geometry
circumcircle
incenter
trapezoid
conics
ellipse
power of a point

Let $ I$ be the incenter of $\triangle ABC$ and $\omega$ the circle passing through $ A$ and tangent to $ BC$ at the foot $ W$ of the internal bisector of $\angle A.$ Perpendiculars to $BC$ through $ B,C$ cut  external bisector of $ \angle A$ at $ B', C',$ respectively and the lines $ CB'$ and $ BC'$ meet the circle $ \odot (IBC)$ again at $ M,N.$ Show that the perpendicular lines through $ W$ to $ CB'$ and $ BC',$ the tangent to $\omega$ at $ A$ and the radical axis of $\omega , \odot (IBC)$ bound a complete quadrangle whose Newton line passes through the circumcenter of $ \triangle AMN.$

Let $ P \equiv BC' \cap CB'.$ From trapezoid $ BB'C'B,$ $ \frac {BP}{PC'} \equal{} \frac {BB'}{CC'} \equal{} \frac {AB}{CA} \equal{} \frac {BW}{WC}$ $ \Longrightarrow$ $ PW \perp BC.$ Let $ W' \equiv PW \cap B'C',$ then $ P$ is midpoint of $ WW'.$ Let $ AI$ cut circumcircle $ (O)$ of $ \triangle ABC$ again at $ X.$ Let $ X' \equiv OX \cap B'C',$ then $ \triangle AWW' \sim \triangle AXX'$ are centrally similar with center $ A.$ Their circumcircles $ w \equiv (P), (O)$ are therefore internally tangent at $ A.$ Let $ t \perp PA$ be their single common tangent at $ A.$

Let $ (X)$ be circumcircle of $ \triangle IBC.$ Let $ BB'$ cut $ (X)$ again at $ B''$ and let $ M \equiv B''W \cap CB'.$ $ W$ is diagonal intersection of $ BMCB''$ and $ B' \equiv BB'' \cap CM$ is on polar of $ W$ WRT $ (X)$ $ \Longrightarrow$ $ M \in (X)$ and $ MW \perp CB'.$ Similarly, if $ N$ is foot of perpendicular from $ W$ to $ BC',$ then $ N \in (X).$

Let $ (Q)$ be circle with diameter $ PW,$ then $ M, N \in (Q).$ Let $ (S)$ be circumcircle of $ \triangle AMN.$ $ \{M, N\} \equiv (Q) \cap (S)$ $ \Longrightarrow$ $ P$ is inside of $ (S).$ Inversion in $ (X)$ takes $ BC$ to $ (O),$ $ W$ to $ A$ and $ (Q)$ to $ (S).$  Let $ K$ be inversion image of $ P$ in $ (X),$ which lies on the radical axis $ r$ of $ (X), (P)$ and $ PK \perp r.$ Since $ P \in (Q),$ it follows that $ K \in (S).$ Since $ A, K, M, N \in (S)$ are concyclic pedals of $ P$ in the quadrilateral $ (t, r, WM, WN),$ $ (S)$ is pedal circle of an ellipse $ \mathcal E$ with one focus $ P$  inscribed in this quadrilateral. Projecting this ellipse to its pedal circle $ (S)$ by a parallel projection keeps $ S$ in place and takes $ (t, r, WM, WN)$ with its Newton line $ n$ to a quadrilateral tangential to $ (S)$ and its Newton line $ n'.$ Since $ S \in n',$ it follows that $ S \in n.$

Thanks dear Vladimir for such an elegant proof. I knew that you were going to take this problem!. My solution is very similar, I'll post it next time. :)

Let $ D, F$ be the points where $\omega$ cuts $ AB, AC.$ Simple angle chasing yields

$ \angle ADF \equal{} \angle AWF \equal{} \angle ABC \ , \ \angle AFD \equal{}\angle AWD \equal{} \angle ACB$

Thus, $\triangle ADF \sim \triangle ABC,$ which implies that $\omega$ and the circumcircle $ (O)$ of $\triangle ABC$ are internally tangent through $ A.$ Antipode of $I$ WRT $ (K) \equiv \odot (BIC)$ is the A-excenter $I_a$ and due to $ (A,W,I,I_a)=-1,$ it follows that $ (K) \perp \omega.$ As result,  radical axis $ p_a$ of $ \omega, (K)$ is the polar of $ K$ WRT $\omega.$

On the other hand, center $ J$ of $\omega$ is the intersection of the lines $ BC'$ and $ CB'.$ For a proof see the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265876]If and only if[/url]. Inversion WRT $ (J)$ takes  $ (K)$ into itself  due to the orthogonality. Thus, it takes $ (O)$ into the circle $ (O') \equiv \odot (AMN).$ Polars of $ B,C$ with respect to $\omega$ are the perpendiculars $ p_b, p_c$ through $ W$ to $ BC',CB'$ and  tangent $\tau$ to $\omega$ at $ A$ is the polar of $A.$ Since $ A,$ $B,$ $C,$ $K$ lie on $ (O),$  their polars $ \tau,p_b,p_c,p_a$ are tangent to the polar conic $ \mathcal{E}$ of $ (O)$ WRT $\omega,$ which in this case is an ellipse since $ J$ lies inside $ (O).$ The center of $ \mathcal{E}$ is precisely the center $ O'$ of the inverse circle of $ (O)$ in the inversion WRT $ (J),$ but according to Newton's theorem, the center of all inscribed conics in the quadrangle $\tau,p_b,p_c,p_a$ lies on its Newton line.


Let $a,b,c>0$ such that $a^2b^2 +b^2c^2 +c^2a^2=1$.Prove that

                     $a+b+c+ab+bc+ca\geq3$    ;)

It looks elegant Cezar!! :)

Could you give a hint?? ;)

Are you sure your ineq true ,Cezar lupu ?  :huh:

I know Schur's inequality, but I don't know Vornicu-Schur's inequality. Please, if you know, tell and me. Thank.

Well,nttu I haven't checked it in particular cases,but I think it is true.If you find a counter example :clap2:

Levi,I'm sure Vornicu-Schur's inequality is at Inequalities Theorems and Formulas ;) .By the way Vornicu-Schur's
inequality is an extension of Schur's inequality.I think mr. Sorin Radulescu and mr. P.Alexandrescu told me about
this inequality a few years ago,but I'm not sure.I will ask them once again  ;)

[quote="harazi"]Of course it is not a new inequality, Cezar. Don't imagine that such results are unknown in the literature... ;)[/quote]

 Well,did you see it somewhere? :huuh:

I was not speaking about the inequality you posted, Cezar.  ;)

Sorry,...about Vornicu-Schur you were talking about.I agree with you on that.There aren't new inequalities in literature,especialy elementary ones. ;)

cezare, i thing the ineq is strictly grater than 3

For Socrates:

I think the crucial fact is the following surprising, but easy inequality:

[color=blue][b]Theorem 1.[/b] If a, b, c are three positive reals, then $\left(a+b+c\right)^4>16\left(b^2c^2+c^2a^2+a^2b^2\right)$.[/color]

[i]Proof.[/i] Since the inequality in question is symmetric, we can WLOG assume that c is the greatest among the numbers a, b, c. Then, $c\geq a$ and $c\geq b$, so that $c^2\geq ab$.

Now, by the AM-GM inequality, applied to the numbers a + b and c, we have $\left(\left(a+b\right)+c\right)^2\geq 4\left(a+b\right)c$. In other words, $\left(a+b+c\right)^2\geq 4\left(a+b\right)c$. Squaring this, we obtain $\left(a+b+c\right)^4\geq 16\left(a+b\right)^2c^2$. But

$\left(a+b\right)^2c^2=\left(a^2+2ab+b^2\right)c^2=b^2c^2+c^2a^2+2abc^2$,

and $c^2\geq ab$ yields $2abc^2\geq 2abab=2a^2b^2>a^2b^2$, so that $\left(a+b\right)^2c^2>b^2c^2+c^2a^2+a^2b^2$, and thus $\left(a+b+c\right)^4\geq 16\left(a+b\right)^2c^2>16\left(b^2c^2+c^2a^2+a^2b^2\right)$. Thus, Theorem 1 is proven.

Now, in order to solve the problem, we note that Theorem 1, applied to the numbers a, b, c, yields $\left(a+b+c\right)^4>16\left(b^2c^2+c^2a^2+a^2b^2\right)=16\cdot 1=16$, so that $a+b+c>2$. On the other hand, the condition $b^2c^2+c^2a^2+a^2b^2=1$ implies that the numbers $b^2c^2$, $c^2a^2$, $a^2b^2$ are all < 1; hence, their square roots bc, ca, ab are also < 1, and thus $bc>\left(bc\right)^2$, $ca>\left(ca\right)^2$ and $ab>\left(ab\right)^2$, so that $bc+ca+ab>\left(bc\right)^2+\left(ca\right)^2+\left(ab\right)^2=b^2c^2+c^2a^2+a^2b^2=1$.

Hence, $a+b+c+bc+ca+ab=\left(a+b+c\right)+\left(bc+ca+ab\right)>2+1=3$, and the problem is solved.

  Darij

Indeed, very surprising this "theorem 1". I did not post it about one year and half before in the general case with $ a^n b^n+b^n c^n+c^n a^n$...


Pick two subset $ A_1$ and $ A_2$ each has $ m$ elements from set $ P$ with $ n$ elements, $ A_1 \cap A_2 \equal{} B$, how many $ B$ have exactly $ k$ elements?
Generalized version
Pick $ j$ subset $ A_1,A_2....A_j$ each has $ m$ elements from set $ P$ with $ n$ elements, pick $ c$ sets from those subsets form $ S$. $ \bigcap S \equal{} B$. how many $ B$ have exactly $ k$ elements?
$ j \le {n \choose m}$
$ j,m,n,c,k$ are parameters.

Part (1)
The set $ B$ with $ k$ elements can be selected in $ n\choose k$  ways. We shall reconstruct the ordered pair $ (A_1,A_2)$ in the follwing way: we pick $ 2(m\minus{}k)$ elements among left out $ n\minus{}k$ elements and partition that set into two parts. Hence,

\[ {n\choose k} {{n\minus{}k}\choose {2(m\minus{}k)}} {{2(m\minus{}k)}\choose {m\minus{}k}}\]
Part(2)
This takes a messy form.
If $ c$ of the subsets among $ j$ subsets have some $ k$ elements common, then we have to distribute(a element can be more than a set) rest $ n\minus{}ck$ elements among $ c$ sets so that no element is present in every set and every set recieves exactly $ m\minus{}k$ elements. This can be done in the following number of ways.
\[ {n\choose k}\bigg( \sum_{i\equal{}0}{(\minus{}1)^i{{n\minus{}ck\minus{}i\choose m\minus{}ck\minus{}i}\choose c}}\bigg)\]
We are left to reconstruct the rest $ j\minus{}c$ subsets which can be done in $ {{{n\choose j}\minus{}c}\choose {j\minus{}c}}$ ways. Hence,
\[ {n\choose k}{{{n\choose j}\minus{}c}\choose {j\minus{}c}}\bigg( \sum_{i\equal{}0}{(\minus{}1)^i{{n\minus{}ck\minus{}i\choose m\minus{}ck\minus{}i}\choose c}}\bigg)\]


It takes Amelia five hours to mow the yard and it takes her brother Tom 7.5 hours to mow the same yard. If they have two lawn mowers, how many hours will it take for them to mow the yard together?

In one hour, they do $ \frac{1}{5}\plus{}\frac{2}{15}\equal{}\frac{1}{3}$ of the yard. So 3 hours.


Prove that for natural numbers $a < b < c < d$ with $ad=bc$ we have $\frac{(a-d)^{2}}{2}\ge a+2$  :)

What do $b$ and $c$ have to do with this?

It is not my own problem, i took it from a book of Valentin Vornicu. :)

As far as i can see, $\frac{(a-d)^{2}}{2}\ge a+2$ is supposed to be true only for a specific set of numbers, and that's $a < b < c < d$ with $ad=bc$.

For example, the inequality isn't valid for $a=3$ and $d=6$, but for $a=2$, $d=6$ (and obviously $b=3$ and $c=4$) everything's fine.

Still no answer? I think it's in the wrong section since it's from an olympiad level book :maybe:


How many perfect square with 200 digits that the first 99 digits are 9?

Its factorial + 1 is a perfect square.

There are only 3 such numbers so it should be easy to guess.   (and it is not 4)   [hide](and not 7 either)  so I would guess it is five   [/hide]

{ Hint Hint above is the world's most useless hint }


algebra
polynomial
linear algebra
linear algebra unsolved

Hi,

in a paper I found a claim I somehow do not see a reason for. Does anyone of you?

Let $ A\in \mathbb{R}^{n\times n}$ be a symmetric matrix.
Let the trace of $ A$ be zero. Then the smallest (remember
that all eigenvalues are real) of the eigenvalues of A is not
strictly positive.

I know that the trace appears as the coefficient of the second-to-largest
power in the characteristic polynomial of A. But I do not see how
to deduce the claim.

The only theorem about signs of real zeroes of univariate polynomials
with real coefficients that I know is:

Let $ f\in \mathbb{R}[x]$ have only real roots, i.e. let it factor entirely
over $ \mathbb{R}$. Then
All roots of $ f$ are strictly negative $ \Longleftrightarrow$ All coefficients of $ f$ are strictly positive

and

All roots of $ f$ are strictly positive $ \Longleftrightarrow$ None of the coefficients of $ f$ vanishes and they alternate in sign.

This does not help me in this case.

Does anyone of you see the reason?

Salut,

Sébastien

The sum of the eigenvalues is equal to the trace.


What is the remainder when $22^{50}+15^{50}$ is divided by $13$?

Well...

[hide]eli mods say 3, but i doubt thats right.  cause i didnt make them for this stuff.[/hide]


Let polynomial $ p(x)$ have just real zeros, and $ a\in R$ is multipli zero of $ p'(x)$.
Prove that $ p(a)\equal{}0$.

[quote="suadnovic"]Let polynomial $ p(x)$ have just real zeros, and $ a\in R$ is multipli zero of $ p'(x)$.
Prove that $ p(a) \equal{} 0$.[/quote]

Let $ p(x)\equal{}c(x\minus{}a_1)^{p_1}(x\minus{}a_2)^{p_2}...(x\minus{}a_n)^{p_n}$ , where $ a_1,a_2,...,a_n$ are distinct real numbers and let $ p_1\plus{}p_2\plus{}...\plus{}p_n\equal{}m$ . Clearly $ \exists a_1<b_1<a_2 , a_2<b_2<a_3 , ... , a_{n\minus{}1}<b_{n\minus{}1}<b_n$ so that $ b_1, b_2, ... , b_{n\minus{}1}$ are roots of $ p'(x)$. Comparing the degree of $ p'(x)$ we have $ (b_1,b_2,...,b_{n\minus{}1})$ is unique and $ b_i$ are all single roots of $ p'(x)$

Now assume that $ p(a) \neq 0$ then $ a\equal{}b_k$ for some $ k \in \{1,2,...,n\minus{}1 \}$
Then $ p'(a)\equal{}0$ and $ p''(a)\equal{}0 \Rightarrow a$ is not a single root of $ p'(a)$ , contradiction.

So $ p(a)\equal{}0$


Prove that among the numbers of the arithmetic progressions 3,7,11,15,19,23,... and 5,11,17,23,29,35,... there are an infinite number of primes.

In other words, that there are an infinite number of primes of the form $4k+3$ and $6k+5$.  Both well-known exercises in similar veins.

[hide="Huge hint"] Any number of the form $4k+3, 6k+5$ must have a divisor of that form. [/hide]

proof by contradiction will be a good way to approach this right?

Yes.  In fact, both proofs have essentially the same form.  Now, a harder question:  Can you prove that there are also an infinite number of primes of the form $6k+1$?  (The question of the primes of the form $4k+1$ was addressed in a recent thread.)

[hide="slightly better (12k+7)"]
We know there is a prime that is 7 mod 12 (7, for instance). Now suppose there are finitely many such primes. Let $P$ be their product and consider
\[Q=3P^{2}+4\]
We have $Q\equiv 7\pmod{12}$. On the other hand, all prime divisors of $Q$ must be 1 mod 6. (For those of you who know about Legendre Symbols, this is simple enough. To do it without this...
[hide]
let $p$ be a prime divisor of $Q$. Clearly $p=2$ and $p=3$ can't happen. Then $3P^{2}\equiv-4\pmod{p}$ so letting $m=2P\cdot 3^{-1}$, we have $m^{2}\equiv-3\pmod{p}$. Now let
\[x=2^{-1}(1+m)\]
We get $x^{2}-x+1\equiv 0\pmod{p}$. Therefore $x^{6}\equiv 1\pmod{p}$. Also $x^{p-1}\equiv 1\pmod{p}$ since $m\not \equiv-1\pmod{p}$. If $p\not\equiv 1\pmod{6}$, then the order of $x$ modulo $p$ divides both $6$ and $p-1$, so it divides 2. Therefore $x^{2}\equiv 1\pmod{p}$, then $x^{2}-x+1\equiv 0\pmod{p}$ implies $x\equiv 2\pmod{p}$, and then $2^{2}\equiv 1\pmod{p}$. This is bad![/hide]

In addition, not all of $Q$'s prime divisors can be 1 mod 12, since that would imply $Q\equiv 1\pmod{12}$. Therefore, $Q$ has a prime divisor $p\equiv 7\pmod{12}$. But this means $p|P$ and $p|Q$, so $p|4$, a contradiction! This means infinitely many primes of form $12k+7$.
[/hide]


geometry
conics
parabola
search
Putnam
calculus
calculus computations

A square has side length $ 2$. Find the area of the region consisting of all points closer to the center of the square than to any of its sides.

Is the answer $ \frac{4}{3} (4\sqrt{2}\minus{}5)$ ?

[quote="Nora.91"]Is the answer $ \frac {4}{3} (4\sqrt {2} \minus{} 5)$ ?[/quote]That is correct. May I see your solution?

Actually i am also getting the same answer ie $ \frac{4}{3}(4\sqrt{2}\minus{}5)$ but while doing this i took the sides of the square as ( -1,1) , (1,1) , ( -1 , 1) , ( 1 , -1) . so how can we actually do this without assuming any of these vertices .

[quote="vijaymenon"]Actually i am also getting the same answer ie $ \frac {4}{3}(4\sqrt {2} \minus{} 5)$ but while doing this i took the sides of the square as ( -1,1) , (1,1) , ( -1 , 1) , ( 1 , -1) . so how can we actually do this without assuming any of these vertices .[/quote]

It's perfectly fine to place the square that way. No matter where you place the square, it doesn't change the outcome.

you can use symmetrical arguments

The 4 parabola that close up the region are

$ x^2\pm2y\minus{}1\equal{}0$

$ y^2\pm2x\minus{}1\equal{}0$

If you graph the 4 lines and $ x\equal{}\pm1$, $ y\equal{}\pm1$

You'll see the complete diagram

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=141840028&t=253140
and linked from there is
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1095827834&t=106690

(I found that by searching for "putnam parabola" without quotes.)


I am getting 253.Hell lotsa fight for 2 parents to handle I say. :roll:

A perfect example as to why the population of India is increasing at such an alarming rate  :lol:

Anyway, the answer is 191 i think?

Take the number of children conceived by Pramila "thanks to" narendra = $ y$  :rotfl:  :rotfl:

$ x\plus{}x\plus{}1\plus{}y \equal{} 24$

also, total number of fights :
$ F \equal{} x(x\plus{}1) \plus{} x.y \plus{} y(x\plus{}1)$

substitute $ y$ in terms of $ x$ and solve $ \frac {dF}{dx}$ = 0. $ x$ turns out to be 7.5, which is not possible (unless, of course, there are siamese twins :D)
So, substitute $ x\equal{}7$ and $ x\equal{}8$. Both cases give the answer $ 191$


geometry
inequalities proposed
inequalities

If K' is the area of the triangle with sides  \sqrt a, \sqrt b, \sqrt c  prove that

        4K' \geq  \sqrt 3 K

where K is the area of the triangle with sides a,b,c.

Its  well  known  that  [tex] 16K^2=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4) [/tex]Aply  this  formula  to  the  triangle  with sides  [tex] \sqrt{a} , \sqrt{b} , \sqrt{c}  [/tex] and  find [tex] 16K'^2 = 2(ab+bc+ca)-(a^2+b^2+c^2)[/tex] . So [tex]4K'^2\geq\sqrt{3}K [/tex]  after  computations  gives  [tex]a^4+b^4+c^4+abc(a+b+c)\geq\
a^3(b+c)+b^3(c+a)+c^3(a+b) [/tex]  which  is  Shur.


algebra polynomial limit inequalities logarithms calculus trigonometry

[quote="Lovasz"]Compare the two number: $e^{\pi}$ and $\pi^e$.[/quote]

It would be interesting to find solution without calculus.....
Cause I have not seen a such one yet....... :D

[quote="Lovasz"]Compare the two number: $e^{\pi}$ and $\pi^e$.[/quote]
That remember me the trascendence of $e^e$ adnd or similar (above all about polynomials but i think that series could be too advanced for this section :D )

 can we use the definition of $e=\lim_{x\to 0} {(x+1)}^{\frac{1}{x}}$ or use complex numbers?
if we can't,
since $e<\pi$ and both the numbers are \geq e, the one with the least base is the greatest one (that's simple to show that by the rearrangement inequality), so $e^\pi$ is the highest one.

 if we can't use neither these arguments, let's try to mayorize $\pi^e$ and minimize $e^{\pi}$ taking the logarithm of the two sides and dividing by $\pi$: show that $\ln \pi \leq \frac{e}{\pi}$

Define $x\ .s.s.\ y\ (x,y$ have the same sign $)\Longleftrightarrow$ $sgn\ x=sgn\ y$ $\Longleftrightarrow$ $x=y=0\ \vee\ xy>0\ .$
$\left(e^{\pi}-\pi ^e\right)\ .s.s.\ \left(\pi \ln e-e\ln \pi \right)\ .s.s.\ \left(\frac{\ln e}{e}-\frac{\ln \pi}{\pi }\right)\ .s.s.\ \left[f(e)-f(\pi )\right]$,
where $f: (0,\infty )\rightarrow R,\ f(x)=\frac{\ln x}{x}\ .$ But $f'(x)\ .s.s.\ 1-\ln x\ .s.s.\ e-x\ .$
Therefore, $e<\pi\Longrightarrow f(e)>f(\pi)\Longrightarrow \boxed {\ e^{\pi}>\pi ^e\ }\ .$

Virgil Nicula, the question asked for NON caluculus proof, if possible :)

Why ? I don't agree with you say. I don't give indications, I solve the proposed problem clearly, rigorously and utterly. I suppose that Lovasz has no against to this form of the reply. You can ask him. I didn't calculate $e^{\pi}$ and $\pi ^e$ (with the computer or ... with my pen) ! I wish my proposed problems to be solved like these solutions ... which you don't agree.

[b]Other exercises from the same family.[/b] ([u]without calculus ![/u] )
$\sin 50^{\circ}<\frac{\sqrt 2}{2}+\frac{\pi}{36}\ \ ;\ \ 3^{\sqrt 5}<5^{\sqrt 3}\ ;\ \boxed {\ \mathrm {(own)}\ \ \frac{\pi}{2}+\sqrt 2<3\ \ \ ;\ \ \pi +2<3\sqrt 3\ \ \mathrm {(own)}\ }\ .$

$\sqrt [3] 2+\sqrt [3] 5<\sqrt [3] 3+\sqrt [3] 4\ ;\ \sqrt [3] 5+\sqrt 2<\sqrt [3] 4+\sqrt 3\ ;\ \sqrt [3] 3+\sqrt [5] 5<\sqrt [3]5+\sqrt [5] 3\ .$

$\ln(n-1)\ln (n+1)<\ln ^2n\ \ ;\ \ (e+2)^{e-2}>(e-2)^{e+2}\ \ ;\ \ \ln$$\left( e^{\pi}-1\right) \cdot \ln \left( e^{\pi} +1\right) <\pi ^2\ .$

[quote="socrates"]It would be interesting to find solution without calculus.....
[/quote]

Actually, this is what I referred to. And even for a calculus solution, yours is way too complicated. It's enough to examine the function $f(x)=x^{1/x}$. Its derivative is $f'(x)=x^{1/x}{{1-\ln x}\over{x^2}}$ and it's obvious that it is negative for $x>e$. Hence $\pi>e\implies \pi^{1/\pi}<e^{1/e}\implies\pi^e<e^{\pi}.$

I have no commentary w.r.t. the comparison between these two solutions. And yet, why $x^{\frac 1x}$ and not $\frac{\ln x}{x}$ ? It is same thing. And yet, for my function, the derivative is more easy calculated than your. Serves you right ! Sold again ! You didn't understand why my solution shows such. [b]Only one way ?[/b]

I'll repeat, use the rearrangement inequality that says that $\prod x_i^{a_i}$ arrive to the maximum when $x_i>x_j\rightarrow a_i>i_j$ and the fact that $x>y\rightarrow y^x\geq x^y$. If you want to proceed to calculus, show that $x^\frac{1}{x}$ reach the maximum when $x=e$

The rearranjement inequality (if is truly what you say) is proved with derivatives, i.e. ... with calculus !

Indeed, Virgil Nicula and Farenhajt both showed exactly when $x^y\geq y^x$.

...and solution without calculus is a trouble...
happy asian new year  :lol:  :D

[quote="ThAzN1"][quote="JohnConstantine"]the fact that $x>y\rightarrow y^x\geq x^y$[/quote]

Umm? $x = 3, y = 2$ gives $8 = 2^3 \geq 3^2 = 9$[/quote]

But $2<e$...an intuitive proof of the rearrangement inequality exixts, i know :?


Is there any unconventional method to find out the relation between group velocity and phase velocity? know there is a method employing Fourier tarnsforms and another easier method as well.


Let $ x,y,z>0$,such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx,t\geq0$,prove that:
$ (x \plus{} y \plus{} z)^2 \geq \frac {3}{1 \plus{} t}(\sqrt {(x^2 \plus{} t)(y^2 \plus{} t)} \plus{} \sqrt {(y^2 \plus{} t)(z^2 \plus{} t)} \plus{} \sqrt {(z^2 \plus{} t)(x^2 \plus{} t)}).$

BQ

thank you chan_...
BQ

[quote="xzlbq"]Let $ x,y,z > 0$,such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx,t\geq0$,prove that:
$ (x \plus{} y \plus{} z)^2 \geq \frac {3}{1 \plus{} t}(\sqrt {(x^2 \plus{} t)(y^2 \plus{} t)} \plus{} \sqrt {(y^2 \plus{} t)(z^2 \plus{} t)} \plus{} \sqrt {(z^2 \plus{} t)(x^2 \plus{} t)}).$

BQ

thank you chan_...
BQ[/quote]

$ t\geq \frac {4}{5}$

can use the power mean inequality proof.we need only prove

$ (x \plus{} y \plus{} z)^4 \geq \frac {27}{(1 \plus{} t)^2}\sum (x^2 \plus{} t (\frac {xy \plus{} yz \plus{} zx}{x \plus{} y \plus{} z})^2)(y^2 \plus{} t(\frac {xy \plus{} yz \plus{} zx}{x \plus{} y \plus{} z})^2).$

 :lol:


calculus integration trigonometry geometry perimeter conics ellipse

evaluate
$\int\frac{dx}{1+\sin(x)+\cos(x)}$

also
find the value of $\int_{0}^{2\pi}\sqrt{a^{2}\sin(x)+b^{2}\cos(x)}dx$

The second one is the perimeter of an ellipse, and doesn't have a simple form.

The first is tricky, but I think I've done it before.

[hide="Hint"] Consider multiplying numerator and denominator by expressions of the form $1 \pm \sin x \mp \cos x$. [/hide]

The second one (as written) is not the perimeter of an ellipse. Note that it is $\sin x$ and $\cos x,$ not $\sin^{2}x$ and $\cos^{2}x.$

That means that the integral is not defined as a real integral, since the function under the radical is negative over half the period.

For the first one, just use the (standard) substitution $y = \tan\frac{x}2.$

[quote="maokid7"]Evaluate $F=\int\frac{1}{1+\sin x+\cos x}\ dx\ .$[/quote][color=darkblue]Come on gambol without substitutions and only with the primitive relation $\int f(\phi (x))\phi '(x)\ dx=F(\phi (x))+C$, where $F=\int f(x)\ dx$
and the function $f$ is remarkablely. Suppose $\sin x+\cos x\ne 1$ for any $x$ from the integration domain of the function $f(x)=\frac{1}{1+\sin x+\cos x}\ .$ Thus,
$f(x)=\frac{(\sin x+\cos x)-1}{(\sin x+\cos x)^{2}-1}=$ $\frac{\sin x+\cos x-1}{2\sin x\cos x}=$ $\frac{(\sin x-\cos x)'}{1-(\sin x-\cos x)^{2}}-\frac{1}{2\sin x\cos x}=$ $-\frac{(\sin x-\cos x)'}{(\sin x-\cos x)^{2}-1}-\frac{1}{2}\cdot \frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x}=$ $-\frac{(\sin x-\cos x)'}{(\sin x-\cos x)^{2}-1}+\frac{1}{2}\cdot \left[\frac{(\cos x)'}{\cos x}-\frac{(\sin x)'}{\sin x}\right]\ .$ Therefore, $\boxed{\ F=\frac{1}{2}\cdot\ln\left|\frac{\cos x(1+\sin x-\cos x)}{\sin x(1+\cos x-\sin x)}\right|+C\ }\ .$

[b]Remark.[/b] Denote $\tan\frac{x}{2}=t\ .$ Then $\sin x=\frac{2t}{1+t^{2}}$, $\cos x=\frac{1-t^{2}}{1+t^{2}}$ and $\frac{\cos x(1+\sin x-\cos x)}{\sin x(1+\cos x-\sin x)}=$ $\frac{(1-t^{2})(1+t^{2}+2t-1+t^{2})}{2t(1+t^{2}+1-t^{2}-2t)}=$
$\frac{1}{2}\cdot (1+t)^{2}=$ $\frac{1}{2}\cdot\left(1+\tan\frac{x}{2}\right)^{2}\ .$ Therefore, $F=\frac{1}{2}\ln\left[\frac{1}{2}\cdot\left(1+\tan\frac{x}{2}\right)^{2}\right]+C=$ $\ln\left|1+\tan\frac{x}{2}\right|+C\ .$[/color]

[color=darkred]Like you the entertainment ? Was an amusing game ? For me, yes.[/color]

For de second one, consieder that

$a \sin x+b\ cos x \, =\, \sqrt{a^{2}+b^{2}}\, \sin \left(x+\arctan \frac{b}{a}\right)$

and you will get

$\int \sqrt{a \sin x+b\ cos x}\, \cdot dx \, = \,-2 \sqrt{2 \sqrt{a^{2}+b^{2}}}\, E \left( \frac{1}{\sqrt{2}}\, , \, \arccos \sqrt{\frac{a \sin x+b \cos x}{\sqrt{a^{2}+b^{2}}}}\right) \,+\, 2 \sqrt{2 \sqrt{a^{2}+b^{2}}}\, F \left( \frac{1}{\sqrt{2}}\, , \, \arccos \sqrt{\frac{a \sin x+b \cos x}{\sqrt{a^{2}+b^{2}}}}\right)$

or, if you don´t have probles with elliptic functions witn moludule greater than one,

$\int \sqrt{a \sin x+b\ cos x}\, \cdot dx \, = \,-2 \sqrt{\sqrt{a^{2}+b^{2}}}\, E \left( \sqrt{2}\, , \, \frac{\pi-2x}{4}-\frac{1}{2}\arctan \frac{b}{a}\right)$


If a convex polygon has 90 distinct diagonals, how many sides does it have?

$ \frac{n(n\minus{}3)}{2}\equal{}90$

$ n^2\minus{}3n\equal{}180$

$ n^2\minus{}3n\minus{}180\equal{}0$

$ (n\minus{}15)(n\plus{}12)\equal{}0$

As n is positive, we have 15.


Can somebody solve this:

Prove that natural numbers $m$ and $n$ such that

$m(m+1)(m+2)(m+3)=n(n+1)^2(n+2)^3(n+3)^4$

do NOT exist.

Edited by pbornsztein to make it readable.

Thanks to pbornsztein cos i cant still use latex


abstract algebra
group theory
superior algebra
superior algebra unsolved

Maybe this is trivial :
Let $G$ be a group such that $G \approx \mbox{Aut } ( G )$.
Does it follow that there is an injective homomorphism $\pi : G \longrightarrow S_n$ with $n$ a positive integer such that $n! > |G|.$ ?

Something seems to be wrong here (I assumed that $G$ is supposed to be finite): such homomorphism exists for any group (with e.g. $n=|G|$)  ...  :?

Heh, sorry. I tried to generalize, and as you pointed out, obtained a (really) trivial statement
Let $G$ be a group such that $G \approx \mbox{Aut } ( G )$ and $|G|=n!$ for some positive integer $n$.
Does it follow that there is an isomorphism $\pi : G \longrightarrow S_n$.
I guess this is also quite trivial. I'm asking since I'm really bad when it comes to dealing with $S_n$.
(In particular I want to prove that a group $G$ with $|G|=6$ and $Z(G)={e}$ (then $G \approx \mbox{Inn } (G)$ hence $G \approx \mbox{Aut }(G)$) is isomorphic to $S_6$).

[quote]
(In particular I want to prove that a group $G$ with $|G|=6$ an $Z(G)={e}$ (then $G \approx \mbox{Inn } (G)$ hence $G \approx \mbox{Aut }(G)$) is isomorphic to $S_6$[/quote]

how would being isomorphic with Inn(G) explain being isomorphic with Aut(G)

Anyway, the best and much more general way to do this is learning semidirectproducts :

[url]http://www.mathreference.com/grp,sdp.html[/url]

[quote="{x}"]Thanks ! I'll read through all of that.
I profoundly believe that $|\mbox{Aut }(G)| \leq |G|$, am I wrong ? (I guess yes :))[/quote]

That is definitely incorrect : consider any prime p and the abelian group

$\mathbb{Z}_p\times\mathbb{Z}_p\times \mathbb{Z}_p$

the number of automorphisms is the number of linearly independent triples over $\mathbb{Z}_p$ which is $(p^3-1)(p^3-p)(p^3-p^2)$

that is always more than $p^3$

automorphisms usually have a high number compared to the number of elements

 If it is of interest to you, the groups of form $\mathbb{Z}_p\times\cdots\times\mathbb{Z}_p$  for a prime p , are the only finite groups were all nontrivial elements are 'equivalent' : the automorphism group works transitive on the nontrivial elements


in masale ro chand vaght pish tooye roozname khoondam:(dar mosahebe ba Mr parvis shahriari :D )
$4$ noghteye motamayez bar rooye yek khat darim morabayi ra biabid ke emtedade azlaash az in $4$ noghte bogzarand. :roll:

[quote="fatemeh"][quote="mohamadhosein"][quote="fatemeh"]farz konid in noghat be tartib A,B,C,D bashand. hala farz konid AB=x, CD=y va zavieie f=arctan(y/x) bashe.
hala age az A,C khotooti ba zavieie f (nesbat be khat) va az B,D khotooti ba zavieie 90-f kharej konim, charzeleeie hasel az barkhorde khotoot, ye morabba'e :D[/quote]
hala mitooni ino ba khata kesh va pargar rasm koni? :roll:[/quote]
too sooratesh nagofte boodi faghat az khatkesho pargar mishe estefade kard  :?  :D[/quote]

manzoor az tarsim hendesi yani estefade az khatkesh va pergar.

dayere mosalasati ro dar nazar begir. roye mehvare tangent-ha tooli mosavi ba $\frac{y}{x}$ joda kon.darnatije $\angle AOB=\text{Arc}\tan\frac{y}{x}$

[quote="amir2"][/quote][quote="fatemeh"][quote="mohamadhosein"][quote="fatemeh"]farz konid in noghat be tartib A,B,C,D bashand. hala farz konid AB=x, CD=y va zavieie f=arctan(y/x) bashe.
hala age az A,C khotooti ba zavieie f (nesbat be khat) va az B,D khotooti ba zavieie 90-f kharej konim, charzeleeie hasel az barkhorde khotoot, ye morabba'e :D[/quote]
hala mitooni ino ba khata kesh va pargar rasm koni? :roll:[/quote]
too sooratesh nagofte boodi faghat az khatkesho pargar mishe estefade kard  :?  :D[/quote][quote="amir2"]

manzoor az tarsim hendesi yani estefade az khatkesh va pergar.

dayere mosalasati ro dar nazar begir. roye mehvare tangent-ha tooli mosavi ba $\frac{y}{x}$ joda kon.darnatije $\angle AOB=\text{Arc}\tan\frac{y}{x}$[/quote]
avvalan tarsime hendesi ma'nish in nist. ma'nish ine ke hamechizhaii ke mikhaim bekeshim ma'loom bashan. darzemn age faghat khatkesh va pargaar dashte bashi nemitooni andazeie y/x ro joda koni !!! (y/x mitoone kharej az mahdoodeie darajebandie khatkesh bashe masalan gong bashe)  :roll:

manzooram inbood ke baraye rasmhaye hendesi faghat pargar va [b]khatkeshe gheire modaraj[/b] dar ekhtiyar darim.

mohem ham nist ke $\frac{y}{x}$ gong bashe ya na.

khob age khatkeshet modarraj nabashe, che joori andazeie y/x ro joda mikoni?

dar tarsimhaye hendesi farz bar in ast ke toole vahed ra niz hamishe dar ekhtiyar darim.

ba tavajoh be shekle zir $z=\frac yx$
($AC||BD$)

baba asan man hamoon harfe avvalamo kamel mikonam ke in moshkelat pish naiad:
az noghteie B ye amood bar khate ABCD kharej mikonim. ba'd be andazeie y azash joda mikonim va noghteie hasel ro P minamim. hala age az A be P vasl konim, zavieie BAP, hamoon f ast.(chon AB=x va tan(PAB)=y/x) ke hameie in kara ro ba khatkesh va pargar (albatte gheire foroorikhtani) mishe kard.hala az B ye khat movazie hamin khat kharej mikonim. hala khate d ro khati amood bar AP dar nazar begirid. az C,D do khat movazi ba khate d rasm mikonim.
hala hameie in karharo ba khatkeshe GHEIRE MODARAJ va pargar anjam dadim :D

va in ham rahe hale man:

farz konid $EFGH$ morabae mafrooz bashad.$A\in EH\ ,\ B\in EF\ ,\ C\in FG\ ,\ D\in HG$ morabae $EFGH$ ra nesbat be markaze an $90^{\circ}$ davaran bedahid.
oonvaght khate $AC$ be khate $A'C'$($A'\in HG\ ,\ B'\in EF$) va amood bar an tabdil khahad shod.hal az $D$ khati movazi ba $A'C'$ bekeshid ta $EF$ ra dar $D'$ ghat konad.darim $DD'=AC\ ,\ DD'\perp AC$
pas ma noghteye $D'$ darim ,faghat kafist $D'$ ra be $B$ vasl konim .va az $D$ movaziye $D'B$ va az A va C amood bar an rasm konim.

niazi ham nist ke $A,B,C,D$ rooye yek khat bashand.


Very easy. If we put $f(a,b)=LHS-RHS$, we have $f(a,b)=f(b,a)$ and $f(-a,b)=f(a,b)$. So we can assume wlog $a\ge b\ge0$. So we have just to prove $2(1+a^2)(1+b^2) \ge 2a\cdot2$ for $ab\le1$ and $2(1+a^2)(1+b^2) \ge 2a\cdot2ab$ for $ab\ge1$. Both obvious.


linear algebra
matrix
vector
LaTeX
linear algebra unsolved

Let $A \in \mathcal M_3 \left( \mathbb R \right)$. Prove that there is $n \in \mathbb N^{\ast}$ s.t. $\left( A + A^{\ast} \right)^{2n} = A^{2n} + \left( A^{\ast} \right)^{2n}$ iff $\det A = 0$.

In the above, $A^{\ast}$ reffers to the "romanian" adjoint, i.e. the transpose of the matrix of the cofactors, which verifies $A A^{\ast} = \left( \det A \right) I_3$.

[i]Romanian Mathematical Olympiad, 2003, 11th Grade, Problem 4, by Laurentiu Panaitopol[/i]

The Romanian adjoint is the french comatrice ,l'adjoint in french has an other meaning and it's relating to linear algebra too:).
If det(A)=0 then the conclusion is directly deduced from Newton formulae.
Now we are working in M_3(R) thus the matrix A has at least one real eigenvalue "r",let's suppose that det(A)<>0 then r is must be different from  0 .
For X a vector of R^3 such that AX=rX we have A^(-1)X=(1/r)X so A*X=(det(A)/r)X
The equality of the exercice means that there is a natural number n such that (r²+det(A))^2n=r^(4n)+det(A)^2n s=r² so s>0 and (s+det(A))^2n=s^2n+det(A)^2n
so we deduce that s and det(A) have different signs wich means det(A)<0
But |s-|det(A)||<=max(s,|det(A)|)  and for s & |det(A)|>0  s^2n+|det(A)|^2n>max(s,|det(A)|) ^2n  Thus we have max(s,|det(A)|) ^2n>max(s,|det(A)|) ^2n :CONTRADICTION so det(A)=0.

[quote="MoHaMeD"]The Romanian adjoint is the french comatrice ,l'adjoint in french has an other meaning and it's relating to linear algebra too:).[/quote]

On this forum, I've seen many people call the adjoint the conjugate of a transpose of a matrix.

I $\LaTeX$ed your solution. It's very hard for me to follow it, in the way you wrote it.

[quote="MoHaMeD"]If $\displaystyle \det \left( A \right) = 0$ then the conclusion is directly deduced from Newton formulae.

Now we are working in $\displaystyle \mathcal M_3 \left( \mathbb R \right)$ thus the matrix $\displaystyle A$ has at least one real eigenvalue $\displaystyle r$, let's suppose that $\displaystyle \det \left( A \right) \neq 0$ then $\displaystyle r$ is must be different from  $\displaystyle 0$.

For $\displaystyle X$ a vector of $\displaystyle \mathbb R^3$ such that $\displaystyle AX=rX$ we have $\displaystyle A^{-1} X = \frac{1}{r} X$ so $\displaystyle A^{\ast} X= \frac{\det \left( A \right)}{r} X$.

The equality of the exercice means that there is a natural number $\displaystyle n$ such that $\displaystyle \left( r^2 + \det \left( A \right) \right)^{2n} = r^{4n} + \left( \det \left( A \right) \right)^{2n}$. $\displaystyle s=r^2$ so $\displaystyle s>0$ and $\displaystyle \left( s + \det \left( A \right) \right)^{2n} = s^{2n} + \left( \det \left( A \right) \right)^{2n}$,
so we deduce that $\displaystyle s$ and $\displaystyle \det \left( A \right)$ have different signs wich means $\displaystyle \det \left( A \right) < 0$.

But $\displaystyle \left| s - \left| \det \left( A \right) \right| \right| \leq \max \left( s , \left| \det \left( A \right) \right| \right)$  and for $\displaystyle s \, \& \, \left| \det \left( A \right) \right| >0$, $\displaystyle s^{2n} + \left| \det \left( A \right) \right|^{2n} > \max \left( s , \left| \det \left( A \right) \right| \right)^{2n}$.

Thus we have $\displaystyle \max \left( s , \left| \det \left( A \right) \right| \right)^{2n} > \max \left( s , \left| \det \left( A \right) \right| \right)^{2n}$ :CONTRADICTION so $\displaystyle \det \left( A \right)=0$.[/quote]

The solution is nice ... :thumbup:

And I think that we could replace $3$ with any other $2k+1$.

The problem given in the olympiad had also a point (a), which leads you to a nice solution to point (b). Namely, if $f \in \mathbb R[X]$ has no real roots, then $f \left( A \right) \neq 0$. This is a pretty unfair problem to give on an olympiad. I solved point (a) pretty easily using eigenvalues, but those are not in the syllabus, and the "non-eigenvalues" solution is very hard to find. (at least for me)


prove that if $ m,n>1$ have the same parity then $ 2^m\minus{}1 \nmid 3^n\minus{}1$.

http://www.mathlinks.ro/Forum/viewtopic.php?p=1148144#1148144

( This is for $ m,n$ odd, for $ m,n$ even the problem is obvious since $ 2^m\minus{}1$ is divisible by $ 3$)


AoPS community member Mildorf will be hosting a second Mock AIME contest on Sunday, August 8th, at 2:00 PM Eastern time. If you are not available for this date, you may still take the test earlier by letting Mildorf know via PM. Further information about this contest can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14772]here[/url].

I encourage all problem solvers of the Intermediate forum to check out this contest! It will be great practice for the real thing next winter.

Hi,

 I just found out about this site yesterday, and I'm being flooded with more math than ever!
 I'm really interested in this mock AIME, but I obviously just missed it.  So is it possible to still get it "PM"ed?  and what is "PM" anyways?

Go and look around in the AMC forum.  The questions should still be lying around in the Mock AIME thread.

PM = private message. All users have some sort of message box on this site.


induction
combinatorics unsolved
combinatorics

Let $t(A)$ denote the sum of elements of a nonempty set $A$ of integers, and define $t(\emptyset)=0$. Find a set $X$ of positive integers such that for every integers $k$ there is a unique ordered pair of disjoint subsets $(A_{k},B_{k})$ of $X$ such that $t(A_{k})-t(B_{k}) = k$.

The set $X$ is the set of the powers of $3$. Let's prove by induction that this set satisfies the conditions required. I'll let the base of induction to the end of the proof.
Suppose that we have done for each positive integer less than or equal to $\frac{3^{m}-1}{2}$ for some $m$ positive integer. Let $k$ be a positive integer such that $\frac{3^{m}-1}{2}< k \leq 3^{m}$. If $l =3^{m}-k$ then there exists $(A_{l}, B_{l})$ such that $t(A_{l})-t(B_{l}) = l$, for $l \leq \frac{3^{m}-1}{2}$.  Then let $A_{k}= (B_{l})U(3^{m})$ and $B_{k}= A_{l}$. So $t(A_{k})-t(B_{k}) = k$.

Suppose that there exist others sets $C_{k}$ and $D_{k}$ such that $t(C_{k})-t(D_{k}) = k$. Then now it's suficient to prove that $3^{m}$ belongs to $C_{k}$. But $t(C_{k}) \geq k > \frac{3^{m}-1}{2}> 3^{m-1}$, so there will     exist a power of $3$ greater than $3^{m}$ on $C_{k}$. It's easy to see that $t(D_{k}) \leq \frac{3^{m}-1}{2}$, so $k \geq 3^{m+1}-\frac{3^{m}-1}{2}> 3^{m}$, contradiction. Then $3^{m}$ belongs to $C_{k}$, and $t(D_{k})-t(C_{k}-3^{m}) = 3^{m}-k = l$ which implies $A_{l}= B_{k}$ and $B_{l}= C_{k}-3^{m}$. So $A_{k}= C_{k}$ and $B_{k}= D_{k}$. So to each positive integer less than or equal to $3^{m}$ we have done. But if $n$ is such that $3^{m}< n \leq 3^{m}+\frac{3^{m}-1}{2}= \frac{3^{m+1}-1}{2}$, then call $l = n-3^{m}$, and by the uniqueness of $A_{l}$ and $B_{l}$ we may find unique $A_{n}$ and $B_{n}$, following the same steps of the last proof. And the induction is done.

Base of induction: For $k = 1$, $A_{1}= [1]$ and $B_{1}= [0]$. For $A_{2}= [3]$ and $B_{2}= [1]$.

PS: I know that I didn't complete this proof, but with a little bit of comprehension you would understand that I would repeat the first proof.

If for finding only one set $X$, the set of powers of 3 clearly works (just consider digits of $k$ in base 3). The more challenging question is to find [b]all[/b] such set $X$.


geometry
3D geometry
sphere
induction
algebra solved
algebra

This is a solid geometry problem :

On a sphere, 2003 great circles are drawn (so these are circles which are the intersection of the sphere with a plane passing trough the center of the sphere), no three of which are concurrent. In how many parts the surface of the sphere is divided like this ?

I think this cannot be too difficult (it's from the Belgian mathematics olympiad 1998, and the belgian level is not very high I must admit - i'm Belgian myself), but I don't find the solution since I'm bad at solid geometry. I never manage to visualize the situation ... Please help !

Since nobody comes up with a solution this problem seems to be quite difficult ...

Just some ideas

a(n) the number of parts

compute a(1), a(2), a(3), try to find an recurrence relation

What do you think ?

I also thought about a recurrence ... but I don't manage to realoze the idea.

a(1) = 2
a(2) = 4
a(3) = 8 ? (-> I'm even not sure)

An educated guess :

a(n) = 2^n

but I'm not sure whether this is correct, and if this would be correct, it can be proven by induction probably. But I don't manage to visualize it, and I don't know anything about lines of a sphere and so on. So maybe a hint or so ? This should be the idea but probably my guess (a(n) = 2^n) is just false. :?

My guess is : n(n-1) + 2   :?

Let's lay on one of the plane going through the center and one big circle.  Zzzz.
Let's look at the ceiling.
We see a disk D (= our plane) with (n-1) arcs of circle cutting it.
By hypothesis no two arcs of circle intersect on the border of D.

So the problem is to count the number of partitions of a disk by (n-1) arcs of circle.
This is the "projective" problem of counting the number of zones defined by (n-1) lines, no 2 parallel and no 3 intersecting.

Let a(n) be the number of zones defined by n lines, no 2 parallel and no 3
intersecting.

Let's have n such lines and let's add a new line. Each time we hit one of the n lines a new zone is created and the last hit add one more zone : a(n+1) = a(n) + (n+1), a(1) = 2 so a(n) = n(n+1)/2 + 1

The two hemisphere are symetrical relatively to the center of the sphere hence my guess is 2*a(n-1)

What do you think of this ? does it make sense ?


$ a,b,c,d\geq 0 and a^{2}\plus{}b^{2} \plus{}c^{2} \plus{}d^{2} \equal{}4$
prove that :
$ a^{3} \plus{}b^{3} \plus{}c^{3} \plus{}d^{3} \plus{}abc\plus{}bcd\plus{}cda \plus{}dab \leq 8$

[quote="iqtimo"]$ a,b,c,d\geq 0 and a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \equal{} 4$
prove that :
$ a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} d^{3} \plus{} abc \plus{} bcd \plus{} cda \plus{} dab \leq 8$[/quote]

I remenber this is darij's inequlity....and it had been posted already in the forum.

[quote="iqtimo"]$ a,b,c,d\geq 0$ and $ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \equal{} 4$
prove that :
$ a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} d^{3} \plus{} abc \plus{} bcd \plus{} cda \plus{} dab \leq 8$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99927


induction
algorithm
combinatorics solved
combinatorics

Let G be a graph such that there are no 3 vertex mutually joined. Prove one can color the vertex of G with no more than $2\sqrt{n}$ colors, such that if two vertex are joined, they have different colors.

We will use induction on $k= [ \sqrt n]$ where $[.]$ denote the integer part.

First, for $k=1$, it means that we have a graph of at most $3$ vertices with no triangle, and we want to color it we at most two colors. Direct checking leads to that obvious result.

Now, let $k-1 \geq 1$ be given, and assume that the result holds for any graph with no triangle and $N$ vertices where $k-1 \geq [ \sqrt N ]$.
Now let's give a graph $G$ with no triangle and $n$ vertices where $k = [\sqrt n ]$.

Let $\Delta$ be the maximum degree in $G$.
It is a very well-known result that $\chi (G) \leq \Delta +1 $ for any finite graph (1).
The problem is to prove that for our graph we have $\chi (G) \leq 2\sqrt n $. (2)

Therefore, we may assume that $\Delta > 2 \sqrt n - 1$ otherwise (1) leads immediately to (2).
Let $A$ be a vertex with degree $\Delta$.
We assign color $c_1$ to $A$ and $c_2$ to all the vertices adjacent to $A$. Since there are no triangle, two vertices adjacent to $A$ are not adjacent with each other, so that there is no two adjacent with the same color until now.
Now delete $A$ and the vertices adjacent to $A$ (and all the edges with endpoint one of these vertices) and consider the resulting graph $G'$.
Then $G'$ has no triangle and has $n'$ vertices with $n' < n - 2\sqrt(n) + 1 = (\sqrt n - 1)^2$. It follows that $k' = [\sqrt {n' }] < [\sqrt n ] - 1 = k$. Then, using the induction hypothesis, we may properly color $G'$ with at most $2 \sqrt {n'} - 2$ colors, that we choose different from $c_1,c_2$.
Then, we have properly color $G$ with at most $2 \sqrt {n'} - 2 + 2 \leq 2 \sqrt n $  colors, which ends the induction steps and the proof.

Pierre.

by inductive for n proof this problem.
for n=3 is trivial, suppose n=m is true now we want to prove for n=m+1 is true.

phrase star* n+2 :sqrt:  n   +  1 =  :sqrt: n +1.
according to phrase 1 until the number of vertex has not reached for the value of 2 n +1+n the number
of used colors should be equal to the time we had n vertex.

becuase t, :sqrt: 2  :sqrt: n +1 ---->  n+ :sqrt: t <  :sqrt:  n +1

i have an algorithm for proving this problem. we must paint this graph with the same number colors
that we used for n vertex.
choose one vartex in graph g with maximum degree.we name this vertex funny.if it's degree was less than 2 n then
it' s trivial that degree of adjacent vertex of funny are not bigger than 2 n.
so for each adjacent vertex of funny and funny at least one color that sa not used in the adjacent of adjacent of funny because the value of the is lessthan 2 n.

now we paint each adjacent of vertex funnywith the colors there were not used for each of them.
now we omit funny and adjacent of funny vertex which are painted in the wanted form.
so according to inductiv hypothesis we have painted the remaind graph with less than 2 n color.
because we havent used new colorto paint funny and adjacent so we can paint it entire graph with less than 2 n color..

now we suppose the degree funny to be more than 2 n or equal to it. in this time the number of the memebers of this sub is morethan 2 n.

t<2  :sqrt: n +1   :sqrt:  :P n+t < n +1  .

so we can use a new color now we paint funny with one of the colors which we used to paint the n-vertex graph.and we paintall adjacent of funny with the new color.and remind graph painting with 2 n color.
my proove is complete.

Myth asked: "Is 2\sqrt{n} is an exact estimate?"

No, the best estimate is a constant times sqrt{n} divided
by log(n)^2/3.
This follows by combining Turan's theorem with the results
of:

Nilli Alon:
Triangle-free graphs with large chromatic numbers
Discrete Mathematics, Volume 211, Number 1, 28 January 2000, pp. 261-262(2).


function
calculus
derivative
Pythagorean Theorem
geometry
calculus computations

Hi again,

Here's the problem:

A man starts walking north at 4 ft/sec from a point P. Five minutes later a woman starts walking south at 5 ft/sec from a point 100 feet due east of P. At what rate are the people moving apart 15 minutes after the woman starts walking?

Im having a little trouble putting this all together. Here's what I have:

So I should draw a diagram. It will be a right angle triangle with point "P" at the right angle. I'll have to find the distance between the two people after 15 minutes. Remembering that the man will have travelled for 20 minutes, and the woman for 15. Converting these times to seconds and multiplying by the speed determines the distance that each of them has travelled. So:

The man has traveled for 1200 seconds at 4 ft/sec = 4800 ft.
The woman has traveled for 900 seconds at 5 ft/sec = 4500 ft.

Now where do I go from here?  I figure that I need to use the pyth. theorem to find the distance they are apart. Then what? The deriv of pyth theorem?  Thanks for your help.
Chris

Use pythagorean theorem to create a function of the distance the two people are apart in terms of the time.  Take the derivative of the distance apart with respect to time to determine the change in distance apart per time, then plug in the time you wanted (15 min, but remember t is in sec.) into your derivative.


hello!

i just lost like 5 videos!

i had some more and they have vanished

From the [url=http://www.artofproblemsolving.com/Alcumus/FAQ.php]FAQ[/url]:

How come I can no longer see subjects on my report page that I use to be able to see?

Alcumus uses a sophisticated algorithm to determine which subject topics you are able to receive problems from and for which you can see the status of on the report page. [b]This algorithm will also affect which videos you can see[/b]. If you begin to answer problems incorrectly in the subject or related subjects, or others start answering problems correctly that you previously answered incorrectly, it may be determined that you are no longer ready to do problems in a particular subject. At that time you will no longer be able to see the subject in your report. Once you have sufficiently answered problems in related subjects, you will again be flagged as "ready" to do that subject and it will re-appear on your report.

[quote="mathwiz85"]that was in the FAQlike a direct copy[/quote]

Obviously. Is that a problem? Also, levans probably wrote that.

is that why his little smily character is wearing dark glasses? :agent:


calculus
integration
calculus computations

Make the substitution $x^{2}=y$ and the integral transforms to a Beta function.

[quote="dyta"]Can we have a general problem ?[/quote]

For a general solution, use the Beta as suggested.

For $\int_{0}^{1}x^{2n+1}(1-x^{2})^{n}dx$

the general result ends up being:

$\frac{(n!)^{2}}{2(2n+1)(2n)!}$

This can be derived from the Beta as shown in previous posts.


I have a research paper coming up in a few weeks on a mathematician (past or present) of my choice.  I usually choose someone who deserves more credit or attention than he or she has actually gotten.  Any suggestions?

You can look up a nice formula and see who first came up with it. They're like the "inventors" of it. I think. It helps me. Or you can write about Dr. Kofman. She was my old math teacher and she had a phd in physics. JK :lol:

Thank you.  That has started me thinking about some possibilities.  I'm also very interested in science, so I might do someone who has a strong connection to both.


calculus
derivative
algebra
function
domain
search
calculus computations

If f is atleast 3 times differentiable in its domain and f" be continous.Prove that there exists a real number satisfying
f(a).f'(a).f"(a).f'''(a)>=0

Also,what is the geometrical meaning of third order derivative?

The geometric meaning of a third-order derivative is complicated.  The third derivative of displacement is known as "jerk"; I don't think it has a simple "geometric" meaning.

http://en.wikipedia.org/wiki/Jerk

This problem has been discussed at least three other times (possible exaggeration). You can search for it, here is (I believe) the most recent http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=differentiable+three&t=129827


geometry
circumcircle
parallelogram
geometry unsolved

Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.

(2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.

AP is the A-symmedian of the triangle $\triangle ABC.$ Let O be the triangle circumcenter and K the symmedian point.

[color=blue][b](1)[/b][/color] AEDF is a parallelogram, hence its diagonals AD, EF cut each other in half. Since the midpoint of EF lies on the A-symmedian AD, EF is antiparallel to BC with respect to the angle $\angle A,$ wich means that the points B, C, E, F are concyclic.

[color=blue][b](2)[/b][/color] Let parallels to the B-, C-symmedians BK, CK through the foot $D \in BC$ of the A-symmedian $AK \equiv AD \equiv AP$ meet the rays (AB, (AC at B', C'. The triangles $\triangle AB'C' \sim \triangle ABC$ are centrally similar with the similarity center A and D is the symmedian point of the triangle $\triangle AB'C'.$ It immediately follows that the circumcircle $(A_{1})$ of the quadrilateral BCEF is the 1st Lemoine circle of the triangle $\triangle AB'C'$ centered at the midpoint X' of the segment DO', where O' is the circumcenter of this triangle. Therefore, $AA_{1}$ intersects the segment KO of the original triangle $\triangle ABC$ also at its midpoint X, the center of the 1st Lemoine circle of the original triangle. Simiarly, $BB_{1}, CC_{1}$ cut KO at its midpoint X, hence all three are concurrent at X.

$ \mbox{The three parallels to the sides of a triangle ABC through the Lemoine point of the triangle ABC, }$
$ \mbox{ determine on the sides of triangle ABC, 6 concyclic points.}$
$ \mbox{The circle of the 6 points is the 1\minus{}st Lemoine circle of triangle ABC.}$

[quote=alpha-beta]can someone define 1st Lemoine circle or give some links?[/quote]

can you explain me about Lemoin circle please

Here is my solution with some angle and length chasing

[b]Disclaimer: [/b] This is definitely not as elegant as yetti's beautiful  solution (:thumbup:), but it is much neater than I had originally expected it to be, which is the reason I decided to mention it anyway.

[b]$(1):$[/b]

By Thales' Theorem, $\frac{BD}{BC} = \frac{BF}{BA}$ and $\frac{CD}{CB} = \frac{CE}{CA}$. As $ADP$ is the symmedian, $\frac{BD}{DP} = \frac{AB^2}{AC^2}$ (as the symmedian is the reflection of the median over the angle bisector).

This yields the following, where $a=BC$ etc. (we will use these in part [b]$(2)$[/b] as well):
$BF=\frac{c^3}{b^2+c^2} ...(1)\\ \\AF=\frac{cb^2}{b^2+c^2} ...(2)\\ \\AE=\frac{bc^2}{b^2+c^2} ...(3)\\ \\CE=\frac{b^3}{b^2+c^2}...(4)$

From here we get $AF.AB = AE.AC = \frac{b^2c^2}{b^2+c^2}$ and concyclicity follows.

[b]$(2):$[/b]

Let the radius of circle $BFEC$ be $r$.

Let $\angle BCF=\alpha \implies \angle BA_1F=2\alpha \implies \angle A_1BF=\angle A_1FB=90-\alpha \implies \angle A_1BC = B+\alpha-90 = \angle A_1CB \implies \angle A_1CE= 90-\alpha-B+C = \angle A_1EC \implies \angle CA_1E = 2(\alpha+B-C)$.

Now, in $\Delta sA_1BF$ and $A_1CE$ we get, using equations $(1)$ and $(4)$ above,
$2r sin \alpha = \frac{c^3}{b^2+c^2}$ and $2r sin (\alpha+B-C) = \frac{b^3}{b^2+c^2}$
$\implies \frac{sin \alpha}{sin (\alpha+B-C)} = \frac{c^3}{b^3}    ...(5)$

Now we observe that,
$\frac {[ABA_1]}{ACA_1]} = \frac{\frac{1}{2}AB. AA_1 sin \angle BAA_1}{\frac{1}{2}AC.AA_1 sin \angle CAA_1} = \frac{c}{b}.\frac{sin \angle BAA_1}{sin \angle CAA_1} ...(6)$
and
$\frac {[ABA_1]}{ACA_1]} = \frac{\frac{1}{2} AB. BA_1 sin \angle ABA_1}{\frac{1}{2} AC.CA_1 sin \angle ACA_1} =  \frac{c}{b}.\frac{sin(90-\alpha)}{sin(90-\alpha-B+C)}  = \frac{c}{b}.\frac{cos \alpha}{cos (\alpha-B+C)}  ...(7)$

$(6)$ and $(7)$ together imply
$\frac{sin \angle BAA_1}{sin \angle CAA_1} = \frac{cos \alpha}{cos (\alpha-B+C)}  ...(8)$

Now after some elementary manipulations on relation $(5)$ we get,
$\frac{cos \alpha}{cos (\alpha-B+C)} = \frac{\frac{b^3}{c^3} - cos (B-C)}{\frac{b^3}{c^3}cos (B-C) - 1}    ...(9)$

Finally we use $cos \theta = cos^2 \frac{\theta}{2} - sin^2 \frac{\theta}{2} = \frac{1-tan^2 \frac{\theta}{2}}{1+tan^2 \frac{\theta}{2}}$ (on $\theta = B-C$ duh  :P ) and $tan \frac{B-C}{2} = \frac{b-c}{b+c}.cot\frac{A}{2}$ in relations $(8)$ and $(9)$ to finish the proof by the trigonometric form of Ceva's theorem.


In how many ways can 5 identical balls be arranged in 10 bowls (labelled 1 to 10)? Generalize to N (identical or numbered) balls in M (numbered or identical) bowls.

How many sequences $\{a, b, c, d, e\}$ are there satisfying $0\leq a\leq b\leq c\leq d\leq e\leq 9$? Generalize to sequences of N nonnegative non-decreasing integers less than or equal to M.

[hide="Short answer to the first one."]This is equivalent to the number of solutions there are in nonnegative integers there are to the equation
\[x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}=5\]
which is $\textstyle{{{14}\choose{9}}=2002}$.  In general, for $n$ balls and $m$ bowls, the answer is $\textstyle{{{m+n-1}\choose{n}}}$.[/hide]

[hide="Longer answer to the second one."]Any such sequence can be expressed in the following alternate format: a sequence of nine A's and five x's where $a$ is the number of A's before the first x, $b$ is the number of A's before the second x (including those A's that were before the first x), $c$ is the number of A's before the third x, and so on.  For example, AAxAxAAAxxAxAA represents the sequence $2, 3, 6, 6, 7$.  Thus, the number of sequences is the number of ways to arrange the "word" xxxxxAAAAAAAAA, which is $\textstyle{{{14}\choose{9}}=2002}$ (same as before... did you get these from some 2002 test? ;)).  In general, when you have $n$ integers and $m$ as the maximum, the answer is $\textstyle{{{n+m}\choose{n}}}$ (or $\textstyle{{{n+m}\choose{m}}}$, whichever you prefer).[/hide]


yes! it really was fun.
Thanks for everybody who [u]listened[/u] to the announcements and came to help the team.

But not that many people showed up.
Mr. Linker and I would like this to be handed down to the future new york city math team.
:help::help: :help:

[hide="Time"] 12:00 - 1:00 PM ET, 1:50 - 3:50 PM ET (3hrs total - lunch break in between)[/hide]
[hide="Participants"]
Alan [i]A[/i]
Maria [i]A[/i]
Kevin [i]A[/i]
Stan [i]A[/i]
Peak [i]A[/i]
Hanpin [i]A[/i]
Danny [i]A[/i]
Zuoyu [i]A[/i]
Wilson [i]C[/i]
Justin [i]C[/i]
Alex [i]C[/i]
Yuri [i]N[/i]
Mark [i]N[/i]
Brady [i]N[/i]
Tao [i]N[/i]
Ben [i]R[/i]
Amin [i]R[/i]
Chun [i]R[/i]
Steven [i]R[/i]
Danny [i]Y[/i]
Tony [i]Y[/i]
Edward [i]Y[/i]
[/hide]
[hide="plus"]Geehoon from A team[/hide]

[hide="scores for the first round"][u]Team 1 [/u]
Justin Wilson Alex
2 0 4 0 $6$

[u]Team 2[/u]
Kevin Alan Stan
4 0 4 0 $8$

[u]Team 3[/u]
Ben Steven Amin
0 0 4 4 $8$

[u]Team 4 :first:[/u]
Peak Maria Hanpin
2 4 4 2 $12$

[u]Team 5[/u]
Tao Yuri Edward
4 0 4 2 $10$

[u]Team 6[/u]
Brady Danny Mark
- - 4 0 $4$
[/hide][hide="scores for the second round"][u]Team 1 :first:[/u]
Kevin Alan Stan
4 4 4 4 2 4 4 $26$

[u]Team 2[/u]
Danny B./Peak(from 10th round) Mark Alex
0 0 4 2 0 2 4 $12$

[u]Team 3[/u]
Ben Steven Amin
0 4 2 0 0 4 2 $12$

[u]Team 4[/u]
Tao Yuri Edward
0 0 0 0 0 0 2 $2$

[u]Team 5[/u]
Tony Maria Hanpin
4 4 4 2 0 4 2 $20$

[u]Team 6[/u]
Danny Z. Zuoyu Chun(from 11th round)
- - 0 0 0 0 0 $...$
[/hide]

Once again, thank you all!
it really was my pleasure.

Thanks and great job to Geehoon for organizing the relays and dealing with the little problems that came up (like the people with no teams  :blush: )
I had so much fun today and at the other practices, hope ARML goes well :)
best of luck to NYC at ARML :D

you're very welcome.
I had a lot of fun today as well.

Yes, let's do better than last year :icecream:

these were alot of fun, and provided great practice...were they all from actual ARML's???


limit
real analysis
real analysis solved

OK, let's see if my second attempt will be better.

Suppose that $f\in C^1(0,+\infty)$ and $\lim_{x\to\infty}(f(x)+f'(x))=L$.
Prove that $\lim_{x\to\infty}f(x)=L$ and $\lim_{x\to\infty}f'(x)=0$.

No, that one is too classical. I will let for another member to post the solution.  :D

And it's been posted before :). I think I posted it, but I'm not sure. We can look at $e^x f(x)$ and use L'Hospital. We can use $f(x)+1$ instead of $f(x)$ so that we don't have problems, and $e^x(f(x)+f'(x))=(e^xf(x))'$ tends to $\infty$.


Find all natural numbers $ n$ for which the inequality $ a(a \minus{} b)^n \plus{} b(b \minus{} c)^n \plus{} c(c \minus{} a)^n\ge 0$ holds for all real numbers $ a,b,c$.

1,3,5?

Obviously,n must be odd number.Now,let
$ a\equal{}4,b\equal{}3,c\equal{}2$,we have
$ 4\plus{}3\minus{}2.2^{n} \geq 0$
$ \Leftrightarrow 7 \geq 2^{n\plus{}1}$.
Thus $ n< 2$.
For $ n\equal{}1$,ineq is true.Thus $ n\equal{}1$ is the only answer.

[quote="andreikoo"]Find all natural numbers $ n$ for which the inequality $ a(a \minus{} b)^n \plus{} b(b \minus{} c)^n \plus{} c(c \minus{} a)^n\ge 0$ holds for all real numbers $ a,b,c$.
[/quote]
I think, the following very known problem is more interesting:
Find all natural numbers $ n$ for which the inequality $ a(a \plus{}b)^n \plus{} b(b \plus{} c)^n \plus{} c(c \plus{} a)^n\ge 0$ holds for all real numbers $ a,b,c$. :wink:

[b]arqady [/b], i saw the answer from SII-Pham Kim Hung , $ n\equal{}3$ or $ n\equal{}5$

yes that is the problem with +. and it is true for n=1 too. can somebody proove it

andreikoo, see here the "hard" case:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55723

For $ n\equal{}3$

Let $ x\equal{}a\plus{}b,y\equal{}a\plus{}c,z\equal{}b\plus{}c$ we must prove that :

$ x^4\plus{}y^4\plus{}z^4\plus{}x^3y\plus{}y^3z\plus{}z^3x\ge xy^3\plus{}y^3z\plus{}z^3x$

We prove

$ x^4\plus{}y^4\plus{}2x^3y\ge 2xy^3$

-If $ xy\ge 0$ by AM-GM we have :

$ x^4\plus{}y^4\plus{}2x^3y\ge y^4\plus{}2xy^3\frac{3}{\sqrt[3]{2}}\ge 2xy^3$

-If $ xy\le 0$ simirlary:

$ x^4\plus{}y^4\plus{}2x^3y\ge x^4\plus{}2|xy^3|\ge 2|x^3y|$

Are $ n\equal{}1,3,5$ the only answers??
If so,how to prove it?? :?:

for n =3, can we show that   $ x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \minus{} xy^3 \plus{} y^3z \minus{} yz^3 \plus{} z^3x \plus{}zx^3 \equal{} sum of squares$ ?

For $ n\equal{}5$ we have a same solution with $ n\equal{}3$  :D


[b]if a,b,c,d$/geq$0 and a$/leg/$1     a+b$/leg/$5 , a+b+C$/leg/$14  ,  a+b+c+d$/leg/$30      prove that   ${sqrta}$+${sqrtb}$+${sqrtc}$+${sqrtd}$ $/leg/10[/code]

Can you write it in Latex correctly? I don't understand what you have written.

May be this sol is better  :D

 Let $x,y,z,t$ be a permutation of $(a,b,c,d)$ such that $x \leq y \leq z \leq t$

 Easy to see

 $x \leq a\leq 1$
 $x+y \leq a+b \leq 5$
 $x+y+z \leq a+b+c \leq 14$
 $x+y+z+t =a+b+c+d \leq 30$

 Now we have
$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}-10$

$ =\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}-10$

$ =\frac{x-1}{\sqrt{x}+1}+\frac{y-4}{\sqrt{y}+2}+\frac{z-9}{\sqrt{z}+3}+\frac{t-1}{\sqrt{t}+4}$

$ =(x-1).\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{y}+2}\right)+(x-1+y-2)\left(\frac{1}{\sqrt{y}+2}-\frac{1}{\sqrt{z}+3}\right)$+$(x-1+y-2+z-3)\left(\frac{1}{\sqrt{z}+3}-\frac{1}{\sqrt{t}+4}\right)+(x-1+y-2+z-3+t-4)\left(\frac{1}{\sqrt{t}+4}\right)$

$ \leq 0$

  :D  :D  :D

[quote="keira_khtn"][quote="siuhochung"][quote="keira_khtn"]We can apply Karamata's theorem!![/quote]
I don't think so, you cannot order a,b,c,d at all.[/quote]
We can !! ;) (I expect so)[/quote]
how??? I've said that we cannot order a,b,c,d at all. Ordering is important in Karamata.

I do as :
[quote="Viet Math"]May be this sol is better  :D

 Let $x,y,z,t$ be a permutation of $(a,b,c,d)$ such that $x \leq y \leq z \leq t$

 Easy to see

 $x \leq a\leq 1$
 $x+y \leq a+b \leq 5$
 $x+y+z \leq a+b+c \leq 14$
 $x+y+z+t =a+b+c+d \leq 30$
[/quote] and we can apply Karamata's ineq

I suppose in majorization, we should first compare the largest term instead of the smallest one?


Can anyone solve this problem? in term of d1,d2, a and b where a and b are angle


geometry
perimeter
conics
analytic geometry
linear algebra
parallelogram

Let the circle $ (I; r)$ be inscribed in the triangle $ ABC$. Let $ D$ be the point of contact of this circle with $ BC$. Let $ E$ and $ F$ be the midpoints of $ BC$ and $ AD$, respectively. Prove that the three points $ I$, $ E$, $ F$ are collinear.

[quote="thanhnam2902"]Let circle $ \omega (I;r)$ be incircle of the $ \triangle ABC$.Let $ D$ be the point of tangency $ \omega$ with $ BC$. Let $ E, F$ be the midpoints of $ BC$ and $ AD$ respectively.Prove that the points $ I, E, F$ are collinear.
[/quote]
Proof:
Let $ K$ be the point of intersection $ ID$ with $ \omega$ and $ G$ be the point of intersection $ AK$ and $ BC$.Then it is well-known that $ GE \equal{} ED$ and $ ID \equal{} IK \equal{} r$,hence $ IE\|AG$,but at the same time $ EF\|AG$,so $ I,E,F$ are collinear.
Best regards,Erken.

Hi!!
I think this problem has a mass center´s solution. Is anybody know that?

Another hint:
Use
$ 2\overrightarrow{IE} \equal{} \overrightarrow{IB} \plus{} \overrightarrow{IC},2\overrightarrow{IF} \equal{} \overrightarrow{IA} \plus{} \overrightarrow{ID}$
And
$ a\overrightarrow{IA} \plus{} b\overrightarrow{IB} \plus{} c\overrightarrow{IC} \equal{} \overrightarrow{0}$
$ \overrightarrow{ID} \equal{} \frac {p \minus{} c}{a}\overrightarrow{IB} \plus{} \frac {p \minus{} b}{a}\overrightarrow{IC}$

Hi $ Erken$, I think you not true. Can you answer me why $ GE \equal{} ED$ ?
Thank you very much.

[quote="thanhnam2902"]Hi $ Erken$, I think you not true. Can you answer me why $ GE \equal{} ED$ ?[/quote]
At first,if the fact,that was written by me,is wrong,then your problem is wrong,which is obviously not so.
Here is the proof of this [size=150]well-known[/size] fact:
[img]http://img232.imageshack.us/img232/3120/obviousonewg7.png[/img]
Let $ l$ be a line through $ K$,which is parallel to $ BC$,then it's tangent to $ \omega$ at $ K$.So the point $ K$,is the point of tangency $ \omega$ with $ MN$(see the image),but $ \omega$ is $ A$-exincircle of $ \triangle AMN$,hence $ G$ is the tangency point of $ A\minus{}exincirlce$ of the $ \triangle ABC$ with $ BC$,now it is obvious that $ CG \equal{} BD \equal{} s \minus{} c$,where $ s$ is a semi-perimeter,and $ c$ is a length of the side $ AB$.So from $ CG \equal{} BD$ and $ BE \equal{} EC$,we conclude that $ DE \equal{} EG$.

Oh! I see. Because your last post was very sort. Now I understand. Thank you very much. I wish you good luck.

[quote="thanhnam2902"][color=darkred]Let $ w\equal{}(I,r)$ be the incircle of the triangle  $ ABC$ . Let $ D$ be the contact point of the circle $ w$ with the side $ [BC]$ .

Let $ E, F$ be the midpoints of the segments $ [BC]$ and $ [AD]$ respectively . Prove three point $ I\in EF$ .[/color] [/quote]
[color=darkblue][b][u]Remark.[/u][/b] The line $ \overline {EIF}$ is the [b]Newton's line[/b] of the degenerate convex quadrilateral $ ABDC$ circumscribed to the circle $ w$ .[/color]

[size=150][color=darkblue]Thank [b]Erken, mihai miculita, Virgil Nicula, tdl, conejita [/b]for your reply. I wish you good luck.[/color][/size]

 :lol:  :rotfl: . Good bye! See you again.

Hi [color=red][b][size=150]tdl[/size][/b][/color] ! Can you post full your slove by vecto? Thank you very much.

$ 2\overrightarrow{IE} \equal{} \overrightarrow{IB} \plus{} \overrightarrow{IC},2\overrightarrow{IF} \equal{} \overrightarrow{IA} \plus{} \overrightarrow{ID}$
And well-known that:
$ a\overrightarrow{IA} \plus{} b\overrightarrow{IB} \plus{} c\overrightarrow{IC} \equal{} \overrightarrow{0}\Leftrightarrow \overrightarrow{IA} \equal{}\minus{}\frac {b}{a}\overrightarrow{IB} \minus{}\frac{c}{a} \overrightarrow{IC}$
$ \overrightarrow{ID} \equal{} \frac {CD}{BC}\overrightarrow{IB} \plus{} \frac {BD }{BC}\overrightarrow{IC}\equal{}\frac {p \minus{} c}{a}\overrightarrow{IB} \plus{} \frac {p \minus{} b}{a}\overrightarrow{IC}$
Then $ 2\overrightarrow{IF} \equal{} \overrightarrow{IA} \plus{} \overrightarrow{ID}\equal{}\frac{p\minus{}b\minus{}c}{a}(\overrightarrow{IB} \plus{} \overrightarrow{IC})\equal{}2\frac{p\minus{}b\minus{}c}{a}\overrightarrow{IE}$
Then OK!

[quote="thanhnam2902"][color=darkred]Let $ w \equal{} (I,r)$ be the incircle of the triangle  $ ABC$ . Let $ D$ be the contact point of the circle $ w$ with the side $ [BC]$ . Let $ E, F$ be the midpoints of the segments $ [BC]$ and $ [AD]$ respectively . Prove three point $ I\in EF$ .[/color] [/quote]
[color=darkblue][b][u]Remark.[/u][/b] The line $ \overline {EIF}$ is the [b]Newton's line[/b] of the degenerate convex quadrilateral $ ABDC$ circumscribed to the circle $ w$ .

[b][u]Direct proof of the proposed problem.[/u][/b] Observe easily that $ [EBD] \plus{} [EAC] \equal{}$ $ [FBD] \plus{} [FAC] \equal{}$ $ [IBD] \plus{} [IAC] \equal{} \frac S2$ .   The geometrical locus of a interior point $ X$ for which $ [XBD] \plus{} [XAC] \equal{} k$ (constant) is a segment (well-known). Thus, $ I\in EF$ .[/color]

=============================================================
[quote="mihai miculita"][color=darkred]Let $ P$ be a point in the plane of triangle $ ABC$ and points $ \{ P_a \} \equal{} AP \cap BC$, $ \{ P_b\} \equal{} BP \cap AC$, $ \{ P_c \} \equal{} CP \cap AB$.  Let $ M_a$, $ M_b$, $ M_c$ the midpoints of
[$ AP_a$], [$ BP_b$], [$ CP_c$], respectively; and $ N_a, N_b, N_c$ the midpoints of [$ BC$], [$ AC$], [$ AB$], respectively. Prove that $ M_aN_a$, $ M_bN_b$ and $ M_cN_c$ are concurent!
The point $ \{ O \} \equal{} M_aN_a \cap M_bN_b \cap M_cN_c$ is the centre of conic touches the lines BC, CA and AB at points $ P_a, P_b, P_c$ respectively.[/color] [/quote]
[color=darkblue][b]Prove [u]easily[/u] (with the [u]barycentrical coordinates[/u]) that[/b]

$ P(a,b,c)$ $ \Longrightarrow$ $ O(x,y,z)$ , where $ \frac {x}{a(b \plus{} c)} \equal{} \frac {y}{b(c \plus{} a)} \equal{} \frac {z}{c(a \plus{} b)}$ .[/color]

[quote="mihai miculita"][color=darkred]Let $ X$ be a point in the plane of triangle $ ABC$ . Denote the intersections

$ \|\begin{array}{c}
D\in AX\cap BC\\\
E\in BX\cap CA\\\
F\in CX\cap AB\end{array}$ , the middlepoints $ M$ , $ N$ , $ P$ of the sides $ [BC]$ , $ [CA]$ , $ [AB]$

and the middlepoints $ U$ , $ V$ , $ W$ of the segments $ [AD]$ , $ [BV]$ , $ [CW]$ respectively.

Prove that the lines $ UM$ , $ VN$ , $ WP$ are concurently.[/color] [/quote]
[color=darkblue][b]Proof[/b] (with the [u]barycentrical coordinates[/u]).

$ \|\begin{array}{ccc}
A(1,0,0) & B(0,1,0) & C(0,0,1)\\\\
M(0,1,1) & N(1,0,1) & P(1,1,0)\end{array}$ and $ \begin{array}{c}
\blacktriangleright\ \ X\ (\ a\ ,\ b\ ,\ c\ )\ \ \blacktriangleleft\\\
==================\\\
\|\begin{array}{ccc}D(0,b,c) & , & U(b+c,b,c)\\\\
E(a,0,c) & , & V(a,c+a,c)\\\\
F(a,b,0) & , & W(a,b,a+b)\end{array}\|\\\
==================\end{array}$ .

$ UM\cap VN\cap WP\ne\emptyset$ $ \Longleftrightarrow$ $ (\exists )\ Y(x,y.z)\in UM\cap VN\cap WP$ so that

$ |\begin{array}{ccc}
x & y & z\\\
0 & 1 & 1\\\
b+c & b & c\end{array}|=|\begin{array}{ccc}
x & y & z\\\
1 & 0 & 1\\\
b & c+a & c\end{array}=|\begin{array}{ccc}
x & y & z\\\
1 & 1 & 0\\\
a & b & a+b\end{array}|=0\ \Longleftrightarrow$

$ \|\begin{array}{ccc}
c-b & b+c & -(b+c)\\\
-(c+a) & a-c & c+a\\\
a+b & -(a+b) & b-a\end{array}\|\bullet\|\begin{array}{c}
x\\\
y\\\
z\end{array}\|=\|\begin{array}{c}
0\\\
0\\\
0\end{array}\|$ . Observe that

(in the upper system) the matrix $ \mathcal A$ of the coeficients is singularly, i.e.  $ \det \mathcal A=0$

because the sum of the its lines is equally to $ 0$ . Thus this system is compatiblely and

$ \frac {x}{(b+c)(c+a)+(b-c)(b+c)}=$ $ \frac {y}{(b+c)(c+a)+(b-c)(b+a)}=$

$ \frac {z}{c+a)(c+b)+(c-b)(a-c)}$ , i.e. $ \boxed {\ \frac {x}{a(b+c)}=\frac {y}{b(c+a)}=\frac {z}{c(a+b)}\ }$ . Thus

the lines $ UM$ , $ VN$ , $ WP$ are concurrently in $ Y[\ a(b+c)\ ,\ b(c+a)\ ,\ c(a+b)\ ]$ .[/color]

See also
http://www.mathlinks.ro/Forum/viewtopic.php?t=5811
http://www.mathlinks.ro/Forum/viewtopic.php?t=112509
http://www.mathlinks.ro/Forum/viewtopic.php?t=106337
http://www.mathlinks.ro/Forum/viewtopic.php?t=15932 for the same question with excircle instead of incircle

  darij

Thanks [b]darij grinberg [/b]very much. I wish you good luck.

Here is my solution using complex numbers:
Let the incircle of the triangle be the unit circle and let $P$ and $Q$ be the points of tangency to sides $AB$ and $AC$. Now, we have $a=\frac {2pq}{p+q}$, $b=\frac{2pd}{p+d}$ and $c=\frac {2qd}{q+d}$. We can easily continue our computations: $e=\frac{dq}{d+q}+\frac{dp}{d+p}$ and $f=\frac{pq}{p+q}+\frac{d}{2}$. Now, we need ony to show that $E, I, F$ are collinear, which is equivqlent to the condition $e\bar{f}=f\bar{e}$, or $(\frac{dq}{d+q}+\frac{dp}{d+p})(\frac{1}{p+q}+\frac{1}{2d})=(\frac{pq}{p+q}+\frac{d}{2})(\frac{1}{d+q}+\frac{1}{d+p})$. Multipying all by $(d+q)(d+p)(p+q)$, and expanding, we get $4dpq+2p^2q+2q^2p+2d^2p+p^2d+dpq+2d^2q+q^2d+dpq$ $=2d^2p+4dpq+2d^2q++dp^2+dpq+2p^2q+2q^2p+dpq+dq^2$, which is obvious.

[quote="jayme"]Dear Mathlinkers,

1. R the point of intersection of EI and the A-altitude.

Prove : AR = r

Sincerely
Jean-Louis[/quote]
My solution:

Let $ D' $ be the antipode of $ D $ in $ \odot (I) $ .

Since $ I $ is the complement of the Nagel point of $ \triangle ABC $ ,
so $ EI $ is parallel to the $ A- $ Nagel line $ AD' $ of $ \triangle ABC \Longrightarrow AD'IR $ is a parallelogram $ \Longrightarrow AR=D'I=\text{r} $ .

Q.E.D


[quote="Arrange your tan"][quote="undead"]find all values of n $ \in N$ show that: n can express as
              n=[a,b]+[b,c]+[c,a]                      (a,b,c $ \in N$)[/quote]

In general, what are you defining [x,y] as?  It's the what of x and y?[/quote]
it is smallest common multiple

[hide="A start"]
For arbitrary m,k:
(a,b,c) = (mk,k,1)
N=(2m+1)k

This takes care of all numbers containing at least one odd factor.

Left to examine is the attainability of powers of 2.[/hide]


integration
algebra
polynomial
calculus
calculus computations

Are you sure that you typed it correctly?
I don't think it has an elementary primitive...In the denominator there is a quartic polynomial under the radical sign...which leads us to an elliptic integral, I think...

[quote="Myth"]Why did he wrote $x>1$?[/quote]

 May be he meant $\int_{1}^{\infty}$

No, I don't think so, since $6x^2-(x-1)^4<0$ for a sufficiently large $x$.


induction
advanced fields
advanced fields unsolved

If $ m,n\geq 0$ then every point $ z\in S^{m\plus{}n\plus{}1}\subset\mathbb{R}^{m\plus{}n\plus{}2}\equal{}\mathbb{R}^{m\plus{}1}\times \mathbb{R}^{n\plus{}1}$ can be represented in the form $ z\equal{}cos(t).x\plus{}sin(t). y$ with $ x\in S^m, y\in S^n$, and $ t\in [0,\pi/2]$, and this representation is unique except that x resp. y is undetermined when $ t\equal{}\pi/2$ resp. t=0. Given $ f: S^m\to S^m$ and $ g : S^n\to S^n$, we define their join$ f*g : S^{m\plus{}n\plus{}1}\to S^{m\plus{}n\plus{}1}$ by $ (f*g)(z)\equal{}cos(t). f(x)\plus{}sin(t). g(y)$

prove that $ deg(f*g)\equal{}deg(f)deg(g).$

There is one hint that, first, we easy to prove f*g=(f*id)(id*g). Then prove deg(f*id)=deg f by induction on n. But I get stuck in the later.

Thanks in advance.

The idea is that $ S^0 * X \equal{} \Sigma X$ for any space $ X$ ( the join operator has a general definition. ); furthermore, the canonical map $ H_n(X) \to H_{n\plus{}1}(\Sigma X)$ is an isomorphism for every $ n$.


http://www.mathlinks.ro/Forum/viewtopic.php?t=60595

If this turns out to be the actual polish problem, please a mod. delete my post ;)

$\binom{n}{k} = \frac{n!}{(n-k)!k!}$ So when the number of factors of 2 in both $n!$ and $(n-k)!k!$ is the same, it's odd. Else, even. Using Legendre's (?) formula, if $\displaystyle \sum_{i=1}^{\infty} [\frac{n}{2^i}] - \sum_{i=1}^{\infty} [\frac{k}{2^i}] - \sum_{i=1}^{\infty} [\frac{n-k}{2^i}] = 0$ then $_nC_k \equiv 0 (\bmod 2)$
(Note [$x$] is the integer part of $x$, or the floor function.)
Maybe we can go from here?

That seems a little overkill.  Start with your original idea and work from there.

Oh and a really good related problem:

Prove that $C^{n}_k$ is an integer.


Please post transcript from Saturday 4/17/04 if you have them.

Thanks!

Please leave your e-mail and I will send it to you.

Can't you just post them in here?

I don't feel like receiving them through e-mail so..

Thanks!


Let:$ a;b;c>0;ab\plus{}bc\plus{}ca\equal{}1$.Prove that:
$ \sum \frac{1}{a^2\plus{}2bc} \ge 2\plus{}\frac{27}{(a\plus{}b\plus{}c)^6}$

[quote="caubetoanhoc94"]Let:$ a;b;c > 0;ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:
$ \sum \frac {1}{a^2 \plus{} 2bc} \ge 2 \plus{} \frac {27}{(a \plus{} b \plus{} c)^6}$[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt3}.$ :wink:

[quote="arqady"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:
$ \sum \frac {1}{a^2 \plus{} 2bc} \ge 2 \plus{} \frac {27}{(a \plus{} b \plus{} c)^6}$[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt3}.$ :wink:[/quote]

is there a problem with that? you get $ 3\ge3$.

We use two inequalities.
\[ \sum {\frac{1}{{{a^2} \plus{} 2bc}}}  \ge \frac{2}{{ab \plus{} bc \plus{} ca}} \plus{} \frac{1}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}\], which is equivalent to \[ \frac{{{{\left( {a \minus{} b} \right)}^2}{{\left( {b \minus{} c} \right)}^2}{{\left( {c \minus{} a} \right)}^2}\left( {ab \plus{} bc \plus{} ca \plus{} 2{a^2} \plus{} 2{b^2} \plus{} 2{c^2}} \right)}}{{\left( {{a^2} \plus{} 2bc} \right)\left( {{b^2} \plus{} 2ca} \right)\left( {{c^2} \plus{} 2ab} \right)}} \ge 0\]
and
\[ {\left( {a \plus{} b \plus{} c} \right)^6} \ge 27\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)\] which is obtained by using AM-GM inequality with three variables \[ {\left( {{a^2} \plus{} {b^2} \plus{} {c^2} \plus{} 2ab \plus{} 2bc \plus{} 2ca} \right)^3} \ge 27\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right){\left( {ab \plus{} bc \plus{} ca} \right)^2}\]
The conclusion follows immediately

[quote="caubetoanhoc94"][quote="arqady"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:
$ \sum \frac {1}{a^2 \plus{} 2bc} \ge 2 \plus{} \frac {27}{(a \plus{} b \plus{} c)^6}$[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt3}.$ :wink:[/quote]

It's true arquady  :)[/quote]
OK. My mistake.
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, your inequality is equivalent to
$ \frac{(2u^2\minus{}v^2)v^2}{w^6\plus{}2u(2u^2\minus{}3v^2)w^3\plus{}2v^6}\geq\frac{2}{3v^2}\plus{}\frac{v^4}{3u^6}.$
Id est, we need check only $ w^3\equal{}0$ and $ b\equal{}c,$ which is not hard.


linear algebra
matrix
abstract algebra
vector
rotation

I was wondering if my work in doing the following problem is correct. Any help is greatly appreciated!

Consider the linear transformation L: M_22 -> R^4 defined by

L of
[a b
c d]

=
(-10 a - c + 2d, 5a + b - d, -2c + d, 9a + 17b + 2d)

a) What is the matrix (namely A_L) corresponding to L with respect to the standard bases for M_22 and R^4?

Okay, so I basically put the coefficients into a matrix like so:
First row:  -10, 0, -1, 2
2nd row:     5, 1, 0, -1
3rd row:      0, 0, -2, 1
4th row:     9, 17, 0, 2

b) Compute the determinant of A_L.

Basically, I did this out by cofactor expansion and got 152. However, the second part to the question is, based on the result you get for the determinant, is L an isomorphism? Explain.

A linear transformation with nonzero determinant is an isomorphism. (Because it is invertible: you can write down the inverse in terms of the cofactor matrix and the reciprocal of the determinant.)

Thanks!  I think I know what to do now.  I have another question, though with regards to L.  How do I compute the ker(L)?  I've looked in my textbook but it doesn't explain it very well.

So far I've put the coefficients of L into a matrix (same one I used to find the determinant) and row reduced the matrix to get

1st row:  1, 0, 0, 0
2nd row: 0, 1, 0, 0
3rd row: 0, 0, 1, 0
4th row: 0, 0, 0, 1

But I'm unsure of what to do from there.

A linear transformation is an isomorphism if and only if its kernel is $\{0\}$.

Thanks, but how can I tell what the kernel is by looking at the row-reduced echelon form of the matrix?

The kernel consists of all vectors $x$ such that $Ax=0$ (use can use a row-reduced form of $A$ here). With the matrix you got (ones on diagonal, zeros elsewhere) you have $Ax=x$ for any $x$. So, the kernel consists only of the vector $x=0$.

Okay, thanks!  Is there a specific way to notate the kernel or would it be okay if I just wrote

ker(L) = {0}?


hello,
i just wanted to ask if anyone here can give a solution for the money problem of the usaco december 04 bronze division contest:
you have coins of 1, 5, 10, 20, 50, 100 cents and he asks you how many ways you can make a sum of A cents, with 1 <= A <= 1,000,000,000

thank you..

Hmm... What is the problem? Dynamic programming. Just keep $B[n]$, $B[n+1]$, ..., $B[n+99]$, where $B[i]$ is a number of possibilities to make sum $i$. Then you can calculate $B[n+100]$ and then alter $B[n]$, so you will again have array with 100 numbers. Take into consideration, for $B[i]$ you need to use long arithmetic.

i know that dp is used to solve these types of problems, but the problem here, is that A can be as big as 1,000,000,000
so i think that the dp solution will timeout for the last cases (when becomes A very big).. the time limit for this problem is 0.3 seconds

thank you..

Ops.. You are right. Time limit.
I think, I know approach how to accelerate calculationes, but in this case solution becomes too complicated, in my opinion.

I don't know if this is the best way, but there are closed-form formulas for the number of ways to make a particular sum.  If you restrict to any particular A mod 100, then I believe the number of ways is a polynomial in A of degree 5.

For example, the book [i]Concrete Mathematics[/i] by Graham, Knuth, and Patashnik gives a formulas for the case when the coins are 1, 5, 10, 25, and 50 cents.  Their solution uses generating functions.

[quote="Ravi B"]Also, I assume that the order of the coins does not matter.  (That is, 5 + 10 is the same as 10 + 5.)  In that case, Myth, I think your solution needs modification.[/quote]
Oh! You are right!

[quote="Ravi B"]If you restrict to any particular A mod 100, then I believe the number of ways is a polynomial in A of degree 5.[/quote]
At first sight, this approach would mean hard-coding 100 different polynomials into the program.  (We could reduce the number to 20 polynomials, because without loss of generality, A is a multiple of 5.)

But we can avoid hard-coding any polynomials.  Instead, we could have the program compute B[0..599] using any method.  Then we could compute any higher value by polynomial extrapolation.


I read somewhere (not sure where/when) that if one were to go into a teaching or government position for a mininum of 3 years after college, then the tuition would be paid for by the government. Is there any truth to this?

Maybe [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33638]this[/url] is what you are talking about (or at least it is similar).

wow, free tuition. i don't think the government is expecting many people to go into math or science with an offer like that. where would they get all the money from? don't they need it for other things?

Hmm, not exactly what i was considering, but i think its also quite interesting. I dont believe, though, that what i read was found on aops (hmm, it just occured to me when my inner voice was talking while my outer hands were typing that i now pronounce it ay-ops, whereas i used to say ay-oh-pee-es. I wonder when this happened.).

Actually, if you read the thing its only 10,000 dollars, which a lot of times isn't even half of a year's worth of college expenses.


A finite set [tex]S=\{a_1,...,a_n\}[/tex] positive integers is called a [i]unique sum set[/i] if for each sequence x1,...,xn of nonnegative integers the following implication holds:
[tex]a_1x_1+\cdots+a_nx_n=a_1+\cdots+a_n \to x_1=\ldots=x_n=1.[/tex]
In other words, if the only way to write the sum of all elements of [tex]S[/tex] as a sum of elements of [tex]S[/tex], where we may use some elements more than once, is the sum of all elements itself.

First an easy question:
show that
[tex]\{2^n-2^0, 2^n-2^1,\ldots,2^n-2^{n-1}\}[/tex]
is a unique sum set for each [tex]n[/tex].

Now a challenging one:
is the unique sum set given in the previous question the one with the smallest sum if the number of elements is [tex]n[/tex]?

[quote="thabit"][quote="pbornsztein"]Is the challenging one an open question? Or do you have the answer (and then we have to move this post into the 'proposed and own problems' section?

Pierre.[/quote]
To my knowlegde, it is open.[/quote]

Indeed, it is problem C19 (attributed to John Selfridge) in "Unsolved Problems in Number Theory" (3rd edition) by Richard Guy (Springer, 2004).  Incidentally Selfridge offers $10 for a solution.


The sequences $ (a_n),(b_n),(c_n)$ are given by the conditions: $ a_1\equal{}4, a_{n\plus{}1}\equal{} a_n(a_n\minus{}1),2^{b_n} \equal{}a_n, e^{n\minus{}c_n} \equal{}b_n$ for $ n\equal{}1,2,3,...$. Prove that the sequence $ (c_n)$ is bounded.

You are wrong, because $ c_n\equal{}n\minus{}ln(log_2(a_n)),a_n\le 2^{2^n}$. It give $ c_n\ge n\minus{}nln 2$.


[quote="Smartguy"]Does anyone know of any good introduction NT books?
besides AoPS?
and intermediate books? besides 104 NT??

Thanks[/quote]
The theory of algebraic numbers
...never read it though

I have "Number Theory" by George E. Andrews. It's a Dover book, so you may be able to find it at Borders. I'm not sure how good it is, relatively speaking.

It covers a fairly diverse range of topics, from basic congruences to quadratic reciprocity to geometric number theory and more. It has a "combinatorics approach" to solving the problems, so you may be able to learn new ways of looking at problems (like the stellated p-gon approach to Wilson's theorem).

What are you looking for? If you want a purely contest-oriented book, get the 104 Problems. Otherwise, the Andrews is good, albeit rather unorthodox for an introductory text. Hardy and Littlewood is a fairly comprehensive classic. Stark is good too, I'm told, but I don't actually have it.

From experience I can strongly recommend An Introduction to the Theory of Numbers by Ivan Niven, et al. It can be quite expensive therefore I suggest you look for a second hand copy.

what about the books Kiran Kedlaya recommends on the AMC website?

Ivan Niven et al. is good but for a more general overview try David Burton's.


The way you guys pose this formula is not only confusing, but hard to comprehend.
Basically, emphasize the point that you're trying to find square inches, [color=#f00]NOT CUBIC INCHES. [/color]
10x10=100    Simply add these numbers to get 340. [color=#0ff]Then multiply this by 0.05 to get 34.[/color]
12x10=120
10x12=120


Let $x_0=\theta$ and $x_{n+1}=\frac{4}{7}\left(x_n^2+\frac{\theta}{x_n^2}\right)$.
Does $\displaystyle\lim_{n\to\infty}x_n$ exist, and if so what is its value.

I'll assume that $\theta$ is real, and we are working in the real line.

Here's a start: if this sequence were to converge, it would have to converge to a root of the equation $x^4-\frac74 x^3+\theta=0.$

This equation only has real roots if $\theta\le\frac13\left(\frac{21}{16}\right)^4,$ so that becomes a necessary condition.

The question of what would be a sufficient condition would depend on stability; I haven't investigated that yet.


calculus
integration
trigonometry
calculus computations

Set $ \tan x\equal{}\sqrt{2}u$

$ \int \sqrt{2\plus{}\tan ^{2}x}dx$

$ \equal{}\allowbreak \allowbreak 2\int \left( \frac{1}{\left( 1\plus{}2u^{2}\right) \sqrt{1\plus{}u^{2}}}\plus{}\,\frac{1}{\sqrt{1\plus{}u^{2}}}\right) du$

$ \equal{}\allowbreak 2\tan ^{\minus{}1}\frac{\tan x}{\sqrt{2\plus{}\tan ^{2}x}}\plus{}2\sinh ^{\minus{}1}\left( \frac{1}{\sqrt{2}}\tan x\right) \plus{}C$


i noticed that there are a lot more scary movies coming out after The Grudge, is this true?

[quote="plokoon51"]lol, i thought this topic was about the Scary Movies (Scary Movie 1,2,3...) :D[/quote]Lol, me too :)

I'm allergic to scary movies. I don't get nightmares, but still...urgh...

[quote="djimenez"][quote="juicybooty911"][quote="furious"]the scariest movie ever is Finding Nemo.[/quote]

i found Nemo in my fish tank[/quote]
I found Nemo in my sushi roll!  :D[/quote]

which part did you eat? :P


geometry
rectangle
absolute value
number theory unsolved
number theory

A table with an infinite number of fields contains numbers.Empty fields are considered to contain zeros.In any square with sides along the grid lines is the absolute value of the sum of the numbers equal or less 1.Regarding these conditions, determine the maximum value of the sum of all elements in a rectangle 5 x 1.

The problem is called "Der Burische Hammer" after Wolfgang Burmeister

Did you by any chance post it in this subforum because the numbers in the table are integers? (you just mentioned they were "numbers")

Good question grobber!

The numbers are real numbers!

I don't quite get this? Are there multiple numbers per square? Or if there is just 1 what do you mean by "in any square is the absolute value of the sum of the numbers"? Adjacent squares? I don't understand the question sorry :(

You can have squares formed by many small squares. For example, if you label the small squares $a_{ij},\ i,j\ge 1$ in the obvious way, the cells $a_{11},a_{12},a_{21},a_{22}$ for a square of side $2$. The sum of the four numbers in these four cells has absolute value $\le 1$.

so you can divide up each square? :?

Then I still don't get why it isn't infinite, like replace 10^9 by whatever in the below example
(where we split each square in 9 squares so all border squares fullfil the requirement):

$\begin{array}{ccccccccccccccc}
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\
0&0&0&0&0&0&10^9&0&0&0&0&0&0&0&0\\
0&0&0&0&0&0&0&0&0&0&0&0&0&0&0
\end{array}$

Perhaps can you reformulate it and/or make a drawing, cuz my english is too bad to understand this I'm afraid :P

No, man, you cannot divide each square. You have an infinite chess board (without any coloring, though :)). This board is made up of small squares of fixed dimension. However, from these small squares you can form [b]larger[/b] squares, made up of a certain number of small squares. Is it that hard? :D

For example:

$\begin{array}{cc}a&b\\c&d\end{array}$.

$a,b,c,d$ are small squares (you can't find smaller squares on your board). However, the array formed by $a,b,c,d$ is another (larger) square. The problem says that, for example, $|a+b+c+d|\le 1$ in addition to $|a|,|b|,|c|,|d|\le 1$.

then what does he mean by: 'an infinite field' and 'squares along the grid lines'?

If the fields is infinite, and you may just group together to larger squares, where are the grid lines then??? :?

Gosh I'll better just quit this problem, i don't get the heck of it  :blush:

Dude, think about the integer lattice. That's naturally divided in small squares of side $1$ and it has grid lines of equations of the types $x=k$ and $y=t$. Let $a_{mn}$ be the number written in the cell numbered $m,n$. The problem says that (I'm just giving some examples, I can't write all of the relations because there are infinitely many :)) $|a_{mn}|\le 1$. It also says that $|a_{m\ n}+a_{m\ n+1}+a_{m+1\ n}+a_{m+1\ n+1}|\le 1$. It also says that $|a_{m\ n}+a_{m\ n+1}+a_{m\ n+2}+a_{m+1\ n}+a_{m+1\ n+1}+a_{m+1\ n+2}+a_{m+2\ n}+a_{m+2\ n+1}+a_{m+2\ n+2}|\le 1$, and so on (for all integers $m,n$).

How on Earth can it still be unclear?


How many 4 digit postive numbers are divisible by 3 and have a last digit of 1?

Let the number be $ \overline{abc1}$ , then we have, by divisibility rule of 3,

$ a \plus{} b \plus{} c \plus{} 1 \equiv 0 \mod 3$

$ a \plus{} b \plus{} c \equiv 2 \mod 3$

We have 9 possible residues for $ (a,b,c)$:

$ (0,0,2)$ There are $ 3 \cdot 4 \cdot 3 \equal{} 36$ numbers
$ (0,2,0)$ There are $ 3 \cdot 3 \cdot 4 \equal{} 36$ numbers
$ (2,0,0)$ There are $ 3 \cdot 4 \cdot 3 \equal{} 36$ numbers
$ (1,1,0)$ There are $ 3 \cdot 3 \cdot 4 \equal{} 36$ numbers
$ (1,0,1)$ There are $ 3 \cdot 4 \cdot 3 \equal{} 36$ numbers
$ (0,1,1)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers
$ (1,2,2)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers
$ (2,1,2)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers
$ (2,2,1)$ There are $ 3 \cdot 3 \cdot 3\equal{}  27$ numbers

In total there are $ 36(5) \plus{} 27(4) \equal{} \boxed{288}$ numbers.

EDIT: THanks palytoxin. Why is it still wrong?

You get the numbers from 1001 to 9991.

So that's $ \frac{8990}{10}\equal{}899$ four-digit numbers which end in 1.
Because $ 1001\equiv 2\mod3$ and $ 9991\equiv 1\mod3$, add 1 and divide by 3 to get all the four digit numbers which are divisible by 3. $ \frac{900}{3}\equal{}\boxed{300}$

[quote="modularmarc101"]
EDIT: THanks palytoxin. Why is it still wrong?[/quote]
Check your third case  :)

[quote="itzmiko"]how does that work, adding 1 then dividing by 3? how did you get that by observing 1 mod 3 and 2 mod 3?[/quote]

Well, if you look at a smaller example, you would ideally want a full cycle. Look at 1 through 6. There are obviously 2 numbers divisible by three.
$ 1\equiv 1\mod3$
$ 6 \equiv 0\mod3$
So this case goes through the full cycle. However, the numbers are inclusive, so $ 6\minus{}1$ will only give 5, but you have to add 1 because subtracting only gives the spaces in between the numbers.

[quote="modularmarc101"]Let the number be $ \overline{abc1}$ , then we have, by divisibility rule of 3,

$ a \plus{} b \plus{} c \plus{} 1 \equiv 0 \mod 3$

$ a \plus{} b \plus{} c \equiv 2 \mod 3$

We have 9 possible residues for $ (a,b,c)$:

$ (0,0,2)$ There are $ 3 \cdot 4 \cdot 3 \equal{} 36$ numbers
$ (0,2,0)$ There are $ 3 \cdot 3 \cdot 4 \equal{} 36$ numbers
$ (2,0,0)$ There are $ 3 \cdot 4 \cdot 3 \equal{} 36$ numbers<----Wrong ($ 3\cdot4\cdot4$)
$ (1,1,0)$ There are $ 3 \cdot 3 \cdot 4 \equal{} 36$ numbers
$ (1,0,1)$ There are $ 3 \cdot 4 \cdot 3 \equal{} 36$ numbers
$ (0,1,1)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers
$ (1,2,2)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers
$ (2,1,2)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers
$ (2,2,1)$ There are $ 3 \cdot 3 \cdot 3 \equal{} 27$ numbers

In total there are $ 36(5) \plus{} 27(4) \equal{} \boxed{288}$ numbers.

EDIT: THanks palytoxin. Why is it still wrong?[/quote]

[quote="calculatorwiz"]well a faster way to do it is to find the first one

the first one has to be

$ 1_ _1$
so to be a multiple of $ 3$ it has to add up to $ 3$ or its multiple so the smallest one would be $ 1011$

now you do $ 10000 - 1011 = 8989$

so we also know that everye thirty numbers there will be a multiple of three with the digit 1

so now we do $ \frac {8989}{30}$

if you compute it you get $ 299$r$ 11$

so the answer would be $ 299$

and you guys if you read the question it says that it has one as the units digit not a remainder of $ 1$

you do not need to use modular arithmitec[/quote]Unfortunately, that's the wrong answer.

Clearly, the sum of the first three digits must yield 2 mod 3. These possible residues:

$ (2,0,0): 3\cdot4\cdot4 \equal{} 48$
$ (0,2,0),(0,0,2): 2\cdot3\cdot3\cdot4 \equal{} 72$
$ (1,1,0),(1,0,1): 2\cdot3\cdot3\cdot4 \equal{} 72$
$ (0,1,1): 3\cdot3\cdot3 \equal{} 27$
$ (1,2,2),(2,1,2),(2,2,1): 3\cdot3\cdot3\cdot3\cdot \equal{} 81$.

I count $ \boxed{300}.$

Alternatively, we know $ 1011$ is divisible by 3. Keep on adding 30 to get multiples of 3. The greatest such four-digit number should be $ 9981$. There are $ 9000/30 \equal{} 300$.

sorry I accidentally did not count 1011

sorry


The area between two varying concentric circles is always $ 9\pi in^{2}$. The rate of change of the area of the larger circles is $ 10\pi in^{2} /sec$ How fast is the circumference of the smaller circle changing when it has area $ 16\pi in^{2}$?

I would just like to have some hints, not the answer. But if you really want to, you can put the answer under a hide tag.

Thank you very much in advance.

[hide="Hint"]
$ \pi R^2 \minus{} \pi r^2 \equal{} 9 \pi$
Then differentiate.
[/hide]

[hide="Solution"]
From beforehand,
$ 2 R \frac{dR}{dt} \equal{} 2 r \frac{dr}{dt}$
We are given that $ r \equal{} 4$ and $ \frac{d[\pi R^2]}{dt} \equal{} 10 \pi \Rightarrow R \frac{dR}{dt}\equal{}5$. So,
$ 5 \equal{} 4 \frac{dr}{dt}$
$ \frac{dr}{dt} \equal{} \frac{5}{4}$
Now, we find the rate of change of the circumference, or $ d[2 \pi r]/dt \equal{} 2 \pi \frac{dr}{dt}$
So, plugging in $ \frac{dr}{dt}$, we get  $ d[2 \pi r]/dt \equal{} \frac{5 \pi}{2}$
[/hide]

[quote="stuck"][hide="Hint"]
$ \pi R^2 \minus{} \pi r^2 \equal{} 9 \pi$
Then differentiate.
[/hide][/quote]

What are you differentiating in respect to? As in $ \frac {d?}{d?}$

Can someone come up with an more understandable hint/solution?

Thank you.

For some reason, I end up with $ 10 \pi$ as my answer...

Can anybody help?


Find an integer $n$ ,where $100 \leq n \leq 1997$ ,such that
$\frac{2^n +2}{n}$ is an integer.
And hence,prove that it is unique.

I think this problem must be a national(international) olympiad in 1997  :P  (blind guess)

It's APMO 1997.

I believe the integer is 946 :)

Thanks
Do you have a solution for this Q?

Well, I have seen the solution before.

I am pretty sure you can find it on this forum, right here :)

I tried typing APMO 1997 in the searching bar,
but i can't get the related topic =(