谢谢。看上去觉得很简单，但不知是不是最近大脑短路，狂K但K不出来.

[quote="siuhochung"][quote="Skyward_Sea"]用这个吧  Holder的特例
$(a_1+b_1)...(a_n+b_n)\geq(\sqrt[n]{a_1...a_n}+\sqrt[n]{b_1...b_n})^n$
左边先AM-GM  再用它[/quote]
you can't do that. Remember the index is -0.5. If you use huygens, the sign is reversed.[/quote]
看错了，不好意思...还是你接那个引论有用 :oops:

[quote="siuhochung"][quote="Skyward_Sea"]用这个吧  Holder的特例
$(a_1+b_1)...(a_n+b_n)\geq(\sqrt[n]{a_1...a_n}+\sqrt[n]{b_1...b_n})^n$
左边先AM-GM  再用它[/quote]
you can't do that. Remember the index is -0.5. If you use huygens, the sign is reversed.[/quote]
Hi. Does "huygens" equal "Holder"? How to pronuonce it?

[quote="siuhochung"][quote="dingdongdog"][quote="siuhochung"]Is it really that easy...  :?
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=beautiful+inequality&t=5911[/quote]
But the two ineq are different.[/quote]
no, the same.[/quote]
why？How to prove it？

[quote="dingdongdog"][/quote][quote="siuhochung"][quote="dingdongdog"][quote="siuhochung"]Is it really that easy...  :?
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=beautiful+inequality&t=5911[/quote]
But the two ineq are different.[/quote]
no, the same.[/quote][quote="dingdongdog"]
why？How to prove it？[/quote]
Ahh,我说一说好了。（希望这次不要错了...）
首先 $\frac{1}{\sqrt{1+x}}=\frac{n}{\sqrt{n^2(1+x)}}\ge\frac{2n}{n^2+1+x}=\frac{\frac{2n}{n^2+1}}{1+\frac{x}{n^2+1}}$
以$C_i$代入x，再利用香港朋友的那个不等式：
$\frac{1}{1+C_1}+...+\frac{1}{1+C_1}\ge\frac{n}{1+\sqrt[n]{C_1C_2...C_n}}$, 其中
$\sqrt[n]{C_1C_2...C_n}\ge n-1$

[quote="zhaobin"]首先skyward_sea我不认为你的做法是对的...
我的思路是归纳法,有点麻烦.
加强命题
$(1+C_1)^{-0.5}+(1+C_2)^{-0.5}+\cdots + (1+C_n)^{-0.5} \geq n(1+(C_1 \cdots C_n)^{1/n})^{-0.5}$
when $C_i>0,C_1 \cdots C_n \geq (n^2-1)^n$
不妨设$C_1 \geq \cdots \geq C_n$
可由归纳假设$(1+C_1)^{-0.5}+(1+C_2)^{-0.5}+\cdots + (1+C_{n-1})^{-0.5} \geq (n-1)(1+(C_1 \cdots C_{n-1})^{1/{n-1}})^{-0.5}$
剩下的就是时间问题了 :)[/quote]
用归纳当然是最简单的办法了。我只是想试试可不可以由那个式子直接导出结果。
可能我理解错了人家说的“the same”，他或许是指思想相同，而非可以套用。 :blush:

我發現現在我根本想不出來為甚麼我那時說它們是一樣的呢..

突然发现这道题目是42IMO的推广  可以用反证法，和零件不等式的方法

[quote="zhaobin"]突然发现这道题目是42IMO的推广  可以用反证法，和零件不等式的方法[/quote]

零件不等式???

[quote="siuhochung"]原來這叫零件不等式.. 我還真不知道這個名稱呢
但42imo 那個方法怎麼用呢？我一向用holder 來解那題，這題不見得有用[/quote]
Can you post the problem?

$a,b,c>0$, then $\sum \frac{a}{\sqrt{a^2+8bc}} \ge 1$

好像就是蒙 $k$, 使得
$\frac{a}{\sqrt{a^2+8bc}} > \frac{a^k}{a^k+b^k+c^k}$...

这个k应该可以算的  假如存在的话
是4/3

但我覺得這題好像這樣不太可行..
要是這樣做你要怎樣用題目給的條件呢

$(\sum \frac{a}{\sqrt{a^2+8bc}})^2(\sum a(a^2+8bc)) \ge (a+b+c)^3$
只剩下展開和amgm

[quote="dingdongdog"]
事实上原题就是要证
$A = \sum \sqrt{\frac{a_1^{n-1}}{a_1^{n-1}+(n^2-1)a_2...a_n}} \ge 1$
看来我把它“简单化”了。[/quote]
Ahhh 如果直接看到原来的形式，多半会去想Holder的，只是我不熟悉Muirhead，自己能否做出也不知道... :roll:


Given $ 2k\minus{}1$ distinguishable boxes, how many ways are there to place $ k$ indistinguishable balls into them?

[hide="Solution"]Well the title says it all...

We need to partition $ k$ balls into $ 2k \minus{} 1$ boxes, which means we need $ 2k \minus{} 2$ dividers.

$ \binom{k \plus{} 2k \minus{} 2}{2k \minus{} 2} \equal{} \binom{3k \minus{} 2}{3k \minus{} 2 \minus{} 2k \plus{} 2} \equal{} \boxed{\binom{3k \minus{} 2}{k}}$.[/hide]

[hide="Solution"]By [url=http://www.artofproblemsolving.com/Wiki/index.php/Balls_and_urns]Balls and Urns[/url], we find the answer to be $ \dbinom{k \plus{} 2k \minus{} 1}{k \minus{} 1} \equal{} \boxed{\dbinom{3k \minus{} 1}{k \minus{} 1}}$.[/hide]

grn_turtle- I don't understand your solution. Are both of ours essentially the same or is it one of us's that's incorrect?

Let's see...

Let's try the small case where $ k\equal{}2$, so that we have $ 3$ boxes and $ 2$ objects.

1:0, 2:0, 3:2
1:0, 2:1, 3:1
1:1, 2:1, 3:0
1:2, 2:0, 3:0
1:0, 2:2, 3:0
1:1, 2:0, 3:1

Clearly those are the only cases, and there are 6.

Plugging $ k\equal{}2$ into my formula gives 6. Plugging it into yours gives 5.

[hide="Why you are incorrect."]
Well, in the previous example we had $ 2$ objects ($ \Box$) and $ 3$ boxes:
\[ \Box \Box\]

Here's how we could split this up into three boxes:
\[ \Box | \Box |\]

This corresponds to the solution 1:1, 2:1, 3:0 (do you see why?)

Notice how many dividers there are. There are 2, not three. It's not hard to see that in general, given $ b$ boxes, you need $ b\minus{}1$ dividers (in this case, $ 2k\minus{}1\minus{}1\equal{}2k\minus{}2$).
[/hide]

[hide="Why you both are incorrect"]The number of ways to distribute $ n$ balls among $ r$ urns is $ {n\plus{}r\minus{}1\choose r\minus{}1}$.
Thus, the number of ways to distribute $ k$ balls among $ 2k\minus{}1$ urns is $ {(k)\plus{}(2k\minus{}1)\minus{}1\choose(2k\minus{}1)\minus{}1} \equal{} {3k\minus{}2\choose2k\minus{}2}$

grn_turtle, the only reason yours worked was because $ k$ happened to equal $ 2k\minus{}2$ in this case.
[/hide]

[hide="Um, dude...?"]
$ \binom{n}{k}\equal{}\binom{n}{n\minus{}k}$, remember?

$ \binom{3k\minus{}2}{2k\minus{}2}\equal{}\binom{3k\minus{}2}{3k\minus{}2\minus{}2k\plus{}2}\equal{}\binom{3k\minus{}2}{k}$.

I just used this at the end to simplify things down.
[/hide]


calculus
integration
quadratics
number theory unsolved
number theory

Find all integers $ a$ for which the cubic equation has $ 3$ integers roots
$ x^3 \minus{} x \plus{} a \equal{}0$

Suppose the equation has an integer root $ r$.  Then $ a \equal{} r \minus{} r^3$, so we can write the equation as

$ (x^3 \minus{} r^3) \minus{} (x \minus{} r) \equal{} (x \minus{} r)(x^2 \plus{} xr \plus{} r^2 \minus{} 1) \equal{} 0$

and the resulting quadratic has integer roots if and only if $ r^2 \minus{} 4(r^2 \minus{} 1) \equal{} 4 \minus{} 3r^2$ is a perfect square, which occurs if and only if $ r \equal{} 0, \pm 1$.  In each case this gives $ a \equal{} 0$.


The surface area remains constant (can u see why?) so its 1.5^2x6=13.5

Yeah, it's constant, but it's:
$ 6\cdot {4.7}^2\equal{}132.54$

[quote="GameBot"]If a cube whose edge measures 1.50 centimeters is cut out of the cube whose edge measures 4.70 centimeters, how many square centimeters are in the surface area of the resulting figure?
[asy]pair A,B,C,D,E,F,G,H,L,M,N,O,P,Q,R;
A = (0,0); E = (0,2); D = (2,0); B = (1,1);
C = B+D; F = E+B; G = E+B+D; H = E+D;
L = (2*H+A)/3; M = (2*H+B)/3; N = (2*H+C)/3;
O = (2*H+D)/3; P = (2*H+E)/3; Q = (2*H+F)/3;
R = (2*H+G)/3;
draw(A--E--F--G--C--D--A);
draw(E--P--Q--R--G--R--N--O--D--O--L--M--N--M--Q--P--L);
label("4.70", A--D, S);
label("1.50",Q, N);[/asy][/quote]

Yes, the surface area remains constant. This is because of the cube that was cut out, the sides you can see, which previously couldn't be seen, are the same as the sides taken away.

Oh yeah whoops total side is 4.7  :blush:


I gave the link some time ago, and I saw that some people here already knew that site (you can read their comments to the problems online). This is a very interesting resource !

that's right. do you have any other good links?
I found Parabla as a good publication http://endeavor.macusa.net:591/macusa/mathpropress/FMPro with 500 problems and at my level. do you have any other good publications not too difficult?


algebra polynomial topology inequalities function limit

Let $ P\left(x\right)\equal{}x^n\plus{}a_1x^{n\minus{}1}\plus{}a_2x^{n\minus{}2}\plus{}...\plus{}a_{n\minus{}1}x\plus{}a_n$ be a polynomial with integer coefficients such that $ a_n$ is non-zero.

If $ \left|a_1\right| > 1 \plus{} \left|a_2\right| \plus{} \left|a_3\right| \plus{} ... \plus{} \left|a_{n\minus{}1}\right| \plus{} \left|a_n\right|$, then prove that $ P\left(x\right)$ is irreducible in $ \mathbb{Z}\left[x\right]$.

[Moderator edit: This post referred to a wrong version of the problem, where instead of $ \left|a_1\right| > 1 \plus{} \left|a_2\right| \plus{} \left|a_3\right| \plus{} ... \plus{} \left|a_{n\minus{}1}\right| \plus{} \left|a_n\right|$ the condition was (incorrectly) written $ \left|a_1\right| > 1 \plus{} \left|a_2\right| \plus{} \left|a_3\right| \plus{} ... \plus{} \left|a_{n\minus{}1}\right|$.]

I think something is wrong here:

If we put n=3, a2=0, a1=2 then a1>1+|a2| because a2=an-1 but then we can take any a3 that we want, so if we have a3=-1 then the plymiomial=
x^3+2x^2-1=(x+1)*(x^2+x-1) so the problem is wrong. This is because we can choose any an that we want.

Maybe it's like this:a1>1+|a2|+..+|a(n-1)|+|an|. What do you think?

[quote="Namdung"]Hi all,

I will correct this problem (posted by hxtung):

P(x) is polynomial with integer coefficients, an is non zero
P(x) = x^n + a_1x^(n-1) + ...+ a_(n-1)x + a_n.
If |a1| > 1 + |a_2| + ... +|a_n|,
prove that P(x) is irreducible over Z[x]

It's well-known Perron critery. We also can show that if |a1| = 1 + |a_2| + ... +|a_n| and f( \pm 1) <> 0 then P(x) is irreducible over Z[x].

I have the proof of this theorem, but it is not very elementary (using Rushe theorem). Anyone can find elementary solution?[/quote]

Could you please tell us more about Perron's criterion ?

I hardly see the link between the inequality $|a1| > 1 + |a_2| + ... +|a_n|$
and the irreducibility of $P$. How do you prove the unsolvability
of the galois group of $P$ (I mean $Gal(\mathbb{Q}[R]/\mathbb{Q})$)
where $\mathbb{Q}[R]$ is the decomposition field of $P$ .
Is the fact that $R \subset ]-1,0[$ relevant ?

[quote="fmathurin"]
I hardly see the link between the inequality $|a1| > 1 + |a_2| + ... +|a_n|$
and the irreducibility of $P$.[/quote]

The link is pretty simple: if the inequality holds, then the polynomial $P(z)$ has [i]exactly[/i] one root outside the unit circle, and such a polynomial is obviously irreducible.

Namdung, the correct spelling is 'Rouche's theorem.'  :) (those who hear it for the first time can Google it). It is a standard result in complex analysis stating that if $f,g \in \mathbb{C}[x]$ are two polynomials and $\Gamma$ is a simple closed curve in the plane such that $|g(z)|<|f(z)|$ for all $z \in \Gamma$, then the polynomials $f(z)$ and $f(z)+g(z)$ have the same number of roots (counted with their multiplicities) inside $\Gamma$. For this problem, the assumed inequality allows us to apply Rouche's theorem in the special case when $\Gamma$ is the unit circle to conclude that all but exactly one root of $P$ lie strictly inside the unit circle. Another immediate corollary of Rouche's theorem is the fundamental theorem of algebra (and in fact the stronger statement that if $f(z)=z^n  + c_{n-1}z^{n-1} + \cdots + c_0$, then the disc $|z|<1 + \max_i |c_i|$ contains precisely $n$ roots of $f$).

--Vesselin

[quote="vess"][quote="fmathurin"]
For this problem, the assumed inequality allows us to apply Rouche's theorem in the special case when $\Gamma$ is the unit circle to conclude that all but exactly one root of $P$ lie strictly inside the unit circle. Another immediate corollary of Rouche's theorem is the fundamental theorem of algebra (and in fact the stronger statement that if $f(z)=z^n  + c_{n-1}z^{n-1} + \cdots + c_0$, then the disc $|z|<1 + \max_i |c_i|$ contains precisely $n$ roots of $f$).

--Vesselin[/quote][/quote]

Sorry, but I'm really lost here. Can you please elaborate on how you prove perron's criterion with rouche's theorem? What are your choices of f, and g, for example. Your help would be deeply appreciated.

Indeed, it is a well know result.
Consider $f(x)=a_1x^{n-1}$ and $g(x)=x^n+a_2x^{n-2}+...+a_n$. Then $|f(z)|=|a_1|$ for all $z$ with $|z|=1$, and in the same time $|g(z)|\leq |z^n|+|a_2z^{n-2}|+...+|a_n|=1+|a_2|+...+|a_n|$ if $|z|=1$. So $|f(z)|>|g(z)|$ for all $z$ on the unit circle.
Due to Rouche's theorem we conclude $f(z)=a_1z^{n-1}$ and $f(z)+g(z)=z^n+a_1z^{n-1}+...+a_n=P(x)$ have the same number of complex roots inside the unit circle. But $f(z)=a_1z^{n-1}$ has exactly $n-1$ roots there. It follows that $P(x)$ has exactly 1 root outside of $|z|<1$.

Suppose that $P(x)$ is a reducable polynomial. Then $P(x)=u(x)v(x)$, there $u,v\in \mathbb{Z}$.  And we know $u(0),v(0)\neq 0$. But each such polynomial has at least one root outside of $|z|<1$ (otherwise integer nonzero $u(0)$ is a product of complex numbers, where each numbers is less than 1 in magnitude). Therefore, $P(x)$ has at least two roots outside of $|z|<1$. Contradiction.

Valentin said me some time ago that he knows a proof without use of complex analysis (Rouche's theorem).

[quote="vess"]A proof that $P(x)$ has precisely $n-1$ roots inside the unit circle [i]without complex analysis[/i]??!!!??[/quote]Yes. Why is that so hard to belive? :D

So, let's take $P(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_1X + a_0$, $a_k \in \mathbb{Z}$, $k=\overline{0,n-1}$ such that $|a_{n-1}| > 1 + |a_{n-2}| + \cdots + |a_1| + |a_0|$.

We can obviously suppose that $a_0 \neq 0$ because otherwise we just extract the largest $X^j$ type factor we can find out of $P(X)$. Let's first prove that there is no root $\alpha$ of $P(X)$ with absolute value $|\alpha |=1$.

We have that \[  -a_{n-1}\alpha^{n-1} = \alpha^n + a_{n-2} \alpha^{n-2} + \cdots + a_1 \alpha + a_0 \] thus
\[ |a_{n-1} \alpha^{n-1} | = | \alpha^n + a_{n-2} \alpha^{n-2} + \cdots + a_1 \alpha + a_0 | \]
\[ |a_{n-1} \alpha^{n-1} | \leq | \alpha^n | + | a_{n-2} \alpha^{n-2}| + \cdots + | a_1 \alpha | + | a_0 |  \]
and because $|\alpha=1|$ it follows that $|a_{n-1}| \leq 1 + |a_{n-2}|+\cdots + |a_1|+|a_0|$ which is a contradiction.

Let's denote with $\alpha_1$, $\alpha_2$, $\ldots$, $\alpha_n$ be the roots of $P(X)$. Because $|\alpha_1 \cdot \alpha_2 \cdots \alpha_n| = a_0$ it follows that at least one of the roots is larger than 1 in absolute value.

Let $|\alpha_1 | >1$ and let $Q(X) = X^{n-1} + b_{n-2}X^{n-2} + \cdots + b_1X + b_0$ polynomial with roots $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_n$. We have that

$P(X) = (X-\alpha_1)Q(X) = X^n + (b_{n-2}-\alpha_1)X^{n-1} + (b_{n-3}- b_{n-2}\alpha_1)X^{n-2}+$[tex]\cdots + X(b_0 - b_1\alpha_1)-b_0\alpha_1[/tex]

It follows that $a_k = b_{k-1} - b_k \alpha_1$, for all $1 \leq k \leq n-1$, $b_{n-1}=1$, and $a_0 = - b_0 \alpha_1 $, therefore the condition in the statement becomes

\[ |b_2 - \alpha_1 | > 1 + |b_{n-3} - b_{n-2}\alpha_1 | + \cdots + |b_0\alpha_1 | \geq \]
\[ \geq 1 + |b_{n-2}||\alpha_1|  - |b_{n-3}| + |b_{n-3}||\alpha_1| - |b_{n-4}| + \cdots + |b_1||x_1| - |b_0| + |b_0||x_1| = \]
\[ = 1 + |b_{n-2} | + (|\alpha_1| -1 ) (|b_{n-2} | + |b_{n-3} | + \cdots + |b_1 | + |b_0| ). \]

But as $|b_{n-2}  - \alpha_1 | \leq |b_{n-2} | + |\alpha_1| $ it follows that

\[ |b_{n-2} | + |\alpha_1 | > 1  + |b_{n-2}| + (|\alpha_1| -1 ) \sum^{n-2}_{k=0} b_k \ \Rightarrow 1 > \sum^{n-2}_{k=0} |b_k| \eqno (*) \]

Suppose now that there exists an $j>1$ such that $|\alpha_j |>1$. As $Q(\alpha_j) = 0$ we have
\[ \alpha_j^{n-1} = - \sum^{n-2}_{k=0} b_k \alpha_j^k  \ \Rightarrow \ 1 = - \sum^{n-2}_{k=0} b_k \frac 1{\alpha_j^{n-1-k}} \ \Rightarrow \]
\[ 1 = \left| \sum^{n-2}_{k=0} b_k \frac 1{\alpha_j^{n-1-k}} \right| \leq \sum^{n-2}_{k=0} |b_k| \frac 1 {|\alpha_j^{n-1-k}|} \leq \sum^{n-2}_{k=0} |b_k| \] which is a contradiction with (*). Because $|\alpha_j |\neq 1$ it follows that $|\alpha_j| < 1$, and as $j$ was chosen arbitrarily, all $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_n$ are smaller than 1 in absolute value.

No complex analysis whatsoever. Vess I'd expect you of all people never to say never in mathematics ;)

[quote="Valentin Vornicu"]No complex analysis whatsoever. Vess I'd expect you of all people never to say never in mathematics ;)[/quote]

 :)  :)  :)  You're right --- never say 'never.'  :D

Does this solution belong to Laurentiu Panaitopol? It's very beautiful.
 :lol:

--Vesselin

There are also non-elementary but interesting ways of avoiding complex analysis, I think.

It's easier for me to write it as $P(x)=a_nx^n+x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_1x+a_0$, with $|a_n|+|a_{n-2}|+\ldots+|a_1|+|a_0|<1$.

First we show, just as Valentint did, that there are no roots on the unit circle. Then, we construct the homotopy $h(x,t)=x^{n-1}+(1-t)\cdot(P(x)-x^{n-1})$. We have $h(x,0)=P(x),\ h(x,1)=x^{n-1}$. The important thing here is to notice that, just like we showed that there are no roots on the unit circle for $P$, we can show that there are no roots on the unit circle for $h(x,t)$, for any $t\in[0,1]$. This means that it makes sense to talk about $H(x,t):S^1\to S^1,\ H(x,t)=\frac{h(x,t)}{|h(x,t)|}$ (here $S^1$ is the unit circle). Since the order of a curve wrt a point does not change with a homotopy if the curve does not cross that point during the process, it means that $\frac{P(x)}{|P(x)|}=H(x,0)$ has the same order as $H(x,1)=\frac{x^{n-1}}{|x|^{n-1}}=x^{n-1}$ (remember that $x$ takes values on the unit circle) wrt the origin of the plane, and this order must then be $n-1$.

However, there is a theorem stating that the number of roots inside the unit disk for a polynomial $P$ with no roots on the unit disk is precisely the order of $H(x,0)$ defined as above wrt the origin (this isn't hard to show either).

P.S. I really hope there are no mistakes. You see, I'm a newbie in the field, so be forgiving :)

This, in essense, immitates the proof of Rouche's theorem, so it's basically the same idea.

I can see that, but there's still no complex analysis involved :).

i did not quite understand it,what am i supposed to do?

Is complex analysis the non-real part of calculas?

in its roots, complex analysis is the analysis of the functions $f$ that are 'complex derivable' = analytical (meaning that the complex differential quotient $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$  exists and is the same independently 'in which direction' $h$ tends to $0$ in the complex numbers) on all it's points of definition.
But there are other 'definitions', since you can start looking at equivalent properties of these functions.

Alright. I think I understand. Probably take the class in college.

Do you have a source for the above proof credited to Laurenţiu Panaitopol ?

I have a student who would like to use the proof in his Senior Paper.

[quote="DLF"]Do you have a source for the above proof credited to Laurenţiu Panaitopol ?[/quote]It's an article in Gazeta Matematica 10/1993. I'm not sure about its name though ...

[quote] The link is pretty simple: if the inequality holds, then the polynomial $P(z)$ has [i]exactly[/i] one root outside the unit circle, and such a polynomial is obviously irreducible. [/quote]

I understand that this proof shows that under this criteria, there are no
roots that are equal to 1 in absolute value.  Furthermore, I also understand
that it shows there is exactly one root that is greater than 1 in absolute
value.  My question is, how does this imply irreducibility?  It is not
obvious to me why a polynomial (with leading coefficient 1
and integer coefficients and constant) with only one root greater than 1 in absolute
value is irreducible.  Please explain why this is the case!  Any help would
be greatly appreciated!

Assume there is some factorisation $(x^m + ... + a_1 x + a_0 ) (x^n + ... + b_1 x + b_0)$. Then the roots of one factor would all have modulus $<1$ and $\neq 0$, thus the constant coefficient, being their product, would be an integer $c$ with $|c| <1$ and $c \neq 0$, clearly impossible.

I'm not sure this completely answers the question.  Your comment explains why there must be at least one root whose absolute value is larger than one.

We are wondering why having EXACTLY one root larger than one is enough to conclude the polynomial is irreducible over the rationals.

Thanks for the help!

No it shows that there must be at least two roots with modulus greater than $1$ since every factor must have one.

Sorry if I'm missing some important idea here.

I thought that it could be proven like this:

If P(x) has exactly 1 root whose absolute value is greater than 1 and is reducible, then it can be written as P(x)=p1(x)*p2(x).  But the polynomials p1 and p2 have exactly 1 root whose absolute value is greater than 1, forcing the original P(x) to have two such roots, which would obtain the contradiction.

But then,  I thought that p1 and p2 would not necessarily have integer coefficients, so they wouldn't be of the form which forced them to have exactly one such root.

What am I missing?

Gauß' Lemma: when an integer polynomial is reducible as rational polynomial, then it is reducible as integer polynomial (so rational factorisation implies integral factorisation). More exactly, the integral factors are up to multiplication of some (rational) constant the same as in the rational factorisation.

[quote=Valentin Vornicu][quote="vess"]Does this solution belong to Laurentiu Panaitopol? It's very beautiful.
--Vesselin[/quote]Yes, it does.[/quote]

Obviously a super ancient thread, but it's worth noting that Perron also proved a multi-variate version of this result and the method of the late Laurentiu Panaitopol was generalized to this setting by Nicolae Ciprian Bonciocat in 2010.

[url=https://www.worldcat.org/title/on-an-irreducibility-criterion-of-perron-for-multivariate-polynomials/oclc/6733580644]link[/url]


geometry
circumcircle
incenter
inradius
geometric transformation
homothety
ratio

Prove that radii r1 and r2 of two circles improperly touching the circumcircle of ABC
      and decently touching each other at the incenter of ABC are tied into relationship

      2/r = 1/r1 + 1/r2    with inradius r.

      Lt. Col.   Pestich

Apply an inversion of pole $I$ (the incenter) and power equal tt $-4r_1r_2$. This turns the circumcircle into a circle tangent to two parallel lines with distance $2(r_1+r_2)$ between them. This means that it turns the circumcircle into a circle with radius $r_1+r_2$, which means that $\frac{r_1+r_2}R=\frac{4r_1r_2}{O(I)}$, where $O(I)$ is the power of $I$ wrt the circumcircle $(O)$. Since that power is $2Rr$, we get the required result.

I used the fact that if an inversion of pole $I$ has power of absolute value $k^2$, then it turns a circle $(O)$ into its image through the homothety of center $I$ and ratio $\frac{k^2}{O(I)}$.


limit
function
inequalities
parameterization
real analysis
real analysis unsolved

a) Prove that the equation
$t^{2}+1+\frac{1}{t}=3(\frac{t^{\alpha}+2}{3})^{\frac{1}{\alpha}}$
has only one real positive root which is $\alpha (t)$  $\forall t>1$
b)Prove that $\exists lim_{t\rightarrow 1^{+}}\alpha (t)=\alpha_{0}$ and find it.
 :(  I only unsolve part b)

What is $\alpha$ ?

because when $\alpha =8$ ,$t=1,0.8681,1.664$(from mathematica) are three positive roots for the solution ....

[quote="shyong"],$t=1,0.8681,1.664$(from mathematica) [/quote]
I don't understand it

[quote="jmerry"]A restated version of the problem, which should clear up the confusion:

(a) For any fixed $t>1$, show there is exactly one positive $\alpha$ satisfying $t^{2}+1+\frac{1}{t}=3\left(\frac{t^{\alpha}+2}{3}\right)^{\frac{1}{\alpha}}$. Let $\alpha(t)$ be this value.
(b) Show that $\lim_{t\to1^{+}}\alpha(t)$ exists and find this limit.[/quote]
Yes,thank you! :oops:
Please hep me to solve part b)

[quote="fedja"]Apparently, people who could solve the problem didn't get interested and people who got interested couldn't solve it :). Indeed, it is fairly standard if you are comfortable with Taylor expansions and almost impossible to solve if you aren't. This is a big hint by itself. If it is not enough, here is the [hide="solution."] Note that the right hand side is an increasing function of $\alpha$ (by Holder inequality, say). Now fix $\alpha>0$ and write $t=1+x$. The left hand side equals $3+x+2x^{2}+O(x^{3})$ while the right hand side equals $3\left[1+\frac{1}{3}\left(\alpha x+\frac{\alpha(\alpha-1)}2x^{2}\right)+O(x^{3})\right]^{1/\alpha}$, which is $3+x+\frac{\alpha-1}3 x^{2}+O(x^{3})$. Thus, if $\alpha>7$, the left hand side gets eventually less than the right hand one as $x\to 0+$ and if $\alpha<7$, it gets greater. From here it follows that the limit exists and equals $7$. I hope I haven't made a stupid arithmetic mistake somewhere...
[/hide][/quote]
Thank you,but I think you don't understand my Pro :(
$\alpha$ is the variable number, t is fixed.

fedja knows what he's doing; there's nothing wrong with looking at what happens when we change a parameter to get a handle on the solution of an equation involving that parameter.
[quote="fedja"]Note that the right hand side is an increasing function of $\alpha$[/quote]
This is why the solution $\alpha(t)$ exists and is unique- use the Intermediate Value Theorem with this fact (the left-hand side is constant in $\alpha$).

The remainder of the argument is the calculation of the limit. If $t^{2}+1+\frac{1}{t}>3\left(\frac{t^{s}+2}{3}\right)^{\frac{1}{s}}$, $\alpha(t)>s$ by the increasing function argument. Given any fixed $s<7$, this is true for all $t$ in some interval $(1,1+\epsilon)$.
The upper bound is a similar argument, so by squeezing we have that $\lim_{t\to1^{+}}\alpha(t)=7$.


a) Prove that $10201$ is composite in all bases greater than 2.

b) Prove that $10101$ is composite in all bases.

[hide]
$101^2 = 10201$
and the other one is divisible by $111$ because $10101=111(101-10)$   :)
(In the base of  your choosing , of course!)
[/hide]

[hide]
part a)
basically we have $10201_n$ where $n$ is the base.
in base 10 this is equal to $n^4+2n^2+1=(n^2+1)^2$ thus it is always composite.

part b)
we have $10101_n$ in base $n$.
in base 10 this is $n^4+n+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$
Q.E.D.
[/hide]


Demonstraţi că patru numere pozitive $a,b,c,d$ pot fi ariile feţelor unui tetraedru dacă şi numai dacă
\[a<b+c+d, \ b<a+c+d, \ c<a+b+d, \ d<a+b+c .\]
Am mai postat această problemă în precambrianul forumului (secţiunea "Geometry Unsolved Problems"), dar vreau să văd şi un "second opinion".

PS: Problema din Shortlist a lui maky m-a făcut să-mi amintesc de aceasta (e tot dintr-un Shortlist, da' pentru naţională).

[quote="mychrom"]Fie $E$ proiectia lui $A$ pe $(BCD)$. Evident ca :
$[BCD]=[BCE]+[BDE]+[CDE]$.
Dar stim ca au loc si inegalitatile stricte:
$[BCE]<[ABC]$ si analoagele. Prin adunarea relatiilor se ajunge la concluzie.
Reciproc se poate demonstra prin reducere la absurd.
[/quote]
Poti să-mi arăţi cum se demonstrează cealaltă implicaţie? Că tocmai asta mi se pare greu la problema asta.

la cealalta implicatie nu am nicio idee (nu merge in niciun caz cu reducere la absurd cum am scris in postul precedent).

Daca notam cu S1, S2, S3, S4 ariile fetelor opuse virfurilor A1, A2, A3 si A4 ale tetraedrului cu virfurile in aceste puncte; iar cu alfa(ij) diedrul determinat de fetele opuse varfurilor Ai si Aj, atunci are loc relatia:
S1=S2.cos[alfa(12)]+S3.cos[alfa(13)]+S4.cos[alfa(14)]{asta este teorema cosinusului in tetraedru}>S2+S3+S4.
Pentru alte referinte referitor la tetraedru, se poat consulta cartile(in lb.romana):
[1]. Branzei sa.: "PLANUL SI SPATIUL EUCLIDIAN", Ed. Academiei-1986;
[2]. M.Miculita & D. Branzei: "ANALOGII TRIUNGHI_TETRAEDRU", Ed. Paralela-45(Pitesti)-2000.


Find or prove that lim doesn't exist   ;)

\[ \lim_{n\rightarrow \infty}\frac{n+1}{2^{n+1}}(\frac{2^{1}}{1}+\frac{2^{2}}{2}+...+\frac{2^{n}}{n}) \]

Denote $(a_n)_{n\geq 1}$ your sequence and $b_n=\frac{2^n}{n}$. Then

   $a_n=\frac{b_1+...+b_n}{b_{n+1}}$

Apply Stolz-Cesaro ($b_n$ is unbounded and strictly increasing for $n\geq 2$). The limit we have to analyse is

   $\lim_{n\rightarrow \infty}\frac{b_{n+1}}{b_{n+2}-b_{n+1}}=\\=\lim_{n\rightarrow \infty}\frac{1}{\frac{b_{n+2}}{b_{n+1}}-1}=1$,

so the original limit has the same value.

What is Stolz-Cesaro (in general terms)?


My teacher didn't give me back my booklet, so I was wondering if anyone posted the questions somewhere. If so, can you post the link? Thanks.

yeah. I wanted to know the same thing. I lost my booklet(and i lost).

116 people viewed and only one response - and that of someone who wondered the same thing as I did!! I guess that just means lots of people would LIKE to see all the answers online, but that they just aren't ... oh, well.  I wonder when the results themselves will be posted ...

YAY. My teacher got an email today. I got 25 :)


If S is the subspace of M6(R) consisting of all upper triangular matrices, find the dimension of S.

[quote="maokid7"]could you please explain this in more detail to me. i dont quite understand what the question is asking.[/quote]

Well, in fact, I don't quite think it belongs to the pre-olympiad forum :)


can someone post a book which has SELFRIDGE's theorem?

Man, do you think that such a great mathematician has only one theorem? Be more precise in your request, there are hundreds of results that can be called Selfridge's theorem. My guess is that you are looking for that one with collections of numbers $a_{i}$, and $b_{j}$ such that the collections $a_{i}+a_{j}$ and $b_{i}+b_{j}$ are identical. But there can be very well something else.

i think it true.Harazi has book about set theory?


I guess math jams? Most math jams are for contest reviews or class previews, not for concepts.

Speaking of that, does AoPS plan to offer any more Math Jams about particular topics, such as that one with inversion, and the other one with infinite sets, etc. ?

You can see what the upcoming Math Jams will be about on the top left tab that says "Math Jams" (next to Foum and My Classes)
 :)

Yup; there are no upcoming "topic" Math Jams, I was just wondering if they will offer them ever.  :maybe:

[quote="centry57"]I hope there will be a place talk about special functions！[/quote]

The "advanced topics" and "calculus and analysis" fora (specifically the "unsolved and proposed problems" subfora) are the appropriate place for this.

You can just study general functions in a book.
If you have specific questions you can ask in the appropriate forum

The phrase "special functions" refers to a specific object of study, including the gamma and beta functions, among many others.  It's a college-level subject matter  :wink:

Here's an article.  I know nothing about place functions and I'm not even close to learning calculus but to those who are, look here:
[url]http://en.wikipedia.org/wiki/Function_(mathematics)[/url]


hi,
Calculate the Sum
\[ S_{n} \equal{} \sum_{k \equal{} 1}^{n\minus{}1}{\sin(kx)\cos(n\minus{}k)x}.\]

hello, do you mean $ S_{n} \equal{} \sum_{k \equal{} 1}^{n\minus{}1}{\sin(kx)\cos\left((n\minus{}k)x\right)}$.?
Sonnhard.

$ sin (kx) cos (n\minus{}k)x \plus{} sin (n\minus{}k)x cos (kx)\equal{}sin nx$
$ S_n\equal{}[sin x cos (n\minus{}1)x \plus{} sin (n\minus{}1)x cos x]\plus{}[sin 2x cos (n\minus{}2)x \plus{} sin (n\minus{}2)x cos 2x]\plus{}\dots$
When $ n$ is odd , $ S_n\equal{}\frac {n\minus{}1}{2} sin \;nx$
When $ n$ is even , $ S_n\equal{}\frac {n\minus{}2}{2} sin \;nx \plus{} sin \frac {nx}{2} cos \frac {nx}{2}\equal{}\frac {n\minus{}2}{2} sin \;nx \plus{}\frac {sin \;nx}{2}\equal{}\frac {n\minus{}1}{2} sin \;nx$

[quote="Dr Sonnhard Graubner"]hello, do you mean $ S_{n} \equal{} \sum_{k \equal{} 1}^{n \minus{} 1}{\sin(kx)\cos\left((n \minus{} k)x\right)}$.?
Sonnhard.[/quote]
yes

[quote="stephencheng"]$ sin (kx) cos (n \minus{} k)x \plus{} sin (n \minus{} k)x cos (kx) \equal{} sin nx$
$ S_n \equal{} [sin x cos (n \minus{} 1)x \plus{} sin (n \minus{} 1)x cos x] \plus{} [sin 2x cos (n \minus{} 2)x \plus{} sin (n \minus{} 2)x cos 2x] \plus{} \dots$
When $ n$ is odd , $ S_n \equal{} \frac {n \minus{} 1}{2} sin \;nx$
When $ n$ is even , $ S_n \equal{} \frac {n \minus{} 2}{2} sin \;nx \plus{} sin \frac {nx}{2} cos \frac {nx}{2} \equal{} \frac {n \minus{} 2}{2} sin \;nx \plus{} \frac {sin \;nx}{2} \equal{} \frac {n \minus{} 1}{2} sin \;nx$[/quote]

Simplify this:

$ 2S_n\equal{}\sum_{k \equal{} 1}^{n \minus{} 1}{\sin(kx)\cos\left((n \minus{} k)x\right)} \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}{\sin(kx)\cos\left((n \minus{} k)x\right)}$

$ \equal{}\sum_{k \equal{} 1}^{n \minus{} 1}{\sin(kx)\cos\left((n \minus{} k)x\right)} \plus{}\sum_{k \equal{} 1}^{n \minus{} 1}{\cos(kx)\sin\left((n \minus{} k)x\right)}$

$ \equal{}\sum_{k \equal{} 1}^{n \minus{} 1}{\sin(kx)\cos\left( (n \minus{} k)x\right) \plus{}\cos(kx)\sin\left((n \minus{} k)x \right) }$

$ \equal{}\sum_{k \equal{} 1}^{n \minus{} 1}{\sin(nx) } \equal{} (n\minus{}1)\sin(nx)$.

So

$ S_n\equal{}\frac{(n\minus{}1)\sin(nx)}{2}$.


geometry
rectangle
combinatorics unsolved
combinatorics

There is $N*N$ board. We want to paint squares in this board such that there is no any 4 squares that if we connect them we get a rectangle.
What is the maximum number of squares we can paint and on which places do we need to paint ?
For $n=4$ a good painting:(1-painted, 0-not painted)
1101
1010
0101
0011
A bad painting:
[color=red]11[/color]01
1010
[color=red]11[/color]00
0011

Is the forbidden rectangle parallel to the grid?
In that case, as far as I know, only bounds are known in the general case. The equality case is related to the existence of projective planes of given order, which is an open question.

Pierre.


Was it some place like guam or a department?

Uh, I doubt an answer to this question will be, or should be, publically stated. That would be disrespectful to the team that got the lowest. Also, I don't think anyone should be guessing as to who did, as that's similar. This topic should be locked.


Note: the "goose" in the title exists because it wouldn't let me post without lowercase letters in the title.

So, the second Johns Hopkins Math Tournament of 2006 is tomorrow.  Who here is going?

chess64 and I are on the Fighting 42's, coached by CatherineAsaro, and we are going to sweep the team awards, just like last year.  :D And of course I will win first place individual again, unless chess64 does.

For the non-Marylanders who have appeared in our forum recently, we will post the problems here after the competition.

Well you never know... like maybe you'll randomly change all your answers from right to wrong and so will I and so will everyone else on our team what then :P i'm gonna go read that Combinatorics thing by J. Steinhardt... :ninja:

As expected, I got

First Place Algebra
Second Place Combinatorics
First Place Individual

and our team got

First Place Team Round
First Place Scientific Math
First Place Overall Team

I had good reason to be cocky  :wink:.

Anyway, the competition was lots of fun.  I'll post the tests later; they're good problems.


integration
calculus
calculus computations

Let W be the solid bounded by the paraboloid $ x \equal{} y^2 \plus{} z^2$ and the plane $ x \equal{} 9$. Let $ \vec{F}\equal{} 5x\vec{i}  \plus{} y\vec{j}  \plus{} z\vec{k}$

$ div\vec{F} \equal{} 7$

$ \int\limits_W 7 dV$

my question, what is dV here? What is the volume of the bounded region?

[quote="Fermat's Little Turtle"]Are you familiar with setting up multiple integrals?
If you can visualize the section of the parabolloid over which you are integrating it should work out well.
it looks like a parabolloid opening along some axis (in this case the x) and being closed off by a wall (in this case a plane).
It looks like the rays of a flashlight onto a wall (that is something opening up but going only so far).
Does this help in setting up the integral?

The bounds for the $ y$ and $ z$ integrals should work out easily by considering $ (y,z) \equal{} (0,3)$ or $ (y,z) \equal{} (3,0)$[/quote]

is it something like the following:

$ \int_{\minus{}3}^{3} \int_{ \minus{} \sqrt(9 \minus{} z^2)}^{\sqrt(9 \minus{} z^2)}\int_{y^2 \plus{} z^2}^9 7 dx dy dz$


Functional Analysis
advanced fields
advanced fields unsolved

Assume D and F are self adjoint positive semidefinite (possibly unbounded) operator on a Hilbert Space
Assume D is invertible (i.e. spectrum of D is separated from zero)

Prove or disprove: A= D^-1 - (D+F)^-1 is positive semidefinite.

Furthermore, if F is invertible, is A invertible also?

(If F and D commute, then the conclusion follows easily from the spectral theorem)

1) $ D\plus{}F$ is invertible, then $ A\geq0$ iff $ (D\plus{}F)A(D\plus{}F)\geq0$. $ (D\plus{}F)A(D\plus{}F)\equal{}F\plus{}FD^{\minus{}1}F\geq0$ because $ F\geq0,D^{\minus{}1}\geq0$.
2) If $ x$ is a column and $ x\equal{}(D\plus{}F)y$, then $ Ax\equal{}0$ implies $ (y,(F\plus{}FD^{\minus{}1}F)y)\equal{}0$; then $ (y,Fy)\equal{}0$ and $ y\equal{}0$ (if $ F$ is invertible then $ \sqrt{F}$ exists). Thus $ x\equal{}0$ and $ A$ is one to one.
I don't know if $ A$ is onto.

Thanks for 1)
for 2), A is self adjoint means injective <=> surjective. But I think we also need to show $ A$ has spectrum separated from 0, so that the inverse is a bounded operator...


Line $ m$ has the equation $ y \equal{} 3x \plus{} 5$.  Line $ n$ has the equation
$ y \equal{} kx \minus{} 7$.  Lines $ m$ and $ n$ intersect at the point $ (\minus{}4, \minus{}7)$.  What is the value of $ k$?

If they intersect at $ (\minus{}4, \minus{}7)$, that means that both equations are true for those values.

$ y \equal{} kx \minus{} 7$

$ \minus{}7 \equal{} \minus{}4k \minus{} 7$

$ 0 \equal{} \minus{}4k$

$ \boxed{0} \equal{} k$


function
geometry
geometric transformation
reflection
real analysis
real analysis unsolved

This took me pretty long time and maybe a hard problem:

You have some increasing continuous function f:[0,1]->[0,1] satisfying f(x)<=x for x in [0,1]. What is the class of functions g:[0,1]->[0,1], which satisfy f(g)=g(f)?

Comment: if f(x)=ax^b, it is easy. But whatever else seems dificult. Thanks for any help or ideas.

if the function is monotonically increasing then the inverse will do. :)
otherwise, it seems to me that the given condition seems to imply that the graph of the function
lies below the line y=x. may be reflection about y=x can be used in some way...... :maybe:

Examples of functions $ g$ that work:
 :!:  As already noted, the inverse of $ f$, let's write it $ f^{\minus{}1}$.
 :!:  $ f \circ f$, where $ \circ$ means function composition.
 :!:  $ f \circ f \circ f$
 :!:  $ f^{\minus{}1} \circ f^{\minus{}1}$
and so on


Can someone explain how many people are selected for the Olympiad camp after USAMO. Also i heard that there are red MOSP , black MOSP etc. Can anyone explain what these are and what the cutoffs are to qualify to these MOSP's of various levels.

red mop: 9th graders only, about 10+ needed
blue: everyone allowd: 18+ usually
black: everyone allowed: about 25 needed

This should be mooved to the AMC forum

... ummm

Red mop: Top 30 freshmen... or around there, with cutoff score around 10.
Blue: All non-seniors are allowed; around 15-20 people in here.  Cutoff score is around 18 usually?
Black: Top 12 in all grades. Cutoff score is around 25.

i thought blue was around 20-24. black was ~28

umm cutoffs change from year to year. try using the search function, guys... I really think some mods or someone should make a thread in the AMC forum about this

So do only the people that are in one of these mops go to the Olympiad camp?

You mean IMO?

Yes, only people in black mop can go...top 6 on the TST

So the top twelve go to the camp, and then the top six there go to the IMO? What happens there?


Hi,

I couldn't figure out how to do this problem. Could anyone help me?

Bill is sent to a donut shop to purchase exactly six donuts. If the shop has four kinds of donuts and Bill is to get at least one of each kind, how many combinations will satisfy Bill's order requirements?

Thanks!

[hide]cALL th 4 typed of donutsa,b,c,d
We have

a+b+c+d=6

so the possible combinations for (a,b,c,d) are

(3,1,1,1)=Can be arranged in 4 ways

(2,2,1,1)=Can be arranged in 6 ways

tHere no orher ones where a,b,c or d is not 0

so the is 10.[/hide]

Since he needs to get six donuts with at least one of each kind, his first four donuts must be D1, D2, D3, and D4. Then the last two donuts can be:
D1 with another D1 or a D2, D3, or D4= 4 possibilities
D2, D2 (don't count D2, D1 since we already counted it before in the D1 row.) D2, D3; D2, D4
=3 possibilities
D3, D3 and D3, D4= 2 poss.
D4, D4=1 poss.
4+3+2+1=[color=green]10 Possibilities[/color]

Or another method is:

First assign a donut to each kind.

Now with the remaining 2, insert 3 dividers seperating them into groups. Now its just the number of ways to arrange 5 objects with a repeated 2 and 3.

$ \frac{5!}{2!*3!}\equal{}\boxed{10}$

Another way to calculate above, and faster, is just $ \binom{5}{2}\equal{}\boxed{10}$


inequalities
trigonometry
inequalities proposed

Let $ 0\leq x_1 \leq x_2 \leq ...\leq x_n$ and $ x_{n\plus{}1}\equal{}x_1$
Prove that
$ \sum_{k\equal{}1}^{n} \frac{x_k\minus{}x_{k\plus{}1}}{1\plus{}x_kx_{k\plus{}1}} \geq 0$

\[ \sum\limits_{k \equal{} 1}^n {\frac{{x_k }}
{{1 \plus{} x_k x_{k \plus{} 1} }}}  \geqslant \sum\limits_{k \equal{} 1}^n {\frac{{x_{k \plus{} 1} }}
{{1 \plus{} x_k x_{k \plus{} 1} }}} \]
it is just rearrangement, or it seems like that.
Am I wrong?
 :maybe:

[quote="theSA"]
\[ \sum\limits_{k \equal{} 1}^n {\frac {{x_k }} {{1 \plus{} x_k x_{k \plus{} 1} }}} \geqslant \sum\limits_{k \equal{} 1}^n {\frac {{x_{k \plus{} 1} }} {{1 \plus{} x_k x_{k \plus{} 1} }}}
\]
[/quote]

This holds just when $ x_{k} \geq x_{k\plus{}1}$ ,but that's not true by conditions of the problem.  :maybe:

[quote="enndb0x"][quote="theSA"]
\[ \sum\limits_{k \equal{} 1}^n {\frac {{x_k }} {{1 \plus{} x_k x_{k \plus{} 1} }}} \geqslant \sum\limits_{k \equal{} 1}^n {\frac {{x_{k \plus{} 1} }} {{1 \plus{} x_k x_{k \plus{} 1} }}}
\]
[/quote]

This holds just when $ x_{k} \geq x_{k \plus{} 1}$ ,but that's not true by conditions of the problem.  :maybe:[/quote]

No, by conditions of the problem, we have $ x_{n} \geq x_{n\plus{}1}\equal{}x_{1}$  :rotfl:

ohh ..my bad.
However you should write that $ k\geq n$

[quote="xxx001"]Let $ 0\leq x_1 \leq x_2 \leq ...\leq x_n$ and $ x_{n \plus{} 1} \equal{} x_1$
Prove that
$ \sum_{k \equal{} 1}^{n} \frac {x_k \minus{} x_{k \plus{} 1}}{1 \plus{} x_kx_{k \plus{} 1}} \geq 0$[/quote]
Chevbyshev?

[quote="xxx001"]Let $ 0\leq x_1 \leq x_2 \leq ...\leq x_n$ and $ x_{n \plus{} 1} \equal{} x_1$
Prove that
$ \sum_{k \equal{} 1}^{n} \frac {x_k \minus{} x_{k \plus{} 1}}{1 \plus{} x_kx_{k \plus{} 1}} \geq 0$[/quote]
Let $ x_i \equal{} \tan{\theta_i}$.so that $ \theta_i \ge 0$
Then to prove that
$ \tan{(\theta_n \minus{} \theta_1)} \ge \tan{(\theta_n \minus{} \theta_{n \minus{} 1})} \plus{} \cdots \plus{} \tan{(\theta_2 \minus{} \theta_1)}$. which needs essentially to prove that
$ \tan{(a \plus{} b)}\ge\tan{a} \plus{} \tan(b)$. which is true for tan a and tan b positive.
Here $ tan(a)tan(b)<1$, since $ tan(a\plus{}b)\ge0$

[quote="ith_power"][quote="xxx001"]Let $ 0\leq x_1 \leq x_2 \leq ...\leq x_n$ and $ x_{n \plus{} 1} \equal{} x_1$

$ \tan{(a \plus{} b)}\ge\tan{a} \plus{} \tan(b)$. which is true for tan a and tan b positive.
Here $ tan(a)tan(b) < 1$, since $ tan(a \plus{} b)\ge0$[/quote][/quote]
Why?I don't think it true?

[quote="manymany"][quote="ith_power"][quote="xxx001"]Let $ 0\leq x_1 \leq x_2 \leq ...\leq x_n$ and $ x_{n \plus{} 1} \equal{} x_1$

$ \tan{(a \plus{} b)}\ge\tan{a} \plus{} \tan(b)$. which is true for tan a and tan b positive.
Here $ tan(a)tan(b) < 1$, since $ tan(a \plus{} b)\ge0$ [/quote]
Why?I don't think it true? [/quote] [/quote]
See that $ tan(a \plus{} b)$ is in the problem $ tan(x_i \minus{} x_j)$ for some $ i > j$.Hence $ tan(a \plus{} b)$ is positive in this context. So when $ tan (a)$ and $ tan (b)$ are positive$ 1 \minus{} tan[a]tan[b]$ is positive . so it is less than 1. Hence $ \frac {tan(a) \plus{} tan(b)}{1 \minus{} tan{a}tan{b}} > tan(a) \plus{} tan(b)$. :)
Is anything wrong in that argument? :maybe:  if yes, please clarify. :)

[quote="Stephen"]Now, we have to prove tan(x1-x2)+tan(x2-x3)+...+tan(x(n-1)-xn)+tan(xn-x1)≥0.[/quote]
correct.
And we have tan(xn-x1)>tan(x_n-x_{n-1})+tan(x_{n-1}-x_1)>tan(x_n-x_{n-1})+tan(x_{n-1}-x_{n-2})+tan(x_{n-2}-x_1)>...>..
by "tan(a+b)>tan(a)+tan(b)" as i said earlier.

[quote="ith_power"][/quote][quote="manymany"][quote="ith_power"][quote="xxx001"]. Hence $ \frac {tan(a) + tan(b)}{1 - tan{a}tan{b}} > tan(a) + tan(b)$. :)
Is anything wrong in that argument? :maybe:  if yes, please clarify. :)[/quote][/quote][/quote][quote="ith_power"][/quote]
If $ a = 179^o,b = 5^o$ fail!

[quote="xxx001"][/quote][quote="ith_power"][/quote][quote="manymany"][quote="ith_power"][quote="xxx001"]. Hence $ \frac {tan(a) + tan(b)}{1 - tan{a}tan{b}} > tan(a) + tan(b)$. :)
Is anything wrong in that argument? :maybe:  if yes, please clarify. :)[/quote]
If $ a = 179^o,b = 5^o$ fail![/quote][/quote][quote="ith_power"][/quote][quote="xxx001"][/quote]
please check my posts correctly.
I mentioned that Tan a and Tan b MUST BE POSITIVE. :mad:
ANYWAY ALL x_i ARE POSITIVE. :wink:
Hence $ x_i-x_{j}>0$ if i>j. And if tan a= x_i, then $ 0\le a \le \pi/2$.


induction
number theory solved
number theory

Let p > 3 be prime. A sequence of positive integers is defined by
an = n for n = 0, 1, 2, ..., p - 1
an = an - 1 + an - p for n \geq p
Set q = p3. Determine aq (mod p).
[Poland 1995/3/3.]

Use induction on k to prove that, for all integers n,k such that n \geq kp \geq p, we have :
a_n =  \sum (i = 0,...,k) C(k,i)*a_(n-ip-k+i),  (1)
where C(k,i) is the binomial coefficient.

It is well-known that, since p is prime then C(p,i) = 0 mod[p] for each i = 1,...,p-1.
From (1), we deduce that for each n  \geq p 2 :
a_n = a_(n-p) + a_(n-p 2 ) mod [p]
Using that a_n = a_(n-1) + a_(n-p), it follows that for n \geq p 2  : a_(n-1) = a_(n-p 2 ) mod[p].
Then the sequence (a_n) is periodic modulo p with period p 2  - 1.
Since p 3  = p(p 2 -1) + p, we deduce that a_p 3  = a_p mod[p].
Since a_p = a_(p-1) + a_0 = p-1, it leads to a_p 3  = p-1 mod[p].

Pierre.


Combinatoricsman wrote:Here's a neat olympiad problem that I found

Show that there is a multiple of 2004 whose binary expression contains 2004 0's and 2004 1's

Hint:[hide] n*2004 = something that contains k ones and j zeros, where k,j divides 2004 (base 2) and k<j. Try multiplying this by 100000....00001. [/hide]

Masamune, [hide]i don't see how ur hint helps. Multiplying it by 100...001 still leaves it with more 0's than 1's. Further explanation would be appreciated.[/hide]

Anyway, here's how I thought about it.

[hide]If we could find a number with k 0's and k 1's where k is a divisor of 2004, then we'd be happy because stringing a bunch of these together would give us the desired number. Furthermore, in our search for this magic number, we don't need k 0's to begin with... all we really need is a number with k 1's and at most k 0's, cause we can just add 0's on the end.

When writing small multiples of 2004 in binary, i found that the only numbers with 6 1's have more than 6 0's, so the next k we can hope for is k=12. With (alot of) patience, i came up with

47*2004 = 10110111111101100,

which has 12 1's and 5 0's. So, multiplying this number by 2^7 will add 7 0's on the end, giving us the magic number with 12 1's and 12 0's. Then, stringing 167 of these together gives us a multiple of 2004 which has 2004 1's and 2004 0's, as desired.[/hide]

NOTE: [hide]this is my thought process, NOT a formal solution! Maybe I'll come back later and write up a solution as i would on a contest.[/hide]

zabelman wrote:[hide]
When writing small multiples of 2004 in binary, i found that the only numbers with 6 1's have more than 6 0's, so the next k we can hope for is k=12. With (alot of) patience, i came up with

47*2004 = 10110111111101100

[/hide]

Just curious, do you know if there's any quicker way to do this step than by testing?

Hi, by the way.  Haven't talked to you since camp.  =)

my thoughts on an easier way:

hint:[hide]do you have to use a string of the same multiple of 2004?[/hide]

solution (i hope):

[hide]

lets take 2004 in binary -

11111010100

and then 3*2004 in binary -

1011101111100

we know we can add any amount of 0's on the end creating a binary number that is still a multiple of 2004.

so with 7 1's in 2004 and 9 1's in 3*2004, we need to make the number of 1's a divisor of 2004 (since there will be a total 2004 1's) and we can make the number of 0's of 2004 and 3*2004 the same as the number of 1's by multiplying by a 2^n.  We could even find a number with 2004 1's and do it (add 0's) at the end!

So, since the factors of 2004=2*2*3*167, and 7+9 is already greater than 12, choose 167.

Anything here works, i chose 2*7+17*9=167 as an easy one.

so 2 "strings" of 2004 followed by 17 strings of 3*2004 will have 167 1's and will also have 4*17+4*2=76 0's, so we can tack on 91 to get 167 0's by multiplying by 2^91.

Now we have a string of 167 0's and 167 1's....the desired number is going to be 12 of those.

Sorry if it is a bit confusing

[/hide]

i like that much better than my way... good job!


quadratics
search
number theory solved
number theory

hi,

i will be really happy if someone helps me to undertand the

underlined part of this exercice's solution  :

[u]problem :[/u]
find all integers $x$ such that 7 devides $x^{2}+x+1$

[u]solution:[/u]
$x^{2}+x+1 = 0(mod 7)$ [u]after trying the seven residus

[/u] ,we find that the solutions are : 2 and 4

why did they test only those numbers ?

thanks in advance.

$(7k+r)^{2}+(7k+r)+1 \equiv r^{2}+r+1 (mod\ 7)$, so we only have to check all the residues mod 7
 ($r$ is such that $r^{2}+r+1\equiv0 (mod\ 7)\Leftrightarrow$ all the numbers of form $7k+r$ have this property).

you have made me happy ! thank you so much !

In general, if the modulus is prime a quadratic has only 2 roots.

[quote="hassenhk"]In fact 11 is a also  a root
because 11^2 +11+1= 133 = 19*7
so you must search more ![/img][/quote]

Huh ?

Obviously, as it was written, solutions are $ 2$ and $ 4$ [u]modulus 7[/u] : so $ 2,4,9,11,16,18, ...$


inequalities
number theory
relatively prime
number theory unsolved

[b]Problem:[/b]For a positive integer $ n$,let $ V(n; b)$ be the number of decompositions of $ n$ into a
product of one or more positive integers greater than $ b$. For example,$ 36 \equal{} 6.6 \equal{}4.9 \equal{} 3.12 \equal{} 3 .3. 4$, so that $ V(36; 2) \equal{} 5$.Prove that for all positive integers $ n$; b it holds that $ V(n;b)<\frac{n}{b}$.  :)

Is it too hard? It's really interesting. Can anybody solve it?  :(

If $ b\equal{}1$ We have $ V(n;1)\equal{}d(n)\minus{}1$ and the inequality holds trivially.
Now suppose $ b > 1$  We obviously have $ V(mn;b)\equal{}V(m;b)V(n;b)$ for relatively prime m,n; Suppose $ n\equal{}p^{a}$ where p is a prime then if $ p \leq b$ then $ V(n;b)\equal{}0$ and we have the inequality.If $ p > b$ then $ \frac{n}{b} \geq a > V(n;b)$ .
Now by the multiplicativity we have the inequality $ V(n;b) < \frac{n}{b}$ for all natural number n


Let $f \in End( C_{n}[x])$defined by
$(f(T))(x) = T(x+1)-T(x)$ ,  $\forall T \in C_{n}[x]$ . Find Kerf và Imf.

[quote="jmerry"]A nice basis for looking at the action of $T$: $f_{k}(x)=x(x-1)(x-2)\cdots (x-k+1)$ for $0\le k\le n$, with the empty product $f_{0}$ interpreted as 1. Can you give the matrix of the difference operator $T$ in this basis? It will make the answer obvious.

The standard basis isn't bad either; the computations are a little uglier, but the essential nature of the matrix is the same.[/quote]

Oh, yes.
$Ker f = C$ and $Im f = C_{n-1}[x]$ . Right?


why do we equate the rate obtained by law of mass action to obtain the equilibrium conc. and not use the more perfect rate laws

What do you mean by "obtain the equilibrium concentration by equating the rate..."?

i mean using the rate law expression obtained from experiments

Equilibrium concentrations are usually computed by using equilibrium constants, not rate laws.

but those equilibrium constants are derived from law of mass action expressions

That's right: equilibrium constants are related to thermodynamics, which don't have nothing to do with kinetics.

well then if u derive them here i think my confusion will be gone


algorithm
number theory
Euclidean algorithm
relatively prime

Prove that the fraction $ \dfrac{21n\plus{}4}{14n\plus{}3}$ is irreducible for any natural number $ n$.

[hide="What I think might be a quick solution"]Using the Euclidean algorithm, we get:
$ gcd(21n\plus{}4, 14n\plus{}3) \implies gcd(7n\plus{}1, 14n\plus{}3) \implies gcd(7n\plus{}1, 7n\plus{}2) \implies gcd(7n\plus{}1, 1)$.
Therefore, the two expressions are relatively prime and thus irreducible.
[/hide]

Considering that that was an IMO problem I just solved, I probably missed something.  Is that enough for a solution?

correct. though to be mathematically precise, the arrows should be replaced with $ \equal{}$ signs.

and yes, the IMO was much much easier back then. problem solving has come a long way...

Yeah, thats fine. But remember, in 1959 a lot of the topics that we know now (especially in NT) were not known back then, so its not like a lot of people in 1959 saw this problem and immediately knew what to do.

Hm, I see.  It's interesting how problem solving has progressed over 50 years.

Faster solution:

$ 3(14n\plus{}3)\minus{}2(21n\plus{}4)\equal{}1$ so the fraction is irreducible.

And also, look at the resources. Every IMO problem is there. http://www.mathlinks.ro/viewtopic.php?p=341470#341470


$f(x)$= $\frac{e^{x}-1}{x}$ if x is a real number not equal to $0$ or
 $f(x)= 1$ if $x=0$;
compute :
$\lim_{n\to\infty}\sum_{k=1}^n\frac{f^{(k)}(0)}{k}$

Nice problem. Yes, it can be explicitly and exactly computed, and fairly quickly at that.

Yes.  Did you try the hint I posted?  How far did you get?

and by the way:
how can I determine the
$ +... + \frac{{(x-x_{0})^{n} f^{(n)} (x_{0})}}{ n!} $

now I know how to solve it:
$\lim_{n\to\infty}\sum_{k=1}^n\frac{f^{(k)}(0)}{k} = f(x_{0}) = 1 $

That is not correct.

Let's back up a step. We don't want power series centered at 1; we want series centered at 0.

What is the power series centered at 0 for $f(x)=\frac{e^x-1}x$ ?

Huh?  Let's back up one more step.  What is the power series of $e^x$ centered at $x = 0$?

[color=blue][b]
Let $f\left( x \right) = \left\{ \begin{array}{l}
 \frac{{e^x  - 1}}{x},x \in \mathbb{R} - \left\{ 0 \right\} \\
 1,x = 0 \\
 \end{array} \right.$ be a real function.

Find $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{f^{\left( k \right)} \left( 0 \right)}}{k}} $
(autor prof. Mircea Ivan, UT Cluj-Napoca, Teste grila, 2005)
[/b][/color]

[b][u]Solution:[/u][/b]

We have:
$e^x  = 1 + \frac{x}{{1!}} + \frac{{x^2 }}{{2!}} + \frac{{x^3 }}{{3!}} + \frac{{x^4 }}{{4!}} +  \cdots  + \frac{{x^n }}{n} +  \cdots  \Rightarrow $

$ \Rightarrow \frac{{e^x  - 1}}{x} = 1 + \frac{x}{{2!}} + \frac{{x^2 }}{{3!}} + \frac{{x^3 }}{{4!}} +  \cdots  + \frac{{x^{n - 1} }}{{n!}} +  \cdots  \Rightarrow $

$ \Rightarrow \left\{ \begin{array}{l}
 f'\left( 0 \right) = \frac{1}{2} \\
 f^{\left( 2 \right)} \left( 0 \right) = \frac{1}{3} \\
 f^{\left( 3 \right)} \left( 0 \right) = \frac{1}{4} \\
  \cdots  \\
 f^{\left( k \right)} \left( 0 \right) = \frac{1}{{k + 1}} \\
  \cdots  \\
 \end{array} \right. \Rightarrow $
$ \Rightarrow\sum\limits_{k = 1}^n {\frac{{f^{\left( k \right)} \left( 0 \right)}}{k}}  = \sum\limits_{k = 1}^n {\frac{1}{{k\left( {k + 1} \right)}}}  = 1 - \frac{1}{{n + 1}} \Rightarrow $

$\Rightarrow\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{f^{\left( k \right)} \left( 0 \right)}}{k}}  = 1$

Thanks, metru. As far as I can see, that's the only way to solve it. Ravi B. and I were trying to talk nemesis into seeing this, but I guess it was going to take too long.


why the sky become red when the weather is snowy? :)

I guess its cos the sunrays are reflected by the snow and so we see the sky red. Actually I got this idea when I was in Pokhra(nepal), The whole mountain seems to be on fire when the sun rises and the snow begins to give all this wonderful colours of a burning mountain.

[quote="wizard2378"]It's not just snow.

Sometimes in the summer there are marvelous red sun rises or sun sets.

You know the saying red at night, sailor's delight (good weather), red in the morning, sailor's take warning (bad weather)?

See Matthew 16:2 where Jesus is quoted "When evening comes, you say, 'It will be fair weather, for the sky is red,' and in the morning, 'Today it will be stormy, for the sky is red and overcast.'"  (It's not a new observation or one linked to snow.)  (He lived in an area where snow was and is not normal).[/quote]

I remember the saying having to do with fronts bringing in dust particles. Anyway, the sky is probably red because of dust particles that the snow storm brings in.

The red and blue thing of the sky is due to rayleigh's law

Somwething like scattering is inversely ptoprtional to ${\text{wavelenght}^4}$ so during daytime,blue is scqattered more and by the setting sun,is $\text{scattered out}$ leading in predominancy of red colour.Too much red indicates pollution.


function
real analysis
calculus
calculus computations

This probably has been posted before but I couldn't find the link so here we go...

What are all Holder Continuous Functions with $ \alpha > 1$, given that a Holder Continuous Function is defined as a function $ f$ such that:

$ |f(x) \minus{} f(y)| \le C|x\minus{}y|^\alpha$

with $ C > 0$.

Here's my solution, although I'm not sure it's correct.

[hide]
Replace $ x$ with $ x\plus{}h$ and $ y$ with $ x$. Then, our goal is to find a $ \delta > 0$ such that if $ |x\plus{}h \minus{} x| < \delta$ then $ \Big|\frac{f(x\plus{}h) \minus{} f(x)}{h}\Big| < \epsilon$ for some $ \epsilon > 0$. Let $ \delta \equal{} \frac{C|h|^{\alpha \plus{} 1}}{|f(x\plus{}h) \minus{} f(x)|} > 0$. Then
$ |h| < \delta \implies |h| < \frac{C|h|^{\alpha \plus{} 1}}{|f(x\plus{}h) \minus{} f(x)|}$
$ \implies \Big|\frac{f(x\plus{}h) \minus{} f(x)}{h} \Big| <C|h|^{\alpha \minus{} 1} \equal{} \epsilon$.

Hence $ f'(x) \equal{} 0$ and the only possible functions are constant functions $ f$.

[/hide]

[quote="gotztahbeazn"]well the reason i'm skeptical with my solution is that we never use the fact that $ \alpha > 1$

for example, why wouldn't this be true for $ 1 > \alpha > 0$? clearly holder continuous functions are not constant for all $ \alpha > 0$[/quote]

If $ \alpha<1$, $ |x\minus{}y|^{\alpha\minus{}1}$ can be very large when $ x$ and $ y$ are close together :)

The proof is not 'fine'. It makes no sense.

$ \left| \frac {f(x \plus{} h) \minus{} f(x)}{h} \right| \leq C|h|^{\alpha \minus{} 1} \to 0$ when $ h \to 0$ is all that is needed. The first step about a $ \delta$ and stuff doesn't imply a good understanding of what you're doing since you're letting $ \delta$ depend on $ h$.

I just looked at the last line and figured that it was alright.


Consider the two dimensional (x,y)-plane. Let L1 be a straight line which does not pass through the origin. Let L2 be the shortest straight line joining L1 to the origin. Prove that L1 and L2 meet at right angles.

Hello,

I'm rather stuck on the above question, but I would really prefer a hint as to what path to go down rather than the full answer.

Thanks very much.

J.

Suppose L2 does not meet L1 at a right angle and draw L3 through the origin that does meet L1 at a right angle.  L1, L2, and L3 form a triangle...

That was an amazingly quick reply. Still stumped though...

Have you drawn the triangle?  Because we specified that L3 meets L1 at a right angle, this triangle is a [i]right[/i] triangle.  Its hypotenuse is L2, but...

I feel like an idiot; I can't see it.

I have drawn the triangle and all I can think of is 'well mod L2 > mod L1 or mod L3

...

Absolutely correct.  Now, L2 is [i]supposed[/i] to be the shortest line from the origin to L1, but L3 is a line from the origin to L1 that is [i]shorter[/i] than L2.  What does this mean?

Contradiction, therefore angle where L2 and L1 meet must be right as we assumed it was not right in the first place? proof by contradiction?


For $x,y,z$ positive reals such that $xy+yz+zx=\frac{1}{3}$ prove that

$\frac{x}{x^2-yz+1}+\frac{y}{y^2-zx+1}+\frac{z}{z^2-xy+1} \geq \frac{1}{x+y+z}$

From Cauchy we have  :Sigma: x^2/x^3-xyz+x >= (x+y+z)^2/(x+y+z)(x^2+y^2+z^2-xy-yz-xz+1)=1/x+y+z


probability
search
advanced fields
advanced fields theorems

We have n envelopes where are written n different addresses. In those envelopes we put n letters in an aleator way( the letters are one for each address). What is the probability that at least one of the letters it will be in the envelpe with the right address. What can we say if n->infinity.

[quote="blahblahblah"]It should go to $1-\exp(-1)$.  Google 'derangements'.[/quote]

Thank uuuuuu! I find it!  :D


A football game was played between two teams, the Cougars and the Panthers.  The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points.  How many points did the Panthers score?

$ \textbf{(A) } 10 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 24$

[quote=ythomashu][quote=MAA]A [url=https://www.google.com/?ion=1&espv=2#q=2016%20super%20bowl]football game[/url] was played between two teams, the [color=#f00][b]Broncos[/b][/color] and the Panthers.  The two teams scored a total of 34 points, and the [b][color=#f00]Broncos[/color][/b] won by a margin of 14 points.  How many points did the Panthers score?

$ \textbf{(A) } 10 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 24$[/quote]

 :o :o :o :o :o[/quote]

Even more freaky, the answer is $\boxed{A}$, $10 \text{ points}$ (well this is really obvious so maybe not that much more freaky but...). The total is also $34$ points, and the Broncos won by a margin of $14$.

:wacko: :wacko: :wacko:

It is all a conspiracy of the Mathinati (which is a subsidiary of the illuminati) !

[quote=Carrots101]It is all a conspiracy of the Mathinati![/quote]

Remember, math (especially geo (all circles and triangles and stuff)) was invented by the illuminati. But how the heck was MAA able to predict the future!?!?! :o

Wait this is trivial.  As Obama said on the CBS late show after the superbowl, all the scores for the superbowl are written on the back of the constitution.


The updated mathleague.org website has launched.  i invite everyone to check it out.  (Please let me know if you find anything broken, too)..

hmmm then what is mathleague.com? two totally different things?


x,y,a,b are integers that are greater than 1. Solve
${x^a}-{y^b}=1$
(Show the methods)

Man,  you can wait here years and nobody will write a proof, since I don't know if anyone on the forum is capable to understand all the things necessary to prove Catalan's conjecture. It is an extraordinary difficult problem, solved very recently by Preda Mihailescu with very advanced tools and using work of generations of mathematicians who struggled to solve it. And you post it and ask to indicate methods.  :D

Well...I really don't know Catalan conjecture is this...
All my mistakes...
P.S.  I thought CC was wrong when I was a child.(I should have known it!)


I believe that the given equation should be $ (x^2\plus{}4x\minus{}2)^3\plus{}(2x^2\minus{}7x\plus{}4)^3\equal{}(3x^2\minus{}3x\plus{}2)^3.$

[quote="kunny"]I believe that the given equation should be $ (x^2 \plus{} 4x \minus{} 2)^3 \plus{} (2x^2 \minus{} 7x \plus{} 4)^3 \equal{} (3x^2 \minus{} 3x \plus{} 2)^3.$[/quote]

Good remark , kunny but not there was my wrong . I edited . If you observed that , i think the equation is solved .

[quote="alex2008"]Solve in the set of reals numbers the following equation :

$ (x^2 \plus{} 4x \minus{} 5)^3 \plus{} (2x^2 \minus{} 7x \plus{} 4)^3 \equal{} (3x^2 \minus{} 3x \minus{} 1)^3$[/quote]

$ a\plus{}b\plus{}c\equal{}0\Longrightarrow a^3\plus{}b^3\plus{}c^3\equal{}3abc$ gives the [hide="Answer"]$ x\equal{}\minus{}5,\ 1,\ \frac{7\pm \sqrt{17}}{4},\ \frac{3\pm \sqrt{21}}{6}$[/hide].


This is probably going to be somewhat trivial, since it's one of my first geometry problems.  :)

Let $O$ be the circumcenter of acute triangle $ABC$ and let the altitude from $A$ meet the circumcircle of $ABC$ at $A_1$.  Let the perpendicular to $AA_1$ at $A_1$ meet the circumcircle of $ABC$ again at $A_2$, let $A_3$ be the midpoint of $A_1A_2$, and let $AA_3$ meet $OA_1$ at $A_4$.  Points $B_4$ and $C_4$ are similarly defined.  Prove that $ABC$ and $A_4B_4C_4$ have the same circumcenter.

Let $R$ be the circumradius of the triangle $\triangle ABC$. Since the angles $\angle AA_1A_2 = \angle BB_1B_2 = \angle CC_1C_2 = 90^o$ are right, the segments $AA_2, BB_2, CC_2$ are diameters of the circumcircle $(O)$ and the circumcenter $O$ is their common midpoint. Hence, $AA_3, A_1O = R$ are medians of the triangle $\triangle AA_1A_2$ intersecting at its centroid $A_4$, $BB_3, B_1O = R$ are medians of the triangle $\triangle BB_1B_2$ intersecting at its centroid $B_4$ and $CC_3, C_1O = R$ are medians of the triangle $\triangle CC_1C_2$ intersecting at its centroid $C_4$. Consequently, $A_4O = B_4O = C_4O = \frac R 3$ and $O$ is the circumcenter of the triangle $\triangle A_4B_4C_4$. Nice problem.

Nice problem!  It's a very interesting fact although it isn't too hard to prove once you've drawn the diagram as someone else said.


An 800-kg sports car with a 700-N driver hits a dip in the road.  The dip has a vertical radius of curvature of 60 m.  This causes the car's springs to "bottom out," or compress completely for an instant at the bottom of the dip.  When standing still, it takes 5000 N of weight in addition to the car's weight to fully compress the springs.  How fast was the car going when it encountered the dip?  What was the apparent weight of the driver at that instant?

Okay, so I desperatly need help with this problem.  I have no idea how to even start it.  I know that I need to use a = v^2 / r  and other circular motion equations but how do I account for the spring?  And what is a vertical radius of curvature.  Any help would be appreciated.  Thanks!

Well, it says it needs 5000 newtons in addition to the car's weight.

You take away 700 because of the driver

And the rest has got to be supplied by the centrifugal-normal force from the car going in the dip.
So set that $4300 Newtons=m\frac{v^2}{60m}$
and solve for $v$

Thank you so much.  That helped a lot.  Just to clarify:

1.  The mass would be the sum of the mass of the car and the mass of the person, right?

2.  Would the apparant weight of the driver simply be 700/9.8, or am I reading the question wrong?

yeah, the mass would be combined mass

The apparent weight of the driver would be his original weight(700 newtons) plus the extra force from the centripetal motion--- $his weight*(v^2/r)$

Great.  Thanks.

One more question:  "hisweight"  would have to be in kilograms right?  Otherwise we would not have Newtons....

oh, yeah, that's what I meant.  His Mass, not weight.


geometry
rectangle
function
complex numbers
complex analysis
complex analysis unsolved

For a fixed complex number $ \lambda$, show that if $ m$ and $ n$ are large integers, then the equation $ e^z\equal{}z\plus{}\lambda$ has exactly $ m\plus{}n$ solutions in the horizontal strip $ \{ \minus{}2\pi i m < \text{Im}(z)<2 \pi i n \}$.

The hint says to consider $ f(z)\equal{}e^z \minus{}z\minus{}\lambda$ and sketch the image of the boundary of $ \{ \minus{}R< x<R, \minus{}2 \pi m < y < 2 \pi n \}$. Keep in mind $ |\lambda | << m, n << R$.

I do not see how to prove this now. I need some help on this one. Thank you.

This is just the argument principle, though you should be a bit careful. Look at the rectangle $ \minus{}2\pi m\le y\le 2\pi n$, $ \minus{}R\le x\le R$. The right vertical side is easy: the function $ f$ there is essentially $ e^z$, so the increment of the argument is $ 2\pi(m\plus{}n)$ with very high precision. Similarly, the increment over the left vertical side is almost $ 0$ (only $ \minus{}z$ matters there). The difficult thing is horizontal sides. To handle the top one, note that $ e^z\equal{}e^x\in\mathbb R$ on both of them and $ \minus{}z\minus{}\lambda\equal{}\minus{}x\minus{}\mu\plus{}i(\minus{}2\pi n\minus{}\nu)$ where $ \lambda\equal{}\mu\plus{}i\nu$. Now, $ f(z)\equal{}e^x\minus{}x\minus{}\mu\plus{}i(\minus{}2\pi n\minus{}\nu)$, so the real part of $ f(z)$ (the only one that changes) runs from a big positive number $ e^R\minus{}R\minus{}\mu$ to a big positive number $ e^{\minus{}R}\plus{}R\minus{}\mu$ as $ z$ runs over the top side from the right to the left while the imaginary part stays fixed and negative. This means that the argument gets almost no increment when you go from the right top corner to the left top one. The bottom side can be treated similarly. The rest should be clear.


probability
counting
distinguishability
combinatorics proposed
combinatorics

I have a solution to the following problem but I am not sure about this one .Please try it
We have $k$ girls and $a$ boys sitting in a row at $k+a$ seats .
If $a\leq k+1$ find the propability such that two boys are sitting at adjacent seats .

Justify your answer .

We first place 1 girl between consecutive boys.  This leaves $k-a+1$ girls, and we need to distribute them into the $a+1$ "boxes," where the boxes are "left of the leftmost boy," "to the right of exactly one boy," ..., "to the right of the rightmost boy."  There are $m+n-1 \choose n-1$ ways to distribute $m$ identical things into $n$ distinguishable boxes, so this gives us ${k-a+1+a+1-1 \choose a+1-1}={k+1\choose a}$ ways so that no two boys are adjacent.
(Having got that answer, an alternative method: add one girl, but make her sit at the far left.  This obviously doesn't change the number of desireable arrangements.  Now, every boy will have a girl immediately to his left, so we need to choose $a$ of the $k+1$ girls to be to the left of boys.)

So, the answer is $\frac{{k+a \choose a}-{k+1 \choose a}}{{k+a \choose a}}$.

[quote="silouan"]Thank you .But here
[quote="JBL"]  There are $m+n-1 \choose n-1$ [/quote]

you took the formula for the combinations with repetitions ?
If yes I think we have
 $m+n-1 \choose n$

Am I wrong ?[/quote]

I had $m$ things, $n$ boxes ... I think I am right.  Maybe you are thinking of $m$ boxes, $n$ things?  In any event, my second argument (which gives the same result) seems good, doesn't it?


Let $\gamma$ be circle and let $P$ be a point outside $\gamma$. Let $PA$ and $PB$ be the tangents from $P$ to $\gamma$ (where $A, B \in \gamma$). A line passing through $P$ intersects $\gamma$ at points $Q$ and $R$. Let $S$ be a point on $\gamma$ such that $BS \parallel QR$. Prove that $SA$ bisects $QR$.

$AS$ bisects $RQ$ if and only if $\dfrac{\triangle{RSA}}{\triangle{QSA}}=1$.

Since $\angle{RSA}+\angle{QSA}=180^o$, so we have:
\[ \dfrac{\triangle{RSA}}{\triangle{QSA}}=\dfrac{RS \cdot RA}{QS \cdot QA}=\dfrac{RS}{QS}\cdot \dfrac{RA}{QA} \]
Because $\dfrac{RS}{QS}=\dfrac{BQ}{BR}=\dfrac{PB}{PR}$, $\dfrac{RA}{QA}=\dfrac{PR}{PA}$

Hence: $\dfrac{RS}{QS}\cdot \dfrac{RA}{QA}=\dfrac{PB}{PR}\cdot \dfrac{PR}{PA}=1$

Therefore: $\triangle{RSA}=\triangle{QSA}$, thus $AS$ bisects $RQ$.

Nice solution !

 [color=red]Second Solution[/color]

[b] HINT[/b]

If $ M$ is the common point of $ AS,QR$, then $ PBMA$ is inscribed(produce PB to PBB', so $ \angle MPB= \angle SBB' = \angle MAB$). Now playing a little with angles , we find that $ \triangle MSB$ is isosceles ect.

 [u] Babis[/u]

[color=red]Third solution.[/color]

We have $ AB$ is $ A$ - symmedian line of triangle $ AQR.$
On the other hand, $ \widehat{QAB}=\widehat{RAS},$ so $ AS$ is median of triangle $ AQR.$
Hence $ SA$ bisects $ QR$

[u]April[/u] (Style like Babis') :D
[color=red]
[b]P.S.[/b] Thank you Mr. Babis for your remark.  :) [/color]

[quote="April"][color=red]Third solution.[/color]

We have $ AB$ is $ A$ - symmedian line of triangle $ AQR.$
[/quote]

 Hallo April !

 I see that  '' $ RP$ is $ R$ - symmedian line of triangle $ ABR.$ ?

 Can you explain a little more your  arguments ? Probably I do not see something.

 Thank you - Babis

[quote="April"]Hi, Mr. Babis!

We have $ AQBR$ is a harmonic quadrilateral, so $ A,\,B$ and $ T$ the intersection of tangent lines at $ Q$ and $ R$ of the given circle are collinear.
And you known that $ AT$ is $ A$ - symmedian line of triangle $ AQR,$ i.e. $ AB$ is $ A$ - symmedian line of triangle $ AQR.$

Good luck.

[u]April.[/u]  :D[/quote]

 Thank you for your nice explanation. Of course , I would not ask you this , but in the first message I did not read carefully  the image. In our language letter $ P$ is our $ R$ , and  by reading $ P$ as $ R$, I did not see the right triangle.
 A!( a typo): I think that at the end of the second  sentence to the first message you wanted to write ''.....median of triangle $ AQR$ ''instead of ''triangle $ APQ$''.

 Now , I have read carefully your solution and I see that [color=red]it is really very nice , short and genious [/color]!

 Sorry , and again many thanks.
[u]
 babis[/u]

denote $ M$ as the intersection of $ SA,QR$.
$ PA,PB$ are tangent to the circle so we get that $ P(AB,QR)=-1$ hence $ S(AB,QR)=-1$ now if we take the intersections with line $ QR$ we get that:
$ (MB',QR)=-1$ where $ B'$ is the intersection of $ SB,QR$ but we had $ BS \parallel QR$ so $ B' \rightarrow \infty$ so we get that:
$ (M\infty,QR)=-1$ so $ M$ is the midpoint of $ QR$ as wanted...

[quote="shobber"]
Since $\angle{RSA}+\angle{QSA}=180^o$, so we have:

\[ \dfrac{\triangle{RSA}}{\triangle{QSA}}=\dfrac{RS \cdot RA}{QS \cdot QA}=\dfrac{RS}{QS}\cdot \dfrac{RA}{QA} \]

[/quote]

Hi shobber i have some question, can you explain me :

- How we have [img]http://data.artofproblemsolving.com/images/latex/0/a/d/0addb3650a467dff4f44094604eaf51796804d5a.gif[/img] from [img]http://data.artofproblemsolving.com/images/latex/d/e/2/de26302451dfb5257ed6439d058e6603fc1fbf24.gif[/img] ?

- How we get [img]http://data.artofproblemsolving.com/images/latex/d/6/e/d6e33665c63e0a6db8baa42bc91ee93c6a9a6540.gif[/img] from  [img]http://data.artofproblemsolving.com/images/latex/a/9/e/a9e741999575c3318ef8d9ee12ea9f314b7ed0eb.gif[/img] ?

shobber meant $\triangle RSA$ to be Area of $\triangle RSA$
Now,FOR THE FIRST PART $\frac{\triangle RSA}{\triangle QSA}=\frac{\frac{1}{2}RS\times RA \times \sin(\angle {ARS}}{\frac{1}{2}QS \times QA \times \sin(\angle{SQA}}$
FOR the second part,
first observe,
$PR \times PQ=(PA)^2=PB \times PB$
WHICH GIVES,
$\frac{PQ}{PB}=\frac{PB}{PR}$......(2)
then observe that $\triangle RSB$ IS SIMILAR TO $\triangle PQB$
WHICH GIVES,
$\frac{RS}{PQ}=\frac{BR}{PB}=\frac{QS}{PB}$         (SINCE, BR=QS )
so $\frac{RS}{QS}=\frac{PQ}{PB}=\frac{PB}{PR}$                 (USING (2))
simillarly,
$\frac{RA}{QA}=\frac{PR}{PA}=\frac{PR}{PB}$

my solution (using angle chasing)

let SA intersects QR at X

<AXP = <ASB = <ABP so, A,B,P,X cyclic.

also, denote O is the circumcentre of the circle(gamma).

<OAP = <OBP = 90

<OAP + <OBP = 180

so, A,B,O,P cyclic.

so, A,O,X,P cyclic.

<OXP + <OAP = 180

so, <OXQ = <OXP = 90 thus QX = RX
                                                                  (proved)
any nicer solution

:o [url=https://artofproblemsolving.com/community/c5h404355p2254813]USAJMO 2011 P5[/url]

Rename some points and use phantom points. We will solve the following equivalent problem: [quote]In $\triangle ABC$ with circumcircle $\Omega$, let $D$ be a point on $\Omega$ such that $AD$ is the $A$-symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$. Show that $E$ lies on $\Omega$.[/quote]

We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Since $D$ is on the $A$-symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$. Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$, so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$. So$$D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$$ Thus, the point at infinity along the line $A$-symmedian has coordinates $$P_\infty=(b^2+c^2:-b^2:-c^2).$$

Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$. The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$. Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$. Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$. Hence, their intersection must have coordinates $$E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$$ It remains to check that $\text{Pow}_\Omega (E)=0$. Plugging $E$ into the circumcircle equation yields $$a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$$ which is clearly true after factoring.

[hide=smooth]Take the markups i did in figure.
claim:-$\angle{PMO}=90^{\circ}$\\
Notice $QR\cap AS=M$\\
also we have $QBSR$ to be a isosceles trapezoid which gives $QB=RS\implies \widehat{QB}=\widehat{RS}$\\
also notice we have $\angle{PMA}=\frac{\widehat{RS}+\widehat{BQ}}{2}=\frac{\widehat{QB}+\widehat{BQ}}{2}=\frac{BA}{2}=\frac{\angle{AOM}}{2}=\angle{POB}$\\
which implies points $P,A,O,M$ are conyclic. also we have $OA \perp PA\implies \angle{PAO}=90^{\circ}\implies \angle{PMO}=90^{\circ}$\\
Hence claim follows $\square$\\
from claim $OM \perp QR\implies M$ is midpoint of $QR$ and hence $SA$ bisects $QR$ as desired $\blacksquare$  [/hide]


Hi, there is a brute inequality

Prove that for any $ a,b,c>0$ it's true that

$ (\displaystyle \sum a^2b)^2\ge 2abc\displaystyle \sum a^3\plus{}4abc\displaystyle \sum a^2b$

where the sums are symmetric

[quote="BlAcK_CaT"]Hi, there is a brute inequality

Prove that for any $ a,b,c > 0$ it's true that

$ (\displaystyle \sum a^2b)^2\ge 2abc\displaystyle \sum a^3 \plus{} 4abc\displaystyle \sum a^2b$

where the sums are symmetric[/quote]

$ \iff \sum_{sym}{\left(a^4b^2\plus{}a^3b^3\plus{}a^2b^2c^2\right)} \ge \sum_{sym}{\left(a^4bc\plus{}2a^3b^2c\right)},$

$ \sum_{sym}{a^4b^2} \ge \sum_{sym}{a^4bc} \iff \sum_{cyc}{a^4(b\minus{}c)^2} \ge 0$;

$ \sum_{sym}{\left(a^3b^3\plus{}a^2b^2c^2\right)} \ge \sum_{sym}{2a^3b^2c} \iff 2\sum_{cyc}{b^2c^2(a\minus{}b)(a\minus{}c)} \ge 0$.


geometry
trapezoid
ratio
parallelogram
similar triangles

This is a problem from the Canadian Waterloo Pascal Contest.

I have tried solving it but I'm not that great at solving problems.  I need your help.  Thanks.

This is a cheap way to get the answer, but since the problem is so general and is asking you to find the ratio of two things, you can assume that AD is perpendicular to CD and the problem becomes a lot easier. This trick wouldn't work though if you were asked on an olympiad to prove that the ratio is always equal to something, but since this is multiple-choice... :D

It's a good idea to take advantage of the format of the contest whenever you can.

I haven't taken a close look, but I'm almost positive it will be similar triangles that solves this one....

Ok I got it... The important thing to notice is thatt the two parallelograms are equal in area and thus their bases both equal 2.

Solution in spoiler:

[hide]

AB=2  and DC=5

Because ABDY is a parallelogram, DY is 2.

ABCX is also a parallelogram which shares AB, XC=2

XY=DC-DY-XC=1

Triangle BCY is similar to Triangle ZAB in the ratio of 3/2 thus side BC=3/2AZ;  because These sides are in the ratio of 2/3, and BCW is similar to ZWA, the ratio between BCW and ZWA is also 3/2.

Let us set the height of WYC as L, then the height of WAB is 2/3L

If the total height is H, H=L+2/3L or L=3/5H.

The area of triangle WYC is 1/2(3*3/5H)

The area of triangle BYC is 1/2(3*H)

The area of triangle BWC must be H3/2-9H/10=6H/10

The area of AZW must be (2/3)^2*6/10H or 24H/90

The area of ABCD is 7H/2

The ratio is (24H/90)/(7H/2) or 8/105. B.  [/hide]

Am I right?

Would it be important to calculate the area of the parallelogram?

[hide]I assume from the multiple choices that the area of the trapezoid is 105 so if that assumption is true, then the height would be 30 therefore the area of the parallelogram would be 60.  However, I'm not sure if this point is helpful in solving this problem.[/hide]

I also want to know how it would be easy to solve this problem if I assumed that AD is perpendicular to CD. (Like I said, I'm not that great in problem solving... )

Thanks

kool_dudy: the key to this problem is that you dont need to (and you cant) calculate the area of the trapezoid.  Nor do you need to calculate the area of the parallelogram.  The both the area of triangle AZW and the area of the trapezoid can be expressed in terms of h -- and when you take the ratio between them you can cancel the h.

Ill try and explain my thought process as I looked at this problem:

[hide]Because the problem asks for a ratio it is natural to start looking for similar triangles.  Also the abundance of congruent angles created by the parallel lines makes this course even more enticing.

There are so many similar triangles in the diagram that it is hard (at least for me) to discern which ones would be helpful.  We are looking for which triangles would allow us to find the area of triangle AZW.

In this case, I think triangles ABZ and ABW are the quickest ones to use.

We know that ABZ is similar to XYZ with a ratio of 2:1 (because XY = 5 - 2 - 2 = 1).  Together their areas are (3*h)/2, where h is the height of the trapezoid.  The ratio of the height of ABZ to that of XYZ is 2, and they add to the height of the trapezoid (which I will call h) so the height of ABZ must be 2h/3.  This means that the area of ABZ is (1/2) * 2 * (2h/3) = 2h/3.

We also know that ABW is similar to CYW with a ratio of 2:3.  By noting as we did above that their heights add to the height of the trapezoid we find that the area of ABW is 2h/5.

The area of AZW is ABZ - ABW = 2h/3 - 2h/5 = 4h/15.

The area of the trapezoid is h * (3+2)/2 = 5h/2.

The ratio, then, between the area of AZW and the area of the trapezoid is 4h/15:5h/2 which simplifies to 8:105.[/hide]

After looking at Ragingg's solution I realized that mine is exactly the same.  Oh well :P

Thanks for all your help.

B is the correct answer.

Ian : How come your answer 4h/15:5h/2 doesn't result in 8:105...
or is it just me?

[quote]The area of the trapezoid is h * (3+2)/2 = 5h/2. [/quote]

lol -- that should be h*([i]5[/i] + 2)/2 = 7/2.  Thanks for catching that.

This is sort of silly, but I would have guessed B right away based on the fact that it's the only fraction there that doesn't simplify ...


calculus
integration
algebra
polynomial
function
domain
Ring Theory

Let $R$ be an integral domain. Prove that $Ra=Rb$ iff $a=ub$ for some
  unit $u$ in $R$. Try to construct a commutative ring $R$ containing
  elements $a,b$ such that $Ra=Rb$ but there does not exist a unit $u$
  such that $a=ub$.

2. Compute $Z[i]/(3+i)$. Is $3+i$ prime in $Z[i]$?
  (Z denotes the ring of integers; i a root of $X^2+1$.)

3. Let $k$ be a field. Show that the rings $k[X,Y]/(X^3-Y^2)$
  and $k[T^2,T^3]$ are isomorphic. Is the image of $X$
  (i) an irreducible element (ii) a prime element in the first ring?

4. Let $|R$ denote the field of real numbers. Show that
  $|R [X,Y]/ (X^2+Y^2-1)$ is an integral domain.

Aww guys I think I need some help with the 3rd problem.I think I am missing somthing

Just show that the polynomial is irreducible. This means that the quotient ring corresponding to the ideal generated by it is an integral domain, because the ideal generated by it is prime.

Grobber are you sure that you gave me the hints for the 3rd one & not the 4th one.Because the 4th problem I have done by showing that the ideal is  a prime.
Could you please elaborate your idea on the 3rd problem.
Also i dont know how to show whether  the image of  $R$  is(i) an irreducible element (ii) a prime element in the first ring?

Oh, sorry. I did indeed look at the fourth problem instead of the third.

Hmm all right I am able to solve the problem using the 1st isomorphism theorem.But as I was saying,the 2nd part of the problem is totally unclear to me.I didnt understand what the problem was trying to say.


Prove that the polynomial $Q(x) = x^{p-1}+x^{p-2}+...+x+1$, where $p$ is prime, cannot be written as a product of two nonconstant polynomials with rational coefficients.

The rest s fairly obvious i think. Roots of unity

I was thinking of that, before i realised i completely forgot what it was! :rotfl: What is it again?

[quote="warut_suk"]Can we use Eisenstein's criterion to prove it?[/quote]

Yes, using Eisenstein's seems the easiest. Substitute $x=y+1$ into $\displaystyle x^{p-1}+x^{p-2}+...+x+1=\frac{(x^p-1)}{x-1}$ and you will get a new polynomial in $y$ satisfying all the conditions for Eisenstein's criterion

try here http://mathworld.wolfram.com/EisensteinsIrreducibilityCriterion.html

the site's quite good

[quote="amirhtlusa"]how can you use the Eisenstein's criterion, on this, because all the coefficients are $1$... the Eisenstein's criterion  says:
let: $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ then $P(x)$ is irreducable if some prime number $p$ divides all coefficients: $a_0,a_1,...,a_{n-1}$, but not the leading coefficient $a_n$, also $p^2$ does not divide the constant term $a_0$, but you can not use the Eisenstein's criterion here because all the coeffiecents are $1$ and no prime number devides $1$.
 ;)[/quote]
 :huuh:  Have you even looked at my post #6 of same topic? It tells you to substitute $x=y+1$, not to apply Eisenstein's directly.  :ninja:


P2

Prove that in the sequrnce of complex numbers z_n=(1+i)(2+i)..(n+i) there are infinitely many numbers with imaginary part >0 and infinitely many with imaginary part <0

I don't have time to write a full solution but I think it enough to show that

S(n) =  :Sigma: arcsin (1/sqrt(k+1)) -> +oo

This is clear because 0 < arcsin(1/sqrt(k+1)) ~ 1/k

So we have infinitely many S(n) such as 0 < S(n) < Pi (mod 2:pi:) and infinitely many such as :pi: < S(n) < 2:pi: (mod 2:pi:).

That's definitely the right idea. I think it might have been tougher for young students who participated in that test. I mean the ones who don't know complex numbers or limits :)


geometry
inequalities
inequalities proposed

[quote="xzlbq"][size=150]Let $ ABC$ is a triangle,$ w_a, w_b, w_c$ are bisectors of angles,$ h_a, h_b, h_c$ are altitudes respectively, $ r$ is radius of the incircle,prove that:

$ \frac {w_bw_c}{w_a} \plus{} \frac {w_cw_a}{w_b} \plus{} \frac {w_aw_b}{w_c} \geq \frac {3}{4}(\frac {a^2w_a}{w_bw_c} \plus{} \frac {b^2w_b}{w_cw_a} \plus{} \frac {c^2w_c}{w_aw_b}) \geq \sqrt {3}s$
BQ[/size]

[size=150]another,I have to tell you,known inequality:

$ r_a \plus{} r_b \plus{} r_c \geq \frac {w_bw_c}{w_a} \plus{} \frac {w_cw_a}{w_b} \plus{} \frac {w_aw_b}{w_c}$
BQ[/size][/quote]

easy

$ \frac {w_bw_c}{w_a} \plus{} \frac {w_cw_a}{w_b} \plus{} \frac {w_aw_b}{w_c} \geq \frac {3}{4}(\frac {a^2w_a}{w_bw_c} \plus{} \frac {b^2w_b}{w_cw_a} \plus{} \frac {c^2w_c}{w_aw_b})$

$ \Longleftrightarrow 4w_b^2wc^2 \plus{} 4w_c^2w_a^2 \plus{} 4w_a^2w_b^2 \minus{} 3a^2w_a^2 \minus{} 3b^2w_b^2 \minus{} 3c^2w_c^2\geq 0$

$ \Longleftrightarrow \sum x^3(y \plus{} z)(x \minus{} y)(x \minus{} z) \plus{} 11(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2$$ \plus{} 40\sum y^2z^2(x \minus{} y)(x \minus{} z))$

$ \plus{} 4xyz\sum x(x \minus{} y)(x \minus{} z) \plus{} 9xyz\sum (y \plus{} z)(x \minus{} y)(x \minus{} z) \geq 0$

$ \frac {3}{4}(\frac {a^2w_a}{w_bw_c} \plus{} \frac {b^2w_b}{w_cw_a} \plus{} \frac {c^2w_c}{w_aw_b}) \geq \sqrt {3}s$

$ r_a \plus{} r_b \plus{} r_c \geq \frac {w_bw_c}{w_a} \plus{} \frac {w_cw_a}{w_b} \plus{} \frac {w_aw_b}{w_c}$

use [size=200]Muirhead theorem[/size]  can prove it :lol:


For positive real numvers x,y,z prove:

$ \sum_{cyc}{\frac{xy}{xy+x^2+y^2^}} \leq \sum_{cyc}{\frac{x}{2x+z}}$

[quote="MathUniverse"]For positive real numvers x,y,z prove:

$ \sum_{cyc}{\frac {xy}{xy + x^2 + y^2^}} \leq \sum_{cyc}{\frac {x}{2x + z}}$[/quote]
Let $ \frac {x}{y} = a,\frac {y}{z} = b$ and $ \frac {z}{x} = c$,then,It seems we'll obtain an inequality that I have already created.See here :

http://www.mathlinks.ro/viewtopic.php?t=302320

Can anybody post all 4 problem in the Mediterranean mathematics olympiad 2009?
Thanks a lot. :)

See here for another approach:

http://gbas2010.wordpress.com/2009/12/17/inequality-11mediterranean-mathematical-olympiad/

I think that you have mistake in the last line of your proof. There should be:
$ \sum(x\minus{}y)^2\ [\frac{1}{3(x^2\plus{}y^2\plus{}xy)}\minus{}\frac{1}{4(x^2\plus{}y^2\plus{}z^2)\plus{}2(xy\plus{}xz\plus{}yz)}] \ge 0$ instead of
$ \sum(x\minus{}y)^2\ [\frac{1}{3(x^2\plus{}y^2\plus{}xy)}\minus{}\frac{1}{2(x^2\plus{}y^2\plus{}z^2)\plus{}xy\plus{}xz\plus{}yz}] \ge 0$.

But why is this true?
$ \frac{1}{3(x^2\plus{}y^2\plus{}xy)}\minus{}\frac{1}{4(x^2\plus{}y^2\plus{}z^2)\plus{}2(xy\plus{}xz\plus{}yz)} \ge 0$
$ \Leftrightarrow$ $ 4(x^2\plus{}y^2\plus{}z^2)\plus{}2(xy\plus{}xz\plus{}yz)\minus{}3(x^2\plus{}y^2\plus{}xy) \ge 0$
$ \Leftrightarrow$ $ x^2\plus{}y^2\plus{}z^2\minus{}xy\plus{}2xz\plus{}2yz \ge0$ ?
 :?:

Another elementary solution:

We have:
$ \sum_{cyclic}{\frac{x}{x\plus{}2y}}\equal{}\sum_{cyclic}{\frac{x^{2}}{x^{2}\plus{}2yx}}\geq\frac{(x\plus{}y\plus{}z)^{2}}{x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2(xy\plus{}xz\plus{}yz)}\equal{} 1$

Hence

$ \sum_{cyclic}{\frac{x}{2x\plus{}z}}\equal{}\frac{1}{2}\sum_{cyclic}{\frac{2x}{2x\plus{}z}}\equal{}\frac{1}{2}\sum_{cyclic}{\frac{2x\plus{}z\minus{}z}{2x\plus{}z}}\equal{}\frac{1}{2}\left(3\minus{}\sum_{cyclic}{\frac{z}{2x\plus{}z}}\right)\leq\frac{3\minus{}1}{2}\equal{} 1$

Hence

$ \sum_{cyclic}{\frac{x}{2x\plus{}z}}\equal{}\sum_{cyclic}{\frac{x}{2x\plus{}z}}\plus{}1\minus{}1\geq\sum_{cyclic}{\frac{x}{2x\plus{}z}\plus{}\frac{z}{2z\plus{}x}}\minus{}1$

Now we need to prove:

$ \sum_{cyclic}{\frac{x}{2x\plus{}z}\plus{}\frac{z}{2z\plus{}x}}\minus{}1\geq\sum_{cyclic}{\frac{xz}{x^{2}\plus{}z^{2}\plus{}xz}}\Leftrightarrow\sum_{cyclic}{\frac{x}{2x\plus{}z}\plus{}\frac{z}{2z\plus{}x}}\geq\sum_{cyclic}{\frac{xz}{x^{2}\plus{}z^{2}\plus{}xz}\plus{}\frac{1}{3}}\Leftrightarrow\sum_{cyclic}{\frac{x^{2}\plus{}z^{2}\plus{}4xz}{2(x^{2}\plus{}z^{2})\plus{}5xz}}\geq\sum_{cyclic}{\frac{x^{2}\plus{}z^{2}\plus{}4xz}{3(x^{2}\plus{}z^{2}\plus{}xz)}}$

Which is obvious using AM-GM at the denominators of the three fractions.


In a bag there are 10 marbles, 3 red ones, 3 green ones, and 4 blue ones. I will randomly pick marbles from the bag without replacement. I lose if I pick out all of the green marbles OR all red marbles before I pick out all the blue marbles. If I pick out all the blue marbles first, then I win. What is the probability that I win?

Let us denote $ p(m,n)$ the number of ways we can randomly pick  $ 3$ blue marbles , $ m$ green marbles and $ n$ red marbles ; and with $ s(m,n)$ the number of ways we can randomly select $ m$ green marbles and $ n$ red marbles.Then the number of cases in wich we win is:
\[ \sum _{0\leq m,n\leq 2} p(m,n)s(3\minus{}m,3\minus{}n)\]
But $ p(m,n)\equal{}\binom{3\plus{}m\plus{}n}{3,m,n}\equal{}\frac{(3\plus{}m\plus{}n)!}{3!m!n!}$ and $ s(m,n)\equal{}\binom{m\plus{}n}{n}\equal{}\frac{(m\plus{}n)!}{m!n!}$ , so:
\[ \sum _{0\leq m,n\leq 2} p(m,n)s(3\minus{}m,3\minus{}n)\equal{}\sum _{0\leq m,n\leq 2} \frac{(3\plus{}m\plus{}n)!(9\minus{}m\minus{}n)!}{3!m!n!(3\minus{}m)!(3\minus{}n)!}\]
Because the total number of cases in wich we can pick $ 3$ green marbles ,$ 3$ red marbles and $ 4$ blue marbles is $ \binom {3\plus{}3\plus{}4}{3.3,4}\equal{}\frac{10!}{(3!)^{2}4!}\equal{}4200$ we conclude that the probability $ P$ equals:
\[ P\equal{}\frac{1}{4200} \sum _{0\leq m,n\leq 2} \frac{(3\plus{}m\plus{}n)!(9\minus{}m\minus{}n)!}{3!m!n!(3\minus{}m)!(3\minus{}n)!}\]

I prefer to take a "dynamic programming" approach to such problems.

At each stage of the game you have $ r$ red marbles, $ g$ green marbles and $ b$ blue marbles left. Let $ p(r,g,b)$ be the probability the you eventually win the game starting from that position. For example, we have:

$ p(1,1,0) \equal{} 1$ since if there are no blue marbles left but there's still a red and a green one, we have already won!

$ p(0,1,1) \equal{} 0$ since we're out of reds while still having some blues - we lost.

$ p(1,1,1) \equal{} 1/3$ since the only way to win is to pick the one blue marble.

In general, what we need here is two things: a recurrence relation and some boundary conditions. Both are not hard to find. To discover the recurrence relation, ask yourself: starting from $ (r,g,b)$, what are the possible next positions, and what are the probabilities of arriving at them?

Note that from $ (r,g,b)$ the next possible positions are $ (r\minus{}1,g,b)$ if we pick a red marble, $ (r,g\minus{}1,b)$ if we pick a green one and $ (r,g,b\minus{}1)$ if we pick the blue. The key observation is that the probability of winning starting from $ (r,g,b)$ is precisely the combined probabilities of winning from each of those three possible next positions, weighted by their probabilities.

Specifically, letting $ m \equal{} r\plus{}g\plus{}b$, we have

$ p(r,g,b) \equal{} \frac{r}{m} p (r\minus{}1, g, b) \plus{} \frac{g}{m} p(r, g\minus{}1, b) \plus{} \frac{b}{m} p (r,g,b\minus{}1)$

[b]provided[/b] that all of $ r, g, b$ are positive. If one of them is zero we have our boundary conditions:

$ p(0,g,b) \equal{} 0$ if $ g>0$, $ b>0$
$ p(r,0,b) \equal{} 0$ if $ r>0$, $ b>0$
$ p(r,g,0) \equal{} 1$ if $ r>0$, $ g>0$

It's now a slightly tedious but totally straightforward exercise to calculate $ p(3,3,4)$. When I have a few minutes I'll add the actual number to this post :-) Of course this is also easy to program.

- R


What do you guys think you'll get on this round?

I'm kind of posting this ahead of time so that I get to be the one who posts it.  :P

For me,

Problem 1- 5
Problem 2- 5
Problem 3- 1 or 2  :(
Problem 4- 5
Problem 5- 5

So a total of about 21 or 22 points.

well...for number 1, I didn't show the grids, so maybe deduct one point for that.  Else, I think I got 24-25. I hope I get some commended

[quote="towersfreak2006"][quote="joml88"]Hopefully a 25 :) Maybe 0 to 2 commended.[/quote]Agreed.

This is so that I can't be shot for making one word posts.[/quote] Bang! :rotfl:

Oh great, I made a one word post also.  ;)

I'm hoping 5/5/5/5/2, couldn't figure out the last one.  Hopefully I will get one or two commended this time around!

22-25, the range cuz i don't like one of my solutions, tho its the right answer and the same as what some people have posted, so maybe 25.

almost definitely 0 commended, cuz i typed all solutions at the last minute and I hate them all :ewpu:    must get crackin on round 3 earlier this time around. . .

As far as I know all my answers are corect, and I tried to make all of my solutions commendable. I typed them up weeks before the deadline and spent the rest of that time revising them and making parts clearer. If I get less than 25 I will be rather disappointed, and I certainly hope for at least one commendation, if not two or three.

I'm hoping 5/5/2/5/5

I felt really really stupid for my solution to #3 after i read the ones in the other topic. I didn't get the right answer but i hope i get 1,2 points for gettin somewhere...

I got the right answer on 1, but I didn't prove that every configuration worked so I don't know if I'll get full credit on that. 2 was correct, people seem to disagree about 3 so my answer may or may not be right, 4 I think is correct but I haven't checked carefully, 5 I'm hoping for 1 point or so on. So...er, I'll probably be somewhere between 14 and 21 points. I think.

I did really bad...... :lol:

4-5/5/1/5/1-2
Total: 16-18

Next round I better do the problems right away.

I'm hoping 4/5/5/5/4, possibly a commended solution for #4, maybe. I noticed that everyone else whose solutions I've read got the answer in two equations, but I condensed it to one, so maybe that'll get me something. :lol:

my comp crashed so ihandwrote it at the last second...
so:
#1 - 2
#2 - 5
#3 - 4/5
#4 - 3/4/5
#5 - 5

Thats about 19-22 points...
ugh i could've easily had 25... i thought the problems were pretty easy

I would say...

#1: 5 i think
#2: 5
#3: 5
#4: probably a 4, I made a careless stupid mistake at the end, but it was minor, and fortunately, it wasn't one of those which can be mistaken for a math error...yet it's possible that they'll give me a 3...yet based on the criteria for grading, I'm going with 4
#5: I used a calculator for my final calculations, but I explained it clearly enough so I hope for a 5.

Total: 24, adding wtih the previous round to make 46.  Heck, I might even make silver...

I got all the solutions correct, and my methods are (hopefully) rigorous, so I'm hoping for a 25.  Whether I get commended is completely up in the air - I've written solutions I considered very nice with no commendation, and on the other hand some of my less creative or interesting solutions have been commended.

I procrastinated :wallbash_red:

#2 - Solved a few weeks before deadline - 5pts.
#1 - Solved $\approx$ 3 hours before deadline - 5pts.
#4 - Solved $\approx$ 1.5 hours before deadline - 4pt.
#3 - Read my "solution", it's so funny :D - 0 pt.

:( :(

Total: $\approx$ 14 pts.........

Hmm ... hopefully all 5's, but possibly a four or on a couple.  I'm guessing 23-25...

JB