I encountered this problem.In fact,when i tried this I did get close to the solution but it seems, I have to seek some help.

$ a\plus{}b\plus{}c\equal{}0$

$ a^5\plus{}b^5\equal{}(a^3\plus{}b^3)(a^2\plus{}b^2)\minus{}a^2b^2(a\plus{}b)$

$ a^3\plus{}b^3\plus{}c^3\equal{}3abc$

Find $ \frac{a^5\plus{}b^5\plus{}c^5}{(a^3\plus{}b^3\plus{}c^3)(a^2\plus{}b^2\plus{}c^2)}$

Thank you.

Why have you given the second and third conditions??

$ (a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3) \equal{} a^5 \plus{} b^5 \plus{} c^5 \plus{} a^2b^2(a \plus{} b) \plus{} b^2c^2(b \plus{} c) \plus{} c^2a^2(a \plus{} c) \equal{} a^5 \plus{} b^5 \plus{} c^5 \minus{} a^2b^2c \minus{} ab^2c^2 \minus{} a^2bc^2 \equal{} a^5 \plus{} b^5 \plus{} c^5 \minus{} abc(ab \plus{} bc \plus{} ca) \equal{} a^5 \plus{} b^5 \plus{} c^5 \minus{} \frac {1}{6}(3abc)(2ab \plus{} 2ac \plus{} 2bc) \equal{} a^5 \plus{} b^5 \plus{} c^5 \minus{} \frac {1}{6}(a^3 \plus{} b^3 \plus{} c^3)( \minus{} a^2 \minus{} b^2 \minus{} c^2)$
$ \Rightarrow \frac {5}{6}(a^2 \plus{} b^2 \plus{} c^2)(a^3 \plus{} b^3 \plus{} c^3) \equal{} a^5 \plus{} b^5 \plus{} c^5$

So the answer is $ \frac {5}{6}$

Another way to do such problem(the method is general and works well)
let $ a,b,c$ be roots of the equation $ x^{3} \plus{} px \minus{} q \equal{} 0$. It sure does satisfy an equation of this form as the sum of $ a,b,c$ is given to be zero and hence $ x^{2}$ doesnt appear in the equation.

Now $ \sum ab \equal{} p$ and $ abc \equal{} q$.
further sums of higher powers of $ a,b,c$ can be got easily:
$ x^{4} \equal{} \minus{}px^{2} \plus{}qx$
Since $ a,b,c$ satisfy this equation u can substitute it here and all three equations, to get something like
$ \sum a^{4} \equal{} \minus{}p\sum a^{2}$
and $ \sum a^{2} \equal{} 0 \minus{} 2p$
Similarly find $ \sum a^{5}$. and divide ur expression in $ p, q$ by the expressions u get for $ \sum a^{2}$ and $ \sum a^{3}$
and u r done  :lol:

[quote]
Thank you.
But, I did not understand the last post.Can you please explain what you meant?[/quote]
which part did u not understand? it is using polynomial algebra.

[quote="muks"]Thank you.
But, I did not understand the last post.Can you please explain what you meant?[/quote]
If you have any Olympiad book like V. Seshan, you'll prolly find a lot of problems like that. Basically, you know, that there is nothing wrong with taking a,b,c as roots of a polynomial equation. Now anything you prove with those a,b,c will be valid for ANY a,b,c. Because here, we are taking the coefficients in the polynomial equation as variables NOT constants. Now what happens normally is, the use of the sum on n roots taken together at a time formulae greatly reduce the amount of work you need to put in to solve the problem. Coz you can get parts of the final expression asked in the question in the form of p,q etc. i.e. the coefficients of the polynomial equation. Finally you'll see that they'll tend to cancel out each other leaving a constant as an answer in this case or evolve into whatever algebraic form you need to bring stuff into to solve the question. It will appear a bit wierd reading me say this. But once you try a few problems of this kind you'll understand more clearly how and when this happens and also where this will turn out to be a good strategy.
Hope that clears up your doubts a bit. :)

@Euclidean Geometer: muks is a VIII th or IX th student. He probably didn't understand $ \sum$ notations

@muks: I'm not sure if you'll understand this, but I'll try:

$ \sum_{cyc} f(a,b,c) \equal{} f(a,b,c) \plus{} f(b,c,a) \plus{} f(c,a,b)$
where $ f$ is any function of $ (a,b,c)$

For example:
When $ f(a,b,c) \equal{} (a^2b \plus{} 2c^3)$

$ \sum_{cyc} f(a,b,c) \equal{} \sum_{cyc} (a^2b \plus{} 2c^3) \equal{} (a^2b \plus{} 2c^3) \plus{} (b^2c \plus{} 2a^3) \plus{} (c^2a \plus{} 2b^3)$

and similarly $ \sum_{cyc} ab \equal{} ab \plus{} bc \plus{} ca$

And $ \sum_{cyc} a^4 \equal{} a^4 \plus{} b^4 \plus{} c^4$

This is just a shorthand notation to save writing similar terms. Sometimes $ cyc$ is omitted as in his post.

Also you might have trouble understanding the first line of that post. Check out the following Wiki article:
http://www.mathlinks.ro/Wiki/index.php/Vieta's_Formulas
(copy the whole link and pest in the url, the hyper link won't work coz it has an apostrophe)

[quote="Akashnil"]@Euclidean Geometer: muks is a VIII th or IX th student.[/quote]
So sorry bt that. since i ve not beeen very active on mathlinks over the last 6 months i dont really no the new ppl out here. (there were green old days when i was very active here  :) )

Well fine thread to see Soumya, Euclid, Akashnil posting.
Akashnil, when did u come to know about the sigma cyclic and symmetric expressions? $ \sum_{cyclic}$, $ \sum_{symmetric}$.

[quote="Akashnil"]
$ \sum_{cyc} f(a,b,c) \equal{} \sum_{cyc} (a^2b \plus{} 2c^3) \equal{} (a^2b \plus{} 2c^3) \plus{} (b^2c \plus{} 2a^3) \plus{} (c^2a \plus{} 2b^3)$
[/quote]

this needed a little correction

Well yes I didnt notice it. 1st time Akashnil has gone wrong in latexing.

Ah yes... Thanks Anu. :)
And sorry muks. It'd be very misleading for a new learner. I've edited.

[quote="Aravind Srinivas L"]Akashnil, when did u come to know about the sigma cyclic and symmetric expressions? $ \sum_{cyclic}$, $ \sum_{symmetric}$.[/quote]Don't remember.

Wat would be the symmetric sum for the same function...??? Even i find these a bit confusing sometimes.....  :blush:

And that makes it pretty clear...
Thanks

[quote="nuclear_alchemist"]So basically cyc= all permuts of a,b,c and sym= all possible permuts of a,b,c taken in the same order?[/quote]
The other way round

I understood nothing.either it's too tough for me or i am making things complicated;can anyone post the solution using the method specified by Euclidean Geometer?

Denote $ p\equal{}ab\plus{}bc\plus{}ca,q\equal{}abc$

$ a,b,c$ satisfies the equation:
$ (x\minus{}a)(a\minus{}b)(x\minus{}c)\equal{}0$
$ \iff x^3\minus{}(a\plus{}b\plus{}c)x^2\plus{}(ab\plus{}bc\plus{}ca)x\minus{}abc\equal{}0$
$ \iff x^3\plus{}px\minus{}q\equal{}0\cdots (i)$

$ a^2\plus{}b^2\plus{}c^2\equal{}(a\plus{}b\plus{}c)^2\minus{}2(ab\plus{}bc\plus{}ca)\equal{}0\minus{}2p\equal{}\minus{}2p\cdots (1)$

From $ (i)$:
$ x^3\equal{}q\minus{}px$
Since all of $ a,b,c$ satisfy this equation, we can plug them in:
$ a^3\equal{}q\minus{}pa$
$ b^3\equal{}q\minus{}pb$
$ c^3\equal{}q\minus{}pc$
Adding the last 3:
$ a^3\plus{}b^3\plus{}c^3\equal{}3q\minus{}p(a\plus{}b\plus{}c)\equal{}3q\cdots (2)$

From $ (i)$:
$ x^5\equal{}qx^2\minus{}px^3$
Since all of $ a,b,c$ satisfy this equation, we can plug them in:
$ a^5\equal{}qa^2\minus{}pa^3$
$ b^5\equal{}qb^2\minus{}pb^3$
$ c^5\equal{}qc^2\minus{}pc^3$
Adding the last 3:
$ a^5\plus{}b^5\plus{}c^5\equal{}q(a^2\plus{}b^2\plus{}c^2)\minus{}p(a^3\plus{}b^3\plus{}c^3)$
Plugging in the values for $ a^2\plus{}b^2\plus{}c^2$ and $ a^3\plus{}b^3\plus{}c^3$ from $ (1)$ and $ (2)$:
$ a^5\plus{}b^5\plus{}c^5\equal{}q(a^2\plus{}b^2\plus{}c^2)\minus{}p(a^3\plus{}b^3\plus{}c^3)\equal{}q(\minus{}2p)\minus{}p(3q)\equal{}\minus{}5pq\cdots (3)$

From $ (1)$,$ (2)$ and $ (3)$ we have
$ a^2\plus{}b^2\plus{}c^2\equal{}\minus{}2p$
$ a^3\plus{}b^3\plus{}c^3\equal{}3q$
$ a^5\plus{}b^5\plus{}c^5\equal{}\minus{}5pq$

Hence $ \frac{a^5\plus{}b^5\plus{}c^5}{(a^3\plus{}b^3\plus{}c^3)(a^2\plus{}b^2\plus{}c^2)}\equal{}\frac{(\minus{}5pq)}{(3q)(\minus{}2p)}\equal{}\boxed{\frac{5}{6}}$

Man u gotta open a coaching institute!!!! You will be a HIT!!!!!!!!  :D

Ok, if it is $ a+b+c=0$, find  $ \frac{  \sum_{cyc} a^6}{( \sum a^4)( \sum a^2)}$

q=$ abc$ , p= $ \sum ab$

$ x^3=-px^2+q$

Hence,$ \sum a^3 =3q$

$ \sum (ab)^2 = 0$

$ \sum a^4 = 4p^2$

$ \sum a^6= -p(\sum a^4) + q( \sum a^3)
= -4p^3 + 3q^2$

$ \sum a^2= -2p$

Thus, $ \frac{  \sum_{cyc} a^6}{( \sum a^4)( \sum a^2)}$

 = $ \frac {1}{2} + \frac {q^2}{8p^2}$

I do not know whether the sum is valid as I made it up.But is my method correct?

And THANK YOU all of you.

You got the idea correct. But messed up a bit.

[quote="muks"]$ x^3 \equal{} \minus{} px^2 \plus{} q$[/quote]
It should remain $ x^3 \equal{} \minus{} px \plus{} q$

[quote="muks"]$ \sum (ab)^2 \equal{} 0$
$ \sum a^4 \equal{} 4p^2$[/quote]
How did u get these?

To evaluate $ \sum a^4$ use $ \sum a^4 \equal{} \minus{} p(\sum a^2) \plus{} q( \sum a) \equal{} \minus{} p( \minus{} 2p) \plus{} q( 0) \equal{} 2p^2$

[quote="muks"]$ \sum a^6 \equal{} \minus{} p(\sum a^4) \plus{} q( \sum a^3)$[/quote]
This is fully correct.

Plug in the correct value for $ \sum a^4$ here.

Also they wudn't have given this problem actually since the answer is dependent on $ p,q$
But this shouldn't happen. You defined $ p,q$ in the soln. They shouldn't be in the answer.
And they also wouldn't want the answer to be dependent on $ a,b,c$

Sorry, I committed certain errors while evaluating $ \sum (ab)^2$. Then i tried to find $ \sum a^4$ by squaring $ \sum a^2$. Since that involved using $ \sum (ab)^2$, I got it wrong.
But $ THANK YOU$ again. I think I would be more confident while doing sums of this type.


An organic compound A $ (C_2H_7NO_2$) on slow distillation loses water and gives B which on further distillation with $ P_2O_5$ forms C. The compound C on hydrolysis gives D which on treatment with $ P\plus{}Cl_2$ followed by hydrolysis forms glycolic acid. What are A, B, C, D?

Another strange scheme..

A=$ OHCH_2CH(OH)NH_2$
B=$ OHCH_2CONH_2$
C=$ OHCH_2C\equiv N$
D=$ OHCH_2COOH$ (=glycolic acid  :o )

Hell_ever, your first compound is too unstable to be isolable.

[hide]A = ammonium acetate

B = acetamide

C = acetonitrile

D = acetic acid.[/hide]

How does chlorination and hydrolysis of acetic acid give glycolic acid?


geometry
rectangle
Intersection
combinatorics
graph theory
IMO Shortlist

In the plane we have $n$ rectangles with parallel sides.  The sides of distinct rectangles lie on distinct lines.  The boundaries of the rectangles cut the plane into connected regions.  A region is [i]nice[/i] if it has at least one of the vertices of the $n$ rectangles on the boundary.  Prove that the sum of the numbers of the vertices of all nice regions is less than $40n$.  (There can be nonconvex regions as well as regions with more than one boundary curve.)

This is a pretty ambiguous problem. "Each rectangle has parallel sides and the sides of distinct rectangles lie on distinct lines." I'll take to mean that any two rectangles have two pairs of respectively parallel sides and that no four vertices are collinear, but I need a few other clarifications as well:

- Are we counting the "unbounded" region which is not included in any rectangle?
- Are we only counting regions which don't have any lines passing through their interiors, or all regions?

i cannot understand this question?! what does region mean? for example if the rectangles ABCD and EFGH  have intersection and their intersection (that is a rectangle) named KFLD, then ABCD and KFLD both are region or only KFLD is region? can any body help me to understand... (i dont need solution!!)(sorry for my poor english!!)

I think the problem statement is pretty clear. "Regions" are the regions into which the plane is cut by the rectangles (no more inner lines), including the infinite one. Note that the sum only counts the regions that contain vertices of the rectangles, e.g. for the four rectangles in the attached image, only the grey regions (and the surrounding infinite one) are counted. (and the sum would be here $12\cdot4+28=76$.)

Here's a solution that cwein3 came up with.

Let $R=\{R_1,R_2,\cdots,R_{|R|}\}$ be the regions satisfying the condition, and let $c(R_i)$ be the number of corner vertices that region $R_i$ has on its border. Notice that since each exterior angle of a region is $\pm \frac{\pi}{2}$ and the sum must be $2\pi$, we have that $v(R_i)$ is always even.

[b]Lemma:[/b] For a given $R_i$, let $v(R_i)=2k$. Then $c(R_i)\ge k-2$.
[b]Proof:[/b] First, realize that every angle of a region is either $\frac{\pi}{2}$ or $3\frac{\pi}{2}$; note that any $\frac{3\pi}{2}$ angle must be the corner of a rectangle, as if the edges leading up to it corresponded to different rectangles and stopped at that vertex their sides would lie on the same lines; contradiction. This gives us that the number of square corners is at least the number of obtuse angles.

Thus, we show that for $v(R_i)=2k$, the number of such obtuse angles is at least $k-2$. In fact, it is equal to $k-2$, which can be seen again by the exterior angles argument: each obtuse angle contributes $-\frac{\pi}{2}$ and each right angle contributes $\frac{\pi}{2}$. As the sum is $2\pi$, there must be exactly four more right angles than obtuse ones, and the result follows. $\square$

From this lemma, it is immediately clear that $c(R_i)\ge \frac{v(R_i)}{2}-2$.

Let $X,Y$ be the set of non-rectangular and rectangular regions in $R$. Using the lemma, it is apparent that $\sum_{R_i\in X} \left (\frac {v(R_i)} {2} - 2 \right) + |Y| \le 8n$, or $\sum_{R_i\in X} v(R_i) \le 16n+4|X|-2|Y|\Rightarrow \sum v(R) \le 16n+4|X|+2|Y|$. Note a region is a member of $X$ iff it has an obtuse angle, each of which requires a corner, so $|X|\le 4n$. Then $16n+4|X| + 2|Y| \le 16n+8n+2|R|\le 40n$, as desired.

Let $r(R)$ be the number of "real" vertices of a region $R$. We are asked to prove $ \sum v(R)-5r(R) \le 0$. For any $R$ that isn't a hexagon, this is quickly verified to be negative, since a $270º$ angle vertex must always be a real vertex. When $R$ is a hexagon, look at its one $270$º angle vertex, which also belongs to another region. Let $r_1,...,r_{\ell}$ be these "another regions", and let $r_i$ be the "counterpart" of $a_i$ hexagons. Then since $v(R)-5r(R)=1$ for $R$ hexagon, and there are $a_1+...+a_{\ell}$ hexagons, it is enough to prove

$\sum_{i=1}^{\ell} v(r_i)-5r(r_i)+a_i \le 0$.

Let $r_i$ have $2k_i$ sides, then it has $k_i-2$ $270º$ angle vertices (which must all be real), and also at least $a_i$ extra real vertices. Thus it suffices to prove (if $K=\sum k_i, A=\sum a_i$) that

$\sum_{i=1}^{\ell} 2k_i -5(k_i-2+a_i)+a_i \le 0$, i.e.
$-3K-4A + 10 \ell \le 0$,

whch is easy, since $4A+3K = \sum (4a_i+3k_i) \ge \sum (4+3(2)) = 10 \ell$.

Funny enough, my proof doesn't work for $(40-\epsilon)n$, even though it seems $40$ is kinda arbitrary.

Note that each region's boundary is a polygon (or, in the case of the infinite region, many polygons) with vertices having angles either $90$ or $270$ (we use degrees throughout). Furthermore, each vertex can only be either a vertex of a rectangle (which we color red) or the intersection of two sides of different rectangles  (we color these blue).

Note that each red vertex has a $90$ angle and a $270$ angle, while blue vertices are $4$ $90$ angles:

[asy]
size(8cm);

dotfactor *= 2;

fill((0,0)--(0,2)--(2,2)--(2,2.2)--(-.2,2.2)--(-.2,0)--cycle,pink);
fill((0,0)--(.2,0)--(.2,1.8)--(2,1.8)--(2,2)--(0,2)--cycle,palecyan);
draw((0,0)--(0,2)--(2,2));
dot((0,2),red);

fill((5,1)--(5,.8)--(5.8,.8)--(5.8,0)--(6,0)--(6,1)--cycle,lightyellow);
fill((5,1)--(5,1.2)--(5.8,1.2)--(5.8,2)--(6,2)--(6,1)--cycle,lightred);
fill((7,1)--(7,1.2)--(6.2,1.2)--(6.2,2)--(6,2)--(6,1)--cycle,lightblue);
fill((7,1)--(7,.8)--(6.2,.8)--(6.2,0)--(6,0)--(6,1)--cycle,lightgreen);
draw((5,1)--(7,1));
draw((6,0)--(6,2));
dot((6,1),blue);
[/asy]

Now, suppose a closed portion of a region's boundary has $v$ vertices, $r$ of which are red and form a $270$ angle (when measured by the angle in the region in question). If the region is in the interior of this part of the boundary, then we have \[
90(v-r) + 270r = 180(v-2) \implies 2v-4 = v+2r \implies v = 2r+4.
\]Otherwise, if the region is exterior to this part of the boundary:
[asy]
fill((0,0)--(5,0)--(5,5)--(0,5)--cycle,pink);
filldraw((1,2)--(2,2)--(2,1)--(3,1)--(3,2)--(4,2)--(4,3)--(3,3)--(3,4)--(2,4)--(2,3)--(1,3)--cycle,orange);
[/asy]
We have, then, that $v-r$ have an angle of $90$ (measured from the exterior) and $r$ have an angle of $270$ (measured from the exterior). Thus, measured form the interior, they become $270$ and $90$, respectively; for reference, in the above image, the pink is the exterior region, and the orange is the interior. So, \[
270(v-r) +90r = 180(v-2) \implies 3v-2r = 2v-4 \implies v = 2r- 4 \leq 2r+4.
\]Summing over all connected boundaries of the region gives \[
\sum v \leq \sum(2r+4),
\]Summing over all nice regions gives \[
\textnormal{\# of vertices over all regions}\leq 2\sum r + 4(\textnormal{\# of nice regions}) .
\]But $\sum r$ counts each red vertex once (the pink region in the first diagram), so it equals $4n$. Furthermore, the number of nice regions is at most $8n$, so we get \[
\textnormal{\# of vertices over all regions} \leq 2(4n)+4(8n) = 40n.
\]For equality to hold, we need $2r+4  = v$ for every connected part of every region, i.e. every region needs to be interior to every connected, closed part of its boundary, which is false for the infinite exterior region outside of all rectangles. So, equality cannot hold, and the number of vertices is less than $40n$.

We define the [i]leakiness[/i] of a nice region to be one less than the number of rectangle vertices it contains. We will prove the stronger statement that $$\sum_{R\in\text{nice}}4\cdot\text{Leakiness}(R)+\text{\#Vertices}(R)\le 40n$$
To do this, we use induction. The base case $n=1$ is trivial. Now, suppose we have an arrangement of $n$ rectangles and we add the $n+1$st. First, consider where the vertices land. We will only consider the case where vertices land in different regions, since it is not hard to show that vertices landing in the same region is strictly worse.

In this case, the addition of each vertex divides the region it landed in into two parts, and since each part contains it, both regions are nice. As we add three new vertices to each region created (ie the vertex, and the intersection of the sides incident upon it with the boundaries of the region) the vertex sum increases by $6$ per vertex. Now, we claim that each vertex can increase leakiness by at most 1. This is pretty easy to show - if the region the vertex landed in had $a+b$ rectangle vertices and the vertex split it into groups of $a,b$, then the leakiness sum increases from $a+b-1$ to $(a+1-1)+(b+1-1)=(a+b-1)+1$ as desired. So, each vertex increases our sum by $6+4\cdot 1=10$, which is $40$ in total.

Now, we consider how the edges of the new rectangle affect our leakiness/vertex sum. If an edge passes through a mean (ie not nice) region, then it partitions it into two mean regions, which contributes nothing to the sum. On the other hand, if it goes through a nice region, it can partition it into a nice region and a mean region, or two nice regions. In the former case, leakiness sum doesn't change, and it is clear that the vertex sum cannot increase. On the other hand, if we form two nice regions, note that each nice region gets two more vertices which increases the vertex sum by $4$. However, since the vertices of the nice region were split, the sum of the leakinesses decreased by $1$, meaning our sum changes by $4-4\cdot 1=0$.

So, we have shown that the vertices can increase the sum by at most $40$ and the sides can change the sum by at most $0$, and the induction is complete.

Color a point nice-inducing if it is the vertex of a rectangle. A region has an outer boundary and perhaps inner boundaries, each of which are polygons. We assume the unbounded region to be bounded by some large imaginary rectangle, which we will never interact with. Note that this unbounded region has at least one inner boundary. By exterior angle theorem, in any outer boundary of a region there must be $2c+4$ vertices where $c$ is the number of concave vertices, and in any inner boundary this number is $2c-4$.

Let $m$ be the number of nice regions. Each nice-inducing point can appear only once in a boundary as convex, so summing over all nice regions, the number of vertices in nice regions is at most $2(4n)+4m$ with equality if and only if every boundary is an outer one. In other words, equality is impossible. There are at most $8n$ nice regions since each nice-inducing vertex and induce at most two nice regions, so we are done.


geometry
3D geometry
algebra
factorization
sum of cubes

Any one remember Team round 2:
[quote]Select any three-digit multiplier of 3. Calculate the sum of the cubes of the digits of that number. This is now your new number. Now calculate the sum of the cubes of the digits of this new number.  Continue this process of adding the cubes of the digits of the resulting number until you arrive at a number that is equal to the sum of the cubes of its digits. What is this number? [/quote]

The answer was  153...It's the smallest number which can be expressed as  sum of the cubes of its digits.
They restricted you with 3 digit number and if you (or your team) were really unlucky you would select  171 and it would take 13 cycles...but it is said that even if they allowed 20 digit numbers within  15 cycles you would have reached 153. (Worse known case , I was told , is to start with  $44499999999999999999$

 For fun, here are    some other magic properties of 153.

$153 = 1^3+5^3+3^3$  (Sum of cubes of its digit)
$153=1!+2!+3!+4!+5!$
$153=1+2+3+4+5+\dots +17$  (Triangular number and 17 is see the next line...)
$\frac {153}{(1+5+3)} = 17$
$1+5+3 = 3^2$  (Sum of the digits is a perfect square)
$1 + 3 + 9 + 17 + 51 = 81 = 9^2$   ( Sum of the factors is a perfect  square too!)
$153 = (3)(51)$   (Product of two of its decomosition )
$135=1^1+3^2+5^3$ (Rearrangement of digits are fun too!)

Not to mention:
 $  1^0 + 5^1 + 3^2 = 15 $
$ 1^1 + 5^2 + 3^3 = 53$

$115153$ is prime but $153153$  has lot of factors, in fact the smallest odd abudant number, ending in 3
And if you like permutations ... permute all 6 possibilities and you will see:
$153 + 315 + 531 = 351 + 135 + 513$

[size=150]WOW[/size]

Here is the unlucky cycle ..
(UNLUCKY) $771 => 687 => 1071 => 345 => 216 =>225 => 141 => 66 => 432=> 99 => 1458=> 702 => 351 =>153$

Yes that was an irritating problem to look at... I read it three times and realized that the answer was facing me on my own calculator... it ends up that nearly every trial ends up somehow at 153... That was probably the only one I used luck to get... It's kinda funny... I was pissed that 153 kept popping up on my calc screen, and then re-read t he question and we got the question right!...

BTW that list is really interesting...

I think Gyan meant sum of proper factors

Also, even though it isn't amazing: (1+5)/2 =3

[quote="baseballfreak2"][quote="Gyan"]$1 + 3 + 9 + 17 + 51 = 81 = 9^2$   ( Sum of the factors is a perfect  square too!)
[/quote]

How about 153? Thats a factor too.[/quote]
Yes  I meant, as  Hexahedron said, proper facotors only..above,  but now since you mention it:

$1 + 3 + 9 + 17 + 51 + 153 = 234 $   ( *  The sum of all divisors of 153 is 234.)

And The product of all divisors of 153 is 23409:

$1 \times  3\times   9 \times   17 \times   51 \times   153 = 23409$

Note that the product of all  divisors of 153 contain the sum of all divisors of 153 juxtaposed:

$23409$ or   [size=150] [color=blue]234[/color] 09 [/size]
 :)!!!

Also, I am sure some one will mention it here:  the New Testament, where in the net Simon Peter drew from the Sea of Tiberias held 153 fishes. :)

If you haven't noticed yet... 153 is SPECIAL...

[quote="baseballfreak2"]What do you mean?[/quote]
Well... aside from being the answer to number two on the team round :D , it has tons or properties that I don't feel like listing since they already have been listed...(see above)

For that question I made some mistake and got 371 as the answer. That works too, right? probably that isn't valid for starting with a multiple of 3...

But it's not a multiple of three!!! Although 343+1+27 does equal 371

I know, I brought down my team with me :(

[quote="236factorial"]New York, and in one of the worst chapters.[/quote]
Did you make states?

I would've if I didn;t have to go somewhere just on that day (March 19)! In the school competition I got 1st, and then in the chapters I bombed... My teacher is sad..

I know...

Here's how it works out:

The USA was no. 25 out of 29 countries in a recent test. Then, New York is one of the worst states for education in the nation. after that, our chapter is almost last in the state, and finally, our team is second in our chapter. What must my team place be at the international level?! Number 3^6756th?


modular arithmetic
algebra
polynomial
Vieta
symmetry
search
induction

Let $p$ be an odd prime. If $g_{1}, \cdots, g_{\phi(p-1)}$ are the primitive roots $\pmod{p}$ in the range $1<g \le p-1$, prove that \[\sum_{i=1}^{\phi(p-1)}g_{i}\equiv \mu(p-1) \pmod{p}.\]

It's a well known fact of live, that in any field, the roots of $\Phi_n(x)$, the $n$-th cyclotomic polynomial, are exactly the primitive $n$-th roots of unity.
Especially the roots of $\Phi_{p-1}(x) %Error. "modp" is a bad command.
$ are exactly the primitive roots $\mod p$.

And by Vieta, the sum of the roots of a polynomial is the coefficient of $x^{n-1}$, $n$ the polynomial's degree. Thus we look for the coefficient of $x^{\varphi(p-1)-1}$ of $\Phi_{p-1}(x)$, but by it's symmetry we can also look for the coefficient of $x$.
So we search for $\Phi_n^\prime(0)$ (we take arbitrary $n$ instead of $p-1$).

Deriving the recursive formulas $\Phi_{pn}(x) = \frac{\Phi_n(x^p)}{\Phi_n(x)}$ iff $p\nmid n$ and $\Phi_{pn}(x) = \Phi_n(x^p)$ iff $p|n$, we get our result $\mu(n)=\Phi_n^\prime(0)$ by induction on prime divisors.

Another way to conclude: let $f(n)$ be the coefficient of $x^{n-1}$ in $\Phi_n(x)$, or the sum of the roots of $\Phi_n(x)$.

We know that $\prod_{d \mid n} \Phi_d(x) = x^n-1$. If we just compare the sum of the roots of the polynomials in LHS and RHS, we get:
$\sum_{d \mid n} f(d) = 0$ if n > 1, and $f(1) = 1$.
But this is a definition of $\mu$!!  :cool:

Are there any elementary solutions? I guess this isn't well known to high school students. :P

I don't see anything nonelementary (the polynomials pop out if you start thinking about it). And to be known for high school students can't be a good way to judge as even Fermat's little theorem isn't known to them...

FLT is standard part of any olympiad training. This isn't, at least it's not in any of the standard olympiad books I have here...
Seriously, I heared about this for the first time in second year of university... :?

Not the cyclotomic polynomials itself occure, but the polynomials with roots the $ g_i$ (and they are the same). After some more work, one considers the other ones, too.
I agree that the problem gets a lot easier if you know them, but primitive roots are in their nature [i]primitive roots[/i] of unity.
And if some training doesn't consider them, then they should be added  :P

Filling in an elementary solution void here...:P

$ \sum_{(a,p \minus{} 1) \equal{} 1,1\le a\le p \minus{} 1}g^a$ is what we have to evaluate, where $ g$ is one of the primitive roots. Let $ k$ be the number of distinct prime factors of $ p \minus{} 1$.

Now, by PIE (Principle of Inclusion Exclusion) this is equivalent to

$ 1 \plus{} g \plus{} g^2 \plus{} ... \plus{} g^{p \minus{} 1} \minus{} \sum_{a|p \minus{} 1, a prime} (1 \plus{} g^a \plus{} g^{2a} \plus{} ... \plus{} g^{p \minus{} 1}) \plus{} \sum_{a,b|p \minus{} 1, a < b primes} \minus{} ... \plus{} ( \minus{} 1)^{k}(1 \plus{} g^{\Pi prime factors} \plus{} g^{2\Pi prime factors} \plus{} ... \plus{} g^{p \minus{} 1})$.

All but the last term definitely evaluate to 0 by the sum of a geometric series (using $ g^p\equiv 1$ mod p), and the final term turns out to be $ \mu(p \minus{} 1)$ mod p upon inspection, hence the desired result.

Cheers,

Rofler


Find all $ (x;y)$ such that:
$ \frac{x^2}{x^2\plus{}y^2\plus{}2x}\plus{}\frac{y^2}{x^2\plus{}y^2\minus{}2y}\equal{}\frac{\sqrt{2(x^2\plus{}y^2)}}{4}$

The curve represents an animal which shows its tongue; and then?


Let $ I_m$ is $ m$x$ m$ identity matrix  ($ m \in \mathbb{N^{*}}$) .
 Calculate
1) $ \lim_{x \to 0} \begin{pmatrix}{\frac {1}{x}} \lim_{n \to \plus{} \infty}\begin{pmatrix}\begin{bmatrix}1 & 0 & 0 & \minus{} \frac {x}{n} \\
0 & 1 & \minus{} \frac {x}{n} & 0 \\
0 & \frac {x}{n} & 1 & 0 \\
\frac {x}{n} & 0 & 0 & 1\end{bmatrix}^n \minus{} I_4\end{pmatrix}\end{pmatrix} \equal{} ?$
2) $ \lim_{x \to 0}\begin{pmatrix} {\frac {1}{x}} \lim_{n \to \plus{} \infty} \begin{pmatrix}\begin{bmatrix}1 & 0 & 0 & 0 & 0 & \minus{} \frac {x}{n} \\
0 & 1 & 0 & 0 & \minus{} \frac {x}{n} & 0 \\
0 & 0 & 1 & \minus{} \frac {x}{n} & 0 & 0 \\
0 & 0 & \frac {x}{n} & 1 & 0 & 0 \\
0 & \frac {x}{n} & 0 & 0 & 1 & 0 \\
\frac {x}{n} & 0 & 0 & 0 & 0 & 1\end{bmatrix}^n \minus{} I_6\end{pmatrix}\end{pmatrix} \equal{} ?$
 [hide="A small example again and answer of it"]General it is very interesting  :play_ball:
     $ A(x,y,z) \equal{} \lim_{n\to \plus{} \infty}\begin{pmatrix}\begin{bmatrix}1 & 0 & 0 & 0 & 0 & \minus{} \frac {z}{n} \\
0 & 1 & 0 & 0 & \minus{} \frac {y}{n} & 0 \\
0 & 0 & 1 & \minus{} \frac {x}{n} & 0 & 0 \\
0 & 0 & \frac {x}{n} & 1 & 0 & 0 \\
0 & \frac {y}{n} & 0 & 0 & 1 & 0 \\
\frac {z}{n} & 0 & 0 & 0 & 0 & 1\end{bmatrix}^{n}\end{pmatrix}$ = $ \begin{bmatrix}cosz & 0 & 0 & 0 & 0 & \minus{} sinz \\
0 & cosy & 0 & 0 & \minus{} siny & 0 \\
0 & 0 & cosx & \minus{} sinx & 0 & 0 \\
0 & 0 & sinx & cosx & 0 & 0 \\
0 & siny & 0 & 0 & cosy & 0 \\
sinz & 0 & 0 & 0 & 0 & cosz\end{bmatrix}$ ,Here  $ x,y,z \in \mathbb{C}$
Again  we have :
$ A(x_1,y_1,z_1)A_(x_2,y_2,z_2) \equal{} A(x_1 \plus{} x_2,y_1 \plus{} y_2,z_1 \plus{} z_2)$,$ A(mx,my,mz) \equal{} A^m(x,y,z) ,\forall m \in \mathbb{N^{*}}$ I hope that you can define $ A(x,y,z)$ or $ A(x_1,x_2,....,x_m)$ ....[/hide]

[quote="loup blanc"]$ \lim_{n\rightarrow\infty}(I \plus{} \dfrac{xU}{n})^n \equal{} e^{xU}$.
$ \lim_{x\rightarrow{0}}(e^{xU} \minus{} I)/x \equal{} U$.[/quote]
  You use it, is  strong  . Proof it isn't easy . But it is very nice , I remember proof it ...


algebra
system of equations
algebra proposed

Let $ n$ be a positive integer.Solve the system of equations
$ x_1\plus{}{x_2}^2\plus{}...\plus{}{x_n}^n \equal{} n$
$ x_1\plus{}2x_2\plus{}...\plus{}nx_n\equal{}\frac{n(n\plus{}1)}{2}$
for $ n$-tuples $ (x_1,x_2,...x_n)$ of nonnegative real numbers.

By AM-GM we have $ a^k\plus{}(k\minus{}1)\geq ka\forall a\geq 0\forall k\in\mathbb{N}$.  :)


a) Consider the function from $ R^2$ to $ R^2$ defined by

$ f\left(\begin{array}{cc}x \\
\ y\end{array}\right) \equal{} \left(\begin{array}{cc}x \minus{} y \\
\ 2x \plus{} y\end{array}\right)$

Is $ f$ invertible? If so, find a formula for the inverse.

b) For any numbers $ a$ and $ b$, let

$ A \equal{} \left(\begin{array}{cc}1 & a \\
\ 0 & 1\end{array}\right)$

Show that $ A$ is always invertible.(without using determinants)

[hide="Part a Solution"]The function is invertible. Let $ x\minus{}2y\equal{}a$, and $ 2x\plus{}y \equal{} b$. Solving for $ x$ and $ y$ in terms of $ a$ and $ b$ yields $ x \equal{} \frac{a\plus{}2b}{5}$ and $ y \equal{} \frac{b\minus{}2a}{5}$.[/hide]

[hide="Part b Solution"]The determinant of $ A$ is equal to $ 1\times 1 \minus{} 0 \times A \equal{} 1$, so since the determinant is nonzero, $ A$ is always invertible.

If you make the $ a$ negative, you get the inverse of $ A$.

[/hide]

I should have stated that I wanted part b to be solved without using determinants.

For part a, where did you get the 5s from?

By the way, darkdieuguerre, the determinant of that matrix is $ 3,$ not $ 1.$ Or $ 5.$

The matrix for the transformation in part (a) is $ \begin{bmatrix}1&\minus{}1\\2&1\end{bmatrix}.$

Matrix inverses can be computed by row elimination:

$ \left[\begin{array}{cc|cc}1&\minus{}1&1&0\\2&1&0&1\end{array}\right]\ \sim \ \left[\begin{array}{cc|cc}1&\minus{}1&1&0\\0&3&\minus{}2&1\end{array}\right]\ \sim \ \left[\begin{array}{cc|cc}1&\minus{}1&1&0\\0&1&\minus{}\frac23&\frac13\end{array}\right]$

$ \sim \ \left[\begin{array}{cc|cc}1&0&\frac13&\frac13\\0&1&\minus{}\frac23&\frac13\end{array}\right]$

So the inverse matrix is $ \begin{bmatrix}\frac13&\frac13\\\minus{}\frac23&\frac13\end{bmatrix}\equal{}\frac13\begin{bmatrix}1&1\\\minus{}2&1\end{bmatrix}.$

Hence the inverse function is $ f^{\minus{}1}\begin{pmatrix}x\\y\end{pmatrix}\equal{}\frac13\begin{pmatrix}x\plus{}y\\\minus{}2x\plus{}y\end{pmatrix}.$

There is an easily remembered formula for the inverse of a $ 2\times 2$ matrix:

$ \begin{bmatrix}a&b\\c&d\end{bmatrix}^{\minus{}1}\equal{}\frac1{ad\minus{}bc}\begin{bmatrix}d&\minus{}b\\\minus{}c&a\end{bmatrix}.$ Of course, you can say that that does involve determinants.

As for part (b): that's an elementary matrix, which is to say a matrix that performs an elementary row operation. In particular, it's a "type 3" row operation: add $ a$ times the second row to the first row. The way to invert that is to undo that row operation: subtract $ a$ times the second row from the first row. This gives an inverse matrix of
\[ A^{\minus{}1}\equal{}\begin{bmatrix}1&\minus{}a\\0&1\end{bmatrix}.\]


Well , there is a conceptual doubt .

I was reding the introduction to the denstiy functional theory .
when I came across this.

it says that a important basic concept is the concept of " ELECTRONEGATIVITY EQUALIZATION"

IT SAYS THAT THE ELECTRON CLOUD REACHES A POSITION WHERE THE NET FORCE ON IT IS ZERO .
I'VE GOT  DOUBT . PLEASE CLARIFY.

IN HCl MOLECULE.  THE e-s are near Cl  , going by the electrostatic forces formula . F = kq1q2 / r^2  .

the force experienced by electrons by Cl must be more no???   coz of this , the net charge on it is greater than on H . so , dist b/w the e-s and Cl must be greater . But it's opposite.

I know my understanding is wrong.  but where ???/
please tell.

Carcul  Sir .
help needed badly.

please help . is my doubt so foolish  :blush:  :oops: ???? I know there's some conceptual mistake but, where????

[quote="lordvoldemort"]
I was reding the introduction to the denstiy functional theory .
when I came across this.

it says that a important basic concept is the concept of " ELECTRONEGATIVITY EQUALIZATION"

[/quote]

Never even heard of those. Any links or a book you are referring that you can share?

Density functional theory is a method for quantum chemistry calculations that gained popularity since 1996 or so. For a start, look [url=http://en.wikipedia.org/wiki/Density_functional_theory]here[/url].


greatest common divisor
number theory
least common multiple

What is the product of the greatest common factor and the least common multiple of 10 and 35?

The product of the GCF and LCM of two numbers is always the product the two numbers themselves.

10*35 = 350


Find all triples $a,b,c$ of positive integers such that \[ ( 1 + \frac{1}{a} ) ( 1 + \frac{1}{b}) ( 1 + \frac{1}{c} ) = 3. \]

AMGM says $3 = \prod (1 + \frac{1}{a}) \le (\frac{3 + \sum \frac{1}{a}}{3})^3$

So it remains to find a,b,c so that $\sum \frac{1}{a} \ge 3\sqrt[3]{3} - 3 > 1.3$

Obviously, smaller a,b,c makes $\sum \frac{1}{a}$ bigger.  Take a>b>c.  Now, 2,3,4 is too small, so we must have a = 1.

Then $2(b+1)(c+1) = 3bc$ so that $bc=2b+2c+2$, or $(b-2)(c-2) = 6$.  Its now easy to solve.

Sorry to bring up an old post but I am not satisfied by Singular's solution.

I provide my solution instead:

Because the equation is [i]homogenously symmetric[/i] with respect to multiplication, suppose WLOG that $0<a\le b\le c$.

This implies that $1+\frac1{a}\ge1+\frac1{b}\ge1+\frac1{c}$.

Then, $\prod_{a,b,c}\left(1+\frac1{a}\right)=3\le\left(1+\frac1{a}\right)^3\Rightarrow a\le\frac1{\sqrt[3]{3}-1}$

So $a\le2$, we have 2 cases:

[b]Case 1[/b]: Let $a=2$.

Then, $\prod_{b,c}\left(1+\frac1{b}\right)=2\le\left(1+\frac1{b}\right)^2\Rightarrow b\le\frac1{\sqrt2-1}$

So $b\le2$. Since $a=2$, $b=2$, and $1+\frac1{c}=\frac43$ gives $c=3$.

[b]Case 2[/b]: Let $a=1$.

Then, $\prod_{b,c}\left(1+\frac1{b}\right)=\frac32\le\left(1+\frac1{b}\right)^2\Rightarrow b\le\frac1{\sqrt{\frac32}-1}$

So $b\le4$.

If $b=1$, $1+\frac1{c}=\frac43\Rightarrow c=-4<0$, not a possible triple.

If $b=2$, $1+\frac1{c}=1$, is unsolvable.

If $b=3$, $1+\frac1{c}=\frac98\Rightarrow c=8$.

If $b=4$, $1+\frac1{c}=\frac65\Rightarrow c=5$.

The possible unordered triples are $(2,2,3),(1,3,8),(1,4,5)$.


the sum S=2001+2(1+2+....+2000) is the square of what number?


integration
induction
logarithms
algebra proposed
algebra

$ 0<c<e$
$ \{a_{n}\}$ is a positive sequence .
Prove that there exist infinite $ n\in N$ such that
$ 1\plus{}a_{n\plus{}1}>\sqrt[n]{c}a_{n}$

Assume there exist only finitely many $ n$ with the required property. This implies the existence of some $ N$ such that $ \forall n\geq N: 1\plus{}a_{n\plus{}1}\leq c^{\frac{1}{n}}a_{n}$ holds. The case $ c\leq 1$ yields $ a_{n\plus{}1}\leq a_{n}\minus{}1$ and more generally, $ a_{N\plus{}k}\leq a_{N}\minus{}k$, a contradiction to the positiveness of the sequence.

From now on assume $ c>1$ and with $ \frac{1}{n}<\int_{n\minus{}1}^{n}\frac{dx}{x}$ observe that $ a_{n\plus{}1}\leq a_{n}\,c^{\int_{n\minus{}1}^{n}\frac{dx}{x}}\,\minus{}1$. Similarly to the previous case we get by induction:

$ a_{N\plus{}k}\leq a_{N}\,c^{\int_{N\minus{}1}^{N\plus{}k\minus{}1}\frac{dx}{x}}\minus{}\sum_{t\equal{}1}^{k}c^{\int_{N\plus{}t\minus{}1}^{N\plus{}k\minus{}1}\frac{dx}{x}}\equal{}a_{N}\,c^{\ln\left(\frac{N\plus{}k\minus{}1}{N\minus{}1}\right)}\minus{}\sum_{t\equal{}1}^{k}c^{\ln\left(\frac{N\plus{}k\minus{}1}{N\plus{}t\minus{}1}\right)}$

Since the $ a_{n}$ form a positive sequence we must have $ a_{N}\geq\sum_{t\equal{}1}^{k}c^{\ln\left(\frac{N\minus{}1}{N\plus{}t\minus{}1}\right)}$ . Now, the subsitution $ x\equal{}e^{\minus{}\beta}$ with $ \beta\geq 0$ requires $ \forall k: a_{N}\geq\sum_{t\equal{}1}^{k}\left(\frac{N\plus{}t\minus{}1}{N\minus{}1}\right)^{\beta}\geq k$, a contradiction.

spotted a typo in my final substitution step:

let $ c=e^{\beta}$ with $ \beta >0$, then a substitution yields

$ \forall k : a_{N}\geq\sum_{t=1}^{k}\left(\frac{N-1}{N+t-1}\right)^{\beta}\geq (N-1)^{\beta}\sum_{t=1}^{k}\int_{N+t-1}^{N+t}\frac{dx}{x^{\beta}}=\left\{\begin{array}{ll}(N-1)^{\beta}\ln(1+\frac{k}{N}) &\mbox{if }\beta =1\\[0.1cm]\frac{(N-1)^{\beta}}{1-\beta}\left[(N+k)^{1-\beta}-N^{1-\beta}\right] &\mbox{otherwise}\end{array}\right.$

i.e. for $ c\leq e\Longleftrightarrow\beta\leq 1$ we get a contradiction due to the limit behavior for $ k\rightarrow\infty$.


inequalities
induction
inequalities unsolved

Let $ x_1,x_2,...,x_n$ such that
$ 0<x_1\leq x_2\leq x_3\leq...\leq x_n$
 and $ P\equal{}x_2^k(x_1\minus{}x_3)\plus{}x_3^k(x_2\minus{}x_4)\plus{}.....\plus{}x_1^k(x_n\minus{}x_2)$
Prove that
$ P\geq 0$ if $ k>1$
$ P\leq 0$ if $ 0<k<1$

[quote="sylvect"]Let $ x_1,x_2,...,x_n$ such that
$ 0 < x_1\leq x_2\leq x_3\leq...\leq x_n$
 and $ P \equal{} x_2^k(x_1 \minus{} x_3) \plus{} x_3^k(x_2 \minus{} x_4) \plus{} ..... \plus{} x_1^k(x_n \minus{} x_2)$
Prove that
$ P\geq 0$ if $ k > 1$
$ P\leq 0$ if $ 0 < k < 1$[/quote]
Induction over $ n$.

For $ n \equal{} 3$ let $ a < b < c$. ($ a \equal{} b$ or $ b \equal{} c$ trivial) Then

$ b^k(a \minus{} c) \plus{} c^k(b \minus{} a) \plus{} a^k(c \minus{} b)$

$ \equal{} (b \minus{} a)(c^k \minus{} b^k) \minus{} (c \minus{} b)(b^k \minus{} a^k)$

$ \equal{} (c \minus{} b)(b \minus{} a)\left(\frac {c^k \minus{} b^k}{c \minus{} b} \minus{} \frac {b^k \minus{} a^k}{b \minus{} a}\right)$.
\[ As \ t\mapsto t^k \ is \begin{cases} \mathrm{convex\ } & if\ k > 1 \\
\mathrm {concave\ } & if\ 0 < k < 1 , \end{cases}
\]
the result follows, e.g. from Rolle's theorem.  :)

For the induction step, assume it is proven for $ P_n(x_1,x_2,...,x_n)$ and insert an additional variable $ x$ between $ x_1$ and $ x_2$. Now $ P_{n \plus{} 1}(x_1,x,x_2,...,x_n) \minus{} P_n(x_1,x_2,...,x_n) \equal{} P_3(x_1,x,x_2)$.  :)


geometry
incenter
trigonometry
circumcircle
parallelogram
angle bisector
geometry proposed

[color=darkblue] Let $ABC$ be an $A$- isosceles triangle and let $M\in (BC)$ be a mobile point for which $MB<MC\ .$ Construct the circles $w_{1}=C(I_{1},r_{1})$ , $w_{2}=C(I_{2},r_{2})$ so that $\{M,B\}\subset w_{1}$ , $\{M,C\}\subset w_{2}$ and the lines $AB$ , $AC$ are tangent to the circles $w_{1}$ , $w_{2}$ respectively. Denote the intersection $L\in I_{1}I_{2}\cap BC$ and the points $\{D,E\}\subset AB$ for which $LD=LE=LM\ .$ Prove that the sum $r_{1}+r_{2}$ is constant and the point $M$ is the incenter of the triangle $CDE\ .$

[b]Remark.[/b] See the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=113634 .[/color]

I proved the first .
Let $\angle{BI_{1}M}=2k$  Now we have that $\angle{ABC}=k$ and $\angle{ACB}=\angle{ABC}=k$  .
But $\angle{MI_{2}C}=2\angle{ACB}=2k$  so $\angle{BI_{1}M}=2k=\angle{MI_{2}C}$
By the cosine low we have
$BM=r_{1}\sqrt{2}\sqrt{1-cos\ 2k}=2r_{1}sink$ (1) and similarly
$MC=2r_{2}sink$ (2)

(1)+(2) => $BC=2sink(r_{1}+r_{2})$  so $r_{1}+r_{2}=\frac{BC}{2sink}=constant$

The first claim is simple - see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113634]"the line from centers passes through fixed point"[/url]. If AZ is a diameter of the circumcircle (O) of $\triangle ABC,$ then the isosceles triangles $\triangle BI_{1}M \sim \triangle MI_{2}C \sim \triangle BZC$ are similar, $I_{1}ZI_{2}M$ is a parallelogram, $I_{1}Z = I_{2}M = r_{2},$ and $BZ = I_{1}B+I_{1}Z = I_{1}B+I_{2}M = r_{1}+r_{2}.$ As for the second claim, L is the similarity center of the centrally similar isosceles triangles $\triangle BI_{1}M \sim \triangle MI_{2}C$ and consequently, L is also the external similarity center of the circles $(I_{1}),\ (I_{2}).$ A circle (L) with radius LM intersects AB at D, E. Since L is on the center line $I_{1}I_{2}$ of the circles $(I_{1}),\ (I_{2}),$ it also passes through their other intersection N different from M. Power of A to the circle (L) is $AD \cdot AE = AM \cdot AN = AC^{2},$ hence a circle (K) tangent to AC at C and passing through D also passes through E. Let this circle meet the line $LM \equiv BC$ again at L' different from C. Let M' be the other end of the diameter of (L) from M. Power of B to the circle (L) is $BD \cdot BE = BM \cdot BM' = BM \cdot (BL+LM),$ while the power of B to the circle (K) is $BD \cdot BE = BL' \cdot BC = BL' \cdot (BM+MC).$ Thus we have

$BL' \cdot (BM+MC) = BM \cdot (BL+LM)$

Substituting $MC = BM \cdot \frac{r_{2}}{r_{1}},$ $LM = BL \cdot \frac{r_{2}}{r_{1}},$ and reducing by $BM \left(1+\frac{r_{2}}{r_{1}}\right)$, we obtain BL' = BL. Thus the points $L' \equiv L$ are identical and the circumcircle (K) of triangle $\triangle CDE$ passes through L. Since LD = LE, L is the midpoint of the arc DE of (K) opposite to C, so that CL bisects the angle $\angle DCE.$ The circle (L) centered at this arc midpoint and passing through the vertices D, E cuts the angle bisector CL at the incenter M of the triangle $\triangle CDE.$

Very interesting results in this figure .Yetti could we find another interesting properties of this figure ?
Thank you


trigonometry
logarithms
IMC
college contests

Compare $\tan(\sin x)$ with $\sin(\tan x)$, for $x\in \left]0,\frac{\pi}{2}\right[$.

[quote="zhaobin"][quote="Frozen"]Compare $tan(sinx)$ with $sin(tanx)$, for $x\in (0,\frac{\pi}{2})$.[/quote]
I don't think they can compare on $x \in (0,\frac{\pi}{2})$
Is there anything wrong?
Or it is  $x \in (0,\frac{\pi}{4})$?[/quote]

I have checked with mathematica and for $x\in (0,\pi)$ ,

$\tan (\sin x)>\sin (\tan x)$

but still cannot find a proof .....

[quote="zhaobin"][quote="Frozen"]Compare $tan(sinx)$ with $sin(tanx)$, for $x\in (0,\frac{\pi}{2})$.[/quote]
I don't think they can compare on $x \in (0,\frac{\pi}{2})$
Is there anything wrong?
Or it is  $x \in (0,\frac{\pi}{4})$?[/quote]
What is the problem with it?
[hide="Hint"]Just calculate $(\tan(\sin x)-\sin(\tan x))'$ and show that it is positive in $(0,\frac{\pi}{2})$ :) [/hide]

[quote="zhaobin"][quote="rogue"]
[hide="Hint"]Just calculate $(\tan(\sin x)-\sin(\tan x))'$ and show that it is positive in $(0,\frac{\pi}{2})$ :) [/hide][/quote]
But it seems the derivability will be negative when $x \to \frac{\pi}{2}$.
Am I wrong?[/quote]

but for all $x\in \left(\sin^{-1}\left(\frac{\pi}{4}\right),\frac{\pi}{2}\right)$

$\tan (\sin x)>1>\sin (\tan x)$

so I think that $\to \frac{\pi}{2}$ wouldn't affect much

As Shyong mentioned if $\sin x>\frac{\pi}{4}$ then everything is OK.
Let $\sin x\le\frac{\pi}{4}.$ Then $\tan x<\frac{\pi}{2}$ (else $\tan x\ge\frac{\pi}{2}$ which imply $\cos x\le\frac{1}{2}$ and $\sin^{2}x+\cos^{2}x\le\frac{\pi^{2}}{16}+\frac{1}{4}<1.$)
And now apply the hint here:
$(\tan(\sin x)-\sin(\tan x))'=\frac{\cos x}{cos^{2}(\sin x)}\ge \frac{\cos(\tan x)}{\cos^{2}x}\ge0, x\in (0,\arctan\frac{\pi}{2})\Leftrightarrow$
$\cos^{3}x\ge cos^{2}(\sin x)\cos(\tan x), x\in (0,\arctan\frac{\pi}{2})$
Since $\ln\cos x$ is concave in $(0,\frac{\pi}{2})$ it remains to apply
$2\sin x+\tan x\ge3x, x\in(0,\frac{\pi}{2})$(well-known) to get
$\cos^{3}x\ge\cos^{3}(\frac{2sin x+\tan x}{3})\ge cos^{2}(\sin x)\cos(\tan x).$   :)

It looks like more high school problem than competition problem for
university student.
It is not a good choice for IMC.


inequalities
geometry
inradius
calculus
geometry proposed

Let $ a$ and $ b$ sides ,  $ r$ - inradius of $ \triangle ABC$ . Find minimum : $ \boxed{\frac{ab}{r^2}}$ and then does minimum holdes?

hello, with the help of calculus i have got
$ \frac{ab}{r^2}\geq\frac{1}{2}(11\plus{}5\sqrt{5})$ and the minimum will be attained for $ a\equal{}1,b\equal{}1,c\equal{}\minus{}1\plus{}\sqrt{5}$
Sonnhard.

please post your solution.i want to see how calculus can help


geometry
trapezoid
rectangle
ratio
number theory
relatively prime
AMC

Let $ ABC$ be a triangle with sides $ 3,4,$ and $ 5$, and $ DEFG$ be

a $ 6$ by $ 7$ rectangle.  A segment is drawn to divide triangle

$ ABC$ into a triangle $ U_1$ and a trapezoid $ V_1$, and another segment is

drawn to divide rectangle $ DEFG$ into a triangle $ U_2$

and a trapezoid $ V_2$ such that $ U_1$ is similar to $ U_2$ and $ V_1$ is similar

to $ V_2$.  The minimum value of the area of $ U_1$ can be written in the form

$ m/n$, where $ m$ and $ n$ are relatively prime positive integers.  Find $ m \plus{} n$.

How do you find the lengths of triangle $ U_2$?  Let $ s_1$ be the line segment

drawn in triangle $ ABC$. How do you know that "

were $ s_1$ parallel to the side of length $ 4$, the corresponding acute angles in

$ V_1$ and $ V_2$ would not be congruent, therefore trapezoids $ V_1$ and $ V_2$ would not be similar"?

Thanks!

havent really read the problem but the ratios of the sides probably dont match...

anyways i think the point is to try both cases..


algebra
polynomial
function
functional equation

Does there exist a polynomial f(x) for which $ xf(x\minus{}1)\equal{}x\plus{}1f(x)$?

is that really $ x\plus{}1f(x)$ or $ (x\plus{}1)f(x)$?

Well if it is the second edited version
[hide]We can do
$ x\equal{}0$
so $ f(0)\equal{}0$
$ x\equal{}1$, plugging in known values
so $ f(1)\equal{}0$
and if we continue, we see that $ f(x)$ has a root at all integer values.
Thus $ f(x)\equal{}0$ since it is a polynomial
[/hide]
Not sure if it is valid proof

You could make it a more rigorous proof by using induction.

[hide="My Solution"]
We claim that the function $ f(x)$ that satisfies $ xf(x \minus{} 1) \equal{} (x \plus{} 1)f(x)$ also satisfies $ f(x) \equal{} 0$ for all $ x \in \mathbb{Z}_0$. Checking the base case, we have $ (0)f(\minus{}1) \equal{} (1)f(0)\implies f(0) \equal{} 0$. We assume that $ f(k) \equal{} 0$ for some nonnegative integer $ k$. Then letting $ x \equal{} k \plus{} 1$ in the functional equation, we have $ (k \plus{} 1)f(k) \equal{} (k \plus{} 2)f(k \plus{} 1)$. We know that $ f(k) \equal{} 0$, so we have $ (k \plus{} 2)f(k \plus{} 1) \equal{} 0\implies f(k \plus{} 1) \equal{} 0$. Therefore, our claim holds for all nonnegative integers.

Now we know that the only polynomial that has an infinite number of roots is $ \boxed{f(x) \equal{} 0}$, hence that is the only solution.
[/hide]


Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \sqrt {\frac {a^2}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2}{a^2 \plus{} b^2}} \ge \frac {1}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {1}{\sqrt {b^2 \plus{} ca}} \plus{} \frac{1}{\sqrt {c^2 \plus{} ab}}.
\]
Equality holds if and only if $ (a,b,c) \equal{} (1,1,1)$ or $ (a,b,c) \equal{} \left( \frac {3}{2},\frac {3}{2},1\right).$
:)

Sorry, I misstyped. The second eqaulity point must be $ (a,b,c)\equal{}\left( \frac{3}{2},\frac{3}{2},0\right).$  :blush:


Good problem.
[hide="Solution"]
This can be written as
$\frac{1}{\frac{4!}{0!}}+\frac{1}{\frac{5!}{1!}}+\ldots+\frac{1}{\frac{31!}{27!}}$

These can be re-written as the sum of the reciprocals:
$\frac{0!}{4!}+\frac{1!}{5!}+\ldots+\frac{27!}{31!}$
The common denominator must be $31!$, so we multiply each fraction by $\frac{31!}{N!}$ where N is the denominator of the fraction.

We then get a long and messy thingy, which I could simplify, but it's late, and I don't feel like it.
I might edit this later.
[/hide]

I think the best way to approach this problem is with [hide="a blow-dart gun"] partial fractions.  :wink: [/hide]

I tried doing that but I coulndt get a partial fraction sequence that would cancel out terms..

Yes, I get the same answer as you.  Yours is the method I was trying to think of because I think I saw kunny do it once, but I couldn't quite recall...

[quote="t0rajir0u"][hide="Solution"] $\frac{1}{k(k+1)(k+2)(k+3)}= \frac{1}{(k^{2}+3k+2)(k^{2}+3k)}= \frac{1}{2}\left( \frac{1}{k^{2}+3k}-\frac{1}{k^{2}+3k+2}\right)$

Letting the original summation be $S$, we have

$2S = \sum_{k=1}^{28}\frac{1}{k(k+3)}-\sum_{k=1}^{28}\frac{1}{(k+1)(k+2)}$

The first summation can be written $\frac{1}{3}\left( \frac{1}{k}-\frac{1}{k+3}\right)$, which telescopes to

$\frac{1}{3}\left( \frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{29}-\frac{1}{30}-\frac{1}{31}\right)$

And the second can be written as $\frac{1}{k+1}-\frac{1}{k+2}$ which telescopes to

$\frac{1}{2}-\frac{1}{30}$

So... whatever the difference of those is, divided by two!  :D [/hide][/quote]

This is actually a pretty deep method... I think it works for any sum of the form: \[\frac{1}{(k+w)(k+x)(k+y)(k+z)}\] such that $w+z=x+y$

That really reduces an entire class of problems!!!

Well, partial fraction decomposition is even more general, so...  :|


calculus
integration
number theory proposed
number theory

Prove that there exists $ m \in N$ such that there exists an integral sequence $ \{a_n\}$ which satisfies:
$ i.a_0 \equal{} 1,a_1 \equal{} 337$
$ ii.(a_{n \plus{} 1}a_{n \minus{} 1} \minus{} a_n^2) \plus{} \frac {3}{4}(a_{n \plus{} 1} \plus{} a_{n \minus{} 1} \minus{} 2a_n) \equal{} m,\forall n\geq 1$
$ iii.\frac {1}{6}(a_n \plus{} 1)(2a_n \plus{} 1)$ is perfect square $ \forall n\geq 1$.

No one is interesting in this problem ?  :(


(1) For $ f: R\rightarrow R$ is a monotonic increasing/decreasing function. Supposed $ a,b > 0$ is a real constant that statisfies $ f\left( x\right)\plus{}f\left( y\right)\minus{}a\leq f\left( x\plus{}y\right)\leq f\left( x\right)\plus{}f\left( y\right)\plus{}b$ for all $ x,y > 0$
Prove that there exist any reals $ r$ so that $ f\left( x\right)\minus{}rx$ is a bounded function
(2) For $ \frac{a}{b}\in\left( 0,1\right)$ and $ a$ is a natural n umber, $ b$ is an odd natural number $ a$ and $ b$ relatively prime
Prove that there exist integers $ x,y,z$ that statisfies $ \frac{a}{b}\equal{}\left\{\frac{xyz}{x^{2}\plus{}y^{2}\plus{}z^{2}}\right\}$
NB: $ \left\{ x\right\}$ is the fractional part of the real number $ x$

(3) For $ a,b,c\in R^{\plus{}}$. Find the smallest possible value of $ A$ that statisfies $ A\geq\frac{\left|\left( a\minus{}b\right)\left( b\minus{}c\right)\right|^{3}\plus{}\left|\left( b\minus{}c\right)\left( c\minus{}a\right)\right|^{3}\plus{}\left|\left( c\minus{}a\right)\left( a\minus{}b\right)\right|^{3}}{\left( a^{3}\plus{}b^{3}\plus{}c^{3}\right)^{2}}$

(4) Supposed the distance of any two points in a plane is an integer (positive absolutely). Prove that all the points is in one line.

(5) For $ a,b\in R^{\plus{}}$. Prove that
(a) $ \frac{\left( 1\plus{}r\right)\left( a\minus{}b\right)^{2}\plus{}8ab}{4\left( a\plus{}b\right) }\geq\left(\frac{a^{r}\plus{}b^{r}}{2}\right)^{\frac{1}{r}}$ for all $ r\geq 2$ is a real number
(b) $ \frac{\left( 1\plus{}r\right)\left( a\minus{}b\right)^{2}\plus{}8ab}{4\left( a\plus{}b\right) }\leq\left(\frac{a^{r}\plus{}b^{r}}{2}\right)^{\frac{1}{r}}$ for $ 1\leq r\leq\frac{3}{2}$ or $ r\leq\minus{}1$ is a real number

(6) For $ a,b,c\in R^{\plus{}}$ that statisfies $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$. Prove that $ 2abc\left( a\plus{}b\plus{}c\plus{}9\right)\plus{}30\geq a^{2}b^{2}c^{2}\plus{}17\left( ab\plus{}bc\plus{}ca\right)$

In (4), are there an infinite number of points? (3, 4, 5 triangle...)

Is there a source for these problems? With exactly six of them, it looks suspiciously like an Olympiad. If so, it'd probably be a good idea to get it in the Resources section. And to split these into multiple posts.

[hide]For (3), I claim $ A \equal{} 1$. WLOG suppose $ a\ge b\ge c$. Then

$ (a^{3}\plus{}b^{3}\plus{}c^{3})^{2}\ge a^{6}\plus{}2a^{3}b^{3}\ge |(a\minus{}b)(a\minus{}c)|^{3}\plus{}|(a\minus{}b)(b\minus{}c)|^{3}\plus{}|(a\minus{}c)(b\minus{}c)|^{3}$,

which is exactly what we want. Equality occurs at $ b \equal{} c \equal{} 0$.[/hide]

[quote="Altheman"][quote="MellowMelon"][quote="t0rajir0u"]I believe (4) is valid for a certain number of points, possibly $ n\ge 4$.[/quote]
I think it has to be infinite. Pick an even number $ x$ with an arbitrarily high number of divisors. Now pick $ m,n$ such that $ m > n$ and $ x \equal{} 2mn$ to generate a pythagorean triple ($ m^{2}\minus{}n^{2}$,$ 2mn$,$ m^{2}\plus{}n^{2}$). Since the number of ways to pick $ m,n$ is half of the number of divisors of $ x$, we can get as many pythagorean triples as we want by picking certain $ x$.

Then just take $ (0,0)$, $ (0,x)$, $ (0,\minus{}x)$, and put points on the $ x$-axis to form all the right triangles generated by the pythagorean triples, so then we could have a set of points, not all collinear, with any finite number of elements.[/quote]

That is not the same thing as infinite.[/quote]

He's showing that the problem doesn't hold for a finite number of points.

[quote="MellowMelon"]Then just take $ (0,0)$, $ (0,x)$, $ (0,\minus{}x)$, and put points on the $ x$-axis to form all the right triangles generated by the pythagorean triples, so then we could have a set of points, not all collinear, with any finite number of elements.[/quote]

Ah, true.  Sorry, I was thinking about $ n \equal{} 4$ points, [i]none[/i] of which are collinear, which is a different problem.


hallo,
Find all funktions $f: R\to R$ s.t.
$f(x+y)=f(x)f(a-y)+f(y)f(a-x)$ for $(x,y) \in R \times R$ and $a \in R$. :)

I find that $\ f(x)=0$ or $\ f(x)=\frac{1}{2}$

First, with $\ y=0$ we have: $\ f(x)=\frac{f(0)f(a-x)}{1-f(a)}$  (1)

So $f(0)\neq 0$ because if it is the case then $\boxed{f(x)=0}$ and we are done.

It fallows that $\ f(a)=\frac{1}{2}$ and $\ f(0)=\frac{1}{2}$ with equation (1)$\ \to$ $\ f(x)=f(a-x)$ and $\ f(x)=f(x+a)$     (2)

We have $\ f(x)=f(a-x)$ $\ \to$ $\ f(-x)=f(x+a)=f(x)$ So $\boxed{f(x)=f(-x)}$ (3)

We can wright the original equation as fallow:$\boxed{f(x+y)=2f(x)f(y)}$ (4)

Equations (3) and (4) and replacing $\ y=-x$ gives: $\ f(0)=2f(x)f(-x)=2f^{2}(x)$

Finally $\boxed{f(x)=\frac{1}{2}}$

[quote="El fisico"]So $f(0)\neq 0$ because if it is the case then $\boxed{f(x)=0}$ and we are done.[/quote]
I think that you should consider the case $f(0)=0$ and $f(a)=1$.

[quote="El fisico"]It follows that $f(a)=\frac{1}{2}$ and $f(0)=\frac{1}{2}$[/quote]
I think that you should consider the case $f(0)=-\frac{1}{2}$.


function
algebra
rational function
complex analysis
complex analysis unsolved

Suppsose that $R(z)$ is a rational function (i.e. there exist $f,g\in\mathbb C[x],\ \ h=\frac fg$) which for each $|z|=1$ then $|R(z)|=1$. Prove that there exist $\alpha_1,\dots ,\alpha_n$ that are not on unit circle, and $m\in \mathbb Z$ and $a$ on unit circle which : \[ R(z)=az^m\prod_{i=1}^{n}\frac{z-\alpha_i}{1-\overline{\alpha_i}z} \]

almost the same solution that http://www.mathlinks.ro/Forum/viewtopic.php?highlight=entire&t=78814 works.


How can draw geometric figures by using Flash?
The figure that I'm going to draw is too complex.

What I usually do is create the image in other, more advanced art programs, then import it to flash and work with it like that. If needed, you can draw a few frames of it.


1) Say we have a homogeneous function f(x, y, z) in an inequality. Can we normalize and say that x/y = 1? I can't seem to come up with the proper proof for this case like I can with other equalities like x+y+z=1, xyz=1, x=1, or xy=1.

2) On an olympiad, can we normalize a homogeneous function in an inequality without proof (i.e. say that "since this function is homogeneous, we can normalize and assume that xyz = 1")?

Thanks.

im not an expert, but i dont think you can normalize with x/y=1, because then you are just proving the inequality for x=y, that wouldn't prove it for all (x,y,z)

btw in an olympiad you can say "because this inequality is homogeneous we may assume xyz=1" without proof.

zscool is right. You cannot assume that x=y but you can assume that xyz=1 IF IT IS HOMOGENEOUS!

[quote="JBL"]The reason that you can't make that substitution is actually stronger, I believe: You've set equal two expressions of the same degrees (x/y and 1 both have degree 0).  However, the homogenity of the expression only allows you to set equal things of different degrees.  That is, because it is homogeneous, if you have two expressions of different degree, you can multiply everything through by a constant until you get the two expressions are equal (since they must cross somewhere).  If both expressions are of the same degree, however, this does not apply.[/quote]

Hmm, Joel, so you're saying that the difference in degree is a necessity, but can we set ANY two expressions of different degrees equal? For example, if f(x,y,z) is homogeneous, can we set x^3 + xy^2 + 3z^3 = 1? How about x^3 + y = 1 (I assume not, since the LHS is not homogeneous)?

Well, here is my thinking.  Let's work with x, y, z on the domain of the reals.  Say we have some inequality we are trying to prove true, f(x, y, z) >= 0 where f is homogenous in some degree.

Take any two expressions g(x, y, z) and h(x, y, z) where both are homogeneous, but of different degrees.

We can break every triple (x, y, z) into equivalence classes so that (x, y, z) :equiv: (ax, ay, az) for all a in the reals.  Every solution is in exactly one equivalence class except for (0, 0, 0), but it will obey the inequality trivially anyhow.

Now, consider h and g acting on triples within one equivalence class.  Since they are of different degrees, taking a very large positive a and a very large negative a give us two points at which the parity of g - h is different.  So there must be some point in between where g - h = 0 (since they are continuous functions of a, after all).  Thus, for each equivalence class there exists a point at which the thing holds.  Thus, we can set it as a condition with no problems.

Here are possible problems with what I've just done:

1) The point (0, 0, 0) is a problem -- what if we only have that the condition is met when a = 0?  This seems to me like it is an avoidable problem, but I don't know really what to do with it.

2) It only works if our numbers can be on all the reals.  If they are restricted, for example to the positive reals, I'm not sure what happens.  I think this is avoidable in most cases, too, but I also don't know what to do about it.

As for your two examples, if I'm right then yes, you could assume the first condition, but you would need perhaps more complicated work to assume the second.  At least, you might be able to assume the second, but I'm not positive.


$\mathbb{N}$ se partitioneaza in $n$ progresii aritmetice, cu ratiile $r_{1},r_{2},...r_{n}$.Sa se arate ca:

$\frac{1}{r_{1}}+\frac{1}{r_{2}}+...+\frac{1}{r_{n}}=1$

Daca nu exista un numar care sa apartina la 2 progresii, atunci toate ratiile sunt egale.

Sun mai multe solutii deja postate. Iata una simpatica:
probabilitatea ca alegand un numar natural la intamplare el sa apartina primei progresii este $\frac{1}{r_{1}}$...
Apropo, se poate demonstra ca macar doua dintre ratii sunt egale.

da, asta era rezolvarea care mi s-a dat si mie la problema, cea cu probabilitatea, mi s-a parut insa interesant rezultatul

http://www.mathlinks.ro/Forum/viewtopic.php?p=148351 pentru problema cu ratiile egale.

De altfel, un exercitiu interesant: nu exista nici o masura de probabilitate pe N astfel incat probabilitatea ca un numar sa fie multiplu de $k$ este $1/k$. Bine, problema de fata se poate face evident fara a invoca probabilitati, ci lucrand mai degraba cu densitati.

[quote="perfect_radio"][quote="enescu"]Probabilitatea ca alegand un numar natural la intamplare el sa apartina primei progresii este $\frac{1}{r_{1}}$...[/quote]
Cum definim alegerea unui număr natural la întâmplare?[/quote]
A se vedea "Probleme neelementare tratate elementar", A.M. Iaglom, I.M. Iaglom, Editura tehnică, 1983, pagina 30.


Fiind the period of the linear relativistic oscillations is the amplitude is A(in laboratory system).You can use the following approximation.

I got \[T = 2\pi \sqrt{\frac{m}{k}}+\frac{A^{2}\pi}{2c^{2}}\sqrt{\frac{k}{m}}.\] Is that ok?

first termen of the sum is ok  but you should have 3pi/8 instead of pi/2 in ht second one

No, I can't get $3\pi/8$ :( But, on the other hand, I didn't use the approximation you gave (not in that form). I got \[\gamma = 1+\frac{k(A^{2}-x^{2})}{2mc^{2}},\] from c.o.e, and then said that \[T = 4\int_{0}^{A}\frac{\text dx}{v},\] after which it's straightforward. Do I have a mistake somewhere?

you  are correct this far.maybe your answer is different because of the use of a different approx

The equation of motion:
$\frac{d}{dt}\bigg ( \frac{m_{0}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\bigg ) =-Dx$
Lets assume that at $t=0$ time $x=A$, $v=0$.
using the $\omega_{0}=\sqrt{\frac{D}{m}}$ substitution:
$c\sqrt{c^{2}-v^{2}}-\frac12{\omega_{0}}^{2}x^{2}=c^{2}-\frac12{\omega_{0}}^{2}A^{2}$
From here the square of the velocity:
$v^{2}={\omega_{0}}^{2}(A^{2}-x^{2}) \bigg (1-\frac{{\omega_{0}}^{2}A^{2}}{4c^{2}}+\frac{{\omega_{0}}^{2}x^{2}}{4c^{2}}\bigg )$
using the :
$y=\frac{x}{A}$ and $\varepsilon=\frac{\omega_{0}A}{2c}$
substitutions, the period of the oscillator:
$T=\frac{4}{\omega_{0}}\frac{1}{\sqrt{1-{\varepsilon}^{2}}}\int_{0}^{1}\frac{dy}{\sqrt{(1-y^{2}) \bigg (1-\frac{{\varepsilon}^{2}}{{\varepsilon}^{2}-1}y^{2}\bigg )}}=\frac{4}{\omega_{0}\sqrt{1-{\varepsilon}^{2}}}K \bigg (\frac{\varepsilon}{\sqrt{{\varepsilon}^{2}-1}}\bigg )$
Where $K(k)$ is first order whole elliptic integral
This integral can be written in taylor series (now I write it till the first piece):
$T=\frac{4}{\omega_{0}}\frac{1}{\sqrt{1-{\varepsilon}^{2}}}\frac{\pi}{2}\bigg [1-\bigg (\frac12 \bigg )^{2}\frac{{\varepsilon}^{2}}{1-{\varepsilon}^{2}}+... \bigg ] =\frac{2 \pi}{\omega_{0}}\bigg (1+\frac{{\varepsilon}^{2}}{4}\bigg )$
writing here $\varepsilon$ We get:
$T=2\pi \sqrt{\frac{m}{D}}+\frac{\pi}{4}\frac{A^{2}}{c^{2}}\sqrt{\frac{D}{m}}$
now that must be right.

you must have sqr(c) in your second term.

Ok, I've been more careful with my approximations, and here it is:

The integrand is $\gamma/c\sqrt{\gamma^{2}-1}$, so from my previous post we get
\[\frac{\gamma}{c\sqrt{\gamma^{2}-1}}= \frac{1+\frac{\omega^{2}(A^{2}-x^{2})}{2c^{2}}}{c\sqrt{\left (1+\frac{\omega^{2}(A^{2}-x^{2})}{2c^{2}}\right)^{2}-1}}= \frac{1+\frac{\omega^{2}(A^{2}-x^{2})}{2c^{2}}}{c\frac{\omega\sqrt{A^{2}-x^{2}}}{c}\sqrt{1+\frac{\omega^{2}(A^{2}-x^{2})}{4c^{2}}}}\approx \]

\[\approx \frac{1+\frac{\omega^{2}(A^{2}-x^{2})}{2c^{2}}}{\omega\sqrt{A^{2}-x^{2}}}\left(1-\frac{\omega^{2}(A^{2}-x^{2})}{8c^{2}}\right) \approx \frac{1}{\omega\sqrt{A^{2}-x^{2}}}+\frac{3\omega\sqrt{A^{2}-x^{2}}}{8c^{2}}. \]
Now integrating this we get what we need:
\[T = 2\pi\sqrt{\frac{m}{k}}+\frac{3A^{2}\pi}{8c^{2}}\sqrt{\frac{k}{m}}. \]
:)

The answer is just a question of how you approximate :D

You just need to be consistent and keep all the terms of the same order of magnitude :)


In the figure below, $ AB, CD,$ and $ EF$ are parallel. If $ AB \equal{} 300$ and $ EF \equal{} 200$, then $ CD \equal{}$

[img]http://i204.photobucket.com/albums/bb97/hasan4444/Problem2.jpg[/img]

$ \textbf{(A)}\ 100 \qquad \textbf{(B)}\ 120 \qquad \textbf{(C)}\ 140 \qquad \textbf{(D)}\ 160 \qquad \textbf{(E)}\ 180$

Please could you clarify more how you make those ratios because I'm pretty bad at Geometry  :)

Triangles $ ABC$ and $ FEC$ are similar

$ AC: CF\equal{}AB: EF\equal{}300: 200\equal{}3: 2$

Triangles $ FCD$ and $ FAB$ are similar

$ FD: FB\equal{}2: (3\plus{}2)\equal{}CD: AB$

$ CD\equal{}300\times \frac{2}{5}$


[quote="shfdfzhjj"]Equation
Can you?[/quote]

No one?


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167109
reading this topic I thought about the following generalisation:

consider the sum

$ \sqrt {1} \plus{} \sqrt {2} \plus{} \dots \plus{} \sqrt {n}$

of course it is 1 - rational number for $ n \equal{} 1$.

for $ n \equal{} 2$ we have $ 1 \plus{} \sqrt {2}$ irrational

the number obtained for $ n \equal{} 3$ will be irrational as  shown there,

is it true than $ \sqrt {1} \plus{} \sqrt {2} \plus{} \dots \plus{} \sqrt {n}$  is rational iff n=1 :?:

There are verious ways to prove it. Short but not so elementary one:

One of them: let $ p$ be the biggest prime $ \leq n$. Then $ p$ occurs only once in $ \{ 1,2,3,...,n\}$ (by Bertrand's postulate), thus $ p$ is multiplicatively independent from the other numbers. By Kummer theory, the desired sum is irrational.

One can avoid Bertrand's theorem rather easily. Avoiding Kummer theory is a bit harder, here one needs often other theorems of "Pre-Class-Field-Theory".

thanks for your answer, but I'm afraid it's too advanced for me :(

is there any proof of this fact which can be read with understanding by first-year student? :wink:


conics
hyperbola
Asymptote
function
geometry
trigonometry
analytic geometry

[quote="anirudh"]Find the equation and asymptotes of:

[img]http://i96.photobucket.com/albums/l167/adasarathy/graphhyperbola.gif[/img][/quote]

[hide]Just by [i]looking[/i] at the graph: $y=-4$, $x=-5$

It [i]looks[/i] like it goes through (-6,0), so the equation would be $(x+5)(y+4)=-4$.

But seriously, this is a horrible question. You don't know if each mark on the axes on the labels represents 1 unit or not or if the hyperbola really goes through some of the points and not just near it. [/hide]

[hide="Looking at the graph, I would conjecture that..."]
... the asympotes are $x=-5$ and $y=-4$.

Would the equation then be $y=\frac{-1}{x+5}-4$?[/hide]

I agree with Splash: this question is foul.

It was a question on Green Globs.

I got the part about $(x+5)(y+4)$, but how would I know what the value on the other side is? Is it guess and check

What's Green Globs?

I don't think its guess and check as much as just guess. Splash's criticisms (does it pass through $(-6,0)$ or go close, does each unit marked equal 1...?) basically show that the question is pretty much invalid.

I gave a value of $-1$ on the RHS of $(x+5)(y-4)$ because $1$ is a round number, but $-4$ is probably the closest integer.

[quote="tim1234133"]Splash's criticisms (does it pass through $(-6,0)$ or go close, does each unit marked equal 1...?) basically show that the question is pretty much invalid.[/quote]
Keep in mind that their diagram might have actually been good.
EDIT: Never mind, it seems that their diagram was this indeed so this question is invalid

I used printkey to get the diagram. I don't know how you get the RHS fast?

Green Globs is the BEST GAME EVER

you hit GREEN GLOBS with graphs and you use random trigonometric functions that you don't even understand in order to get obscenely high scores

Get a book about pre-calc and learn about it.

[quote="Ubemaya"]where do you learn about hyperbolas and stuff? I don't get any of this...[/quote]

you take a analytic geometry class. Also green globs is fun. Once I was messing around in expert mode (because I like trig functions) and I got 128 in one move or something.


Prove for $ x,y,z\in\mathbb{R}^ \plus{} ,x \plus{} y \plus{} z \equal{} \frac {1}{2}$ that
\[ \left(\sum_{sym}\frac {x}{y}\right)\left(\sqrt {2x^2 \plus{} 2y^2 \plus{} 2z^2}\right)\ge 1
\]

The first parenthesis is greater than 6 by am-gm and the second parenthesis we can use RMS-AM...

My solution was that $ \sum_{sym}\frac {x}{y}\ge\frac {1}{\sum\limits_{sym}x\sqrt {y}}$ by Jensen's and that $ \frac {1}{\sum\limits_{sym}x\sqrt {y}}\ge \frac {1}{\sqrt {(2x^2 \plus{} 2y^2 \plus{} 2z^2)(2x \plus{} 2y \plus{} 2z)}}$ by Cauchy.


Έστω μια ακολουθία συνεχών συναρτήσεων $ \{f_n : [0,1] \to [0,1] \}_{n\equal{}1}^\infty$. Αν $ f_n \to 0$ κατά σημείο, τότε φυσικά θα έχουμε και $ \int_{[0,1]} f_n \to 0$ κατά Lebesgue, αλλά μπορούμε αναγκαστικά να πούμε το ίδιο και κατά Riemann?

Επειδή έτυχε να είμαι online:
[hide]Απαρίθμηση των ρητών και όριο στην Dirichlet... ή εννοείς κατά σημείο [i]παντού[/i] (και όχι σχεδόν παντού)?[/hide]

Cheerio,

Durandal 1707

Κατά σημείο παντού.

[quote="cmad"]Κατά σημείο παντού.[/quote]

Ah, ok! Μου φαίνεται ότι παραέχω συνηθίσει με Lebesgue και ξεχνάω ότι καμιά φορά οι συναρτήσεις μας είναι τόσο καλές που είναι ακόμα και συνεχείς :lol:

Δε βάζω λύση για να μη γίνουν ταλάντωση μεταξύ μας τα προβλήματα, πες μου μόνο αν σωστά σκέφτομαι ότι [hide]τελικά πέφτουμε στο κλασικό παιχνίδι ομοιόμορφης σύγκλισης.[/hide]
Cheers mate, ωραίο πρόβλημα!

Durandal 1707

[quote="Durandal"]πες μου μόνο αν σωστά σκέφτομαι ότι [hide]τελικά πέφτουμε στο κλασικό παιχνίδι ομοιόμορφης σύγκλισης.[/hide][/quote]

Έχεις κάποια συγκεκριμένη ακολουθία στο μυαλό σου; Δοκίμασε να κατασκευάσεις κάποια, και αν θελεις το συνεχίζουμε σε PM για να μην το κάνουμε ping pong.

[quote="cmad"]Έχεις κάποια συγκεκριμένη ακολουθία στο μυαλό σου; Δοκίμασε να κατασκευάσεις κάποια, και αν θελεις το συνεχίζουμε σε PM για να μην το κάνουμε ping pong.[/quote]

Με παρεξήγησες, the other way around πήγαινε το σχόλιό μου ;)

Anyway, μιας και έχουν περάσει δυο-τρεις μέρες και δεν έχει δείξει κάποιος άλλος ενδιαφέρον, να βάλω hidden αυτό που είχα σκεφτεί (πολύ βιαστικά ασχολήθηκα, οπότε είναι σχετικά πιθανό να μου έχει ξεφύγει κάτι...)

[hide]Λοιπόν, αν βλέπω σωστά, η σύγκλιση που περιγράφεις είναι ομοιόμορφη. Πράγματι, έστω ότι υπάρχει $ \varepsilon > 0$ τέτοιο ώστε να υπάρχει ακολουθία $ n_k$ με $ f_{n_k}(x_{n_k}) > 2\varepsilon$ για κάθε $ n_k$. Εφόσον το $ [0,1]$ είναι συμπαγές, μπορούμε να περάσουμε σε μία υπακολουθία και να υποθέσουμε ότι $ x_{n_k}\to x\in[0,1]$. Τότε όμως, για αρκετά μεγάλα $ n_k$ θα έχουμε λόγω συνέχειας:
\[ 2\varepsilon < f_{n_k}(x_{n_k}) \equal{} f_{n_k}(x_{n_k}) \minus{} f_{n_k}(x) \plus{} f_{n_k}(x) < f_{n_k}(x) \plus{} \varepsilon
\]
οπότε $ f_{n_k}(x) > \varepsilon$, contradiction αφού $ f_n\to 0$ παντού.

Συνεπώς, η σύγκλιση πρέπει να είναι ομοιόμορφη και νομίζω ότι τότε το όριο περνάει και μέσα στο ολ/μα Riemann (αν θυμάμαι καλά... έχω να κάνω ο,τιδήποτε με Riemann αρκετό καιρό :P).[/hide]

Cheerio,

Durandal 1707

[quote="Durandal"]Με παρεξήγησες, the other way around πήγαινε το σχόλιό μου ;)[/quote]

Ooops, sorry about that  :oops:

Συγχώρεσέ με in advance για μια ακόμα πιθανή παρεξήγηση (είναι 3:40am, after all), αλλά έχω την εντύπωση ότι η ακολουθία $ f_n : x \mapsto \frac{1}{n^3\left(x \minus{} \frac{1}{n}\right)^2 \plus{} 1}$, ικανοποιεί τις συνθήκες του προβλήματος, και συγκλίνει, αλλά όχι ομοιόμορφα, στο 0 (είναι ενα pinched curve που συρρικνώνεται και συγκλίνει προς τον άξονα των $ y$, αλλά έχει σταθερό ύψος $ 1$). Το πρόβλημα στο συλλογισμό σου αν δεν κάνω λάθος είναι εκεί που ισχυρίζεσαι ότι λόγω συνέχειας αναγκαστικά έχουμε $ f_{n_k}(x_{n_k}) \minus{} f_{n_k}(x) < \varepsilon$ for sufficiently large $ n_k$.

[quote="cmad"]Συγχώρεσέ με in advance για μια ακόμα πιθανή παρεξήγηση (είναι 3:40am, after all), αλλά έχω την εντύπωση ότι η ακολουθία $ f_n : x \mapsto \frac {1}{n^3\left(x \minus{} \frac {1}{n}\right)^2 \plus{} 1}$, ικανοποιεί τις συνθήκες του προβλήματος, και συγκλίνει, αλλά όχι ομοιόμορφα, στο 0 (είναι ενα pinched curve που συρρικνώνεται και συγκλίνει προς τον άξονα των $ y$, αλλά έχει σταθερό ύψος $ 1$). Το πρόβλημα στο συλλογισμό σου αν δεν κάνω λάθος είναι εκεί που ισχυρίζεσαι ότι λόγω συνέχειας αναγκαστικά έχουμε $ f_{n_k}(x_{n_k}) \minus{} f_{n_k}(x) < \varepsilon$ for sufficiently large $ n_k$.[/quote]

Bleh, ναι, προφανώς :( Serves me right για να είμαι τσαπατσούλης :oops:

Three cheers!

Durandal 1707

Λοιπόν, για να ξεκαθαρίσει το τοπίο, η άσκηση είναι αστεία με τη χρήση εργαλείων απο τη θεωρία Lebesgue -- όπως θα σας υπογράψει και ο Παναγιώτης. Το interesting part είναι να δοκιμάσετε να τη λύσετε με θεωρία Riemann αφού στο κάτω κάτω η άσκηση αυτή είναι το dominated convergence theorem στα ολοκληρώματα [u]Riemann[/u].


[b]ABCD [/b]is a parallelogram. [b]P [/b]is a point on the ray [b]DA[/b]. construct a line form [b]P [/b]to [b]C [/b]such that [b]P [/b]meets [b]AB [/b]at [b]Q [/b]and diagonal [b]DB [/b]at [b]R[/b]. [b]PQ = 735[/b], [b]QR=112[/b], find the distance of [b]RC[/b].
plz help me!  :D

[hide="Solution"]
Since $PD\parallel BC$, $\triangle PDR\sim\triangle CBR$.
So $\frac{DR}{RB}=\frac{PR}{RC}$

Similarly, $QB\parallel DC$, $\triangle QBR\sim\triangle CDR$.
So $\frac{DR}{RB}=\frac{CR}{RQ}$

Let $x=RC$.
Combining the two equations, we obtain $\frac{PR}{RC}=\frac{CR}{RQ}$ $\Longleftrightarrow\frac{735+112}{x}=\frac{x}{112}$

We can simplify the equation as $x^2=847\cdot 112=4^2\cdot 7^2\cdot 11^2$, so $x=4\cdot 7\cdot 11=\boxed{308}$
[/hide]


Hi!It's my first topic on ML.
So:
Given a point P inside the triangle ABC such that APB+ABC=APC+ACB, D the incircle in triangle APB, E the incircle in APC, proove that BD, CE and AP are concurrent.
I couldn't solve it... :blush:

I guess you mean IMO 1996 problem 2. Solutions can be found at http://www.kalva.demon.co.uk/imo/isoln/isoln962.html or at http://www.mathlinks.ro/Forum/viewtopic.php?t=1108 .

  Darij


function
algebra
polynomial
functional equation
complex analysis

Hello, could anybody help me to solve the following question: Let $ f$ be an entire non-constant function that satisfies the functional equation $ f(1 \minus{} z) \equal{} 1 \minus{} f(z)$ for all $ z \in \mathbb C$. Show that $ f(\mathbb C) \equal{} \mathbb C$.

Besides, I need to how which tricks we usually use to prove something like $ f(\mathbb C) \equal{} \mathbb C$. Thank you.

Oops, please move this topic to the section entitled Unsolved and Proposed Problems in the Complex Analysis category.

Looks like homework to me, because this is a trivial consequence of a well known theorem.

[quote="1234567a"]Aryabhatta : are you talking about Picard's theorem?
I cannot see why? can you explain?[/quote]

Thanks Aryabhatta, I got my own way. Assume by contradiction, then W.L.O.G. by the Picard's Theorem we can assume that $ f$ misses $ a \in \mathbb C$. By Hadamard's Theorem,
\[ f(z) \minus{} a \equal{} e^{p(z)}\]
for some polynomial $ p$. Therefore, $ f(z) \equal{} a \plus{} e^{p(z)}$ for all $ z \in \mathbb C$. From the fact that $ f(1 \minus{} z) \equal{} 1 \minus{} f(z)$ we get
\[ \underbrace {a \plus{} {e^{p\left( {1 \minus{} z} \right)}}}_{f\left( {1 \minus{} z} \right)} \equal{} \underbrace {1 \minus{} \left( {a \plus{} {e^{p\left( z \right)}}} \right)}_{1 \minus{} f\left( z \right)}\]
which yields
\[ {e^{p\left( z \right)}} \equal{} 2a \minus{} 1 \plus{} {e^{p\left( {1 \minus{} z} \right)}}.\]
Put $ z \equal{} 0$ and $ z \equal{} 1$, we obtain
\[ {e^{p\left( 0 \right)}} \equal{} 2a \minus{} 1 \plus{} {e^{p\left( 1 \right)}}, \quad {e^{p\left( 1 \right)}} \equal{} 2a \minus{} 1 \plus{} {e^{p\left( 0 \right)}}.\]
Hence
\[ {e^{p\left( 0 \right)}} \equal{} 2\left( {2a \minus{} 1} \right) \plus{} {e^{p\left( 0 \right)}}\]
which implies $ a \equal{} \frac {1}{2}$.
From the identity $ f(1 \minus{} z) \equal{} 1 \minus{} f(z)$ put $ z \equal{} \frac {1}{2}$ we then deduce that
\[ f\left( {\frac {1} {2}} \right) \equal{} \underbrace {\frac {1} {2}}_a\]
a contradiction.

Yes, Picard's theorem. The little one to be precise.

Basically $ f(z)$ takes all values except possibly some $ a$. Also note that $ f(1/2) \equal{} 1/2$. So we can assume $ a \neq 1/2$. It does take the value $ 1\minus{}a$, and hence takes the value $ a$.


Please help me with the following easy problem: The sum of 49 positive integers is 624. Prove that three of them are equal to one another.

Note that 1+1+2+2+3+3......+24+24+25=2(1+2+3+4...24)+25=625. So to make the sum less, you'd have to have three of the same number.


floor function
inequalities
graph theory

prove that in any bicolored complete graph with 14 vertices, there exists a monochromatic quadrangle, and similarly, in any tricolored complete graph with 80 vertices, there exists a monochromatic quadrangle.

quadrangle = ?  Does that mean we need the K4 to me monochromatic, or only a C4?  (I hope that terminology is correct -- haven't done graph theory in a while.  C4 means a cycle on 4 vertices, right?)

It means C4 otherwise it would be R(4,4)=18. (I'm not sure about that notation for Ramsey numbers.) I'll trust your notation for a cycle on 4 vertices. It's good enough to convince me.

So does this mean we need 4 vertices and 4 lines connecting them and forming a cycle to be monochromatic, or all 6 lines connecting them to all be the same color?

We can use the following known result:  If a graph with n vertices has more than n (1 + :sqrt:(4n - 3)) / 4 edges, then it has a 4-cycle.  Challenge: Can you prove that result?  Hint:  Count the number of simple paths of length 2, in two different ways.

With that result, we can prove stronger Ramsey bounds than you asked for.  For example, if we two-color a complete graph on 7 vertices, then there exists a monochromatic 4-cycle.  If we three-color a complete graph on 14 vertices, then again there exists a 4-cycle.  Do you see how to prove these results?

[color=green][b]Rieman's Theorem.[/b][/color] [i]If the graph $ G$ on $ n$ vertices contains no $ 4$-cycles, then:

$ |E|\leq\left\lfloor \frac {n}{4}\cdot (1 + \sqrt {4n - 3}\right\rfloor.$[/i]

[b]Proof.[/b] Denote by $ d(u)$ the degree of $ u.$ Let $ S$ be the set of pairs $ (u,\{v,w\}),$ where $ u$ is adjacent to $ v$ and $ w,$ with $ v\neq w.$
We count the following set $ S$ in two ways. Summing over $ u,$ we find $ |S| = \displaystyle\sum_{u\in V} \dbinom {d(u)}{2}.$ On the other hand, every pair $ \{u,w\}$ has at most one common neighbor, by the $ 4$-cycle condition. Hence $ |S|\leq\dbinom{n}{2},$ and we conclude:

$ \displaystyle\sum_{u\in V} \dbinom {d(u)}{2}\leq\dbinom{n}{2},$ or

$ \displaystyle\sum_{u\in V} d(u)^2\leq n(n - 1) + \displaystyle\sum_{u\in V}d(u).$ Using [color=blue]Cauchy-Schwarz inequality[/color] we have:

$ \left (\displaystyle\sum_{u\in V}d(u)\right )^2\leq n\cdot\left (\displaystyle\sum_{u\in V} d(u)^2\right ),$ therefore $ n\cdot\left (\displaystyle\sum_{u\in V} d(u)^2\right )\leq n^2 (n - 1) + n\cdot\left (\displaystyle\sum_{u\in V}d(u)\right ).$

Because $ 2|E| = \displaystyle\sum_{u\in V}d(u)$ we get:

$ |E|^2 - \frac {n}{2}\cdot |E| - \frac {n^2 (n - 1)}{4}\leq 0.$ Solving the quardatic equation we obtain the desired result.


probability
function
integration
advanced fields
advanced fields unsolved

Let $\Omega \subset \mathbb{R}$ be an open subset and $u_{n}$ a bounded sequence in $L^{\infty}(\Omega)$. Prove that one can extract a subsequence $v_{n}$ such that for almost every $x \in \Omega$ one can find a borel probability measure $\nu_{x}$ on $\mathbb{R}$ such that for every continuous function $f : \mathbb{R}\to \mathbb{R}$, $f(v_{n})$ tends $L^{\infty}$ weak* towards $\overline{f}$
where
$\overline{f}(x) = \int_{\mathbb{R}}f(y) d \nu_{x}(y)$

What can one say if $\nu_{x}$ is a Dirac measure for almost every $x$ ?

Sorry I just thought this was funny: is $u_{n}$ a sequence of bounded sequences whose bounds are bounded? (If not, then I don't understand the question. So I'm being silly and asking a serious question at the same time.)

edit: Okay, now I think maybe $u_{n}$ is a sequence of functions, and by $f(v_{n})$ you meant function composition? Then, I think if $\nu_{x}$ is a Dirac measure for a.e. $x$ then the sequence $v_{n}$ is pointwise convergent a.e. But I still didn't prove the main part. :|

you're right: $u_{n}$ is a sequence of functions, and there exists a constant $M$ such that $\|u_{n}\|_{\infty}< M$ for every $n$.

[quote="alekk"]$f(v_{n})$ tends $L^{\infty}$ weak* towards $\overline{f}$[/quote]
Sorry, what does it mean to tend $L^\infty$ weak*? Does that mean that for any bounded continuous function $g$, $\sup_{x}f\circ v_{n}(x) \cdot g(x) \to \sup_{x}\overline{f}(x) \cdot g(x)$?

$L^{\infty}$ is the dual of $L^{1}.$ To say that $f_{n}\to f$ weak* in $L^{\infty}$ is to say this:

For all $g\in L^{1},\ \int f_{n}g\to\int fg.$


the graders are so careless, they don't carefully read your solution at all.

http://68.38.137.202/round2.pdf

Look at question #5.
All they did was see the 13 at the end (which was not even my answer to the question) and gave a 4/5, saying the correct answer was 14, with the comment "You missed an edge. The answer is 14."

Unbelievable. Even when I explicitly stated in the first sentence, we show that the minimum is FOURTEEN. And I provided one of the two possible correct polyhedrons (up to an isomorphism in the polyhedral graph) WITH FOURTEEN edges.

Now, I guess this is forgiveable since the graders are not paid, but that is NO excuse to be negligent, even in volunteer work!

They did something similar to mine.
Except it helped me.
I got a totally wrong answer for number one or two, and they gave full credit.
I think they look for the number of pages your solution takes up.
For the previous two rounds I never answered one problem in each.
Now I think I should write something- I think they will give me a poit, or something.

well, they should give you a point or so, no? it shows you tried to at least comprehend the problem.

if you are headed in the right direction with at least some work they give you 2.

i got number 5 completely wrong and got 3 points

Update: I emailed Mr. Kelly about this problem 2 weeks ago and still I have not received a response. The USAMTS has basically lost my respect. Clearly I don't need to spend the time to write rigorous proofs, because they will give full credit anyway because they don't feel like reading the proof, and even if the proof is rigorous, it will be marked wrong because the person did not feel like reading it! And complaints about the grading are not even addressed!

I will make sure to \mathbf my answers this time in Latex, but it is truly a pity that I have to do so.

Before you totally lose respect - keep in mind the two weeks you're talking about are Christmas/New Years.  And that the USAMTS is free, and these graders have to look over a large number of papers.  Mistakes happen in grading, just like mistakes happen while you're taking a test; hopefully Dr. Kelly will be able to address this one.

Mr. Kelly seems to be a nonresponder in general, as I emailed him a personal question regarding my entries two weeks ago and he has not yet answered.

that seems weird. on the website it says,"In spite of the number of participants in the USAMTS, we aim to maintain a close, personal relationship with all USAMTS participants. In particular, we wish to serve as your mentors, your mathematical friends, and your advisors within areas of our expertise. Hence we welcome your correspondence, your asides, your comments. Write to us if you have complaints about the grading, if you wish to call some shortcomings to our attention, or if you have any suggestions for improving the program." Apparently this is not happening; I sent an email to mr. kelly also and he never responded.

Hmm... it's been another month since my 1/1/04 post. STILL no response. Now that's just poor.

(Dies laughing)

Sorry, but those claims on the USAMTS website are simply ridiculous.

[quote="JBL"]I've always found Dr. Kelly to be an excellent responder -- he's never taken more than 2 days to respond to any e-mail I've sent him.  Perhaps you should try again, [i]politely[/i] asking him about it.[/quote]

My reaction to this thread, as a busy adult who naturally favors

1) my own family time

and

2) my paying work

to my volunteer activities, was to wonder what the emails to the USAMTS graders REALLY said. It's one thing to say, "I wrote to the guy, and he hasn't answered me yet," and another to say, "Here is what I wrote [quoting full text of email] to [name of person] on [date]" and then let busy adults like me with off-line personal lives assess how rapidly WE would answer such an email, given the high-priority claims on our time from our work and our families.

We haven't participated in USAMTS yet, but it's on our list for next school year. I hope that the busy volunteers who make that program possible get enough appreciation to make up for the lack of pay and personal free time inherent in grading USAMTS contest papers. I view the various math contests available for young people as learning experiences--and one thing for young people to learn is how to deal with the other claims on adults' time that make the young learner's priorities less than the top priority.

I did politely ask him, and my questions were like clarifications for the problems. However, since he never responded, I had to guess the problem's meaning all by myself. But Mr. Kelly personally knows you, right JBL?

I met him once, but I wouldn't call us personally acquainted or anything.  He wouldn't recognize me if he saw me in the street.  If it makes you feel any better about the question clarification question, I once e-mailed him for clarification as to whether the question was asking for one of two things, and the response amounted to "Well, part of the question is to decide what it asks."  Like most other contests, I don't think they give "hints" or clarification.

Okay, here goes:

12/22/03:

email sent to usamts@nsa.gov

Dear Mr. Kelly,

My control number is 7023, and my name is Yifei Chen, Grade 11, from NJ. On problem #5, I received a 4/5 with the comment "You missed an edge. The answer is 14." However, in my solution, I provided a polyhedron with 14 edges and proved that 14 was the minimum. Can you look into this? Thank you.

Yifei Chen

well, i've asked him questions before and he's responded pretty quickly, so i don't know what to say.  But did you send in a note of inquiry with your solutions to round 3?  I think that's part of the intent of the 'note of inquiry' - to deal with grading problems like yours.

I sent him an email and he replyed like 3 months later so...


geometry
function
analytic geometry
calculus
integration
search
algebra

Officially this problem comes from algebraic geometry, but I'm posting it here since I'll try to reduce it to a purely algebraic problem.

If $ X$ is a variety (of dimension $ \geq 2$), $ P\in X$ is a normal point, and $ f$ is a regular function on $ X\setminus P$ then $ f$ extends to be regular on $ X$.

The definition of normal point is that $ \mathcal{O}_P$ is integrally closed (the coordinate ring localized at the maximal ideal that vanishes at $ P$, so $ \mathcal{O}_P\equal{}(k[x_1, \ldots, x_n]/I(X))_{\frak{m}_P}$).

These are just rational functions whose denominators don't go to zero at $ P$. So the idea is that if you have a regular function on $ X\setminus P$ we must show that it is an integral element over $ \mathcal{O}_P$ since this would mean that it actually is in $ \mathcal{O}_P$ which would give the extension.

It seems this question has been asked at least twice on the forum before this with no success...  :(

For other interpretations see http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1431815286&t=110506

and http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1431815286&t=232159

Is your question related to the fact $ k[\mathbb{A}^n\minus{}\{0\}] \equal{} k[\mathbb{A}^n]$ for $ n\ge2$?

Steve

Well, being able to extend a function on a neighborhood of $ P$ and being able to extend on all $ X$ are equivalent, so technically our first assumption can be that $ X$ is affine by just doing the problem on an affine neighborhood of $ P$.

There is probably some relation to the fact you wrote, but isomorphic coordinate rings isn't quite enough. My guess on the $ n\geq 2$ condition is that polynomials in 2 or more variables can't have zeros at single point, so a rational function can't blow up just at $ P$. I'd like to use this fact to show that the element is integral, but nothing comes to mind.

OK, perhaps you're right.  But note I didn't write the rings were isomorphic, they are in fact equal.  I thought something similar would happen since $ P$ was normal, regular functions on an affine patch would be quotients of functions in the local ring at $ P$, so your case would proceed similarly to the case I mentioned.  But it might not be enough.

Steve

Actually, maybe I dismissed this idea too quickly. It didn't occur to me to use the "normal" condition to get into the situation $ \mathbb{A}^n\setminus 0$, I was convinced that the localized coordinate ring being integrally closed meant that the best technique would be to show that a general rational function (in the right place) was integral and hence actually didn't blow up at $ P$.

This might be better suited for the "Advanced fields". I talked with some people today, and didn't realize it would turn into something this complicated.

The ideas I received were along the following lines.

$ X\setminus P\hookrightarrow X$, so we get an injective map $ \phi: \mathcal{O}_X\to \mathcal{O}_{X\setminus P}\simeq \mathcal{O}_P$.

Since the maximal ideal associated to the closed point $ P$ is has height $ \geq 2$ (since $ \dim X\geq 2$), we get by Serre's criterion for normality that depth$ (\mathcal{O}_P)\geq 2$.

Now consider the exact sequence $ 0\to \mathcal{O}_X\to \mathcal{O}_P\to coker \phi\to 0$.

Try to use the depth condition to get $ coker \phi\equal{}0$ and then $ \mathcal{O}_X\to \mathcal{O}_P$ is surjective (which says precisely that all rational functions whose denominators don't go to zero at $ P$ has a natural extension on $ \mathcal{O}_X$).

No one knew how to proceed to actually get this.

My guess is that we can now use the equivalent form of depth in terms of the non-vanishing of Ext groups and show that the long exact sequence in Ext induced from that given short exact sequence somehow says that $ \phi$ is surjective or that $ coker \phi\equal{}0$ or something.

Any other guesses on how to proceed or does anyone see that this might work?

Alright. I finally got this, so the next time someone asks it there will actually be an answer somewhere.

Let $ Y$ be a variety of dimension $ \geq 2$ and let $ P\in Y$ be a normal point. Let $ f$ be a regular function on $ Y\setminus P$. Claim: There is a regular function on $ Y$, say $ \hat{f}$ such that $ \hat{f}\big|_{Y\setminus P}\equal{}f$.

First note that this is a local property, so we may assume $ Y$ is affine.

Now we just rephrase the question in terms of algebra. Our assumption is that $ f\in\mathcal{O}_{Y\setminus P}$. We must show that really $ f\in \mathcal{O}_Y$. Note that it is sufficient to show that $ f\in \mathcal{O}_P$, since we know that away from $ P$, $ f$ is regular and so if $ f$ can be expressed as a rational function whose denominator is never $ 0$ at $ P$ we will be done.

The fact that $ P$ is normal tells us that $ \mathcal{O}_P$ is integrally closed. But then since it is also a Noetherian domain $ \displaystyle\mathcal{O}_P\equal{}\cap_{ht(q)\equal{}1}\mathcal{O}_{Z(q)}$ where $ q$ ranges over all height 1 prime ideals contained in $ m_P$. But each $ \mathcal{O}_{Z(q)}\hookrightarrow \mathcal{O}_{Y\setminus P}$, so by restriction ($ f\mapsto f\big|_{Z(q)}$) we get that $ f$ is in each $ \mathcal{O}_{Z(q)}$. Thus $ f\in \mathcal{O}_P\Rightarrow f\in \mathcal{O}_Y$, which completes the proof.


What is the sum of all values of $ x$ that are solutions to the equation $ x^2 \equal{} 7x \minus{} 12$?

$ x^2\equal{}7x\minus{}12 \iff x^2\minus{}7x\plus{}12$.
Thus,
The sum of all values of $ x$ is $ \frac{7}{1}\equal{}\boxed{7}$.
The product of all values of $ x$ is $ \frac{12}{1}\equal{}12$.
by Vieta's Formulas.


Find integers satisfying $ x^{2} \plus{} y^{2} \equal{} z^{2} \plus{} 1$.

You can let either x or y equal one.  If you let y equal one then you could make infinitely many pairs by having x = z and the opposite is true if you let x equal one.

$ x\equal{}2mn\plus{}1, y\equal{}mn^{2}\minus{}m\plus{}n, z\equal{}mn^{2}\plus{}m\plus{}n$ for integers $ m, n$.


inequalities
real analysis
real analysis unsolved

$ f: R^m\minus{}>R^m$ is a mapping satisfying
$ |f(x)\minus{}f(y)|\equal{}|x\minus{}y|$, where $ | |$ represents the Euclidean norm.
Prove that f is surjective.

$ \mathbb{R}^m$ is an affine space; we prove easily (draw circle arcs) that $ f$ transforms the barycenter of a couple of points in the barycenter of the transformed couple. Then $ f$ is an affine application. Moreover $ g$, the linear application associated to $ f$, is a linear isometry then is invertible. $ f(M)\equal{}f(O)\plus{}g(OM)$; then $ f\equal{}f_2.f_1$ where $ f_1$ is the affine bijection associated to $ g$ which fixes $ 0$ and $ f_2$ is a translation. $ f_1,f_2$ are bijections then $ f$ also.

[quote="loup blanc"]$ \mathbb{R}^m$ is an affine space; we prove easily (draw circle arcs) that $ f$ transforms the barycenter of a couple of points in the barycenter of the transformed couple. [/quote]

how to prove this? I cannot reach this assertion.

We use the notation $ U'\equal{}f(U)$. Let $ G$ be the barycenter of $ A(\alpha),B(\beta)$ and $ G_1$ be the barycenter of $ A'(\alpha),B'(\beta)$. $ d(G,A)\equal{}d(G',A'),d(G,B)\equal{}d(G',B')$; moreover $ d(A',B')\equal{}d(A,B)$ implies that $ d(G,A)\equal{}d(G_1,A'),d(G,B)\equal{}d(G_1,B')$. Thus $ d(G',A')\equal{}d(G_1,A'),d(G',B')\equal{}d(G_1,B')$ and $ G_1\equal{}G'$ because the triangular inequality is an equality (it's a property of the euclidean norm).


under each one is my answer,just check it
2006, S27

In a regular polygon there are two diagonals such that the angle between them is 50^{o}. What is the smallest number of sides of the polygon for which this is possible?
my answer12

2006, S22

The vertices of a cube are each labelled with one of the integers 1, 2, 3, ..., 8. A face-sum is the sum of the labels of the four vertices on the face of a cube. What is the maximum number of equal face-sums in any of these labellings?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
E
2006, S14

In a triangle PQR, \sin P = \frac{1}{3} and \sin Q = \frac{1}{4}. How many different values can the size of R have?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
D
2006, S15

How many different pairs of 2-digit numbers multiply to give a 3-digit number with all digits the same?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
C

where are these from?

also if you want to see $\frac{a}{b}$ you have to put dollar signs ($\$\$$) around \frac{a}{b}, and \sinQ, and whatever.b

these comes from 2006 australian math competion, as it is copyrighted. no reference is available

The S15 should be 2.

One angle is P and other is Q.  P and Q are acute angles.

From the resulting sines, it can be a triangle with

P and Q
180-P and Q
180-Q and P
180-Q and 180-P

The fourth one obviously doesnt work since they are both obtuse and the thrid angle would have to be negative.   Since P is larger than Q, 180-Q+P is also greater than 180 so it doesnt work.  The first two's sums are less than 180 so there can be a third angle, giving two different values for the thrid angle.


In a normed linear space, if $x_n \rightarrow x$, then $||x_n|| \rightarrow ||x||$.

by assumption we have that for all $\epsilon>0$ there is an $N$ such that for all $n \geq N$ we have $||x_n-x|| < \epsilon$.

then we look at

$\left| \, ||x_n|| - ||x|| \, \right| \leq ||x_n|| - ||x||$

but since $||x_n|| - ||x|| \leq ||x_n-x|| < \epsilon$ we have $||x_n|| \leq ||x|| + \epsilon$, hence

$\left| \, ||x_n|| - ||x|| \, \right| \leq ||x_n|| - ||x|| \leq ||x|| + \epsilon - ||x|| = \epsilon$

does this work for a proof?

oops, I dont think it works... I switched an inequality around on accident... is there a way to repair this argument?


algebra
polynomial
linear algebra
linear algebra unsolved

Suppose $ T$ is a linear map from $ \mathbb{C}^n$ to itself. Let $ p$ be a polynomial in $ \mathbb{C}$ and $ a\in C$. Show that if $ a$ is an eigenvalue of $ p(T)$, then $ a\equal{}p(\lambda)$ for some eigenvalue $ \lambda$ of $ T$.

I can show this by considering the Jordan normal form of $ T$.  But was wondering if there's another way around this problem.

Factorize $ p(T) \minus{} aI \equal{} (T\minus{}r_1I)...(T\minus{}r_kI)$. If $ p(T)\minus{}aI$ is not invertible, at least one of $ T\minus{}r_iI$ is not invertible, and note that $ a \equal{} p(r_i)$, so we are done.


pigeonhole principle
geometry unsolved
geometry

In a disk with center $ O$,there are four points such that the distance between every pair of them is greater than the radius of the disk.Prove that there is a pair of perpendicular diameters such that exactly one of the four points lies inside each of the four quarter disks,formed by the diameters.

Nobody?Please share your ideas with me if it is really useful.Thank you.

By pigeonhole pick two points $ A$ and $ B$ with central angle greater or equal to $ \frac {\pi}{2}$ .
Divide the disk in $ 12$ central sectors passing through $ A$ and number them with $ S_1,S_2,\cdots, S_{12}$
Then $ B \in S_i$ with $ i \ge 4$
and there are at least one sector between any two points then posible cases are:
$ (i,j,k) | (B,C,D) \in (S_i,S_j,S_k)$

does it help?


Prove that if $a, b, c, d$ are integers such that $d=( a+\sqrt[3]{2}b+\sqrt[3]{4}c)^{2}$ then $d$ is a perfect square.

Expanding the LHS, we get

\[ d=a^2+b^2\sqrt[3]{4}+2c^2\sqrt[3]{2}+2ab\sqrt[3]{2}+4bc+2ac\sqrt[3]{4}\hspace{1.5cm}(*)\]

Because $ \sqrt[3]{2},\sqrt[3]{4}$ are irrational, the coefficients of each must be 0; that is,

\begin{eqnarray}
b^2+2ac=0 \\
2c^2+2ab=0
\end{eqnarray}

We wish to show that $ b=c=0$, from which $ d=a^2$, as desired, would follow.

Note that $ a\neq 0$ because the RHS of (*) would be irrational.

Assume for contradiction that neither $ b$ nor $ c$ are zero. Then, moving the squares of (1) and (2) to one side and dividing, we have
\[ \frac{b^2}{c^2}=\frac{2c}{b} \implies 2c^3=b^3 \implies c=b\sqrt[3]{2}\], which contradicts $ b,c\in\mathbb{Z}$. Therefore, at least one of them is zero. Then, (1) implies that the other is also zero, as desired.