algebra
polynomial
vector
linear algebra
matrix
calculus
integration

The sum and product of two algebraic numbers are again algebraic. This should be obvious but I don't see why.  :?:

If $ \alpha$ and $ \beta$ are algebraic, then $ \mathbb{Q}(\alpha, \beta)$ is a finite extension of $ \mathbb{Q}$. But any finite extension is algebraic, so in particular the product and sum of $ \alpha$ and $ \beta$ are algebraic, since they lie in $ \mathbb{Q}(\alpha, \beta)$.

Is there a more elementary way? This is not so "obvious".

I don't know what theory you know. I think any argument will be theoretical in nature, not obvious by inspection. I mean, I don't see how to find a polynomial having $ 2^{1/3} \plus{} 13^{1/7}$ as a zero, so certainly it's hard to do it directly.

It's possible but notoriously complicated to construct polynomials with the sums and products as roots. The advanced tricks are far nicer in this case.

For Kalle's argument, interpret these extensions as vector spaces over $ \mathbb{Q}$. Suppose $ p(\alpha)\equal{}0$ and $ q(\beta)\equal{}0$, where $ p(x)\equal{}x^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}a_1x\plus{}a_0$ and $ q(x)\equal{}x^m\plus{}b_{m\minus{}1}x^{m\minus{}1}\plus{}\cdots\plus{}b_1x\plus{}b_0$; in addition, choose these to be the minimal polynomials.

Now, we can write $ \mathbb{Q}(\alpha)$ as a vector space with basis $ 1,\alpha,\alpha^2,\cdots,\alpha^{n\minus{}1}$ and $ \mathbb{Q}(\beta)$ with basis $ 1,\beta,\beta^2,\cdots,\beta^{m\minus{}1}$. The matrix of multiplication by $ \alpha$ is the companion matrix $ \begin{bmatrix}0&0&0&\cdots&0&\minus{}a_0\\ 1&0&0&\cdots&0&\minus{}a_1\\ 0&1&0&\cdots&0&\minus{}a_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&\minus{}a_{n\minus{}2}\\ 0&0&0&\cdots&1&\minus{}a_{n\minus{}1}\end{bmatrix}$.

Similarly, $ K\equal{}\mathbb{Q}(\alpha,\beta)$ is a vector space. It's dimension is at most $ mn$, with $ \alpha^i\beta^j$ for $ 0\le i< n,0\le j<m$ a spanning set Choose a basis $ B$ from among these elements. Consider multiplication by $ \alpha\plus{}\beta$ (or the product, or any other combination). This is a $ \mathbb{Q}$-linear map from $ K$ to itself, so it has a matrix- with coefficients in $ \mathbb{Q}$. By Cayley-Hamilton, it satisfies its characteristic polynomial $ g$. Now multiplication by $ g(\alpha\plus{}\beta)$ is multiplication by the zero matrix, and $ g(\alpha\plus{}\beta)$ must be zero as a number. That makes $ \alpha\plus{}\beta$ algebraic.

The constructive methods amount to actually calculating those characteristic polynomials. It gets seriously unwieldy for large $ m$ and $ n$, since we're dealing with $ mn\times mn$ matrices.

You might want to try it explicitly for a small case: can you find the fourth-degree polynomial that $ \sqrt{3}\plus{}\frac12\sqrt{5}\plus{}\frac12$ satisfies?

Two generalizations:
We can use the same argument to show that the algebraic integers are a ring; if all of the coefficients of the polynomial are integers, all entries in the matrices for sums and products will be integers as well.
Other than the mention of integers I just made, there's nothing special about $ \mathbb{Q}$. This applies to extensions over any field: any element in a finite extension is algebraic over the base field.

[quote="jmerry"]
Similarly, $ K \equal{} \mathbb{Q}(\alpha,\beta)$ is a vector space. It's dimension is at most $ mn$, with $ \alpha^i\beta^j$ for $ 0\le i < n,0\le j < m$ a spanning set Choose a basis $ B$ from among these elements. Consider multiplication by $ \alpha \plus{} \beta$ (or the product, or any other combination). This is a $ \mathbb{Q}$-linear map from $ K$ to itself, so it has a matrix- with coefficients in $ \mathbb{Q}$. By Cayley-Hamilton, it satisfies its characteristic polynomial $ g$. Now multiplication by $ g(\alpha \plus{} \beta)$ is multiplication by the zero matrix, and $ g(\alpha \plus{} \beta)$ must be zero as a number. That makes $ \alpha \plus{} \beta$ algebraic.
[/quote]
For this bit it can be done with more simple linear algebra by saying that if $ \gamma \in K$, then $ {1,\gamma, \gamma^2, \ldots \gamma^{mn}}$ can not be linearly independent, so there is a nontrivial linear relation (with rational coefficients) between them, which is the same as having a non-zero polynomial in $ \mathbb{Q}[x]$ with $ \gamma$ as a zero.

OK, that works too. I prefer the matrices here, since it indicates a constructive approach and also handles the integer case. A generic linear dependence argument won't give a monic polynomial there.

I constructed the polynomials in the last post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=194797]here[/url] using the [url=http://en.wikipedia.org/wiki/Resultant]Resultant[/url].
This actually shows, that the integral elements from any extension of commutative rings form again a ring.

Yes, that's true. My argument gives the same result, combining the two extensions; we don't need to actually calculate the entries of the matrices to see that they're integers, and characteristic polynomials are naturally monic.
Over in the linked topic, darij was using matrices; that's a quantified version of what I said, adapted to the specific problem. The resultant is either the obvious $ \prod_{i,j}(x\minus{}r_i\minus{}s_j)$ with coefficients computed by nonobvious symmetric polynomial tricks, or the characteristic polynomial of a fairly sparse matrix. Concrete examples with high degree are a lot of work no matter what you do.

Why answer a nine-month old rhetorical question?

Also, there's a mistake in that. The minimal polynomial should be monic. I haven't worked out the whole thing, but the constant term is $ \minus{}2^7\minus{}13^3\equal{}\minus{}2325$.

Hi jmerry,
"Why answer a nine-month old rhetorical question?"
 is an incredible remark !! It is not the first time that you answer in an unpleasant way to my posts.
Have you a problem with me ?
    You have a curious behavior but you are a good mathematician; for the second point: "The minimal polynomial should be monic", you are right because $ 2^{1/3}$ and $ 3^{1/7}$ are algebraic integers.
I have chosen $ 60$ digits for the calculation with Maple but it is not sufficient; the correct choice is $ 300$ digits. This choice is very important when you use the LLL method.
The true solution is (I hope so):
p := -2325+14196*_X-17472*_X^2+448*_X^3-35490*_X^4-82992*_X^5-672*_X^6+507*_X^7
-55692*_X^8+560*_X^9-6006*_X^11-280*_X^12-39*_X^14+84*_X^15
-14*_X^18+_X^21

Working entirely by hand, I get
$ x^{21}\minus{}(2\cdot 7)x^{18}\plus{}(2^2\cdot  21)x^{15}\minus{}(13\cdot 3)x^{14}\minus{}(2^3\cdot 21)x^{12}\plus{}(2\cdot 13\cdot 231)x^{11}$
$ \plus{}(2^4\cdot 35)x^9\minus{}(2^2\cdot 13\cdot 1071)x^8\plus{}(13^2\cdot 3)x^7\minus{}(2^5\cdot 21)x^6\plus{}(2^3\cdot 13\cdot 738)x^5$
$ \minus{}(2\cdot 13^2\cdot 105)x^4\plus{}(2^6\cdot 7)x^3\minus{}(2^4\cdot 13\cdot 84)x^2\plus{}(2^2\cdot 13^2\cdot 21)x\minus{}(2^7\plus{}13^3)$
Your revised answer is confirmed.

Switching out the $ 2$s and $ 13$s would give you the polynomial for any other $ a^{\frac13}\plus{}b^{\frac17}$ The technique would also work for other sums of roots; the only tricky bit is the combinatorics for some of those coefficients in the middle.

The other point is that there was no reason to answer that question. The topic was already complete, and that question was merely asked as a way of making the point that the polynomials here aren't obvious.
On the attitude- I just looked at every thread we ever both posted in, and can't find any examples of that unpleasantness. I've been provoked to unpleasantness by other posters, but I mostly don't respond at all to you.

See http://www.mathlinks.ro/viewtopic.php?t=249755 for another approach that also works for integers.


geometry
geometric transformation
homothety
circumcircle
geometry proposed

Given two triangles ABC and DEF, draw a triangle KLM congruent to DEF, such that $K\in AB$, $L\in BC$,$M\in AC$.

i think its not hard kondr.
we can solve it by homothecy.
draw aline paralell to BC wich goes through F.
draw aline paralell to AC wich goes through E.
draw aline paralell to AB wich goes through D.
they make another triangle wich is homothecy with triangle ABC.

lets name it A'B'C'.
then to find the center of homothecy draw lines AA', BB', CC' and let they meet each other in point O.

then draw line OD and let it intersect AB at K.
       draw line OF and let it intersect BC at L.
       draw line OE and let it intersect AC at M.

KLM is that triangle . ;) .

Ehr ... well, it says [i]congruent[/i] to DEF, not similar ...
And, by the way, th points K,L,M must be on the segments AB, BC, CA or on the lines on which the sides are?

Kondr's problem was discussed and solved in [url=http://groups.yahoo.com/group/Hyacinthos/messages/8185?xm=1&viscount=32&m=e]Hyacinthos messages #8185, #8189, #8194, #8195, #8197, #8198, #8203, #8205, #8206, #8208, #8210, #8212, #8215, #8218[/url]. (Mainly, I recommend you to read JPE's solution in message #8194.)

  darij

yes yes .i didnt pay attention to it.but i finally solved it.

but some times kondrs problem is wrong .for example triangle ABC be small enough that we cant draw KLM is it.

then i prove the question if we suppose that its possible to draw KLM in ABC. ;)

we can change the problem to another lproblem wich is equal to u problem.

suppose we have KLM and we want to draw triangle ABC in wich L be on AB.and M be on BC.and K be on AC.

draw KLM and ABC in paper.then draw aline paralell to AB wich goes through L.

draw aline paralell to BC wich goes through M.draw aline paralell to AC wich goes throughK.

let the lines that go through K and L ,intersect each other in point A'.let the lines that go through L and M ,intersect

each other in point B'.and let the lines that go thorough K and M intersect each other in point C'.

these three lines will make a triangle A'B'C' wich is simillar to triangle ABC.but we arnt sure that its equal to ABC.

then we have to find the locus of the points A' and B' and C'.

the locus of A' will be arc KA'L of the circumcircle of the triangle KA'L.

the locus of B' will be arc LB'M of the circumcircle of the triangle LB'M.

the locus of C' will be arc KC'M of the circumcircle of the triangle KC'M.

then the problem will change to another one.

problem:let two circle (C) and (C') intersect each other in points P and Q.

draw aline wich goes through P and is equal to a desired line .by (desired), i mean for each sum of X(wich is

possible ),draw a line wich goes through P.

then if we can this problem ,kondrs problem will solved because :

imagin the circles KA'L and LB'M .one of they meet each other in two points, in wich one of them is point L.

then if we draw a line wich goes through L and is equal to segment AB,the problem will be solved.

because by drawing this line A'B' will be equal to AB. and the angle B'A'C'  = the angle BAC

and the angle A'B'C'= the angle ABC. and it means that the triangles ABC and A'B'C' are equal.

so try to solve the problem that i said.its a well known problem and i think it doesnt mather wether i solve it or not .

but if u want the solution just tell me. ;)  :D

according to evaristeGs question, my answer is:

drawing triangle KLM in triangle ABC,isnt out of three kind.

1)K is on AB.L is on BC.M is on AC.

2)K is on BC.L is on AC.M is on AB.

3)K is on AC.l is on AB.M is on BC.

and for each kind if we know its possible to draw triangle KLM such that the problem wanted,we can draw it. ;)

is it clear or not ?

To Asegh: Your solution is the same as mine, you don't have to post the remaining part.
To alll: thank you for your replies.

Surely my poor english made me misunderstood : I asked whether the points K,L,M should be in the interior of the side (i.e., if K is on AB, the points are ordered as A,K,B) or, more generally, on the straight lines containing the sides.
Even if the triangle ABC is too small wrt triangle DEF, there could exist points K,L,M on the lines containing the sides of ABC such that KLM is congruent to DEF.

but i want to talk about the problem that i said in sixth post.

let O be the center of (C).and O'be the center of (C').and AB be aline inwich goes through P,inwich A be on (C),and

B be on (C').

then the triangles QOO' and QAB are simillar .because :  angle QO'O=arcQR/2=angle QBR.

and : angle QOO'=arcQR/2=angle QAB.

we have the sums of QO,QO'and OO'.and again we have the sum of AB.

QO'/QB=OO'/AB

then we can find the sum of QB.and draw a circle wich its center is Q.and its radius is QB.

then find the inter section of this circle and circle (C').and name it B.then draw line BP and let it intersect (C)in point A.then the segment AB will be equal to side AB in triangle ABC.  ;)

ure right evaisteG .if the problem said K,L,M can be on line AB,BC,AC,it would be true.

but it said it should be on sides BC,AC,AB,

and i think we can draw each triangle in another triangle with this fact.


inequalities
invariant
inequalities theorems

What are Vornicu-Schur ineq and SOS ineq?

As for the Vornicu-Schur inequality, it is mainly (in its most general form) the following assertion:

[color=blue][b]Theorem 1.[/b] Let a, b, c be three reals and x, y, z three nonnegative reals such that $ a\geq b\geq c$, and such that $ x\geq y$ or $ z\geq y$. Then,

$ x\left(a - b\right)\left(a - c\right) + y\left(b - c\right)\left(b - a\right) + z\left(c - a\right)\left(c - b\right)\geq 0$.[/color]

[hide="Proof of Theorem 1"][i]Proof of Theorem 1.[/i] First consider the case $ x\geq y$. In this case, we have the inequalities $ x\geq y$ and $ a - c\geq b - c$ (that's because $ a\geq b\geq c$); note that all four sides of these two inequalities are nonnegative (in fact, both a - c and b - c are nonnegative since $ a\geq b\geq c$), so that we can multiply these inequalities and get

$ x\left(a - c\right)\geq y\left(b - c\right)$.

Since $ a - b\geq 0$ (again from $ a\geq b\geq c$), we can multiply this with a - b and get

$ x\left(a - b\right)\left(a - c\right)\geq y\left(b - c\right)\left(a - b\right)$,

so that

$ x\left(a - b\right)\left(a - c\right) - y\left(b - c\right)\left(a - b\right)\geq 0$;

in other words,

$ x\left(a - b\right)\left(a - c\right) + y\left(b - c\right)\left(b - a\right)\geq 0$.

Also, $ a\geq b\geq c$ yields $ c - a\leq 0$ and $ c - b\leq 0$, so that $ \left(c - a\right)\left(c - b\right)\geq 0$, and since $ z\geq 0$, we get

$ z\left(c - a\right)\left(c - b\right)\geq 0$.

Adding this to the inequality $ x\left(a - b\right)\left(a - c\right) + y\left(b - c\right)\left(b - a\right)\geq 0$, we get $ x\left(a - b\right)\left(a - c\right) + y\left(b - c\right)\left(b - a\right) + z\left(c - a\right)\left(c - b\right)\geq 0$, as intended to prove.

Now remains the case $ z\geq y$. This time, have the inequalities $ y\leq z$ and $ a - b\leq a - c$ (that's because $ a\geq b\geq c$); note that all four sides of these two inequalities are nonnegative (in fact, both a - b and a - c are nonnegative since $ a\geq b\geq c$), so that we can multiply these inequalities and get

$ y\left(a - b\right)\leq z\left(a - c\right)$.

Since $ b - c\geq 0$ (again from $ a\geq b\geq c$), we can multiply this with b - c and get

$ y\left(a - b\right)\left(b - c\right)\leq z\left(a - c\right)\left(b - c\right)$,

what rewrites as

$  - y\left(b - c\right)\left(b - a\right)\leq z\left(c - a\right)\left(c - b\right)$.

Hence,

$ y\left(b - c\right)\left(b - a\right) + z\left(c - a\right)\left(c - b\right)\geq 0$.

Also, $ a\geq b\geq c$ yields $ a - b\geq 0$ and $ a - c\geq 0$, so that $ \left(a - b\right)\left(a - c\right)\geq 0$, and since $ x\geq 0$, we get

$ x\left(a - b\right)\left(a - c\right)\geq 0$.

Adding this to the inequality $ y\left(b - c\right)\left(b - a\right) + z\left(c - a\right)\left(c - b\right)\geq 0$, we get

$ x\left(a - b\right)\left(a - c\right) + y\left(b - c\right)\left(b - a\right) + z\left(c - a\right)\left(c - b\right)\geq 0$.

And thus Theorem 1 is proven.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=382086#p382086]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64059 post #10[/url] for a further discussion of the Vornicu-Schur inequality. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=271175#271175]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41284 post #4[/url] Theorem 3 for the Schur inequality in its strongest form. And see [url=http://www.artofproblemsolving.com/Forum/portal.php?t=162684]my Vornicu-Schur article[/url] for a few applications and a number of variations.

As for SOS (short for: sum of squares), it is a generally used term for a way of proving inequalities: If you want to prove an inequality $ a\geq b$, you write a - b as a sum of squares or of other trivially nonnegative terms, so that it follows that $ a - b\geq 0$ and thus $ a\geq b$.

However, the word SOS is also used for a particular case of this technique, namely the application of the following theorem:

[color=blue][b]Theorem 2.[/b] Let a, b, c, x, y, z be six reals satisfying the following conditions:
[b](1)[/b] first, the number b lies "between" the numbers a and c; that means, we have either $ a\geq b\geq c$ or $ a\leq b\leq c$;
[b](2)[/b] then, $ y\geq 0$;
[b](3)[/b] either $ x\geq 0$ and $ z\geq - y$, or $ z\geq 0$ and $ x\geq - y$ (or both).

Then,

$ x\left(b - c\right)^{2} + y\left(c - a\right)^{2} + z\left(a - b\right)^{2}\geq 0$.[/color]

[hide="Proof of Theorem 2"][i]Proof of Theorem 2.[/i] By condition [b](3)[/b], we have either $ x\geq 0$ and $ z\geq - y$, or $ z\geq 0$ and $ x\geq - y$. We can WLOG assume that $ z\geq 0$ and $ x\geq - y$.

Since $y\geq 0$, the number $-y$ is nonpositive.

Now, from [b](1)[/b], we have $ \left|b - c\right|\leq\left|c - a\right|$, so that $ \left(b - c\right)^{2}\leq\left(c - a\right)^{2}$. Multiplying this inequality with the nonpositive number $-y$, we must reverse its sign: $ -y\left(b - c\right)^{2}\geq -y\left(c - a\right)^{2}$. Since $ x\geq - y$ and $\left(b-c\right)^2\geq 0$, we have $ x\left(b - c\right)^{2}\geq -y\left(b - c\right)^{2}\geq - y\left(c - a\right)^{2}$. Thus,

$ x\left(b - c\right)^{2} + y\left(c - a\right)^{2} + z\left(a - b\right)^{2}\geq - y\left(c - a\right)^{2} + y\left(c - a\right)^{2} + z\left(a - b\right)^{2}$
$  =  z\left(a - b\right)^{2}\geq 0$,

since $ z\geq 0$. Thus, Theorem 2 is proven.[/hide]

Here are, by the way, two other useful results of a similar kind. The first one is a very trivial fact, and generalizes, among other things, Lemma 1 from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=204446#p204446]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6138 post #2[/url] and Lemma 1 from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=204800#p204800]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3547 post #10[/url]:

[color=blue][b]Theorem 3.[/b] Let a, b, c be three nonnegative reals satisfying $ a\geq c$ and $ b\geq c$, and let u, v, w be three reals such that $ u\geq 0$, $ v\geq 0$ and $ u + v + w\geq 0$. Then, $ au + bv + cw\geq 0$.[/color]

[hide="Proof of Theorem 3"][i]First proof of Theorem 3.[/i] Since $u\geq 0$ and $a\geq c$, we have $au\geq cu$. Since $v\geq 0$ and $b\geq c$, we have $bv\geq cv$. Thus,

$\underbrace{au}_{\geq cu}+\underbrace{bv}_{\geq cv}+cw \geq cu+cv+cw = \underbrace{c}_{\geq 0}\underbrace{\left(u+v+w\right)}_{\geq 0} \geq 0$.

Thus, Theorem 3 is proven.

[i]Second proof of Theorem 3.[/i] If $ w\geq 0$, then (combined with $ u\geq 0$, $ v\geq 0$ and the nonnegativity of the numbers a, b, c) this yields $ au + bv + cw\geq 0$. Thus, we only have to consider the case when w < 0.

We can WLOG suppose that $ a\geq b$; else, we just interchange a and b (of course, interchanging u and v at the same time). Now, since w < 0, the inequality $ c\leq b$ becomes $ cw\geq bw$, and since $ u\geq 0$, the inequality $ a\geq b$ becomes $ au\geq bu$. Hence, $ au + bv + cw\geq bu + bv + bw  = \underbrace{b}_{\geq 0}\left(\underbrace{u + v + w}_{\geq 0}\right)\geq 0$. Thus, Theorem 3 is proven.[/hide]

And the last one is Lemma 1 from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=386442#p386442]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64654 post #7[/url]:

[color=blue][b]Theorem 4.[/b] Let a, b, c be three reals; let x, y, z, u, v, w be six nonnegative reals. Assume that the number arrays (a; b; c) and (x; y; z) are equally sorted, and

$ u\left(a - b\right) + v\left(b - c\right) + w\left(c - a\right)\geq 0$.

Then,

$ xu\left(a - b\right) + yv\left(b - c\right) + zw\left(c - a\right)\geq 0$.[/color]

[hide="Proof of Theorem 4"][i]Proof of Theorem 4.[/i] Since the number arrays (a; b; c) and (x; y; z) are equally sorted, we have either ($y\geq z$ and $b\geq c$) or ($y\leq z$ and $b\leq c$). In other words, either ($y-z\geq 0$ and $b-c\geq 0$) or ($y-z\leq 0$ and $b-c\leq 0$). Thus, $\left(y-z\right)\left(b-c\right)\geq 0$. Similarly, $\left(z-x\right)\left(c-a\right)\geq 0$ and $\left(x-y\right)\left(a-b\right)\geq 0$. Thus,

$\left(y-z\right)\left(a-b\right)\left(z-x\right)\left(b-c\right)\left(x-y\right)\left(c-a\right)$
$=\underbrace{\left(y-z\right)\left(b-c\right)}_{\geq 0}\underbrace{\left(z-x\right)\left(c-a\right)}_{\geq 0}\underbrace{\left(x-y\right)\left(a-b\right)}_{\geq 0}\geq 0$.

Hence, at least one of the numbers $\left(y-z\right)\left(a-b\right)$, $\left(z-x\right)\left(b-c\right)$ and $\left(x-y\right)\left(c-a\right)$ must be $\geq 0$ (because otherwise, each of these numbers would be $< 0$, so that their product would also be $< 0$, but this contradicts the fact that their product is

$\left(y-z\right)\left(a-b\right)\left(z-x\right)\left(b-c\right)\left(x-y\right)\left(c-a\right)\geq 0$

).

Since the statement of Theorem 4 is invariant under cyclic permutations (of course, when these are performed for the number arrays (a; b; c), (x; y; z) and (u; v; w) simultaneously), we can thus WLOG assume that $\left(y-z\right)\left(a-b\right)$ is $\geq 0$. Assume this. Then,

$\left(x-z\right)\left(a-b\right) = \underbrace{\left(x-y\right)\left(a-b\right)}_{\geq 0} + \underbrace{\left(y-z\right)\left(a-b\right)}_{\geq 0} \geq 0$.

Now,

$ xu\left(a - b\right) + yv\left(b - c\right) + zw\left(c - a\right)$
$  =  u\underbrace{\left(x - z\right)\left(a - b\right)}_{\geq 0} + v\underbrace{\left(y - z\right)\left(b - c\right)}_{\geq 0} + z\underbrace{\left(u\left(a - b\right) + v\left(b - c\right) + w\left(c - a\right)\right)}_{\geq 0}$
$ \geq 0$,

and Theorem 4 is proven.[/hide]

  Darij

I am curious about S.O.S technique.
  But how can u rewrite a homogenize inequality to the form
  $\sum f(a,b,c)(a-b)^2$


linear algebra
matrix
vector
algebra
function
domain
linear algebra unsolved

Let $A\in M_{n}(\mathbb{R})$ such that $det(A)=0.$ Prove that there is a matrix $B\in M_{n}(\mathbb{R})$ such that $AB=O_{n}.$ ;)

one of the many many ways to do this:
otherwise the linear map $f: M_n(R) \mapsto M_n(R)$ defined by $f(M)=AM$ would be injective. But $M_n(R)$ is a finite dimensional vector space. So $f$ would be bijective, hence there would exist $P$ such that $AP=I$.

To avoid trivialities, the original problem should have specified that $A\ne0$ and that we are looking for a solution $B$ such that $B\ne0.$

If $\det A=0,$ then there exists a vector $x\ne 0$ such that $Ax=0.$

Now pick any $y\ne 0.$ Let $B=xy^T.$ $B$ is a rank one matrix and clearly $AB=A(xy^T)=(Ax)y^T=0y^T=0.$

As is well-known, if $AB=0,\,A\ne0,\,B\ne0,$ then both $\det A=0$ and $\det B=0.$

use row-reduction to A so that the reduced row echelon form is reached. Suppose the first r diagonal entries of the form are 1 and the rest are 0. We take B' to be the matrix with first r diagonal entries to be 0 , rest of diagonal entries be 1 and others be 0. Then we can take some inverse gaussian elimination process from A (multiply B' by elementary matrices), and we can get a B satisfying AB = O

There are of course many such matrices $B.$ If I'm not mistaken, if $\text{rank}(A)=r,$ then the set of all $B\in M_n$ such that $AB=0$ is a vector space of dimension $(n-r)^n.$

It makes no difference whether the scalars are $\mathbb{R}$ or any other field.

[quote="Kent Merryfield"]If $\det A=0,$ then there exists a vector $x\ne 0$ such that $Ax=0.$[/quote]

Why is that?

Again sorry for lack of knowledge in this domain

Call it the first fundamental theorem of linear algebra. David Lay, in his textbook, calls it the "Invertible Matrix Theorem." Now, I've got to say that Lay goes in for extreme overkill with this statement, (and spreads it over three chapters, 70 pages apart) but sometimes extreme overkill can be fun. Here it is:

[u][b]The Invertible Matrix Theorem[/b][/u]

Let $A$ be a square $n\times n$ matrix. Then the following statements are equivalent. That is for a given $A,$ the statements are either all true or all false.

a. $A$ is an invertible matrix.
b. $A$ is row equivalent to the $n\times n$ identity matrix.
c. $A$ has $n$ pivot positions.
d. The equation $Ax=0$ has only the trivial solution.
e. The columns of $A$ form a linearly independent set.
f. The linear transformation $x\mapsto Ax$ is one-to-one.
g. The equation $Ax=b$ has at least one solution for each $\mathbb{R}^n.$
h. The columns of $A$ span $\mathbb{R}^n.$
i. The linear transformation $x\mapsto Ax$ maps $\mathbb{R}^n$ onto $\mathbb{R}^n.$
j. There is an $n\times n$ matrix $C$ such that $CA=I.$
k. There is an $n\times n$ matrix $D$ such that $AD=I.$
l. $A^T$ is an invertible matrix.
m. The columns of $A$ form a basis of $\mathbb{R}^n.$
n. $\text{Col} A=\mathbb{R}^n.$
o. $\text{dim Col} A=n.$
p. $\text{rank} A=n.$
q. $\text{Nul} A=\{0\}.$
r. $\text{dim Nul}A=0.$
s. $\text{det}A\ne 0.$

So let's say I was using the equivalence of (s) with (d).

In this massive recitation of conditions, Lay was assuming this was a real matrix, and saying $\mathbb{R}^n$ in a number of cases. That's for purposes of pedagogical familiarity. None of the proofs use any properties of $\mathbb{R}$ other than that it is a field. Hence, this theorem is also true for matrices with entries in the field $K,$ as long as you replace $\mathbb{R}^n$ with $K^n,$ where appropriate.

[quote="Kent Merryfield"][quote="Soarer"][quote="cimabue"]how about the matrix $B=(b_{ij})$ where $b_{ij}=(-1)^{i+j}\det A_{ij}$ etc[/quote]
i'm not sure, but doesn't this make AB=I?[/quote]
No, it makes $AB=(\det A)I,$ and remember that we were assuming that $\det A=0.$ cimabue's example may be more specific than is entirely needed, but it is correct.[/quote]

oh, sorry for my stupidity. :P

[quote="perfect_radio"]Thanks, Kent. For the equivalence you used, at least, again we can use row-reduction.

[quote="alekk"]the linear map... would be injective. But $M_n(R)$ is a finite dimensional vector space. So $f$ would be bijective[/quote]

Injective, I understand why. I don't get it why it is also surjective. Could you explain?[/quote]
dimension theorem.
T is injective <=> nullity T = 0
dimension theorem states that if T:U->V is linear, nullity T + rank T = dim U
and you see that rank T = dim U <=> T is surjective.


geometry
ratio
inequalities
number theory
relatively prime
triangle inequality

In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD=2$ and $DB=5,$ point $E$ is on $\overline{AC}$ with $CE=1$ and $EA=3,$ $AB=8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Umm if $EA = 32$ then Triangle $ABC$ is a $7: 8: 33$..which isnt possible because of the triangle inequality theorem.. $a+b > c$ ...in this case it isnt...

Is there something wrong with the question?

If AE=3  I think I solve without "AB=8"

By menelaus(?) in triangle ACD and line(?) BPE
 PD/AP=5/21
 So we can get AP/AD=21/26
 And PR=5x21/26=105/26 (since ARP and ABD is similar(?))

  PQR/ABC = (PR/BC)^2 = (15/26)^2 = 225/676

  ,,,, Something wrong???

That is correct, you do not need AB=8, I solved it the same way using menelaus.

i have a great approach in applying mass points to find AP PD ration and would get 901 as answer only

Easy with barycentrics.

Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then clearly $D=(0:2:5)$ and $E=(1:0:3)$ so we deduce $P=(5:6:15)$. Since $R$ lies on $AB$, it can be parameterized by $R=(t:1:0)$. However, it also lies on line $PP_\infty$ where $P_\infty=(0:-1:1)$ is the point at infinity along line $BC$. Hence, $$\begin{vmatrix}t&1&0\\ -1&0&1\\ 5&6&15\end{vmatrix}=0\iff t=\frac{10}3$$ so $R=(10:3:0)$. Similarly, $Q=(5:21:0)$. Now just remark $$\frac{[PQR]}{[ABC]}=\frac{1}{(5+6+15)(10+3+0)(5+21+0)}\begin{vmatrix}5&6&15\\ 10&3&0\\ 5&21&0\end{vmatrix}=\frac{15(210-15)}{13\cdot 26^2}=\frac{225}{676}$$ and we are done.

We will use barycentric coordinates, We have $D=(0,2/7,5/7)$ and $E=(1/4,0,3/4)$ The equation for line $AD$ is $5y-2z=0$, while the equation for $BE$ is $3x-z=0$, This gives that $$P=(5/26,6/26,15/26).$$ Noting that $Q$ has the same second coordinate as $P$ since it is translated parallel to $AC$, we have $$Q=(20/26,6/26,0).$$ Similarly, $$R=(5/26,21/26,0).$$ Rather than using area formula now which is quite messy, note that $\triangle PQR$ and $\triangle ABC$ are similar due to the parallel, and the first coordinates of $Q$ and $R$ differ by $15/26$, so $QR=(15/26)AB$, so our answer is $$(15/26)^2\rightarrow 901.$$

[quote=OlympusHero]Obviously $\triangle PQR \sim \triangle CAB$, so denote $PQ = 4x, PR = 7x, QR = 8x$. But we also have $\triangle APR \sim \triangle ADB$ and $\triangle BPQ \sim \triangle BEA$, so $AR = \frac{56x}{5}$ and $QB = \frac{32x}{3}$. This means $AQ = \frac{16x}{5}$ and $RB = \frac{8x}{3}$, so $\frac{16x}{5}+8x+\frac{8x}{3} = 32 \implies x  = \frac{15}{26}$. This is the ratio of sides, so squaring gives $\frac{225}{676} \implies \boxed{901}$.[/quote]

same as this post
$Nice!!!!!$


inequalities
LaTeX
inequalities unsolved

Let $k \geq 2, 1 < n_1 < n_2 < \ldots < n_k$ are positive integers, $a,b \in \mathbb{Z}^+$ satisfy \[ \prod^k_{i=1} \left( 1 - \frac{1}{n_i} \right) \leq \frac{a}{b} < \prod^{k-1}_{i=1} \left( 1 - \frac{1}{n_i} \right)  \]
Prove that: \[ \prod^k_{i=1} n_i \geq (4 \cdot a)^{2^k - 1}.  \]

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

[quote="hxy09"]solution :)[/quote]

nice solution :lol:

[quote="Jianxing113725"][quote="hxy09"]solution :)[/quote]

A fomous university in China...
The paper is your examination test paper... :D[/quote]
Very sorry,the paper isn't mine,my mother is a teacher in this university.She gave me these papers to write my solution down :blush:
I am only a high school student :)

Anyone have different ideas?Solution posted by hxy09 is the official solution...It's unimaginable I think....

We will prove the result by strong induction on $k$. When $k = 1$, the result is obvious. Now, suppose the result holds for $k < t.$ When $k = t,$ we consider two cases.

[b] Case 1. [/b] There exists an $1 \le i \le k-1$ such that $n_1 n_2 \cdots n_i \le (4a)^{2^i-1}.$

Then, observe that:

$$\prod_{j = i+1}^{k} \left(1 - \frac{1}{n_j}\right) \le \frac{a n_1 n_2 \cdots n_i}{b (n_1 - 1)(n_2-1)\cdots (n_i-1)} < \prod_{j=i+1}^{k-1} \left(1- \frac{1}{n_j}\right).$$
Hence, since $k - i \ge 1$, the inductive hypothesis implies that:

$$(4an_1 n_2 \cdots n_i)^{2^{k-i}-1} \ge \prod_{j= i+1}^{k} n_j. \qquad (1)$$
Therefore, observe that from our assumption that $(4a)^{2^i-1} \ge n_1n_2 \cdots n_i$, we get:

$$((4a)^{2^i})^{2^{k-i}-1} \cdot n_1n_2 \cdots n_i \ge \left(\prod_{j=i+1}^{k} n_j\right) \cdot n_1n_2 \cdots n_i,$$
so that

$$(4a)^{2^k-1} \ge ((4a)^{2^i})^{2^{k-i}-1} \cdot n_1n_2 \cdots n_i \ge \prod_{i=1}^{k} n_i,$$
as desired.

[b] Case 2. [/b] There does not exist an $1 \le i \le k-1$ such that $n_1n_2 \cdots n_i \le (4a)^{2^i-1}.$

Then, from the fact that $n_1 < n_2 < \cdots < n_k$ we get that:

$$n_i > \left ((4a)^{2^i-1} \right)^{\frac1i}, \forall 1 \le i \le k-2.$$
Since $\frac{2^i-1}{i} \ge i-1$ for $i \ge 2$, we get that $n_1 > 4a$ and $n_i > (4a)^{i-1}$ for $2 \le i \le k-1.$ We also clearly have $n_k > n_{k-1} > (4a)^{k-2}.$

For $k \le 4$, this already yields from $4a < n_1 < n_2 < \cdots < n_k$ and telescoping that:

$$\prod_{i=1}^{k} \left( 1 - \frac{1}{n_i} \right) \ge \frac{4a+1}{4a+k+1} < \frac{a}{a+1} < \frac{a}{b},$$
which is clearly absurd.

Hence, we can assume that $k \ge 5$ from now on. Observe from telescoping  that:

$$\left(1-\frac{1}{4ax-4a+1}\right)\left(1 - \frac{1}{4ax-4a+2}\right)   \cdots \left( 1 - \frac{1}{4ax}\right) = \frac{x-1}{x},$$

and so since $1 - \frac{1}{4ax}$ is greater than or equal to all of the multiplicands on the LHS we obtain that

$$1 - \frac{1}{4ax} \ge \left(1 - \frac{1}{x}\right)^{\frac1{4a}}, \forall x \in \mathbb{N}.$$
A simple induction then yields that:

$$1 - \frac{1}{(4a)^{i}} \ge \left(1 - \frac{1}{4a}\right)^{\frac{1}{(4a)^{i-1}}}, \forall i \in \mathbb{N}$$ This yields for $k \ge 5$ that:

$$\prod_{i=1}^{k} \left(1 - \frac1{n_i}\right) \ge \left (1 - \frac1{4a}\right )^{\frac{1}{(4a)^0} + \frac{1}{(4a)^0} + \frac{1}{(4a)^1} + \frac{1}{(4a)^2} \cdots + \frac{1}{(4a)^{k-4}} + \frac{1}{(4a)^{k-3}} + \frac{1}{(4a)^{k-3}}}.$$
For $k \ge 5$, it's easy to verify that the exponent is at most $2.5$. Therefore, we have that if $a \ge 2$, then:

$$\prod_{i=1}^{k} \left(1 - \frac1{n_i}\right) \ge \left(1-\frac{1}{4a}\right)^{2.5} > \frac{a}{a+1} \geq \frac{a}{b},$$
where the second inequality used above is easily verified. This is a clear absurdity.

It remains only to tackle the case where $a = 1.$ To do so, we will use the improvements $1 - \frac{1}{n_1} \ge 1 - \frac{1}{4a+1} = \frac{4}{5}$ and $1 - \frac{1}{n_2} \ge 1 - \frac{1}{4a+2} = \frac{5}{6}.$ Implementing these then yields the stronger bound of:

$$\prod_{i = 1}^{k} \left(1 - \frac{1}{n_i}\right) \ge \frac23 \cdot (1 - \frac14)^{0.5} > \frac12 \ge \frac{a}{b}.$$
This is clearly absurd.

As we've reached absurdity in all cases, we in fact obtain that Case $2$ is not even possible.

Hence, the induction is complete, and the problem is solved.

$\square$

Sorry but shouldnt this problem be in chinese if it is a china test


Show that for any positive integer $ n$, the polynomial $ f(x) \equal{} (x^2 \plus{} x)^{2^n} \plus{} 1$ is irreducible over $ \mathbb{Z}$.

[quote="Moonmathpi496"]Show that for any positive integer $ n$, the polynomial $ f(x) \equal{} (x^2 \plus{} x)^{2^{n}} \plus{} 1$ is irreducible over $ \mathbb{Z}$.[/quote]

You meant this, I suppose  :wink:

It's probably among the top 5 of "Problems posted $ n$ times on Mathlinks, $ n \ge 5$" chart:

http://www.mathlinks.ro/Forum/viewtopic.php?t=49799
http://www.mathlinks.ro/viewtopic.php?t=186280
http://www.mathlinks.ro/viewtopic.php?t=70003
http://www.mathlinks.ro/viewtopic.php?t=80920


geometry
3D geometry
vector
symmetry
analytic geometry
combinatorics unsolved
combinatorics

Hello all! I know that the hypercube graph (aka k-cube) is symmetric. I've succeeded proving that it's vertex transitive (with the "decomposition" into $ K_2 \times ... \times K_2$) - but I haven't been able to prove that it's also edge-transitive. Does anybody know how to do it and can give me a hint please?  :blush:

Thank you all for your help.

[quote="Nukular"]You can use the $ K_2 \times K_2 \times \dots \times K_2$ decomposition.

In fact, let's do a little better; instead label the vertices as $ \left(\mathbb{Z}/2 \right)^k$.
(k-tuples of mod-2 integers). Two vertices are adjacent iff they differ in exactly one
coordinate. However, this vertex set is a mod-2 vector space; we can "add" two
vertices by $ (a_1, \dots, a_k) \plus{} (b_1, \dots, b_k) \equal{} (a_1 \plus{} b_1, \dots, a_k \plus{} b_k)$
(taken all mod-2 of course). It's not too hard to show that the map
$ f_w : v \mapsto v \minus{} w$ is a graph automorphism for any $ w \in \left(\mathbb{Z}/2 \right)^k$.
Additionally, if $ \omega$ is a permutation of $ 1,2,\dots,k$, then $ g_\omega : (a_1,\dots,a_k) \mapsto (a_{\omega 1}, \dots, a_{\omega k})$ is also a graph automorphism.

From there, you get any symmetry you want "for free". In particular, the edge $ (v,w)$ is
taken to the edge $ (0,w \minus{} v)$ (where 0 = (0,0,...,0)) by $ f_v$. We must have that
$ w \minus{} v \equal{} (0,0,0,\dots,0,1,0,\dots, 0)$ (with the 1 in the $ j$th coordinate); a permutation automorphism $ g_\omega$, where
$ \omega j \equal{} 1$ then takes $ (0,w \minus{} v)$ to $ (0,(1,0,0,\dots,0))$. Therefore the graph is edge-transitive.

(Perhaps this last argument is a bit messy, but when the graph is easily reduced to vector space
identities you get basically any symmetry you want directly from that).

EDIT: In general, you can do this by two steps: 1) Show vertex-transitivity (comes from the $ f_w$ automorphisms),
2) Show edge transitivity *on any individual vertex* (comes from the $ g_\omega$ automorphisms). I suppose
that's what I did above, but the separation in these two steps is not as apparent as it could be.[/quote]

Thanks! I had to read your message a couple of times before I finally understood the process you've suggested. However, I've filled in all the details and it works beautifully. Thank you very much for your help, I don't know how many more hours it would have taken me to find similar process myself :)


I need help with the following problems:

[b]#16, #22, and #23[/b]

If AOPS members could help me out, I'd greatly appreciate it.  Thanks for your time!

[img]http://img178.imageshack.us/img178/6134/problemset9am9.jpg[/img]

[hide="16."]
Since you know $-1$ is a root, $(x+1)$ is a factor. Use synthetic division to find that the quotient is $(x+4)(x-3)$. So the other roots are $-4$ and $3$.[/hide]
[hide="22."]
Note that the degree is even and the leading coefficient is positive.

This means $f(x)\rightarrow+\infty$ as $x\rightarrow\pm\infty$.[/hide]
[hide="23."]
Note that the degree is odd and the leading coefficient is negative.

This means $f(x)\rightarrow-\infty$ as $x\rightarrow+\infty$ and $f(x)\rightarrow+\infty$ as $x\rightarrow-\infty$[/hide]

[quote="ckck"]First of all Welcome to Aops!
Second what is left end behavior and right end behavior?
Third what does it mean when it says find all zeros?[/quote]
Thanks for the welcome!  This is my username, I registered "iMath" because I forgot my password.

Left and right end behavior describes how the graph looks at "plus infinity" and "minus infinity".

When it says find all zeros, I need to find all the zeros of the function.  Meaning the value that you put for x that makes the equation equal to zero.

i don't know waht left and right hand behavior is but #16:

$x^{3}+2x^{2}-11x-12=0$

$x^{3}+x^{2}+x^{2}-11x-12=0$

$x^{2}(x+1)+(x+1)(x-12)$

$(x+1)(x^{2}+x-12)$

solve either $x+1=0$ or $x^{2}+x-12=0$

the  zeroes are:

-1,-4, and 3

EDIT: sorry if i gave you any confusion. now all the x's are lower case

By the way, this looks suspiciously like Algebra 2 homework.  :)

@rnwang2:  Yes, this is Algebra 2 homework.  This is part of my problem set, which is graded on accuracy! It's probably a piece of cake for you AOPS people.  My brain is just slow. :(

click "online school" on the left and for each class there is a:

1. Do You need this class?

2. are you ready for this class?

An easier way to do #24 is factoring:

$x^{4}-5x^{2}-36 = 0$

This is the same as a quadratic in $x^{2}$:

$(x^{2})^{2}-5x^{2}-36 = 0$

Through guess-and-check/experience, notice that $-9$ and $4$ are factors of $-36$ that add up to $-5$.  Then factor appropriately:

$(x^{2}-9)(x^{2}+4) = 0$

Thus:

$x^{2}-9 = 0$ or $x^{2}+4 = 0$

Solving each...

$x^{2}= 9$ or $x^{2}=-4$

Take the square root of both sides in each equation:

$\sqrt{x^{2}}= \pm \sqrt{9}$, $\sqrt{x^{2}}= \pm \sqrt{-4}$

Thus, the solutions are:

$x = \pm 3, \pm 2i$.


[quote="KMK00009"]a,b,c are real numbers where a^2+b^2+c^2=669
find the maximum value of (a-b)^2+(b-c)^2+(c-a)^2[/quote]

We have $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2$ $ \equal{}3(a^2\plus{}b^2\plus{}c^2)\minus{}(a\plus{}b\plus{}c)^2\leq 3(a^2\plus{}b^2\plus{}c^2)$ $ \equal{} 3*669 \equal{} 2007$

And this number may be reached for example with :

$ a\equal{}\sqrt{\frac{669}{2}}$

$ b\equal{}\minus{}\sqrt{\frac{669}{2}}$

$ c\equal{}0$

So the requested value is $ 2007$


POW 7: One Mile at a Time

You may have noticed that many highways have signs along the side of the road that list your milage from some place behind you. This POW concerns a situation involving these mileposts.

The Situation

You are driving down a highway and notice a milepost showing a distance that is a two-digit number. Exactly one hour later, you notice a milepost that shows the same two digits as the first, but in opposite order.

Then, exactly one hour after seeing the secong milepost, you notice a third one that shows a three-digit number. The middle digit on this sign is 0. The other two digits are the same as those on the first milepost and are in the same order as on that first milepost.

Your Task

The basic question for the situation is, "How fast have you been traveling?" This question has a simple numerical answer that you may be able to find by guess-and-check. (Assume that you have been traveling at the same speed throughout your trip.)

Your task is to go beyond finding the answer, by expressing the situation in algebraic terms and proving using algebra, that your solution to the problem is correct. (Hint: Recall that a number like 386 can be written as 3 x 100 +8 x 10 +6 and apply this idea to express an arbitrary two- or three-digit number in terms of its digits).

ok. thats the problem.

the simple numerical answer i got is 16. so, the car is going at 45 miles and hr, and so its 16, 61, 106.

but then, i dont know how to put in algebraic terms.

Please help me out!
Thanks!

Well, I have a few questions:

What is POW?
Is POW 7 over?
When was POW 7 over?

Let the first sign be $\overline{ab}=10a+b$, the second sign is $\overline{ba}=10b+a$ and the third sign is $\overline{a0b}=100a+b$. Since there is exactly one hour between the time you saw each sign and you are travelling at a constant speed, $(10b+a)-(10a+b)=(100a+b)-(10b+a)$
$20b+2a=110a+2b$
$10b+a=55a+b$
$9b=54a$
$b=6a$.
Now since $a$ and $b$ are single digit integers, the only solution can be $b=6$,$a=1$.

Let's call the three digits $a,\ b,$ and $c.$

We know that $b=0.$

The two differences are the same, so $(10c+a)-(10a+c)=(100a+10b+c)-(10c+a),$ or $c=6a.$

This has only one solution, $a=1,\ b=6.$

$61-16=\boxed{45}$


Prove that no ordered pair of integers $ (x, y)$ satisfies the equation $ x^{2}+3xy-2y^{2}=T$

Sorry, if I take any x and y and calculate $ T=x^{2}+3xy-2y^{2}$ it wouldn't be counterexample?
So I believe T is some definite integer?

Assuming you want the general form of of $ T$ for which this equation is satisfied, you might consider that solving this equation modulo $ p$ yields $ x=y(\frac{-3 \pm \sqrt{17}}{2}$, so for $ p$ such that $ \left(\frac{17}{p}\right)=1$, this equation has solutions modulo $ p$, or otherwise we have $ x, y \equiv 0\mod{p}$: from there, you can take one solution and set up a system to which Minkowski's theorem can be applied, giving you all primes for which $ x^{2}+3xy-2y^{2}$ has a solution.  Solving this equation for general $ n$ gives the same thing: $ 17$ is a quadratic residue mod $ n$.  Then for all $ p$ dividing $ n$, $ 17$ is a quadratic residue mod $ p$.  Then $ p \equiv \pm 1, \pm2, \pm4, \pm8 \mod{17}$.  I'm fairly certain you can use Minkowski's Theorem to show that if the congruence is solvable, the equation will be solvable, and so $ T$ will be of the form that for each $ p|T$, if $ 17$ is a quadratic nonresidue mod $ p$, then $ p$ divides $ T$ an even number of times.

Just a stab in the dark, but can you do something like infinite descent?
I read about that being used to solve these kinds of problems.


linear algebra
matrix
algebra
polynomial
number theory
complex numbers
linear algebra unsolved

$A$ is with entries that are integers and $p>2$ is a prime. If the entries of $A-E$ are all divisible by $p$ and for some $k<p$  $A^{k}=E$ then $A=E$.

solution: the minimal polynomial of A divides $x^{k}-1$, which had different complex roots and therefore A is diagonalisable with entries on the main diagonal eventually $1$ and eventually complex numbers(some of the roots of $x^{k}-1$). But for making $A$ diagonalisable (A is with whole entries) we don't need multiplying,dividing by a complex(not real) number so on the diagonal of $A$ there are only 1s.(so all we needed was A with whole entries and $A^{k}=E$).

No, you can't say that about the diagonalization. There are many examples of matrices with integer entries that have some power equal to the identity.

You need number theory to do this problem.

but where is the mistake in this solution then? i only used that to make $A$ diagonalisable we don't need multiplying by a complex number.i think the last is true so the mistake is somewhere else.

i know now to solve it without eigenvalues

You do need complex numbers to diagonalize it- consider something like $\begin{bmatrix}0&1&1\\0&0&1\\1&0&0\end{bmatrix}$. The basic premise of your attempted solution was nonsense.

We have that $A = I+pB$, where $B \in \mathcal M_{n}\left( \mathbb Z \right)$.
Trivially,
\[p^{n}\cdot \det \left( B-\frac{x-1}{p}\cdot I \right) = \det \left( pB+I-x I \right) . \]
This equality of polynomials means that if $\lambda_{i}$ are the eigenvalues of $pB+I$, then $\frac{\lambda_{i}-1}{p}$ are [i]exactly[/i] (i.e. with multiplicity) the eigenvalues of $B$. If $\lambda$ is an eigenvalue of $pB+I$, then $\lambda^{k}$ is an eigenvalue of $\left( pB+I \right)^{k}= A^{k}= I$, i.e. $\lambda^{k}= 1$. Hence, $\left| \lambda_{i}\right| = 1, \, \forall i$. Therefore,
\[\left| \frac{\lambda_{i}-1}{p}\right| \leq \frac{\left| \lambda_{i}\right|+1}{p}= \frac2{p}< 1 . \]
Consequently, $\left| \prod_{i=1}^{n}\frac{\lambda_{i}-1}{p}\right| < 1$. But that's the product of the eigenvalues of an integer matrix, so it must be an integer. The only possibility is that the product is $0$. Hence, $\lambda_{j}= 1$ for some $j$. Now put this eigenvalue of $B$ aside and look at the product of others. Its absolute is of course $<1$. Moreover, it's an integer (this follows from the fact that the characteristic polynomial of $B$ has, after all, integer coefficients). So we get another eigenvalue equal to $0$. Inductively, all are $0$, so all the $\lambda_{i}$'s are equal $1$.

Let $m_{A}$ be the minimal polynomial of $A$. We all know that $m_{A}$ divides any polynomial that annihilates $A$. $X^{k}-1$ is one such polynomial, as mentioned in the hypothesis. By Cayley-Hamilton, $(X-1)^{n}$ is another one. Hence, \[m_{A}\mid \gcd \left( X^{k}-1, (X-1)^{n}\right) = X-1 ,\]
so $A = I$.

Is this correct? I'm wondering, because what I proved is stronger than the original problem. Among others, I do not seem to be using the fact that $p$ is prime :blush:

It's a very nice and clever solution- definitely not something the proposer thought of.

$A=I_{n}+pB$
$A^{k}=I_{n}+C_{k}^{1}pB+p^{2}(......)=I_{n}$ taking this modulo $p^{2}$ and using the fact that $k<p$ and $p$ is prime it follows that
$pB\equiv_{p}0$. So $B=pB_{1}$.
Inductively $p^{n}|B, n\ge 1$ so $B=O_{n}$, hence $A=I_{n}$.
The solution of perfect_radio shows that the hypothesis of the problem can be stronger.


Let $n=p^{a}m$ where $p$ is a prime. prove:
$\binom{n}{p^{a}}\equiv m\pmod{p}$

[hide="hint"]
consider $(x+1)^{p^{2}}\pmod{p}$.[/hide]

[hide="solution"]
note $(x+1)^{p^{2}}\equiv (x^{p}+1)^{p}\equiv x^{p^{2}}+1 \pmod{p}$

by the same argument we have:
$(x+1)^{p^{a}}\equiv x^{p^{a}}+1\pmod{p}$
and
$(x+1)^{n}=(x+1)^{p^{a}m}\equiv(x^{p^{a}}+1)^{m}\pmod{p}$
comparing the coefficient of $x^{p^{a}}$ on both sides yields
$\binom{n}{p^{a}}\equiv\binom{m}{1}=m\pmod{p}$

as desired.[/hide]


induction
pigeonhole principle
function
algorithm
geometry
trigonometry
calculus

Dear All,

Math-M-Addicts are back, this time with 3 separate groups -- Number Theory, Intermediate, and Advanced.

The Number Theory group will become very comfortable with proof by induction, proof by pigeonhole principle and other mathematical techniques.  The goal is also to become very comfortable with modular arithmetic, Euler's Phi function, Euclid's Algorithm, Chinese Remainder Theorem, and other topics in Number Theory.  The second semester for this group will concentrate on geometry -- triangles, similar shapes, circles, etc.  Because we did not have topical focus last year, this class is recommended for most of the returning students from group I.  The ages are 6th - 12th grade, no exceptions.

The Intermediate group will feature topics from various areas of mathematics; extreme familiarity with mathematical induction, pigeonhole principle, and other mathematical techniques is recommended.  Certain areas, such as trigonometry and calculus, will not be included.  The ages are 6th - 10th grade, though we may admit some older students upon request.

The Advanced group will feature what the Intermediate group will feature and more.  People who on a good day will be qualifying for the USAMO belong to this group.  The ages here are 9th -- 12th grade, though we may admit some younger students upon request.

To apply, please fill out the 3 forms attached; Email the "Application 2008 - 2009", and bring the other 2 forms, as well as the check made out to "Pushkin Academy of Russian Heritage", in the first week of classes (the checks are due the second week, but the sooner I can be done with the paperwork, the happier I will be -- I promise not to cash the checks till after week 2).  The classes will be at the same time and location and will start on Saturday, September 27th.  The details are in the "Application 2008 - 2009".

Consent form: In order for you to participate in the program, we need
to ascertain that parents / legal guardians are aware that kids
are attending the program.  Therefore, we need one of the following to
take place:

A) A parent notarizes the signature on the parent consent form              OR
B) A parent provides a copy of their picture ID with the signature
matching the one on the parent consent form   OR
C) A parent provides payment by check with signature matching the one
on the parent consent form     OR
D) A parent arrives with you to the first meeting of the group.
Sorry for the inconvenience, but I am sure that at least one of these
is not going to be too difficult for anyone involved.

For reference, here is our website (still being changed from last year as I am sending this Email): http://www.pushkinacademy.com/en/pamm -- and here is the calendar: http://www.pushkinacademy.com/en/calendar20082009.

Please let your friends and math teachers know about Math-M-Addicts -- the more people find out about the program, the larger we can make the classes, which I think will result in more diversity and more fun among students.

Should you have any questions, please don't hesitate to ask.

Good luck and see you soon!
Ruvim

The program sounds great, and I have a few questions.

How many weeks will the class last?  What topics do each of the classes cover?  If we take the Advanced class this year, will there be a suitable course next year?

Thanks to all.

Alex, I know you have already joined the program, but for the benefit of other prospective students, I will try to answer your questions.  (Note: I'm only a parent, not an instructor of the class.)

The class runs almost every Saturday morning from the end of September to June.  The topics vary depending on which group you are in, but are chosen from the usual subjects that arise in math competitions: Algebra, Geometry, Number Theory, and Combinatorics.

Even though you are taking the advanced class, I don't think you will run out of topics anytime soon.  Math is infinite, and so is apparently the number of things the instructors know about.  Some students have been in the program now for 3 years, and are still learning new things.


[size=150][b]Give a method and mechanism of dehydration of alcohol without rearrangement.[/b][/size]

Hmmmm... I hope this helps:

To dehydrate an alcohol, you will need to use mineral wool to soak up the alcohol. Then, as some aluminium oxide catalyst in the form of pellets in a boiling tube along with the soaked mineral wool.

Using a bunsen burner, warm up the wool (make sure that alcohol does not catch fire) and then continue to heat the aluminium oxide. The gas needs to be collected in a downward displacement apparatus. The first test tube taken should be thrown away, since it will contain only air, the next test tubes will contain the alkene that you are probably looking for.

Hope this helps!

use TsCl to convert -OH to -OTs
Then use any ordinary base to eliminate the tosylate to get alkene without rearrangement.


function
real analysis
absolute value
number theory theorems
number theory

Prove (using elementary techniques) that the Dirichlet convolution inverse of the Liouville function is the absolute value of the Mobius function. Symbolically: $\lambda*|\mu|=\delta_{1}$.

$\lambda$ and $|\mu|$ are multiplicative, and therefore so is their convolution. So, we need only show that $\lambda\ast|\mu|$ takes the value zero at any prime power $n=p^{a}$ (where $a>0$).
\[(\lambda\ast|\mu|)(p^{a})=\lambda(n)|\mu(1)|+\lambda(n/p)|\mu(p)|+\text{ terms equal to 0}\]
\[=\lambda(n)+\lambda(n/p)\]
\[=(-1)^{a}+(-1)^{a-1}= 0\]


[quote="arqady"][quote="DucHuynh"]Given $\left(c^{2}+d^{2}\right) = \left(a^{2}+b^{2}\right)^{3}$, where $a,b,c,d>0$,

prove that $\frac{a^{3}}{c}+\frac{b^{3}}{d}\geq 1$.[/quote]
$\frac{a^{3}}{c}+\frac{b^{3}}{d}=\frac{a^{4}}{ac}+\frac{b^{4}}{bd}\geq\frac{(a^{2}+b^{2})^{2}}{ac+bd}\geq\frac{(a^{2}+b^{2})^{2}}{\sqrt{(a^{2}+b^{2})(c^{2}+d^{2})}}=1.$ :)[/quote]

Sorry to ask, but how is the first inequality accomplished?

[quote="arqady"][quote="DucHuynh"]
Sorry to ask, but how is the first inequality accomplished?[/quote]
It's cauchy-schwarz's inequality:
$\frac{a^{4}}{ac}+\frac{b^{4}}{bd}\geq\frac{(a^{2}+b^{2})^{2}}{ac+bd}\Leftrightarrow(ac+bd)\left(\frac{a^{4}}{ac}+\frac{b^{4}}{bd}\right)\geq(a^{2}+b^{2})^{2}\Leftrightarrow$
$\Leftrightarrow\left(\left(\sqrt{ac}\right)^{2}+\left(\sqrt{bd}\right)^{2}\right)\left(\left(\frac{a^{2}}{\sqrt{ac}}\right)^{2}+\left(\frac{b^{2}}{\sqrt{bd}}\right)^{2}\right)\geq(a^{2}+b^{2})^{2}.$ :wink:[/quote]

Thanks. I'm very much new to inequalities. :-)


[quote="littletush"]let us suppose that f is non-constant($f(x)\equiv 0 $is a trivial solution)let $x=0$,$f(0)+f(f(y)+f(0))=yf(0)+f(f(y))$
let $y=0,f(xf(0))+f(f(x)+f(0))=f(x+f(0))$
it's trivial that f is injective.
let $y=1$,then $f(f(1)+f(0))=f(f(1))$,hence $f(0)=0$
so $f(f(x))=f(x)$ hence $f(x)=x$.[/quote]
Very interesting brand new solution !
Congrats !

is this solution ok?
Let $P(x,y)$ be assertion of given equation.We can easily prove that the function is an injection(like in the post before).
$P(x,x)$ ->$f(xf(x))+f(2f(x))=xf(x)+f(x+f(x))$
$P(f(x),x)$ ->$f(f(x)x)+f(2x)=f(x)x+f(f(x)+x)=f(xf(x))+f(2f(x))$ which implies $f(2x)=f(2f(x))$ so since the function is an injection we get $f(x)=x$ which is solution.
Hence the only solutions are $f(x)=0,f(x)=x$

[quote=littletush]let us suppose that f is non-constant($f(x)\equiv 0 $is a trivial solution)let $x=0$,$f(0)+f(f(y)+f(0))=yf(0)+f(f(y))$
let $y=0,f(xf(0))+f(f(x)+f(0))=f(x+f(0))$
it's trivial that f is injective.
let $y=1$,then $f(f(1)+f(0))=f(f(1))$,hence $f(0)=0$
so $f(f(x))=f(x)$ hence $f(x)=x$.[/quote]

How did u get that $f$ is injective?

Let $P(x,y)$ be the assertion of the given equation. Suppose that $f(a)=f(b)$.  Then, $P(x,a)$ and $P(x,b)$ gives $af(x)=f(xf(a))+f(f(x)+f(a))-f(x+f(a))=f(xf(b))+f(f(x)+f(b))-f(x+f(b))=bf(x)$.  We have $a=b$ if there is some value of $x$ such that $f(x)\neq 0$.  If an value of $x$ doesn't exist, then we have $\boxed{f(x)=0}$ as a solution, which indeed works.  If an value of $x$ does exist, now we have $a=b$ or $f$ is injective.  Let $P(0,1)$ so that $$f(0)+f(f(0)+f(1))=f(0)+f(f(1))\implies f(f(0)+f(1))=f(f(1)).$$ Since $f$ is injective, we have $f(0)+f(1)=f(1)\implies f(0)=0.$  Now, let $P(x,0)$ so $$f(f(x))=f(x)$$ and because $f$ is injective, we have $\boxed{f(x)=x}$, which is another solution. $\blacksquare$

[quote=pco][quote="FelixD"]Find all functions $ f: \mathbb{R} \to \mathbb{R}$, such that
\[ f(xf(y)) \plus{} f(f(x) \plus{} f(y)) \equal{} yf(x) \plus{} f(x \plus{} f(y))\]
holds for all $ x$, $ y \in \mathbb{R}$, where $ \mathbb{R}$ denotes the set of real numbers.[/quote]

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

$ f(x)\equal{}0$ $ \forall x$ is a solution. Let us from now consider $ \exists c$ such that $ f(c)\neq 0$

If $ f(y_1)\equal{}f(y_2)$, subtracting $ P(c,y_1)$ from $ P(c,y_2)$ implies $ 0\equal{}f(c)(y_2\minus{}y_1)$ and so $ y_1\equal{}y_2$ and $ f(x)$ is injective.

$ P(0,1)$ $ \implies$ $ a\plus{}f(a\plus{}f(1))\equal{}a\plus{}f(f(1))$, so $ f(a\plus{}f(1))\equal{}f(f(1))$ and so, since $ f(x)$ is injective, $ a\equal{}0$

Then $ P(x,0)$ $ \implies$ $ f(f(x))\equal{}f(x)$ and, since $ f(x)$ is injective, $ f(x)\equal{}x$

Hence two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$[/quote]

Could you explain some motivation,for such problems. this part (Let us from now consider $ \exists c$ such that $ f(c)\neq 0$)

[quote=pco][quote=itslumi]Could you explain some motivation,for such problems. this part (Let us from now consider $ \exists c$ such that $ f(c)\neq 0$)[/quote]
I claimed that $f(x)\equiv 0$ is a solution
So from then, I search only non allzero solutions. So it's normal to consider that $\exists c$ such that $f(c)\ne 0$[/quote]

Yes i undertand but how to now intuitivly as a costant to try this method at a right time and at a right problem

I dont know what to explain exactly ...
I wanted to test / prove injectivity and comparing $P(x,y_1)$ with $P(x,y_2)$, it was obvious that finding an $x$ such that $f(x)\ne 0$ was enough to conclude injectivity.

Hence the process :
If $f(x)=0$ $\forall x$, we have a solution (non injective :) ).
If $f\not\equiv 0$, then using an $x$ such that $f(x)\ne 0$ allows to prove injectivity. Hence my choice about $c$.

If $f\not\equiv 0$, it's clear that $f$ is injective.

If we set $x=y=0$, we find that $f(2f(0))=0$. Then setting $x=y=2f(0)$ gives $f(0)=0$.

Now by putting $y=0$, we find that $f(f(x))=f(x)$, from which we conclude $f(x)=x$ by injectivity.

So the only functions are $f(x)=0$ and $f(x)=x$, which clearly work.


limit
inequalities
triangle inequality
real analysis
real analysis unsolved

this is an interesting theorem, but i just don't get the last line of the pf. Help pls!

 :(

Suppose $ \sum{c_n}$ converges. Put  f(x) = $ \sum^{\infty}_{n\equal{}0} {c_n x^n}$ (-1<x<1)
Then   $ \lim_{x\minus{}>1} {f(x)} \equal{} \sum^{\infty}_{n\equal{}0} {c_n}$

pf : let $ s_n\equal{} c_0 \plus{} ...\plus{}c_n$, then,

$ \sum^{m}_{n\equal{}0} {c_n x^n}\equal{} \sum^{m}_{n\equal{}0} (s_n \minus{}s_{n\minus{}1})x^n \equal{} (1\minus{}x)\sum^{m\minus{}1}_{n\equal{}0} {s_n x^n} \plus{}s_m x^m$
For |x| <1, we let m->$ \infty$ and obtain

(9) f(x) = $ (1\minus{}x)\sum^{\infty}_{n\equal{}0} {s_n x^n}$

Suppose s = $ \lim_{n\minus{}>0} {s_n}$. Let $ \epsilon > 0$ be given. Choss N so that n > N implies |s - $ s_n$| <$ \epsilon /2$

Then since (1-x)$ \sum^{\infty}_{n\equal{}0}{x^n} \equal{} 1$ (|x| <1)
we obtain from (9)

$ |f(x) \minus{} s| \equal{} | (1\minus{}x) \sum^{\infty}_{n\equal{}0} {(s_n \minus{}s) x^n} | \leq{\epsilon}$ if $ x >1\minus{}\delta$,
 for some suitably chosen $ \delta$ >0.
end of pf.

I understand why $ |f(x) \minus{} s| \equal{} | (1\minus{}x) \sum^{\infty}_{n\equal{}0} {(s_n \minus{}s) x^n} |$
but I don't get why $ | (1\minus{}x) \sum^{\infty}_{n\equal{}0} {(s_n \minus{}s) x^n} | \leq{\epsilon}$ if $ x >1\minus{}\delta$,

It looks like it's just using the triangle inequality to split up the sum so that you get two sums that are both $ \leq\varepsilon/2$, so when you add them, you get $ \varepsilon$. Typically, when one uses a $ \varepsilon/n$ trick, the end goal is to split up the desired expression into $ n$ parts by the triangle inequality.  Then since each individual part is $ <\varepsilon/n$, you end up with $ n\times\varepsilon/n\equal{}\varepsilon$.

The proof skips one step: you split the sum into $ \sum_0^N \plus{} \sum_{N \plus{} 1}^\infty$. The second sum is less than $ \varepsilon/2$ (can you see why?). To handle the first one, notice that it doesn't exceed $ (1 \minus{} x)\sum_0^N|s\minus{}s_n|\le \delta\sum_0^N|s\minus{}s_n| < \varepsilon/2$ if $ \delta > 0$ is small enough.


If $ xyz \equal{} 1$ and $ x,y,z > 0$ then $ x^{x \plus{} y \plus{} z} \plus{} y^{x \plus{} y \plus{} z} \plus{} z^{x \plus{} y \plus{} z} \ge 3$.

By AM_GM:

$ LHS\geq x^3\plus{}y^3\plus{}z^3\geq3$

CLAIM: if $ x,y,z>0$ $ xyz\equal{}1$, $ a \ge b$, then $ \sum x^{a} \ge \sum x^{b}$. (this is easily proven by AM-GM)

$ \Rightarrow  x^{x\plus{}y\plus{}z}\plus{}y^{x\plus{}y\plus{}z}\plus{}z^{x\plus{}y\plus{}z}\ge x^{3}\plus{}y^{3}\plus{}z^{3} \ge 3$  :D

hello, by the AM-GM inequality we get
$ \frac{x^{x\plus{}y\plus{}z}\plus{}y^{x\plus{}y\plus{}z}\plus{}z^{x\plus{}y\plus{}z}}{3}\geq\sqrt[3]{(xyz)^{x\plus{}y\plus{}z}}\equal{}1$.
Sonnhard.


If sqrt (10-2(sqrt(24)))=sqrt(x)-sqrt(y),
what is x?
8
12
6
4
none of the above

[hide="tentative solution"]
square both sides.
we get
$ 10 \minus{} 2 \sqrt{24} \equal{} x \plus{} y \minus{}2\sqrt{xy}$
since the values given are integral(rational), we can compare the rational and irrational parts:
$ \equal{}> x \plus{} y \equal{} 10$ and $ xy \equal{} 24$
2 solutions exist: x=6, y=4 and x=4, y=6
x=4 and y=6 can't be possible as the RHS in the question will become negative.

Hence,  $ x \equal{} 6$

[/hide]


In quadrangle ABCD,$ A \plus{} C \equal{} 90^0$

PROVE:$ (AB*CD)^2 \plus{} (AD*BC)^2 \equal{} (AC*BD)^2$

Inversion with center $ D$ and arbitrary power $ r^2$ takes $ A, B, C$ into $ A', B', C'$ and

$ B'C' \equal{} BC \cdot \frac {r^2}{DB \cdot DC},\ C'A' \equal{} CA \cdot \frac {r^2}{DC \cdot DA},\ A'B' \equal{} AB \cdot \frac {r^2}{DA \cdot DB}$

$ \angle A'B'C' \equal{} \angle A'B'D \plus{} \angle DB'C' \equal{} \angle DAB \plus{} \angle BCD \equal{} 90^\circ$ $ \Longleftrightarrow$

$ A'B'^2 \plus{} B'C'^2 \equal{} C'A'^2$ $ \Longleftrightarrow$

$ \frac {AB^2}{DA^2 \cdot DB^2} \plus{} \frac {BC^2}{DB^2 \cdot DC^2} \equal{} \frac {CA^2}{DC^2 \cdot DA^2}\ \Longleftrightarrow$

$ AB^2 \cdot DC^2 \plus{} BC^2 \cdot DA^2 \equal{} CA^2 \cdot DB^2.$

Denote $ AB\equal{}a$ , $ BC\equal{}b$ , $ CD\equal{}c$ , $ DA\equal{}d$ , $ AC\equal{}e$ , $ BD\equal{}f$ $ \implies$ ([u]generalized [b]Ptolemeu[/b]'s relation[/u])

$ \boxed {\ e^2f^2\equal{}a^2c^2\plus{}b^2d^2\minus{}2abcd\cos (A\plus{}C)\ }$ (well-known !). For $ A\plus{}C\equal{}90^{\circ}$ obtain the required relation.

Dear Mathlinkers,
in: Sharygin I. F., Problemas de geometria, Editions Mir, Moscou, 1986, problem II. 236 p. 117-118
you can found this theorem called under the name of Bretschneider, the proof p.328-329.
This book is in Spain...I think that the same book in English is also available.
Sincerely
jean-LOuis

[quote][color=darkred][u][b]Bretschneider[/b]'s relation[/u] ([u]generalized [b]Ptolemeu[/b]'s relation[/u]) $ \boxed {\ e^2f^2 \equal{} a^2c^2 \plus{} b^2d^2 \minus{} 2abcd\cos (A \plus{} C)\ }$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] Construct the point $ E$ for which the line $ CD$ separates the points $ A$ , $ E$ and $ \triangle ABC\sim\triangle EDC$ . Therefore,

$ \frac {a}{ED} \equal{} \frac bc\ \implies$ $ \boxed {\ ED \equal{} \frac {ac}{b}\ }$ . Observe that $ \triangle BCD\sim\triangle ACE$ from where obtain $ \frac be \equal{} \frac {f}{AE}$ , i.e. $ \boxed {\ AE \equal{} \frac {ef}{b}\ }$ .

Apply now the [u]generalized [b]Pitagora[/b]'s theorem[/u] in $ \triangle ADE\ : \ AE^2 \equal{} DA^2 \plus{} DE^2 \minus{} 2\cdot DA\cdot CE\cdot \cos\widehat {ADE}\ \implies$

$ \frac {e^2f^2}{b^2} \equal{} d^2 \plus{} \frac {a^2c^2}{b^2} \minus{} 2d\cdot\frac {ac}{b}\cdot$ $ \cos (B \plus{} D)\ \implies\ e^2f^2 \equal{} a^2c^2 \plus{} b^2d^2 \minus{} 2abcd\cdot\cos (A \plus{} C)$ .[/color]

[quote="yetti"]Inversion with center $ D$ and arbitrary power $ r^2$ takes $ A, B, C$ into $ A', B', C'$ and

$ B'C' \equal{} BC \cdot \frac {r^2}{DB \cdot DC},\ C'A' \equal{} CA \cdot \frac {r^2}{DC \cdot DA},\ A'B' \equal{} AB \cdot \frac {r^2}{DA \cdot DB}$

$ \angle A'B'C' \equal{} \angle A'B'D \plus{} \angle DB'C' \equal{} \angle DAB \plus{} \angle BCD$ ...[/quote]

$ A'B'^2 \plus{} B'C'^2 \minus{} 2 \cdot A'B' \cdot B'C' \cos \widehat{A'B'C'} \equal{} C'A'^2$ $ \Longleftrightarrow$

$ \frac {AB^2}{DA^2 \cdot DB^2} \plus{} \frac {BC^2}{DB^2 \cdot DC^2} \minus{} 2 \cdot \frac {AB \cdot BC \cdot \cos (\widehat{DAB} \plus{} \widehat{BCD})}{DA\cdot DB^2 \cdot DC} \equal{} \frac {CA^2}{DC^2 \cdot DA^2}\ \Longleftrightarrow$

$ AB^2 \cdot DC^2 \plus{} BC^2 \cdot DA^2 \minus{} 2 \cdot AB \cdot BC \cdot DC \cdot DA \cdot \cos (\widehat{DAB} \plus{} \widehat{BCD}) \equal{} CA^2 \cdot DB^2.$


Let quadrangle ABCD be inscribed in a circle with center O, radius $ R\equal{}\sqrt{5}$. Two diagonals of it are perpendicular each other at I, $ OI\equal{}1$. Denote $ S$ be the area of the triangle ICD. Prove that
\[ 1\le S\le 4
\]

Call $ M,N$ are midpoints of $ AC, BD$ respectively and denote $ S_1,S_2,S_3,S_4$ for $ S_{AIB},S_{BIC},S_{CID},S_{DIA}$. We have:
-) $ S_1.S_3\equal{}S_2.S_4\equal{}\frac{IA.IB.IC.ID}{4}\equal{}\frac{(IA.IC)^2}{4}\equal{}\frac{(R^2\minus{}OI^2)^2}{4}\equal{}4$
-) $ AC^2\plus{}BD^2\equal{}4(AM^2\plus{}BN^2)\equal{}4(R^2\minus{}OM^2\plus{}R^2\minus{}ON^2)\equal{}4(2R^2\minus{}OI^2)\equal{}36$
Then $ S_1\plus{}S_2\plus{}S_3\plus{}S_4\equal{}S_{ABCD}\equal{}\frac{AC.BD}{2}\leq \frac{AC^2\plus{}BD^2}{4}\equal{}9$
But $ S_2\plus{}S_4\geq 2\sqrt{S_2.S_4}\equal{}4$ then $ S_1\plus{}S_3\leq 5$
It mean $ \frac{4}{S_3}\plus{}S_3\leq 5\Leftrightarrow S_3^2\minus{}5S_3\plus{}4\leq 0\Leftrightarrow 1\leq S_3\leq 4$


If $ \frac{1}{4}$ of $ 2^{30}$ is $ 4^x$, then what is the value of $ x$ ?

$ \frac {1} {4}$ of $ 2^{30}$ would be $ 2^{28}$ which i just know to be true, so would someone explain why

therefore $ 2^{28} \equal{} 4^{x}$ since four is twice the value of two, then the exponential value has to be half of the original one
therefore $ 2^{28} \equal{} 4^{14}$

[quote="vallon22"]$ \frac {1} {4}$ of $ 2^{30}$ would be $ 2^{28}$ which i just know to be true, so would someone explain why[/quote]

$ \frac {1}{4}$ is the same thing as $ 2^{ \minus{} 2}$, which, when multiplied by $ 2^{30}$, yields $ 2^{30 \minus{} 2}$, and that is the same as $ 2^{28}$


superior algebra
superior algebra unsolved

I see the sentence in a book:` The flop is a simple algebraic manipulation'. what is `flop' ? thanks

Try this link
[url]http://books.google.com/books?id=7GDyRMrlgDsC&pg=PA166&lpg=PA166&dq=flop+algebraic+manipulation&source=bl&ots=2e9D_BS8wT&sig=Ytr84sXwhsqPim5xpGL3mHOdGFg&hl=en&ei=MAe2SrblF4KTsAb9wPTuDA&sa=X&oi=book_result&ct=result&resnum=5#v=onepage&q=flop%20algebraic%20manipulation&f=false[/url]

or try looking on Google (that's what I did, using "flop algebraic manipulation" keywords).

this is the book I am just reading .thanks


Suppose that $ \sin a \plus{} \sin b \equal{} \sqrt {\frac {5}{3}}$ and $ \cos a \plus{} \cos b \equal{} 1.$ What is $ \cos(a \minus{} b)?$

$ \textbf{(A)}\ \sqrt {\frac {5}{3}} \minus{} 1 \qquad \textbf{(B)}\ \frac {1}{3}\qquad \textbf{(C)}\ \frac {1}{2}\qquad \textbf{(D)}\ \frac {2}{3}\qquad \textbf{(E)}\ 1$

[hide]
If we square both equations and add them, we get

$\sin^{2}a+2\sin a\sin b+\sin^{2}b+\cos^{2}a+2\cos a\cos b+\cos^{2}b=\frac{8}{3}$.

Using the facts that $\sin^{2}x+\cos^{2}x=1$ and $\cos (a-b)=\cos a\cos b+\sin a\sin b$, we find

$2+2\cos (a-b)=\frac{8}{3}\Rightarrow \cos (a-b)=\frac{1}{3}$.

The answer is [b]B[/b].
[/hide]

Not that I did this, but if you are [b]really[/b] lazy, or don't know how to do this, you can simply type the system into a TI-89 and it will give the correct answer. And this is a #17 on the AMC12, which is kind of sad...

That's mathematica, and you KNOW that's cheating alex  :P

Seriously, this felt way too easy for a number 17.

uhhh...no

if you take out the input, output, it is TI-89 syntax

Aaaah sorry!
I don't have one of those, and that looked suspiciously like Mathematica, which we all know is not allowed on the AMCs.

So sorry for assuming that.

[quote="JavaMan"]Are ti-89s allowed on the AMC 12?[/quote]
I would assume so, since they are accepted for use on the SAT.

[quote="Altheman"]input: solve(cos(x)+cos(y)=(5/3)^(1/2) and sin(x)+sin(y)=1 and cos(x-y)=z,z)
output:z=1/3

SNAP[/quote]

Hey, that looks just like how my calculator looked after I did this problem...

i got a nice solution... but i dun kno how to post  :blush:

im kinda new to this

LaTeX code is a way to make nice looking math. You can learn LaTeX code in the LaTeX page on the navigation bar to your left. Enclose the code between two dollar signs, like
[code]$\frac{1}{2}$[/code]
becomes $\frac{1}{2}$.

Also use hide tags, so to give other people a chance to do a problem without seeing the solutions.

[quote="Altheman"]input: solve(cos(x)+cos(y)=(5/3)^(1/2) and sin(x)+sin(y)=1 and cos(x-y)=z,z)
output:z=1/3

SNAP[/quote]

I typed that in and my the output was the same as my input. Am I on a wrong setting?

[quote="Altheman"]either estimate, or do

,{x,y,z}) instead of ,x)[/quote]

{x,y,z} did the trick. By estimate, you mean the approx() function, right?

i did not guess, but i had a hard time with this problem :maybe:


Do anybody have a list of high school honor societies that look good on college apps? Stuff like national honor society or something like that.

I know that National Honor Society looks good on college apps, and probably Mu Alpha Theta, but you have to be careful about other honor societies, because several of them are basically vanity publishers.

I'm pretty sure getting into a "good" (HYPMS) college has absolutely nothing to do with the number of honor societies you're in. From what I know, a vast majority of upper-tier students get involved in NHS. But there's no need to go hunting for clubs that look good on a college app. Do MAO, do NHS, StuCo, Orch, volunteering (Amnesty Intl maybe?), MUN, Debate, FBLA, varsity sports, whatever you're interested in...and stick to it.

[quote="Xevarion"]At my school many of the top students considered NHS to be a joke (which it is). All you need is halfway-decent grades and to get "points" through various means (like community service). But your points form was filled out by another student at the school, so there was rampant corruption. So a lot of us didn't join in a weak form of protest/expression of disdain.[/quote]

Yeah, same here =P In my school NHS and MAO are jokes, and I'm very suspicious of other honor societies being the same. And yeah, I could have applied to NHS but chose not to (somewhat out of apathy; somewhat out of principle), and I was in MAO for its one and only year (soon died as people found out that they were actually supposed to be doing math).

The reason I say they're jokes is because not only of what Xevarion said -- the unimpressive prerequisites -- but also because a (rather depressingly vast) majority of people in honor societies don't actually do academics in their spare time, or moreover, they don't do it because they love it -- they only do it because it's good for colleges and because their parents and society expect them to. For example, in MAO, there were only two kids who actually did math in their spare time (my friend and me).


probability
combinatorics unsolved
combinatorics

There is a biased coin, with 2/3 probability that it will land on heads.
What is the probability of an even number of heads in 50 throws of the coin.

(I think there is a simple method for approaching it, but i think im over-complicating it)

$ Pr \equal{} \sum_{i \equal{} 0}^{25} \frac {1}{3}^{2i} \frac {2}{3}^{50 \minus{} 2i}\binom{50}{2i}\equal{} \frac{(\frac {1}{3} \plus{} \frac {2}{3})^{50} \plus{} (\frac {1}{3} \minus{} \frac {2}{3})^{50}}{2} \equal{} \frac {1}{2} \plus{} \frac {1}{2 *3^{50}}$

I just summed over all the options (2i heads and 50-2i tail, and we can choose any 2i coins to be heads). I used the identity $ \sum_{i \equal{} 0}^{n} x^{2i}y^{2n \minus{} 2i}\binom{2n}{2i} \equal{} \frac {(x \plus{} y)^{2n} \plus{} (x \minus{} y)^{2n}}{2}$

The general formula for 2n flips is therefore $ \frac {1}{2} \plus{} \frac {1}{2 *3^{2n}}$. It works for n=1 ($ \frac {5}{9} \equal{} \frac {1}{3}*\frac {1}{3} \plus{} \frac {2}{3}*\frac {2}{3}$), so I guess it is right.


USAMTS
combinatorics unsolved
combinatorics

Determine the smallest number of squares into which one can disect a 11 by 13 rectangle. Squares can be of different size but should be integer and no overlap.

I think it is 6 with sides 1,4,4,5,6,7 the squares with sides 4,5,6,7 are in the corners of the rectangle.

This is from the USAMTS, right?  (As have been some of your other recent posts.)  What year/problem?


If (a^2).b + (b^2).c + (c^2).a = a.(b^2) + b.(c^2) + c(a^2), prove that at least two of a,b and c are equal.

Can a,b and c be the sides of triangle and satisfy this equation?

take all summands to one side and then it can be factored as
$(a-b)(a-c)(b-c)=0$ and the conclusion follows!

Yes, it does indeed factor to $(a-b)(b-c)(a-c) = 0$.
If $a,b,c$ are sides of a triangle, then $a = x+ y$, $b= x + z$ and $c = y + z$ where $x,y,z$ are the distances from vertices to tangency points of the incircle. Plugging them in we get:

$((x+y)-(x+z))((x+z)-(y+z))((x+y)-(y+z)) = (y-z)(x-y)(x-z) = 0$

Thus, we know $y=z$, $x=y$ or $x=z$ which tells us the triangle is at least isosceles.
**I just realized we could tell that without adding anything to socrates' answer other than $a=b$ implies isosceles or more.


guys please help me and give an efficient way to solve this problem

150. Where to Drink the Plonk?
Problem code: PLONK

Consider a city bounded by a square, whose n horizontal and n vertical streets divide it into (n+1)2 square blocks. However, in tribute to the ancient traditions of the first dwellers (who tended to overindulge in alcohol), all the inhabitants live at crossroads. A group of friends would like to meet for an evening of merriment at the place of residence of one of them. With a good deal of foresight, anticipating the difficulties they might have getting back to their respective homes, they would like to meet in the house of the friend which minimises the total walking distance for all of them. Assume that everybody walks along the streets, turning only at crossroads, and the distance between any pair of adjacent crossroads is 1.

Input
The input begins with the integer t, the number of test cases. Then t test cases follow.

For each test case the first line of input contains the integer n - the number of friends who want to meet (1<=n<=10000). The next n lines contain two integers each, the i-th being xi yi, standing for the x and y coordinates of the crossroads at which the i-th friend lives (0<=xi,yi<=100000).

Output
For each test case output the total distance covered by all friends when walking to the meeting place.

Example
Sample input:
1
7
1 3
3 2
3 5
6 9
10 1
12 4
5 7

Sample output:
39

i cant come up with any idea other than brute forcing over all points and finding their distances and suming them up...
one more way could be to sort the points according to their x coordinates...find the distances for the x direction for each point... and then compute the minimum y distances....and choose the overall minima....
what is the best way to solve this problemm???

best regards

Sort the houses in order of x coordinate.

Step through the houses in order, storing how many houses have x coordinate <= current one, and how many have x coordinate > current one. That only takes linear time, since they're sorted.

Calculate total x distance if you choose the first house. Then move through the houses in order - moving from an x coordinate of x1 to an x coordinate of x2 adds a total of (x2-x1)*(num <= current house - num > current house) to what you have so far. That only takes linear time.

Do the same for the ys, and choose the house with minimal x and y total.


Carachterise all integer solutions of the dyophantine equation \[ 5x^2\plus{}4\equal{}y^2\]

Good work :wink:

Who know how to solve this equation ？
\[ x^2 \plus{} 4 \equal{}5 y^4
\]  :)

First equation remains me Bulgarian Math Olympiad IV round problem ... 1979

[quote="salva90"]Carachterise all integer solutions of the dyophantine equation
\[ 5x^2 \plus{} 4 \equal{} y^2
\]
Good work :wink:[/quote]

It's Pell's equation.

[quote="shfdfzhjj"]Who know how to solve this equation ？
\[ x^2 \plus{} 4 \equal{} 5 y^4
\]
:)[/quote]

It's Pell's equation too in the shape of $ x^2 \minus{} 5(y^2)^2 \equal{} \minus{} 4$

http://www.imomath.com/tekstkut/pelleqn_ddj.pdf

[quote="Ramanujan47"][quote="shfdfzhjj"]uot;]Who know how to solve this equation ？
\[ x^2 \plus{} 4 \equal{} 5 y^4
\]
:)[/quote]

It's Pell's equation too in the shape of $ x^2 \minus{} 5(y^2)^2 \equal{} \minus{} 4$

http://www.imomath.com/tekstkut/pelleqn_ddj.pdf[/quote]

Yes,I know.and then? :maybe:

[quote="salva90"]Carachterise all integer solutions of the dyophantine equation
\[ 5x^2 \plus{} 4 \equal{} y^2
\]
Good work :wink:[/quote]
I have an ideal$ 5x^2 \plus{} 4 \equal{} y^2<\equal{}>5x^2\equal{}(y\minus{}2)(y\plus{}2)$
Hence $ y\minus{}2\equal{}5a^2,y\plus{}2\equal{}b^2$or$ y\minus{}2\equal{}a^2,y\plus{}2\equal{}5b^2$ there $ ab\equal{}x$
The last is solving two system

(1,3)is one of the solution, and

$ (20x\plus{}9y)^2 \minus{}5(9x\plus{}4y)^2 \equal{}y^2\minus{}5x^2$ must be useful!

[quote="shfdfzhjj"]Who know how to solve this equation ？
\[ x^2 \plus{} 4 \equal{} 5 y^4
\]
:)[/quote]

Who has idea about this? :)

The solutions $ (y, x)$ to the original problem are given by $ y_n \plus{} x_n \sqrt{5} \equal{} (3 \plus{} sqrt{5})(9 \plus{} 4 \sqrt{5})^n, n \in \mathbb{Z}$.  This is a standard application of the Pell's technique.


function
number theory proposed
number theory

Hello,

The problem is to find all the positive ingeters $x$ and $y$ such that
\[x^{x+y}=y^{y-x}\]

I haven't solve it yet,  but it doesn't seem to be that hard.

Best

First note that if $x=y$, then $x^{2x}=1$ so $x=1$ and we have a solution.

Take $(x,y)=d$, so that $x=dp, y=dq, (p,q)=1$.

Then we have
\[(dp)^{dp+dq}=(dq)^{dq-dp}\]
Since both exponents are multiples of d you can take dth roots and get
\[(dp)^{p+q}=(dq)^{q-p} \\
d^{2p}p^{p+q}=q^{q-p}\]
Then since $p$ and $q$ are coprime, and the equation implies that $p$ divides $q^{q-p}$, we must have $p=1$. Then
\[d^2=q^{q-1}, d=q^{ \frac {q-1}{2}}\]
So since p=1, x=d and y=dq, so
\[x=q^{ \frac {q-1}{2}}, y=q^{ \frac {q+1}{2}}, \] for any q which is either odd or a perfect square. It's easy to check that such $x, y$ satisfy the conditions.

[quote="djimenez"] find all positive integers $x$ and $y$ such that \[x^{x+y}=y^{y-x}\] [/quote]
[hide="The method"] (to let $k = x/y$ and parametrize all rational solutions as a function of $k$)[/hide] is the same as for solving $x^y = y^x$.


The algebraic sum of the perpendicular distances from [b]A(x1,y1),B(x2,y2) and C(x3,y3)[/b] to a variable line is zero, then the lines passes through
[b](a) the orthocentre of triangle ABC
(b) the centroid of triangle ABC
(c) the circumcentre of triangle ABC
(d)none of these.[/b]

Its answer is [b](b)[/b]. [u]And I got its solution[/u]. [u]But the problem is that [/u]I want to know how can the sum of some distances whether perpendicular distances or non-perpendicular distances is zero?If the sum of some distances is zero then all the distances must be zero. Am I wrong?And if distance between two points is zero then both the points are same. I am very much confused.I know that my problem may seem to you silly,but that's my problem. I need your help. Thanks a lot.

This might be wrong; I'm not sure.

Instead of considering distances, consider drawing vectors from the centroid of the triangle perpendicular to the three sides. If you add the three vectors, you get the zero vector.


The first inequality follows from the fact that each of the terms is greater than $\frac{x_1}{x_1+...+x_n}$.

The second is a consequence of the first since if we add the expression $\sum \frac{x_i}{x_i+x_{i-1}}$ we get $n$.

@Armo: Our posts are sounding eerily similar...

Oh, I am very sorry [b]Fiachra[/b] .I just haven't seen the second part of your proof. :blush:  :blush:  :blush:

[quote="Armo"]Now the second side. It is equvalent to $(1-\frac{x_{1}}{x_{1}+x_{2}})+
(1-\frac{x_{2}}{x_{2}+x_{3}})+...+(1-\frac{x_{n-1}}{x_{n-1}+x_{n}})+(1-\frac{x_{n}}{x_{n}+x_{1}})>1$ or $\frac{x_{2}}{x_{1}+x_{2}}+\frac{x_{3}}{x_{2}+x_{3}}+...+\frac{x_{n}}{x_{n-1}+x_{n}}+\frac{x_{1}}{x_{n}+x_{1}}>1.$
And this is the same as the first part.
What is Shapiro?[/quote]

Thank you for your reply,Armo.

[b]Shapiro's cyclic inequality[/b]!

kunny

For relevant informations about Shapiro's inequality, you may look at :
http://www.mathsoft.com/article/0,,2144,00.html

Pierre.

Shapiro's inequality is the statement that the
  cyclic sum over a1...an , of  a1/(a2+a3),
is at least  n/2.  It turns out to be very subtle -
true for small n, false for large n, and with some
interesting work on the optimal constant replacing
1/2 done by Drinfeld (one of 2-3 top living mathematicians)
when he was around 15 years old.


limit
real analysis
real analysis unsolved

Find the limit:

$\lim_{x\rightarrow \infty}\left(\zeta(x+1)\right)^{2^{x}}$,

where $\zeta$ is the famous Riemann-Zeta function.

As $x\to\infty,\ \zeta(x+1)=1+\frac1{2^{x+1}}+O(3^{-x}).$

From that, it is clear that your limit equals $e^\frac12.$


The equation $ x^{x^{x}}...\equal{}2$ is satisfied when $x$ is equal to:

$\textbf{(A)}\ \infty \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ \sqrt[4]{2} \qquad
\textbf{(D)}\ \sqrt{2} \qquad
\textbf{(E)}\ \text{None of these}$

[quote="Smartguy"]The equation $ x^{x^{x}}... \equal{} 2$ is satisfied when x is equal to:
(A) infinity
(B) 2
(C) $ \sqrt [4] {2}$
(D) $ \sqrt {2}$
(E) none of these[/quote]
[hide]$ x^{x^{x\ldots}}\equal{}x^{\left(x^{x^{x\ldots}}\right)}\equal{}x^2\equal{}2 \Rightarrow x\equal{}\sqrt{2}$. Actually, a similar thing works for $ x^{x^{x\ldots}}\equal{}4$, paradoxically yielding the exact same answer, but in this case, it actually is correct... :P [/hide]

[quote=m5s][quote="Brut3Forc3"]$ x^{x^{x\ldots}}\equal{}x^{\left(x^{x^{x\ldots}}\right)}\equal{}x^2\equal{}2 \Rightarrow x\equal{}\sqrt{2}$. Actually, a similar thing works for $ x^{x^{x\ldots}}\equal{}4$, paradoxically yielding the exact same answer, but in this case, it actually is correct... :P [/quote]
So, does $ \sqrt {2}^{ \sqrt {2}^{ \sqrt {2}\ldots}}$ equal 2 or 4 ?

It seems to me that the meaning of an expression like $ x^{x^{x}}... $ (or $ x^{x^{x\ldots}}$ ) needs to be clarified, in order for the question to even make sense.
Presumably, one possible definition could be: the limit (if it exists) of a sequence, defined recursively: $t_0\equal{}a, t_n\equal{}x^{t_{n-1}}$.
However, the limit (and even its existence) would depend on the choice of the sequence's initial term, $a$. For example, when $x\equal{}\sqrt{2}$, the sequence will converge to 2 if $a<4$, converge to 4 if $a=4$, and diverge if $a>4$.

Curiously enough, the technique used by Brut3Forc3 above could be applied to any equation of the type $ x^{x^{x\ldots}}=b$, producing the solution $x=b^{b^{-1}}$, but this solution will be "unstable" (in the same way as $\sqrt{2}$ (that is, $\sqrt[4]{4}$) is an "unstable" solution of $ x^{x^{x\ldots}}=4$) unless $b<e$. For example, $x=\sqrt  [3] {3}$ is an "unstable" solution of $ x^{x^{x\ldots}}=3$ (it "works" only if the sequence's initial term is 3), but $x=2.5^{0.4}$ is a "stable" solution of $ x^{x^{x\ldots}}=2.5$ (because $3>e$, but $2.5<e$).[/quote]

This is a massive bump, but I wanted a clarification on this point.

Although this method of finding $x$ seems to provide a solution(namely $\sqrt{2}$), is it really valid in this case?  $\sqrt{2} > 1$ and raising a real $x > 1$ to another real $y>1$ will output a larger number.  For this reason, it seems that the sum diverges.

Brut3Forc3 also brought up a good point that this solution of $\sqrt{2}$ outputs multiple values, so is this problem broken?

[quote=RamanConjecture]
$\sqrt{2} > 1$ and raising a real $x > 1$ to another real $y>1$ will output a larger number. For this reason, it seems that the sum diverges.
[/quote]
This doesn't imply anything about divergence. Another way of thinking about it; consider the sequence $a_1 = \sqrt{2}$ and $a_n = \sqrt{2}^{a_{n-1}}$. Even though it is increasing, $\lim_{n \to \infty} a_n = 2$.

Also, see [url=https://artofproblemsolving.com/community/q1h1522041p9098016]this[/url] thread, which is very similar.

It is satisfied when the answer is $\sqrt{2}$, but it technically isn't correct. You start off by assuming there is a value of x that works, which isn't necessarily true.

[quote=m5s][quote="Brut3Forc3"]$ x^{x^{x\ldots}}\equal{}x^{\left(x^{x^{x\ldots}}\right)}\equal{}x^2\equal{}2 \Rightarrow x\equal{}\sqrt{2}$. Actually, a similar thing works for $ x^{x^{x\ldots}}\equal{}4$, paradoxically yielding the exact same answer, but in this case, it actually is correct... :P [/quote]
So, does $ \sqrt {2}^{ \sqrt {2}^{ \sqrt {2}\ldots}}$ equal 2 or 4 ?

It seems to me that the meaning of an expression like $ x^{x^{x}}... $ (or $ x^{x^{x\ldots}}$ ) needs to be clarified, in order for the question to even make sense.
Presumably, one possible definition could be: the limit (if it exists) of a sequence, defined recursively: $t_0\equal{}a, t_n\equal{}x^{t_{n-1}}$.
However, the limit (and even its existence) would depend on the choice of the sequence's initial term, $a$. For example, when $x\equal{}\sqrt{2}$, the sequence will converge to 2 if $a<4$, converge to 4 if $a=4$, and diverge if $a>4$.

Curiously enough, the technique used by Brut3Forc3 above could be applied to any equation of the type $ x^{x^{x\ldots}}=b$, producing the solution $x=b^{b^{-1}}$, but this solution will be "unstable" (in the same way as $\sqrt{2}$ (that is, $\sqrt[4]{4}$) is an "unstable" solution of $ x^{x^{x\ldots}}=4$) unless $b<e$. For example, $x=\sqrt  [3] {3}$ is an "unstable" solution of $ x^{x^{x\ldots}}=3$ (it "works" only if the sequence's initial term is 3), but $x=2.5^{0.4}$ is a "stable" solution of $ x^{x^{x\ldots}}=2.5$ (because $3>e$, but $2.5<e$).[/quote]

There's a video on youtube that explains this "paradox". It uses minimum and maximum values related to $e$, and shows that 2 is in the range.
Here's the link: https://www.youtube.com/watch?v=DmP3sFIZ0XE

Nice problem boss
Got another one for you:
Find 1-1+1-1+1-... (infinetly many ones with alternating signs)

[quote=RadianEleven]And also 1+n+n^2+n^3+...(infinitely terms)[/quote]

that's a generalization of the previous one

[quote=mathleticguyyy][quote=RadianEleven]And also $1+n+n^2+n^3+...$(infinitely terms)[/quote]

$\LaTeX$[/quote]

Use $\ldots$ or $\cdots$ for dots. Also, it is just a geometric series, which equals $\frac{1}{1-n}$ if $|n| < 1$, and does not converge otherwise.

[quote=JohnHankock]@scrabbler94 it kind of does. $2^{\sqrt{2}}>2$, and thus the solution is broken (possible with one way of looking at it, impossible another).[/quote]

Are you referring to my post #5? How does that fact "break" the solution?

The inverse function of $f(x)=x\uparrow\uparrow\infty$ on $\frac{1}{e^e}\leq x\leq e^{1/e}$ is $f^{-1}(x)=x^{1/x}$ on $\frac{1}{e}\leq x\leq e$.

We know thus that $f^{-1}(2)=\sqrt{2}$

[quote=JohnHankock]@scrabbler94 it kind of does. $2^{\sqrt{2}}>2$, and thus the solution is broken (possible with one way of looking at it, impossible another).[/quote]

[pywindow]
import math

a = math.sqrt(2);
b = a;
for i in range(0, 100):
  print (b);
  b = a**b;
print (b);
[/pywindow]
Here's convergence for you.

Edit: why doesn't it work?
Edit2: ok fixed

@3above (yes post 5) the idea is that, according to @bengabriel's proof, $x=\sqrt{2}$. However, infinity isn't one decided value. As such, raising that to the power of $\sqrt{2}$ should yield the same result, but it in fact doesn't.

@above it should be $\sqrt{2}^2$ which indeed does equal $2$

Yes, but raising that to another power of $\sqrt{2}$ changes the result.

[quote=Radiate]@RadianEleven - The first one is 1/2. The second one is the Maclaurin series for 1/(1-n).[/quote]

How come $1+1-1+1-1+ \cdots \, \infty =\dfrac 12 \, ?$
I have found the topic in brilliant wiki. But, AoPS doesn't allow me to post a link as I am a new user. Search for "1+1-1+1 ..." in Google and one should get the link.

Euler did 'prove' it, but he was actually wrong. It is not $\frac{1}{2}$.


the function $f$ is defined by $f(n)=f(n-1)+f(n-2)$. it's also true that $f(1)=3$ and $f(3)=10$. wat's the value of $f(6)$

[hide]For n plug in three so you get f(3)=f(2)+f(1). Since you know    f(3) and f(1) solve for f(2) getting 7. Now plug in 4 for n. So its f(4)=f(3)+f(2). f(4)=17. Now plug on 5 for n. f(5)=f(4)+f(3). Solve for that getting f(5)=27. Finally plug on 6 to get our answer. f(6)=f(5)+f(4). Solving for that gives f(6)=44.[/hide]


Spirit and Opportunity have been collecting very good evidence that there was once water on Mars, and astronomers are beginning to discover more "earthlike" planets than the hot gas giants that were first found. Since it seems that there may be a possibility for life (in general) on other planets, how long do you think it will take for us to find it?

I think scientists could discover evidence of past life on Mars in less than ten years, and we might find life such as bacteria on another planet or moon within the next few decades. As for intelligent life, I believe it will find us before we find it. Maybe it already has found us. Who knows?

Voted a decade, as KrazyNerd said with the Mars Rovers, evidence should show up soon. :)

As for us finding other intelligent life or it finding us first, of course it will be possible with enough time but, if that were to happen imagine the people from those cults that worship "aliens" or just the public's, in general, reaction to that occasion.  Mass hysteria.

Well I just saw that a new planet has been found in the habitable zone. And there is a very big chance that there may be life there. Here are more details:
[url]http://en.wikipedia.org/wiki/Gliese_581_c[/url]

Astronomers have many planets like that already. :)
They have also found planets that appear to be not within the habitable zone.

Yeah, that's what I was talking about with the hot gas giants, AstroPhys. Some of the first extrasolar planets found were called "hot Jupiters" because they were large gas giants that orbited very close to their stars. Now that detection methods are increasing in sensitivity, less massive (and perhaps more earthlike) planets are being found.

I was very excited when I read about Gliese 581 c, and I was just talking about it to someone the other day. That's what prompted me to make this poll. :)

If no one has posted this before:

there's methane on Mars and Titan, and that could lead to life. But if there will be life, it will be very small, so no one will have to worry about being eaten up or anything.

For the whole story, get the May issue of Scientific American year '07. It's front page material, so I don't want to type it ALL out.

There is life on mars, but not like on earth. We cant see them because they are in a different dimension.
Why cant it be possible, that for xmple a lifeform breaths cyan gas in and nitrous oxide out, or whatever.
The problem is that we are so isolated in thought that we can never think of this stuff, like only the lifeform which is on earth can be anywhere else.  :idea:

I have noticed that you all have considered primitive life on such planets and bodies like Mars. At 1=2, correct, it should be of nothing significant, unless it is a possible disease or organism that will affect us.

However, we are considering only our solar system. What about outside our solar system? Outside or galaxy? Maybe even in another universe or dimension! There are billions of stars in a galaxy, and billions of galaxies, and with so many planets and possibilities, it is almost guaranteed that there is life somewhere else, even if it is small.

With an infinite universe, and possibly more, there must be life!

The universe is incredibly large.  I agree.  If intelligent life is out there, we will need to wait lightyears for it to get here.  What about the Fermi paradox.  It is possible that aliens will visius somehow.  Who knows?

I voted a few. I don't think there is intelligent life, but I think there could be micro bacteria and stuff beneath the icy surface of europa.

I don't think there is life anywhere else, so I voted likewise.

Why...
There is a good chance there might be microorganisms of other planets

Well, there are so many planets though...

Have scientists actually [i]seen[/i] salt water forming usable proteins?

On a microscopic level, probably. It takes a really long time tho so I don't think they have that kind of patience :P

Maybe if they did, they'd have better evidence...

You little genius.
You should actually send an email to this guy:Professor Michael Blaber from florida state univertisy telling him to do just that and put a screenshot of this. Credit's all yours.

[size=200]I really mean it[/size]

Is that sarcasm, or do you really mean it? :huh:

Yeah! I have the same doubt. Regarding the poll- do you know something called the Arecibo message?

[quote=rainbowcharizard12]What does that have to do with anything, though?[/quote]

"Extraterrestrial life"

But it didn't have anything to do with extraterrestrial life.

Evidence should come up soon, because the probability of extraterrestrial life is pretty high.

I don't mean to be a conspiracy theorist but the chances that we will be the ones finding is is much lower than being the ones found to think about it we have been in this universe for 0.00145985401% of it and thats including extremely primate humans if you notice that we've been looking for only 1% that in that 0.001% and only broadcasting 3.75% of the time since then we are extremely young evolutionary and only 0.7 on the Kardashev scale we will probably be the ones found.

on the other hand if other planets had life long before us then light from other plants would reach us sooner than our light reaching theirs, right


Arrange the digits 1 to 9 inclusive in a 3x3 array such that the sum of a number's immediate neighbors, including diagonals, is a multiple of that number.

By immediate neighbours you mean the numbers in squares sharing a vertice and/or an edge with the given square?


geometry
analytic geometry
similar triangles

Given a triangle $ ABC$ and an arbitrary point $ P$ internal to it, let the line through $ P$ parallel to $ BC$ meet $ AC$ at $ M$. Similarly let the line through $ P$ parallel to $ AC$ meet $ AB$ at $ N$, and let the line through $ P$ parallel to $ AB$ meet $ BC$ at $ L$.

Prove that

$ \frac{BL}{LC} \times \frac{CM}{MA} \times \frac{AN}{NB} \le \frac{1}{8}$

and locate the position of $ P$ for which equality occurs.

I have a pretty long solution to this problem.
[hide="Solution"]
Let LP intersect AC at S, MP intersect AB at T, NP intersect BC at R.
Let AP intersect BC at X, BP intersect AC at Y, CP intersect AC at Z. Then
$ \frac {BL}{LC} \equal{} \frac {BL}{LX} \cdot \frac {LX}{XR} \cdot \frac {XR}{RC} \cdot \frac {CR}{CL} \equal{} \frac {AP}{PX} \cdot \frac {BX}{XC} \cdot \frac {PX}{AP} \cdot \frac {SP}{SL} \equal{} \frac {BX}{XC} \cdot \frac {SP}{SL}$.
But by Ceva $ \frac {BX}{XC} \cdot \frac {CY}{YA} \cdot \frac {AZ}{ZB} \equal{} 1$, so the product we seek is $ \frac {SP}{SL} \cdot \frac {RP}{RN} \cdot \frac {TP}{TM}$.
Note that $ SL \equal{} SP \plus{} PL$, $ RN \equal{} RP \plus{} PN$, $ TM \equal{} TP \plus{} PM$, so we have to show $ \frac {PS}{PS \plus{} PL} \cdot \frac {PT}{PT \plus{} PM} \cdot \frac {PR}{PR \plus{} PN} \le \frac {1}{8}$
Let $ \frac {PL}{PS} \equal{} x$, $ \frac {PM}{PT} \equal{} y$, $ \frac {PN}{PR} \equal{} z$.
By similar triangles, $ PN \equal{} \frac {AC}{BC}PT$, $ PM \equal{} \frac {BC}{AB}PS$, and $ PL \equal{} \frac {AB}{AC}PR$, so $ PN \cdot PM \cdot PL \equal{} PT \cdot PS \cdot PR$, so $ xyz \equal{} 1$. We have to show that
$ \frac {1}{1 \plus{} x} \cdot \frac {1}{1 \plus{} y} \cdot \frac {1}{1 \plus{} z} \le \frac {1}{8}$, or $ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \ge 8$. But $ x \plus{} 1 \ge 2 \sqrt {x}$ by AM-GM so $ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \ge 8 \sqrt {xyz} \equal{} 8$.
Equality holds when $ x \equal{} y \equal{} z \equal{} 1$, or $ PL \equal{} PS$, $ PN \equal{} PR$, $ PM \equal{} PT$, or P is the centroid of $ \triangle ABC$. [/hide]

[hide="My solution"]Using areal coordinates:

Let the areal coordinates of $ P$ be $ (p,q,r)$. The line $ PM$ has equation $ x\equal{}p$. The point $ M$ also lies on the line $ y\equal{}0$, so $ M$ has coordinates $ (p,0,1\minus{}p)$.

$ \frac{CM}{AM} \equal{} \frac{[BCM]}{[BMA]} \equal{} \frac{[BCM]}{[ABC]} \div \frac{[BMA]}{[ABC]} \equal{} \frac{p}{1\minus{}p}$

Similar working for the other two gives $ \frac{AN}{NB}\equal{}\frac{q}{1\minus{}q}$ and $ \frac{BL}{LC} \equal{} \frac{r}{1\minus{}r}$

So the question becomes $ \frac{pqr}{(1\minus{}p)(1\minus{}q)(1\minus{}r)} \le \frac{1}{8}$ for $ p\plus{}q\plus{}r\equal{}1$.

Multiplying out: $ \frac{pqr}{1\minus{}p\minus{}q\minus{}r\plus{}pq\plus{}qr\plus{}rp\minus{}pqr}$.

But $ p\plus{}q\plus{}r\equal{}1$ so this becomes:

$ \frac{pqr}{pq\plus{}qr\plus{}rp\minus{}pqr} \le \frac{1}{8}$.

Cross multiplying we are left to prove that $ 9pqr \le pq\plus{}qr\plus{}rp$.

Multiply the RHS by $ (p\plus{}q\plus{}r)\equal{}1$ and we must prove:

$ p^2q\plus{}q^2p\plus{}p^2r\plus{}r^2p\plus{}q^2r\plus{}r^2q \ge 6pqr$

Since $ P$ is internal to the triangle, $ p,q,r > 0$ so we can use AM-GM.

By AM-GM we get: $ p^2q\plus{}q^2r\plus{}r^2p \ge 3\sqrt[3]{(pqr)^3} \equal{} 3pqr$

$ q^2p\plus{}r^2q\plus{}p^2r \ge 3pqr$.

Adding these two we get the desired result.

Equality occurs iff $ p\equal{}q\equal{}r\equal{}\frac{1}{3}$, ie $ P\equal{}(\frac{1}{3},\frac{1}{3},\frac{1}{3})$, ie $ P$ is the centroid of ABC.[/hide]