trigonometry
inequalities
geometry proposed
geometry

One more a equivalence for "[b]A=60[/b]" in the triangle ABC: "[b]$r_a=R+r$[/b]".
$(r_b+r_c=3R)$

[quote="levi"]One more a equivalence for "[b]A=60[/b]" in the triangle ABC: "[b]$r_a=R+r$[/b]".[/quote]
 I think this works by calculation.Well $r_a=S/p-a $and $ R=abc/4S $ and cosine law.I'm not sure,but when
 I finish my solution I'll post it.I hope very soon.

I'm courious in your solution,levi? can you post it?

[quote="levi"]Ce faci, Cezar ? Putem comunica prin intermediul Forumului. Ai grija, aici ori raspunzi complet si (evetual) corect, ori nu apari. Salutari de pe coasta insorita a Californiei, [i]V.N.[/i][/quote]Please restrain from using other languages than English in this forums. The Romanian community forums are located [url=http://www.mathlinks.ro/Forum/index.php?f=250]here[/url].

Look at the one of the solutions :
$r_a=R+r$  iff  $p.r_a=p.R+p.r$  iff  $p.r_a=p.R+(p-a).r_a$  iff
$a.r_a=p.R$  iff  $2R\sin A.p\tan\frac{A}{2}=pR$  iff  $2\sin A.\tan\frac{A}{2}=1$
 iff  $4\sin^2\frac{A}{2}=1$  iff  $2(1-\cos A)=1$  iff  $\cosh=\frac12$  iff  $A=60^{\circ}$.

I think the formula:[tex]r_a-r=4R(sin\frac{A}{2})^2[/tex]is well-known

Yes,Keira,you're right.I can't believe I did not remember this formula,although it's a well known one.
Ok,I have a problem concerning the formula remembered:

   Prove that in any acute triangle the following inequality holds:

              $\frac{r_a-r}{2R}\leq\frac{a^2}{b^2+c^2}$ ;)


function
limit
calculus
derivative
calculus computations

If $f$ is a function satisfying

$|f(x)-f(y)|\leq M|x-y|^{\alpha}$

$\forall x$ and $y$ and some fixed $M$ and $\alpha>1$, prove that $f$ is constant.

Stick $n$ copies of this together: $|f(x+nh)-f(x)|$
$\le |f(x+h)-f(x)|+|f(x+2h)-f(x+h)|+\cdots+|f(x+nh)-f(x+(n-1)h)|$
$\le Mh^{a}+Mh^{a}+\cdots+Mh^{a}=M(nh^{a})$.
If $nh=y-x$, we get $|f(y)-f(x)|\le M\frac{n}{n^{a}}|y-x|^{a}$.
Let $n\to\infty$ in this last relation.

That's different than the proof I know. For utter simplicity, just take $\lim_{x\to y}\frac{f(x)-f(y)}{x-y}.$ We have that $f'$ exists everywhere and $f'(x)=0$ everywhere.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=813852#p813852]here[/url] for a related TST problem.
(Give or take an argument involving Cauchy sequences.)

And, of course, I cannot understand what an analysis problem is doing on a TST for IMO.

The Romanians have a different idea of what should be on high school contests than anyone else- that TST covers the same material as the tests before it.

The $f'=0$ argument conceals important details- you need local uniform continuity of the difference quotient (or a compactness argument) to get from a zero derivative to a constant function.

Assuming we're talking about a function defined on an interval in $\mathbb{R},$ I don't see that at all. The only hypotheses of the Mean Value Theorem are $f$ continuous on $[a,b],$ differentiable on $(a,b),$ and the Mean Value Theorem is all we need to go from zero derivative to constant function.

Ah- I was thinking of a different proof, not using the Mean Value Theorem. There is still compactness in the background- we use the extreme value theorem to prove Rolle and the MVT.


Euler
trigonometry
geometry solved
geometry

I is the isogonic point of a triangle [tex]ABC[/tex] (the point in the interior of the trangle for which [tex]\angle{AIB}=\angle{BIC}=\angle{CIA} =120^0[/tex] ).Prove that the Euler line of the trangles [tex]ABI , BCI[/tex] and [tex]CAI[/tex] are concurrent.

The point [tex]D[/tex] line inside the acute trangle [tex]ABC[/tex] .Three of the circumscribed circles of the trangles [tex]ABC,ABD,BCD[/tex] and [tex]CAD[/tex] have equal radii.Prove that the fourth circle has the same radius

Number 2: Use the formula $l=d\sin{\theta}$ and the acute angles, we can easily get the answer.

For the first problem, see http://www.mathlinks.ro/Forum/viewtopic.php?t=47955 (of course, isogonic point = Fermat point).

For the second problem, see part 2) of http://www.mathlinks.ro/Forum/viewtopic.php?t=2033 .

  darij


I am aware that the range are the y-values.

Given the function:

y = 1 + 8/(x-4) - 4/(x-3)

What would the range be?

I graphed it on my calculator and I am not getting very precise answers. Please help.

THanks!

Hi !
choose window  for $y$ between $-100$ and $100$ and see what's happened ! Make the variations of the functions (with derivatives...) and you'll get a maximum for $x=2+\sqrt2$. Calculate this maximum and you'll have the answer !

Actually you see that $y \to \pm \infty$ as $x \to \pm 4$, so you'd better not speak of global minima and maxima :wink: From this you see that the range must be of the form $(-\infty, a) \cup (b,+\infty)$. (It might jolly well happen that $a>b$; in that case, the range would be all of $\mathbb{R}$...) It remains to find the values of a and b, and for this you need to do a little bit of calculus.

It works best if you construct the graph by hand - you find the asymptotic behaviour of $y(x)$ as $x \to 3$ and $x \to 4$, and then you find, where the local extremes are, by solving $y'=0$ w.r.t x.

If I did not miscalculate, then the loc. extr. pts are solutions to $x^{2}-4x+2=0$, i.e., $x=2 \pm \sqrt{2}$. It is left as an exercise to find whether these are local maxima or minima, to evaluate $y$ at these points and find the range  :D

After simplifying the equation
I got

y=(x^2-3x+4)/(x^2-7x+2)
I don't know where to go from here on
(though I'll keep on trying)

anyone care to help?

and

chickendude
can you please explain how you got from step 3 to step 4?

Edit:
nvm i get it
you subtracted
alright

Edit2:
OK I think i've solved it
after much help from chickendude's unfinished solution
[hide]
y>-2.05011111...
y<-20.79499999...
[/hide]


search
inequalities
inequalities proposed

By AM-GM, we have $LHS \geq  \frac{2}{(a+1)(b+1)}\geq  \frac{2}{(\sqrt{a}\sqrt{b}+1)^{2}}$, by Cauchy.

Now, since $2(\sqrt{a}\sqrt{b})^{2}+2*1^{2}\geq  (\sqrt{a}\sqrt{b}+1)^{2}$, by an application of AM-GM,
we have the previous expression $\geq  \frac{2}{2ab+2}= \frac{1}{ab+1}$.

Therefore, we have $\frac{1}{(a+1)^{2}}+\frac{1}{(b+1)^{2}}\geq  \frac{1}{(a+1)(b+1)}$.

[quote="PTynan89"]By AM-GM, we have $LHS \geq \frac{2}{(a+1)(b+1)}\geq \frac{2}{(\sqrt{a}\sqrt{b}+1)^{2}}$, by Cauchy.

.[/quote]

i think you had a mistake

[quote="xXx_MrRomano_xXx"]i think this problem is familiar with many yours but can you post a nice solution

a,b>0
prove  1/(1+a)^2  +1/(1+b)^2 >=1/(1+ab)[/quote]
[hide="hint"]
clear the denominators[/hide]

[quote="xXx_MrRomano_xXx"]i think this problem is familiar with many yours but can you post a nice solution

a,b>0
prove  1/(1+a)^2  +1/(1+b)^2 >=1/(1+ab)[/quote]
I think it not difficult  :D Above inequality $\to (ab+1)(a-b)^{2}\ge 0$
This lemma used to prove China TST's inequality  :wink:

i know it lemma of china'tst2004 and vietnam'tst2005 but i like a nnice solution,
 zaizai, can you post a detail solution?

[quote="xXx_MrRomano_xXx"]$\{a,b\}\subset (0,\infty )\Longrightarrow$ $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}\ge \frac{1}{1+ab}\ \ (1)\ .$[/quote]
[color=darkblue][b]Proof 1.[/b] $\blacktriangleright$ I'll use the [u]Pohoatza's hint[/u]. Denote $a+b=S\ ,\ ab=P\ .$ Therefore,

$(1+ab)\cdot[(1+a)^{2}+(1+b)^{2}]-(1+a)^{2}(1+b)^{2}=$

$(1+P)(2+2S+S^{2}-2P)-(1+S+P)^{2}=PS^{2}-3P^{2}-2P+1=$

$P(S^{2}-4P)+(P-1)^{2}=\boxed{\ ab(a-b)^{2}+(ab-1)^{2}\ge 0\ }\ \ (2)\ .$

We 'll have equality in the relation $(1)$ if and only if $a=b=1\ .$

In conclusion, the proposed inequality $(1)$ is equivalently with the inequality $(2)\ .$

[b]Proof 2.[/b] $\blacktriangleright$ I"ll use the identity $\boxed{(a+b)(1+ab)=a(b+1)^{2}+b(a-1)^{2}}\ \ (3)\ :$

$\{\begin{array}{c}(a+b)(1+ab)=a(b+1)^{2}+b(a-1)^{2}\ge a(b+1)^{2}\Longrightarrow\frac{1}{(b+1)^{2}}\ge \frac{a}{a+b}\cdot\frac{1}{1+ab}\\\\ (a+b)(1+ab)=b(a+1)^{2}+a(b-1)^{2}\ge b(a+1)^{2}\Longrightarrow\frac{1}{(a+1)^{2}}\ge\frac{b}{a+b}\cdot\frac{1}{1+ab}\end{array}\|\bigoplus$

[b]Proof 3.[/b] $\blacktriangleright$ Again I"ll use use the identity $(3)\ :$

$\{\begin{array}{c}\frac{a(b-1)^{2}}{(a+1)^{2}}+\frac{b(a-1)^{2}}{(b+1)^{2}}\ge 0\Longleftrightarrow \frac{(a+b)(1+ab)-b(a+1)^{2}}{(a+1)^{2}}+\frac{(a+b)(1+ab)-a(b+1)^{2}}{(b+1)^{2}}\ge 0\Longleftrightarrow\\\\ \frac{(a+b)(1+ab)}{(a+1)^{2}}+\frac{(a+b)(1+ab)}{(b+1)^{2}}\ge a+b\Longleftrightarrow\frac{1}{(a+1)^{2}}+\frac{1}{(1+b)^{2}}\ge \frac{1}{1+ab}\end{array}$

[b]Remark.[/b] $\{a,b,c\}\subset (0,\infty )\Longrightarrow$ $\boxed{\frac{1}{(a+c)^{2}}+\frac{1}{(b+c)^{2}}\ge \frac{1}{c^{2}+ab}}\ \ (4)\ .$

We"ll have equality in the relation $(4)$ iff $a=b=c\ .$[/color]


vector
function
calculus
integration
linear algebra

Let V be the real functions y=f(x) satisfying d^2(y)/(dx^2) + 9y=0.

a. Prove that V is a 2-dimensional real vector space.
b. In V define (y,z) = integral (from 0 to pi) yz dx. Find an orthonormal basis in V.

part A:
I integrated and got that f(x)= (-3/2)y^3 + Cy, C is a real number.
It seems like I need to use dot product here. I don't know how, though.

part B:
Completely lost.

Finding solutions to this differential equation has nothing to do with part a. Also, that is not a solution.
A more general result here:
Let $ V$ be the set of (real) solutions to the $ n$th order linear differential equation $ f^{(n)}(x)\equal{}g_{n\minus{}1}(x)f^{(n\minus{}1)}(x)\plus{}\cdots\plus{}g_1(x)f'(x)\plus{}g_0(x)f(x)$, on some interval $ I$ on which all of the $ g_i$ are nice. This $ V$ is a $ n$-dimensional vector space. You will need the existence-uniqueness theorem.

For part b, you do have to correctly find the solutions. The most natural choice of basis is already orthogonal, so you just have to scale it to make it orthonormal.


Is this the correct version of the following problem?  :?

If x, y, z are real and x + y + z = xyz, minimise \sumcyclicx7(yz - 1).

[i]China 2003, TST 1, Problem 1[/i]

Well, if we consider just the case x,y,z>=0 (we hope this is the right problem), then the problem is is simple:
  sum x^7(yz-1)=xyzsum x^6-sum x^7=(x+y+z)(x^6+y^6+z^6)-sum x^7=sum x(y^6+z^6)>=2sum xy^3z^3>=6(xyz)^7/3 and of course xyz>=3sqrt(3).
  And I strongly believe that the problem is not correctly stated, since we could put x=-a, y=-b, z=-c with a,b,c>0 such that a+b+c=abc. Then we would need to maximize sum a^7(bc-1) and this is of course unbounded.

If x, y, z \geq 0 the problem is too simple. I think there must be another condition.

I also suspect that. After all, it's China Imo Selection Test, not a joke.


algebra
polynomial
quadratics
modular arithmetic
number theory unsolved
number theory

show that for $n\in\mathbb{Z}$, $n^{2}+n+1$ does not have any divisors of the form $6k-1$.

[b]Lemma.[/b] $-3$ is a quadratic residue of primes of the form $6n+1$ and a quadratic non-residue of primes of the form $6n+5$.

[i]Proof.[/i] See in G. H. Hardy and E. M. Wright page 75, theorem 96.

Now if exist $n$ such that $n^{2}+n+1$ have divisor form $6k-1$ then $n^{2}+n+1$ have prime divisor $p$ form $6k+5$. Then $4(n^{2}+n+1)\equiv 0\pmod{p}$ therefore $-3\equiv (2n+1)^{2}\pmod{p}$, so $-3$ is a quadratic residue of $p$. Contradiction!

This is the cyclotomic polynomial $\Phi_{3}(n)$. For prime $p$ and any divisor $d$ of $\Phi_{p}(n)$; $d\equiv~1(\mod~p)$ or $d\equiv~1(\mod~p)$.
In our case, suppose $6k-1|\Phi_{3}(n)$. Then $6k-1\equiv~2(\mod~3)$. Contradiction. $6k-1$ is not a divisor of $\Phi_{3}(n)$.


function
integration
calculus
real analysis
real analysis unsolved

Suppose that $ f: \mathbb{R}\to \mathbb{R}$ is such that $ |f(x)\minus{}f(y)|\le c|x\minus{}y|$ for all $ x,y\in \mathbb{R}$ for some $ c>0$.
Then it is clear that $ f$ is uniformly continuous.
Can we say anything about its differentiability?

Thanks.

The name of the hypothesis: $ f$ is Lipschitz, or $ f$ satisfies a Lipschitz condition.

$ f$ need not be differentiable everywhere. The simplest example would be $ f(x)\equal{}|x|,$ which fails to be differentiable at $ 0.$

Saying more about the differentiability of $ f$ requires quite a bit of analysis theory and machinery. $ f$ being Lipschitz implies that $ f$ is absolutely continuous on each bounded interval. That means that there is a locally integrable $ g$ such that we can write $ f(x)\equal{}a\plus{}\int_b^xg(t)\,dt$ and then $ f$ is differentiable almost everywhere and $ f'(x)\equal{}g(x)$ almost everywhere. We can also show that $ g\in L^{\infty}.$ In fact, that is the complete characterization: a function on the line is Lipschitz if and only if it is the indefinite integral of an $ L^{\infty}$ function.


function
real analysis
real analysis unsolved

$ \displaystyle f: [0,1]\to R$ is a twice differentiable function , such that $ f''_{(1)}\equal{}0$.  Prove that there is $ c\in ]0,1[$ such that $ f_{(0)}\equal{}f_{(c)}\minus{}cf'_{(c)}\plus{}c^2f''_{(c)}$
 :cool:

[quote="Diogene"]$ \displaystyle f: [0,1]\to R$ is a twice differentiable function , such that $ f''_{(1)} \equal{} 0$.  Prove that there is $ c\in ]0,1[$ such that $ f_{(0)} \equal{} f_{(c)} \minus{} cf'_{(c)} \plus{} c^2f''_{(c)}$
 :cool:[/quote]

[color=red]what dose $ c\in]0,1[$ mean?[/color]

[quote="limsup"][quote="Xingyuan"][color=red]what does $ c\in]0,1[$ mean?[/color][/quote]
I hope it is $ c\in(0,1)$[/quote]

[b]limsup [/b]is right  .. Now I hope that the proof will be there  :D
 :cool:


geometry
Gauss
topology
3D geometry
sphere
search
Functional Analysis

Sorry for old problem, but I need to learn some things I don't know, please
How many ways to define $ S^{\infty}$?, can we contruct some basic concept of differentiall geometry on $ S^{\infty}$ as on $ S^{n}$ maybe as geodesic, curvature, can we apply Gauss-Bonet theorem on $ S^{\infty}$, is $ S^{\infty}$ manifold?

[url=http://en.wikipedia.org/wiki/Banach_manifold]Banach manifolds[/url]

For example, a sphere in a Hilbert space is a smooth Hilbert manifold: any point has a neighborhood diffeomorphic to the Hilbert space itself. Geodesics are the arcs of great circles. Sectional curvature (using two-dimensional sections) is $ 1/R^{2}$. There are also ways to introduce "curvature bounds" (from above and below) on rather general metric spaces.

In a general Banach space the sphere is a topological (and Lipschitz) manifold, but it does not have to be smooth, and its curvature may be unbounded. Geodesics should exist, though. You may want to consult "Geometry of spheres in normed spaces" by Schaffer.

oh thank you for your link I did't know it under the name Banach manifolds, about the book Geometry of spheres in normed spaces" by Schaffer  do you have ebook? please I see the title is very interesting but [url=http://www.alibris.com/search/books/isbn/0824765540]its cost[/url] is so expensive.  :(

I don't have Schaffer's book in electronic form, and frankly speaking, it's a somewhat obscure subject. If you search Google Scholar for "Banach manifolds", you'll find more interesting sources. To me, the most interesting example of an infinite-dimensional manifold is the set of (smooth/Lipschitz/continuous...) maps $ f\colon M\to N$, where $ M$ and $ N$ are finite-dimensional manifolds.

Ok thank you for your helping, I am a student in mathematics but I am really intertested in physics and diffrential geometry is really as language of mathematics in physics.


There are only six red, four blue and eight brown buttons in a
jar. A button is randomly selected from the jar. What is the
probability that the button selected is $ \textit{not}$ blue? Express your answer as a common fraction.

The probability that it is not blue is $ 1\minus{}\frac{4}{18}\equal{}\boxed{\frac{7}{9}}$.

Ill answer this in a series of steps.

[hide="THIS IS MY CAREFULLY EXPLAINED ANSWER TO THIS MATH PROBLEM"]1.we first know that there are 18 marbles all together

2.and 4 of them are blue so we have a ratio of picking blue of 4/18

3.then we need to count the ones that we dont want to count and we get the equation 1-4/18

4.[hide="ANSWER"]7/9[/hide][/hide]

[hide="."][size=200][color=#808040][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!![/u][/i][/b][/color][/size][/hide]

[size=50]please post me if i was wrong thank you so much!!![/size]


geometry
invariant
parallelogram
power of a point
radical axis
geometry solved

Hi, this is a geometry conjecture I discovered a pair of weeks ago; but I know that it's  true because I checked it in my computer and by myself of course :D  So I would like that you help me to prove it... (I've tried the enough amount of time :blush:) So, here it goes:

Let ABC be a triangle, and let D, E, F be the feet of the altitudes from A, B, C, respectively. P and Q are the respective feet of the perpendiculars from D to AB and AC. Let R be the intersection of BE and DP, and let S be the intersection of CF and DQ. If M is the intersection of BQ and CP, and N the intersection of RQ and PS, prove that M, N and the orthocenter H of ABC are collinear.

Also, If you notice it's a problem already known, say, used in an olympiad or something, please tell me!!! Thanks

Since the radical axis of two circles having cevians as diameters passes through the orthocenter it is enough to prove that [tex]M,N[/tex] lie on the radical axis of the circles [tex]\Gamma_1, \Gamma_2[/tex] having [tex]CP[/tex], respectively [tex]BQ[/tex] as diameters.
First let's show that [tex]B,C,P,Q[/tex] are concyclic. Since [tex]A,D,P,Q[/tex] are concyclic ([tex]\angle DPA= \angle AQD= \frac{\pi}{2}[/tex]) we get [tex]\angle PQC=\pi-\angle AQP =\angle ADP =\angle CBA[/tex].
Thus [tex] QM\cdot MB= PM\cdot CM [/tex] and [tex]M[/tex] lies on the radical axis of [tex]\Gamma_1, \Gamma_2[/tex].

Now let [tex] S', R'[/tex] be the intersections of [tex]PS[/tex] with [tex]\Gamma_1[/tex] respectively of [tex]QR[/tex] with [tex]\Gamma_2[/tex]. Since [tex]Q,S',C,S[/tex] are concyclic (right angles at [tex]Q,S'[/tex]) we get [tex]\angle QS'P=\angle QS'S=\angle QCS =\frac{\pi}{2}- \angle BAC[/tex]. Similarly we get [tex]\angle QR'P=\frac{\pi}{2}- \angle BAC[/tex]. Thus [tex]P,Q,R',S'[/tex] are concyclic and [tex]QN\cdot NR'=PN\cdot NS'[/tex]. So [tex]N[/tex] lies on the radical axis of [tex]\Gamma_1, \Gamma_2[/tex] as well and the conclusion follows.

[quote="pestich"]This property is obviously not exclusive to orthocenter,
    it stands for any point X, the lines thru D have to be parallel to the cevians
    of X. The radical line proof is not applicable then.
    We need another proof.

    Maj.  Pestich[/quote]

Why the royal we? The proof is fine. Just project 2 of the cevians into altitudes by a parallel projection and the 3rd one follows the suit. Simple, dear Watson.

Actually, it's not that useless to consider the affine generalization of the problem. One can even go a step further and arrive at the following, projective generalization:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle, and H a point in its plane. Let the line AH meet the line BC at a point D. Let P and Q be arbitrary points on the lines AB and CA, and let the lines DP and DQ intersect the lines BH and CH at the points R and S, respectively. If M is the point of intersection of the lines BQ and CP, and N is the point of intersection of the lines RQ and PS, then the points M, N and H are collinear.[/color]

[i]Proof of Theorem 1.[/i] Since Theorem 1 is a pure incidence theorem, its statement is invariant under any non-degenerate projective transformation. Hence, we can WLOG assume that the line BC is the line at infinity. Then, the points B, C and D, all lying on this line BC, must be infinite points.

The lines AP and HR are parallel, since they intersect at the infinite point B. The lines AH and PR are also parallel, since they intersect at the infinite point D. Hence, the quadrilateral APRH is a parallelogram; therefore, $\overrightarrow{PR}=\overrightarrow{AH}$. Similarly, the quadrilateral AQSH is a parallelogram, and $\overrightarrow{QS}=\overrightarrow{AH}$. Thus, $\overrightarrow{PR}=\overrightarrow{QS}$, and it follows that the quadrilateral PRSQ is a parallelogram. Thus, PQ || SR.

Since the lines QM and RH have a common infinite point (namely, the point B), they are parallel. Similarly, the lines MP and HS, having a common infinite point (namely, the point C), must be parallel, too. Thus, altogether, we have PQ || SR, QM || RH and MP || HS. It follows that the triangles PQM and SRH are homothetic; thus, the lines PS, QR and MH concur. In other words, the point $N=QR\cap PS$ lies on the line MH. In other words, the points M, N and H are collinear. Theorem 1 is proven.

This is, by the way, a very nice example on how generalizing a problem simplifies its solution (although I find Christian Hirsch's solution of the original problem very beautiful).

  Darij

Thank you by your help, guys :) Ehhhhh, well, there's actually a "part b", so if you wanna try...

With the same notation of the original problem, let W=BE :cap: PS, X=CP :cap: RQ, Y= BQ :cap: PS, and Z=RQ :cap: CF. Show that BC, WX and YZ are concurrent.

Thank you in advance.

Part b, follows from part a, by the application of [url=http://planetmath.org/encyclopedia/DesarguesTheorem.html]Desargue's Theorem[/url].
From part a, we know that [tex]XZ\cap WY=N[/tex], [tex]BY\cap XC=M[/tex] and [tex]BW\cap CZ=H[/tex] are collinear.
Thus triangles [tex]\Delta XZC[/tex] and [tex]\Delta WYB[/tex] are perspective and the three lines [tex]BC, ZY, WX[/tex] are concurrent


Euler
combinatorics proposed
combinatorics

On the edges of a convex polyhedra we draw arrows such that from each vertex at least an arrow is pointing in and at least one is pointing out.
Prove that there exists a face of the polyhedra such that the arrows on its edges form a circuit.

[i]Dan Schwartz[/i]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31951]All the questions of the test[/url]

I think it was an easiest problem in the set, wasn't it?  :?

[quote="Myth"]I think it was an easiest problem in the set, wasn't it?  :?[/quote]Most likely problem 1 will be the one solved by the most contestants. This problem proved to be rather difficult, since most solutions are based on case argumentations and thus spread over many pages.

But maybe you have a short solution? :)

Consider the smallest cycle in our oriented graph (it is more convinient to consider it as a planar graph), i.e. the cycle, inside of which has the minimal number of facets. Then, if the inside of this cycle however has at least two facets, we go to this inside, and, by procceding the path, we form a smaller cycle.

I think my idea is the same as that of Fedor Petrov, but I think that when he says "a minimal cycle" he means a cycle that doesn't contain any other cycles inside of it (inside - geometrically speaking). Now, if it would not be a facet, it would be easy to take a vertex, say X, and then go forward from it and somewhere hit the cycle, at point say A. Also, by going back from X, we could hit the cycle at another point say B. It is easy now to obtain another cycle inside our cycle, so a minimal cycle is a facet....
I hope it was as clear as possible, and maybe even correct, although I suspect it is the same idea as that of Feodor Petrov!
A note : X could be a vertex of the cycle!!!! Even simpler I think would be to take X to be a vertex of a cycle, that has an edge, because if all the verices of the cycle wouldn't have any edges, it would be a facet!!!!
Now it became even easier I think!

:( You wrote it before me. (I received the notification only after 50 minutes Fedor wrote it)

That's is what I meant. A very easy problem.

Yes, DusT, it is exactly what I meant, but could not say in English.

i tried to find out what "facet" means on the net, but i didn't really get it

can anyone explain?

Fedor means "face" instead of "facet" :) (in Romanian "faţă").

[quote="DusT"]It is easy now to obtain another cycle inside our cycle, so a minimal cycle is a facet....
[/quote]

i don't see the way to construct a smaller cycle in our cycle

Well, let our cycle $\gamma$ be $A_1A_2\dots A_k$, and let $A_1B_1$ be an edge such that $B_1$ lies inside $\gamma$. Wlog, it is oriented as $A_1-B_1$. Next, we find edges $B_1-B_2$, then $B_2-B_3$ and so on, until some $B_k$ equals some $A_i$. So, we get a cycle: either $A_1B_1B_2\dots B_{k-1}A_iA_{i+1}A_{i+2}\dots A_nA_1$, or $A_1B_1B_2\dots B_{k-1}A_iA_{i-1}A_{i-2}\dots A_2A_1$. Is it clear now?

now i understand how to build the new cycle and what "inside" means

Let $F$ be the set o faces.Denote by $V$ the set of vertices and by $E$ the set of directed edges.By Euler relation we know that $|V|+|F|-2=|E|$.Now,suppose that there are no cycles in $F$.Then,for each face there must exist a vertice $v$ such that $\deg^+(v)\ge 2$.Denote by $M$ the set of this vertices,such that $|M| \le |F|$(if there are at least 2 vertices,we will choose only one to be element of $M$).Then,it is pretty simple that $\sum _{v \in M}\deg^+(v)\ge 2|F|$.But $\sum _{v \in V}\deg^+(v)= \sum _{v \in M}\deg^+(v)+ \sum _{v \in V - M}\deg^+(v)=|E|=|V|+|F|-2\Rightarrow  \sum _{v \in V - M}\deg^+(v) \le |V|-|F|-2$.But we know that$ \sum _{v \in V - M}\deg^+(v) \ge |V|-|M| \ge |V|-|F|\Rightarrow |V|-|F|-2 \ge |V|-|F|$ which is false.

Please check my solution :help:

[hide = Solution] Assume the polyhedron is made up of rubber sheet. If you have a directed cycle in the convex polyhedron, then call [i] atlas [/i] as the plane figure obtained by cutting the polyhedra along the directed cycle and flattening the cut out portion. Note that the boundary of an atlas, by definition is a directed cycle, and every vertex in the atlas has an inedge and outedge.

[b] Lemma [/b]: If an atlas $A$ has $F \geq 2$ faces, then it's possible to cut out a new atlast $A'$ from $A$ having $F' < F$ faces.
[hide = Case bash]
[b] Case - 1 [/b]: There's a vertex $v$ in $A$ not in the boundary
[hide = Proof] Start from the vertex $v$ and go to a vertex $u$ such that $\vec{vu}$ exists. Then go from $u$ to $w$ such that $\vec{uw}$ exists. Keep going that until you reach the boundary or visit a vertex already visited. If you visit a vertex already visited, then you're done since you form a directed cycle and can cut out the atlas along that cycle and form a smaller atlas. So assume you reach the boundary vertex $x$ through a directed path $v \overset{P_1}{\rightarrow} x$.

Start from the vertex $v$ and go backward to a vertex $u$ such that $\vec{uv}$ exists. Then go from $u$ to $w$ such that $\vec{wu}$ exists. Keep going that until you visit a vertex already visited (same handling as above) or a vertex in the path $P_1$ (again it's easy to see a directed cycle is formed so same as above) or to a boundary vertex $y$. So assume you reach the boundary vertex $y$ backward through a directed path $y \overset{P_2}{\rightarrow} v$.

Since both $x$ and $y$ are on the directed cycle forming the boundary, there's a directed path $P$ on the boundary from $x$ to $y$ (and to $y$ to $x$ also, but that's not needed). So we can now cut out the smaller atlas with the boundary $v \overset{P_1}{\rightarrow} x \overset{P}{\rightarrow} y \overset{P_2}{\rightarrow} v$ and this case is handled. [/hide]

[b] Case - 2 [/b] There's no vertex in $A$ not on the boundary
[hide = Proof] Then since the number of faces is greater than one, there's two different vertices $x,y$ on the boundary such that there's an $\bar{xy}$ edge. Assume $x \overset{P_1}{\rightarrow} y \overset{P_2}{\rightarrow} x$ is the boundary cycle.

If the edge $xy$ is from $x$ to $y$, then cut out the smaller atlas having the boundary $x \rightarrow y \overset{P_2}{\rightarrow} x$, or if the edge $xy$ is from $y$ to $x$, then cut out the smaller atlast having the boundary $y \rightarrow x \overset{P_1}{\rightarrow} y$. Either way, we're done. [/hide] [/hide]

Now since it's well know that a directed graph with edge having an inedge and outedge has a directed cycle, we first pick such an cycle and cut out it to form an atlas. Now we apply Lemma repeatedly to reduce the number of faces in it until it's one, when we're done. [/hide]


[quote="jli"]hey classroom math people.

theres this guy called interval who keeps posting kinda easy questions but ive been told not to move them so you guys could also try tackling those if you wanted.[/quote]
Not all the problems are easy for classroom people. Some are actually hard. Well, not really. OK, then maybe semi-hard but not all easy.

some of his problems are not fit for middle school
others are outrageously basic that they need to make an elementary school section for them.........
then there's the in-between problems that could go here.

The one about the Pythagorean one may be moved to MC I guess, but definitely not classroom math.

[quote="jli"]hey classroom math people.

theres this guy called interval who keeps posting kinda easy questions but ive been told not to move them so you guys could also try tackling those if you wanted.[/quote]

yeah i believe they are some easy and basic homework problems but it does fall under Algebra 2 and that does go with 9th grade curriculm (Even though im in 8th and our curriculm says we take it before geometry) and therefore worthy of HSB

i believe the following types of math may be considered for

HSB- geometry/alg 2
Classroom-anything lower than geometry
Intermediate- trig/extreme basic calc
Pre-olmyp-calc

I'm only in geometry. They seem too easy for me, so I guess it should DEFINITELY be in MC and below.

I'm in 8th grade and I'm in geometry (I've already done algebra) and some of these problems are kinda easy.
Also, for the classroom math, it says Middle School students & teachers only so maybe he didn't want to put them there.

There's no CALCulus in the Preolympiad fourms :!:


The smallest three digit number is $ 100$.

This was the FUNNIEST Alcumus problem I encountered.

I got 128
:D

Note to self: Think outside the box

The box is your friend.  Remember that $\binom{n}{1} = n$.  Or just look at that nice diagonal in Pascal's triangle.  The "first" diagonal is all $1$, but the second diagonal gives you the counting numbers.  When you count, you don't skip any numbers, so the answer is $100$.

For fun, the third diagonal gives triangular numbers, and the fourth diagonal gives tetrahedral numbers.  I suppose the next diagonal would give the 4D-hedral numbers.

[quote="GameBot"]What is the smallest three-digit number in Pascal's triangle?[/quote]
[hide="Solution"]Every number appears in pascals triangle (Hint: Second forward diangonal in triangle). So the answer is 100.[/hide]

HAHAHAHAHAH i wish that was on a math test.

:) XD got tricked and put 126. I thought they meant FIRST. Then I got it on my second try.

It is level 17 Counting+Probability. All you have to do is think.

There is every positive number in the pascal's triangle. How could it not be 100?

I got this problem in Probability - Think About It! and I was like "this is perfectly placed" then I realized it was a REVIEW PROBLEM :wacko:

I made this problem so complicated on the first try, but I should have saw a pattern  :noo:

[quote=gonoahgo][quote=GameBot]What is the smallest three-digit number in Pascal's triangle?[/quote]

so who cares[/quote]

I don't care; do you?

[quote=Jwenslawski][quote=Shadowlight]This would be a good alcumus problem for April Fools Day[/quote]

Why?[/quote]

its a troll problem

hmm looking at friends posts interesting problem

[quote=bluehurricane][quote=ArsenalFC]There is every positive number in the pascal's triangle.[/quote]

Wait, how exactly??[/quote]

$\dbinom{n}{1}=n,$ so $\mathbb{Z}\subseteq\{x\in\mathbb{Z}\,|\,x\text{  is a binomial coefficient}\}$. So it is just the smallest three-digit positive integer which is $\boxed{100}$.

[hide=Solution] 100 because every whole number is in Pascal's triangle.[/hide]

[quote=Andrew2019]how is this level 17 though
it's literally asking "what's the smallest three digit number?"[/quote]

yeah i think it should be around level 7

wow one of my first posts is here


function
limit
calculus
calculus computations

Find a discontinuous function on $ \mathbb{R}$ such that

$ \lim_{h\rightarrow 0} \frac{f(x\plus{}h)\minus{}f(x\minus{}h)}{h} \equal{} 0$ for every $ x\in\mathbb{R}$.

I was wondering whether a good answer to this problem would a step function of some kind?

Thanks.

Your suspicions are well-founded, my friend.

Consider the following function:

$ f(x) \equal{} 1$, if $ x \equal{} 0$
$ f(x) \equal{} 0$, otherwise.

It certainly satisfies the conditions that you require.


[quote="Anh Cuong"]Uhm.., this is not true for all $a,b,c >0$, simply take $a=b$ and judge the coefficient of $a^5$[/quote]
You're right, my ineq is wrong  :blush: .Just take c=0,a=1,b=2


The sequence 2, 3, 5, 6, 7, 10, 11, $ \ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes nor perfect fifth powers (in form of $ x^{5}$, $ x$ is an integer). What is the $ 1000^{\mathrm{th}}$ term of the sequence?

It's mainly just estimation: we know that we have to be above $ 1000$ and that there are $ 31$ perfect squares we already have to discount, so $ 32^2\equal{}1024\le x<33^2\equal{}1089$ is a nice place to start.

From the solution to this problem, where do they 1080 from. I thought they were dealing with 1089

Sorry to revieve but I don't understand the solution given by alcumus and in this thread.

[quote]The solution: We see that it's easier to count the number of perfect squares, perfect cubes and perfect fifth powers less than $ 33^2 \equal{} 1089$. We see there are 32 perfect squares less than 1089, which are $ 1^2$,$ 2^2$,...,$ 32^2$ and then there are 10 perfect cubes which are $ 1^3$, $ 2^3$,..., $ 10^3$. There are 4 perfect fifth powers less than 1089 which are $ 1^5$,..., $ 4^5$. Then notice there are 3 numbers that are both perfect squares and perfect cubes which are $ 1^6$, $ 2^6 \equal{} 64$  and $ 3^6 \equal{} 729$. There are also 2 numbers that are both perfect squares and perfect fifth powers which are $ 1^{10} $ and $ 2^{10} $ $ \equal{} 1024$. The only number which is both a perfect cube and a perfect fifth power is $ 1^{15} $. The only number which is a perfect square, perfect cube, and perfect fifth power all at the same time is $ 1^{30} $ $ \equal{} 1$. So within the first 1080 positive integers we need to get rid of $ 32 \plus{} 10 \plus{} 4 \minus{} 3 \minus{} 2 \minus{} 1 \plus{} 1 \equal{} 41$ integers which means the $ 1000th$ term is $ 1000 \plus{} 41 \equal{} 1041$. [/quote]

I don't understand the addition involved in this problem. Since there are 3 perfect sixth powers, 1 perfect tenth power, 1 perfect fifteenth power, and 1 perfect thirtieth power, shouldn't the answer be $1000 \plus{} 32 \plus{} 10 \plus{} 4 \minus{} 3 \minus{} 2 \minus{} 1 \minus{} 1 \equal{} 1039$ integers ???? why would the thirtieth power be added in the solution, instead of subtracted as in my solution?

The thirtieth power is added a total of three times: once for the squares, once for the cubes, and once for the fifth powers. It is then subtracted three times: once for the sixth powers, once for the tenth powers, and once for the fifteenth powers. This means it was counted a total of 0 times, so it must be added in once.

For more information, see here: [url=http://www.artofproblemsolving.com/Wiki/index.php/Principle_of_Inclusion-Exclusion]Principle of Inclusion-Exclusion[/url].

But isn't 1024 32^2? So that number shouldn't be included, right?

I started with 1089 and subtracted 41 to get 1048. What does this mean? Also, why do they just decide to add 41 to 1000?

Since 32^2 = 4^5 = 1024, 1024 is a good place to start. There are 32 perfect squares (1^2,.... 32^2), 10 perfect cubes (1^3,...,10^3), and 4 perfect fifth powers (1^5, ..., 4^5). However we have to be careful as we've overcounted some of them. In particular we've overcounted the number 1 two extra times, and we've also overcounted possibilities like x^6 = (x^2)^3 = (x^3)^2  (e.g. 4^3 = 8^2, and 9^3 = 27^2), so each of these was overcounted once. We also have the possibility x^10 = (x^5)^2 = (x^2)^5 (e.g. 32^2 = 4^5), this is another overcount by one. Hence we have 32+10+4-5 = 41 numbers that we should exclude from the first 1024 numbers. This means our sequence has 1024-41 = 983 numbers. We need 17 more numbers to make it into 973+17 = 1000 number sequence. Therefore from 1024 if we add the next 17 numbers, we get 1041

Guys... Y there is a plus one in the end?? (32+10+4-3-2-1[color=#f00]+1[/color]=41)
The solution says it was because $1^{30}=1$.........But i don't really get it.
Someone, Pls help

[quote=Kwow]Have you read all the solutions in this thread?  I think they explain it all.[/quote]

I got it~  :)

[hide][quote=Truffles]The solution says : We see that it's easier to count the number of perfect squares, perfect cubes and perfect fifth powers less than $ 33^2 \equal{} 1089$. We see there are 32 perfect squares less than 1089, which are $ 1^2$,$ 2^2$,...,$ 32^2$ and then there are 10 perfect cubes which are $ 1^3$, $ 2^3$,..., $ 10^3$. There are 4 perfect fifth powers less than 1089 which are $ 1^5$,..., $ 4^5$. Then notice there are 3 numbers that are both perfect squares and perfect cubes which are 1,$ 2^6 \equal{} 64$  and $ 3^6 \equal{} 729$. There are also 2 numbers that are both perfect squares and perfect fifth powers which are $ 1^19$ and $ 2^10 \equal{} 1024$. The only number which is both a perfect cube and a perfect fifth power is . The only number which is a perfect square, perfect cube, and perfect fifth power all at the same time is $ 1^15 \equal{} 1$. So within the first 1080 positive integers we need to get rid of $ 32 \plus{} 10 \plus{} 4 \minus{} 3 \minus{} 2 \minus{} 1 \plus{} 1 \equal{} 41$ integers which means the $ 1000th$ term is $ 1000 \plus{} 41 \equal{} 1041$.

Why decide to count the number of perfect squares, cubes, and fifth powers below 33^2?  I know that it would be easier to count the squares, cubes, and fifth powers and then subtract.  But what made them decide below 33^2?[/quote][/hide]

You gotta hide it!!!!!!!!

[quote=Truffles]The solution says : We see that it's easier to count the number of perfect squares, perfect cubes and perfect fifth powers less than $ 33^2 \equal{} 1089$. We see there are 32 perfect squares less than 1089, which are $ 1^2$,$ 2^2$,...,$ 32^2$ and then there are 10 perfect cubes which are $ 1^3$, $ 2^3$,..., $ 10^3$. There are 4 perfect fifth powers less than 1089 which are $ 1^5$,..., $ 4^5$. Then notice there are 3 numbers that are both perfect squares and perfect cubes which are 1,$ 2^6 \equal{} 64$  and $ 3^6 \equal{} 729$. There are also 2 numbers that are both perfect squares and perfect fifth powers which are $ 1^{19}$ and $ 2^{10} \equal{} 1024$. The only number which is both a perfect cube and a perfect fifth power is . The only number which is a perfect square, perfect cube, and perfect fifth power all at the same time is $ 1^{15} \equal{} 1$. So within the first 1080 positive integers we need to get rid of $ 32 \plus{} 10 \plus{} 4 \minus{} 3 \minus{} 2 \minus{} 1 \plus{} 1 \equal{} 41$ integers which means the $ 1000th$ term is $ 1000 \plus{} 41 \equal{} 1041$.

Why decide to count the number of perfect squares, cubes, and fifth powers below 33^2?  I know that it would be easier to count the squares, cubes, and fifth powers and then subtract.  But what made them decide below 33^2?[/quote]

LaTeX errors

[quote=Mathcounts829][quote=sw993][quote=Truffles][hide]The solution says : We see that it's easier to count the number of perfect squares, perfect cubes and perfect fifth powers less than $ 33^2 \equal{} 1089$. We see there are 32 perfect squares less than 1089, which are $ 1^2$,$ 2^2$,...,$ 32^2$ and then there are 10 perfect cubes which are $ 1^3$, $ 2^3$,..., $ 10^3$. There are 4 perfect fifth powers less than 1089 which are $ 1^5$,..., $ 4^5$. Then notice there are 3 numbers that are both perfect squares and perfect cubes which are 1,$ 2^6 \equal{} 64$  and $ 3^6 \equal{} 729$. There are also 2 numbers that are both perfect squares and perfect fifth powers which are $ 1^{19}$ and $ 2^{10} \equal{} 1024$. The only number which is both a perfect cube and a perfect fifth power is . The only number which is a perfect square, perfect cube, and perfect fifth power all at the same time is $ 1^{15} \equal{} 1$. So within the first 1080 positive integers we need to get rid of $ 32 \plus{} 10 \plus{} 4 \minus{} 3 \minus{} 2 \minus{} 1 \plus{} 1 \equal{} 41$ integers which means the $ 1000th$ term is $ 1000 \plus{} 41 \equal{} 1041$.[/hide]

Why decide to count the number of perfect squares, cubes, and fifth powers below 33^2?  I know that it would be easier to count the squares, cubes, and fifth powers and then subtract.  But what made them decide below 33^2?[/quote]

LaTeX errors[/quote]

LaTeX errors? There aren't any LaTex errors...[/quote]

I fixed the LaTeX errors.

We see that it's easier to count the number of perfect squares, perfect cubes and perfect fifth powers less than $33^{2}=1089$. We see there are 32 perfect squares less than 1089, which are $1^2$, $2^2$, $\ldots$, $32^2$ and then there are 10 perfect cubes which are $1^3$, $\ldots$, $10^3$. There are 4 perfect fifth powers less than 1089 which are $1^5$, $\ldots$, $4^5$. Then notice there are 3 numbers that are both perfect squares and perfect cubes which are 1, $2^{6} = 64$ and $3^{6} = 729$. There are also 2 numbers that are both perfect squares and perfect fifth powers which are $1^{10} = 1$ and $2^{10} = 1024$. The only number which is both a perfect cube and a perfect fifth power is $1^{15}=1$. The only number which is a perfect square, perfect cube, and perfect fifth power all at the same time is $1^{30}=1$. So within the first 1089 positive integers we need to get rid of $32+10+4-3-2-1+1 =41$ integers which means the $1000^{\text{th}}$ term is $1000+41 = \boxed{1041}$.

If you're interested, I wrote the following C++ program to check my answer to this problem (somewhat inefficient, can be optimized a lot further, but who cares lol...C++ is really fast):
[hide=C++ Code]
[cpp]
#include <iostream>
#include <vector>
using namespace std;

bool power(int n, int k){
    bool b=false;
    for(int i=1; i<=n; i++){
        int prod=1;
        for(int j=1; j<=k; j++){prod*=i;}
        if(prod==n){b=true; break;}
    }
    return b;
}

int main(){
    vector <int> v;
    int i=1;
    while(true){
        if(v.size()==1000){
            break;
        }
        if(!(power(i, 2) || power(i, 3) || power(i, 5))){
            v.push_back(i);
        }
        i++;
    }
    cout << v.at(999) << "\n";
    return 0;
}
[/cpp][/hide]

[quote=OronSH]i thought 1042[/quote]

me too  :(

[quote=Gerard.123][quote=Benq]@TMM There are 41 terms under 1089 which fit into one of the above categories. Therefore, 1088 is the 1088-41=1047th term. Subtract 47 from each sides to get the answer.[/quote]

I would like to point out that, while the solution is correct, it would not work if there were any powers of 2, 3 or 5 between 1041 and 1089, and I think that should be addressed in the official solution.[/quote]

I reported this as a bug and the solution has been rewritten.

first power problem!

[hide=insert shruggie]
we first guess 1050 as the number
so there are ;;

32 squares
10 cubes
4 perfect fifth powers

in the range 0-1050, to get 46 to subtract

but there are overlaps, we compensate for this by subtracting the overlaps ;;

3 perfect 6th powers (n^2 and n^3 overlap)
2 perfect 10th powers (n^2 and n^5 overlap)
1 perfect 15th power (n^3 and n^5 overlap)

to get 40 to subtract

but there is an overlap in the subtraction :|

so we add back this overlap ;;

1 perfect 30th power (all three overlap)

to get 41 to subtract

since we want 1000th term, we add 1000+41, to say that [answer] - 41 = 1000, to get 1041
[/hide]

[quote name="OronSH" url="/community/p10751988"]
i thought 1042
[/quote]

omg sameeeeeeee

(oop user is inactive for 2 days but wtv)

I did this problem today (maybe yesterday) on FTW.

Also, sorry to bump with something off-topic.

[hide=easier sol]do PIE on the number of unique squares, cubes, powers of 5 until 1000. Computing gives 41. Adding that to 1000 gives $ \boxed{1041}.$  [/hide]


number theory
least common multiple
greatest common divisor
modular arithmetic
Euler
relatively prime

Find the last digit of $ \text{lcm}(F_2,F_3,\cdots,F_{1985})$

Note that $ F_n$ is a Fermat number not a Fibonacci number and by the way shouldn't they have different letters at least to differentiate between them??

If you mean GCD, the answer should be $ 1$ as any two Fermat numbers are relatively prime.

Well, the question statement is to find the last digit

I think $ 1$ is the last digit of $ 1$. :wink:

$ 0\le 1\le 9$ and $ 1\equiv 1\pmod{10}$. Hence the conclusion.

[quote="hasan4444"]LOOL I didn't understand anything :D sorry I'm very bad at NT but what do you mean by $ 0 \le 1 \le 9$[/quote]
Hasan you're really the limit... How can you [size=200][b]NOT[/b][/size] know the last digit of the number "$ 1$"
I think that you're REALLY spamming asking such things which an infant is taught...

[quote="hasan4444"]LOOL sorry about that but this is not your business. So stop the spam and answer me.

LOOL I'm sorry now I just get it but I think the problem is harder than just finding that GCD of them is 1 and say so  :|
[b]
Edit: [/b]I saw a very long solution of it in the past but it was ended with 1 also that's why I got confused[/quote]
First of all its not LOOL but LOL with a single O in it (It stands for "Laugh Out Loud  :D ")

And you can check that two of these nos. ($ F_2,F_3$ are)are prime acc to me...so the result of the GCD being 1 immediately follows...

It's reasonably clear that the question [i]isn't[/i] meant to be about the GCD.  (How do I know?  Because that's a silly and not very interesting question to ask.)  What's not clear is what the question actually [i]is[/i] meant to ask.  Hasan, since you're the one who provided the question, perhaps you could clarify what it is supposed to mean -- what is meant by $ (F_2, \ldots, F_{1985})$?  (In case it wasn't clear to you, everyone else on the thread has been treating this as if it means the GCD of the $ F_i$.)

[quote="hasan4444"][code]\lcm(F_{2},F_{3},\cdots,F_{1985})[/code][/quote]

Looking at the code is the first step...

(Note to hasan4444: The "\lcm" command does not exist in LaTeX. Instead, you should type \text{lcm} to get the following)
\[ \text{lcm}(F_{2},F_{3},\cdots,F_{1985}).\]
And [b]that[/b] is what the question is meant to ask.

Wow.
Well, now that we know it's LCM...

[hide="Solution"]Going off the fact that any two Fermat numbers are relatively prime, the LCM of those numbers is just the product, so we have $ \prod_{n \equal{} 2}^{1985} 2^{2^n} \plus{} 1\bmod 10$.

Suppose $ 2^{2^n}\equiv 2x\bmod 10$, so $ 2^{2^n \minus{} 1}\equiv x\bmod 5$.  Using FLT, we need $ 2^n \minus{} 1\bmod 4$, which is clearly $ \minus{} 1\equiv 3$ for $ n\geq 2$. So $ 2^{2^n \minus{} 1}\equiv 2^3\equiv 8\equiv 3\bmod 5$, and $ x \equal{} 5k \plus{} 3$.

Therefore, $ 2^{2^n}\equiv 6\bmod 10$ for $ n\geq 2$, so $ F_n\equiv 7\bmod 10$.

So the product is $ 7^{1984}\bmod 10$. $ 1984\equiv 0\bmod\varphi(10)$, so $ 7^{1984}\equiv\boxed{1}\bmod 10$.[/hide]

Oh Thanks a lot guys I didn't notice that it wasn't there. Thanks a lot

Well firstly a Fermat number is a number that is in the form 2^2^n+1. So for all n>=2 2^n is divisible by four. So then that would make the units digit of 2^2^n+1 7. There are 1984 numbers in the set, and since all Fermat numbers are relatively prime, the last digit of 7^1984 would be one.

for all n>1,the last digit of 2^(2^n) is 6 because6^m is 6 mod 10 for any m>0
so we need to just count 7^1984,by euler
7^4 is 1 mod10 and 4 divides 1984,so its 1


Find all positive integers x and y such that $x^2+xy=y^2+1$

Deleted: wrong

I'm sorry i did it in a hurry

That's not correct, you factored the equation incorrectly.  In fact there are other solutions such as $\ (x,y) = (2,3)$

I think there are infinitely many solutions.  Consider the given equation as a quadratic in $\ x$.  The discriminant becomes $\ 5y^2 + 4$ which must be a perfect square in order to satisfy the condition of integer solutions.

You should actually look at the Fibonacci numbers :) and maybe you know how to solve IMO 1988 Problem 6?
Well, that technique might be useful as well :D

This is actually quite hard for Pre-Olympiad but I'll leave it here since it has already been discussed in the Advanced section, and that would spoil your problem solving fun :)

incidentally im solving a similar question and notice some pattern (which I am latter told that it is cassini idenity)where i posted few days ago in

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89971[/url]

notice that

$F_{n-1}F_{n+1}-F^2_n=F^2_{n-1}+F_{n-1}F_n-F^2_n$ and using the identity in above post and we can solve the above question . The problem is that I dont know how to prove that " this pattern" is the unique solution for above equation and no others.

$x^2+xy=y^2-1$
$(x+y)x=(y-1)(y+1)$
$\Rightarrow x\mid(y+1)\Rightarrow x=1;y^2-y=2\Rightarrow y=2$ :|

Here's a proof that there are no other solutions.

Let's suppose that there are other solutions, and choose such $(x, y)$ that $x$ is minimal.

First we prove that $y>x$. If $y\leq x$ then $y^2+1\leq x^2+1 < x^2+xy$ (unless $xy=1$, giving as solution $(1, 1)$ which is of the form $(F_{2n-1}, F_{2n}$). Then we prove that $2x>y$ in a similar manner.

Now we observe that $(2x-y, y-x)$ is also a solution of the original equation. Since $x<y$, $2x-y<x$, hence $2x-y=F_{2n-1}$ and $y-x=F_{2n}$ for some positive integer $n$. But now $x=2x-y+y-x=F_{2n-1}+F_{2n}=F_{2n+1}$ and $y=y-x+x=F_{2n}+F_{2n+1}=F_{2n+2}$. Contradiction.


[quote="Mark1"]The pole is 28 feet high.  Point A is at a 60 degree angle (connecting to point B) away from the pole on the ground.  Point B is at the top of the pole, point C is at the bottom of the pole.

What is the length from point A to point B?

Thank you.[/quote]

Anyone?

Thank you.

[quote="demji"]easy
[hide]
since we can make it into a 30-60-90 triangle with one leg 28, across from angle A=30 since angle B=60.  Thus $AB=2*BC=2*28=56$
[/hide][/quote]

Thank you, demji  :)

its quite a simple triangle identity, i reccommend you learn these.  You can also use trigonometry to solve the problem.  Also i reccommend posting problems like this in getting started.


[img]http://img148.imageshack.us/img148/60/sepppnm3.jpg[/img]

What is the solution?
-*geniuises is not the answer to all question but the question to all answer*

This is highly trivial to be in Olympiad Section, I suspect it's your HW in Math 9.

You should find all natural solutions to: |(88+n)/(19+n)-88/19|<2.

It isn't as hard as you think.
Because $ 4<\frac{88}{19}<5$,       $ 2<\frac{88\plus{}n}{19\plus{}n}<3$. It means that $ 38\plus{}2n<88\plus{}n \Leftrightarrow n<50$ and $ 88\plus{}n<57\plus{}3n \Leftrightarrow n>15,5$. That is $ 15,5<n<50$. So $ n \in [16; 49]$. There are $ 49\minus{}16\plus{}1\equal{}33$ of these positive integers.

No, really, this is extremelly easy, if I understood it correctly.

[hide="For those who are startled by the contrast of the title and the content"]Check out the other topic titles and contents of hotgal24:
http://www.mathlinks.ro/search.php?search_author=hotgal24
 :P  :P  :rotfl: [/hide]


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ A\in\mathcal{M}_{n,k}(\mathbb{C})$ and $ B\in\mathcal{M}_{k,n}(\mathbb{C})$ such that $ rank(AB)\equal{}k$ ,$ (AB)^2\equal{}AB$.Prove that $ BA\equal{}I_k$

$ \text{rank}(BA)\equal{}\text{rank}(AB)\equal{}k.$

$ (BA)^2\equal{}B(AB)A\equal{}B(AB)^2A\equal{}(BA)^3.$

So, $ BA$ satisfies the polynomial equation $ x^3\minus{}x^2\equal{}0$ and all of its eigenvalues are either $ 0$ or $ 1.$ But $ BA$ is of full rank, hence invertible, as a $ k\times k$ matrix, so $ BA$ cannot have $ 0$ as an eigenvalue.

$ BA$ has $ 1$ as an eigenvalue of algebraic multiplicity $ k.$ Since the geometric multiplicity of an eigenvalue is greater than or equal to its algebraic multiplicity, we have that $ BA$ has $ 1$ as an eigenvalue of geometric multiplicity $ k.$ So $ BA\minus{}I$ has nullity $ k,$ hence rank $ 0,$ hence it is the zero matrix. We conclude that $ BA\equal{}I.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=86370]This[/url] is basically the same problem.


Estoy colgando el examen de las olimpiadas de un nivel aceptable para esta seccion
 que es de nive de inciacion en olimpiadas .
espero les agrade...

Carlos Bravo ;)
----
[b][size=150][color=blue]VII Olimpiada del Cono Sur
PERU 1996[/color][/size] [/b]

$\blacktriangleright$  [color=blue]Problema 1.[/color]
En la figura siguiente, el cuadrado mayor esta dividido en dos
cuadrados y tres rect\'angulos, conforme se muestra:
\includegraphics{conosur96_prob1.eps}
El area de cada cuadrado menor es igual a $a$ y el area de
cada rectangulo menor es igual a $b$. Si $a+b = 24$ y la raiz
cuadrada de $a$ es un numero natural, halle todos los posibles
valores para el area del cuadrado mayor.

$\blacktriangleright$  [color=blue]Problema 2.[/color]
Considere la secuencia de numeros reales positivos $a_0, a_1, a_2,...$ definida por:
$a_{n+1} = a_n + \frac{1}{a_n}, \forall n\geqslant 0.$
Muestre que, respecto al valor de $a_0$, siempre se tiene
$a_{1996} > 63$.

$\blacktriangleright$ [color=blue]Problema 3.[/color]
Una tienda vende botellas con las siguientes capacidades: $1l, 2l, 3l, ..., 1996l$.
Los precios de las botellas satisfacen las siguientes condiciones:
(a) Dos botellas tienen un mismo precio si y solo si sus
capacidades $m>n$ son tales que $m-n=1000$.
(b) Cada botella de $m$ litros ($1\leqslant m \leqslant 1000$)
cuesta $1996 - m$ soles.
Determine todos los pares de capacidades $(m,n)$ de las botellas
tales que:
  (i) $m+n = 1996$.
 (ii) El costo total del par sea el menor posible.
(iii) Con este par de botellas, podamos medir $k$ litros, para
cada entero $k$ desde $1$ hasta $1996$.
(Nota: Las operaciones permitidas son: echar o sacar cualquiera de
las botellas; pasar liquido de una botella para otra. Habremos
conseguido medir $k$ litros cuando la cantidad de litros en una
botella mas la cantidad de litros en otra, fuera igual a $k$).

$\blacktriangleright$ [color=blue]Problema 4.[/color]
La secuencia $0, 1, 1, 1, ..., 1$ contiene $1996$ n\'umeros,
siendo el primero igual a $0$ y todos los demas iguales a $1$.
Escogemos dos o mas numeros de la secuencia (mas no todos
ellos), y sustituimos cada uno de esos numeros por la media
aritmetica de los numeros escogidos, se obtiene asi una
nueva secuencia de $1996$ mumeros. Muestre que existe una
secuencia de esas operaciones que nos permite obtener una
secuencia con $1996$ n'umeros iguales.

$\blacktriangleright$ [color=blue]Problema 5.[/color]
Pretendemos cubrir totalmente un cuadrado de lados $k (k>1$ entero
$)$ con los siguientes rectangulos: $1$ rectangulo $1\times1$,
$2$ rectangulos $2\times1$, $4$ rectangulos $3\times1$, ...,
$2^n$ rectangulos $(n+1)\times1$, de manera que los
rectangulos no se superpongan ni excedan los limites del
cuadrado. Halle todos los valores de $k$ para los cuales esto es
posible y, para cada uno de tales valores dise\~ne una soluci\'on.

$\blacktriangleright$  [color=blue]Problema 6.[/color]
Hallar todos los enteros $n>2$ tales que exista un conjunto $S_n$
formado por $n$ puntos en el plano, satisfaciendo las siguientes
condiciones:
(a) Tres puntos cualesquiera no son colineales.
(b) Ningun punto es interior a cualquier circulo que tenga por
diametro dos de los puntos de $S_n$.

$\text{\LaTeX}{}ed$ by Carlos Bravo 2006.

Haber .. aki la version en pdf ..

les sugieron que los intenten y cuelguen sus soluciones .....
o comentarios de soluciones previas ..

Graciias .

Carlos Bravo
Lima -Peru

¿Cómo se podría hacer para que los problema de la olimpiada del Cono Sur salgan en $Resources$ ?, para que se pueda comentar la solución de cada uno de los problemas por separado.

$Tipe$

Haber , debemos contactarnos con ....
 la persona encargada ..
en este caso es :

orl

msn: orl@gmx.de

Pero debemos presentarlo en ingles ..
obviamante ..

Carlos  Bravo ;)


inequalities
factorial
calculus
integration
real analysis

proove: n! > (n/e)^n
where e is the base of a natural log.
and n is the nth riemann sum

[quote="Chris Redifield"]
and n is the nth riemann sum[/quote]

what does it mean?

[quote]
Chris Redifield wrote:

and n is the nth riemann sum
[/quote]

Hm...I think you should ignore it, Liyi...
That looks like a problem I've recently come across:
Prove the following inequality:
[tex]\Big(\frac{n}{e}\Big)^n<n!<e\Big(\frac{n}{2}\Big)^n, n\in\mathbb{N}[/tex]

Shall I place my solution for it here?

i meant to say using the nth riemann sum as a hint

sorry about that i meant to say just prove n! (n factorial) is greater than (n/e)^n (the e is 2.718281828...) and ignore the riemann sum statement.

Stirling's Approximation

\[\sqrt{2\pi n} (\frac{n}{e})^n \leq n! \leq \sqrt{2\pi n}(\frac ne)^n (1+\frac{1}{12n-1})\]

Look at the graph of ln n and  notice:
Integral (from 1 to n) ( ln x ) dx    < ln 1 + ln 2 + ... ln n

or n ln n -n < ln (n!)

or (n/e)^n < n!

OK...right...but what if it's asked to solve this problem without knowing what an integral and the Stirling's approximation are?...Because I found my problem in a problembook in the chapter on sequences and limits... :?

do this without stirling's approximation

Gyan gave a proof using integration...
i think the proof probably uses some calculus, since there is $e$ which is a limit...

Same thing  as integral, if you prefere:

Start with

(1+1/n)^n  < e

or  ((n+1)^(n+1))/(n^n) < e (n+1)

so we have
e* n>   (n^n/((n-1)^(n-1))
e* (n-1) >( (n-1)^(n-1))/((n-2)^(n-2))

e *2 > 2^2/1^1)

Mulityply all togather, and you get:

e^n  * n! >   n^n
QED


Let $ n$ be a $ 5$-digit number, and let $ q$ and $ r$ be the quotient and remainder, respectively, when $ n$ is divided by $ 100$. For how many values of $ n$ is $ q \plus{} r$ divisible by $ 11$?

$ \textbf{(A)}\ 8180 \qquad
\textbf{(B)}\ 8181 \qquad
\textbf{(C)}\ 8182 \qquad
\textbf{(D)}\ 9000 \qquad
\textbf{(E)}\ 9090$

[quote="xpmath"][hide="hint"]We have n=100q+r, so n=99q+q+r. Now, 11|q+r, so we have 11|n.[/hide][/quote]
Too lazy to type that, I'll just quote it.
Next:
[hide]
There are 90000 5 digit numbers, so there are floor(99999/11)-floor(9999/11) numbers that satsify the condition which is 8181.[/hide][/hide]

$n=100q+r$ which would be. $99q+(q+r)$. $99$ is already divisible by $11$. So for the $n$ to be $0mod11$ (since we want $(q+r)$ to be divisible by $11$), we can use floor functions. Using division we find minimum of $n$ is $10010$, smallest $5-digit$ multiple $11$ and $99990$, large $5-digit$ multiple $11$. These are $910, 9090$ respectively. $9090-910+1=8181, B.$

We have $n=100q+r$, and $q+r=11k$. Substituting, we get $n=99q+11k=11(9q+k)$. Thus, we need the $5$-digit multiples of $11$. The largest is $99990$ and the smallest is $10010$, so there are $$\frac{99990-10010}{11}+1=\boxed{8181}$$ such $n$.

Let $n=100q+r$. Then, if $11\mid q+r$ then $q+r+11(99q)\equiv 0\pmod{11}\to n\equiv 0\pmod{11}$. This is basically asking for the number of 5 digit numbers that are divisible by $11$. The first one is $10010=910\cdot 11$. Then, the last one is $99990=9090\cdot 11$. Thus, the answer is $9090-910+1=\boxed{8181}$.

Note, that $n=100a+b,$ then if $11|(a+b),$ then we have that $11|n,$ because $100a+b=a+b+11(9a).$ So, this is basically the number of $5,$ digit numbers that are divisible by $11.$ Now, we have $9090-910+1=\boxed{8181}.$

:D :yup: :)
:D :yup: :)
:D :yup: :)


calculus
derivative
calculus computations

could someone please explain how to find the second derivative of

$\sin(y)-xy+x^{3}=0$.

[hide=" i get"]
 $y'=-\frac{3x^{2}-y}{\cos(y)-x}$
so taking the derivative again i get

$y''=-\frac{(\cos(y)-x)(6x-y')-(3x^{2}-y)(-\sin(y)y'-1)}{(\cos(y)-x)^{2}}$

and then simplify. i am wondering if i did this correctly and if there is a faster way to solve this. thanks
[/hide]

Might be easier to take two implicit derivatives as the beginning, then substitute for $y'$ and solve for $y''$.  No promises, though.

[quote="maokid7"]
i am wondering if i did this correctly and if there is a faster way to solve this. thanks[/quote]

Yeah, I got the same answer. I tried both ways, and neither seems to be particularly faster. Matter of taste I suppose *shrug*


function
probability
calculus
derivative
LaTeX
real analysis
algebra

Suppose there is a probability W(n,m), n and m are non-negative integers, and satisfying W(n,0)=0, W(n,m)=1 if n<=m-1. When n>=m, the following recurrence equation holds: W(n,m)=W(n-2,m)*n/(n+m)+W(n-1,m-1)*m/(n+m).
My question is: Does there exist a real smooth function W1(x,y) defined on the interval [0,Infinity)*[0,Infinity), and satisfying W1=W on the integer points, i.e., W1 is the continuous and smooth generalization of W. If W1 exists, please give your construction, thanks.

Yes, but you're asking the wrong question. We could fit a smooth function to any values on a discrete set, and it doesn't tell you anything about those values.

An explicit construction is given by a sum of [url=http://en.wikipedia.org/wiki/Bump_function]bump functions[/url] and reveals nothing about the original set of values.

Yes, I asked the wrong question. My original question is : can we generalize W to a smooth function W1 and satisfying W1(x,y)=W1(x-2,y)*x/(x+y)+W1(x-1,y-1)*y/(x+y)?

Even there, "smooth" is not really enough- given any solution, we could add a bump in one place and translate it with the functional equation. I think you really want an analytic function satisfying the functional equation $ (x\plus{}y)W(x,y)\equal{}xW(x\minus{}2,y)\plus{}yW(x\minus{}1,y)$ and the chosen initial condition, which could lay claim to being "the" extension. For that question, you have the problem that you're specifying too many initial conditions, and some of them don't satisfy the functional equation. In particular, the functional equation isn't satisfied when $ n\equal{}m$ and both are integers greater than or equal to $ 2$. You can't keep a problem like that down in one place, so you need to change some of those initial conditions.

Yes, I  really want an analytic function satisfying the functional equation:

(x+y)W(x,y)=x*W(x-2,y)+y*W(x-1,y-1)

and the chosen initial conditions (on integer points). If we remain the initial conditions the same, but consider a W with continuous derivatives of second order, then can we construct the W?

Relaxing the smoothness condition won't help. The problem is the initial conditions, which force the functional equation to be false at some points. If it's false at those points, it's false on an open set. If we have analyticity, it's false on an open set which is dense in the domain.

Sorry, I do not understand why when n=m  and both are integers greater than or equal to 2, the functional equation isn't satisfied. I think for all integers n>=2 and m>=1, W(n,m) is obtained by the recurrence equation. The boundary are n<=m-1, or m=0.

Oops. I misread the functional equation- you should have noticed the error in what I wrote. I also misread $ (n,0)$ as $ (n,n)$ in the initial conditions- which are really boundary conditions, now that I see the system i the proper light.

Now, a triangular wedge is a fairly awkward domain. How about a change of variables?
Let $ u=\frac{x-y}{2}$, $ v=y$, and set $ F(u,v)=W(x,y)$. The functional equation $ (x+y)W(x,y)-x\cdot W(x-2,y)-y\cdot W(x-1,y-1)=0$ becomes $ (2u+2v)\cdot F(u,v)-(2u+v)F(u-1,v)-v\cdot F(u,v-1)=0$, or $ (2u+v)[F(u,v)-F(u-1,v)]+v[F(u,v)-F(u,v-1)]=0$, with boundary conditions that $ F$ is zero on lattice points on the horizontal axis and one at $ (-\frac12,n)$ and $ (-1,n)$ (Why not the vertical axis $ x=y$/$ u=0$?). Now this looks like a discretized partial differential equation, of which I sadly know relatively little. There likely is an analytic solution (except for isolated singularities), but I don't know much about how to find it.

A side note: the pretty math is $ \text{\LaTeX}$, for which there are various guides on the site. Math mode is enabled on the forum, by putting the code between dollar signs.

Sorry, I thought it was only your written mistake :)  The following is the origin of my problem:
In research I came with a probability W(n,m), n and m are non-negative integers, and satisfying W(n,0)=0, W(n,m)=1 if n<=m-1. When n>=m, the following recurrence equation holds: \[ W(n,m)\equal{}W(n\minus{}2,m)\times \frac{n}{n\plus{}m}\plus{}W(n\minus{}1,m\minus{}1)\times \frac{m}{n\plus{}m}\]. I want to analyze the property of W(n,m) when n and m approach Infinity using the technique of continuous analysis, so the continuous generalization (with at least continuous second order derivate) of W(n,m) is needed.

If you want approximations, don't interpolate. Instead, use the PDE form $ (2u\plus{}v)F_u\plus{}v\cdot F_v\equal{}0$ (after the change of variables I gave in my last post), with boundary conditions 0 on the $ u$-axis and $ 1$ on the $ v$-axis.
Is this form simpler and easier to understand? Yes.
Is the solution a decent approximation? Maybe.
Is the solution equal to the original solution anywhere? No.

So, we try to solve this PDE- and we run into trouble, because the solution is not unique if there is a smooth one. If $ f(u,v)$ is a solution, so is $ f(tu,tv)$ for any fixed $ t>0$. Looking for a purely radial solution to avoid this leads to the fact that $ uf_u\plus{}vf_v\equal{}0$ for any radial function $ f$, so $ (u\plus{}v)f_u\equal{}0$, and the only radial solution is $ F(u,v)\equal{}\begin{cases}1&v>u\\0&v<u\end{cases}$
This is not close to the discrete version. While it's accurate on the $ v>u$ side, it's nowhere close below.


algebra
polynomial
calculus
quadratics
real analysis
real analysis unsolved

3x^2-2x-8.....want whole worked solution step by step

Why is this calculus? Are you asking for the value of x such that the polynomial equals zero? Because if you are, it can be done by simply using the quadratic formula...

Unless I'm missing something.


A straw is placed into a rectangular box that is 4 inches by 3 inches by 8 inches. If the straw fits exactly into the box diagonally from the bottom left front corner to the top right back corner, how long is the straw, to the [i]nearest tenth of an inch[/i]?

THanks for replies.

In the wrong section.  Should be in Classroom Math or thereabouts.

[hide="Solution"] $\sqrt{ 3^{2}+4^{2}+8^{2}}\approx 9.4$ [/hide]


number theory
prime numbers
number theory proposed

Show that there are infinitely many prime numbers $p$ having the following property: there exists a natural number $n$, not dividing $p-1$, such that $p|n!+1$.

Take an odd prime $q$. Let $p \equiv 3 \ mod \ 4$ be a prime divisor of $(2q)!-1$. We have $(p-2q-1)! \equiv (p-1)! \equiv-1 \ mod \ p$. So we have to check that $p-2q-1 \not |p-1$. Assume to the contrary that it holds. As $p-1 \equiv 2q \ mod \ p-2q-1$ so $p-2q-1 |2q$. Hence $p-2q-1=2q$ or $p-2q-1=2$. In both cases we get a contradiction looking modulo $4$ (as $p\equiv 3 \ mod 4$). Now it's easy to see that there are infinitely many numbers $p$ with wanted property. QED

Hello, where does the $ (p \minus{} 2q \minus{} 1)! \equiv (p \minus{} 1)!$ come from

we have $ (2q)!\equiv 1 (mod p)$so we have $ (p \minus{} 2q \minus{} 1)!\equiv (p \minus{} 1)!$
the same as mine.

My solution:
Lemma: let $p$ be a prime number and $k<p$ be a positive integer then $(k-1)!(p-k)!\equiv (-1)^k\pmod{p}$.
Solution to lemma:
From wilson's theorem we have $(p-1)!\equiv -1\pmod{p}\Longrightarrow (p-1)(p-2)\dots (p-(k-1))(p-k)!\equiv -1\pmod{p}\Longrightarrow (-1)^{k-1}(k-1)!(p-k)!\equiv -1\pmod{p}\Longrightarrow (k-1)!(p-k)!\equiv (-1)^k\pmod{p}$.
Back to the main problem:
Consider an arbitrary even positive integer $k\ge 4$ such that $k\equiv 2\pmod{3}$ now let $p$ be a prime number such that $p\mid (k-1)!+1\longrightarrow (k-1)!\equiv -1\pmod{p}$(1). From the lemma $(k-1)!(p-k)!\equiv 1\pmod{p}$(2) now multiplying $(p-k)!$ to the sides of (1) and using (2) we get $1\equiv –(p-k)!\pmod{p}\Longrightarrow p\mid (p-k)!+1$ thus:
$p\mid (p-k)!+1,p\mid (k-1)!+1$
So $p-k\mid p-1\longrightarrow p-k\mid k-1$(3)(otherwise we are done)
And $k-1\mid p-1\longrightarrow k-1\mid p-k$(4)(otherwise we are done)
From (3),(4) we deduce that $p-k=k-1\longrightarrow p=2k-1\equiv 0\pmod{3}\longrightarrow p=3,k=2$ contradiction.
DONE


function
limit
complex analysis
calculus
integration
real analysis
abstract algebra

Fatou's theorem can be generalized by allowing a point to approach the boundary in larger regions, as follows.
For each $0 < s < 1$ and a point $z$ on the unit circle, consider the region $\Gamma_s(z)$ defined as the smallest closed convex set that contains $z$ and the closed disc $D_s(0)$. In other words, $\Gamma_s(z)$ consists of all lines joining $z$ with points in $D_s(0)$. Near the point $z$, the region $\Gamma_s(z)$ looks like a triangle.
We say that a function $F$ defined in the open unit disc has a [b]non-tangential limit[/b] at  a point $z$ on the circle, if for every $0 < s < 1$, the limit

$\lim_{{w\to z}_{w \in \Gamma_s(z)}} F(w)$

exists.
Prove that if $F$ is holomorphic and bounded on the open unit disc, then $F$ has a non-tangential limit for almost every point on the unit circle.
[Hint: Show that the Poisson integral of a function $f$ has non-tangential limits at every point of the Lebesgue set of $f$.]

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

In the text it was proven that under the same conditions, $F$ has a radial limit for almost every $\theta$, and that limit is $F(e^{i\theta})$. It seems like we ought to use that to get this theorem. If $F$ is holomorphic in the unit disc, then is $F(e^{i\theta})$ a continuous function? Do we even need that, or is $L^2$ enough?

In short, we have a vague idea of what to do, but it isn't definite enough for us to get a proof out of it. Should we just be estimating some integrals? Any suggestions?

Thanks.

We finally got this (hopefully). The proof is rather lengthy and requires a lot of approximations. We write $F$ as a convolution with an $L^2$ function on the unit circle with the Poisson kernel, and then we approximate $f$ by continuous functions, and then we do two more approximations (using the theorem from the text and continuity) and the triangle inequality, and finally things looked right.

But feel free to take a crack at it if you have some time and like challenging problems. :D


invariant
geometry
3D geometry
sphere
function
Euler
topology

I wonder if there is any Euler's formula V+E-F = 2 in higher dimension?

Any sources on the web I can consult?  :?:

someone told me there's one...
and working on symplexes, you get it's something like $\sum (-1)^{k}S_{k}= 1-(-1)^{n}$, where $S_{0}, S_{1}, \dots$ are the number of vertices, edges, etc...

ma_go's formula is correct (for suitable interpretation of n), but the reason it works is far more interesting:  The LHS of the formula is a topological invariant, and the RHS just happens to be its value for polytopes describing a sphere.  Therefore the formula generalizes not just for higher dimensional objects but also for objects with more complicated topology.  (For example, a torus gives you RHS=0.)

Furthermore, it is not necessary to use simplices; it isn't even necessary to use polytopes.  For example, the two sphere can be thought of as one vertex with one "face" glued to it, giving the usual answer V-E+F=2.

To learn more about this stuff, you can study homology theory (the last bit I mentioned is cellular homology), but to understand the case of polytopes, it is probably possible to generalize an elementary proof of the usual Euler's formula, but it probably gets messy.

So, for example an $n-$dimensional cubiod, how can we count the numbers $S_{0}$, $S_{1},...S_{n}$

Certainly we know for $n=2$ $S_{0}=4,~S_{1}=4$
for $n=3$, $S_{0}=8$, $S_{1}=12$, $S_{2}=6$

How about the rest? :(

The generating function for the $i$-facets of an $n$-cube is $(x+2)^{n}$.  This actually is pretty simple to figure out -- the $n+1$-cube is what you get if you drag the $n$-cube along via a new axis (the same way you get a line segment from a point, a square from a line segment and a cube from a square).  Each $i$-face of the original cube picks up a new dimension this way, so it becomes an $i+1$-face.  In addition, we have 2 copies of the cube we started with (pre-drag and post-drag).  And that's all we need.

[quote="Ottem"]http://www.applet-magic.com/eulerpoincare.htm
http://www.math.ohio-state.edu/~fiedorow/math655/HyperEuler.html[/quote]

For the sources, we know
(1)  Volume of k-hypercube in $\Re^{n}$ is $C^{k}_{n}2^{n-k}$
(2)  Volume of k-simplex in $\Re^{n}$ is $C^{k+1}_{n+1}{\sqrt{k+1}\over{k!\sqrt{2^{k}}}}$
(3)  How about the volume of k-hyperoctahedron which has a number of  ${{n-k}\over{k+1}}2^{k+1}C^{k}_{n}$  faces?

[quote="Omid Hatami"]Look at this [url=http://www.ics.uci.edu/~eppstein/junkyard/euler/]link[/url]. It has 19 proofs for Euler's formula. You see proof 14 (Noah's Ark) is one of its most beautiful proofs. Proof 17 also proves Euler's Formula for higher dimensions.[/quote]

Well. Euler-Poincare formula holds for all objects with "corners" and we count the number of their edges. Are there any generalization of the formula to smooth objects, such as spheres, torus and ellipsoids? :?:

Well, let's say the Poincare-Euler Equation

${\sum_{k=0}^{n-1}(-1)^{k}F_{k}=1-(-1)^{n}}$

where $F_{k}$ is the number of $k$ faces of the convex manifold in $n$ dimensional.

The formula only gives information about the number of faces $F_{k}$ but has no references to the volumes and areas etc. Could there be a formula which brunch all the "number of faces" and their "areas" altogether  :?:

I don't think the latter is possible: you can have alot of different arrangements with same numbers but with different areas; also the question of "what is area" occurs (these properties depend only on the topology, not on the exact sizes; one could slightly deformate it to get another area even in normal space).
There are relations for simplices (e.g. Heron's formula), but It already fails in quadrilaterals.

For the formula: a standard result from homological algebra is:
Let $S$ be a finite dimensional CW-complex (if you are not known to the term, imagine it as segments, triangles, tetrahedrons, in general as simplexes that are deformated and glued together; instead of simplexes, one could for example also use balls of the same dimension).

Let $c_{n}$ be the number of $n$-dimensional simplexes/balls used.
Then $\sum_{n}(-1)^{n}c_{n}$ does only depend on the space. In fact, the sum equals the Euler-characteristic \[\chi(S) : = \sum_{n}(-1)^{n}\mathrm{rank}(H_{n}(S)),\] where $H_{N}$ represents the homology groups.
As a result, everything with the shape of a $n$-dimensional ball (which is also a CW-complex) has the same such sum. Same holds e.g. also for all things of torus-like shape, and etc..

I mean, say for a unit cube in ${\mathbb R}^{3}$ which has a number $S_{0}=8$ vertices, $S_{1}=12$ edges and $S_{2}=6$ faces. Its total volume is $V_{3}=1$, total surface is $V_{2}=6$, total edge lengths is $V_{1}=12$ and total verttices is $V_{0}=8$

In this case of course we have $V_{i}=S_{i}$ which satisfies Euler Poincare Equation.
How about non-unit cubes or other convex objects?
Should we possibly understand $k-$ faces are the boundary of the $k$ faces. Or else should we take the union set of the $0,1,2,....k-1$ faces as the boundary of the $k-face$?


videos
Alcumus
factorial
Pascal\u0027s Triangle
algebra
binomial theorem
binomial coefficients

Can somebody give me a brief rundown on Pascal's triangle?  I was never taught it in school.................

Pascal's Triangle is a very useful tool in combinatorics as well as in algebra. Your question is a little too broad to answer. But here are a few things in which the pascals triangle can be useful.

It can be used to prove many combinatorial identities by methods of block-walking.

It can be used when expanding something like: $ (x \plus{} y)^{64},$ or any binomial basically.

It can also be useful in combinatorics problems, such as in: In how many ways can 12 pieces of candy be distributed among 4 kids?

There are many more uses of the Pascal's Triangle, I suppose.

EDIT: I'd recommend that you look over chapter 12 onwards in Introduction to Counting and Probability. It is a GREAT book for such concepts.

There are also some good videos about Pascal's triangle on Alcumus. That is where I learned all about Pascal's triangle.

Yeah, Alcumus hasn't covered that with me in the questions.  Ditto with the binomial theorem.  Two things I really want to learn about...........:(.  I think I'm doing OK with the problems, but they won't let me go on yet.  Any free sources available?

To quickly generate a chunk of Pascal's triangle on yuor computer screen so you can look at it (and there are many patterns worth looking for):

Open a spreadsheet program, such as MS Excel.

Put $ 1$ in cell A1.

Select a chunk of the column A, including cell A1; issue a "Fill Down" command.

Go to cell B2. Enter this formula: =A1+B1.

Select a large group of cells with B2 as its upper left corner.

Issue a "Fill Down" command, and without letting go of that selection, issue a "Fill Right" command.

Once you've done all that you can extend selections and fills into territory you haven't covered, you can delete zeros if you want to, you can play with the formatting, including making columns wide enough to display integers as large as you have.

One variant worth trying: instead of "=A1+B1" in cell B2, try this command: "=mod(A1+B1,2)". And do the "fill down" and "fill right" with that.

If you know programming, I believe you can write a program to calculate pascal's triangle to a certain row.

Are there any free websites available that I could learn from?

@powerofpi: How did you get that awesome image?

@powerofpi(again :P ): I think he meant programming, not Pascal's triangle.

@maybach: There are plenty of online programming tutorials, just check the web!

Regarding the Binomial Theorem and Pascal, the entries of each row $ n$ are considered binomial coefficients for $ (x\plus{}y)^n$ (where the first row is row $ 0$.
$ (x\plus{}y)^0\equal{}1, (x\plus{}y)^1\equal{}x\plus{}y, (x\plus{}y)^2\equal{}x^2\plus{}2xy\plus{}y^2, (x\plus{}y)^3\equal{}x^3\plus{}3x^2y\plus{}3xy^2\plus{}y^3$...take the groups of coefficients $ 1; 1, 1; 1, 2, 1; 1, 3, 3, 1$ and note they form the first 4 rows of pascal's triangle. And the pattern continues on, and that is what the Binomial theorem is.

[quote="Falcon1"][quote="Kent Merryfield"]Published more than 300 years before Pascal was born. Note the typographical error in the $ \binom{7}{3}$ position - that looks like 34 when of course it's supposed to be 35.[/quote]

Is that in chinese[/quote]

yeah

http://ptri1.tripod.com/
Very helpful site!

here is what it says!:

How Pascal's Triangle is Constructed
At the tip of Pascal's Triangle is the number 1, which makes up the zeroth row. The first row (1 & 1) contains two 1's, both formed by adding the two numbers above them to the left and the right, in this case 1 and 0 (all numbers outside the Triangle are 0's). Do the same to create the 2nd row: 0+1=1; 1+1=2; 1+0=1. And the third: 0+1=1; 1+2=3; 2+1=3; 1+0=1. In this way, the rows of the triangle go on infinitly. A number in the triangle can also be found by nCr (n Choose r) where n is the number of the row and r is the element in that row. For example, in row 3, 1 is the zeroth element, 3 is element number 1, the next three is the 2nd element, and the last 1 is the 3rd element. The formula for nCr is:

n!
--------
r!(n-r)!

! means factorial, or the preceeding number multiplied by all the positive integers that are smaller than the number. 5! = 5 × 4 × 3 × 2 × 1 = 120.

[img]http://ptri1.tripod.com/ptreal1r.gif[/img]

The Sums of the Rows
The sum of the numbers in any row is equal to 2 to the nth power or 2n, when n is the number of the row. For example:

20 = 1
21 = 1+1 = 2
22 = 1+2+1 = 4
23 = 1+3+3+1 = 8
24 = 1+4+6+4+1 = 16

Prime Numbers
If the 1st element in a row is a prime number (remember, the 0th element of every row is 1), all the numbers in that row (excluding the 1's) are divisible by it. For example, in row 7 (1 7 21 35 35 21 7 1) 7, 21, and 35 are all divisible by 7.

Hockey Stick Pattern
If a diagonal of numbers of any length is selected starting at any of the 1's bordering the sides of the triangle and ending on any number inside the triangle on that diagonal, the sum of the numbers inside the selection is equal to the number below the end of the selection that is not on the same diagonal itself. If you don't understand that, look at the drawing.
1+6+21+56 = 84 (red)
1+7+28+84+210+462+924 = 1716 (green)
1+12 = 13 (blue)

	[img]http://ptri1.tripod.com/ptreal1h.gif[/img]

...You can find the rest on the website just scroll down to "Magic 11's"


Prove that:
$ (a\minus{}\frac{1}{b})(b\minus{}\frac{1}{c})(c\minus{}\frac{1}{a})\geq (a\minus{}\frac{1}{a})(b\minus{}\frac{1}{b})(c\minus{}\frac{1}{c})$
with $ a;b;c>0$

[quote="toan hoc"]Prove that:
$ (a\minus{}\frac{1}{b})(b\minus{}\frac{1}{c})(c\minus{}\frac{1}{a})\geq (a\minus{}\frac{1}{a})(b\minus{}\frac{1}{b})(c\minus{}\frac{1}{c})$
with $ a;b;c > 0$[/quote]
I think, it's wrong. Try $ a\equal{}1$ and $ b\equal{}c\equal{}0.9.$ :wink:

To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now.


limit
logarithms
trigonometry
calculus
calculus computations

Find $ \lim_{x\rightarrow 0}\left(\frac{tanx\plus{}1}{sinx\plus{}1}\right)^{\frac{1}{sinx}}$.

Thank.

$ L\equal{}    \lim_{x\rightarrow 0}\left(\frac{tanx\plus{}1}{sinx\plus{}1}\right)^{\frac{1}{sinx}}$
$ \ln L \equal{}     \lim_{x\rightarrow 0}\frac{\ln \left(\frac{tanx\plus{}1}{sinx\plus{}1}\right)}{sinx}$

$ \ln L\equal{} \lim_{x \to 0} \frac{ \tan x \sec x \plus{}\sec^2 x \minus{}\tan x\minus{}1}{\cos x(\sin x \plus{}1)(\tan x\plus{}1)}\equal{}0$

so $ L\equal{}1$ ?

hello
 $ \lim_{x\rightarrow 0}\left(\frac{tanx\plus{}1}{sinx\plus{}1}\right)^{\frac{1}{sinx}}$
=$ e^{\lim_{x\rightarrow 0} \frac{1}{sin(x)}(\frac{tanx\plus{}1}{sinx\plus{}1} \minus{}1)}$
=$ e^{\lim_{x\rightarrow 0}\frac{1\minus{}cos(x)}{cos(x)(sin(x)\minus{}1)}}$
=$ e^{0}$
=$ 1$
thank u


inequalities
trigonometry
Putnam
college contests

$ \cos(\cos x) \geq \sin x , \forall x \in [0,\frac {\pi}{2}$

Let $ f(x)\equal{}cos(cosx)\minus{}sinx$.

We have: $ f(0)\equal{}sin1$ and $ f(\frac{\pi}{2})\equal{}0$,
$ f'(x)\equal{}sin(cosx)sinx\minus{}cosx\leq cosxsinx\minus{}cosx\leq0$ for $ x\in[0,\frac{\pi}{2}]$ hence $ f(x)\geq 0$.

hello, do you mean that $ \cos(\cos x)\geq\sin x ,\forall x\in [0,\frac{\pi}{2}]$
Sonnhard.

$ \cos{(\cos{x})} \ge \sin{x} \Leftrightarrow$
$ \cos{(\cos{x})} \ge \cos{(\frac{\pi}{2} \minus{} x)} \Leftrightarrow$
$ \cos{x} \le \frac{\pi}{2} \minus{} x$
Now just sketch the unit circle, and you should see.


trigonometry
integration
calculus
calculus computations

The problem is as follows, and we need to use a power reducing formula right? $\sin^{2}x = \frac{1-\cos{2x}}{2}$

$\int{\sin^{2}(2x)}dx$

would we use this in evaln'g the integral?

Yes, you should use this formula.

Philosophically speaking, some formula like that must be used, because the antiderivative will involve something else besides trig functions. Indeed, the integrand is nonnegative, which means that the antiderivative is non-decreasing, and in particular non-periodic.

[b]The usual linear substitution :[/b] $\int f(ax+b)\ dx=\frac{1}{a}F(ax+b)+C$, where $F(t)=\int f(t)\ dt\ .$


For the first and second we use $ \displaystyle \frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b} $

For the first and third$ \sum\frac{1}{a}\geq\frac{(1+1+1)^2}{a+b+c}=\frac{9}{a+b+c} $

And for the second and third the same$ \sum\frac{2}{a+b}\geq\frac{(3\sqrt{2})^2}{2(a+b+c)}=\frac{9}{a+b+c} $

My proof:
Let $\frac{1}{a}=x$, $\frac{1}{b}=y$ and $\frac{1}{c}=z$. Hence the inequality becomes:

\[x+y+z \geq \frac{2xy}{x+y}+\frac{2yz}{y+z} +\frac{2xz}{x+z}\]

Since $x+y \geq \sqrt{xy}$, so:
\[ \sum \frac{2xy}{x+y} \leq \sum \sqrt{xy}\]

Now it becomes: $x+y+z \geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$, which is obviously true.

The next sign is a simple application of AM-HM. ( That's why i used this title.)

to prove the 1st and 3rd, i used AM-HM:

$\displaystyle\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9abc}{ab+ac+bc} \ge \frac{9}{a+b+c}$

this is because $\displaystyle\frac{abc}{ab+bc+ac} \ge \frac{1}{a+b+c+}$

since:

$(abc)(a+b+c) \ge ab+bc+ac \Rightarrow a^2bc + ab^2c +abc^2 \ge ab+ac+bc \Rightarrow abc^2 \ge ab$

and goes the same for the rest of them

Your last inequality, that $abc^2\geq{ab}$ is only true if c is greater than 1.

However, comparing the first and the third is immediate from AM-HM.  You seemed to have forgotten to take the reciprocal.  You get
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq{\frac{9}{\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}+\frac{1}{\frac{1}{c}}}}$

For the second and third, simply substitute $ x\equal{}a\plus{}b$, etc. to get $ \sum 1/x \ge 9/(x\plus{}y\plus{}z)$ which is just direct AM-HM.

Hello.

[hide="second"]
$ \frac {2}{a \plus{} b} \equal{} \frac {2^2}{2(a \plus{} b)}$

$ \frac {2}{b \plus{} c} \equal{} \frac {2^2}{2(a \plus{} b)}$

$ \frac {2}{c \plus{} a} \equal{} \frac {2^2}{2(c \plus{} a)}$

By Titu's Lemma

$ \frac {2}{a \plus{} b} \plus{} \frac {2}{b \plus{} c} \plus{} \frac {2}{c \plus{} a} \equal{} \frac {2^2}{2(a \plus{} b)} \plus{} \frac {2^2}{2(b \plus{} c)} \plus{} \frac {2^2}{(c \plus{} a)} \geq \frac {(2 \plus{} 2 \plus{} 2)^2}{4(a \plus{} b \plus{} c)} \equal{} \frac {9}{a \plus{} b \plus{} c}$ [/hide]

umm... Jensen sorta kills this problem without any algebraic manipulation... Just set in $ f(x)\equal{}\frac{1}{x}$, which is convex in the interval $ (0,\infty)$.

 $ \frac{f(a)\plus{}f(b)}{2}\plus{}\frac{f(b)\plus{}f(c)}{2}\plus{}\frac{f(c)\plus{}f(a)}{2}\geq f(\frac{a\plus{}b}{2})\plus{}f(\frac{b\plus{}c}{2})\plus{}f(\frac{c\plus{}a}{2})\Rightarrow \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq \frac{2}{a\plus{}b}\plus{}\frac{2}{b\plus{}c}\plus{}\frac{2}{c\plus{}a}$

For the RHS we use $ \frac{f(\frac{a\plus{}b}{2})\plus{}f(\frac{b\plus{}c}{2})\plus{}f(\frac{c\plus{}a}{2})}{3}\geq f\left(\frac{a\plus{}b\plus{}c}{3}\right)$ and the result follows.

[quote=shobber]For any positive real numbers $a,b$ and $c$, prove:
\[ \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \geq \dfrac{2}{a+b} + \dfrac{2}{b+c} + \dfrac{2}{c+a} \geq \dfrac{9}{a+b+c} \][/quote]


Μετά από πολύ καιρό βάζω μια σχετικά εύκολη αλλά αρκετά ενδιαφέρουσα γεωμετρία.

Εστω 3 τυχαίοι κύκλοι (ας τους πάρουμε καλύτερα μη τεμνόμενους, χωρίς να αλλάζει βέβαια τίποτα). Εστω τώρα $ X_1, X_2, X_3$ τα σημεία στα οποία τέμνονται οι εξωτερικές εφαπτομένες (ανά δυο) των κύκλων. Να αποδείξετε ότι $ X_1, X_2, X_3$ είναι συνευθειακά!

Το θεώρημα είναι γνωστό ως θεώρημα Monge-D'Alembert
Έστω $ R_1,R_2,R_3$ οι ακτίνες των τριών κύκλων . Τα σημεία $ X_1,X_2,X_3$ είναι τα εξωτερικά κέντρα ομοιοθεσίας των τριών κύκλων .
Η ομοιοθεσία με κέντρο το $ X_1$ και λόγο $ \frac {R_2}{R_1}$ στέλνει τον πρώτο κύκλο στον δεύτερο . Η ομοιοθεσία με κέντρο το $ X_2$ και λόγο $ \frac {R_2}{R_3}$ στέλνει τον  δεύτερο κύκλο στον τρίτο και η ομοιοθεσία με κέντρο $ X_3$ και λόγο $ \frac {R_1}{R_3}$ τον τρίτο κύκλο στον πρώτο .

Επομένως η σύνθεση των ομοιοθεσιών αυτών είναι ταυτοτική και με λόγο το γινόμενό τους δηλαδή 1 (και κάθε φορά δείτε όταν κάνουμε ομοιοθεσία που πάνε τα άλλα δύο κέντρ ομοιοθεσίας  :wink: ) . Επομένως τα κέντρα ομοιοθεσίας είναι συνευθειακά σημεία .

(μπορεί να γίνει απόδειξη και με το θεώρημα Μενελάου στο τρίγωνο $ O_1O_2O_3$)

Δηλαδή Σιλουανέ αυτό είναι το θεώρημα που χρησιμοποιούνταν στο 6ο πρόβλημα IMO? Εχει πολύ πλάκα γιατί ένας φίλος απ'τη Βουλγαρία μου είχε βάλει αυτό το πρόβλημα όταν ήμουν στην Αμερική, μία μέρα πριν την ΙΜΟ!
Btw εγώ το είχα λύσει χρησιμοποιώντας Μενέλαο σε συνδιασμό βέβαια με ότι το γινόμενο αυτό ισούται με 1 (λόγω ομοιοθεσίας όπως είπες).

Ναι ουσιαστικά μπορούσες να το χρησιμοποιήσεις και αυτό σε εκφυλισμένες περιπτώσεις  (αν και έβγαινε και πιο εύκολα)


in Vietnam, I'm a member of Team Vietnam 1; we had 4 rounds of competition.
But I do not find the results team Vietnam.
Can anyone help me??

thank you very much.
But I still do not understand all.
Can you help me?


Find the rate of change of the distance btwn the origin and a moving point on the graph of $ y=x^{2} +1$ if $ \frac{dx}{dt} =2$ cm per sec
I don't understand how to do the problem, but I know it has to do something with the distance formula.
The answer is, $ \frac{2(2x^{3}+3x)}{\sqrt^{x^{4}+3x^{2}+1}}$

Let $ P(x,y)$ is on the point we observe. We want the rate of $ \overline{OP}\equal{}\sqrt{x^2\plus{}y^2}$ as time goes by.

$ \frac{\mathrm{d} }{\mathrm{d} t}\sqrt{x^2\plus{}y^2}\equal{}\frac{\mathrm{d}}{\mathrm{d}t}\sqrt{x^2\plus{}(x^2\plus{}1)^2}\equal{}\frac{\mathrm{d} }{\mathrm{d} t}\sqrt{x^4\plus{}3x^2\plus{}1}$

=$ \frac{\mathrm{d} }{\mathrm{d} x}\sqrt{x^4\plus{}3x^2\plus{}1}\cdot\frac{\mathrm{d}x }{\mathrm{d} t}\equal{}\equal{}\frac{4x^3\plus{}6x}{2\sqrt{x^4\plus{}2x^3\plus{}1}}\cdot2\quad(\because\frac{\mathrm{d} x}{\mathrm{d} t}\equal{}2)$

=$ \frac{4x^3\plus{}6x}{\sqrt{x^4\plus{}2x^3\plus{}1}}$.


AMC USA(J)MO USAMO geometry inequalities Putnam calculus

Are there any plans to come up with a new text on problem solving from the AoPS peeps <Geared towards USAMO problems specifically> ? Like an AoPS Vol. III or maybe USAMO: Unleashed.

I really enjoy the AoPS books and find that their tactic of teaching within the text is exquisite, however much of the material is still basic. Perhaps a book where the USAMO problems are grouped into their own subject <IE Geometry, inequalities etc.> and then solved by one of the author explained step-by-step in terms of their thought process? That would really help me as a competitor. What do you guys think, agree? disagree?

Appreciate the input,
The Idiot without a village.

Agreed that the AoPS books are more better for AMC/AIME than for the USAMO.  They merely lay the foundation for the latter, but that training comes better elsewhere.

We don't have more books in the pipes at this point, but we will have numerous online classes along the lines that you suggest - grouped into specific subjects, targeted to problems as difficult as the USAMO.  The two classes we offer now are not nearly at this level, but we will be offering some this fall that will be...

The Zeitz book listed on our resources page is a good stand-in for AoPSIII.  Even has nearly the same title & a nature-inspired cover.

I would like to recommend two books that may be useful for the USAMO. The first is a book called [u]Problem-Solving Strategies[/u], written by the coach for the German IMO team. It contains a plethora of problems with solutions to help people prepare for the IMO, but I think it is useful for the USAMO as well.

The other book is called [u]Problem-Solving Through Problems[/u], written by Loren Larson. It is designed to help students do well on the Putnam exam I think, so it contains some calculus, but if you ignore the chapter on real analysis (chapter 6 I think) then it is almost right for USAMO training. There is some calculus in some of the other problems, but you can ignore them if you do not know calculus or if you specifically want USAMO training.

Also, to study for the USAMO, you probably should not study only USAMO-like problems. Any (fairly challenging) proofs you do will help you on the USAMO even if they are not USAMO-like problems.

Well, I appreciate your input and time. Apparently, it's a long road ahead but knowing which roads to take in getting to this goal is indeed a time-saver. I, unfortunately, have also fallen victim to this "being-notified-too-late" syndrome. I didn't find out about this USAMO or any other olympiad for that matter until Junior year and from what I have seen, there are 8th graders already in preparation. AMC is doing a great job however, if people spread the word more and moreover, if people had more interest then, perhaps, we may have something going.

Thanks for the input!
The Idiot without a village.


trigonometry
calculus
derivative
function
limit
inequalities
calculus computations

can someone offer a proof on this? I can't seem to prove it.

Use the unit circle and the definitions of sine , tan, and radian

Edit: Ya I overlooked it , but the result is obviously true in $ (0,\frac{\pi}{2})$

hello, why should be $ x<\tan(x)$? Do you have something forgotten to say, i think.
Sonnhard.

[quote="Xaenir"]Edit: Ya I overlooked it , but the result is obviously true in $ (0,\frac{\pi}{2})$[/quote]

Yeah, that.

Ok.
You can look at the sine graph.  You know that sine is concave down from $ (0,\frac{\pi}{2})$.  In addition, the tangent line at x=0 is y=x (it's not that hard to prove).  Because we want to prove that $ sin(x)<x$, then y=sin(x) should be less than y=x at all points where we want to prove it.  Combine all the facts you have, and the result is proved.

For tangent, on the other hand, the first derivative is $ sec^2x$ which is positive on the interval $ (0,\frac{\pi}{2})$, in addition, the second derivative, $ 2sec^2xtanx$ is positive, so it is concave up, making it increase even more and making it greater than y=x, so we are done.

The logic is backwards in that, probably - all that information about concavity and derivatives comes from knowing the derivatives of the trigonometric functions. But to know the derivatives of the trigonometric functions, we must first prove that $ \frac{\sin x}x\to 1$ as $ x\to 0,$ and to know that limit, we first need this inequality.

Someone draw the right picture, please! It should have a wedge of a circle and more than one triangle.

[u][b]Note[/b][/u]

$ \sin{x}<x$ if $ x >0$

[u][b]Solution[/b][/u]

Let $ y\equal{}x\minus{}\sin{x}, x>0$

We get $ y'\equal{}1\minus{}\cos{x} \ge 0, \forall x>0$

Hence $ y(x)>y(0) \Longrightarrow x\minus{}\sin{x}>0 \Longrightarrow \sin{x} < x, \forall x>0$

The point is still being missed. It's one thing if we're defining the sine to be the power series (or defining it through the exponential, which is itself defined by the power series), but let's say we define the sine and cosine geometrically, by reference to the unit circle. In that case, we don't know the derivatives of the trigonometric functions until we know $ \lim_{x\to 0}\frac{\sin x}{x},$ and this inequality, $ \cos x<x<\tan x$ for $ x$ in the first quadrant, is an intermediate step toward proving that limit.

So let's assume that you don't already know the derivatives of the trigonometric functions. Hence we reject the posts by thanhnam2902 and centry57 as representing circular reasoning, as we previously rejected zapi2007's post for the same reason.

Who is going to draw the appropriate picture?

I have been searching t0rajir0's post, I can't find it.

Here is a MathLinker's text.


ABCD is a square side 1. Points A'B'C'D' are taken on the sides AB BC CD DA respectively so that AA'/AB = BB'/BC = CC'/CD = DD'/DA = 1/n. The strip bounded by the lines AC'and A'C meets the strip bounded by the lines BD' and B'D in a square area 1/1985. Find n.

hello its my first time posting a problem hopefully i posted it in the right place  ;)

Hmmm....this is a problem from some old AIME if I recall correctly, right?

a.. not sure i think so. my teacher likes to give us problems form contests

Yeah, this is from '85 AIME (#4) :)

[url=http://www.kalva.demon.co.uk/aime/aime85.html]kalva[/url]


algebra
difference of squares
special factorizations

Source: 2005 AMC-10B #24

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$. The integers $x$ and $y$ satisfy $x^2-y^2=m^2$ for some positive integer $m$. What is $x+y+m$?

$\text{(A)}\ 88 \qquad\text{(B)}\ 112 \qquad\text{(C)}\ 116 \qquad\text{(D)}\ 144 \qquad\text{(E)}\ 154$

[hide="hint"]let x=10a+b where a and b are the digits[/hide]

[hide="Solution"]As Altheman said, let one be 10a+b and the other be 10b+a.

Also, $(x+y)(x-y)=m^2$.

So substituting them in, $(11a+11b)(9a-9b)=m^2$

$99(a+b)(a-b)=m^2$

So, for $m$ to be an integer, $(a+b)(a-b)$ needs to have a factor of 11 and another perfect square (in this case 1 because it's the easiest).  Thus, $a=6, \ b=5$.

$56+65+33=154 \implies \boxed{E}$ (I hope).   [/hide]

Why the factor have to be 11? How did you get 11?

99 is made up of 9 and 11.

9 is already a perfect square, so $99 \times \text{something}$ is only a perfect square there is another 11.  Did that help? (probably not, as I am a bad explainer)  :P

[hide="My solution"]

First, I realize how similiar (or probably is the same) this solution is to Robin's

$x=10a+b$ and $y=10b+a$ and $(x+y)(x-y)=m^2$

Substitute in to get $(10a+b)(10b+a)=(100a^2+20ab+b^2)-(100b^2+20ab+a^2)$

It all cancels out to $99a^2-99b^2=m^2$

Factor out a $99$ and use difference of squares to get $99(a+b)(a-b)=m^2$

Factor the $99$ into $3^2*11(a+b)(a-b)=m^2$ and then take the square root of both sides to get $\sqrt{3^2*11(a+b)(a-b)}=\sqrt{m^2}$

Simplify that to get $3\sqrt{11(a+b)(a-b)}=m$

Because m is an integer, that means that $\sqrt{11(a+b)(a-b)}$ is also an integer, but in order for that to be true, there must be another $11$ in order for you to take out an $11$.

Since $a$ and $b$ are both digits, $(a+b)$ must equal $11$, and therefore $(a-b)$ is $1$.

Solve that to get $a=6$ and $b=5$ (since $a$ is the larger one)

Plug it all back in to get $65^2-56^2=m^2$ and solve to get $m=33$

Therefore $x=65$, $y=56$, and $m=33$

$65+56+33=\boxed {154}$

So $\boxed {E}$[/hide]


A cubic tank holds 1000kg of water. What are the dimensions of the tank in meters.  I totally do not understand this. Any and all help is greatly appreciated. Thank you

[quote]1 kg of water is supposed to have dimensions 10cm*10cm*10cm
so 1000kg has dimensions 1m * 1m * 1m

this is not exact, but it is extremely accurate
it is in reality a few millionths off  :D[/quote]

It would be exact if the temperature is 4 degrees Centigrade in both times.  :)

indeed water does chenge its density with respect to temperature and other stuff
BUT the density of water is NEVER $ 1 \frac{g}{cm^3}$
but it is very often a few millionths off

im sure given the proper mineral concentration and temperature it could be... unless we're talking about pure water.

And the conversion thing can be found on thousands of sites, just google metric system.


A container in the shape of a right circular cylinder with no top has surface area 3pi ft.^2    .  What height (h) and base radius (r) will maximize the volume of the cylinder?  (Hint: area of the curved side of the cylinder=2pi rh)

Maximize $V=\pi r^2h$ subject to $A=\pi r^2+2\pi rh=3\pi.$

This is precisely the sort of problem for which Lagrange multipliers is the most efficient method:

$\frac{\partial V}{\partial r}V=\lambda \frac{\partial A}{\partial r}$ or $2\pi rh=\lambda(2\pi r+2\pi h)$
$\frac{\partial V}{\partial h}V=\lambda \frac{\partial A}{\partial h}$ or $\pi r^2=\lambda 2\pi r$

From the second equation $\lambda = \frac r2.$ Put that into the first equation to get

$2\pi rh=\pi r^2+\pi r h$ or $\pi rh=\pi r^2.$

So $h=r.$ The radius of the cylinder must be the same as its height.

Plug this into $\pi r^2+2\pi rh=3\pi$ to get the actual dimensions of $r=h=1.$