trigonometry
inequalities
inequalities proposed

Prove that : In the triangle ABC :
 $ m_a.cos\frac{A}{2} +  m_b.cos\frac{B}{2} +  m_c.cos\frac{C}{2} \geq \frac{3}{4}(a+b+c) $  :)

Let $M$ be the midpoint of $BC$ and let $x_1$ and $x_2$ be the angles made by the median $AM$. $(x_1+x_2=A)$
Then $m_{a}.cos\frac{a}{2} \ge $ $ m_{a}.cos\frac{a}{2}.cos\frac{x_2-x_1}{2} $ $ = m_{a}.\frac{1}{2}(cosx_1 + cosx_2)$. Now let $P$ and $Q$ be the ortogonal projections of the point M on the sides $AB$ and $AC$ respectively. Then it's easy to notice that $m_{a}.cosx_2 = c- \frac{a}{2}.cosB$. Summing all the terms in $LHS$ and taking into consideration that $a.cosb+b.cosa=c$ we get the desired inequality.

Nice solution , Andrei  :lol:
 I solved this problem by applying the lemma :
 " Let ABC be a triangle , a point I is in the triangle ABC . $ <BIC = x , <AIC = y < AIB = z $ , M is an arbitrary point in the plane " , we have :
 $ MA.sinx + MB .siny + MC .sinz \ geq IA.sinx + IB .siny + IC .sinz $

 Let $ I\equiv I $ be the center of the inscribed circle , $ M\equiv G $ be the centroid of the triangle ABC , we can solve  my problem

Both solutions are great!
And here comes an ugly one:
We pass to dual principle.
It's easy to see that $m_a^2=x(x+y+z)+\frac {(y-z)^2}4$.
$\cos {\frac A2} =\sqrt {\frac {x(x+y+z)}{(x+y)(x+z)}}$
Now let's prove the following:
$m_a \cos {\frac A2} \geq p (\frac x{x+y}+ \frac x{x+z})$.
The inequality would follow then by summing the desired terms.
So, let's go.
1) $m_a-\sqrt{x(x+y+z)}=\frac {m_a^2-\sqrt{x(x+y+z)}^2}{m_a+\sqrt{x(x+y+z)}}=\frac {(y-z)^2}{4(m_a+\sqrt{x(x+y+z)})}$. Denote this expression by $E$.
2) Then we must have
$(\sqrt{x(x+y+z)}+E) \frac {\sqrt{x(x+y+z)}}{\sqrt{(x+y)(x+z)}}-p(\frac x{x+y}+\frac x{x+z})$ $=\frac {E \sqrt {x(x+y+z)}} {\sqrt {(x+y)(x+z)}} - $ $\frac 12 p(\sqrt{\frac x{x+y}}-$ $\sqrt{\frac x{x+z}})^2$.
We are to prove this expression is greater than zero.
3)However $(\sqrt{\frac x{x+y}}-\sqrt {\frac x{x+z}})=$ $\frac {\sqrt{x(x+z)}-\sqrt{x(x+y)}}{\sqrt{(x+y)(x+z)}}=$ $\frac
{x(x+z)-x(x+y)}{ (\sqrt{x(x+z)}+\sqrt{x(x+y)})sqrt{(x+y)(x+z)}}$.
Squaring we get $\frac {x^2 (y-z)^2}{ (\sqrt{x(x+z)}+\sqrt{x(x+y)})^2 (x+y)(x+z)}$
Substituting 1) and 3) into 2) we receive the monster
$\frac {\sqrt{x(x+y+z)}(y-z)^2}{4(m_a+\sqrt{x(x+y+z)})} $ $\geq \frac {x^2 (x+y+z)(y-z)^2}{2 (\sqrt{x(x+z)}+\sqrt{x(x+y)})^2 (x+y)(x+z)}$.
Now cancelling common terms and grouping leads to
$2 (m_a +\sqrt{x(x+y+z)}) $ $\sqrt {x(x+y+z)}$ $ \leq \sqrt {(x+y)(x+z)} $ $(\sqrt {x+y}+\sqrt {x+z})^2$.
However $sqrt{(x+y)(x+z)}\geq $ $\sqrt{x(x+y+z)}$ thus is suffices to prove that
$2(m_a+\sqrt{x(x+y+z)}) $ $\leq (\sqrt {x+y}+\sqrt {x+z})^2=$ $2x+y+z + 2\sqrt {(x+y)(x+z)}$.
This breaks into the inequality $2m_a<2x+y+z$ and $\sqrt{x(x+y+z)} \leq $ $\sqrt {(x+y)(x+z)}$ which are true (the first one is the well known $m_a\leq \frac {b+c}2$.
QED


The last one does not converge.  (It is the harmonic sequence without the 1/1 term.)


[quote="ABRORBEK"]If $ a$, $ b$, $ c$, $ d$are real numbers, then
$ 6(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} (a \plus{} b \plus{} c \plus{} d)^2\ge12(ab \plus{} bc \plus{} cd)$.[/quote]
$ LHS\minus{}RHS\equal{}(2a\plus{}2d \minus{}b\minus{}c)^2\plus{}3(a\plus{}c \minus{}b\minus{}d)^2\plus{}3(b\minus{}c)^2$ :D
[hide="how did I obtain that"]Was easy once I had found the equality case $ (1,2,2,1)$  :jump: [/hide]

it's really easy if you normalize to $ a\plus{}b\plus{}c\plus{}d\equal{}6$

Then you can rewrite it as $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}6\geq 2bc$
this is the same as $ (b\minus{}a)^2\plus{}(c\minus{}d)^2\plus{}6\geq 2bc$

If you substitute $ a\minus{}b\equal{}x, d\minus{}c\equal{}y$ you now have the new condition $ x\plus{}y\plus{}2b\plus{}2c\equal{}6$

Optimal case is when $ x\equal{}y, b\equal{}c$ from now its really easy.


[quote="anorithdialga"]
\[ \begin{array}{l} If \\
{\rm{a}} \plus{} b > 5c \\
in\_\Delta ABC,show\_c\_is\_the\_sm{\rm{a}}llest\_side \\
\end{array}
\]
[/quote]
if c is the greatest side , we can easily see the contradiction
WLOG a>c>b then a+b > 5c > 5(a-b) -> 3b > 2a
on the other hand , a+b > 5c > 5b -> 2a>8b (contradiction)
so c is the smallest side


Determine maximum real $ k$ such that there exist a set $ X$ and its subsets $ Y_{1}$, $ Y_{2}$, $ ...$, $ Y_{31}$ satisfying the following conditions:
(1) for every two elements of $ X$ there is an index $ i$ such that $ Y_{i}$ contains neither of these elements;

(2) if any non-negative numbers $ \alpha_{i}$ are assigned to the subsets $ Y_{i}$ and $ \alpha_{1}+\dots+\alpha_{31}=1$ then there is an element $ x\in X$ such that the sum of $ \alpha_{i}$ corresponding to all the subsets $ Y_{i}$ that contain $ x$ is at least $ k$.

If any element belongs to at most $ 25$ of the sets we can choose $ \alpha_{i}=\frac 1{31}$ and the sum calculated for each element will be at most $ \frac{25}{31}$.

Otherwise, there exist an element $ x$ that belongs to all sets except $ m$ of them where $ 1\leq m \leq 5$. In this case, we may assign $ \alpha_{i}=\frac{1}{m}$ if $ x\notin A_{i}$ and $ \alpha_{0}=0$ otherwise. Every element will not belong to at least one of the $ m$ sets (because otherwise there is no set containing neither it neither $ x$), so the sum associated to it is at most $ 1-\frac{1}{m}\leq 1-\frac{1}{5}<\frac{25}{31}$. We have proven that $ k\leq \frac{25}{31}$.

Now we prove $ k$ can be $ \frac{25}{31}$. $ 31=5^{2}+5+1$ where $ 5$ is prime.

Therefore by considering a projective plane for $ p=5$ we get a set of $ 31$ elements anf $ 31$ subsets $ B_{1}, B_{2}, \ldots, B_{31}$ of it such that each pair of elements belongs to (exactly) one set and each set consists of $ 6$ elements while each element belongs to $ 6$ sets (and also any two sets have intersection one element, so there is a complete correspondence between elements and sets). Now just let $ A_{i}$ be the complement of $ B_{i}$, having $ 25$ elements, and each element belonging to $ 25$ sets.

Then no matter how we assign the $ \alpha_{i}$, if we sum all the sums obtained for the $ 31$ elements, we get each $ \alpha_{i}$ $ 25$ times, so a total of $ 25$. Then at least one of the assigned sums is at least $ \frac{25}{31}$, as desired.

Note: The projective plane for a prime $ p$ is a well-known example and is constructed as follows: Consider a set of $ p^{3}-1$ triplets $ (x,y,z)\neq (0,0,0)$ where $ x,y,z\in \mathbb{Z}_{p}$. We consider two triplets $ (a,b,c)$ and $ (d,e,f)$ equivalent if and only if $ (a,b,c)=(\lambda d, \lambda e, \lambda f)$ for some $ \lambda \in \mathbb{Z}_{p}\setminus \{0\}$. Each class of equivalency consists of $ p-1$ elements so we get $ p^{2}+p+1$ classes of equivalency. Now let the element associated to the class of equivalency of $ (a,b,c)$ belong to the set associated to the class of equivalency of $ (d,e,f)$ if and only if $ ad+be+cf=0$ (in $ \mathbb{Z}_{p}$). It is easy to check the definition is proper. Also each elee
ment belongs to $ p+1$ sets, each set contains $ p+1$ elements, for every pair of elements there is exactly one set containing them, and every pair of sets intersect in exactly one element.


We know that $R\geq2r$ in every triangle.

For right triangles we have in addition a sharper ineq, this one
$R\geq(1+\sqrt{2})r$

Using this formula $\sum cosA =1+\frac{r}{R}$
and the fact that $cosB=sinC$ it can be proven by $Caushy$ ineq.

There is also a pure geometric solution.

I am interested in an algebrical one. Could anyone help??

I didn't quite understand your question. What exactly do you want???

an [u]algebrical [/u]proof of the statement

For right triangles we have $\displaystyle R\geq(1+\sqrt{2})r$

ok? can help?

Since it is a right triangle, so we have the followings:
$a^2+b^2=c^2$
$R=\frac{c}{2} $
$r=\frac{ab}{a+b+c}$

So...
$\frac{R}{r}=\frac{c(a+b+c)}{2ab}=\frac{ac+bc+a^2+b^2}{2ab}=\frac{a+c}{2b}+\frac{b+c}{2a}=\frac{a}{2b}+\frac{b}{2a}+\frac{c}{2a}+\frac{c}{2b} \geq 1+\sqrt{\frac{a^2+b^2}{ab}}= 1+\sqrt{\frac{(a-b)^2+2ab}{ab}} \geq 1+\sqrt{\frac{2ab}{ab}} = 1+ \sqrt{2}$

We have equality if $a=b$.

What are the other two proofs you mentioned?

Using this formula $\sum cosA =1+\frac{r}{R}$ (take $A=90^0$)
and the fact that $cosB=sinC$ it can be proven by $Caushy$ ineq.

There is also a pure geometric solution:
$OI\geq r$ (it folows immediately from the sketch)
and $OI=\sqrt{R^2-2Rr}$ etc

Ok???

Actually, this is Emmerich's inequality for right triangle. :P


When $ (0.2)^4$ is written as a decimal, what is the product of the last two digits?

You can ignore the decimal and think of the problem as 2 to the fourth, which is 16. 1 times 6= the answer, 6.

what threw me off is that it asked for the last TWO digest, but once i plug it in it said incorrect and that the answer is 6; which i get but then why ask for the last TWO digest?


A flag is made of three horizontal strips of fabric, each of a solid color, either red, white, blue or green. If no two adjacent strips can be the same color, how many distinct flags are possible? These two flags are different.

[asy]draw((0,0)--(0,6),linewidth(1));
filldraw((0,3)--(4,3)--(4,4)--(0,4)--(0,3)--cycle,white);
filldraw((0,4)--(4,4)--(4,5)--(0,5)--(0,4)--cycle,red);
filldraw((0,5)--(4,5)--(4,6)--(0,6)--(0,5)--cycle,green);
filldraw(circle((0,6),1/6),white);
label("Green",(2,5.5));
label("Red",(2,4.5));
label("White",(2,3.5));

draw((5,0)--(5,6),linewidth(1));
filldraw((5,3)--(9,3)--(9,4)--(5,4)--(5,3)--cycle,green);
filldraw((5,4)--(9,4)--(9,5)--(5,5)--(5,4)--cycle,red);
filldraw((5,5)--(9,5)--(9,6)--(5,6)--(5,5)--cycle,white);
filldraw(circle((5,6),1/6),white);
label("White",(7,5.5));
label("Red",(7,4.5));
label("Green",(7,3.5));[/asy]

We have $ 4$ choices for the top color, $ 3$ choices for the middle color because it can't be the same as the top one, and $ 3$ choices for the bottom color because it can't be the same as the middle one. Multiplying these, we get $ 4\times3\times3\equal{}\boxed{36}$.

[quote=hoyam]I did $4\cdot3\cdot4$!  :wallbash_red: :wallbash_red: :wallbash_red: :wallbash_red:[/quote]

oof

oof i thought there were only 3 colors. Didn't read the problem correctly.
:(

Um this is definitely not a Level 19? I thought I sillied for a sec
[hide = sol]$4$ options for top, $3$ for middle bc it cant be the top color, and $3$ for bottom bc it can be anything but the middle color. $4 \cdot 3 \cdot 3 = \boxed{36}$.[/hide]


geometry
geometric transformation
rotation
reflection
ratio
invariant
vector

In a triangle with vertices $A_0$ $B_0$ and $C_0$, each side is divided by k evenly spaced points. The point closest to $A_0$ on side $A_0$ $C_0$ is called $B_1$. The point closest to $B_0$ on side $A_0$ $B_0$ is called $C_1.$ The point closest to $C_0$ on line $B_0$ $C_0$ is called $A_1$. The same operation is continued for triangle $A1$ $B1$ $C1$, and for $A2$ $B2$ $C2$, and so forth.

prove that the centroid (intersection of the 3 medians) of $A_0$ $B_0$ $C_0$ is inside of triangle $A_n$ $B_n$ $C_n$ for any value of n.

[hide="Intro to Affine Transformations"]

Introduction to Affine Transformations:

You may remember the familiar geometric transformations such as translation, rotation, reflection, etc. Each of these transformations has its own balance of freedom and structure. For example, translation has little freedom because you are very limited with how you can morph your figures with it. However, it has a lot of structure because almost everything is preserved (parallelism, angles, lengths, areas, etc.)

The [i]affine[/i] transformations are quite advanced; they have a lot of freedom while having very little structure. Nevertheless, they often come extremely useful. If you take a triangle (three non-collinear points), you can map it to ANY OTHER triangle in the same plane by a UNIQUE affine transformation.

Affine transformations preserve the following:
(1) ratios of lengths along a segment
(2) ratios of areas
(3) collinearity of points
(4) concurrency of lines (three or more lines meeting at a point)

However, make sure you know that affine transformations DO NOT preserve the following:
(1) lengths of segments (as opposed to their ratios)
(2) areas
(3) angles
(4) parallel lines
(5) orientation

Affine transformations cover a lot; I believe that all the ways to play with your monitor settings and completely warp your monitor are affine transformations.

So remember, if you see a problem that deals a lot with RATIOS, but not LENGTHS or ANGLES, keep affine transformations in the back of your mind!

To learn more about them, go here: http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html
Click on Packet of Geometry; I believe affine transformations are in Chapter 3.

[/hide]

EDIT: Try to use these to solve this problem!

[hide="Solution"]

We will prove the stronger statement that the centroid of $A_0B_0C_0$ is in fact also the centroid of $A_1B_1C_1$, and so on.

We will perform an affine transformation on $A_0B_0C_0$, turning it into an equilateral triangle.

Now the ratios $\frac{B_0C_1}{B_0A_0}$, $\frac{C_0A_1}{C_0B_0}$, $\frac{A_0B_1}{A_0C_0}$ are all preserved under the affine transformation. Also, the medians are still medians (they divide the opposite side in the ratio 1:1). They still concur at the centroid, because affine transformations preserve concurrency.

So we have all of the givens in our original problem, BUT WE GET TO ASSUME THAT $A_0B_0C_0$ IS EQUILATERAL!!!

After drawing the diagram, the lengths clearly indicate that we can rotate the whole figure (both $A_0B_0C_0$ and $A_1B_1C_1$) around the centroid by 120 degrees and we will have the same image. It is not hard to see that $A_1B_1C_1$ must also be equilateral.

But if we rotate an equilateral triangle 120 degrees about a point while leaving it invariant, then that point must be ..... ITS CENTROID!!!

So we have now shown that the centroid of $A_0B_0C_0$ is in fact also the centroid of $A_1B_1C_1$.

The centroid of a triangle is always inside it, so if we keep applying this operation, the centroid of $A_nB_nC_n$ will also be the centroid of $A_0B_0C_0$ and thus be inside $A_nB_nC_n$.

I understand that this subject is quite difficult, so if you have any questions about it or you think I made a mistake somewhere in this proof, please feel free to PM me!

[/hide]

another solution could use vectors...(this all uses stuff from aops vol 2, ch 12)

the vertices of the triangle are the vectors [b]a[/b],[b]b[/b],[b]c[/b]

the centriod is ([b]a[/b]+[b]b[/b]+[b]c[/b])/3

suppose that the ratio $A_nB_{n+1}$ to $B_{n+1}C_n$ is $d: e$

the new verticies are given by {[b]a[/b]d+[b]b[/b]e}/{d+e}, doing the similar for the others, then adding, and dividing by 3, for the centriod

centroid new= {[b]a[/b]d+[b]b[/b]e+[b]b[/b]d+[b]c[/b]e+[b]c[/b]d+[b]a[/b]e}/{3(d+e)}=([b]a[/b]+[b]b[/b]+[b]c[/b])/3, so the centriods are the same...

Yeah, vectors are also really good when dealing with lengths and their ratios.  :)

When you do affine transformations to simplify the problem, it is often a good idea to try to finish the proof with vectors or complex numbers because they have the inherent machinery of lengths and their ratios. (And, like affine transformations, they are not very good with angles.) However, if you are dealing with ratios of AREAS, then vectors and complex numbers won't be as useful; and remember, affine transformations are still legal for ratios of areas.

Remember to make sure that all your invariants are affine-legal; they can be more error-prone if you accidentally discount an invariant. So when your answer turns out to be stronger than what the question is asking, you should start worrying and check your invariants (such as 90 degree angles and the like). That is what happened here, but luckily the second approach confirmed my stronger statement.  :lol:


quadratics
number theory proposed
number theory

with out using Dirichlet's theorem.prove that there are infinity prime in the form 8k+5.
how to prove it? I have one solution,but,I don't sure it's true.
[hide="My solution"]Let $ p_1, p_2, ... , p_n$ be any finite collection of primes of the form
$ 8n \plus{} 5$.  Consider the number $ (2*p_1*p_2*...*p_n)^2 \plus{} 1$.  It cannot be
divisible:by $ 2$, because it's odd;by any $ 4n \plus{} 3$ prime, because $ \minus{} 1$ is a quadratic residue;solely by $ 8n \plus{} 1$ primes, because it's of the form $ 8n \plus{} 5$.Therefore it has a prime factor of the form $ 8n \plus{} 5$.$ \therefore$ contradiction[/hide]

With the same method we can prove that there exist infinite prime in the form $ 8k\plus{}1,8k\plus{}3,8k\plus{}5,8k\plus{}7$

Do not double post! ( http://www.mathlinks.ro/Forum/viewtopic.php?t=169527 )
Locked!


Find all pairs (x, y), where x, y are integers such
$ x^3 \plus{} 11^3 \equal{} y^3$

[b]My solution:[/b] By Fermat's last theorem, the only solutions are (-11, 0) and (0, 11). Am I right?

[quote="thomaskurian89"]
By Fermat's last theorem[/quote]

Well, only a special and quite classical case of it...

The real question is : why is this problem in the algebra section? :wink:

Pierre.

$ (y\minus{}x)(x^2\plus{}xy\plus{}y^2)\equal{}11^3$

since $ 11$ is prime so :

$ y\minus{}x\equal{}11$ and  $ x^2\plus{}xy\plus{}y^2\equal{}11^2$

$ \iff$   $ (y\minus{}x)^2\equal{}x^2\plus{}xy\plus{}y^2$ and $ y\minus{}x\equal{}1$

so the solutions are : $ (0,11)$ and $ (\minus{}11,0)$

or : $ x^2\plus{}xy\plus{}y^2\equal{}11$ and $ y\minus{}x\equal{}11^2$

and in this case we don't have solutions in integers ;)


algebra
polynomial
number theory
greatest common divisor
superior algebra
superior algebra solved

Let $F$ be a field, and $R$ a ring included in(sub-ring of) $F[X]$ and such that $F\subset R$. Prove that there exist $f_1,f_2,...,f_n$ such that $R=F[f_1,f_2,...,f_n]$.

The condition $F\subset R$ is necessary. For example take $\mathbb Q\subset \mathbb R[X]$.

If $R=F$, then take $f_1=1$ and $n=1$.
If $R\neq F$, then take $f_1$ a monic non-constant element of $R$ of minimal degree such that $f_1(0)=0$. $f_1$ is unique with these properties. Let $R_1=R[f_1]$.
While $R_n\neq R$, take $R_{n+1}=R_n[f_{n+1}]$ with $f_{n+1}$ chosen such that it is monic of minimal degree in $R\setminus R_n$.
What we want to prove is that this construction ends after a finite number of steps.

Let $d_i=deg f_i$.

I will first prove that $(d_i)_i$ is a strictly increasing sequence. The sequence is clearly increasing, but not clearly strictly increasing. Actually I will prove the stronger fact that $d_n$ is not divisible by $d_i$ for any $i<n$. Assume $d_n$ is divisible by $d_m$ for some $m<n$. Let $g=f_n-f_m^{\frac{d_n}{d_m}}$. Since $f_k$ are all monic polynomials, deg $g\leq d_n-1$. Since $f_n\not\in R_{n-1}$ and $f_m\in R_{n-1}$, it follows that $g\not\in R_{n-1}$. These contradict the choice of $f_n$ as a monic polynomial of minimal degree in $R\setminus R_{n-1}$.
The same ideas can be used to prove that $d_n$ is not in the additive submonoid of $\mathbb N$ generated by $d_1,\ldots,d_{n-1}$.

Denote by $D_n$ the additive sub-monoid of $\mathbb N$ generated by $d_1,\ldots,d_n$. By what we have proved, $D_n\subsetneq D_{n+1}$
Let $\bar d_n=gcd(\{k|k\in D_n\})$. Clearly $\bar d_{n+1}|\bar d_n$ for all $n$. In particular Since $\bar d_n$ is a decreasing sequence of positive integers, there exists $N$ such that $\bar d_n=d$ for all $n\geq N$ for some integer $d$. It also follows that $d|\bar d_n$ for all $n$. There exists $M>0$ such that $d\cdot r\in D_N$ for all $r>M$ i.e. $D_N$ contains all the multiples of $d$ greater than $d\cdot M$. But then since $d|\bar d_{n}|d_n$ for all $n\geq N$ and since $d_n\not\in D_N$ for $n>N$, we get that $d_n<dM$ for all $n>N$.
Hence $d_n$ is a striclty increasing bounded sequence of positive integers. This is not possible, thus the construction stops and $R=R_n=F[f_1,\ldots,f_n]$ for some $n$.

Very nice approach. Thank you! I think that this problem is from G.M. 198?..


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

We have $n$ numbers. Each number consist of four digits. Possible dgits are $0$ or $1$.
We are interested in [b]all[/b] possible numbers. Even like $0010$ or $0101$ and so on.
For each pair of numbers there are two digits situated on the same position in both numbers satisfying:
$A_1 \not =B_1$
$A_2 \not =B_2$
$A_1$ and $A_2$ are the two digits of number $A$.
$B_1$ and $B_2$ are the two digits of number $B$.
Remember that digits $A_1$ and $B_1$ are situated on the same position.
The same is in $A_2$ and $B_2$
For example in numbers
$0110$
$1100$
$A_1$ and $B_1$ are the first digits
and $A_2$ and $B_2$ are on the third position.
This pair fulfill the conditions of the task, but pair $0001$   $0000$ does not.

Evaluate the maximum value of $n$.

I hope I've explained it in an understandable way.

I know that we can co it in a mechanical way, but I would be grateful if you solve it by a method.
For example try to do it when number of digits is six and possible digits are: $0$, $1$, $2$, $3$.
In that situation we can't do it by writing all possibilities.

This problem can be reformulated as [url=http://www.mdstud.chalmers.se/~md7sharo/coding/main/node29.html#SECTION02055300000000000000]packing spheres[/url] of radius 1 with respect to Hamming distance.

Sorry if I'm annoying, but I don't quite understand this proof.
Could you explain it to me.....or give some more simple  :)

It was not a proof but a direction for search.
Likely, the problem you are asking is NP-complete and there is no much better solution than bruteforce.

I have doubts whether we can do it bruteforce. In case when we have six task and four possibilities of markink one task. There are $4^6$ possibilities of creating a number.


trigonometry
AoPS Books
complex numbers
\/closed

Would you recommend the Complex Numbers/Trig class or should I just work through the respective chapters in AoPS II?  I know the class will obviously be more in-depth, but I am wondering whether the things that are taught can just be learned over time by doing problems.

i would suggest taking classes. Though books are always helpful, there are always some parts where you can't just learn by doing problems and reading. take a look at the course outline for the online class. if you can not fully understand any of the sections of the class by using AoPS books, definitely take online class.

I would suggest take the online class and read the chapters thoroughly before each class. This will help you to concentrate and make the most out of the class and understand the materials way better  :) .

The trigonometry and complex numbers covered in AoPS 2 is nowhere near the depth in which they cover in the class. It really isn't that neccessary to read through the text, because they cover all the stuff in the text in the class anyway. If you really feel the compulsive need to read it, fine. Also most of the problems appear in the class as well in some form (example, message board, test, etc)

Thank you!
I think (if I do the class)  I'll just use the books if I get confused on a particular topic.

The class goes to much greater depth than AoPS Vol 2.  At the end of 2008 or beginning of 2009, we expect to have an Intermediate Trig/Complex Numbers (not sure yet exactly what the title will be) text that will be a companion to the class, but right now, AoPS Vol 2 would be a helpful supplement to the class, but not an effective complete replacement for it.

well you took the class  :D

-Pi-dimension


Express $ \dfrac{\sqrt{0.44444\cdots}}{2}$ as a common fraction.

Since $ .44444\cdots \equal{} \frac49$, we have $ \dfrac{\sqrt {\frac49}}{2} \equal{} \dfrac{\frac23}{2} \equal{} \boxed{\frac13}$.

This might help you problem solvers for future problems.
Whenever you see a number that has a terminating decimal such as 0.444444 Its fraction form is 44/99, not 44/100.
I wont give the proof but I will tell you that this will come in very handy for some problems. For instance, if we didnt know that 0.444444 is just 4/9, we could have found that out by the method above.

Good Day.

In general, $0.\overline{a_1a_2\cdots a_k}$ = $\dfrac{\overline{a_1a_2\cdots a_k}}{10^k-1}$.

Proof of AkshajK's statement:

Let $n=0.\overline{a_1a_2\ldots a_k}$. Then, we have $10^kn=a_1a_2\ldots a_k.\overline{a_1a_2\ldots a_k}$. Subtracting the first equation from the second gives $\left(10^k-1\right)n=a_1a_2\ldots a_k$, so therefore $n=\frac{a_1a_2\ldots a_k}{10^k-1}$.


algebra
polynomial
calculus
integration
algebra solved

Suppose f(x) is a polynomial with integral coefficients. If f(x) = 2 for three different integers a,b and c, prove that f(x) can never be equal to 3 for any integer x.

This one is easier than the previous one about polynomial...
Suppoqe that f(a) = f(b) = f(c) = 2, for a,b,c pairwise distinct integers.
Then, f(x) - 2 = (x-a)(x-b)(x-c)g(x), where g is a polynomial with integer coefficients.
Now suppose that f(t) = 3, for some integer t. Then (t-a)(t-b)(t-c)g(t) = 1, and each of the 4 factors in the LHS is an integer. Thus, each of them is equal to -1 or 1. By the pigeon-hole principle, it follows that at least two among the numbers t-a,t-b,t-c are eaqual, contradicting that a,b,c are pairwise distincts.

Pierre.


inequalities unsolved
inequalities
algebra


A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long. What is the smallest number of blocks needed to build the wall?

A. 344
B. 347
C. 350
D. 353
E. 356

[quote="i_like_pie"][hide]$\text{C}$

$350\,\text{blocks}$[/hide][/quote]

That's exactly what I thought but that's not correct.

Are you sure that you are looking at the correct space at the answers?

Each are 1 foot above and 2 acroos. Then you need 50*7=350.

Here is the exact problem.

A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 1 feet or 2 feet long (no blocks may be cut).  The vertical joins in the block must be staggered, and the wall must be even on the ends.  What is the smallest number of blocks needed to bulid this wall?

a) 344
b) 347
c) 350
d) 353
e) 356

[quote="mathgeniuse^ln(x)"]Here is the exact problem.

A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 1 feet or 2 feet long (no blocks may be cut).  [b]The vertical joins in the block must be staggered,[/b] and the wall must be even on the ends.  What is the smallest number of blocks needed to bulid this wall?[/quote]
What does that mean?

It means the edges of the blocks must not coincide with the edges of the blocks directly above and below any given block...

ya, that sounded way too easy at first
-jorian

[quote="anirudh"]Oh...So then it would be E, right?[/quote]

OOPS :(  :oops_sign:

I forgot that the 1, 3, 5, and 7'th row could be only 50. The answer I wish to change Is D.  :blush:

Yes, that is the right answer...

goldendomer, reading the question helps...

[hide]
The bottom row can be made of 50 2 feet blocks.
The second to bottom row can be made of 1 one foot block, then 47 2 feet blocks, then one 1 foot block, then a 2 foot block, then another 1 foot block.
Then, keep alternating this pattern.
So, there are 4 rows of 50 blocks, and 3 rows of 51 blocks.
So, the answer is 353, D.[/hide]


[quote]I think this is a hard trig question[/quote]

You thought right.

I concur. *sigh* i'm going to use calculus soon if i can't find a better way. *sigh*

by the way, which contest is this from? Is a knowledge of calculus expected or is there an elementary way?

[quote="darkquantum"]Let a=(secx)^2

If y=(a+tanx-1)/(a+1)

Then find the minimum of y.[/quote]

Writing everything in terms of tan x=a yields (a:^2:+a)/(a:^2:+2)=1+(a-2)/(a:^2:+2). Therefore, to find the minimum of the original expression, we find the minimum of (a-2)/(a:^2:+2) and add 1. Okay now, (a-2)/(a:^2:+2)=1/((a:^2:+2)/(a-2)). Thus, I will find the maximum of (a:^2:+2)/(a-2) and invert it to find the mininum of the other equation. (a:^2:+2)/(a-2)=a+2+(6/(a-2)).  erg I gotta go bye.

err
just use the tan-sec identity, and things cancel out fairly nicely...

err
i'll tell u one thing tho
there is a minimum


induction
combinatorics unsolved
combinatorics

Some pairs of towns in a country are joined by roads, so that there is a unique route from any town to another which does not pass through any town twice. Exactly 100 of the towns have only one road. Show that it is possible to construct 50 new roads so that there will still be a route between any two towns even if any one of the roads (old or new) is closed for maintenance.

This one is cute, and I don't recall seeing it on the forum, but I could be wrong :).

We can prove by induction on the number $k$ the following proposition:

Given a tree with $2k$ leaves, we can add $k$ edges with endpoints among the leaves to the tree such that it become $2$-edge-connected (i.e. the removal of any edge will leave it connected after the addition).

The base case ($k=1$) is simple, so let's assume it's proven for values smaller than $k$, which we assume to be $>1$. If all leaves are connected to the same non-leaf, the assertion is obvious: we just pair them up two by two, and connect the leaves in the same pair. On the other hand, if there are at least two non-leaves $x,y$, start going from $x$ towards $y$ and continue until you hit a leaf $a_{xy}$, and then start from $y$ towards $x$, and continue, again, until you hit another leaf $a_{yx}$. Connect $a_{xy}$ and $a_{yx}$ with an additional edge. Because our graph was a tree, this extra edge creates precisely one cycle. Collapse this cycle inti a single vertex, identifying all its vertices and edges to a single point. Because $a_{xy},a_{yx}$ are not connected to the same non-leaf, after this operation we're left with a tree with precisely two fewer leaves than the previous one, and we can apply the induction hypothesis.


Two angles are supplementary. The measure of one angle is  three times that of the other. What is the number of degrees in measure of the greater angle?

Oops, read the problem.  The answer is $X=135$, because it asks for the greater angle.


Hey,

I am currently studying for AIME.

I have AoPS Volumes 1 and 2. I am studying from AoPS Volume 2 right now and it's not far beyond what I know. I got a 132 on the AMC 10B, and I missed the other points because I was rushed towards the end not because I didn't know how to do them.

My question is if I need to study Volume 1.

Thank you for your help.

I don't think you need to. It'll be useful as a reference, but otherwise you should be fine.

You should be fine plowing ahead with Volume 2.


Express $\frac{\frac{1}{2} - \frac{1}{3}} {\frac{1}{3} - \frac{1}{4}} \cdot \frac{\frac{1}{4} - \frac{1}{5}} {\frac{1}{5} - \frac{1}{6}} \cdot \frac{\frac{1}{6} - \frac{1}{7}} {\frac{1}{7} - \frac{1}{8}} ... \cdot \frac{\frac{1}{2004} - \frac{1}{2005}} {\frac{1}{2005} - \frac{1}{2006}}$ as a rational number in lowest terms.
So far I have:
[hide] $\displaystyle\prod_{n=2}^{2004} \frac{\frac{1}{n} - \frac{1}{n+1}} {\frac{1}{n+1} - \frac{1}{n+2}}$
$\rightarrow \displaystyle\prod_{n=2}^{2004} \frac{n+2}{3n}$[/hide]
I don't know much about the properties of the product thing... thx in advance...

[hide]Do the actual subtracting. For the first term:

$\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$, $\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$.

$\frac{\frac{1}{6}}{\frac{1}{12}}=2$.

The second:

$\frac{1}{4}-\frac{1}{5}=\frac{1}{20}$, $\frac{1}{5}-\frac{1}{6}=\frac{1}{30}$.

$\frac{\frac{1}{20}}{\frac{1}{30}}=\frac{3}{2}$.

If we had continued, we would have $2\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{1003}{1002}=1003$.

How do we find the last term? Notice that for each term, the numerator comes out to be the smallest fraction's denominator, and the denominator is the biggest fraction's denominator (this can be shown, but I don't want to do that).

For the last term, the biggest fraction is $\frac{1}{2004}$, with denominator 2004, and the smallest is $\frac{1}{2006}$, with denominator 2006. Thus the last term is
$\frac{2006}{2004}=\frac{1003}{1002}$.
[/hide]

[hide="Click Me"]
Notice that for every $n$, $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{(n)(n+1)}$.

Now everything is easy and becomes ${\frac{\frac{1}{2}\cdot\frac{1}{3}\cdot...\cdot\frac{1}{2005}}{\frac{1}{3}\cdot\frac{1}{4}\cdot...\cdot\frac{1}{2006}}}=\frac{3\cdot4\cdot...2006}{2\cdot3\cdot...2005}$.

Canceling things, you have $\frac{2006}{2}=1003$.
[/hide]

[quote="Apprentice"]...
So far I have:
[hide] $\displaystyle\prod_{n=2}^{2004} \frac{\frac{1}{n} - \frac{1}{n+1}} {\frac{1}{n+1} - \frac{1}{n+2}}$
$\rightarrow \displaystyle\prod_{n=2}^{2004} \frac{n+2}{3n}$[/hide]
I don't know much about the properties of the product thing... thx in advance...[/quote]

This is not exactly correct, you have to notice that the first number in each term of the product changes by 2's i.e. it is 2,4,6,8,... and you did 2,3,4,5,...


rotation
integration
symmetry
geometry
3D geometry
trigonometry

The MOI of the cylinder about its symmetry axis is $\dfrac{1}{2} MR^2$ ( you can get this easilt from integration). Now we want the MOI about the axis perpendicular to the symmetry axis . Now we consioder the perpendicular axis theorem.

By symmetry, the MOI About any axis $\perp$ to the symmetry axis is the same say = $I$
Thus, from perpendicular axis theorem , $2I = \dfrac{1}{2} MR^2$ thus $I = \dfrac{1}{4} MR^2$....

I think this may be faulty for I read that the $\perp$ axis theorem works only for planar figures!!!!

What exactly does the perpendicular axis theorem state?

And I see you got the 1/4MR2 but what abuot the 1/12ML2?

The $\perp$ axis theroem states that for a planar body, $I_x + I_y = I_z$ where $I_x$ is the MOI about some chosen x-axis and so on...

it should be noted that in the perpendicular axis theorem, x and y are the two axis that make up the plane that contains the planar face of the object

Wait a minute...
[quote]about its central diameter through the center[/quote]
what do you mean by that?  is it like the thin rod case, but with a big cylinder?

couldnt you cut up the cylinder into slices, find the MOI w/ the Perpendicular and then parallel axis theorem, and integrate?

Hm, I integrated it and came to a different formula - I thought that I don't understand the "axis about its central diameter through the center" part properly so I tried the geometrical center (1/2 L), center of mass (1/4 L) and both ends (1 L, 0 L) but still not what you wrote up there. Perhaps I made some stupid mistake but to clear things up, where exactly is the axis again?

Here is a picture of what I am trying to describe:

[url]http://library.thinkquest.org/16600/advanced/moment-cylinder-diameter.gif[/url]

Whoops, I thought you were talking about a cone, not a cylinder. I need to practise my English... :oops:

Well, then it should be even easier. Too bad it's so late, I'll have to post my solution tomorrow  :P.

Lets give the cylinder the following dimensions, then.  It's got radius $R$ and length $l$.  so lets call it's mass $M=\rho*\pi*R^2l$, where $\rho$ is the cylinder's density.

Let's take any random slice of this cylinder, now.  It's got a cross-sectional area of $\pi*R^2$ and a thickness $dx$, where $x$ is the distance away from the central axis that you were asking about in the beginning

Then the moment of inertia of this guy(with axis through its center) would be $1/2*\rho*\pi*R^4*dx$

Use the perpendicular axis theorem, and you get that the MOI of this slice is $1/4*\rho*\pi*R^4*dx$ when rotated about a diameter

Then, knowing that each slice is a distance $x$ away from the central axis, you use the parallel axis theorem for each of these slices.  So you get that the MOI of each slice is
$1/4*\rho*\pi*R^4*dx + \rho*\pi*R^2*x^2*dx$
And you would integrate this from 0 to l/2 (doing this from the center) and multiply it by 2, because the complete cylinder is 2 of them.

That should be the solution.  There might be some mistakes, but I think the idea is correct

or its completely wrong.  go figure

"lo, pranay.  I'm on your noml team.

When I integrated it, it did come up with the right answer, but I am still confused.

How did you get the $1/2*\rho*\pi*R^4*dx$?  Did you use $I=1/2*MR^2$? And then why did you suddenly divide it in half?

You use $I=1/2MR^2$ and then from the perpendicular axis theorem:

$I_x+I_y=I_z=1/2MR^2$

and $I_x=I_y$ from symmetry so you get $1/4MR^2$. I wonder what do you do next though. What doest the parallel axis theorem state?

Just that I don't forget, here's my solution by integrating for those real physicists out there :) :

$I=\int_{-L/2}^{L/2}\int_0^R\int_0^{2\pi}(x^2+r^2\sin^2\alpha)\rho r\,\text{d}\alpha\,\text{d}r\,\text{d}x=$
$=\rho\int_{-L/2}^{L/2}\int_0^R[x^2r\alpha+r^3(\dfrac{\alpha}{2}-\dfrac{\sin{2\alpha}}{4})]_0^{2\pi}\,\text{d}r\,\text{d}x=$
$=\rho\int_{-L/2}^{L/2}\int_0^R(2\pi x^2r+\pi r^3)\,\text{d}r\,\text{d}x=$
$=\pi\rho\int_{-L/2}^{L/2}[x^2r^2+\dfrac{1}{4}r^4]_0^R\,\text{d}x=$ :dry:
$=\pi\rho\int_{-L/2}^{L/2}(R^2x^2+\dfrac{1}{4}R^4)\,\text{d}x=$
$=\pi\rho R^2[\dfrac{1}{3}x^3+\dfrac{1}{4}R^2x]_{-L/2}^{L/2}=$
$=\pi\rho R^2L(\dfrac{1}{4}R^2+\dfrac{1}{12}L^2)=\dfrac{1}{4}MR^2+\dfrac{1}{12}ML^2$ :whistling:

Parallel axis theorem states that if you have the Moment of Inertia$(I)$ of an object about its center of mass, you can find it's moment of inertia about a parallel axis anywhere else by adding $I+MR^2$, where $M$ is the mass and $R$ is the distance from the center of mass of the object to the axis

[quote="pkothari13"]Does anyone know a proof for the moment of inertia of a solid cylinder being rotated on the axis about its central diameter through the center?

The formula for it is $I=\displaystyle\dfrac{1}{4}MR^2 + \displaystyle\dfrac{1}{12}ML^2$, where L is the length (or you could say height) of the cylinder and R is the radius.

Recall that the moment of inertia for a continuous object is $I=\displaystyle\int r^2\,dm$ .

I thought maybe using the parallel axis theorem might work, but I cuoldn't find an efficient solution.[/quote]The MOI for a cylinder about its central diameter is a superposition of the MOI of a thin rod about an axis through it's center, and the MOI of a disc about an axis in the plane of the disc.

The second part is harder to get. By the perpendicular axis theorem $I_z = I_x + I_y$ ([url=http://hyperphysics.phy-astr.gsu.edu/hbase/perpx.html#ppx2]A proof of which is available here.[/url]) For a disc, $I_z = \frac12MR^2$; this is derivable from the expression $I =\int r^2\; \mathrm{d}m$ evaluated for a planar figure in polar coordinates. Since the disc is circular, completely symmetric, $I_x = I_y = I$. Therefore, $I =\dfrac{I_z}{2} =\frac14MR^2$.

The first part, the long thin rod is the simplest MOI there is, $I =\int_{\textstyle{\dfrac{-L}{2}}}^{\textstyle{\dfrac{+L}{2}}} \rho x^2\; \mathrm{d}x =\dfrac{2\rho L^3}{24} = \dfrac{ML^2}{12}$

Now just add them together, $I_{\text{total}} = \dfrac{1}{12}ML^2 +\frac14MR^2$.


Cara is sitting at a circular table with her five friends as shown to the right. How many different possible pairs of people could Cara be sitting between?

[asy]draw(circle((0,0),1));
label("$\_$",1.5dir(0));
label("$\_$",1.5dir(60));
label("Cara",1.5dir(120));
label("$\_$",1.5dir(180));
label("$\_$",1.5dir(240));
label("$\_$",1.5dir(300));[/asy]

We only care about the people sitting next to her. There are 2 people sitting next to her out of 5 total people, giving us an answer of $ \binom{5}{2} \equal{} \boxed{10}$.


Given $ x,y,z \in R$ such that $ xyz\equal{}\minus{}1$,show that
$ x^4\plus{}y^4\plus{}z^4\plus{}3(x\plus{}y\plus{}z)$≥$ \sum_{cyc}x^2(\frac{1}{y^2}\plus{}\frac{1}{z^2})$

$ x\equal{}\minus{}1$, $ y\equal{}\minus{}1$, $ z\equal{}\minus{}1$ says this inequality is false =[


limit
integration
trigonometry
real analysis
real analysis unsolved

Compute
$ \lim_{n\rightarrow \infty} \int_{0}^{\pi} \frac {\sin{x}}{1 \plus{} \cos^2 nx}dx$ .

The limits equals 0. We have, by using the substitution $ nx\equal{}y$, that

$ |I_{n}|\equal{}\biggm|\frac{1}{n}\int\limits_{0}^{n^{2}}\frac{\sin y/n}{1\plus{}\cos^{2}y}dy\biggm|\leq \frac{1}{n}\int\limits_{0}^{n^{2}}\frac{dy}{1\plus{}\cos^{2}y}dy.$

Using the substitution $ \tan y\equal{}t$ we get that

$ |I_{n}|\leq \frac{1}{n}\int\limits_{0}^{\tan n^{2}}\frac{dt}{2\plus{}t^{2}}\equal{}\frac{1}{n\sqrt{2}}\arctan(\frac{\tan n^{2}}{\sqrt{2}})<\frac{\pi}{2\sqrt{2}n}$,

and the result follows.

I am sorry, there was a typo in the beginning. Look at this version now.

Use the same substitution and you get that

$ I_{n}\equal{}\frac{1}{n}\int\limits_{0}^{n\pi}\frac{\sin y/n}{1\plus{}\cos^{2}y}dy\equal{}\frac{1}{n}\sum\limits_{k\equal{}0}^{n}\int\limits_{k\pi}^{(k\plus{}1)\pi}\frac{\sin y/n}{1\plus{}\cos^{2}y}dy.$

Use the substitution $ y\minus{}k\pi\equal{}z$ and you get that

$ I_{n}\equal{}\frac{1}{n}\sum\limits_{k\equal{}0}^{n}\int\limits_{0}^{\pi}\frac{\sin (\frac{k\pi\plus{}z}{n})}{1\plus{}\cos^{2}z}dz\equal{}\frac{1}{n}\int\limits_{0}^{\pi}\left(\sum\limits_{k\equal{}0}^{n}\sin(\frac{k\pi \plus{}z}{n})\right)\frac{dz}{1\plus{}\cos^{2}z}$

$ \equal{}\frac{1}{n}\int\limits_{0}^{\pi}\frac{\cos(z/n)\cos(\pi/(2n))}{\sin(\pi/(2n)}\frac{dz}{1\plus{}\cos^{2}(z)}$

And this converges to $ \frac{2}{\pi}\int\limits_{0}^{\pi}\frac{dz}{1\plus{}\cos^{2}z}\equal{}\frac{2}{\pi}\cdot \pi\sqrt{2}/2\equal{}\sqrt{2}$.

I hope that I made no computational mistakes.


[quote="K Sengupta"]I am looking for a solution to the puzzle furnished hereunder, which I found in a Puzzle periodical.

A man returns from a long swim at the beach and takes a trip to the dry cleaners.
Thereafter, he goes to a restaurant and orders his food. When he finishes his food, he realizes something, and runs out into the street, gets hit by a car, and dies.

Why did the man run into the street?[/quote]
There are many of these probelms; here is a similar one:

A man walks into a restaurant, orders albatros, eats it, and immediately kills himself. Why?

The man, his wife and his friend were on a ship a few weeks ago, but they ran out of food, so the chef served them "albatros." At the same time, his wife disappeared. When they got back, the man ordered some albatros, and when he realized that he had eaten his own wife, he killed himself.

I think that can be a solution to yours, also.

You mean you walked to the restaurant naked? :huh:

Please tell me that abcdefghijklmnopq is lying, that must have been horribly humiliating.

[hide] Maybe he had realized something bad happened in the restaurant to him before and he was very superstitious, so he went out and INTENTIONALLY GOT HIT BY A TRUCK![/hide]

 Sounds pretty stupid, doesn't it?

[quote]You mean you walked to the restaurant naked? [/quote]

Actually, I left my thong on, since I usually wear one  for a week and this happened after I had only worn itfor one day.  However, i did throw my shirt and pants(which unfortunately had my wallet) into the laundry machine before going to the restaurant.

[quote="abcdefghijklmnopq"]he left his wallet in his pants which he took off toput in the washing machine and then walked to the restaurant naked. This exact scenario has happened to me numerous times.[/quote]

uhh what? why though?

Reveal and be scared by a quote from abcdefghijklmnopq:[hide]Actually, I left my thong on, since I usually wear one for a week and this happened after I had only worn itfor one day. However, i did throw my shirt and pants(which unfortunately had my wallet) into the laundry machine before going to the restaurant.[/hide]

[hide]ABCDEFGHIJKLMNOPQ! YOU DO NOT SHARE THAT WITH OTHER PEOPLE! ALLOW YOURSELF SOME PRIVACY! GOD FORGIVE YOU FOR SAYING SUCH IRRATIONAL THINGS! YOU ARE SCARING ME! NOW I AM RUNNING AROUND IN CIRCLES SCREAMING LIKE A MANIAC! :maybe: [/hide]

[hide]the man runs out into the street because the restaurant does not have a trash can. he didnt wait a long time before eating so he will throw up, but there is no trash can. so then the man knows a trash can across the street, but he dies in the process.[/hide]


Just read today's newspaper regarding the Indian Institute of Science and Educational  Research (IISER).Sounds quite nice.

I couldn't understand what they meant by KVPY 2006 SA batch.I got thru the the KVPY in 2005 but was  given the scholarship amount only for March 2006 - March 2007.So does that mean that I am eligible...

I think they mean those who got KVPY this year will be qualified for applying next year i.e. our batch.

But the admissions are starting right from this year ,Possibly ,they'll be starting the first batch of students from this year's may or june.

if you look at the last few lines of the ad then ashwin probably u will have to contact the director IISER, KOLKATA .once u do the entire process this year please do write about it here ,that will surely help us :P  :D

Hello,
    The IISER Admissions were started from last year.It is a great place if you are interested in the Pure Sciences.
Sathej

[quote="arijit91"]what exactly does INDIAN NATIONAL OLYMPIAD mean.does it require participation in an international olympiad or going to a national camp?[/quote]
well itmeans u and ur brother can both make it 2 the IISER(if u are interested) by virtue of ur INMO-07 merit list .Anyways he also has his INAO certificates...

I had been to the website of IISER and I saw that the admissions for the year 2007 are being made from the KVPY 2006 SA batch.Last year they had selected from the KVPY 2004 batch it seems...Then what about we KVPY 2005 batch students?
And also it seems odd to select KVPY 2006 batch students Who have now just finished their 11th to apply for the year 2007...

SOMEONE PLEASE CLARIFY ME ON THIS.

Opps sorry... I made a mistake !
We're the KVPY 2007 batch since we gave the exam this year.

Don't know about the 2005 batch though.

ashwin r u sure the mean '2006' SA batch or they have some kind of misconception as vihang...

[quote="ashwinkrishnan"]Well it would be wrong to suspect them of having misconceptions(they are after all eminent people and the whole idea is theirs...),but still that could be another possibility as well.[/quote]

no it is not exactly a misconception but a conception that 2006 SA batch refers to urs

[quote="ashwinkrishnan"]Just a sec,Vihang...u know Varun Suhas Jog right...He is two years elder to u( i dont know which class u study in) and a year elder to me and he was from the 2004 batch(his enrolment no. is 10400...something got it from the IISER website itself)mine is 1050026 which means that I am from 2005 batch.Now assuming that you should be in the 2006 batch.....have we been forgotten[/quote]

Yup I checked my card and my enrolment number is 1061016.
And yes I'm in Eleventh right now.

Well it must be a printing mistake in the newspaper then.
I think you should contact some IISER representative and get it confirmed.

Maybe KVPY refers to us as 2006 batch but IISER think of us as the 2007 batch because we actually gave the exam (and also got the scolarship) in 2007.

just follow this link

http://www.iiserpune.ac.in/files/admission/imp/IISER_Admissions_2007_08_BATCH.pdf

Ashwin in the link you gave ....

[quote]Students wishing to pursue Science as a career with scientific research and teaching and have passed (10+2) or equivalent qualifying examination in 2006 or will be appearing in 2007 may apply in plain paper giving the following information[/quote]

I'll be appearing for my HSC exam next year so clearly I'm not allowed to apply this year. (And anyway that would be wierd IISEr after eleventh !)

You should just contact them and get this cleared.


How can I convert an image .bmp to .eps?, is there any software?

.png to .eps?

Eps is a vectorial encapsulator, whereas bmp is not. If you could convert it, I would not see the point.

I assume you want free software that others have used as Googling for "convert bmp to eps" gives lots of links.
[url=http://www.openoffice.org/]OpenOffice[/url] makes some of the best quality eps files I have seen using OpenOffice Draw. You can either open the bmp/png file or copy and paste it in then export as EPs.

[url=http://www.imagemagick.org/script/index.php]ImageMagick[/url] is fast (and is used by AoPS behind the scenes) but uses the command line where you say [i]convert image.bmp image.eps[/i] and it's done instantly.

I'm sure others will come up with their favourite software but meanwhile, are you aware that you can use png, jpeg and gif images in latex documents when converting to pdf?

To answer Valentin's point, it is necessary to use eps when compiling latex files to dvi or ps but most people use pdf these days where it isn't necessary.

I have another problem

Below I attached the file (.eps) that I want to put in the PDF

My LaTeX code is

[color=indigo]\documentclass{article}
\usepackage[latin1]{inputenc}
\usepackage{principal}
\usepackage{multicol}
\usepackage{color}
\usepackage{cancel}
\textheight=24cm
\definecolor{rojo}{rgb}{1,0,0}
\definecolor{verde}{rgb}{0,0.7,0}
\definecolor{azul}{rgb}{0,0,1}
\definecolor{naranjo}{rgb}{1,0.5,0}
\definecolor{cyan}{rgb}{0,0.5,1}
\definecolor{fucsia}{rgb}{1,0,0.5}
\definecolor{negro}{rgb}{0,0,0}
\definecolor{marron}{rgb}{0.8,0,0}
\begin{document}
\end{document}[/color]

I don't know what I must add to the code to be able to generate the image.

If you read [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]Images in LaTeX[/url] you'll see you need \usepackage[pdftex]{graphicx} in the preamble and \includegraphics{myimage.png} to get an image.
BUT you can't include eps images in this way as compiling to pdf requires images in gif, png, jpg or pdf format.
You said you started with bmp and had to convert it to eps - convert the bmp to png instead and then you can use:
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
% put your other \usepackage and \definecolor stuff here
\begin{document}

\includegraphics{Ilustraciones.png}

\end{document}[/code]

Look at

[img]http://img247.imageshack.us/img247/1179/figxy2.png[/img]

There are the diverse qualities about the drawing in different formats. Indeed .eps format appears, this means that the graph could be inserted.

Sorry, but I still don't know how to make this thing.

Are you using Windows? If so, then use Paint (Start->All Programs->Accessories->Paint). Open the bmp image and save as png and that's all that's needed.


calculus
integration
calculus computations

Summation (of n = 4 to 19) of 1/sqrt(n + 5.5)

Please show your work.

You haven't shown me anything (for this challenge).


Let M be a point inside a triangle ABC. Prove the inequality
$AM+BM+CM+\min{\{AM,BM,CM\}} \leq AB+BC+CA.$

I really don't know how to solve it.

This is ISL 1999 geometry problem 1, and one solution can be found on http://www.kalva.demon.co.uk/short/soln/sh99g1.html . The problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?p=131828 but the discussion got lost.

  Darij

I had this problem in 2000 IMO TST :) I just posted the simplest solution possible http://www.mathlinks.ro/Forum/viewtopic.php?p=145946#p145946

Try to prove the lemma separate :P

Thank you very much. But I know a person from Kazakhstan who gave me this problem. He told me that it had been proposed at Kazakhstan olympiad (where he took part). Could it be true?

It's a problem from the ISL, and many countries are tempted to use these in their olympiads.


Find the least positive integer n such that 1/(sin 45 sin 46) + 1/(sin 47 sin 48) + ... + 1/(sin 133 sin 134) = 2/sin n.
No solution on kalva. Help!

$\cot(n) - \cot(n+1) = \frac{\sin(1)}{\sin(n) \sin(n+1)} \Rightarrow$ ${\text{blah} = \frac{1}{\sin(1)} \cdot (\cot(45) - \cot(46) + \cot(47) - ... - \cot(134))}$ but $\cot(180 - x) = -\cot(x)$, so $\text{blah} = \frac{1}{\sin(1)} \cdot \cot(45) = \frac{1}{\sin(1)}$. Er yea, so I'm getting a $1$ instead of a $2$ in the numerator, but I bet the answer's 1...now where did I screw up...

I think you forgot a 1/2 in one of your equations (probably for sinasinb). Could you be slightly more verbose?

[b]MysticTerminator[/b], no, your solution is completely correct. I checked myself, and also ran a computation through Maple, and indeed, the answer is $1/\sin(1)$.

[b]Kierenjl[/b], my guess is Kalva's site has a small typo for the problem. The numerator should be $1$ instead of $2$. :lol:


hey guys. i'm rachel.  first time on this forum.  i'm obsessed w/ math and the amc, i'm pretty good but i'm trying to get better.  kind of intimidating reading all the posts about people who got perfect or close-to-perfect scores!

can anyone remind of the numbers problems for #1? something like, CDs are $\$$ 20 each and Pam buys 5 at 30% off, and Sue buys 4 at 25%, what is the difference between the amount Pam paid and the amount Sue paid?

In that case the answer was 10 (C)? is this right or am i not remembering correctly?


In a circle with center O, lines WZ and XY are diameters with lengths of 12. What is the area of the shaded region? (see attached)

I'm lost. I am not doing this right at all. I know that my answer and calculations are way off. Please steer me in the right direction. I have the wrong values. Here's what I've done:

The triangles are special right triangles.

<WXO = 45°   <WOX = 45°  <YOZ = 45°   <ZOY = 45°

The hypotenuse of both triangles = 6 which is also √2s.

6=√2s--->s = 18--->The other 2 sides = 18 each.

area = 1/2 base * height

1/2 (9) * 18 = 81

The other triangle is identical. The total area of the shaded region = 162

I also have:

c^2 = a^2 + b^2

c^2 = 36 = a^2 + b^2 which my calcs don't indicate - I have a + b =36

4.2476^ + 4.2476^ = 36

1/2(4.2476) * 18 = 38

[quote="Your Mistake"]The hypotenuse of both triangles = 6 which is also √2s.
[color=red]6=√2s--->s = 18[/color]--->The other 2 sides = 18 each. [/quote]

$ 6 \equal{} s \sqrt {2} \implies s \equal{} \frac {6}{\sqrt {2}} \neq 18$  :P

Then, use $ A_{triangle} \equal{} \frac{1}{2} (s) (s)$, and double it.

[b]Visual Solution:[/b] Imagine line $ \overline{XY}$ as a ramp, and slide the lower triangle up to the top of the ramp. Now note that the two triangles together form a square (consider the angles -- you've already calculated these). Now, just find the area of this square.

WZ = 12, so WO = 6, and OX = 6 as well. The sidelength of the square is then 6 / sin(45) = 6/sqrt(2). Then, the area of the square is (6/sqrt(2))^2 = 18.

[quote="The Zuton Force"]Imagine line $ \overline{XY}$ as a ramp, and slide the lower triangle up to the top of the ramp. Now note that the two triangles together form a square (consider the angles -- you've already calculated these). Now, just find the area of this square.

WZ = 12, so WO = 6, and OX = 6 as well. The sidelength of the square is then 6 / sin(45) = 6/sqrt(2). Then, the area of the square is (6/sqrt(2))^2 = 18.[/quote]Wow-I didn't even think to put the two triangles together to make a square. You explained it very clearly.
Thank you so very much.

:)

You dont even need to put them together to make a square. They are 45-45-90 triangles with hypotenuse 6. The answer proceeds very easily from there xd


$a,b,c,d>0$ find the minimum of $\frac{cd}{a^2+b^2}+\frac{da}{b^2+c^2}+\frac{ab}{c^2+d^2}+\frac{bc}{d^2+a^2}$
I guess it is $2$ but I can't prove it. :(

Take a = c = 1 and b = d = 0 (or, if you, want $b=d=\epsilon$ for a very small $\epsilon$). Then, your expression is 0. So the minimum (or infimum) is 0.

  Darij

you are right.
but what about the minimum of $\frac{cd}{a^2+b^2}+\frac{da}{b^2+c^2}+\frac{ab}{c^2+d^2}+\frac{bc}{d^2+a^2} +\frac{ac}{b^2+d^2}+\frac{bd}{a^2+c^2}$

[quote="darij grinberg"] ...So the minimum (or infimum) is 0...
  Darij[/quote]
Sorry, Darij. Infinum! ;)

Well, this follows directly from cauchy-schwarz in engel-form:

$\sum_{sym} \frac{ab}{c^2+d^2} = \sum_{sym} \frac{a^2b^2}{abc^2+abd^2}\geq \frac{(\sum_{sym} ab)^2}{2\sum_{sym} abc^2}\geq 12$.

The last inequality follows from expanding the square and using $a^2b^2 + a^2c^2\geq 2a^2bc$ and stuff like that.

Peter


calculus
integration
geometry
derivative
function
calculus computations

Prove that $ \int_a ^b f(x) dx$ is the area under the curve $ f(x)$ between the points $ a$ and $ b$.

That is [u]not[/u] the Fundamental Theorem of Calculus.

What you just said is not the Fundamental Theorem of Calculus.

What did you say?

$ \int_a^bf(x)\,dx$ is simply a symbol for the integral of $ f.$ To give it any meaning, we must supply a careful mathematical definition of the integral (as, for instance, the appropriate limit of Riemann sums, or else perhaps as the coinciding supremum of lower sums and infinum of upper sums).

"The area under the curve ..." is not a careful mathematical statement at all. It's an informal heuristic, a motivation for what we would like to see in the definition of the integral. In the end, it's valid in some circumstances but requires modification in other circumstances, such as $ f$ taking on negative values, or such as the graph of $ f$ not being reasonably described as a "curve."

To give a careful definition of the integral, and to show that it meets the reasonable expectation of that heuristic -- that would be a reasonable exercise. In fact, it's something you have to do to develop the theory of the integral. But that exercise does not go by the "Fundamental Theorem" name.

The actual Fundamental Theorem of Calculus is the statement that integration and differentiation are inversely related.

As preliminaries before the FTC, we must have a theory that includes the following:

• A theory (Riemann or Darboux) of integration, sufficiently developed to include the following theorems:

• $ \int_a^cf(x)\,dx\equal{}\int_a^bf(x)\,dx\plus{}\int_b^cf(x)\,dx$ whenever all of those make sense.

• If $ a<b$ and $ f(x)\le g(x)$ on $ [a,b]$ with $ f,g$ both integrable, then $ \int_a^bf(x)\,dx\le\int_a^bg(x)\,dx.$

• The definition of continuity, and the basic $ \epsilon$-$ \delta$ tools for dealing with limit definitions.

• The definition of the derivative.

• The Mean Value Theorem.

Once we have all of those pieces in place, we can state the FTC. Some textbooks organize it into two separate theorems, and I will do so here.

1. The Derivative of the Integral version of the FTC:

Suppose $ f$ is integrable over an interval that contains $ a,$ and suppose $ f$ is continuous at $ x_0.$ Define $ F(x)\equal{}\int_a^xf(t)\,dt.$ Then $ F$ is continuous on the interval in question, and $ F$ is differentiable at $ x_0,$ with $ F'(x_0)\equal{}f(x_0).$

2. The Integral of the Derivative version of the FTC:

Suppose $ f$ is integrable on $ [a,b]$ and $ G$ is a function continuous on $ [a,b]$ and differentiable on $ (a,b)$ such that $ G'(x)\equal{}f(x)$ everywhere on $ (a,b).$ Then $ \int_a^bf(x)\,dx\equal{}G(b)\minus{}G(a).$

You can prove #2 as a corollary of #1, or you can give it its own independent proof. Either way, you're going to need the Mean Value Theorem, if just for the statement that two functions with the same derivative on an interval must differ by a constant.

I just watched a lecture that stated that as the fundamental theorem of calculus.  I apologize, I may have misunderstood.

(I edited my first post.)

As I said, that wouldn't so much be a proof as it would be showing that a formal definition satisfies certain informal expectations.


inequalities
algebra
polynomial
three variable inequality

Let a,b,c be any real number and a,b,c>0.prove that:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq\frac{a\plus{}b}{b\plus{}c}\plus{}\frac{b\plus{}c}{c\plus{}a}\plus{}1$

A counterexample : $ c\equal{}a\equal{}2,b\equal{}3$.

[quote="xiaohui"]Let a,b,c be any real number and a,b,c>0.prove that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} 1$[/quote]
I think
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a}\plus{}\frac{a\plus{}c}{a\plus{}b}$
Here is a problem from mathlinks contest :lol:

[quote="xiaohui"]Let a,b,c be any real number and a,b,c>0.prove that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} 1$[/quote]
I thinks should be rewrite below
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a\plus{}b} \plus{} 1$
and I had slution for this probelem

[quote="lenhamquy"]Let $ x,y,z > 0$, prove that

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}\geq\frac {x \plus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{x \plus{} y} \plus{} 1.$[/quote]
$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \minus{} \frac {x \plus{} y}{y \plus{} z} \minus{} \frac {y \plus{} z}{x \plus{} y} \minus{} 1$

$ \equal{} \frac {y^4x \plus{} x^3z^2 \plus{} y^3z^2 \plus{} y^3x^2 \plus{} z^3y^2 \minus{} 2y^3zx \minus{} 2xy^2z^2 \minus{} zx^2y^2}{xyz(y \plus{} z)(x \plus{} y)}$

$ \equal{} \frac {2y^3(z \minus{} x)^2 \plus{} x(y^2 \minus{} zx)^2 \plus{} y^4x \plus{} 2x^3z^2 \plus{} z^3y^2 \minus{} 4xy^2z^2}{2xyz(y \plus{} z)(x \plus{} y)}\geq0$,

which is true by the arithmetic mean - geometric mean inequality.

[quote="lenhamquy"][quote="xiaohui"]Let a,b,c be any real number and a,b,c>0.prove that:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} 1$[/quote]
I thinks should be rewrite below
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} 1$
and I had slution for this probelem[/quote]
The inequality is equivalent to
\[ (a\plus{}b\plus{}c)\left( \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{}3\right) \ge \frac{(a\plus{}b\plus{}c)(a\minus{}c)^2}{(a\plus{}b)(b\plus{}c)}\]
We have
\[ (a\plus{}b\plus{}c) \left( \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{}3\right) \equal{}\frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \plus{}\frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b} \minus{}2(a\plus{}b\plus{}c) \ge \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \minus{}a\minus{}b\minus{}c \equal{}\frac{(a\minus{}b)^2}{b}\plus{}\frac{(b\minus{}c)^2}{c}\plus{}\frac{(a\minus{}c)^2}{a} \ge \frac{(a\minus{}c)^2}{b\plus{}c} \plus{}\frac{(a\minus{}c)^2}{a}\]
It suffices to prove that
\[ \frac{1}{a} \plus{}\frac{1}{b\plus{}c} \ge \frac{a\plus{}b\plus{}c}{(a\plus{}b)(b\plus{}c)}\]
\[ \Leftrightarrow \frac{1}{a} \ge \frac{c}{(a\plus{}b)(b\plus{}c)}\]
which is obviously true because
\[ \frac{c}{(a\plus{}b)(b\plus{}c) } \le \frac{c}{ac} \equal{}\frac{1}{a}\]
We have done.
:)

Beautiful proof, can_hang2007! :lol:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107534

[quote="Math pro"][quote="arqady"][quote="Ji Chen"]

$ \equal{} \frac {2y^3(z \minus{} x)^2 \plus{} x(y^2 \minus{} zx)^2 \plus{} y^4x \plus{} 2x^3z^2 \plus{} z^3y^2 \minus{} 4xy^2z^2}{2xyz(y \plus{} z)(x \plus{} y)}\geq0$,

which is true by the arithmetic mean - geometric mean inequality.
[/quote]
I don't see it.  :maybe: Explain please. Thank you![/quote]

$ y^4x \plus{} x^3z^2 \plus{} z^3y^2 \plus{} x^3z^2 \minus{} 4xy^2z^2\geq0$
it is true by AM-GM! :D[/quote]
Sure?
;)

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \minus{} \frac {x \plus{} y}{y \plus{} z} \minus{} \frac {y \plus{} z}{x \plus{} y} \minus{} 1$

$ \equal{} \frac {2y^3(z \minus{} x)^2 \plus{} x(y^2 \minus{} zx)^2 \plus{} y^4x \plus{} x^3z^2 \plus{} 2z^3y^2 \minus{} 4xy^2z^2}{2xyz(y \plus{} z)(x \plus{} y)}\geq0$,

which is true by the arithmetic mean - geometric mean inequality.

In fact, the generalized inequality

$ \frac {x^p}{y^p} \plus{} \frac {y^p}{z^p} \plus{} \frac {z^p}{x^p}\geq\left(\frac {x \plus{} y}{y \plus{} z}\right)^p \plus{} \left(\frac {y \plus{} z}{x \plus{} y}\right)^p \plus{} 1$

holds for any real number $ p$.

If $ a,b,c>0$, then

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} k$,

where $ k\equal{}0.90993888\cdots$ is a root of the irreducible polynomial

$ k^{14}\plus{}16k^{13}\plus{}82k^{12}\minus{}29k^{11}\minus{}2598k^{10}\minus{}19736k^9\minus{}100825k^8\minus{}348832k^7$

$ \minus{}867328k^6\minus{}1981184k^5\minus{}2965984k^4\minus{}1805568k^3\plus{}1314304k^2\plus{}3594240k$

$ \plus{}1000192$;

with equality if $ b\equal{}1$,$ c\equal{}0.70196811\cdots$ is a root of the irreducible polynomial

$ c^{14}\plus{}9c^{13}\plus{}28c^{12}\plus{}24c^{11}\minus{}51c^{10}\minus{}135c^9\minus{}131c^8\minus{}80c^7\minus{}33c^6\minus{}9c^5\plus{}10c^4$

$ \plus{}24c^3\plus{}19c^2\plus{}7c\plus{}1$,

and $ a\equal{}0.76314238\cdots$ is a root of the irreducible polynomial

$ a^{14}\minus{}4a^{13}\plus{}6a^{12}\minus{}18a^{11}\plus{}32a^{10}\plus{}7a^9\minus{}35a^8\plus{}36a^7\minus{}27a^6\minus{}70a^5\plus{}28a^4$

$ \plus{}38a^3\minus{}6a^2\minus{}5a\plus{}1$.

[quote="can_hang2007"][quote="shfdfzhjj"][quote="Ji Chen"]
In fact, the generalized inequality

$ \frac {x^p}{y^p} \plus{} \frac {y^p}{z^p} \plus{} \frac {z^p}{x^p}\geq\left(\frac {x \plus{} y}{y \plus{} z}\right)^p \plus{} \left(\frac {y \plus{} z}{x \plus{} y}\right)^p \plus{} 1$

holds for any real number $ p$.

[/quote][/quote]
Yeah, I think this is a nice lemma. It is known, and we can prove it by Abel's summation. :)[/quote]

Yes,this lemma is known,I mean I get the inequality Ji Chen notice last summer.
Sorry for my bad English. :(


After this one

http://www.mathlinks.ro/viewtopic.php?t=319530&sid=c068f5cadd6caea70bb1dc3e256ae1db

you also have

$ A,B$ in $ M(n,C)$ such that $ AB^2\minus{}B^2A\equal{}B$
then $ B$ is nilpotent with odd order

[quote="Moubinool"]
$ A,B$ in $ M(n,C)$ such that $ AB^2 \minus{} B^2A \equal{} B$
then $ B$ is nilpotent with odd order[/quote]
 The same  we have  : $ B^{k\minus{}1}(AB^2\minus{}B^2A)\equal{}B^k ,\forall k \geq 1 ,k \in \mathbb{N}$
 Use $ Tr(XY)\equal{}Tr(YX) ,\forall X,Y \in M(n,C)$ then
  $ Tr(B^k)\equal{}Tr(B^{k\minus{}1}AB^2)\minus{}Tr(B^{k\plus{}1}A)\equal{}0$, $ \forall k \geq 1 ,k \in \mathbb{N}$
 Easy ....

you have to prove $ B^{2m\plus{}1}\equal{}0$ for some positive integer

By a recurrence reasoning:  $ AB^{2k}\minus{}B^{2k}A\equal{}kB^{2k\minus{}1}$.
Thus if $ B^{2k}\equal{}0$ then $ B^{2k\minus{}1}\equal{}0$ and the index of nilpotence of $ B$ is odd.


Η γεωμετρία που πρότεινε ο κύριος Στεργίου είναι πολύ καλή και θα ασχοληθώ με αυτή σήμερα.

Για όσους όμως τελειώνουν γρήγορα με αυτές (που υπάρχουν αρκετοί στο ελληνικό φόρουμ... :lol: ) να άλλη μία.

Έχω δύο λύσεις γι αυτήν. Μία που προτείνεται και μία δικιά μου.
Δική σας:

Δίνεται τρίγωνο $ ABC$ και έστω ο εγγεγραμμένος με κέντρο $ I$ που εφάπτεται της $ AC$ στο $ E$, της $ AB$ στο $ F$ και της $ BC$ στο $ D$. Η διχοτόμος της $ <BIC$ τέμνει την $ BC$ στο $ M$. Οι $ AM$ και $ EF$ τέμνονται στο $ P$. Να δείξετε ότι η $ DP$ είναι διχοτόμος της $ <FDE$.

Exei syzhththei arketes fores sto forum auti i askisi...

http://www.mathlinks.ro/Forum/viewtopic.php?t=24699

aaa kai gia ti gonia boreis na xrisimopoieis to \angle


Can someone please post the names of the individuals at the California and Penn State sites who got 7's on the individual?  I would also be interested in knowing a rough guess as to how many 6's there were.  Thanks!

At Iowa, the two 7's were Zach Abel (Texas) and Laurence Chan (Chicago).  There were probably 20-30 6's for the site tiebreaker.

I remember hearing the number 18 somewhere, although that may have had nothing to do with sixes. Yeah, I know I'm helpful.  :D

At San Jose, there was 1 7's and that was Curtis Liu.

There were 11-12 6's

I think there were 12 at PS, 3 elsewhere for 15 total. The 7s I remember were Linda Liu, Brian Lawrence, Arnav Tripathy, Tom Belulovich, yea mainly the top placers whose names you could look up anyway

[quote="MysticTerminator"]Linda Liu[/quote]

Dude...the one at MOP last year? I didn't know she pwned... O.O

I know some of the other PS 7's... HyunSoo Kim and Ryan Ko.

BTW, didnt Richard McCutchen also score a 7?

[quote="xxreddevilzxx"]BTW, didnt Richard McCutchen also score a 7?[/quote]As far as I know, he [url=http://oldweb.uwp.edu/academic/mathematics/usaco/2005/camp05/camp05.htm]wasn't[/url] at ARML.

I think Ameya Velingker also had a 7.

i got a 7 =D at penn state, kinda slow on the tiebreakers though....on our bus ride i had to listen to ryan mocking me about how he was 6x faster than me....

[quote="MysticTerminator"]...Brian Lawrence, Arnav Tripathy...[/quote]
weird... Two of the top 10 or so people at Mathcounts about 3 years ago

Also, from our Penn State Site

one person from the Florida Team: Gabor Debreczeni

[quote="nat mc"][quote="Iversonfan2005"][quote="MysticTerminator"]Arnav Tripathy[/quote]

I wonder who that could be :roll:[/quote]
Oh so you're Arnav! Whenever I looked at your username and avatar I thought that you would look like AI and I never would have thought you were Arnav (yourself)  :lol:[/quote]Actually I think Arnav is MysticTerminator.

[quote="nat mc"][quote="Iversonfan2005"][quote="MysticTerminator"]Arnav Tripathy[/quote]

I wonder who that could be :roll:[/quote]
Oh so you're Arnav! Whenever I looked at your username and avatar I thought that you would look like AI and I never would have thought you were Arnav (yourself)  :lol:[/quote]

oh no no, i'm not arnav, I was just laughing at the fact that mysticterminator always refers to himself (Arnav) in third person.  :P


If $ z$ satisfies $ \frac{i\plus{}z}{i\minus{}z}\minus{}\frac{i\minus{}z}{i\plus{}z}\equal{}\minus{}2i$ , prove that $ z\in \mathbb{R}$

[quote="Pain rinnegan"]If $ z$ satisfies $ \frac {i \plus{} z}{i \minus{} z} \minus{} \frac {i \minus{} z}{i \plus{} z} \equal{} \minus{} 2i$ , prove that $ z\in \mathbb{R}$[/quote]
What is $ i$ $ ?$
If you are mean that $ i^2\equal{}\minus{}1$ then, $ z\equal{}1\in\mathbb R.$

hello, we have
$ \frac{i\plus{}z}{i\minus{}z}\minus{}\frac{i\minus{}z}{i\plus{}z}\plus{}2i\equal{}\minus{}\frac{2i(z\minus{}1)^2}{(i\minus{}z)(i\plus{}z)}\equal{}0$ and from here we get $ z\equal{}1$
Sonnhard.

Let $ t=\frac{i+z}{i-z}$.
So
\begin{align*}
\frac{i+z}{i-z}-\frac{i-z}{i+z}=-2i
&\iff t-\frac{1}{t}=-2i \\
&\iff t^2+2it-1=(t+i)^2=0 \\
&\iff t=-i
\end{align*}

Thus
\begin{align*}
i = \frac{i+z}{i-z}
&\iff -zi-1=i+z \\
&\iff (1-i)z=-(1-i) \\
&\iff \boxed{z=-1}
\end{align*}

hello, correct your solution please, the right answer is $ z\equal{}1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, correct your solution please, the right answer is $ z = 1$.
Sonnhard.[/quote]

Sorry.
I'm typo :(

[quote="Kouichi Nakagawa"]Thus
\begin{align*}
i = \frac {i + z}{i - z}
& \iff - zi - 1 = i + z \\
& \iff (1 \color{red}+\color{black} i)z = - (1 \color{red}+\color{black} i) \\
& \iff \boxed{z = - 1}
\end{align*}

(This post is typo, too. But I did a bigger typo.)[/quote]

Thus
$ \begin{align*}
\color{red}-\color{black}i = \frac {i + z}{i - z}
& \iff zi + 1 = i + z \\
& \iff (1 - i)z = 1 - i \\
& \iff \boxed{z = 1}$

hello, thank you, everybody can make a mistake or errare humanum est.
Sonnhard.


MATHCOUNTS
articles
ratio
number theory
prime numbers

I will be compiling articles, problems, concepts, etcetera to help any beginning "Mathlete", or anyone needing help in a topic, perform better in Mathcounts, along with solidifying a base somewhat.  Please post any topics you feel that should be covered, and help each other out.

Websites:

http://www.purplemath.com/modules/index.htm
      This is a great site that teaches many basic to advanced things in Algebra.

http://www.mathcounts.org/Program/toolbox.pdf
      Definitions, formulae, random things that should be known, you name it (including common fractions!!!)

Basic Definitions:

Rational Numbers - the ratio of integers formed by dividing one by another

Natural Numbers - the set of the counting numbers. They can be classified as odd or even. {1,2,3,4,5,6...}

Whole Numbers - the set of natural numbers and zero. {0,1,2,3,4,5...}Integers-the set of the natural numbers, their opposites, and zero. {-4,-3,-2,-1,0,1,2,3,4...}

Irrational Numbers - the set of all the numbers that can be expressed in the form a/b where a and b are integers. b cannot equal zero. this is because anything divided by 0 is undefined.  The set of the numbers that cannot be written as terminating or repeating decimals.

Real Numbers - rational and irrational numbers

Prime Numbers - Integers that only have divisors of one and itself.  {2,3,5,7,11,13,17,19,23...}

Well, so far there's basically nothings, but there's more to come soon!

If you have any questions on anything, please post here.

oooo, good idea. i think common fraction should be on that list because it's something you really don't wanna mess up on...

Not necessarily but it's good if you can do some basic logarithrms or maybe decimal conversions like [tex]\frac {1}{11}[/tex]is 0.0909090....


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finitely generated group. Prove that for each $n$ number of subgroups of index $n$ in $G$ are finite.

It's quite straightforward, isn't it? Any subgroup of index $n$ induces a homomorphism from $G$ to $S_{n}$ and any such morphism is determined by its values on the generators of the group, thus there are finitely many such morphisms.

[quote="harazi"] Any subgroup of index $n$ induces a homomorphism from $G$ to $S_{n}$ and any such morphism is determined by its values on the generators of the group, thus there are finitely many such morphisms.[/quote]
I have a problem!
we know that there finite Homomorphism that sends $G$ to $S_{n}$ but it's not enough ,maybe one
this kind of Homomorphism sends $G$ to $S_{n}$  for infinitley subgroups.
Could you say an example about this kind of homomorphism for this problem?

As harazi has said it is quite straightforward to show that there are finitely many such morphisms. And sth easier is what you have stated. Prove that these homomorphisms are distinct for $H\neq K$ subgroups. You may need to see the proof of Cayley's theorem.

i think it's a serious problem which payman_pm has mentioned.

if $H$ has index $n$, we get a homomorphism $G \to Sym(G/H)$, but then we have to choose a bijection $G/H \cong \{1,...,n\}$ to get homomorphisms $G \to S_{n}$. this choice makes it impossible to get distinct homomorphisms ... unless we specify it.

For every $H<G$ we difine $\phi_{H}: G\longrightarrow S_{n}$ such that
$\forall x \in G : \phi_{H}(x)= yH\longmapsto xyH$ and suppose
$H_{1}$,$H_{2}$,...,$H_{n}$ are left cosets of $H$ which $H_{1}$=$H$ and $H_{i}$ is the same as $i$ in $S_{n}$.
we can simply show that $\forall H<G$  $\phi_{H}$ is a homomorphism,but i want to show that if $\phi_{H}=\phi_{K}$ $\longrightarrow$  $H=K$
$\forall$ $h\in H$ $\phi_{H}(h)$ sends$1\longmapsto 1$ thus $\phi_{K}(h)$ sends $1\longmapsto1$ thus $h\in K$ thus $H\subseteq K$ similary we can show that $K\subseteq H$ thus  $K=H$


This is from the 2005 Mock AMC A. I couldn't find it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15513]here[/url].

Compute $\tan{2A}$ if $\sin{A}\cos{A}\cos{B}^2 + \sin{A}\cos{A}\sin{B}^2 - \sin{B}\cos{B}\cos{A}^2 - \sin{B}\cos{B}\sin{A}^2$ = 1.

Working from that, I got $\sin{2A} - \sin{2B} = 2$. However, this does not seem to get me to the answer.

The choices are 0, 1, $\sqrt{3}$, $\frac{\sqrt{3}}{3}$ and undefined.

That equation $\sin 2A-\sin 2B=2$ certainly can be used to arrive at the answer.

[hide="hint"]think range of sine[/hide]

Edit: You're right, oops. LOL. It's TAN 2A... not $\tan{a}$.

[hide]Anyway, so the period of that is $\pi$, so it's $1$ at $A = \frac{\pi}{4}$, and $\tan{\frac{\pi}{2}}$ is undefined. [/hide]

Actually, that doesn't give you the right answer, since you answered the wrong thing. ;)

[hide="Answer"]First two terms simplify to $\sin A\cos A$, second two simplify to $\sin B\cos B$, so we have $\sin A\cos A-\sin B\cos B=1\Rightarrow 2\sin A\cos A-2\sin B\cos B=\sin 2A-\sin 2B=2$. Therefore, we must have $\sin 2A=1$ and $\sin 2B=-1$, so $2A=\frac{\pi}{2}$, and $\tan 2A$ is undefined. $\boxed{E}$[/hide]


I'm currently reading Euclid's "The Elements," and I had some questions which might be useful to others who are also studying the book.

First of all, what is Euclid meant to teach me?  I see that most of his proofs show concepts we learned in basic geometry, although I would never have thought of them myself.  Is the importance of Euclid in the way he reached his solutions (proofs)?

Second, I was wondering what I could do to get the most out of the book.  One of my friends said that he tried to independently prove each proposition before looking at the so-called "solution," but I was wondering whether this was efficient.  I know many of you have found the solution of a problem obvious, but only after you see it.  That happens to me a lot, so I'm kind of hesitant about reading the proofs straight.

Thank you for considering this post.  Any suggestions would be greatly appreciated.

I read the Elements in around April. And I think that the most important thing the book gave me is its logic system. I was really astonished to see that Euclid implied everything from just 5 axioms. The problems aren't so difficult but it is very helpful to read how he built up the classical geometry.

I agree that Euclid's work is a supurb reference for geometry.  However, it is incomplete logically.  Also it has five "postulates" and five "common notions", for a total of ten.  Furthermore, in the first book's definitions, sneaky assumptions are made :  in definition 17, he assumes that a diameter bisects a circle, which can be proved using simpler axioms; and in definition 21, he assumes without due argument that a triangle can obtain a right angle, an obtuse angle, or three acute angles, which in reality requires justification, with help from the parallel postulate.   In the first five propostions of the first book we can find fault with propositions 1 (how do we know that the circles intersect? (cliché)), 4 (the controversial principle of superposition), and 5 (he assumes without justification that isosceles triangles are possible, and also that we can select a "random point" on a line).

Nonetheless, as I said above, it is a very nice work, and the objections I listed were considered obvious in those days.  Of course, if we can declare something "obvious" from a diagram, then that defeats the notion of axioms.  There are some very elegant proofs in there (for instance, that there are infinitely many primes, which I had not seen before).

However, the supreme master of axiomatic reasoning and logic is the great DAVID HILBERT.  I have looked at his [i]Foundations of Geometry[/i] (tenth German edition/second English edition).  The entire book looks like a very tall order, but the first chapter is true light and splendor. In this, he gives nineteen axioms (yes, you really do need that many), not counting the parallel axiom, which, coupled by beautiful proofs that overawe me and subtle, precise definitions, reduce every other basic geometric system to ashes.  With Hilbert's system, one dous not need to draw a diagram to determine anything; using only words and those wonderful axioms, one can justify [i]everything[/i].

[quote="Boy Soprano"]I agree that Euclid's work is a supurb reference for geometry.  However, it is incomplete logically.  Also it has five "postulates" and five "common notions", for a total of ten.  Furthermore, in the first book's definitions, sneaky assumptions are made :  in definition 17, he assumes that a diameter bisects a circle, which can be proved using simpler axioms; and in definition 21, he assumes without due argument that a triangle can obtain a right angle, an obtuse angle, or three acute angles, which in reality requires justification, with help from the parallel postulate.   In the first five propostions of the first book we can find fault with propositions 1 (how do we know that the circles intersect? (cliché)), 4 (the controversial principle of superposition), and 5 (he assumes without justification that isosceles triangles are possible, and also that we can select a "random point" on a line).[/quote]That's a very good and often ignored point. The Elements is in no way a "perfect" logical system. In fact it was Bertrand Russell's horror upon realizing this (when he was younger he had believed The Elements was such) that ultimately led to his become probably the greatest philosopher of mathematics.

I may have exaggerated, but I read something like that in a short introduction to Russell and math philosophy.

I have been trying to write an automated theorem prover for easy geometry problems, and although Euclid's Elements are great for human proofs, they just aren't logically sound enough for a computer to use. I have been using the book mentioned above: [i]Foundations of Geometry[/i] as my source of axioms, and it works beautifully. That really is a fantastic book, and as it's only $10 on http://www.amazon.com, it's definitely worth buying if you're interested in taking the subject of geometry to it's foundations and basically turning it into an algebra.

10 dollars is too much for me.  :P
I read some of the Axioms of Hilbert today. They are indeed more logical than Euclid's. But always remember that Euclid was thousands years of earlier than Hilbert...


logarithms
calculus
calculus computations

Prove that
\[ \sum_{n\equal{}2}^{\infty}\frac{1}{(\ln{n})^{\ln{n}}}\]
converges, but
\[ \sum_{n\equal{}2}^{\infty}\frac{1}{(\ln{n})^{\ln({\ln{n}})}}\]
diverges

Use Cauchy condensation test (see http://mathworld.wolfram.com/CauchyCondensationTest.html). You obtain in fact a new infinite series but the convergence character is unchanged.

Then apply the root test (the limit-version) to obtain the desired results.


find all function $ f: \mathbb R \to \mathbb R$ that satisfy the following condition for all reals $x$ and $y$:
\[f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y).\]

[quote="bjh7790"]find all function $ f: \mathbb R \to \mathbb R$ that satisfy the following condition.

$ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$[/quote]

This problem has already been posted and I answered with a different demo (but same result :) )

Let $ P(x,y)$ be the assertion $ f(x\plus{}yf(x))\equal{}f(x)\plus{}xf(y)$

(i) $ f(x)\equal{}0$ $ \forall x$ is a trivial solution. So we'll consider from now that $ \exists c$ such that $ f(c)\ne 0$

(ii) $ P(1,0)$ $ \implies$ $ f(1)\equal{}f(1)\plus{}f(0)$ and so $ f(0)\equal{}0$

(iii) If $ f(a)\equal{}0$ then $ P(a,c)$ $ \implies$ $ 0\equal{}af(c)$ and so $ a\equal{}0$

(iv) $ P(\minus{}1,\minus{}1)$ $ \implies$ $ f(\minus{}1\minus{}f(\minus{}1))\equal{}0$ $ \implies$ (from iii) $ f(\minus{}1)\equal{}\minus{}1$

If $ f(1)\ne 1$, $ P(1,\frac 1{1\minus{}f(1)}$ $ \implies$ $ f(\frac 1{1\minus{}f(1)})\equal{}f(1)\plus{}f(\frac 1{1\minus{}f(1)})$ and so $ f(1)\equal{}0$ which is impossible (see iii). So $ f(1)\equal{}1$

(v) $ P(1,x)$ $ \implies$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$ and so $ f(x)\equal{}f(x\plus{}1)\minus{}1$ and so $ f(x\minus{}1)\equal{}f(x)\minus{}1$ and so $ f(n)\equal{}n$ $ \forall n\in\mathbb Z$

(vi) Let $ n\in\mathbb Z$ : $ P(n,x\minus{}1)$ $ \implies$ $ f(nx)\equal{}n\plus{}nf(x\minus{}1)\equal{}n(f(x\minus{}1)\plus{}1)\equal{}nf(x)$ and so $ f(nx)\equal{}nf(x)$ and so $ f(x)\equal{}x$ $ \forall x\in\mathbb Q$

(vii) If $ f(a)\equal{}a$ for $ a\ne 0$, then $ P(a,x\minus{}1)$ $ \implies$ $ f(ax)\equal{}a\plus{}af(x\minus{}1)\equal{}af(x)$ and then $ P(a,y)$ may be written $ f(a\plus{}ya)\equal{}a\plus{}af(y)\equal{}a\plus{}f(ay)$; Setting then $ y\equal{}\frac xa$, we get $ f(a\plus{}x)\equal{}a\plus{}f(x)$, still true for $ a\equal{}0$

(viii) $ P(x,1)$ $ \implies$ $ f(x\plus{}f(x))\equal{}x\plus{}f(x)$. Using then $ a\equal{}x\plus{}f(x)$ in (vii), we get the new assertion $ Q(x,y)$ : $ f(x\plus{}y\plus{}f(x))\equal{}x\plus{}f(x)\plus{}f(y)$

(ix) $ Q(x,\minus{}x)$ $ \implies$ $ f(f(x))\equal{}x$

(x) For $ x\ne 0$, $ P(x,1\plus{}\frac y{f(x)})$ $ \implies$ $ f(x\plus{}y \plus{} f(x))\equal{}f(x)\plus{}xf(1\plus{}\frac y{f(x)})$ $ \equal{}x\plus{}f(x)\plus{}xf(\frac y{f(x)})$
Comparing this with $ Q(x,y)$, we get $ x\plus{}f(x)\plus{}f(y)\equal{}$ $ x\plus{}f(x)\plus{}xf(\frac y{f(x)})$ and so $ f(y)\equal{}xf(\frac y{f(x)})$
Using $ y\equal{}uv$ and $ x\equal{}f(u)$, we get then $ f(uv)\equal{}f(u)f(v)$ $ \forall uv$

(xi) from (x) we get that $ f(u^2)\equal{}f(u)^2$ and so $ f(x)>0$ $ \forall x>0$ and $ f(x)<0$ $ \forall x<0$
So, for $ x\ne 0$ $ P(x,\frac y{f(x)})$ $ \implies$ $ f(x\plus{}y)\equal{}f(x)\plus{}xf(\frac y{f(x)})$ and $ xf(\frac y{f(x)}$ has same sign as $ y$. So $ f(x)$ is increasing.
And since $ f(x)\equal{}x$ $ \forall x\in\mathbb Q$, we get that $ f(x)\equal{}x$ $ \forall x\in\mathbb R$

Hence the two solutions :
1) $ f(x)\equal{}0$ $ \forall x$
2) $ f(x)\equal{}x$ $ \forall x$


calculus
derivative
limit
calculus computations

I was asked this question recently:

let f be differentiable for all real numbers, suppose (a) is a real number, and:

(lim x --> infinity)(f(x)) = a;

assuming that the limit of

(lim x-> infinity)( x * f '(x)) : {IMPORTANT f'(x)  is the derivative of f(x), (it may be hard to visually spot on your computer screen) }
exists,

find this limit,

is the answer

limit = (number of roots of f(x)) * a ?

Im suspecting its actually zero, can someone show me the answer?

1) Please use Latex.

2) What limit are one supposed to find?

$ \lim_{x\to\infty}f(x) \equal{} \lim_{x\to\infty}\frac {f(x)x}{x} \equal{} (L'Hospital) \equal{} \lim_{x\to\infty}(f'(x)x \plus{} f(x))$
so the limit that you ask is $ 0$


number theory
greatest common divisor
number theory proposed

Sequence $ a_n$ of positive integer numbers satisfies the followin relation: This sequence is determined by its two first terms. And $ (a_(n\plus{}2)) GCD(a_(n\plus{}1),a_n) \equal{} a_n \plus{} a_(n\plus{}1)$.
Find all values for $ a_1 and a_2$ for which this sequence is bounded.

Since the sequence is bounded, we can only see finitely many different pair $ (a_n,a_{n + 1})$. Hence, the sequence is periodic.

It is not hard to see $ \gcd(a_n,a_{n + 1})$ is non-increasing. But since the sequence is periodic, there exists $ m > 0$ such that $ \gcd(a_n,a_{n + 1}) = t$ is constant for $ n\ge m$.

$ a_{n + 2} = \frac {a_n + a_{n + 1}}t,\forall n\ge m$.
Obviously $ t = 1$ will lead a unbounded sequence, $ t > 2$ will lead non-integer term in the sequence, So $ t = 2$.
$ a_{n + 2} - a_{n + 1} = - \frac12(a_{n + 1} - a_n)$. To guarantee all numbers to be integer, $ a_n$ is constant $ c$ for $ n\gem$.
Hence, $ c = \frac {c + c}{\gcd(c,c)} = 2$.
It is not hard to see the whole sequence is constant $ 2$.

Sorry but Could You explain because I don't understand why:

 there exists $ m > 0$ such that $ gcd(a_{n \plus{} 1} , a_{n})$  is constant for ALL $ n > \equal{} m$ ? Why this gcd could not be smaller than $ t$ for some pair $ (a_{i},a_{i\plus{}1})$ where $ i>\equal{}n$?

thanks

I will give more details.
Suppose $ c\equal{}\frac{a\plus{}b}s$, where $ s\equal{}\gcd(a,b)$.
$ \gcd(\frac as,\frac bs)\equal{}1\Rightarrow \gcd(\frac{a\plus{}b}s,\frac bs)\equal{}1\Rightarrow \gcd(c,b)\equal{}\gcd(\frac{a\plus{}b}s,s*\frac bs)\le s$.
Hence, $ \gcd(a_n,a_{n\plus{}1})$ is non-increasing in $ n$.
But we know $ a_n$ is periodic. say $ a_k\equal{}a_{p\plus{}k},\forall k\ge m$.
Then you look at the series $ b_n\equal{}\gcd(a_n,a_{n\plus{}1})$. It is also periodic for $ n\ge m$. A periodic non-increasing series have to be all constant.
Hopefully it is clear now.


Let a,b,c,d be real number such that
i)$ 0<a,b,c<d$;
ii)$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \minus{} \sqrt{\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}}\equal{}\frac{2}{d}$
Prove that
 $ ab\plus{}bc\plus{}ca<\sqrt{3}d^2$

[quote="FibonacciFan"][hide]
Multiply ii) by abc and rearrange:

$ ab \plus{} bc \plus{} ac \equal{} \frac {2abc}{d} \plus{} \sqrt {a^2 b^2 \plus{} b^2 c^2 \plus{} a^2 c^2}$

Then use i) inside the square root:

$ \frac {2abc}{d} \plus{} \sqrt {a^2 b^2 \plus{} b^2 c^2 \plus{} a^2 c^2} < \frac {2abc}{d} \plus{} \sqrt {3 d^4} \equal{} \frac {2abc}{d} \plus{} d^2\sqrt {3}$

So, RHS of ii) could really have been $ 0$ instead of $ 2/d$.
[/hide][/quote]
Your solution only prove $ ab \plus{} bc \plus{} ac <\frac {2abc}{d} \plus{} d^2\sqrt {3}$
but my my problem is $ ab \plus{} bc \plus{} ac < d^2\sqrt {3}$ (note that $ \frac{2abc}{d} >0)$  :rotfl:

[quote="8826"]Let a,b,c,d be real number such that
i)$ 0 < a,b,c < d$;
ii)$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \sqrt {\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}} \equal{} \frac {2}{d}$
Prove that
 $ ab \plus{} bc \plus{} ca < \sqrt {3}d^2$[/quote]
Since ii) we have :
$ \sqrt{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}\equal{}ab\plus{}bc\plus{}ca\minus{}\frac{2abc}{d}\equal{}ab\plus{}bc(1\minus{}\frac{a}{d})\plus{}ca(1\minus{}\frac{b}{d})$  ,,, (*)
Since i) and (*) we have:$ \sqrt{a^2b^2\plus{}b^2c^2\plus{}c^2a^2} < ab\plus{}bd\minus{}ab\plus{}ad\minus{}ab \equal{}d^2\minus{}(d\minus{}a)(d\minus{}b) <d^2$
But $ ab\plus{}bc\plus{}ca \ge \sqrt{3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}$
Then.we have Q.E.D


Which of the following can be (and is) the sum of 5 consecutive integers?

(F)  43,210
(G) 43,211
(H) 43,212
(J) 43,213
(K) 43,214

Let the consecutive numbers be $ x \minus{} 2, x \minus{} 1, x, x \plus{} 1, x \plus{} 2$, then the sum of these numbers is $ 5x$. So the question becomes which of the numbers is divisible by $ 5$, which is clearly $ \boxed{F) \ 43,210}$.


vector
superior algebra
superior algebra unsolved

Let (A,+,*) be a ring with 5k+2 elements s.t. 1+1=0. Show that the eqn x4+x3+1=0 has no solutions in A.

If $1+1=0$ then $A$ contains $k = GF(2)$.  If $x^4 + x^3 + 1$ has a solution $a$ in $A$ then $A$ contains all powers of $a$, which is a primitive element of $K = GF(16)$.  So $A$ contains $K$ and hence is a finite-dimensional vector space over $K$.  So $\vert A \vert$ is a power of 16 and so $\equiv 1 \bmod 5$.


There are a few problems on FTW like,

[b]1.If you flip 5 coins, what is the probability that you will get more than two heads?[/b]

Wouldn't this just be $ \dfrac{1}{2} \plus{} \dfrac{1}{4} \plus{} \dfrac{1}{8} \equal{} \dfrac{7}{8}$ ?

You have a $ \dfrac{1}{2}$ probability of flipping a head once.

You have a $ \dfrac{1}{4}$ probability of flipping a head twice.

You have a $ \dfrac{1}{8}$ probability of flipping a head three times.

[b]2.If you flip a coin three times, what is the probability that you will get tails fewer than twice?[/b]

Isn't this just $ \dfrac{1}{2}^3 or \dfrac{1}{8}$

$ \dfrac{1}{2}$ probability of flipping a head.

$ \dfrac{1}{2}^2$ probability of flipping two tails.

Can someone tell me what I'm doing wrong?

Problem 1.
The probability that you get 5 heads is the same as the probability that you get 0 heads (5 tails).
The probability that you get 4 heads is the same as the probability that you get 1 heads (4 tails).
The probability that you get 3 heads is the same as the probability that you get 2 heads (3 tails).
So the answer is 1/2.
Problem 2.
Same reasoning gives 1/2.

1/2 is the probability of flipping a tail on one particular turn. (1/2)^2 is the probability of flipping two heads on two particular turns. Since you can get HHT, HTH, THH all with probability (1/2)^3, and also HHH with the same probability, you get (1/2)^3 *4=4/8=1/2, as before.

I suggest that you use Alcumus if you don't already.

[quote="AIME15USAMO"](stuff)

Can someone tell me what I'm doing wrong?[/quote]

May you explain your reasoning that got you those answers?  It would make it much easier to understand.

Ok, number one for example, there is a $ \dfrac{1}{2}$ probability of getting one head, $ \dfrac{1}{2}^2$ of getting 2 heads and so on. I know I'm doing it wrong but instead of telling me I'm doing it wrong and giving me the answer, can u tell me what to do?

I think that the main error is that when trying to compute the probability of getting $ n$ heads, the above (OP) method just finds the probability of the first $ n$ flips being heads and leaves off the rest.  However, we then need the rest of the coins to flip tails, with probability $ (1/2)^{\text{however many left}}$.  The final probability would then be $ (1/2)^{\text{total number of flips}}$.  As mentioned before, this assumes the first $ n$ flips are heads, but any set of $ n$ could be heads.

For example, suppose 3 coin tosses and you want exactly 2 to be heads (probability of).  The total number of ways to flip 3 coins should be simple, and you want 2 of the 3 to be heads.  In how many ways can you do this?

Hopefully this helps.


let
p(m,x)=Product (i=1 to m) x/(1-x^i)
q(m,x)=Product (i=1 to m) x^m/(1-x^i)^2

please prove:

Sum (i=1 to infinity) p(i,x)=Sum (i=1 to infinity) q(i,x)

The identity is false?

Take x=3.
\sum_{i=1}^{\infty} p(i,x)= -1.
\sum_{i=1}^{\infty} q(i,x) = 1.23862

??

i think I proved it using generating functions by considering two different ways to express a partition of n (parition function P) and establish a bijection.  the formula is up in sloane's integer sequences database but without proof: G.f.: Product_{k>0} 1/(1-x^k)=1+Sum_{k>0}
              x^(k^2)/(Product_{i=1..k} (1-x^i))^2.
there is also a proof using ferrers diagram

Just curious-- I have seen people write "bump" before. What does that mean?


Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca} \geq\ \frac{a}{a^2\plus{}2bc}\plus{}\frac{b}{b^2\plus{}2ca}\plus{}\frac{c}{c^2\plus{}2ab}$

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \geq\ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ca} \plus{} \frac {c}{c^2 \plus{} 2ab}$[/quote]
\[ \Leftrightarrow \sum {\frac {{a\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}}{{a^2 \plus{} 2bc}}} \ge 0
\]
Without loss of generality, we can assume $ a \ge b \ge c$, then we have
\[ \frac {{c\left( {c \minus{} a} \right)\left( {c \minus{} b} \right)}}{{c^2 \plus{} 2ab}} \ge 0
\]

\[ \frac {{a\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}}{{a^2 \plus{} 2bc}} \plus{} \frac {{b\left( {b \minus{} a} \right)\left( {b \minus{} c} \right)}}{{b^2 \plus{} 2ca}} \equal{} \frac {{\left( {a \minus{} b} \right)^2 \left( {2a^2 c \plus{} 2b^2 c \plus{} abc \minus{} ac^2 \minus{} bc^2 } \right)}}{{\left( {a^2 \plus{} 2bc} \right)\left( {b^2 \plus{} 2ca} \right)}} \ge 0
\]
Thus the inequality has been proved. Equality holds if and only if $ a \equal{} b \equal{} c$ or $ abc \equal{} 0$, :).

Nice proof, my friend.
Besides, expand solve it easily  :)


I was doing Chapter Sprint, and I got and understood all but these 2, so can you guys help?

10. A four-digit number uses each of the digits 1,2,3, and 4 exactly once. What is the probability that the number is a multiple of 4? Express your answer as a common fraction.

[hide] I got 1/2 but the answer is 1/4.[/hide]
[b]1st was COMPLETED[/b]

15. Right Triangle ABC is inscribed in a circle, the radius of the circle is 5 centimeters, and BC = 5 cm. AB is the hypotenuse. What is the number of centimeters in the length of (arc)BC? Express you answer as a common fraction in terms of pi.

I don't know the equation for how to find the degree of an arc, so plz try to explain that too.

[hide] all I could get was the sides of the triangle were 10, 5, and 5*sqrt3.

The answer is 5pi/3 (centimeters)

[/hide]

Thanks in advance[/b]

1.
Think like this
$ 0\pmod{4}\\
1\pmod{4}\\
2\pmod{4}\\
3\pmod{4}$
So out of the 4 things the number could be, and only one is a multiple of 4. $ \frac{1}{4}$

There may be a solution that way, but the solution I posted is the one I remember.

[hide]15. Let point O be the center of hte circle. OC and OB equal 5, since they are radii of circle O. You are given the BC is also 5, making triangle COB equilateral. Since Triangle COB is equilateral, angle COB is 60 degrees, making arc CB
(60/360)(diameter*pi)
1/6(10)pi
5/3pi[/hide]

To ihatepie:

About the 4 divisibility rule, it's because 100 is a multiple of 4 but 10 is not. By looking at only the ones digit of the number, you are saying that adding ten does not matter.
It is obvious that adding 100 doesn't change the divisibility of a number by 4 since 4x25=100

andrew, it is 5pi/3. Anyways, I forgot to say earlier that AB was the hypotenuse.

@dark I wasn't saying that adding by 10 doesn't matter, that's what hunter said.

My problem is I found all possible last two numbers,

12,13,14,21,23,24,31,32,33,34,41,42,43 which is 12.

I found the ways that it was divisible by 4 which was 12, 24, 32.

Which is 3/12 or 1/4.

Woah, I just realized now I made a mistake on the test. I got total possible as 3! on the test, not 3!2!.

[quote="darkmatter47"][quote="Ihatepie"]what I don't get about that, is that it is the last 2 digits that count for determining if it is divisible by four, not the last digit.[/quote]

I was refering to this statement.[/quote]

In that statement, I was talking about hunter's previous post.

[quote="hunter34"]1.
Think like this
$ 0\pmod{4} \\
1\pmod{4} \\
2\pmod{4} \\
3\pmod{4}$
So out of the 4 things the number could be, and only one is a multiple of 4. $ \frac {1}{4}$[/quote]

I said that I don't understand how his technique works because to find if a number is divisible by 4, you have to know the last two digits, not just the last digit.

Sorry for not being clear.

Oh...I see... Pardon my ignorance.

BTW- Where did you get the 2000 competition? I can only get 2001-2007.

I think it had stutz on the name of the website. It only 2000 Chapter Sprint, nothing else from 2000.

Edit: I looked and this is the site.

http://www.stutzfamily.com/mrstutz/mathcounts/index.html


A star-polygon is drawn on a clock face by drawing a chord from each
number to the fifth number counted clockwise from that number.  That
is, chords are drawn from 12 to 5, from 5 to 10, from 10 to 3, and
so on, ending back at 12.  What is the degree measure of the angle
at each vertex in the star-polygon?

First, we find the arc measure between two adjacent numbers.  That is simpily $ \frac {360}{12} \equal{} 30$.  If you draw out the star, you see that each vertex angle cuts off an arc of $ 2\cdot30 \equal{} 60$, since it is an arc made by $ 3$ consecutive numbers (an adjacent pair of $ 2$ adjacent numbers).  But we're not finished.  Since this is an inscribed angle, the angle measure is half of the intercepted arc.  Therefore, our answer is $ \frac {60}{2} \equal{} \boxed{30}$.  If anyone would like to post a picture, it might help...


function
real analysis
real analysis unsolved

if $ f(x,y)$ is a function, such that the x-sections and the y-sections of $ f$: $ f_{x}$ and $ f_{y}$ are measurable

what's a counterexample that $ f(x,y)$ is not a measurable function

(If we add the condition that $ f_{y}$ is continuous then it is true)

I think this is related to the continuum hypothesis; see [url=http://www.math.wvu.edu/~kcies/STA/STRAsurv/node2.html]here[/url].