[quote="Pain rinnegan"]If $ a_1,a_2,a_3 > 0$ prove that $ \sum_{k \equal{} 1}^n(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\ge 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)$ , with $ a_4 \equal{} a_1$ .[/quote]
I think $ n\equal{}3$ then $ LHS \le RHS$

[quote="Pain rinnegan"]If $ a_1,a_2,a_3 > 0$ prove that $ \sum_{k \equal{} 1}^3(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\le 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)$ , with $ a_4 \equal{} a_1$ .[/quote]

Must be : $ \boxed{\sum_{k \equal{} 1}^3(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\le 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)}$

For any nonnegative $ a_1,a_2,a_3 > 0$  we have triangle  $ ABC$  with point  $ T$ inside  , so that :
 $ TA\equal{} a_1$ , $ TB \equal{} a_2$ , $ TC\equal{}a_3$ and $ \angle ATB \equal{} \angle BTC \equal{} \angle CTA \equal{} \frac {2\pi}{3}$.

Point   $ T$ for $ \triangle ABC$ is $ Torricelli$ Point.

Then  :  $ AB^2 \equal{} a_2^2 \plus{} a_2a_3\plus{}a_3^2$ , $ BC^2 \equal{} a_3^2 \plus{} a_3a_1\plus{}a_1^2$ , $ CA^2 \equal{} a_1^2 \plus{} a_1a_2\plus{}a_2^2$

So we must prove : $ \boxed{AB^2\plus{}BC^2\plus{}CA^2 \le 3(TA^2\plus{}TB^2\plus{}TC^2)}$

if $ M$ - median point $ \triangle ABC$ then :

$ \boxed{TA^2\plus{}TB^2\plus{}TC^2 \equal{} AM^2\plus{}BM^2\plus{}CM^2 \plus{} 3TM^2\equal{} \frac{1}{3}(AB^2\plus{}BC^2\plus{}CA^2)\plus{}3TM^2}$ :)

[quote="Sergic Primazon"][quote="Pain rinnegan"]If $ a_1,a_2,a_3 > 0$ prove that $ \sum_{k \equal{} 1}^3(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\le 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)$ , with $ a_4 \equal{} a_1$ .[/quote]

Must be : $ \boxed{\sum_{k \equal{} 1}^3(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\le 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)}$

For any nonnegative $ a_1,a_2,a_3 > 0$  we have triangle  $ ABC$  with point  $ T$ inside  , so that :
 $ TA \equal{} a_1$ , $ TB \equal{} a_2$ , $ TC \equal{} a_3$ and $ \angle ATB \equal{} \angle BTC \equal{} \angle CTA \equal{} \frac {2\pi}{3}$.

Point   $ T$ for $ \triangle ABC$ is $ Torricelli$ Point.

Then  :  $ AB^2 \equal{} a_2^2 \plus{} a_2a_3 \plus{} a_3^2$ , $ BC^2 \equal{} a_3^2 \plus{} a_3a_1 \plus{} a_1^2$ , $ CA^2 \equal{} a_1^2 \plus{} a_1a_2 \plus{} a_2^2$

So we must prove : $ \boxed{AB^2 \plus{} BC^2 \plus{} CA^2 \le 3(TA^2 \plus{} TB^2 \plus{} TC^2)}$

if $ M$ - median point $ \triangle ABC$ then :

$ \boxed{TA^2 \plus{} TB^2 \plus{} TC^2 \equal{} AM^2 \plus{} BM^2 \plus{} CM^2 \plus{} 3TM^2 \equal{} \frac {1}{3}(AB^2 \plus{} BC^2 \plus{} CA^2) \plus{} 3TM^2}$ :)[/quote]
Your solution very long,this inequality is equivalent to:$ (a_1\minus{}a_2)^2\plus{}(a_2\minus{}a_3)^2\plus{}(a_3\minus{}a_1)^2 \ge 0$

[quote="quykhtn-qa1"][quote="Sergic Primazon"][quote="Pain rinnegan"]If $ a_1,a_2,a_3 > 0$ prove that $ \sum_{k \equal{} 1}^3(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\le 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)$ , with $ a_4 \equal{} a_1$ .[/quote]

Must be : $ \boxed{\sum_{k \equal{} 1}^3(a_k^2 \plus{} a_k\cdot a_{k \plus{} 1} \plus{} a_{k \plus{} 1}^2)\le 3(a_1^2 \plus{} a_2^2 \plus{} a_3^2)}$

For any nonnegative $ a_1,a_2,a_3 > 0$  we have triangle  $ ABC$  with point  $ T$ inside  , so that :
 $ TA \equal{} a_1$ , $ TB \equal{} a_2$ , $ TC \equal{} a_3$ and $ \angle ATB \equal{} \angle BTC \equal{} \angle CTA \equal{} \frac {2\pi}{3}$.

Point   $ T$ for $ \triangle ABC$ is $ Torricelli$ Point.

Then  :  $ AB^2 \equal{} a_2^2 \plus{} a_2a_3 \plus{} a_3^2$ , $ BC^2 \equal{} a_3^2 \plus{} a_3a_1 \plus{} a_1^2$ , $ CA^2 \equal{} a_1^2 \plus{} a_1a_2 \plus{} a_2^2$

So we must prove : $ \boxed{AB^2 \plus{} BC^2 \plus{} CA^2 \le 3(TA^2 \plus{} TB^2 \plus{} TC^2)}$

if $ M$ - median point $ \triangle ABC$ then :

$ \boxed{TA^2 \plus{} TB^2 \plus{} TC^2 \equal{} AM^2 \plus{} BM^2 \plus{} CM^2 \plus{} 3TM^2 \equal{} \frac {1}{3}(AB^2 \plus{} BC^2 \plus{} CA^2) \plus{} 3TM^2}$ :)[/quote]
Your solution very long,this inequality is equivalent to:$ (a_1 \minus{} a_2)^2 \plus{} (a_2 \minus{} a_3)^2 \plus{} (a_3 \minus{} a_1)^2 \ge 0$[/quote]

I only want to show geometrical solution for : $ a_1a_2 \plus{} a_2a_3 \plus{} a_3a_1 \le a_1^2 \plus{} a_2^2 \plus{} a_3^2$


calculus integration limit trigonometry logarithms function algebra

Evaluate the following limit.

\[\lim_{n\to\infty}\int^{\pi}_0 |\sin nx|\ln (1+x) dx\ \ (n=1,2,\cdots)\]

We were discussing similar problems for some times.

For any continuous function $f$ following equality holds:
\[\lim_{n\to\infty}\int_a^b |sin(nx)|f(x)dx=\frac{2}{\pi}\int_a^b f(x)dx.\]

Though I am sure Kent will comment it. ;)

Yes we talked about this one

If $f,g:R \rightarrow R$ continuous, $g$ is T-periodic then

$\lim_{n \rightarrow +\infty} \int_{0}^{T}g(nx)f(x)dx=1/T.\int_{0}^{T}g.\int_{0}^{T}f$

here $g(x)=|\sin x|$ is $\pi$ periodic, $f(x)=\ln(x+1)$

And we have $\hbox{Fej\'er}$'s formula.
$g(x)$ is a measurable function with period of $T$. $\{\lambda_n\}$ is a sequence of reals. For $f\in L(\mathbb{R})$ it holds
$
\lim\limits_{n\to\infty} \int_\mathbb{R} f(x) g(nx + \lambda_n) dx = \left(\frac{1}{T}\int_0^T g(x)dx\right)\left(\int_\mathbb{R} f(x)dx\right)
$

[quote="Myth"]Though I am sure Kent will comment it.[/quote]
While I was away today, Moubinool and liyi have already said what I was going to say. But since I had the answer prepared, I'll run it anyway:

Suppose $f$ is periodic of period $P$ and that $f$ and $g$ satisfy certain other sufficient conditions. Then

$\lim_{n\to\infty}\int_0^Pf(nx)g(x)dx = \left(\frac1P\int_0^Pf(x)dx\right) \left(\int_0^Pg(x)dx\right).$

The additional sufficient conditions might be $f\in L^1(0,P),g\in L^{\infty},$ or that $f,g\in L^2(0,P),$ among other possibilities.

Assume that $f$ is a trigonometric polynomial. The finish of the proof will come from the fact that trignometric polynomials are dense in $L^1$ or $L^2$ or whatever space we're using. That is,

$f(x)=\sum_{k=-N}^N\hat{f}(k)e^{2\pi ikx/P}.$ Then,

$\lim_{n\to\infty}\int_0^Pf(nx)g(x)dx=\lim_{n\to\infty}\int_0^P \sum_{k=-N}^Ng(x)\hat{f}(k)e^{2\pi inkx/P}dx.$

But $\lim_{n\to\infty}\int_0^Pg(x)e^{2\pi inkx/P}dx=0$ for $k\ne0$ by the Riemann-Lebesgue Lemma. That leaves as the overall limit $\hat{f}(0)\int_0^Pg(x)dx,$ where $\hat{f}(0)=\frac1P\int_0^Pf(x)dx.$

[quote="koo"]You don't need to use trigonometric polynomials, using that the simple functions are dense in $L^1$ suffices.[/quote]
But how are we going to bring periodicity into this? Suppose that $f$ is the characteristic function of a set $A\subset[0,P],$ extended to be periodic of period $P.$ Suppose $g$ is a function satisfying whatever hypotheses we need in what follows. Now try to prove that

$\lim_{n\to\infty}\int_0^Pf(nx)g(x)dx=\frac{|A|}P\int_0^Pg(x)dx.$

This doesn't look obvious to me at all, not without bringing in some Fourier tools.

First off if you're going to do liyi's problem then approximating $g$ (his notation) with step functions won't do since they are not dense in ${L^\infty(\mathbb{R}})$. With that into account it's even more obvious that we should approximate $f$ instead, going over to the simple functions case is trivial in the $\mathbb{R}$ setting.

So assume $f$ is the step function $\chi_{[a,b]}$ and $g \in L^\infty$ then
\[\int f(nx)g(x)\, dx = \frac{1}{n}(\frac{n(b-a)}{T} + O(1))\int_0^T g(x)\, dx.\] What's left is to pass through the limit, and approximate with Fatou's lemma.


can someone post the address to it? i just can't seem to find it. thanks


vector
rotation
analytic geometry
graphing lines
slope
limit
complex numbers

Let $ x_1,z_2,...z_n$ be complex numbers such that
\[ |z_1| \plus{} |z_2| \plus{} ... \plus{} |z_n| \equal{} 1.
\]
Prove that there exists a subset $ S$ of $ {x_1x_2,...,x_n}$ such that
\[ \left| {\sum\limits_{z \in S} z } \right| \ge {1 \over \pi }.
\]
and prove that the lowest bound is $ {1 \over \pi }$.

[quote="HelloWorld"]Let $ z_1, z_2, ..., z_n$ be complex numbers such that
\[ |z_1| \plus{} |z_2| \plus{} ... \plus{} |z_n| \equal{} 1.
\]
Prove that there exists a subset $ S$ of $ {z_1, z_2, ..., z_n}$ such that
\[ \left| {\sum\limits_{z \in S} z } \right| \ge {1 \over \pi }.
\]
and prove that the lowest bound is $ {1 \over \pi }$.[/quote]

If anyone could fix the "obvious" gap in my proof, that would be excellent. My calc knowledge is virtually nonexistent.

EDIT: Scratch that, I suck at math. No calc needed. Solution now complete.

[hide="The complete solution"]Here, we will use "vector" and "complex number" interchangeably, whatever is most convenient at the time. The vectors will be $ r cis \alpha$ and the complex numbers will be $ a \plus{} bi$, although they really are the same thing.

Let's say we have our set $ S$ with the maximum sum. Assume for the sake of contradiction that $ {\left| \sum\limits_{z \in S} z \right|} < {1 \over \pi}$. Let $ x \equal{} \sum\limits_{z \in S} z$. WLOG, rotate all the vectors so that $ x \equal{} a \plus{} 0i$, where $ a$ is a positive real number.

Now, if there's a number $ z_k$ with a positive $ a$, putting that element in $ S$ will increase the absolute value of the sum of the elements in $ S$. So all elements with a positive $ a$ value must already be in $ S$.

Now designate all vectors to have an "On/Off" switch. This switch changes depending on whether each number $ z_k$ is included in $ S$. Have all switches be on the "Off" position.

For each element not in $ S$, turning the switch to the "On" position will include that element in $ S$.

For each element in $ S$, turning that switch to the "On" position will take that element out of $ S$, thereby subtracting that element from the total sum. This is equivalent to adding the opposite vector to $ x$.

Now we consider the set of vectors not in $ S$ and the opposite of the vectors in $ S$. Term this set $ A$. All vectors in $ A$ have a negative $ a$ value. When we have all the switches of $ A$ off, we end up at $ x \plus{} 0i$.

We must not have a series of "On/Off" switches that lets us increase $ x$, which means that we must not have a series of "On/Off" switches that gives us a value that lies outside the circle of radius $ x$, centered at $ 0 \plus{} 0i$. Our starting point for the On/Off switches with the set $ A$ will be at $ x \plus{} 0i$.

We turn on all the switches one by one, counterclockwise using the angle of the vectors. The path delineated must never exit the top edge of the circle, and will end up at a point $ M$. The length of this path is equal to the sum of the magnitudes of our vectors.

We turn off all the switches, and turn them back on one by one, clockwise using the angle of the vectors. The path delineated must never exit the bottom edge of the circle. We rotate this diagram 180 over the origin and translate it so that it fits on top of our first path. (see diagram)

Assume that the longest path does not lie on the arc shaped by the two semicircles. Then, between every two adjacent slope changes that do not lie on the arc, we must have a vector. If we "push" this vector up to become two smaller vectors, we will get a longer path by the triangle inequality. We have a contradiction, so the longest path must lie on the arc.

The longest arc occurs when $ M$ is at $ \minus{} x \plus{} 0i$, because moving it to the right will chop off sections off the top of the arcs, and it is not possible to move it to the left. The greatest arc is therefore $ 2x\pi \over 2$, which is less than 1. The magnitude of the vectors in $ A$ and the magnitude of the original vectors are equal (we only flipped some). Therefore, $ |z_1| \plus{} |z_2| \plus{} ... \plus{} |z_n|$ cannot be 1, and we have a contradiction. So $ x \ge \frac {1}{\pi}$.

For the second part, we have $ z_n \equal{} \lim_{x\to\infty}\frac {1}{x} \left({cis \frac {2\pi n}{x}}\right)$, where $ n \equal{} 1, 2, 3, \cdots, x \minus{} 1, x$. Choosing the vectors with positive $ a$ value gives us a semicircle with length 1/2. That means the diameter must be $ \frac{1}{\pi}$, and so the bound is completed.[/hide]


What is the therom or deffinition called that states when several things are spinnig around an epicenter and one breaks off it has to spin in the same direction?

Are you sure that's math? IF it isnt', the physics forums can help...

I found it out
It's called "The Conservation of Angular Momentum"

Please read the posting guidelines before posting in the middle school forums.


inequalities
search
LaTeX
inequalities theorems

I need some inequaliti is prove by Chebyshev's inequality
Who can help me ? Thanks a lots!

[quote="N.T.TUAN"]Ok, this is one :D

''Prove that $\sum\frac{a}{b+c}\geq\frac{3}{2}\forall a,b,c>0$''[/quote]
This is Nesbit's inequality

We can use Chebyshev's inequality :
 for $a\geq b\geq c$ We have :
 $a+b\geq a+c\geq b+c$ Use Chebyshev's ineq :
 $2(a+b+c)(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b})\geq 3(a+b+c)$
               :wink:  :wink:  :wink:  :wink:

[quote="chienthan"]Nobody can prove it?[/quote]
We can prove :
 $2(a^{2}+b^{2}+c^{2}+ab+bc+ca)(\sum\frac{a^{2}+bc}{(b+c)^{2}})\geq 3(a^{2}+bc+b^{2}+ac+c^{2}+ab)$
          ([b]Chebyshev's inequality [/b])
chienthan:Em đừng post những bài thế này nữa nhé (Anh chỉ góp ý thôi _ Anh không muôn em bước vào vết xe đổ của anh)

How about:

$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+a}+\frac{e}{a+b}\ge \frac{5}{2}$

It looks similar to the above question...just a bit more complicated  :roll:

[quote="problem_solver"]How about:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} d} \plus{} \frac {c}{d \plus{} e} \plus{} \frac {d}{e \plus{} a} \plus{} \frac {e}{a \plus{} b}\ge \frac {5}{2}$

It looks similar to the above question...just a bit more complicated  :roll:[/quote]

Nesbitt's Type.

[quote="Probably the anser is:"]
Let left hand side be $ I$.
By using Chebyshev's Inequality, $ 2(a\plus{}b\plus{}c\plus{}d\plus{}e)I\ge5(a\plus{}b\plus{}c\plus{}d\plus{}e)$
Done. :P [/quote]

[quote="mathboy_cntHDVN"]
chienthan:Em đừng post những bài thế này nữa nhé (Anh chỉ góp ý thôi _ Anh không muôn em bước vào vết xe đổ của anh)[/quote]
Yes I think that .

This one is very nice application of Chebyshev inequality:
For $ a\leq b\leq c\leq d\leq e$ and $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}1$,prove the following inequality:
    $ ad\plus{}dc\plus{}cb\plus{}be\plus{}ea\leq\frac{1}{5}$.

Why don't you people make a search for the keyword "Chebyshev" in the Inequalities section? It is not far fetched to expect such a search to bring up some examples of applying Chebyshev's inequality.

zeroεinteger: your proof is wrong. Chebyshev's inequality has conditions. ;)

  darij

[quote="chien than"]Continue...
Let $ a;b;c$ be positive real numbers .Prove that:
$ \frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{a \plus{} b \plus{} c} \plus{} \frac {a^{3} \plus{} b^{3} \plus{} d^{3}}{a \plus{} b \plus{} d} \plus{} \frac {a^{3} \plus{} c^{3} \plus{} d^{3}}{a \plus{} c \plus{} d} \plus{} \frac {b^{3} \plus{} c^{3} \plus{} d^{3}}{b \plus{} c \plus{} d}\geq a^{2} \plus{} b^{2} \plus{} c^{2}$[/quote]
By chebyshev,we have
$ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{a\plus{}b\plus{}c} \geq \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{3}$.
Summin these inequality,we get the desired result.

[quote="stephen38"][quote="chien than"]Continue...
Let $ a;b;c$ be positive real numbers .Prove that:
$ \frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{a \plus{} b \plus{} c} \plus{} \frac {a^{3} \plus{} b^{3} \plus{} d^{3}}{a \plus{} b \plus{} d} \plus{} \frac {a^{3} \plus{} c^{3} \plus{} d^{3}}{a \plus{} c \plus{} d} \plus{} \frac {b^{3} \plus{} c^{3} \plus{} d^{3}}{b \plus{} c \plus{} d}\geq a^{2} \plus{} b^{2} \plus{} c^{2}$[/quote]
By chebyshev,we have
$ \frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{a \plus{} b \plus{} c} \geq \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{3}$.
Summin these inequality,we get the desired result.[/quote]
Looks like my [url=http://www.mathlinks.ro/viewtopic.php?t=161284]solution[/url] :wink:

[quote="darij grinberg"]Why don't you people make a search for the keyword "Chebyshev" in the Inequalities section? It is not far fetched to expect such a search to bring up some examples of applying Chebyshev's inequality.
[/quote]
Maybe because they are too lazy,or if you try search,it might be too many posts,hence it would be much easier just to collect them here,isn't it?

i dont know is here the place or not , but i have a question about chebyshev's ineq:
can we say it's a particular case of karamta's ineq when it can be proven with rearrangement ,
but rearrangement is a particular case of karamata  :maybe:

[quote="BG Yoda"]but rearrangement is a particular case of karamata  :maybe:[/quote]

How?

  darij

edit : i'm sorry for the doc file , but i dont know latex  :blush:

edit2 : now i realize that it's true only for positive numbers  :( sorry

:roll: that from the previous post is absolutely wrong  :mad:
the real reason for the majorization is that :
 $ \sum^{k}_{i \equal{} 1} (t_i \plus{} k_i)$ is possibly the greatest , because $ t_1\geq t_2\geq ...\geq t_n$ and $ k_1\geq k_2\geq ...\geq k_n$
and so $ \sum^{k}_{i \equal{} 1} (t_i \plus{} k_i) \geq \sum^{k}_{i \equal{} 1} (t_{j(i)} \plus{} l_{j(i)})$  for $ 1\leq k\leq n \minus{} 1$
$ \sum^{n}_{i \equal{} 1} (t_i \plus{} k_i) \equal{} \sum^{n}_{i \equal{} 1} (t_{j(i)} \plus{} l_{j(i)})$
$ \{j(i)\}$ is such a permutation of $ \{i\}$  that  $ (t_{j(1)} \plus{} l_{j(1)})\geq (t_{j(2)} \plus{} l_{j(2)})\geq ...\geq (t_{j(n)} \plus{} l_{j(n)})$
in this way $ [(t_1 \plus{} k_1),(t_2 \plus{} k_2),...,(t_n \plus{} k_n)]\succ [(t_{j(1)} \plus{} l_{j(1)}),(t_{j(2)} \plus{} l_{j(2)}),..., (t_{j(n)} \plus{} l_{j(n)})]$

it seems that when I was typing the doc file , I should have looked at sth similar to this now , which is on bulgarian and finally I posted that nonsense from the previous post :blush:

sorry for my english but it's the same as my mathematical acknowledges . i hope i speak at least a little clearly :huh:

Nice proof, though it needs an additional argument in order to be extended to arbitrary reals (and not just positive ones).

By the way, your post proving the Fuchs inequality is correct. I don't know why the $ p_i$ are required to be positive in the paper.

  Darij

btw if $ a_i$ and $ x_i$ are all negatives we'll have :
$ a_1\equal{}\minus{}e^{t_1}> ... >a_n\equal{}\minus{}e^{t_n}$ : $ t_1<t_2<...t_n$
$ x_1\equal{}\minus{}e^{k_1}> ... >x_n\equal{}\minus{}e^{k_n}$ :  $ k_1<k_2<...k_n$
$ a_i.x_i\equal{}\minus{}e^{t_i}.\minus{}e^{k_i}\equal{}e^{t_i\plus{}k_i}$
and $ [(t_n \plus{} k_n),(t_{n\minus{}1} \plus{} k_{n\minus{}1})...(t_1 \plus{} k_1)]\succ [(t_{j(1)} \plus{} l_{j(1)}),(t_{j(2)} \plus{} l_{j(2)})...(t_{j(n)} \plus{} l_{j(n)})]$

so $ \sum^{1}_{i\equal{}n} e^{t_i \plus{} k_i} > \sum^{1}_{i\equal{}n} e^{t_i\plus{}l_i}$
$ \equal{}> \sum^{n}_{i\equal{}1} a_ix_i > \sum^{n}_{i\equal{}1} a_iy_i$

however this isn't a proof for arbitrary reals again but for all positives or all negatives only :(

[quote=chien than]I have inequality:
Let $a;b;c$ are positive real numbers.Prove that:
$\frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq \frac{3}{2}$[/quote]

One more approach..

Observe that the function $$f(x)=\frac{x^2+\frac{abc}{x}}{(a+b+c-x)^2}$$
is convex as $$f''(x)\geqslant 0$$.
Now the given inequality follows by application of Jensen's inequality.

[quote=chien than]I have inequality:
Let $a;b;c$ are positive real numbers.Prove that:
$\frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq \frac{3}{2}$[/quote]

Does anyone know this source of this problem? :(

[quote=chien than]I have inequality:
Let $a;b;c$ are positive real numbers.Prove that:
$\frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq \frac{3}{2}$[/quote]
We have
\[4(a^2+bc)(a^2+b^2+c^2)-(b+c)^2[6a^2-a(b+c)+2bc] = a(a+b+c)(2a-b-c)^2+(a^2+2bc)(b-c)^2 \geqslant 0.\]
Therefore
\[\sum \frac{a^{2}+bc}{(b+c)^{2}} \geqslant \frac14\sum \frac{6a^2-a(b+c)+2bc}{a^2+b^2+c^2} = \frac32.\]

$\sum{\frac{a^2 +bc}{(b+c)^2}} \geq \frac{3}{2}$ iff

$\sum{\frac{2a^2 +2bc}{(b+c)^2}} \geq 3$ iff

$\sum{\frac{2a^2 +2bc}{(b+c)^2}} -1 \geq 0 $ iff

$\sum{\frac{2a^2-b^2-c^2}{(b+c)^2}} \geq 0$...................(*).This inequality is symmetric in a, b and c. We can prove it using Chebyshev's inequality .It suffices to prove it for the ordering $ a \leq b \leq c $.In which case
$\frac{1}{b+c} \leq \frac{1}{c+a} \leq \frac{1}{a+b} $, and
$2a^2 -b^2-c^2 \leq 2b^2-a^2-c^2 \leq 2c^2-a^2 -b^2 $. And Chebyshev's inequality gives

$\sum{\frac{2a^2-b^2-c^2}{(b+c)^2}} \geq \frac{1}{3}.(\sum{2a^2-b^2 -c^2}).(\sum{\frac{1}{(b+c)^2}})$ = 0, as desired.

[quote=chien than]I have inequality:
Let $a;b;c$ are positive real numbers.Prove that:
$\frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq \frac{3}{2}$[/quote]

For $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\frac{a^{3}+abc}{(b+c)^{2}}+\frac{b^{3}+abc}{(c+a)^{2}}+\frac{c^{3}+abc}{(a+b)^{2}}\geq \frac{3}{2}$$

[quote=chien than]I have inequality:
Let $a;b;c$ are positive real numbers.Prove that:
$\frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq \frac{3}{2}$[/quote]

We have:
$$\sum_{cyc}\frac{a^2+bc}{(b+c)^2} \geq \sum_{cyc}\frac{\frac{b^2+c^2}{2}+bc}{(b+c)^2}=\frac{1}{2}\sum_{cyc}\frac{b^2+c^2+2bc}{(b+c)^2}=\frac{3}{2}$$


I'm sorry if this had been asked before.

What year's chapter; states; nationals competition was the hardest?

What I've heard is that 2005 states and nats were pretty hard and that the cutoff for national countdown in 2005 was 30...then again I got like a 12 on just the target a while back and I could've gotten a 16 without too much trouble. It didn't seem that hard, but that's just target. 2005 states definitely wasn't hard compared to some other ones, as I had my highest score on 2005.

Now, are you asking for hardest overall, or just hardest to get a certain score in? For example, several have only a few difficult questions in them so it's not too hard to get 30+ but difficulty to get in the 40's...

Looking at my most recent chapter and state scores (only from 2000-2007, excluding 2004 since that's all I kept track of) the chapter that I scored lowest on was 2007, but that's because there was this annoying grid question where you had to count and I fail at organized counting. The lowest state score I got was in 2001 because there were two annoying target questions and then a couple sprints I can't remember right now.

So I'd say 2005 overall was the hardest year according to people, but I'm not sure about the specifics.

I agree that the general consensus is that 2005 Nationals was brutal.

I'm not sure about Chapter tests, because they're basically too easy to compare, so there's not too much that would make one test significantly more difficult than others.

For state tests, I think 2008 was the hardest. :P Probably because it was in actual competition. Otherwise, I think 2006 was sort of difficult (if I remember correctly).

I remember some year there were weird and ambiguous questions on the Target, but I forget which...

OK, here's a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98871&vote=viewresult&search_id=1384474796]poll[/url] and here's some [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1384474796&t=116597]discussion[/url].

BOGTRO, we were talking about the 2006 State tests.

I haven't actually done many MATHCOUNTS problems since last year, so that's why I don't really remember much about the difficulties, other than the fact that 2004-2007 were generally harder than the rest.

[quote="xpmath"]Well, the target was pretty easy in 2008 except for #8. The sprint did seem a bit difficult...I guess you could argue that it was the hardest state. I don't think 2006 was that hard, but I can't seem to find my sprint score.[/quote]

#8 actually wasn't that bad.  Without looking at the official solutions, I solved this in about 4 minutes, but #7 was abnormally easy.

And 2006 was actually pretty easy, in my opinion.  I got 20 sprint and 12 target, so you would get somewhere in the mid-30s. :P

The only person I know when I went to state last year who was able to solve #8 target was "**stevenmeow**". For me it was ridiculously hard. I did miserablely in state last year; only 18 in sprint! My only excuse for that is: "well, I'm only a 6th grader..." which is a ridiculous excuse.
I'm not participating this year though.

Meh, #8 wasn't too hard with casework.  And #7 was uber-easy.

It seems that we are discussing different tests after each post... :wink:

Back in my day, I think 2000 state had a reputation for being quite difficult.  If I remember correctly, 4th place in Minnesota that year was a 27.  It might be considered easier by current standards, though.

My MC coach gave me a bunch of MC problems and the lowest score that I ever got on any of the national sprint rounds was 8/30, but that's just me, but it was the 1996 National Sprint. So I think 1996 was a hard year. I've never seen 2005 National problems, so I don't know about that.

where do u guys get the national rounds?

I got all mine by trading :D

To Wickedestjr, 1996 wasn't that bad...


hi there's a possiblitly that i might be moving to gainsville so i was wondering what math team tournys that buchholtz goes to. thanks:)

well, for the past two years, they've gone to vestavia.  :P
and just all the usual stuff FL schools do (all the MAO comps, florida math league, AMC/AIME, etc)

well i go to buchholz and walk around the river is pretty much right (hey eli)...except when u go bak to vestavia everyone will glare @ u with ur buchholz math team shirt...so be prepared!

bahaha i know but i won't wear them:) haha and ive probably met some of yall at that picnic thingy:)


calculus
integration
trigonometry
calculus computations

try to find some recurent formula
$I_{n}=\int(\frac{\sin\frac{x-a}{2}}{\sin\frac{x+a}{2}})^{n}dx$ where $n$ is natural.  :)

Ok
$t=\frac{\sin\frac{x-a}{2}}{\sin\frac{x+a}{2}}$. then $x=2arctan(\frac{(1+t)tan(a/2)}{(1-t)}$ and derivetive is $\frac{4tan(a/2)}{(1+tan^{2}(a/2))(1+2At+t^{2})}$ where $A=\frac{tan^{2}(a/2)-1}{tan^{2}(a/2)+1}$ and after few calculus we find that $I_{n}=(2sina)(\frac{t^{n+1}}{n+1}-\frac{I_{n+1}A}{sina}-\frac{I_{n+2}}{2sina}+C$ Done. I only wonted looked some different idea


function
search
algebra unsolved
algebra

Find all functions $ f: R\rightarrow R$ such, that for every real $ x,y,z$ we have:
\[ f(x+f(y+f(z)))=f(x)+f(f(y))+f(f(f(z))).\]

[quote="N.T.TUAN"]If $ f$ is continuous then here is answer http://www.mathlinks.ro/Forum/viewtopic.php?t=132174  :wink:[/quote]
Hard to find something new for this forum  :lol:

Yes, I think so! Hence, Use function search before posting!  :wink:


Prove

$4(a+b+d)(a+c+d)\geq (2a+3d+c)(2a+2b+c+d)$  for $a\leq b\leq c\leq d$ or $d\leq c\leq b\leq a$

After expansion the inequality is equivalent to $(2b-c-d)(c-d) \ge 0$.


Let $ b \in \mathbb{N}$. For which $ a \in \mathbb N$ the number $ a^2\plus{}b$ is a perfect square? That is, when do $ a,c$ exist such that $ (c\minus{}a)(c\plus{}a)\equal{}b$?

[quote="Stewie"]Let $ b \in \mathbb{N}$. For which $ a \in \mathbb N$ the number $ a^2 \plus{} b$ is a perfect square?[/quote]

[quote="Stewie"]That is, when do $ a,c$ exist such that $ (c \minus{} a)(c \plus{} a) \equal{} b$?[/quote]

Are these problems equivalent? If so, for the second problem we need to find $ b$ that can be expressed as a difference of two squares, which happens iff $ b\not\equiv 2\pmod 4$.

$ b \equal{} 4k \plus{} 1\implies (2k)^2 \plus{} b \equal{} (2k \plus{} 1)^2$
$ b \equal{} 4k \plus{} 3\implies (2k \plus{} 1)^2 \plus{} b^2 \equal{} (2k \plus{} 2)^2$
$ b \equal{} 4k\implies (k \minus{} 1)^2 \plus{} b \equal{} (k \plus{} 1)^2$

If $ b\equiv 2\pmod 4$, we have $ c^2 \minus{} a^2\equiv 2\pmod 4$, which is impossible as $ x^2\equiv 0,1\pmod 4\, \forall x$.

I'm not sure if this is what you intended though...

[quote="Stewie"]Let $ b \in \mathbb{N}$. For which $ a \in \mathbb N$ the number $ a^2 \plus{} b$ is a perfect square? That is, when do $ a,c$ exist such that $ (c \minus{} a)(c \plus{} a) \equal{} b$?[/quote]

For the question "For which $ a \in \mathbb N$ the number $ a^2 \plus{} b$ is a perfect square?" :

If $ b\equal{}1\pmod{2}$, We find all the values of $ a$ as following : $ a\equal{}\frac{1}{2}(\frac{b}{d}\minus{}d)$ for any divisor $ d$ of $ b$ such that $ d<\sqrt{b}$

If $ b\equal{}2\pmod{4}$, no such $ a$ exist.

If $ b\equal{}0\pmod{4}$, We find all the values of $ a$ as following : $ a\equal{}\frac{1}{2}(\frac{b}{2d}\minus{}2d)$ for any divisor $ d$ of $ \frac{b}{4}$ such that $ d<\sqrt{\frac{b}{4}}$

[quote="Stewie"]It's not clear to me the last case. I mean, why the denominator $ 2d$ and the condition $ d|b/4$?[/quote]

We have $ a^2\plus{}b\equal{}c^2$ and so $ (c\minus{}a)(c\plus{}a)\equal{}b$ and so $ c\plus{}a\equal{}d_1$ and $ c\minus{}a\equal{}\frac{b}{d_1}$. and so $ a\equal{}\frac{1}{2}(d_1\minus{}\frac{b}{d_1})$ and $ c\equal{}\frac{1}{2}(d_1\plus{}\frac{b}{d_1})$

But then, in order $ a$ and $ c$ be integers, we need $ d_1$ and $ \frac{b}{d_1}$ have same parities (both even or both odd). So, if $ b$ is even, we need $ d_1$ and $ \frac{b}{d_1}$ be even. So we need $ b\equal{}0\pmod{4}$ and $ d_1\equal{}2d$ and $ \frac{b}{2d}$ even. So we need $ d|\frac{b}{4}$

And so the result.

[quote="Stewie"]Let $ b \in \mathbb{N}$. For which $ a \in \mathbb N$ the number $ a^2 \plus{} b$ is a perfect square? That is, when do $ a,c$ exist such that $ (c \minus{} a)(c \plus{} a) \equal{} b$?[/quote]
$ \{a\in\mathbb{N}|\ a^2 \plus{} b\ perfect\ square\}$
$ \equal{} \{a\in\mathbb{N}|\ \exists c\in\mathbb{N},b \equal{} (c \minus{} a)(c \plus{} a)\}$
$ \equal{} \{\frac {|d \minus{} \frac {b}{d}|}{2}: \ d|b,d \minus{} \frac {b}{d}\ even\}$

My apologies since I have missed something, but wont this be true for all a and odd b? I mean

Let $ f(x)$ be the $ x^{th}$ odd number then we have

$ a^2\plus{}f(a)\equal{}(a\plus{}1)^2$

Hence for all a, there exists b such that $ a^2\plus{}b\equal{}c^2$?

Have I missed something here?

Thanks

[quote="SimonM"]My apologies since I have missed something, but wont this be true for all a and odd b? I mean

Let $ f(x)$ be the $ x^{th}$ odd number then we have

$ a^2 \plus{} f(a) \equal{} (a \plus{} 1)^2$

Hence for all a, there exists b such that $ a^2 \plus{} b \equal{} c^2$?

Have I missed something here?

Thanks[/quote]

What you have shown is that for all $ a$ $ \exists b\ |\ a^2 \plus{} b$ is a perfect square, which is quite trivial. But the problem asks to find all $ a$ for a given $ b$ such that $ a^2 \plus{} b$ is a perfect square.

PS. 700th post! :D

Thank you, can't believe I didn't see that

Congrats on you're 700 posts

No problem. We all make mistakes. (especially me! :P ) And thank you.

[quote="nayel"]
What you have shown is that for all $ a$ $ \exists b\ |\ a^2 \plus{} b$ is a perfect square, which is quite trivial. But the problem asks to find all $ a$ for a given $ b$ such that $ a^2 \plus{} b$ is a perfect square.

PS. 700th post! :D[/quote]
i posted that
[quote="aviateurpilot"]
$ \{a\in\mathbb{N}|\ a^2 \plus{} b\ perfect\ square\}$
$ \equal{} \{a\in\mathbb{N}|\ \exists c\in\mathbb{N},b \equal{} (c \minus{} a)(c \plus{} a)\}$
$ \equal{} \{\frac {|d \minus{} \frac {b}{d}|}{2}: \ d|b,d \minus{} \frac {b}{d}\ even\}$[/quote]
then $ \{a\in\mathbb{N}|\ a^2 \plus{} b\ perfect\ square\} \equal{} \{\frac {|m \minus{} \frac {b}{m}|}{2}: \ m|b\}\cap \mathbb{N}$
so if $ b\equal{}k2^n$ where $ k$ is odd
 $ \{a\in\mathbb{N}|\ a^2 \plus{} b\ perfect\ square\} \equal{} \{\frac {|m \minus{} \frac {b}{m}|}{2}: \ m|b,m\neq 2^k\}$


algebra
polynomial
calculus
integration
quadratics
modular arithmetic
induction

Determine a monic polynomial $f(x)$ with integral coefficients such that $f(x) \equiv 0 \bmod p$ is solvable for every prime $p$ but $f(x) = 0$ is not solvable with $x$ an integer.

[hide="Hint"] Quadratic residues ;) [/hide]

Don't have a complete solution yet, but for all $p \equiv 1 \pmod{4}$, we know there exists an integer $x$ such that $x^{2}\equiv-1 \pmod{p}$, so the polynomial $x^{2}+1$ suffices. Not sure how to get the $p \equiv 3\pmod{4}$ case.

[hide="Hint 2"] Split it up into $p \equiv 3 \bmod 8, p \equiv 7 \bmod 8$. [/hide] [hide="Hint 3"] Quadratic reciprocity might be useful here!  Well, unless you already know common quadratic residues  :D [/hide]

$f(x) = (x^{2}+1)(x^{2}-2)$ can cover the $p \equiv 7 \pmod{8}$ case now... I'm thinking $x^{2}+3$ for the last factor. I don't remember quadratic residues, though.

You're really close, so I'm just going to [hide="tell you the last factor..."] $x^{2}+2$ covers $p \equiv 3 \bmod 8$  :wink:

Edit:  $x^{2}+1$ also covers $p = 2$, which is a case you neglected ;) ;) ;)  [/hide]

Easy to those who knows [hide="dont touch"]quadratic residues[/hide]
[hide="for instance"]$(x^{2}-2)(x^{2}-3)(x^{2}-6)$[/hide]

there's also
[hide="blah"]
$(x^{2}+1)(x^{2}+2)(x^{2}-2)(x^{2}-17)$
[/hide]
which has solutions to $f(x)\equiv 0\pmod{n}$ for [b]any[/b] positive integer $n$ (!)

[hide="sketch of proof"]
We only need to show it for prime powers, then CRT will do the rest.

For the odd prime powers, we can prove by induction that if $r$ is a square root of $a$ modulo $p^{n}$ where $p$ is an odd prime and $n$ is a positive integer, then there exists a number $s$, congruent to $r$ modulo $p^{n}$, such that $s^{2}\equiv a\pmod{p^{n+1}}$. This isn't hard; just write $s=r+kp^{n}$ and solve for $k$.

This leaves the powers of 2, and the really strange factor $x^{2}-17$ deals with numbers of the form $2^{n}$. It's trivial for $n\leq 3$, then for the inductive step, show that if $r^{2}\equiv 17\pmod{2^{n}}$ for $n\geq 3$, then either $r^{2}$ or $(r+2^{n-1})^{2}$ is congruent to 17 modulo $2^{n+1}$.
[/hide]

I need not to see what  Scorpius hides.
[hide="I bet I know"]To cover all odds it is sufficient to consider polinom of the kind $f(x)=(x^{2}-a)(x^{2}-b)(x^{2}-ab)$ for any integers $a$ and $b$. For odd prime $p$ it follows from that: if both $a$ and $b$ are not  quadratic residues then $ab$ is. Now if $x^{2}-c \equiv 0 (mod p^{k})$ solvable then it is solvable modulo $p^{k+1}$ - each solution might be uniquely expanded  because the derivative of $x^{2}-c$ is $2x$ and so couldn't be congruent to $0$ modulo odd $p$ exept $x\equiv 0$ which is neglected.
Now we only need to choose $a$ and $b$ to cover $2^{k}$ and of course have no integer roots for $f(x)=0$
Just because it's more easy to seek under lamp and also to make simple 3d factor $x^{2}-ab$ choose $a=-1$ It remains to take $b$ such that $x^{2}-b$ covers all $2^{k}$. As $x^{2}\equiv 1 (mod 8)$ for odd $x$ the first candidat for this is $17$. Now it is easy that all $2^{k}$ are covered indeed because if $x^{2}\equiv 17 (mod 2^{k}$ but fails to $\equiv (mod 2^{k+1}$ we can improve it: $x: =x+2^{k-1}$
So should I open hidden text I must discover somewhat of $f(x)=(x^{2}+1)(x^{2}-17)(x^{2}+17)$[/hide]
P.S. After looking: No principal difference exept Scorpius have not care of the economy.  :lol:

[hide="On second thoughts"]When I've exit I realized what I've missed.Of course $b=17$ is the 1st [b]positive[/b] candidat but we can take negative $b=-7$ which is also work. And generelising all the polinoms of the kind $f(x)=(x^{2}-a)(x^{2}-b)(x^{2}-ab)$ are required when at least one of $a, b, ab$ is congruent $1 (mod 8)$ and no of them are perfect square.
For instance we can take $f(x)=(x^{2}+1)(x^{2}+7)(x^{2}-7)$ [/hide]
It looks like true that $6$ is the least degree for such polinom but at the moment I have no idea to prove it or affirmate.


Which absorbs gaseous carbon dioxide most effectively?

(a) solid potassium hydroxide

(b) solid silicon dioxide

(c) aqueous hydrogen chloride

(d) aqueous sodium fluoride

[hide="Click here"]In my opinion it is (a), because hydroxide ions are better able to react with carbon dioxide by carbonyl attack - for those who work in the laboratory, it is known that solid sodium or potassium hydroxide pellets are covered by a white layer made up of sodium (or potassium) carbonate and bicarbonate.[/hide]

[hide="me too"]
yes it is (a).Caustic Potash is used to absorb carbon di oxide[/hide]


let a1,a2,...,an  are positive numbers that :
a1+a2+...+an=3

a1^2 + ...+ an^2 >1
prove : there are three numbers like Ai,Aj,Ak that : Ai+Aj+Ak>1

it's wrong. take $ n\equal{}6$ and $ a_1\equal{}a_2\equal{}\ldots\equal{}a_6\equal{}\frac12$.

No,Im sure that it is  correct completely

i think Extremal can be useful for this problem :maybe:

From QM-AM we obtain $ \sum_{i\equal{}1}^{n}a_{i}^{2}\ge\frac{\left(\sum_{i\equal{}1}^{n}a_{i}\right)^{2}}{n}$ so if $ n>9$ we get contradiction.


What you do is for each choice, plug $ \minus{}3$ and see if it evaluates to $ 1$ as expected. If it doesn't, then that cannot be $ f(x)$.

[quote="Yongyi781"]What you do is for each choice, plug $ \minus{} 3$ and see if it evaluates to $ 1$ as expected. If it doesn't, then that cannot be $ f(x)$.[/quote]

I think that's what I did above. I only had to find the 1 that does not work. Did I do the math correctly?

[quote="Yongyi781"]Yes, you are correct.[/quote]

Thank you.

:)


i got this problem:
 Prove that  t(1)+t(2)+...t(n)= nlogn +O(n).

 ¿is solving the above equivalent to proving that g(n)= t(1)+t(2)+...t(n)-nlogn=O(n)?

ok so i think i got this one :

 First we prove that t(1)+t(2)+...t(n)= [n/1]+[n/2] + ... [n/n].

 This is true because m is a divisor of exactly [n/m] numbers between 1 and n. ok now we see that
  log(n)> 1/2+1/3+...1/n.

  so we have t(1)+t(2)+...+t(n) - nlog(n)< t(1)+t(2)+...+t(n) - (+n/2+n/3+...+n/n)=

  [n/1]+[n/2]+...[n/n] -(+n/2+...+n/n)< n.

 So we see that t(1)+t(2)+...+t(n) - nlogn < n for every so we have M=2 AND N=0

 meaning for every n>0  we have t(1)+t(2)+...t(n)-nlogn < n so we have t(1)+t(2)+...+t(n)= nlog(n)+O(n).

  Am i right or is there something wrong here?


geometry
3D geometry
pyramid
MATHCOUNTS
probability
Gauss
blogs

I'm designing a Chapter/State level mock test for you to do. Right now I have completed typing up the Target Round. If you're interested in doing it, you don't need to signup, all you do is PM your answers to me. Please be on the honor system with six minutes per pair of problems!!! You will have until [b]AUGUST 10TH [/b]to PM your Target answers! GOOD LUCK!

Here's the Target Round: (no signups needed,  If you wanna do it please PM me the answers and I'll score it)

1. Callie bought six ounces of coffee cream and 10 ounces of coffee. She first mixes both the cream and the coffee in one large pitcher, then pours 3/5 of the coffee mixture into her cup. What fraction of the coffee mixture in Callie's coffee cup is cream? Express your answer as a common fraction.

2. How many integers between 1 and 100 inclusive are divisible by 4 but not divisible by 5?

3. Bobby has a square based toolshed measuring 15 feet on each side. He tethers his dog to one vertex of the shed with a 10-foot leash. What is the area of the region where his dog can roam freely? Express your answer in terms of pi.

4. On GINORMO Sword, Lawrence kills 5 Golems in 2 minutes. If he played for 1.75 hours, how many Golems can he murder?

5. Paul decides to make a 4 feet by 6 feet montage of smilies, either with Starwars or Spam emoticons. Starwars emoticons cover 2 square feet of space and Spam emotes cover 1 square foot of space. What is the maximum number of smilies Paul can fit on the montage?

6. Xiaoyu is collecting a pyramid structure of red and green apples. On the top level, she puts 1 red apple. On the second level, she puts 4 green apples, and on the third level she puts 9 red apples. If she follows this pattern and makes the pyramid 20 rows high, how many green apples would be in her pyramid?

7. While staying at the Denver Marriott Tech Center for the 2008 Lockheed Martin MATHCOUNTS Competition, Darryl enters an elevator that travels through 20 floors. He enters the elevator on the second floor. Darryl tosses a fair, six-sided die. If the number appearing is prime, Darryl moves down one floor, otherwise, he goes up one floor. What is the probability that after 5 die rolls, Darryl would be on the 5th floor or higher? Express your answer as a common fraction?

8. Kevin leaves his pet monkey on the computer to spam in gauss1181's blog. The monkey is ordered to use a 26 letter keyboard to type "HEY GAUSS WANNA SPAM WITH ME SPAM SPAM SPAM SPAM SPAM". If the monkey does not know how to type in English letters and instead just strikes random keys on the keyboard, what is the probability that the monkey would spell GAUSS correctly AND spell SPAM correctly AT LEAST ONCE? Express your answer as a common fraction.

[quote="shentang"]On question number 7, what if Darryl rolls primes in his first two rolls? Does he go to the 0th floor, or stay where he is?
Edit: Now I see, it doesn't even matter.

And on number 8, I got a rediculously small number that will not fit on my calculator, so what do I do?[/quote]
stays where he is.

on 8, just express in scientific notation :P

Just use rounding, whatever feels best for you. I'll accept any closely rounded answers because I'm a nice kid :D

Oh and also note this target round is on my blog too :)

Guyz come on! Let's get more people involved.

MOCK TEST ANSWERS FOR TARGET ROUND DUE TOMORROW AT 11:59 PM CENTRAL TIME!! NO LATE ONES ACCEPTED!!!!!!!!

So far, we've only had one submission:

Submitted:
1. FantasyLover
Waiting:
1. shentang
2. Poincare

Please also note you CAN STILL take my mock test sprint round if you haven't submitted your target answers in time, it's just that you'll receive a 0 on the target round if not turned in on time.

Again, people! MOCK TESTT TARGET ANSWERS DUE TODAY AT 11:59 PM Central TIME!!!!
PLEASE SUBMIT ASAP FOR THE SCORE YOU WANT!!!!

im thinking that theres gonna be only 3 ppl advancing to sprint round...

Why?

There's no advancement qualification, people just can participate whenever they want to. If they don't submit target answers, automatic 0 for target.

ATTENTION EVERYONE:

Since there has only been 2 submissions of answers, I will extend the date to August 15 (Friday) since many of you might be watching the Olympics! (perhaps)

Therefore, [size=150][b]ALL TARGET ANSWERS ARE DUE FRIDAY AUGUST 15 AT 10 PM CENTRAL TIME. NO EXCUSES/EXCEPTIONS!!!!!!!!!!!!!!!!!!!!![/b][/size]

thanks! I forgot about this mock test  :P  :D

*BUMP* Again, deadline changed to [b]Friday August 15 10 PM CENTRAL TIME!!!!!!![/b]

Send in your answers--we need more peeps on this!

this isn't hard guyz

this is just target. sprint would be posted next week.

Your sprint round looks so much more like the real thing than mine, in terms of cover page, etc... :(

I think that you got your template from Mr. Frost? If so, then thats like the exact copy of the real tests...

Actually, I got my template, but I changed it a little, using the 2006 National Sprint round cover page as a guide :D

Number 29, every odd prime up to 10,000? That's nuts!

Also, my name is not shen like tinytim's isn't tim. :P

You might think it's nuts... but... it's tricky... :D
It becomes easy once you think about it for a while. :D

Yeah, I finally got it....or at least I think. :ninja:

You actually already submitted your answers, and the answers aren't due till September 12.

Please refrain from discussing what you put as answers until I tell you it's okay. If you want to post here, please keep it only to stuff related to questions about the wording of the problem and/or any typos you see. Thanks. :)

Heres an update on submissions of sprint answers:

Submissions:
1. chenhsi
2. shentang

Cmon we need more participation! Answers due Sep. 12!

Another Update!
Heres an update on submissions of sprint answers:

Submissions:
1. chenhsi
2. shentang
3. FantasyLover

Cmon we need more participation! Answers due Sep. 12! ONLY A WEEK AWAY! (also my birthday is on the day the answers are due!) :D

Good for you. I'm glad. :D

thank you
that is all
 :)

um gauss... results??

its 9/17 now...

EDIT:lol, nvm.

She gave the results in the discussion thread thingy, with the poll.

sorry for the massive bump but are there answers to this?
thanks


trigonometry
inequalities proposed
inequalities

Prove that  $2(sinx_1+sinx_2+....+sinx_n)(cosx_1+cosx_2+....+cosx_n)\leq n^2$

ain't this quite easy?

$2\sum_{i=1}^n \sin x_i \cdot \sum_{i=1}^n \cos x_i=\sum_{i=1}^n 2\cos x_i\sin x_i+\sum_{i<j}2\cos x_i\sin x_j+2\cos x_j \sin x_i \\\leq \sum_{i=1}^n \cos^2x+\sin^2x+\sum_{i<j}\cos^2x_i+\sin^2x_j+\cos^2x_j+\sin^2x_i \\=n+2 {n \choose 2}=n^2$


vector
analytic geometry
inequalities
number theory proposed
number theory

A very nice problem:
  Let $2^n-1$ vectors in $Z^n$. Prove the existence of a vector $x\in Z^n$ such that the open segment joining this vector to each of the $2^n-1$ vectors does not contain any laticial point.

[quote="harazi"]A very nice problem:
  Let $2^n-1$ vectors in $Z^n$. [/quote]

Let these vectors what?

I just wanted to underline they are arbitrary. There is nothing else to add to the statement of the problem.

At first, there exists a vector $(\epsilon_1,\,\epsilon_2,\,\dots,\epsilon_n)$ with each coordinate 0 or 1 such that no of our $2^n-1$ vectors coincides with it modulo 2.

Then we take large $N$ and consider $N^n$ vectors $(\epsilon_1+2x_1,\,\epsilon_2+2x_2,\dots,\,\epsilon_{n}+2x_n)$ for $x_i=1..N$. Assume that an open interval between an endpoint $(a_1,\,a_2,\,\dots,\,a_n)$ of one of our vectors and a point $(\epsilon_1+2x_1,\,\epsilon_2+2x_2,\dots,\,\epsilon_{n}+2x_n)$ contains a lattice point. It means that for some prime $p$ we have $a_i\equiv \epsilon_i+2x_i$ modulo $p$. Clearly, we may assume that such $p$ is odd and not greater that $2N+1+\max |a_i|<3N$ for large $N$. For any given $p$, numbers $x_i$ are uniquely determined modulo $p$, and so there are at most $(N/p+1)^n$ such vectors. Sum up by all $p$, we get at most $N^n(1/3^n+1/5^n+\dots)+o(N^n)$ "bad"
vectors. Multiplying it by $2^n-1$ (number of initial vectors), we get less then

$N^n((2/3)^n+(2/5)^n+(2/7)^n+\dots)+o(N^n)$.

It suffices to verify that $(2/3)^n+(2/5)^n+\dots\le (2/3)^2+(2/5)^2+(2/7)^2+\dots<1$

Last inequality is clear: replace primes to all odds and use $2/p^2<1/(p-1)-1/(p+1)$ for $p\ge 5$. Everything cancels, and we get $4/9+1/2<1$.


Pythagorean Theorem
geometry
angle bisector

If two circles with centers $U$ and $V$ and radii $u$ and $v$ touch each other externally, and one of the two external common tangents of these two circles touches them at the points $P$ and $Q$, respectively, then $PQ=2\sqrt{uv}$.

This is a rather known fact and far from being difficult; I hope this is the right section for it.

  Darij

[hide]WLOG, $u\gev$ With some manipulation, you get a triangle with a hypotenuse of length $u+v$, a leg of length $u-v$ and another leg of length $PQ$.

Applying the Pythagorean theorem,
$PQ^{2}+(u-v)^{2}=(u+v)^{2}$
$PQ^{2}=4uv$
$PQ=2\sqrt{uv}$[/hide]

Yeah, this solves it. The reason why I posted the problem here was, however, to strike the existence of a different, completely synthetic solution:

[hide="Synthetic solution"]Let the two circles touch each other at the point T, and let their common tangent at the point T meet the line PQ at a point G. Since the two tangent segments from a point to a circle are equal, GP = GT and GQ = GT, so that GP = GQ. Thus, the point G is the midpoint of the segment PQ, so that $PQ=2\cdot GP$.

Since the circle with center U and radius u touches the line PQ at the point P, we have $UP\perp PQ$ and UP = u, and similarly, $VQ\perp PQ$ and VQ = v.

Now, since the circle with center U and radius u touches the segments GP and GT, its center U must lie on the angle bisector of the angle PGT, so that $\measuredangle UGP=\frac{\measuredangle TGP}{2}$. Similarly, $\measuredangle VGQ=\frac{\measuredangle TGQ}{2}$. Thus, $\measuredangle UGP+\measuredangle VGQ=\frac{\measuredangle TGP}{2}+\frac{\measuredangle TGQ}{2}=\frac{\measuredangle TGP+\measuredangle TGQ}{2}=\frac{180^{\circ}}{2}=90^{\circ}$, so that $\measuredangle VGQ=90^{\circ}-\measuredangle UGP$. But $UP\perp PQ$ yields that the triangle UPG is right-angled at P, so that $\measuredangle GUP=90^{\circ}-\measuredangle UGP$. Thus, $\measuredangle VGQ=\measuredangle GUP$. Also, $\measuredangle VQG=\measuredangle GPU$ (because actually $\measuredangle VQG=\measuredangle GPU=90^{\circ}$, since $UP\perp PQ$ and $VQ\perp PQ$). Hence, the triangles GUP and VGQ are similar. Consequently, UP : GP = GQ : VQ, what rewrites as u : GP = GP : v. Thus, $GP^{2}=uv$, so that $GP=\sqrt{uv}$, and therefore $PQ=2\cdot GP=2\cdot\sqrt{uv}$, and the problem is solved.[/hide]

  Darij


hi, please may i know the formula of distance of 2 lines please ?

$y= a x+b$
$y= c x+d$

Supose a point on the first line and a point on the second line

so pitagoras

if you want the minimum distance

$a=c$  so  $|b-c|$

se a different of c  the minimum distance is zero

if you want in R^3   or someting else this can be done with vectors


geometry
trapezoid
geometric transformation
homothety
parallelogram
geometry proposed

Trapezium ABCD is given (AB||CD). On the segment AB is chosen random point M. On AD and BC are chosen points K and L so MK||BD and ML||AC. KL meets BD and AC at points N and P. Prove that KN=LP.

I think it is not difficult, but it is not as easy as it seems also. The problem was proposed at "Vasil Levski" High School competition /Roman - Bulgaria/ for 9-th grade. Maybe some Russian will write a solution :-).

$ E: AC\cap BD$, $ S: BD\cap ML$ and $ T: AC\cap KM$; for an homothety with center in A and one in B we have $ \frac{KT}{TM}\equal{}\frac{DE}{EB}\equal{}\frac{EC}{AE}\equal{}\frac{SL}{MS}$
so $ ST\parallel KL$ for taletes and the triangle $ \triangle KTP$ and $ \triangle MLS$ are congruent so $ KP \equal{} NL\ \Longleftrightarrow\ KN \equal{} PL$

[color=darkblue][b]A little error (in a hurry)[/b] in the [b]¬[ƒ(Gabriel)³²¹º]¼[/b]'s proof ! Is correctly such (see the box from quote) :[/color]

[color=darkred]Denote $ E\in AC\cap BD$, $ S\in BD\cap ML$ and $ T\in AC\cap KM$. We have $ \frac{KT}{TM}\equal{}\frac{DE}{EB}\equal{}\boxed{\frac{CE}{EA}\equal{}\frac{LS}{SM}}$ $ \implies$

$ \frac{KT}{TM}\equal{}\frac{LS}{SM}$ $ \implies$  $ ST\parallel LK$ $ \implies$ the quadrilaterals $ KTSN$, $ LSTP$ are parallelograms $ \implies$ $ KN \equal{} TS \equal{} PL$.[/color]

I'm happy you liked the problem. I said that maybe some russian will solve the problem because there was a problem in Prasolov's textbook saying that: If in triangle AM, BN are medians in triangle ABC, then MN is divided in three equal parts by AM and BN. I saw the idea for my solution from this book. Indeed I think proposed from me problem is very good for early rounds in math olympiads.

[quote="Virgil Nicula"][color=darkblue][b]A little error (in a hurry)[/b] in the [b]¬[ƒ(Gabriel)³²¹º]¼[/b]'s proof ! Is correctly such (see the box from quote) :[/color][/quote]

 :blush: thank you Virgil Nicula :blush:


Suppose ABC is a triangle with M the midpoint of BC.
Suppose that AN intersects the incircle at K,L.
We draw parallel line from K and L to BC and name their second intersection point with incircle X and Y. Suppose that AX and AY intersect BC at P and Q.
Prove that BP=CQ.   :oops:  :oops:

I think it is hard  :D  Let's take a look at http://www.mathlinks.ro/viewtopic.php?p=463068#463068


inequalities
algebra
polynomial
inequalities proposed

let $ a,b,c$  [b]real numbers[/b] such that : $ a\plus{}b\plus{}c\equal{}1$ , Prove that :
$ 2\plus{}32(ab\plus{}ac\plus{}bc)^2 \geq 14(ab\plus{}ac\plus{}bc)\plus{}24abc$
I hope it's not trivial  :D  :maybe:

[quote="anas"]let $ a,b,c$  [b]real numbers[/b] such that : $ a + b + c = 1$ , Prove that :
$ 2 + 32(ab + ac + bc)^2 \geq 14(ab + ac + bc) + 24abc$
I hope it's not trivial  :D  :maybe:[/quote]

[Edit: My solution is wrong ! :) ]

your problems look like a nice lema of "pqr technique".

However, it is followed directly Vasc's lema at http://www.mathlinks.ro/viewtopic.php?t=6026&start=60

Basically, homogenize the inequality yield:
$ \Sum a^4 + 8\sum a^2b^2 - 3 \sum_{cyc} (a^3b) - 3\sum_{cyc} (ab^3) - 3abc(a + b + c) \geq 0$

Apply Vasc's lema for $ p = q = 3$ then

$ \sum a^4 + 5\sum a^2b^2 - 3 \sum_{cyc} (a^3b) - 3 \sum_{cyc} ab^3 \geq 0$
and $ 3\sum a^2b^2 \geq 3 abc(a + b + c)$

Can I see your solution ? [ I guess it comes from the identity ?]

Inequality is equivalent to:

$ 1\plus{}16(ab\plus{}bc\plus{}ca)^2 \ge 7(ab\plus{}bc\plus{}ca)\plus{}12abc$

this is all I can do  :lol:

[quote="Gibbenergy"][quote="anas"]let $ a,b,c$  [b]real numbers[/b] such that : $ a + b + c = 1$ , Prove that :
$ 2 + 32(ab + ac + bc)^2 \geq 14(ab + ac + bc) + 24abc$
I hope it's not trivial  :D  :maybe:[/quote]

your problems look like a nice lema of "pqr technique".

However, it is followed directly Vasc's lema at http://www.mathlinks.ro/viewtopic.php?t=6026&start=60

Basically, homogenize the inequality yield:
$ \Sum a^4 + 8\sum a^2b^2 - 3 \sum_{cyc} (a^3b) - 3\sum_{cyc} (ab^3) - 3abc(a + b + c) \geq 0$

Apply Vasc's lema for $ p = q = 3$ then

$ \sum a^4 + 5\sum a^2b^2 - 3 \sum_{cyc} (a^3b) - 3 \sum_{cyc} ab^3 \geq 0$
and $ 3\sum a^2b^2 \geq 3 abc(a + b + c)$

Can I see your solution ? [ I guess it comes from the identity ?][/quote]

I have another aproach , but I can't post it here , I'm waiting for more ideas  :wink:

[quote="Gibbenergy"]
Apply Vasc's lema for $ p \equal{} q \equal{} 3$ then

$ \sum a^4 \plus{} 5\sum a^2b^2 \minus{} 3 \sum_{cyc} (a^3b) \minus{} 3 \sum_{cyc} ab^3 \geq 0$
[/quote]
Try $ a\equal{}2$ and $ b\equal{}c\equal{}1.$ :wink:

nice inequality Anas  :lol:

after normalisation we get :

$ \sum_{cyc}a^4\plus{}8\sum_{cyc}a^2b^2\minus{}3\sum_{cyc}a^3b\minus{}3\sum_{cyc}ab^3\minus{}3\sum_{cyc} a^2bc\geq 0$

wich is equivalent with :

$ \sum_{cyc}((a\minus{}b)(a\plus{}b\minus{}3c))^2\geq 0$

:)

[quote="mehdi cherif"]nice inequality Anas  :lol:

after normalisation we get :

$ \sum_{cyc}a^4 \plus{} 8\sum_{cyc}a^2b^2 \minus{} 3\sum_{cyc}a^3b \minus{} 3\sum_{cyc}ab^3 \minus{} 3\sum_{cyc} a^2bc\geq 0$

wich is equivalent with :

$ \sum_{cyc}((a \minus{} b)(a \plus{} b \minus{} 3c))^2\geq 0$

:)[/quote]

that's it Mehdi  :lol:  :)

[quote="arqady"][quote="Gibbenergy"]
Apply Vasc's lema for $ p \equal{} q \equal{} 3$ then

$ \sum a^4 \plus{} 5\sum a^2b^2 \minus{} 3 \sum_{cyc} (a^3b) \minus{} 3 \sum_{cyc} ab^3 \geq 0$
[/quote]
Try $ a \equal{} 2$ and $ b \equal{} c \equal{} 1.$ :wink:[/quote]

I see your point Arqady. The identity does not true.
For the original inequality:
\[ \sum_{cyc}a^4 \plus{} 8\sum_{cyc}a^2b^2 \minus{} 3\sum_{cyc}a^3b \minus{} 3\sum_{cyc}ab^3 \minus{} 3\sum_{cyc} a^2bc\geq 0
\]
is equivalent with
\[ \frac {1}{4} \left(2a^2 \minus{} b^2 \minus{} c^2 \minus{} 3ab \plus{} 6bc \minus{} 3ac \right)^2 \plus{} \frac {1}{12} \left(3b^2 \minus{} 3c^2 \minus{} 9ab \plus{} 9ac \right)^2 \geq 0
\]
[quote="mehdi cherif"]nice inequality Anas

after normalisation we get :
\[ \sum_{cyc}a^4 \plus{} 8\sum_{cyc}a^2b^2 \minus{} 3\sum_{cyc}a^3b \minus{} 3\sum_{cyc}ab^3 \minus{} 3\sum_{cyc} a^2bc\geq 0
\]
which is equivalent with :
\[ \sum_{cyc}(a \minus{} b)^2(a \plus{} b \minus{} 3c)^2\geq 0
\]
[/quote]
Very nice solution  :lol:

nice identity Gibbenergy  :lol:  , My solution is the same as Mehdi's one  :)

[b]Lemma.[/b]Let $ p,q,r$ and $ a,b,c$ be real numbers. If $ 3(1 \plus{} r)\ge p^2 \plus{} pq \plus{} q^2$, then

$ \sum a^4 \plus{} r\sum a^2b^2 \plus{} (p \plus{} q \minus{} r \minus{} 1)abc\sum a \ge p\sum a^3b \plus{} q\sum ab^3$.

For the main case, $ 3(1 \plus{} r) \equal{} p^2 \plus{} pq \plus{} q^2$, the inequality is equivalent to each of the following:

$ \sum (2a^2 \minus{} b^2 \minus{} c^2 \minus{} pab \plus{} (p \plus{} q)bc \minus{} qca)^2 \ge 0$

and

$ \sum (3b^2 \minus{} 3c^2 \minus{} (p \plus{} 2q)ab \minus{} (p \minus{} q)bc \plus{} (2p \plus{} q)ca)^2 \ge 0$

and

$ 3(2a^2 \minus{} b^2 \minus{} c^2 \minus{} pab \plus{} (p \plus{} q)bc \minus{} qca)^2 \plus{} (3b^2 \minus{} 3c^2 \minus{} (p \plus{} 2q)ab \minus{} (p \minus{} q)bc \plus{} (2p \plus{} q)ca)^2 \ge 0$.

[quote="Vasc"][b]Lemma.[/b]Let $ p,q,r$ and $ a,b,c$ be real numbers. If $ 3(1 \plus{} r)\ge p^2 \plus{} pq \plus{} q^2$, then

$ \sum a^4 \plus{} r\sum a^2b^2 \plus{} (p \plus{} q \minus{} r \minus{} 1)abc\sum a \ge p\sum a^3b \plus{} q\sum ab^3.$

For the main case, $ 3(1 \plus{} r) \equal{} p^2 \plus{} pq \plus{} q^2$, the inequality is equivalent to each of the following:

$ \sum (2a^2 \minus{} b^2 \minus{} c^2 \minus{} pab \plus{} (p \plus{} q)bc \minus{} qca)^2 \ge 0$

and

$ \sum (3b^2 \minus{} 3c^2 \minus{} (p \plus{} 2q)ab \minus{} (p \minus{} q)bc \plus{} (2p \plus{} q)ca)^2 \ge 0$

and

$ 3(2a^2 \minus{} b^2 \minus{} c^2 \minus{} pab \plus{} (p \plus{} q)bc \minus{} qca)^2 \plus{} (3b^2 \minus{} 3c^2 \minus{} (p \plus{} 2q)ab \minus{} (p \minus{} q)bc \plus{} (2p \plus{} q)ca)^2 \ge 0$.[/quote]

Thanks Vasc for a very nice lemma  :) . I try to prove your statement and I get:
Consider the 4-degree polynomial:
\[ \sum_{cyc} a^4 \plus{} \left ( \frac {\Delta_1 \plus{} \Delta_2}{2} \minus{} 1 \right) \sum_{cyc}a^2b^2 \plus{} m_1 \sum_{cyc}a^3b \plus{} m_2\sum_{cyc}ab^3 \plus{} m_3 \sum abc (a \plus{} b \plus{} c)
\]
where
\[ \Delta_1 \equal{} m_1^2 \plus{} m_1m_2 \plus{} m_2^2
\]

\[ \Delta_2 \equal{} m_1 \plus{} m_2 \plus{} m_3
\]
[b]If $ \Delta_1 \plus{} 3\Delta_2 \geq 0$ then $ P(a,b,c) \geq 0$  for all real numbers $ a,b,c,m_1,m_2,m_3$ [/b]

[u]Proof:[/u]  $ P(a,b,c) \equal{} \sum_{cyc}(a^2 \minus{} c^2 \plus{} pab \plus{} qbc \plus{} rca)^2$
where
\[ p \equal{} \frac23 m_1 \plus{} \frac {1}{3}m_2 \plus{} \frac {1}{3}\sqrt {\Delta_1 \plus{} 3\Delta_2}
\]

\[ q \equal{} \minus{} \frac {2}{3}m_2 \minus{} \frac {1}{3} m_1 \plus{} \frac {1}{3}\sqrt {\Delta_1 \plus{} 3\Delta_2}
\]

\[ r \equal{} \minus{} \frac {1}{3}m_1 \plus{} \frac {1}{3}m_2 \plus{} \frac {1}{3}\sqrt {\Delta_1 \plus{} 3\Delta_2}
\]


Has anyone read any good biographies, if so please post the title here.

The absolute "King Royale" of Mathematics Biographies is E.T. Bell's book [u]Men of Mathematics[/u]. Once you pick it up you won't want to put it down.

As for books devoted to individual mathematicians, [u]A Mathematician's Apology[/u] by Hardy is one of the most famous. I also found [u]Genius[/u] by James Glieck especially engaging. It's about Richard Feynmann, who was technically a physicist, but I still think would be compelling to those interested in math.

There are also a lot of books that aren't pure biographies, but that have a lot of interesting biographical information in them. For example, Peter Beckmann's, [u]The History of Pi[/u] and Moir's [u]e: The Story of a Number[/u]. and Marcus Du Sautoy's [u]The Music of the Primes[/u] - all have very interesting biographical tidbits about various mathematicians. I have a lot of books like these, but I think these three are good because they 1) span different eras of time, and 2) introduce you to some cool areas of mathematics that you can understand.

Have you ever seen the St. Andrew's Biographies of Mathematicians site? It has a lot of very good biographiesl. http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html

And here's a link to a collection of books of Mathematics Biogragraphies - http://www.ericweisstein.com/encyclopedias/books/MathematicsBiography.html

This may be more information that what you were looking for, but if I had to pick just [i]one[/i] I might start with Beckmann's book about Pi or Moir's book about e.

The two best math biographies I've read are The Man Who Knew Infinity by Robert Kanigel (about Ramanujan) and The Man who Loved only Numbers by Paul Hoffman (about Erdos -- the title makes him sound more 1-d than he was).  Also I read a good biography of Sonya Kovalevskaya (or however you want to spell her name) a few years back.

Thanks for those books gauss202, but i am actually doing this for a school project (should have said this ealier), so i actually need to read a biography about someone in particular (not neccesarly math realated, just interesting). Do you have any suggestions for that?

Both of the books that Alison suggested are good - as is the book A Beautiful Mind. And there's a good autobiography by Stanislaw Ulam called [u]Adventures of a Mathematician[/u]. He imigrated to America from Poland just before WWII, and ended up being a major contributor to the Manhattan Project.

The books about Ramanujan, Erdos, and Nash might be the most interesting because they were such unique personalities - and those books should all be easy to find.

I don't know if there are specific books on them, but a few other great people to do a biographies on would be Hypatia of Alexandria, Gregor Cantor, Evariste Galois, Andre Weil, and Alexander Grothendieck.

Particularly Grothendieck, who I consider to be the greatest mathematician of the 20th Century - as well as one of the most interesting. I bet that no one else would think to do a paper on him (as opposed to Ramanujan, Erdos, or Nash).

I know that Grothendieck wrote an autobiography that's about 1000 pages. I would like to read it at some time when I have less schoolwork and more shelfspace.

[quote="Alison"]The two best math biographies I've read are The Man Who Knew Infinity by Robert Kanigel (about Ramanujan) and The Man who Loved only Numbers by Paul Hoffman (about Erdos -- the title makes him sound more 1-d than he was).[/quote]

Hi, Alison, what does "1-d" mean? If it's not printable on a public forum, could you PM me about it?

I like the book by Kanigel too. I haven't tried out the Hoffman book yet, but I've seen another Erdos biography--isn't there one called My Brain Is Open, or something like that? When Erdos was still alive, there was an article about him in Atlantic magazine that made for very interesting reading.

My favorite biography is probably Mornings on Horseback, a biography of President Theodore Roosevelt, which won the Pulitzer prize in 1981 or so. My favorite biography of a mathematician is probably Paul Halmos's "automathography," I Want to Be a Mathematician.

The Atlantic Monthly article about Erdos was written by Paul Hoffman, who later wrote the book [i]The Man who Loved only Numbers[/i].

I think "1-d" just meant "one-dimensional".

[quote="Alison"]Yes.  *amused*  I've also read My Brain is Open, but I think the Hoffman book did a better job -- it also spends more time on his math community.  There's a lot of overlap between the two books though.

[quote="Ravi B"]
I think "1-d" just meant "one-dimensional".[/quote][/quote]

Thanks for the further information. Hmmm, I never would have guessed "one-dimensional" in description of as colorful a character as Erdos. I was thinking along the lines of "whole but for derangement," but I missed the obvious parallelism with 3-d, 4-d, etc.

I had met Erdos a few times while he was alive.  He was indeed a colorful character.  My former colleagues Joel Spencer and Noga Alon had written many papers with him.  Since I wrote a handful of papers with them, that makes my Erdos number 2. :)  Um, I guess that's the closest to fame I'll get.


algebra
polynomial
ceiling function
Support
search
probability
absolute value

all polynomial with degree greater than 1 with all the coefficients being positive integer $ a_{i}\in{N^{+}}$ from the set{$ 1,2,3,\cdots$}and all coefficients different to each other $ {a_{n}}\neq{a_{n-1}}\neq{a_{n-2}}\cdots\neq{a_{0}}$, provided that $ a_{0}\neq0$ has atleast one root that is $ \left\lceil{|x|}\right\rceil\geq degf(x)+1$
i hope i am not wrong again

$ P(x) = 1000x^{2k+1}+1$.

I don't understand why you would think a relation exists between the degree of a polynomial and the size of its roots at all.

Something tells me you didn't read that little blurb I left you last time  :o
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156242&start=21

Go ahead and read it if you haven't. In particular, look at 5. If you think you're on to something, you should consider exploring why that might be true.

I AM NOT MAKING THIS UP WITH NO REASON !!!!!! dont think i am foolish,stupid, i have a reason for writing it.
t0rajir0u you think there is no relation. if all the mathematicians who discovered something were like you, no theorem,formula,conjecture or hypothesis would exist now!.you are always dissin, you are telling me that there is no relation between the degree of a polynomial and the size of its roots at all. u need to put self in my shoes then say that. you dont know why i am saying that. The Zuton Force, i know this very clearly, that i have to consider these Is it true? Must it be true? Why is it true? What does it mean if it is true? What kinds of interpretations of it are there? What does it suggest/imply?. i have considered more than those questions,
I know this theorem exists because i have found  this much with logic and reasoning, but still something is missing.

[quote="binomial_4eva"]t0rajir0u you think there is no relation. if all the mathematicians who discovered something were like you, no theorem,formula,conjecture or hypothesis would exist now!.[/quote]

As someone who is currently working on a mathematical research project, this is unfair.  I've spent a lot of time this past week asking [i]meaningful[/i] mathematical questions based on some kind of [i]motivation.[/i] I have in fact made several conjectures which have all turned out to be false, but in each case I had a [i]reason[/i] to suspect that they could be true, and in each case I broadened my understanding of my subject through exploring that conjecture.

You haven't given any of your reasons for suspecting this conjecture.  Part of being a mathematician is being able to communicate and express your mathematical ideas.  If nobody understands what you're talking about, then what good is your mathematics?

[quote="binomial_4eva"]you are telling me that there is no relation between the degree of a polynomial and the size of its roots at all. u need to put self in my shoes then say that. you dont know why i am saying that. [/quote]

No, I don't know why you're saying that.  Why don't you tell me?  Here is a polynomial of degree $ 2k+1$ with an arbitrarily large (negative) root:

$ P(x) = x^{2k+1}+N, N >> 1$

And here is a polynomial of degree $ 2k+1$ with an arbitrarily small root:

$ P(x) = Nx^{2k+1}+1, N >> 1$

Because you've restricted the coefficients to being positive, no positive roots can exist, but within that bound for any fixed $ k$ the roots of $ P(x)$ span the range $ (-\infty, 0)$, so no bound of any kind can exist, much less one based on degree.

When you say you've thought about it a lot you seem to be implying that I haven't.

@binomial_4eva: All mathematical theorems are based upon rigorous proofs. You cannot go around claiming things if you do not have enough evidence to support your claim. Maybe you are claiming a new conjecture but still you have to search for counter-examples or evidences to support your conjecture. Maybe t0rajir0u is not claiming that there are no reltationships what so ever between degree of a polynomial or the size of their roots maybe we just haven't discovered it yet.

But still I find the way that you have treated t0rajir0u to be quite unfair because most (maybe all) of his posts are relevant and good information for all of us. I am sure t0rajir0u is a very experienced problem solver who has much to contribute.

So, I think that it is good that you like mathematics but just remember that one has to start from the basics and build your way upwards. :)

Totally agree with the other two previous posters. This isnt science. But making conjectures is important, but you cant just write them you need to prove them [url=http://en.wikipedia.org/wiki/Rigorous#Mathematical_rigour]rigorusly[/url]

By the way, how old are you?

I agree with both of you BanishedTraitor and sylow_theory,but i never said  t0rajir0u isnt a very experienced problem solver who has much to contribute.Of course he is, even like my teacher.but after reading his posts i feel like quiting,forgetting the thing i am doing, because i think it is either impossible or i am very stupid.
Sylow_theory: i love maths only because "All mathematical theorems are based upon rigorous proofs".and i didnt say its science.
how old am i ? I am under 20, how old ru sylow_theory ? how old is t0rajir0u?
i am gonna make a poll soon about ages in Aops
i dont know how to make one.

[quote="binomial_4eva"]but after reading his posts i feel like quiting,forgetting the thing i am doing, because i think it is either impossible or i am very stupid.[/quote]

It would be instructive to look at what can be learned from considering this conjecture (and this is what we are all here to do, is learn):

- The restriction to integers does nothing.  For arbitrary real integers, multiplying by a suitable real should get each coefficient arbitrarily close to an integer (this is obvious for a single coefficient, but can you prove or disprove it for multiple coefficients?), which means that in terms of absolute value, the roots of integer polynomials are pretty much the same as the roots of real polynomials.

- The restriction that the coefficients are different from each other does almost nothing.  Rather than saying $ P(x) = x+1$ we could simply let $ P(x) = 1000x+999$.  The coefficients can be arbitrarily close without being equal; this is related to the above point, and acts on the roots in a similar manner (which is to say, hardly at all).

- The restriction that the coefficients are positive does do something.  It's clear that all the roots are negative.  In fact, let me venture a guess as to the source of your conjecture:

If a polynomial $ P(x)$ can only have distinct positive integer [b]roots[/b] (which has little to do with having distinct positive integer [b]coefficients[/b]), then such a polynomial of degree $ n$ has to have at least a root of size $ n$.  That's because the minimal such polynomial is

$ P_{n}(x) = (x+1)(x+2)...(x+n)$

This is close to what you're talking about, but the relationship falls apart when we move from talking about the roots themselves to talking about the coefficients.

[quote="binomial_4eva"]
how old am i ? I am under 20, how old ru sylow_theory ?.[/quote]
Im 19. Im not an experienced problem because when I was younger I used to do advanced math. So for my age I know way more math than other people my age but when I comes to problem solving Im not a newbie but Im not that great, thats one of the things I started practicing during my summer. I guess Im getting better.

By the way why dont you like probability theory? I can tell you now, it is rigourous. Perfectly rigorous.


Find all the possible values of the sum of the digits of all the perfect squares.

[Commented by djimenez]
[b]Comment: [/b]I would rewrite it as follows:
Let $f: \mathbb{N}\rightarrow \mathbb{N}$ such that $f(n)$ is the sum of all the digits  of the number $n^2$. Find the image of $f$ (where, by image it is understood the set of all $x$ such that exists an $n$ with $f(n)=x$).

Yes, but how to prove that all these values can be attained?

[quote="Rust"]f(n)=0,1,4,7 (mod 9).[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=86823:
[quote="Rust"]It is equavalent $x^8-y^8=10(z^4-y^8)$. It had onlu solutution x=y=z=1.[/quote]

:what?:
[u]Very[/u] good and [u]very[/u] complete solutions indeed.

Just for your (and all others, there are many people doing that way) information: The next post (in Number Theory) only containing the trivial answers - or even more badly: only one special answer out of many - without any further statement or a proof (or a reasoning) or at least anything usefull will be deleted without any more comment. :bomb:

@all: at this case there is at least a small bit of information given, but answering to
"find all solutions of $x^{1337}+y^{1337}=z^{1337}$"
with
"one answer is $x=y=z=0$"
is pure spamming!
[sarcasm] Note: such a very ingenious remark is surely much too complicated to be found out by others themself... [/sarcasm]

It's understandable that there are users new to the forum not knowing about how,where and what to post. But from someone having more than 400 posts, I think one can expect to know that (and it was told to you before; I would not complain if this would be the first time). I think you can solve it, but either post a solution/a sketch of it or don't post.

A remark to the problem: showing that $451132$ and $451332$ are in the image was the hardest problem in the third round of the german mathematical olympiad, grades 11/12/13.

we can also prove it by using some well-known result .
we know that

$11\cdots 1122\cdots 225$ ($n$ times $1$ ,$n+1$ times $2$) is a perfect square . And the sum of digit is

$n+2(n+1)+5=3n+7$ .

Also from $(10^n-1)^2=100^n-2\cdot 10^n+1=99\cdots 9800\cdots 001$ ($n-1$ times $9$ ,$n-1$ times $0$) , we know that the sum of digit is $9(n-1)+8+1=9n$ .

Hence $f(n)=0,1,4,7\mod 9$


I don't know what the heck you guys are thinking, but there will be NO tolerance of AMC discussions until after it is authorized, which is Thursday or so. Any subjects containing any information about either AMC test will immediately be deleted.
[color=darkred]
The time has passed. Feel free to talk about it.[/color]

[quote="words from adminster"]This means no talking about specific problems, about how hard you think it was, about whether or not you passed, etc.[/quote]

DON'T EVEN THINK ABOUT CHEATING!!!!!!!!!!

THIS IS NOT A CHEATING WEBSITE..

THIS SITE IS WHERE WE ARE SUPPOSED TO LEARN, NOT CHEAT!!!

AGAIN,,

NO TELLING ANSWERS OR DISCUSSIONS ABOUT IT UNTIL THE DATE!!
 :mad:

good thing that lasts only until thursday...

Thursday at noon, specifically.

If there was a warn system/suspension system, I'd put a three day suspension on anybody who discusses the AMCs before then...


Well, now that most scores are available by phone (but you have to pay :( ) I thought I would start up a thread for people to post their scores.  I'll probably wait until I get my scores in the mail as opposed to paying the $\$8.

Already? I thought it was July 1. Oh well, though. My parents might call... I'd rather wait.l

Don't know if I'll wait or not.  Usually my parents have a harder time waiting than I do.  :roll:

come on guys, ya'll all know that service is just a scam for collegeboard to get more money. eight dollars for one itty bitty phone call? i think i can wait two weeks.

[quote="furious"]i'll wait as well. i took 3 exams so i don't feel like dishing out $24 to find out what i will know in like 3 weeks anyway.[/quote]
I believe it's $8 for one call, not for each test.

[quote="joml88"]I believe it's $8 for one call, not for each test.[/quote]
o.  :blush:  well, either way, i'm still not doing it.

I wanted to find out the AP English Score (just in case i get a 5, and not have to take the MIT essay evaluation), so i called and dutifully paid the $8.
i got 5 on BC Calc, 5 on AP Chem, and a 3 on English :(

[quote="Mariag"]i got 5 on BC Calc, 5 on AP Chem, and a 3 on English :([/quote]
Good job! English AP Exam is very very hard I heard. so no need for :(
I'm in Korea so I guess there is no way for me to find it out unless I call them from here, which is clearly more than $8..
I took 5 exams, but I only care about bio score.

I got 5s in English, US & European History, Chem, and Physics B and C.

Let the buyer's remorse begin.

No need to see the actual score, I already knew I got and 1 in AP CS AB and another 1 in AP WH.

Not that my AP scores mean anything, but I hope I did good in Calculus for personal satisfaction

Physics I bombed....

ok, im going to go ahead and make a prediction on mine. lets see how i measure up come... probably (hopefully!) tuesday.
bio- 5. i felt REALLY confident about that one.
cal bc- 4. im not going to hold out for the coveted 5 on that one, just on sheer confidence level.
us history- PLEASE PLEASE GOD BE A THREE. at least. please please PLEASE!!!! i worked my buns off in that class all year, i really want a good AP score. a four would be phenomenal.

I'll wait to find out what I got, but as a prediction, I'm going to award myself a 5 on Calc AB  (the only AP I took).  I'd be very surprised if I got anything else.

Chemistry: 4
Calculus BC: 5
European History: 5
World history: 5

I'm a sophomore who never took any AP courses and who self-studied them a little, although i was already very fluent in history. I really could have studied more (but played pokemon instead) and got a 5 on Chemistry....

I only took 1 AP this year... and i got a 4 on it... it was European History  :( ... I'm so poor at history it's crazy  :(

U.S. History, English, Physics C Mechanics, Stats: 5
Chemistry: 4

I live in New England, the scores just came today...

not if you are in an international school.
in china there're international schools for American kids and they're just like(well actually almost identical) the ones here.

any jabronis still not get their AP scores?

[quote="1+1"]I thought you had to live in USA to take the tests.(?)

How do you take the tests if you live in another country?
What's the max and min ages?

How much does it cost?[/quote]
http://www.collegeboard.com/student/testing/ap/about.html

Calc AB/BC=5/5
Human Geography=4...I completely made up one of my free response  :lol:

US History-3   :(  :stink:
Statistics-5
Calc BC-5/5
Physics B-5

Needless to say, I am very disappointed with History. I'm so, SO glad that I managed to pull out a 5 on stats by self studying it.  Physics was also very much a relief given our school's poor history with it.

[quote="joml88"]I'm so, SO glad that I managed to pull out a 5 on stats by self studying it.  [/quote]

Same. I was more worried about that than World History, which is saying something :D

Euro Hist: 5
Stats: 5
Physics C: 5/5

I was worried that scores would take forever to get to Hawaii...but they got here...

i took the class and got a 4.  i thought for sure i got a 5.  my actual scores were pretty much all off of what i predicted...

Calc BC - 5 (5 on AB composite)
English Lang and Comp - 5

got mine a while ago...

sophomore year

Statistics - 5
Computer Science A - 4

this year...

Chemistry - 5
Calculus BC - 5
Calculus ABSub - 5
Computer Science AB - 4
Macroeconomics - 4
Government - 4
Physics B - 5


A bottling plant fills 0.5-liter bottles from a 56-liter tank. A flat consists of 24 bottles. How many complete flats can be filled from a single tank?

the info says that in a 56 liter tank, 0.5 liter bottles are filled and a flat has 24 bottles.

Therefore,

0.5*24=12, so you use 12 liters per flat. then you calculate how many FULL flats can be filled.

56 divided by 12 is bigger than 4, so you can fill

4 flats

1 flat has 24 bottles which can contain 12 litres.And as you know 56/12 gives the answer which is more than 4 but lesss than 5.So a maximum of 4 flats can be filled from one single tank.


algebra
polynomial
number theory
relatively prime
superior algebra
superior algebra unsolved

Does all polynomial map from $ \mathbb{Q}$ to  $ \mathbb{Q}$ which is surjective must be degree 1??(Linear)

Polynomial map means it can be expressed by a polynomial.

Yes.

First, you need to argue that the coefficients must be rational numbers.  One way to do this is write down $ n \plus{} 1$ linear equations $ \sum_i c_i k^i \equal{} f(k)$ and solve for the $ c_i$ using Gaussian elimination.

Next, write $ f(x) \equal{} 1/n \cdot \sum_i c_i x^i$, where the $ b_i$ are integers.  We can assume $ n \equal{} 1$, because $ f$ is surjective if and only if $ nf$ is surjective.  Choose some prime $ p$ that is relatively prime to $ c_n$, and suppose that $ f(a/b) \equal{} 1/p$, with $ a/b$ in lowest terms.  This gives $ p \sum_i c_i a^i b^{n \minus{} i} \equal{} b^n$, implying that $ p | b$.  But $ p$ divides every term in the sum except $ c_n a^n$, so the let side is divisible by $ p$ exactly once.  Hence the right side is as well, implying that $ n \equal{} 1$.

You can strengthen this a little bit: instead of assuming $ f$ is surjective onto $ \mathbb{Q}$ we can assume that every rational in the image of $ f$ is mapped to by some rational.  Then we can take $ N$ relatively prime to $ p$ and large enough so that $ \pm N/p$ is in the image and apply the same argument.

Thank you very much.

But I don't know what you mean by " every rational in the image of $ f$ is mapped to by some rational."?
I don't know the means of words.

I mean: $ (\forall x \bigm| f(x)\in\mathbb{Q})\ (\exists y\in\mathbb{Q} \bigm| f(x)\equal{}f(y))$.

I originally saw this problem in the following form: Prove that if $ f$ is a polynomial with real coefficients such that $ x\in\mathbb{Q}\iff f(x)\in\mathbb{Q}$, then $ {\rm{deg}}\ {f}\leq 1$.  This is the same idea but requires the little trick at the end, because $ 1/p$ may not be in the image of $ f$.


You are on death row, you are set to die today, but the warden decides to play a game on you -As so many wardens seem to do in Mensa land =P

The Warden tells you, I will bring you into a room with 2 electric chairs.

One Chair is broke, One works 100% of the time.

Each chair is guarded by 1 guard.

1 Guard always tells the truth, 1 guard always lies.

You have no clue which chair is broke, which works, you have no clue which gaurd tells the truth, which lies.

You can ask 1 guard, 1 question.

If you choose the chair that is broken, you will be set free so you can murder more people.

What is the question you ask?

yay. the other one is nice too but i already know the solution. but this is nice. just because of the...er...closing remark.

[hide]ask one of the guards, "Which chair would the other guy point to if I asked him, 'Which chair is working?'?". Go sit in the chair that he points to. Murder some more peeps.[/hide]

I think this also works:

You ask a guard: "Which chair would YOU show me if I asked which was the broken chair?" Then you go seat in the chair he points to.
There are more questions of this type. I can only remember one right now:

You come to a crossroad and one road takes you to certain death while the other one takes you to eternal bliss ( :) ). Guarding the roads are 3 guards: guard A always tells the truth, guard B always lies, guard C lies or tells the truth randomly. Unfortunately, I can't remember how many questions you were allowed to ask (it's 1 or 2)
to find out which road is the correct one. Try to solve it with 1 question (I'll try to solve it too, as I don't have a soln). I rember solving it once, but I completely forgot how (you can see I can't even rember the right number of questions).

[quote="grobber"]
You come to a crossroad and one road takes you to certain death while the other one takes you to eternal bliss [/quote]
doesn't that sound familiar...

probably 2 questions; i can't think of how you would do it w/ 1 since you have that guy that tells the truth randomly...

I also recall a really easy one: You are in a city, but you can't remember if you're in Truthville or Liartown. There are people who says the truth in Liartown and vice versa since there are tourists. If you can ask one question to anyone in the town that you're currently in, what would you ask?

This is easier compared to the first two.

I don't think it's obvious that you can't do it in 1 q. What if there is a strange question involving logical relations, like iff and stuff like that. Here's another one which I kind of remember: you have the same prb, the only difference being that C answers "yes" or "no" randomly (I forgot to say that you're only supposed to ask yes-no questions, but that's not a big problem, as you can easily do for the problem posted here "another prison game", you can adapt the questions to make them yes-no type). For this new problem you definitely need 2 questions.

I think they can be found on a site having puzzle-type prbs (rec puzzles or sth like that, try writing "rec puzzle archive" in Google).

I checked it on the Internet: you only need 1 question for the first problem I posted here. I must add that the 3 guards all know each-other, meaning that A knows whichof the other ones tells the truth or lies and so on.

For the second one you need 2 questions.

I still don't know the answers, but at least I know the questions now! :D

[color=cyan]I think you can do it in one question:  for example, if you are trying to get to eternal bliss, you might ask, "Hey, did you guys hear that one of these roads leads to eternal bliss?  Let's go!"  Or, for the Liartown and Truthville, "Hey, I hear they're giving away free beer in Truthville!"

More seriously:
Given three guys and the knowledge that one of them tells the truth always, one lies always, and one of them tells the truth or lies at random.  In three questions, determine which one is which.  (Actually give a list of 3 questions as your answer -- who you ask the questions to will depend on the answers, but the questions tehmselves should not.)  It might be useful, in solving questions like this, to refer to the characters as the truthteller, liar, and fool, and to the individuals as 1, 2, and 3.
Oh, and questions must be yes-no only.
Grobber's question definitely requires 2 questions, and if you solve it you will be well on your way to solving mine.
[/color]

Hey, did you solve that problem or were you just joking? :D

I just wanted to make it clear: when I say one question I mean you ask ONLY ONE OF THEM that question!

I don't know if I've mentioned this already, but I have a weird kind of soln, but I didn't know if it was correct because I was assuming that the one which both lies and tells the truth knows whether he's gonna lie or not on his next answer (I mean the one he's going to give me for the 1 question I'm asking him).

I couldn't help it and I looked at the soln on the internet (rec puzzle archive) and I saw that they had a soln using the same thing, so I guess I have a soln for the first question.

Really sorrry for boring you with this... :lol:

It can be done without future or past knowledge of the fools answers:

[hide]

Ask person #1, "What would you say if I asked you, 'Is person number 2 the fool?'"

If he says "yes:"

He is the liar or the truthteller and person number 2 is the fool.  OR he is the fool.  Either way, we know person number 3 is NOT the fool.

If he says "no:"

He is the liar or the truthteller and person number 2 is NOT the fool.  OR he is the fool.  Either way, we know person number 2 is NOT the fool.

Now, ask the person we determined is not the fool, "What would you say if I asked you, 'Which is the road that leads to eternal bliss?'"[/hide]

So knowledge of what the fool will actually answer to such a "double question" is irrelevant.

This can be modified slightly to deternmine what type all three individuals are.

But all those questions don't exactly help you find the right way, do they? :D

[color=cyan]I do not know to what you are referring.[/color]

I meant that you can find out who is who, but then you need a second question to find out which road is the correct one. Remember, that is what the prb asks.

[color=cyan]Ah, yes, I know.  But I asked a seperate problem, after you did, which was "how can you find out in three questions who all of the people are given that one is a truthteller, one is a liar, and one is a fool?"  And you need the same method to approach that problem as to approach yours.

Here is a question:  Given two guys standing around, and given that each is either a truthteller or a liar (making 4 possible combinations), how many questions are necessary to figure out what each one is?

Here is a harder question, whose answer I don't know:  Given three guys standing around, each of whom is either a liar, truthteller, or fool, how many questions will you need to be perfectly sure what each one is?  (There are 26 valid combinations for this one -- we discard FFF, since you cannot gain any information from fools alone.)

In fact, before tackling that one, you might want to try it where you are limitted to 0 fools, exactly 1 fool, at most 1 fool, exactly 2 fools, and then finally at most 2 fools.[/color]

Oh.. I get it now. Sorry abt not paying attention.


hi friends..
after helping me so much with my poker questions here is another one,hope you can help me again..
i play a lot of [url=http://www.888.com]online casino[/url] games..i read in other forums about softwares that can caunt the card when you play blackjack
can you explain me how its working?

This is your fourth post, and they have all been advertisements for online casino sites. Administrators / moderators, please ban him!

i don't think that he needs to be banned. He just needs to be told that AoPS is not for advertising. :)

what is the point of banning someone with 4 posts they could just make another account


Should move to FF, and I voted mechanical pencil.

Mechanical Pencil:

Reasons:

* smooth, sharp writing
* pencil gives some fat lines
* i dont think you can use pen in mathcounts

I've always preferred a regular pencil, since it's heavier than a mechanical pencil so it makes me less delusional and slightly more conscious, and also because it feels good to be a nerd and have $ 30$ sharpened pencils in front of me. I like spinning regular pencils better as well. :P

Yeah math...

Whenever we have a math test at school, there is a pile of regular pencil in front of me.

[quote="math154"]I like spinning regular pencils better as well. :P[/quote]

Lol, I can't agree more  :rotfl:

Dang, nothing like 50 or so Dixon Oriole Number 2 pencils at a MathCounts comp. I find they write well and are really really really good for phailage-spinning. I often spin them during school and then suddenly find I'm not holding the pencil anymore...and see it's pinning some poor kid to the wall.

If you know me in real life, I have this pocket and there's n pencils in that pocket... where n is an integer between 1 and 3, inclusive...

Yes. 24/7.

[quote="BOGTRO"]At mathcounts, the person says "Stop, pencils down", so if you use a pen you can keep working  :rotfl:  :rotfl:  :rotfl:[/quote]

rofl i love your theory bogtro... :lol:

[quote="nikeballa96"]keep it simple: BIC mechanical pencils, 10 or so...too bad i cant spin.[/quote]

Same, for both the BIC and the can't spin part...except I only have 2.  Plus a bottle of Gatorade or Propel. :)

I'm kinda split between mechanical and wood. I just voted for wood because it writes nicely with my hand for some reason, but i use mechanical all the time at school.

Who doesn't want to walk into MC with 50 something "Mirado Black Warriors"?

[quote="gauss1181"]I always use a mechanical pencil! Never breaks, good eraser, no need to sharpen![/quote]
"never breaks"?!

I mean, all u need to do is release new lead, and no sharpening is needed.

mechanical pencils can be pretty annoying. like when theres still lead in it but it won't come out. Or when you put new lead in and when you try to put the eraser back in the lead starts coming out and becomes really long.

[quote="3.141592653589"]Pen is the BEST. Looks "professional." :starwars:[/quote]

yeah, and pens are easier to write with

What I do is take the first two pencils I see, and take them. (No Pens) [color=#F8F8F8]One of the things I hate about USAMO...[/color]

I don't care what type of pencil I use. As long as it's not a pen (xD pencil =/= pen)

I find mechanical pencils easier for writing for a long time without sharpening but I need a lotta lead. OR i can just bring 487239 pencils and lose all of them :)

[quote="mathemagician1729"]If I use mechanical, then my favorite is Clearpoint the only thing is some people *Cough Cough * like to break them.  :wink:[/quote]

Hey, it was an accident.  And i gave you  new pencil (with 3+ pieces of lead) AND an pencil top eraser.

[i]Tip for everyone else:  Do not mess with mathemagician1729 and his Papermate Clear Point pencils.[/i]

Ballpoint pens have less friction so it is much easier to write with
Try to get an erasable one
$ BIC$ $ FTW$

Very cool, interesting, different type of poll. I voted pencil. In a competition I always like to have multiple pencils (4-5), and a separate bar of eraser.

mechanical pencils, because I'm left handed, so I don't like using a pen where the whole paper has ink smudged all over it, and I just don't like having to sharpen my pencil all the time.


algebra
polynomial
function
calculus
integration
domain
Ring Theory

If $ R$ is a finite ring, can $ R[x]\cong P(R,R)$, where $ P(R,R)$ is the set of all polynomial functions $ R\to R$?

Also, let $ r\mapsto\bar{r}$ denote a ring homomorphism $ \theta: R\to S$.  Let $ \overline{f(x)}\equal{}\bar{a}_{0}\plus{}\cdots\plus{}\bar{a_{n}}x^{n}$.  If $ R\cong S$ (as rings), then there is some isomorphism $ \alpha: R\to S$.  Does it necessarily follow that $ \bar{\alpha}: R[x]\to S[x]$ defined by $ \bar{\alpha}(f(x))\equal{}\overline{f(x)}$ is a ring isomorphism between $ R[x]$ and $ S[x]$?  If not, how can I construct one?

HINT (a) Compare cardinatlities. (b) Use the universal mapping property (or direct verification).

I assume that for (a), the cardinality of $ P(R,R)$ exceeds that of $ R[x]$?  But I know for a fact that the result holds if $ R$ is an infinite integral domain-where is the difference?

(b) Direct verification of...?

No, $ P(R,R)$ is necessarily [i]finite[/i], where $ |P(R,R)| \le |R|^{|R|}$; however $ x^n \in R[x]$ for every $ n$ and so $ R[x]$ is infinite.

For (b):

If $ \alpha : R \to S$ is a homomorphism then there is a homomorphism $ \alpha_* : R[x] \to S[x]$. This is functorial in the sense that if $ \beta : S \to T$ then $ (\alpha\beta)_* \equal{} \alpha_*\beta_*$. Furthermore $ (id)_* \equal{} id$. Then if $ \alpha$ is an isomorphism it [i]necessarily[/i] follows that $ \alpha_*$ is an isomorphism, simply by the functoriality.

Thanks a lot.  What does "functorial" mean?

Above, simply that $ (\alpha\beta)_* \equal{} \alpha_*\beta_*$. In general, something similar about "induced" morphisms (category theory).


Bumping this for the person [url = https://artofproblemsolving.com/community/c4h2457550]here[/url]

[quote=mathpk]Give $a,b,c \geq 0$ prove that,
$a^{12}(a^{18}+26b^9c^9)^{2/3}+b^{12}(b^{18}+26a^9c^9)^{2/3}+c^{12}(c^{18}+26a^9b^9)^{2/3}\leq(a^8+b^8+c^8)^3$[/quote]
By Holder $$\left(\sum_{cyc}a^{12}\left(a^{18}+26b^9c^9\right)^{\frac{2}{3}}\right)^3\leq\left(\sum_{cyc}a^9\left(a^{18}+26b^9c^9\right)\right)^2\sum_{cyc}a^{18}.$$
Thus, it's enough to prove that:
$$\left(\sum_{cyc}a^9\left(a^{18}+26b^9c^9\right)\right)^2\sum_{cyc}a^{18}\leq\left(\sum_{cyc}a^8\right)^9.$$
But, by Holder again we obtain:
$$\left(\sum_{cyc}a^8\right)^9\sum_{cyc}a^{18}\geq\left(\sum_{cyc}a^9\right)^{10}.$$
Now, let $a^9=x$, $b^9=y$ and $c^9=z$.
Thus, we need to prove that:
$$(x+y+z)^5\geq(x^3+y^3+z^3+78xyz)(x^2+y^2+z^2),$$ which is $f(w^3)\leq0,$ where $f$ increases.
Id est, it's enough to prove the last inequality for an equality case of two variables and $y=z=1$ gives:
$$(x-1)^2(5x^2-10x+14)\geq0.$$
Done!


I saw somewere on forum:

Set of tangents on fixed circle is projective line.

How do we define cross-ratio of 4 tangents on circle?

projective line is the polar of a fixed point to the circle...


induction
number theory unsolved
number theory

Find all natural numbers $n$ and $s$ such that \[n^{\varphi(s)}=s\]

Suppose $\phi(s) \ge 4$.  If $p^k \mid n$ then $s$ is divisible by $p^{k\phi(s)}$ for some prime $p$.  So $\phi(s)$ is divisible by $p^{k\phi(s)-1} - 1$ and so $\phi(s) \ge 2^{\phi(s)-1} - 1$.  But $2^n > 2n+1$ for $n \ge 3$.  So $\phi(s) \le 2$.

Yes those are the solutions Ghang Hwan, and rgep nice sol. mine is similar I just prove that $3^{\varphi (s)}>s$ induction, induction, induction....


Find all functions $ f: \mathbb{R}-\{1\}\to\mathbb{R}$ such that $ f\left(\frac{x+1}{x-1}\right)=2f(x)+\frac{3}{x-1}\forall x\in\mathbb{R}-\{1\}.$

Replacing $ x$ with $ \frac{x+1}{x-1}$ we obtain
$  f(x)=2f\left(\frac{x+1}{x-1}\right)+\frac{3(x-1)}{2}$
Now, from the two equalities we get $  f(x)=-\frac{x^{2}-2x+5}{2(x-1)}$


geometry
Support
rotation
geometric transformation

A uniform round tabletop of diameter 4.0 m and mass 50.0 kg rests on massless, evenly spaced legs of length 1.0 m and spacing 3.0 m. A carpenter sits on the edge of the table. What is the maximum mass of the carpenter such that the table remains upright? Assume that the force exerted by the carpenter on the table is vertical and at the edge of the table.

The center of mass has to stay within the area of support; letting the leg be at position $ 0$, we have the center of the table is $ 1.5$ away, edge $ 0.5$ away, and $ 50\,\text{kg} \cdot 1.5\,\text{ft} \ge m \cdot 0.5\,\text{ft} \Longrightarrow m \le 150\,\text{kg}$.

Can you ignore the normal force exerted on the legs when you take the net torque of the system to be equal to 0? If so, why?
thanks.

No - but in this case, there is no normal force exerted on the other leg, as a little thinking will show.  When enough force is applied to the other side, the other leg is barely in contact with the floor.

I don't understand how we can just assume all of 50 kg of the table to be at the center of the table, since part of the table is on the other side of the leg and must be added to the weight of the person. the center of mass of the side with the majority of the table should have center of mass slightly more than 1.5m away.

There's a pretty good discussion of this problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321591
Maybe this can clear things up.

The torque about the bottom of the table leg on the carpenter's side is zero. Because if the table were to rotate, it would rotate about that point.

Yes that is what I thought for about half an hour after reading the discussions about this problem. However, I think what the others are saying is that since it does not rotate yet in the problem along with the fact that the mass of the small portion of the table to the right of the right leg is almost negligible, we can make any point we want to be the center of rotation. However, in fact that small part is not negligible, so the answer of 150 is off by a little.

If you read the other discussion in the link, then you can set the the center of rotation right under where the person is sitting, so you do not need to make any assumptions about negligible masses, and therefore perfectly comes out to 150kg.