geometry
circumcircle
radical axis
geometry proposed

Let $ ABC$ be a triangle and a point $ P$, $ A'B'C'$ is Cevian triangle of $ P$, $ PA\cap B'C'\equal{}A''$, $ M$ is projection of $ A''$ on $ BC$ circumcricle $ (ANC)\cap AB\equal{}\{E\},(ANB)\cap AC\equal{}\{F\}$, let circumcircle $ (BME)\cap (CMF)\equal{}\{M,P'\}$ prove that $ M,P,P'$ are collinear.

There is a typo in your problem, N is not previously defined, but I believe N=M.

By Miquel theorem, $AEP'F$ is cyclic, thus $AE,FP',CM$ concur at the radical center $B$ of $\odot(AEP'F),$ $\odot(CFP'M)$ and $\odot(AEMC),$ i.e. $B,F,P'$ are collinear. Hence $\angle P'MC=\angle P'FA \equiv \angle BFA=\angle AMB$ $\Longrightarrow$ $MA',MA''$ bisect $\angle AMP'$ $\Longrightarrow$ $M(A,P',A'',A')=-1.$ But $M(A,P,A'',A')=-1$ $\Longrightarrow$ $M,P,P'$ are collinear.

[quote="encyclopedia"]Let $ ABC$ be a triangle and a point $ P$, $ A'B'C'$ is Cevian triangle of $ P$, $ PA\cap B'C' \equal{} A''$, $ M$ is projection of $ A''$ on $ BC$ circumcricle $ (ANC)\cap AB \equal{} \{E\},(ANB)\cap AC \equal{} \{F\}$, let circumcircle $ (BME)\cap (CMF) \equal{} \{M,P'\}$ prove that $ M,P,P'$ are collinear.[/quote]

Sorry it must be

Let $ ABC$ be a triangle and a point $ P$, $ A'B'C'$ is Cevian triangle of $ P$, $ PA\cap B'C' \equal{} A''$, $ M$ is projection of $ A''$ on $ BC$ circumcricle $ (AMC)\cap AB \equal{} \{E\},(AMB)\cap AC \equal{} \{F\}$, let circumcircle $ (BME)\cap (CMF) \equal{} \{M,P'\}$ prove that $ M,P,P'$ are collinear.

Let $ (O_{1}),\ (O_{2})$ be, the circumcircles of the triangles $ \bigtriangleup AMB,\ \bigtriangleup AMC$ respectively and also let $ (O_{3}),\ (O_{4})$ be, the ones, of the triangles $ \bigtriangleup BME,\ \bigtriangleup CMF$ respectively.

It is easy to show that the quadrilateral $ AEP'F,$ where $ P'\equiv (O_{3})\cap (O_{4}),$ is cyclic, from $ \angle BEP' \equal{} \angle CMP' \equal{} \angle AFP'$ $ ,(1)$

Also it is easy to show that the line segment $ EP',$ passes through the vertex $ C$ of $ \bigtriangleup ABC,$ from $ \angle FP'P'' \equal{} \angle A \equal{} \angle FMC$ $ ,(2)$ where $ P''$ is an arbitrary point on the extension of the line segment $ EP',$ towards $ P'.$

Similarly, the line segment $ FP',$ passes through the point $ B.$

$ \bullet$ We have now, the configuration of the complete quadrilateral $ AEP'FBC,$ where $ AEP'F$ is cyclic with circumcircle so be it $ (O)$ and let be the point $ N\equiv EF\cap AP'.$
The point $ M,$ as the common point of the circumcircles $ (O_{1}),\ (O_{2}),\ (O_{3}),\ (O_{4})$ of the triangles $ \bigtriangleup ABF,\ \bigtriangleup ACE,\ \bigtriangleup BEP',\ \bigtriangleup CFP'$ respectively, is the [b][size=100]Miquel point[/size][/b], of the complete quadrilateral $ AEP'FBC,$ which lies on $ BC$ as well, because of cyclic $ AEP'F.$

Also, we have the result that the points $ O,\ N,\ M,$ are collinear and $ ONM\perp BC$ as well and easy to prove elementary or shortest by polar theory ( the line segment $ BC,$ is the polar of $ N$ with respect to circumcircle $ (O),$ of $ AEP'F$ ).

So, we conclude that the point $ A''\equiv B'C'\cap AP,$ lies on the line segment $ ONM\perp BC.$

$ \bullet$ Let be the point $ N'\equiv BC\cap AP'$ and from the complete quadrilateral $ AEP'FBC$ we have that the points $ A,\ N,\ P',\ N',$ are in harmonic conjugation and then, the pencil $ M.ANP'N',$ is also in harmonic conjugation.

Similarly, from the complete quadrilateral $ AB'PC'BC$ we have that the points $ A,\ A'',\ P,\ A',$ are in harmonic conjugation and then, the pencil $ M.AA''PA',$ is in harmonic conjugation too.

So, because of the harmonic pencils $ M.ANP'N',\ M.AA''PA',$ have the line segment $ MA$ as their common ray and also they have $ MN\equiv MA''$ and $ MN'\equiv MA',$ we conclude that they have also $ MP'\equiv MP$ $ ,(3)$

Hence, from $ (3)$ we conclude that the points $ M,\ P,\ P',$ are collinear and the proof is completed.

Kostas Vittas.

Nice solution dear vittasko, here is my solution

Because $ (APA''A')$ is harmonic and $ MA''\perp MA'$ thus $ MA''$ is bisector of $ \angle PMA$ similar if $ B'C'\cap BC\equal{}A'''$ then $ (BCMA''')$ is harmonic then $ MA''$ is bisector of $ \angle B'MC'$, this is succifient to show $ \angle AMB\equal{}\angle PMC$, let $ BF\cap (MFC)\equal{}\{P'\}$ we have $ \angle BMA\equal{}\angle BFA$ but $ \angle BMA\equal{}\angle PMC$ so that $ MP'FC$ is cyclicmso we can prove $ BEP'M$ is cyclic, too. We are done. It is the way I created this problem!

First I want to sorry if my solution below is same as anyone's :D .
Here is my solution:
Let D be the intersection of B'C' with BC. It is easy to see that (D,A',B,C)=-1.
Therefore (D,A",C',B')=-1. Then M(D,A",C',B')=-1, but MA" is perpendicular with MD. So MA" is the bisector of (MC',MB'). With the same argument, we are also able to prove that A"M is the bisector of  (MA,MP). As the result, we have (MC',MA)=(MP,MB').
Now it is easy to prove that E, M', C are collinear (Directed angle can kill it easily :lol: ).
Therefore, (MM',MB')=(MM',MF)+(MF,MC)+(MC,MB')=(CM',CA)+(AC,AB)+(MC',MB)=(ME,MA)+(MB,ME)+(MC',MB)=(MC',MA). Then, (MM',MB')=(MP,MB').
This prove that M, M', P are on the same line.
Our proof is completed. :D  :lol:  :blush:


geometry
geometric transformation
reflection
linear algebra
matrix
search
linear algebra unsolved

One of the problems published in Mathematical Reflections asserts that if $A$ is a commutative subalgebra of $M_{n}(R)$ such that all matrices in $A$ have nonnegative determinant, the same property holds for all matrices whose elements are actually elements of $A$ (that is, block matrices with blocks belonging to $A$). Is the same property true for noncommutative subalgebra?

Tried to construct a counterexample based on the quaternion algebra \[\begin{pmatrix}a&b&c&d\\-b&a&-d&c \\-c&d&a&-b \\-d&-c&b&a\end{pmatrix}\] but failed. Perhaps block-quaternion matrices must have a nonnegative determinant.  :maybe: Still think that the answer to the general question is no.

Any idea, guys? I also believe that the answer is negative, but how to prove (or disprove) it?

I gave this problem to my algebra professor and he spent the first half of today's lecture explaining the basics of matrix algebra so that he could tell the problem to the class. He offered a dollar to anyone who could prove the Mathematical Reflections problem by elementary means (he destroyed it with some machinery that I haven't heard of).

He said that the question about noncommutative subalgebra is vacuous because there is no noncommutative subalgebra of $M_{n}(R)$ whose matrices all have nonnegative determinant.

(Just so you know, I have currently had about 4 lectures' worth of algebra, so I have rather little idea of what's going on.)

You should tell your professor about [url=http://en.wikipedia.org/wiki/Quaternion#Representing_quaternions_by_matrices]representing quaternions by matrices[/url].

He knows about that. He was talking about it.

edit: I should specify that the reason for my last comment about my own lack of knowledge about algebra is that I may not be reliably relaying what the professor said, since I didn't understand any of it. But he said that the noncommutative version is vacuous.

[quote="Xevarion"]
He said that the question about noncommutative subalgebra is vacuous because there is no noncommutative subalgebra of $M_{n}(R)$ whose matrices all have nonnegative determinant.
[/quote]

Here's a counterexample to that:

Let $n=2k,\ k\ge 2$, and consider all matrices in $\mathcal M_{n}(\mathbb R)$ of the form $\begin{pmatrix}A&B\\0&A\end{pmatrix}\ (*)$, where $A,B,0$ are $k\times k$ matrices. These form a subalgebra of $\mathcal M_{n}(\mathbb R)$, which is clearly non-commutative (for $k\ge 2$, as specified), and the determinant of the typical member $(*)$ is $(\det A)^{2}$.

[quote="grobber"]Here's a counterexample to that:[/quote]
I'll try to figure out what he actually said, then.

Xevarion, did you find out what he actually wanted to say?

Sorry, he misunderstood me and though we want determinants positive, not nonnegative. :P [probably my fault :|] I think in this case his statement is true. Anyway he has not yet answered me regarding the actual problem.

Even in that case, he is wrong. Look at the quaternions algebra.

Come on, I thought he was serious enough not to consider the zero element!!!

That is my question not his. :P I am sorry, I have only had 2 weeks of algebra, I hardly understand what is happening...  :(  :(

It's ok, I just insisted in order to see what your teacher knows about this fact, because the problem seems quite difficult (at least for me) and I have no clue where to search in the literature for it.


function
LaTeX
limit
trigonometry
algebra
rational function
calculus

[cos(sin(x))-cos(x)]/x^4
limt x tends to zero

A bad solution will use L'Hopital repeatedly as long as the resultant rational function satisfies the conditions.
A more natural approach would be to approximate the top portion.

In $ LaTeX$:

$ \lim_{x \to 0} \frac{\cos \sin x \minus{} \cos x}{x^4}$

Near zero we have:

$ \cos x \sim 1 \minus{} \frac{x^2}{2} \plus{} \frac{x^4}{24}$

$ \sin x \sim x \minus{} \frac{x^3}{6}$

Just it's just the matter of simple expression manipulation..

[quote="milin"]In $ LaTeX$:

$ \lim_{x \to 0} \frac {\cos \sin x \minus{} \cos x}{x^4}$

Near zero we have:

$ \cos x \sim 1 \minus{} \frac {x^2}{2} \plus{} \frac {x^4}{24}$

$ \sin x \sim x \minus{} \frac {x^3}{6}$

Just it's just the matter of simple expression manipulation..[/quote]

hello
$ \lim_{x\to 0}\frac{\cos\sin x\minus{}\cos x}{x^{4}}$
= $ \lim_{x\to 0}\frac{2sin(\frac{x\plus{}sinx}{2})sin(\frac{x\minus{}sinx}{2})}{x^{4}}$
=$ \lim_{x\to 0}\frac{2(\frac{x\plus{}sinx}{2})(\frac{x\minus{}sinx}{2})}{x^{4}}$
using the series of cos and sin .After simplification
=$ \lim_{x\to 0}(2\minus{}\frac{x^{2}}{3!}\plus{}.........)(\frac{1}{3!}\minus{}\frac{x^2}{5!}.........)\frac{1}{2}$
=$ \frac{1}{6}$
thank u

[quote="kabi"]hello
$ \lim_{x\to 0}\frac {\cos\sin x \minus{} \cos x}{x^{4}}$
= $ \lim_{x\to 0}\frac {2sin(\frac {x \plus{} sinx}{2})sin(\frac {x \minus{} sinx}{2})}{x^{4}}$
=$ \lim_{x\to 0}\frac {2(\frac {x \plus{} sinx}{2})(\frac {x \minus{} sinx}{2})}{x^{4}}$
using the series of cos and sin .After simplification
=$ \lim_{x\to 0}(2 \minus{} \frac {x^{2}}{3!} \plus{} .........)(\frac {1}{3!} \minus{} \frac {x^2}{5!}.........)\frac {1}{2}$
=$ \frac {1}{6}$
thank u[/quote]

A substitution up front to make the power series solution easier: let $ u\equal{}\sin x$ on a suitable neighborhood of $ x\equal{}0.$ The limit then becomes:

$ \lim_{u\to0}\frac{\cos u\minus{}\sqrt{1\minus{}u^2}}{(\arcsin u)^4}$

We have $ \cos u\equal{}1\minus{}\frac12u^2\plus{}\frac1{24}u^4\plus{}O(u^6)$ and $ \sqrt{1\minus{}u^2}\equal{}1\minus{}\frac12u^2\minus{}\frac18u^4\plus{}O(u^6).$

Subtracting, and using $ \frac1{24}\plus{}\frac18\equal{}\frac16,$ we get that the numerator is $ \frac16u^4\plus{}O(u^6).$

The denominator is easy, since we're not asking for much: $ (u\plus{}O(u^3))^4\equal{}u^4\plus{}O(u^6).$

Divide to get that the limit is $ \frac16.$

elementary solution ?

is there an elementary solution other than power series ?  maybe which involves only basically
 limit sin x / x = 1  when x is near 0.

It's enough to consider $ x > 0$. The mean value theorem gives ${ \cos{\sin x} - \cos x} = \sin{c_x}(x - \sin x)$, $ \sin x < c_x < x$. We know $ (x - \sin x)/x^3 \rightarrow 1/6$, so we just need to show $ \sin{c_x}/x \rightarrow 1$. But $ \sin{\sin x}/x < \sin{c_x}/x < \sin x/x$. Here the right side $ \rightarrow 1$, and multiplying and dividing by $ \sin x$ shows the left side $ \rightarrow 1$, hence $ \sin{c_x}/x \rightarrow 1$ as desired.

i think this may help u .

$ \lim_{x\to 0}\frac{\cos\sin x\minus{}\cos x}{x^{4}}$

= $ \lim_{x\to 0}\frac{2sin ( {\frac{x\plus{}sinx}{2} })\sin({ \frac{x\minus{}sinx}{2} })}{x^{4}}$

here multiply and divide by $ \frac{\sin x \plus{} x }{2}$ and $ \frac{\sin x \minus{} x }{2}$

hence it reduces to

= $ \lim_{x\to 0}\frac{2\sin({\frac{x\plus{}sinx}{2} })\sin({\frac{x\minus{}sinx}{2}})\frac{\sin x \plus{} x }{2} \frac{\sin x \minus{} x }{2}
}{ \frac{\sin x \plus{} x }{2} \frac{\sin x \minus{} x }{2}x^{4}}$

= $ \frac{1}{2} \lim_{x\to 0}\frac{x^2 \minus{} sin^{2} x }{x^4}$  this limit by l'hospital turns out to be $ \frac{1}{3}$ . hence answer is  $ \frac{1}{6}$


[quote="mdk"]Let f: R to R be a differentiable function such that f(x+h)-f(x)=hf'(x), for all x,h in R.
Prove that f is a linear function[/quote]

$ x=0\implies f(h)=f'(0)h+f(0)=ah+b,\; \forall h\in \mathbb{R}$


How did you do on the fall start up? I got 50 for 4th place in the middle school division. They put my name as Jeny Li when my name is Jerry. -.-
i need to work on my handwriting....

My coach only gave me like 15 minutes to work on the test (Said I started at 9:25 when I came to the buliding at 9:35...) so I only got 40 (83 last year)

69 according to the website, though my teacher had said 68. worse than last year as expected :) cuz i didnt reach the end. Interesting how no one got in the 90's this year.

Ouch BOGTRO. That stings.

I got 52, hope to get better next year. Are the scores usually a lot higher than this?

Got 65 officially, but I was told that I had 72.  Ah well, first middle school, although BOGTRO would have beat me if he had enough time...

@hm29168: Scores were higher last year, although I heard last years' test was easier.

Got 58- not bad, but could have done alot better.

I got 5/6 of the problems I answered right and ranked (6+7)th place nationally.


What is the remainder when X^1982 +1 is divided by X −1?

[hide]just plug in 1, so it should be 2?[/hide]

Answer (using congruences):

$ x-1 \equiv 0(mod\,(x-1))$
$ \Longrightarrow x \equiv 1(mod\,(x-1))$
$ \Longrightarrow x^{1982}\equiv 1(mod\,(x-1))$
$ \Longrightarrow x^{1982}+1\equiv 2(mod\,(x-1))$

Hence remainder is 2 if $ x \geq 4$,

There is no remainder if $ x = 3$ (since $ 2(mod\,2) \equiv 0(mod\,2)$)
There is no remainder if  $ x = 2$ (since $ 2(mod\,1) \equiv 0(mod\,1)$).
x cannot take the value 1.
There is no remainder if $ x = 0$ (can be checked by substitution).
There is no remainder if $ x =-1$ (can be checked by substitution).
If $ x \leq-2$ then there is remainder 2.

$ x$ is being considered as a variable and is not assigned a value.

t0rajir0u...do you agree with my solution? Or is there an easier way?

Your solution is correct, although you could've safely stated $ x^{1982}+1 \equiv 1^{1982}+1 \equiv 2 \bmod (x-1)$.  But the reasoning after that is unnecessary since we are considering polynomial division and polynomial remainders.


linear algebra
matrix
vector
inequalities

Prove that frobenius norm ( matrix norm) is compatible  with 2-norm ( vector norm)

Do you want an inequality between these 2 norms ?


Is there any way to create a flowchart in latex?

You could try:
1. flow package available at http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=nassflow
2. pstricks see http://tug.org/PSTricks/main.cgi?file=psmatrix/psmatrix#flowchart for an example (clciking on the tex or pdf icon to the right will give you the latex code used
3. draw the flowchart elsewhere and import the image


Let $ ABC$ be a triangle and an arbitrary point $ M$ on its side $ BC$. Let $ \mathcal{C}(O)$, $ \mathcal{C}(O_1)$, $ \mathcal{C}(O_2)$ be respectively the inscribed circles of triangles $ ABC$, $ ABM$, $ ACM$. Denote by $ T_1$, $ T_2$ the tangent point of $ \mathcal{C}(O_1)$, $ \mathcal{C}(O_2)$ with $ AM$, respectively, and $ D$ the tangent point of $ \mathcal{C}(O)$ with $ BC$. Let $ T$ be the foot of the perpendicular line to $ O_1O_2$ from $ D$. Prove that $ \angle T_1TT_2 \equal{} 90^{\circ}$.

It is well-known that I1,I2,D,M are concyclic  => ∠I1DI2=90   => ∠I1DF+∠I2DE=90
I2TDE are concyclic   =>  ∠I2TE+∠I1I2M=∠I2DE+∠I1DF=90
=>  TE⊥MI2   but T2E⊥MI2  =>  T2,T,E are collinear
analogously , T1,T,F are collinear
done!

Dear plane geometry:
In your figure, I found $ DE\equal{}MF$, how to show it?
thank you!

[quote="Zhang Fangyu"]Dear plane geometry:
In your figure, I found $ DE \equal{} MF$, how to show it?
thank you![/quote]
MF=(BM+AM-AB)/2

DE=ME-MD=(AM+MC-AC)/2-(BD-BM)=(AM+MC-AB-BC+2BM)/2=(BM+AM-AB)/2

So DE=MF

[hide]See for a familiar problem, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=223789]Here[/url][/hide]

Kostas Vittas.


trigonometry
geometry
circumcircle
trig identities
Law of Sines

Find the length of the longest diagonal of a regular dodecagon of side length 1.

I divided the dodecagon into twelve congruent isosceles triangles (30-75-75) and applied the extended law of sines, but I ended up getting 2R (where R is the circumradius)=1/(sin 30) or R=1, which made no sense because if all the side lengths were 1 then the triangle would be equilateral.

Sorry about the horrible format. I'm new to this forum.

I am going to move this to Intermediate topics.

You were right to divide the dodecagon into (30-75-75) triangles.  The Law of Sines states $\frac{sin{A}}{a}=\frac{sin{B}}{a}=\frac{sin{C}}{c}$.  This gives $\frac{\sin{30}}{1}=\frac{sin{75}}{r}$ and $2r=2\frac{sin{75}}{\sin{30}}$ If you work this out with a calculator you get $2r\approx 1.932$
Did this help?

[quote="ineedmathhelp"]... I ended up getting 2R (where R is the circumradius)=1/(sin 30) or R=1, which made no sense because if all the side lengths were 1 then the triangle would be equilateral. [/quote]
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
In the extended law of sines formula, $R$ is the circumradius of the triangle, not of the dodecagon. The circumradius of the isosceles triangle is indeed 1, but that is not the same as the circumradius of the dodecagon. We are trying to find the circumradius of the dodecagon, not of the triangle.
diophantus correctly applied the law to get the desired length.

Oops. I realized that I was finding the wrong circumradius. Sry.

if you then divide those triangles into 2 right triangles, you can solve for the hypotenuse = circumradius


need help -  :P Express 4.134134134...5 as a common fraction in base 10.

:thumbup: thank you

i knew that

not. :roll:


quadratics
algebra
polynomial
pigeonhole principle
number theory theorems
number theory

I don't know whether my topic belong to here..
Would you like to give me no standard examples of applications of the quadratic reciprocity theorem?
Standard applications are; solving x^2 = p mod q  with p,q odd primes, calculating legendre symbole.......

Even a olympiad problem where this theorem is very usefull to solve with is welkom!!
thanks

An example, in my opinion.  :blush:

The sequences $ a_0, a_1, a_2,...$and $ b_0, b_1, b_2,...$ are defined by $ a_0 = 1, b_0 = 4, a_{n+1} = a{}^n{}^{2001} + b_n, b_{n+1} = b_n^{2001} + a_n$. Show that no member of either sequence is divisible by $ 2003$.

[hide="Solution"]http://www.kalva.demon.co.uk/ibero/isoln/isol036.html[/hide]

Another nice application (local square => global square):

If $ a$ is an integer such that $ a$ is a quadratic residue $ \mod p$ for all primes $ p$ (we count $ 0$ as a quadratic residue), then $ a$ is a perfect square. This still holds if we exclude finitely many primes.

Okey thank u both of you..
but how can we get a local->global effect if we need to check  mod p for all primes!?

By the way, I heard of another application but I only remember that if we take a monic polynomial f in Z[x].
let's calculate f mod p for a lot of primes: 2,3,5,7,...........,q.
Somehow we will see that f splits in linear factors or ifor some primes. This has also to do with the discriminant of f.
I do not exactly know the connection between these primes and the discriminant.. may be you know something about this..

THank you again

Local to global means:

Let $ S$ be some property (e.g. being a square). We say (here) that $ S$ satisfies the local-to-global-principle if the following is true:
$ S$ holds in $ \mathbb Z / p \mathbb Z$ for most (or all) primes $ p$ (a local property), then $ S$ holds in $ \mathbb Z$ (a global property).

There are different types of local-to-global-principles. But often it's easier to check the local properties than to check the global ones.
One example is the theorem of Hasse-Minkowski, literally saying that if a quadratic form $ q(x_1, ... , x_n)$ (this is by definition a homogeneous polynomial of degree $ 2$) over the integers has a nontrivial (meaning: not all $ x_i \equal{} 0$) zero $ \mod n$ for any integer and in the reals, then the same holds in the integers.
One corollary of this: every positive integer not of type $ 4^s(8k \minus{} 1)$ is a sum of three perfect squares.

Why is something local sometimes easier: You work in some finite set, thus you can use counting arguments. Or you may be able to use additional structures you gained (like Fermat's theorem, quadratic reciprocity, or else). And there is Hensel's Lemma: a polynomial $ f(x)$ with a root $ f(a) \equiv 0 \mod p$ fulfulling $ f^\prime (a) \not\equiv 0 \mod p$ has a root $ \mod p^k$ for all $ k$. Thus to use Hasse-Minkowski's theorem, you only have to check a couple of prime powers (there is a version covering even those with $ f^\prime (a) \equiv 0 \mod p$).
Not that it would be good example for the principle itself, but for why local properties can be easier: Show that $ x^2\plus{} y^2 \plus{} 1 \equiv 0 \mod p$ has a solution, e.g. use pidgeonhole principle.

Splitting of polynomials $ \mod p$ is a bit trickier. Some facts:
- If $ p$ divides the discriminant, only then the polynomial may have a double root $ \mod p$.
- There are polynomials that cannot be factored in the rationals, but $ \mod p$ for all primes $ p$. For example $ x^4 \plus{} 1$.
- There are polynomials that have no rational root, but one $ \mod p$ for all $ p$. For example $ (x^2 \minus{} 2)(x^2 \minus{} 17)(x^2 \minus{} 34)$.
- For any polynomial $ f(x)$, there are infinitely many primes $ p$ such that $ f \mod p$ splits completely into linear factors.
- More general, for irreducible polynomials, there is Chebotarev's theorem, a mighty (but nonelementary) theorem describing how the polynomial can split into factors $ \mod p$ and how "many" such primes $ p$ with that splitting there are. See http://mathworld.wolfram.com/ChebotarevDensityTheorem.html or http://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem .


Trapezoid ABCD has vertices A (0,0), B (6,0), C (12,8) and D (0,8). A line through the origin intersects DC at E, dividing the trapezoid into two regions with equal perimeters. What is the number of square units in the area of quadrilateral ABCE?

[hide]By the Pythagorean Theorem, we know that $BC=10$.  Now we can set up equations:

$8+DE+AE=6+10+EC+AE$  and
$DE+EC=12$.
$AE$ cancels out, and we solve for $DE$ and $EC$ and we get
$DE=10$ and $EC=2$.  Now we find the area of trapezoid $ABCE$, which comes out to be $32$.[/hide]


Hi,.. I gave this problem to my friend, carlosbr. we really want to know how we can use the books hint?  the problem is:
Prove that the only positive integer solutions of
y^2=1+x+x^2+x^3 where  (x,y)=(1,2),(7,20)
Hint: use that x^4+y^4=z^2 has no positive integer solutions

it's posted only 3 topics below yours
http://www.mathlinks.ro/Forum/viewtopic.php?t=14926
you should check stuff before you post ;)


Express the number 14 as the sum of fewer than 4 positive integers. What is the largest possible product resulting from these integers?

[quote="sup3rcrash3r"][hide="Solution"]
We want to find four integers that are as close to each other as possible. This is done by using two 3's and two 4's. The product of these integers is 144.[/hide][/quote]

Not quite.

[hide="Solution"]It's asking for fewer than 4.  We get two 5s and 4, and therefore the greatest product is 100.[/hide]

[quote="sup3rcrash3r"][hide="Solution"]
We want to find four integers that are as close to each other as possible. This is done by using two 3's and two 4's. The product of these integers is 144.[/hide][/quote]

I think you misread the question. It asks for [b]less than[/b] 4 numbers.

Case 1: 1 number. Clearly, the number must be 14.

Case 2: 2 numbers. By AM-GM, the maximum product is 49.

Case 3: 3 numbers. The maximum is $ 4 \cdot 5 \cdot 5\equal{}100$

Our answer is 100.

Oops...  :P

In that case, I agree with the above two posts.

[quote="AdithyaGanesh"][quote="BOGTRO"][quote="sup3rcrash3r"][hide="Solution"]
We want to find four integers that are as close to each other as possible. This is done by using two 3's and two 4's. The product of these integers is 144.[/hide][/quote]

I think you misread the question. It asks for [b]less than[/b] 4 numbers.

Case 1: 1 number. Clearly, the number must be 14.

Case 2: 2 numbers. By AM-GM, the maximum product is 49.

Case 3: 3 numbers. The maximum is $ 4 \cdot 5 \cdot 5 \equal{} 100$

Our answer is 100.[/quote]

Correct answer, but you really don't need to make three cases.  It is clear that 3 numbers multiply to more than 2 or 1 right?[/quote]

It may be clear to you and me, but to someone starting out in this forum or unfamiliar with the type of problem, it may not be.


calculus
integration
vector
trigonometry
analytic geometry
parameterization
calculus computations

Evaluate the surface integral $ \int\int_{S} \mathbf{F} \cdot d\mathbf{S}$ for the given vector field $ \mathbf{F}$ and the oriented surface $ \textit{S}$. In other words, find the flux of $ \mathbf{F}$ across $ \textit{S}$. For closed surfaces, use the positive orientation.

Problem: $ \mathbf{F}(x, y, z) \equal{} y \mathbf{j} \minus{} z \mathbf{k}$. $ \textit{S}$ consists of the paraboloid $ y \equal{} x^{2} \plus{} z^{2}$, $ 0\leq y \leq 1$, and the disk $ x^{2}\plus{}z^{2}\leq 1$, $ y\equal{}1$.

My Work:
$ \int\int_{S} \mathbf{F} \cdot d\mathbf{S}$ = $ \int\int_{S_{1}}\mathbf{F} \cdot d\mathbf{S}\plus{}\int\int_{S_{2}}\mathbf{F} \cdot d\mathbf{S}$.

$ S_{1}:$ $ x\equal{}x$, $ z\equal{}z$, $ y\equal{}x^{2}\plus{}z^{2}$
$ \vec{r_{x}}\equal{} \mathbf{i}\plus{}2x\mathbf{j}$
$ \vec{r_{z}}\equal{} 2z\mathbf{j}\plus{}\mathbf{k}$

Crossing these two vectors, we have $ 2x\mathbf{i}\minus{}\mathbf{j}\plus{}2z\mathbf{k}$.
So, $ \mathbf{F}\cdot\mathbf{n} \equal{} \minus{}y\minus{}2z^{2}$.

$ \int\int_{S_{1}}\mathbf{F} \cdot d\mathbf{S} \equal{} \int\int_{D} \minus{}y\minus{}2z^{2} dA$.

$ \int\int_{S_{2}}\mathbf{F} \cdot d\mathbf{S} \equal{} \int\int_{D} y dA$
$ \equal{}\int\int_{D} x^{2}\plus{}z^{2} dA$.

Because $ \int\int_{S} \mathbf{F} \cdot d\mathbf{S}$ = $ \int\int_{S_{1}}\mathbf{F} \cdot d\mathbf{S}\plus{}\int\int_{S_{2}}\mathbf{F} \cdot d\mathbf{S}$. This equals:

$ \int\int_{D} \minus{}x^{2}\minus{}3z^{2}\plus{}x^{2}\plus{}z^{2} dA$
$ \equal{}\int\int_{D} \minus{}2z^{2} dA$
$ \equal{}\minus{}\int^{2\pi}_{0}\int^{1}_{0} 2r^{3}\sin^{2}\theta dr d\theta$ (switch to polar coordinates)
$ \equal{}\minus{}\int^{2\pi}_{0} \frac{1}{2}\sin^{2}\theta d\theta$
$ \equal{}\minus{}\frac{\pi}{2}$.

The answer is 0, but I keep on getting $ \minus{}\frac{\pi}{2}$. Where did I mess up?

[quote]For closed surfaces, use the positive orientation. [/quote]
Which orientation did you put on $ S_1$?

The normal vector for $ S_{1}$ points left, and the normal vector for $ S_{2}$ points right.

I see. The mistake is elsewhere, in a part that you omitted: the parametrization of $ S_2$. It is $ x\equal{}x$, $ y\equal{}1$, $ z\equal{}z$.


The given diagram represents the layout for a poster that has been formed by 6 congruent rectangular sections. If the area of the layout is 600 square inches, what is the perimeter?

Answer & Explain, plz :-)

Er...for each of the rectangles, let the shortest side be $ w$. Then the longer side is $ 4w$, since you can see in the diagram that...4 widths makes up one length. There are 6 rectangles, so the area of each is 100. Thus we have $ 4w^2\equal{}100\implies{w\equal{}5}$. We can see that the perimeter is made up of 4 lengths and 4 widths, for a total of 20 widths, or $ 20\cdot5\equal{}100$.


Compute: $ \sqrt [3]{cos \frac {2 \pi}{7} } \plus{} \sqrt[3]{cos \frac {4 \pi}{7} } \plus{} \sqrt[3]{cos \frac {6 \pi}{7} }$

Let $ x \equal{} \sqrt [3]{cos\frac {2\pi }{7}},y \equal{} \sqrt [3]{cos\frac {4\pi}{7}},z \equal{} \sqrt [3]{cos\frac {6\pi}{7}}$ and $ u \equal{} x \plus{} y \plus{} z,v \equal{} xy \plus{} yz \plus{} zx,t \equal{} xyz$.
We had, that $ x^3,y^3,z^3$ are roots $ a^3 \plus{} \frac 12 a^2 \minus{} \frac 12 a \minus{} \frac 18 \equal{} 0$.
Therefore $ t \equal{} \frac 12 , u^3 \minus{} 3uv \plus{} t \equal{} \minus{} \frac 12 , v^3 \minus{} 3uvt \plus{} t^2 \equal{} \minus{} \frac 12 .$ It give $ u^3 \minus{} 3uv \equal{} \minus{} 1,2v^3 \minus{} 3uv \equal{} \minus{} \frac 32$ or $ v^3 \equal{} \frac {u^3}{2} \minus{} \frac 14.$ Therefore $ u \equal{} \sqrt [3] b$, were b is negative (<0) root of $ 4b^3 \minus{} 42b^2 \plus{} 39b \plus{} 4 \equal{} 0$.

Can use complex:
$ z \equal{} \cos{\frac {2\pi}{7}} \plus{} i\sin{\frac{2\pi}{7}}$

TShpn, could you explain your solution in detail? I don't know much about complex numbers...


Could you guys please give me some examples of measure $ \nu$ on $ \mathbb{R}$ which is (mutually ) singular with respect to the Lesbeques measure and $ \nu(I) > 0$ for any non-empty interval $ I$ ?

many thanks

Reusing an example I gave a week ago: counting measure on the rationals. Every nontrivial interval has infinite measure, and we can write $ \mathbb{R}$ as the union of a set with Lebesgue measure zero ($ \mathbb{Q}$) and a set with $ \nu$-measure zero ($ \mathbb{R}\setminus\mathbb{Q}$).

Whatever else you build, you should probably start with a dense set of measure zero. The rationals are the standard example.

That example gives every interval infinite measure. If you'd prefer to avoid such infinities, we can tweak that a little. For instance: let $ \{r_j\}$ be an enumeration of the rationals and let $ \nu(E) \equal{} \sum_{\{j\,|\,r_j\in E\}}2^{ \minus{} j}.$

It's still basically the same idea.


I see on the forums that many people are saying they ran out of problems but still have problems missed.

Does that mean if you get fed an old problem and get it right, it is taken out of the "wrong"?

If the answer to the previous question is yes, is there a difference between missing a problem, then getting it right again and missing no problems?

[quote="sup3rcrash3r"]I see on the forums that many people are saying they ran out of problems but still have problems missed.

Does that mean if you get fed an old problem and get it right, it is taken out of the "wrong"?[/quote]

Yup

[quote]If the answer to the previous question is yes, is there a difference between missing a problem, then getting it right again and missing no problems?[/quote]

Nope. Not currently. Future updates may take this into account.

Wait...but if future correct answers overwrite previous incorrect answers, shouldn't everybody who finished all of the questions have an alcumus score of 100.00?

I'm not sure. The developers who know the algorith would be able to answer that. Unfortunately, they don't come here.


geometry
power of a point
radical axis
Pythagorean Theorem
geometry unsolved

5 circumferences of different radiuses touch each other. 4 of them are circumscribed around 5th.
Find formula, expressing the dependency of the radius of the 5th circumference from the other 4 circumferences.

P.S. This problem was proposed on the Russian scientific forum (dxdy), however, no one could solve it.

When 4 circles are tangent in pairs the radical axis of one non-tangent pair passes thru

   the similitude center of the other non-tangent pair.    The rest is the difficult part of

   the arithmetic of proportions.

   M.T.

Maybe this can help: http://mathworld.wolfram.com/InnerSoddyCircle.html

Is this kind of a problem which solution is worth 1000s of dollars? Is it really so difficult for mathematicians?

Given a central circle and n circles tangent to it and pairwise to each other, the last radius is easy to calculate, if n is odd. When using the same method for even n, like n = 4, the last radius cancels out. Instead, the calculation yields that the cyclic 2n-gon of the tangency points with the central circle is harmonic and this does not make calculation of the last radius for n = 4 any simpler.

Let $ r$ be radius of the central circle $ (O)$ and $ a, b, c, d$ radii of circles $ (A), (B), (C), (D)$ externally tangent to $ (O)$ at $ H, I, J, K.$ Suppose $ r, a, b, c$ are given. Invert the figure in a circle $ (I)$ with radius $ 2 \sqrt {rb}$; power of inversion is $ 4rb.$ Circles $ (O), (B)$ go to parallel lines $ o' \parallel b',$ perpendicular to $ OB,$ at distances $ 2b, 2r$ from $ I,$ respectively. Circles $ (A), (C)$ go to congruent circlres $ (A'), (C')$ tangent to $ o', b',$ with radii $ a' = c' = r + b.$ By Pythagorean theorem, $ IA'^2 = \delta^2(A', OB) + (r - b)^2,$ where $ \delta(A', OB)$ is distance of $ A'$ from $ OB.$ Inversion center is similarity center of $ (A), (A'),$ with similarity coefficient equal to power of inversion divided by power of inversion center to the inverted circle.

$ \frac {a}{a'} = \frac {a}{r + b} = \frac {4rb}{IA'^2 - (r + b)^2} = \frac {4rb}{\delta^2(A', OB) - 4rb}$

so that distances of $ A', C'$ from $ OB$ are

$ \delta_a = \delta(A', OB) = \sqrt {\frac {4rb(r + b + a)}{a}},\ \ \delta_c = \delta(C', OB) = \sqrt {\frac {4rb(r + b + c)}{c}}.$

$ A, C$ and $ A', C'$ have to be on the opposite sides of $ OB,$ otherwise tangencies of $ (A), (C)$ with $ (O), (B)$ would not be external, so that $ H'J' = A'C' = \delta_a + \delta_c.$ Tangency point $ K'$ of $ (D')$ with $ o'$ is midpoint of $ H'J'$ and tangent length $ K'H'$ of $ (D'), (A')$ yields radius of $ (D')$ as

$ d' = D'K' = \frac {K'H'^2}{4(r + b)} = \frac {(\delta_a + \delta_c)^2}{16(r + b)}.$

Distance of $ D', K'$ from $ OB$ is $ \delta_d = \delta(D', OB) = \frac {_1}{^2}\|\delta_a - \delta_c|.$ This allows to calculate power of $ I$ to $ (D')$ by Pythagorean theorem:

$ ID'^2 - d'^2 = \frac {_1}{^4}(\delta_a - \delta_c)^2 + (2b - d')^2 - d'^2 =$

${ = \frac {_1}{^4}(\delta_a - \delta_c)^2 + 4b^2 - \frac {b}{4(r + b)}(\delta_a + \delta_c)^2}.$

As before, inversion center is similarity center of $ (D), (D')$ with similarity coefficient

$ \frac {d}{d'} = \frac {4rb}{ID'^2 - d'^2},$

${ d = \frac {4rb \cdot d'}{ID'^2 - d'^2}} = \frac {rb (\delta_a + \delta_c)^2}{(r + b)(\delta_a - \delta_c)^2 + 16b^2(r + b) - b(\delta_a + \delta_c)^2}.$

$ \delta_a, \delta_c$ are given in terms of $ r, a, b, c$ and I do not see how to simplify the formula significantly.
[hide]
$ d = \frac {r^2b \left(\sqrt{\frac{r + b + a}{a}} + \sqrt{\frac{r + b + c}{c}}\right)^2}{r(r + b)\left(\sqrt{\frac{r + b + a}{a}} - \sqrt{\frac{r + b + c}{c}}\right)^2 + 4b(r + b) - rb\left(\sqrt{\frac{r + b + a}{a}} + \sqrt{\frac{r + b + c}{c}}\right)^2}$[/hide]


Given $ a,b,c \ge 0$ and abc=1.Prove that:
(a) $ (a\plus{}1)^4(b\plus{}1)^4(c\plus{}1)^4 \ge 512(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)$
(b)$ (x\plus{}1)^3(y\plus{}1)^3(z\plus{}1)^3 \ge 64(x\plus{}y)(y\plus{}z)(z\plus{}x)$
(c)$ 64(x\plus{}1)(y\plus{}1)(z\plus{}1)(x\plus{}y)(y\plus{}z)(z\plus{}x) \ge (x\plus{}1)^4(y\plus{}1)^4(z\plus{}1)^4$
I hope you like them,my friend.
(a) is nicer than (c),but (c) is stronger than (a)

[quote="secrets"]Given $ a,b,c \ge 0$ and abc=1.Prove that:
(a) $ (a \plus{} 1)^4(b \plus{} 1)^4(c \plus{} 1)^4 \ge 512(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)$
(b)$ (x \plus{} 1)^3(y \plus{} 1)^3(z \plus{} 1)^3 \ge 64(x \plus{} y)(y \plus{} z)(z \plus{} x)$
(c)$ 64(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(x \plus{} y)(y \plus{} z)(z \plus{} x) \ge (x \plus{} 1)^4(y \plus{} 1)^4(z \plus{} 1)^4$
I hope you like them,my friend.
(a) is nicer than (c),but (c) is stronger than (a)[/quote]
$ x,y,z?$ my friend :wink:I think $ a\equal{}x,b\equal{}y,c\equal{}z : maybe:$ (a) we have: $ (a\plus{}1)^4 \ge 8a(a^2\plus{}1)$
Similar, we have Q.E.D :)

[quote="secrets"]Given $ a,b,c \ge 0$ and abc=1.Prove that:
(b)$ (x \plus{} 1)^3(y \plus{} 1)^3(z \plus{} 1)^3 \ge 64(x \plus{} y)(y \plus{} z)(z \plus{} x)$
(c)$ 64(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(x \plus{} y)(y \plus{} z)(z \plus{} x) \ge (x \plus{} 1)^4(y \plus{} 1)^4(z \plus{} 1)^4$
I hope you like them,my friend.
(a) is nicer than (c),but (c) is stronger than (a)[/quote]
If $ xyz\equal{}1$ then
(b) $ (x\plus{}1)(y\plus{}1)\equal{}(xy\plus{}1)\plus{}(x\plus{}y) \ge 2\sqrt{(xy(z\plus{}1)(x\plus{}y)}$
Similar ,we have Q.E.D :)
(c) this inequality is false (see(b)) :wink:

[quote="secrets"][quote="quykhtn-qa1"][quote="secrets"]Given $ a,b,c \ge 0$ and abc=1.Prove that:
(b)$ (x \plus{} 1)^3(y \plus{} 1)^3(z \plus{} 1)^3 \ge 64(x \plus{} y)(y \plus{} z)(z \plus{} x)$
(c)$ 64(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(x \plus{} y)(y \plus{} z)(z \plus{} x) \ge (x \plus{} 1)^4(y \plus{} 1)^4(z \plus{} 1)^4$
I hope you like them,my friend.
(a) is nicer than (c),but (c) is stronger than (a)[/quote]
If $ xyz \equal{} 1$ then
(b) $ (x \plus{} 1)(y \plus{} 1) \equal{} (xy \plus{} 1) \plus{} (x \plus{} y) \ge 2\sqrt {(xy(z \plus{} 1)(x \plus{} y)}$
Similar ,we have Q.E.D :)
(c) this inequality is false (see(b)) :wink:[/quote]
 But (c) is true,try it.
[hide] bài này hình như đúng đấy anh ạ,em cũng check thử rùi[/hide][/quote]
Are you sure? :wink:
Since (b),we have: $ (x\plus{}1)^3(y\plus{}1)^3(z\plus{}1)^3 \ge 64(x\plus{}y)(y\plus{}z)(z\plus{}x)$
(c) is equivalent to $ 64(x\plus{}y)(y\plus{}z)(z\plus{}x) \ge (x\plus{}1)^3(y\plus{}1)^3(z\plus{}1)^3$
Then, it is wrong :wink:
@secrets:[hide] em hoc lop may' nhi?[/hide]

[quote="secrets"][quote="quykhtn-qa1"][quote="secrets"]Given $ a,b,c \ge 0$ and abc=1.Prove that:
(b)$ (x \plus{} 1)^3(y \plus{} 1)^3(z \plus{} 1)^3 \ge 64(x \plus{} y)(y \plus{} z)(z \plus{} x)$
(c)$ 64(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(x \plus{} y)(y \plus{} z)(z \plus{} x) \ge (x \plus{} 1)^4(y \plus{} 1)^4(z \plus{} 1)^4$
I hope you like them,my friend.
(a) is nicer than (c),but (c) is stronger than (a)[/quote]
If $ xyz \equal{} 1$ then
(b) $ (x \plus{} 1)(y \plus{} 1) \equal{} (xy \plus{} 1) \plus{} (x \plus{} y) \ge 2\sqrt {(xy(z \plus{} 1)(x \plus{} y)}$
Similar ,we have Q.E.D :)
(c) this inequality is false (see(b)) :wink:[/quote]
 But (c) is true,try it.
[hide] bài này hình như đúng đấy anh ạ,em cũng check thử rùi[/hide][/quote]

$ (c)$ is FALSE, try the values:
$ x \equal{} 2, y \equal{} 1, z \equal{} \frac {1}{2}$ :wink:
Then we have from (c) that:
$ 720\ge729$ which is false. :roll:
[edit] Sorry to secrets, I did not reload the page before replying so I did not know you had corrected the error.
Extremely sorry! :(

[quote="secrets"]Sr,it must be:
$ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(x \plus{} y)(y \plus{} z)(z \plus{} x) \ge 8(x^2 \plus{} 1)(y^2 \plus{} 1)(z^2 \plus{} 1)$
@quykhtn:[hide]Sorry anh,em gõ nhầm đề,thế mà không để ý hic,em học lớp 10,không rõ là anh hay là bạn nữa =))[/hide][/quote]]
Nice inequality,secrets :lol:
Use $ xyz\equal{}1$,then this inequality equivalent to: $ (x\plus{}y)(y\plus{}z)(z\plus{}x)(yz\plus{}1)(zx\plus{}1)(xy\plus{}1) \ge 8(x^2\plus{}1)(y^2\plus{}1)(z^2\plus{}1)$
We have: $ (x\plus{}y)(xy\plus{}1)\equal{}y(x^2\plus{}1)\plus{}x(y^2\plus{}1) \ge 2 \sqrt{xy(x^2\plus{}1)(y^2\plus{}1)}$
Similar,we have Q.E.D :)

[quote="secrets"]Sr,it must be:
$ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(x \plus{} y)(y \plus{} z)(z \plus{} x) \ge 8(x^2 \plus{} 1)(y^2 \plus{} 1)(z^2 \plus{} 1)$
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262448


AMC
AIME
USA(J)MO
USAMO
AMC 8
AMC 10
AIME I

Can we order online? Or do we have to print that order form and send it to the address?

Print the order form and mail in along with payment, or
call 800-527-3690, between the hours of 9:00 am - 4:00 pm (Central time)
Monday through Friday, and order using a credit card.  Orders are usually filled in no more than 2 business days and mailed via U.S. postal mail.

Steve Dunbar
MAA Director, American Mathematics Competitions

Yup. I ordered a copy of that CD. The only ways you can order it is by mail or fax.

It contains all the AMC 8 from 2001-2006, and the AMC 10 A&B, AMC 12 A&B, AIME I & II, USAMO and the IMO Team tests from 2001-2007.

Does it have the solutions or do we have to buy them separte

As well as the contests themselves, the complete and full solution booklets for all contests, exactly as printed, are on the CD.  This includes the solutions for the USAMO problems too.

-- Steve Dunbar
MAA Director, American Mathematics Competitions


When $ 2^{100} \plus{} 3^{100}$ is divided by $ 5$, what is the remainder?

$ 2^4\equiv 1(mod 5)\rightarrow 2^{100}\equiv 1(mod 5)$
$ 3^4\equiv 1(mod 5)\rightarrow 3^{100}\equiv 1(mod 5)$
Remainder: 2

Alternatively, note that $2^2\equiv3^2\equiv -1 \pmod5$.

This becomes $1+1\equiv2\pmod5$ due to even powers.


search
geometry
circumcircle
geometry unsolved

[color=green]Let $ ABCDEF$ be the convex hexagon,such that $ \frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{EF}{FA}\equal{} 1$ and $ \angle ABC\plus{}\angle CDE\plus{}\angle EFA \equal{} 2\pi$.Prove that $ \angle FAE\plus{}\angle ECD \equal{}\angle FBD$.[/color]
I'm sure that this is very famous problem.So if anyone knows where i can find a solution,please give me a link.

Hello all

This looks like a generalization of  Bellavitis'  theorem for balanced quads

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1252704492&t=88077

Seems that for n>4 the description word 'balanced' has a slightly different meaning.

M.T.

[quote="pohoatza"]I guess here the famous problem is the ISL 98, G6 posted here: [url]http://www.mathlinks.ro/viewtopic.php?p=3488#3488[/url].

Now, this similar problem can be done using the same idea of the official solution of G6, i.e. consider $ P$ a point such that $ \triangle{CDP}$ and $ \triangle{CBA}$ are similar and equally oriented and the rest is angle-chasing.[/quote]
Yes,this problem is very similar with ISL 98,but i don't think that this problem could be solved in such easy way,it is much harder.But,anyway thanks  :) (but if you solve it,please post your solution).
P.S.Using my problem you could solve ISL 98. :wink:

[quote="Erken"]Does someone have any ideas or complete solutions?[/quote]
well,as said in the link given,the circumcircles of $ \triangle ABC,\triangle CDE,\triangle EFA$ have a common point $ P$...

now the problem asks us to prove that:

$ \angle FAE \plus{} \angle ECD \equal{} \angle FBD$ which is equivalent to:

$ \angle PAF \minus{} \angle PAE \plus{} \angle PCD \minus{} \angle PCE \equal{} \angle PBF \minus{} \angle PBD$

consider all the angles directed...

so after inversion with center $ P$ we have to prove that:

$ \angle PF'A' \minus{} \angle PE'A' \plus{} \angle PD'C' \minus{} \angle PE'C' \equal{} \angle PF'B' \minus{} \angle PD'B'$

or equivalently:

$ (\angle PF'A' \minus{} \angle PF'B') \minus{} (\angle PE'A' \plus{} \angle PE'C') \equal{} (\angle PD'C' \plus{} \angle PD'B')$

$ \Rightarrow \angle A'F'B' \minus{} \angle C'E'A' \equal{} \angle B'D'C'$

which is true because after inversion we get that $ A',B',C'$ and $ C',D',E'$ and $ E',F',A'$ are colinear and as showed in the link we conlude that $ F',D',B'$ are colinear too...
now the problem becomes trivial,since we have:

$ \angle C'D'B' \equal{} \angle E'D'F'$

and also according to $ \triangle E'D'F'$ we get that:

$ \angle A'F'D' \equal{} \angle A'E'D' \plus{} \angle F'D'E'$

which solves the problem...


function
algebra
polynomial
rational function
complex analysis
complex analysis unsolved

A rational function R is defined to be the quotient of two polynomial functions in which the denominator is not the zero polynomial function.
Prove that if the polynomials in a rational function R have real coefficients, then (R(z))*=R(z*)  where *=conjugate

Monomials $ cz^n$ with $ c\in \mathbb R$ have this property. By summation, the property passes to polynomials, and by division, to rational functions.


calculus
integration
trigonometry
calculus computations

Hello ;) !!!

we have : $ \int \frac{dx}{(x\plus{}1)^5 \sqrt{(x\plus{}1)^2\minus{}1}}$

$ Let$   $ (x\plus{}1)\equal{}y$

than the integral become:

$ \int \frac{dy}{y^5 \sqrt{y^2\minus{}1}}$

than by changing variable that's be very easy to calcul  it ...
______________________________________________________
lahoucine

hello, to derive $ \int\frac{1}{t^5\sqrt{t^2\minus{}1}}\,dt$ we setting $ z\equal{}\frac{1}{t}$ and we get $ \minus{}\int\frac{z^4}{\sqrt{1\minus{}z^2}}\,dz$, setting $ z\equal{}\sin(u)$ we get the integral $ \minus{}\int\sin^4(u)\,du$. This is easy, when we know that
$ \sin^4(u)\equal{}\frac{1}{8}(3\minus{}4\cos(2u)\plus{}\cos(4u))$.
Sonnhard.


Find all triples (x, y, z) of positive integers such that \[ x | y^2 + y + 1,\;y | z^2 + z + 1,\;z | x^2 + x + 1.\]

Did you find it somewhere or were you just wondering about it? I failed to find any triplets except for $(1,1,1),(1,1,3),(1,13,3),(13,3,61)$ and their cyclic permutations, but this was just a small manual search. I was merely trying to find a pattern in case there were infinitely many solutions. There might be more and they might be relatively small numbers, but not as obvious as these ones.


group theory
abstract algebra
superior algebra
superior algebra solved

Let N be a subgroup of the center of G. Show that if G/N is a cyclic group, then G must be abelian

Let $N=\{g_{1},g_{2},\cdots,\}$
Because G/N is cyclic.
We can write:
$G=N \bigcup Ng \bigcup Ng^{2}\bigcup \cdots$
So for every $a,b \in G$
we can write:$a=g_{k}g^{m},b=g_{s}g^{m}$
Because $g_{k},g_{s}\in C(G)$
thus $ab=g_{k}g_{s}g^{m+n}=ba$
Right?


Let $a,b,c>0$ such that $a+b+c+abc=4$.Prove the following inequality:

       $\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq a+b+c$ ;) ;)  ;)  ;)

[quote="cezar lupu"]Let $a,b,c>0$ such that $a+b+c+abc=4$.Prove the following inequality:

       $\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq a+b+c$ ;)[/quote]

 to prove     $\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq a+b+c$

adding both side by $abc$ ,
$\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}+abc\geq 4$ which is equivalent to

$a^3c^2+a^2b^3+b^2c^3+(abc)^3\geq 4(abc)^2$ which is AM-GM


Let a,b,c,d,e be reals such that any three form triangle side lengths.
Prove that at least one triangle is acute angled.

Bomb

Suppose the contrary, i.e., all the triangles formed are non-acute.
WLOG, assume $a\ge b\ge c\ge d\ge e$, then $a^2\ge b^2+c^2\ge 2c^2+d^2\ge 3d^2+2e^2>(d+e)^2$. Contradiction.


analytic geometry
Functional Analysis
topology
advanced fields
advanced fields solved

Let $X$ be a Banach space, and let $M$ be a bounded set in $X$.
Define \[ \alpha(M)=\inf\{\varepsilon>0|\exists\ n\in\mathbb N\mbox{ and }M_1,\ldots,M_n\subset M\mbox{ such that }M\subset\cup_{i=1}^nM_i\mbox{ and the diameter of }M_i\mbox{ is }\varepsilon\} \].

Assume that $X$ is infinitely dimensional. Prove that $\alpha(B(0,1))=2$.

Dude, this problem is so cool! It totally made my day :).

Suppose we have $n$ sets with diameters $<2$ which cover the unit ball of $X$. We may as well assume that the sets are closed, since their closures have the same diameters. Then we can restrict our attention to an $n$-dimensional subspace $X^n$ of $X$, and conclude that the unit ball of $X^n$ is covered by $n$ closed sets of diameter $<2$. Suppose we have shown that $(P)$ holds:

$(P)$ The unit ball $\mathbb D^n$ of $\mathbb R^n$ (with the usual, Euclidean norm) cannot be covered by $n$ closed sets of diameter $<2$.

The contradiction we're after then follows, since $(P)$ basically says that given any symmetric convex body centered at the origin of $\mathbb R^n$ (such as the unit ball of $X^n$, for instance) and any $n$ closed subsets of $\mathbb R^n$ covering it (such as our $n$ closed sets), there is a set which contains two antipodes.

Assume the contrary of $(P)$. For each closed set $M_i,\ i=\overline{1,n}$ among the $n$ which cover $\mathbb D^n$, consider a small enough open set $V_i$ around $M_i\cap\mathbb S^{n-1}$ which still has diameter $<2$. The $V_i$'s are $n$ open subsets of $\mathbb S^{n-1}$, which cover $\mathbb S^{n-1}$, none of which contains two antipodes. We derive a contradiction from this.

Define $\phi_i: \mathbb S^{n-1}\to\mathbb R$ by $\phi_i(x)=d(x,\mathbb S^{n-1}\setminus V_i)$. Each $\phi_i$ is non-negative, and positive exactly on $V_i$. Finally, define $\varphi: \mathbb S^{n-1}\to\mathbb R^{n-1}$ by $\varphi(x)=\left(\frac{\phi_i(x)}{\sum_{j=1}^n\phi_j(x)}\right)_{i=1}^{n-1}$. For any $x\in\mathbb S^{n-1}$ there is an $i\in\overline{1,n}$ such that $x\in V_i\Rightarrow -x\not\in V_i$. If $i\in\overline{1,n-1}$, then the $i$-coordinate of $\varphi(x)$ is non-zero, while the $i$-coordinate of $\varphi(-x)$ is zero. On the other hand, if $i=n$, the coordinates of $\varphi(x)$ do not sum up to $1$, while those of $\varphi(-x)$ do. In any case, $\varphi$ is a continuous map from $\mathbb S^{n-1}$ to $\mathbb R^{n-1}$ which does not map any pair of antipodes onto the same point, and this contradicts the [url=http://en.wikipedia.org/wiki/Borsuk-Ulam_theorem]Borsuk-Ulam Theorem[/url].

[quote="grobber"]Dude, this problem is so cool! It totally made my day :).[/quote]

I thought the same thing the first time I've read the proof. It is one of those problems that bring out beauty in Mathematics.

The proof I know is about just like yours, but it uses (P'): If $\partial\mathbb D^n$ is covered by $n$ closed sets, $F_1,\ldots,F_n$, then at least one of the $F_i$s contains antipodal points.

This proof is also a consequence of Borsuk-Ulam.
Consider $f: \partial\mathbb D^n: \mathbb R^{n-1}$ defined by $f(x)=(d(x,F_1),\ldots,d(x,F_{n-1}))$.
By Borsuk-Ulam, there exists $x\in\partial\mathbb D^n$ such that $f(x)=f(-x)$.
Since the $F_i$s cover $\partial\mathbb D^n$, $x\in F_i$ for some $i=\overline{1,n}$.
If $i<n$, then $x\in F_i\Rightarrow d(x,F_i)=0$. On the other hand, $f(x)=f(-x)\Rightarrow d(x,F_i)=d(-x,F_i)\Rightarrow d(-x,F_i)=0\Rightarrow -x\in F_i$, so $F_i$ contains the pair of antipodal points $\pm x$.
If $i=n$, we can assume then $x\in F_n$ and $x\not\in F_j$ for $j<n$ (otherwise we can apply the previous case).
For $j<n$, $x\not\in F_j\Rightarrow d(x,F_j)>0$. $f(x)=f(-x)\Rightarrow d(-x,F_j)=d(x,F_j)>0\Rightarrow -x\not\in F_j$. So $-x\not\in F_j$ for all $j<n$. Using again that the $F_i$s cover $\partial\mathbb D^n$, we get $-x\in F_n$. So $F_n$ contains the antipodal points $\pm x$.

[b]Remark:[/b] In the proof of (P') we haven't used that $F_n$ is closed, so we can improve the hypothesis by assuming that $\partial\mathbb D^n$ is covered by $n$ subsets, of which at most one is not closed.


If CosA +CosB +CosC +CosACosBCosC =0

Prove Tan(A/2).Tan(B/2).Tan(C/2) = 1 or -1

Note that A,B,C are not the angles of a triangle

This is not difficult, but I think it is still rather Advanced Level than Pre-Olympiad.

In any case, here is a solution:

You consider three angles A, B, C satisfying

  cos A + cos B + cos C + cos A cos B cos C = 0,

and ask for a proof that $\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}$ is either 1 or -1.

In order to prove this, it will be obviously enough to show that

$\left( \tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\right) ^{2}=1$.

But for every angle x, we have

$\tan ^{2}\frac{x}{2}=\frac{\sin ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}=\frac{1-\cos ^{2}\frac{x}{2}}{\cos ^{2}\frac{x}{2}}=\frac{1}{\cos ^{2}\frac{x}{2}}-1=\frac{2}{1+\cos x}-1$
$=\frac{2-\left( 1+\cos x\right) }{1+\cos x}=\frac{1-\cos x}{1+\cos x}$.

Hence,

$\left( \tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\right) ^{2}=\tan ^{2}\frac{A}{2}\cdot \tan ^{2}\frac{B}{2}\cdot \tan ^{2}\frac{C}{2}$
$=\frac{1-\cos A}{1+\cos A}\cdot \frac{1-\cos B}{1+\cos B}\cdot \frac{1-\cos C}{1+\cos C}=\frac{\left( 1-\cos A\right) \left( 1-\cos B\right) \left( 1-\cos C\right) }{\left( 1+\cos A\right) \left( 1+\cos B\right) \left( 1+\cos C\right) }$.

And thus, instead of proving

$\left( \tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\right) ^{2}=1$,

I will show

$\frac{\left( 1-\cos A\right) \left( 1-\cos B\right) \left( 1-\cos C\right) }{\left( 1+\cos A\right) \left( 1+\cos B\right) \left( 1+\cos C\right) }=1$.

This is obviously equivalent to

(1 - cos A) (1 - cos B) (1 - cos C) = (1 + cos A) (1 + cos B) (1 + cos C).

And this expands to

1 - (cos A + cos B + cos C) + (cos B cos C + cos C cos A + cos A cos B) - cos A cos B cos C
= 1 + (cos A + cos B + cos C) + (cos B cos C + cos C cos A + cos A cos B) + cos A cos B cos C,

or, equivalently,

- (cos A + cos B + cos C) - cos A cos B cos C = (cos A + cos B + cos C) + cos A cos B cos C.

But this is clear, since we are given that cos A + cos B + cos C + cos A cos B cos C = 0.

  Darij


[quote="Ragingg"]Your homepage is http://www.artofproblemsolving.com.[/quote]

I like this one. It has been literally true of me for more than a year now.

[quote="Valentin Vornicu"]Great ones frankthetank, but I'd modify some of them as follows:

- you often say AM-GM instead of AM-FM or AM-PM
- apple $\pi$ reminds you of math
- you clean with a mop instead of a vacuum cleaner hoping people will start calling you a mopper
- there's someone on your buddy list whose screen name ends in "ahedron"

:D[/quote]

Props Valentin. Haha...those are great. I'd capitalize MOP in mopper, but otherwise it's awesome.

[quote="MithsApprentice"]Props Valentin. Haha...those are great. I'd capitalize MOP in mopper, but otherwise it's awesome.[/quote]Thanks, but I just made some minor adjustments. 90% of the credit goes to frankthetank :)

It's kind of scary so many of the supposedly-jokes actually describe me..
Is anybody else feeling that way?

[quote="jpark"]You know you're a aopser if...

... you try to write QED at the end of your English paper.
(which is exactly what I did on today's English exam)[/quote]

heh, I think I'm going to start doing that on purpose  :ninja:

On the other hand, in my Lit book they use Halmos symbols ($\blacksquare$) to denote the end of a passage.  I thought that was just a math thing but i must be wrong.

(sorry for going off topic :blush: )

[quote] you think that iff is a word[/quote]
But that is a word.  Ask any Scrabble player (specially those who  challanged me and lost  :) ) Take it from the one who   has been chanlanged  (more than once) by English majors while playing Scrabble! (When I first arrived in US, they never understood how come a non native speaker of English can beat them in Scrabble -  And all funny words like  phi, xi ,  , mu, nu , mho  etc suddenly appear in  Scrabble dictinaries..:) ( I  looked up and learned by heart the spelling of  [i]all[/i] two and three letter words - but that is besides the point..)

Oops.  I looked in an ordinary dictionary, which didn't have iff.  But you're right -- it's in the Scrabble dictionary.  So I take back the entry.

[quote="mathfanatic"][quote="white_horse_king88"]You know you're a aopser if...

You spell aim AIME.[/quote]

Heh, I've done that a lot.  The E seems natural...[/quote]

LOL. Now that's good. I love puns.  :D

[quote="zscool"]...if you consistently generalize homework problems

...you correct your teacher on a regular basis[/quote]

my teacher isn't too happy with me when i correct him. so, now, i just let him screw it up and pretend he has the right answer. :)

So... When does this contest end?  What are these going to be used for? lol

[quote="filletwho"]
...you've spent hours trying to come up with a good answer to this topic.
[/quote]

That's a good one, :lol:


calculus
integration
real analysis
real analysis unsolved

Integrate the following.
$ \int (1\plus{}x^{2}\plus{}x^{4}) \cdot e^{\frac{x^2}{2} \plus{} \frac{x^4}{4}} dx$


Is there a function $f: N \rightarrow N$ such that $f(f(n)) = n^2$?

Yes,i think it is,but to find one on $N$ is a little bit hard.I post a function as soon as I find one.

Clearly,$f$ is injective.we put $f(0)=0,f(1)=1$.Clearly $f(4)$ is defined corectly.We put $f(2)=3$ and we deduce that
 $f(3)=4$ and from here $f(4)=9$.We put $f(5)=6$.Well we assume that $f$ is defined for $k=1,...,n$ and we shall
 define the function for $n+1$.We have two cases:

 In the $ first case $ if $n+1=f(m)$ with $m\leq n$ we have $ f(n+1)=f(f(m))=m^2 $.
 In the $ second case $ if $n+1$ is not equal with $f(m)$ then we consider $p$ with the properties:
 $p\>n+1$ and $p$ isn't equal to $f(m)$ for any $m=1,...n$.We put $f(n+1)=p$ and the coclusion follows.
 Clearly,$f$ is well defined. :P

i've already posted this problem and  Noname did a good solution

if your solution is exactly my solution,then I apologise. :blush:

Its not my solution but Noname solution  :)
And there are infinite solutions!

I don't really understand your solution. Can you define $f(n) = ?$

[quote="Cezar Lupu"]Clearly,$f$ is injective.we put $f(0)=0,f(1)=1$.Clearly $f(4)$ is defined corectly.We put $f(2)=3$ and we deduce that
 $f(3)=4$ and from here $f(4)=9$.We put $f(5)=6$.Well we assume that $f$ is defined for $k=1,...,n$ and we shall
 define the function for $n+1$.We have two cases:

 In the $first case$ if $n+1=f(m)$ with $m\leq n$ we have $f(n+1)=f(f(m))=m^2$.
 In the $second case$ if $n+1$ is not equal with $f(m)$ then we consider $p$ with the properties:
 $p\>n+1$ and $p$ isn't equal to $f(m)$ for any $m=1,...n$.We put $f(n+1)=p$ and the coclusion follows.
 Clearly,$f$ is well defined. :P[/quote]

You make mistake. 0 isn't a positive integer.

The problem didn't mention "positive integers" but some set $\mathbb{N}$. But it's different from country to country if $0$ is in that set or not.
So a definition of it should be given everytime it's used (but well, the solution doesn't make any difference here).


ratio
geometry
similar triangles
angle bisector
geometry proposed

Two different points $A$ and $B$  and a circle $\omega$ that passes through $A$  and $B$  are given. $P$  is a variable point on $\omega$  (different from $A$  and $B$). $M$ is a point such that $MP$  is the bisector of the angle $\angle{APB}$  ($M$ lies outside of $\omega$) and $MP=AP+BP$. Find the geometrical locus of $M$.

Take point $K$ on half of arc $AB$. We have that $KP=(AP+BP)\frac{AK}{AB}$. So we see that locus of the points M is the arc (if point P is only on the one arc).

According to the Ptolemy's Theorem $KP.AB=AK.PB+KB.AP\Rightarrow KP=(AP+PB).\frac{AK}{AB}=MP.\frac{AK}{AB}$ ok.
now, why locus of $M$ is an arc?

Let $B'$ and $A'$ be on $AX$ and $BX$ beyond $X$ such that $AB'=BA'=AB$ and $X$ is the middle of arc $AB$ without $P$
locus of $M$ is the arc $A'B'$ on circle $A'B'X$ without $X$.
hint:
i did it by using the centres  $O_{1},O_{2}$ of the circles $PAB$ and $XA'B'$ and point $D$ which is the intersection of $PX$ and $AB$. It's not so hard it's just some ratio chase that uses similar triangles and $AX=BX$.

Not a different solution, but the complete result is a pair of arcs.

We have $AP+BP$, angle bisector, a circumcircle. This is the special case of Ptolemy.
Let $N$ be the midpoint of $\overarc {AB}$. By Ptolemy,
$PA\cdot BN + PB\cdot AN = PN\cdot AB$
$\Rightarrow (PA+PB)\cdot AN = PN \cdot AB$

$\Rightarrow \frac{PA+PB}{PN} = \frac {AB}{AN} = \text{Constant.}$

Let $A'$ be a point on the extension of the ray $[NA$ such that $AB=A'A$. Since $A,N,B$ are constants, $A'$ is a constant point. We have

$\frac{MP}{PN}=\frac{AA'}{AN} = \frac {AB}{AN}$

So $A'M \parallel AP$, and $\angle A'MN = \angle APN$.
Similarly, construct $B'$ which is constant. We have $\angle A'MB' = \angle APB = \text{Constant}$.
So $M$ is on the arc $%Error. "widearc" is a bad command.
{A'B'}$.
Since there is another $N$, call it $N'$, such that $N'$ is the midpoint of the arc $\overarc{AB}$. Let those points be $A''$ and $B''$.
The complete solution is that the loci are the arc $A'B'$ of $(A'NB')$ not containing $N$ and the arc $A''B''$ of $(A''N'B'')$ not containing $N'$.

Why,   $M$  lies  outside  of  $ \omega $ ?
There are  $A,B$ and $P$ such that  $PA+PB<2R$ and  $ M(.)\in \omega $.


Sassy Fashions buys dresses at wholesale and then marks them up for retail sale. They recently sold a dress at a 40% discount off of their marked-up price. What percent mark-up did they originally apply to the dress if they broke even on the sale? Express your answer to the nearest whole percent.

[hide] $ x(.6)\equal{}1 \implies x\equal{}\frac{5}{3}$ which implies a $ 66\frac23\%$ increase.[/hide]


$k \in \mathbb{N}$
Let $S=k^8+(k+1)^8+(k+2)^8+...+(k+99)^8$
Evaluate two last digits of number $S$

I don't quite understand your solution. Could you explain it more clearly?

I reduced the calculation to some smaller sums (but even a little computation).
[b]1)[/b] You just sum up $100$ consecutive integers (so all residue classes $\mod 100$ once) and so you can take every $100$, especially the smallest ones.
[b]2)[/b] By the chinese remainder theorem it's enough to find it's values $\mod 4$ and $\mod 25$
[b]3)[/b] You can again by the same idea as in 1) reduce the number of calculations you need (I think it can even be done better)


geometry
trapezoid
projective geometry
geometry unsolved

Let $ ABCD$ be a trapezoid such that $ AB$ is perpendicular to both $ BC$ and $ AD$. The circle with $ AB$ as a diameter cut $ CD$ at $ R$ and $ S$. Let $ l_1$ be the line passing through $ D$ and parallel to $ BR$. Let $ l_2$ be the line passing through $ C$ and parallel to $ AR$. Prove that $ l_1$ and $ l_2$ intersect on $ AB$.

The solution is quite straightforward using Pappus theorem.  :P

Lines $ AD$ and $ BC$ intersect at the infinite point of $\perp AB,$ lines $ \ell_1,RB$ intersect at the infinite point of $RB$ and lines $\ell_2,RA$ meet at the infinite point of $RA.$  If $ R ,C, D$ are collinear, then  $ A,B$ and the intersection $ \ell_1 \cap \ell_2$ are collinear too.

Yes, that's a nice solution!

What about without Pappus?

Let $ l_1$ intersect $ AB$ at $ T$ and $ AR$ at $ K$. Extend $ AR$ to meet $ BC$ at $ Q$.
Since $ TD\parallel{}BR$ we have that $ AK\perp TD$. Moreover $ \triangle DAT \sim \triangle ABQ$ thus
\[ \frac {DK}{KT} \equal{} \frac {AR}{RQ} \equal{} [\textrm{since } AD\parallel{}BC] \equal{} \frac {DR}{RC},
\]
which proves that $ CT\parallel{}AR$, as desired.


can anyone write the derivation of f=ma

and explain why the constant after derivation is 1

[quote="bkunchanapalli"]can anyone write the derivation of f=ma
[/quote]

Well what can we assume to prove F=ma?  F=ma inan inertial frame is one of the most fundamental concepts in physics.

Actually, one of Newton's laws says $\vec F = \frac{d \vec p}{dt}$. When $m$ is constant, you get the formula $\vec F = \frac{d \left( m \vec v \right)}{dt}= m \cdot \frac{d \vec v}{dt}= m \vec a$.

But I don't know what constant equal to 1 he's talking about.

I think they defined the constant to be $1$ when they made $Newton$ the unit of force.

[quote="bkunchanapalli"]the formula of force
is
f=kma actually
but why k=1 in si system[/quote]

I believe as Newton orignally stated it, $a\propto F$.

The constant of proportionality is $m$.

[quote="perfect_radio"]Actually, one of Newton's laws says $\vec{F}=\frac{d \vec{p}}{dt}$ ....[/quote]
This is Euler's formula for Newton's 2nd law; useful in fluid mechanics where mass flow is possible.

[quote="Franz Joseph"]Newton found that $F \propto ma$ mass was not the constant of proportionality.[/quote]
The question is not worth quibbling over. After you've done physics for awhile you'll discover that the values of such proportionality constants are irrelevant (they can be scaled to any number you please). The important part is the behavior of the function itself.

For example, one can write Coulomb's law as: $F = \frac{q_{1}q_{2}}{R^{2}}$, the value of $k$ that normally appears is only necessary to reconcile the formula with the system of units.

That said, mass is a multiplicative constant that may be universally eliminated from equations of motion in Newtonian mechanics without affecting the character of said equations, i.e. [b][color=blue]Mass is a constant of proportionality.[/color][/b]


[quote="Abdek"]Let $ a,b$ and $ c$ be  the sides of a triangle $ ABC$ such that $ a^2 \plus{} b^2 > 5c^2$.
Prove that $ c$ is the smallest side of $ \Delta ABC$[/quote]
It is easy
If c is not the smallest side, We can assume that c>b
Then $ a^2 > 4c^2$
Thus, a>2c.Therefore, a>b+c(which is impossible)


geometry
circumcircle
vector
ratio
trigonometry
perpendicular bisector
geometric transformation

$ABC$ is acute-angled. $D$ s a variable point on the side BC. $O_1$ is the circumcenter of $ABD$, $O_2$ is the circumcenter of $ACD$, and $O$ is the circumcenter of $AO_1O_2$. Find the locus of $O$.

Perpendicular bisector of AC e BC are fixed, so the variable is the perpendicular bisector of AD: O_1 moves on a line, such as O_2 and D; I'd like to show thet they move with the proportional velocity, so we can find an omothety...should we use vectors?

let $X$ denote the circumcenter of $ABC$ and $R$ the length of $AX$.

notice that in every position $AO_1O_2$ is spirally similar to $ABC$ through a spiral similarity centered at $A$. let $\theta$ be the angle of the spiral similarity at one of these arbitrary positions. the ratio of that similarity is $\frac{1}{2\cos\theta}$. this makes $AO = R \cdot \frac{1}{2\cos\theta}$. the projection of $AO$ on $AX$ equals (because the angle between $AO$ and $AX$ is $\theta$)

$AO \cos\theta = \frac{R}{2}$.

the locus is therefore the perpendicular bisector of $AX$.

Here's a slightly different way to finish...

Since $\triangle{AO_1O_2}\sim\triangle{ABC}$, we also have $\triangle{AO_1B}\sim\triangle{AO_2C}\sim\triangle{AOX}$ (where as in the previous solution, $X$ is the circumcenter of $\triangle{ABC}$). But $AO_1=BO_1$, so $AO=XO$ as well. Thus $O$ lies on the perpendicular bisector of $AX$.

Conversely, for any point $O$ on the perpendicular bisector of $AX$, the unique spiral similarity about $A$ taking $X$ to $B$ also takes $O$ to a point $O_1$ with $AO_1=BO_1$. This uniquely determines $D$, which uniquely determines the circumcenter $O_2$ of $\triangle{ACD}$.

How do you prove that there is a spiral similarity?


Not only have you posted this before, it's been answered:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=235766[/url]


calculus
integration
geometry
3D geometry
calculus computations

[b]IS THIS CORRECT ?[/b]
[i]please comment[/i]

Calculate the integral
\[\iiint_{V}x^{2}+y^{2}dxdydz \]
where $V$ is bounded by cone $(1-z)^{2}=x^{2}+y^{2}$ and plane  $z=0$

[hide="my try"]

$V$ is a cone with vertex at $(0,0,1)$ and its base is the unit circle in the $xy$-plane.
[i]Method 1:[/i]
\[\iint_{E=\mbox{proj}_{xy}V}\left(\int_{0}^{1-\sqrt{x^{2}+y^{2}}}x^{2}+y^{2}dz\right)=-\iint_{E}(x^{2}+y^{2})(\sqrt{x^{2}+y^{2}}-1)dxdy= \]
\[\ldots={polar subst.}=-\int_{0}^{2\pi}d\varphi\int_{0}^{1}r^{2}(r-1)rdr=-2\pi\left(\frac{1}{5}-\frac{1}{4}\right)=\boxed{\frac{\pi}{10}}\]

[i]Method 2:[/i]
Slicing the cone we have discs with radius $(1-z)$ and the "mass" (contribution) of the scalar field on each disk (there are many  :P ) is integrated with respect to $z$:
\[\int_{0}^{1}\left(\iint_{x^{2}+y^{2}\leq(1-z)^{2}}x^{2}+y^{2}dxdy\right)dz=\int_{0}^{1}\left(2\pi\int_{0}^{(1-z)}r^{3}dr\right)dz= \]
\[\ldots=-2\pi\int_{0}^{1}\left(\frac{(1-z)^{4}}{4}\right)dz=\frac{2\pi}{4}\frac{(1-0)^{5}}{5}=\boxed{\frac{\pi}{10}}\]

[/hide]

Your computations are correct. I would use method 1 myself, but both are reasonable.

Thank you for answering  :) .
By the way, I prefer the first method too.


I am in 8th grade and this is the last year for MATHCOUNTS for me. I have simply one question that my coach brought up last year:

ARe we allowed to use programs on our TI calculators? My friend and I have programmed ours to do calculations we haven't been taught yet (my friend is like 10th grade so he knows more stuff than i do). My coach said it was illegal and demanded to delete it immediately (about 2 weeks hard work) but I was hesitant.

Is it or is it not illegal?

Think about it-how could proctors check anyway? I've always used them.

In fact, they actually allow calculator programs to be used (even at Nationals)!
I know someone who used one of his calculator programs to solve a National Target problem in 2005 (if I remember the year correctly). I thought it was a little unfair, but I guess that's just the way it is, so it's fine for you to use them.  :)
But, of course, don't get too dependent on your calculator; it's not a good idea.  :wink:

I believe that eventually they're gonna tell you no graphing calculators. In like 20 years, you will be able to have counting programs that you can write in under a minute, you can just program some restrictions and it will count for you. That will just completely defeat the purpose of MC.

I don't use calculator programs.

Too bad George isn't here-he'd enjoy this topic...

He only wrote about 6 calculator programs on the free one he got between 2 PM and 9 PM, then he lost the calculator 3 days later  :P

Every year at the coaches dinner the night before Nationals some coach brings this up.  The answer is always the same.  There are too many different types of calculators to assume that the proctors would even know how to erase all the memory.  If students have learned how to program them than good for them.

Mathcounts has publicly stated though that they are trying to write target and team questions for which advanced calculators and programs would not be a major advantage.  For the most part I have found the newer tests to be more thought provoking and less equation or computation heavy.

You'd be better served spending the hours you took writing programs and practicing more problems!

I just don't use them. Its faster just to solve it mentally, than to use a calculator program. But, they are supposed to erase the memory...

It's not faster if you already have a program for something long.

jw, what are good programs to put in if you had that capability?
-jorian

[quote="7h3.D3m0n.117"]I don't use programs a lot; I only use them for problems I haven't been taught how to do (my friend puts them on there).[/quote]
Wouldn't learning how to do them work better for problems that are similar to those kinds?

math has a lot....learning every aspect might take a while

[quote="anirudh"]so what. better than getting caught using one[/quote]
Um, programs are allowed...

how would i get my calc. to do these if i had one of the TIs? what probs do they help on?

Use the TIConnect Software. You need to connect it to your computer.

They help most with combinatorics problems, but George Silvis, I remember, wrote a program that tested for primes.

You can write programs to plug numbers in to solve for things in algebra, like heron's or the quartic or cubic formula or something.

I have a TI-89, and despite it's immense power, The only two special functions I ever really use in contests are one that will factor numbers, which is extremely useful both as a way to factor integers (duh), and as a prime tester (I often find it useful to check both  at the same time.)  And one that will solve equations, which often are most of a problem in the target or team rounds.

If your calculator doesn't already have a function for testing for primes, I would recommend writing one for it.  Writing a function to solve any equation is probably more work than it's worth.  Otherwise, I never really use any of the other (read 400+) special functions on my calc.

As was stated earlier, there isn't much use for functions in MATHCOUNTS.

I mostly use them for plugging in #s for formulas(quartic=helpful). Also, I have a prime tester. That's basically it. And I only put on programs that I understand.

btw-remember when George wrote a program to find squares? That was a pointless program.

I need a TI-89 I'd be so much further in my list of primes. Wait a minute does this only work for the TI-89s of the TIs?


calculus
inequalities proposed
inequalities

a^b//geq//1/2 I can solve it but I want to know other ways to solve :D

sorry for a ,b //geq//0 and kill it without calculus!!! ;)


Given a $ m*n$ typed chessboard. Paint each square black or white such that for all black squares, the number of black squares neighbouring to it is odd.
Show that the total number of black is even. Note that two squares are neighbourhood means that they are distinctive and have a common side.

[hide="solution"]Let $ f$ represent the number of neighboring black squares of a black square. If $ a$ adjacent to $ b$, then $ b$ is adjacent to $ a$. So the value of $ f$ is even.

Assume that the number of black squares is odd. But, $ f$ of a black square is odd. So the total number of $ f$ would be odd. Contradiction, so the number of black squares is even. Q.E.D.[/hide]


calculus
integration
calculus computations

Evaluate the following integral.
$\int_{1}^{\infty}\frac{dx}{e^{x+1}+e^{3-x}}$.

I have an [hide="answer"] $\frac{\pi}{4e^{2}}$[/hide] but am looking to see if people solved it differently and if they got the same answer.

For this problems there is a general technique which is classical:

$\int_{1}^{\infty}\frac{1}{e^{x+1}+e^{3-x}}dx=\int_{1}^{\infty}e^{-x-1}\frac{dx}{1+e^{2-2x}}dx=\int_{1}^{\infty}e^{-x-1}\sum_{n=0}^{\infty}(-1)^{n}e^{(2-2x)n}dx$
$=\sum_{n=0}^{\infty}(-1)^{n}e^{2n-1}\int_{1}^{\infty}e^{-(2n+1)x}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}e^{-2}}{2n+1}=\frac{1}{e^{2}}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}=\frac{\pi}{4e^{2}}$
the last summation equality follows from the expansion

$\arctan(x)=\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)x^{2k+1}}$, by setting $x=1$.
Didi

Just make the substitution $y = e^{x}$ :)
You'll get that the integral is equal to $\frac1{e}\cdot \int_{e}^\infty \frac{dy}{y^{2}+e^{2}}$. We can find an antiderivative of this one, $\frac1{e^{2}}\arctan \left( \frac{x}{e}\right)$.
Hence, the result.


trigonometry
inequalities unsolved
inequalities

Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$\sin\frac{2AB}{A+B}+\sin\frac{2BC}{B+C}+\sin\frac{2CA}{C+A}\leq\frac{3\sqrt{3}}{2}$

[quote="hien"][quote="cefer"]
$\frac{2AB}{A+B}+\frac{2BC}{B+C}+\frac{2CA}{C+A}\leq A+B+C$
[/quote]

It's very easy by using fact: $\frac{2AB}{A+B}\leq\frac{A+B}{2}$[/quote]

You are right... :blush:


Outline methods for the conversion of propyne into:

1) $ CH_3C \equiv C\minus{}CH(CH_3)_2$

2) $ CH_3CH_2CH_2C \equiv CH$

3) $ CH_3C \equiv C\minus{}C_6H_5$

4) 1,2-Dimethyloxirane

why the first two induces the ethyl group and the third is the alkyne zipper reaction

The alkyne zipper reaction cannot be made with NaHO, I thought that was already understood. A stronger base is needed.

The so called "Acetylene Zipper Reaction" is the migration of an internal triple bond to a terminal position on a linear chain, by the action of metal amide as base.

The best base for the Zipper reaction is reported to be $ H_2NCH_2CH_2CH_2NHK$, but sodium amide can also be used.

Thanks. One more question:

The alkyne in question. Can it be any form or is there a specific type of alkyne that works best?

There is no special type of internal alkyne; naturally, however, it cannot contain functional groups that interfere in the reaction (remember we are using a very strong base).

Try to think by yourself a little bit on that last question, thinking in relevant examples. Then if there are still doubts, post them.

To help, I have just posted an example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=182058]here[/url], as the form of an exercise, which shows how the zipper reaction can be used.

my book says to convert a 2-Alyne to 1-alkyne NaOH can be used

Then your book is wrong (one more time). The [i]reverse[/i] process - conversion of a terminal alkyne into an internal one - is best done with a weaker base like NaHO (and even so, its not easy because the reaction can stop at the allene stage, or there can be more than one product). However, conversion of internal triple bonds to terminal ones [i]must[/i] be done with stronger bases - because then the terminal alkyne hydrogen can be removed and that's what makes the reaction favourable.

sorry Carcul i am wrong my book says the same as u but for 2-Alkyne to 1-Alkyne it is NaNH2

Yes, that's right. But it was found that $ H_2NCH_2CH_2CH_2NHK$ in $ H_2NCH_2CH_2CH_2NH_2$ gives the best results.

for the second conversion,

1.treat with NBs
2.perform wurtz reactionwith ethyl chloride.

Is this correct :wink: ????????????????

I think Wurtz rn is seldom used in synthesis as you get many side products.  :)

[quote="Carcul"]Yes, Madness is correct.[/quote]
hey if wurtz reaction cant be done you can always replace it by house corey reaction no??? :D


geometry
3D geometry
sphere
calculus
integration
Pythagorean Theorem
calculus computations

Proof that V = 4/3 :pi: R^3 for a sphere. (In wich V is the volume and R is the radius)
Use pythagorean theorem, (a triple) integral and infinitesimal.

[quote="Myth"]Hmm... I think there are much more appropriate sections for such problems.[/quote]

Like what section? Geometry? You can move it if you like, I don't mind.
I thought since it obtained integral and infinitesimal that it belonged here.

[quote="rhynoceros"][quote="Myth"]I think like this.[/quote]

Did I make a mistake in my sentence....I'm quite bad at English...[/quote]

u can make it by integrals!

[quote]u can make it by integrals![/quote]

Then show us how.


geometry algebra polynomial incenter circumcircle inequalities function

Please post this topic in the right area :blush:
Some problems are easy but some are very hard and nice :D Please post your solutions.
First day
1. All positive integers 1,2,...,121 are arranged in the squares of a 11*11 table. Dima found the product of numbers in each row and Sasha found the product of the numbers in each column. Could they get the same set of 11 numbers?
2. Six members of the team of Fatalia for the International Mathematical Olympiad are selected from 13 candidates. At the TST the candidates got a1,a2,...,a13 points with ai not equal aj if i not equal j.
     The team leader has already 6 candidates and now wants to see them and nobody other in the team. With that end in view he constructs a polynomial P(x) and finds the creative potential of each candidate by the formula ci=P(ai).
For what minimum n  can he always find a polynomial P(x) of degree not exceeding n such that the creative potential of all 6 candidates is strictly more than that of the 7 others?
3. Organizers of a mathematical congress found that if they accomodate any participant in a room the rest can be acomodated in double rooms so that 2 persons living in each room know each other.
    Prove that every participant can organize a round table on graph theory for himself and even number of other people so that each participant of the round table knows both his neigbours.
4. In a triangle ABC A1, B1, C1 are the points where excircles touch the sides BC, CAand AB respectively. Prove that AA1,BB1 and CC1 are the sides os a triangle.

 Second day
1. Sorry if i cant explain very well the problem :blush:  :blush:  :blush:
Is a picture... You have 2 columns of 11 squares in the middle,in the right and in the left  you have columns of 9 squares  (centered on the ones of 11 squares), then columns of 7,5,3,1 squares. Several rooks stand on the table and beat all the squares ( a rook  beats the square it stands in too). Prove that one can remove several rooks such that not more than 11 rooks are left and still beal all the table. :D
2. Given are positive integer n and infinite sequence of proper fractions  x(0)=a(0)/n, ... x(i)=a(i)/(n+i), a(i) is less than n+i. Prove that there exist a positive integer k and integers c(1),....., c(k) such that
 c(1)x(1)+.....+c(k)x(k)=1.
3. I is the incenter of triangle ABC. A circle containing the points B and C meets the segments BI and CI in P and Q respectively. It is known that BP*CQ=PI*QI. Prove that the circumcircle of triangle PQI is tangent to the circumcircle of ABC.
4. For every positive a,b and c such that a^2+b^2+c^2=1 prove the inequality:
Sum a/(a^3+bc) >3 (the inequality is strict)
This was the hardest...all the participants got 0 points :?

The time is 5h for each day.

[quote="cezar lupu"]About the 4-th problem here is a sharper inequality:

    if $a,b,c>0$ such that $a^2+b^2+c^2=1$.Prove that the following inequality:

                $\frac{a}{a^3+bc}+\frac{b}{b^3+ca}+\frac{c}{c^3+ab}\geq\frac{9}{1+\sqrt3}$ ;)[/quote]
[hide="This is wrong"]
After Modifying each term and apply Cauchy-schwartz we get
$\frac{a^4}{a^6+a^3bc}+\frac{b^4}{b^6+b^3ca}+\frac{c^4}{c^6+c^3ab}\geq \frac {(a^2+b^2+c^2)^2}{a^6+b^6+c^6+(abc)(a^2+b^2+c^2)}$

Substitue $a^2+b^2+c^2=1$ we get
$LHS \geq \frac {1}{a^6+b^6+c^6+abc}$

since $a^2b^2c^2 \leq (\frac{a^2+b^2+c^2}{3})^3 = \frac{1}{27}$  so  $abc \leq \frac{\sqrt3}{9}$

and from the  inequality  below (Power Mean)

          if $x,y,z \geq 0$ then $9(x^3+y^3+z^3) \leq (x+y+z)^3$

we have $9(a^6+b^6+c^6) \leq (a^2+b^2+c^2)^3$  so $a^6+b^6+c^6 \leq \frac{1}{9}$  and then
$ a^6+b^6+c^6+abc \leq \frac{1+ \sqrt3}{9}$ so
$LHS \geq \frac {1}{a^6+b^6+c^6+abc} \geq \frac{9}{1+\sqrt3}$
[/hide]

Ciprian, can you tell us the results of the romanian team?

[quote="harazi"]It is wrong. For example, if $b,c$ are very close to 0, the inequality you proved is false (that one obtained after applying Cauchy). Cezar, can you post a solution or as usual you say you're here but you're not? ;)[/quote]
You're Right Harazi. THank very much to notice me:). Actually,I didn't believe that the solution can obtained easily like that. I'll edit my works, hope its right.  Fell free to correct me.. I got some lesson from you.  :)
As a counter-example for the problem take
$a=0.9999999999999849999999999998875$
$b=0.1414213562373095048801688724210\,10^{-6}$
$c=0.1\,10^{-6}$
(using 31 digits rounding aritmatics)
then the inequality in the problem will give $3.121320343559645250872062750576 \geq 3.294228634059947820873508536776
$ which is false.
Cezar maybe you should verify that inequality once again, I have working more than one hour with that and I obtained nothing.

Beginner in your post you said that $9(x^3+y^3+z^3)\leq (x+y+z)^3$,I think it is $\geq$

[quote="cezar lupu"]About the 4-th problem here is a sharper inequality:

    if $a,b,c>0$ such that $a^2+b^2+c^2=1$.Prove that the following inequality:

                $\frac{a}{a^3+bc}+\frac{b}{b^3+ca}+\frac{c}{c^3+ab}\geq\frac{9}{1+\sqrt3}$ ;)[/quote]

That's false ! Take for ex.  $c$ very close to $1$ and $a=b$ ,  so $S = \frac 4{1-c^2 + 2c}+ \frac{2c}{2c^3+1-c^2}$, so $c=1 \Longrightarrow S=3$ :D
 :cool:  Manifestly no  sharper inequality for  the 4-th problem  :cool:

Can you post solutions for problems 4 and 8? I know the solutions of the jury...are short but very hard to find in a contest so i would like to see your solutions :D

ciprian what are the results of the Romanian team at the competition?

Ganea Octavian 29 pct silver
Talau Cristian 27 pct  silvert
Ionescu Ciprian 21 pct bronz

[quote="harazi"]Ciprian, here is my solution for problem 8. It is very ugly (but natural), so I would be glad if you posted the official solution.
  Take $P=abc$ and $x=a^2, y=b^2, z=c^2$. We must prove that $\sum{\frac{x}{x^2+P}>3}$. Clear denominators (sorry, but I had to) and always use that $x+y+z=1$. Thus we must prove that $\sum{x(P+y^2)(P+z^2)}>3(P+x^2)(P+y^2)(P+z^2)$ or equivalently $ P^2+P\cdot \sum{x(y^2+z^2)}+xyz S> 3P^3+3P^2 (1-2S)+3P(x^2y^2+y^2z^2+z^2x^2)+3P^2$ where $S=xy+yz+zx$. Next, observe that $ x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)=xy(1-z)+yz(1-x)+zx(1-y)=S-3P$ and $ x^2y^2+y^2z^2+z^2x^2=S^2-2P$. Thus we must prove that $ P^2+PS-3P^2+PS>3P^3+3P^2-6P^2S+3PS^2-6P^2+3P^2$. But this easily becomes $ 3P^2-6PS+3S^2<2(S-P)$, or $ (S-P)^2<\frac{2}{3} (S-P)$. SInce of course $S>P$ (because $S>xy>xyz=P$) it remains to prove that $S<P+\frac{2}{3}$, which follows by Schur immediately or which can be proved in the following more elegant way: write it as $ (1-x)(1-y)(1-z)<\frac{2}{3}$. It is immediate from AM-GM.
   I hope it's not disgusting as solution.[/quote]

harazi, i think you made a mistake here: you took first $P=abc$; but then you also took $P=xyz$; but $abc$ and $xyz=(abc)^2$ are not quite the same    :(    (to me the soln looked wrong when i first read it because of the incrediblely weak inequality at the end)

i hope i didn't write nonsense above   :P

anyway, congrats to the romanian team; (especially to ciprian for his devilish accomplishment)     :D

and i'd also like to see the official solution to this blooody ineq

Oh, thanks for noticing that, I was also wondering why it is so simple... :blush:

An attempt to a solution of problem $8$..
By letting $x=a^2,y=b^2,z=c^2$ and homogenysing we have to prove the following
$\sum \frac {x(x+y+z)}{x^2+t}\geq 3$ where $t=\sqrt{xyz(x+y+z)}$. which is in turn equivalent to
$(x+y+z) \sum x(y^2+t)(z^2+t)-3(x^2+t)(y^2+t)(z^2+t) \geq 0$ .
Now let $u=x+y+z,v=xy+yz+zx,w=xyz$.
we transform our inequality to
$g(x,y,z)=u(vw+(uv-3w)\sqrt{uw}+u^3w)-3w^2-3(v^2-2uw)\sqrt{uw}-3(u^2-2v)uw-3uw\sqrt{uw}\geq 0$.
Now let's assume that this inequality fails for some $x,y,z$  thus $g(x,y,z)<0$.
Now keep $u,w$ fixed and let's try to vary $v$ in order to minimise $g$. It can run on a set intervals of reals that ensures that the polynomial $p(x)=x^3-ux^2+vx-w=0$ has solutions or equivalently that the line $y=-v$ intersects the graph of function $x^2-ux-\frac  wx$ in three points. Now $g(x,y,z)$ transforms to a quadratic expression in $v$ with leading coefficient negative thus it takes minimum at some interval. However the extremum for $v$ realises when the line $x=-v$ touches  the graph of the function somewhere which implies that the equation has a double root.
This  means that two of $x,y,z$ are equal. So we need to check the relation only for $x=y, z=1$.
This is quite simple to check:
after clearing denominators this expression transforms to
$(x^2+4x-2)\sqrt{2x+1}\geq 4x^2+x-2$.
This inequality can be easily proved as follows:
If $4x^2+x-2\geq 0$ then we shall have  $x\geq \frac {\sqrt 33-1}8>0.6$ and necessarily $x^2+4x-2\geq 0$ and thus by squaring we deduce that we have to prove $2x^4+x^3+24x^2-5x-4>0$ which is clearly true for $x>0.6$.
if $4x^2+x-2<0$ we must look only at the case $x^2+4x-2<0$ which is $x<\sqrt 6-2$.
By squaring we must prove the inequality $2x^4+x^3+24x^2-5x-4<0$ which again easily follows from $x<\sqrt 6-2$.
QED?

I have posted the problems in separate topics (they look to nice to be in the same topic :D) :

[b]Day 1:[/b]

[i] Problem 1[/i] http://www.mathlinks.ro/Forum/viewtopic.php?t=46290

[i] Problem 2 [/i] http://www.mathlinks.ro/Forum/viewtopic.php?t=46293

[i] Problem 3 [/i] http://www.mathlinks.ro/Forum/viewtopic.php?t=63070

[i] Problem 4 [/i] http://www.mathlinks.ro/Forum/viewtopic.php?p=292643#292643

[b]Day 2:[/b]

[i] Problem 1[/i] http://www.mathlinks.ro/Forum/viewtopic.php?t=46291

[i] Problem 2[/i] http://www.mathlinks.ro/Forum/viewtopic.php?t=46295

[i] Problem 3[/i] http://www.mathlinks.ro/Forum/viewtopic.php?t=45653

[i] Problem 4 [/i] http://www.mathlinks.ro/Forum/viewtopic.php?p=292646#p292646

Also, one can download [url=http://www.mathlinks.ro/Forum/resources/files/international_competitions/98/International_Competitions-Tuymaada_Yakut_Olympiad-2005-98-1.pdf]PDF Version[/url] of the problems.

What's the "international olympiad tuymaada" ? I mean who are the contestants and so on.

[quote=ciprian]
 Second day
4. For every positive $a,b$ and $c$ such that $a^2+b^2+c^2=1$ prove the inequality:
$\frac{a}{a^3+bc}+\frac{b}{b^3+ca}+\frac{c}{c^3+ab}>3$[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $w^3$ and $\sum_{cyc}\frac{a}{a^3+bc}>3\Leftrightarrow\sum_{cyc}(a^3b^3c+a^4b^2+a^4c^2+a^3bc)>\sum_{cyc}(a^3b^3c^3+a^2b^2c^2+3a^4b^4+3a^5bc)$.
Easy to see that
$\sum_{cyc}a^3b^3c=-6uw^6+A_2(u,v^2)w^3+B_2(u,v^2)$;
$\sum_{cyc}(a^4b^2+a^4c^2)=-3w^6+A_1(u,v^2)w^3+B_1(u,v^2)$;
$\sum_{cyc}3a^4b^4=3((a^2b^2+a^2c^2+b^2c^2)^2-2a^2b^2c^2(a^2+b^2+c^2))=3(9v^4-6uw^3)^2-6(9u^2-6v^2)w^6=(54u^2+36v^2)w^6+A_3(u,v^2)w^3+B_3(u,v^2)$;
$\sum_{cyc}3a^5bc=36uw^6+A_4(u,v^2)w^3+B_4(u,v^2)$;
Thus, our inequality is equivalent to $f(w^3)<0$, where
$f(w^3)=3w^9+6w^6+(54u^2+36v^2)w^6+42uw^6+A(u,v^2)w^3+B(u,v^2)$.
But it's obvious that $f$ is a convex function.
Hence, $f$ gets a maximal value for an extremal value of $w^3$, which happens in two following cases.
1. $c\rightarrow0^+$. We need to prove that $\frac{1}{a^2}+\frac{1}{b^2}>3$, where $a^2+b^2=1$, which follows from C-S:
$\frac{1}{a^2}+\frac{1}{b^2}=(a^2+b^2)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\geq4>3$;
2. $b=a$, $c=\sqrt{1-2a^2}$, which gives $\frac{2}{a^2+\sqrt{1-2a^2}}+\frac{\sqrt{1-2a^2}}{\sqrt{(1-2a^2)^3}+a^2}>3$, which is also easy.
Indeed, let $a^2=x$. Hence, we need to prove that $\sqrt{1-2x}(6x^2-x+2)>15x^2-12x+2$, where $0<x<0.5$,
which is obvious for $15x^2-12x+2\leq0$. But for $15x^2-12x+2>0$ or $x\in\left(0,\frac{2-\sqrt{\frac{2}{3}}}{5}\right)$
we need to prove that $(1-2x)(6x^2-x+2)^2>(15x^2-12x+2)^2$, which is true for $x\in\left(0,\frac{2-\sqrt{\frac{2}{3}}}{5}\right)$.
Done!  :)


[quote="M4RI0"]$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\geq 0$

$(a,b,c,d>0)$[/quote]
The inequality is equivalent to:
$\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\geq 4$
Maybe this is simpler to prove...

$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\geq \frac{4(a+c)}{(b+c)+(d+a)}+\frac{4(b+d)}{(c+d)+(a+b)}-4=0$

by Cauchy inequality .

PS .The above problem belongs to Vasc .Also the following conjecture belongs also to Vasc.

$a,b,c,d,e>0$  then prove or disprove that
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq 0$

$a,b,c,d,e>0$  then prove or disprove that
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq 0$[/quote]
we can use Cauchy inequality :
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq$ $\frac{((a-b)+(b-c)+(c-d)+(d-e)+(e-a))^{2}}{ac+bd+ce+ad+eb-a^{2}-b^{2}-c^{2}-d^{2}-e^{2})}$ $=0$

Sorry but you used Cauchy with denominators non specified sign and the solution is wrong  :(

[quote="M4RI0"] Prove the inequality for nonnegative reals $a,b,c,d$: \[{a-b \over b+c}+{b-c \over c+d}+{c-d \over d+a}+{d-a \over a+b}\ge 0.\] [/quote]

[hide=" :!: "]
Terrible bad, I solved this problem long time ago (maybe last year). I tried it last night, but failed. Luckily, a look to the original helped. [/hide]

[b]Solution:[/b] \[{a-b \over b+c}+{b-c \over c+d}+{c-d \over d+a}+{d-a \over a+b}\ge 0 \\ \iff{a+c \over b+c}+{b+d \over c+d}+{c+a \over d+a}+{d+b \over a+b}\ge 4.\] Since the wellknown ${1 \over x}+{1 \over y}\ge{4 \over x+y}$, we obtain ${a+c \over b+c}+{c+a \over d+a}\ge{4(a+c) \over a+b+c+d}$ and ${b+d \over c+d}+{d+b \over a+b}\ge{4(b+d) \over a+b+c+d}$. Take sum of these two inequalities, we get the desired one.
Done.

PS: there was so much solutions here!
Silouan problem is Rearrangement?

[quote="silouan"]Also the following conjecture belongs also to Vasc.

$ a,b,c,d,e > 0$  then prove or disprove that
$ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} e} \plus{} \frac {d \minus{} e}{e \plus{} a} \plus{} \frac {e \minus{} a}{a \plus{} b}\geq 0$[/quote] For $ (a,b,c,d,e)\to(1,0,2,0,2), LHS\to \minus{} \frac16$ if I'm right.

[quote="spanferkel"]For $ (a,b,c,d,e)\to(1,0,2,0,2), LHS\to \minus{} \frac16$ if I'm right.[/quote]

a,b,c,d,e [b]>[/b]0

Yes, but the LHS is a continuous function, so you can replace 0 with some small $ \varepsilon$ without changing the value by more than, say $ \frac{1}{12}$.  But in fact the arithmetic is wrong: $ \frac{1 \minus{} 0}{0 \plus{} 2} \plus{} \frac{0 \minus{} 2}{2 \plus{} 0} \plus{} \frac{2 \minus{} 0}{0 \plus{} 2} \plus{} \frac{0 \minus{} 2}{2 \plus{} 1} \plus{} \frac{2 \minus{} 1}{1 \plus{} 0} \equal{} \frac{5}{6}$.

[quote=M4RI0]$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\geq 0$

$(a,b,c,d>0)$[/quote]
[url=https://artofproblemsolving.com/community/c6h1983621p13795466]Dutch IMO TST1 2012 p2[/url]
[url=https://om.mimuw.edu.pl/static/app_main/camps/oboz2017.pdf]Poland 2017, MO Science Camp[/url]


Hi, I have a question about showing the following, could anyone help? Thx!!

Let $X$ be a subset of vector space $V$.
Show that $\langle x \rangle = \langle\langle x \rangle\rangle$

Is that any other way to show by not using the terms of bases?

I guess no, because the definition of the generated vector spaces uses linear combinations, and in order to talk about linear combinations, you need to use a base I guess ...

You do mean $\langle X\rangle$ right?

Well what is the definition of $\langle A\rangle$?  If it is what I think it is : the smallest subspace containing $A$, then I think this is obvious and requires no bases? :maybe:

Well, true. As a definition I took the set of all linear combinations of $X$. But actually, I think your definition is the one currently used.


[hide]we know that Delia sits at one end. Since E (I'll use initials) sits two seats away and has an arrangement AB between the two, we can conclude that D has to be sitting at the left end.

DABE_ _
This should be easy, Abraham sits next to Delia[/hide]

[quote="hwenterprise"]Mr. Cortes is trying to determine where each student in his class should sit. He knows that he wants Abraham, Brian, Cynthia, Delia, Eugene and Faye to sit in the six chairs in the first row. [color=blue]Using the following criteria, determine who sits next to Delia:[/color]

-Eugene sits next to Cynthia but not Faye
-Abraham's nickname, ABE, appears in the seating chart if you only use the first letters of each person's name.
-Faye and Delia sit at each end.
-There are two seats between Delia and Eugene.[/quote]

Abraham. [url]http://www.phpbbforfree.com/forums/aplethora-post-42.html#42[/url]

[size=75][color=darkblue]Misleading tag removed by moderator RCV.[/color][/size]


logarithms
inequalities proposed
inequalities

$ a,b,c \in ]0,1[$ and $ x,y,z \in ]0,\plus{}\infty[$ such that $ a^x\equal{}bc$, $ b^y\equal{}ac$, $ c^z\equal{}ab$

Then prove:

$ \frac {1}{2\plus{}x} \plus{} \frac {1}{2\plus{}x} \plus{} \frac {1}{2\plus{}x} \le \frac{3}{4}$

some one knows where it was proposed, please?

[quote="FOURRIER"]$ a,b,c \in ]0,1[$ and $ x,y,z \in ]0, \plus{} \infty[$ such that $ a^x \equal{} bc$, $ b^y \equal{} ac$, $ c^z \equal{} ab$

Then prove:

$ \frac {1}{2 \plus{} x} \plus{} \frac {1}{2 \plus{} x} \plus{} \frac {1}{2 \plus{} x} \le \frac {3}{4}$

some one knows where it was proposed, please?[/quote]

With $ a,b,c \in \left( {0,1} \right)$ then

$ x \equal{} \frac {{\ln b \plus{} \ln c}}{{\ln a}},y \equal{} \frac {{\ln a \plus{} \ln c}}{{\ln b}},z \equal{} \frac {{\ln a \plus{} \ln b}}{{\ln c}}$

now we have to prove the follow ineq with

$ \frac {a}{{2a \plus{} b \plus{} c}} \plus{} \frac {b}{{a \plus{} 2b \plus{} c}} \plus{} \frac {c}{{a \plus{} b \plus{} 2c}} \le \frac {3}{4}$

with $ a,b,c \in \left[ {0, \plus{} \infty } \right]$

And it is easy  :)


geometry
geometric transformation
rotation
trigonometry

Given a point $P$, inside an equilateral triangle $ABC$, such that $|AP|=3, |BP|=4$ and $|CP|=5$. find the length of the side of the triangle ABC.

[hide]
try rotation NOT herons formula for areas[/hide]

Rotate the triangle around $A$ for $60^\circ$ so that $C$ falls onto $B$. If the image of $P$ is $P'$, then $P'A=3, P'B=5,\angle PAP'=60^\circ$, meaning that $\triangle PAP'$ is equilateral, and $\triangle BPP'$ is $3-4-5$, meaning that $\angle BPP'=90^\circ$. Hence $\angle APB=\angle APP'+\angle P'PB=150^\circ$. By Cosine Law in $\triangle APB$, $a^{2}=3^{2}+4^{2}-2\cdot 3\cdot 4\cdot\cos 150^\circ=25+12\sqrt{3}$, giving $a=\sqrt{25+12\sqrt{3}}$


for $n\in \mathbb N$ define $f(n)$ to be the number of ways $n$ can be expressed as an ordered sum of at least one positive integer. for example $f(4)=8$, since $4=1+3=3+1=2+2=2+1+1=1+2+1=1+1+2=1+1+1+1$
find, with proof, a closed expression for $f(n)$.

Isn't this just the number of compositions of $n$?

$f(n) = 2^{n-1}$

See http://en.wikipedia.org/wiki/Composition_%28number_theory%29

my solution is like this:
denote the "dot-representation" of $n$ by n dots $***\cdots**$
and we have $1+2=*+**=***=3$ etc.
now $n=***\cdots**$
call the free space between each consecutive pair of $*$s a "[i]gap[/i]"
there are $n-1$ "[i]gap[/i]"s in $n$'s dot representation. and we would like to fill the "gap"s by "+"s. so how many ways can we choose at least $0$ "gap"s from a set of $n-1$ gaps? the ans is simply $\binom{n-1}{0}+\binom{n-1}{1}+\cdots+\binom{n-1}{n-1}=2^{n-1}$

$f(n)$ is equivalent to the number of ways one can place $)+($ in the $n-1$ empty spaces of 1_1_..._1.  You either put one in an empty space or you don't; there are 2 choices.  As a result, $f(n)=2^{n-1}$.  This problem is not at the Olympiad level.  A better problem is to consider a permutation of the sums equivalent rather than distinct, that is, $1+3,3+1$ are counted once.


[hide]
Let $a = p_{1}p_{2}\cdots p_{m}$ and $d = q_{1}q_{2}\cdots q_{n}$ where all those $p$'s and $q$'s are prime. Then $b$ and $c$ each contains some $p$'s and some $q$'s and they are disjoint. Let $r_{1}, r_{2}, \ldots, r_{k}$ be the $p$'s that $b$ has and $s_{1}, s_{2}, \ldots, s_{l}$ be the $p$'s that $c$ has. Finally, let $R = \prod r_{i}$ and $S = \prod s_{i}$. Then

$a+b+c+d = \left(R+\frac{c}{S}\right)\left(S+\frac{b}{R}\right)$

since $RS$ has all the $p$'s so $a = RS$ so $\frac{bc}{RS}= \frac{ad}{a}= d$. Since the terms are all at least $1$, the sum is not prime. [/hide]

The same idea in somewhat more accessible manner:

[hide]${a\over b}={c\over d}$. Let that ratio be equal to ${p\over q}$, where $p,q$ are coprime. Then for some $k$ we have $a=kp, b=kq$, and for some $l$ we have $c=lp, d=lq$. Hence $a+b+c+d=kp+kq+lp+lq=(k+l)(p+q)$. Since both factors are $\geqslant 2$, the conclusion follows.[/hide]


how to prove sin4^x = 3/8 - 4/8 cos2x + 1/8 cos4x

Your formula isn't correct.  Plugging in $x=0$ gets you a false equality.

Perhaps you mean something like $(sin x)^4$

[quote="fourierseries"]Perhaps you mean something like $(sin x)^4$[/quote]

That is what that means.


prove that circumcenters of six triangles that formed by three medians of triangle lie on a circle.

There is a solution to your problem... I think it is very interesting magazine. Look at the first problem in this section. I think you will like it more...


I've got the following problem in mathematica and wonder if anyone could help:

I want to write a function with vector valued argument in mathematica. For example

$ f(x,y) = x\times y$ where ${ x = {x_1,x_2,x_3},y = [y_1,y_2,y_3}]$ are vectors

What should be the codes in mathematica?

Hi;

Simplest seems:

f[x_?ListQ,y_?ListQ]:=Cross[x,y]

As you know Mathematica uses a list to represent a vector. So enter x and y accordingly.

The syntax should be the same as if you were taking scalar variable inputs:

f[x_,y_]:=Cross[x,y];

To call the function, you would have to put in appropriate inputs, without which the function will result in an error.

[b]Input:[/b] f[{1,2,3},{3,4,5}]
[b]Output:[/b] {-2, 4, -2}