vector
combinatorics unsolved
combinatorics

Let $ A\equal{}\left\{1,2,...,n\right\}$, and let $ a\left(1\right)$, $ a\left(2\right)$, ..., $ a\left(k\right)$ be pairwisely distinct subsets of $ A$ such that for every $ i$ and $ j$, the number $ \left|a\left(i\right)\cap a\left(j\right)\right|$ is odd. Find the maximum possible value of $ k$.

It is on the forum. Pierre (grobber, someone else?) wrote very nice solution for it.

Are you sure, Myth? I don't remember this one, although there are lots of problems similar to this one on the forum.

I think I have an almost complete solution. In fact, I believe it's complete if in the we allow for $i$ and $j$ to be equal, or, in other words, if all the sets have an odd number of elements. I'll assume this is the case.

Identify each set with its characteristic vector with elements $0$ or $1$, and we assume we work in the vector space $V=\mathbb Z_2^n$ over the field $\mathbb Z_2$. For $v,w\in V$ we define $v^t\cdot w$ in the same way as the usual scalar product.

If $v_i$ are our vectors, we know that for all $i,j$ we have $v_i^t\cdot v_j=1$. Let $U$ be the dimension of the vector space generated by $v_i$. For all $j\ne 1$ and all $i$, we have $(v_1+v_j)^t\cdot v_i=0$, so the vector space spanned by $v_1+v_j,j\ne 1$ is included in $U^{\perp}$. Since $\dim(U)+\dim(U^{\perp})=n$, we get $\dim(<v_1+v_j|j\ne 1>)\le n-\dim(U)$. On the other hand, it's not hard to see that $\dim(<v_1+v_j|j\ne 1>)=\dim(U)-1$, so we have $\dim(U)\le\frac{n+1}2\ (*)$.

Choose a basis for $U$ with vectors among the $v_i$. It's clear that all the other vectors we have considered are sums with an odd number of terms among the elements of the basis. This means that the maximal number of vectors we can get for a fixed dimension of $U$ is $\binom{\dim(U)}1+\binom{\dim(U)}3+\ldots=2^{\dim(U)-1}$. Together with $(*)$, this gives a maximum number of sets (vectors) equal to $2^{\frac{n-1}2}$.

It's easy to actually reach the bound if $n$ is odd.

I hope it's correct.

I think Myth reffers to the attached file in it :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=caruso&t=13941

Pierre.


inequalities
function
inequalities proposed

[quote="Peter VDD"]Let $a,b,c$ be positive reals with $a+b+c=1$. Show that \[ \prod_{cyc}\left(\frac{(1-a)^2}{1+a}\right) \ge abc.  \][/quote]

$\left(\frac{(1-a)^2}{1+a}\right) \ge -\frac{5}{4}a+\frac{3}{4}\Leftrightarrow (3a-1)^2\ge 0$
Summing over $a,b,c$ we get  \[ \prod_{cyc}\left(\frac{(1-a)^2}{1+a}\right) \ge 1\ge abc  \]

[quote="hardsoul"][quote="Peter VDD"][quote="hardsoul"]I think...[/quote] Can you prove that? Your solution looks nice. (and completely different from mine)[/quote]

I`ve edited.[/quote]

I think you have confused $\sum$ with $\prod$.

Doing omogenization:
$((a+b)(b+c)(a+c))^2\geq abc(a+(a+b+c)))(b+(a+b+c))(c+(a+b+c))$

$((\sum a)(\sum ab)-abc)^2\geq abc(abc+(\sum ab)(\sum a)+2(\sum a)^3)$

$(\sum a)^2(\sum ab)^2\geq 2abc(\sum a)(\sum ab)+abc(\sum a)(\sum ab)+2abc(\sum a)^3)$

$(\sum a)^2(\sum a^2b^2)\geq 3abc(\sum a)(\sum ab)$

But $\displaystyle (\sum a^2b^2)\geq \frac{(\sum ab)^2}{3}$ and

it  is remained to prove  $(\sum a)^2(\sum ab)^2\geq 9abc(\sum a)(\sum ab)$

or $(\sum a)(\sum ab)\geq 9abc$ - easy by AM-GM inequality.

[quote="son of Skanderbeg"][/quote][quote="hardsoul"][quote="Peter VDD"][quote="hardsoul"]I think...[/quote] Can you prove that? Your solution looks nice. (and completely different from mine)[/quote]

I`ve edited.[/quote][quote="son of Skanderbeg"]

I think you have confused $\sum$ with $\prod$.[/quote]

Oops  :roll:

[quote="Peter VDD"]Let $a,b,c$ be positive reals with $a+b+c=1$. Show that \[ \prod_{cyc}\left(\frac{(1-a)^2}{1+a}\right) \ge abc.  \][/quote]

Anyways another solution:
By Popoviciu`s inequality to the concave function $lnx$ we get
$\prod_{cyc}\left(\frac{a+b}{2}\right)^2\ge \left(\frac{a+b+c}{3}\right)^3abc$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19097]www.mathlinks.ro/Forum/viewtopic.php?t=19097[/url]
After slight simplifications it reduces to $\\(a+b+c)^3\ge 27/64(2a+b+c)(2b+c+a)(2c+a+b)\Leftrightarrow 64/27\ge (1+a)(1+b)(1+c)$ which is now trival. ;)

I think there is no need for popoviciu here.
Here's my solution:
$\frac{abc(2a+b+c)(a+2b+c)(a+b+2c)}{[(a+b)(b+c)(c+a)]^2} = \frac{a(2a+b+c)}{(a+b)(a+c)} \frac{b(2b+c+a)}{(b+c)(b+a)} \frac{c(2c+a+b)}{(c+a)(c+b)}$
$= (\frac{a}{a+b}+\frac{a}{a+c})(\frac{b}{b+a}+\frac{b}{b+c})(\frac{c}{c+b}+\frac{c}{c+a}) \le (\frac{\sum_{cyc}\frac{a}{a+b}}{3})^3 = 1$
by AM-GM obviously
now we substitute $a+b+c=1$
 ;)

Indeed.

My own solution is based on a pretty simple trick: rewrite the LHS as \[ \prod_{cyc}\left(\frac{(1-b)(1-c)}{1+a}\right) = \prod_{cyc}\left(\frac{a+bc}{1+a}\right).  \]

Assuming $a\ge b\ge c$ and therfor $ab\ge ac\ge bc$, we can now rewrite the ineq as $\prod_{cyc} (a+bc) \ge \prod_{cyc}(a+ab)$, which is a trivial product rearrangement.


$ m_b\cos C \equal{} m_c\cos B \Longleftrightarrow \sqrt{2(a^2 \plus{} c^2) \minus{} b^2} \cos C \equal{} \sqrt{2(a^2 \plus{} b^2) \minus{} c^2} \cos B \Longrightarrow$

$ \Longrightarrow[2(b^2 \plus{} 2ac \cos B ) \minus{} b^2]\cos ^2 C \equal{} [2(c^2 \plus{} 2ab \cos C ) \minus{} c^2]\cos ^2 B \Longleftrightarrow$

$ \Longleftrightarrow b^2 \cos ^2 C \plus{} 4ac \cos ^2 C \cos B \equal{}  c^2 \cos ^2 B \plus{} 4ab \cos ^2 B \cos C \Longleftrightarrow$

$ \Longleftrightarrow \frac{b}{c} \equal{} \frac{\cos C}{\cos B} \Longleftrightarrow \frac{\sin B}{\sin C} \equal{} \frac{\cos C}{\cos B} \Longleftrightarrow \sin {2B} \equal{} \sin {2C}$

then or $ B \equal{} C$ or $ 2B \plus{} 2C \equal{} 180 \Longleftrightarrow A \equal{} 90$

But it's olso easy to prove that if $ A\equal{}90$, $ b \equal{} c$:

take $ a\equal{}1$
$ m_b\cos C \equal{} m_c\cos B \Longleftrightarrow \frac{b}{c} \equal{} \frac{m_c}{m_b} \Longleftrightarrow \frac{b^2}{c^2} \equal{} \frac{1\plus{}3b^2}{1\plus{}3c^2}  \Longleftrightarrow$

$ \Longleftrightarrow b^2 \plus{} 3b^2c^2 \equal{} c^2 \plus{} 3b^2c^2 \Longleftrightarrow b \equal{} c$

[quote][color=darkred]$ 1\blacktriangleright$ In any $ A$ - right or acute $ \triangle ABC$ , $ m_b\cos C \equal{} m_c\cos B\Longrightarrow B \equal{} C$ .

$ 2\blacktriangleright$ Exists triangles $ ABC$ with $ A > 90^{\circ}$ and $ B\ne C$  so that  $ m_b\cos C \equal{} m_c\cos B$ ?

[b]Notation.[/b] $ m_a$ is the length of the $ A$ - median of the triangle $ ABC$ .[/color][/quote]
[color=darkblue][b][u]Remark.[/u][/b]   $ \boxed {m_b\cos C \equal{} m_c\cos B\Longleftrightarrow\sin (B \minus{} C)\left(1 \plus{} 4\cos A\cos B\cos C\right) \equal{} 0}$

[b]Proof.[/b] I"ll use the well-known relation $ 4m_a^2 \equal{} a^2 \plus{} 4bc\cdot\cos A$ a.s.o. Thus,

$ m_b\cos C \equal{} m_c\cos B$ $ \Longleftrightarrow$ $ 4m_b^2\cos^2C \equal{} 4m_c^2\cos^2B$ $ \Longleftrightarrow$

$ \left(b^2 \plus{} 4ac\cdot\cos B\right)\cos ^2C \equal{} \left(c^2 \plus{} 4ab\cdot\cos C\right)\cos^2B$ $ \Longleftrightarrow$

$ b^2\cos^2C \minus{} c^2\cos^2B \equal{} 4a\cos B\cos C(b\cos B \minus{} c\cos C)$ .

But $ a \equal{} b\cdot\cos C \plus{} c\cdot \cos B$ . Therefore,

$ b\cdot\cos C \minus{} c\cdot\cos B \equal{} 4\cos B\cos C(b\cos B \minus{} c\cos C)$ $ \Longleftrightarrow$

$ \sin B\cos C \minus{} \sin C\cos B \equal{} 4\cos B\cos C(\sin B\cdot\cos B \minus{} \sin C\cdot\cos C)$ $ \Longleftrightarrow$

$ \sin (B \minus{} C) \equal{} 2\cos B\cos C(\sin 2B \minus{} \sin 2C)$ $ \Longleftrightarrow$

$ \sin (B \minus{} C) \plus{} 4\cos B\cos C\sin (B \minus{} C)\cos A \equal{} 0$ $ \Longleftrightarrow$

$ \sin (B \minus{} C)\left(1 \plus{} 4\cos A\cos B\cos C\right) \equal{} 0$ .

[u]Case I[/u] - $ \triangle ABC$ is no obtuse. Then  $ 1 \plus{} 4\cos A\cos B\cos C\ge 1 > 0$ .

Thus, in this case $ m_b\cos C \equal{} m_c\cos B$ $ \Longleftrightarrow$ $ B \equal{} C$ .

[u]Case II[/u] - $ \triangle ABC$ is obtuse and $ \cos A\cos B\cos C \equal{} \minus{} \frac 14$ . This is the true problem !  [/color]


limit
trigonometry
real analysis
real analysis unsolved

i can't find the limit
\[\lim_{n\to\infty}\ \frac{1}{n}\sum_{k=1}^{n}\left(\frac{2+\sin k}{3}\right)^{k}\]
could you help?

Since for positive $a_{k}$ the convergence of $\sum \frac{a_{k}}{k}$ implies $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}{a_{k}}=0$, the limit in question is $0$. On the other hand, reducing this question to the one about the convergence of the corresponding series is a clear overkill. Here it is enough to know that $\pi$ is irrational, not that it is non-Liouvillian. Indeed, let us fix some integer $A>0$ and look at the set of indices $k$ such that $\sin k\ge 1-\frac{A}{k}$ (all other terms are small if $A$ is large). Suppose that the portion of such terms is not small, say $1/A$. Then one can find two indices $k'>k''>\frac 1{2A}n$ with difference $k'-k''\le 2A$ in this set. But this will mean that one of the numbers $1,2,\dots,2A$ is an integer multiple of $\pi$ with precision about $A^{2}/n$, which is false for large $n$. The rest should be clear.


Given a graph $ G\equal{}(V,E)$, say that a set $ S \subset V$ is an edge cover if every edge $ e \in E$ has some endpoint in $ S$. Let $ \psi(G)$ be the minimum edge cover in $ G$. Prove that for any matching $ M\subset E$ in $ G$, we must have $ \mid M\mid \leq \psi(G)$

I think you have confused the terms "edge cover" (a set of edges covering all vertices) and "vertex cover" (a set of vertices covering all edges).

Choose any matching $ M$ and any vertex cover $ C$.  By the definition of vertex cover, at least one endpoint of each element of $ M$ belongs to $ C$.  By the definition of matching, all these endpoints are distinct.  It follows that every matching is no larger than any vertex cover, and so in particular the largest matching is no larger than the smallest vertex cover, as desired.

The inequality is sharp for bipartite graphs (this is the Konig-Egervary; e.g. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43344 or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154989 ) but not in general (e.g., it fails for a triangle).

In the future, questions of this sort should be placed in the Combinatorics section of the forum.  (No sense in reposting this one, of course.)


For what value of the constant k does y=x^2+2x-k satisfy the differential equation xy'-y=x^2+1?

Just plug y=x^2+2x-k into the equation and see what happens... naturally, y'=2x+2.


[quote="tobeno_1"]let $ a,b,c \geq 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$, try to prove that
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq \frac {5}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31377
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476

first square both sides of the inequality after that use IRAN's (1996) inequality then multiply and let [b]ab=x[/b] and [b]bc=y[/b] and [b]ca=z[/b]


larry owes 5000 dollars on his credit card. he is charged 1.75% interest on the remaining balance every month, which means 1.75% of the amount he owes is added to the amount he owes at the end of each month. suppose that he makes no new purchases with his credit card and pays 200 dollars on the first day of every month.

write a recursive equation describing his balance just after he makes each payment.

how long will it take him to pay off his credit card debt?

how much will he pay, total, during that time?

[hide]
Amount he owes just after he makes his nth payment:
$5000\cdot101.75^{n}-200-200\cdot101.75-200\cdot101.75^{2}\cdots-200\cdot101.75^{n-1}$
[/hide]

If $a_{n}$ is how much money he owes at the end of $n$ months, and $a_{0}=5000$, the recursive formula is:
[hide]a_n=1.0175\left(a_n-200\right)[/hide]

It will take him:
[hide]33 months[/hide]
to pay it off, over which he will pay
[hide]$\$6480.18$[/hide]


6. Many television screens are rectangles that are measured by the length of their
diagonals. The ratio of the horizontal length to the height in a standard television screen is 4:3. The horizontal length of a "27-inch" television screen is closest, in inches, to which of the following?

$ (A) 20 \qquad (B) 20.5 \qquad (C) 21 \qquad (D) 21.5 \qquad (E) 22$

[hide]We have a side of 4x, a side of 3x, and a side of 27 in a right triangle. This is 3 4 5 right triangle, so we have 27=5x, and $ x \equal{} \frac {27}{5}$. Since we are solving for 4x (the horizontal length), we see the length is $ \frac {108}{5}$ or 21.6. [b]This is closest to 21.5, or [size=150]D[/size][/b] [/hide]


Given n numbers a[size=59]1[/size], a[size=59]2[/size], ... , a[size=59]n [/size]> 0 (constant).

With n number of x > 0, find the max value of:

$(x_1^{a_1})(x_2^{a_2})...(x_n^{a_n})$

 or (x[size=59]1[/size]^a[size=59]1[/size])(x[size=59]2[/size]^a[size=59]2[/size])...(x[size=59]n[/size]^a[size=59]n[/size])

such that

x[size=59]1[/size] + x[size=59]2[/size] + ... + x[size=59]n[/size] = 1. (xs are the variables)

The code you want is x_1^{a_1} = $x_1^{a_1}$.

[hide="the first step"]For the case $n = 2$ we can find the maximum at $x_1 = \frac{a_1}{a_1+a_2}$ (I used calculus to do this -- there may be a better way.)  [/hide]

Agggghhhh! All those powers! Take the log!

By weighted AM-GM (or plain old Jensen's inequality), if the sum of the $a_i$'s is $1$, we get

$\prod_i x_i^{a_i}\leq \sum_i a_ix_i$

Was this what you were looking for?

The sum of the $x_i$ is one, not the $a_i$

Id say its the maximum in terms of the $a_i$

The value of $a_n$ is given first, then you can select x as you want to max the value of the product.  I have restated the question to make it clearer.

[quote="blahblahblah"]By weighted AM-GM (or plain old Jensen's inequality), if the sum of the $a_i$'s is $1$, we get

$\prod_i x_i^{a_i}\leq \sum_i a{}_ix{}_i$

Was this what you were looking for?[/quote]

this idea isn't so bad

put $y_i=x_i/a_i$ and use Jensen inequality
you find :$\prod{}_a{}_i^{a_i}/(\sum{}_a{}_i)^{\sum{}_a{}_i}$

and the equality hold if $y_i=y_j$

souhail.
[code][/code]

lol my first time with latex
:)
this idea isn't so bad

put $y_i=x_i/a_i$ and use Jensen inequality
you find :${\prod a_i^{a_i}}/{(\sum a_i)^{\sum a_i}}$

and the equality hold if $y_i=y_j$

souhail.
[code][/code][/quote]

[quote="kenshin_rurounie"]lol my first time with latex
:)
put $y_i=x_i/a_i$ and use Jensen inequality
you find :${\prod a_i^{a_i}}/{(\sum a_i)^{\sum a_i}}$

and the equality hold if $y_i=y_j$

souhail.
[code][/code][/quote][/quote]

Sorry, I didn't get your solution.  Explain please..

[quote="Laputa"]Hm.. I still didn't get it.  What happen if $a_1 = 2, a_2 = 3, a_3 = 4$, can you find the specific value of x for each a?[/quote]

if so then we have  $x_1/2=x_2/3=x_3/4=(x_1+x_2+x_3)/9=1/9$

[quote="kenshin_rurounie"]put $y_i=x_i/a_i$

then we have to maximise  $\prod a_i^{a_i}*\prod y_i^{a_i}$
forget $\prod a_i^{a_i}$ and maximise $\prod y_i^{a_i}$

so bye the Jensen Inequality :$\log(\prod y_i^{a_i})=\sum a_i\log(y_i) \leq \sum a_i*\log(\frac{\sum a_i*y_i}{\sum a_i}) = ...$

so we find ${\prod a_i^{a_i}}/{(\sum a_i)^{\sum a_i}}$[/quote]

You did a good job of figuring out the solution but I quite can't understand your solution, what is the $...$?


hello, the coefficient of $ x^{2009}$ is $ 2147483648$.
Sonnhard.

[hide="Solution"]This problem basically asks, how can we make 2009 by using powers of 3 from $ 3^0$ to $ 3^6$, each being used at most twice?

This is the same thing as representing 2009 in ternary. Express 2009 in ternary; we get $ 2202102_3$. Therefore, the desired coefficient is
\[ 2^{2\cdot 6 \plus{} 2\cdot 5 \plus{} 2\cdot 3 \plus{} 2} \equal{} 2^{30} \equal{} 1073741824.\]But wait! The $ 1$ in the ternary representation indicates that there are two ways to choose that particular coefficient of $ x^9$, since there are two of them. Hence the answer is $ 2^{31}\equal{}2147483648$, as desired.
[/hide]

Yongyi, your solution confuses me. I don't see how the problem basically asks how can we make 2009 by using powers of 3 from 3^0 to 3^6 , each being used at most twice. I can't quite put my hands around it.

Also, how is this the same thing as representing 2009 in ternary?

The product is $ \prod_{k \equal{} 0}^6 (1 \plus{} 2^kx^{3^k})^2$, which is equivalent to $ (1 \plus{} x)^2(1 \plus{} 2x^3)^2(1 \plus{} 4x^9)^2(1 \plus{} 8x^{27})^2(1 \plus{} 16x^{81})^2(1 \plus{} 32x^{243})^2(1 \plus{} 64x^{729})^2$.

[hide="Old way I explained it"]
This is a product of 14 binomials, consisting of 2 of each binomial of the form $ (1 \plus{} 2^kx^{3^k})$ when $ k$ ranges from 0 to 7. I will assume you already know how to expand products of binomials, so I will get straight to the point.

The coefficient of $ x^k$, for any $ k$, is determined by the number of combinations of choices for each binomial such that the product of the choices is in the form $ kx^{2009}$.

For example, the coefficient of $ x^2$ is just $ 1$ since there is only one way to get an exponent of $ 2$ from the terms, namely just choosing $ x$ for both of the $ (1 \plus{} x)$ terms and $ 1$ for the rest of the terms.

Likewise, the coefficient of $ x^{87}$ can only be obtained by choosing $ 2x^3$ for both of the $ (1 \plus{} 2x^3)$ terms and $ 16x^{81}$ for both of the $ (1 \plus{} 16x^{81})$ terms, and choosing $ 1$ for the rest of the terms.  This is the only combination that yields a sum of 87. This yields a coefficient of $ 2\cdot 2\cdot 16\cdot 16 \equal{} 1024$.

In other words, the coefficient of $ x^k$ is the sum of the values of $ a_0 \plus{} 2a_1 \plus{} 4a_2 \plus{} 8a_3 \plus{} 16a_4 \plus{} 32a_5 \plus{} 64a_6$ for each possible way to express $ k$ as $ a_03^0 \plus{} a_13^1 \plus{} a_23^2 \plus{} a_33^3 \plus{} a_43^4 \plus{} a_53^5 \plus{} a_63^6$ where $ a_i\in \{0,1,2\}$ for all $ i$. But from ternary representation, we know there is only one unique way to express any number in the above form. Also, since there is a 1 in the ternary representation, we multiply the coefficient by 2 to account for the two different choices of the monomial that corresponds to the $ 1$ in the ternary representation.[/hide]

Edit: I've found a better way to explain it. Expand the squares of each term; we get $ \prod_{k\equal{}0}^6 (1\plus{}2^{k\plus{}1}x^{3^k}\plus{}2^{2k}x^{2\cdot 3^k})$, or $ (1\plus{}2x\plus{}x^2)(1\plus{}4x^3\plus{}4^6)(1\plus{}8x^9\plus{}16x^{18})\cdots(1\plus{}128x^{729}\plus{}4096x^{1458})$. Take the ternary representation of 2009, choose the appropriate term from each monomial, and multiply.

[quote="bigman"]I think I understand everything until you start talking about x^87

I see how you choose 2x^3 from both of the (1+2x^3) terms but not 16x^81 from both (1+16x^81) terms. Wouldn't you just pick 16x^81 from 1 of the terms?[/quote]

Yes, that was a mistake, and I apologize. I intended for the exponent to be 168 instead of 87.

cool. Now I understand everything until "From ternary representation..."

I guess I also don't understand how you said in your 1st post, "This problem basically asks, how can we make 2009 by using powers of 3 from 3^0 to 3^6, each being used at most twice? "

and then, how do you get the desired coefficient?

Look at this product:
\[ (1 \plus{} 2x \plus{} x^{2})(1 \plus{} 4x^{3} \plus{} 4^{6})(1 \plus{} 8x^{9} \plus{} 16x^{18})\cdots(1 \plus{} 128x^{729} \plus{} 4096x^{1458})\]
We know there is only one way to represent 2009 in ternary; it is $ 2202102$. In other words, $ 2009\equal{}2\cdot 3^0\plus{}0\cdot 3^1\plus{}1\cdot 3^2\plus{}2\cdot 3^3\plus{}0\cdot 3^4\plus{}2\cdot 3^5\plus{}2\cdot 3^6\equal{}2\plus{}0\plus{}9\plus{}54\plus{}0\plus{}486\plus{}1458$. How does that correspond with the exponents in the product above?

uh each of the summands is a power (1) of the numerically corresponding product of polynomials

Yes. Do you get it now? So the coefficient of $ x^{2009}$ is simply the product of the coefficients from each of the factors.


Wow , I don't have a slightest clue. Please give us a hint.

The question essentially is:
There is a function f(x) from N to R
We are given f(1),f(2),...f(10).
We have to prove that f(11) can be anything.ie,we have to find f(x) which satisfies 1,2,..10 and f(11) should be something else.

[hide="hint"]
f(x)=2x-1 + g(x)

g(x) should be 0 for x=1,2,...10
and should be some real no for x>10.
We can then vary g(x) to get any real for x>10.
The essential part is that it should be 0 for x<=10.
[/hide]

[quote="arjunarul"]The first ten terms of an infinite sequence are 1,3,5,7,9,11,13,15,17,19 in the given order.If the sequence is not defined,prove that the 11th term of the sequence can be a term of any real value.(INMOTC-08)[/quote]

Define : $ f(n) \equal{} (n\minus{}1)(n\minus{}2)...........(n\minus{}10)/k$ + $ 2n\minus{}1$ where k is any constant real number.


geometry
geometric transformation
reflection
geometry unsolved

Two perpendicular lines divide a square into four parts. Three of these parts have area $1$.
Show that the fourth part also has area $1$.

I don't have a proof, but I'll tell you what I've been trying to do.

Some notations:

Let's take square ABCD s.t. AB and CD are horizontal. Let's also work with N, S, E and W (cardinal points). On my picture AB is the N side and BC is the W side. Now let P be the point of intersection between the 2 perpendicular lines d and d'. We tak the 2 perpendicular lines passing through O (the center of the square) and having the same directions as d and d'. Let the first such line, the one parallel to d, cut AB in X and CD in Z, and let the second one cut BC and DA at Y and T.

We name the 4 regions formed by XZ, YT and the sides of the square according to their position (for example, the region XBYO is NW, coming from North-West).

The whole problem revolves around the position of P in one of these regions: NE, NW, SW or SE. The lines d and d' also define some regions, so let's call these NW', SW' etc.

Here are some conjectures which (I think) might help:

(1) if P is in the region XY (wit X being N or S and Y being E or W) then XY' is th smallest among the 4 regions defined by d and d', so the 3 regions of equal areas are the other 3;

(2) if P is a point on (YT) then S(SW')>=S(NW') with equality iff P=O;

I don't really know whether these are true or not, but if they are, I think we can use them to show that we can never have 3 regions of equal areas except when P=O, and in this particular case all 4 regions have equal areas, so we would be done.

Comments?

Let $L_1,L_2$ the lines.

If the two lines intersect the area of square, then the problem is almost obvious.

If one of lines form a triangle, say $L_1$, reflect this triangle in the same line $L_1$.Then this reflected triangle is in the other part of the square, except in the case of $L_1$ is a diagonal.Then in its case the problem is finished.

If the lines form four quadrilaterals, take the auxiliar construction:
$L_1;L_2$ cut the sides of squares in $M,Q;N,P$ respectively.
$P'N'$ and $M'Q'$ are perpendiculars of PN and MQ respectively, and they contains the center of square.
Then, prove that the two formed paralellograms are congruent and the central quadrilateral in your intersections is a square, and use the first case for calculate the areas.

I sent a solution to john Scholes site like a month ago but they havent published it. I am going to describe briefly the key to the proof because i gotta go to school :D  :D

1: the idea is to show that the intersection of the 2 lines must be the center of the circle.

2: if both lines intersect in the center the result is obvious.

3: if the lines doesnt intersect in the center, draw 2 lines, each one paralel to one of the original lines, so that the new lines intersect in the center and then give a name to the area of every region and write down equations and you ill get to a contradiction.......I am sure because i did it, maybe later i'll tell you what the equations are, but from here it is just work!!!

I solved almost all the unsolved MO problems in the world. These solutions are in my books.


Hey all how u doin. i need help with the following 2 questions.

1. A mass m is given an initial velocity of 'u' in the x-direction. It moves in a medium that resists its motion with a force proportional to the velocity, F = -kv. Find v(x) and how far the far the mass travels before stopping. Find v(t)

2. A ball is thrown horizontally from the top of 150-ft cliff at A with a speed of 50ft/s and lands at point C. Because of a strong horizontal wind, the ball has a constant acceleration in the negative x-direction. Determine the radius of curvature 'p' of the path of the ball at B, where its trajectory makes an angle 45 degrees with the horizontal. i got "normal acc" to be 38. need to check on tht.

p.s. the attached pic is for question 2
thnk u

[quote="glitch"]what about v(x) ?[/quote]
the given facts say

$\frac{d\vec v}{dt}$ $=$ $-\frac{k\vec v }{m}$

since the body moves in st. line with decellaration
 we have ,

$|\frac{d\vec v}{dt}|$ $=$ $-\frac{d|\vec v|}{dt}$

so that gives by taking modulus on the sides

$\frac{d|\vec v|}{dt}$ $=$ $-\frac{k|\vec v| }{m}$
but ,

$\frac{d|\vec v|}{dt}$ $=$ $(\frac{d|\vec v|}{dx})(\frac{dx}{dt})$
that gives
$(\frac{d|\vec v|}{dx})(\frac{dx}{dt})$ $=$ $-\frac{k|\vec v| }{m}$
this gives us
$d|\vec v|$ $=$ $-\frac{k{dx}}{m}$
this gives

${v(x)}^{2}$ $=$ $u^{2}-{\frac{2kx }{m}}$
this gives1

${v(x)}$ $=$ $\sqrt{u^{2}-{\frac{2kx }{m}}}$

[quote="sos440"][quote="sos440"]1. Instead of considering the 2D displacement vector, we express the displacement of a mass $s$ with complex number. Also we denote $v = v(t) = \frac{ds}{dt}$. Then

$mv' =-img-kv$, $v(0) = u$

where $i$ is the imaginary unit. This initial value problem yields the solution

$v(t) = ue^{-\frac{kt}{m}}-\frac{img}{k}\left( 1-e^{-\frac{kt}{m}}\right)$

If you specify what $x$ means, then you may also find the expression of $v$ in terms of $x$.[/quote]

Sorry, I didnt know there was no gravitatioal force...  :blush:[/quote]

well since the particle is supposed  to move in the $x$ axis and assuming the g.force if any acts on the y axis .it gets balanced down by other forces.i think it's clear from the question that only 1 net force acts on the body


integration
calculus
calculus computations

The velocity is given by $ v(t)\equal{}84\minus{}14t$, so it will stop when the velocity is zero, or $ 84\minus{}14t\equal{}0\Rightarrow t\equal{}84/14$ s. The position is given by $ r(t)\equal{}84t\minus{}7t^2$, so you can plug in that $ t$ value to find how far it went.

Note that you can integrate to get those equations. $ v(t)\equal{}\int a(t)dt$ and $ r(t)\equal{}\int v(t)dt$.

We have $ a(t)\equal{}\minus{}14\implies v(t)\equal{}\minus{}14\int dt\equal{}\minus{}14t\plus{}C$. We already know that the velocity is 84, so we have $ v(t)\equal{}84\minus{}14t$.
$ p(t)\equal{}84\int dt\minus{}14\int t\ dt\equal{}84t\minus{}7t^2\plus{}C$ (but $ C\equal{}0$ because the initial position is taken to be $ 0$).

So if $ v(t)\equal{}0$, we have $ \minus{}14t\equal{}\minus{}84\implies t\equal{}\boxed{6\text{ seconds}}$.
And $ p(6)\equal{}\boxed{252\text{ feet}}$

[quote="math92"]A car traveling at 84 ft/sec begins to decelerate at a constant rate of 14 ft/sec^2.  After how many seconds does the car come to a stop and how far will it travel before stopping?[/quote]
[hide]We have $ s \equal{} ut \plus{} \frac{1}{2}at^2$ and $ v \equal{} u \plus{} at$, here $ v\equal{}0$.

I think this should've belonged to the Introductory Physics forum.[/hide]

[quote="Rajiv"][quote="math92"]A car traveling at 84 ft/sec begins to decelerate at a constant rate of 14 ft/sec^2.  After how many seconds does the car come to a stop and how far will it travel before stopping?[/quote]
[hide]We have $ s \equal{} ut \plus{} \frac {1}{2}at^2$ and $ v \equal{} u \plus{} at$, here $ v \equal{} 0$.

I think this should've belonged to the Introductory Physics forum.[/hide][/quote]

you can solve it with basic physics or integrals.  either one.  the first two peeps chose integrals and you chose physics.

[quote="math92"]you can solve it with basic physics or integrals.  either one.  the first two peeps chose integrals and you chose physics.[/quote]

Depends on individual choice!

[quote="I"]
This should've belonged to the Introductory Physics forum

[/quote]

My suggestion was a joke and not a serious comment although it may have been devoid of any grinning smileys!!!   :D  :D

Of course, many of the formulas that exist in physics are largely derived from calculus, so we're both doing the same thing in effect.
Case in point: $ s\equal{}ut\plus{}\frac{1}{2}at^2$. Differentiating, we have $ \frac{\mathrm{d}s}{\mathrm{d}t}\equal{}v\equal{}u\plus{}at$.


quadratics
modular arithmetic
Divisibility Theory
pen

Let $d$ be any positive integer not equal to 2, 5, or 13. Show that one can find distinct $a$ and $b$ in the set $\{2,5,13,d\}$ such that $ab - 1$ is not a perfect square.

[quote="nicetry007"]Let  $ 2d-1 = x^{2}$, $ 5d-1 = y^{2}$ and $ 13d-1 = z^{2}$.

Suppose $ d$ is even. This would imply $ 2d-1 = 3 ( mod \; 4)$. But a perfect square is either $ 0$ or $ 1 (mod \; 4)$. Hence, $ d$ is odd.

This implies that $ x$ is odd and $ y$ and $ z$ are even.

$ y^{2}-z^{2}= 8d \Rightarrow y^{2}= z^{2}(mod \;8)$.

Since the perfect square of an even number is either $ 0$ or $ 4 (mod \; 8)$,  $ y^{2}$ and $ z^{2}$ are either both $ 0(mod\; 8)$ or $ 4 (mod \; 8)$.

Suppose $ y^{2}= 0 (mod \; 8)$ and $ z^{2}= 0 (mod \; 8)$.

If the square of an integer is divisible by $ 8$, then it is also divisible by $ 16$. Hence, it follows that $ y^{2}= 0 (mod \; 16)$ and $ z^{2}= 0 (mod \; 16)$.

This, in turn, implies that $ y^{2}-z^{2}= 0 (mod \;16)$.

But  $ y^{2}-z^{2}= 8d$ and $ d$ is odd.  Hence, we arrive at a contradiction.

Suppose $ y^{2}= 4 (mod \; 8)$ and $ z^{2}= 4 (mod \; 8)$.

Let $ y^{2}= 4p^{2}$ and  $ z^{2}= 4q^{2}$ where $ p,q$ are odd.

$ y^{2}-z^{2}= 4(p^{2}-q^{2}) = 4(p-q)\;(p+q)$. Since $ (p-q)$ and $ (p+q)$ are both even, $ y^{2}-z^{2}$ is divisible by $ 16$.

But  $ y^{2}-z^{2}= 8d$ and $ d$ is odd.

We observe that in both cases we arrive at a contradiction. Hence,  the result follows.[/quote]

[hide="A52"]We consider it modulo 16. It's trivial observation that 0, 1, 4, 9 are the quadratic residues modulo 16. We use contradiction method.

$2d-1 \equiv 0, 1, 4, 9 \Rightarrow d \equiv 1, 9, 5, 13 \pmod{16}$.
$5d-1 \equiv 0, 1, 4, 9 \Rightarrow d \equiv 13, 10, 1, 2 \pmod{16}$.

Therefore, $d \equiv 1, 13 \pmod{16}$. But then, $13d-1 \equiv 12, 8 \pmod{16}$, so it's impossible. [/hide]

We see that $y$ and $z$ have the same parity since $8$ divides $y^2-z^2$.But we see that $y$ must be of the form $5k+/-2$ and $z$ must be of the form $13k+/-5$.Here we see that $y$ and $z$ have different parity.Hence

[hide=Solution]It suffices to show that one of $2d-1, 5d-1, 13d-1$ is not a perfect square. The quadratic residues modulo $16$ are $0, 1, 4, 9$. We will prove this result by contradiction. Suppose that all subsets of cardinality $2$ of $\{2,5,13,d\}$ satisfy that it's two elements, $a$ and $b$, satisfy that $ab-1$ is a perfect square. It follows that $2d, 5d, 13d$ must all be either $1, 2, 5, 10$ $(\text{mod}$ $16)$. Solving these congruences give us the following results: $$2d\equiv 1,2,5,10: d\equiv 1,5,9,13$$ $$5d\equiv 1,2,5,10: d\equiv 1,2,10,13$$ $$13d\equiv 1,2,5,10: d\equiv 2,5,9,10$$ Clearly no residue in each of the three congruences satisfies all $3$ of them, thus we have our contradiction.[/hide]

let assume $2d-1=x^2,5d-1=y^2,13d-1=z^2 \implies$

$x^2$ is odd so $x^2=2d-1 \equiv 1 \mod 8 \implies d \equiv 1 \mod 4$

now let $d=4k+1$ which implies that $8k+1=x^2,20k+4=y^2,52k+12=z^2$ now let:

$z=2z_0,y=2y_0 \implies 5k+1={y_0}^2,13k+3={z_0}^2$, we shall assert $5k+1,13k+3$ cannot be perfect square at the

same time and for proving this we check four cases:

[color=#f00]case 1:[/color] $k \equiv 0 \mod 4 \implies {z_0}^2 \equiv 3 \mod 4$ which is impossible

[color=#f00]case 2:[/color] $k \equiv 1 \mod 4 \implies {y_0}^2 \equiv 2 \mod 4$ which is impossible

[color=#f00]case 2:[/color] $k \equiv 2 \mod 4 \implies {y_0}^2 \equiv 3 \mod 4$ which is impossible

[color=#f00]case 2:[/color] $k \equiv 3 \mod 4 \implies {z_0}^2 \equiv 2 \mod 4$ which is impossible


The topic pretty much says it all, but where do moved topics go to? I have had a recent post moved, and yet I can still reply on it. Does it actually get moved to a different forum group, or is it just deleted? :huh:

They go to where they are moved. If you click on a "moved" topic in a certain forum, it takes you to the topic in the destination forum.

Oh....that actually makes sense. Never noticed. Apparently the old topic still works. Thanks!(prob should not post this message here, but I think it's ok....not too much spam)

That forwarding alias is only there if the moderator who moved the post specifically activated it; many times it's not. If you can't find a post where you put it, one thing you could try is going into your own profile in "Memberlist" and then clicking on the tag for seeing your own posts. That should show where the one you're interested is.

Or just click "View your posts" at the top right corner of the main frame :P.

I would add that if your posts are being moved, then you're posting them in the wrong places to begin with. Take a moment to consider that, and learn from it. Put the next one in the right place the first time.


4 of her daughters must have daughters, and 2 have none.  So the answer is 2+24=26.

[font=TIMES NEW ROMAN][hide=Oops... Complementary counting is not my best topic]It is easy to see that 4 of Bertha's daughters have 6 daughters each.

[hide=Diagram. Press the expand button at the top of the feed box to view it] [center][size=50][b]D[/b] = Daughter, [b]G[/b] = Granddaughter[code]
---------------------------- BERTHA ----------------------------
/            /            /            \            \            \
D            D            D             D            D            D
/ / / \ \ \  / / / \ \ \  / / / \ \ \  / / / \ \ \  ----- NO DAUGHTERS -----
G G G  G G G G G G  G G G G G G  G G G G G G  G G G  ----- NO DAUGHTERS -----
-------------------------------- NO DAUGHTERS --------------------------------[/code][/center][/hide]

We can see that all 24 G's plus two D's don't have daughters. Therefore, $24+2=\boxed{26}$ of Bertha's daughters and granddaughters don't have daughters. $$\definecolor{color}{rgb}{0.6, 0.6, 0.7} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

[quote=pifinity][font=TIMES NEW ROMAN][hide=Oops... Complementary counting is not my best topic]It is easy to see that 4 of Bertha's daughters have 6 daughters each.

[hide=Diagram. Press the expand button at the top of the feed box to view it] [center][size=100][b]D[/b] = Daughter, [b]G[/b] = Granddaughter[code]
---------------------------- BERTHA ----------------------------
/            /            /            \            \            \
D            D            D             D            D            D
/ / / \ \ \  / / / \ \ \  / / / \ \ \  / / / \ \ \  ----- NO DAUGHTERS -----
G G G  G G G G G G  G G G G G G  G G G G G G  G G G  ----- NO DAUGHTERS -----
-------------------------------- NO DAUGHTERS --------------------------------[/code][/center][/hide]

We can see that all 24 G's plus two D's don't have daughters. Therefore, $24+2=\boxed{26}$ of Bertha's daughters and granddaughters don't have daughters. $$\definecolor{color}{rgb}{0.6, 0.6, 0.7} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$[/quote]

Why not make the font size bigger?

[quote=hansenhe][quote=mikekongmike]no \;;;;;;;;;;;;;;;;;;;;;;[/quote]

What do you mean by this?[/quote]

Probably the 8-char requirement

This problem is 2004 AMC 10A Problem 6/2004 AMC 12A Problem 4.

[quote=GameBot]Bertha has 6 daughters and no sons. Some of her daughters have 6
daughters, and the rest have none. Bertha has a total of 30 daughters
and granddaughters, and no great-granddaughters. How many of Bertha's
daughters and granddaughters have no daughters?[/quote]

I will tackle this problem by sorting out info
Daughters: x
Grandaughters: 30-x
Great-grandaughters: 0 (not needed).
Well, we know that the total daughters are 6, so granddaughters are 24.
We know that each daughter who has children has 6, which means that 4 have daughters.
That means that 24 grandaughters+2 daughters don't have children, or [size=150]28[/size] do not have children.
Sorry if this was confusing, but I hope it helped.

[quote=angrybird_r][quote]Your First Answer: 2
Your Second Answer: 26[/quote]
*sigh*[/quote]

Wow, there is like a small chance that a family can only have a daughter lol alcumus defied human nature

[quote=Lamboreghini]As this was an AMC question, it was MAA that defied human nature, not Alcumus.[/quote]

Bertha must have a hard time caring for her daughters.
[hide][center]Bertha
Bertha's daughter                Bertha's daughter                Bertha's daughter                Bertha's daughter                Bertha's daughter                Bertha's daughter
Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               Bertha's Granddaughter               [/hide]


Prove that:
$ x^4\plus{}y^4\plus{}z^4\equal{}2x^2y^2\plus{}2y^2z^2\plus{}2z^2x^2\plus{}24$ have no integer solutions.


linear algebra
matrix
invariant
ratio
parameterization
linear algebra unsolved

How do I randomly generate a real matrix that is positive-definite?

A standard method: Generate a random matrix $ A$- using independent identically distributed Gaussians for each entry is most common.
The desired matrix is $ A^{T}A$ if $ A$ is real, and $ A^{*}A$ if $ A$ is complex.

If we use Gaussians, these distributions are invariant under orthogonal/unitary similarity.

Another note: $ A$ does not have to be square. A tall and skinny $ A$ works as well, and the ratio of width to height is a parameter we have to track.


Blizi nam se BMO, tako da sam odlucio da otvorim i ovaj topic. Na njemu cemo za par sedmica iscekivati rezultate, komentirati zadatke, a zasad mozemo objaviti sastave ekipa :)

Sto se BiH tice, ovaj post ce biti pravovremeno editovan kad saznam cijelu ekipu, a zasad tu su:

1. Armin Besirevic
2. Vedran Karahodzic
3. Salem Malikic
4. Jelena Radovic
5. Vlado Uljarevic
6. xxxxxxxxxxxxx

Kako to da ekipu saznajes " u delovima " ?  Jeste li je izabrali ili ...?
( Inace, cula sam neku cudnu pricu o izboru pa me zanima je li stvarno tako  )

Upravo zahvaljujuci cudnoj prici koju si vec cula (i koja je nazalost istinita, ako mislimo na isto: sifra "3+2+1"), ekipu saznajem u dijelovima.

konačni sastav ekipe:
BOSNIA AND HERZEGOVINA TEAM

Leader: Vidan Govedarica

Deputy leader: Amer Krivosija

Contestants:

   1. Besirevic Admir
   2. Karahodzic Vedran
   3. Malikic Salem
   4. Radovic Jelena
   5. Sarcevic Franjo
   6. Uljarevic Vladimir

 vidimo se u ohridu! :D


2 circles [b]O[/b] and [b]O'[/b] are internally tangent at point [b]P[/b].[b]O[/b] is the larger one]AB is a chord of [b]O[/b] and a tangent of [b]O'[/b] at point [b]C[/b].Prove that PC bisects the angle APB.

[hide="Hint"]
draw a tangent line to the circles at point $ P$.[/hide]

On that given lemma, could somebody explain exactly where to place the points X and Y? I'm having trouble comprehending it.

X is the intersection of the common tangent and the tangent at the point N.Y is any point that lies on the otherside of of the common tangent(the side not containing X).
Take a look at the diagram I included.

Though I am not sure about the NA=Nb part.Could someone explain that?

Ok.I got how NA=NB comes.But how does it prove the statement?Could someone please explain?
And X & Y can be the same points, right?

It's not letting me view your attachment, could you repost it please?

And if NA = NB, then the angles extending to those chords are equal, which are the angles that form APB (according to the notation in which you originally posted the problem), and are divided by PC.


if P=NP then every problem in P (except $ \emptyset$ and $ \Sigma^*$) is in NPC. How do I have to proceed? Thanks

Fill in the blanks:

A problem Y is NPC if and only if for every problem X in NP, there exists __________.  If P = NP, then for any problem Y in NP there exists ____________.  Therefore, if P = NP, for any problem X in P and any problem Y in NP, we can build ____________ by using ____________ directly, without any additional machinery needed.

I have no idea what those exceptions are doing there: if P = NP then every problem in P (including $ \emptyset$ and $ \Sigma^*$) are in NPC.

The first one is just the definition, that is iff $ X \le Y$. But I have the impression that the rest of the blanks involve some equivalent definition of NPC, which makes things easier..

@JBL:  The exceptions are there for a reason.  It has to do with the definition of polynomial-time reductions that underlies the definition of NP-completeness.

@Stewie:  Let $ L$ be a language in $ P$ besides the empty language and the full language.  Let $ M$ be an arbitrary language in $ P$.  Show that there is a polynomial-time reduction from $ M$ to $ L$.  You don't even need the assumption $ P \equal{} NP$ for this step.

Ravi B, I don't know how to deal with this problems, so any intelligente hints will slip over me..

Stewie, writing "$ A \leq B$" is using the first sentence to define notation.  That completely misses the point, which was to write down [i]what that notation means[/i].

[quote="Ravi B"]@JBL:  The exceptions are there for a reason.  It has to do with the definition of polynomial-time reductions that underlies the definition of NP-completeness.[/quote]

Okay, fair enough.  It's been a little too long since my Intro to Computation course, I'm afraid.  It seems to me (not having thought about it very hard) that the exceptions are an indispensible part of the definition of [i]reduction[/i], but in special cases like polynomial-time reduction (where the nature of the reducing function is limited, esp. in ways that make the most sense when we need to introduce some computational model) one could rework the definition in such a way as to avoid these exceptions without changing the sets P, NP, or NPC, right?  This would have the disadvantage that a polynomial-time reduction would no longer be a reduction, of course.

So ...if for every problem X in NP, there exists a polynomial-time reduction from Y to X.

But I'd really don't know how to use this..

Joel, I should have mentioned that the definition of NP-complete depends on whether we are using (polynomial-time) "Turing reductions" or "many-one reductions".  If we are using Turing reductions, then the exceptions are not needed.  With many-one reductions, the exceptions are needed.  The most common definition of NP-complete uses many-one reductions.

Stewie, let's pick two concrete languages.  Suppose that $ L$ is the set of all strings that begin with a $ 0$.  Suppose that $ M$ is the set of nonnegative integers (in binary notation) that are multiples of $ 3$.  Show that there is a polynomial-time, many-one reduction of $ M$ to $ L$.  You will have to look up the definition of reduction to proceed.  Remember that $ M$ itself is a polynomial-time problem.


algebra
polynomial
inequalities
number theory solved
number theory

let $f(x)= x^{100} +600x^{99} + \sum{ a_{k}.x^k} $wherek<=99
 and f(7)>1 have 100 a real proof .
            prove that :  $\exists x >7  satisfying f(x)=0$

[quote="tranthanhnam"]let $f(x)= x^{100} +600x^{99} + \sum{ a_{k}.x^k} $wherek<=99
 and $f(7)>1 $have 100 a real proof .
            prove that :  $\exists x >7$  satisfying$ f(x)=0$[/quote]

[quote="tranthanhnam"][quote="tranthanhnam"]let $f(x)= x^{100} +600x^{99} + \sum{ a_{k}.x^k} $where$k\leq99$
 and $f(7)>1 $have 100 a real proof .
            prove that :  $\exists x_{o} >7$  satisfying$ f(x_{o})=0$[/quote][/quote]

Do you know what the EDIT button under your posts stands for? ;)
And I don't understand the problem itself also, even after your reposts.

I think it's asking to show that if a degree 100 polynomial of the given form (i.e. coefficient of $x^{100}$ is $1$, coefficient of $x^{99}$ is $600$) has all real roots and $f(7)>1$, then it must have a root $\alpha>7$.

(Although I'm assuming where it says $k\le99$ beside the sum that it should say $k<99$, otherwise I don't know what the point of the $600x^{99}$ term is.)

There is an error in the problem. Letting $f(x) = (x+6)^{100}$, we have $f(x) = x^{100} + 600x^{99} + \ldots$, and $f(7) = 13^{100} >1$, yet all the roots of the equation are $<7$.

I'm guessing the problem should be as follows:

Given a polynomial $f(x) = x^{100} - 600x^{99} + \ldots$ having $100$ real roots and satisfying $f(7) > 1$, prove that at least one of the roots is greater than $7$.

If this is the problem, it becomes a beautiful inequality question in disguise.

Let $a_1, a_2, \ldots , a_{100}$ be the $100$ real roots of polynomial. Note that by theory of equations, we have $\sum_{i=1}^{100}a_i = 600$. Suppose that all the roots of the polynomial are $<7$, then $f(7) > 1 \Rightarrow (7-a_1)(7-a_2) \cdots (7-a_{100}) > 1$. Note that $\frac{(7-a_1)+ (7-a_2) + \ldots + (7-a_{100}) }{100} = 1$.
$\Rightarrow \sqrt[100]{(7-a_1)(7-a_2) \cdots (7-a_{100})} > \frac{(7-a_1)+ (7-a_2) + \ldots + (7-a_{100}) }{100}$, which contradicts AM-GM inequality, since the $100$ terms $(7-a_1), (7-a_2),  \ldots, (7-a_{100})$ are all positive, from the assumption that all of the $a_i$ are $<7$. Done!  :lol:


rotation
LaTeX
geometry
geometric transformation

I would like to know if it is possible to rotate something in LaTeX without having to use an external library (graphicx for example). In particular, I'm wondering if you can rotate something using the forum's BBCode-LaTeX setup.

I can't see that being possible. The nearest you can get to it is to use the rotating package. This doesn't need the graphicx package and will rotate text and pictures. But you would have to persuade Valentin to include it in the forum code or else import a picture from elsewhere.
You could I suppose simulate rotation using the picture environment so $\setlength{\unitlength}{1cm}\begin{picture}(5,2) \put(0,0){\vector(1,0){3}}\end{picture}$ becomes $\setlength{\unitlength}{1cm}\begin{picture}(5,2) \put(0,0){\vector(2,1){2}}\end{picture}$ for example.


Find $ f(x): R\minus{}>R$ so that:
$ f(x)\plus{}f( \frac{x\minus{}1}{x})\equal{}x\plus{}1, x\neq0,1$

[hide]Plugging in $ x$, $ \frac {x \minus{} 1}{x}$, and $ \frac {1}{1 \minus{} x}$ yields the following:

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

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

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

Adding the three equations and dividing by $ 2$ yields:

$ f(x) \plus{} f\left(\frac {x \minus{} 1}{x}\right) \plus{} f\left(\frac {1}{1 \minus{} x}\right) \equal{} \frac {1}{2}\left(x \plus{} \frac {x \minus{} 1}{x} \plus{} \frac {1}{1 \minus{} x} \plus{} 3 \right)$

Subtracting the second equation gives:

$ f(x) \equal{} \frac {1}{2}\left( x \minus{} \frac {x \minus{} 1}{x} \plus{} \frac {1}{1 \minus{} x} \plus{} 1 \right)$[/hide]

EDIT: BTW, I believe that is Day 1, Problem 1 of [url=http://www.artofproblemsolving.com/Forum/resources.php?c=64&cid=54&year=2004]Germany TST 2004[/url]

we can do it byreplacing x by 1-1/x , 1/x , 1/1-x.................but oops............my answer doesnt match with penguins :maybe:  :maybe:  :maybe:

i did the same way penguin did.................sorry i over looked it :(  :(  :( .......and answer...........i got it now


This is a simple task for students who've just learnt continuity:

A function $f : \mathbb{Q} \rightarrow \mathbb{R}$ is called $\mathbb{Q}$-continuous iff for any $a,(a_n) \in \mathbb{Q}$ with $\lim_{n\rightarrow \infty} a_n=a$ we have $\lim_{n\rightarrow \infty} f(a_n)=f(a)$.
Also we must have for any $a \in \mathbb{R}$ and $(a_n)\rightarrow a$, $(a_n) \in \mathbb{Q}$ $|a_n|$ is bounded
A function $f: \mathbb{Q}\rightarrow \mathbb{R}$ is called $\mathbb{R}$-continuous iff for any $a \in \mathbb{R}$ and $(a_n),(b_n) \in \mathbb{Q}$ with
 $\lim_{n\rightarrow \infty} a_n=\lim_{n\rightarrow \infty} b_n=a$ we have
$\lim_{n\rightarrow \infty} f(a_n)=\lim_{n\rightarrow \infty} f(b_n)=l$ where $l$ exists and is finite.
(so that actually, $f$ can be uniquely be extended from $\mathbb{Q}$ to $\mathbb{R}$ to be continuous)

Prove that any $\mathbb{Q}$-continuos function is $\mathbb{R}$-continuous

[Edit: edited according to Pierre's observation below, indee the limit must be finite]

Isn't there some hypothesis missing about the existence of the limits or about their 'values'?
I mean : consider the function $f(x) = \frac 1 {x - \sqrt 2}.$
It is clearly $\mathbb {Q}$-continuous.
But if $(a_n),(b_n)$ both have limit $\sqrt 2$ with $a_n < \sqrt 2 < b_n$ for all $n$ the $f(a_n)$ has limit $- \infty$ and $f(b_n)$ has limit $+ \infty.$ Thus it is not $\mathbb {R}$-continuous.

Pierre.

It doesn't work.

Consider $f(x)=\{1\,\text{if}\,x>\sqrt{2}\atop{ 0\,\text{if}\,x<\sqrt{2}}$

The only way to get continuous extensions in general is to start with a closed domain. The rational numbers have far too many holes for this to work.

Here's an example to really kill the idea:

Let $g$ be the Cantor "devil's staircase" function, extended to a function from $\mathbb{R}$ to $\mathbb{R}$ by the relation $g(x+n)=g(x)+n$ for all integers $n$.
Define $f(x)=g^{-1}(\pi x)$ for $x$ rational. We can show that $f$ is $\mathbb{Q}$-continuous everywhere, but cannot be extended to an $\mathbb{R}$-continuous function on any interval. We even have that $f$ is strictly increasing, but its natural extension has a dense set of jump discontinuities; $g^{-1}$ is discontinuous at every non-integer dyadic rational and continuous elsewhere.

The key to this topologically is that the closure of the image of $f$ is a totally disconnected set; the Intermediate Value theorem says that a continuous function from a connected set such as an interval of $\mathbb{R}$ has a connected image.


prove that in any triangle:

p*sqrt(3)>=la+lb+lc.

Solution:

You have that la=2sqrt(bc)/(b+c)*sqrt(p*(p-a))<=sqrt(p(p-a)) and then apply Cauchy.

cheers! :D :D

This problem was also given at the contest : "SIMION STOILOW" this year in Romania! cheers


Constructive Counting Part 1

We solve two problems using "constructive counting", which means finding a strategy for counting by considering how to construct an example of what we want to count.

[quote="minerva"]What if Ann chose another color other than black, say blue. Doesn't that mean Bob would only have 2 choices, rather than 3?[/quote]

No, Bob would still have three choices because he is only restricted by one person (i.e. Cam), leaving three possible shirt colors. If one chose to start with Ann, it would become a messy casework counting problem and we would still get the same answer (try it to prove it to yourself), but to make life much easier, we begin with Bob.

Just a point of interest, if the third and the fifth were kind enough to choose the same color, then the one sandwiched in the middle would have three choices. Nevertheless, Bob is only concerned with what color Cam wears and would still have three choices, so the net result would be that there 3^4 or 81 possibilities instead of 54.

Your drawings are actually really good. I can tell that you're good at Geometry.


Which of these compounds is not adequately represented by a valence bond model?

I - CO2, II - SO2, III - SIO2

A. I
B. II
C. I and III
D. II and III

How do you do this?

My answer would be D

CO2 is a standard linear molecule
SO2 has resonance, which valence bond theory cannot completely explain (MO theory explains the 1.5 bond order)
SiO2 has a network covalent shape because its p orbitals cannot effectively overlap, and so valence bond theory fails there too.

Thank you.  :)

I've a few other ones I'm having trouble with:

#35 Carbonic Acid H2CO3, is a diprotic acid for which K1 = 4.2x10^-7 and K2 = 46.x10^-11. Which solution will produce a pH closest to 9?

A) .1 M H2CO3
B) .1 M NaHCO3
C) .1 M Na2CO3
D) .1 M Na2CO3

#39 For which of these oxidation/reduction pairs will the reduction potential vary with pH?

I. AmO2^(2+)/AmO2^(+) II. AmO2^(2+)/Am^(4+) III. Am^(4+)/Am^(2+)

A) I
B) II
C) I and II
D) I, II, and III

hmm err.. the actual answer to 54 was B?!? I'm confuddled on that as well then. somebody explain!

#35:
The actual problem I believe is slightly different. I'll solve for the actual.

pK1=6.38
pK2=10.33

Hence A is already not plausible as it will be too acidic
For B, set up the equilibrium CO32- + H2O -> HCO3- + OH- and solve for OH-, get pH from that and you get pH 8.34
For C, this is (pK1+pK2)/2=8.36
For D, this is pK2=pH=10.33

Closest one is clearly C

#39:
If you wish for the potential to be affected by pH, then the overall Q_eq expression must have an H+ in it. For I and III, only electrons are needed to balance the equation so those are out. For II, The balanced reduction equation is $ 2e^\minus{} \plus{} 4H^\plus{} \plus{} AmO_2^{2\plus{}} \rightarrow Am^{4\plus{}} \plus{} 2H_2O$ So pH would indeed affect that. The answer is then B. (II only)

54 is B because SiO2 can be drawn as O-Si-O, with 2 lone pairs on Si.

$ \ce{SiO2}$ is nothing like $ \ce{CO2}$ because, as someone else mentioned, silicon dioxide forms a covalent network. Each $ \ce{Si}$ atom is $ sp^{3}$ hybridized because it is present in a tetrahedral environment.

let me explain myself...

in the problem, it's just not the answer

but, maybe they need to define adequately...


a)Aratati ca orice matrice $A \in M_{4}(C)$ se poate scrie ca suma a patru matrice
din $B_{i} \in M_{4}(C)$,i={1,2,3,4} de rang 1.
b)Aratati ca $I_{4}$ nu se poate scrie ca suma a mai putin de 4 matrici de rang 1.

Sursa: problema cat se poate de cunoscuta. Dar oricum, foarte frumoasa.

Totusi care este ideea pentru rezolvarea ei? :)

Hai sa o rezolvam pe cazul general: un endomorfism de rang r este suma a r endomorfisme de rang 1 ( identific aici matricile cu endomorfismele canonic asociate lor). Avem $ dim Im f=r$ unde f este endomorfismul nostru. Luand $ e_1,...,e_r$ o baza in imaginea lui $f$ pentru orice x putem scrie $ f(x)=x_1 e_1+...+x_r e_r$ cu $x_i$ din corpul in care consideram ca are elemente matricea. Atunci nu avem decat sa definim $ f_i(x)=x_i e_i$ si avem r endomorfisme de rang 1 cu suma f. Cred ca se poate deduce destul de repede si din faptul ca o matrice de rang r este echivalenta cu o matrice ce are r de 1 pe diagonala si restul elementelor 1. Punctul al doilea al problemei este imediat din faptul ca rangul sumei este cel mult egal cu suma rangurilor.


[quote="vess"]Let $f(x) = x - \frac{1}{x}$. For which $a \in \mathbb{R} \setminus \{0\}$ is the sequence $f^{(n)}(a)$ of iterations bounded? What happens for $a=2$?[/quote]
If you draw the function you can see that the only posible limits are the solutions of $2L^2-1=0$, I mean $L=\pm \frac 1 {\sqrt 2}$, and this ones are the only posible limits, because if $x_0>1$ then $0<f(x_0)<x_0$ and if $x_0<-1$ tehn $0>f(x_0)>x_0$ and if $|x_0|<\frac 1 {\sqrt 2}$ then $|f(x_0)|>1$ hence the only posibility is $f^2(x_0)=x_0$, and this is eaxctly the values I wrote before, so of course is bounded for every $a\ne 0$.

Best regards.

RX TCM

:?  :?  :?  I don't agree. Yes, of course $\lim_{n \to \infty} f^{(n)}(x)$ exists iff $x = \pm \frac{1}{\sqrt{2}}$, but how is this relevant? Why is, for instance, $\{f^{(n)}(2)\}_{n \geq 1}$ bounded? By [i]what[/i] is it bounded?

To [i]me[/i], it's an open question (I have absolutely no idea what is going on in this problem  :( ), and I don't know if this has actually been solved ... (though I think I have seen it formulated (but not solved) on the www in the case $a=2$), so this may well be an open question! (however strange that seems in such a simply looking problem).


Show that if 2 + 2 √(28 n2 + 1) is an integer, then it is a square (for n an integer).

If 2 + 2\sqrt(28n + 1) is an integer, then 28n + 1 must be an odd square.
Hence there exists a positive integer k such that 28n + 1 = (2k + 1).
Hence 7n = k(k + 1). Now k and k + 1 are relatively prime.
So there are two possibilities:
(1) k = a, k + 1 = 7b for certain positive integers a, b.
Then a + 1 = 7b. Hence a = 6 (mod 7) which is impossible.
(All squares equal 0, 1, 2, 4 mod 7.)
(2) k = 7a, k + 1 = b.
Since (1) is impossible, we must have case 2.

Hence 2 + 2\sqrt(28n + 1) = 2 + 2(2k + 1) = 4k + 4 = 4b = (2b).
So we are done.


Let a circle be with equations $ x^2 \plus{} y^2 \equal{} 1$ and $ P$ be a point  on the circle in the first quadrant be with coordinates $ (a,b)$. Do $ c \equal{} \frac {a}{b}, b \neq 0$ constitute all rational numbers?

Yes.  Positive rational numbers.

This is of course provided that $ a$ and $ b$ are rational. Is there a proof for this or this is simply accepted as it is. I came across with an article but it is quite more complicated, so I simplified it.

[quote="guile"]This is of course provided that $ a$ and $ b$ are rational. [/quote]

No; for example, try achieving $ c \equal{} \frac ab \equal{} 1$.

Parameterizing in terms of the angle with the x-axis, we have $ c \equal{} \cot \theta, 0 < \theta < \frac {\pi}{2}$, and all positive real numbers can be achieved (provided $ a$, $ b$ are positive reals).

By Pythagoren triples, a and b can be written as $ \frac{2pq}{p^2\plus{}q^2}$ and $ \frac{p^2\minus{}q^2}{p^2\plus{}q^2}$, so $ \frac{a}{b}\equal{}1$ (from azjps) $ \implies 2pq\equal{}p^2\minus{}q^2 \implies 2q^2\equal{}(p\minus{}q)^2$, but obviously one square can't be twice another.

Definitely $ a$ and $ b$ should be rational since our radius is only 1 unit. Does this mean that if $ a$ and $ b$ are rational, it constructs all rational numbers except 1? Or are there any other that is inconstrible?

Letting $ t \equal{} \frac{q}{p}$ gives $ \frac{a}{b} \equal{} \frac{2t}{1 \minus{} t^2}$.  Suppose that this is equal to some rational number $ r$.  Then

$ 2t \equal{} (1 \minus{} t^2)r \Leftrightarrow$
$ rt^2 \plus{} 2t \minus{} r \equal{} 0$

has discriminant the square of a rational; the discriminant is given by $ 4 \plus{} 4r^2$, so $ r$ is a possible value of $ \frac{a}{b}$ if and only if there exists some rational $ s$ such that $ 1 \plus{} r^2 \equal{} s^2$.  In particular, no integer values of $ r$ except $ 0$ are possible.


First Day – Budapest, April 18
Find all sequences $x_1,\cdot\cdot,x_n$ of distinct positive integers such that $\frac{1}{2}=\frac{1}{x_1^2}+\frac{1}{x_2^2}+\cdot\cdot+\frac{1}{x_n^2}$.


induction
geometry
rectangle
combinatorics proposed
combinatorics

Several rooks are arranged on a $8 \times 8$ chessboards in such a way that every rook attacks even number of other rooks. Find the maximal number of rooks.

You can put $64$ rooks on the board (one on each square), and every rook would attack $14$ others, so an even number. Or maybe by "attack" you mean something else? Perhaps a rook attacks another rook iff they are on the same row-column but no other rooks are between them? :?

You should specify stuff like this :).

Umm, in chess, rooks cannot attack over pieces, so no 64 is wrong...

[hide="answer?"]
32?
[/hide]

I think the answer is 40. Constellation: ('x' are the rooks)
[b]
oooxxooo
ooxxxxoo
oxxxxxxo
xxxxxxxx
xxxxxxxx
oxxxxxxo
ooxxxxoo
oooxxooo
[/b]

But I don't see how to prove that's best...

[quote="grobber"]Perhaps a rook attacks another rook iff they are on the same row-column but no other rooks are between them? :?[/quote]
Yes.

I think we can show by a sort of induction that for a maximal constellation, for $k\le n/2$, the $k$th row (or column), counting from either side of an nxn board, cannot have more than $2k$ rooks in it. The idea is as following: Suppose say the uppermost row has $m>2$ rooks, then the $m-2$ middle ones cannot have any other rook in their respective columns beneath. So those columns are "wasted." Likewise more generally,if in a maximal constellation a row (or column) has $m$ rooks, the adjacent one(s) must have between $m-2$ and $m+2$ rooks each.
The numbers $2k$ can be attained by the diamond-like constellation as indicated for $n=8$. For odd $n$, the best constellation would be similar to the following for $n=5$:
[b]
ooxxo
oxxxx
xxxxx
xxxxo
oxxoo
[/b]
There are rooks everywhere except two triangles of side $\frac{n-1}2$ on two opposite corners of the board and two of side $\frac{n-3}2$ on the other two opposite corners.
i.e. for an $n$x$n$ board, best possible is $\frac{n(n+1)}2$ for $n$ even and $\frac{n^2+2n-1}2$ for $n$ odd.

Now let's ask the same question for an nxnxn cube. I have the impression we cannot put significantly more rooks there then in a nxn square  :?  Still working on that one.  :)

And the same question for an $m$x$n$ rectangle??? And for a $m$x$n$x$r$ cuboid :?:


Solve the system of the equations:
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}9$
$ x^{4}\plus{}y^{4}\plus{}z^{4}\equal{}33$
$ xyz\equal{}\minus{}4$

We can use elementary symmetric equations: $ \sigma_{1}^{4}\minus{}4\sigma_{2}\sigma_{1}^{2}\plus{}4\sigma_{3}\sigma_{1}\plus{}2\sigma_{2}^{2}\equal{}33$, $ \sigma_{1}^{2}\minus{}2\sigma_{2}\equal{}9$, $ \sigma_{3}\equal{}\minus{}4$.

Trivially, we can find a 4th degree polynomial in $ \sigma_{1}$. We get that $ \sigma_{1}\equal{}3,1,\minus{}5$.
It follows that $ (\sigma_{1},\sigma_{2},\sigma_{3})\equal{}(3,0,\minus{}4),(1,\minus{}4,\minus{}4),(\minus{}5,8,\minus{}4)$.

Note that the polynomial: $ t^{3}\minus{}\sigma_{1}t^{2}\plus{}\sigma_{2}t\minus{}\sigma_{3}\equal{}0$ has roots $ x$, $ y$, and $ z$.

We get the solutions $ (x,y,z)\equal{}(2,2,\minus{}1), (2,1,\minus{}2), (\minus{}1,\minus{}2,\minus{}2)$ and permutations.


The person who has them doesn't have an AoPS account, but I'm emailing him now.

Semi-on topic: what did people think of the Penn State song contest?  I thought Central Jersey was great on Friday night, but they seemed to lose the audience on Saturday.  AAST did have a good parody, though. And although they weren't going to win, why was Maine trying to sing in a huge basketball arena without a microphone?

Does the singing at ARML happen every year?

Yes, it does.  Although I heard that last year they canceled it at Las Vegas because it was too hot (am I right about that?)

It wasn't on the schedule this year. We talked Paul Dreyer into it when he said it would be another 5 minutes before they had results. I said, "Well, in 5 minutes, we could have Jacob sing." And last year, after we retreated to the relative cool of the dining commons, some songs happened unofficially.

Okay, I don't know what I'm talking about.  Penn State is the only site I have experience with, and there it is a fairly prominent event.

[quote="MysticTerminator"]dude AAST got booed pretty badly

I suspect it is because everyone else was like wtf they are only winning because their school brought like 929834928347091834130487 teams[/quote]

yea,

our team was like VOTE FOR THE WORST so we were going to vote for AAST cuz they were the worst by far but we realized by the coaches' awards adn stuff already that only AAST votes for AAST so we voted for maine but that wasn't enough i guess :( but all three songs were pretty bad, the first one had no creativity, second one had no rhythm, third one was a trainwreck

but seriously, that coach of AAST had a huge cult following

Iowa has never had singing.  I don't really know why not, although the ballroom where the individuals and relays are might not be the greatest for song contest finals.  There is a room in Currier (dorm next to Burge) that would be perfect for Friday night prelims.  Penn State has had it for a long time, and it's a pretty big deal.  I think at Vegas it's been fairly informal in the past.  Next year, though, it might be made into a bigger event, complete with a preliminary round on Friday.

There were better acts on Friday than the 3 picked.  It seems like that guy picked them completely on his own which is a pretty horrible system.

I can't remember who did what, but one of them was decent and the other two were pretty bad.  The one that won was in the latter category.

Iowa might not have singing, but it does have a T-shirt contest (which as far as I know neither Penn State nor Las Vegas has).

I was pretty unhappy with the whole Penn State song contest.  First, the preliminaries started late and no one knew who was in charge (at least 6 people asked me where in the Commons it would be or who would start it off).  It wasn't clear who was doing the judging or what the criteria were. (Personally, I disagreed with the choice of teams to make it into the finals.)  Worst of all, for the preliminaries there was nothing approaching a sound system or any attempt to compensate for the lack of a sound system.

I'm sort of thinking of contacting ARML and volunteering to take charge of the song contest, but I don't want to step on anyone's toes---for all I know, the person who organizes it now really likes doing it, and was just frazzled this time.

[quote="Phoenix257"]"It ain't gonna be me" was stinkin brilliant!!!!  Can someone from that team please post the lyrics?[/quote]
albert's teapot song pwned that  :P

were we supposed to have voting to decide the winner? because i don't think we did at las vegas

[quote="MysticTerminator"]dude AAST got booed pretty badly

I suspect it is because everyone else was like wtf they are only winning because their school brought like 929834928347091834130487 teams[/quote]

For once, I agree. AAST didn't do too well at all; and they did bring a lot of team members (I saw an entire row from top to bottom of width of about 7 people of only AAST). And their song wasn't too good either; they kept messing up. The group with the YMCA wasn't too shabby though.

The traditions of the song contest are very different between the western ans eastern site- it's always been very informal at Las Vegas/San Jose, with just a judge or two to decide a winner.

There is also a "Best T-shirt" award at the western site, handled the same way.

[quote="smbelcas"]Iowa might not have singing, but it does have a T-shirt contest (which as far as I know neither Penn State nor Las Vegas has). [/quote]
When the west was in San Jose, we used to have both a song contest and a T-shirt contest. While San Jose had certain problems having to do with transportation and dorm cost, what it did have was a spectacular room for the Saturday events. If you go to http://www.arml.com and click on the "About ARML" button, you'll get a rotating slide show. The very first picture shows that auditorium at San Jose. That, and the fact that we'd only get 10-15 teams, seemed to lend itself to us having lots of time for the super relay, a song contest, and a T-shirt contest, as well as such things as a "most spirited team" award, which was awarded on entirely arbitrary criteria.

The one year we should have won the best T-shirt contest, they'd already given away that particular bucket of candy to Macky who did his juggling and cubing show during the song contest time.

Since we've moved back to Las Vegas, we still haven't had a really good room for the Saturday events (although this year's room was much, much better than last year's!) I think by next year, they'll have the rest of the student union finished and we might finally have enough space.

Edit: much of jmerry's post should be read in the past tense - the last ARML he came to was in San Jose.

[quote="junggi"]were we supposed to have voting to decide the winner? because i don't think we did at las vegas[/quote]

No, it was just to fill time.  Next year, though, we might.

The Saturday venue this year felt a lot like state Mathcounts.  It was great, especially for the number of students we had.  But it didn't really have the feel of competing for the national championship.  Sounds like that will change next year, though.

[quote="Sly Si"][quote="junggi"]were we supposed to have voting to decide the winner? because i don't think we did at las vegas[/quote]

No, it was just to fill time.  Next year, though, we might.[/quote]
we really should...then we could annually get the our team to sing the teapot song and win every year

[quote="Ubemaya"]What was the best song from the Las Vegas site?[/quote]
yours, probably

The songs at Penn State weren't as good as last year's.

The YMCA one was creative, but it didn't really hook me.  The second one had a lot of pauses, possibly from forgetting the lyrics.  And I couldn't hear the third one too well, despite the fact that I was sitting relatively close to the "stage".

I can now finally answer the original poster of this topic.

[quote="Jacob Levine"]It Ain't Gonna Be Me (sung to the tune of It's Gonna Be Me)

 I hate equations
 It ain't no lie
 With numbers I'm just so slow
 I wasn't kidding
 When I said that math makes me want to die
 Maybe that's why

 Chorus:
 Even adding two plus two
 Tends to get me so confused
 I don't want to fail math again
 But I can barely count to ten
 Baby if you want a guy
 Who can multiply
 Guess what?
 It ain't gonna be me

 Don't mention fractions
 Or even pi
 Unless you want to bore
 'Cause I'm just too dumb to see
 That 3! doesn't mean THREE!
 I can't deny
 I'm not a smart guy

 (Chorus)
[/quote]
Of course Jacob also had some visual aids to go along with it, and Paul Dreyer (the site coordinator) was getting into the spirit of the gestures somewhere behind him.

[quote="E^(pi*i)=-1"]Actually, I thought that the songs picked for finals at Penn State were reasonable.  "These Are a Few of My Favorite Theorems", by a team I forget, was pretty good.  And Lehigh Valley's parody of "White and Nerdy" was clever, though no one understand what the guy was saying.  One of them could have replaced Maine.  But Central Jersey and AAST were the two best in my opinion.  While the huge AAST contingent (8 teams?!?!) helped, Central Jersey's song was so much worse on Saturday that I'm not sure it mattered.

I'm not sure what would have happened with a Penn State T-shirt contest.  Upstate New York's "Armlican Idol" shirt was certainly the corniest, though.[/quote]

 :D  :D I wrote "my favorite theorems" for the most part (we did one verse at HMMT as a team, then I wrote the rest minus a bit).  It was Upstate NY.  And our shirt design was...yeah.  Apparently there were 5 designs submitted, one from me, one from Mike Gottlieb (both to the effect of "We're number 1", which is what we did last year), and three from our coach.  The first email he sent to us to say that we were on the team had subject line "This...is ARMLican Idol" and somehow that became our t-shirt.

I was cheering for Maine in the song contest finals just because I didn't really like Central Jersey's song all that much and because I couldn't understand a bit of what AAST was singing.  But we should have won :D

If Andre and Chris sang, we (Upstate New York) could have won! AAST's song was good, I guess, but Maine should have won.

I was going to sing a song by myself, which Tim wrote, but we decided not to. Maybe next year.

[quote="E^(pi*i)=-1"]Actually, I thought that the songs picked for finals at Penn State were reasonable.  "These Are a Few of My Favorite Theorems", by a team I forget, was pretty good.  And Lehigh Valley's parody of "White and Nerdy" was clever, though no one understand what the guy was saying.  One of them could have replaced Maine.  But Central Jersey and AAST were the two best in my opinion.  While the huge AAST contingent (8 teams?!?!) helped, Central Jersey's song was so much worse on Saturday that I'm not sure it mattered.

I'm not sure what would have happened with a Penn State T-shirt contest.  Upstate New York's "Armlican Idol" shirt was certainly the corniest, though.[/quote]

umm wtf??
please learn a little bit about music before commenting
im pretty sure the songs were
YMCA (WOW REPLACING YMCA WITH ARML, CREATIVE!!) - Doesn't need an artist
Remember the Name - Fort Minor (uh yea. lets be uhhhh damn i forget my lyrics off beat and stuff)
Bad Day - that Powter dude (YAY FOR THE AAST CULT)

so yeah Maine (YMCA??) won by far. AAST was the worst but their cult saved them

I don't know how the songs were on friday (my team arrived at midnight) but the songs were not too great on saturday. My team cheered for Maine and they would have been  better with a microphone. (or maybe it doesn't matter since everyone knows the YMCA song...)
Central Jersey team forgot the lyrics and the AAST was not vert good either, and it was almost impossible to hear the lyrics so well.... whatever, hopefully it's better next year.

hmm i think the songs at arml were kinda bad y'know but they were ok for math songs i guess...

first of all i couldn't hear a word they were saying
second of all they were out of tune

and 3rd

NC LOST

WHATTTTTTTTTTTTTTTTTTTTTTT


LaTeX
limit
trigonometry
Euler
real analysis
real analysis unsolved

hello, in LaTeX looks your problem like
Evaluate the limit $ \lim_{n \to \infty}\sum_{k\equal{}1}^n\sin\left(\frac{k}{n^2}\right)$.
Sonnhard.

Not very hard.
Answer...
[hide]0.5[/hide]

You may explicitly evaluate the given sum via trigonometry, Euler formula, etc. This reduces the problem to a simple limit.

Or, note that $ \sin x \equal{} x \plus{} O(x^3)$ as $ x \to 0$. This shows $ \sum_{k \equal{} 1}^{n} \sin \left( \frac {k}{n^2} \right) \equal{} \sum_{k \equal{} 1}^{n} \frac {k}{n^2} \plus{} O \left( \frac {1}{n^2} \right)$, which immediately leads to the answer.


Acabo de recibir la informacion de los resultados de
la seleccion de PERU:

PER1: Daniel Soncco -------- ORO-------- 57 ptos. (máx. puntaje)
PER2: Franco Vargas -------- ORO ------- 52 ptos.
PER3: Maria Huanuco -------- PLATA
PER4: Freddy Ochoa --------- PLATA

PD: subir informacion de los otros equipos....
y cuadros de merito generales ....
gracias

Carlos Bravo ;)

Lima - peru

last year i was participating in the cono sur, and Daniel Soncco was the first gold medal (with 53 points)
congratulations to him, and to the peruvian team!

you know about the brazilian results? (last years we win with 2 golds and 2 silver medals  :lol: )

(sorry, but i don't speak spanish very well  :blush: )

Ah si ...
Daniel Soncco...
 un muy buen estudiante ...
pero no es el unico eh??

Saludos para los chicos de Brasil ....
un pais con mucho empuje ;)

Carlos Bravo ;)
Lima peru

[quote="e.lopes"]

(sorry, but i don't speak spanish very well  :blush: )[/quote]
I think that you can post here in portuguese... the spanish communitie will understand you.... portuguese and spanish are so similar.

$Tipe$

[quote="e.lopes"]yes, all of peruvian guys was medallists in last IMO! (Daniel Soncco was bronze medal, with one point to the silver  :( )

congratulations to them!  :)

and, about the brazilian results in cono sur, you know anything?

thanks![/quote]

BUENO ESO ES CIERTO, E.LOPES..
PODES ESCRIBIR EN PORTUGUESE SIN PROBLEMA ..

AH Y RESPECTO A TU PREGUNTA SOBRE LOS RESULTADOS DE BRAZIL ..
BUENO AUN NO TENEMOS LA INFORMACION DE LOS OTROS EQUIPOS ....

PERO NO DUDES QUE SI OBTENEMOSA ALGUNA  INFORMACION LA
POSTEAREMOS DE IGUAL MANERA ;)

SALUDOS

CARLOS BRAVO ;)

[quote="carlosbr"][quote="e.lopes"]agora eu tenho informações sobre o resultado:

           BRA1  10  10  7  10   9  10     56  Ouro
           BRA2   1  10   5  10   9   2     37  Prata
           BRA3  10   0   4  10  10  2     36  Prata
           BRA4  10  10  4  10   9   0     43  Prata

 :lol:[/quote]

Muy buen trabajo de parte del equipo Brasileiro ;)

Saludos ........[/quote]

gracias, e mais uma vez, saludos à equipe peruana!

[quote="adanluna"]SALUDOS A TODOS ESPERO QUE JOSE CUELGUE LOS RESULTADOS DEL EQUIPO ARGENTINO EN ESTAS ULTIMAS OLIMPIADAS  DEL CONO SUR  EN (ESCOBAR -ARGENTINA)[/quote]

Acá está lo que sé!!!

Fernando M. Vidal,  Villa Cañás, Santa Fe - Medalla de Plata
Mauro Schilman, Cdad de Córdoba - Medalla de Plata.
Pablo Nicolás Zimmermann, Rosario - Santa Fe - Medalla de Bronce.
Nicolás Ponieman,  Cdad de Buenos Aires - Medalla de Bronce

[quote="Guía Rojo"][quote="José"][quote="M4RI0"]Alguien sabe donde se puede encontrar los resultados completos?[/quote]

El torneo tiene página oficial?[/quote]
no...  :(

me tuve que comunicar con Chen para saber cómo anduvo la cosa por ahí...

felicitaciones al equipo peruano...  :)[/quote]

entonces la verdad que no sé de donde sacar los resultados oficiales


calculus
integration
function
geometry
3D geometry
sphere
calculus computations

I dont know how to start with this problem,  :blush: any help would be appreciate. :)

Suppose that f(X,Y,Z)=g((X^2+Y^2+Z^2)^(1/2)) ,where g is a function of one variable such that g(2)=-5. Evaluate the surface integral of f over the surface S, where S is the sphere X^2+Y^2+Z^2=4

Note that you've been told that the function you want to integrate is constant (and takes the value -5) on the set you're integrating over. The integral of a constant should always be that constant times the "measure" of the set you're integrating over. What do I mean by "measure"? If it were an ordinary integral over an interval, that would be the length of the interval. In this case, since it's a surface integral, the "measure" of the set you're integrating over should be the surface area of that set.

That should be enough for you to figure it out. No actual integration should be necessary.


induction
number theory unsolved
number theory

[u][b]Problem:[/b][/u]A large number of cards are each given a number in the set {1, 2, ... , n} in such a way that the sum of the numbers on all the cards is k·n! for some positive integer k. Show that the cards can be divided into k groups, so that the sum of the numbers on the cards in each group is n! .

I don't know why this problem took so much time from me :(

The main difficulty of this problem is to check the base of induction $n=1$. :lol:

Place cards with numbers 1, 2, ..., $n-1$ into a row. Consider groups which consists of 1, 2, ..., $n$ first cards (if we have so many cards, of course ;) ). Then we will find two groups which sum up to the same number modulo $n$. It means that we can find group of at most $n$ cards which sums up to $mn$, where $m\leq n-1$. Join correspondig cards into one mega card $n\cdot m$ and exclude them from consideration :) We will continue this process while we have cards with numbers 1, 2, ..., $n-1$. Since the whole sum is divisible by $n$, finally we will join all cards into mega cards $n\cdot 1$, $n\cdot 2$, ..., $n\cdot n-1$. Let $a_i$ be the number of mega cards $n\cdot i$ then $\sum_{i=1}^{n-1}a_ini=k\cdot n!$, therefore $\sum_{i=1}^{n-1}a_ii=k\cdot (n-1)!$. Hence, applying induction, we obtain the result :)


Express the following summation in terms of rational numbers:

\[ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{6^n}{\left( 3^n\minus{}2^n \right) \left( 3^{n\plus{}1}\minus{}2^{n\plus{}1} \right)} \]

\[ \sum_{n\equal{}1}^m\frac{6^n}{\left( 3^n\minus{}2^n \right) \left( 3^{n\plus{}1}\minus{}2^{n\plus{}1} \right)}  \equal{}  \sum_{n\equal{}1}^m\left(\frac{2^n}{3^n\minus{}2^n}\minus{}\frac{2^{n\plus{}1}}{3^{n\plus{}1}\minus{}2^{n\plus{}1}}\right)
 \equal{} 2\minus{}\frac{2^{m\plus{}1}}{3^{m\plus{}1}\minus{}2^{m\plus{}1}}\]
Now, sending $ m$ to infinity we obtain that the limit is $ 2$.


Take a random irrational number such as $\pi=3.141596535...$, and add up the first $k$ digits after the decimal point. For example, with $k=5$, 1+4+1+5+9=20. What is the probability that, for some $k$, this sum is 2005? The answer is closest to:

A)0
B) 1/10
C) 1/9
D) 1/5
E) 2/9

I think zero. proof I made was too wordy...

basically if you add up any number k ordered decimals of Pi, to form the sum n, the series is divergent as the decimals are always positive digits, and are never ending so cannot be simplified to end with 000000.... (which would end the summation). since k can be any of an infinite number of choices, you can never randomly get a number k which will allow n = 2005.

Either that or I am misinterpreting the question, but I doubt you'd be required to show that the sum of the first kth decimals cannot equal 2005.

I think the wording of this problem isn't that great.  I think something like the following would have been a clearer problem statement:

Take a random irrational number such as $\pi=3.141592653\ldots.$  Take the sum of the first digit after the decimal place, the first two digits after the decimal place, the first three digits after the decimal place and so on.  So, for example, we would take the sums $1$, $1+4=5$, $1+4+1=6$, $1+4+1+5=11$, etc.  What is the probability that one of these sums is equal to 2005?

At first, I interpreted it like degreestudent did.  But that's not what they want.

Take the last sum less than 2005, call it n.  The next digit can't be 0, because then the next sum would also be less than 2005. The difference between 2005 and n is a number 1-9 and the probability is 1/9 that the next digit will be this number.

Unfortunately, that idea doesn't work.  For example, suppose that $n$ is 2000.  Then the next digit is not uniformly distributed from 1 to 9; in particular, the next digit can't be 1, 2, 3, or 4.

Here's the sketch of another approach:

[hide="Hint"]First, we can ignore 0's.  Let's look at the first million nonzero digits.  What do they add up to (on the average)?  How many different partial sums do they generate?  From the answers to those two questions, we can estimate what fraction of integers are equal to a partial sum.[/hide]

[quote="Ravi B"]Unfortunately, that idea doesn't work.  For example, suppose that $n$ is 2000.  Then the next digit is not uniformly distributed from 1 to 9; in particular, the next digit can't be 1, 2, 3, or 4.[/quote]Um....why not?  The digits aren't random?  :?:

No, he's not saying that the digits aren't random.  He is saying that all of the values of $n$, the last sum before the sum is $\ge 2005$, are not equally likely to result in a sum of 2005 on the next addition of a digit.

Right.  For example, if $n$ (the last sum before 2005) happens to be 2000, then the next digit can't be 1, or else 2000 wouldn't be the last sum before 2005.  Once we condition on a particular value of $n$, the next digit is no longer uniformly random.


You are given a soap solution together with the usual apparatus found in the lab. You are also given 3 samples of water. Design an experiment that allows you to to compare the degree of hardness. Explain your procedure

The hardest the water, less foam we will obtain and more precipitate we will see.

Your concept is right...
But give me  the procedure

Isn't the procedure obvious? Just add some soap solution to each sample of water and shake it. Then you already know what observations you might expect.

But i want you to compare the relative degree of hardness

Sorry, but I cannot understand no longer where your doubt is.

that is comparing the relative hardness........
this is done by taking the soap solution in the burette and water in the conical flask....
one which consumes greater amount of soap solution to form the permanent l;ther is the hardest...
in this way we can compare the relative hardness(only approximately)


Please, I really need this answer by today (before 12 hrs from now), for my take-home-quiz..
-----------------------------------------------------------------

[size=100]A quadrilateral has vertices K(-1,4) L(2,2) M(0,-1) and N(-3,1). Verify that:
[b]
a) a quadrilateral is a square[/b]

[b]b) each diagonal of the quadrilateral is the perpendicular bisector of the other diagonal[/b]

[b]c) the diagonals of the quadrilateral are equal in length[/b][/size]

(1): use $|P_1P_2|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ to find the side length, then use $k_1k_2=-1$ to prove that the angles are $90^o$.

(2)(3): When we know the quadrilateral is a square, we should imply these two properties immediatly.

Welcome tamilan.

Just to tell you that from now on, be aware that AoPS is not a place you ask for your homework or take-home quiz. This is a place for enthusiastic math students, professors, and coaches and please DO NOT post homework problem.

I notice this is your third post so please don't do it again.

Thanks.

HI shobber, I have never learned those formulas in my math class, do you know any easier or pre-teached forumals i can work with for this problem?

Use: $a^2+b^2=c^2$ to prove the distance formula.

you guys should not be doing his quiz...and you need to learn how to do these problems if you are going to do well in class, but the easiest way to show this would be to show that all the sides are equal in length, and one of the angles is a right angle, then once you have a, b and c follow immediately (unless your teacher has a different style than mine did)


Opposite sides of a regular hexagon are 12 inches apart.  What is the length of each side?

[hide]First, we need to find the length of the diagonal. Create a 30,60,90 triangle in the hexagon with 12 as the longer leg.

Because we know that ratios of the sides of a 30,60, 90 triangle, the hypotenuse is $8\sqrt{3}$.

Now, divide the hexagon into 6 equilateral triangles with side length of half the hypotenuse, or $4\sqrt{3}$. Finally, find the area of one of the equilateral triangles and multiply by 6.

The answer is $\72\sqrt{3}$

[/hide]


algebra
function
domain
calculus
calculus computations

If f'(x)=0 for all x in the domain of f, then f is a constant function.

Is the above statement true or false, and why?

Thanks!

False, if the domain of $ f$ is not an interval.

Let $ f(x)\equal{}\begin{cases}1&x>0\\0&x<0\end{cases}$ with $ x\equal{}0$ lying outside the domain of $ f.$

If the domain of $ f$ is an interval, then see what the Mean Value Theorem can do for you.

[quote="Kent Merryfield"]False, if the domain of $ f$ is not an interval.[/quote]

[quote="Kent Merryfield"]If the domain of $ f$ is an interval, then see what the Mean Value Theorem can do for you.[/quote]

I get the example case you gave me, but I don't get what's above.  Could you please explain it a bit more?  Thanks a lot!

Suppose the function is defined on an interval but is not constant.  Then, by the definition of non-constant, [fill in the blank].  Then by the Mean Value Theorem, [fill in the blank].


Given that $ \minus{}\pi < x < \pi$, solve each of the following inequalities.
1) $ \cos{\left(2x\minus{}\frac{\pi}{6} \right)} > \frac{\sqrt{3}}{2}$
2) $ \cos{\left(\frac{x}{2} \plus{} \pi \right)} \plus{} \frac{1}{\sqrt{2}} > 0$

Without graphing everything, is there a general way to do these types of problems?

Thanks in Advance

Sure; figure out where equality holds, and think about whether cosine is increasing or decreasing near the point of equality.


topology
advanced fields
advanced fields unsolved

(I don't have the English book, so I translate the problem from Chinese back to English)

$X=A_1\cup A_2\cup \cdots$, where $A_n\subseteq A^\circ_{n+1}$ for all $n$.
$f: X\to Y$ is map, satisfying $f|_{A_n}: A_n\to Y$ is continuous with regard to the subspace topology on $A_n$ for all $n$.
Show that $f$ is continuous.

Notations: $S^\circ$ means the interior of $S$; $f|_{A_n}$ means the restriction of $f$ on $A_n$.

Consider an open set $V$ of $Y$, then $f_{A_{n}}^{-1}(V)=A_{n} \cap U_{n}$for some open set $U_{n}$ of $X$ (this is just the definition of the induced topology). Now consider the open set $U=\cup_{n \geq 1} {U_{n}}$, then $f^{-1}(V)=\cup_{n \geq 1} {f_{A_{n}}^{-1}(V)}=\cup_{n \geq 1}{(A_{n} \cap U_{n})}=\cup_{n \geq 1}{(A_{n} \cap U)}=U$ since $U_{n} \cap A_{n} = U \cap A_{n}$ $(*)$. Whence $f$ is continuous on $X$.

N.B: I only used the fact that $A_{n} \subseteq A_{n+1}$ by asserting $(*)$ so maybe there's a flaw.

I thought I posted something earlier- oh well.

If you leave out the mention of the interior, this is false.

We actually need to show that if $U_n=U\cap A_n$ is open in $A_n$ for all $n$, $U$ is open.
Now, let $B_n=\mathrm{Int }A_n$ and $V_n=U_n\cap B_n=U\cap B_n$. We clearly have that $V_n$ is open in $B_n$. Since $B_n$ is open, this means that $V_n$ is open. Now $\cup V_n\subset \cup U_n=U\subset\cup V_{n+1}$, so $U=\cup V_n$ is a union of open sets and hence open.

[quote="liyi"][quote="julien_santini"]since $U_{n} \cap A_{n} = U \cap A_{n}$[/quote]
Why this?[/quote]

That's what I suspected there's a flaw here ( I found it out when I tried to prove that argument ). jmerry's proof seems fine to me.


Thanks. I know only A.M.-G.M. Cauchy-Schwarz Jensen' Tchevichev Jacobstahl's inequalities. :lol:

[quote="kunny"]Let $a>0,\ b>0,\ c>0.$ Prove that $\frac{1}{a^{2}(b+c)}+\frac{1}{b^{2}(c+a)}+\frac{1}{c^{2}(a+b)}\geq \frac{3}{2abc}.$[/quote]

It is not difficult either.
$\sum \frac{bc}{a(b+c)}\geq \frac{3}{2}$
$\sum \frac{b^{2}c^{2}}{abc(b+c)}\geq \frac{3}{2}$
$\sum \frac{b^{2}c^{2}}{abc(b+c)}\geq \frac{(ab+bc+ca)^{2}}{2abc(a+b+c)}\geq \frac{3}{2}$
which is equivalent to
$(ab+bc+ca)^{2}\geq 3abc(a+b+c)$
but $(ab+bc+ca)^{2}= (ab)^{2}+(bc)^{2}+(ca)^{2}+2abc(a+b+c) \geq 3abc(a+b+c)$
 :)

Yes, my solution is exactly same as yours, Gibbenergy. :lol:

[quote="Albanian Eagle"]Isn't this just Nesbitt inequality ??? :wink:[/quote]

Wow, you are absolutely correct. Nesbit is with $x=ab$, $y=bc$, $z=ca$. You are so fast! :)


salam dustan.
ye khabare jadid:CD e olampiade riazi ham darumad.
tuye in CD ghesmathaye zir vojud dare:
1-soalate olampiade iran(moghaddamati,marhaleye avval va marhaleye dovvom)+rahe halle masael+soalate olampiade jahani(IMO)
2-amuzeshe kolliyeye matalebe morede bahs dar olampiade riazi be surate tabaghe bandi shode(tarkibiat:shomaresh,esteghra,navardayi,...,hendese,....)
3-hezaran soale olampiad dar shakhehaye mokhtalef hamrah ba hal ba ghabeliate tabaghe bandi(az nazare mozu, sathe soalat,manbae soalat,...)
4-ghashangtaribn ghesmate in CD ine ke soalate olampiade riazie 32 mantaghe az jahan az ebtedaye bargozari ta ebtedaye sale 2006 jamavari shode.mesle amrica,england,canada,IMO shortlist,irland,... .
az oonjayi ke kheili az soalate olampiadhaye iran az keshvarhaye dige entewkhab mishe, in CD mitune marjae monasebi baraye amadegi dar marhaleye avval va dovvome olampiad bashe.
5-vaje nameye takhassosie riazi(englisi be farsi va bar aks)
6-moarrefie manabe olampiade riazi ,tarikhcheye olampiad,bargozidegane olampiad,...
7-bish az 300 bazie fekri va sargarmi ke aksaran baziaye ghashangi hastand.
esme in narmafzar ((daneshpajooh)) hastesh va gheymatesh ham 10000 tomane ke bazi jaha ba takhfife 20%, 8000 toman midan.
in CD ro mitunid az banke ketab(tel:66410000) va entesharate khoshkhan va bazi jahaye dige tahiyye konid.narmafzar ham ruye XP bayad nasb beshe.
ruye in CD ham dabir haye matrahe olampiad kar kardan.masle aghayan servati,ahmadpur,alipur,sharifi,hajizade,safa,... .
man tosie mikonam kesayi ke mikhan olmpiad sherkat konan in CD ro bekharan.

ye sari ham be in link bezanid:
[url]http://www.havaasil.com[/url]


[hide="Very interesting"]I get $\frac{1}{2^{k-2}}$ (for $k \ge 2$ since $k=1$ gives 1). It's really weird and interesting. I don't know how to best explain how I get this, other than it can be illustrated like so:
$\frac12 + \frac14 + \frac18 + \cdots = 1$
$\ \ \ \ \ + \frac14 + \frac18 + \cdots = \frac12$
$\ \ \ \ \ \ \ \ \ \ \ + \frac18 + \cdots = \frac14$
And of course on the RHS we get 2 because of geometric series. When we use this method, the infinite stepping up and layering of sequences, it all depends where our starting point is, which is $\frac{1}{2^k}$. Thus we use the model above and our successive sums come out to be $\frac{1}{2^{k-1}}$, $\frac{1}{2^k}$, etc. which gives a final sum of $\frac{1}{2^{k-2}}$.[/hide]

Wait, I got something different. I did it like this.
[hide]Start with the inner summation: $\sum_{a_k=1}^{\infty} \frac{1}{2^{a_1+a_2+\ldots+a_k}}=\frac{1}{2^{a_1+a_2+\ldots+a_{k-1}+1}}+\frac{1}{2^{a_1+a_2+\ldots+a_{k-1}+2}}+\ldots=\frac{\frac{1}{2^{a_1+a_2+\ldots+a_{k-1}+1}}}{1-\frac{1}{2}}=\frac{1}{2^{a_1+a_2+\ldots+a_{k-1}}}$

So now we have $\sum_{a_1=1}^{\infty}\sum_{a_2=1}^{\infty}\sum_{a_3=1}^{\infty} \ldots \sum_{a_{k-1}=1}^{\infty} \frac{1}{2^{a_1+a_2+\ldots+a_{k-1}}}$, which becomes $\sum_{a_1=1}^{\infty}\sum_{a_2=1}^{\infty}\sum_{a_3=1}^{\infty} \ldots \sum_{a_{k-2}=1}^{\infty} \frac{1}{2^{a_1+a_2+\ldots+a_{k-2}}}$. The "collapse" of the summations leaves us with $\sum_{a_1=1}^{\infty} \frac{1}{2^{a_1}}=1$.[/hide]

That does make sense. Can you point out what I did wrong, because I don't see it. I would think that the terms would be something like what I wrote out, but maybe that was a mistake in the first place.

Yeah I'm trying to understand your method and see where it goes wrong

[hide="let me try..."]
for a given summation sign, all the other variables are const., so let's collectively call their sum $S$, the most-inner sum becomes $\frac{1}{2^S}$ by simple geometric sum formula, therefore we deduce that each summation eliminates the variable from the sum.  The answer is $1$.
[/hide]


Compute the limits:

a) lim(x-->oo) [(ax+b)/(ax-b)]
b) lim(x-->oo) [(ax-b)/(ax+b)], a,b pozitive reals.

where [y]=y-{y}.

are these limits 1?

thank you very much!

The first one is 1 since the stuff in the floor brackets will always exceed 1. The second one is 0 since the stuff in the floor brackets will eventually go below 1 and approach 1 from below.

Hmm... I can't really see how it goes to 0, can you explain it again?

... Never mind, I see.


[quote="Davron"]Lets talk about the Geometry problems of an IMO level please it will be very usefull.

In my opinion the geo problems of an IMO level are can be solven by finding one usefull point .

What about the others! What do you think it will be very useful.
Please imo contestants answer ... Darij Zetax Soarer Tarlan Pavel Myth campos ... all of you ...

Davron[/quote]

I doubt you're quite right :)
There are irregular solutions sometimes ;)

I think the IMO's geometry problems are quite nice usually :thumbup:

But there wasn't very difficult geometry problems on the IMO(except Emel'yanov's problem №6 on the IMO 2000)-max 2nd or 5th place. Main of them rather easy.


In $\triangle ABC$, let $I$ be the incenter, $O$ be the circumcenter, $H$ be the orthocenter, $R$ be the circumradius, $E$ be the midpoint of $OH$, $r$ be the inradius, and $s$ be the semiperimeter.

[b][color=blue](a)[/color][/b] Find the distance $IH$ in form of $R,r,s$
[b][color=blue](b)[/color][/b] Find the distance $IE$ in form of $R,r$

Using Leibniz theorem for the circumcenter $O,$ we obtain the relation

$OG^2=\frac{_1}{^3}(OA^2+OB^2+OC^2)-\frac{_1}{^9}(a^2+b^2+c^2)=R^2-\frac{_1}{^9}(a^2+b^2+c^2)$

Since $OG=\frac{_1}{^3}OH,$  it follows that $OH^2=9R^2-(a^2+b^2+c^2).$

Incircle $(I)$ and 9-point circle $(E)$ of $\triangle ABC$ are internally tangent $\Longrightarrow$ $IE=\frac{_1}{^2}R-r.$ Notice that $IE$ becomes the I-median of $\triangle IOH,$ therefore

$IE^2=\frac{_1}{^2}(IO^2+IH^2)-\frac{_1}{^4}OH^2 \Longrightarrow \ IH^2=2IE^2+\frac{_1}{^2}OH^2-IO^2$

$IH^2=2\left(\frac{R}{2}-r \right)^2+\frac{_9}{^2}R^2-\frac{_1}{^2}(a^2+b^2+c^2)-(R^2-2Rr)$

$IH^2=4R^2+2r^2-\frac{_1}{^2}(a^2+b^2+c^2)$

Because of $a^2+b^2+c^2=2s^2-2r^2-8Rr$ $\Longrightarrow$ $IH=\sqrt{4R^2+3r^2+4Rr-s^2}.$


parameterization
vector
linear algebra
advanced fields
advanced fields unsolved

Hello, can anyone help me with that ?

Let M^n be an oriented surface of class C^k. For each p in M define w(p) to be the element of volume from the tangent space of M at p. Show that if M is a hipersurface and phi:U->V is a parametrization around p, with p=phi(u), then w(p)=a(u)du1 V .. V dun
where a(u)=volume(dphi/du1,..,dphi/dun) and V denotes the wedge produt.

One def of a volume form is an n-form on M with magnitude one at each point, with respect to the metric.  Since du_1...du_n is an n-form on M, you just need to find the right normalization factor by computing <du_1...du_n,du_1...du_n>_g.

How to do this.

Some simplifying notation.  Let pu_i=\partial/\partial u_i, that is, the vector dual to du_i.  Then the right normalization will be the square root of <pu_1...pu_n,pu_1...pu_n>_g.  Now use pu_i=(d\phi/du_i).  (Actually, those are partials, of course.)  The rest follows from linear algebra (volume of a parallelpiped).

As usual with elementary diff geom, the hardest part is understanding how to abuse notation correctly.