I have been ignoring this type of problem but since i see it often, i would like to know how it is solved.

Determine the are of the region in the Cartesian plane satisfying the equation $|x|+|y| < 20$[/quote]

well, draw a cartesian plane. mark the dots $(20,0), (0,20), (-20,0)$ and $(0,-20)$. conect that dots. every dot $T(x,y)$ in that square (not including the lines) is the solution.

later, if i found some time, i'll make it on a paper and upload it on the site[/quote]

Is there an algebraic solution to this?

Yes, that's correct nrz. I just found the equation for the square centered at the origin and shifted it 2 units right. :D

[quote="guile"]
Is there an algebraic solution to this?[/quote]

i won't recomand you to do it on that way 'cause there are so many various cases (when is $x>0$ and $y<0$; ...). in this case graph is easier


calculus
integration
logarithms
algebra
partial fractions
calculus computations

Let $a,\ b$ be constant numbers such that $a^{2}\geq b.$
Find the following definite integrals.

(1) $I=\int \frac{dx}{x^{2}+2ax+b}$

(2) $J=\int \frac{dx}{(x^{2}+2ax+b)^{2}}$

Your answer is almost correct, Please correct $\ln A\rightarrow \ln\ |A|.$ :)


Hmm.. I'm not sure which forum to ask this but how do you put the circle degree symbol in Latex? 45 degrees, etc.?

$4^\circ$
cool! I never knew you could do that!

[quote="jeez123"]yeah, THANKS.

How do you put the fractions bigger? When I do fractions, the num and denom skrinked to 1/2 size. My font is 12, so the num would be like 6 and the denom 6?

And also, slashing the numbers? (r top, r bottom, canceling out slash)[/quote]
I think doing /displaystyle before the fraction makes the fraction a bit bigger.

but how do you post a vertical fraction(as in num on top and denom on bottom) instead of horizontal(num on left and denom on right(as in $4/3$))?

Random $\LaTeX$ tip: $\varnothing$ looks better than $\emptyset$.

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php[/url]
Note that on AoPS, if you have a sum or product (such as $\sum_{k=1}^{\infty}\frac{k}{2^k}$), things are automatically put in display mode.  This also happens if you simply have a complicated fraction (such as $\frac{k}{2^k}$).  I'm not sure exactly how it works, though.  Also, note that you must put \begin{eqnarray*}...\end{eqnarray*} in dollar signs.

[quote="K81o7"]\frac{4}{3}
$\frac{4}{3}$
\[ \frac{4}{3}  \]
Double dollar signs make the 4/3 bigger (as well as centering it and giving it its own line).  Using double dollar signs is an equivalent to displaystyle i think, but also centers.[/quote]

I went through so many \displaystyle in my document! Thanks for the advice.

Back to my other question: How do you slash out (canceling things) in LATEX?


Find all functions $ f(x)$ defined on reals except $ \{0,1\}$ and taking real values such that:

$ f(x) \plus{} f(\frac {1}{1 \minus{} x}) \equal{} \frac {1}{x}$

holds for all real numbers $ x$ except $ \{0,1\}$.

[hide="hint"]Let $ A \equal{} \{ x\in \mathbb{R}: x < 0\}$, $ B \equal{} \{x\in \mathbb{R}: 0 < x < 1\}$, $ C \equal{} \{x\in \mathbb{R}: x > 1\}$, and let $ g(x) \equal{} \frac {1}{1 \minus{} x}$.  Note that $ f(A) \equal{} B$, $ f(B) \equal{} C$, $ f(C) \equal{} A$, and $ g(g(g(x))) \equal{} x$.[/hide]

put$ x,\frac {1}{1 \minus{} x},\frac {x \minus{} 1}{x}$ in equation:
$ f(x) \plus{} f(\frac {1}{1 \minus{} x}) \equal{} \frac {1}{x}$
$ f(\frac {1}{1 \minus{} x}) \plus{} f(\frac {x \minus{} 1}{x}) \equal{} 1 \minus{} x$
$ f(\frac {x \minus{} 1}{x}) \plus{} f(x) \equal{} \frac {x }{x\minus{}1}$
solving these 3 equation we have:$ f(x) \equal{} \frac {x^3 \minus{} x^2 \plus{} 2x \minus{} 1}{2x(x \minus{} 1)}$

[quote="sumita"]put$ x,\frac {1}{1 \minus{} x},\frac {x \minus{} 1}{x}$ in equation:
$ f(x) \plus{} f(\frac {1}{1 \minus{} x}) \equal{} \frac {1}{x}$
$ f(\frac {1}{1 \minus{} x}) \plus{} f(\frac {x \minus{} 1}{x}) \equal{} 1 \minus{} x$
$ f(\frac {x \minus{} 1}{x}) \plus{} f(x) \equal{} \frac {x }{x \minus{} 1}$
solving these 3 equation we have:$ f(x) \equal{} \frac {x^3 \minus{} x^2 \plus{} 2x \minus{} 1}{2x(x \minus{} 1)}$[/quote]

You should make a confirmation the answer satisfies the given functional equation. :wink:

[quote="kunny"]
You should make a confirmation the answer satisfies the given functional equation. :wink:[/quote]

You are right kunny. :)
It is often forgotten.


Prove that for postive real numbers a,b,c,d,

$ \frac{a}{b+2c+3d} + \frac{b}{c+2d+3a} + \frac{c}{d+2a+3b} + \frac{d}{a+2b+3c} \geq \frac{2}{3}. $

Well, i couldn't find another solution!
....
only the solution in Kim L.' book...

could you help me?

I do not know that book, but the solution using Cauchy ( call it as you wantn
, Andreescu lemma, I prefer Cauchy) is really straightforward.

Just in case Carlos wants some details :

By Cauchy-Schwarz :  :Sigma:  ai/bi * :Sigma:  ai*bi >= (:Sigma:  ai)

Then apply this with a1=a, ... and b1=b+2c+3d, ...

End using :Sigma: a*b <= :Sigma: a to prove :Sigma: (symetric) a*b <= 3/8 (:Sigma: a)

Thanks a lot!!!

I TRIED........THAT'S RIGHT!!

----------------------------------------
Where can i find information about "elementary inequalities" like these ??

for example, any web site where....i can download papers like *.pdf or *.ps...

thanks  for all!!! :)
-----------------------------------------


If we're in 7th grade, are we allowed to grade papers at the 6th grade competition if the coach of the 6 grade team lets us?

What is the sixth grade competition? Isn't everyone in the same (chapter) competition at the same time?

Yes. There's no different tests or times for different grades in MC.

Where I live, there is an unofficial competition for 6th grade only at the beginning of March which is run by the same people who run the official competition. :)

sorry... but all us NORMAL ppl dont have that... :P jk  thats cool... having a 6th grade comp, but who DOES grade our tests anyways

In my chapter, 6th graders can compete can get awards and stuff, though either sixth graders aren't skilled enough or they're not allowed to do the countdown round or go to state.

whoa... actually on our team...(even they're on the B team) we have two REALLY SMART GENIUS 6th graders...they got great ranks on individual... but one is EXTREMELY annoying which is a problem when u get 6th graders who are smarter than some of the A team and get cocky
but they are REALLLLLLLY smart

In my area 6th graders are in elementary schools, not middle/jr. high, so you can't really have them on the same team as 7th and 8th graders. And it's in two other chapters too, though all three of the mentioned chapters have their competitions in the same place.

at our school... 6th graders on
mahcounts was unheard of... until I showed up
so I was first 6th grader ever to even go to chapter
but now our coach looks at a couple of sixth graders... generally they are (mostly) good at basic arithmetic, but terrible at abstract math (algebraic reasoning)... so they have not yet made it on the team
as to who grades our tests: volunteers, mostly professional engineers (at least in Texas, the texassociety of professional engineers coordinate all chapter and state events... so they get to grade them too)


Dose there exist a positive integer a such that in the sequence of numbers $(x_{n})$ defined by $x_{n}=x^{3}+n^{3}$ for all $n=1;2;3;...$ every two consecutive terms are coprime integers?

This is a problem discussing in "Youghth and mathematic" magazine in VietNam. Don't discuss any more, please Sup lock it!

I think x=a.
[quote="chien than"]Dose there exist a positive integer a such that in the sequence of numbers $(x_{n})$ defined by $x_{n}=a^{3}+n^{3}$ for all $n=1;2;3;...$ every two consecutive terms are coprime integers?[/quote]
$(x_{n},x_{n+1})=(x_{n},x_{n+1}-x_{n})=(a^{3}+n^{3},3n^{2}+3n+1)$.
If $7|not a$, then $a^{3}=1(mod 7)$ or $a^{3}=-1(mod 7)$. Therefore $7|(x_{n},x_{n+1})$, when $a^{3}=1(mod 7),n=5(mod 7)$ or $a^{3}=-1(mod 7),n=1(mod 7)$


algebra
polynomial
trigonometry
geometry
3D geometry
blogs
complex numbers

[quote="vijaymenon"]Find the roots of the equation  $ z^6 \equal{} 64$ where the real part is positive .[/quote]

The six complex roots of $ z^6\equal{}2^6$ are $ 2e^{ik\frac{\pi}{3}}$ with $ k\in\{0,1,2,3,4,5\}$

The constraint "positive real part" reduces this set to $ k\in\{0,1,5\}$

And this is surely not an olympiad problem, IMHO.

But how to do u get that that is how do u get $ 2e^{ik\frac{\pi}{3}}$. can u please explain

[quote="vijaymenon"]But how to do u get that that is how do u get $ 2e^{ik\frac {\pi}{3}}$. can u please explain[/quote]

Huh... The $ n$ $ n^{th}$ roots of $ \rho e^{i\theta}$ ($ \rho>0$) are $ \rho^{\frac 1n}e^{i\frac{\theta\plus{}2k\pi}{n}}$ with $ k\in\{0,1,...,n\minus{}1\}$

According to me, this is a part of the basic knowledge for students involved in olympiad contests.
IMHO.

Any proof for this ?? i can't still get it

Check that they work; then recall that a degree $ n$ polynomial over a field (reals, complex numbers, for example) has at most $ n$ roots, so you must have found them all.

Eh, it's a reasonable guess, and this form gets used often enough that you use
it without giving it a second thought.

Idea being that the $ n$th roots of $ 1$ are $ \exp(\frac{2 \pi i k}{n})$. (Intuition: looking at the complex plane, these are the $ n$ points equally distributed around the unit circle.) You can get everything else from there. This is a fact you should learn and become comfortable with.

[quote="vijaymenon"]My question is how do u find those roots[/quote]

Do you know the de Moivre formula: $ (\cos \theta \plus{} i\sin\theta)^n \equal{} \cos(n\theta) \plus{} i\sin(n\theta)$ ?
Use it to solve $ z^n \equal{} 1$ .

Although that is somewhat redundant since it brings in $ \cos$ and $ \sin$, which are unnecessary here; all you need is that $ 2 \pi i$ is the smallest value of $ z$ such that $ e^z \equal{} 1$ (depending
on your source, this is an excellent definition of $ \pi$ up to sign); and of course the additional fact $ (e^x)^y \equal{} e^{xy}$.

If you don't understand DeMoivre's formula, then here is a more tedious factoring technique.

$ z^6 \equal{} 64 \Longleftrightarrow z^6\minus{}64 \equal{} 0$

Use the difference of squares formula: $ A^2\minus{}B^2 \equal{} (A\minus{}B)(A\plus{}B)$. Here $ A\equal{}z^3$ and $ B\equal{}8$:
$ z^6\minus{}64 \equal{} (z^3\minus{}8)(z^3\plus{}8)$

Use the difference of cubes formula: $ A^3\minus{}B^3 \equal{} (A\minus{}B)(A^2\plus{}AB\plus{}B^2)$. Here $ A\equal{}z$ and $ B\equal{}2$:
$ z^3\minus{}8 \equal{} (z\minus{}2)(z^2\plus{}2z\plus{}4)$

Use the sum of cubes formula: $ A^3\plus{}B^3 \equal{} (A\plus{}B)(A^2\minus{}AB\plus{}B^2)$. Here $ A\equal{}z$ and $ B\equal{}2$:
$ z^3\plus{}8 \equal{} (z\plus{}2)(z^2\minus{}2z\plus{}4)$

Therefore:
$ z^6\minus{}64 \equal{} (z^3\minus{}8)(z^3\plus{}8) \equal{} (z\minus{}2)(z^2\plus{}2z\plus{}4)(z\plus{}2)(z^2\minus{}2z\plus{}4) \equal{} 0$

Setting each factor equal to $ 0$ and solving for $ z$ yields the following solutions:
$ z \equal{} 2$
$ z \equal{} \minus{}1 \pm i\sqrt{3}$
$ z \equal{} \minus{}2$
$ z \equal{} 1 \pm i\sqrt{3}$

The solutions with a positive real part are: $ z \equal{} 2, 1 \pm i\sqrt{3}$.

This problem should probably be moved to the high school section.

The math behind Euler's and DeMoivre's is very beautiful. [url=http://fermatslasttheorem.blogspot.com/2006/02/eulers-formula.html]This[/url] blog post is very informative on that topic.

My favourite derivation is as follows:

$ f(\theta) \equal{} \cos(\theta) \plus{} i\sin(\theta) \implies f'(\theta) \equal{} if(\theta) \implies \frac {df}{d\theta} \equal{} if(\theta)$

Then moving the differentials (though can be justified :P) we get $ \frac {df}{f(\theta)} \equal{} id\theta$

This is known as the world's simplest differential equation with solution $ f(\theta) \equal{} e^{i\theta}$.

That is, Euler's Formula =) Then using (as already posted) $ (e^x)^y \equal{} e^{xy}$ you can see how DeMoivre's formula comes into play. However I just got that from this topic, as I am familiar with its proof through induction.

Now to find the roots suppose for an easy example we have (z-complex) $ z^n \equal{} a$ and wish to find all values of $ z$. $ a$ should be rewritten in complex-polar form.

Let $ z^n \equal{} r^n[\cos(\theta \plus{} 2k\pi) \plus{} i\sin(\theta \plus{} 2k\pi)]^n$ $ \implies$ $ z \equal{} r[\cos(\frac {\theta \plus{} 2k\pi}{n}) \plus{} i\sin(\frac {\theta \plus{} 2k\pi}{n})]$

Now the magic of this really comes from congruency. This is valid for $ \{0,1,...,k \minus{} 1\}$. Prove that you have congruency classes for each $ (a\mod k)$, and each class represents one of the n solutions to the above equation.
That is if $ i \equiv j$ then $ z(i) \equal{} z(j)$. Here the $ z(k) \equal{} r[\cos(\frac {\theta \plus{} 2k\pi}{n}) \plus{} i\sin(\frac {\theta \plus{} 2k\pi}{n})]$
Also to be clear, show that if $ i \ncong j$ then they produce different values. Thus you get the $ n \minus{}$roots.

Hopefully that made it really clear. Night! 4am  :rotfl:

Plug in values for $ k \equal{} 0,1,2,3,4,5$ and see if you get what Xantos C. Guin got.

[quote="Xantos C. Guin"] This problem should probably be moved to the high school section.[/quote]

Kind of harsh, eh?

And what an heretic thought! Olympiad section is the place where everyone can post problems difficult for them, whatever level is the problem.

Dont be afraid, vijaymenon, by those rude opinions, and dont hesitate to post your problems here.

[quote="pco"]
Kind of harsh, eh?

And what an heretic thought! Olympiad section is the place where everyone can post problems difficult for them, whatever level is the problem.

Dont be afraid, vijaymenon, by those rude opinions, and dont hesitate to post your problems here.
[/quote]

You need to lay off the sarcasm. There are many ways to say the same thing; it's fine
to suggest (note "probably") that a problem belongs somewhere else, its far different to [i]laugh[/i] at the original poster. Like

[quote="pco"]
According to me, this is a part of the basic knowledge for students involved in olympiad contests.
IMHO.
[/quote]

Which is tactless, and wasn't really necessary. As a filter, pretend you're saying that to
their face, before you post it.

If you can't understand the difference, I'm sorry for your social skills.

[quote="Nukular"] You need to lay off the sarcasm. [/quote]

Oh, I'm very sorry but you misunderstood. There was no sarcasm at all.

Please accept all my apologizes.


summer program
Mathcamp
linear algebra
matrix
search

Ah so it's about a week (1.5 for me) until MathCamp starts. Hope everyone's excited.

For everyone with digital cameras, I thought it would be a good idea to do a cumulative album after camp is over. The obvious reasons are that generally the photographers don't get to be in any picturs, so this will be a big advantage for them. We'll definitely have many times the amount of pictures that normal albums would. And also, since no one will hang out with everyone, with lots of different pictures from different people we have optimal photo opportunities for everyone. :)

Haha so if you've got a digi, please bring it to camp. And don't forget the spare batteries, memory cards, lenses, etc. :) Oh and I guess you can post here if you plan on helping, just so I can get an idea of how many people might be taking pictures (i.e. how many cameras to dodge).

[quote="MithsApprentice"](i.e. how many cameras to dodge).[/quote]
Ditto.

I'll be helping out with my three-and-a-half-year-old-but-still-functioning digital camera. Just in case I take more than 400 pictures, I'll be bringing my connection cord as well ;)

Can't wait!

Sounds like a good idea.  Is anybody watching this anymore (yes there do exist STUPID questions)?  I bet a lot of people would like the talent show.  I wonder if the mathcamp site would host some such thing.  BTW, who is ztbb?  I need the Matrix parody, or at least the script.  Alex Schwender wrote it right?

ztbb is Dave (Savitt), the guy on staff.

The Matrix script is at http://web.mit.edu/~alexrs/www/matrix.pdf

About a cumulative album:
Interesting, it seems we came up with the same idea. :)

Anyway...

As you probably know if you've read the messages from the mailing list (so this message is really for those who aren't on the list for whatever reason),  for the past couple of weeks Andrew and I (I do the code and Andrew does the webdesign) have been working on a password-protected photo album for Mathcamp. It's been taking a while because
1) this is the first project of this scale I've done and my second time working with MySQL
2) I've got a huge bunch of features I have planned to make this thing, for example usuable for multiple years using the same database, etc. and let you search for photos by the main subject (as in person) of the photo, etc.

But it is nearly done. The database design is finalised, the login system is finished, the admin cp and user cp is finished, and, the thing that took me ages to fix due to random annoyances (chmod'ing the wrong folder, etc.): the uploading and thumbnail generating page are finished! So all that's left are a few moderator cp pages that are just copies of my old pages, and the search and browsing pages, which should be fairly simple.

Well, and then get everyone's accounts set up, but bleh.


[quote="loup blanc"]Necessarily $ rank(ABC)\leq2$; in your $ ABC$ the blockmatrix of dimension $ 3$ which is up and left is invertible. You are not carefully.[/quote]
$ (ABC)^{2} \equal{} (19^{2} \plus{} 5^{2})I_{4}$
 then $ (ABC)^{4} \equal{} (19^{2} \plus{} 5^{2})^{2}(ABC)^{2}$ from here
$ (BCA)^{5} \equal{} BC(ABC)^{4}A \equal{} (19^{2} \plus{} 5^{2})^{2}(BCA)^{3},$ and $ rank(BCA) \equal{} 4$
and $ (BCA)^{2} \equal{} (19^{2} \plus{} 5^{2})^{2}.I_{4}$........

[quote="loup blanc"]Hi studen,
I read your post "Nice  inequality" which is locked now (not by me); what I may write ? Naughty boy ?[/quote]
Yes !!! " $ det(19A^{2}\plus{}5AB\plus{}1890B^{2}) \geq 0$
[b]"..Do math , love math to love native place , love country ... Who do You think ?"  [/b]
 [b]...Name show to call ..." [/b]:P
[b]...iron-steel technique  don't very math ....[/b]

There are no such matrices. This is not a valid question.


algorithm
number theory
greatest common divisor
least common multiple
Euclidean algorithm

[hide]
1. I used the euclidean algorithm for gcd, which is 180. Then 1440*900/180 = 7200. 7200 - 180 = 7020.
2. $42 - 3 = 39 = 100111_2$
3. 11, since the number would be in the form, abba, using the divisible by 11 trick you get a+b-a-b = 0. Which means its divisible by 11.

Took me about 2 minutes.
[/hide]

[hide]about 3 mins 20 secs

1) 900 = 2 * 2 * 3 *3 * 5 * 5

1400 = 2 * 2 * 2 * 2 * 3 *3 * 5

GCF = 2^2 * 3^2 * 5^1 = 180

LCM = 2^4 * 3^2 * 5^2 = 3600

Difference = 3420

2) 101010 - 11 = 100101

3) 4 digit palindromes can be separated by multiples of 110 (i.e. 2442 --> 2552). Thus, the prime number in question must divide 110. 110 = 2 * 5 * 11, and not all 4 digit palindromes are divisible by 5 or are even. Hence, the solution must be 11.[/hide]

EDIT: wow... and i made simple arithmetic errors on 2 of them...  :blush:

[hide]0 min 55 sec (I knew some of these problems)

1. LCM (1440, 900)=7200; GCF (1440, 720)=180. 7200-180=7020.

2. $101010_2-11_2=100111_2$ (since $3=11_2$).

3. All 4-digit palindromes must be divisible by [color=red]11[/color] by the divisibility rule of 11.   [/hide]

[quote="lkryptonicl"]Three relatively easy problems with time limit of 3 minutes. Be honest post solutions [hide]and time.[/hide]
1. What is the positive difference between the least common multiple and the  greatest common factor of 1440 and 900?

2. Write in base 2, the result of  $(101010)_{2}-(3)_{10}$.

3. What prime number is a divisor of every four-digit palindrome?[/quote]
[hide]
1)
1440=144*2*5=2^2*72*5=2^3*36*5=2^4*18*5=2^5*3^2*5
900=2^2*3^2*5^2
LCM=2^5*3^2*5^2=7200
GCF=2^2*3^2*5=4*45=180
Difference=7020.

2)101010_2-3_10=2+8+32-3=39=100111_2

3)101.

4 min. I stink at calculation.

[/hide]

[quote="lkryptonicl"]Three relatively easy problems with time limit of 3 minutes. Be honest post solutions and time.

1. What is the positive difference between the least common multiple and the  greatest common factor of 1440 and 900?

2. Write in base 2, the result of  $(101010)_{2}-(3)_{10}$.

3. What prime number is a divisor of every four-digit palindrome?[/quote]

[hide="answers"]
1. 7020? i'm not very good at subtracting in my head
2. 100111 (in base 2)? 42 - 3 = 39, that turns into 100111
3. 11? just a random guess by looking at 1001 and 1111

around 28 seconds
[/hide]

i don't have a solution nobody else used, but i had them all in 1:56


For all you chess players who play online, what is your favourite chess server/why?

I like kurnik.org - not a very mainstream site, but very nice, easy, no-nonsense layout with reliable servers (and many other games too).

(lately, my computer has been acting up so I've been having some problems with the javascripts used to run the site)

I usually play at yahoo because thats where most of the people play at. The servers arn't too bad either.  Whats your rating? Mine's around 1650.

Thanks a lot.

My rating's very low... around 1350... =\
(I'm getting better though  :) used to be below 1000)

It depends what you're looking for.  Yahoo is easy for a quick game, but if you want to play consistently it is not so good as there are some people who just sit there if they are losing.  Free servers like FICS (at http://www.freechess.org : "Free Internet Chess Server") are better for the long term - but not worth setting up an account just to play one quick game.

Yahoo is what I use.  However, I hate that there is no friends list, like there is on XBOX Live.  If anyone wants to play me on Yahoo, IM me and we'll play right then, rahter than setting something up beforehand.  My Yahoo ID is [b]seans41k[/b], by the way.


(i) Given real numbers $ a, b, c$, with $ a\plus{}b\plus{}c\equal{}0$, prove that

$ a^{3}\plus{}b^{3}\plus{}c^{3} >0$ if and only if $ a^{5}\plus{}b^{5}\plus{}c^{5} >0$.

(ii) Given real numbers $ a, b, c, d$, with $ a\plus{}b\plus{}c\plus{}d\equal{}0$, prove that

$ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3} >0$ if and only if $ a^{5}\plus{}b^{5}\plus{}c^{5}\plus{}d^{5}>0$.


plz do Assigment 2 as soon as possible thank you........

when u have answer plz show ur setps..and if u can plz explain it.....thnx very much

[img]http://i74.photobucket.com/albums/i249/wenlong104/scan0002.jpg[/img]

this isnt a forum for getting homework answers.  if you dont understand a concept or need help on a problem or two, thats ok.  but no one here will be doing this for you.

and no chatspeak please.

Lol, do you really expect us to do your homework for you? We have enough of that as it is...

Locked.  :mad:


[quote="GameBot"]How many terms of the arithmetic sequence 88, 85, 82, ...
appear before the number -17 appears?[/quote]

Just find the number of terms in the sequence 88, 85, 82, ..., -14.  If you apply +, -, *, or / to all the terms, the number of terms is the same, and it's still a sequence.  So if you mess with it until you get 1, 2, 3, ..., n, the sequence counts itself.  First, add 17 to all the terms.

105, 102, 99, ..., 3

Now, divide by 3.

35, 34, 33, ..., 1

It counted itself! There are 35 terms.

Or to find the difference betweent the largest and smallest you could just take 88 - (-17) = 105.  Then divide by the distance(3) to get 105/3 = 35

Uh... you are both wrong...

The sequence is

$ 88, 85, 82, \cdots, \minus{}17$. So, adding $ 17$ gives us

$ 105, 102, 99, \cdots 0$.

Dividing by $ 3$ gives us

$ 35, 34, 33, 32, \cdots, 0$.

So there are $ 35\minus{}0\plus{}1\equal{}36$ terms.

[quote="BOGTRO"]Uh... you are both wrong...

The sequence is

$ 88, 85, 82, \cdots, \minus{} 17$. So, adding $ 17$ gives us

$ 105, 102, 99, \cdots 0$.

Dividing by $ 3$ gives us

$ 35, 34, 33, 32, \cdots, 0$.

So there are $ 35 \minus{} 0 \plus{} 1 \equal{} 36$ terms.[/quote]

the problem says 'before -17 appears'.

so it's 36-1.

the answer is 35.

the way i did it was just using the equation

(85--5)/3 =35

so the answer $\boxed{35}$

P.S. you can PM me for more INFO and ill be happy to tell you how i got that ! :)

 :wallbash_red:  :wallbash_red:  :wallbash_red:
:

Hello,
It says before -17 appears.

So,
(85 5) devided by 3= 35
Or 36-1= 35
Which means the answer would be [size=150]35[/size]

How do you find the distance in the problem??? What does the distance mean???

The distance is the largest subtracted from what you need.
$88-(-17)$ which equals to $105$.
Since you're counting by 3, you divide $105$ by $3$ to get $35$. But because you didn't count 1 of them, your answer is $35+1=36$


trigonometry
college
calculus
calculus computations

Prove that\[ \begin{gathered} \left( {\cos \theta } \right)^p \leqslant \cos p\theta \hfill \\
for0 \leqslant \theta \leqslant \frac {\pi } {2} \hfill \\
and0 < p < 1 \hfill \\
\end{gathered}
\]

You can look for this problem is Berkeley problems in mathematics


This is a more comprehensive version of the USAMO index poll

I can't see an index going above 240. It doesn't affect me personally but it just seems so harsh haha.

241 here... but I'm a soph so it doesn't matter.

I'm a freshman.  I scored a 222 on the index; how's that?  (This is my first year, so excuse my ignorance)

222 is quite admirable for a 1st-year. Congrats.

254.5

lol, SOMEHOW, the distrubution of votes is like a bell curve.

Excuse me, what is USAMO index ?
   By the way, may I ask you how they use the AMC and the AIME score to qualify for the USAMO ? Thank you in advance.

USAMO selection has been one of those interesting topics because it changes every year. I anticipate we'll be discussing it more thoroughly later on (especially after the second AIME). For now, your USAMO index is AMC + 10(AIME) out of a possible 300. Last year's cutoff was 210, which is in range. However, this year many people (as you can see) anticipate a cutoff around 230.

[quote]257. I wonder how many 300s this year...[/quote]
No more than 6, I think.  Only 6 got 150 on AMC.[/quote]

[quote="ednerd"]257.  I wonder how many 300s this year...[/quote]

None probably.

Zach and Adam came soooo close. 290. Wow...

I tied Zach? Wow, yay! Congratulations, Zach, which one did you miss?

-Adam Hesterberg

He missed 13...or at least I think that's what he told me O.O


[quote=Twins]Find the minimal $k$ fits that
For $a,b,c>0$, $\frac{a^{2}}{\sqrt{a^{4}+3b^{4}+3c^{4}}}+\frac{k}{a^{3}}(\frac{c^{4}}{b}+\frac{b^{4}}{c})\geq \frac{2\sqrt{2}}{3}$[/quote]
Any one?

Just curious, where did this problem come from?

Also, a ridiculously old post lol I probably won't get an answer.

[quote=sqing][hide=Thanks.][quote=dragonheart6][quote=Twins]Find the minimal $k$ fits that
For $a,b,c>0$, $\frac{a^{2}}{\sqrt{a^{4}+3b^{4}+3c^{4}}}+\frac{k}{a^{3}}(\frac{c^{4}}{b}+\frac{b^{4}}{c})\geq \frac{2\sqrt{2}}{3}$[/quote]

WLOG, assume $a = 1$.

By the power mean inequality, we have
$$\frac{b^4}{c} + \frac{c^4}{b} = \frac{b^5 + c^5}{bc} \ge
\frac{2(\frac{b^4 + c^4}{2})^{5/4}}{\sqrt{\frac{b^4 + c^4}{2}}} = 2\left(\frac{b^4 + c^4}{2}\right)^{3/4}.$$
Then, take derivative.[/quote]
[/hide]
$$k_{min}=?$$[/quote]

$$k_{\min} = \frac{1}{6}2^{1/4}3^{3/4}$$


Let n be a natural number.[tex]Sn=\sum_{k=1}^{n} [\frac{n}{k}]-[\sqrt{n}][/tex].Prove that Sn is even


ok, so facing the facts, I have come the realise that in the current development way MathLinks needs a dedicated server to work perfectly. unfortunately I do not posses the resources myself to arange this, so I am asking for any help from you guys.

I am also trying to obtain some sponsorships, but this is quite hard in Romania, with all the sponsorships laws being so not welcoming for the sponsors.

Well I tell about mathlinks everytime I see a mathematicien!

And I think I do know some way to get more money: use Google Ads :) that's very easy and has probably no law problem stuff

And you could ask very fund members of the site to sponsor :D

And I may be able to arrange some hosting for mathlinks if that'd appear necessary. Currently everything here works fine though...


The pairs of values of $ x$ and $ y$ that are the common solutions of the equations $ y\equal{}(x\plus{}1)^2$ and $ xy\plus{}y\equal{}1$ are:

$ \textbf{(A)}\ \text{3 real pairs} \qquad
\textbf{(B)}\ \text{4 real pairs} \qquad
\textbf{(C)}\ \text{4 imaginary pairs} \\
\textbf{(D)}\ \text{2 real and 2 imaginary pairs} \qquad
\textbf{(E)}\ \text{1 real and 2 imaginary pairs}$

hello, inserting the first equation in the second one we get $ x(x\plus{}1)^2\plus{}(x\plus{}1)^2\equal{}1$ simplifying and factoring we get $ x(x^2\plus{}3x\plus{}3)\equal{}0$, from here we have $ x_1\equal{}0,y_1\equal{}1$, $ x_2\equal{}\minus{}\frac{3}{2}\plus{}\frac{\sqrt{3}}{2}i,y_2\equal{}\minus{}\frac{1}{2}\minus{}\frac{\sqrt{3}}{2}i$, $ x_3\equal{}\minus{}\frac{3}{2}\minus{}\frac{\sqrt{3}}{2}i,y_3\equal{}\minus{}\frac{1}{2}\plus{}\frac{\sqrt{3}}{2}i$, so we have one real solution and two imanigary solitions, and the answer is $ \text{E}$.
Sonnhard.

[hide="Alternative Solutions"]No doubt that the equation simplifies to $ (x\plus{}1)^3\equal{}1$. One way of looking at it is to consider the cube roots of unity (1 real and 2 complex roots). Another way is to see that $ x^2\plus{}3x\plus{}3>3\minus{}\frac94\equal{}\frac34>0$.[/hide]


summer program
MathPath
geometry
calculus
number theory

Okay, so getting your thread locked up feels real bad. (see http://www.mathlinks.ro/Forum/viewtopic.php?t=192846 to get what I mean.) So, back to the topic. Who's qualified already, and where are you from? And, please, no more discussion of test questions. Not even the question number. Not even what question you have done, have difficulty with or have not attempted to solve. No more kind souls trying to help others allowed. Understand?

I'm from Singapore, am 13 7/12 years (which kind of means that I might be able to sign up for MP09) and is a math person. Prefers pure maths over applied maths. Prefers algebra and number theory over geometry. I've just sent my application over, although my 2nd Recommendation Form was sent over by the person who filled it up two days ago. I hope that doesn't void my application. If that got snipped out I swear I'll never post again ever.

[quote="ZhangPeijin"]Eh, ok no discussing about the test anymore then I guess.[/quote]

um, you "guess"?

I answered __ questions so far.  I'm pretty sure I'm going, since I went last year and for some reason I feel like going again.

Sorry to revive, but I MADE IT!!

I'm from NJ, am 5 and 7 months old (jk, 11), I love math.

I also like pure math more than applied math. l believe its more fun

Geometry is my favourite subject, and I'm getting better at it.

I love Number Theory, which I'm getting pretty good at.

I sent my applicastion about a week ago, and got an e-mail

telling me that I've been accepted. Adrenaline rush.

Well, if you haven't noticed, there are some other threads on this (although I don't exactly object to reviving my own topic :P )

A recent turn of events might just allow me to join you guys at MP'08. (Applause please.)

Almost forgot about my math preferences.

Favourite subfield: Number Theory (just because I'm good at it)
Likes: Calculus, challenging (well-defined please) questions, etc.
Hates: Geometry (actually, I'm just bad at it)


Compound R-X is a molecule in which X is a good leaving group (and a very poor nucleophile) attached to a primary carbon. When R-X is dissolved in a polar aprotic solvent with equimolar concentrations of KI and KCN the following is observed:

1) If a portion of the reaction mixture is taken after 5 minutes and rapidly evaporated, and the residue is taken up in ether, the only organic product detected is R-I.

2) If a portion of the reaction mixture is taken after 2 hours and subjected to the same procedure, the vast majority of the organic product is R-CN.

Explain these observations.

Hmmm.... I have sources differing on the relative nucleophilicity of $ CN^{ \minus{} }$ and $ I^{ \minus{} }$ but I think if the iodide ion is a stronger nucleophile then quite clearly it should attack the primary carbon attached to $ X$ first and hencing giving an initially high concentration $ R \minus{} I$.

Once $ [I^{ \minus{} }]$ and $ [RX]$ decreases, the rate of this reaction also decrease and once $ [RI]$ is sufficiently high $ S_N2$ reaction occurs between $ CN^{ \minus{} }$ and $ RI$ giving $ RCN$ (NOTE: $ I$ is a very good leaving group). Now once again over time cyanide ion concentration decreases and $ [RI]$ also decrease and $ I^{ \minus{} }$ (a better nucleophile than $ X^{ \minus{} }$ will attack the primary carbon of $ RCN$ but $ \minus{} CN$ is a very poor leaving group (I think, I can't find sources on this) hence this reaction is slow.

So over long period of time we have a major amount of $ RCN$.

I hope this explanation is correct.


algebra
polynomial
calculus
calculus computations

Let $n\in \mathbb{N}$ and $P\in \mathbb{R}[x],deg ~P=n$ such that $P(k)=2^{k}, \forall k=1,2,\dots, n+1$ .Calculate $P(n+2)$


Its probably not original, but unique enough so that few has heard of them. Comes from some sort of scam/free money ad site.

Heres how game #1 goes:
-You and an opponent have 6 stones of magnitude 1, 2, 3, 4, 5, and 6.
-The center also has the same set.
-The game last for 6 rounds with each round, the center randomly selects a stone.
-Both player sees the stone and decide which stone they will pit against each other.
-The player who plays the stone with greater magnitude inherits all the stones. These stones may not be used again but are added up and counted towards your score
-In the events of a tie, both players' stones are put up for grabs in addition to the stones up for grabs last round and the victor is determined by the magnitude of the next set of stones.
-The player with the most points wins.
-A tie of 0-0 happens when both player matches all 6 rounds.

Game #2
-This rock-scissor-paper variation lasts 6 rounds.
-Each round one of the 3 elements is chosen at random and announced and if you play that element, you receive 0 bonus points, otherwise you receive 2 bonus points.
-The victor receives 3 points.
-In case of ties, 3 additional points are added to the reward of next round.

So my question to you is, what strategies will have the best probabilty of defeating a completely random opponent? Does these strategies work in general or is it sometimes better to adapt a worser strategy?  Please share your analysis.

If there are any ambiguity, i'll edit this post and clarify.

for gold rush i pick 3 first, then use 4 for winning smaller numbers

basically i try to lure out the opponent's 5 and 6 and then go for the kill, or if possible, win early by using my 5 and 6 unexpectedly

The strategy to beat #2 against a completely random opponent is to always choose the element that gives you bonus points ....

A common element in this sort of game: the best strategies are not pure deterministic strategies, but instead rely on randomness.

Consider the following tweak to #2: there are no carryover points in case of a tie, and the number of games played is large rather than the fixed 6 (This simplifies the analysis).
A strategy for this tweak: If rock is the non-bonus element this round, choose paper with probability $\frac19$, rock with probability $\frac39$, and scissors with probability $\frac59$. Rotate accordingly for different element choices.
That is an equilibrium strategy; all other strategies have exploitable weaknesses, where an opponent can choose a strategy that wins in the long run. This particular strategy has no weaknesses: if you play a large number of games against any opponent with it, you can expect wins and losses to balance no matter what the opponent does.
It also has no strengths: even the dumbest opponent strategies have a fair chance against it.

Yes, we all love to get AoPSers to find us winning strategies for http://www.moola.com

Now just hope you don't go up against one.


MCrawford wrote:(a) What is the sum of all odd integers between -10 and 12?

(b) What is the sum of all odd integers between -100 and 102?

[hide]

a)1-9 and -1 -  -9 cancel each other out... so it only leaves 11

b) Same thing... 101[/hide]

[hide]a.) 11

b.)101

All the terms cancel each other out except for the 10^x+1[/hide]

[quote="Tare"]There must be a more clever way to do this...[/quote]

The "clever" way was simply to notice the cancellation of terms.  Such a collapsing sum is pretty clever relative to a MATHCOUNTS sprint round problem.

i cancelled, but make sure to read that it says [i]odd[/i]


A mathematical organization is producing a set of commemorative
license plates. Each plate contains a sequence of five characters
chosen from the four letters in AIME and the four digits in 2007. No
character may appear in a sequence more times than it appears among
the four letters in AIME or the four digits in 2007. A set of plates
in which each possible sequence appears exactly once contains $ N$
license plates. Find $ \frac{N}{10}$.

[quote=RunpengFAILS]Case 1: no 0's

$ 6\times{5}\times{4}\times{3}\times{2}\equal{}720$

Case 2: one 0

$ 5\times{6}\times{5}\times{4}\times{3}\equal{}1800$

Case 3: 2 0's

$ 5C2\times{6}\times{5}\times{4}\equal{}1200$

$ 720\plus{}1800\plus{}1200\equal{}3720$

$ 3720/10\equal{}\boxed{372}$[/quote]

I think it might just be easier to combine the cases where the number of 0's is 0 or 1.

i don't really get how to do this problem, still.

[hide=Alternate Solution with permutations]
Case 1:(0 or 1 0's)
Since there are 7 distinct characters and we are choosing 5 we have 21 ways to arrange them. Since order does matter we multiply by 5! so the number of license plates in this case is 2520
Case 2:( 2 0's)
In this case there are 6 distinct characters(we can't use 0's) and we are choosing 3( we already have 2 0's). Since order does not matter we do 5!/2! to account for permutations so the number of license plates in this case is 1200
Adding the cases up we get the answer as 3720/10=372
[/hide]

Then it is not a problem. $N$ is defined to be the number of possible license plates.

Got this on second try, wrote 3 as 4 :(, got 384


[quote="BanishedTraitor"]$n$ is an odd positive integer. Find the largest $k$ (another positive integer) such that $2^{k}$ always divides

$n^{12}-n^{8}-n^{4}+1$[/quote]

Notice that $(n^{8}-1)(n^{4}-1)=(n^{4}+1)(n^{2}+1)^{2}(n+1)^{2}(n-1)^{2}$

and $n^{4}+1\equiv 2(\mod 4)$ , $n^{2}+1\equiv 2(\mod 4)$ and
$8\parallel n^{2}-1$

[quote="BanishedTraitor"]So what's your answer? This is what I got:

[hide=" Answer"] $k=9$[/hide][/quote]

you are right


solve in real number the equation :

\[ 6^x \plus{} 1 \equal{} 8^x \minus{} 27{^{x \minus{} 1}}\]

Is there any other solutions other than $ x\equal{}1$ ?

hello, $ x\equal{}2$ is another solution.
Sonnhard.


Let $ a,b,c>0$ and
$ \ S\equal{}\frac{a^2}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^2}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^2}{c^2\plus{}ca\plus{}a^2}$
Show that

$ \ 1\le S\le 1\plus{}\frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{2(a^2\plus{}b^2\plus{}c^2)}$

The left hand ineq is easy after let $ x \equal{} \frac {a}{b}$...and expand.
With the right ineq, I think it's hard.
I have seen a nice and hard proof in Pham Van Thuan's book:
$ 2 \minus{} \sum\ \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \minus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \equal{} \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)(\sum\ ab^5 \minus{} abc\sum\ ab^2)}{(a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)(a^2 \plus{} b^2 \plus{} c^2)} \geq\ 0$
I havent' found another proof yet  :wink:


integration
probability
function
expected value
advanced fields
advanced fields solved

I don't know if this is the right place to post this, but I'm sure it will be moved if it's not :).

A player wins at each turn an amout uniformly distributed in [$0,1$], and stops when he has won a cumulative total of at least $1$. What's the expected number of turns that the game lasts?

I doubt this is correct but..
Let $f(b)$ = the expected number of turns to get a cumulative total at least b. Then

$f(b)=(1-b) + \int^b_0 (f(b-x) + 1) dx$

$= 1+\int^b_0 f(b-x) dx$

$= 1+\int^b_0 f(x) dx$

So
$F'(b) = 1 + F(b) - F(0)$
$F(b) = e^b + C$

So $f(b) = e^b$ and the answer is $f(1) = e$.

I got the same answer, but I'll have to read this carefully. What is $f$ exactly? The expected number of turns when you start from $0$ and stop at $b$, or the expected number of turns when you have $b$ and you want to reach $1$?

The latter. The first formula for f just comes (hopefully) from the fact that you have a (1-b) probability of getting over 1 next turn, otherwise you get a value of x, and the expected value will be 1 (this turn) + f(b-x).

But if you have the probability $1-b$ of getting over $1$, why do you add this probability to the expected number of turns?

Besides, if you define $f(b)$ as the expected number of turns if you have $b$ and want to reach $1$, then $f(1)$ should be $0$, don't you think?

Gah, edited. I did mean that f(b) is the expected number of turns to get a total of b, not the other one. Then expected total is (1-b) * 1 + other probabilities * turns in those cases.

But with this new definition of $f$, why are you interested in the probability of getting over $1$, which is $1-b$? Aren't you interested in getting a total of at least $b$? What does $1$ have to do with this case? I'm sorry if I start to ask stupid questions, I'm a bit tired right now (it's past my bed time :)).

If b = 0.7, say, then you get over b if you get anything from 0.7 to 1, which is a probability of 1-b. Note that I'm "translating everything backwards" after each turn, ie if you first turn is 0.3, then you assume after that you start from 0 and are aiming to get to 0.7.

I got it. Yes, everything seems to be falling into place now :).

I have an extension of this, and will start a new thread for it, which will be titled "Expected number of turns (2)."

As for the right forum for this? This seems as good as any, so let's keep it here.

I'm going to go ahead and move this to "solved." TripleM's approach is correct, and his function, $f(b)$ is indeed $e^b.$ The answer to the original question is $e.$ This approach can be generalized considerably, which is what the "Number of turns (2)" topic is about.


Prove that it is impossible for three consecutive squares to sum to another perfect square.

[hide="Answer"]  Let our 3 perfect squares be designated by:

  $n^2 + (n+1)^2 + (n+2)^2$

  Among any three consecutive integers: n, n+1, n+2 we will always obtain the modular 3 residues: 0, 1 , 2.  Hence we re-write our equation to match:

  $0^2 + 1^2 + 2^2 = 5 = 2$ ( mod 3 )

  As we know, perfect squares can only have residues of 0 and 1 mod 3.  Hence, the original equation cannot equal a perfect square.[/hide]

BTW, proofs are beyond getting started level

But this problem would fit better in the intermediate section :)


In a right angled triangle ABC ( $ \angle A \equal{} 90$ ) we draw the altitude $ AH$. If $ I,K$ are the incenters of triangles

  $ ABC , AHC$ respectively and the circle $ (I)$ touches side $ BC$ at $ P$, prove that $ KP \parallel AB$

 Babis

$ D,E$ are the tangency points of the incircles $ (K),(J)$ of $ \triangle AHC,\triangle AHB$ with $ BC$

$ PD \equal{} PC \minus{} DC \equal{} \frac {_1}{^2}(BC \plus{} AC \minus{} AB \minus{} AC \minus{} HC \plus{} AH)$

$ PD \equal{} \frac {_1}{^2}(BC \minus{} AB \minus{} HC \plus{} AH)$

On the other hand, we have:

$ 2JE \plus{} AB \equal{} BH \plus{} AH \Longrightarrow JE \equal{} \frac {_1}{^2}(BH \plus{} AH \minus{} AB)$

$JE \equal{} \frac {_1}{^2}(BC \minus{} AB \minus{} HC \plus{} AH) \equal{} PD$

Similarly, we have $ KD \equal{} PE$ $\Longrightarrow$ $ \triangle KPD \cong \triangle PJE.$ Then it  follows that $ \angle KPJ \equal{} 90^{\circ},$ which implies that $KPHJ$ is cyclic $ \Longrightarrow$ $ \angle KPD \equal{} \angle KJH.$

Since $ \triangle AHC \sim \triangle BHA$ and $ \triangle KHD \sim \triangle JHE$ $ \Longrightarrow$ $ \frac {KD}{JE} \equal{} \frac {AC}{AB} \equal{} \frac {HK}{HJ}$

Thus, $ \triangle ABC \sim \triangle HJK$ $ \Longrightarrow$ $ \angle KJH \equal{} \angle ABC \equal{} \angle KPD$ $ \Longrightarrow$ $ KP\parallel AB.$

We denote the points $ E\equiv (O)\cap CI$ and $ F\equiv (O)\cap EP$ and $ K\equiv CI\cap AF$ and we will prove that $ K$ is the incenter of the right triangle $ \bigtriangleup DAC,$ where $ AD$ ( instead of $ AH$ ) is the altitude of the given right triangle $ \bigtriangleup ABC$ and $ (O)$ its circumcircle.

The quadrilateral $ BPIE$ is cyclic, because of $ \angle CEB \equal{} 90^{o} \equal{} \angle BPI$ and so, we have that $ \angle IBP \equal{} \angle CEF \equal{} \angle CAF$ $ ,(1)$

Because of now, $ \angle CAD \equal{} 90^{o} \minus{} \angle C \equal{} \angle B$ and $ \angle IBP \equal{} \frac {\angle B}{2}$ we conclude that $ \angle CAF \equal{} \frac {CAD}{2}$ $ ,(2)$ and then, the point $ K\equiv CI\cap AF,$ is the incenter of $ \bigtriangleup DAC.$

$ \bullet$ It is easy to Show now, that the quadrilateral $ CKPF$ is cyclic, because of $ \angle KFP \equal{} \angle KCA \equal{} \angle KCP$ and so, we have $ \angle KPC \equal{} \angle KFC$ $ ,(3)$

But, $ \angle KFC\equiv \angle AFC \equal{} \angle B$ $ ,(4)$

From $ (3),\ (4)$ $ \Longrightarrow$ $ \angle KPC \equal{} \angle B$ $ \Longrightarrow$ $ KP\parallel AB$ and the proof is completed.

Kostas Vittas.

Let J be the incenter of triangle $ AHB.$
This result is well-known: $ \Delta JHK\sim \Delta BAC$
Let P' be a point such that $ JKP'$ is a right isoscele triangle. Since $ JHP'K$ is cyclic we get $ \angle KHP'\equal{}\angle KJP'\equal{}45^o\equal{}\angle KHC$
Therefore P' lies on BC.
$ \angle JIK\equal{}135^o\equal{}180^o\minus{}1/2\angle JP'K$.
So  $ P'I\equal{}P'K\equal{}P'J$
Note that $ AC\equal{}AL$ and $ CK$ is the bisector of angle $ ACB$ thus $ KI\perp AJ$, similar we obtain I is the orthocenter of triangle $ AJK.$
$ \angle P'IJ\equal{}\angle SIK\equal{}\angle C/2\plus{}45^o$
$ \Rightarrow \angle JP'I\equal{}\angle B$
But $ \angle KP'C\equal{}\angle KJH\equal{}\angle B$ hence $ \angle IP'C\equal{}90^o \Rightarrow P'\equiv P \Rightarrow KP//AB$

Dear Stergiu and Mathlinkers,
I think that this problem is a consequence of
http://www.mathlinks.ro/viewtopic.php?t=300127.
Sincerely
Jean-Louis

note that $ CK \equal{} (c(b^2)\div (a))/((a \plus{} b \plus{} c)sin(C/2))$ and $ CP \equal{} (bc\div (a \plus{} b \plus{} c))/tan(C/2)$
$ CK/CP \equal{} b/(a cos(C/2)) \equal{} sin KPC/ sin CKP \equal{} sin KPC/ sin((45\minus{}B/2 \plus{} KPC) \equal{} (sin B / cos (45\minus{}B/2))$
let $ x \equal{} KPC$
hence $ cos(x \plus{} 45 \minus{} 3B/2) \equal{} cos(45 \plus{} B/2 \minus{} x)$
which implies that $ B \equal{} x$
therefore $ KP//AB$

$ (K)$touches $ AC$ at$ M.$
$ MK \cap BC \equal{} P'$
We have :
$ \frac {CP'}{CB} \equal{} \frac {CM}{CA} \equal{} \frac {CA \plus{} CH \minus{} AH}{2 .CA} \equal{} > CP' \equal{} \frac {CB.CA \plus{} CB.CH \minus{} CB.AH}{2.CA} \equal{} \frac {CB.CA \plus{} CA^2 \minus{} CA.AB}{2.CA} \equal{} \frac {CB \plus{} CA \minus{} AB}{2} \equal{} CP \equal{} > P'\equiv P$and the proof is completed. :)

[quote="stergiu"] [color=darkred]Let $ ABC$ be an $ A$-right triangle with the incircle $ w$ . Denote the point $ P\in BC\cap w$ , $ H\in BC$

for which $ AH\perp BC$ and the incircle  $ \theta \equal{} C(K)$ of $ \triangle\ AHC$ . Prove that $ KP \parallel AB$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ L\in AC\cap \theta$ . Observe that $ \triangle ABC\stackrel{(*)}{\sim}\triangle HAC$ $ \implies$ $ \boxed {\ \frac {LA}{LC} \equal{} \frac {PB}{PC}\ }$ because the points $ L$ , $ P$ are

analogously in a similarity $ (*)$ $ \implies$ $ PL\parallel AB$ $ \implies$ $ PL\perp AC$ . Since $ KL\perp AC$ obtain $ K\in PL$ , i.e. $ KP\parallel AB$ .[/color]

[quote] [color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle with the incircle $ w$ . Suppose $ a>b$ . Denote the point $ P\in BC\cap w$ ,

$ H\in (BC)$ for which $ \widehat {HAC}\equiv\widehat {ABC}$ and the incircle  $ \theta$ of $ \triangle\ AHC$ , $ L\in AC\cap \theta$. Prove that $ PL \parallel AB$ .[/color] [/quote]

Dear Mathlinkers,
a synthetic proof of this result can be see
on: http://perso.orange.fr/jl.ayme    vol. 5  Le théorème de Feuerbach-Ayme p. 19
Sincerely
Jean-Louis

See that $ K$ and $ I$ are analogous in the similarity $ \triangle AHC \sim  \triangle BAC$, hence $ \frac{CK}{CI}=\frac{AC}{BC}$ ( 1 ).
If $ { D } = CI \cap AB$, then $ \frac{CI}{CD} = \frac{a+b}{a+b+c}$ ( 2 ),  from (1) & (2) we get $ \frac{CK}{CD}= \frac{b\cdot (a+b)}{a\cdot(a+b+c)}$ ( 3 ), but $ \frac{PC}{BC} = \frac {s-c}{a}$ ( 4 ), where $ s= \frac{a+b+c}{2}$, from (3) & (4) we see that $ \frac{CK}{CD} = \frac{CP}{BC}$, i.e. $ PK \parallel AB$.

Best regards,
sunken rock


this is a easy inequality

let $ x$ , $ y$ , $ z$ $ \in ]0;\plus{}\infty [$ , prove that :

$ \frac{x^2\plus{}y^2}{x^2\plus{}xy\plus{}y^2}\plus{}\frac{y^2\plus{}z^2}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{z^2\plus{}x^2}{z^2\plus{}zx\plus{}x^2} \geq \frac{3}{2}$

huh

\[ x^2\plus{}y^2 \ge 2xy \implies x^2\plus{}y^2\plus{}xy \le \frac{3}{2}(x^2\plus{}y^2) \implies \frac{x^2\plus{}y^2}{x^2\plus{}xy\plus{}y^2}\ge \frac{2}{3}\]

... so the sum is at least 2...?

thank's

the minimum that you find is  good . :)


MATHCOUNTS geometry probability AMC USA(J)MO USAMO function

[color=red][Joedit:  This was split from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84288]this[/url] topic.] [/color]

This goes on my all-time favorite problems list.  It was a MathCounts problem, yet fooled all the smartest AoPSers (Joe!).

[quote="chess64"]Joe is all the smartest AoPSers? I mean, he is smart, but I don't think he counts as all the smartest AoPSers.[/quote]

i think he means for SC AOPSers :D

[quote="chess64"]Joe is all the smartest AoPSers? I mean, he is smart, but I don't think he counts as all the smartest AoPSers.[/quote]

lol...I am not even on the same level as many AoPSers much less the smartest.

[quote="mysmartmouth"]Errr...I have half a mind to split this topic into a thread entitiled "Is Joe Smart(er than most AoPSers)?"[/quote]

No, don't do that...I think we've already settled the issue :)

[quote="mysmartmouth"]Errr...I have half a mind to split this topic into a thread entitiled "Is Joe Smart(er than most AoPSers)?"[/quote]

uh..what? ;)

[quote="chess64"]lol, Joe is smart, but you said he is [i]all[/i] the smartest AoPSers. I have to disagree with that one. Even rrusczyk isn't [i]all[/i] the smartest AoPSers. :lol:[/quote]

agreed. However, if I was to pick the smartest active HIGHSCHOOL aopsers (atleast the ones that I know of), i would have to say they would be Zabelman (Zach) as the smartest, and probability1.01 (Alex) /MysticTerminator (Arnav), not far behind.

And if I were to pick the smartest aopser (regardless of age), it would definitely be Myth.  :D

What about David "Perfect Index" Benjamin (Dnas)? :D And ThAzN1?

(Oh BTW most people don't know that Alex skipped a grade!)

oh, i didn't know David Benjamin was on AOPS :)

I split it since we were getting off topic.

But what is this now?  How can you leave out Peter Scholze?  Anyways, going by precedent ( http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58218 ) this topic should be locked but I'll leave it open a little longer...if it starts to get hot at ALL I will lock it immediately.

(Oh yeah, I assume we're talking about math smartness only)

My list:

1. Myth (by far)
2. Darij Grinberg (I'm going w/ my geometry bias)
3. Peter Scholze
4. rrusczyk
5. DPatrick

Err... though I know Mr. Rusczyk is exceptionally smart, I seem to recall he himself saying that many AoPSers were at a far higher level than him.

Well for me its(in no order)

Farenhajit

mathmanman

riddler

Arne

paladin8

yetti

kubus

I don't want to leave out many names, but a couple of people jump out if you post in the olympiad/advanced sections often enough, namely (and in no particular order!)

Peter Scholze
myth (well, in the past)
pbornzstein
jmerry
grobber(!)
iura
vess(!)
Ravi B

among semi-regular to regular posters.  There are many other IMO medalists and whatnot who do not post regularly.

you definitely have to include yetti and probability1.01

idk, if you include probability1.01, you'll have to include Mildorf for obvious reasons and then you have to include a bunch more people that hang out in the Olympiad section, i think the list that blablahblah gave is good (athough where is Darij!)

Speaking of inactive...Dnas....

Mildorf's active!  He doesn't post 10 times a day but he still posts a little.  For example, he left some comments about USAMO in the AMC forum and he also posted in the college forum in the past couple days.  To top it off...he's employed by AoPS and helps out with classes and such.

[quote="joml88"][quote="mysmartmouth"]Maybe we should have a thread for worst AoPSers,,,just kidding!  Anyway, I wanted to add Nathan Benjamin to this list.  Not sure of his screenname, but I know he has one...  He's not on a high IMO level yet, but he's only in 8th grade.  He'll make IMO by 10th.[/quote]

His sn is [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=2471]Dnas[/url].

Why do you have to include Mildorf if you include probability?  probability got a 41 on the USAMO in 10th grade when he was 14.[/quote]
Dnas is his brother, David Benjamin. Nathan is I_Would_Say

[quote="chess64"]Mildorf isn't active. Probability1.01 should definitely be on there. Youngest student to get a 41 on USAMO ever.[/quote]

Age doesn't matter for this thread, why do you think that most of the people on our lists are college students/math proffesors, you don't think Mildorf is better than Alex at this current moment? ;)

Shobhit...I believe you were the first person in this thread to select your list of top HIGH SCHOOLERS.  I thought it was implied thereafter that we would be naming the smartest people from the two sets {still in high school} and {not in high school anymore}.

Of course I would not say that Zhai is smarter than Mildorf at this very moment.

Shobhit: Smartness is not defined as a function of knowledge. By that token a current middle/high schooler would be smarter than Albert Einstein. ;) And I'm certainly not one to decide whether Mildorf or Zhai is smarter.

Oh, and I didn't know Mildorf was active because I don't read the college forum :P

[quote="joml88"][quote="Iversonfan2005"][quote="chess64"] And I'm certainly not one to decide whether Mildorf or Zhai is smarter. [/quote]

yes, i agree with you chess64. none of us are on their level so I don't think it's fair for any of us to decide who's smarter. :)[/quote]

Exactly...so what's the point of this thread?[/quote]

Err...you're the one who split the topic...And I agree, this is pointless, and will make people feel sad.  LOCKED.


$ a,b,c$ positive reals with $ ab\plus{}bc\plus{}ca\equal{}1$. Prove
\[ (a\plus{}b)(b\plus{}c)(c\plus{}a)\ \leq\ \frac{8\sqrt{3}}{27}(a\plus{}b\plus{}c)^2
\]

[quote="Jane_Bennet"]$ a,b,c$ positive reals with $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove
\[ (a \plus{} b)(b \plus{} c)(c \plus{} a)\ \leq\ \frac {8\sqrt {3}}{27}(a \plus{} b \plus{} c)^2
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188529


function
algebra proposed
algebra
Functional Equations

Let $f$ a fuction $\large f: R\to R$ sastifying $\large |f(a)-f(b)|<|a-b| \forall a,b \in R; a\neq b$. Prove that if $f(f(f(f....f(x)))...))=x$ ($n$ letter $fs$) then $f(x)=x$

Why you post it twice? :maybe:
Please delete my post!

there is a problem when i click the Submit button  :D
Please del my post and the same topic

Yeah, I wonder if where you find this? Is it from "Cac bai toan ham so qua cac ki thi Oplympiad"?

oh no!!! this is my problem  :)  (tổng quát từ bài tập của tớ mà Bách  :D )
there are some problem else, i will post later (until this problem is solved)


prove that there exists a positive number k (k<2001) such that $A/k!$ is a perfect square
A=1!2!...2000!

Sorry so much if it has been posted.

That is the first question.
Here is the second one:
Do there exist k if $A=1997!$?

Do you mean A=1!2!...1997! ?
If so, then we just need to check whether 1!2!...1996! is a perfect square, since 1997 is a prime.

[quote="Collins"]:?[/quote]
was a NICE jokeeee but it's true, come onn!!! ;)