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Hello!
To be easy for you - just see the attachment: There are given six points on a circle BHCFEG (in this order). KLMNIJ are the vertices of the hexagon formed by the intersecting lines. Prove that the opposite diagonals of KLMNIJ - KN, JM, IL intersects at a common point.

I discovered this statement using Geogebra. The software says it it is true. My questions are:
1. Is it a famous theorem/problem?
2. How to prove it.

I would like to ask you - if it is possible please send me the proofs in a private message and if it possible, please use as simple techniques as possible. If you would like to share your proofs to all the members of the forum - it is not a problem.

Thank you very much!

I don't know if it is a famous theorem dear [b]Borislav[/b], but surely it is coming from the past.

So, there is another more geberal configuration of an arbitrary convex hexagon, as follows :

[b][size=100][color=DarkBlue]GENERAL THEOREM. - Let $ ABCDEF$ be, an arbitrary convex hexagon and we denote the points $ K\equiv AB\cap CD$ and $ L\equiv BC\cap DE$ and $ M\equiv CD\cap EF$ and $ N\equiv DE\cap AF$ and $ P\equiv EF\cap AB$ and $ P\equiv AF\cap BC$. Prove that :

(a) - If the diagonals $ AD,\ BE,\ CF$ of the given hexagon, are concurrent at one point then, the line segments $ KN,\ LP,\ MQ,$ taken per two of them, intersect each other at three points $ X,\ Y,\ Z,$ lie on the above diagonals. The inverse is also true ( [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=23602]schema t=310049[/url] ).

(b) - If the line segments $ KN,\ LP,\ MQ$ are concurrent at one point then, the diagonals $ AD,\ BE,\ CF$ of the given hexagon taken per two of them, intersect each other at thre points $ R,\ S,\ T,$ lie on the above line segments. The inverse is also true ( [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=23601]schema t=310049(a)[/url] ).[/color][/size][/b]

$ \bullet$ Return now in to your problem ( [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=23600]schema [b]t=310049(b)[/b][/url] ), based on the [b]Pascal theorem[/b], it is easy to show that the points $ A,\ R,\ D,$ are collinear (consider the non-convex cyclic hexagon $ PKMQNL$ ).

That is, the point $ R\equiv LP\cap MQ,$ lies on the diagonal $ AD$ of the convex hexagon $ ABCDEF$ and similarly, the points $ S\equiv KN\cap LP$ and $ T\equiv MQ\cap KN$, lie on the diagonals $ CF,\ BE,$ respectively.

Hence, based on the part [b](a)[/b] of the above general theorem, we conclude that the diagonals $ AD,\ BE,\ CF$ of $ ABCDEF$, are concurrent at one point and the proof of the proposed problem is completed.

Best regards, Kostas Vittas.

PS. I will search my archive and I will post here next time the ( not difidult ) proofs I have in mind, of the parts [b](a), (b),[/b] of the General theorem I mentioned.

[quote="borislav_mirchev"]Very nice proof dear Kostas!

Thank you very much!
Do you like the problem?
What is the level of difficulty of that problem?
Everyone says it is probably a result from the past but no one said who is the inventor. Probably it is not very famous result.
What if we don't know these properties of the hexagon and Pascal theorem and we start from school level /we know similarities, inscribed angles, chord proportions, Sine law and at most Ceva and Menelaus/?[/quote]
Dear [b]Borislav[/b], I have also (re)discovered the general theorem some years ago and I don't know any reference about it.

But, I think that it is not a new result.

Best regards, Kostas Vittas.

Dear Kostas,

It is not my first interesting result i rediscovered ... for example I rediscovered a generalization of the Butterfly theorem and Romania TST 1993 problem /the equilateral triangle/ as well as the formulae sin(a+b). Anyway I'll be happy the people spent good time in solving the problem.

The problem is probably true if the circle is an ellipse as the Pascal theorems. It probably can be solved in many ways - with analytical geometry / complex numbers / trigonometry, similarities. Any other solutions are appreciated and aswers of the questions above.

[quote="borislav_mirchev"]I'm sorry for my double post but can we see the proofs of the lemas you use? I like these statements.[/quote]
Dear [b]Borislav[/b], I searched a little today morning but, I didn't find the proofs ( I don't remember them ) and I am too lazy to try again to solve the [b]Lemmas[/b] I mentioned.

Give me please a little time and I hope that I wiil find them.

Kostas Vittas.

see here:
http://www.mathlinks.ro/viewtopic.php?t=295936
(note that the proof that i posted works also for this).

Dear Kostas,

Thank you for the nice materials ... there is not need to hurry ... I'll try to solve the original problem. I hardly start anything but when I start I hardly stop ... who know maybe I can prove the lemmas Smile ... but if you find the materials, please post for all mathlinkers or send me them in a PM.

In the weekend I'll try to find some time for thinking. Why you write my name in bold?

Shoki, your solution is simpler. It is a nice application of the Pascal theorem. Thank you! Indeed may I ask you to post a picture with exact letters on it, to replace the word "obvious" with your exact arguments and every time when you apply Pascal theorem to say the vertices of the hexagram you apply it for.

I like both solutions.

[quote="borislav_mirchev"]There are given six points on a circle BHCFEG (in this order). KLMNIJ are the vertices of the hexagon formed by the intersecting lines. Prove that the opposite diagonals of KLMNIJ - KN, JM, IL intersects at a common point[/quote]
Due to Brianchon's theorem the problem is equivalent to show that that the convex hexagon LMNIJK is circumscribed in a conic. Take a look at the topic [url=http://www.artofproblemsolving.com/viewtopic.php?t=233430]Two triangles on a circle[/url].

[quote="borislav_mirchev"]
....
to replace the word "obvious" with your exact arguments and every time when you apply Pascal theorem to say the vertices of the hexagram you apply it for.
...
[/quote]
i used papus theorem twice in the second line.

shoki's and yetti's solutions use similar arguments. yetti's solution is very clear and elegant with excellent descriptions it makes the problem to looks as a simple game. i think the result i posted is great and beautiful and i had the luck to find it on my own. in the weekend i'll try to solve Kostas' lemmas and two more people promised me to show their solutions.

[quote="vittasko"][b][size=100][color=DarkBlue]GENERAL THEOREM. - Let $ ABCDEF$ be, an arbitrary convex hexagon and we denote the points $ K\equiv AB\cap CD$ and $ L\equiv BC\cap DE$ and $ M\equiv CD\cap EF$ and $ N\equiv DE\cap AF$ and $ P\equiv EF\cap AB$ and $ P\equiv AF\cap BC$. Prove that :

(a) - If the diagonals $ AD,\ BE,\ CF$ of the given hexagon, are concurrent at one point then, the line segments $ KN,\ LP,\ MQ,$ taken per two of them, intersect each other at three points $ X,\ Y,\ Z,$ lie on the above diagonals. The inverse is also true ( [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=23602]schema t=310049[/url] ).[/color][/size][/b][/quote]
$ \bullet$ [b]PROOF[/b]. - Let be the points $ O\equiv AD\cap BE\cap CF$ and $ R\equiv CP\cap EQ$ ( [b]schema t=310049(c)[/b] ).

Applying the [b]Pappos theorem[/b], for the triads of points $ P,\ F,\ E$ and $ Q,\ B,\ C,$ on the line segments $ PM,\ QL$ respectively, we have that the points $ A\equiv PB\cap QF$ and $ R\equiv PC\cap QE$ and $ O\equiv FC\cap BE,$ are collinear.

Because of now the collinearity of the points $ A,\ O,\ D,$ we conclude that the $ R,$ lies on the diagonal $ AD$ of the given hexagon $ ABCDEF.$

$ \bullet$ Applying again the [b]Pappos theorem[/b], for the triads of the points $ P,\ E,\ M$ and $ Q,\ C,\ L,$ we have that the points $ R,\ X\equiv PL\cap QM$ and $ D\equiv EL\cap CM,$ are collinear and so, we conlude that the point $ X,$ lies also on $ AD.$

Similarly, the points $ Y\equiv PL\cap KN$ and $ Z\equiv QM\cap KN,$ lie on the diagonals $ CF,\ BE$ and the proof of the part [b](a)[/b] of the general theorem, is completed.

Kostas Vittas.

Dear Kostas,

Thank you very much! You always show us the beauty of the clear geometry approaches.
These theorems are one of the many proofs of my words. A person can learn many things from you and the other mathlinkers.
You were right - it was an old problem but maybe a little forgotten.

[quote="vittasko"][b][size=100][color=DarkBlue]GENERAL THEOREM. - Let $ ABCDEF$ be, an arbitrary convex hexagon and we denote the points $ K\equiv AB\cap CD$ and $ L\equiv BC\cap DE$ and $ M\equiv CD\cap EF$ and $ N\equiv DE\cap AF$ and $ P\equiv EF\cap AB$ and $ P\equiv AF\cap BC$. Prove that :

(b) - If the line segments $ KN,\ LP,\ MQ$ are concurrent at one point then, the diagonals $ AD,\ BE,\ CF$ of the given hexagon taken per two of them, intersect each other at thre points $ R,\ S,\ T,$ lie on the above line segments. The inverse is also true ( [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=23601]schema t=310049(a)[/url] ).[/color][/size][/b][/quote]
$ \bullet$ - ( [b]schema t=310049(d)[/b] ) - Because of the concurrency of $ O\equiv PL\cap KN\cap MQ,$ based on the [b]Desarques theorem[/b], we conclude that the triangles $ \bigtriangleup PKM,\ \bigtriangleup LNQ$ are perspective and so, we have that the points $ S\equiv PK\cap LN$ and $ U\equiv PM\cap LQ$ and $ V\equiv KM\cap NQ$ are collinear.

We consider now these points as the points of intersection of the opposite sidelines of the given hexagon $ ABCDEF$ and based on the [b]Pascal theorem[/b] and because of their collinearity, we conclude that this hexagon is inscribed in a conic.

Applying again the [b]Pascal theorem[/b], in the non-convex hexagon $ ADEFCB$ inscribed in that conic, we conclude that the points $ P\equiv EF\cap BA$ and $ X\equiv AD\cap FC$ and $ L\equiv DE\cap CB$ are collinear.

That is, the point $ X,$ as the point of intersection of the diagonals $ AD,\ CF$ of the given convex hexagon $ ABCDEF,$ lies on the line segment $ PL.$

Similarly, the points $ Y\equiv BE\cap CF$ and $ Z\equiv AD\cap BE,$ lie on the line segments $ KN,\ MQ$ respectively and the proof of the part [b](b)[/b] of the general theorem is completed.

Kostas Vittas.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14447[/img]

 :(

[url]http://www.polarprof.org/geometriagon/wesercizio.asp?c=545437&n=431[/url]


I want your opinion. Which do you think is better.
Doing graduation in India or abroad.
I want to know whether it is better to study in IIT or in foreign universities like MIT etc.
I know that if I have a fixed goal and study hard I can get into IIT or even MIT. But now I have no fixed goal.
I want advice on which is more advisible in terms of quality of education.
Also will I get financial aid if I study abroad?

well according to me mit isTHE best but getting admission there isnt easy besides academic achievements u have to show some extracurricular skills
so if you are in 8th and 9th standard start working on your extacurricular achievements and if u are in class10th or above iit is a better option
ps if u get into iit or mit there would a wonderful feeling in your a life and a sense of achiement
to achieve this feeling work hard :lol:

for more info visit the official site of mit

Unless you are like a genius or extremely well-rounded applicant or in class 11 or lower, I would strongly suggest against MIT. You don't need to be the best in the world to get there, but yeah, the standards are pretty high and you'll need yoru SAT subject test scores, SAT reasoning scores, extracurriculars and weird activities to set you apart from the crowd.
Also, you cannot define if IIT is better or MIT, they are good int heir own departments. But for my purpose, IITs are worse than eeven IISERs. Also, MIT is not the best in the US. It depends on your preferences. Its not even in my 'top 3' American universities.
And for the record, its called 'undergraduation' in the US.

What kind of extracurricular skills are required?
What are the subjects tested in SAT?
Which is the best university for undergraduation in the world (tell me for the 5 most popular departments)?

What kind of extracurricular skills are required?
Anything you are passionate about
What are the subjects tested in SAT?
Whatever subjects you chose in the Subject tests; English and Maths(10th grade) in SAT Reasoning test.
Which is the best university for undergraduation in the world (tell me for the 5 most popular departments)?
It is IMPOSSIBLE to say that. Its always about being the right 'fit' with a college. All colleges have there own qualities, and its up to you to decide. If it were not so, they wouldn't be really in the class they are in. But still, if you live by the famed numbers, the most reliable rankings are here- http://colleges.usnews.rankingsandreviews.com/best-colleges/engineering

And please mention your class/rough profile before asking questions next time. I might be able to help you better.


(as posted earlier by Surreality)
[quote]
For those of you who didn't take the AIME (especially beta, who asked), here are the questions:

1. The digits of a positive integer n are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 37?

2. Set A consists of m consecutive integers whose sum is 2m and set B consists of 2m consecutive integers whose sum is m. The absolute value of the difference between the greatest element of A and the greatest element of B is 99. Find m.

3. A convex polyhedron P has 26 vertices, 60 edges, 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does P have?

4. A square has sides of length 2. Set S is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set S enclose a region whose area to the nearest hundredth is k. Find 100k.

5. Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta, who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success ratio (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could have achieved is m/n where m and n are relatively prime positive integers. What is m + n?

6. An integer is called snakelike if it's decimal representation a(subscript 1)a(subscript 2)a(subscript 3)......a(subscript k) satisfies a(subscript i) < a(subscript i+1) if i if i is odd and a(subscript i) > a(subscript i+1) if i is even. How many snakelike integers between 1000 and 9999 have 4 distinct digits?

7. Let C be the coefficient of x  in the expansion of the product:
(1 - x)(1 + 2x)(1 - 3x)...(1 + 14x)(1 - 15x). Find |C|.

I'll put up the other 8 in a separate post, because this one's getting long.[/quote]

(again, as posted by Surreality...)
[quote]
here are the other 8. (sorry if i mess up the list in #8, i've never used that function before)

8. Define a regular n-pointed star to be the union of n line segments P(subscript 1)P(subscript 2), P(subscript 2)P(subscript 3),......, P(subscript n)P(subscript 1) such that:

the points P(subscript 1), P(subscript 2),......, P(subscript n) are coplanar and no three of them are collinear,
each of the n line segments intersects at least one of the other line segments at a point other than an endpoint
all of the angles at P(subscript 1), P(subscript 2),......P(subscript n) are congruent,
all of the n line segments P(subscript 1)P(subscript 2), P(subscript 2)P(subscript 3),......, P(subscript n)P(subscript 1) are congruent, and
the path P(subscript 1)P(subscript 2)......P(subscript n)P(subscript 1) turns clockwise at an angle of less than 180  at each vertex.

There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are 2 non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?

9. 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  ABC into a  U(subscript 1) and a trapezoid V(subscript 1), and another segment is drawn to divide rectangle DEFG into a  U(subscript 2) and a trapezoid V(subscript 2) such that U(subscript 1) is similar to U(subscript 2) and V(subscript 1) is similar to V(subscript 2). The minimal value of the area of U(subscript 1) can be written in the form m/n where m and n are relatively prime positive integers. Find m + n.

10. A circle of radius 1 is randomly laced in a 15-by-36 rectangle ABCD so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal AC is m/n, where m and n are relatively prime positive integers, find m + n.

11. A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid C and a frustum-shaped solid F, in such a way that the ratio between the areas of the painted surfaces of C and F and the ratio between the volumes of C and F are both equal to k. Given that k = m/n, where m and n are relatively prime positive integers, find m + n.

12. Let S be the set of ordered pairs (x,y) such that 0 < x < 1, 0 < y < 1, and |log(subscript 2)*(1/2*x)| and |log(subscript 5)*(1/y)| are both even. Given that the area of the graph of S is m/n where m and n are relatively prime positive integers, find m + n. The notation |z| denotes the greatest integer that is less than or equal to z.

13. The polynomial:
P(x) = (1 + x + x  + ... + x^17)^2  - x^17
has 34 complex zeros of the form z(subscript k) = r(subscript k)*[cos(2*  *a(subscript k)], k = 1,2,3...,34, with 0 < a(subscript 1)  a(subscript 2)  a(subscript 3)  ...  a(subscript 34) and r(subscript k) > 0. Given that a(subscript 1) + a(subscript 2) + a(subscript 3) + a(subscript 4) + a(subscript 5) = m/n, where m and n are relatively prime positive integers, find m + n.

14. A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from its nearest point on the tower, and the length of rope that is touching the tower is (a- :sqrt: b)/c feet, where a, b, and c are positive integers, and c is prime. Find a + b + c.

15. For all the positive integers x let:
f(x) =
1
if x = 1
x/10
if x is divisible by 10,
x+1
otherwise

and define a sequence as follows: x(subscript 1) = x and x(subscript n+1) = f(x(subscript n)) for all positive integers n. Let d(x) be the smallest n such that x(subscript n) = 1. (For example d(100) = 3 and d(87) = 7.) Let m be the number of positive integers x such that d(x) = 20. Find the sum of the distinct prime factors of m.

Good luck working on these.
[/quote]

[edit]All fixed! Thanks guys!

Problem 14 should read   (a -  :sqrt: b)/c

   in case square root emoticons didn't work: (a - squareroot(b))/c

#13: the geometric series is supposed to be squared

#15: d(x) = 20, not 120.


calculus
integration
calculus computations

Calculate $ \int_0^1 \int_y^1 x^2 e^{xy} dx dy$.

The point of this problem is that sometimes switching the order of integration simplifies things.  Use a geometric argument to see why:
\[ \int_0^1\int_y^1x^2e^{xy}dxdy \equal{} \int_0^1\int_0^xx^2e^{xy}dydx\]
The algebra for the integral on the right hand side is significantly easier to carry out, and does not require integration by parts.

So, can I do it like this? (Switching the order is using Fubini's Theorem, right?)
$ \int_0^1\int_y^1 x^2e^{xy} dx dy \equal{} \int_0^1 \int_0^x x^2e^{xy} dy dx \equal{} \int_0^1 \left( \int_0^x x^2e^{xy} dy\right) dx \equal{} \int_0^1 \left[xe^{xy}\right]_0^x dx \equal{} \int_0^1 xe^{\left(x^2\right)} \minus{} x dx \equal{} \int_0^1 x(e^{x^2}\minus{}1) dx \equal{} \left[\frac{1}{2}\left(e^{x^2}\minus{}x^2\right)\right]_0^1 \equal{} \frac{e\minus{}1}{2} \minus{} \frac{e^0}{2} \equal{} \frac{e\minus{}2}{2}$


Punctele $\ E,F$ sunt mijloacele laturilor $\ AB,AD$ ale paralelogramului $\ ABCD$ iar segmentele $\ CE,BF$
se intersecteaza in punctul $\ K$.Punctul $\ M$ se ia pe $\ (EC)$ iar $\ BM$ e paralela cu $\ KD$.Demonstrati ca
$\ S_{KFD}=S_{KMBD}$.

intai alegem $U$ intersectia dintre $CE$ si $AD$, apoi $V$ intersectia dintre $BF$ si $DC$, $S$ intersectia dintre $DK$ si $AB$.
menelaus pe acolo de muuulte ori (chiar ca imi e lene sa scriu) si se obtin rapoarte :
$KF/KB=3/2$, $KC/KA=4$, apoi $SA/SB=3$, deci $SA=3SB$, deci $AB=4SB$. iese imediat $K$ mijlocul lui $ME$, si apoi iar cu menelaus $DK/SK=4$, deci $DK=4SK=2BM$. ducem inaltimea din $F$ la $DK$ o notam $3y$. cu thales avem ca distanta de la $K$ la $BM$ o sa fie $2y$. ramane sa aratam ca $DK\cdot 3y=(DK+BM)\cdot 2y$, sau $DK=2BM$, care este adevarat conform relatiei de mai sus.
stiu ca am omis o groaza de pasi dar sunt doar menelausuri si (desi nu imi aduc aminte sa fi folosit) umpic de calcul cu proportii...

Eu am rezolvat-o ducand o paralela prin $\ D$ la $\ EC$ pana la intersectia ei cu $\ AB$ in $\ N$.Apoi segmentul $\ BF$ il prelungim pana cand intersecteaza pe $\ DN$ in $\ P$.....e greu dupa aceea.Trebuie sa vezi niste asemanari, de exemplu $\ BCK$,$\ FPD$...calculezi rapoarte si ajungi la concluzie.E totusi o problema mai greuta, dar frumoasa. :D

cand am vazut despre ce e vorba ma gandeam "sa aplic menelaus, sa nu aplic menelaus"... la primele doua menelausuri a mai mers da deja pe la al cincilea nu prea mai imi ardea chiar asa...


[quote=dduclam]For $ a,b,c$ be nonnegative real numbers, we have
\[ \frac1{\sqrt{a^2\plus{}bc}}\plus{}\frac1{\sqrt{b^2\plus{}ca}}\plus{}\frac1{\sqrt{c^2\plus{}ab}}\ge\frac{\sqrt2}{\sqrt{a^2\plus{}b^2\plus{}c^2}}\plus{}\frac2{\sqrt{ab\plus{}bc\plus{}ca}}
\][/quote]
Any one?
[url=https://artofproblemsolving.com/community/c6h1312855p7039304]h [/url] [url=https://artofproblemsolving.com/community/c6h1085761p6026769]h[/url] [url=https://artofproblemsolving.com/community/c6h1293262p6916809]h[/url] [url=https://artofproblemsolving.com/community/c6h1333249p7200047]h[/url]

[quote=dduclam]For $ a,b,c$ be nonnegative real numbers, we have
\[ \frac1{\sqrt{a^2\plus{}bc}}\plus{}\frac1{\sqrt{b^2\plus{}ca}}\plus{}\frac1{\sqrt{c^2\plus{}ab}}\ge\frac{\sqrt2}{\sqrt{a^2\plus{}b^2\plus{}c^2}}\plus{}\frac2{\sqrt{ab\plus{}bc\plus{}ca}}
\][/quote]
I agree, it's really nice. There is a nice proof by C-S, AM-GM  and SOS.


do you know what$u_{2},u_{3}$is,it is important that find the number $i$satisfy $u_{i+1}-u_{i}=1$


geometry
geometric transformation
rotation
calculus
calculus computations

Okay I am learning Double and my book starts using these terms. The only explanation given for a Lamina is that it's a flat surface???  and it keeps saying "the moments about a particular axis".

Thanks!

A lamina is a closed two-dimensional planar surface with a given mass and density: http://mathworld.wolfram.com/Lamina.html, and I'm pretty sure that "moment" means moment of inertia.

Thanks!

Okay....what's a moment of inertia?

The moment of inertia of an object is its resistance to a change in its state of motion.

Here is a sight that has a few examples of laminas and their moments: http://ltcconline.net/greenl/courses/202/multipleIntegration/MassMoments.htm

to understand, it would help to have taken physics, but the idea is that just as objects will not accelerate with out a force exerted on it, we say that the resistance to that is somethings mass, and when one does rotations, then the object stays with a constant angular velocity


calculus
derivative
limit
calculus computations

How to compute the Taylor series of $ (x\plus{}1)^{x\plus{}1}$ about the point $ 0$?

Can someone please help. Thank you very much. :P

You would have to follow the Taylor formula and keep taking the derivative.  As to how to take that derivative, I'm really sure.

I would compute the first few derivatives:

$ f(x) \equal{} (x \plus{} 1)^{(x\plus{}1)}$

$ f'(x) \equal{} (x \plus{} 1)^{(x\plus{}1)}[ln(x\plus{}1) \plus{} 1]$

$ f''(x) \equal{} (x \plus{} 1)^{x}[(x\plus{}1)(ln(x\plus{}1))^{2 }\plus{} 2(x\plus{}1)ln(x\plus{}1) \plus{} (x\plus{}1) \plus{} 1]$

$ f'''(x) \equal{}(x \plus{} 1)^{x\minus{}1}[(x\plus{}1)^{2}(ln(x\plus{}1))^{3} \plus{} 3(x\plus{}1)^{2}(ln(x\plus{}1))^{2} \plus{} 3(x\plus{}1)(x\plus{}2)ln(x\plus{}1) \plus{} (x\plus{}1)^{2} \plus{} 3(x\plus{}1) \minus{} 1]$

So that we have :
$ f(0) \equal{} 1; f'(0) \equal{} 1 ; f''(0) \equal{} 2 ; f'''(0) \equal{} 3$

I am suspecting $ f^{n}(0) \equal{} n$ (this may need verification)
We will then have
$ f(x) \equal{} 1 \plus{} x \plus{} x^{2} \plus{} \frac{x^{3}}{2!} \plus{} ....\plus{} \frac{x^n}{(n\minus{}1)!}$

akech, the series you have written down is $ f(x) \equal{} 1 \plus{} xe^x.$ That can't be right, since $ (1 \plus{} x)^{1 \plus{} x}\ne1 \plus{} xe^x.$

In fact, the correct series should only have a radius of convergence of $ 1,$ since $ (1 \plus{} x)^{1 \plus{} x}$ has a singularity of some sort at $ x \equal{} \minus{} 1.$

Trying another approach:

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

$ \equal{} x \plus{} \frac {x^2}2 \minus{} \frac {x^3}6 \plus{} \frac {x^4}{12} \plus{} \cdots \plus{} \frac {( \minus{} 1)^nx^n}{n(n \minus{} 1)} \plus{} \cdots$

Now, we try to take the exponential of this series - a tedious process. I got this far:

$ f(x) \equal{} 1 \plus{} x \plus{} x^2 \plus{} \frac {x^3}2 \plus{} \frac {x^4}3 \plus{} O(x^5).$

Thank you. I first thought of your approach, but then I thought taking the exponential was going to look complicated.

Kent Merryfield is right .

I want to say that sometimes take exponential is bad , I mean it's don't save aproximation. but $ log$ save aproximation:
for example
for big $ n$
we have $ n^{2}$~$ n^{2} \plus{} n$ its true because
$ \lim_{n \minus{} > \plus{} oo}\frac {n^{2}}{n^{2} \plus{} n} \equal{} 1$
but if you take exponenta you have
$ e^{n^{2}}$~$ e^{n^{2} \plus{} n}$ but it's not true because
$ \lim_{n \minus{} > \plus{} oo}\exp({n^{2} \plus{} n \minus{} n^{2}}) \equal{} \lim_{n \minus{} > \plus{} oo}\exp (n) \minus{} > \plus{} oo$
remember

The sequence of the coefficients times $ n!$ is [url=http://www.research.att.com/~njas/sequences/A005727]A005727[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url].


function
algebra proposed
algebra
Fixed point

Suppose that $f$ is real continuous function satisfying for each real number $x$, there is an integer $n=n(x)$ such that
$f_n(x)=1$ where $f_n(x)=f(f_{n-1}(x))$
Determine $f(1)$

$f(1)=1$.  Suppose $f(1)\neq 1$; then the function has no fixed points, because if $a\neq 1$ is a fixed point then $f_n(a)=a\neq 1$ for any $n$.  As $f(x)$ never equals by $x$, by continuity either $f(x)>x$ for all $x$, or $f(x)<x$ for all $x$.  If the former, the sequence $2, f(2), f(f(2)), \ldots$ is increasing and $f_n(2)>1$ for all $n$.  If the latter, the sequence $0, f(0), f(f(0)), \ldots$ is decreasing and $f_n(0)<1$ for all $n$.


Let $\alpha ,\ \beta \ (\alpha < \beta )$ be the roots of the quadratic equation $x^{2}-x-1=0,$ we have $\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=\frac{2-\beta}{2-\alpha}\cdot \frac{a_{n}-\alpha}{a_{n}-\beta},$ thus we obtain $\frac{a_{n}-\beta}{a_{n}-\alpha}=\frac{\beta}{\alpha}\left(\frac{2-\beta}{2-\alpha}\right)^{n},$ yielding $a_{n}=\frac{(2-\beta)^{n}-(2-\alpha)^{n}}{\alpha (2-\alpha)^{n}-\beta (2-\beta)^{n}}.$

A trick for easier calculation: Put $a_{n}={p_{n}\over q_{n}}$ and create a system in $p_{n}, q_{n}$


Let $ \minus{} 1\leq a,b,c\leq 1$ s.t. $ \quad \frac {1}{1 \minus{} a} \plus{} \frac {1}{1 \minus{} b} \plus{} \frac {1}{1 \minus{} c} \equal{} \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b}\plus{}\frac {1}{1 \plus{} c}$. Prove that
\[ \frac {1}{3 \minus{} a} \plus{} \frac {1}{3 \minus{} b} \plus{} \frac {1}{3 \minus{} c}\geq 1
\]

[quote="Lippilus"]Let $ \minus{} 1\leq a,b,c\leq 1$ s.t. $ \quad \frac {1}{1 \minus{} a} \plus{} \frac {1}{1 \minus{} b} \plus{} \frac {1}{1 \minus{} c} \equal{} \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c}$. Prove that
\[ \frac {1}{3 \minus{} a} \plus{} \frac {1}{3 \minus{} b} \plus{} \frac {1}{3 \minus{} c}\geq 1
\]
[/quote]
See here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=183247
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176808
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172516


Ross Mathematics Program
email
summer program
PROMYS
number theory

Who is applying to the Ross Program for 2006?

Wow......no one?  I thought Ross was awsome! :roll:

Thanks for your reply. I think Ross is a great program too (I know an alumnus from the 1970s locally, and have email correspondence with a math professor who knows the program well), but it seems that its daughter (sister?) program PROMYS gets all the electrons in online discussion.

Yep, Ross is awsome!  Everyone apply now!!! :lol:

I wish I could go back again to study there.  It has an extremely positive effect on me in terms of learning math.  It's just the atmosphere.  However, for those who don't know much about rigor, it is a great place to learn how to start formulating your own systems starting from the axioms.  It took me a whole week to include an axiom that was basically an awkward form of WOP for my integers.  Unfortunately, this summer, I am succumbing to the excitement of the TST and I don't think Ross will want to pay me to be a counselor for only part of the summer.

Also, if you go to Ross, you can get those great QR shirts and wear them all the time!

Heh, too bad they probably won't invite me back again... :roll: those celebrations... :roll:

and peter: Of course they will hire you for the last 5 weeks.  :)

I went to Ross last year too and had possibly the best time (and most educational) time of my life.  Highly recommend it.

Thanks for checking into this thread with reports on your experiences at Ross.

I have noticed that Ross is very underrepresented here.  I think it is because it tends to be a little secretive and for good reason.  When I was there, people who went in knowing very little actually gained much more and learned much more.  This is because they didn't spend all of their time whining about rigor or simply quoting theorems they had to learn before.  It was much more fun rediscovering CRT or fermat's little theorem with no prior background.  Most of the time we proved something and didn't even know it was a named theorem.  Conjecturing and self-exploration are so heavily emphasized that often counselors do not like to reveal all of the information to incoming students.  I remember a kid trying to post information about Ross in wikipedia and all of us counselors and jc's getting angry and editting it heavily.  The self-exploration factor really is key although highly highly highly qualified and knowledgeable counselors can always guide you.  I remember one particular counselor challenging me until late into the night about what axioms to use for what I now recognize to be set theory.

any french speaking guy going to the Ross? It will be a pleasure for me to talk to him before meeting him in Columbus.

I really agree with Pdiao in the whole notion of the self-exploration.  That was definitely a huge part of my experience at Ross and was really helpful in truly making the connections myself, rather than having things shown to me.  And yes, the counselors/jc's are EXTREMELY qualified and are really amazing.

[quote="kevivek"] And yes, the counselors/jc's are EXTREMELY qualified and are really amazing.[/quote]
As in they usually only have one response to any of your questions: "I don't know, what do you think?"  :lol:  :D  :rotfl:

Well I suppose...I was thinking more along the lines of they really know their stuff and are able to effectively help you.  They seem to really understand what theyre teaching to the point where they can subtly prod you in the right direction without giving much away and still leaving room for self-discovery.  I suppose giving only one answer falls under that though.

A member of the math team I coach is on his way to Ross 2006. I'll be eager to hear all about it.

i'm going to promys this year, but i'll probably apply to ross next year (its being longer: this year, I couldn't because of my orchestra's trip to france from june 15-24) along with others (ie promys, rsi etc).

Sounds like a really good program though...

I'll be at Ross this summer and I'm ridiculously excited about it. Everything posted about it so far makes me glad that I chose to go there.

Et j'essaie de me convaincre que je peux parler francais, mais je sais que je suis nul et que j'ai rate l'examen d'AP.

Washington representin'!

I'm really not one of those Asian kids who pretends to be black...

I'll be a parent at the Ross Program orientation meeting on Sunday evening. I look forward to saying hello to the AoPSers there. We're driving over from Minnesota, with some stops to see relatives and friends along the way.

Have a safe trip there, everyone. It sounds like Ross 2006 will be an intense experience.

unfortunately i will not be at the orientation on the 18th.

i will arrive on the 21st; due to the University Advising Sessions for fall.

I'll be at Ross as a Junior Counselor this year.  I was a first year student last year.

haven't been to this forum in a while...but I'll be a counselor for the second time this summer.

I'm looking forward to meeting all of you this weekend, my name is Zach.


geometry
perpendicular bisector
Pythagorean Theorem
geometry solved

[quote="cosinerburc"]Let a circle with center H intersects AB,BC andAC at F_1,F_2;D_1,D_2;E_1,E_2 respectively,where H is
the orthocenter of ABC.Prove that AD_1=AD_2=BE_1=BE_2=CF_1=CF_2.[/quote]

Wrong. I guess you meant the following result (discovered by Jakob Steiner, proved by Arnold Droz-Farny, cf. http://www.pandd.demon.nl/drozf.htm#4 ):

[color=blue][b]Theorem 1.[/b] Let H be the orthocenter of a triangle ABC, and let A', B', C' be the midpoints of its sides BC, CA, AB. Some circle with center H meets the line B'C' at the points $D_1$ and $D_2$, the line C'A' at the points $E_1$ and $E_2$, and the line A'B' at the points $F_1$ and $F_2$.

Then, $AD_1=AD_2=BE_1=BE_2=CF_1=CF_2$.[/color]

[i]Proof.[/i] Since B' and C' are the midpoints of the sides CA and AB of the triangle ABC, we have B'C' || BC. The line AH, being an altitude of triangle ABC, is perpendicular to BC, and thus to B'C' as well (since B'C' || BC). In other words, the line AH is the perpendicular to the line B'C' through the point H.

Since the points $D_1$ and $D_2$ lie on one circle with center H, we have $HD_1=HD_2$, so that the perpendicular bisector of the segment $D_1D_2$ passes through the point H; obviously, this perpendicular bisector is perpendicular to $D_1D_2$, i. e. to B'C'. Hence, the perpendicular bisector of the segment $D_1D_2$ is the perpendicular to the line B'C' through the point H. Thus, the perpendicular bisector of the segment $D_1D_2$ coincides with the line AH. From this, we can conclude that the line AH passes through the midpoint X of the segment $D_1D_2$, and that the point A lies on the perpendicular bisector of the segment $D_1D_2$, so that $AD_1=AD_2$.

The Pythagorean theorem for the right-angled triangles B'XA and $D_1XA$ yields $B^{\prime }A^2=B^{\prime }X^2+AX^2$ and $AD_1\;^2=D_1X^2+AX^2$, so that

$B^{\prime }A^2-AD_1\;^2 =\left( B^{\prime }X^2+AX^2\right)-\left( D_1X^2+AX^2\right) =B^{\prime }X^2-D_1X^2$
$= \left( B^{\prime }X-D_1X\right) \cdot \left( B^{\prime }X+D_1X\right)$.

Since X is the midpoint of the segment $D_1D_2$, we have $D_1X=XD_2$, and thus we obtain

$B^{\prime }A^2-AD_1\;^2=\left( B^{\prime }X-D_1X\right) \cdot \left( B^{\prime }X+XD_2\right) =B^{\prime }D_1\cdot B^{\prime }D_2$.

Similarly, $B^{\prime }C^2-CF_2\;^2=B^{\prime }F_1\cdot B^{\prime }F_2$. But as the points $D_1$, $D_2$, $F_1$, $F_2$ lie on one circle, the intersecting chords theorem yields $B^{\prime }D_1\cdot B^{\prime }D_2=B^{\prime }F_1\cdot B^{\prime }F_2$, so that $B^{\prime }A^2-AD_1\;^2=B^{\prime }C^2-CF_2\;^2$. Since B' is the midpoint of the segment CA, we have $B^{\prime }A^2=B^{\prime }C^2$; therefore, this becomes $AD_1\;^2=CF_2\;^2$, and $AD_1=CF_2$. Together with $AD_1=AD_2$, this becomes $CF_2=AD_1=AD_2$. Analogous reasoning shows that $AD_2=BE_1=BE_2$ and $BE_2=CF_1=CF_2$. Altogether, we get $AD_1=AD_2=BE_1=BE_2=CF_1=CF_2$, and Theorem 1 is proven.

  Darij


group theory
abstract algebra
pigeonhole principle
superior algebra
superior algebra unsolved

Let G be a transitive subgroup of S_n where n>1. Show that G contains a permutation without fixpoints, i.e., where all cycles has length >1.

The number of orbits is equal to the average number of fixed points.  Apply the pigeonhole principle.


calculus
conics
geometry
quadratics
articles
ARML
geometric transformation

Hi, I'm wondering who among the math-aware people here is able to study math books in languages other than English. I have a good number of Chinese-language math books, and what limits my reading ability in them is more my limitations in math knowledge than my limits in Chinese, strictly speaking. (After edit:) My first, and very satisfying, attempt to read a math book in German was reading Edmund Landau's Grundlagen der Analysis in the original German in the Chelsea reprint edition now published by AMS, and recommended by Michael Spivak as a good first book for studying mathematical German. I have a few other classic German-language books about math at home.

There are many children of first-generation immigrants on this forum, and some participants in countries outside the United States, so I'm wondering what other languages you might read about math in.

Having spent a reasonable amount of time in the Mbaracayu reserve in Paraguay with the indigenous Ache and Guarani peoples, I would probably be able to decipher a math text written in Guarani. However, finding one would probably be difficult...

Edit: I've also taken latin for several years and was attempting to read portions of Newton's Principia. In particular, I wanted to see his demonstration (without the multivariable calculus, which I find not too satisfying) that all planets orbit in conic sections. However, I got lost near the end because of his use of, what I might consider, obscure properties of conic sections. The latin the the text was pretty simple, and so I'd bet most people having taken >3 years would probably be able to read most of it.

I can read anything in Croatian without any difficulty (its my first language). I could probably read some books in other Slavic languages (as long as they are not written in Cyriliic) because of the likeness between them. Also I could likely read them in French thanks to studying the language from grade 1 to 9 (I live in Canada so that was required).

I can read math in Romanian (which I'm fluent in) and have a few times. I also have a cool Analytical Geometry book, written in french, which I mostly understand.

I have taken 6 years of Jr High/HS french, so I could probably read a french text (maybe not learn anything new from one, but I would probably be able to read a Precalculus text).  We don't learn words like "quadratic" and "combinatorics" but usually the really complicated words look similar to english.

The best I've done is read one french article with the help of an online translator. I know no french. In fact, the only language other than english that I partially know is latin, which I have been taking for 4 years.

well, I was going to select French and Spanish at the same time... but I couldn't.  So I guess there should be a vote for Spanish as well...

I took four years of French in high school and can read math books in French pretty fluently. I also know no Chinese whatsoever, but I have read a few Chinese language math books. (I was preparing for ARML 2003, and a friend had a set of books designed for training the Chinese olympiad team, and he let me borrow them. He helped me a bit with understanding the problems that I couldn't understand on my own.)

i voted for Spanish because it is the language i know best apart from English. i'm sure my observation is not unique, but i find it easier to read math in a foreign language than any other subject, largely because it is a context with which i am familiar and because there are so many universal symbols.  that said, i am confident based on my own language training that i could read math in any western European language (most of you probably could as well assuming you have studied any Indo-European languages).  the only foreign languages in which i have experience proofreading mathematics are Portuguese and Afrikaans, due to my time spent in Southern Africa this year teaching and overseeing translations of math contest material..

I can read math books in Hindi, but very few math books in India beyond calculus are written in Hindi; most of them are in English.

I've taken 5 years of Spanish, so I might be able to manage some books but nothing with too much new material.  I'm not very good with languages.  I also took Hebrew for eight years but it was in elementary and JR high so it's pretty weak now.  Also, since the Hebrew alphabet is completely different, I'm not sure if abbreviations (cos(x), for example) would be the same.

Just because you are fluent in a language does not necessarily mean you can read about math in it. Even if you know how to read The Three Musketeers in French and understand almost everything does not mean that you would understand the French word for tetrahedron or for orthogonal conjugate. That's why I have a nice Russian-English-German-French mathematical dictionary.

I could read most Russian math texts through high school, and with this dictionary I would probably be able to read other texts.

I think math in Hebrew is more similar to math in English than you think. I have an Israeli friend who showed me some Israeli math exams or contests, and I could (sort of) read them. (My Hebrew isn't nearly as good as it should be.)

[quote="Yasha"]Just because you are fluent in a language does not necessarily mean you can read about math in it. Even if you know how to read The Three Musketeers in French and understand almost everything does not mean that you would understand the French word for tetrahedron or for orthogonal conjugate. That's why I have a nice Russian-English-German-French mathematical dictionary.[/quote]

Too true... I don't know the Russian for "orthogonal conjugate" since I didn't even know what one was before MOP. I haven't had problems reading math texts in Russian so far though...

If you were to start reading about math in a language from a middle school math level (assuming you could read that), I'd imagine you'd be able to understand everything when you got to it. I could probably read an elementary French math book if I saw one, but I haven't tried.

I've lived for 17 years in Vietnam so I actually can easily read it, and use books in VIetnamese to compare with what I've learned so far in English

Nowadays, some Chinese universities begin to use the textbook in English...

I'm quite surprised at the number of people that can read chinese :?

tokenadult, I think you should divide the choice "chinese" into two parts: traditional chinese and simplified chinese :D

[quote="mecrazywong"]tokenadult, I think you should divide the choice "chinese" into two parts: traditional chinese and simplified chinese :D[/quote]

I've always (= for almost thirty years) had to read both "traditional" Chinese characters and "simplified" Chinese characters. (I put both terms in quotation marks because neither is wholly accurate.) The overlap in character forms between the two forms of script is quite large, and these days rather few readers of Chinese are wholly unacquainted with one form or the other. Because my wife grew up in Taiwan, I at one time had an advantage over her in reading "simplified" script, even though she continues to have a HUGE advantage over me in general Chinese-language reading ability, but all it takes is practice, and anyone who reads Chinese at all can read either form of script.

Welcome to the combined forum. I was last in Hong Kong in 1987, so I am curious about how things are there these days.

I can read simplified Chinese, but I have some trouble understanding traditional Chinese.

what if one can read in more languages? :)
for example I can read in Romanian (obviously :P), English (well that's obvious too), French (that is not that clear! :)) and Italian ...

[quote="mecrazywong"]I'm quite surprised at the number of people that can read chinese :?
[/quote]
I'm surprised, too.

Apart from English I can read obviously in Polish [my mother's tongue]  :)

I can read dutch :)

though I still have to see the first good dutch maths book... :?

English & Polish. (I invite all to http://www.matematyka.pl )
OffTopic- see http://grad.exo.pl/files/letni_hicior_o_zone_lotr.avi

Quote:what if one can read in more languages?

That depends on how many languages one can read -

If (s)he can read two,  (s)he is called "Bilingual"

If (s)he can tead three,  languages (s)he is called "Trilingual"

And if (s)he can read only one language (s)he is called :

[hide]An American[/hide]

[hide](Sorry could  not resist it ... It was told by one of my Europian Friends)   [/hide]


integration
calculus
logarithms
calculus computations

$\frac{d}{dx}(xlogx)=1+logx$

$\int 1+logx\,dx = xlogx+C$
$\int logx\,dx = xlogx-x+C$

--------------------------------------

Differentiate $x^{2}logx$ and hence determine
$\int^{e}_{e}xlogx\,dx$

where the bottom limit e is $sqrt{e}$. I didn't know how to write it.
--------------------------------------
Differentiate $\sqrt{x}logx$ and hence determine $\int \frac{logx}{\sqrt{x}}\,dx$

You have just found out the famous integral formula, bos1234. :)

You can type the square root of $e$ as [code]\sqrt{e}[/code]

$\left(\frac{x^{2}}{2}\ln x\right)'=.....\Longrightarrow \int x\ln x\ dx=.....$.


Find the compounds A - E bellow:

1) $ 2\, CH_3COCH_3 \plus{} A \plus{} NaNH_2 \longrightarrow B$

2) $ B \plus{} H_2/Pd \longrightarrow C$

3) $ C \plus{} HCl \longrightarrow D$

4) $ D \plus{} C_6H_5Br/AlCl_3 \longrightarrow E$

5) $ E \plus{} Mg/THF, \,\,then \, H_3O^\plus{} \longrightarrow \,1,1,4,4\minus{}tetramethyltetralin$.

B seems to be the aldol condensation product of acetone but doesnt it give a very poor yield. (unless you use ba(oh)2 and a special apparatus)

sorry dint see A. thought it was just acetone + base.


inequalities
Diophantine equation
number theory unsolved
number theory

Has $x^2-2y^2 =-1$ any integer solutions ?

I used Gica's algorithm(see http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Panaitopol/panaitopol41.pdf -> theorem 1.2 ).
i read it not from the article but from another book called
"Introducere in aritmetica si teoria numerelor" by L. Panaitopol and Gica Alexandru.
I'm sure there's an english version "Introduction into arithmetics and number theory".
I made some rough approximations on those square roots.But the basic ideea is...to find a minimal not trivial solution to the pell equation,then find the solutions with the constraints
$x<\sqrt{|k|\epsilon}$ and $y<\sqrt{\frac{|k|\epsilon}{d}}$
and from those solutions (x,y) arise an infinty of solutions wich are described
by that $\pm\mu^n$ stuff.

This is the minimal solution, generating all others by $x+\sqrt{2} y = \pm (1+\sqrt{2})^{2n+1}$ (the $2n+1$ because only odd powers work).
E.g. $x=7, y=5$ from $n=1$.

it gave me that same minimal $(7,5)$ solution ZetaX.
but could you show how you got $1+\sqrt{2}$ there ?

The minimal solution is not $x=7 , y=5$ but $x=y=1$, which give that $1+\sqrt{2}$ by looking at $x+\sqrt{2}y$.

sorry,it seems i calculated wrong the minimal sol.
but now one must find those $\mu_1,\mu_2,..,\mu_r$ satisfying those
inequalities i wrote above

this is prolly so well known that nobody posted it. i post it for clarification anyway ... ;)

if one found a solution $(x_0; y_0)$ (we only consider positive solutions) to the equation $x^2-Dy^2= \pm 1$ then $(x_1;y_1)$ with $x_1+\sqrt D y_1 = ( x_0 + \sqrt D y_0 )^k$ for all $k\in \mathbb N$ is also a solution, and this is easily shown by induction. the other way around, that if one found the minimal solution $(x_0;y_0)$ (note that $x<x' \iff y<y'$, when $(x;y)$ and $(x';y')$ are positiv solutions), for all positive solutions $(x;y)$ there is a natrual $k$ so that $x+ \sqrt D y = ( x_0 +\sqrt D y_0)^k$, is proven a little harder but can be done by inequalities. so one has all the solution by this method.


number theory
greatest common divisor
number theory unsolved

Let $ n$ be a positive integer and $ \delta (n): \equal{} \{ m \in \mathbb{Z}_{ > 0}: m\mid n\}$. Let $ W\subset \delta (n)$ and let $ a(n,W): \equal{} \{m \in \mathbb{Z}_{ > 0}: \operatorname{gcd} (m,n) \in W\}$.

Prove that
1. If $ n\mid n'$ then $ a(n,W) \equal{} a(n',W')$ where $ W': \equal{} \{d \in \mathbb{Z}_{ > 0}: d|n' \text{ and } \operatorname{gcd}(n,d) \in W\}$.
2. $ \mathbb{Z}_{ > 0}\setminus a(n,W) \equal{} a(n,\delta (n)\setminus W)$.
3. $ a(n,W) \cup a(n,W') \equal{} a(n,W \cup W')$.

nice problem!

1) $ n\equal{}\prod_{i\equal{}1}^{n}{p_i^{\alpha_i}} \text{and} n'\equal{}(\prod_{i\equal{}1}^{n}{p_i^{\beta_i}})(\prod_{i\equal{}n\plus{}1}^{k}{p_i^{\gamma_i}}) \text{with} \beta_i \ge \alpha_i \forall i \in [1,n]$, and we name $ p\equal{}\prod_{i\equal{}n\plus{}1}^{k}{p_i^{\gamma_i}}$.
we have $ \delta(n)\equal{}\{c_1, c_2, c_3, ..., c_h, c_{h\plus{}1},...c_{\prod_{i\equal{}1}^{n}{\alpha_i\plus{}1}}\}$ and let $ W\equal{}\{c_1,...c_h\}$
so $ a(n,W)\equal{}\{c_jT, \text{where} j \in [1,h], \forall T \in \mathbb{N}, p_i \nmid T, i \in [1,n]$.
anyway $ W'\equal{}\{c_j F \text{where} F \in \delta(p), j\in [1,h]\} \implies a(n',W')\equal{}\{c_j F G \text{where} p_i \nmid G,$$ i \in [1,k]\}\equal{}a(n,W)$

2)$ Z_{>0}\minus{}a(n.W)\equal{}Z_{>0}\minus{}\{c_j T, j \in [1,h]\}\equal{}\{c_j T, j \in [h\plus{}1, \prod_{i\equal{}1}^{k}{\alpha_i\plus{}1}]\}\equal{}a(n,\delta(n) \minus{}W)$

3) it is trivial since $ W \cup W' \equal{} W' \text{and} a(n,W) \le a(n,W')$


Can someone who's done it explain the general idea? (And if it actually works...)

I trieed this in 03 and it was dently succefull, and will probably have to do it again this year if I want to go back to mathcamp(aka will probably have to do it again this year.). Basically I wrote a couple of paraphras discribing my interest in math and mathcamp. Then me and my family took that, added some things like my sucess and past mathconests and all that, and contacted a person with the "right connections" and all that to be our representative. After that it was pretty much sitting back and letting her do most of the work. It was really quite easy, and if you know someone whod be willing to be your reprsentaive, I would recommend it if mathcamp would either possible but a fincial strain or totally impossible without outside help. You never know how succesful its gonna be, so it seems to me a risk worth taking :)

When you apply, I belive you can check a box that says "please send me information about local fundrasing." when they send you the camp imformation they send you the fundrasing stuff as well :)

the checkbox is on the registration form that we send you once you're admitted, not on the application form.


superior algebra
superior algebra solved

Is it true that $\overline{\mathbb{Q}}(x)$ is an algebraic extension of $\mathbb{Q}(x)$, where $\overline{\mathbb{Q}}$ is the field of algebraic numbers?

Excuse me if I am posting a silly question but I can't see why, because I am a starter in algebra.
Isn't it $\overline{\mathbb{Q}}$ that is the algebraic extension of $\mathbb{Q}$(of course it is the algebraic closure)? But why is any element of $\overline{\mathbb{Q}}(x)$ algebraic over $\mathbb{Q}(x)$, where $x$ is transcendental over $\mathbb{Q}$.

Oh- I completely misread your question.

Actually, it is still essentially trivial; just use the fact that $\overline{\mathbb{Q}}(x)$ is the join of $\overline{\mathbb{Q}}$ and $\mathbb{Q}(x),$ interpreting $x$ as a transcendental element. By standard theorems, the lift of an algebraic extension is algebraic. After all, the field is generated by $\overline{\mathbb{Q}}$ and $x,$ so it is generated over $\mathbb{Q}(x)$ by $\overline{\mathbb{Q}}.$

Abstract nonsense is very useful.


analytic geometry
graphing lines
slope
trigonometry

When I read colors post, the following problem came to my mind:
We have a slope with incline angle angle $30^{\circ}$. At the bottom end of the slope a circular loop-the-loop is connected with a radius of $R$. We put a solid ball of radius $r$ and mass $m$ at height $h=3R$ on the slope. ($R\gg r$)At least how big should the coefficient of friction be so that the ball goes through the whole path without sliding?

Are you sure that for the ball to go through the whole path without sliding there must be a minimum coefficient of friction? If yes, then explain why the following reasoning is wrong.

Although there is friction between the ball and the surface, the friction force doesn't do work and so we can use conservation of energy. Let 1 be the position of the ball at height 3R and 2 the position when the ball is at the highest point of the circular path (height = 2R). The minimum velocity at 2 is such as $N = 0 \Rightarrow  F_{c}= P \Rightarrow  v_{min}^{2}= gR$ where N is the reaction of the surface on the ball. Now we can write:

$E_{m,1}= E_{m,2}\Rightarrow  mg3R = mg2R+\frac{1}{2}mv_{2}^{2}+\frac{1}{2}I\omega^{2}$
$mgR = m\frac{v_{2}^{2}}{2}+\frac{1}{2}\cdot \frac{2}{5}mr^{2}\cdot \frac{v_{2}^{2}}{r^{2}}\Rightarrow  gR = \frac{7v_{2}^{2}}{10}\Rightarrow  v_{2}^{2}= \frac{10gR}{7}$.

As $v_{2}> v_{min}$, the ball is able to make the whole path.

[b]@carcul[/b]
without friction how wil the ball move without sliding. since the c.m of the ball is accelarating so there must be a net torque for angular  accleration .Yes the necessary speed is much more than the min but how does it tell u that ther is no  friction. friction is there but during motion it does no work and it's just not any arbitrary value of friction but a fixed one. see u have assumed while conserving energy that the ball rolls without slipping but this can only happen when there is certain minimum frictional coefficient.

[hide="ans"]
$max(\frac{2}{7\sqrt{3}}, \frac{2 \sin(\ arctan \frac{\sqrt{23}}{3})}{\frac{16}{7}+\cos ( \ arctan \frac{\sqrt{23}}{3})})$ $\approx$ $0.486$[/hide]

[quote="Pardesi"]without friction how wil the ball move without sliding.(...) the ball rolls without slipping but this can only happen when there is certain minimum frictional coefficient.[/quote]

Yes, you are right. I missunderstood the question. Thanks.

Pardesi i dont understand your answer.

why is the coma in the middle?

[quote="pardesi"]

$max(\frac{2}{7\sqrt{3}}, \frac{2 \sin(\ arctan \frac{\sqrt{23}}{3})}{\frac{16}{7}+\cos ( \ arctan \frac{\sqrt{23}}{3})})$ $\approx$ $0.486$[/quote]
thsi means that the answer is the maximum between $\frac{2}{7\sqrt{3}}$
and $\frac{2 \sin(\ arctan \frac{\sqrt{23}}{3})}{\frac{16}{7}+\cos ( \ arctan \frac{\sqrt{23}}{3})}$ which turns out to be $0.486$

nope unfortunately thats not the correct answer :(

[b]@pardesi[/b]

Perhaps it is better if you show your solution, so one can see in detail or in principle how you thought?
Then any eventual corrections are done more easily.

[quote="Thaakisfox"]$\mu_{min}= \frac{2}{\sqrt{111}}$[/quote]
Yes, I also got that answer :)

Ok golduck you can post your solution if you want. :D  If not I will post mine.

On the loop:

$\mu \ge \frac{2\sin\theta}{17\cos\theta+20}$

Is that what you have found? It gives the same answer.  :lol:

i got on the loop
$\mu$ $\geq$ $\frac{14sin \theta}{16+7cos\theta}$
on the slope

$\mu$ $=$ $\frac{2}{7\sqrt{3}}$

On the slope:
The linear acceleration of the ball equals $a=g\sin 30^{o}-\frac{T}{m}$ and the angular acceleration of the ball equals $\epsilon=\frac{5T}{2mr}$ (from Newton's second law). As the ball does not slide we have got $a=\epsilon r$, from which we obtain:
$T=\frac{2}{7}mg\sin 30^{o}$
The ball does not slide when $\frac{T}{N}\leq\mu$ ($N=mg\cos 30^{o}$ - normal force)
$\frac{2}{7}\tan 30^{o}\leq\mu$
$\mu_{min}=\frac{2}{7\sqrt{3}}$

On the loop:
Let $\alpha$ be the angle between the position of the ball and the vertical. From conservation of energy we get:
$3mgR=\frac{mv^{2}}{2}+\frac{\frac{2}{5}mr^{2}\omega^{2}}{2}+mgR(1-\cos\alpha)$
As the ball is rolling without sliding we get $v=\omega r$, so:
$v^{2}=\frac{10gR(2+\cos\alpha)}{7}$
The difference between the normal force and the component of the gravitational force perpendicular to the loop equals the centripedal force:
$N-mg\cos\alpha=\frac{mv^{2}}{R}$ and by putting $v^{2}$ we get:
$N=\frac{mg(20+17\cos\alpha)}{7}$

The linear acceleration of the ball equals $a=g\sin\alpha-\frac{T}{m}$ and the angular acceleration of the ball equals $\epsilon=\frac{5T}{2mr}$ (from Newton's second law). As the ball does not slide we have got $a=\epsilon r$, from which we obtain:
$T=\frac{2}{7}mg\sin\alpha$

The ball does not slide when $\frac{T}{N}\leq\mu$
$\frac{2\sin\alpha}{20+17\cos\alpha}\leq\mu$

By derivating we find that the maximum of $\frac{2\sin\alpha}{20+17\cos\alpha}$ is for $\cos\alpha=\frac{-17}{20}$, $\sin\alpha=\frac{\sqrt{111}}{20}$, which gives $\frac{2}{\sqrt{111}}\leq\mu$ so:
$\mu_{min}= \frac{2}{\sqrt{111}}$

As $\frac{2}{7\sqrt{3}}<\frac{2}{\sqrt{111}}$, the final answer is $\mu_{min}= \frac{2}{\sqrt{111}}$.


inequalities
quadratics
inequalities proposed

Let $a,b,c>0$ and $x,y,z$ be some real numbers which verify:
   a) $ ax^2+by^2+cz^2\leq abc+2xyz$.
   b) $ ab+bc+ca\geq x^2+y^2+z^2$.
     Then prove that for all $n\geq 0$ we have $ x^{2n}+y^{2n}+z^{2n}\leq (ab)^n+(bc)^n+(ca)^n$.
    At least Muierhead and other "mathlinks" origine principles don't work here.  :D

We may assume $x,y,z\geq 0$ otherwise replacing them by their absolute values.
Looking the first relation as a quadratic expression in $z$ the associated equation must have positive discriminant which is $4(y^2-ac)(z^2-ab)$. Analogously for $y,z$ we conclude that $x^2-bc,y^2-ac,z^2-ab$ must have the same sign.
The second condition implies that this sign is not $>0$ hence $x^2\leq bc,y^2\leq ca,z^2\leq ab$ and hence the conclusion


Find "x" if:
[2Sin18*Sin18] / [2Cos42º - Sin36º] = Tan x

hello, you have to prove that
$ \sin(18^\circ)\equal{}\frac{1}{4}(\sqrt{5}\minus{}1)$
$ \sin(36^\circ)\equal{}\frac{1}{4}\sqrt{2}\sqrt{5\minus{}\sqrt{5}}$
$ \cos(42^\circ)\equal{}\frac{1}{8}\left(\sqrt{15}\minus{}\sqrt{3}\plus{}\sqrt{2}\sqrt{5\plus{}\sqrt{5}}\right)$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, you have to prove that
$ \sin(18^\circ) \equal{} \frac {1}{4}(\sqrt {5} \minus{} 1)$
$ \sin(36^\circ) \equal{} \frac {1}{4}\sqrt {2}\sqrt {5 \minus{} \sqrt {5}}$
$ \cos(42^\circ) \equal{} \frac {1}{8}\left(\sqrt {15} \minus{} \sqrt {3} \plus{} \sqrt {2}\sqrt {5 \plus{} \sqrt {5}}\right)$
Sonnhard.[/quote]

Do you prove following? :D
You [b]mustn't[/b] calculate $ \tan 12^\circ$ !! (Because we don't yet know that a calculation result becomes $ 12^\circ$.)
\[ \arctan \frac{3\minus{}\sqrt{5}}{\minus{}\sqrt{10\minus{}2 \sqrt{5}}\plus{}\sqrt{3}\left(\minus{}1\plus{}\sqrt{5}\right)\plus{}\sqrt{2\left(5\plus{}\sqrt{5}\right)}} \equal{} 12^\circ\]


What are bad boxes? In most of my documents I have 6 or 7 errors but like 70-80 bad boxes.

As long as the document prints out fine you don't have to worry about those.

A "bad box" basically means that your text didn't fit.  "Overfull" is the most common, meaning that your text extended into the margin.  "Underfull" means LaTeX thinks there's too much whitespace in your document.

If your output looks OK, just ignore them.


Hi everyone,

Could someone please help me on this problem?

A spring has a natural length of 20 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

I know that work=(force)(distance)

Force=25 N
distance=25-20=5
work=(25N)(5cm)

Is this correct so far? If it isn't, could you please show me what to do?

Thank you very much

The force is not a constant- it requires $ 25$ N at length $ 30$, but less below that.
What do you know about springs and the forces involved?


Prove that $(\frac{1}{2})(\frac{3}{4})....(\frac{2n-1}{2n}) \leq \frac{1}{\sqrt{3n}}$ using maths induction

This is no longer tricky if we prove the following stronger version:

$(\frac12)(\frac34)...(\frac{2n-1}{2n}) \leq \frac1{\sqrt{3n+1}}.$

[quote="eddie77"]nice job ...fieryhydra!!  :lol: ...would you like to show others why??[/quote]

Base case: $\frac12 \leq \frac1{\sqrt{3+1}}.$ So, the statement is true for $n=1.$

Induction case: Let the statement be true for $n=k,$ where $k>1$.

So, we have $(\frac12)(\frac34)...(\frac{2k-1}{2k}) \leq\frac1{\sqrt{3k+1}}$
$\therefore (\frac12)(\frac34)...(\frac{2k-1}{2k})\cdot(\frac{2(k+1)-1}{2(k+1)}) \leq \frac1{\sqrt{3k+1}}\cdot(\frac{2(k+1)-1}{2(k+1)})$

Now, $\frac1{\sqrt{3k+1}}\cdot(\frac{2k+1}{2(k+1)}) \leq \frac1{\sqrt{3(k+1)+1}}$
$\Leftrightarrow (3k+4){(2k+2)}^{2}\leq (3k+1){(2k+2)}^{2}$
$\Leftrightarrow 12k^{3}+28k^{2}+19k+4 \leq 12k^{3}+28k^{2}+20k+4$
$\Leftrightarrow 0 \leq k,$ which is obviously true.
So, the statement is true for $n=k+1$ as well. And, we are done.

That's pretty obvious from geometric sum, but here:
[hide="the stronger version"]
$\sum_{i=1}^{n}{\frac{1}{2^{i}}}= 1-\frac{1}{2^{i}}$.
[/hide]
With that, the rest should be quite obvious...

[quote="FieryHydra"][quote="maokid7"]how did u know to try and solve the stronger case instead of the original problem.
I saw the solution in ACoPS but dont really know why he did that.[/quote]
The idea is if we stick with $\frac1{\sqrt{3n}},$ then $\frac12 \leq \frac1{\sqrt{3}}$ is a very weak inequality. It is not good enough to lead us to the case when $n=2$. Note that induction relies on showing that a statement true for $n=k$ (say) is also true for $n=k+1.$ If the case when $n=k$ is weak, then we might not be able to prove the case when $n=k+1.$ This is what happens when we try to prove the weaker version in our example.

To illustrate this further, try solving the following problem using [b]only[/b] induction.

[b]Problem[/b]: Show that $\frac12+\frac1{2^{2}}+...+\frac1{2^{n}}< 1$, for all $n \in \mathbb{N}.$

[b]Hint[/b]: You will have to look for a stronger version.[/quote]
How does this require a stronger version? The base case is simple, and then you just need to prove that if
$ \sum_{j=1}^{k}{\frac{1}{2^{j}}}=1-\frac{1}{2^{k}}$, then
$ \sum_{j=1}^{k+1}{\frac{1}{2^{j}}}=1-\frac{1}{2^{k+1}}$,
which just requires adding $\frac{1}{2^{k+1}}$.

[quote="nebula42"][quote="FieryHydra"][quote="maokid7"]how did u know to try and solve the stronger case instead of the original problem.
I saw the solution in ACoPS but dont really know why he did that.[/quote]
The idea is if we stick with $\frac1{\sqrt{3n}},$ then $\frac12 \leq \frac1{\sqrt{3}}$ is a very weak inequality. It is not good enough to lead us to the case when $n=2$. Note that induction relies on showing that a statement true for $n=k$ (say) is also true for $n=k+1.$ If the case when $n=k$ is weak, then we might not be able to prove the case when $n=k+1.$ This is what happens when we try to prove the weaker version in our example.

To illustrate this further, try solving the following problem using [b]only[/b] induction.

[b]Problem[/b]: Show that $\frac12+\frac1{2^{2}}+...+\frac1{2^{n}}< 1$, for all $n \in \mathbb{N}.$

[b]Hint[/b]: You will have to look for a stronger version.[/quote]
How does this require a stronger version? The base case is simple, and then you just need to prove that if
$\sum_{j=1}^{k}{\frac{1}{2^{j}}}=1-\frac{1}{2^{k}}$, then
$\sum_{j=1}^{k+1}{\frac{1}{2^{j}}}=1-\frac{1}{2^{k+1}}$,
which just requires adding $\frac{1}{2^{k+1}}$.[/quote]
My point was suppose if someone didn't know the formula for the sum of a geometric series or didn't know how to compute the same -- all hypothetical cases, then he/she would have to look for a stronger version of the equality, as induction would not work on the weaker version. Of course, the example problem I gave was all too simple but it was meant as an illustration.

Hope this clarifies what I meant to imply in my earlier post.


yek khate mafrooz darim va dar yek tarafe an do dayere joda az ham vojood darand.az noghte p rooye khat bar har dayere yek momas rasm mikonim(yek momas bar dayere aval va yek momas bar dayere dovom).farz konim  ke do  khati ke az p bar har kodam az dayere ha momas karde boodim dar noghat K va T bar har dayere momas shode bashnad.
noghte P ra tori biyabid ke PK+PT barabare adade mafrooze L shavad. :)

(1)yeki az davayer ro nesbat be khat gharine mikonim noghteye barkhorde momas moshtarak ba khat hamooniye ke behesh niaz darim :)

ma ino az asatid yad gereftim:
noghate $A,A'$ ro tori tain konid ke khate $\Delta$ m3hvar aslie $A,C$ va hamchenin mehvar aslie $A',C'$ bashe
pas $PT=PA,PK=PA'$
hal mikhaim noghte $P$ ro rooye khate $\Delta$ tori tain konim ke $PA+PA'=L$
bahjaiasham ke hame baladid dige:
dayerei be markaze $A$ va shoae $L$ rasm mikonim
hal makan hendesie nofghati manande $P$ ra mikhahim ke dayere $K(P,PA')$ bar dayere aval momas bashad
amma dayere $K$ az noghate $A_1,A'_1$ ke gharine noghate $A,A'$ nesbat be khate $\Delta$ hastand migozarad
pas ma bayad dayerei bekeshim ta az noghate $A.A'$ begozarad va bar dayuere avalie momas bashad
halle??
 ;)  :D
sharmande az asatid age eshtebah goftam


function
logarithms
complex analysis
complex analysis unsolved

I ran into the following problem on page 42 of Conway's "Functions of One Complex Variable":

If $ f: G\minus{}>\mathbb{C}$ is a branch of $ z^a$ and $ g: G\minus{}>\mathbb{C}$ is a branch of $ z^b$, prove that $ fg$ is a branch of $ z^{a\plus{}b}$.

The book defines a branch of $ z^a$ as $ e^{a l(z)}$, where $ l$ is a branch of the logarithm.

Now, this problem seems blatantly false to me.  For example if $ \ln$ is the principal branch of the logarithm, then $ f(z)\equal{}e^{{1\over 2} (\ln{z}\plus{}2\pi i)}$ is a branch of $ z^{1\over 2}$, and so is $ g(z)\equal{}e^{{1\over 2} \ln{z}}$.  But $ f(z)g(z) \equal{} e^{\ln{z} \plus{} \pi i}\equal{}\minus{}z$ is most certainly not a branch of $ z^1$.

If both branches arise from the same branch of the logarithm, then it's true.  And the second part of the exercise, to show that $ g\circ f$ is a branch of $ z^{ab}$, is correct.

If another pair of eyes could let me know whether I'm crazy or there's actually something wrong with the problem, that'd be swell.  Maybe there's some interpretation that I'm just not seeing?

You are right. There's no interpretation of the problem to make it right, as you demonstrated. Since $ z^1$ has an entire function element (the identity map) all branches are restrictions of it to a region, none of which correspond to $ \minus{}z$


geometry
incenter
projective geometry
geometry solved

How to solve this problem with projective geometry?

ABC has incenter I and the incircle touches BC, CA, AB in K, L, M. The line through B parallel to MK meets LM, LK at R, S. Prove that RIS is acute.

You might want to take a look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=beautiful&t=17895]this discussion[/url]. Look at sprmnt21's last post. He states a generalization of the problem, and the idea of the proof, using the idea of pole.


Let a,b,c are positive real number. Prove that:
$\frac{a^}{a^3+b^3}+ \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{(a+b+c)^3}{6(ab+bc+ca)}$
I think this problem very difficult if it's true. But my solution is incomplete. I used SOS method.
 I hope anyone can prove that it.
The following similar this inequality, but much more difficult (I think so :D)
$\frac{a^}{a^3+b^3}+ \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{3(a^3+b^3+c^3)}{2(ab+bc+ca)}$
Easy , you can see this inequality is stronger than inequality:
$\frac{a^}{a^3+b^3}+ \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}{2}$

Now who can give me a nice solution ? In diffirent case you can disprove it. But I think this inequality is true.

[quote="zaizai-hoang"]Let a,b,c are positive real number. Prove that:
$\frac{a^}{a^3+b^3}+ \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{(a+b+c)^3}{6(ab+bc+ca)}$

$\frac{a^}{a^3+b^3}+ \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{3(a^3+b^3+c^3)}{2(ab+bc+ca)}$
[/quote]
These are not true.
Try $c\rightarrow0^+$ and $a=b=1.$ ;)

\frac{a^}{a^3+b^3}+ \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{(a+b+c)^3}{6(ab+bc+ca)}
i see that the inequality is not homogenous ....there's any thing missed???!!!!
like a^4!!!! :huh:

Okie,  :(
Now who can give me solution that inequality:
$\frac{a^4}{b^4+c^4} + \frac{b^4}{c^4+a^4} + \frac{a}{b} \ge \frac{3(a^3+b^3+c^3)}{2(ab+bc+ca)}$

 :D  :roll:

[quote="justdudxit"]What's the "SOS" method?  I must learn...[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127


I searched on AoPS but couldn't find the answer... :oops:

no, i think the show where they announce the winner will be on sept. 12th or something like that. so, yeah, i don't think they have announced the winner.

Oh, thanks for letting me know -- I was just curious ...

Kevin and his mom were in New York this past weekend to do some kid of taping for the show.  I do not know if it was just Kevin or if they flew everyone out to be taped for the awards show.  they should announ ce is somewhere around the 20th......

Jeff Boyd
Kevin's coach

which is friday as people said................

I didn't see the show, but the Exec Director of MATHCOUNTS emailed me this morning and told me he won.  Nice job, all you ballot box stuffers!

Each category has a list of nominees under it, with one of the nominees in red and the others in black.  Presumably the red names are the winners.

Kevin's name is in red under the "Best Junior Achiever" category.

Yes, he did win : )

I really cannot believe that, but I guess all of the math geeks swung the vote : )

[quote="13375P34K43V312"]Wow, I am actually very surprised. Not because I doubt Kevin's ability and his achievements, I certaily think he deserved to win, but because of today's mainstream and how all teenage girls worship Miley Cirus (who is a very unskilled lip singer and has very corny lyrics), but I guess we (as in you guys) infiltrated the system quite effectively.

If we (again, you guys) did not infiltrate the system quite effectively, however, I'm happy that the mainstream does not exclusively worship people that do "popular" or more socially acceptable things (don't get my wrong, I'm not saying that math is not socially acceptable).

Congratulations, Kevin.

Actually, thinking about it, on a sidenote, I am surprised that Evan O'Dorney did not get nominated, but that's another matter.[/quote]
I don't think you understand, probably about 5000 people voted outside of those voting for Kevin Chen, I mean did you hear anything at all about this award anywhere?

I personally think it'd be very humorous if they released the number of votes, each of his opponents got about 500 votes and he got like 20000


I'd like to welcome everyone to the (new) New Jersey sub-community of AoPS. Feel free to post problems and solutions here just like you would in any other forum.

Already mcrawford has posted several problem practice topics for everyone to work on.

If you have any questions about this sub-forum please post them here.

[quote="yugrey"]Hello.

Please be active.  Our forum is dead.[/quote]

There hadn't been much activity previously on this subforum -- if the demand exists, I will start being more active :).


trigonometry
integration
calculus
calculus computations

3. Integrate by parts. $u=\frac1{t+1},\ dv=\sin t\,dt,$ so $du=-\frac{dt}{(t+1)},\ v=1-\cos t.$ Note the slightly unusual choice for $v.$ We're allowed to do that because we can add any constant we want to to $v.$ My reasons for doing that will become apparent.

$\int_{0}^{x}\frac{\sin t}{t+1}\,dt= \left.\frac{1-\cos t}{t+1}\right|_{0}^{x}+\int_{0}^{x}\frac{1-\cos t}{(t+1)^{2}}\,dt$

$=\frac{1-\cos x}{x+1}+\int_{0}^{x}\frac{1-\cos t}{(t+1)^{2}}\,dt$

Since $1-\cos x\ge0$ for all $x,$ this is clearly nonnegative.

Incidentally, we have also shown that $\int_{0}^{\infty}\frac{\sin t}{t+1}\,dt$ converges, to a positive number.

[quote="sinajackson"]nice soluTions guys,and sorry , i edited the first problem,It is not $f^{'}(x)=....$ but $f(x)=.....$ :) :

$1$.Find $f(x)$ such that : $f(x)=\sin x+\int^{x}_{0}f^{'}(t)(2\sin t-\sin^{2}t)dt$[/quote]
In the current form, the problem is much more tractable; earlier, I thought the problem was somewhat difficult.

A sketch of the soution: differentiate both sides w.r.t. $x.$ We get a first order linear ODE with variables separable. After integrating and noting that $f(x) = 0$, we obtain $f(x) =-1+\frac1{1-\sin x}.$

[Edit: I had made an error while computing the integral and this was pointed out by Ramanujan. I have edited my answer now.]

[quote="Ramanujan"]Check it again FieryHydra, i think it is $f(x) =-1+\frac{1}{1-\sin x }$[/quote]
Thanks, Ramanujan, for pointing out the error. I was indeed sloppy with my calculations.

[quote="maokid7"]I was wondering.
if the problem were actually $f'(x)=\sin(x)+\int_{0}^{x}f'(t)(2\iint-\sin^{2}t)dt$

couldnt u try and solve it the same way as if it were $f(x)$?
i know u would get  second degree equation (which i currently dont know how to solve) but it is possible to solve right?[/quote]
You can still convert the linear second degree ODE to a first degree ODE by using the substitution $w(x) = f'(x).$ It just means  more work.


calculus
integration
number theory unsolved
number theory

show there are atleat n 1's in binary representation of integral multiple of 2^n-1,where n a natural number. :wink:

This is a special case of an IMO ISL 1993 problem.
You may find a solution, in french, at the IMO ISL problem here, as problem #164

http://www.animath.fr/cours/arithm.pdf

Pierre.


geometry
3D geometry
algebra
polynomial
LaTeX
linear algebra

Use the fact that $ x\plus{}a$ is always a factor of $ x^{3}\plus{}a^{3}$ to find a formula for the sum of perfect cubes.

(I'm trying to use polynomial long division.)

Thanks in advance!

If you know polynomial long division, just divide $ x \plus{} a$ into $ x^3 \plus{} a^3$.
I don't know if this is how you're supposed to do it, but I see that there are at least $ x^2$ number of $ x\plus{}a$ in $ x^3\plus{}a^3$. Then, you are left with a remainder of $ a^3\minus{}ax^2\equal{}a\left(a^2\minus{}x^2\right)\equal{}a\left(a\plus{}x\right)\left(a\minus{}x\right)$. This remainder divided by $ x\plus{}a$ is $ a\left(a\minus{}x\right)$, so the final answer is $ x^2\plus{}a\left(a\minus{}x\right)\equal{}x^2\minus{}ax\minus{}x^2$. The factorization of $ x^3\plus{}a^3$ is thus $ \left(x\plus{}a\right)\left(x^2\minus{}ax\minus{}x^2\right)$.

[quote="123456789"]If you know polynomial long division, just divide $ x \plus{} a$ into $ x^3 \plus{} a^3$.
I don't know if this is how you're supposed to do it, but I see that there are at least $ x^2$ number of $ x \plus{} a$ in $ x^3 \plus{} a^3$. Then, you are left with a remainder of $ a^3 \minus{} ax^2 \equal{} a\left(a^2 \minus{} x^2\right) \equal{} a\left(a \plus{} x\right)\left(a \minus{} x\right)$. This remainder divided by $ x \plus{} a$ is $ a\left(a \minus{} x\right)$, so the final answer is $ x^2 \plus{} a\left(a \minus{} x\right) \equal{} x^2 \minus{} ax \minus{} x^2$. The factorization of $ x^3 \plus{} a^3$ is thus $ \left(x \plus{} a\right)\left(x^2 \minus{} ax \minus{} x^2\right)$.[/quote]

I got the stuck with the polynomial long division.  Could someone show or describe to me the steps with this particular problem?  Thanks!

The long division process is exactly the same as with integers. We ask ourselves how many times $ x+a$ divides each term of $ x^3+a^3$, then multiply by the  quotient, subtract, and repeat the process. A picture is worth a thousand words, though.

Please excuse my ridiculous attempt to show long division without $ \text{\LaTeX}$:

[code]

       x^2 - ax + a^2
     ________________________
 x+a ) x^3 + 0x^2 + 0x + a^3
       x^3 + ax^2
      ______________________
            -ax^2 + 0x
            -ax^2 - a^2x
           __________________
                    a^2x + a^3
                    a^2x + a^3
                   ___________
                             0

[/code]

From this, we conclude that $ (x+a)(x^2-ax+a^2)=x^3+a^3$.

[hide="Another way..."]

Another way we can find the formula for the some of cubes, which I find to be much easier to handle in the general sense, is to "assume" that there is a quartic polynomial $ P(n)$ that returns the sum of the first $ n$ perfect cubes. That is:

$ P(n)=an^4+bn^3+cn^2+dn+e$

As you probably know, we need only five points to determine this polynomial, so we use the first 5 cubes: $ {(0,0),(1,1),(2,1+8),(3,1+8+27),(4,1+8+27+64)}$.

$ 0=0a+0b+0c+0d+e$
$ 1=a+b+c+d+e$
$ 9=16a+8b+4c+2d+e$
$ 36=81a+27b+9c+3d+e$
$ 100=256a+64b+16c+4d+e$

This can easily be solved with some algebra, or, even easier, Gaussian elimination or other linear algebra methods, to find $ {a,b,c,d,e}$, which would give us the desired polynomial.

You may have noticed that the sum of the first $ n$ cubes is the square of the sum of the first $ n$ integers. This is no coincidence, and in fact:

$ \sum_{i=1}^ni^3=\left(\sum_{i=1}^ni\right)^2$

is something that can be proved easily by induction.
[/hide]

[quote="jvenezuela716"]The long division process is exactly the same as with integers. We ask ourselves how many times $ x + a$ divides each term of $ x^3 + a^3$, then multiply by the  quotient, subtract, and repeat the process. A picture is worth a thousand words, though.

Please excuse my ridiculous attempt to show long division without $ \text{\LaTeX}$:

[code]

       x^2 - ax + a^2
     ________________________
 x+a ) x^3 + 0x^2 + 0x + a^3
       x^3 + ax^2
      ______________________
            -ax^2 + 0x
            -ax^2 - a^2x
           __________________
                    a^2x + a^3
                    a^2x + a^3
                   ___________
                             0

[/code]

From this, we conclude that $ (x + a)(x^2 - ax + a^2) = x^3 + a^3$.

[hide="Another way..."]

Another way we can find the formula for the some of cubes, which I find to be much easier to handle in the general sense, is to "assume" that there is a quartic polynomial $ P(n)$ that returns the sum of the first $ n$ perfect cubes. That is:

$ P(n) = an^4 + bn^3 + cn^2 + dn + e$

As you probably know, we need only five points to determine this polynomial, so we use the first 5 cubes: $ {(0,0),(1,1),(2,1 + 8),(3,1 + 8 + 27),(4,1 + 8 + 27 + 64)}$.

$ 0 = 0a + 0b + 0c + 0d + e$
$ 1 = a + b + c + d + e$
$ 9 = 16a + 8b + 4c + 2d + e$
$ 36 = 81a + 27b + 9c + 3d + e$
$ 100 = 256a + 64b + 16c + 4d + e$

This can easily be solved with some algebra, or, even easier, Gaussian elimination or other linear algebra methods, to find $ {a,b,c,d,e}$, which would give us the desired polynomial.

You may have noticed that the sum of the first $ n$ cubes is the square of the sum of the first $ n$ integers. This is no coincidence, and in fact:

$ \sum_{i = 1}^ni^3 = \left(\sum_{i = 1}^ni\right)^2$

is something that can be proved easily by induction.
[/hide][/quote]

Thanks a lot!  I got it now!

The problem already gives you that $ x \plus{} a$ divides $ x^3 \plus{} a^3$.  What it's asking you to do from there is to write

$ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} ... \plus{} n^3$

as

$ (1^3 \plus{} n^3) \plus{} (2^3 \plus{} (n\minus{}1)^3) \plus{} (3^3 \plus{} (n\minus{}2)^3) \plus{} ...$.

:)


MATHCOUNTS
modular arithmetic
AMC
USA(J)MO
USAMO
geometry
3D geometry

As the title says, discuss and post your solutions to the problems here

[u]My personal thoughts[/u]
-Upon taking this test, I realized that this test is obviously harder than an actual mathcounts one.
-Partially because many questions ended up being trick questions (oops), particularly #15 and #27
-The problem with the prime factors of $ 2^{22}\minus{}1$ has been nominated by me as the stupidest problem on the test, since originally I somehow thought that 2731 is divisible by 7.  It turns out to be prime.
-I lost
-shentang suggested a few problems, which I then modified.  Most of those problems he got wrong anyways...

[hide="solution to number 15"]
Make an 8x8 square.  Look at it, and you'll see 7x7s, 6x6s, etc.  There are then obviously $ \boxed{8}$ distinct squares.

The comment about $ n$x$ n$ squares requiring $ n^2$ unit squares was to prevent the answer of $ 15$, see if you can figure out how to get that
[/hide]

The rest of the solutions are for you guys to talk about

I phailed terribly on that so how do you do #26?

I'd like to know what level you all thought this was? Especially people who've gone to nationals before.

I think this was atleast a hard state, considering I think #14 was the exact same as an old #30 on a state sprint, except with 7 instead of 4.

[hide="26"] Let it be n. We know that $ n\equiv1\pmod{8}$, $ n\equiv{2}\pmod{9}$ and $ n\equiv3\pmod{5}$. One possible way would be to actually write out in terms of variables and do a bunch of random manipulation. Or, you could write $ n\equiv3\equiv\minus{}2\pmod{5}$ and see that $ n\equiv{x\minus{}7}\pmod{x}$ for x=5, 8, and 9. Hence, it is 7 less than the lowest common multiple of 5, 8, and 9, which is 360-7=353.[/hide]

[quote="5849206328x"]
[hide="solution to number 15"]
Make an 8x8 square.  Look at it, and you'll see 7x7s, 6x6s, etc.  There are then obviously $ \boxed{8}$ distinct squares.

The comment about $ n$x$ n$ squares requiring $ n^2$ unit squares was to prevent the answer of $ 15$, see if you can figure out how to get that
[/hide]

[/quote]

blah... didnt read the part 'positive integer length'...  :(

Hmm....

I tihnk it was hard state

I found it way too calculation intense. But maybe thats because i'm used to things like USAMO with not a lot of calculation?

This was a nice mock mathcounts... even if it was a little hard and some problems were bad...

Are you sure there is only one solution to 29?

I will post my solutions:

1. pythag theorem
2. brute force... ugh... this took me a long time... and it was #2!!!!
3. divisibility by 9 test
4. guess & check
5. pretty easy
6. the variables are independent
7. pie/sieve formula
8. brute force
9. too lazy to do, too much calculation.
10. f(a,b)=a^3-b^3
11. all the centers are the same
12. its c(x-3)^2
13. sum from 1 to n formula
14. cube
15. brute force... this took me a long time...
16. well known problem, from 1 to sum of total
17. power of points
18. brute force!
19. x has to be > 6, other things work
20. am-gm
21. pie
22. 2-1 = 1, 8-4=4...
23. some brute force... and Fermat-Euler
24. area bashing
25. ugh...
26. BRUTE FORCE! (Chinese Remainder THeorem)
27. DISTINCT roots, vieta's
28. Dirichlet Series Factorization
29. hmm... I get that there is no answer
30. balls & urns.

Do you realize how incredibly hard it is to write the first few sprint problems? They're supposed to be easy, but we can't just give you like 1+2. Plus, Mathcounts often has a lot of time waster problems, which I hate too. Also, I suppose you are used to USAMO, considering that a lot of things that you said are brute force are really pretty common for Mathcounts. Like, #18 is pretty easy after you factor and see that for x>8, both factors are positive, and so it's 8.

Maybe I should learn what Dirichlet Series are, but #28, which a bunch of people missed (or so it seemed like) is easy if you notice that 496 is a perfect number, so the sum of its factors is 2 (an interesting result). Then, if you remember what a proper factor is, which I always forget, you just subtract.

Anyway, I'd just like to thank Alan (the number person) for writing such a beast sprint+target.

EDIT: Ok, so maybe I everything I wrote wasn't the exact truth. Sorry.

parts of the test phailed

pythag - for #29, can you post how you get no answer?

[quote="Temperal"][quote]Do you realize how incredibly hard it is to write the first few sprint problems?[/quote]

I don't - writing the first few is far easier than the rest.[/quote]

Also, I must mention that it is VERY hard to write good USAMO questions: Most of the problems I write are either

1.) Way too hard
2.) Way too easy
3.) Not very nice
4.) Way too similar to an existing problem

For writing the first few: Unfortunately, I can't see how to do this except for using brute force questions (Which apparently are standard on mathcounts.)

Personally, I wish that the sprint round had only 10-15 problems, and they were all hard.

Here's my solution for $ 29$.

 [geogebra]7008a0fb1aab8e96019297bfe7b290b5912366b4[/geogebra]

So by triangle similarities, we get most of the diagram up there.

Also by triangle similarities, we have $ \frac {5a}x \equal{} \frac {3a}{6 \plus{} 12} \equal{} \frac a6\implies\frac5x \equal{} \frac16\implies x \equal{} 30$ and $ \frac {8a}6 \equal{} \frac {4a}3 \equal{} \frac x{12}\implies a \equal{} \frac x{16} \equal{} \frac {15}8$, so

$ [ABC] \equal{} [AGB] \plus{} [BGC] \plus{} [AGC] \equal{} 8a \plus{} x \plus{} 18 \equal{} 15 \plus{} 30 \plus{} 18 \equal{} \boxed{63}$.

I didn't participate, though, but whatever.

Edit: Sorry, I misread it.

[quote="math154"]Here's my solution for $ 29$.

 [geogebra]7008a0fb1aab8e96019297bfe7b290b5912366b4[/geogebra]

So by triangle similarities, we get most of the diagram up there.

Also by triangle similarities, we have $ \frac {5a}x \equal{} \frac {3a}{6 \plus{} 12} \equal{} \frac a6\implies\frac5x \equal{} \frac16\implies x \equal{} 30$ and $ \frac {8a}6 \equal{} \frac {4a}3 \equal{} \frac x{12}\implies a \equal{} \frac x{16} \equal{} \frac {15}8$, so

$ [ABC] \equal{} [AGB] \plus{} [BGC] \plus{} [AGC] \equal{} 8a \plus{} x \plus{} 18 \equal{} 15 \plus{} 30 \plus{} 18 \equal{} \boxed{63}$.

I didn't participate, though, but whatever.[/quote]
I just have one question: How do you know that the area ratio between ADG and DGB is 3/5??? The problem states that AD:BC=3/5, which I can't connect to any theorems or area ratio(I took the test and left it blank because of this ratio!). If [b]AD:BD[/b]=3:5, it can be killed easily by Menelaus's theorem. :wink:

[quote="5849206328x"]hmm, #16 is well known?

@ghjk: did I type AD:BC?  phail...[/quote]

And thats how I got no answer.

yeah, pretty much
just wondering, how many people took note of the change on #19 which was posted here?

Hmmm... are we allowed to post our answers and figure out our scores?

[quote="mathemagician1729"]I beg of your assistance to #30 since I can't count for my life.[/quote]

I [tryed] the problem and got 55.  Is this problem basically 9 indistinguishable things to 3 distinguishable things?  If so, 55 should be correct...

[hide="hint 1"] If you've done a lot of problems, balls and urns should immediately pop into your mind. (If you don't know what balls and urns is, try searching it, or ask).[/hide]

[hide="hint 2"] Distribute the apples and oranges (or whatever the two things were, I forget) seperately.[/hide]

EDIT: Ah, didn't see the post. That doesn't work directly, since they're not exactly indistinguishable, since apples are different from oranges. The [i]apples[/i] are indistinguishable from other apples, and [i]oranges[/i] are indistinguishable from other oranges, but together, not everything is indistinguishable. Person A getting two apples is different from person A getting two oranges. The clever part is seeing that this could be solved if the apples were by themselves and oranges were by themselves. Refer to hint 2.

Ahh...

So to distribute the apples is $ \binom{4 \plus{} 3 \minus{} 1}{3\minus{}1}$ and to distrubute the oranges it is $ \binom{5 \plus{} 3 \minus{} 1}{3\minus{}1}$, so $ \binom{4 \plus{} 3 \minus{} 1}{3\minus{}1}\binom{5 \plus{} 3 \minus{} 1}{3\minus{}1} \equal{} 315$

That's correct.

I dunno if I should be saying that, but oh well. I'm still not going to just tell the answers if someone asks, just give hints.

For #4 on the target, it seems there were quite a few even answers.  there's 13 odd numbered terms, so their sum can't be even :wink:

anyone want to post a solution?

I will.

Also, I'm willing to write a solution guide, if there isn't one already. should I write one?

We prove by induction that $ a_n \equal{} 2^{n \plus{} 1} \minus{} 1.$ This is true when n = 0. Assume this is true when n = k. When n = k+1, $ a_n \equal{} a_{k \plus{} 1} \equal{} 2a_k \plus{} 1 \equal{} 2(2^{k \plus{} 1} \minus{} 1) \plus{} 1 \equal{} 2^{k \plus{} 2} \minus{} 2 \plus{} 1 \equal{} 2^{k \plus{} 1} \minus{} 1.$ Therefore, $ a_0 \plus{} a_1 \plus{} \ldots \plus{} a_{12} \equal{} 2^1 \plus{} 2^2 \plus{} \ldots \plus{} 2^{13} \minus{} 13.$ Using geometric sequences, we have that $ 2^1 \plus{} 2^2 \plus{} \ldots 2^{13} \equal{} 2^{14} \minus{} 2.$ Therefore,  $ 2^1 \plus{} 2^2 \plus{} \ldots \plus{} 2^{13} \minus{} 13 \equal{} 2^{14} \minus{} 15$. We can calculate 2^14 using the fact that 2^10=1024.

Um does anyone know how to do number 26 using variables and manipulations?

I tried and I keep getting it wrong.

Is it true that if 8b = 32 (mod 9)
then b = 4 (mod 9) ?

Well, you COULD use lagrange interpolation... but thats messy and not mathcounts level.

Yes I think I have tried Chinese Remainder Theorem but I keep getting the answer wrong. Maybe I'm dividing wrong in mods.  :maybe:

[quote="pythag011"][quote="ghjk"][quote="mathemagician1729"]sorry to double post, but it seems no one has responded to my question on the last page.[/quote]
Use Chinese Remainder Theorem! You'll immediately get a form of x, like ak+b, then the rest is just plug and chug to find the smallest one.[/quote]

Right, but you can generalize it using Lagrange interpolation.[/quote]

right; a vital concept to understand for all mathcounts competitions.

[quote="dafjbomb"]hey pythag do you want to do cd later today?[/quote]

maybe...

btw, this is probably wron place to put this, but because I absolutely hate computatoin, I have started writing a computation-free mathcounts...


[quote="GameBot"]What is the least common multiple of 45,60, and 75?[/quote]

$ 15 \cdot 3 \cdot 4 \cdot 5\equal{}900$.


For every $ n$ the sum of $ n$ terms of an arithmetic progression is $ 2n \plus{} 3n^2$.  The $ r$th term is:

$ \textbf{(A)}\ 3r^2 \qquad \textbf{(B)}\ 3r^2 \plus{} 2r \qquad \textbf{(C)}\ 6r \minus{} 1 \qquad \textbf{(D)}\ 5r \plus{} 5 \qquad \textbf{(E)}\ 6r \plus{} 2 \qquad$

The $ r^{th}$ term of the arithmetic progression?

$ a_1 \equal{} 2(1) \plus{} 3(1)^2 \equal{} 5$
$ a_1 \plus{} a_2 \equal{} 2(2) \plus{} 3(2)^2 16$
$ a_1 \plus{} a_2 \plus{} a_3 \equal{} 33$

$ \Rightarrow a_1,a_2,a_3... \equal{} 5, 11, 17...$

$ \Rightarrow a_r \equal{} 5 \plus{} 6(r\minus{}1) \equal{} 6r \minus{}1 \Longrightarrow \boxed{C}$


[quote="hucht"]A ver a todos los que les gusta mucho el libro "los elementos" y que se rasgan las vestiduras cuando ven soluciones con otras erramientas

[b]Cuantos axiomas tiene la geometria euclidea????[/b][/quote]

Creo que tiene 5, el conflictuoso es el famoso "5º postulado", que define y separa la euclidiana de la no euclidiana

[quote=hucht]A ver a todos los que les gusta mucho el libro "los elementos" y que se rasgan las vestiduras cuando ven soluciones con otras erramientas

[b]Cuantos axiomas tiene la geometria euclidea????[/b][/quote]

no se cuantos axiomas tiene ;¨=


number theory
prime numbers
number theory unsolved

Show that:there exists infinitely many positive integer $n$ such that the largest prime divisor of $n^4+1$ is $>2n$

[b]Lemma:[/b] There is infinitely many primes which divides numbers of a type $n^4+1$.
[i]Proof:[/i] It's easy. Else if $p_1, p_2,...,p_n$ are all these prime numbers, then number $(p_1p_2...p_n)^4+1$ isn't divisible by $p_i$ for each $i=1,2,...,n$.
Let's take $p$ - one of such prime numbers. There is $n$: $p|n^4+1$. Let $n=ps+r$, where $s,r\in\mathbb{Z}$, $0\leq r<p$. Then both numbers $r^4+1$ and $(p-r)^4+1$ is divisible by $p$, but one of numbers $r$ or $p-r$(we name it as $t$) less, then $\frac{p}{2}$. Also $t^4+1$ is divisible by $p$, but $p>2t$.
So fof each such prime $p_i$ we built $t_i$, satisfying condition of our problem.
But if set $(t_1,t_2,...)$ is finite, then there is $t^4_k+1$ having infinitely many different prime divisors from $p_1,p_2,..$ - it'sn't true.
It means, that we got what we need.

[quote="pavel kozlov"][b]Lemma:[/b] There is infinitely many primes which divides numbers of a type $n^4+1$.
[i]Proof:[/i] It's easy. Else if $p_1, p_2,...,p_n$ are all these prime numbers, then number $(p_1p_2...p_n)^4+1$ isn't divisible by $p_i$ for each $i=1,2,...,n$.[/quote]

How does this prove the lemma?

Assume that there are only finetely many primes that can divide $n^4+1$ and call them $p_1,p_2,...,p_n$. Now read the proof again and you will see the contradiction ;)


integration
calculus
trigonometry
calculus computations

Find the volume of the region in $ \mathbb R^3$ lying above the $ xy$-plane and below the suface with equation $ z\equal{}4\minus{}x^2\minus{}y^2$.

Double integral.  We need $ \iint_{S}4\minus{}x^{2}\minus{}y^{2}\ dy\ dx$ where $ S$ is the surface.  If we draw a picture, we see that on the $ xy$-plane, it forms a circle of radius $ 2$.  Then we may write
$ \int_{\minus{}2}^{2}\int_{\minus{}\sqrt{4\minus{}x^{2}}}^{\sqrt{4\minus{}x^{2}}}4\minus{}x^{2}\minus{}y^{2}\ dy\ dx$.
Now this is a messy integral due to trig substitutions so if we instead perform a change of variables, it becomes
$ \int_{0}^{2\pi}\int_{0}^{2}r(4\minus{}r^{2})\ dr\ d\theta\equal{}\int_{0}^{2\pi}\left(2r^{2}\minus{}\frac{r^{4}}{4}\right)_{0}^{2}\equal{}\int_{0}^{2\pi}4\ d\theta\equal{}8\pi$
But for some reason, this doesn't look right to me.  Check it.


Can somebody solve this problem for me?

" Prove that there exist 2004 points on a plane such that if they are taken by three they define a triangle with whose sides are irrational numbers ".
bye and thank u everyone

Hint: Actaully you can have 2004 (or any nmber > 2) points in a plane where  all triangles have integer lengths.

Now if you  "enlarge" so that one side is irrational, all sides would be irrational.

Proof; <spoiler start>

[hide]Note that: (Let m,n be integers)

(2*m*n)^2 + (m^2-n^2)^2  = (m^2+n^2)^2

Now take  pairs  (m_1,n_1),(m_2,n_2) such that   all ratios m_k/n_k are distinct and m_k/n_k > 1  (You can have as many such pairs as you need, select 2002 such pairs)

Now   consider a  semicircle  with diameter   A B (lenght AB=x)   and  2002  points on the semicircle P_1,P_2 such that for all P's

AP_k  =  (2*m_k*n_k)/(m_k^2+n_k^2) * x

          = (rational_number)*x

This will gve BP_k =( (m^2-n^2)/(m^2+n^2)) * x

          = rational_naumber*x

Also AB = x = rational_number*x

  Traingle  (A P_1 B)   is right angle  and  sin (and cosine) of all angles are  rational.

That will be case for ANY   angle between any thee points ( Remember if   sin (R), cos (R), sin (S), cos (S) are rational so are sin  (R+S) or  sin (R-S) etc... since sin of all angles are rational .. all sides are tational multiple of each other...

All you have to do is now take x as any irrational number and you are done.

Hope it helps.[/hide]

Thanks for helping me in solving this problem... i guess it wasn't so hard finally!!
 :)  Au revoir or better au re - ecrire.


inequalities
topology
inequalities proposed

[quote="tnk"]Given a,b,c,d , x,y,z,t > 0 and a + b + c + d = x + y + z + t = 1

Prove that:

 $\sum  \frac{xyzd}{(d+a+b)(d+b+c)(d+c+a)}   \leq   \frac{1}{27}$

What about a general inequality for x1,...xn and a1,...an???[/quote]
Yes, it can be generalize to

$
\sum\frac{a_ix_1x_2...x_n(1-a_i)}{x_i(1-a_1)(1-a_2)...(1-a_n)}\leq\frac{1}{(n-1)^{n-1}}
$

if $x_1+...+x_n=a_1+...+a_n=1$

My proof...well...it makes use of Lagrange multiplier...

[quote="tnk"]Does the equality happen when t = (a+b+c)/3 ....???

I like to see algebraic tranformations in the proof !! Please help me![/quote]
Yes, the equality occurs iff $x_i=\frac{1-a_i}{n-1}$
Here is my proof:
Fixed $a_1,...,a_n$. Consider the closed set $\{(x_1,x_2,...,x_n)|x_1+...+x_n=1,x_1,x_2,...,x_n\geq 0\}$. The left term of the inequality occurs its maximum at somewhere in this set, say $(y_1,y_2,...,y_n)$
Let $z_i=\frac{1-a_i}{n-1}$. What we want to do is to prove that $y_i=z_i$
It can be proved that if $y_i=0$ for some i then the left term $\leq\frac{1}{(n-1)^{n-1}}$. So we ignore that case.
Suppose that $0<y_1,...,y_n<1$. Apply Lagrange multiplier, we have
$
\frac{\delta\sum\frac{a_iy_1y_2...y_n(1-a_i)}{y_i(1-a_1)(1-a_2)...(1-a_n)}}{\delta y_1}=...=\frac{\delta\sum\frac{a_iy_1y_2...y_n(1-a_i)}{y_i(1-a_1)(1-a_2)...(1-a_n)}}{\delta y_n}
$
That is
$
\sum_{i!=1}\frac{a_iy_1...y_n(1-a_i)}{y_1y_i(1-a_1)...(1-a_n)}=...=\sum_{i!=n}\frac{a_iy_1...y_n(1-a_i)}{y_ny_i(1-a_1)...(1-a_n)}
$
That is
$
\frac{1}{y_1}\sum_{i!=1}\frac{a_i(1-a_i)}{y_i}=...=\frac{1}{y_n}\sum_{i!=n}\frac{a_i(1-a_i)}{y_i}
$
Suppose that $(y_1,...,y_n)!=(z_1,...,z_n)$. WLOG assume that $\frac{z_1}{y_1}\leq ...\leq\frac{z_n}{y_n}$ and $\frac{z_1}{y_1}<\frac{z_n}{y_n}$. Now we have

$\frac{1}{z_1}\sum_{i!=1}\frac{a_i(1-a_i)}{z_i}=\frac{1}{z_n}\sum_{i!=n}\frac{a_i(1-a_i)}{z_i}$(this can be verified.)

Which implies

$\frac{1}{y_1}\sum_{i!=1}\frac{a_i(1-a_i)}{y_i}<\frac{1}{y_n}\sum_{i!=n}\frac{a_i(1-a_i)}{y_i}$, contradiction!

Thank for your solution. But what I mean is a elementary proof, not using a tool like Lagrange multiplier for undergraduates.

If you are interested in solving the inequality in that way. Try this one,

Find the maximal value (I can give u the result, but if I do that, the result may hint you the solution) of :

$\sum_{i<j} (a_{i} + a_{j}) x_{i} x_{j}$

Given

$1< a_{i} < 2 ,  \sum x_{i} = 1$

In this problem, my solution requires only Cauchy's inequality for 2 numbers!


Let $k$ be a circle, $AB$ and $CD$ are parallel chords and $l$ is a
line from C, that intersects $AB$ in its middle point $L$ and $l\cap k=E$. $K$ is the middle of $DE$. Prove that $\angle AKE=\angle BKE.$

$AB\parallel CD$, thus direct computation of the angles in terms of the arcs shows that $\angle ALE=\angle EBD$. But $L$ is the midpoint of $AB$ and from the previous equality of angles we infer that $AEBD$ is a harmonic quadrilateral.  Since $K$ is the midpoint of $ED$ this implies immediately $\angle AKE=\angle ADB=\angle EKB$. Hence the result.