inequalities
function
algebra unsolved
algebra

Find all quadruples of real numbers [tex]x,y,z,t[/tex] which satisfy : [tex]x+y+z+t=x^7+y^7+z^7+t^7=0[/tex]

i think the only two solutions are $(0,0,0,0)$ and $(-0,-0,-0,-0)$ ;)

[hide="heres my proof"]hölders inequality says [by [url=http://mathworld.wolfram.com/HoeldersInequalities.html]wolfram[/url]]:
$\sum_{k=1}^n |a_kb_k| \le \left(\sum_{k=1}^n |a_k|^p\right)^{\frac1p}\cdot \left(\sum_{k=1}^n |b_k|^q\right)^{\frac1q}$
with $\frac1p + \frac1q = 1$

now setting $a= (x,y,z,t)$, $b = (1,1,1,1)$ and $p=7$, $q=\frac76$
we get:
$0=x+y+z+t \le 4\cdot \sqrt[7]{x^7+y^7+z^7+t^7} = 0$

now wolfram says that equality is when $|b_k|=c|a_k|^{p-1}$
that would mean that they are all the same because $b_k=1$ for all $k$
hence $0=4\cdot  x= 4\cdot x^7$

so $x=\pm 0$ and we arrive at and only at the given solutions[/hide]

Answer: $\left(  x,y,z,t\right)  =\left(  a,b,-a,-b\right)  $ with $a,b\in R$ (and permutations).

Discarding the trivial solution $\left(  0,0,0,0\right)  ,$ observe that $x,y,z,t$ cannot have the same sign. Thus, modulo some permutations, we have
to analyze two cases:

1. $x,y,z\geq0\geq t;$

2. $x,y\geq0\geq z,t.$

In the first case, replacing $t$ by $-t,$ we have
\[
x+y+z=t
\]
\[
x^{7}+y^{7}+z^{7}=t^{7},
\]
so
\[
\left(  \frac{x}{x+y+z}\right)  ^{7}+\left(  \frac{y}{x+y+z}\right)
^{7}+\left(  \frac{z}{x+y+z}\right)  ^{7}=1.
\]
But
\[
\left(  \frac{x}{x+y+z}\right)  ^{7}\leq\frac{x}{x+y+z},
\]
\[
\left(  \frac{y}{x+y+z}\right)  ^{7}\leq\frac{y}{x+y+z},
\]
\[
\left(  \frac{z}{x+y+z}\right)  ^{7}\leq\frac{z}{x+y+z},
\]
therefore the above equality holds only if one of the fractions equals $1$ and the other two equal $0.$ We obtain the solution $\left(  1,0,0,-1\right)  $
(and permutations).

In the second case, we have, for the nonnegative numbers $x,y,z,t:$
\[
x+y=z+t,
\]
\[
x^{7}+y^{7}=z^{7}+t^{7}.
\]
Since the function $f:[0,+\infty)\rightarrow R,$ $f\left(  x\right)  =x^{7}$ is convex, the above equalities can hold only if $\{x,y\}=\{z,t\}$. We obtain the solution $\left(  a,b,-a,-b\right)  $ (and permutations).

your solution quadruple seems obvious, and is without a doubt correct.
but then were is the mistake in my proof ... ?

peeta

[quote="peeta"]but then were is the mistake in my proof ... ?[/quote]
Holder's inequality holds only for positive numbers, that is your mistake.


Is it just more or did it say Greg Gauthier was an 8th grader in the 2003 Mathcounts ESPN airing? Was that just a mistake or did he actually stay back to redo mathcounts?

I'm 99% sure that he wasn't in 8th that year. Probably a mistake.

Greg won MATHCOUNTS in 2004 as an 8th grader, he was 7th in 7th grade in 2003


Where can i find solutions to HC verma (concepts of physics)?is there any site where they can be found or on any p2p application?if yes pls tell me and if possible send the link.[/b]

hey you use HC verma too?? thats a cool book , isnt it ???i use it too :lol:  which standard are you in ??

i advise you to try resnick halliday krane, its cooooooooooooooooool too!

ummm, i dont know about the solutions, i usually ask them to friends if i dont get some problems..........

How about asking those problems here?there are many people here willing to help you :lol: Feel free.

ya resmick halliday is a gud book but i prefer using hc verma.i was having  a few problems in solving  a few problems of laws of motion.i cud not solve the monkey problems as well as the last 5 problems.if u cud help i wud b great.thanx

if u could mention the chapter then we could help :)

[quote="shadysaysurspammed"][quote="riddler"][quote="eminem"]
i advise you to try resnick halliday krane, its cooooooooooooooooool too![/quote]
 :o  :mad:  :|  :ninja:  :wallbash:  :wallbash_red:  :agent:  :bomb:  Forget it yes u can ask ur questions here![/quote]

 :D  :rotfl: You ain't the only person in the universe who knows it riddler.[/quote]
I know

hey cool dude , about the monkey ones as you are saying in the laws of motion are you talking about the TENSION ones?, did you get that one, the one in which 2 guys(me and Shady ) are pulling out their friend(riddler , ahahahahahahhaha) who's stuck in a ditch???

I don't think HC Verma solutions will be on a P2P server. It's mostly used in Indian high schools and I even doubt whether most of indian universities even have any P2P servers. But as eminem and shady and riddler... said above, you can alwyas discuss your problems here!!! [hide]( hey guys, no new posts?!)[/hide]

no i didnt get the ditch question and the monkey question is the one where one monkey is hanging on the tail of the other.and as far as the solutions story goe they are available in the market i guess.but if u get them online whats better :P

Which problem are you refering to? Please be specific!!!

I NEED HELP  FOR SUMS 57,25-b,c,d,68,48 from the 31st chapter - capacitors.
I would really appreciate ur help.
(P.s: Does ne one the the exact name of the book containing the solns to hc verma co i can't get it newhere..???)[/size]

Yup,good ol days you just revived.

Anyway,They are available here in the market by the name "Solutions of concepts of physics"Depends where you live

could u tell me the author's name or the publication bcoz i asked for it in 2 bookshops in b'mbay but couldn't find it...or could u give me the add. of some book shop from where i could order it??...

H C verma solutions are availble at
http://www.vidyadrishti.com/H%20C%20Verma%20Solutions.html
The most imp point is that problems r solved by iitians.

guys, this thread was made THREE YEARS AGO!!!  :mad:


search
number theory unsolved
number theory

If $a,b$  $(a^2+b^2)/(ab-1)=q$ are integers, then $q=5$

Posted before...
But do you really think that "Nt" as title says something to the readers¿ Use topic names that describe (parts of) the problem!

Tried "a^2+b^2 ab-1", with additionally the search restricted to Number Theory¿


Bobby stands 10 feet from Sam, and Eric stands 8 feet from Bobby. What is the least number of feet that Eric could be from Sam?

Let's think about everything in comparision to Bobby. Sam is somewhere in a 10 foott radius circle from Bobby. Eric can now be anywhere in an 8 foot radius circle from Bobby. For Sam and Eric to be as close together as possible, they will be in the same exact same direction, thus making their distance from each other $ 10 \minus{} 8 \equal{} 2$ feet.


ABCD is a trapezoid.  AB is parallel to CD.  If DA = AB + CD, and AE is the angle bisector of angle A, show that BE = EC.

[hide]Instead I'll prove that if M is the midpoint of BC, then AM bisects angle BAD. Continue DM to hit AB at F, and notice that MCD and MBF are congruent. So AF=AB+BF=AB+CD=AD, meaning triangle DAF is isosceles, and so the median AM of that triangle is also the angle bisector.

Since there is only one point of intersection between the angle bisector of angle A and segment BC, point M must be this point.[/hide]


Let be $ a,b,c\in (0,\infty )$ such that $ a\plus{}b\plus{}c\equal{}1$ . Prove that :

$ a\sqrt{a}\plus{}b\sqrt{b}\plus{}c\sqrt{c}\ge \sqrt{a^4\plus{}b^4\plus{}c^4\plus{}24abc\sqrt[3]{abc}}$

[hide="Ugly"]

By Holder

$ (a \plus{} b \plus{} c)(a\sqrt {a} \plus{} b\sqrt {b} \plus{} c\sqrt {c})(a\sqrt {a} \plus{} b\sqrt {b} \plus{} c\sqrt {c})\ge (a^{\frac {4}{3}} \plus{} b^{\frac {4}{3}} \plus{} c^{\frac {4}{3}})^3$

So the inequality is reduced to

$ (a^{\frac {4}{3}} \plus{} b^{\frac {4}{3}} \plus{} c^{\frac {4}{3}})^3\ge a^4 \plus{} b^4 \plus{} c^4 \plus{} 24a^{\frac {4}{3}}b^{\frac {4}{3}}c^{\frac {4}{3}}$

Which is obvious[/hide]

Any symmetric sum of degree 4 with 24 terms is $ \ge 24abc \sqrt [3] {abc}$ by Muirhead (or AM-GM) since all the exponents on a,b,c are equal.
So squaring, we have to show
$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 2 \sum a^{3/2}b^{3/2})(a \plus{} b \plus{} c) \ge a^4 \plus{} b^4 \plus{} c^4 \plus{} 24abc \sqrt [3] {abc}$
But the LHS contains an $ a^4 \plus{} b^4 \plus{} c^4$ and 24 other terms of degree 4, so it is obviously true


graph theory
Ramsey Theory
combinatorics theorems
combinatorics

Let $ G$ be a graph with $ n$ vertices that does not contain a $ K_k$ as subgraph ($ K_k$ is a complete graph with $ k$ vertices).
Then the maximum number of edges in G is equal to
\[ \displaystyle M(n,k) = \frac {k - 2}{k - 1}\cdot \frac {(n - r)(n + r)}{2} + \frac {r(r - 1)}{2}
\]
if
\[ \displaystyle n = (k - 1)t + r,0\leq r \leq k - 2.
\]

I read tht there is a consecuence of this theorem that is a generalization, it is a fotmula such that f(n,k,r) is the maximun number of edges you can draw and color in r colors in such a way that there is no complete k monocromatic subgraph. I think that furthermore, if A= R(r1,r1,r3,.....,rt) is the minimun number such that in any coloraTion of a Complete graph with A vertices with t colors there is a Kri for some i, then if we chose a quantity of edeges in a graph with n vertices which depends on n and A, and color this edges in t colors, then there  is a Kri.....

The question is, I remember this formula was very similar to turans, does anyone nows it, maybe perhaps Pierre?

The problem you are referring to smells really like Ramsey's theorem, which states that :

Let $k,r \geq 1$ and $m_1,...,m_r \geq k$ be integers.
There is a integer $n_0$ such that, for each $n \geq n_0$, no matter how each $k$-elements subsets of a set $X$ with $|X|$ are colored by $r$ colors (say $c_1,...,c_r$) there exists a subset $X_i$ of $X$ such that $|X_i| = m_i$ and each $k$-subset of $X_i$ has color $c_i$.

It may be useful to get the sense of this theorem by checking small values of $k,r, \cdots $.

In another hand, for $k=2$ the $2$-subsets of a set $X$ can be seen as the edges of the complete graph whose vertices are the elements of $X$. Thus, Ramsey's theorem states that, given $m_1,...,m_r \geq 2$, then for $n$ sufficientely large no matter how you color the edges of $K_n$ each with one from $r$ colors, there will be a complete monochromatic $K_{m_i}$.

The least integer $n_0$ which have the property described in the theorem is called the Ramsey number $R(m_1,...,m_r, k)$.
Note that, the exact value of the Ramsey numbers are unknown, except for some exceptional and small values of $r,k,m_i$

Is that what you were talking about?

Pierre.

:?
can you tel me please somthings about the proof
and thank you johan
:roll:

Turans Theorem talks about having a certain number of vertices, and no clique larger than some K_n. I have an article where it says tells you the amount of vertices  n and edges, and it says you have averege degree no more than 4. Thus Turans theorem implies that G has independence number at least n/5. I have no idea how they could use turans theorem, to calculate the independence number of the graph. Do you have any idea? Maybe i should explain more, i dont know.

The version of Turan's theorem I am familiar with is actually stronger: it gives the unique structure for a graph on $n$ vertices with no $K_k$ subgraph to have a maximal number of edges.  When one considers this version of the theorem it implies much more than the version in the original post.


geometry
quadratics
area of a triangle
Heron\u0027s formula

1. In Triangle ABC, AB=2, BC=3, AC=4. Determine the length of the altitude of BD.

2. Factor $ a^4c^2\minus{}b^4c^2\plus{}d^2b^4\minus{}d^2a^4$

So I have I have gotten this problem down to:
$ (c\plus{}d)(c\minus{}d)(a^4\minus{}b^4)$. I know $ (a^4\minus{}b^4)\equal{}(a\plus{}b)(a\minus{}b)(a^2\plus{}b^2)$ but can someone please post an explanation.

3. Rewrite $ x^4\plus{}4$ as the product of two quadratics. The answer is below but how would you do this.
  [hide="answer"]$ (x^2\plus{}2x\plus{}2)(x^2\minus{}2x\plus{}2)$[/hide]

[hide="1"]
By Heron's ($ s \equal{} \frac {a \plus{} b \plus{} c}2 \equal{} \frac {2 \plus{} 3 \plus{} 4}2 \equal{} \frac92$), the area is

$ \sqrt {\frac92\left(\frac92 \minus{} 2\right)\left(\frac92 \minus{} 3\right)\left(\frac92 \minus{} 4\right)} \equal{} \sqrt {\frac92\cdot\frac52\cdot\frac32\cdot\frac12} \equal{} \sqrt {\frac {135}{16}} \equal{} \frac {3\sqrt {15}}4$

$ \implies\frac {3\sqrt {15}}4 \equal{} \frac12\cdot4\cdot BD\implies BD \equal{} \boxed{\frac {3\sqrt {15}}8}$.

Alternatively,...

 [geogebra]3ea30d445718d6dc1e9a01805a5fb4b37f243ab9[/geogebra]

We have

$ \sqrt {4 \minus{} x^2} \plus{} \sqrt {9 \minus{} x^2} \equal{} 4\implies\sqrt {9 \minus{} x^2} \equal{} 4 \minus{} \sqrt {4 \minus{} x^2}$

$ \implies9 \minus{} x^2 \equal{} 16 \plus{} 4 \minus{} x^2 \minus{} 8\sqrt {4 \minus{} x^2}\implies\frac {11}8 \equal{} \sqrt {4 \minus{} x^2}$

$ \implies\frac {121}{64} \equal{} 4 \minus{} x^2\implies x^2 \equal{} \frac {135}{64},x > 0$

$ \implies x \equal{} \boxed{\frac {3\sqrt {15}}8}$.
[/hide]
[hide="2"]
Repeatedly grouping and using differences of squares, we get

$ a^4c^2\minus{}b^4c^2\plus{}d^2b^4\minus{}d^2a^4$

$ \equal{}c^2(a^4\minus{}b^4)\minus{}d^2(a^4\minus{}b^4)$

$ \equal{}(c^2\minus{}d^2)(a^4\minus{}b^4)$

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

$ \equal{}\boxed{(c\minus{}d)(c\plus{}d)(a\minus{}b)(a\plus{}b)(a^2\plus{}b^2)}$.
[/hide]
[hide="3"]
I will demonstrate two techniques.

[hide="One way"]
From the problem statement, we can set up $ x^4\plus{}4\equal{}(x^2\plus{}ax\plus{}b)(x^2\plus{}cx\plus{}d)\equal{}x^4\plus{}(a\plus{}c)x^3\plus{}(b\plus{}d\plus{}ac)x^2\plus{}(ad\plus{}bc)x\plus{}bd$, and pairing up coefficients, we get

$ a\plus{}c\equal{}0\implies a\equal{}\minus{}c$ (paring $ x^3$ coefficients),

$ b\plus{}d\plus{}ac\equal{}0$ (paring $ x^2$ coefficients),

$ ad\plus{}bc\equal{}0$ (paring $ x$ coefficients), and

$ bd\equal{}4$ (pairing the constants).

From this, we know that it must be of the form $ (x^2\plus{}ax\plus{}b)(x^2\minus{}ax\plus{}d)\equal{}x^4\plus{}4$, and it is not hard to guess the rest of it.

However, this technique doesn't work well when the coefficients of the quadratics aren't integral.
[/hide]
[hide="Another Way"]
This is a special case of the [url=http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie Germain Identity[/url].

We have

$ x^4\plus{}4\equal{}x^4\plus{}4x^2\plus{}4\minus{}4x^2\equal{}(x^2\plus{}2)^2\minus{}(2x)^2\equal{}\boxed{(x^2\minus{}2x\plus{}2)(x^2\plus{}2x\plus{}2)}$.
[/hide]
[/hide]

[hide="1"] Using Heron's formula For $ \triangle ABC$ we get the area of $ \triangle ABC$, take that thing multi by 2 and divide by $ AC\equal{}4$ get the solution.
[/hide]
[hide="2"]
I don't know what things to explain for you $ (c\plus{}d)(c\minus{}d)(a\minus{}b)(a\plus{}b)(a^2\plus{}b^2)$ and that's it, that's right.
[/hide]
[hide="3"]
Let $ x^4\plus{}4\equal{}(x^2\plus{}ax\plus{}b)(x^2\plus{}cx\plus{}d)$
You have $ x^4\plus{}cx^3\plus{}dx^2\plus{}ax^3\plus{}acx^2\plus{}adx\plus{}bx^2\plus{}bcx\plus{}bd$
$ x^4\plus{}(c\plus{}a)x^3\plus{}(b\plus{}d\plus{}ca)x^2\plus{}(ad\plus{}bc)x\plus{}bd\equal{}x^4\plus{}4$
So
$ c\plus{}a\equal{}0$
$ b\plus{}d\plus{}ca\equal{}0$
$ ad\plus{}bc\equal{}0$
$ bd\equal{}4$
$ \equal{}>(a,b,c,d)\equal{}(2,2,\minus{}2,2)$
and you have $ x^4\plus{}4\equal{}(x^2\plus{}2x\plus{}2)(x^2\minus{}2x\plus{}2)$
[/hide]

Thanks so much for the answers and explanation. Hit my self on the head for the Heron's formula one :)


trigonometry
function
geometry
angle bisector
trig identities
Law of Sines

Let ABC be a triangle ,angle <B=80,[BD] is angle bisector of B angle.D point is between A and C point.The length of [AD],[BD] and [BC] segments are a+b, a and b respectively where a,b real numbers.Find angle <C.

Let's find the angle $\measuredangle A$ instead. Using the sine theorem for the triangles $\triangle ACD, \triangle DBC$ and combining,

$\frac{AD}{BD} = \frac{a + b}{a} = \frac{\sin{40^o}}{\sin{\widehat A}}$

$\frac{BC}{BD} = \frac b a = \frac{\sin{(\widehat A + 40^o)}}{\sin{(100^o - \widehat A)}}$

$\frac{\sin{40^o}}{\sin{\widehat A}} = 1 + \frac{\sin{(\widehat A + 40^o)}}{\sin{(100^o - \widehat A)}}$

$\sin{40^o} \sin{(100^o - \widehat A)} = \sin{\widehat A}\ [\sin{(100^o - \widehat A)} + \sin{(\widehat A + 40^o)}]$

We will show that this equation is satisfied only for $\measuredangle A = 20^o$. If this is true, the triangle $\triangle ABC$ is then isosceles. Let $E$ be another point on the side $AC$, such that the angle $\measuredangle CBE = 70^o$. Using the sine theorem for the triangle $\triangle ABC, \triangle AEC, \triangle BEC$,

$\frac{AC}{BC} = \frac{\sin{80^o}}{\sin{20^o}}$

$\frac{AE}{AC} = \frac{\sin{10^o}}{\sin{150^o}} = \frac{\sin{10^o}}{\sin{30^o}} = 2 \sin{10^o}$

$\frac{BE}{BC} = \frac{\sin{70^o}}{\sin{30^o}} = 2 \sin{70^o}$

$\frac{BE}{BC} = \frac{BE}{AC} \cdot \frac{AC}{BC} = \left( 1 - \frac{AE}{AC} \right) \cdot \frac{AC}{BC}$

$2 \sin{70^o} = (1 - 2 \sin{10^o})\ \frac{\sin{80^o}}{\sin{20^o}}$

$\sin{80^o} = 2 (\sin{70^o} \sin{20^o} + \sin{10^o} \sin{80^o})$

Since $\sin{70^o} = \cos{20^o}$ and $\sin{80^o} = \cos{10^o}$,

$\sin{80^o} = 2 (\cos{20^o} \sin{20^o} + \sin{10^o} \cos{10^o}) = \sin{40^o} + \sin{20^o}$

Let's trisect the angle $60^o$:

$\sin{60^o} = \sin{40^o} \cos{20^o} + \cos{40^o} \sin{20^o} = 2 \sin{20^o} \cos^2{20^o} + (\cos^2{20^o} - \sin^2{20^o}) \sin{20^o}$

$\sin{60^o} =  3 \sin{20^o} - 4 sin^3{20^o}$

Add $\sin{80^o}$ to the left side and $\sin{20^o} + \sin{40^o}$ to the right side of the equation and the multiply both sides by $\sin{20^o}$:

$\sin{60^o} + \sin{80^o} = 4 \sin{20^o} (1 - \sin^2{20^o}) + \sin{40^o}$

$\sin{60^o} \sin{20^o} + \sin{80^o} \sin{20^o} = 4 \sin^2{20^o} \cos^2{20^o} + \sin{20^o} \sin{40^o}$

$\sin{20^o} (\sin{60^o} + \sin{80^o}) = \sin{40^o} (\sin{40^o} + \sin{20^o})$

Substitute the previously derived expression for $\sin{80^o}$ to the right side:

$\sin{20^o} (\sin{60^o} + \sin{80^o}) = \sin{40^o} \sin{80^o}$

But this is exactly the equation of the problem with the angle $\measuredangle A$ replaced by $20^o$. Since $BC = b < a + b < AC$, the angle $\measuredangle A < \measuredangle B = 80^o$ and there is no other solution. Hence, $\measuredangle A = 20^o$ and $\measuredangle C = \measuredangle B = 80^o$.

Is there any pure geometrical(synthetic) solution?

You will have to eventually use at least one trig function, and since none of the triangles are right triangles, you are limited to Law of Sines and/or Law of Cosines.

Perhaps we must think the question like this,perhaps using circles...


Does gravity have an affect on light here on earth or in space? If so can this be proved mathmaticly

I hink it does.I read it somewhere.You want more expert opinion,post such questiuons in the advanced physics section.

Well then can one of the mods move this topic please?


linear algebra
matrix
vector
linear algebra unsolved

Hi all!  :)

[tex]A[/tex] is an [tex]n\times n[/tex] diagonal matrix, such that the distinct values on the diagonal are [tex]c_1, c_2, ... , c_k[/tex], where [tex]c_i[/tex] appears di times and [tex]\sum d_i = n[/tex]. Show that the vector space of [tex]n \times n[/tex] matrices [tex]B[/tex] which commute with [tex]A[/tex] has dimension [tex]\sum d_i^2[/tex].

[Copied from Kalva]

Hanno

As far as I know this holds for any matrix, not only diagonal.

Hi Harazi!

Mhh  :? , what do you mean? I don't think that the statement is true for every matrix, otherwise every matrix for which all diagonal elements are equal would commutate with every other matrix - that's not true.

I think I don't understand what you mean.

Hanno

No, I meant that the dimension of the commutant is the sum of the squares of the dimensions of the proper subspaces of the matrix.

i have a solution for this.Ugly but here it goes:

  Let $A=(a{}_{ij}{}_{1\leq i,j \leqq}$.Same for $B=b_{ij}$, $AB=c_{ij}$ and $BA=d_{ij}$ and finally  we have that$a_{ij}=0$ and because $AB=BA$ we deduce that $(a_{ii}-a_{jj})b_{ij}=0$.If $a_{ii}-a_{jj}\neq 0$ then $b_{ij}=0$ so the conclusion follows. ;)

Harazi have you proved it for any matrix? ;)

I see you adore the sign  ;) . No, I haven't, but I think that it should not be hard with Jordan canonical form. Too bad I don't know it well enough.

I learned it from you.I'm talking about the thing with:  ;)

I'm delighted to see that! But note that it should be used when necessary, not all the time.


geometry
circumcircle
geometric transformation
reflection
geometry solved

O is the circumcenter of the obtuse-angled triangle ABC. K is the circumcenter of AOC. The lines AB, BC meet the circumcircle of AOC again at M, N respectively. L is the reflection of K in the line MN. Show that the lines BL and AC are perpendicular.

Why must triangle ABC be obtuse-angled? The problem is true for every triangle ABC.

The proof is angle chasing. We will use directed angles modulo 180°.

Denote by k the circumcircle of triangle AOC. After the conditions of the problem, this circle k passes not only through the points A, O and C, but also through the points M and N. The center of this circle k is, of course, the circumcenter K of triangle AOC.

The [b]central angle theorem[/b] (for directed angles modulo 180°) states that if X, Y, Z are three points on a circle with center S, then < SXY = 90° - < YZX. Applying this to the points C, A, B lying on a circle with center O (remember that O is the circumcenter of triangle ABC), we get < OCA = 90° - < ABC. But since the points O, A, M and C lie on one circle (namely, on the circle k), we have < OMA = < OCA. Thus, < OMA = 90° - < ABC, so that < OMA + < ABC = 90°. Hence,

  < (OM; BC) = < (OM; AB) + < (AB; BC) = < OMA + < ABC = 90°.

Similarly, < (ON; AB) = 90°. Consequently,

  < MON = < (OM; ON) = < (OM; BC) + < (BC; AB) + < (AB; ON)
             = < (OM; BC) + < CBA - < (ON; AB) = 90° + < CBA - 90° = < CBA.

Now, let U be the reflection of the point B in the line MN. Then, < MUN = - < MBN. But < MBN = < ABC; thus, < MUN = - < ABC = < CBA. Consequently, < MUN = < MON. This means that the points M, N, U and O lie on one circle. Since the points M, N and O lie on the circle k, it follows that the point U also lies on k. Hence, the circle k is the circumcircle of triangle MNU. And the center K of this circle k must therefore be the circumcenter of triangle MNU.

Now, since the point U is the reflection of the point B in the line MN, the point B must be the reflection of the point U in the line MN, and thus, the triangle MNB must be the reflection of the triangle MNU in the line MN. Consequently, the circumcenter of triangle MNB is the reflection of the circumcenter of triangle MNU in the line MN. But the circumcenter of triangle MNU is the point K, and the reflection of this point K in the line MN is the point L. Thus, the circumcenter of triangle MNB is the point L.

In other words, the points B, M, N lie on a circle with center L. Thus, applying the central angle theorem, we get < LBM = 90° - < MNB. But clearly < MNB = < MNC, and since the points M, C, N and A lie on one circle (namely, on the circle k), we have < MNC = < MAC; thus, < MNB = < MAC. Hence,

  < LBM = 90° - < MNB = 90° - < MAC = 90° - < (AB; CA) = 90° + < (CA; AB).

In other words,

  < (BL; AB) = 90° + < (CA; AB).

Thus, < (BL; AB) - < (CA; AB) = 90°; equivalently, < (BL; CA) = 90°, what immediately yields $BL\perp CA$. And the problem is solved.

  Darij


abstract algebra
function
Ring Theory
superior algebra
superior algebra solved

Which of the following rings are Artinian? Noetherian? Justify.

1) Z_7_[x,y]

 2) Z[x]

 3) C [0,1] , the ring of continuous real-valued functions on [0,1] with pointwise multiplication

For 1) I think that this is neither Artinian nor Noetherian but I lack a good understanding and can't justify it.

For 2) Z[x] is a PID, so it is finitely generated but not Artinian.  I'm not sure if, for the Noetherian case, that would be adequete justification.

For 3)  I'm not really certain either way.

I'd appreciate any kind of justification or explanation with these.

James

Proofs that a ring is not artian or not noetherian can in easy cases be given by naming an infinite descending or ascending chain of ideals. Proofs that a ring is artian or noetherian are normally more involved or use some theorem.

For 1)
Z_7 [x,y] is not artian, since the powers of x generate a descending chain of ideals. If R is a noetherian ring, then R[x] is also noetherian, by the Hilbert basis theorem. So this and the next ring are noetherian.

For 2)
Z[x] is not a PID: (2,x) is not principal. But it is a UFD.
I agree that Z[x] is not artian, but it is noetherian.

For 3)
Let A be any closed subset of [0, 1]. The continuous functions on [0, 1] which map A to zero form an ideal of C[0, 1]. Take any sequence $(A_{z})_{z \in \mathbb Z}$ of closed subsets of [0, 1] s.th. $A_{z}\subsetneq A_{z+1}$ for all z. The corresponding ideals form the chains you need to prove that this ring is neither artian nor noetherian.

LiMa


Messy inequality, probably arqady in mind for solution :P
Prove: $a^{4}c+a^{3}c^{2}+b^{4}a+b^{3}a^{2}+c^{4}b+c^{3}b^{2}+c^{3}ab-4a^{3}bc-4ab^{3}c-4abc^{3}-2b^{2}c^{2}a-2c^{2}a^{2}b\geq 0$

(it is equivalent to $\sum_{cyc}\frac{a^{2}}{b}\geq a+b+c+\frac{4(a-b)^{2}}{a+b+c}$)

This is from Balkan 05, no? Just Cauchy:

$(b+c+a) \sum \frac{(a-b)^{2}}{b}\geq (2\text{max}(a,b,c)-2\text{min}(a,b,c))^{2}$

Wow that is a very nice solution... (yes that is where the problem was from) btw what is the thought process behind that solution, it seems pretty contrived...
$(b+c+a) \sum \frac{(a-b)^{2}}{b}\geq (|a-b|+|b-c|+|c-a|)^{2}=(2\text{max}(a,b,c)-2\text{min}(a,b,c))^{2}$

It's known as Balkan Olympiad, http://www.mathlinks.ro/Forum/viewtopic.php?highlight=balkan&t=35937

Well, that is exactly the same as the proof that was just posted... anyway does anybody know another way to prove it?


How do you put the box around the answer or just a number someone wants in the box?

You have to use $\LaTeX$ and use the \boxed{} command.  So, for example, put dollar signs around \boxed{4} to get $\boxed{4}$.

By the way, please don't write everything in bold.


ceiling function
floor function
number theory proposed
number theory

Find all natural number $n$ for which the number of all positive divisors of the number $lcm(1,2,..,n)$ is equal to $2^{k}$ for some posive integer $k$

If $n > 124$, then $\sqrt[3]{n}+1 < \frac{\sqrt{n}-1}{2}$, so by Bertrand's Conjecture (Theorem, whatever), $\exists$ a prime $p$ between $\lceil \sqrt[3]{n} \rceil$ and $\lfloor \sqrt{n} \rfloor$. Thus $p^2 \le n$ but $n < p^3$, so if $L = \text{lcm}{1, 2, \cdots, n}$, we have $p^2 || L$, and thus $3 | \tau(L) \Rightarrow \tau(L) \ne 2^k$. Now, for $n \le 124$, if we can find a prime $p$ between $\sqrt[3]{n}$ and $\sqrt{n}$, we can still apply the above argument. For $25 \le n \le 124, p = 5$ works. For $9 \le n \le 25, p = 3$ works. For $4 \le n \le 7, p = 2$ works. So now we just manually check $n = 1, 2, 3, 8$. For $n = 1$, $\tau(1) = 1 = 2^0$. For $n = 2$, $\tau(2) = 2 = 2^1$. For $n = 3, \tau(6) = 4 = 2^2$. For $n = 8, \tau(840) = 32 = 2^5$.


How do you do vertical multiplication and division in LaTeX?

Thanks! :D

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=xlop]xlop package[/url] will do this for you and will also do the calculations. If you prefer a different layout use a [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Layout#Tables]table[/url].

For tables, here is an example code:

[code]\begin{tabular}{ccc}\\
        & 1 & 2\\
\times& 4 & 3\\
\hline
        & 3 & 6\\
     4 & 8 &   \\
\hline
    5 & 1 & 6
\end{tabular}[/code]

\begin{tabular}{ccc}\\
        & 1 & 2\\
\times& 4 & 3\\
\hline
        & 3 & 6\\
     4 & 8 &   \\
\hline
    5 & 1 & 6
\end{tabular}

That is good, but
[list=1][*]it will give 4 errors in a document! The problem is that \times must be in maths mode so should be \$\times\$ (AoPS treats maths and text modes differently so not important on the forum). Alternatively, use array instead of tabular and stick the whole lot in  \$...\$. That's useful if otherwise you would have a lot of dollar signs to put in.
[*]You can make the separation between the columns more or less \setlength{\tabcolsep}{[i]width[/i]}
[*]the \\ on the first line after {ccc} isn't required.[/list]
Here's an example with the tabcolsep changed to 0.5mm
$ \setlength{\tabcolsep}{0.5mm}
\begin{array}{ccc}
 & 1 & 2\\
\times & 4 & 3\\
\hline
  & 3 & 6\\
     4 & 8 &\\
\hline
    5 & 1 & 6
\end{array}$


So, by default, when we use the two-dollar signs to wrap up our math stuff, they will be centered automatically. What should I do if I just want them to be, says, left hand align?

Use one dollar sign on each side- this puts the math stuff inline, and you can simply add line breaks in if you want it to be at the beginning of a line.

No, that's not what I want, because, for example, the $\sum$, $\int$ etc,.. will get smaller also. I want them to look as we use 2 dollars sign.

Use \displaystyle.

You don't need to do that on this forum- displaystyle is automatic here, as you can see by what you posted.


Given a triangle, and an inside point $ P$ of the triangle. Let $ a,b,c$ be the distances from $ P$ to the vortices and $ x,y,z$ be the distances from $ P$ to the sides of the triangle respectively.
Find the minimum of :

$ \mbox{min}\left ( \frac{\sqrt{a^2\minus{}x^2}}{ax}\plus{}\frac{\sqrt{b^2\minus{}x^2}}{bx}\plus{}\frac{\sqrt{a^2\minus{}y^2}}{ay}\plus{}\frac{\sqrt{c^2\minus{}y^2}}{cy}\plus{}\frac{\sqrt{b^2\minus{}z^2}}{bz}\plus{}\frac{\sqrt{c^2\minus{}z^2}}{cz} \right ) \equal{}?$
For which $ x,y,z,a,b,c$ will it be minimal.

I bumped into this in a physics problem, and I am looking for an elegant solution, not Lagrange method.

If someone has any idea for an elegant solution please share it  :)

your expression is homogenous of degree -1 so we can make it as small as we want. can you check it again.

no thats it. I dont understand, why can it be as small as we want, since $ a,b,c,x,y,z$ arent independant variables, but are dependent, and are "parts" of a triangle, so they cant just take up any value. :huh:

I mean that if you take bigger and bigger triangles similar to the initial one this expression goes to 0. and it is clearly non-negative so...  :wink:

Yes clearly if you could just take bigger and bigger triangles then it goes to 0. but then there would be no point of the problem. The lengths of the  sides of the triangle are clearly fixed values, you cant just  change them.


geometry
geometric transformation
reflection
probability
ratio
absolute value

[quote="aidan"]The numbers 1 to 9 are arranged randomly around the circumference of a circle. There is rotational indistinguishability and reflections are considered the same. The absolute value of the differences between adjacent terms around the circle are added together, for a total of 9 sums. What is the probability that this sum is a minimum?[/quote]

obviously in the sums, each number is going to be used twice
so the sum will always be 90

We're not talking about the sums of adjacent numbers.

[hide="Solution"]The minimum of the sums is $ 16$. Suppose a number $ n$ is next to numbers $ k$ and $ m$. WLOG $ k < m$.
1. If $ k < n < m$, then in adding the absolute values of the differences $ n$ cancels out. $ |k \minus{} n| \plus{} |m \minus{} n| \equal{} m \minus{} k$
2. If $ n < k$ and $ n < m$, then $ n$ is subtracted twice in the resulting sum. $ |k \minus{} n| \plus{} |m \minus{} n| \equal{} m \plus{} k \minus{} 2n$
3. If $ n > k$ and $ n > m$, then $ n$ is added twice in the resulting sum. $ |k \minus{} n| \plus{} |m \minus{} n| \equal{} 2n \minus{} m \minus{} k$
$ 9$ will always be added twice and $ 1$ will always be subtracted twice. That's $ 16$ there. To get lower we have to have a number satisfying case 3 less than a number satisfying case 2. But if you partition the 9 number arrangement into monotonic subsequences, it's pretty clear that a number satisfying case 3 is always greater than another number satisfying case 2. (sorry for lack of rigor)

This also tells us the way to construct all arrangements achieving this is to place the $ 9$ and $ 1$ anywhere, then make sure that going from the $ 9$ to $ 1$ in either direction forms a monotonically decreasing sequence of numbers. Fix the position of the $ 9$ and ignore that we can do reflections (probability, being a ratio, will not change). There are $ 8!$ ways to arrange the other numbers now.

Suppose, going clockwise, that there are $ m$ numbers between the $ 1$ and the $ 9$. Then $ 0 \leq m \leq 7$. The number of arrangements given $ m$ is $ {7 \choose m}$, since there are $ 7$ numbers, and once we have decided which $ m$ are between the $ 9$ and $ 1$ going clockwise, there is only one way to arrange them. So the probability is

$ \frac{\sum_{k\equal{}0}^{7} {7 \choose k}}{8!} \equal{} \frac{2^7}{8!} \equal{} \frac{1}{315}$

unless I screwed up.[/hide]


algebra
function
domain
system of equations
algebra unsolved

Solve the system of equations
$\frac{1}{xy}=\frac{x}{z}+1$ , $\frac{1}{yz}=\frac{y}{x}+1$ , $\frac{1}{zx}=\frac{z}{y}+1$
in the domain of real numbers.


Two computers and a number of humans participated in a large round-robin chess tournament (i.e., every participant played every other participant exactly once). In every game, the winner of the game received one point, the loser zero. If a game ended in a draw, each player received half a point. At the end of the tournament, the sum of the two computers' scores was 38 points, and all of the human participants finished with the same total score. Describe (with proof) ALL POSSIBLE numbers of humans that could have participated in such a tournament.

Let $ h$ be the numbers of humans with an equal final score of $ s$.
The amount of games played is $ (h \plus{} 2)(h \plus{} 1)/2$ so computing the sum of all the points given:
$ 38 \plus{} h \times s \equal{} (h \plus{} 2)(h \plus{} 1)/2$
$ 76 \equal{} h^2 \plus{} 3h \plus{} 2 \minus{} 2sh$
$ 74 \equal{} h(h \plus{} 3 \minus{} 2s)$
so $ h|74$ then $ h \equal{} 1,2,37,74$ so the only possible solution is: $ (h,s) \equal{} (37,19)$


we know schur can be write as (3,0,0)+(1,1,1)>=2(2,1,0) and it seems that it can't be proved directlly by muirhead inequality.
of course i kown schur is equivalent to (a+b-c)(b+c-a)(c+a-b)<=abc,and it can be proved using AM-GM,but i want to know whether it can be proved by muirhead directlly.thanks :D

You'll never get it, i think..

Schur :

$ \sum_{cyc} a^t(a\minus{}b)(a\minus{}c) \ge0$

Since there is $ [t,1,1]$ in LHS but $ [t\plus{}1,1,0]$ in RHS, you will never get a proof with Muirhead only..

You'll need some manipulations first..

Schur can be written (I prefer that form) as
$ (a\plus{}2b,0,0)\plus{}(a,b,b)\ge2(a\plus{}b,b,0)$
From Muirhead you can get $ (a\plus{}2b,0,0)\ge(a\plus{}b,b,0)$, but also $ (a,b,b)\le(a\plus{}b,b,0)$, so it can't be proven just with Muirhead.

@Bugi: Your notation is wrong. Schur is $ [a\plus{}2b,0,0] \plus{} [a,b,b] \ge 2[a\plus{}b,b,0]$! It should not be with ordinary parantheses :)

You probably confused it with $ (a\plus{}2b,0,0) \succ (a\plus{}b,b,b)$


geometry
geometric transformation
reflection
circumcircle
trigonometry
projective geometry
cyclic quadrilateral

(A.Zaslavsky, 9--10) The reflections of diagonal $ BD$ of a quadrilateral $ ABCD$ in the bisectors of angles $ B$ and $ D$ pass
through the midpoint of diagonal $ AC$. Prove that the reflections of diagonal $ AC$ in the bisectors of angles $ A$ and $ C$ pass
through the midpoint of diagonal $ BD$ (There was an error in published condition of this problem).

Notice that the quadrilateral also must be cyclic. In this case, denote by $ M$, $ N$ the midpoints of the diagonals $ BD$, $ AC$, respectively, and by $ P$ the intersection point of the diagonals of $ ABCD$. Since the lines $ BD$, and $ BM$ are isogonal with respect to angle $ ABC$, by Steiner's theorem, $ \frac{AB^{2}}{BC^{2}}\equal{}\frac{AP}{PC}.$ Similarly, the lines $ DB$ and $ DM$ are isogonal with respect to angle $ CDA$, and therefore $ \frac{AD^{2}}{DC^{2}}\equal{}\frac{AP}{PC}.$ Hence $ AB/BC\equal{}AD/DC$; in other words, the quadrilateral $ ABCD$ is harmonic. Now, since the pencil $ B(B, A, D, C)$ is harmonic, by intersecting it with the line $ AC$, we deduce that $ T$ (the intersection point of the tangent at $ B$ to the circumcircle of $ ABCD$ and line $ AC$) is the harmonic conjugate of $ P$ with respect to segment $ (AC)$. On other hand, the pencil $ D(D, A, B, C)$ is harmonic, and thus the tangent at $ D$ to the circumcircle of $ ABCD$ passes through $ T$.

It is now well-known that if $ XYZ$ is the tangential triangle of a given triangle $ ABC$, the lines $ AX$, $ BY$, $ CZ$ are the symmedians of $ ABC$. According to this fact, and to the collinearity of the points $ A$, $ C$, and $ T$, the diagonal $ AC$ is the $ A$-symmedian of triangle $ ABD$, and simultanously the $ C$-symmedian of triangle $ BCD$. Therefore, the reflections of $ AC$ in the bisectors of angles $ A$, and $ C$ of the cyclic quadrilateral $ ABC$ pass through the midpoint of $ BD$. This completes our proof.

I have another solution, which uses the following known fact: considering the projective plane with the circumcircle of $ ABC$ as reference, the symmedians of $ ABC$ are obtained connecting the corresponding vertex to the dual of the opposite side. Then the problem asks to prove that if $ AD$ is the $ A$-symmedian of $ ABC$ then $ BC$ is the $ B$ symmedian of $ ABD$.

In order to prove that, let $ P$ be the dual point of $ BC$ and $ Q$ be the dual point of $ AD$. Then $ P$ belongs to $ AD$ and the result follows from the duality theorem: the dual of $ P$, $ BC$ passes through the dual of $ AD$, $ Q$.

[hide="Solution"]Since $BD$ is a symmedian of triangles $ABC$, $ADC$, we obtain $\frac{[ABD]}{[CBD]}=\frac{\frac{1}{2}BA\cdot BD\sin\angle DBA}{\frac{1}{2}BC\cdot BD\sin\angle DBC}=\frac{BA\sin\angle DBA}{BC\sin\angle DBC}=\frac{BA^2}{BC^2}$. Similarly $\frac{[ABD]}{[CBD]}=\frac{DA^2}{DC^2}$; hence $\frac{BA}{BC}=\frac{DA}{DC}$. Let $D'=BD\cap\odot(ABC)$. Since $\frac{D'A}{D'C}=\frac{DA}{DC}=\frac{BA}{BC}$, it follows that $A$, $D$, $D'$ lie on the $B$-Apollonius circle WRT triangle $ABC$. They are also, however, collinear, so $D'\equiv D$. It follows that quadrilateral $ABCD$ is harmonic, and hence we may conclude. $\blacksquare$[/hide]


The side of square $ S$ is $ x$ units in length. The lengths of two
non-adjacent sides of the square are each increased by 1 unit and the
other two sides are each decreased by 1 unit. Square $ S$ has thus been
transformed into rectangle $ R$. What is the number of square units in
the positive difference between the areas of $ S$ and $ R$?

This is a geometric representation and demonstration of the sum and difference algebraic property. $ (x\plus{}1)(x\minus{}1)\equal{}x^2\minus{}1$, so the difference in area between the two rectangles is $ \boxed{1}$.