[quote="4865550150"]Prove that for all positive numbers $ p, q$,

$ \sqrt {pq} \leq \left(\frac {p \plus{} q}{\sqrt {p} \plus{} \sqrt {q}} \right)^2$.

[hide="My solution"] Since $ p^2 \plus{} q^2 \geq \sqrt {pq}(p \plus{} q)$,

adding $ 2pq$ both sides, we have

[list]$ (p \plus{} q)^2 \geq \sqrt {pq}(p \plus{} 2\sqrt {pq} \plus{} q) \newline (p \plus{} q)^2 \geq \sqrt {pq}(\sqrt {p} \plus{} \sqrt {q})^2$[/list] [/hide][/quote]
You shouldn't post this inequality.It very old and easy

hello, the given inequality is equivalent to $ p^2\plus{}q^2\geq\sqrt{pq}(p\plus{}q)$ squaring this we get $ p^4\plus{}q^4\minus{}p^3q\minus{}pq^3\geq0$, this is true because it is equivalent with $ (p\minus{}q)^2(p^2\plus{}pq\plus{}q^2)\geq0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the given inequality is equivalent to $ p^2 \plus{} q^2\geq\sqrt {pq}(p \plus{} q)$ squaring this we get $ p^4 \plus{} q^4 \minus{} p^3q \minus{} pq^3\geq0$, this is true because it is equivalent with $ (p \minus{} q)^2(p^2 \plus{} pq \plus{} q^2)\geq0$.
Sonnhard.[/quote]

nice!!!! :)

Plugging $ p \equal{} p'^2, q \equal{} q'^2$, this becomes $ G^2 \le \frac {A_2^4}{A_1^2}$, or $ A_1 \cdot G \le A_2^2$
($ G$ - geometric mean of $ q', p'$, $ A_1$ - arithmetic mean, $ A_2$ - root-mean square, etc.)
This is immediate since $ A_2 \ge A_1 \ge G$

A few days ago someone posted some generalization of $ AM \minus{} GM$ which resembles this, can someone find it? :(
(something that implies $ A_1^2 \ge GA_2$ for example)
Also, $ p^4 \plus{} q^4 \minus{} p^3q \minus{} pq^3\geq0$ follows from rearrangement :P

very easy, this is my solution,
$ \left(\frac {p \plus{} q}{\sqrt {p} \plus{} \sqrt {q}} \right)^2\geq \left(\frac{\sqrt{p}\plus{}\sqrt{q}}{2}\right)^2\geq \sqrt{pq}$


There is a theorem in Anton's text:

[b]Theorem.[/b] If $ W$ is a subspace of a finite-dimensional vector space $ V$, then $ \dim W\leq\dim V$; moreover, if $ \dim W\equal{}\dim V$, then $ W\equal{}V$.

In his proof, he use the following proposition first and then by the "Plus/Minus Theorem" (which says if $ \mathbf v\in V$ and $ \mathbf v\notin\text{span}(S)$, where $ S$ is a linearly independent subset of $ V$, then $ S\cup\{\mathbf v\}$ is also linearly independent) to prove the first half of the theorem. Then leave the proof of the proposition as exercise. My question is: could I prove the proposition and the theorem at the same time like following?

[b]Proposition.[/b] Any subspace of a finite-dimensional vector space is finite-dimensional.

[i]Proof.[/i] Suppose $ V$ is an $ n$-dimensional vector space, and let $ U$ be a subspace of $ V$. Then for any subset $ S$ of $ U$ with more than $ n$ vectors, $ S$ is linearly dependent. Thus $ S$ must not be a basis for $ U$. Hence $ \dim U\leq n$. Since $ n$ is finite, $ U$ is finite-dimensional.

[quote="JRav"]Why is $ S$ linearly dependent?  Prove it.[/quote]
There is a theorem which says:

Let $ V$ be an $ n$ dimensional vector space. If a set contains more than $ n$ vectors in $ V$, then it is linearly dependent.

But it is too long to type the proof out.

Okay then.  Well read through your proof.  Are there any potential gaps or flaws?  Is there anything that is unaddressed or ambiguous?  If you can definitively say no to all of these questions, you are done.


[quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ a \plus{} b \plus{} c \equal{} 2$. Prove that:

$ \sqrt {a^2b \plus{} b^2c \plus{} c^2a} \plus{} \sqrt {ab^2 \plus{} bc^2 \plus{} ca^2}\leq 2$[/quote]
\[ ab\left( {a \plus{} b} \right) \plus{} bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right) \le 2\]

[quote="Honey_S"][quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ a \plus{} b \plus{} c \equal{} 2$. Prove that:

$ \sqrt {a^2b \plus{} b^2c \plus{} c^2a} \plus{} \sqrt {ab^2 \plus{} bc^2 \plus{} ca^2}\leq 2$[/quote]
\[ ab\left( {a \plus{} b} \right) \plus{} bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right) \le 2
\]
[/quote]

Let me finish it off !:)
Now, by homogeneous method, we easily get the result, which is known as " Schur degree 3"
$ \sum_{cyc} a^3 \plus{} 6abc \geq \sum_{cyc}ab\left(a\plus{}b \right)$

[quote="great math"][quote="Honey_S"][quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ a \plus{} b \plus{} c \equal{} 2$. Prove that:

$ \sqrt {a^2b \plus{} b^2c \plus{} c^2a} \plus{} \sqrt {ab^2 \plus{} bc^2 \plus{} ca^2}\leq 2$[/quote]
\[ ab\left( {a \plus{} b} \right) \plus{} bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right) \le 2
\]
[/quote]

Let me finish it off !:)
Now, by homogeneous method, we easily get the result, which is known as " Schur degree 3"
$ \sum_{cyc} a^3 \plus{} 6abc \geq \sum_{cyc}ab\left(a \plus{} b \right)$[/quote]
I think it should be $ 3abc$
By the way I dont think this way was useful,My hint is using $ f(a,b,c)\le f(a\plus{}c,b.c)\le 2$
The equation hold when $ (1,1,0$

$ 9abc\plus{}(\sum a)^3\geq 4\sum a\sum ab\Leftrightarrow 4\sum(a^2b\plus{}ab^2)\leq 8\minus{}3abc\leq 8$

Yes, allnames, but It is actually true. I meant that the next step is absolutely true by Schur.


It is easy to ask yourself questions or create problems in mathematics. The difficult part is how to solve them.
Do you agree? Any comments are welcome.

Or maybe it is easy to create poor questions and hard to find good questions? I am not referring to those who ask homework question to be solved but to those who genuinely ask themselves questions as they think about math.

creation of good problems is the tughest job any guy could be facing.

now by tough problems i eman not that creating unsolvable ones, but creating those which tend to dislodge the grey cells present in you..

The strangest question.

Can you prove that for all positive integer n, between n and 2n there is a prime number?

I can't believe so many mathematicians (Fermat, Newton, Leibnitz, Bernoulli, Euler, Lagrange, Gauss, Cauchy) before Bertrand didn't discover this simple theorem.

So my answer is no, it's difficult to ask question, otherwise Euler or Gauss or Cauchy should have discovered this theorem before Bertrand.

Well, there is another explanation: neither of those guys was interested in flooding mathematics with wild conjectures and conditional theorems. ;).

My opinion is that it is very easy to ask a question: every mathematician I have met has asked me at least 5 interestingly looking questions I haven't heard before (not all of them being their own, of course).  As far as I can tell, the questions just start jumping at you if you look at anything closely for at least an hour. I wish it were so with the answers :roll:.

The value of a mathematical question is unpredictable and depends not as much on the question itself as on how people try to solve it. Take the FLT, for instance. The birth of the modern algebraic number theory is usually associated with this question but had anybody found a simple elementary proof of it back in the 17th century, the question would be considered marginal at best.

If you want to object saying that the Riemann hypothesis will have value regardless of the simplicity or complexity of its proof, let me remind you that Riemann didn't specially try to create a new question, he just tried to solve the existing problem about the distribution of primes and recorded his failed attempt pinpointing the difficulty with his approach. It was the approach itself that was exciting, not the particular analytic difficulty in it that is now called RH. So, again, the value here came from a new idea for a possible solution of an existing question.

This does not cover questions coming from physics, engineering, computer science, etc. These have some external value independent on how you approach them, of course, but, again, it is not difficult to ask them. It is much more difficult to avoid asking them.

Anyway, if anybody on this forum feels like he wants more questions, he may just click on the "unanswered posts" link and get hundreds of them immediately. But when somebody wants an answer to his question, the story is often not so simple (I do not mean homework assignments and textbook exercises here, of course).


Let $ \mathbb{F(Z)}$ the set of all functions $ f: \mathbb{Z} \to \mathbb{Z}$ such that $ f(x)\equal{}f(x^2\plus{}x\plus{}1)$ for all integers $x \in \mathbb{Z}$.

a) Find all $ f \in \mathbb{F(Z)}$ such that $ f(x)\equal{}f(\minus{}x)$,for any $x \in \mathbb{Z}$.

b) Find all $ f \in \mathbb{F(Z)}$ such that $ f(x)\equal{}\minus{}f(\minus{}x)$, for all $x \in \mathbb{Z}$.

[quote="prester"]Let $ \mathbb{F(Z)}$ the set of functions $ f: \mathbb{Z} \to \mathbb{Z}$ such that $ f(x) \equal{} f(x^2 \plus{} x \plus{} 1)$, $ \forall x \in \mathbb{Z}$

$ a)$ find all $ f \in \mathbb{F(Z)}$ such that $ f(x) \equal{} f( \minus{} x)$, $ \forall x \in \mathbb{Z}$

$ b)$ find all $ f \in \mathbb{F(Z)}$ such that $ f(x) \equal{} \minus{} f( \minus{} x)$, $ \forall x \in \mathbb{Z}$[/quote]

a) $ f(x) \equal{} f(x^2 \plus{} x \plus{} 1) \equal{} f((x\plus{}1)^2\minus{}(x\plus{}1)\plus{}1) \equal{} f(\minus{}(x\plus{}1)) \equal{} f(x\plus{}1)$ and so $ \boxed{f(x)\equal{}c}$ $ \forall x$

b) $ f(x) \equal{} f(x^2 \plus{} x \plus{} 1) \equal{} f((x\plus{}1)^2\minus{}(x\plus{}1)\plus{}1) \equal{} f(\minus{}(x\plus{}1)) \equal{} \minus{}f(x\plus{}1)$ and so $ f(2p)\equal{}a$ and $ f(2p\plus{}1)\equal{}\minus{}a$
But then $ f(0)\equal{}f(1)$ and so $ a\equal{}\minus{}a$ and so $ \boxed{f(x)\equal{}0}$ $ \forall x$


inequalities
LaTeX
inequalities unsolved

[quote="khj920214"]http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=a%2Bb%2Bc%20%3D%201%2C%5C%20%5C%7B%20a%2Cb%2Cc%20%5C%7D%20%5Csubset%20R%5E%7B%2B%7D%20%5Cimplies%20a%5E2%20%2B%20b%5E2%20%2Bc%5E2%20%2B%203abc%20%20%5Cge%20%5Cfrac%7B4%7D%7B9%7D

Hard to me. '-'; Can you solve it?

ps/ I'm sorry, How can I insert the LaTeX image? I'm the new one. '-';[/quote]
It very easy :)

Very easy? orz I should study more, very very harder.

Would you show me how to show it?

[quote="khj920214"]http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=a%2Bb%2Bc%20%3D%201%2C%5C%20%5C%7B%20a%2Cb%2Cc%20%5C%7D%20%5Csubset%20R%5E%7B%2B%7D%20%5Cimplies%20a%5E2%20%2B%20b%5E2%20%2Bc%5E2%20%2B%203abc%20%20%5Cge%20%5Cfrac%7B4%7D%7B9%7D

Hard to me. '-'; Can you solve it?

ps/ I'm sorry, How can I insert the LaTeX image? I'm the new one. '-';[/quote]
You can just write $ \$$  $ \$$ around the text :)
Like $ a \plus{} b \plus{} c \equal{} 1, \{ a,b,c \} \subset R_{ \plus{} } \implies a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc \ge \frac {4}{9}$
You can "homogenize" the inequality by multiplying by $ a \plus{} b \plus{} c \equal{} 1$:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3abc \ge \frac {4}{9}(a \plus{} b \plus{} c)^3 \iff$
$ \sum_{cyc} 9a^3 \plus{} 9a^2b \plus{} 9a^2c \plus{} 9abc \ge 4(a \plus{} b \plus{} c)^3 \equal{} \sum_{cyc} 4a^3 \plus{} 12a^2b \plus{} 12a^2c \plus{} 8abc \iff$
$ \sum_{cyc} 5a^3 \plus{} abc \ge \sum_{cyc} 3a^2b \plus{} 3a^2c$
Using that $ \sum_{cyc} a^3 \ge \sum_{cyc} abc$ (AM-GM) we obtain:
$ \sum_{cyc} 5a^3 \plus{} abc \ge 3 \left ( \sum_{cyc} a^3 \plus{} abc \right ) \ge 3 \left ( \sum_{cyc} a^2b \plus{} a^2c \right )$
The last inequality is known as Schurs inequality.

Btw. $ \sum_{cyc}$ denotes the cyclic sum. That is $ \sum_{cyc} f(a,b,c) \equal{} f(a,b,c) \plus{} f(b,c,a) \plus{} f(c,a,b)$. E.g. $ \sum_{cyc} a^2b \equal{} a^2b \plus{} b^2c \plus{} c^2a$

Ps: Some of the users on the forum uses a lot of time on inequalities, so when they say they are easy, it just means that they think they are easy. It didn't take me much more than 10 minutes to get this one, so it probably was a bit easy.. But whatever :P

it's an other solution, by schur,
$ (a\plus{}b\plus{}c)^3\minus{}4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}9abc\geq0 \Leftrightarrow 9abc\geq 4(ab\plus{}bc\plus{}ca)\minus{}1$
$ \Leftrightarrow 3(a^2\plus{}b^2\plus{}c^)\plus{}9abc\geq a^2\plus{}b^2\plus{}c^2\plus{}2(a\plus{}b\plus{}c)^2\minus{}1\equal{}a^2\plus{}b^2\plus{}c^2\plus{}1$
by caushy, $ a^2\plus{}b^2\plus{}c^2\geq\frac{1}{3}$ then $ 3(a^2\plus{}b^2\plus{}c^2\plus{}3abc)\geq\frac{1}{3}\plus{}1\equal{}\frac{4}{3}$
$ \Leftrightarrow a^2\plus{}b^2\plus{}c^2\plus{}3abc\geq\frac{4}{9}$


Please give the mechanism of the reaction.

I have never heard of this reaction. Are you sure the starting compound is not phenol?

I think you must mean Br in the product. As far as I know (my professor gave the same question in an examination a year ago) it is because of [url=http://en.wikipedia.org/wiki/Ipso_substitution]IPSO substitution[/url] :maybe:

I'm sorry ,  :oops: I should have been more careful . Madness , thanks.


find all functions f with the property
f(x)=f(x/2) for all x

Let $f(x)=g(sign(x),log_{2}|x|)$. work all periodic function $g(sign(x),t+1)=g(sign(x),t)$.

can you do clearly for me?/ i dont understand

define a function on the interval (0.5,1], then f(x) is that function stretched and compressed by 2 for every interval 2^n,2^(n+1). LOL, simple when you think about it.

@highly_effective_teens: This problem is trivial, if you from Vietnam then you can find Mau Nguyen Van's book. I think it is quite good book on function equations.


If $x$ is a real and $4y^2+4xy+x+6=0$, then the complete set of values of $x$ for which $y$ is real, is:

$\text{(A)} \ x\le -2~\text{or}~x\ge3 \qquad \text{(B)} \ x\le 2~\text{or}~x\ge3 \qquad \text{(C)} \ x\le -3 ~\text{or}~x\ge 2$
$\text{(D)} \ -3\le x \le 2\qquad \text{(E)} \ \-2\le x \le 3$

[hide]Since we have a quadratic in $y$, why not just apply the quadratic formula? In order for $y$ to be real, we want the $\sqrt{16x^2-16(x+6)}=4\sqrt{x^2-x-6}$,the discriminant, to be positive.

$x^2-x-6=(x+2)(x-3)$, and is positive over the interval $x\le -2$ or $x\ge 3$, when both factors are negative or positive.

[/hide]

I was wondering if anybody would beat me. Am I right?

[hide]Rewrite as $4x^2+4xy+x^2-x^2+x+6=0$, $(x+2y)^2=x^2-x-6=(x+2)(x-3)$.
If $y$ is real, then $x+2y$ is real and $(x+2y)^2>0$, $x^2-x-6>0$ The solution set for this is described by $\boxed{A}$[/hide]


linear algebra
matrix
function
geometry
geometric transformation
integration
trigonometry

Problem statement: Consider the graph $ U$ where the vertices are points in $ \mathbb{R}^2$ and two points are joined if they are at distance $ 1$ from each other in $ \mathbb{R}^2$. What is the chromatic number $ \chi$ of this graph?

Known results: $ 4 \leq \chi \leq 7$

Proof: See the diagram here which has the $ 7$-coloring as well as a subgraph with chromatic number $ 4$: http://en.wikipedia.org/wiki/Image:Hadwiger-Nelson.svg. The hexagons all have diameter slightly less than $ 1$.

Idea: Let $ A$ be the "adjacency matrix" for this graph (in this case not actually a matrix but rather a linear operator). Let $ \lambda_{max}$ and $ \lambda_{min}$ be the largest and smallest eigenvalues of $ A$. Then we can show that
\[ \chi \geq 1 \minus{} \frac {\lambda_{max}}{\lambda_{min}}
\]
Proof: Let $ c_1,\ldots,c_{\chi}$ be the color classes. Let $ f$ be the largest eigenfunction of $ A$, and let $ f_1,\ldots,f_{\chi}$ be functions such that $ f_i(x) \equal{} f(x)$ if $ x \in c_i$ and $ 0$ otherwise. Then clearly $ f \equal{} f_1 \plus{} \ldots \plus{} f_{\chi}$. Then we have
\[ 2\lambda_{min}(c \minus{} 1)\parallel{}f\parallel{}^2 \equal{} \sum_{i,j} \lambda_{min}\parallel{}f_i \minus{} f_j\parallel{}^2\]
\[ \equal{} \sum_{i,j} \lambda_{min} < f_i \minus{} f_j,f_i,f_j > \leq \sum_{i,j} < f_i \minus{} f_j,A(f_i \minus{} f_j) >\]
But $ < f_i,Af_i > \equal{} 0$ if we have a proper coloring, since no two two points in $ c_i$ can be connected. Thus (combined with the fact that $ A$ is symmetric) the above reduces to
\[ \sum_{i,j} \minus{} 2 < f_i,Af_j > \equal{} \minus{} 2 < f,Af > \equal{} \minus{} 2\lambda_{max}\parallel{}f\parallel{}^2
\]
Thus $ 2\lambda_{min}(c \minus{} 1)\parallel{}f\parallel{}^2 \leq \minus{} 2\lambda_{max}\parallel{}f\parallel{}^2$

Then, since $ \lambda_{min}$ is negative, when we divide by $ 2\lambda_{min}\parallel{}f\parallel{}^2$ we get $ c \minus{} 1 \geq \frac {\lambda_{max}}{\lambda_{min}}$, which is what we wanted.

Note that this logic holds for any linear operator $ B$ where $ A_{ij} \equal{} 0 \implies B_{ij} \equal{} 0$. Now, assuming that $ B$ is an operator that commutes with translation operators, i.e. $ B(f(x \plus{} t)) \equal{} (Bf)(x \plus{} t)$, then the eigenvectors of $ B$ are given by the Fourier coefficients over $ \mathbb{R}^2$ corresponding to $ Bf$, where $ f$ is such that $ f(0) \equal{} 1$ and $ f(x) \equal{} 0$ if $ x \neq 1$. Working this out, we see that they can be parameterized as
\[ \mu(r,s) \equal{} \int_{ \minus{} \pi}^{\pi} f(\theta) e^{i(r\cos(\theta) \plus{} s \sin(\theta))} d\theta
\]
for any $ f$ we choose, provided that $ f(x) \equal{} f(x \plus{} \pi)$ so that the integral evaluates to a real number always. If $ f$ is the constant function, then using this integral to give us a bound on the chromatic number is not quite good enough to show that $ \chi \geq 5$, but it is possible that a better $ f$ will yield a better bound. Can we find such an $ f$, or prove that no such $ f$ exists? Also note that we can re-paramterize with a proper set of substitutions as
\[ \mu(t,\phi) \equal{} \int_{ \minus{} \pi}^{\pi} f(\theta \plus{} \phi) e^{itcos(\theta)} d\theta
\]
So, the basic question is, can we find an $ f$ to get a better bound than the current one on the chromatic number?

Comment: I've used this same idea with seemingly more success on the [i]odd-distance graph[/i], where two points are connected if their distance is an odd integer. There we just need to show that a certain integral is finite, and if I can get the asymptotics down right I think I can show that. I'd prefer not to type the work for that problem up for now, since I think I'm close to solving it. If I get stuck or end up solving it, I'll put it on arXiv and link to it from here.

Also, I apologize if the exposition is a bit terse here. I will try to make the explanations more detailed later.

Quick question:  what vector space is $ A$ as a linear operator operating on?

We can associate the vertex set $ V$ with $ \mathbb{R}^2$, so $ A$ acts on $ L^2(\mathbb{R}^2)$. One thing to note is that this approach actually only tells us about measurable colorings of the plane, but this question is also interesting (I believe that the current best bound for measurable colorings is $ 5$ colors).


[hide]
No such sequence exists. Suppose one did. Then $P(n_{k})\neq 0$ for all $k$, so no $n_{k}$ has any of its digits 0. In particular, letting $a$ be the number of digits of $n_{1}$, for each $d\geq a$, the sequence must go from a number with at most $d$ digits to one with at least $d+1$ digits.

If $n_{k}$ has at most $d$ digits and $n_{k+1}$ has at least $d+1$, then $n_{k}\leq 10^{d}-1$ and $n_{k+1}\geq 10^{d}+10^{d-1}+\cdots+1=\frac{10^{d+1}-1}{9}$ (no digit can be 0). In particular, this means $P(n_{k})\geq \frac{10^{d}+8}{9}$.

However, we also have $P(n_{k})\leq 9^{d}$ (since at most $d$ digits, each at most 9). Therefore, we must have $9^{d+1}\geq 10^{d}+8$ for each $d\geq a$, but this is false for sufficiently large $d$ (actually, $d\geq 21$ is enough), a contradiction.
[/hide]


MATHCOUNTS
Ring Theory
FTW
AMC
USA(J)MO
USAMO
AMC 8

what time will it be aired tomorowed????
is it after 4:00 eastern time? if not, i will be mad because i will have to skip school to watch it and i cant skip school(although i would like to)

Don't worry its 5:00p.m. eastern, 3:00 mountain. And it says on their website.

Edit:@ajain, i didn't feel like posting another post so hopefully your looking at this. It is to cover countdown. I can't wait. I can't wait. Oh shoot, I have track practice.  :(

Is this regarding the webcast of the Nation MathCounts Countdown round?

Here is the webpage:

[url]http://mathcounts.org/NETCOMMUNITY/Page.aspx?pid=1290&srcid=1290[/url]

But, if you do miss it, it will be archived in the same link i provided.

Message from Denver!!!

Both replies are right.  Sprint starts in like 13 hours.  Shentang, Alan is still using your lucky pencil at all our practices and doing well!!

We'll be sure to post results as soon as we can, but hopefully you catch the webcast.

Frost

Its 30 minutes before the announcing of the winners. From what I have heard, 43 was the top score.

i think it was bobby shen..
my coach told me that a 6th grader from the team he proctored got 41

Dang, people are doing good. I wish I was there. Guess what? My track practice was canceled because of the rain, now I can watch the web cast. @frost13, yep, thats the pencil I used to get the cookie and win bingo. Ah, good times.

ahh!
i am using a crap computer right now. i have 2 right next to me. the one i am typing from is my fast one and there is no link showing up.

Should we be able to see it right now? It's not working for me  :( .

Yes, the webcast is working for me right now.

Is the webcast really live? According to tinytim's post, the thing should have started an hour and a half earlier...

Shoot, I really hate this, so many downloads, I am missing all the fun.....even tried my friends computer and my dad's. :(

kyyuan or whoever is whatching can you give me an update. i cant hear a thing. i just know theyre anouncing some teams

Me too please, cause I don't see or hear anything. Only thing I will be doing-refreshing this page over and over again till someone posts.

You admitted Darryl's luck, didn't you? So why were you expecting maximized credit for Darryl? Just because he's a sixth grader, and he made USAMO, and perfect score on AMC 8, and perfect 800 SAT math score, and a USCF top-rated chess player, or something like those?

How did Maryland get 3 people in top 6 and not win 1st team.... hmmm... they must have done (relatively) poorly on the team round... but still, Texas only had one person in cd. Was Texas written something like 1, 13, 14, 15?

And Indiana must be pretty disappointed.

But congrats to all. (I feel brave congratulating people who are 1000000 times better than me...)

What do you mean Darryl almost failed to make finals? Was it because of Evan Miller?

[quote="mathcrazed"]
Things of note:

Yeah... ermm... sorry.
Excuse list: Bloody nose at 2:30 AM - 3:00 AM day of competition.  Bloody nose when I woke up around 7.  Took advil before test (which isn't supposed to do anything anyway XD).  Bloody nose the day before and the night before (something with bloody noses in spring).  Braces adjusted on Tuesday, but then my bond broke on wednesday and started poking into my cheek.  Uber stomachache the night before.

[/quote]

hey i also had a bloody nose that morning. only i woke up three times during the night and my pillow had some stains of blood.  :oops:

[quote="math154"]

@leoxnlin: They screwed up on the team round. Indiana screwed up too. According to tinytim who heard from Kevin Tian, I heard Lyndon Ji and the Kevins missed "every other" question on the Sprint by misreading.

[/quote]

Hmm.... and Texas had a perfect team round?

[quote="ZhangPeijin"]Hmm, it appears that Denver is a really dry place, with all the bloody noses...
Didn't you have a humidifier in your hotel rooms?
Anyways, I guess most of you are just surprised that a 6th grader won and are placing it on luck...
(Just like me)
Anyways, Darryl seemed sort of... unhappy, when he won... (Hunched back, starting at the floor, shuffling his feet, etc.)[/quote]
Unhappy? :huh: No way, you should have said...unenthusiastic. :o

Ha, I DOUBT he was unenthusiastic.  I bet it was just the shock of suddenly becoming National Champion and that the reality of the situation didn't hit him at that very moment.  I'll bet he's feeling EXTREMELY happy, and lucky right now  :D

I have a feeling it's way more complicated than that...

[quote="math154"][quote="Fanatic"]Could anyone clarify "by luck"? I think we're not giving Darryl Wu nearly enough credit here... or did I miss something?[/quote]

Well, I have to admit Darryl is really beast, but there were a lot of things that made it more "by luck". For example, Lyndon Ji, Kevin Tian, Kevin Li, and many other really good people who should've qualified for countdown did not. Darryl got in the "easier" brackets (I think). He almost did not make finals. Bobby made 4 careless mistakes or something like that. Also, you could say that anyone could have won in Darryl's place in the final round--because he won by getting each of the questions he had 44.99999999 (emphasizing Bobby's speed) seconds on. Again, Darryl is really good, but I think many people are still better than him.

@leoxnlin: They screwed up on the team round. Indiana screwed up too. According to tinytim who heard from Kevin Tian, I heard Lyndon Ji and the Kevins missed "every other" question on the Sprint by misreading.

Anyway, I think nationals was really screwed up this year...[/quote]

umm when i used to lose at something i always thought the other person/people ot luckier than me and that was the reason why I lost and they won, but now i realize maybe they are actually better...
i dont get how other good people who shouldve qualified has anything to do with wheter he won or not...maybe you think they are better than him or something just because youve heard of them before and you never heard of darryl...
ok he made 4 careless mistakes, thats kinda the point of the competition, not to make mistakes so you cant say he got lucky just because his opponent made careless msitakes, you have to give him the credit for not making the careless mistakes...and the 45 seconds, thats actually really smart...you want to use up all your time to check your work so you dont make carless mistakes, im sure he couldve gotten some earlier...and the point about he couldve lost earlier and might not have made the finals, well, maybe had he lost, the person that wouldve beat him might of still beat bobby in the finals and you would be making the same argument against that guy...anyways doesnt matter what i think, the fact of the matter is he won so hes the best...besides he also made usamo as a 6th grader so you cant say he just got "lucky" on one day (and no disrespect, but the guy you are trying to defend did not make usamo in 7th grade).

[quote="the future"](and no disrespect, but the guy you are trying to defend did not make usamo in 7th grade).[/quote]

this is when i get ticked off when people spout off facts about people when they don't know the true situation...

Bobby Shen, like Kevin Chen before him, only took the AMC 8 in 6th grade and 7th grade. So, he didn't make the USAMO because he didn't try to make the USAMO.

I just feel like a lot of people wouldn't be saying the things they are if they knew Darryl posted on AoPS. If I was Darryl and I just won Nats, I log onto AoPS and everyone's spouting off about how lucky I got... I'd be pissed. Down 3-0, 3-1, etc. approaching the 45th second of the countdown question... it's insanely hard to keep your cool.

true, the fact is that Darryl beat Bobby, and one can't say that Bobby didn't win because of stress, because whatever bobby was experiencing, darryl was too. Anyway nice job to both of you. i have a bad post count.

OK Guys, I'm gonna go ahead and lock this.

They both did a great job. Congratulations to both Bobby and Darryl. Special super congratulations for beating out the 8th graders. Maybe we'll have a classic rematch next year!


Solve for $ x$ in terms of $ a,b$: $ |x\minus{}a|\equal{}|x\minus{}b|$.

[hide="solution1"]
By the number line, the distance from $ x$ to $ a$ equals the distance from $ x$ to $ b$ (in terms of magnitude). Thus $ x\equal{}\frac{a\plus{}b}{2}$.
[/hide]

How do we do this the "normal way" (not by number line)?

You mean algebraically?

$ x\minus{}a\equal{}\minus{}(x\minus{}b)\equal{}\minus{}x\plus{}b\Rightarrow 2x\equal{}a\plus{}b\Rightarrow x\equal{}\frac{a\plus{}b}{2}$

Just use the rules of absolute value arithmetic.

This is as clearly as I can put it:
$ |x - a| = |x - b|$
Divide both sides by $ |x - a|$:
$ 1 = \frac {|x - b|}{|x - a|}$
This is the same as:
$ 1 = {|}\frac {x - b}{x - a}{|}$
Which means:
$ 1 = \frac {\pm(x - b)}{x - a}$
So, solve the resulting equations:
$ 1 = \frac {x - b}{x - a}$
$ x - a = x - b$
$ \boxed{a = b}$
That's not very informative, but the next equation:
$ 1 = \frac {b - x}{x - a}$
$ x - a = b - x$
$ 2x = b + a$
$ \boxed{x = \frac{b + a}{2}}$
Which is also equal to $ 2b/2$ or $ 2a/2$ by substitution, which simplifies to $ x = a$ or $ x = b$.
That makes the original equation $ x - x = x - x$ by more substitution, which leads to $ 0 = 0$, which signifies that there are infinitely many solutions.


geometry
rhombus
perimeter
analytic geometry
quadratics
geometry unsolved

A circle intersects each side of a rhombus twice, dividing each side into three parts. Starting from any vertex, go around the rhombus in a fixed direction, color the segments red, white, blue, red, white, blue, red, white, blue, red, white, blue. Prove that the sum of the lengths of the red segments is equal to that of the blue ones.

Let $ XYZT$ be a rhombus, let a circle $ (P)$ with radius $ r$ intersect the rhombus sides $ XY, YZ, ZT, TX$ at $ A, B, C, D, E, F, G, H,$ so that the points $ X, A, B, Y, C, D, Z, E, F, T, G, H$ follow on the rhombus perimeter in this order. We have to show (using directed line segments)

$ \Sigma_{RED} \minus{} \Sigma_{BLUE} \equal{} (\overline{AX} \plus{} \overline{CY} \plus{} \overline{ZE} \plus{} \overline{TG}) \minus{} (\overline{YB} \plus{} \overline{ZD} \plus{} \overline{FT} \plus{} \overline{HX}) \equal{} 0$

Put the coordinate origin at the intersection $ O$ of the rhombus diagonals, x-axis along $ OX,$ y-axis along $ OY.$ Equations of the rombus sidelines and of the circle $ (P)$ are

$ XY: \ \ \ \ y \equal{} \minus{} m(x \minus{} q)$
$ YZ: \ \ \ \ y \equal{} \plus{} m(x \plus{} q)$
$ ZT: \ \ \ \ y \equal{} \minus{} m(x \plus{} q)$
$ TX: \ \ \ \ y \equal{} \plus{} m(x \minus{} q)$

$ (P): \ \ (x \minus{} a)^2 \plus{} (y \minus{} b)^2 \equal{} r^2$

where $ \pm q$ are the x-coordinates of $ X, Z.$ Substituting the 4 line equations into the circle equation yields 4 quadratic equations for the intersections:

$ A, B: \ \ (1 \plus{} m^2) x^2 \minus{} 2(a \minus{} mb \plus{} m^2q)x \plus{} ... \equal{} 0$
$ C, D: \ \ (1 \plus{} m^2) x^2 \minus{} 2(a \plus{} mb \minus{} m^2q)x \plus{} ... \equal{} 0$
$ E, F: \ \ (1 \plus{} m^2) x^2 \minus{} 2(a \minus{} mb \minus{} m^2q)x \plus{} ... \equal{} 0$
$ G, H: \ \ (1 \plus{} m^2) x^2 \minus{} 2(a \plus{} mb \plus{} m^2q)x \plus{} ... \equal{} 0$

Putting $ k \equal{} \frac {1}{1 \plus{} m^2},$ sums of the roots are

$ x_{A} \plus{} x_{B} \equal{} 2k(a \minus{} mb \plus{} m^2q)$
$ x_{C} \plus{} x_{D} \equal{} 2k(a \plus{} mb \minus{} m^2q)$
$ x_{E} \plus{} x_{F} \equal{} 2k(a \minus{} mb \minus{} m^2q)$
$ x_{G} \plus{} x_{H} \equal{} 2k(a \plus{} mb \plus{} m^2q)$

It is sufficient to prove the proposition for projections of the line segments to the x-axis. $ O$ is projection of $ Y, T$ on the x-axis. Let $ A', B', ..., H'$ be projections of $ A, B, ..., H$ on the x-axis.

$ (\overline{A'X} \plus{} \overline{C'O} \plus{} \overline{ZE'} \plus{} \overline{OG'}) \minus{} (\overline{OB'} \plus{} \overline{ZD'} \plus{} \overline{F'O} \plus{} \overline{H'X}) \equal{}$

$ \equal{} \left[(q \minus{} x_{A}) \plus{} (0 \minus{} x_{C}) \plus{} (x_{E} \plus{} q) \plus{} (x_{G} \minus{} 0)\right]\ \minus{}$

$ \minus{} \ \left[(x_{B} \minus{} 0) \plus{} (x_{D} \plus{} q) \plus{} (0 \minus{} x_{F}) \plus{} (q \minus{} x_{H}) \right] \equal{}$

$ \equal{} \left[(x_{E} \plus{} x_{F}) \plus{} (x_G \plus{} x_{H})\right] \minus{} \left[(x_{A} \plus{} x_{B}) \plus{} (x_{C} \plus{} x_{D})\right] \equal{} 0$

Q.E.D.


ratio
analytic geometry
rotation
geometry
power of a point
Pythagorean Theorem

If circular arcs $ AC$ and $ BC$ have centers at $ B$ and $ A$, respectively, then there exists a circle tangent to both $ \stackrel{\frown}{AC}$ and $ \stackrel{\frown}{BC}$, and to $ \overline{AB}$. If the length of $ \stackrel{\frown}{BC}$ is $ 12$, then the circumference of the circle is
[asy]unitsize(4cm);
defaultpen(fontsize(8pt)+linewidth(.8pt));
dotfactor=3;

pair O=(0,.375);
pair A=(-.5,0);
pair B=(.5,0);
pair C=shift(-.5,0)*dir(60);
draw(Arc(A,1,0,60));
draw(Arc(B,1,120,180));
draw(A--B);
draw(Circle(O,.375));
dot(A);
dot(B);
dot(C);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);[/asy]$ \textbf{(A)}\ 24 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 28$

[hide]From Power of a Point and substitution, you can find the ratio of the radius as well as circumference of the smaller circle to AB is $ \frac{3}{8}$. ABC is an equilateral triangle as all sides are radii of congruent circles so arc BC is a sixth of the circumference of circle A. This means the circumference of the smaller circle is $ \frac{3}{8}*72\equal{}27$ or D.[/hide]

I bashed with coordinates. Not that quick, but it works.

[quote="Lazarus"]Cutting a string the length of $ \stackrel{\frown}{BC}$, we find that it is pretty much half the circumference of the circle. $ 12\cdot 2 \equal{} 24$; the answer is $ \boxed{A}$.[/quote]
Human error is a terrible thing.  "Pretty much" is not the same as "exactly," especially when the answer choices are rather close to each other numerically.

how is it equilateral if AB is not an arc but rather a line? is it not isoceles?

nvm

GlenthemanN:
Since B is the center of the arc AC, we have AB=BC. Similarly, since A is the center of the arc BC, we have BA=AC. This gives us AB=BC=CA, or ABC being an equilateral triangle.

mathwiz, could you explain how you get $ (36\pi \minus{} R)^2 \equal{} R^2 \plus{} 18^2/\pi^2\implies R \equal{} 972/(72\pi)$

ive tried going the coordinate route to no avail, I have
$ (x\minus{}r)^2 \plus{}y^2 \equal{} r^2$
$ (x\plus{}r)^2 \plus{}y^2 \equal{} r^2$
and for the circle
$ x^2 \plus{} (y\minus{}R)^2 \equal{} R^2$

big thanks to anyone who helps out  :blush:

If one circle is internally tangent to another, then the centers of the two circles and the point of tangency are collinear. This is harder to see in my opinion than power of a point is.

In this case, draw a segment from A to the point of tangency opposite it (T), let it cut the circle at Q, let M be the midpoint of AM, O the center of the inscribed circle. Then A,O, and T are collinear, so by power of a point, $ AQ \cdot AT \equal{} AM^2 \equal{} (\frac {AQ}{2})^2$. Then the problem is easy, since we know AQ.

[quote="GlenthemanN"]yes thats the issue I had. Cant believe I didn't see it, guess the two arcs threw me off. I wasnt sure if the line BO extended would pass through the tangent or not  :ninja: got it not. makes a lot more sense[/quote]
The diagram is a bit misleading in this regard, but it is a fact that if you have an circle internally tangent to another, then the centers of both circles and the point and tangency will be collinear. It's a bit more obvious if you try a few diagrams yourself, and maybe rotate the paper around when you look at it...

Quick question-- why isn't the triangle made by A, the midpoint of AB, and the center of the circle a 30-60-90? Or is it? After finding AB I tried using ratios to find r but it doesn't work.

It's not  30-60-90. Can you elaborate why you think it is?

Denote $AB=r$ and $\stackrel{\frown}{BC}=l$ . Observe that  $\pi r=3l\implies \boxed{r=\frac {3l}{\pi}}\ (*)$ . Denote the length $x$ of the radius for the circle $w$ which

 is tangent to $AB$ and which is interior tangent to the circles $C(A,r)\ ,\ C(B,r)$ . Hence $(r-x)^2=x^2+\left(\frac r2\right)^2$ $\implies $ $x=\frac {3r}{8}$ .

In conclusion, the length of the circumference $L$ for the circle $w$ is $2\pi\cdot\frac {3r}{8}=\frac {3\pi }{4}\cdot r\stackrel{(*)}{=}\frac {3\pi}{4}\cdot\frac {3l}{\pi}=\frac {9l}{4}$ . For $l:=12$ obtain that $L=27$ .

I thought that it would be since triangle ABC is equilateral.

[quote="SalsaOnStrings"][quote="Ignite168"][hide]From Power of a Point and substitution, you can find the ratio of the radius as well as circumference of the smaller circle to AB is $ \frac{3}{8}$. [/quote]
Could you elaborate on where this particular result came from? I'm not seeing where power of a point comes in.[/quote]

I dont understand that either. Where does the PPT come from?

Edit: Nvm. I figured it out.

[asy]
unitsize(2cm);
defaultpen(fontsize(8pt)+linewidth(.8pt));
dotfactor=3;
pair O=(0,.375);
pair A=(-.5,0);
pair B=(.5,0);
pair C=shift(-.5,0)*dir(60);
draw(Arc(A,1,0,60));
draw(Arc(B,1,120,180));
draw(A--B);
draw(Circle(O,.375));
dot(A);
dot(B);
dot(C);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
draw(C--B,blue);
draw(C--(0,0),blue);[/asy]

[quote=macandcheese][quote=mathwizarddude]My solution:

From the equilateral triangle, we have $ 12 \equal{} 1/6 * 2\pi r \implies r \equal{} 36/\pi$
$ (36\pi \minus{} R)^2 \equal{} R^2 \plus{} 18^2/\pi^2\implies R \equal{} 972/(72\pi)$
$ 2\pi R \equal{} 972/36 \equal{} 27$[/quote]

Could someone explain the $(36\pi \minus{} R)^2$ term?[/quote]
It was a typo. It's supposed to be $(36/\pi-R)^2 = R^2 + 18^2/\pi^2$, where $R$ denotes the radius of the circle.

If we let M be the midpoint of AB and O the center of the circle, then it's just an application of Pythagorean theorem on triangle OMB.

Since $\Delta ABC$ is equilateral, $\angle CAB = 60$ degrees, so the circle with center $A$ containing points $B$ and $C$, has circumference $\frac{360}{60} \cdot 12=72$ and radius $\frac{36}{\pi}$.
Draw the altitude from $C$ to $AB$ and let the intersection with $AB$ be point $M$. Let the radius of the circle we are trying to find be $r$ and let it's center be $O$. Then, extend $AO$ until it intersects circle $A$; call this point of intersection $N$. Then, since $AN=\frac{36}{\pi}$, $AO=\frac{36}{\pi}-r$. We also know that $OM=r$ and $AM=\frac{18}{\pi}$, so by the Pythagorean theorem, $\left(\frac{36}{\pi}-r\right)=r^2+\left( \frac{18}{\pi} \right)^2 \implies$ $r=\frac{27}{2\pi}$. Hence, our answer is $\boxed{27\pi}$.


two guys were thrown in jail and as they were sitting they wanted to know each other's sentence. All they know was that their sentences differ by one year. they had the following coversation:

Person1: I don't know how much you got
Person2: I know you don't know how much I got
person1: Then I know how much you got
person2: Then I also know how much you got

what are the sentences of both of these men?

I like this, but it's not calculus. :)

When person A says he doesn't know what the other got, A can't have 1 year. If he did B would have 2 and A would be able to figure it out.

When B says he is aware of this we know that B can't have 2 either. If B had 2 there would be a possibility that A had 1.

When A then says he knows how much B got, we know that A must have 3. (So B is 2 or 4 but we know that B can't be 2). Again, A can't have 1.

So A has 3 and B has 4. I believe.

The jail sentence are as follows:

Person 1 --> n years
Person 2 --> (n + 1) or (n-1) years

Person1: I don't know how much you got
[b]Person1 is not having a sentence for 1 year or else he could have known that Person2 is having it for 2 years (there cannot be a sentence for 0 years)[/b].

Person2: I know you don't know how much I got
[b]This means Person2 is not having the sentence for 2 years. Then, he/she could have been so sure[/b].

person1: Then I know how much you got
[b]Person1 can interpret this message and he/she comes to know how much Person2 has got. This implies that Person1 is having a sentence for 3 years. And, he/she comes to know that Person2 is not having it for 2 years. Thus, Person2 has it for 4 years[/b].

person2: Then I also know how much you got

[b]Person1 ==> 3 years
Person2 ==> 4 years[/b]


For which integers $n\geq 2$, the number $(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}$ is divisible by $n^{n}$ ?


I am surprised because of it was a contest problem !

Any way, we can get a solution of the wanted result, applying the proof of the generalization of a familiar problem in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140191]Interesting and hard[/url], of [b][size=100]cvix[/size][/b].

Kostas Vittas.

Can someone write out a solution?I can't continue [b]vittasko[/b]'s comment...

[quote="Omid Hatami"]Let $ A,B,C,Q$ be fixed points on plane. $ M,N,P$ are intersection points of $ AQ,BQ,CQ$ with $ BC,CA,AB$. $ D',E',F'$ are tangency points of incircle of $ ABC$ with $ BC,CA,AB$. Tangents drawn from $ M,N,P$ (not triangle sides) to incircle of $ ABC$ make triangle $ DEF$. Prove that $ DD',EE',FF'$ intersect at $ Q$.[/quote]

we just need prove $CP,NB$ and $D'D$  are concurrent at $Q$

 Tangents drawn from $ N,P$ (not triangle sides) to incircle of $ ABC$  intersection $AB, AC at J , K$  and  intersection incircle of $ ABC$ at $O,V$

*)  $PKCB$  is a special quadrilateral  ( i don't know this name :( )  and we know following lemma:
$F'O$ intersection $D'E'$ at $X$ then $X$ is pole of $PC.$

*) $NJBC$ is a special quadrilateral  and $E'V$ intersection $D'F'$ at $Y$ then $Y$ is pole of $BN$

*) $OV$ intersection Tangents drawn from $D'$  to incircle of $ ABC$ at $Z$ then $Z$ is pole of $DD'$

and  use Pascal theorem for $(D'D'F'VOE')$ we have $XYZ$ therefore $DD',PC,BN$ are concurrent
[img]http://ns0.upanh.com/b1.s33.d1/eae7bc59240a3ffea1a834c2d9a151b9_50500680.untitled.700x0.jpg[/img]

The idea of using pole-polar and Pascal is good, but additionally I think the whole problem is rested on nothing else than two well known properties of pole-polar: La Hire’s Theorem ([i]Let x and y be the polars of X and Y, respectively; then X is on line y ⇔ Y is on line x[/i]) and this: [i]Let x, y, z be the polars of distinct points X, Y, Z, respectively; then Z = x∩y ⇔ z = XY[/i] (what can be proved easily using La Hire).

Let $E''F''$ meet $BC$ at $G$, $E'E''$ meet $F'D'$ at $H$, $F'F''$ meet $D'E'$ at $K$. From now on, while talking about poles and polars we do so with respect to the incircle of $ABC$.
First notice that $E''F''$ is the polar of $D$, as $G$ lies on the polar of $D$, $D$ must lie on the polar of $G$, but we know that the tangent point $D'$ is also on the polar of $G$, so $DD'$ is the polar of $G$. Then to show that $DD'$ passes through $Q$, it suffices to show that $Q$ lies on the polar of $G$.
Now notice that $F'F''$ is the polar of $P$ and $D'E'$ is the polar of $C$, then $PC$ is the polar of $K\equiv F'F''\cap D'E'$.
$E'E''$ is the polar of $N$ and $F'D'$ is the polar of $B$, then $NB$ is the polar of $H$. Then $HK$ is the polar of $Q\equiv PN\cap BN$.
Using Pascal for $E''F''F'D'D'E'$ we get $G\equiv E''F''\cap D'D', K\equiv F''F'\cap D'E', H\equiv E''E'\cap F'D'$ collinear.
As $G$ lies on the polar of $Q$, $Q$ must lie on the polar of $G$. q.e.d.
By the same way we get $EE'$ and $FF'$ pass through $Q$.

Note: Having a look back at the [b]cvix[/b]'s [i]interesting and hard[/i] [b]vittasko[/b] mentioned above, we see that the problem can be solved in the same way, of course with the aid of a well known lemma (instead of Pascal in our problem).

We can use brianchon's theorem on $BD'CNDP$ then we get Q, D, D' are collinear, it finishes the proof. :D


inequalities
logarithms
real analysis
real analysis unsolved

Let $p,q$ are positive real numbers.
Prove the following inequality.

\[\frac{p^2+q^2}{2}\ln \frac{p^2+q^2}{2}\geqq -\frac{1}{2}(p-q)^2+\frac{p^2\ln p^2+q^2\ln q^2}{2}\]

I will. ;)
let $f(p)=\frac{p^2+q^2}{2}\ln \frac{p^2+q^2}{2} +\frac{1}{2}(p-q)^2-\frac{p^2\ln p^2+q^2\ln q^2}{2} $
we have $f'(p)=\frac{p^2+q^2}{2} \frac{2}{p^2+q^2}p+p ln \frac{p^2+q^2}{2} +(p-q)-pln p^2-p=p ln \frac{p^2+q^2}{2} +(p-q)-pln p^2$
try to show $f'(p) \leq 0,p \geq q$,$f'(p) \geq 0,p \leq q$(by$f\"(p)$
then we are done :)
thank you for your question

Thank you for your reply,zhaobin.

You are right.

kunny


Let there be a polynomial such that
$ P(x)\equal{}x^{4}\plus{}ax^{3}\plus{}bx^{2}\plus{}cx\plus{}d$.
If $ P(1)\equal{}10$,
$ P(2)\equal{}20$, and
$ P(3)\equal{}30$, find $ \frac{P(10)\plus{}P(\minus{}8)}{10}$.

Let $ Q(x) \equal{} P(x) \minus{} 10x$, which is a monic quartic with roots $ x \equal{} 1,2,3,p$. Then $ Q(x) \equal{} (x \minus{} p)(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)$, and $ \frac {P(12) \plus{} P( \minus{} 8)}{10}$ $ \equal{} \frac {Q(12) \plus{} 120 \plus{} Q( \minus{} 8) \minus{} 80}{10}$ $ \equal{} \frac {[12 \minus{} p]\cdot 9 \cdot 10 \cdot 11 \plus{} 120 \plus{} [ \minus{} 8 \minus{} p] \cdot ( \minus{} 1) \cdot 9 \cdot 10 \cdot 11 \minus{} 80}{10}$ $ \equal{} \frac {20 \cdot 9 \cdot 10 \cdot 11 \plus{} 40}{10} \equal{} 1984$.

- edited.

Oh..OOPS I made a mistake in the question.
You didn't make a stupid mistake...my question was wrong.

It's supposed to be
Find $ \frac{P(12)\plus{}P(\minus{}8)}{10}$.
Sorry =\


Let $A$ be a square matrix of order 3. If $A\acute{A}= I$ and $det.A = 1$, then $det.(A-I) = ?$

Also can anyone help me with some important properties of orthogonal and hermitian matrices ?

I think he meant $A A^{T}= I_{n}$. That's the definition of an orthogonal matrix.

$det(A-I)=0$ . we can prove this as follows. $det(A-I)=detA^{T}.det(A-I)$ since$detA^{T}=1$. therefore $det(A-I)=det(A^{T}A-A^{T})=det(I-A^{T})=det(I-A)^{T}=det(I-A)$. since A is of order 3, $det(I-A)=-det(A-I)$ which gives the answer that the value of the determinant is 0 :wink:


A mathematical prodigy wishes to put 2 of his IMO gold medals and 2 of his IPhO gold medals in one row. How many distinct arrangements are possible?

Assuming the medals are indistinguishable, the answer is $ \frac{4!}{2!2!}$, or $ \boxed{6}$.

Or we could do $ \binom{4}{2}$ to pick the two spots to put the IMO medals and the rest must go into their slots.

This kid has 2 IMO gold medals and 2 IPhO gold medals. He can't figure this out himself?

[quote=huili2010]Bruh, he probably already knows how to do it and the problem is not asking if he knows it or not. Plus, he could've stolen them people.[/quote]

good point.

He is a [quote]...mathematical prodigy...[/quote] He should be able to figure it out.

[hide = Solution if distinguishable]There are $ 4! = 24$ ways to order the medals. [/hide]
[hide = Solution if indistinguishable]There are $ \frac{4!}{2!2!}=6 $ ways to order the medals. We divide by $ 2 $ twice because the IMO and IPhO medals are indistinguishable, but IMO and IPhO medals are probably different. [/hide]

[hide=my dumb solution 1: Write down the options]I write them down. There are $\boxed{6}$[/hide]
[hide=my dumb solution 2]We see that there are $4!$ ways if they were distinguishable. We divide by $2\cdot2$ to account for indistinguishability, for an answer of $\boxed{6}$[/hide]


saabet konid ke be ezaaye har adade tabiii mese $n$,
$\sum_{k=0}^{n} \left(\begin{array}{c} {n-k} \\ {[\frac{n-k}{2}]}\end {array} \right)2^k\left(\begin{array}{c} n\\k\end{array} \right)=\left(\begin{array}{c} 2n+1\\n \end{array} \right)$

khodayish poostam kande shod ta neveshtamesh :stretcher:

[color=red]Moderator Edit : LaTeX Formula Corrected[/color]

Man ke nafahmidam shayad manzooret in bashe: $\sum_{k=0}^{n}\binom{n-k}{[\frac{n-k}2]} 2^k\binom nk=\binom {2n+1}n$

Dar zemn mitoonid be jaye in hame array neveshtan az [color=red]\binom{m}{n}[/color] baraye $\binom mn$ estefade konid.


function
real analysis
real analysis unsolved

Can someone give me the answer to this question?

(X,M) is a measurable space. Show that
if f:X -> [-oo, oo] and f^-1((r,oo]) in M for all rational r,
then f is measurable.

The definition of 'measurable function' is that $f^{-1}(B)$ is measurable for every Borel set $B$, right?

So consider $X = \{X\text{ is a subset of }\mathbb{R} \mid f^{-1}(X)\text{ is measurable }\}$. Show that $X$ is an $\sigma$-algebra. If $X$ contains all $(a,\infty]$, we can conclude that all Borel sets are in $X$, and $f$ is a measurable function.

Now we have only that $(r,\infty]\in X$ for all rationals $r$. For a real $a$, we can find a sequence of rationals $r_n\searrow a$, $(a,\infty] = \bigcup (r_n,\infty]$ then $f^{-1}(a,\infty] = \bigcup f^{-1}(r_n,\infty]$. Hence $(a,\infty]\in X$.


function calculus derivative inequalities

Show whether the function $ f(x)\equal{}2x^3\minus{}x\plus{}1$ is convex or concave.

I think that $ f"(x)\equal{}12x$ is enough to solve the problem!! no?

Adding to the post above, that implies that the function is concave on $ ( \minus{} \infty,0)$ and convex on $ (0,\infty)$.

You could prove it with the converse of Jensen's inequality:

Prove that for all nonnegative $ x,y$,

$ \frac{f(x)\plus{}f(y)}{2}\ge f\left(\frac{x\plus{}y}{2}\right)$.

Although taking the second derivative is without a question the easiest way.

[quote="Yongyi781"]Adding to the post above, that implies that the function is concave on $ ( \minus{} \infty,0]$ and convex on $ [0,\infty)$.[/quote]

Those aren't disjoint partitions of the continuum. $ x \equal{} 0$ is an inflection point of the function - you could describe it as nonconvex and nonconcave, but I believe concave and convex are terms limited to a positive second derivative.

Indeed, you are right. I was being absent-minded when I posted that :blush:

[quote="modularmarc101"]How would you do it without the second derivative?[/quote]
we can say that $ f$ is convex if:

$ \forall(x,y)\in{I^{2}}  \forall(\alpha,\beta)\in{[0,1]^{2}}  \alpha\plus{}\beta\equal{}1\Rightarrow f(\alpha x\plus{}\beta y)\le\alpha f(x)\plus{}\beta f(y)$

Are there any times when the second derivative doesn't work?


trigonometry
geometry
trig identities
Law of Sines

Prove the Law Of Tangents

$ \frac{a \minus{}b}{a\plus{}b}$ = $ \frac{tan\frac{(A\minus{}B)}{2}}{tan\frac{(A\plus{}B)}{2}}$

[hide="Proof"]Since $ \tan x \equal{} \frac {\sin x}{\cos x}$, we know that

$ \frac {\tan (\frac {A \minus{} B}{2})}{\tan ( \frac {A \plus{} B}{2})} \equal{} \frac {\sin (\frac {A \minus{} B}{2}) \cos (\frac {A \plus{} B}{2}) } {\sin (\frac {A \plus{} B}{2}) \cos (\frac {A \minus{} B}{2})}$

By the Product-to-Sum identity, we can rewrite this as

$ \frac {\sin A \minus{} \sin B}{\sin A \plus{} \sin B}$

Now notice that by the Law of Sines,

$ \sin M \equal{} \frac {m}{2R}$  so,

$ \frac {\sin A \minus{} \sin B}{\sin A \plus{} \sin B} \equal{} \frac {a/2R \minus{} b/2R}{a/2R \plus{} b/2R} \equal{} \frac {a \minus{} b}{a \plus{} b}$ [/hide]


probability
combinatorics proposed
combinatorics

[quote="moldovan"]In the weekly State Lottery, a sequence of seven numbers from the set $ \{ 0,1,...,9 \}$ is picked at random. Compute the probability that the sequence contains exactly five distinct numbers.[/quote]

is it {10c5 * 7c5*10 }/10^7  ?


Sure, their intersection is the null set.  :) It's sorta like $ \text{GCD}[1,1]\equal{}1$.

Neither of them have any prime factors, but the GCD still exists.


Let $ a,b,c$ be real positive numbers such that absolute difference between any two of them is less than $ 2$. Prove that: $ a \plus{} b \plus{} c < \sqrt {ab \plus{} 1} \plus{} \sqrt {ac \plus{} 1} \plus{} \sqrt {bc \plus{} 1}$

[b][color=darkblue]This problem is from Russia Olympiad 2004 Grade 9.

We have

$ \begin{array}{cc} \\
 \\
 \\
|a \minus{} b| < 2 \\
 \\
a^2 \minus{} 2ab \plus{} b^2 < 4 \\
 \\
a^2 \plus{} 2ab \plus{} b^2 < 4(1 \plus{} ab) \\
 \\
a \plus{} b < 2\sqrt {1 \plus{} ab} \\
 \\
 \\
 \\
\end{array}$.

By adding them we obtain:
 $ a \plus{} b \plus{} c < \sqrt {ab \plus{} 1} \plus{} \sqrt {ac \plus{} 1} \plus{} \sqrt {bc \plus{} 1}$[/color][/b]


inequalities
search
function
rearrangement inequality

These are 3 inequalities that I was not able to prove:

$1) \frac{a^2}{b^5} + \frac{b^2}{c^5} + \frac{c^2}{d^5} + \frac{d^2}{a^5} \geq \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} +\frac{1}{d^3}$

$2) \frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \leq \frac{a+b+c}{2abc}$

$3) \frac{1}{a^3+b^3+abc} + \frac{1}{b^3+c^3+abc} + \frac{1}{a^3+c^3+abc} \leq \frac{1}{abc}$

Update:  I forgot to say that $a,b,c > 0$

The first inequality is not true for all real $a,b,c,d$. (Take $a=-1, b=-1, c=-1, d=-2$, for example.)

For positive $a,b,c,d$ though, you can use Rearrangement on $\{a^2, b^2, c^2, d^2\}$ and $\left\{\frac{1}{a^5},\frac{1}{b^5},\frac{1}{c^5},\frac{1}{d^5}\right\}$.

[quote="towersfreak2006"]you can use Rearrangement on $\{a^2, b^2, c^2, d^2\}$ and $\left\{\frac{1}{a^5},\frac{1}{b^5},\frac{1}{c^5},\frac{1}{d^5}\right\}$.[/quote]

Can you show me how to rearrangment them?

If $a^2$ is the greatest element in the first set, then $\frac{1}{a^5}$ is the least element in the second set, since $a>0$. Hence, the right hand side, which is equivalent to $a^2\cdot\frac{1}{a^5}+b^2\cdot\frac{1}{b^5}+c^2\cdot\frac{1}{c^5}+d^2\cdot\frac{1}{d^5}$, is the sum of the product of corresponding elements in oppositely sorted sets. From Rearrangement Inequality, we know that the sum of the product of corresponding elements in any other orderings of the two sets is greater than or equal to the sum of the product of corresponding elements in oppositely sorted sets. The left hand side is one such sum.

Rearrangement Inequality, thank you towersfreak2006 I just learn that.

So there're 2 questions left. :roll:

[quote]$\frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \leq \frac{a+b+c}{2abc}$[/quote]

Note that the given inequality is homogenous in its variables. Thus, we can normalize to $abc=1$. But first, rewrite as \[ \frac{1}{a^2+\frac{abc}{a}} + \frac{1}{b^2+\frac{abc}{b}} + \frac{1}{c^2+\frac{abc}{c}} \leq \frac{a+b+c}{2abc}, \] \[ \frac{2}{a^2+\frac{1}{a}}+\frac{2}{b^2+\frac{1}{b}}+\frac{2}{c^2+\frac{1}{c}}\le a+b+c. \] From GM-HM on each of the three terms, we have \[ \frac{2}{a^2+\frac{1}{a}}+\frac{2}{b^2+\frac{1}{b}}+\frac{2}{c^2+\frac{1}{c}}\le \sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}. \] Since $abc=1$, then $\sqrt{abc}=1$. Hence, multiply the given inequality by $\sqrt{abc}$, \[ \frac{2}{a^2+\frac{1}{a}}+\frac{2}{b^2+\frac{1}{b}}+\frac{2}{c^2+\frac{1}{c}}\le\sqrt{bc}+\sqrt{ca}+\sqrt{ab}. \] From Rearrangment Inequality, we know that \[ \frac{2}{a^2+\frac{1}{a}}+\frac{2}{b^2+\frac{1}{b}}+\frac{2}{c^2+\frac{1}{c}}\le\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\le\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}=a+b+c, \] as desired. :)

3 can be found a million times on the forum (use the search function)

Here is a solution to 2.

We want to prove
\[ \sum \frac{1}{a^2+bc}\leq \frac{1}{2}\sum \frac 1{bc}  \]
By summing cyclically, it is enough to prove that
\[ \frac{1}{a^2+bc}\leq \frac{1}{4}\left(\frac 1{ab}+\frac 1{ac}\right)  \]
This rearranges to
\[ 4abc\leq (a^2+bc)(b+c)  \]
By AM-GM, we get
\[ (a^2+bc)(b+c)\geq (2a\sqrt{bc})(2\sqrt{bc})=4abc  \]
so the result follows.

Here is a solution to 3

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=1997&p=185470

I've got a question to towersfreak2006. [quote]Note that the given inequality is homogenous in its variables. Thus, we can normalize to $abc=1$ . But first, rewrite as...[/quote].
Why can we assume, that $abc=1$. Could you explain?

If $abc \neq 1$, simply replace $a$ with $ad$, $b$ with $bd$ and $c$ with $cd$.  The inequality will still look exactly the same (if it was true before, it will be true now), and we have changed the product $abc$ to $d^3$ times the previous product -- in particular, we can change it to 1 with a good choice of $d$.  We can do this because every term has the same total power in the variables (it is homogeneous).

Since $a^2+1/a\geq 2\sqrt{a}$, it follows that
\[ \frac{2}{a^2+a^{-1}}\leq \frac{2}{2\sqrt{a}}=\sqrt{\frac{1}{a}}  \]
In any event, I feel that my solution is more natural, since I avoid dehomogenizing, which is an odd step when presented with a homogeneous inequality (unless, of course, you want to apply Jensen's)

[quote="Laputa"]$2) \frac{1}{a^2+bc} + \frac{1}{b^2+ac} + \frac{1}{c^2+ab} \leq \frac{a+b+c}{2abc}$

I forgot to say that $a,b,c > 0$[/quote]

$\frac{a^2+bc}{2}\geq \sqrt{a^2bc}=a\sqrt{bc}$ $\Rightarrow \frac1{a^2+bc}\leq \frac{1}{2a\sqrt{bc}}=\frac{\sqrt{bc}}{2abc}$

$\frac1{a^2+bc}+\frac1{b^2+ac}+\frac1{c^2+ab}\leq \frac{\sqrt{bc}+\sqrt{ac}+\sqrt{ab}}{2abc}\leq \frac{a+b+c}{2abc}$

The last one is true since Cauchy's yields that $(a+b+c)(b+c+a)\geq (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2$

[quote="Laputa"]$3) \frac{1}{a^3+b^3+abc} + \frac{1}{b^3+c^3+abc} + \frac{1}{a^3+c^3+abc} \leq \frac{1}{abc}$

I forgot to say that $a,b,c > 0$[/quote]

Using your favourite inequality (for instance, rearrangement), we have $a^3+b^3\geq a^2b+ab^2$

$a^3+b^3+abc\geq a^2b+ab^2+abc=ab(a+b+c)\Rightarrow \frac1{a^3+b^3+abc}\leq \frac1{ab(a+b+c)}$

$\frac1{a^3+b^3+abc}+\frac1{b^3+c^3+abc}+\frac1{c^3+a^3+abc}\leq \frac{a+b+c}{abc(a+b+c)}=\frac1{abc}$

Thank you to all of you.  Very nice solutions! :)

[quote="blahblahblah"]We want to prove
\[ \sum \frac{1}{a^2+bc}\leq \frac{1}{2}\sum \frac 1{bc}  \]
By summing cyclically, it is enough to prove that
\[ \frac{1}{a^2+bc}\leq \frac{1}{4}\left(\frac 1{ab}+\frac 1{ac}\right)  \]
[/quote]

What is the sigma for?  I never see such notation before.  And what is summing cyclically?

The $\sum$, in this context, is a kind of shorthand for writing symmetric sums, e.g, we use

$\sum a^2b$ to denote $a^2b+b^2c+c^2a$.

By summing cyclically, I mean that we sum that inequality and the two similar ones

see here a solution for 3
http://www.mathlinks.ro/Forum/viewtopic.php?t=58140


Let ABC be a triangle, BM; CN are medians. Assume that $ \angle{B}\leq\angle{C}$. Prove that: $ BM+CM\geq CN+BN$

If  $\hat C \leqslant {90^o}$  , then: (see fig.)
$\begin{gathered}
  \left. \begin{gathered}
  {\rm A}{\rm E}|| = {\rm B}C\mathop {}\limits_{}  \hfill \\
  AZ||BC\mathop {}\limits_{}  \hfill \\
  ZL \bot BC\mathop {}\limits_{}  \hfill \\
  KZ = KL\mathop {}\limits_{}  \hfill \\
  ZI = EC \hfill \\
\end{gathered}  \right\}\mathop {}\limits_{}^{} \mathop {}\limits_{}^{}  \Rightarrow \mathop {}\limits_{}^{} \left\{ \begin{gathered}
  AB = IL = 2BN\mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{} IE = LC = ZC = 2CN \hfill \\
  BZ = BL = 2MC\mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{} BE = 2BM \hfill \\
\end{gathered}  \right. \hfill \\
   \hfill \\
  So:
\mathop {}\limits_{}^{} IL + IE < BL + BE\mathop {}\limits_{}^{}  \Rightarrow \mathop {}\limits_{}^{} 2BN + 2CN < 2MC + 2BM\mathop {}\limits_{}^{}  \Rightarrow \mathop {}\limits_{}^{} BN + CN < MC + BM \hfill \\
\end{gathered} $.

Τhe same conclusion when $\hat C > {90^o}$


probability
geometry
analytic geometry
similar triangles
absolute value

I hated that target about the elevator...

I figured out it was 7th floor FOR EACH TIME about a minute before and then i put 5/32 and then my brain told me it was 3/16 and i was trying to choose with 10 seconds left but i didn't change it... so i got it wrong... even if i did get that target right, i still would be 2nd but still...

Can anyone post a *coherent* solution for the elevator target (#7), the tile target (#6), and the Fibonacci Sprint Problem (#28)?

I stumbled upon the answer for #6, but it was one of those questions where you kind of intuit the answer, so I am looking for a good solution. Also, I guessed correctly on number 28, but I didn't really solve it either. Thanks a bunch.

28
There's a 6-number pattern of remainders 1, 1, 2, 3, 1, 0, so you divide 100 by 6 for remainder of 4 making the answer $3$.

28. 1 1 2 3 5 8 13 21
Write that in mod 4.
1 1 2 3 1 0 1 1 2 3 1 0 You see the pattern 112310 and you can find that 100 is 4 mod 6, so it's the 4th one, with is 3.

Edit: Tarquin beat me to it

[quote="tarquin"]Well first know that there are $2^5 = 32$ ways to flip the coin.
Then the only combinations that work are,
HTHHH
HHTHH
HHTTH
HHHTH
HHHTT
Which makes 6 combinations so the answer is $\frac 6{32} = \frac 3{16}$.[/quote]

The first one (HTHHH) doesn't work because the problems says "what is the probability that [b]each[/b] of her next five stops is on the 7th floor or higher." So if it was HTHHH, it would be 7,6,7,8,9, but they all have to be above 6. Also, the third one (HHTTH) doesn't work for the same reason (you wind up on the 6th floor after 4 flips). I think the ones you are missing are HHTHT and HHHHH.

[quote="Phelpedo"][quote="bob123"]6. You know that the number of bricks can be expressed as 7x2 (7x*x) where x is the length of the width. Then you start adding layers until you get a number like that 34, 65, 93, 118, 140, 159, 175... 175 is 7*52 so the answer is 7 rows of bricks.[/quote]

Yeah, but is there any way to just get 7, or do you have to add? Also, is there any formula that would allow you to find the number of bricks if there were, say, 14 rows, without adding?[/quote]

Sum the first and last number, divide by 2, multiply by the number of terms... That's a way you could do it... Then you have to find two numbers that are both 1 mod 3, and their sum can be expressed as 2x2... I just said the way I did it at state because I had lots of time there.

[quote="chess64"]Me too! :D They have the solutions up there, but even though I know how to do it, I still hate those kinds of problems.[/quote]

Where can you get the solutions?

The first two flips must be heads. The next two flips must not be TT, and the last flip can be anything.

(1/4)(3/4)

3/16

#29[hide]
Call BE and FC both x.  Lable the lengths of AB as 28 and CD as 58 as given.

Then using similar triangles make the proportion $\frac{28}{x} = \frac{x}{58}$ Cross multiply to get $x^2 = (28)(58) = 1624$ since $x^2$ is the area of the square and what you are looking for, the answer is 1624.[/hide]

#30

[hide]
Call the radius of the small circle 2x and the radius of the larger cirlce 4x.

The combined areas of the two semicircles is then

$\frac{16x^2 \pi}{2} + \frac{4x^2 \pi }{2} = 10x^2 \pi$

Since the sector ACP needs to be half that, its area will be $5x^2 \pi$

To find the measure of the angle ACP, call it y, use the equation

$\frac{y}{360} (16x^2)\pi = 5x^2 \pi$ which simplifies to
$\frac{16x^2y\pi}{360} = 5x^2\pi$
Simplify the $x^2$ and $\pi$ out, and reduce to get
$\frac{2y}{45} = 5$
Multiply by 45
$2y = 225$
$y=112.5$[/hide]

#30
Just know that the little semi circle is 1/4 of the bigger semicircle so that makes a total of 5/4, and all you need is 5/8. 5/8 of 180 is 112.5 or something.

[quote="chess64"][quote="frost13"]#30
and I apologize ahead of time for having to use pi instead of the symbol, everytime I clicked on the Emoticon I got 10 lines of nonsense in my post???
[/quote]

Try $\pi$. "\pi."[/quote]

Thanks Chess64, I edited it

[quote="dwx314"]Wanna hear some thing really bad?
I got EVERYSINGLETARGETPROBLEMWRONG (exept for 8)
Stupid runny nose...ran out of tissues after the sprint.

Well, at least this thread got some of my questions done.

How do you do # 29 and 30 on the sprint?[/quote]

Was mathlover happy  :? ?

I didn't get number 29. In fact, I guessed 2005 because it looked like around 2005 :P .

30 was easy, though. I set the area of the smaller semicircle as 1, so the area of the larger one must be 4. 2.5/4 = 5/8. 5/8 = x/180.

[quote="Treething"][quote="dwx314"]Wanna hear some thing really bad?
I got EVERYSINGLETARGETPROBLEMWRONG (exept for 8)
Stupid runny nose...ran out of tissues after the sprint.

Well, at least this thread got some of my questions done.

How do you do # 29 and 30 on the sprint?[/quote]

Was mathlover happy  :? ?
[/quote]
treething, since I don't expect mathlover will be visiting here anytime soon :

Mathlover was happy for her team who took 1st, 4th, 6th, 7th, and 10th at state, but she wasn't too fond of the target round herself :(

I didnt like number 24 the one with all the equations of the lines that took way too long!
Anyone know a quick and painless way of doing that?

find points of intersection, use distance formula, and then heron's formula? hmm... that'd be a lot longer, wouldn't it....

[quote="NoSoupForYou"][quote="frost13"]Mathlover was happy for [b]her team who took 1st, 4th, 6th, 7th, and 10th at state[/b], but she wasn't too fond of the target round herself :([/quote]

Umm... There's only 4 people in a team.[/quote]

One of the "members" must have been an individual not on the team.

Yes, one of those 5, #7 was an individual who was 10th at chapter.  I consider my "team" to be the 12 kids I take to the competition.

Nice solution I forgot about shoestring! :D

[quote="nat mc"]# 24

[hide]Find the points of intersection.
For any polygon on a coordinate plane, there is a way to find the area that is nice and simple. Triangle:
point (a,b)
point (c,d)
point (e,f)

List them in order and then put the first point again at the end.
(a,b)
(c,d)
(e,f)
(a,b)

Then you do a shoe string like multiplication and addition.
ad + cf + eb and the other is
bc +de + fa
Then take the absolute value of the difference of these quantities and divide by 2.

This works on any sided polygon, just list the points in order[/hide][/quote]

This was actually #25.

i hated number 6 i miss interpreted the problem horrible grammer

Which number 6, because neither one seems to have bad grammar to me?


A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon?

A. $1/3(2000)$

B. $2000 (\sqrt{2}-1)$

C. $2000(2-\sqrt{2})$

D. $1000$

E. $1000\sqrt{2}$

Dude, the correct answer is not up there.

[hide]$x=\frac{(2000-x)\sqrt{2}}{2}\Longrightarrow \boxed{x=2000(\sqrt{2}-1)}$.[/hide]

I think he just mistyped. I got the same thing, although I went through a few more steps... Is that a formula you have memorized, or am I missing something obvious?

Let $x$ be a side of the octagon. We can also write it as $\frac{(2000-x)}{2}\cdot\sqrt{2}$ since we were told that it is a regular octagon and a side of the isosceles triangle is just $\frac{(2000-x)}{2}$. :wink:

Okay, now that I look at it, that's exactly what I did. I just used two variables and didn't substitute soon enough.

ok. Here's my solution.

[hide="solution"]$x\sqrt{2}+x=2000$ Since x/rt2 is 1/2 of the complement of the side.

So $x=2000(\sqrt{2}-1)$[/hide]

edit: same as pianofote's.


abstract algebra
algebra
function
domain
real analysis
real analysis theorems

Let $ \equiv$ and $ \sim$ be two equivalence relations on the same set $ A$.  If $ a\equiv a_{1}$ implies that $ a\sim a_{1}$, prove that each $ \sim$ equivalence class is partitioned by the $ \equiv$ equivalence classes it contains.

How does one handle partitions of equivalence classes?  I know the theorem for equivalence classes under $ \equiv$ partitioning $ A$, but what to do for equivalence classes of other equivalence classes...

Another question: let $ \alpha: A\rightarrow B$ be a map, $ \equiv$ denotes the kernel equivalence ($ a\equiv a_{1}$ iff $ \alpha(a)\equal{}\alpha(a_{1})$ for $ a,a_{1}\in A$), and let $ \varphi: A\rightarrow A_{\equiv}$ denote the mapping $ \varphi(a)\equal{}[a]$.  Finally, let $ \sigma([a])\equal{}\alpha(a)$.  If $ \alpha(A)$ is a finite set, prove that $ \left|A_{\equiv}\right|\equal{}\left|\alpha(A)\right|$.  I already proved that $ \sigma$ is one-to-one, and that $ \sigma$ is onto if $ \alpha$ is.  This is a bijection then, because $ \alpha: A\rightarrow\alpha(A)$ is obviously onto, so $ \sigma$ is onto, correct?

For your first question, what you have says that $ \equiv$ is finer than $ \sim$ (or $ \sim$ is coarser than $ \equiv$). The same translates to the partitions by equivalence classes, just follow the definition.

For your second, you are again right. What you have as equivalence classes are the fibres $ \alpha^{\minus{}1}(b)$ of the map $ \alpha$, and of course $ \sigma$ is only onto when $ \alpha$ is, when considering co-domain $ B$. But clearly what matters is the restriction to $ \alpha(A)$, the range of $ \alpha$, so $ \sigma$ is then a bijection between $ A_{\equiv}$ and $ \alpha(A)$.

Maybe it's that the wording is confusing me, but could you elaborate on the first paragraph?  Thanks.

Assume $ a \equiv b$ always implies $ a \sim b$. Take now an element $ y$ in the class $ [x]_{\sim}$. Clearly $ [y]_{\equiv} \subseteq [y]_{\sim} \equal{} [x]_{\sim}$, since $ z \equiv y$ implies $ z \sim y$. If there exists $ y' \in [y]_{\sim} \setminus [y]_{\equiv}$ do the same, and so on. Another way to see it, is to check that $ \equiv$ is an equivalence relation on the set $ [x]_{\sim}$, and partition it accordingly.

This is probably getting annoying, but could you clarify the statement "do the same, and so on"?  Based on what is there, I am thinking that the partition is all of the elements in $ \equiv$ equivalence classes and those just in $ \sim$, but I'm not sure.

Thanks.

Look at one of the $ \sim$ equivalence classes, be it $ [x]_{\sim}$.

1. Take an element $ y \in [x]_{\sim}$. Clearly $ [y]_{\equiv} \subseteq [y]_{\sim} \equal{} [x]_{\sim}$, since $ z \equiv y$ implies $ z \sim y$. If $ [y]_{\sim} \equal{} [y]_{\equiv}$, we are done, having partitioned $ [x]_{\sim}$ in $ \equiv$ equivalence classes (albeit just one in this case!).

2. If not, there exists $ y' \in [y]_{\sim} \setminus [y]_{\equiv}$. Do the same as step one for $ y'$. If $ [y]_{\sim} \equal{} [y]_{\equiv} \cup [y']_{\equiv}$, we are done, having partitioned $ [x]_{\sim}$ in $ \equiv$ equivalence classes (two  in this case!).

3. If not, there exists $ y'' \in [y]_{\sim} \setminus ([y]_{\equiv} \cup [y']_{\equiv})$. Do the same as step one for $ y''$. And so on.

At some point, there are no more elements left, so we succeed partitioning $ [x]_{\sim}$ in $ \equiv$ equivalence classes. Then we do the same for the other $ \sim$ equivalence classes.

[quote="mavropnevma"]Assume $ a \equiv b$ always implies $ a \sim b$. Take now an element $ y$ in the class $ [x]_{\sim}$. Clearly $ [y]_{\equiv} \subseteq [y]_{\sim} \equal{} [x]_{\sim}$, since $ z \equiv y$ implies $ z \sim y$. If there exists $ y' \in [y]_{\sim} \setminus [y]_{\equiv}$ do the same, and so on.[/quote]
We must pay attention here. This does not work if a $ \sim$-equivalence class decomposes into [u]uncountably[/u] many $ \equiv$-equivalence classes. E.g., on $ \mathbb R$ take $ x\equiv y$ iff $ x\equal{}y$ and $ [x]_\sim\equal{}\mathbb R$ for every $ x\in \mathbb R$. Then this constructive method can never exhaust $ \mathbb R$.
[quote]Another way to see it, is to check that $ \equiv$ is an equivalence relation on the set $ [x]_{\sim}$, and partition it accordingly.[/quote]I believe only an abstract argument like that can be given as  a valid proof, since the cardinality of the searched decomposition can exceed every constructive, hence countable, approach.


search
calculus
complex analysis
complex numbers

Hey

I posted this in the Pre-Olympiad Forum but the answer I received didn't make sense and I was hoping for a clarification. Sorry if I am being pushy, if necessary, please delete one of the double posts....

Anyways, my question is to prove

Prove the following formula

$\sum_{k=0}^{n}{\binom{n}{k}}^{2}= \binom{2n}{n}$

The answer that was shown to me used Lemma (Vandermonde's Identity), but I have never seen that nor do I understand how it works. Is there a simpler method?

Any help is appreciated, thanks.

Both this problem and the more general Vandermonde result (of which this is a special case) are on this forum dozens of times.  I suggest you do a search of the forum.  There are doubtless many explanations available elsewhere on the internet, as well.

Good idea. I guess sometimes my brain just doesn;t work.

In fact, there are two additional proofs of it currently near the top of one of the Olympiad combinatorics forums, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121075

As an aside, "Complex Analysis" refers to analysis (formal calculus) over the complex numbers, so this was a poor choice of forum for your question.   :wink:


\[ \begin{array}{l} k \ge 1,prove: \\
\frac {{a_1^k \plus{} a_2^k \plus{} ...a_n^k }}{n} \ge (\frac {{a_1 \plus{} a_2 \plus{} ...a_n }}{n})^k \\
k < 1,prove: \\
\sqrt [m]{{\frac {{a_1 \plus{} a_2 \plus{} .... \plus{} a_n }}{n}}} \ge \frac {{\sqrt [m]{{a_1 }} \plus{} \sqrt [m]{{a_2 }}...... \plus{} \sqrt [m]{{a_n }}}}{n} \\
\end{array}
\]
Can you tell me how to make this post bigger , it's too small .

Just use the power mean inequalities: http://en.wikipedia.org/wiki/Power_mean

I don't know how to use power mean inequalities in this problem . Can you show me in details ?

If you use "Jensen's inequality", it is easy for you to prove it.
please refer to http://en.wikipedia.org/wiki/Jensen%27s_inequality or http://mathworld.wolfram.com/JensensInequality.html.


Find all real polynomials that

$p(x+p(x))=p(x)+p(p(x))$

 :wink:

Fitim Dika

continuos here, please http://www.mathlinks.ro/Forum/viewtopic.php?p=632796#632796  :wink:


Give an example of a function f:R->R s.t. f(x3)=f(x) for all x \in R and which isn't continuous at any point.

This might be kind of easy, but it sure is nice :).

f(x) = 1 if x is algebraic, else f(x) = 0.

sharpened question: find a function f: R->R that does not only satisfy
the original requirements, but is moreover surjective.


inequalities
trigonometry
vector
inequalities unsolved

As Vasc said, it is right to dedicate an own topic to this nice inequality

[quote="Vasc"]This is the Fan's Inequality:

If $x_1+...+x_n=0$, then $\frac{x_1x_2+...+x_nx_1}{x_1^2+...+x_n^2} \leq cos{\frac{2\pi}{n}}$[/quote]

Equality occurs for $x_k=A\cos{\frac{2k\pi}{n}}+B\sin{\frac{2k\pi}{n}}$, where A and B are arbitrary real numbers. That means for $x_k+x_{k+2}=2cos{\frac{2\pi}{n}}x_{k+1}$.

[quote="grobber"]This isn't a proof "per se", since I'm using a strong result, but I thought the approach was nice enough to post :).

For every $i\in\overline{1,n}$, let $v_i\subset\mathbb R^n$ be the vector $(x_i,x_{i+1},\ldots,x_{i+n-1})$ (where $x_{n+1}=x_1$). What we want to prove is that the angle $\theta$ between $v_i$ and $v_{i+1}$ (which is the same for all $i$) is $\ge\frac{2\pi}n$.

Now consider the piecewise linear curve with vertices $0,v_1,v_1+v_2,\ldots,v_1+\ldots+v_{n-1},v_1+\ldots+v_n=0$. We can approximate this curve arbitrarily well with smooth closed curves, which, by Fenchel's Theorem (the result I mentioned above), have total curvature $\ge2\pi$. The total curvature of these curves, however, tends to the sum of the external angles of our curve, which is $n\theta$, so the conclusion is that $n\theta\ge2\pi$, i.e. exactly what we wanted.[/quote]
Very Very nice idea(I think it is a proof).Can you show me a proof for fenchel's theorem or a link?I can't find it on the internet.
Thanks


geometry
area of a triangle
Heron\u0027s formula

Find the length of the altitude to the longest side of a triangle whose sides have lengths 13, 20, and 21.

By [url=http://mathworld.wolfram.com/HeronsFormula.html]Heron's Formula[/url], the area of the triangle is $ \sqrt {27\times{14}\times{7}\times{6} } = \sqrt { {9}\times{3}\times{7}\times{2}\times{7}\times{3}\times{2} } = 3\times{3}\times{7}\times{2} = 126$

Let $ h$ be the altitude to the longest side (21) of the triangle, then we have $ \frac {21h}{2} = 126\implies{h = \boxed{12}}$


[quote="Art of Owna"]How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)[/quote]

[hide="solution"]There are 6+1=7 multiples of 5 in 31!, 3 multiples of 5 in 16!, 1 multiple of 5 in 8! and none in the others.

7-3-1=3 so $\boxed{3}$ zeroes.[/hide]

Are you sure?

For example, using your way, ur inplying there is 1 zero at the end of (5!)/(4!)

Actually, since $10=2\cdot 5$, you must also consider whether or not there is an ample number of two's to combine with the fives to make zeros.

[hide]For speed, we can only consider even numbers and multiples of 5.  At the end, we get $3$ factors of 5 but only $1$ factor of 2.
edit: scratch that, no factors of 2.
Thus, there is $\boxed{0}$ zeros.  [/hide]

[hide="hmmmm"]The number of zeroes in the numerator is 7. THe number in the denominator is 4. So there are 3. That's what I got. :|

Atleast it was right in a previous MC question :huh: [/hide]

[quote]Oh.

But don't the zeroes cancel out...[/quote]
Not necessarily..., when you cancel out the zeros, ur just really subtracting the number of 2's and 5s in the numerator with the number of 2's and 5s in the denominator, so that only works if there is equal amounts of 2s and 5s in both the demoninator and numerator

For example, 10^3/10^2, you can cancel out zeros because there are 3 2s and 5s in the numerator and 2 2s and 5s in the denominator

but like 400/8, you cannot do that

The reason $\frac{7!}{3!4!}$ doesn't work is because $3!4!$ has as many zeroes as $7!$.

$31!$ has $15+7+3+1 = 26$ zeroes, while $16!8!4!2!1!$ has $8+4+2+1+4+2+1+2+1+1 = 26$ zeroes. So the expression has $\boxed{\text{no}}$ zeroes!

Can someone prove for me that an expression of the form $\frac{n!}{a_{1}!a_{2}!a_{3}!...a_{k}!}$ is always an integer where $a_{1}+a_{2}+... a_{k}= n$? I think it's true but I'm not sure...

because you don't have enough 2's

(You need 2's and 5's to make 10) :D

[quote="Art of Owna"]No shes not... just think about it
Lets give an example.. 7!/(3!*4!), using her way, there is 1 multiple of 5 in 7!, and none in 3! or 4!, 1-0 is 1 so thats the correct answer... Nope.. type it in ur calculator..[/quote]
Jli is a guy.
I'm getting 0 for the answer

[quote="Art of Owna"]kyyuanmathcount i think ur reasoning is correct but I think if u calculate it it would be no zeros

#H34N1, you cant just subtract zeros.. from the top and bottom.. whats 400/8- does that have 2-0=2 zeros?[/quote]

Whoops, my bad, i'll edit that.

[quote="usaha"]What are you talking about? 31! does not have 26 zeros, it hsa 26 factors of 2, infact 31! has 7 zeros, the denominator has 4 zeros BUT when you divide them it will have 0 zeros[/quote]

Whoops, my bad... I meant multiples of 2... and you should have been able to figure that out by yourself.

[quote="13375P34K43V312"][quote="Ubemaya"]Can someone prove for me that an expression of the form $\frac{n!}{a_{1}!a_{2}!a_{3}!...a_{k}!}$ is always an integer where $a_{1}+a_{2}+... a_{k}= n$? I think it's true but I'm not sure...[/quote]

[hide="hmm"]Wouldn't that just be ${n\choose a_{1},a_{2},\ldots,a_{k}}={n\choose a_{1}}{n-a_{1}\choose a_{2}}\cdots{a_{k}\choose a_{k}}$, which is always an integer?[/hide][/quote]
Well for example, can you prove that if $n \geq k$ then $\frac{n!}{k!(n-k)!}$ is an integer without combinatorics?

there are 3 multiples of 5 in the end.  i hope you guys know that.

now count the number of powers of 2's.  the top has $15+7+3+1=26$.  The bottom has $8+4+2+1+4+2+1+2+1+1=26$.  So ther are no multiples of 2.  i feel really stupid now. :wallbash:

the answer is 0

oh, and by the way, im a guy. :D  (j=johnny li=last name)


To all who may be confused: Compiz and Beryl are window managers which run in the Gnome/KDE Desktop Environments. The Gnome window manager is metacity and the KDE window manager is kwin.

I'm curious as to who uses which window manager. Ubuntu comes with metacity (Gnome) by default, so I'm expecting that to be the clear majority. Of course, KDE 4 has completely blown it out of the water. Personally, I like the minimalism of fluxbox, but if you go as far as evilwm, it's a bit too little (although the source for evilwm is nice to read to see how you would make a window manager at all).

I'd vote for Compiz Fusion, but I have an ATI card, which kinda sucks...


at first find $ \sum x^{n}n^{2}$ for $ |x|<1$ you will get some differential equation :)

hello, after the hint above we get
$ \sum_{n\equal{}1}^{\infty}\frac{n^2}{e^n}\equal{}\frac{e(e\plus{}1)}{(e\minus{}1)^3}$.
Sonnhard.


Source: ACoPS
(Hungary, 1899) Let $r$ and $s$ be the roots of
$x^{2}-(a+d)x+(ad-bc)=0$.
Prove that $r^{3}$ and $s^{3}$ are the roots of
$y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}=0$.

[quote="drunner2007"]From ?vietta's? sums $r+s=a+d$ and $rs=ad-bc$. Thus $r^{3}+s^{3}=(r+s)^{3}-3rs(r+s)=(a+d)^{3}-3(ad-bc)(a+d)=a^{3}+3a^{2}d+3ad^{2}+d^{3}-3a^{2}d+3abc-3ad^{2}+3bcd=a^{3}+d^{3}+3abc+3bcd$. Also, $(rs)^{3}=(ad-bc)^{3}$. Therefore the roots of $y^{2}-(a^{3}+d^{3}+3abc+3bcd)y+(ad-bc)^{3}$ are $r^{3}$ and $s^{3}$.[/quote]
Yes, that's how i did..this was pretty easy, since all u have to do is just applying Vieta's theorem.


logarithms
real analysis
real analysis unsolved

[quote="Moubinool"]$ x\in R$

$ u_{n}=\frac{(-1)^{n}}{\sqrt{n}+e^{inx}ln(n)}$

study the convergence of $ \sum u_{n}$[/quote]

$ |\frac{(-1)^{n}}{\sqrt{n}+e^{inx}\ln n}| \geq \frac{1}{|\sqrt{n}|+|e^{inx}\ln n|}\geq \frac{1}{\sqrt{n}+\ln n}\geq \frac{1}{2\sqrt{n}}$

Now, $ \sum_{n=1}^{\infty}\frac{1}{2\sqrt{n}}=+\infty$.


Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices?

[hide]All polygons are distinct unless the have the same vertices. Thus, we simply need to count the number of ways to choose distinct vertices, for a total of $\binom{10}{3}+\binom{10}{4}+\cdots+\binom{10}{10}$ polygons. This expression can be simplified as $2^{10}-\binom{10}{0}-\binom{10}{1}-\binom{10}{2}=1024 - 1 - 10 - 45=\boxed{968}$. [/hide]

[quote="Hokkage"]Altheman: 2^{10}  ;)[/quote]

edited...i did not bother to preview that message

first of all 2^10, is more than 999.  and 2^10 includes polygons with 0, 1, and 2 sides, and u can have polygons with 0,1, and 2 sides

What if the ten points are evenly spaced, and polygons are considered the same if they are congruent instead of sharing the same vertices?
How would you approach?

Combo AIME grind: Use complementary counting. The total amount of polygons is $2^{10}$, and the undesirables give $\binom{10}{0}+\binom{10}{1}+\binom{10}{2}=56$ ways. The answer is $2^{10}-56=\boxed{968}$.

[quote=Hokkage]He subtracted off the aforementioned polygons...
Look at altheman's hidden post.[/quote]

Because nC0+nC1+...+nCn =2^n


k,l,m,n are Natural numbers, k<l<m<n and kn=lm Prove that (n/2-k/2)^2>=k+2

It's well-known that there exist natural numbers $ a,b,c,d$ such that $ k \equal{} ab, n \equal{} cd, l \equal{} ac, m \equal{} bd$.
As $ k < l < m < n$, $ ab < ac < bd < cd$, i.e. $ a < d, b < c$, so $ d\geq a \plus{} 1, c \geq b \plus{} 1$.
So $ cd \geq (a \plus{} 1)(b \plus{} 1) \equal{} ab \plus{} a \plus{} b \plus{} 1$, i.e. $ (cd \minus{} ab)^2 \geq (a \plus{} b \plus{} 1)^2$.

Now back to the original inequality, $ (\frac {n}{2} \minus{} \frac {k}{2})^2 \geq k \plus{} 2 \Leftrightarrow (cd \minus{} ab)^2 \geq 4(ab \plus{} 2)$
Divide two cases,
Case 1) $ a \neq b$, as $ a \geq 1, a \plus{} b \geq 1 \plus{} 2 \equal{} 3$
As $ (cd \minus{} ab)^2 \geq (a \plus{} b \plus{} 1)^2$, it's left to prove that $ (a \plus{} b \plus{} 1)^2 \geq 4(ab \plus{} 2)$
Simiplify to get $ (a \minus{} b)^2 \plus{} 2(a \plus{} b \minus{} 3) \geq 0$, this is true as $ (a \minus{} b)^2 \geq 0$ and $ a \plus{} b \geq 3$.

Case 2) $ a \equal{} b$
As $ ac < bd$, so $ c \neq d$, WLOG let $ c < d$.
So $ d > c \geq a \plus{} 1$, i.e. $ d \geq a \plus{} 2$
So $ cd \geq (a \plus{} 1)(a \plus{} 2) \equal{} a^2 \plus{} 3a \plus{} 2$, then $ (cd \minus{} a^2)^2 \geq (3a \plus{} 2)^2$.

So left to prove $ (3a \plus{} 2)^2 \geq 4(a^2 \plus{} 2)$
$ (3a \plus{} 2)^2 \minus{} 4(a^2 \plus{} 2) \equal{} 5a^2 \plus{} 12a \minus{} 4 \equal{} (a \minus{} 1)(5a \plus{} 17) \plus{} 13 > 0$
So we are done.

Seems the equality can't hold...


How many integer solutions are there for: $-6 < 3x-2 < 7$.

well the easiest way would be from

$\frac{4}{3}< x <3 =$
$\frac{4}{3}< x$ and $x<3$.

Graph or imagine the number line and you can find your solutions.

edit: actually to recieve full credit for that problem the answer should be in the form

$\{x|-1\leq x \leq 2 \}$

[quote="vishalarul"][quote="xxxxx"][quote="vishalarul"][hide]$-4<3x<9$

$-\frac{4}{3}< x < 3$

$-1\leq x\leq 2$

$-1,0,1,2$

$4$ solutions[/hide][/quote]

could you explain how you got: $-1\leq x\leq 2$ ?[/quote]
[hide]Well, you need the integer solutions, so the last two numbers at the end do not count... So we can ignore the $3$, and since $-\frac{4}{3}$ is a fraction anyway, we can get the bounds $-1\leq x\leq 2$[/hide][/quote]

i see.


analytic geometry
combinatorics unsolved
combinatorics

You have a chessboard whose dimensions are 15 x 15 (225 squares), and a chess piece that can move a distance of 8 or 9 squares, horizontally or vertically. You place the piece anywhere on the board. How many moves can you make without visiting the same square twice?

Not hard. Labell the squares of the chessboard with ordered pairs $(x,y)$,  $1\leq x,y\leq 15$
First of all, a path that contains at least one cell of the type $(8,y)$ or $(x,8)$ must contain at most  15 squares, since there is a direction along which you cannot move from or to it. On the other hand you cannot reach a cell of one of those two tipes from a cell not belonging to it, so if the longest non self-intersecting path consistent of allowed moves and containing cells $(x,y)$ with   $x,y\neq8$  has length greater than 15, then that will be maximal. There are  $14^{2}=296$  such squares; we will show we can visit all of them.  Consider the sequence  $a_{0}=9,\;\;a_{1}=9-8=1,\;\;a_{2}=1+9=10,\;\;a_{3}=10-8=2,\;\;a_{4}=2+9=11 \;\ldots\; a_{13}=15,\; \; a_{14}=15-8=7$;  all moves of the type  $(a_{x},a_{y})\mapsto(a_{x\pm1},a_{y})$   or  $(a_{x},a_{y})\mapsto(a_{x},a_{y\pm1})$,  with   $1\leq x,y\leq 14$  and such that for the new coordinates $(a_{x'},a_{y'})$, $x'$  and  $y'$  continue to belong to that interval are allowed moves inside our  $15\times 15$ chessboard. So just take a nonintersecting path  $(x_{i},y_{i})$,   $1\leq i\leq 196$  inside a $14\times 14$ chessboard in which all consecutive cells in the path are adyacent, and which goes through all of them (for example, an s-like one) and the path  $(a_{x_{i}},a_{y_{i}})$,  $1\leq i\leq 196$  inside the initial  $15\times15$  chessboard will achieve maximality within the restrictions initially imposed. Since it consists of  $196$ cells, the number of steps will be  $195$.

Very nice, your solution is absolutely correct  :)


For every positive integer $n$ prove the inequality for all $n \geq 1$

$\sqrt{1+\sqrt{3+\sqrt{5+\sqrt{...+\sqrt{2n-1}}}}} < 2$

This inequality holds iff \[ \sqrt{3+\sqrt{5+\sqrt{...+\sqrt{2n-1}}}} < 3 \]
And again, this inequality holds iff \[ \sqrt{5+\sqrt{...+\sqrt{2n-1}}} < 6 \]
And this one holds iff \[ \sqrt{7+\sqrt{...+\sqrt{2n-1}}} < 31 \]
And this one holds iff \[ \sqrt{9+\sqrt{...+\sqrt{2n-1}}} < 954 \]
...

In the end, we will have to prove that $\sqrt{2n-1} < (((((2^2-1)^2-3)^2-5)^2-7)^2-...)^2- 2n-3$

And that's where I'm stuck  :(

I used induction. There is another way ?


All of the students in a history class are juniors or seniors. There are twice as many girls as boys in the class, and there are 3 times as many senior girls as junior girls. If a student is selected at random, what is the probability that the student is a junior girl?

[hide]Let x be the total number of boys in the class. There are twice as many girls as boys, so the number of girls is 2x. Now the total number of students is 3x. Since there are 3 times as many senior girls as junior girls, the number of sen. girls is (3/4)(2x)=3x/2 and the number of junior girls is (1/4)(2x)=x/2. The probablity of choosing a junior girl is x/2 over 3x=[b]1/6[/b][/hide]


From 6 am to noon the temperature rose $8^\circ$ C.  From noon to 3 pm, it rose another $3^\circ$ C, but then dropped $5^\circ$ C to $-3^\circ$ C at 6 pm.  How many degrees Celsius was the temperature at 6 am?

[hide]You have temperature x, + 8 + 3 -5 = -3. The answer is -9. If you work backwards, you would have -3 + 5 - 3 -8 = x.

Therefore, the answer is -9. [/hide]

[hide]let x= temp at 6 am

x+8+3-5=-3
x+11-5  =-3
x+6       =-3
x           =-3-6
x           = -9

therefore, the temp at 6 am is -9 degrees.[/hide]

[size=84][color=green]Yes, I agree.  Please hide your answers.[/color][/size]

[hide="answer"]x+8+3-5=-3
x+6=-3
x=-9[/hide]

by the way jsc hide your answer next time   ;)


inequalities
calculus
quadratics
function
number theory
inequalities proposed

$a,b,c\ge0$. Prove
$2(1+abc)+\sqrt{2(1+a^2)(1+b^2)(1+c^2)}\ge (1+a)(1+b)(1+c)$. :)
Wanted: solutions with as little calculus as possible  :D

$2(1+abc)+\sqrt{2(1+a^2)(1+b^2)(1+c^2)}\geq (1+a)(1+b)(1+c) \leftrightarrow 2(1+a^2)(1+b^2)(1+c^2)\geq [a+b+c+ab+bc+ca-1-abc]^2 \leftrightarrow (1+a^2)(1+b^2)(1+c^2)\geq 2(a+b+c)(ab+bc+ca)+2abc-2(a+b+c)-2abc(ab+bc+ca) \leftrightarrow (ab+bc+ca-a-b-c)^2+(abc-1)^2\geq 2(abc-1)(a+b+c-ab-bc-ca)$
which is true

so,does this solution satisfy you? :D

yes its really nice
and its application to this problem was straightforward.

I didnt notice Gibbenergy s post when I was posting. Our solutions are identical!
Is it also your solution Spanferkel

no  :blush: I didn't want to expand it all and used some calculus  :blush:

[quote="spanferkel"]This is neat. A nice identity:

$\boxed{(-1+ab+ bc+ca)^2+(a+b+c-abc)^2=(1+a^2)(1+b^2)(1+c^2)}$  :)[/quote]

  Very nice identity.
  Your problem is solve by quadratic form. Since after expand, I treat it as the function with a and using discrimination to access that forms. So, my sols and Albani Eagle is quite different.
    But anyway, Very nice problems

[quote="Gibbenergy"] So, my sols and Albani Eagle is quite different.
    But anyway, Very nice problems[/quote] Well maybe your approach was different in the beginning but the solution is exactly the same. Add one more step in yours (putting it all on the LHS) and you have the same as Albanian Eagle :)
Anyway maybe using that identity in a hidden way can help constructing other ineqs harder to solve   :D  :D

Spanferkel !

  Your identity is very very nice.
  Indeed. Here is my conjecture:
$(-1+ab+bc+ca)^k + (a+b+c-abc) ^k$
 1/ $k$ is odd . It can be factor . I mean it can divided by $(ab+b-abc+bc-1+a+c+ac)$
 2/ $k$ is even.
           If $k$ is not the power of $2$ , I mean $k$ not $2^2,2^3,2^4,...2^n$ it can be factor to $(a^2+1)(b^2+1)(c^2+1)$
            If $k$ is the power of $2$. Irreducible form.
I will try to prove them. :D

Nice idea, Gibbs. Obvious for odd $k$.
But the second one doesn't hold for $k=12$.
In fact, $x^k+y^k$ is generally divisible by $x^2+y^2$ only of $k=2r$, with odd $r$.  :(

What I meant was rather finding nice difficult [i]inequalities [/i]that have an elegant solution based on some (hidden) use of the above identity... Less straightforward ones than my original one. :D

[quote="spanferkel"]Nice idea, Gibbs. Obvious for odd $k$.
But the second one doesn't hold for $k=12$.
In fact, $x^k+y^k$ is generally divisible by $x^2+y^2$ only of $k=2r$, with odd $r$.  :(

 :D[/quote]

  In case $k=12$ and also $k=20$ ...you are right, it doesn't have the factor $(a^2+1)(b^2+1)(c^2+1)$ but it still can be factor out. I think it should be a very nice number theory problem.
   :(

So, good luck!! And especially for proving that for $2^n$, it's irreducible... :P

And I hope you have a computer algebra system for trying :play_ball:   :whistling:

Now let's see if we can do something more general:
If $a_i, b_i, c_i \ge0$ for $i=1,...,n$, prove or disprove

$\sum_{i=1}^n a_ib_ic_i+\sqrt{\frac{\sum a_i^2\sum b_i^2\sum c_i^2}n}\ge \frac2{n^2} \sum a_i\sum b_i\sum c_i$ :lol:

As to the identity, I've just found a generalization of it and a *beautiful* proof. Well, once proven it looks so easy that I don't think it is brand new. I've posted it in Combinatorics [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=474387]here.[/url]

Thanks Albnian Eagle for providing me the link of this topic. The following is my solution:

Because $ 2(1\plus{}a^2)\equal{}(1\plus{}a)^2\plus{}(1\minus{}a)^2,$ and $ (1\plus{}b^2)(1\plus{}c^2)\equal{}(b\plus{}c)^2 \plus{}(bc\minus{}1)^2,$ the Cauchy Schwarz Inequality gives us
$ \sqrt{2(1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)} \ge (b\plus{}c)(1\plus{}a)\plus{}(1\minus{}a)(bc\minus{}1),$
and hence, we deduce that
$ 2(abc\plus{}1) \plus{}\sqrt{2(1\plus{}a^2)(1\plus{}b^2)(1\plus{}c^2)} \ge 2(abc\plus{}1)\plus{}(b\plus{}c)(1\plus{}a)\plus{}(1\minus{}a)(bc\minus{}1) \equal{}1\plus{}a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca\plus{}abc\equal{}(1\plus{}a)(1\plus{}b)(1\plus{}c),$
as desired.


limit
function
real analysis
real analysis unsolved

Let $ \tau : \mathbb{N}\to \mathbb{N}$ strictly increasing, $ (a_n) \in (\mathbb{R}_{\plus{}})^{\mathbb{N}}$ decreasing and $ \varepsilon : \mathbb{N}\to \{1,2\}$. If $ \sum (\minus{}1)^{\varepsilon (n)} a_n$ converge,
Does $ \sum (\minus{}1)^{\varepsilon (n)} a_{\tau (n)}$ converge?

$ \lim_{N\to \infty} \sum_{n\equal{}1}^N (\minus{}1)^{\varepsilon (n)} a_n \equal{} L < \infty$

fix $ \delta > 0$, then $ \exists M_\delta$ s.t.

$ \frac{\delta}{C}>|\sum_{n\equal{}1}^{M_\delta} (\minus{}1)^{\varepsilon (n)} a_n \minus{} L | \equal{}|\sum_{n\equal{}M_\delta\plus{}1}^{\infty} (\minus{}1)^{\varepsilon (n)} a_n|$

$ |\sum_{n\equal{}M_\delta\plus{}1}^\infty (\minus{}1)^{\varepsilon (n)} a_{\tau (n)}| \le C|\sum_{n\equal{}M_\delta\plus{}1}^{\infty} (\minus{}1)^{\varepsilon (n)} a_n|$

and we are done? just having problems with that C.... i want to use 2...

I believe there is not enough information to answer. Did you just make this problem up?

Observe what happens when the sign function is constant and we choose different terms of the harmonic series

[quote="§outh§tar"]I believe there is not enough information to answer. [/quote]
Do you mean that you can find $ \tau : \mathbb{N}\to \mathbb{N}$ strictly increasing, $ (a_n) \in (\mathbb{R}_{ \plus{} })^{\mathbb{N}}$ decreasing and $ \varepsilon : \mathbb{N}\to \{1,2\}$ s.t. $ \sum ( \minus{} 1)^{\varepsilon (n)} a_n$ converges and $ \sum ( \minus{} 1)^{\varepsilon (n)} a_{\tau (n)}$ diverges?

[quote="§outh§tar"]Did you just make this problem up?[/quote]
Yes I did

Related question: can you show that for a decreasing sequence $ (a_n)$, $ a_n > 0$ and $ a_n \to 0$, if $ \varepsilon^{\minus{}1}(2)$ has density $ \frac{1}{2}$ in $ \bf N$ then $ \sum (\minus{}1)^{\varepsilon(n)} a_n$ converges?  (This is close to the alternating series test.)

[quote="JBL"]... if $ \varepsilon^{ \minus{} 1}(2)$ has density $ \frac {1}{2}$ in $ \bf N$ ...[/quote]
If $ \left|\sum_{k\equal{}1}^n(\minus{}1)^{\varepsilon(k)}\right|$ is bounded, then there's an easy proof - but you can have density $ \frac12$ without that sum being bounded.

Joel, are you sure about that result?  What about 1 copy of $ \frac {1}{1}$, minus 2 copies of $ \frac {1}{2}$, plus 3 copies of $ \frac {1}{3}$, and so on?  That series does not converge, but I believe that the plus signs have density $ \frac {1}{2}$.  I hope I haven't miscalculated.

[quote="Ravi B"]Joel, are you sure about that result?[/quote]  Sorry, I wasn't claiming it as true: I played around with it far enough that I could have made Kent Merryfield's comment, suspected that it might not be true if the partial sums weren't bounded, and then didn't find a counter-example immediately so put it up here.  That is a counter-example.  I think my original thought was that a counter-example might be used to construct a counter-example to the original question, but my initial attempts based off your example haven't been successful.  I.e., I'm trying to do the problem in reverse by finding a supersequence that does converge when signed in the same way.

Edit: More broadly, I think the statement should be true if $ \varepsilon^{\minus{}1}(2)$ exists but is not $ \frac{1}{2}$, and probably also if $ \varepsilon^{\minus{}1}(2)$ does not exist, so this looked like the best place to look for a counter-example.

I believe the answer to the original question is "not necessarily".  Here is a counterexample.  First, define
\[ x_k \equal{} \frac {1}{k \log(k \plus{} 1)}.
\]
Note that the sum of the $ x_k$ diverges to infinity.

Consider the series consisting of 1 copy of $ x_1$, minus 1 copy of $ x_1$, plus 2 copies of $ x_2$, minus 2 copies of $ x_2$, and so on.  That series converges (to zero).

Now consider the "subseries" consisting of 1 copy of $ x_1$, minus 1 copy of $ x_2$, plus 2 copies of $ x_2$, minus 2 copies of $ x_3$, and so on.  That subseries diverges to infinity, by comparing to the sum of the $ x_k$.

Thanks.  Joel's side question was useful in pointing the way to the counterexample.


Indeed.

And, as always, I recommend [u]General Chemistry [/u]by Linus Pauling.  Besides those three books, studying from past tests is very likely to put you on the road towards camp.

Thanks, inteluser
I ordered the book just now on Amazon. It looks like a very interesting book,  =)

question:  is Mortimer- Robert G or Charles E?

Also, first two books are somewhat expensive so I don't think I'll get both.  Which do you think is more suitable for someone with a pretty solid understanding of AP chem topics (so like, I wouldn't need a complete intro to those topics.. maybe which is harder?  unless they both are pretty well beyond AP.. then which is more "user-friendly?").  thanks

Charles Mortimers book is my all time favorite chem book.  i think it is the best. :)

[quote="saturnlife"]My AP chemistry class is going at a very slow pace, so I am attempting some self-study on my own. I have Chemistry by Zumdahl, and Chemistry by Mortimer. Would studying solely from these two books and doing practice tests be enough preparation for me to have a decent chance at making it past the local level?[/quote]

principles of chemistry, raymond chang


group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ H$ and $ K$ are subgroup of a finite group then $ |HK| \equal{} \frac {|H\parallel K|} {|H\cap K|}$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232063

See the 10th post. (I think this has been done in other threads as well).