geometry
rectangle
symmetry
calculus
calculus computations

1. Find the length of the diagonal of the rectangle with the largest area that can be inscribed in an isosceles triangle of base 24 cm and height 10 cm.

Place the base of the triangle on the x-axis, centered at 0.  Then you can come up with an equation of the lines that form the other two sides of the triangle.  After you do this, it gets easier.

The height of the rectangle is determined by the value of the diagonal at some point.  So $ bh\equal{}2x\cdot f(x)\equal{}A$.  Here, since we have symmetry, I assumed that the height occurs at some position $ x$ so the entire base is $ 2x$.

Then you can proceed like normal.


inequalities
vector
linear algebra
matrix

This can be heavily simplified by writing $ A \equal{} M^TM$ and $ B \equal{} N^TN$, for $ n \times k$ matrices $ M,N$.

Then, for example, the last inequality reduces to showing

$ \det (M^TM)^2 \plus{} \det(N^TN)^2 \ge 2 \det(M^TN)^2.$

Massage:
$ 2\det(M^TN)^2 \equal{} 2\det(M^TM)\det(N^TN).$

Substitute: we then want
$ \det(M^TM)^2 \plus{} \det(N^TN)^2 \ge 2 \det(M^TM)\det(N^TN),$

which is immediate the quite familiar AM-GM inequality.

Your first two questions 1) and 2) are results about positive semidefinite matrices. That context should make things easier.


superior algebra
superior algebra unsolved

R is an infinite ring, and a is an element of R.
Prove or disprove: aR is either infinite or {0}.

False.

If F is any infinite ring and A is any finite ring, then look at an element (0,1) in F x A.

For example, look at the element $ (0, 1) \in \mathbb{Q} \times \mathbb{Z}/p$


pigeonhole principle
function
algebra
domain

I have one question:What is "pigenhole principle"?

[quote="hydro"]pidgeonhole principle states that if you have $k+1$ objects to put in $k$ boxes, you will have [i]at least[/i] 1 box that contains [i]at least[/i] 2 objects.
A generalization is that if you have $nk+1$ objects to put in $n$ boxes, you will have [i]at least[/i] 1 box that contains [i]at least[/i] $k+1$ objects.
A consequence of this principle is, for example that if you consider a function $f:$ $A \rightarrow B$, where A and B are sets containing a finite number of elements, $f$ is injiective iff it's surjective.[/quote]

Sorry, but what does injective and surjective mean?  Are they related to bijective in any way?

a function is called [i]bijective[/i] iff it's injective [b]and[/b] surjective.

A function is [b]injective[/b] if $\forall x_1 \neq x_2$ with $x_1,x_2 \in D$,
$f(x_1) \neq f(x_2)$ (D is the domain of the function)

I don't know english enough to explain surjectivity, but I try. A function is [b]surjective[/b] if each element of the set of the images of the function has a pre-image (does this word exists?). For example, $f(x)=x^2 \mathbb{R}\rightarrow \mathbb{R}$ is not surjective, but $f(x)=x^2 \mathbb{R}\rightarrow \mathbb{R^+}$ does.

Adeel (chess64) made a nice PDF about this on his Mathematical Ideas website.  I can't find the link, but I have the PDF saved to my computer somewhere.....aha!  See the attachment for a very nice explanation.

So is injective like the proper name for saying that a function is one-to-one?

[quote="hydro"]pidgeonhole principle states that if you have $k+1$ objects to put in $k$ boxes, you will have [i]at least[/i] 1 box that contains [i]at least[/i] 2 objects.
A generalization is that if you have $nk+1$ objects to put in $n$ boxes, you will have [i]at least[/i] 1 box that contains [i]at least[/i] $k+1$ objects.
[/quote]

what's then the dirichlet's principle?


geometry
incenter
circumcircle
geometry proposed

Prove that for every triangle with incenter I and sides a, b and c it is true that $ a.AI^2\plus{}b.BI^2\plus{}c.CI^2\equal{}abc.$

Use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275895]Leibniz formula[/url] for the incenter $ I\equiv (a: b: c)$ and the circumcenter $ O$ of $ \triangle ABC$

$ BC \cdot R^2 \plus{} CA \cdot R^2 \plus{} AB \cdot R^2 \equal{} 2p \cdot IO^2 \plus{} BC \cdot IA^2 \plus{} CA \cdot IB^2 \plus{}  AB \cdot IC^2$

$ BC \cdot IA^2 \plus{} CA \cdot IB^2 \plus{} AB \cdot IC^2 \equal{} 2pR^2 \minus{} 2p(R^2 \minus{} 2Rr) \equal{} BC \cdot CA \cdot AB$


Solve the equation: x^2/3+48/x^2=10(x/3-4/x).. I solved it using a factoring machine... You aren't allowed to do it that way.

You have to be careful when doing these problems and make sure you don't divide out any variables. When you do that you lose answer(s). For example, when you have something like 1/x=x2. If you divide by x, you get the answer 1, when the answer is both 1 and -1.

The problem you wrote is strange, because the answer is difficult to express exactly. Notice the answer x=0 does not work because when x=0, you have division by zero in the original equation. So the answer I got is x=-1.3760008500163. (I'm not sure if that's right, I did it in a hurry)

Spike -- the only real solution to 1/x = x:^2: is 1.  There are also two complex-number solutions.  -1 is definitely not a solution.

Also, I believe that by saying you can't use "factoring machines," he meant to say that you were supposed to do it by hand.  That is, don't use calculators or computers, either.

ARrrg!  I apologize for having a wrong number.. it was ^2 not ^3  :evil:  :cry:

Yeah, I was wondering about that.  Excellent.

Solution:

[hide]Multiply through by 3 and rearrange:

x2 + 144 / x2 = 10x - 120 / x.

Subtract 24 on both sides:

x2 - 24 + 144 / x2 = 10x - 120 / x - 24

(x - 12 / x)2 = 10 (x - 12 / x) - 24.

Let k = x - 12 / x, then

k2 - 10k + 24 = 0

So k = 4 or k = 6; then we can solve for x. The four roots of the equation are:

-2, 6, 3 :pm: :sqrt:21 .[/hide]

[quote="JBL"]Spike -- the only real solution to 1/x = x:^2: is 1.  There are also two complex-number solutions.  -1 is definitely not a solution.[/quote]

Sorry, I wasn't thinking when I made that example. That one's incorrect. I think I was trying to say x=x:^2:. That one works. Sorry for the confusion.

[quote="SPiKEzZ"]Sorry, I wasn't thinking when I made that example. That one's incorrect. I think I was trying to say x=x:^2:. That one works. Sorry for the confusion.[/quote]

That doesn't work... x still can't be -1 or else u get -1=1

Posted by MrMcGeek

I think that he was talking about division by 0, so he meant that it has roots 1 and 0, but if you were to divide by x, you would only get the 1 root.

I should point out that that isn't necessarily relevant to this problem, since the divisions took place before the solving steps.  However, it is good form to go back and check that your eventual answers work, especially when you have reciprocals or roots of any type.


[color=darkblue]Denote the incenter $I$ of the triangle $ABC$ and the intersections $E\in BI\cap AC$, $F\in CI\cap AB$.

Prove that $\boxed{\ A=60^{\circ}\ }$ $\Longleftrightarrow$ $b^{2}+c^{2}=a^{2}+bc$  $\Longleftrightarrow$ $(2a+b+c)(a+b+c)=3(a+b)(a+c)$ $\Longleftrightarrow$ $\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ $\Longleftrightarrow$ $\frac{c}{a+b}+\frac{b}{a+c}=1$ $\Longleftrightarrow$ $\frac{IE}{IB}+\frac{IF}{IC}=1$ $\Longleftrightarrow$ $\boxed{\ \frac{1}{[BIF]}+\frac{1}{[CIE]}=\frac{1}{[EIF]}\ }$, where $[XYZ]$ is the area of the triangle $XYZ\ .$[/color]

First note that: $IF=IE=a$, because the cuadrylateral $IFAE$ is cyclic.

Let: $IB=b$, $IC=c$, then: $[EIF]=\frac{a^{2}}{2}sin60^\circ, [BIF]=\frac{ab}{2}sin60^\circ, [CIE]=\frac{ac}{2}sin60^\circ$

Then the statement is equivalent to show: $\frac{1}{ab}+\frac{1}{ac}=\frac{1}{a^{2}}\Rightarrow \boxed{a(b+c)=bc}$

By cosine law: $BF=\sqrt{a^{2}-ab+b^{2}}, CE=\sqrt{c^{2}-ca+a^{2}}, BC=\sqrt{b^{2}+bc+c^{2}}$

Then: $\frac{\sqrt{a^{2}-ab+b^{2}}}{a}=\frac{\sqrt{b^{2}+bc+c^{2}}}{c}\Rightarrow \boxed{\frac{b-a}{a^{2}}=\frac{b+c}{c^{2}}}...(\alpha)$

$\frac{\sqrt{c^{2}-ca+a^{2}}}{a}=\frac{\sqrt{b^{2}+bc+c^{2}}}{b}\Rightarrow \boxed{\frac{b+c}{c^{2}}=\frac{b-a}{a^{2}}}...(\theta)$

$\frac{(\alpha)}{(\theta)}$: $\frac{b-a}{c-a}=\frac{b^{2}}{c^{2}}\Rightarrow \boxed{a(b+c)=bc}$    $\mathbb{QED}$


Let $ ABC$ be a triangle with angle $ A <$ angle $ C < 90^\circ <$ angle $ B$. Consider the bisectors of the external angles at $ A$ and $ B$, each measured from the vertex to the opposoite side (extended). Suppose both of these line-segments are equal to $ AB$. Compute the angle $ A$.

[hide]
Let $ P$ be the point where the bisector of the external angle at $ A$ meets the extension of $ BC$, and let $ Q$ be the point where the bisector of the external angle at $ B$ meets the extension of $ AC$. Then $ \angle Q = 180^{\circ} - (180^{\circ} - \angle ACB) - \frac{1}{2}(180^{\circ} - \angle ABC) = \frac{1}{2}(\angle ACB - \angle BAC)$. Since $ AB=BQ$, $ \angle Q = \angle BAC$. Thus $ 3 \angle BAC = \angle ACB$. Also, $ \angle P = \angle ABP = \angle BAC + \angle ACB$ and $ \angle P = 90^{\circ} - \frac{1}{4}(180^{\circ} - \angle BAC) = 45^{\circ} + \frac{1}{4}\angle BAC$. So $ \angle BAC = 12^{\circ}$.
[/hide]


is summation (-1)^n/root n where n goes from 1 to infinity convergent?
 sorry if its posted before.

Alternating series test. This is exactly the sort of series it was made for.


algebra
polynomial
group theory
abstract algebra
quadratics
Galois Theory
superior algebra

Let $ \alpha$ to be a root of the polynomial $ x^3\plus{}x\plus{}1$. Does the polynomial $ x^3\minus{}x\plus{}1$ always have a root in $ \mathbb Q [ \alpha]$ ?

This problem was at an exercise-list of my algebra(Galois theory part) course. But none of the students could solve it.

Well, of course. It's just too much work.

First show, that these polynomials are irreducible over $ \mathbb{Q}$, both with only one real root.
Next show under the assumption, that these polynomials must have the same
splitting field $ K$ of degree $ 6$ over $ \mathbb{Q}$.
Then show the Galois group $ \mathcal{G}(K/\mathbb{Q})\cong S_3$ has only one subgroup of index $ 2$,
and $ K$ has only one quadratic subfield $ k$.
Next calculate the [url=http://en.wikipedia.org/wiki/Discriminant]discriminants[/url] of these polynomials and conclude $ k\equal{}\mathbb{Q}(\sqrt{\minus{}23})\equal{}\mathbb{Q}(\sqrt{\minus{}31})$.
And finally use the unique non-trivial automorphism of $ k$ to get the contradiction $ \sqrt{\minus{}23}\equal{}r\sqrt{\minus{}31}$ for some $ r\in\mathbb{Q}$.

That's how I would prove this.
Each step gives a nice little exercise on its own.

With Maple, we see easily that $ x^3\minus{}x\plus{}1$ is undecomposable in $ \mathbb{Q}(a)$ where $ a$ is a root of $ x^3\plus{}x\plus{}1$.

alias( a = RootOf(x^3+x+1) ):
factor(x^3-x+1,{a});
answer: $ x^3\minus{}x\plus{}1$


geometry
inradius
AoPS Books
area of a triangle

Although I am not saying I don't know a fair bit of formulas and methods myself, I'm just wondering how some of you (i.e. Most frequest AoPSers) know an encyclopedia worth of formulas. I mean, I don't even know half the formulas that you use existed until somebody somebody posted a solution such as:

[hide] Easy, just use Heron's formula...blah blah blah blah [/hide]

Thanks...?

Well, we don't memorize them. We learn the concept of them.(i think) After that, we know the formula because we know why it works.

Wow, I use that a lot without even thinking about how I learned it  :o

[quote="happyme"]Thats what I do too, but how do you get the idea of the concept itself? As is, how do you know the concept exists. Do you just go through hundreds of problems everyday and then make a formula for each and every one?[/quote]
That's what coaches and/or AoPS books are for.  Once you learn a formula, prove it or reason out how it works if you can.  Then, it sticks in your brain permanently.  And if you do forget, then like ckck said you can remember the general idea and work from there.
A good example of this is when I took the Chapter MC this year and was working on the Venn Diagram problem.  I knew I had read a formula for that type of problem before, but I had dismissed it as a bunch of A, B, C, AB, BC, blah blah blah that I didn't need to memorize.(Bad idea, of course).  I knew the general idea, though:  When you add all three circles A B and C, you overcount some overlapping parts, so you need to subtract them.  Using this, I drew  the diagram and shaded A one way, B another way, and C a third way.  From there I saw how many times each region had been shaded and derived the Venn diagram formula there on the spot.  And the formula's stuck with me since.
Of course, it would have been better if I had known the formula off the top of my head.  But the point is, it really helps if you know the meaning behind a formula.  It becomes much easier to remember once you know WHY it works.

I have the AoPS books and they help TREMENDOUSLY.

My coach...
Lets just say that he should go to high school again.

yeah as whats his face said
its not memorizing the formulas
its understanding them

I just realized my question is actully what I wanted it to be. Let me rephrase it:

How do you guys know what formulas are useful, and are first introduced into formulas? I know how you learn them, so my first question was kinda off my point.

[quote="ckck"][quote="happyme"]Thats what I do too, but how do you get the idea of the concept itself? As is, how do you know the concept exists. Do you just go through hundreds of problems everyday and then make a formula for each and every one?[/quote]
We don't make a formula everyday. And we get the concept of it IF we understand it. Like if you know why the formula A=rs works, then you know that drawing the perpendicular lines from the center is the key. This kind of concept will remind you that some problems require you to draw the perpendicular lines to get the answer.[/quote]

lol. i definatly didnt know about drawing lines from the center. actually i still dont get it once you said it... :|

[quote="Walk Around The River"][quote="Arvind_sn"]All formulas are useful. They are just used for different things. Mostly, they are used the way that you see the ideas manipulated in the proof or some other thing like that. also, it is pretty obvious if you have A=rs. A means area, so it is pretty obvious that you use it to find the area (and sometimes semiperimeter and inradius) of a triangle.  :wink:[/quote]

i know that much.. im talking about
"then you know that drawing the perpendicular lines from the center is the key"[/quote]

No i was answering happyme's question.

however, this is the way that you do that. [b]draw this out; don't be lazy because the extra 2 seconds can help you understand the concept. [/b]first, draw a triangle (preferrably scalene, but ok if equilateral or isosceles). label the triangle ABC, and the sides a, b, and c, where side a is opposite angle A, and b is opposite B, etc.

Now that you've got that done, draw the incircle of the triangle (this is the circle that is inscribed in triangle ABC). First, we will define the center of the incircle to be the intersection of the perpendicular bisectors of the sides of the triangle. Next, we proceed by drawing segments from the center O of the incircle to the corners of the triangle. We get three triangles. as our next step, we can draw perpendicular segments to the sides from the center because as we defined earlier, the center is the intersection of the perpendicular bisectors. these perpendicular segments, you will find, are the radii of the incircle. so, you know that the area of a triangle is base times height divided by 2. we now can find the areas of the little circles, and they are $\frac{ar}{2}$ , $\frac{br}{2}$ , and $\frac{cr}{2}$ .  If we add these together, we get $\frac{a+b+c}{2}*r$. that is $rs$, our final result.

[b]Again, if you don't get this proof, either go through the proof in AoPS I, or read it again and draw it out.[/b]
 :)  :wink:  :ninja:

@ happyme: Your signature has a logical fallacy present... the last step, divide by $a+b-c,$ is invalid, because $a+b-c = 0$ and dividing by $0$ is a task which only Chuck Norris can perform.


trigonometry
algebra
polynomial
integration
inequalities
real analysis
real analysis unsolved

1) Let $ p(x)$ is trigonometry  polynomial of degree n , $ n \in \mathbb{N}^{*}, x \in \mathbb{R}$
Prove that : $ |p(x)| \leq \frac{1}{\pi}\int_{0}^{2 \pi}{|p(t)|^2dt}\plus{}\frac{2n\plus{}1}{8}$
2) Let $ p(x)$ is trigonometry  polynomial of degree $ \leq n$ , $ n \in \mathbb{N}^{*}, x \in \mathbb{R}$ ,
and $ p(0)\equal{}1,|p(x)| \leq 1 , \forall x \in \mathbb{R}$
 Prove that : $ P(x) \geq cos(nx) ,\forall x \in (\minus{}\frac{\pi}{n},\frac{\pi}{n})$

Hint for 1). Think of $ p(x) \equal{} \sum_{ \minus{} n}^n a_ke^{ikx}$, then use Cauchy-Schwarz on this sum. Now let $ t \equal{} (\sum_{\minus{}n}^n |a_k|^2)^{1/2}$; the inequality you want then looks like $ t(2n \plus{} 1)^{1/2} \leq 2t^2 \plus{} \frac {2n \plus{} 1}{8}$.


projective geometry
geometry
incenter
invariant
geometry solved

Lines connecting centroid with ex-centers
   divide sides of triangle into 6 segments that
   make involution.

   Sgt.  Pestich

I am sorry, could you ask me Pestich few questions:
How you define precisely points of the intersection and segments which we must to consider.
How you define involution on segments, if they aren’t on a line (or are they) ?
Thanks.

@ Pestich: I guess that by "ex-centers", you mean the three excenters and the incenter. Actually, you can call them the [i]tritangent centers[/i], or, like John Conway does, the [i]"four incenters"[/i], but the name "excenters" is reserved for the three "real" excenters only, not for the incenter.

@ Prowler: Here is Pestich's problem, rewritten in a slightly easier to understand  ;)  way:

[color=blue][b]Initial problem.[/b] Let ABC be a triangle, I its incenter, and $I_a$, $I_b$, $I_c$ its excenters. Let G be the centroid of triangle ABC. Let the lines GI, $GI_a$, $GI_b$, $GI_c$ meet the line BC at the points J, $J_a$, $J_b$, $J_c$, respectively. Prove that there exists an involution on the line BC such that the pairs of points (B; C), (J; $J_a$) and ($J_b$; $J_c$) are pairs of conjugate points with respect to this involution.[/color]

Actually, this problem can be generalized: The point G needn't be the centroid of triangle ABC - it can be an arbitrary point in the plane; also, the only thing we need to know about the incenter I and the excenters $I_a$, $I_b$, $I_c$ is that the lines $II_a$ and $I_bI_c$ meet at the point A, that the lines $II_b$ and $I_cI_a$ meet at the point B, and that the lines $II_c$ and $I_aI_b$ meet at the point C; in other words, we only need to know that the triangle ABC is the cevian triangle of the point I with respect to the triangle $I_aI_bI_c$. Hence, we can formulate a general version of Pestich's problem:

[color=blue][b]General problem.[/b] Let $I_aI_bI_c$ be an arbitrary triangle, and I an arbitrary point in its plane. Let ABC be the cevian triangle of the point I with respect to the triangle $I_aI_bI_c$. Also, let G be an arbitrary point in the plane, and let the lines GI, $GI_a$, $GI_b$, $GI_c$ meet the line BC at the points J, $J_a$, $J_b$, $J_c$, respectively. Prove that there exists an involution on the line BC such that the pairs of points (B; C), (J; $J_a$) and ($J_b$; $J_c$) are pairs of conjugate points with respect to this involution.[/color]

Now, the solution of this general problem is quite easy:

Let g be the line through the points $I_b$, I and B, and let this line g meet the lines GC, $GI_c$ and $GI_a$ at the points M, N and L, respectively. Now, we have

$B=g\cap I_cI_a$; $M=g\cap GC$; $I=g\cap CI_c$; $L=g\cap GI_a$; $I_b=g\cap CI_a$; $N=g\cap GI_c$.

Thus, the points B, M, I, L, $I_b$ and N are the points where the line g meets the six lines joining all possible pairs of points from the set {$I_c$; $I_a$; G; C}. Consequently, by the Desargues involution theorem, there exists an involution on the line g such that the pairs of points (B; M), (I; L) and ($I_b$; N) are pairs of conjugate points with respect to this involution.

Now, consider the central projection with projection center G from the line g to the line BC. This central projection takes the points B, M, I, L, $I_b$ and N to the points B, C, J, $J_a$, $J_b$ and $J_c$, respectively. Now, the property of three pairs of points to be pairs of conjugate points under one and the same involution is a projective property, and is invariant under central projection; thus, since the pairs of points (B; M), (I; L) and ($I_b$; N) are pairs of conjugate points with respect to one and the same involution, it follows that the pairs of points (B; C), (J; $J_a$) and ($J_b$; $J_c$) are pairs of conjugate points with respect to one and the same involution. And the general problem is solved.

  darij

Thank you, Darij .
Now problem is clear and already solved.
I think we can use same approaching: note $II_a$ x $BC = U$ .
Applying Desargues theorem for $(GII_aI_b)$ and line $BC$ we get
that pairs $(U,J_b), (J,C),(J_a,B)$ are in involution.

Analogously for $(GII_aI_c)$ and line $BC$ we get
$(U,J_c), (J,B),(J_a,C)$ are in involution.

But composition of two involutions is an involution.
I mean $(J_b - U- J_c), (C-J-B) ,(J_a - B - J)$  so

$(J_b ,J_c), (C,B) ,(J_a, J)$ can be in involution.

Hi, I saw that you sometimes solve the problems using involution so I have a question, Does exist a involution that maps a circle into a circle?  :?

Well, you can take the identity. ( $f(A) = A$ for all $A$ on the circle )

Or you can take a point $P$ not on the circle. Then, for all $A$ on the circle we define $f(A)$ as the second intersection of $AP$ with the circle.

And there's theorem telling us that these are the only ones, if my memory doesn't cheat me. (and a proof by grobber somewhere on the forum)


induction
combinatorics unsolved
combinatorics

Let A1, A2, ... , A100 be subsets of a line, each a union of 100 disjoint closed segments. Prove that the intersection of all hundred sets is a union of at most 9901 disjoint closed segments. [A single point is considered to be a closed segment.]

Let $ A,B$ be two subsets of line with $ A$ union of $ m$ disjoint closed segments and $ B$ union of $ n$ disjoint closed segments .We will prove by induction on $ m\plus{}n$ that the intersection of $ A$ and $ B$ is at most a union of $ m\plus{}n\minus{}1$ disjoint closed segments .
That is trivial for $ m\plus{}n \equal{} 2$ since $ m\equal{}1$ and $ n\equal{}1$ .
Suppose that it's true for $ m\plus{}n \equal{} i$ .We want to prove that it holds for $ m\plus{}n \equal{} i \plus{} 1$ .In the segments of $ A$ and $ B$ take the segment $ [a,b]$ such that $ a$ is maximum . Since $ a$ is maximum every segment having a nonempty intersection with $ [a,b]$ contains $ a$ hence there is at most one other segment having an intersection with $ [a,b]$ .
Without considering $ [a,b]$ the intersection of our 2 sets is at most $ i \minus{}1$ closed disjoint segments adding the intersection of $ [a,b]$ with the other segment it will at most increase $ i\minus{}1$ by $ 1$ so the intersection of $ A$ and $ B$  is at most a union of $ i$ disjoint closed segments therefore our induction is complete.

Now it is easy to see that the intersection of A1,A2,...,A3 is at most a union of $ 100 \times 100 \minus{} 99 \equal{} 9901$  disjoint closed segments.


What is the greatest three-digit prime number each of whose digits is also prime?

It must start with a $ 7$ to be the largest ($ 8$ and $ 9$ are composite).  We list primes:

$ 797$: $ 9$ is composite.

$ 787$: $ 8$ is composite.

$ 773$: All digits are prime!  Our answer is $ \boxed{773}$.

Find a good, effective way to find that $ 773$ is prime w/o using a calc.

We need to test all the primes that are less than $ \sqrt{773}$, which is all the primes up to $ 27$, or all the primes up to $ 23$.

We use the divisibility tests for $ 2$, $ 3$, $ 5$, $ 7$, $ 11$, and $ 13$, and it fails all of them.  We just divide out $ 17$, $ 19$, and $ 23$ to finish.  It's a team round, so you have plenty of time.

On a side note, why is there a school team round?

Good job, AIME15.

[quote="AIME15"]On a side note, why is there a school team round?[/quote]

Maybe to teach "math teamwork"? :lol:

There's just a team round to prepare people for future team competitions.


Hello,
Consider the following question:
Let $ \alpha_1, \alpha_2, ... \in (0,1)$ and let $ X_1, X_2, ... : \Omega \rightarrow \mathbb{R}$ be independent random variables, such that $ \mathbb{P} ( X_n \equal{} 1 \minus{} \alpha_n) \equal{} 0.5 \equal{} \mathbb{P} (X_n \equal{} 1 \plus{} \alpha_n)$ for any n. Let $ Y_n \equal{} X_1 \cdots X_n$.
1) Prove that the event $ Y_n \rightarrow 0$ is of probability 0 or 1.
2) Find $ \mathbb{P} (Y_n \rightarrow 0)$ in case $ \alpha_n \equal{} \frac {1}{1 \plus{} n}$ (for n=1, 2, ...).
3) Same as 2, but in case $ \alpha_n \equal{} \frac {1}{2}$.
4) Again, but in case $ \alpha_n \equal{} \frac {1}{\sqrt {1 \plus{} n}}$.

Now what I have:
1) Let's prove that $ Y_n$ is a tail event, which would prove what we need.
[hide="Proof"]
Let A be the event "$ Y_n \rightarrow 0$". For any k, $ A \equal{} \{ Y_n \rightarrow 0 \} \equal{} \{\frac {Y_n}{X_1 \cdots X_{k \minus{} 1}} \rightarrow 0 \}$, hence $ A \in F_k ^{\infty}$, where $ F_k ^{\infty} \equal{} \sigma(X_n, X_{n \plus{} 1}, ...)$ is the sigma-algebra generated by X_n, X_{n+1}, etc..
Thus $ A \in \bigcap _{n \equal{} 1} ^{\infty} F_k ^{\infty}$.
[/hide]

Now I'm not sure how to decide (0 or 1) in particular examples.
Is there one way to decide, or some different techniques? What could be used here?

Thanks!

Hi kar,

First your proof seems correct to me

Second as far as I know there is no universal recipe for proving if the probability is 0 or 1 for tail events and it is a case by case approach that is needed.

for 2/ I would say (but I might be wrong) that proba is 0.
for 3/ I would say it's 1
for 4/ I can't decide really but if I had to bet I would say it's 1.

To prove the statement you might transform $ Y_n$ by log and try to apply some 3 series type theorem to the transformation no waranty on this it('s just an idea (check Kallenberg's book for example chapter 4, for those kind of convergence theorem for independent random variable series)

Ok, thank you!
Actually now I have a solution for this question (not mine), maybe I'll post later. And try to see if other proofs can be found.


[quote="lambern314"]I was just struggling with the following question:
Show that if a and b are real constants and x is a real variable, the fraction:

(ax^2 + bx -1)/(x^2 +x)

can take on all real values if a-b>= 1

(any hints?)[/quote]

how about using the 'reduction to absurdity'??

i mean; if a-b<1 , the fraction can take on real values.
then find the contradiction

good luck :)

i'm not sure, but is the condition a-b>=1 really necessary? :huh:

A not-so-quick approach

EDIT: the original condition seems to be false, consider the case of $ a \minus{} b \equal{} 1$, where $ a$ and $ a \plus{} 1$ can't be taken.

[hide]
Set $ \frac {ax^2 \plus{} bx \minus{} 1}{x^2 \plus{} x} \equal{} k$.
We want for any $ k$ there is at least one real root. Simplifying gives us
$ (k \minus{} a)x^2 \plus{} (k \minus{} b)x \plus{} 1 \equal{} 0$.
For $ k \equal{} a$, if $ a\neq b$, we always have a solution $ x \equal{} \frac {1}{b \minus{} a}$. Note it can't be extraneous, so $ b \minus{} a\neq \minus{} 1$, $ a \minus{} b\neq 1$.
For $ k\neq a$, the quadratic will have two roots, which can't be both extraneous. Dismiss the case of $ x \equal{} 0$, examine when both are $ \minus{} 1$:
$ k \minus{} a \equal{} 1, k \minus{} b \equal{} 2 \iff a \minus{} b \equal{} 1, k \equal{} a \plus{} 1$. Hence $ a \minus{} b\neq 1$.
Now back to the quadratic, we want it always has some real root for any $ k$,
$ \Delta \equal{} (k \minus{} b)^2 \minus{} 4(k \minus{} a) \equal{} k^2 \minus{} (2b \plus{} 4)k \plus{} b^2 \plus{} 4a \geq 0$.
which means the above inequality should hold for any $ k$, so
$ \Delta_1 \equal{} (2b \plus{} 4)^2 \minus{} 4(b^2 \plus{} 4a) \equal{} 16(b \minus{} a \plus{} 1)\leq 0$,
$ \iff a \minus{} b\geq 1$.
Therefore, $ a \minus{} b > 1$.
[/hide]

How did you know that :

 (2b + 4)^2 -4(b^2 +4a) =< 0 ?


Find all functions $f: \mathbb R \to \mathbb R$ satisfying the equality $$ f (y) + f (x + f (y)) = y + f (f (x) + f (f (y)))$$ happens for all real $ x$ and $ y$.

[quote="tdl"]Find all functions $ f : R\rightarrow R$ satisfying the equality $ f (y) \plus{} f (x \plus{} f (y)) \equal{} y \plus{} f (f (x) \plus{} f (f (y)))$ for all real $ x; y$.[/quote]

Let $ P(x,y)$ be the property $ f (y) \plus{} f (x \plus{} f (y)) \equal{} y \plus{} f (f (x) \plus{} f (f (y)))$
Let $ a\equal{}f(0)$

First, if $ f(y)\equal{}f(z)$, looking at $ P(x,y)$ and $ P(x,z)$ shows that $ y\equal{}z$ and so $ f(x)$ is injective.

$ P(0,0)$ implies $ a \plus{} f (a) \equal{} f (a \plus{} f (a))$  and so $ f(u)\equal{}u$ where $ u\equal{}a\plus{}f(a)$

$ P(x,u)$ implies $ u \plus{} f (x \plus{} u) \equal{} u \plus{} f (f (x) \plus{} u)$, so $ f (x \plus{} u) \equal{}f (f (x) \plus{} u)$ and, since $ f(x)$ is injective, $ x\plus{}u\equal{}f(x)\plus{}u$ and so $ f(x)\equal{}x$

And it is easy to check that this necessary condition is sufficient.

So the unique solution is $ \boxed{f(x)\equal{}x}$


inequalities
calculus
inequalities proposed

$ a,b,c,d\in [1,2]$
show that
$ \frac{a\plus{}b}{c\plus{}d}\plus{}\frac{c\plus{}d}{a\plus{}b}\minus{}\frac{a\plus{}c}{b\plus{}d}\leq \frac{3}{2}$
No calculus pls thx for your nice solution  :)
Credit : problem from Greece's Observer A at 50th IMO

Let $ x \equal{} \frac {a \plus{} b}{c \plus{} d}$ and $ y \equal{} \frac {a \plus{} c}{b \plus{} d}.$

Then $ \frac12\leqslant x,y\leqslant2$ and so $ x \plus{} \frac1x\leqslant\frac52$ and $ y \plus{} \frac1y\geqslant2,$ and we can assume WLOG that $ y\geqslant1.$ Hence

[list]$ x \plus{} \frac1x \minus{} y \minus{} \frac1y\ \leqslant\ \frac12$[/list]
$ \Rightarrow\ x \plus{} \frac1x \minus{} y\ \leqslant\ \frac12 \plus{} \frac1y\ \leqslant\ \frac32$

Why can we assume $ y\ge 1$? If $ y\le 1$, your statement won't hold.

Yeah, I realized that after I had posted. I was a little careless about what I could assume in the proof. :ewpu:


Prove that

 $ \lim_{x \to \infty}\left(e^x-e^{\sqrt{x^2+1} \right) = -\infty}$.

hello, writing your term in the form
$ \frac{1-e^{\frac{\sqrt{1+\frac{1}{x^2}}}}}{e^{-x}}$
now we can use the L'Hospital rules
and we get
$ -\frac{e^{x}e^{\sqrt{1+\frac{1}{x^2}}}}{\sqrt{1+\frac{1}{x^2}}x^3}$
and this tends to  minus infinity as $ x$ tends to plus infinity.
Sonnhard.

Hello
We can prove it by using "mean value theorem"

in fact , $ \exists c  \in [x,\sqrt{x^2 +1}] /  e^x-e^{\sqrt{x^2+1}} = x-\sqrt{x^2+1}.e^c = \frac{-1}{x (1+\sqrt{1+\frac{1}{x^2}})}.e^c$

And $ \lim_{x \to +\infty}\left(e^x-e^{\sqrt{x^2+1} \right) = \lim_{x \to +\infty}\frac{-1}{x (1+\sqrt{1+\frac{1}{x^2}})}.e^c = -\infty}$


Here is the IMO shortlist problems 1988, I got it from internet.

[quote=orl]Here is a complete booklet International Mathematial Olympiad (IMO) Longlist 1988 attachment which is a bit more comprehensive: "An Olympiad down under: a report on the 29th International Mathematical Olympiad in Australia" by W. P. Galvin, D. C. Hunt, and Peter Joseph O'Halloran. For Windows there is a [url=http://www.stdutility.com/download/stduviewer.exe]STDU viewer.[/url]. The link to the file is:

- http://www.4shared.com/file/0eFSbIli/IMO_Longlist_IMO_1988.html[/quote]

This link is not working dear orl can you share it again


Diamonds in mathematical inequalities,page 44,45
equality for the maximum occurs iff $ x_4 \equal{} \minus{} x_1 \equal{} \frac {2 \minus{} \sqrt {2}}{4},x_3 \equal{} \minus{} x_2 \equal{} \frac {\sqrt {2}}{4}$
hint:
try to predict the maximum and then assume $ x_1 \le x_2 \le x_3 \le x_4$ and use only AM-GM.

How do you predict maximum?

Where can I find that book that you have written?

here is the solution written in that book:
we predict that the maximum of the product which is denoted by $ P$ can be obtained when $ x_1\equal{}\minus{}x_4;x_2\equal{}\minus{}x_3 \implies x_2\minus{}x_1\equal{}x_4\minus{}x_3$

Letting $ x_3\minus{}x_2\equal{}a(x_4\minus{}x_3)$,$ P$ reaches Max when the variables are satisfied with the conditions:

$ x_2\minus{}x_1\equal{}x_4\minus{}x_3\equal{}\frac{x_3\minus{}x_2}{a}\equal{}\frac{x_3\minus{}x_1}{a\plus{}1}\equal{}\frac{x_4\minus{}x_2}{a\plus{}1}\equal{}\frac{x_4\minus{}x_1}{a\plus{}2}$

From these ,we come to this solution:

WLOG supposing $ x_1 \le x_2 \le x_3 \le x_4$

$ \implies P\equal{}(x_2\minus{}x_1)(x_3\minus{}x_2)(x_4\minus{}x_1)(x_3\minus{}x_1)(x_4\minus{}x_2)(x_4\minus{}x_3)$

Hence $ \frac{P}{a(a\plus{}2)(a\plus{}1)^2}\equal{}(x_2\minus{}x_1).(x_4\minus{}x_3).\frac{x_3\minus{}x_1}{a\plus{}1}.\frac{x_4\minus{}x_1}{a\plus{}2}.\frac{x_3\minus{}x_2}{a}.\frac{x_4\minus{}x_2}{a\plus{}1}$

$ \ \ \ \ \ \ \ \le \left(\frac{1}{6}((x_2\minus{}x_1)\plus{}(x_4\minus{}x_3)\plus{}\frac{x_3\minus{}x_1}{a\plus{}1}\plus{}\frac{x_4\minus{}x_1}{a\plus{}2}\plus{}\frac{x_3\minus{}x_2}{a}\plus{}\frac{x_4\minus{}x_2}{a\plus{}1}) \right)^6$

it folows that:
$ P \le$ $ \frac{1}{2^8}$$ ((x_4\minus{}x_1)(1\plus{} \frac{1}{a\plus{}1}\plus{} \frac{1}{a\plus{}2})\plus{} ( \frac{1}{a} \plus{} \frac{1}{a\plus{}1}\minus{}1)(x_3\minus{}x_2))^6$

Choose $ a$ such that $ 1\plus{} \frac{1}{a\plus{}1}\plus{} \frac{1}{a\plus{}2}\equal{} \frac{1}{a} \plus{} \frac{1}{a\plus{}1}\minus{}1 \leftrightarrow a\equal{} \sqrt{2}\minus{}1$

Now $ 1\plus{} \frac{1}{a\plus{}1}\plus{} \frac{1}{a\plus{}2}\equal{} \frac{1}{a} \plus{} \frac{1}{a\plus{}1}\minus{}1\equal{}\frac{3\sqrt{2}}{2}$

$ \implies (\sqrt{2}\plus{}1)(\sqrt{2}\minus{}1)(\frac{\sqrt{2}}{2})^2 P \le \frac{1}{2^8}(\frac{3\sqrt{2}}{2}(\minus{}x_1\minus{}x_2\plus{}x_3\plus{}x_4)^6)$

now note that $ \minus{}x_1\minus{}x_2\plus{}x_3\plus{}x_4 \le 1$

the rest is done easily!

I've been looking for more beautiful solution, but I can't find it... Can anybody else find it?

Thanks.

Also, I think this proof is wrong, or I don't understand it completel...
How can we choose a and then prove taht we have reached the maximum. What if $a$ is not $\sqrt{2}-1$? How can we prove that for any other $a$, we will not reach that maximum? Also, how can we prove that maximum won't be reached if we don't have two pairs with sum equal to zero?

Thanks. :)


limit
trigonometry
calculus
calculus computations

Consider the sequence of real numbers $(a_{n})_{n\ge1}$ such that $\lim_{n\to\infty}\sum_{i=1}^{n}{a_{i}}=a$. Prove that ${\lim_{n\to\infty}\sum_{i=1}^{n}{a_{i}\cos{\frac{i\pi}{n}}}}=a$.


For every positive integer $n$, determine the biggest positive integer $k$ so that $2^k |\ 3^n+1$

For every n 3^n+1=2 or 4 mod 8; So for even n k is 1 and for odd - 2

a positive integer -2? :) I think you meant it correctly though...

If $n:$odd,then max $k=2$.
If $n:$even,then max $k=1$.

We see that for $k>2$
$$3^n+1\equiv1,3 +1\equiv2,4 (mod8)$$
And so it is clear that $k<3$
For $k=2$,  $$3^n+1\equiv(-1)^n+1 (mod4)$$
So n must be even.
For $k=1$, it is clear that it works for any $n$ but in our case it is the maximum for n odd.


algebra
polynomial
complex analysis
complex analysis unsolved

Let $ P(z)\equal{}a_0 \plus{} a_{1}z \plus{} ... \plus{} a_{n}z^{n}$ be a complex polynomial of degree $ n$. Suppose that $ \max_{0 \le k < n} |a_k| \ne 0$. Prove that all the roots of $ P$ are in the open circle whose center is (0,0) and radius is $ r \equal{} 1 \plus{} \frac{\max_{0 \le k < n} |a_k|}{|a_n|}$

We may assume $ a_n\equal{}1$. Let $ m\equal{}\max_{0\le k<n}|a_k|$. If $ |z|\ge 1\plus{}m$, then $ |P(z)\minus{}z^n|\le m(1\plus{}|z|\plus{}\dots\plus{}|z|^{n\minus{}1})\equal{}m\frac{|z|^{n}\minus{}1}{|z|\minus{}1}\le |z|^n\minus{}1$, hence $ |P(z)|\ge 1$.


[quote="marko avila"]find all possible x such that :

\[sqrt(x^2-1) + 2*sqrt(x^2-p) = x\] for all numbers p[/quote]

We have ${lim_{p->+\inf}(x^2-p}=-\inf$?

hello, it is $\sqrt{x^2-1}+2\sqrt{x^2-p}=x$?
Sonnhard.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346891&sid=f467b96dfb1a779278a7ebf0f929d637#p346891[/url]

an old imo problem
fully correct solution is by enescu.


induction
MATHCOUNTS
geometry
pigeonhole principle
\/closed

I, personally, would like to see more seminars on certain topics, such as invariance principles, pigeon hole principles, fibonacci numbers, golden sections, mathematical inductions,  combinatorial identities, functional equations, and coloring problems.  I also think that if each seminar lasted several hours, and was divided into a few days, followed by a couple problem sets with solutions, it would be very beneficial.  Some people might argue that we do these problems in the forum all the time, so why should we bother with seminars.  My thought is that, after certain stage of math development, young kids need to approach mathematics problems in a more systemic fashion ---- this would be in contrast to the mathcounts approach, which encourges playfulness, fun, and non-systemic teaching in order to stimulate student's intrest in the studies of math.
    For example, if we look at fibonacci numbers. we encounter them from different angles.  It would save young students lots of time if educators could melt "fibonacci number" topics such as geometry, golden sections, recurrences, and algebra into one pot and present it as a seminar.  I believe that this way, students will be enlightened much faster and much deeper.

We are planning to eventually hold classes on problem solving techniques that do not fit nicely into other curriculum areas.  Methods such as the pigeon hold principle, invariance, induction, and many others will be included.

[quote]Most the principles, formulas, etc that you would like to see on seminars are already covered in the classes. More than half the things you just named are in the Intermediate Counting Class. It will probably be rerun soon. If you are interested in these topics, I would recommend you take that class.[/quote]

The intermediate counting class?  I had watched almost every class over my son's shoulder.  I was hidden.

We did combinatorial identities for a day or two in the intermediate counting.  Functional identities are covered in the algebra class.  We expect to offer a course next year that covers items like induction, invariance, pigeonhole, etc.

I realize that the topics you have mentioned have been taught in some of the classes.  I was just trying to emphasize the importance of in-depth, systemic studies of essential subjects.  I truly appreciate your dedication to the younger mathematicians.


let $ a,b,c$ be a nonegative numbers such that : $ abc\equal{}1$
prove that:

$ \sum \frac{ab}{a^5\plus{}b^5\plus{}ab}\,\geq\,1$

It's reserve .
Use this ineq :$ a^5\plus{}b^5 \geq a^2b^2(a\plus{}b)$ .
$ LHS \leq \sum \frac{1}{ab(a\plus{}b)\plus{}1}\equal{} 1$


yeah, there are a bunch of problems like that but a lot of the time you don't have time to report them.

[quote="toadoncart"]There's this one problem that says, "Refer to the stem and leaf plot shown, blah blah blah blah blah," but there's no stem and leaf plot.

infinity4ever and 3.141957man also know about this.

there's another one that says: here's the value of these letters. How much is [i]papa[/i] worth?

sailingdog knows about this.[/quote]
I saw that problem too

I saw a problem that asked for us to count how many possible ways there are to get to a point, but it did not show us anything.

Just to repeat...

[quote="Xevarion"]You can report an error in a question by going to "review problems" or whatever at the end of the game and then go to that problem and report it.[/quote]

This is the best way to tell us about problems, as it tells us which question you are referring to.


trigonometry
logarithms
calculus
derivative
real analysis
real analysis unsolved

consider  on interval $ x\in [0,1]$ following serie:
$ \sum_{n \equal{} 2}^{ \plus{} oo}\frac {\sin nx}{n\ln n}$
this is not easy problem,
I mean if $ f(x) \equal{} \sum_{n \equal{} 2}^{ \plus{} oo}\frac {\sin nx}{n\ln n}$
then can we diferentiate on our interval. or exist there $ f'(x)$

sorry:
My question was not correct, this is simplest problem. my question is :
for this series :
$ \sum \frac{\sin nx}{n\ln nx}$

In order to derivative the series $ \sum\limits_{n \equal{} 1}^{ \plus{} \infty } {u_n \left( x \right)}  \equal{} f\left( x \right)$, you must check the following facts

+ f(x) exists, e.g. exists for all $ x \in (a,b)$.
+ $ u_n$ is differentiable in $ (a,b)$.
+ $ \sum\limits_{n \equal{} 1}^{ \plus{} \infty } {u'_n \left( x \right)}$ uniformly converges in $ (a,b)$.

Then $ \sum\limits_{n \equal{} 1}^{ \plus{} \infty } {u'_n \left( x \right)}  \equal{} f'\left( x \right)$.


trigonometry
linear algebra
calculus
calculus computations

Solve the eigenvalue problem

$ \frac{d^2 \phi}{dx^2} \equal{} \minus{}\lambda \phi$

subject to

$ \phi(0) \equal{} \phi(2\pi)$

and

$ \frac{d \phi}{dx} (0) \equal{} \frac{d \phi}{dx} (2 \pi).$

I had the solution already, but am looking for a much simpler way (perhaps without linear algebra), if any.

[hide="My roundabout solution"]

Here's what I had:

after solving the ODE, we have the general solution

$ \phi \equal{} C_1 \sin(\sqrt{\lambda}x) \plus{} C_2 \cos(\sqrt{\lambda}x)$

applying the conditions we have the system

$ C_2 \equal{} C_1 \sin(\sqrt{\lambda}2\pi) \plus{} C_2 \cos(\sqrt{\lambda}2\pi)$

$ C_1 \sqrt{\lambda} \equal{} C_1 \sqrt{\lambda} \cos(\sqrt{\lambda}2\pi) \minus{} C_2 \sqrt{\lambda} \sin(\sqrt{\lambda}2\pi)$

then using a little knowledge of linear algebra and determinant, we get, in order not to have trivial solutions,

$ C_1 C_2 u^2 \plus{} C_1 C_2 (v\minus{}1)^2 \equal{} 0$

where $ u \equal{} \sin(\sqrt{\lambda}2\pi)$ and
$ v \equal{} \cos(\sqrt{\lambda}2\pi)$

or simply

$ u^2 \plus{} (v\minus{}1)^2 \equal{} 0$

So $ u^2 \equal{} (v\minus{}1)^2 \equal{} 0$ or
$ u \equal{} 0; v \equal{} 1$

$ \sqrt{\lambda_n}2\pi \equal{} 2n\pi$
$ \lambda_n \equal{} n^2$

So the eigenfunction is

$ \phi_n \equal{} C_1 \sin(nx) \plus{} C_2 \cos(nx)$

I don't think this is complete because we haven't determined $ C_1$ and $ C_2$ yet.
[/hide]

We have eugenfunction $ \varphi_0\equal{}1$ for $ \lambda_0\equal{}0$ and [b]two[/b] eugenfunctions $ \varphi_n^{(1)}\equal{}\cos{nx}$ and $ \varphi_n^{(2)}\equal{}\sin{nx}$ for $ \lambda_n\equal{}n^2$.


Points $ M$ and $ N$ are the midpoints of two sides of the square shown. What fractional part of the square's area is shaded? Express your answer as a common fraction.
[asy]size(3cm,3cm);
pair A,M,N,B; M = (0,1); B = (2,0); N = (1,2); draw(A--(0,2)--(2,2)--B--A); fill(A--M--N--B--cycle); label("$M$",M,W); label("$N$",N,N);[/asy]

Splitting the square down the middle, we see that $ \frac12$ of the right side is shaded.
On the right side, $ \frac14$ is not shaded, so there must be $ \frac34$ of the left side shaded.
Because each fraction is for $ \frac12$ of the square, we need to multiply them by $ \frac12$
Adding, we get:
$ \frac12 (\frac12 \plus{} \frac34) \equal{} \boxed{\frac58}$


Could anyone give me problems of Russia TSTs and USA TSTs(with solution is best).

USA TST problems can be found at
http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html

Although I must say that tests from the http://www.unl.edu/amc archives can be successfully opened about 5% of the time, at least for me.  http://www.kalva.demon.co.uk has lots of tests, but the problems are often very strangely worded :(.


induction
advanced fields
advanced fields unsolved

How do I show that H(XxS^n) is isomorphic to the direct sum of H(X) and
H(X x S^n, X x {s})?

Thanks.

yesterday i've found this problem on hatcher's book, so i tackled it again, and this time i seem to have (partly) won :P

first of all, the much more suggestive first (or global) part of the exercise sounds like:
show that $ H_{i}(X\times S^{n}) \simeq H_{i}(X)\oplus H_{i-n}(X)$
which seems to be false at a first sight!

the second part, which leads to the first one, is partly the first post, and then an induction step, that sounds like:
show that $ H_{i+1}(X\times S^{n+1},X\times\{x_{0}\})\simeq H_{i}(X\times S^{n},X\times \{x_{0}\})$.

first, let's talk about this last part: it's the shortest and easiest part of the proof, being a direct consequence of the relative mayer-vietoris exact sequence, considering the correct spaces.
if someone cares about it, just ask :p it's longer than i thought and than i want to write :p

now, let's come to a proof for the original problem: let me call $ X_{*}= X\times\{x_{0}\}$, $ X^{*}= X\times (S^{n}\setminus \{-x_{0}\})$, and $ SX = X\times S^{n}$; consider the couples $ (SX, X_{*})$ and $ (SX, X^{*})$. since $ X^{*}$ retracts onto $ X_{*}$, their homologies are the same, and also relative homologies of $ SX$ with respect to them, while homology of $ X^{*}$ relative to $ X_{*}$ is trivial..
now write down the long exact sequence for this couple:
$ \to H_{k+1}(SX,X^{*})\to H_{k}(X^{*})\to H_{k}(SX)\to H_{k}(SX,X^{*})\to H_{k-1}(X^{*})\to$
our aim is to prove that the map $ \pi_{*}: H(SX)\to H(SX,X^{*})$ is surjective (for any degree, which index i omitted for sake of everyone's eyes).

why? if it's surjective, since the long sequence is exact, the border map $ H_{k}(SX,X^{*})\to H_{k-1}(X^{*})$ is $ 0$, and the other map $ H_{k}(X^{*})\to H_{k}(SX)$ is injective: then we can split the long sequence at $ H_{k}(SX)$ in short exact sequences
$ 0\to H_{k}(X^{*})\to H_{k}(SX)\to H_{k}(SX,X^{*}) \to 0$.
then we'd like to prove that this sequences split.

let's turn to surjectivity: think of relative cycles in $ S_{k}(SX,X^{*})$ in the geometrical way, as generated by maps $ \sigma^{k}: \Delta^{k}\to SX$ (the image of) which boundary lies in $ X^{*}$.
consider a cycle $ \sigma\in S(SX)$: its boundary is a cycle in $ X^{*}$, because $ \partial\partial \sigma = 0 \in SX$, but boundary map on $ X^{*}$ is just the restriction of boundary map on $ SX$, and $ \partial\sigma \in S(X^{*})$.
so, considering the splitting of all maps ($ \sigma$ and $ \partial$) given by direct products ($ X\times S^{n}$ and $ X\times \mathbb{R}^{n}$), we obtain that $ \partial\sigma$ is a cycle in $ S(X^{*})$, hence it's a boundary, since it was a boundary in the $ X$ factor and $ \mathbb{R}^{n}$ has trivial homology (let's disregard the case $ k=0$, which has no practical interest and can be treated as a separate thing, which i will do later).
this gives a preimage of our cycle $ \sigma$, since we've found a cycle in $ S(X)$ having boundary in $ S(X^{*})$ whose induced projection in homology is $ [\sigma]$.

the case $ k=0$ is just a matter of arcwise connection: just count how many arcwise connected components you have, and you've finished.

summing up, we showed that there's a short exact sequence
$ 0\to H_{k}(X^{*})\to H_{k}(SX)\to H_{k}(SX,X^{*}) \to 0$
that we'd like to split.

this is still missing, maybe tomorrow i'll be thinking a bit more about it: i think the proof can be adapted from the one of splitting of sequences hatcher uses to prove universal coefficient theorems.

i'm back!
so.. let's get back to this short sequence
$ 0\to H_{k}(X^{*})\to H_{k}(SX)\to H_{k}(SX,X^{*}) \to 0$ (1)
the map $ i_{*}: H_{k}(X^{*})\to H_{k}(SX)$ is the induced map of $ i: S_{k}(X^{*})\to S_{k}(SX)$, and is in a short exact sequence
$ 0\to S_{k}(X^{*})\to S_{k}(SX)\to S_{k}(SX,X^{*}) \to 0$
which splits since the last nontrivial term is free. then we have a map $ p: S_{k}(SX)\to S_{k}(X^{*})$ such that $ pi ={\rm id}_{S_{k}(X^{*})}$. then the induced map $ p_{*}i_{*}= (pi)_{*}={\rm id}_{H_{k}(X^{*})}$, which gives $ p_{*}$ that makes our original sequence (1) split.


function
logarithms
limit
algebra unsolved
algebra

Given that $ f$ : $ \mathbb{R} \to \mathbb{R}$ which satisfying : \[ f(2x)\equal{}2f(x)\sqrt{1\plus{}f^2(x)}\; , \forall x\in\mathbb{R}\quad (E) \]
$ i)$ Solve the equation $ (E)$ by assuming $ f$ is differentiable at the point $ 0$.

$ ii)$ Solve the equation $ (E)$ in general case.

[quote="palmita"]Hello,

Given that $ f$ : $ \mathbb{R} \to \mathbb{R}$ which satisfying :
\[ f(2x) = 2f(x)\sqrt {1 + f^2(x)}\; , \forall x\in\mathbb{R}\quad (E)
\]
$ i)$ Solve the equation $ (E)$ by assuming $ f$ is differentiable at the point $ 0$.

$ ii)$ Solve the equation $ (E)$ in general case.[/quote]

Notice that $ a=2b\sqrt{1+b^2}$ $ \implies$ $ b=sign(a)\sqrt{\frac{\sqrt{1+a^2}-1}{2}}$ and so $ f(x)$ is fully determined by the knowleged of the function in for example $ (-2,-1]\cup[1,2)$

Notice also that if $ f(x)=\sinh(u)$, $ f(2x)=\sinh(2u)$ and $ f(\frac{x}{2})=\sinh(\frac{u}{2})$

So a general solution of this equation is :

Let $ u_1(x)$ and $ u_2(x)$ any real functions defined in $ [1,2)$. Then :

$ \forall x<0$ : $ f(x)=\sinh(xu_1(2^{\{\log_2 |x|\}}))$ (where $ \{y\}$ is the fractional part of $ y$, so is in $ [0,1)$).
$ f(0)=0$
$ \forall x>0$ : $ f(x)=\sinh(xu_2(2^{\{\log_2 |x|\}}))$ (where $ \{y\}$ is the fractional part of $ y$, so is in $ [0,1)$).

In order to have a solution differentiable at 0 :
1) We  need $ u_1$ and $ u_2$ to have max and min values over $ [1,2)$ (in order to have continuity)
2) We need to have $ \limits_{x\to 0}\frac{f(x)}{x} = c$

$ \lim_{x\to 0}\frac{f(x)}{x}$ =$ \lim_{x\to 0}\frac{\sinh(xu_i(2^{\{\log_2 |x|\}}))}{x}=c$ $ \implies$ $ \lim_{x\to 0}u_i(2^{\{\log_2 |x|\}}) = c$ and so $ u_1(x)=u_2(x)=c$

And so $ f(x)=\sinh(cx)$

Here is another general form for all the solutions, equivalent to the previous one, but simpler to write :

Let $ g(x)$ any real function defined over $ (\minus{}1,1)$. Then $ f(0)\equal{}0$ and $ f(x)\equal{}\sinh(x g(sign(x)\{\log_2(|x|)\}))$ $ \forall x\neq 0$ (with $ \{y\}\equal{}$ fractional part of $ y$).


There are distinct integers $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$, $a_{6}$, $a_{7}$ such that
 [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/b44a1eca802294d64b1075817ac76afe.gif[/img]

where $0 \leq  a_{i}< i$ for $i = 2,3,4,...7$.

Find $a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}$.

[quote="Nplusnequals"]It says in the problem, "distinct integers". You have three that are the same.[/quote]

It says unique in the actual problem.

[quote="Nplusnequals"]It says in the problem, "distinct integers". You have three that are the same.[/quote]
It means "a unique set of integers."
I saw this problem like 100 times already

[quote="Nplusnequals"]So, can anyone find the answer?[/quote]
Treething's answer is correct

Did you even read any of my posts?
The problem is supposed to read "a unique SET of integers" not "distinct integers."

If you don't believe me, try to find a set of distinct integers statisfying treething's equation.
(this $2520a_{2}+840a_{3}+210a_{4}+42a_{5}+7a_{6}+a_{7}= 3600$)

You can clearly see from Treething's solution that $a_{2}$ must equal $1$ because the maximum of the rest of the summation is $2519$ (there is a [b]very nice[/b] identity here involving factorials that will make this clear), and similarly that $a_{3}$ must equal $1$.

[b]Follow-up:[/b]  The result seems to suggest something about expressing numbers in a "factorial base."  Find this generalization, and prove it.

I'm sorry, but I don't understand how treething found out what the numbers $a_{2}$ through $a_{7}$ were.

It's sort of like case work.
If you try to let $a_{2}=0$, then $a_{3}$ becomes too big, which messes with the restriction that $a_{3}<3.$ So $a_{2}=1$. Similarly, you find $a_{3}$ thru $a_{4}$

[quote="Treething"]$2520a_{2}+840a_{3}+210a_{4}+42a_{5}+7a_{6}+a_{7}= 3600$.[/quote]

Remember that $a_{i}$ are integers such that $0 \le a_{i}< i$.  Therefore, $0 \le a_{2}< 2$ which means $a_{2}= 0, 1$.

If $a_{2}= 0$, then as I said before, the rest of the expression has a maximum value of $2519 < 3600$, which means $\boxed{a_{2}= 1}$.  So now

$840a_{3}+210a_{4}+42a_{5}+7a_{6}+a_{7}= 1080$

$a_{3}= 0, 1, 2$.  If $a_{3}= 2$ the rest of the expression is negative, and if $a_{3}= 0$ again the rest of the expression has a maximum value of $839$ (if you've ever changed a number from base $10$ to another base the slow way, this should be [b]very familiar[/b] logic to you!), so $\boxed{a_{3}= 1}$.  We now have

$210a_{4}+42a_{5}+7a_{6}+a_{7}= 240$

And so forth.  At every step we determine another "digit" based on the constraints on $a_{i}$.

[quote="t0rajir0u"]The integer counterpart consists of proving that there exist unique $a_{1}, a_{2}, ...$ such that

$m = \sum_{k=1}^{\infty}k! a_{k}$

Where $a_{k}\le k$.

[hide="The integer counterpart of the very nice identity; spoiler!"] $\sum_{k=1}^{n-1}k! k = n!-1$ [/hide][/quote]

[hide="hint"] It's easy to show that distinct sets of $a_{i}$ produce distinct $m$ because we always have

$a_{1}\cdot 1!+a_{2}\cdot 2!+\cdots+a_{n}\cdot n! < (n+1)!$.

Using contradiction, suppose

$\sum a_{k}\cdot k! = \sum b_{k}\cdot k!$.

Then

$\sum (a_{k}-b_{k}) \cdot k! = 0$.

Using the fact that $|a_{k}-b_{k}| \le k$, finish the proof.

To show that each positive integer has a representation, use strong induction. If $n! \le m < (n+1)!$, then

$m = a_{n}\cdot n!+r$

where $a_{n}\le n$ and $r < n!$. Finish it from here.[/hide]


One  two one plus two one two
 is the number of cats in a tree.”
My pup  tells, me , and sure, do I check.
They are three times forty-three.
So if my pup  asks  you APos’er what’s your answer be?
“If two one two is added to three one three one three? “

As usual hide your answers.

(It makes sense, and please do not peek at the hint, unless you are really stuck)
[hide]I do math in decimal system, but, obviously,   the Pup uses some other system. .. Only then it can make sense! (Besides  Pups  do not  have 10 fingers. :)  )
[/hide]

[hide]Your pup counts in base $6$
So we have that $212 + 31313$ in base $6$ is equal to $4301$ in base $10$.[/hide]


inequalities
quadratics
function
inequalities proposed

Please try to solve this one,I think it is interesting:
If $a,b,c$ are real numbers satyisfying \[a+b+c=5\] then prove that the following inequality occurs: \[a^{4}+b^{4}+c^{4}+64(ab+bc+ac)\geq 18abc+473\]

No replies? :(  Is it too easy?Or maybe I shoud give a hint?

I'll do it the [b]peasant's[/b] way. Don't you throw [i]rotten tomatoes[/i] at me  :D I probably did some mistakes anyway :(

Let $f(a,b,c) = a^{4}+b^{4}+c^{4}+64(ab+bc+ca)-18abc-473$ and $g(a,b,c) = a+b+c-5$.

At an extremum point we must have $\nabla f = \lambda \cdot \nabla g$, i.e. $4a^{3}+64(b+c)-18bc = \lambda$ etc.

Subtracting the equations, we get $(a-b) \left[ 4 (a^{2}+ab+b^{2})-64+18c \right]=0$ etc.

If $a \neq b \neq c \neq a$, then we have that $2(a^{2}+ab+b^{2})+9c = 32$ etc, so by subtracting again we get $(a-c)(2(a+c)+2b-9)=0$, so $a+b+c = \frac92$, contradiction.

Thus we can WLOG assume that $a=b$, so that $2b+c = 5$.

If $b \neq c$, then we solve the system $2b+c=5, \, 2b^{2}+2c^{2}+2bc+9b = 32$. There are two solutions: $b_{1}= 2, c_{1}= 1$ and $b_{2}= \frac32, c_{2}= 2$. We have that $f(2,2,1) = 0$ and that $f \left( \frac32, \frac32, 2 \right) = .125$.

For $a=b=c= \frac53$ we have that $f(a,b,c) = 0.(148)$ (???).

The last case: WLOG $c=0$. Then, we must solve $a^{4}+b^{4}+64 ab-473 \geq 0$ when $a+b=5$.

Now we introduce another Lagrange multiplier:

$4a^{3}+64 a = \mu$;

$4b^{3}+64 b = \mu$.

Subtracting we get that $(a-b)(4a^{2}+4ab+4b^{3}+64) = 0$, so $a-b = 0$ (since the other parenthesis is $>0$). In this case we have $f \left( \frac52, \frac52, 0 \right) = 5.125$.

The last subcase of the last case is $(a,b,c) = (5,0,0)$: $f(5,0,0) = 152$.

All extremum points yield values $\geq 0$, so the inequality is true.

And of course, equality holds if and only if $(a,b,c) = (2,2,1), (1,2,2)$ or $(2,1,2)$.

@perfect_radio:First of all,thanks for trying the ineq.Secondly,I must sadly say that I did not fully understant your solution,because I am only 9th grade[from september] and do not know some of the simbols there..so I let the others judge the possible mistakes..But the equality case is right and I wait other solutions,not necesarily elementary ones[allthough there is one :P ]

[quote="arqady"][quote="andyciup"]
If $a,b,c$ are real numbers satyisfying \[a+b+c=5\] then prove that the following inequality occurs: \[a^{4}+b^{4}+c^{4}+64(ab+bc+ac)\geq 18abc+473\] [/quote]
$a^{4}+b^{4}+c^{4}+64(ab+bc+ac)-18abc-473=$
$=\frac{1}{625}\cdot\sum_{cyc}(12a^{2}+2b^{2}+2c^{2}-11ab-11ac+4bc)^{2}.$ :D :)[/quote]

arqady, you make me so amazing.
How can you come to this solution, or just only you use maths software? I think that no one can use hand to do that(except you). Okie?

[quote="timlaiminh"]
How can you come to this solution, or just only you use maths software?[/quote]
Which maths software? I made it self.
Moreover, for all $\alpha,$ $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma=16$ and $\alpha\beta+\alpha\gamma+\beta\gamma=52$ holds:

$a^{4}+b^{4}+c^{4}+64(ab+bc+ac)-18abc-473=$

$=\frac{1}{625}\cdot\sum_{cyc}\left(\alpha a^{2}+\beta b^{2}+\gamma c^{2}+\frac{(5\gamma-7\alpha-7\beta)ab}{8}+\frac{(5\beta-7\alpha-7\gamma)ac}{8}+\frac{(5\alpha-7\beta-7\gamma)bc)}{8}\right)^{2}.$

I just came up with a solution which is inspired by pvthuan's article on Mathematical Reflections. (Though his original idea isn't applied here.)

Firstly we assume $a$ is the largest among $a,b,c$, so we have $a\geq\frac53$
Using $x^{4}+y^{4}=(x+y)^{4}-2xy[2(x+y)^{2}-xy]$, the original inequality is equivalent to
$(b+c)^{4}-2bc[2(b+c)^{2}-bc]+a^{4}+64(b+c)a+64bc-18abc-473\geq 0$.

Since $a+b+c=5$, it remains to prove
$(5-a)^{4}-2bc[2(5-a)^{2}-bc]+a^{4}+64a(5-a)+64bc-18abc-473\geq 0$

$\Leftrightarrow 2(bc)^{2}+[64-18a-4(5-a)^{2}]bc+(5-a)^{4}+a^{4}+64a(5-a)-473\geq 0$

$\Leftrightarrow 2(bc)^{2}-2(2a^{2}-11a+18)bc+2(a^{4}-10a^{3}+43a^{2}-90a+76)\geq 0$

Now we consider the quadratic function
$f(t)=2t^{2}-2(2a^{2}-11a+18)t+2(a^{4}-10a^{3}+43a^{2}-90a+76)$
By some calculations one have $\Delta = 4(a-2)^{2}(5-4a)\leq 0$.
Therefore $f(t)\geq 0$, and the desired result follows when we replace $t$ by $bc$.

For the equality to hold, we must firstly have $\Delta =0$, i.e. $a=2$. Now the function becomes $f(t)=2t^{2}-8t+8=2(t-2)^{2}\Rightarrow t=2$. Hence $(b,c)=(2,1)$ or $(1,2)$.

[quote="arqady"]
Which maths software? I made it self.
Moreover, for all $\alpha,$ $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma=16$ and $\alpha\beta+\alpha\gamma+\beta\gamma=52$ holds:

$a^{4}+b^{4}+c^{4}+64(ab+bc+ac)-18abc-473=$

$=\frac{1}{625}\cdot\sum_{cyc}\left(\alpha a^{2}+\beta b^{2}+\gamma c^{2}+\frac{(5\gamma-7\alpha-7\beta)ab}{8}+\frac{(5\beta-7\alpha-7\gamma)ac}{8}+\frac{(5\alpha-7\beta-7\gamma)bc)}{8}\right)^{2}.$[/quote]

Dear Arqady,
can you please explain (with some more detail  :) ) the method you have applied to derive this hard identity.

[quote="andyciup"]
If $a,b,c$ are real numbers satyisfying \[a+b+c=5\] then prove that the following inequality occurs: \[a^{4}+b^{4}+c^{4}+64(ab+bc+ac)\geq 18abc+473\] [/quote]
$a^{4}+b^{4}+c^{4}+64(ab+bc+ac)\geq 18abc+473\Leftrightarrow\sum_{cyc}(a-1)^{2}(a-2)^{2}\geq0.$ :D :)


rotation
combinatorics unsolved
combinatorics

Determine the number of ways of building a wall of dimensions 2xn,
using bricks of size

a)1x2 and 1x1

b)1xk and 2xk

c)1xk  (k=1,...,n)

solution

[a] using 1*2 and 1*1
 This is is a classic argument.let us have 2*n wall with 2 along  x-axis and n along y-axis (as matter of convention).let us mark the lower square of a 1*2 block in the left column of the wall.We can select those squares $ s_i$ among {1,2,..n-1} so that $ |s_i\minus{}s_j|>1$. If $ f(n\minus{}1)$ is the number of ways of finding the the set of lower squares with the above mentioned restriction then,
\[ f(n\minus{}1)\equal{}f(n\minus{}2)\plus{}f(n\minus{}3)\textrm{with initial conditions}f(1)\equal{}1;f(2)\equal{}2\]
Hence,$ f(n)$ is the nth fibonacci number, yielding the solution $ {f^2}(n)$ to [a]

[b] using 1*k and 2*k
It is not specified whether k is fixed or not unlike [c].
case1: k is specified.
We can only build the wall if $ k|n$.
In the left column, identify the left lower square of 2*k block. This can be done in $ 2^{\frac {n}{k}}$
case2: k is not fixed.
letus consider the solutions of the equation $ x_1\plus{}\dots\plus{}x_k\equal{}n$ where the $ x_i$ for odd i denotes the run of 1*k(2*k) and even i denotes the run of 2*k(1*k). This can be done in $ 2\sum_{k\equal{}1}^{n} {n\plus{}k\minus{}1\choose k\minus{}1}$

[c] using 1*k (k=1,2,..,n)
A square in the left column may or may not be the lower end of a 1*k block except the lowermost square.So this can be done in $ 4^{n\minus{}1}$ ways.

so far i have noticed that part a of the question is not correct. first of all f(2)=7 because as far as i understand we have the cases that we just use 1x1 blocks or 1x2 block. so this means we can construct a 2x2 wall by 7 different ways with 1x1 and 1x2 bricks. good luck

Yes Blackmoon, you're right. We can use our 1x2 bricks as 2x1. I mean, we can put our 1x2 bricks vertically and also horizontally...
Thanks, good luck...

[quote="hastings"]...We can use our 1x2 bricks as 2x1. I mean, we can put our 1x2 bricks vertically and also horizontally...[/quote]

Let us stick on to a notation to avoid confusions of this sort. A block of 1X2 means 1 unit on horizontal axis and 2 units on vertical axis.

solution to [a] using 1X1 or 1X2 or 2X1.
We get the recursive relation,
\[ f(n)\equal{}2f(n\minus{}1)\plus{}3f(n\minus{}2)\quad \forall n\geq 3;\quad f(1)\equal{}2,f(2)\equal{}5\]
\[ \Rightarrow f(n)\equal{}\frac {7.3^{n\minus{}1}\minus{}(\minus{}1)^n}{4}\quad \forall n\geq 1\]

No, it is more complicated than that: see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187216]here[/url].

For part (b), if $ k \equal{} 1$ we've just answered it, if $ k \equal{} 2$ we get the recursion $ a_n \equal{} a_{n \minus{} 1} \plus{} 2a_{n \minus{} 2}$ with $ a_0 \equal{} a_1 \equal{} 1$.  If $ k > 2$, srikanth's original argument works (because we cannot rotate the blocks in this case).

Thanks Joel, I did look at your solution in 'rectangle'. I made a mistake by putting $ f(2)\equal{}5$ which is 7 instead. Else I don't find any mistake in my recursion. Let me explain it,
[i] We can build a 2Xn from a 2X(n-1) by either placing two 1X1 dominoes or one 2X1 domino. This gives $ 2f(n\minus{}1)$.
[ii] we can build 2Xn from a 2X(n-2) by three ways as I have shown in the picture. Rest of the four ways of constructing 2X2 as shown in rectangle.jpg are not considered as they are already counted in [i].

@Srikanth:
The problem with your solution is that it doesnt count arrangements like this:
[code]+--+--+--+--+--+--+
|     |     |     |
+--+--+--+--+--+--+
|  |     |     |  |
+--+--+--+--+--+--+[/code]


[quote="manlio"]For $a,b,c$ positive reals prove that

$\frac{a^{2001}}{b+c}+\frac{b^{2001}}{c+a}+\frac{c^{2001}}{a+b} \geq \frac{x^{2000}+y^{2000}+z^{2000}}{2}$[/quote]
what are x,y,z?
if they are a,b,c
then i guess, it can be finished by power mean and chebyshev

Reminds me of Nesbitt's inequality, with the exponents 2001 and 2000 instead of 1 and 0. Wonder if anybody can provide a proof simple as the proof of Nesbitt...?

I think working out gives \[2\cdot\sum_{cyc}a^{2001}.(a+c)(a+b) \geq (a+b)(b+c)(c+a)(a^{2000}+b^{2000}+c^{2000})\]

Which -after simplifying- is: \[\sum_{cyc}(2\cdot a^{2003}+a^{2002}(b+c)) \geq \sum_{cyc}(a^{2000}(b^2c+c^2b)+a^{2001}(b^2+c^2)\]

Which can be split into:
$\sum_{cyc}2\cdot a^{2003}\geq \sum_{cyc}a^{2001}(b^2+c^2)$
and
$\sum_{cyc}a^{2002}(b+c)) \geq \sum_{cyc}a^{2000}(b^2c+c^2b)$
where both follow immediately from muirhead.


Find all strictly increasing functions $f : \mathbb Z \to \mathbb Z$ such that \[f \left( x^{2}+y^{2}\right) = f^{2}(x)+f^{2}(y), \, \forall x,y \in \mathbb Z .\] [b][i]Valentin Vornicu[/i][/b]

We know that $f^{2}(x)+f^{2}(y) = f(x^{2}+y^{2}) = f(|x|^{2}+y^{2}) = f^{2}(|x|)+f^{2}(y)$. This means $f^{2}(x) = f^{2}(|x|)$. If $x<0$, this implies that $f^{2}(x) = f^{2}(-x)$. So $f(x)=f(-x)$ or $f(x)=-f(-x)$. First case is impossible because $f$ is strictly increasing, so we know that $f$ is an odd function, and thus, we just have to solve $f(x^{2}+y^{2}) = f^{2}(x)+f^{2}(y)$ for $x,y\in\mathbb{N}$.

As $f$ is odd, $f(0)=0$. Fill $y=0$ in to get that $\boxed{f(x^{2}) = f^{2}(x)}$. Therefore, by filling $x=1$ in, $f(1) = f^{2}(1)$, so $f(1) \in\{0,1\}$. But $f(0)=0=f(1)$ is impossible, because $f$ is strictly increasing, so $f(1)=1$. From now, it's easy.

[u]Conjecture[/u]: $f(x) = x$, for all $x\in\mathbb{N}$.
[u]Proof[/u]: If we can prove that $f(2^{(2^{n})}+1) = 2^{(2^{n})}+1$, we're done because of the fact that $f$ is strictly increasing. Because $f(5)=f(4+1)=f^{2}(2)+f^{2}(1) = f^{2}(1+1)+1 = (f^{2}(1)+f^{2}(1))^{2}+1 = 5$, it's true for $n=1$. And if true for $n=k$, we know that $f(2^{(2^{k})}) = 2^{(2^{k})}$ because of the $f$ being strictly increasing. Therefore, \[{\begin{aligned}f(2^{(2^{k+1})}+1}) & = f(4^{(2^{k})}+1) = f^{2}(2^{(2^{k})})+f^{2}(1) = (2^{(2^{k})})^{2}+1 \\ & = 2^{(2^{k+1})}+1,\end{aligned}\] which completes the proof. $\blacksquare$

Because $f$ is odd, and we just proved that $f(x)=x$, for all positive integers, it follows that $f(x)=x$ also holds for negative integers. Therefore, $f(x)=x$ is the function we were looking for.


geometry
cyclic quadrilateral
geometry proposed

[b]Problem. [/b]Let $ P$ be a point on the plane of $ \triangle ABC.$ $ X,Y,Z$ are the intersections of $ PA,PB,PC$ with $ BC,CA,AB.$ $ M$ is the midpoint of $ AP.$ Line $ YM$ meets $ AB$ at $ N$ and $ XZ$ meets $ BP$ at $ L.$ Then prove that $ NL \parallel AP.$

By Menelaus' theorem for $ \triangle ABP$ cut by the transversal $ \overline{NMY},$ we get

$\frac {\overline{PY}}{\overline{BY}} \cdot \frac {\overline{BN}}{\overline{NA}} \cdot \frac {\overline{AM}}{\overline{MP}} \equal{} 1  \Longrightarrow  \frac {\overline{BN}}{\overline{NA}} \equal{} \frac {\overline{BY}}{\overline{PY}}$

On the other hand, since $ (B,P,L,Y) \equal{} \minus{} 1,$  we have $  \frac {\overline{BY}}{\overline{PY}} \equal{} \frac {\overline{BL}}{\overline{LP}}$

$\Longrightarrow  \frac {\overline{BL}}{\overline{LP}} \equal{} \frac {\overline{BN}}{\overline{NA}}  \Longrightarrow NL \parallel AP.$

When $ P$ is identical to the orthocenter $ H$ of $ \triangle ABC,$ we get $ NL \perp BC.$

[quote="MJ GEO"]1.(easy) in triangle $ ABC$ $ AD,BE,CF$ are altitudes and concur at $ H$.$ M$ is the midpoint of $ AH$.$ EM$ and $ FD$ intersect $ AB$ and $ BE$ at $ Q$ and $ P$.prove that $ PQ$ is perpendicular to $ BC$
2.(hard) if $ AD,CF,BE$ arent altitude problem is corect again. :!:[/quote]

it's quite easy problem. :)

[quote="luisgeometra"][color=darkred]Let $ P$ be a point in the plane of $ \Delta ABC$ and $ X,Y,Z$ be the meetpoints of $ PA,PB,PC$ with $ BC,CA,AB$

respectively. Denote the midpoint $ M$ of $ AP$ , $ N\in YM\cap AB$ and $ L\in XZ\cap BP$ . Prove that $ NL \parallel AP$ .[/color][/quote]

[b][u]Proof[/u][/b] (similarly with [b]luisgeometra[/b]'s). Denote $ N'\in AB$ for which $ LN'\parallel AP$ and $ R\in LN'\cap AC$ . Since the pencil

$ A(B,L,P,Y)$ is harmonically obtain for the transversal $ \overline {RN'L}$ that $ N'L \equal{} N'R$ , i.e. $ M\in YN'$ . In conclusion, $ N'\equiv N$ .

[quote="MJ GEO"]1.(easy) in triangle $ ABC$ $ AD,BE,CF$ are altitudes and concur at $ H$.$ M$ is the midpoint of $ AH$.$ EM$ and $ FD$ intersect $ AB$ and $ BE$ at $ Q$ and $ P$.prove that $ PQ$ is perpendicular to $ BC$
[/quote]

The 1st part has an easy angle chase solution.

 $ EM$ is the median of right triangle $ AEH$. Therefore, $ EM\equal{}MH\equal{}AM$. THis implies triangle $ EMH$ is isoscles. Thus, angles $ EMH\equal{}EHM$
By vertically opposite angles, $ BHD\equal{} MHE$.
Sicne $ BFHD$ is a cyclic quadrilateral, angles $ BFD\equal{} BHD\equal{} MHE\equal{} MEH$.
Also,  by the same cyclic quadrilateral, angles $ FBH\equal{}FDH$. Since angles $ BFD\equal{} PEQ$, it implies that $ FPEQ$ is a cylic quadrilateral.
Therefore, angles $ PQF\equal{} PEF$. This implies that $ FQP\equal{} FDH$.
THis implies $ PQ\parallel{}AD$. Therefore, $ PQ$ perpendicular to $ BC$.

Dear Mathlinkers,
the problem 1 has been posted but where?
I gave a oroof base on an application of Pascal's theorem.
Sincerely
Jean-Louis


Find all real solutions of the equation

$ \sqrt{x\minus{}\frac{1987}{14}}\plus{}\sqrt{x\minus{}\frac{1988}{13}}\plus{}\sqrt{x\minus{}\frac{1989}{12}}\equal{}\sqrt{x\minus{}\frac{14}{1987}}\plus{}\sqrt{x\minus{}\frac{13}{1988}}\plus{}\sqrt{x\minus{}\frac{12}{1989}}$

hello, i think your equation has no real solutions.
Sonnhard.

Hello
u r right Dr.There is no real solution.
Thank u.


The problem is to show that
\[ \sum_{n\in S}x^{n}\equal{}\frac{1\minus{}x^{6}}{(1\minus{}x^{2})(1\minus{}x^{3})}\quad |x| < 1,\]
where $ S \equal{}\{2i\plus{}3j\mid i,j\in\mathbb{N}\}$, where I consider $ 0$ to be a natural number.

So, my first attempt  was to look at
\[ \sum_{i,j}x^{2i\plus{}3j}\equal{}\big(\sum_{i}x^{2i}\big)\big(\sum_{j}x^{3j}\big) \equal{}\frac{1}{(1\minus{}x^{2})(1\minus{}x^{3})},\]
but I am not sure how to account for the fact that I am counting some elements of $ S$ more than once.

Any help would be much appreciated.

$ \left(\sum_{i}{x^{2i}}\right) ( 1\plus{}x^{3})$
$ \left(\sum_{i}{x^{3i}}\right) ( 1\plus{}x^{2}\plus{}x^{4})$

Either work. The first will compute to $ \frac{1\plus{}x^{3}}{1\minus{}x^{2}}$, and you'll get the answer in the form you want by multiplying top and bottom by $ 1\minus{}x^{3}$. The second is similar, just multiply by $ 1\minus{}x^{2}$ instead.


inequalities
logarithms
Putnam
algebra unsolved
algebra

If $ c \in \mathbb{R}$,

How can I solve for x in:  $ \frac{e^c\plus{}e^{\minus{}c}}{2} \leq e^{xc^2}$ in terms of $ c$?

I've tried logarithms, but it doesn't give me a definite solution.

Take the $ ln$ of both sides, and divide by $ c^2$.


I would like to produce a list in enumerate with paragraph text or section titles in between. That is, once in a while, I would like to get out of the enumerate environment and then resume later, with continued numberings.

The only way I know how to do that now is through \setcounter{enumi}{ * }, but that seems tedious. Is there an easier, more automatic way?

Use the resume option in the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=enumitem]enumitem package[/url]. See page 4 of the manual.


The first three terms of an arithmetic progression are $\tan x$, $\cos x$, and $\sec x$, respectively.  If the $k$th term is $\cot x$, compute $k$.

[hide]$\tan{x}+\sec{x}=2\cos{x}$
$\displaystyle\frac{\sin{x}+1}{\cos{x}}=2\cos{x}$
$\sin{x}+1=2(1-\sin^2{x})$
$2\sin^2{x}+\sin{x}-1=0$
$(\sin{x}+1)(2\sin{x}-1)=0\rightarrow \sin{x}=-1,\frac{1}{2}\rightarrow x=30, 150, 270$
plug back into original
$\text{for} x=30, \tan{x}=\frac{\sqrt{3}}{3}, \cos{x}=\frac{\sqrt{3}}{2}, \sec{x}=\frac{2\sqrt{3}}{3} \rightarrow \frac{2\sqrt{3}}{6},\frac{3\sqrt{3}}{6},\frac{4\sqrt{3}}{6}$
$\cot{30}=\sqrt{3}=\frac{6\sqrt{3}}{6}=\frac{2\sqrt{3}}{6}+(5-1)\frac{\sqrt{3}}{6}$
thus $k=\boxed{5}$
$x=150$ yields the same series, with all values negated, so $k$ would remain the same, and $x=270$ yields undefined values[/hide]


search
number theory unsolved
number theory

would it be better to use modulus or another technique?
For either, how would you solve it?

Prove for

(x^2) + m = 5^(4y+3)
m is not 1,2 or 3.

when x,y are non negative integers.

you already posted this, but then erased most of your post  :what?: : http://www.mathlinks.ro/viewtopic.php?t=232449

no it's different and I have no idea what happened to the other post.
Maybe someone deleted it???
I'm currently at internet cafe and put auto log in....

Anyway the m is on the left hand side this time I wrote the question wrong

I suppose a possible approach would be to factor the left-hand side over the ring of integers in $ \mathbb{Q}(\sqrt{\minus{}m})$. This ring is $ \mathbb{Z}[i]$ for $ m\equal{}1$, $ \mathbb{Z}[\sqrt{\minus{}2}]$ for $ m\equal{}2$ and $ \mathbb{Z}[\frac{1\plus{}\sqrt{\minus{}3}}{2}]$ for $ m\equal{}3$. All of those rings have unique factorization so this should be pretty straightforward if you're familiar with such methods. I didn't sit down to go through the details yet.

hmm do you know anywhere on this site or internet that has a similar problem/explanation on those kinds of factorising?
I'm not really sure how to do them

Not really sure about ring of integers and the Z[] values.
Also would negative root factorisation mean using imaginary numbers in the factorisation?

This has been discussed on the forum many times (search for "Gaussian integers", for instance), but I'm not sure I would recommend trying to learn this stuff just from the posts. Perhaps you can start [url=http://www.imocompendium.com/tekstkut/extensions_ddj.pdf]here[/url]. There are tons of books on the subject, one of my favorites is [url=http://www.amazon.com/Problems-Algebraic-Number-Graduate-Mathematics/dp/0387221824]this one[/url].


\[ {\rm{Solve for }}x: 5\left( {x^2  \plus{} \frac{1}{{x^2 }}} \right) \plus{} 14\left( {x \minus{} \frac{1}{x}} \right) \equal{} 2
\]

We have $ 5x^2 \plus{} 5 \frac {1}{x^2} \plus{} 14(x \minus{} \frac {1}{x}) \equal{} 2$.

Subtracting 10 from both sides, $ 5x^2 \minus{} 10 \plus{} 5 \frac {1}{x^2} \plus{} 14(x \minus{} \frac {1}{x}) \equal{} \minus{}8$.

Thus $ 5(x^2 \minus{} 2 \plus{} \frac {1}{x^2}) \plus{} 14(x \minus{} \frac {1}{x}) \equal{} \minus{}8 \Rightarrow 5(x \minus{} \frac {1}{x})^2 \plus{} 14(x \minus{} \frac {1}{x}) \plus{} 8 \equal{} 0$. Substituting $ y \equal{} x \minus{} \frac {1}{x}$, $ 5y^2 \plus{} 14y \plus{} 8 \equal{} 0$.

Thus $ (y \plus{} 2)(5y \plus{} 4) \equal{} 0$ and $ x \minus{} \frac {1}{x} \equal{} \minus{}2$ or $ \minus{}\frac {4}{5}$. Multiplying by $ x$ throughout in these equations and applying the quadratic formula we get $ x \equal{} \minus{}1 \pm \sqrt {2}$ and $ x \equal{} \minus{}2 \pm \sqrt {29}$.


Please excuse my lack of tex. Its been a long time since I've posted and I'm also in a hurry.
I'm really looking for someone to evaluate my proof and point out any problems with anything that you see. I greatly appreciate this service in advance.

This is the problem statement:
Prove that f: A ---> B is one-one and onto iff there exists g: B ---> A with f(g(B_i)) = B_i and g(f(A_i)) = A_i, i.e. g and f are inverse functions.

Proof

Assume the opposite, that f is not 1-1 and onto.
So say there are some A_1, A_2, such that f(A_1) = f(A_2), but A_1 != A_2, but then A_1 = g(f(A_1)) = g(f(A_2)) = A_2, so A_1 = A_2, which violates our premise, so a contradiction has been found and f cannot be 1-1.

Now say f isnt onto. Then let's take B_1 != f(A_k) for any k. But f(g(B_1)) = B_1, so the A_i that is equal to g(B_1), which we know exists  because g: B ---> A, defies B_1 != f(A_k) and is our contradiction. Thus f is onto.

Now that we have shown that f is onto and one-one if the opening statemenet is true, lets show the only if part by assuming WLOG that f(g(B_i)) != B_i.

Assume then that f is 1-1 and proceed by contradiction.
So, for each A_i, there is one and only one B_i that is its image, thus f(A_i) = B_i >> A_i = g(f(A_i)) = g(B_i) >> B_i = f(A_i) = f(g(B_i)), but we said that B_i != f(g(B_i)), so we found a contradiction and thus f isnt 1-1.

Now let's do a contradiction for onto, assuming f is onto. THus every B_i is an image of at least 1 A.
We know that f isnt 1-1, so there must be some B_n which is the image of at least 2 A, A_1 and A_2.
f(A_1) = f(A_2) = B_n and A_1 != A_2, which implies that g(f(A_1)) = g(f(A_2)) = g(B_n)  >> A_1 = A_2 = g(B_n) >> B_n = f(A_1) = f(A_2) = f(g(B_n)), but this implies that f(g(B_n)) = B_n, which contradicts f(g(B_n)) != B_n, so the premise of f being onto isnt true.

You would probably do better if you posted this under the Calculus-Real Analysis forum.


linear algebra
matrix
superior algebra
superior algebra unsolved

Everyone knows that over fields with $char(K)\not=2$ we have that that antisymmetric matrices in $K^{2n+1\times2n+1}$ are singular, i.e. have determinant zero. This because we have $det(A)=det(A^T)=det(-A)$, thus $2det(A)=0$.

For fields with characteristic 2, we also have that det(A)=0. But we cannot conclude it from $2det(A)=0$, since 2=0. So find another way to prove it.

[hide="This is my solution, can someone tell me if this correct and/or provide a better one?"]Every field F with char F = 2 is embedded in a larger field G with char G = 0. Over G we have det=0, (due tot the classical property), and when expanding det(A)=0 (without simplifying anything!!) we get "a sum of products of elements of F" = 0. Since both sides are in F, restriction to F doesn't change this. So the determinant is also 0 when working over F.[/hide]

Thx

First of all, what do you really mean? :? Not all matrices of odd order have determinant zero, as you wrote. I think you left something out - a condition that these matrices need to satisfy. Maybe they're supposed to be antisymmetric ($A^t+A=0$)? In that case, however, it's not true. The identity matrix $I$ (when we're working with characteristic $2$) always satisfies $I^t+I=0$, and its determinant certainly isn't zero.

Second of all, the characteristic of an extension field $L$ of the field $K$ is the same as the characteristic of $K$. Where did you get the idea that "Every field F with char F = 2 is embedded in a larger field G with char G = 0"? :)

I meant anti-symmetric matrices. And for char K =2, they are thus symmetric matrices [b]with all diagonal elements zero[/b].

And for the field... dunno... $GF(2^h)$ is a subfield of $\mathbb{R}^h$, isn't it?

With all diagonal elements 0, I think you can prove that the determinant is 0 by looking at the sum-of-permutations definition of determinant.  Every permutation will cancel out with its inverse permutation.

[quote="Peter VDD"]And for the field... dunno... $GF(2^h)$ is a subfield of $\mathbb{R}^h$, isn't it?[/quote]what about this? Is this wrong? (probably, since grobber knows more about this than me)
But why?

About the sum-of-permutations rule, that seems to work indeed.

@Peter: For which $h$ is $\mathbb{R}^h$ a field? ;)


Prove for $ a,b,c$ positive reals and $ abc\equal{}1$  that $ \frac{a}{(a\plus{}1)(b\plus{}1)}\plus{}\frac{b}{(b\plus{}1)(c\plus{}1)}\plus{}\frac{c}{(c\plus{}1)(a\plus{}1)}\ge\frac {3}{4}$

http://www.mathlinks.ro/viewtopic.php?t=89302&search_id=984113818

Btw, the generalization of this problem is great.


analytic geometry
algebra
polynomial
ARML
AMC
USA(J)MO
USAMO

1.  If $\displaystyle x^yy^x=e^{\frac{2xy}{e}}$, then evaluate x+y.

2.  Find the smallest positive integer n (n>1), such that $\displaystyle \sqrt{\frac{n(n+1)}{2}}=k$, where k is an integer.

3.  The graph of $f(x)=10x^4-32x^3+12x^2-5x+2$ intersects a certain line in 4 points.  What is the average of the x-coordinates of the intersections points?

[hide="Number 2"]

By brute force, I found the answer to be 8. Perhaps someone could show me a method or formula for finding the answer...?
[/hide]

2
[hide]I just said that the numerator must have an odd number of factors of two, and must be paired with another prime square in order to create the product of two consecutive integers. I just took the smallest numbers possible, $2^3$ and $3^2$, and by stroke of luck they worked... so the answer must be 8.[/hide]

[hide="The answer for 2 is"]8.  I used brute force too...nice solution tarquin[/hide]

For 2, the radicand is the formula for finding the sum of the first n integers.  From there, it was pretty easy to brute force it.

I like 3, however I don't know if I got the right answer.
[hide]
What we are trying to solve for x when we have two equations:
$y=10x^4-32x^3+12x^2-5x+2$
$y=mx+b$
We know there are four solutions.  We can substitute for y and subtract to get
$10x^4-32x^3+12x^2-(5+m)x+(2-b)=0$
Again there are four solutions to this equation, but actually solving for those isn't something we need to do.  The formula for the sum of the roots of a polynomial is $\frac{-b}{a}$ where $a$ is the coeffiecient of the leading term and $b$ the coefficient of the term with degree one less than the polynomial's degree.
So the sum of the roots of the equation, regardless of $m$ or $b$ is $\frac{16}{5}$.  But we want the average of the roots, so we merely divide by 4 to get the answer.
$\frac{4}{5}$
[/hide]

Good work on #s 2 and 3 so far.  (A similar trick to that used on #3 was necessary for an ARML #8 a few years ago.)  #1 seems a bit tough for this forum to me, although with a little sitting and staring it is possible to figure out what the answer must be.  (It's harder to show it's the unique answer.)  When anyone does it, I have a follow-up question, however, which should be done after the question above:

[hide]Question #1 has a unique solution.  That ordered pair is also a solution to $x^xy^y=e^{\frac{2xy}{e}}$, by symmetry.  However, this second equation also has a variety of other solutions.  Give a concise, correct, and intuitive explanation of why this second equation is not solvable uniquely.[/hide]

Yes, I liked the solution for 3.  However, if it appeared on ARML I probably would have to prove the formula for $\frac{-b}{a}$ as the sum of the roots, and I don't know where to begin for that.  I just know it because my coach told us; the state academic team tests like asking for the sum/product of roots and those formulas certainly help.

The ARML individual contest is short-answer.  Perhaps you're thinking of USAMO.  The fact that you use, however, is so well known that you wouldn't need to prove it on USAMO, just state it.

[b]Hi, I have the solution to #1[/b]
[hide]A number, the base of the logarithm (in this case, e) to the logarithm of the number's power is equal to the number. For example 7=10^log 7.

Therefore, we have in this equation x^y*y*x=e^2xy/e.
We first begin by changing it to e^(ln x^y)*e^(ln y^x)= e^2xy/e.
Therefore, ln(x^y)+ln(y^x)=2xy/e
There continues:
ln(x^y*y^x)=2xy/e
2xy/ln(x^y*y^x)=e

Note that x and y both have to be e in order for this not to turn out ugly.

Then we have 2e^2/2e=e.

Therefore, x+y=2e[/hide]

Was my solution a good explanation for your problem?

Yes, that was a good explanation...thanks.
The only thing is...
[quote="jimhu"]Note that x and y both have to be e in order for this not to turn out ugly.[/quote]
 :D I'm not sure how we would show that this is the unique solution, though.  Does anyone have any ideas?

"Not turning out ugly" is a good way to guess an answer -- it's not a good way to prove a solution, though.  Try using the following fact: [hide]$e^{w-1} \ge w$ for all real numbers $w$.[/hide]

hm...I tried applying the fact above, but I'm not quite sure I have a solution...has anyone else come up with a solution using the above fact?

[quote="jen7"]1.  If $\displaystyle x^yy^x=e^{\frac{2xy}{e}}$, then evaluate x+y.[/quote]

So, I'm just going to assume x and y are positive so that I don't have to worry about that kind of issue at all.  (We can pick up an extra answer of 0 if we take $0^0 = 1$, but we'll not worry about that here.)

If x and y are positive, we can choose w, z such that $x = e^w, y = e^z$.  Then our equation becomes
$\displaystyle e^{we^z}e^{ze^w}=e^{\frac{2e^we^z}{e}}$
or, simplifying
$\displaystyle e^{we^z + ze^w}=e^{2e^{w+z-1}}$
and taking a logarithm of both sides,
$\displaystyle we^z + ze^w=2e^{w+z-1}$

However,
$t \le e^{t-1}$ for real $t$ with equality only for $t = 1$, so
$w \le e^{w-1}$ and $z \le e^{z-1}$, and thus
$we^z \le e^{w + z - 1}$ and $ze^w \le e^{w + z - 1}$.  Adding these two inequalities, we find
$\displaystyle we^z + ze^w \le 2e^{w+z-1}$, with equality iff $w = z = 1$.  However, we know equality holds; thus, $w = z =1$, $x = y = e$ is the unique solution.

Keep playing with #2.  Try to find a family of solutions.  It's actually a pretty interesting problem if you're into number theory.


graph theory
combinatorics unsolved
combinatorics

The problem is restated here:
A clique of order $ n$ is a subgraph which is isomorphic to the complete graph $ K_{n}$.

For any graph $ G$, if the largest clique of a graph $ G$ has order less than $ p$, is it always true that $ G$ is a $ p$-partite graph?  Could you mind helping me figure this out?

No: there exist triangle-free graphs of arbitrarily large chromatic number.  More generally, for any $ k$ there exists a graph with both girth and chromatic number larger than $ k$.  (This is a theorem of Erdos, originally proven probabilistically.  Some constructions are known.)


Prove that $\sqrt{a}+\sqrt{b}$ will never be an integer, given that a and b are nonsquare integers.

Okay... assume it is an integer 'k'
square both sides:
1.  a + b + 2√ab = k^2
1.5.  if ab is a perfect square, a is of the form m(n^2) and b is m(o^2) where m, n, o are positive integers and m is not a perfect square.
2.  m(n^2)+m(o^2)+2mno = k^2
3.  m((n^2)+(o^2)+2no)=k^2
4.  m(n+o)^2=k^2
5.  (n+o)√m = k
5.5. k is a positive integer and so is n+o, therefore √m is rational.  But we know from line 1.5 that √m is irrational.
6.  CONTRADICTION

[hide="Alternately"] $\sqrt{a}+\sqrt{b}= k \Leftrightarrow$
$\sqrt{a}= k-\sqrt{b}\Leftrightarrow$
$a = k^{2}-2k \sqrt{b}+b \Leftrightarrow$
$\sqrt{b}= \frac{k^{2}+b-a}{2k}$ [/hide]


[quote="t0rajir0u"]For $n = 1, 2, ...$ let $s(n)$ denote the sum of the digits of $2^{n}$.  Thus, for example, as $2^{8}= 256$ we have $s(8) = 2+5+6 = 13$.  Determine all positive integers $n$ such that

$s(n) = s(n+1)$.

[hide="Remark"] This is by far the easiest problem I've seen in either the Green or Red books, which is why I'm posting it here. [/hide]
[hide="Hint"] Recall that the sum of the digits of a number is congruent to that number $\bmod 9$. [/hide][/quote]

[hide]Wait a second.

If s(n) is going to have the same sum of its digits as s(n+1), then they will have to have the same modulo 9.

But if s(n) is congruent to a(mod 9), then s(n+1) is congruent to 2a(mod 9).

So, inorder for these mods to have any chance of equalling each other, a can only be congruent to 0 mod 9, since all the others will produce different mods.

Since no n will allow 2^n be congruent to 0 mod 9, then the asnwer is no solution.

$\text{no solution}$[/hide]

[hide]Sum of digits of a number is conqruent to the number mod 9.
So $s(n)\equiv2^{n}\equiv x \pmod{9}$.
$s(n+1)\equiv2^{n+1}=2x \pmod{9}$.
$s(n)=s(n+1)\implies x\equiv 2x \pmod{9}$. However, there are no  solutions x to $x\equiv 2x\pmod{9}$, so there are no solutions to the problem.[/hide]

oH, YEAH

I also realized that this is periodic

it goes

2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1...

And since no two consecutive have the same modulo 9, then there is no two consecutive with same digit sum.

[quote="pianoforte"][hide]However, there are no  solutions x to $x\equiv 2x\pmod{9}$[/hide][/quote]

[hide="Well, yes there are..."] How about, say, $x \equiv 0 \bmod 9$?  :P

I think you meant there are no solutions to $2^{n}\equiv 0 \bmod 9$. [/hide]


[quote="Sang Hun Song"]Prove the following inequality:

For a,b,c three [b]real numbers[/b], 3(a^4+b^4+c^4)+4(a^3*b+b^3*c+c^3*a)>=0

 :D[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112423

Sorry...  :blush:
Actually, my solution is quite similar with yours, which I took almost one hour...


geometry
3D geometry
tetrahedron
ratio
invariant
geometry unsolved

Let G be median point of the triangle ABC of the tetrahedron ABCD. A plane intersects DA, DB, DC, DG at A’, B’, C’, G’ respectively. Prove that
\[ \dfrac{DA}{DA'} \plus{} \dfrac{DB}{DB'} \plus{} \dfrac{DC}{DC'} \equal{} 3\dfrac{DG}{DG'}.
\]

Transform the 3D basis $ \overrightarrow{DA}, \overrightarrow{DB}, \overrightarrow{DC}$ into an orthonormal basis $ \overrightarrow{DA^*}, \overrightarrow{DB^*}, \overrightarrow{DC^*}.$ This is a linear transformation preserving segment ratios on a line and its image. Centroid $ G$ of $ \triangle ABC$ goes into centroid $ G^*$ of $ \triangle A^*B^*C^*$ and $ |DG^*| = \frac {_1}{^3}\left|\overrightarrow{DA^*} + \overrightarrow{DB^*} + \overrightarrow{DC^*}\right| = \frac {_{\sqrt 3}}{^3}.$ The plane $ (A^*B^*C^*)$ has equation $ x^* + y^* + z^* = 1$ and the line $ DG^*$ perpendicular to the plane $ (A^*B^*C^*)$ has equation $ x^* = y^* = z^*.$ The plane $ (A'B'C')$ is transfomed to a plane $ (A'^*B'^*C'^*)$ with equation $ \frac {x^*}{|DA'^*|} + \frac {y^*}{|DB'^*|} + \frac {z^*}{|DC'^*|} = 1.$ The line $ DG^*$ intersects the plane  $ (A'^*B'^*C'^*)$ at $ G'^*,$ such that $ |DG'^*| = \frac {\sqrt {3}}{\frac {1}{|DA'^*|} + \frac {1}{|DB'^*|} + \frac {1}{|DC'^*|}}.$ Substituting  ${ 3|DG^*| = \sqrt 3}$ and $ |DA^*| = |DB^*| = |DC^*| = 1$ yields

$ 3 \frac {|DG^*|}{|DG'^*|} = \frac {|DA^*|}{|DA'^*|} + \frac {|DB^*|}{|DB'^*|} + \frac {|DC^*|}{|DC'^*|}.$

This equation is invariant under the above linear transformation, so that we can drop the asterisks.


Show that the equation $ 8^x \minus{} 6^x \minus{} 27^{x\minus{}1} \equal{} 1$ may be simplified to  $ 2^x \equal{} 3^{x\minus{}1} \plus{} 1$ and then solve it by inspection or otherwise for all real $ x$.

[quote="xxxyyyy"]Show that the equation $ 8^x \minus{} 6^x \minus{} 27^{x \minus{} 1} \equal{} 1$ may be simplified to  $ 2^x \equal{} 3^{x \minus{} 1} \plus{} 1$ and then solve it by inspection or otherwise for all real $ x$.[/quote]
Maybe written as $ 2^{3x} \plus{} 2^x \cdot 3^x \minus{} 3 ^{3(x\minus{}1)} \equal{} 1$
$ 2^{3x} \equal{} 3^{3(x\minus{}1)} \minus{} 3 \cdot 3^{x\minus{}1} 2^x \plus{}1$
Try converting to a form: $ a^3 \minus{} 3a^2b \plus{} 3ab^2 \minus{}b^3 \equal{} (a\minus{}b)^3 \equal{} (2^x)^3$.

[hide="solution"]
$ 8^x \minus{} 6^x \minus{} 27^{x \minus{} 1} \equal{} 1$

$ 2^{3x} \minus{} 2^x*3^x \minus{} 3^{3(x \minus{} 1)} \equal{} 1$

$ 2^{3x} \minus{} 2^x*3^x \equal{} 3^{3(x \minus{} 1)} \plus{} 1$

$ 2^x(2^{2x} \minus{} 3^x) \equal{} (3^{x \minus{} 1} \plus{} 1)(3^{2(x \minus{} 1)} \minus{} 3^{x \minus{} 1} \plus{} 1) \implies 2^x \equal{} 3^{x \minus{} 1} \plus{} 1$ if and only if $ 2^{2x} \minus{} 3^x \equal{} 3^{2(x \minus{} 1)} \minus{} 3^{x \minus{} 1} \plus{} 1$
____________________

$ 2^{2x} \minus{} 3^x \equal{} 3^{2(x \minus{} 1)} \minus{} 3^{x \minus{} 1} \plus{} 1$

$ 2^{2x} \equal{} 3^{2(x \minus{} 1)} \plus{} 2*3^{x \minus{} 1} \plus{} 1$

$ 2^{2x} \equal{} (3^{x \minus{} 1} \plus{} 1)^2$

$ 2^x \equal{} 3^{x \minus{} 1} \plus{} 1$
____________________

$ 2^x \equal{} 3^{x \minus{} 1} \plus{} 1 \Longleftrightarrow 2^{2x} \minus{} 3^x \equal{} 3^{2(x \minus{} 1)} \minus{} 3^{x \minus{} 1} \plus{} 1 \Longleftrightarrow 2^x \equal{} 3^{x \minus{} 1} \plus{} 1$

so $ 8^x \minus{} 6^x \minus{} 27^{x \minus{} 1} \equal{} 1$ can really be simplified to $ 2^x \equal{} 3^{x \minus{} 1} \plus{} 1$
____________________

We can easily guess that 2 solutions are $ x \equal{} 1$ and $ x \equal{} 2$ and we will prove that those are the only solutions.
Let $ f(x) \equal{} 2^x$ and $ g(x) \equal{} 3^{x \minus{} 1} \plus{} 1$.
$ f(x)$ aproaches $ y \equal{} 0$ asymptotically while $ g(x)$ aproaches $ y \equal{} 1$ asymptotically. Now it is clear that $ g(x) > f(x)$ for every $ x < 1$. Also $ g(x) > f(x)$ for every $ x > 2$ since $ g(x)$ grows much faster than $ f(x)$.
$ f(x) \neq g(x)$ for every $ x \in < 1,2 >$ because both functions are convex on interval [1,2], so $ f(x) \equal{} g(x)$ has only 2 solutions.[/hide]

[hide="solution#2"]
Let's start with expression $ 2^x\equal{}3^{x\minus{}1}\plus{}1$

$ 2^x\minus{}3^{x\minus{}1}\equal{}1$

$ (2^x\minus{}3^{x\minus{}1})^3\equal{}2^x\minus{}3^{x\minus{}1}\equal{}1$

$ (2^x\minus{}3^{x\minus{}1})^3\equal{}8^x\minus{}2^{2x}*3^x\plus{}2^x*3^{2x\minus{}1}\minus{}27^{x\minus{}1}$

                          $ \equal{}8^x\minus{}6^x(2^x\minus{}3^{x\minus{}1})\minus{}27^{x\minus{}1}$

                          $ \equal{}8^x\minus{}6^x\minus{}27^{x\minus{}1}\equal{}2^x\minus{}3^{x\minus{}1}\equal{}1$

so $ 8^x\minus{}6^x\minus{}27^{x\minus{}1}\equal{}1$ can really be simplified to $ 2^x\equal{}3^{x\minus{}1}\plus{}1$[/hide]

[quote="Flame"]
$ f(x) \neq g(x)$ for every $ x \in (1,2)$ because . . .[/hide][/quote]
You meant the open interval as compared to your original notation?

[quote="Flame"]Yes. I don't know how to make bracket like that in latex so I just used < and >.[/quote]
Just use the racket symbols over 9 and 0.

I have no idea what anyone is talking about, but to make attractive-looking angled brackets (i.e., not parentheses (), square brackets [], or curly brackets {} but rather $ \langle \rangle$) use \rangle and \langle.


$ a,b,c,d\ge0$ $ a\plus{}b\plus{}c\plus{}d\equal{}1$
Prove $ ((a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a))^{\frac{3}{2}}\ge 4abcd$

[quote="dima ukraine"]$ a,b,c,d\ge0$ $ a \plus{} b \plus{} c \plus{} d \equal{} 1$
Prove $ ((a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a))^{\frac {3}{2}}\ge 4abcd$[/quote]
Use $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)\geq(a \plus{} b \plus{} c\plus{}d)(abc \plus{} abd \plus{} acd \plus{} bcd)$ and Maclaren's one:
$ \frac {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}}{4}\geq\sqrt [3]{\frac {\frac {1}{abc} \plus{} \frac {1}{abd} \plus{} \frac {1}{acd} \plus{} \frac {1}{bcd}}{4}}.$ :wink:


function
calculus
derivative
calculus computations

[img]http://img21.imageshack.us/img21/3336/33896893.jpg[/img]

to answer this question I know that I do have to take the derivative of the function and see where it's increasing or decreasing. For this particular question I really don't know how to do it, so i would be very appreciated if someone can help

i took the derivative of the function and it turn out to be (x-3)/(x+1)^2, i guess.

This is a ridiculously poorly worded question (if a function is one-to-one on the interval $ (a, b)$ then it's one-to-one on every subinterval of $ (a, b)$).  But, ignoring that for the moment, here are two suggestions:
1) try to sketch the graph to get an idea of what the answer could be
2) check your differentiation again -- what you've written is very not right.

JBL wrote
[quote]2) check your differentiation again -- what you've written is very not right.[/quote]
Not VERY, just the little ^2, but enough  :)


Let d be a fixed line. A and B are fixed points such that d is the midperpendicular of AB. (O1) and (O2) are circles passing through A and B. Let PQ be the common tangent of (O1) and (O2). Let I be the orthocenter of APQ. Find the locus of I.

I'll denote the orthocenter by $H$ instead of $I$ (which is usually the incenter). Let's look at the triangle $APQ$. Let $M$ be the midpt of $PQ$. It's easy to see that $B$ is on $AM$ and that $\angle PBQ=\pi-\angle PAQ$. Then the points $P, H, B, Q$ lie on a circle. Consider the inversion of pole $A$ and power $AH\cdot AD$. It's easy to see that this inversion turns the circle $PHBQ$ into the nine-point circle of $APQ$, which also passes through $M$, so it turns $B$ into $M$. This means that $AH\cdot AD=AB\cdot AM\Rightarrow$ the quadrilateral $HDMB$ is cyclic, so $\angle HBM=\frac \pi 2$, so $H$ is on the line perpendicular to $AB$ at $B$.

I didn't show that the entire line is the locus (in a rigorous way), but it seems pretty obvious, since the only thing the position of $H$ depends on is the angle made by $PQ$ and $AB$, and this can be controlled by the ratio of the radii of $(
O_1)$ and $(O_2)$, which can be any positive real.

What a beautiful solution ! My solution doesn't use inversion, but it's less interesting than you!  :) Congratulations!. :)


Good day gentlewomen and gentlemen.
I'm mathing last years final problems from a Swedish competition and I'm having problems understanding a part of the solution.

"In the number triplet $ p,q,r$, the three numbers are primes. Also, $ q \minus{} 50 \equal{} p$ and $ r \minus{} 50 \equal{} q$. Find all such triplets."

Solution and question:
[hide]
[i]Solution[/i]
Obviously, we have $ q \equal{} p \plus{} 50$ and $ r \equal{} p \plus{} 100$.
As $ 50/3\equal{}16r2$ and, and $ 100/3\equal{}33r1$, all three numbers will have different reminders when divided by $ 3$. In particular, one of the numbers must have the reminder $ 0$, and be dividable by $ 3$. The only prime dividable by $ 3$ is $ 3$.
The only possibility is as such $ p \equal{} 3$, $ q \equal{} 53$ and $ r \equal{} 103$. As both $ 53$ and $ 103$ are primes, this triplet fulfills the requirements.

[i]Question[/i]
So I somewhat get this thing about different reminders. However, I don't understand why they say they only test $ p \equal{} 3$ when the problem explicitly states [i]find all such triplets[/i]. Is it a misprint or am I missing something?[/hide]

Note that: $ p,q,r$ are primes then if $ p,q,r$ divide by $ 3$ then it must be $ 3$, it mean $ p\equal{}3$ because $ q,r>3$


Let $ x,y,z \geq 0$ and $ xyz\equal{}1$.Prove that :

$ \frac{1}{xy\plus{}2}\plus{}\frac{1}{yz\plus{}2}\plus{}\frac{1}{zx\plus{}2} \geq \frac{1}{x\plus{}yz\plus{}1}\plus{}\frac{1}{y\plus{}zx\plus{}1}\plus{}\frac{1}{z\plus{}xy\plus{}1}$

P.S : Solutions with full expansion aren't allowed :P :).I have a one line proof for this inequality  :)

[quote="anas"]$ \frac {1}{xy \plus{} 2} \minus{} \frac {1}{xy \plus{} z \plus{} 1} \plus{} \frac {1 \minus{} z}{3(2 \plus{} z)} \geq 0$

and it is easy to check that : $ \sum{\frac {1 \minus{} z}{2 \plus{} z}} \leq 0$  :P  :D[/quote]
My proof is exactly the same as yours anas.The following easy inequality holds ( a consequence of the first ) :

$ \sum \frac {1}{x \plus{} 2} \geq \sum \frac {1}{xy \plus{} z \plus{} 1}$

For $ x,y,z \geq 0$ and $ xyz \equal{} 1$.

And I think that the following stronger inequality also holds :

$ \sum \frac {1}{x \plus{} y \plus{} 1} \geq \sum \frac {1}{xy \plus{} z \plus{} 1}$

With the same conditions.

Hi, how did you get an idea to add $ \frac {1 \minus{} z}{3(2 \plus{} z)}$ ? How didn't you come up with some other fraction instead this?  :?:

Thanks!