Euler
geometry
circumcircle
geometric transformation
homothety
ratio
projective geometry

ABC is an equilateral triangle,P is in the interior of triangle ABC
Prove:Euler lines of triangle PAB、PBC、PCA  are concurrent

[quote="pohoatza"]This is obviously not true in general. This property holds only for the points $ P$ lying on the [url=http://pagesperso-orange.fr/bernard.gibert/Exemples/k001.html]Neuberg cubic[/url] of triangle $ ABC$.[/quote]
dear pohoatza
my Geometer's Sketchpad told me it is true  :maybe:

[quote="pohoatza"]My apologies. I didn't see the word "equilateral" in your statement. My intention was to mention that given an [i]arbitrary[/i] triangle $ ABC$ with a point $ P$ inside, we have that the Euler lines of triangles $ PBC$, $ PCA$, $ PAB$ are concurrent if and only if $ P$ lies on the Neuberg cubic of $ ABC$.[/quote]
WOW,my dear friend pohoatza,could you tell more details about that , please? :) Thank you in advance!

Call $ O_B$, $ O_B$, $ O_C$ the circumcenter of $ \triangle BCP$, $ \triangle CAP$, $ \triangle ABP$ and $ G_A$, $ G_B$, $ G_C$ the centroid of $ \triangle BCP$, $ \triangle CAP$, $ \triangle ABP$.

$ \triangle ABC$ and $ \triangle G_AG_BG_C$ are homothetic and so $ \triangle O_AO_BO_C$ and $ \triangle G_AG_BG_C$ are orthologic because the perpendicolar from the vertex of the first to the side of the second concur on the centroid $ G$ of ABC. Call M the midpoint of BC, so an homothety with center M and ratio $ \frac{1}{3}$ take $ PA$ on $ GG_A$, so $ GG_A \parallel PA$ then $ GG_A \perp O_BO_C$ and equaly we have $ GG_B \perp O_CO_A$ and $ GG_C \perp O_AO_B$, the two orthologic centers coincide on G, then $ \triangle O_AO_BO_C$ and $ \triangle G_AG_BG_C$ are also perspective.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]Call $ O_B$, $ O_B$, $ O_C$ the circumcenter of $ \triangle BCP$, $ \triangle CAP$, $ \triangle ABP$ and $ G_A$, $ G_B$, $ G_C$ the centroid of $ \triangle BCP$, $ \triangle CAP$, $ \triangle ABP$.

$ \triangle ABC$ and $ \triangle G_AG_BG_C$ are homothetic and so $ \triangle O_AO_BO_C$ and $ \triangle G_AG_BG_C$ are orthologic because the perpendicolar from the vertex of the first to the side of the second concur on the centroid $ G$ of ABC. Call M the midpoint of BC, so an homothety with center M and ratio $ \frac {1}{3}$ take $ PA$ on $ GG_A$, so $ GG_A \parallel PA$ then $ GG_A \perp O_BO_C$ and equaly we have $ GG_B \perp O_CO_A$ and $ GG_C \perp O_AO_B$, the two orthologic centers coincide on G, then $ \triangle O_AO_BO_C$ and $ \triangle G_AG_BG_C$ are also perspective.[/quote]
Thank you ,dear ¬[ƒ(Gabriel)³²¹º]¼
Here is my solution that I find in books:

Lemma:Given an equilateral triangle ABC , P is an arbitrary point , perpendicular bisectors of PA,PB,PC meet subtense at X,Y,Z, then X,Y,Z are collinear
(to prove it we can use Desargues theorem )

Proof:Denote D,E,F the midpoint of PA,PB,PC
△FG1G2∽△FAB
Z= G1G2 ∩ O1O2   X= G2G3∩O2O3  Y= G1G3∩O1O3
X’=X’D∩PA , and X’D⊥PA    Y’,Z’are defined similarly
ZG2/ZG1=Z’A/Z’B,XG3/XG2=X’B/X’C,YG1/YG3=Y’C/Y’A
ZG2/ZG1* XG3/XG2* YG1/YG3= Z’A/Z’B* X’B/X’C * Y’C/Y’A
By using the lemma we conclude X’,Y’Z’are collinear,so X,Y,Z are collinear
But X,Y,Z are intersections of  sides △G1G2G3 and △O1O2O3
According to Desargues theorem we know the three lines are concurrent.


let $ x$ , $ y$ , $ z$ lenghts of triangle , prove that :

$ x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \geq x^3 \plus{} y^3 \plus{} z^3 \plus{} 2xyz$

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

$ (xy)^2\plus{}(yz)^2\plus{}(zx)^2 \geq \frac{x^4\plus{}y^4\plus{}z^4}{2}$

my inequalities

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ , $ z$ lenghts of triangle , prove that :

$ x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \geq x^3 \plus{} y^3 \plus{} z^3 \plus{} 2xyz$

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

$ (xy)^2 \plus{} (yz)^2 \plus{} (zx)^2 \geq \frac {x^4 \plus{} y^4 \plus{} z^4}{2}$

my inequalities[/quote]

[color=darkblue]$ \bullet$ $ x\plus{}y>z\Longrightarrow xz\plus{}yz>z^2$, (1).

$ \bullet$ $ y\plus{}z>x\Longrightarrow yz\plus{}zx>x^2$, (2).

$ \bullet$ $ z\plus{}x>y\Longrightarrow zy\plus{}xy>y^2$, (3).

Since (1), (2), (3) $ \Longrightarrow xy \plus{} yz \plus{} zx >\frac {x^2 \plus{} y^2 \plus{} z^2}{2}$

[u]Note[/u]: Sybol ''='' is not happen.[/color]

This is the solution for problem 1 and 3
We consider the expanding of following inequality:
1/ $ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)>0$
2/ $ S\equal{}(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\plus{}c) >0$
Now we get the result, it's a very beautiful expansion and useful in proving inquality about triangle.

[color=darkblue][u]Note[/u]: Oh! Your problem Sybol ''='' is not happen.[/color]
let $ x$ , $ y$ , $ z$ lenghts of triangle , prove that :

$ x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) >x^3 \plus{} y^3 \plus{} z^3 \plus{} 2xyz$

$ xy \plus{} yz \plus{} zx >\frac {x^2 \plus{} y^2 \plus{} z^2}{2}$

$ (xy)^2 \plus{} (yz)^2 \plus{} (zx)^2 >\frac {x^4 \plus{} y^4 \plus{} z^4}{2}$

thank's all

for the problem 1 i did like you great math to  creat the inequality


I remember a question in the mandelbrot, i think in 2002 or 2003, that asked something like "Evaluate the sum $\tfrac{F_1}{3^1}+\tfrac{F_2}{3^2}+\cdots$." (Those are fibonacci numbers.) Does anyone know what question I'm talking about? Better yet, do you have it on file? If so, could you please tell me what year it comes from and the actual wording of the problem? I know it's a long shot, but I can hope, right? :)

Thanks.

BTW, you're also welcome do discuss the problem here  ;)

Sorry, I can't help you with the wording of the problem.  I recall seeing this problem (perhaps with a different base instead of 3) on the old AoPS site, but I couldn't find it.

One way to solve it is sort of like the usual way we prove the formula for an infinite geometric series.  Let S be the sum.  Then show that 9S = 3S + S + 3 by using the Fibonacci recurrence.

Here is the question as it appeared on the competition:

Let $F_n$ be the nth Fibonnaci number, where as usual $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}.$ Find the value of the infinite sum
\[\frac 13 + \frac 19 +\frac2{27}+\cdots+\frac{F_n}{3^n}+\cdots.\]

Thanks! I hate to ask for more, but what year and round does that come from?

It was February 2002 Round Four Individual.  I just looked at the other tests for the 2001-2002 year and it looks like they really liked Fibonnaci that year.  They had an article on Fibonacci, 2 individual problems on it and a whole team test on it.  I have the Mandelbrot problem book in case you were wondering where I got all this info.


geometry
3D geometry
algebra
polynomial
Ring Theory
factorization
sum of cubes

Is it possible to represent X as a sum of cubes in the polynomial ring IZ[X]?

I know that it is possible in IQ[X] with 7 cubes (actually in any R[X] where 6 is invertible in the ring R, but I don't know if 7 cubes is minimal) and in IR[X] with 3 cubes (using sqrt(6) and cuberoot(2), this number of cubes seems to be minimal).

LiMa

It isn't: consider the linear coefficient of $(a_{n}x^{n}+...+a_{1}x+a_{0})^{3}$, it is $3a_{0}^{2}a_{1}$, thus divisible by $3$.

You are right. That should have been obvious. :-/

Then, is it possible over the ring $\mathbb Z[1/3]$? Or over $\mathbb F_{2}$? (Or any ring where 3 is, but 2 is not invertible.)

Probably I am too much looking for examples, so I wont find a counter argument easily.

LiMa

It can't be done over the field with two elements (or, by extension, anywhere else that 2 isn't invertible). For any sum of cubes, the sum of the coefficients of $x,x^{2},x^{4},x^{5},x^{7},x^{8},\dots$ is even.

Over $\mathbb{F}_{2}$, we have the identity $(p+q+r)^{3}=p^{3}+q^{3}+r^{3}+(p+q)^{3}+(p+r)^{3}+(q+r)^{3}$; this allows us to reduce any sum of cubes to a sum of cubes of the form $(x^{n})^{3}$ or $(x^{m}+x^{n})^{3}=x^{3m}(1+x^{n-m})^{3}$. The first kind does not contribute to the coefficients of non-cubes; the second either does not contribute or contributes 1 to each of two such coefficients, an even total.

This is the only meaningful restriction; the sums of cubes are a subspace of codimension 1 in the space of polynomials over $\mathbb{F}_{2}$.

Thank you, jmerry. Now I'll fail with the generalisation.

Take a ring R where 2 is not invertible, and a maximal ideal J containing 2. Then the field S:=R/J has characteristic 2 and so your identity holds.
Any presentation of x as a sum of cubes over R would give rise to a presentation of x as a sum of cubes over S, which can be decomposed, using the identity, into cubes of the form $(a x^{n})^{3}$ or $x^{3n}(a+b x^{m})^{3}$. The second one gives rise to the not necessarily cubic term $a b x^{3n+m}(a+b x^{m})$.

So the "sum of the interesting coefficients" changes by $a^{2}b+ab^{2}$. But that does not tell me anything in general, since that sum can equal 1, e.g. in F_4.

LiMa

I'll retract my "anywhere else" comment, but it's impossible in $\mathbb{F}_{4}$. Since all cubes in that field are either zero or 1, only polynomials which take only values of zero and 1 can be written as a sum of cubes.
$x$ can be written as a linear combination of cubes $(1+x)^{3}+a(a+x)^{3}+a^{2}(a^{2}+x)^{3}$, where $a$ is a multiplicative generator.

We should probably look at $\mathbb{F}_{2}[x,y]$; every $\mathbb{F}_{2^{k}}[x]$ is a quotient of this ring. We still have the identity allowing us to express any sum of cubes as a sum of cubes of monomials and of binomials; $(x^{a}y^{b}+x^{c}y^{d})^{3}= x^{3a}y^{3b}+x^{3c}y^{3d}+x^{2a+c}y^{2b+d}+x^{a+2c}y^{b+2d}$. This means that if $\sum a_{ij}x^{i}y^{j}$ is a sum of cubes, $\sum_{i\equiv 1,j\equiv 1}a_{ij}=\sum_{i\equiv 2,j\equiv 2}a_{ij}$, $\sum_{i\equiv 1,j\equiv 2}a_{ij}=\sum_{i\equiv 2,j\equiv 1}a_{ij}$, $\sum_{i\equiv 1,j\equiv 0}a_{ij}=\sum_{i\equiv 2,j\equiv 0}a_{ij}$, and $\sum_{i\equiv 0,j\equiv 1}a_{ij}=\sum_{i\equiv 0,j\equiv 2}a_{ij}$.The subspace of sums of cubes has codimension at least 4 by this count. A slight refinement gives us a codimension of 6 (noncube powers of $x$ or $y$ alone can only be paired with each other and must balance on their own). This is all.

In $\mathbb{F}_{64}$, $x$ can be written as a sum of cubes. Let $a$ be a primitive 9th root of unity in this field: $(1+x)^{3}+(a^{-1}+a^{2}x)^{3}+(a^{-2}+a^{4}x)^{3}=x$.


Find the number of roots of the equation:\[ 2x \equal{} 3\pi (1 \minus{} \cos x)
\]
where $ x$ is measured in radians.

This is equivalent to $ \cos x \equal{} \minus{} \frac2{3\pi}x \plus{} 1$.
$ \minus{} 1\le \minus{} \frac2{3\pi}x \plus{} 1\le 1 \implies x\in[0,3\pi]$

Note that equality is obtained at both ends points. There are 2 intersections for each period (a total of 2) between $ \pi$ and $ 3\pi$, plus one more at 0 (counted as half of a period), so the answer is 2*2+1=5.


group theory
abstract algebra
superior algebra
superior algebra solved

Hello out there!  Newbie posting.

For the life of me, I can't solve this problem:

Show that if G (a finite group) has only 1 p-sylow subgroup, the subgroup must be normal.

I know it has something to do with being a conjugate to itself (right coset = left coset?)...but I'm not sure how to show that.

Thanks for being patient with me.

best,
dogma

Not hard:
Let be $H$ your subgroup and let be $x \in G$.
Since $|H|=|xHx^{-1}|$ conjugation of a subgroup leads to another subgroup with the same number of elements. But there is only one subgroup with that number of elements here, giving that $H=xHx^{-1}$ and $H$ is normal

A big thanks to you, ZetaX...I greatly appreciate your help.

dogma


geometry
perimeter
circumcircle
geometry proposed

Let $O$ be the circumcenter of a triangle $ABC$. Points $M$ and $N$ are chosen on the sides $AB$ and $BC$ respectively so that $2\angle MON=\angle AOC$. Prove that the perimeter of $MBN$ is at least the length of $AC$.

(It is not mine solution. I think grobber wrote it before on forum)
Construct points $M'$ and $N'$ s.t. $\angle BOM=\angle AOM'$, $\angle BON=\angle CON'$, $OM=OM'$ and $ON=ON'$. Then $\triangle BON=\triangle AOM'$, $\triangle BON=\triangle CON'$ and $\triangle MON=\triangle M'ON'$ (we use $O$ is the circumcenter). Thus $P_{BMN}=AM'+M'N'+N'C\geq AC$.

P.S. The gree point is point $O$ :P

Noipe, it's not mine either :). I'm thinking that maybe iura (?) wrote it. Anyway, I believe it was a member of the Moldavian team.

It was Sailor: http://www.mathlinks.ro/Forum/viewtopic.php?t=18806 .

  darij

I call this a good question without any extra needs only basic geometry


Given n greater than 4, and there are 2n points on a plane. If among any n points, there are at least n-1 points lying on a circle. Prove that at least 2n-1 points are concyclic.

I proved a better result in Crux vol.29, n°6, October 2003, p.378-381 that :

Let $ n \geq 5$ be an integer, and $ S$ be a set of $ n$ points in the plane such that among any five of them, four (at least) are concyclic. Denote by $ f(n)$ the largest integer $ k$ for which we can correctly say "At least $ k$ of the $ n$ points lie on a circle".
Then $ f(6) \equal{} 4$ and $ f(n) \equal{} n \minus{} 1$ for each $ n \geq 5,n\neq 6.$

The proof is too long to write it here again (4 pages), but the key lemma is that :
"There do not exist seven pairwise distinct points, $ M_1,\cdots,M_7$ in the plane such that each of the following seven sets is concyclic and the seven circles thus defined are pairwise distinct :
$ M_1,M_2,M_3,M_4$
$ M_3,M_4,M_5,M_6$
$ M_1,M_4,M_5,M_7$
$ M_2,M_4,M_6,M_7$
$ M_1,M_2,M_5,M_6$
$ M_2,M_3,M_5,M_7$
$ M_1,M_3,M_6,M_7$".

This was an answer of a problem of the 1997 Iranian Olympiad, second round (it was asked for $ n\equal{}10$). What is your chinese reference?

Pierre.


[quote="sharkman"]James and Rita agree to share the cost of an 18 dollars pizza based on how much each ate.  If James ate (2/3) the amount Rita ate, how much should each pay?[/quote]

[hide]Let $x$ be the fraction that Rita ate. James ate $\frac{2x}{3}$. Thus, since James and Rita ate the entire pizza,
$x+\frac{2x}{3}= 1$

$x = \frac{3}{5}$

Thus, Rita gives $\frac{3}{5}\cdot \$ 18 = \$10.80$ and James gives $\$ 7.20$[/hide]


algebra
partial fractions
calculus
calculus computations

y' + 3y = sin 2t
Given y(PI) = 1.
If the given condition is not y(0), how to perform lapace transform on it. To find F(s).

Here's a suggestion. First, make a change of variable, letting $u=t-\pi.$ (That is, $t=u+\pi.$)

Define $g(u)=y(u+\pi).$ Then the differential equation is

$g'+3g=\sin(u+\pi),\,g(0)=1.$ Make that $g'+3g=-\sin(u).$

Now we can take the Laplace transform:

$sG(s)-g(0)+3G(s)=-\frac1{s^2+1}$

$(s+3)G(s)=-\frac1{s^2+1}+1=\frac{s^2}{s^2+1}$

I'll let you finish the details from here - the remaining algebra, the partial fractions, the inverse Laplace transform. The result should be a formula for $g(u),$ from which you can find $y(t)=g(t-\pi).$


AMC
AIME
geometry
geometric transformation
reflection
calculus
complex numbers

Hi, maybe it's the first passage written by a chinese student .
i'm in grade 11 and me high school is in BEIJING,CHINA.i did the amc 12b in 45 minutes and i didn't know how to solve 8 of the 25 questions.those were too difficult for me. In CHINA ,a student of grade 11 haven't learned complex number so i don't even know the meaning of some questions.when i was in grade 10,i participated amc9 and succeeded,then i participated the AIME .i thought the question was so so easy for me at that time because the questions in amc9(either a and b) were what chinese student learn in grade 7 or grade 8.but how can the students in the U.S learn so fast in 2 years?i can hardly believe that!! :!:
so,if you are a student in the U.S,please tell me something in your school math learning or your thought after reading the words above(or you can check my grammar mistakes :D
                                                                                THANK U

I think it depends on the school.  In my high school, I learned complex numbers in pre-calc (11th grade), and I think that is where a lot of the AMC 12 stuff comes from.

The AMC 10 (or 9?) is designed to be easy. 8th graders with algebra/geometry background shouldn't have too much trouble on it. The AMC 12 has more focus on more advanced mathematics that students learn at around 10-12 grades.

The tests do not reflect exactly what students learn in specific grades; most high scorers on AMC 12/10 have to spend time outside of school to practice their math-solving abilities.

在AMC 10/12其实我觉得学完了Algebra 2都已经足够(如果你在说知事一方面）。Algebra 2常常是在9或则10年级学的。

如果要把AMC10/12做的很好那就要靠自己做很多AMC, AIME的题目或则数学题目的书。很少会有学校在美国会有数学竞赛的课，好多很好的人在家花很多时间做数学竞赛题目。

。。。希望没有错别字。。。

I'm 11th grade and I've just come American for more than one year. Math competition is not something learned at school, you need some books and lots of exercises. (Before I came here I bought a lot of book about math competition in Chinese), they are all quite useful :lol:

U.S school teach broad knowledges (for example, they have calculus and multivariable in high school), but they are not as deep as that of China.

Since you are in Beijing, you must have met a lot of student who are extremely good at math (you know how China school emphasize math competitions). In fact, American math teacher not at all see math competition as something really important (at least in general). In China, you can always ask teachers for lots of help but here in U.S, I don't think that will work.

HS teachers don't tend to emphasize competitions.  Usually, there are math teams or coaches that can train people interested but the curriculum isn't geared to prepare students for math competitions.

one of the biggest problems i have with American curriculum (especially in my school) is that we have 7th to 12th graders who take AP Calculus BC, but have no idea what to do when they face challenging problems.  It seems like most math teachers try to dumb down students to feel they are doing their jobs because when it comes to math competitions, they are just ignorant. This is one of the things i really hate about my school.  :D  (btw, i'm not Chinese)

I guess this would be better if you posted this is your own countries forum?
Because you have other people that can speak your language and you may be able to get more help from them.
And just to let you know, US curriculum isn't the same as the Chinese curriculum so we may learn things in a different order from you guys so yeah.

Strange thing is, my mom taught me multiplication and division in 1st grade. Then I skipped 3rd grade. Then I learned about the high school math competitions in 9th grade, because our middle school wasn't competitive in math. Then I started looking at Art of Problem Solving.

[quote="zephyredx"]What you learn in schools don't help you much for the math contests. You have to learn Problem Solving. Even if someone finishes every school math class, the problem solving type questions may still be hard. They say Chinese students learn math earlier (like division in 1st grade), and zoom through elementary, but in America, the intensity begins at high school.[/quote]

well, kids here have different standards for math education

真是丢人现眼
北京高考压力小，数学竞赛水平远远落在湖北，湖南和广东之后，根本不足以代表中国中学生的水平
我1年半前来美国，去年来的时候读9年级，在国内念完初二，AMC12都没有说8道题不会做 = =||||
这种题给我中国的同学做大概20分钟左右吧。。。
不要以偏概全，跑到AoPS上面说美国学生amazing，真是丢死人了

I think it depends on the school. I am currently in elementary school ( :o , I know), and in my school I have to learn some of what I think is the stupidest math ever. (I have never experienced Chinese math, because I am a native born American and my parents are Chinese, but my friend owns a Chinese math book and the math in there is much harder than what I am learning now.)

In a lot of America's big public high schools, they don't tend to emphasize creative problem solving.  And AP Calculus is a joke-it is a watered-down curriculum and I've seen many people who took AP in high school, but at the college level, struggle a lot with it.


geometry
calculus
derivative
trigonometry
congruent triangles
area of a triangle
calculus computations

I have a problem where the there is a kite with outer sides are w = z and y = t.

I need to maximize the area of the kite, thus i need to determine the length of the diagonal pieces inside the kite, q and p.

I have tried to solve this problem but i always run into a problem where the only critical point of the derivative is 0, which is obviously not the correct number.

I have tried using the equations i found for a kite on this web page:
http://mathforum.org/dr.math/faq/formulas/faq.quad.html

Can someone help me out!

Thank you!

You make the kite cyclic (that is, you arrange it so it can be inscribed in a circle, or so sums of opposite angles equal $180^{\circ}$).  You may use Bretschneider's (sp?) formula to obtain a relatively neat expression for the area, or simple trig will also do in this case.

Here is one last idea.

We have the relations

$2A=a^2\sin\theta+b^2\sin\psi$ and $2a^2-2a^2\cos\theta=2b^2-2b^2\cos\psi$ by the Cosine Law

Hence $2\frac {dA}{d\theta}=a^2\cos\theta+b^2\cos\psi\frac {d\psi}{d\theta}$

Differentiating the Cosine Law relation, we get $a^2\sin\theta=b^2\sin\psi\frac {d\psi}{d\theta}$

Plugging all of this back in and setting $\frac {dA}{d\theta}=0$, you should get (after some algebra which I'll leave to you)

$a^2\sin(\theta+\psi)=0$

which implies that $\theta+\psi=180^{\circ}$

thank you for your reply, however I still seem to be having trouble with this problem.  I have no idea why i cannot get the right answer.  I have checked my algebra numerous times!

The method of inscribing it in a circle had me somewhat confused.  I prefer the method i was using the first time.

Sorry to ask for more help but this problem seems to be driving me nuts, maybe i am just over looking some stupid mistake!

I am trying to maximize area of a kite and then find the length of the diagonals -  not the 4 outer lines rather the 2 corner to corner lengths.

Once again thank you for any help!

For any polygon with fixed sides, the area is maximized by inscribing it in a circle. blahblahblah gives a general method for this theorem in a quadrilateral, but we can do a kite more easily.

A kite can be seen as two congruent triangles joined together by a common side. The area of a triangle with two fixed sides is obviously maximized when the triangle is right; you should be able to do the rest from there.


function
integration
complex numbers
real analysis
real analysis unsolved

Define the essential range of a function $ f\in L^{\infty}(\mu)$ to be the set $ R_f$ consisting of all complex numbers $ w$ such that $ \mu\left(\{x: |f(x)\minus{}w|<\varepsilon\}\right)>0$ for every $ \varepsilon>0$. Let $ A_f$ be the set of all averages $ \frac{1}{\mu(E)}\int_E fd\mu$ where $ \mu(E)>0$. Are there measures $ \mu$ such that $ A_f$ is convex for every $ f\in L^{\infty}(\mu)$?

What is $ R_f$ doing here?

If $ \mu$ is the Dirac measure at $ x_0$, then $ A_f\equal{}\{f(x_0)\}$, which is a convex set.

Thanks.
The problem also asked to prove that $ R_f$ is compact and to find the relations between $ R_f$ and $ A_f$, but I didn't post it and forgot to remove the definition.

I was trying to prove that it is convex for the Legesgue measure. Can we prove that or it's false?


trigonometry
geometry
cyclic quadrilateral

ABCD is a cyclic quadrilateral
AB,CD(when produced) meet at E and AD,BC meet at F.

Prove that the sum of the squares of the tangents from E and F to the circle is equal to $EF^2$

[hide]Construct tangents from EP and FQ. (for sake of my proof, let P be in $\triangle OEF$)

Apply Pythag to $\triangle EOP$ and $\triangle FOQ$. Then $EO^2=r^2+EP^2, FO^2 = r^2 + FQ^2$

Cosine rule in $\triangle OEF$: $EF^2=OE^2+OF^2-2cos\angle FOE.OE.OF$

So it's sufficient to prove $r^2=cos\angle FOE.OE.OF$

Let H be on OE such that FH is perpendicular to OE.

Then $cos\angle FOE.OE.OF = OH.OE$

Extend PH to meet circle at P'. OH is perpendicular to PP' so PH=P'H.  $\triangle PHE \equiv \triangle P'HE (SAS) so \angle POP' = 2\angle POH$

But by alternate segment, $\angle EPP'=\angle EPH = \angle PDP'=\frac{1}{2}\angle POP'=\angle POH$. Thus $\triangle OHP$ is right angled so P lies on FH.  Then $\triangle OPH \simeq \triangle OEP \implies OH.OE=OP^2=r^2$

I know that it's written up badly but you get the method and the proof. I'll fix it up later, cos I'm falling asleep right now.  :| [/hide]


You remember that girl that the main character from the Wonder years liked? Did you know that she majored in mathematics and graduated suma cum laude from UCLA?

http://www.danicamckellar.com/

click on the mathematics section.

What's teh "Wonder years"

And what's the point? I'm sure there are tons of other girls who have also majored in mathematics and graduated summa cum laude.

[quote="drunner2007"] Did you know that she majored in mathematics and graduated suma cum laude from UCLA?
[/quote]
Yes.

[quote="funcia"]Why does it matter that some girl named Danica majored in mathematics and graduated suma cum laude from UCLA? :?:[/quote]
She's famous because this is her, from the Wonder Years :
[img]http://content.answers.com/main/content/img/PosterThumbs/danicamckellar.jpg[/img]

She's really famous and hardly anyone knows that she really likes math. If you never saw the wonder years you really wouldn't understand.

oh that must be why! i thought it was strange!

[quote="fredbel6"]She actually helps people solving some math problems on her website.

[quote="jackdillon"]$I\;never\;new\;that\;about\;her$[/quote]
knew :wink:[/quote]

I'am a badd sppeller....:(

And what's her website?


group theory
abstract algebra
superior algebra
superior algebra unsolved

$ (G, \star )$ is a group.

for  $ a,b \in G$ and $ m,n \in \mathbb{N} \ \ with \ \ gcd(n,m) \equal{} 1$ .

we have   $ \left\{\begin{array}{l} a^m \star b^m \equal{} b^m \star a^m \\
a^n \star b^n \equal{} b^n \star a^n \\
\end{array} \right.$

prove that  $ a \star b \equal{} b \star a$

Could you be more specific? I mean, are $ a$, $ b$, $ m$, and $ n$ fixed? Do some of them vary, etc.?
I mean, because If $ G\equal{}S_3$, $ a\equal{}(12)$, $ b\equal{}(123)$, $ m\equal{}2$, $ n\equal{}3$ are chosen, your conclusion is false.

Steve

@[b]kai_desmond11[/b]: You might have probably meant [b]gcd(m,n)=1[/b], [i]man[/i]... When is the lowest common multiple of two numbers actually equal to [b]1[/b]? Your proposal would be trivial in that case!

m,n and a,b are all fixed numbers.

I am not familiar with the notation $ S_3$, please write the group completely. because i don't understand it.
thanks.

[quote="coquitao"]@kai_desmond11: You might have probably meant [b]gcd(m,n)=1[/b], man... ;)[/quote]
yes , I am sorry. :blush:

$ S_3$ is the symmetric group on 3 elements.  The idea behind my counterexample is for $ a$ to have order $ m$, and $ b$ order $ n$, then your two hypotheses hold, but we can surely make $ ab\neq ba$.

Steve

[quote="coquitao"]Hi there, Mr. [b]Dalton[/b]...

I see your point, [i]man[/i]. Nevertheless, I think that [b]Desmond[/b] means to ask here whether the commutativity of all m-th powers and the commutativity of all n-powers in a given group G are a sufficient condition for G to be itself abelian.[/quote]

That would be a good question it self  :)  .
But I meant only for the specific $ a$ and $ b$ .

thanks for your counter example.

[quote="kai_desmond11"]
That would be a good question it self  :)  .
But I meant only for the specific $ a$ and $ b$ .[/quote]

Really? Because the other version is not that intractable, dear [b]Desmond[/b]. If M is the subgroup generated by all m-th powers and N the subgroup generated by all n-th powers, it follows that both of 'em are normal subgroups of G. The coprimality of m and n gives now that G = MN and $ M \cap N \subseteq \mathbf{Z}(G).$ To show that G is abelian it remains to see that the commutator of an element of M and an element of N is the identity element of G. From the normality of m and n we obtain at once that the commutator of any two such elements is always in the intersection of M and N, whence in the center of G and we're done...

Why does that commutator being in the center imply it's the identity?

And as far as rating your post, I don't think I've ever rated a post in my life.

Steve

[quote="SD5225272"]Why does that commutator being in the center imply it's the identity?[/quote]

If [a,b]=z where a is in M, b in N, and Z in Z(G), it'd follow that $ aba^{ \minus{} 1} \equal{} zb$. Hence, $ ab^{m}a^{ \minus{} 1} \equal{} z^{m}b^{m}$. Since a is the product of m-th powers, we have that $ b^{m}$ commutes with $ a$. Therefore, $ z^{m} \equal{} e$ and same for $ z^{n}$. Since there exists a linear combination of m and n that gives us 1, what is to be concluded?

[quote="SD5225272"]And as far as rating your post, I don't think I've ever rated a post in my life. [/quote]

Please forget about it, man... I know it couldn't have been you...


geometry
circumcircle
search
analytic geometry
perpendicular bisector
radical axis
power of a point

Triangle ABC inscribed the circle (O) whose diameter is BC. The incircle (I) touches AB,AC at D,E. M,N are the midpoints of AB,AC. MN cuts (O) at P,Q. Prove that D,E,P,Q are concylic.

The perpendicular bisector of DE is AI.
The line PQ is the same as MN, which is the mid-parallel of the triangle and is parallel to BC. Therefore the perpendicular bisector of PQ passes through O, and meet AI at the mid point of the arc BC which does not contain A. This is the center of the circle DEPQ.

This solution is a 'proof' from the past,
  a true deja vu!

  M.T.

[hide]The internal bisector $ AI$ of the $ \angle A$ cuts the circumcircle $ (O)$ again at $ K,$ midpoint of the arc $ BC$ opposite to $ A.$ Let $ \mathcal K$ be a circle with center $ K$ and radius $ KB \equal{} KC \equal{} KI.$ Take 2 tangent circles $ \mathcal U, \mathcal V$ with diameters $ AI, IK$ cutting $ (O)$ again at $ P, Q.$ $ \angle IQK$ is right ($ Q$ is the tangency point of the A-mixtilinear incircle with $ (O),$ not needed). $ KQ$ cuts $ BC$ at $ R.$ Inversion in $ \mathcal K$ takes $ (O)$ to the line $ BC$ and the line $ BC$ into $ (O)$ $ \Longrightarrow$ $ R$ is image of $ Q$ in this inversion. Perpendicular to $ KR$ at $ Q$ is a polar of $ R$ WRT the circle $ \mathcal K$ $ \Longrightarrow$ $ IR$ is tangent to $ \mathcal K$ at $ I.$  $ IR, AP, KQ$ are radical axes of the pairs $ (\mathcal U, \mathcal V)$, $ ((O), \mathcal U)$ and $ ((O), \mathcal V),$ they meet at their radical center $ R \in BC.$

Let $ \mathcal K'$ be a circle with center $ K$ and radius $ KE \equal{} KF.$ This circle cuts the circumcircle $ (O)$ at $ M, N$ and $ MN \parallel BC$ is the radical axis of $ (O), \mathcal K'.$ $ EF \parallel IR$ is radical axis of $ \mathcal K', \mathcal U.$ $ MN, EF$ meet at the radical center $ S$ of $ (O), \mathcal K', \mathcal U.$ $ MN$ is midline of the $ \triangle ABC$ $ \Longleftrightarrow$ $ S$ is the midpoint of $ AR$ $ \Longleftrightarrow$ $ EF \parallel IR$ cuts $ AI$ at its midpoint $ U$ (the center of $ \mathcal U$) $ \Longleftrightarrow$ $ EUF$ is a diameter of $ \mathcal U$ $ \Longleftrightarrow$ $ \angle EAF \equal{} \angle A$ is right.[/hide]

Dear Mathlinker,
this problem has been proposed by Toshio Seimiy pb 1671, Crux Mathematicorum 8 vol. 17 (1991) 237 and proved not synthetically.
I have find a synthetical proof (in spain) published in REVISTEOIM directed by prof. Francisco Bellot Rosado. You can also find this proof in french on my site:
http://perso.orange.fr/jl.ayme  (see vol. 1)
Sincerely
Jean-Louis

[quote="28121941"]Although it is true that the mid point K of the arc BC which does not contains A is the center of the searched circle, I have only proved that KP = KQ and KD = KE. So, for the moment I have not solved the problem.[/quote]

Dear F. , KP^2=KQ^2=KD^2=KE^2=\frac{a^2+bc}{2} when triangle ABC is right.

The proof of the above formula was posted by Virgil Nicula (but I am lazy to search the link  :) )
I intended to post my own inversive problem after the original problem have been solved.
[hide][b]Inversive problem:[/b]
The midline of triangle ABC (which parallel to BC) cuts circumcircle (ABC) at P,Q. The incircle of ABC touches AB,AC at D,E. Know that D,E,P,Q are concylic, show that the angle BAC is right.[/hide]

Unfortunately, the proof of yetti solved it.  :(

Thanks, dear mr.danh. Vigil Nicula is also amazing (same as yetti).
I was pretty sure of have read the problem in Crux, but my search do not was succesful. Thanks also to JLAyme for locate the problem and give more links.

An analytical solution is also possible. Taking B(-a/2,0),C(a/2,0) and supposing WLOG b>c, then A is in the second quadrant and its coordinates are A((c^2 - b^2)/2a,  bc/a) Then the distances KD and KP are equal, and the proof is finished.

I am enclosing a pdf file with the details of my analytical solution.
For more informations, the published solution in CRUX of this problem (1992, pp.216-218) was trigonometric.
I have named the file as "Seimiya", to honor one of the most prolific and clever geometers of the last years.


Prove that $\forall x\in (0;\frac{\pi}{2})$ we always have $tg(sinx)>sin(tgx)$
It's an ineq but I don't we can use any classic ineq here, so I post its in Algebra section

In this interval $tgx>x+\frac{x^{3}}{3}, x-\frac{x^{3}}{6}<sinx<1-\frac{x^{3}}{6}+\frac{x^{5}}{120}.$
If $x>arcsin(\pi /4)$ we have $tg(sinx)>1>sin(tgx)$. Let y=tgx. Because $tg(arcsin(\frac{\pi}{4})<1.27$ we can proof, that
$\frac{1}{\sqrt{1+y^{2}}}(1+\frac{y^{2}}{3(1+y^{2})})>1-\frac{y^{2}}{6}+\frac{y^{4}}{120}$ when $y<1.27$
I solve it a few years ago by thise method. But I don't know nice solution.

i have a similar inequality,cos(sinx)>sin(cosX) for all x in R ,but i think this one is easy


geometry
trapezoid
circumcircle
cyclic quadrilateral
geometry unsolved

Prove that from all quadrilaterals with 3 equal sides (whose lenghth is fixed) the trapezoid has largest area

Let a quadrilateral have 3 sides a and one side d, such that 0 < d < 3a. The quadrilateral with these sides and maximum possible area is cyclic. A cyclic quadrilateral with 3 equal sides is an isosceles trapezoid. For example, let a = AB = BC = CD, d = DA and let O be the quadrilateral circumcenter. The central angles $\angle AOB = \angle BOC = \angle COD$ are equal, hence, the central angles

$\angle AOC = \angle AOB + \angle BOC = \angle BOC + \angle COD = \angle BOD$

are equal and the chords AC = BD are also equal.  Therefore, the quadrilateral  internal angles $\angle ABC = \angle BCD$ spanning the equal arcs AC, BD of the quadrilateral circumcircle (O) are equal.


calculus
inequalities unsolved
inequalities

I don't think it is too hard( if it is true)
because if $t=\frac{a}{b}$
then we only need to prove $f(t)\leq \frac32$
I don't think this method is nice,but maybe enough.

Sorry,I don't think you are right.because when $a^{2}=1,b^{2}=3$ the equality hold.You may make some mistakes. :P

hmm...what`s wrong exactly? the equality case results from $AM-PM$ when $\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}=\frac{b}{\sqrt{9a^{2}+b^{2}}}$, make some calculs before talking please :blush:

[quote="pohoatza"][quote="libra_gold"][quote="pohoatza"]ugly thing :)

$\Rightarrow \frac{(3a^{2}+b^{2})^{2}}{3(a^{2}+b^{2})(9a^{2}+b^{2})}\leq \frac{1}{4}$

$\Rightarrow 9a^{4}+12a^{2}b^{2}+3b^{4}\geq 0$

[/quote]

This Step is error![/quote]

oh i`m sorry i misscalculated that step is equivalent with : $(3a^{2}-b^{2})^{2}\geq 0$ and now results imediatly the eq case. sorry for the mistake : :roll:[/quote]

Misscalculated again... It is equivalent to $(3a^{2}-b^{2})^{2}\leq 0$ which is obvious wrong which means your method is not correct.


Yale
college
Columbia
articles
geometry
3D geometry
sphere

Sorry if someone posted this already.

Elusive Proof, Elusive Prover: A New Mathematical Mystery

By DENNIS OVERBYE
Published: August 15, 2006
New York Times
[EXCERPT]

Three years ago, a Russian mathematician by the name of Grigory Perelman, a k a Grisha, in St. Petersburg, announced that he had solved a famous and intractable mathematical problem, known as the Poincaré conjecture, about the nature of space.

After posting a few short papers on the Internet and making a whirlwind lecture tour of the United States, Dr. Perelman disappeared back into the Russian woods in the spring of 2003, leaving the world’s mathematicians to pick up the pieces and decide if he was right.

Now they say they have finished his work, and the evidence is circulating among scholars in the form of three book-length papers with about 1,000 pages of dense mathematics and prose between them.

As a result there is a growing feeling, a cautious optimism that they have finally achieved a landmark not just of mathematics, but of human thought.

“It’s really a great moment in mathematics,” said Bruce Kleiner of Yale, who has spent the last three years helping to explicate Dr. Perelman’s work. “It could have happened 100 years from now, or never.”

[quote="Saul"]Looks like he got a perfect score at the IMO too.[/quote]

True!!! http://www.imo.org.yu/index.php?options=gl|imotres&p=kLLi3

New York Times....

Perelman was considered for the Fields Medal this year wasn't he?

Also, another proof was claimed by 2 mathematicians from China earlier this year.

[quote="yenlee"]I apologize in advance because I cannot help myself...

\begin{sarcasm}
Wow, this guy got a perfect score at the IMO?  That must mean that this Poincare thing is really hard!
\end{sarcasm}[/quote]

It's like noting that Serre won the Concours.  Not his biggest achievement, but looking back, it was an indicator.

Perelman got a perfect score at the historically difficult IMO of 1982.
The number of contestants was reduced from 8 to 4 per country that year, and the problems were made more difficult than ever before.
Very few people got high scores, and Perelman beat out several clever people who became well-known later.   He was 16 at the time, and so probably also quite young within the pool of top finishers.

It is very good as single-year IMO performances go, even if he went on to bigger and better things.

more about this [url]http://www.rambler.ru/db/news/msg.html?s=12&mid=8479771[/url](in russian).When coming to IMO the school he has finished(Number 239 in ST Petersburg) is still famous.This year in IMO 2oo6 [b]4[/b] from 6 participants from Russia were from that(his) school.

I just wanted to note that there is a discussion about [url=http://www.math.columbia.edu/~woit/wordpress/archives/000077.html]Perelman and the Poincare Conjecture at Columbia University.[/url]

[hide="Local Copy Of The Article"]One of the great stories of mathematics in recent years has been the proof of the Poincare conjecture by Grisha Perelman.  This has been one of the most famous open problems in mathematics and has been around for about one hundred years.  In technical terms the conjecture is that if a space is homotopically equivalent to a three-dimensional sphere it is homeomorphic to the three-sphere. In less technical terms it says that if you have a bounded three-dimensional space in which all loops can be shrunk down to points, it has to be the three-sphere.  In dimensions other than three the analog conjecture has been proved, but the case of three dimensions has resisted all attempts to solve it.

Perelman spent time as a visiting mathematician at Stony Brook, Berkeley and NYU, then went back to St. Petersburg where for eight years he seemed to disappear from mathematics research. In November 2002 he posted a [url=http://arxiv.org/abs/math.DG/0211159]preprint[/url] on the arXiv, which quickly drew a lot of attention.  He seemed to be claiming to have a proof of an even more general conjecture than Poincare, known as the Thurston Geometrization conjecture, but the way his preprint was written, it wasn't clear whether he was claiming to really have a proof.  The method he was using was one pioneered by my Columbia colleague Richard Hamilton, called the "Ricci flow method". This involves something like a renormalization group flow to a fixed point (for more about this, see the talks by Ioannis Bakas at a recent [url=http://www.forth.gr/onassis/lectures/2004-07-05/programme.html]conference[/url] in Crete). If you start with an arbitrary metric on a space you think might be a three-sphere, the hope was that Hamilton's Ricci flow would take you to the standard metric for the three-sphere.  Hamilton had made a lot of progress using his techniques, but as far as pushing them through to give a proof of Poincare, he was stuck.

In the spring of 2003, Perelman traveled to the US and gave talks at several places, including a long series at Stony Brook. By then he was explicitly claiming to have a proof, but few of the details were written down, although he did post two more preprints to the arXiv. His talks were major events in the math community, and at them he was able to answer anyone who asked for details on specific points of his argument.  He gave a somewhat informal talk at Columbia one Saturday, a talk that I attended sitting next to Hamilton, who was hearing Perelman speak for the first time. Hamilton was clearly very impressed, and soon thereafter he and most other experts began to become convinced that Perelman really did have a way of proving the conjecture. By now the situation seems to be that the experts are pretty convinced of the details of Perelman's proof for the Poincare conjecture.  The full Geometrization conjecture requires some more argument and I gather that Perelman is supposed to at some point produce another preprint with more about this.   A workshop was held a couple weeks ago about Perelman's work at Princeton and several people have been carefully working through the details needed to be completely sure the proof works. For this material, see a [url=http://www.math.lsa.umich.edu/research/ricciflow/perelman.html]web-site[/url] maintained at Michigan by Bruce Kleiner and John Lott.

One interesting part of this story is that the Poincare conjecture is one that the Clay Mathematics Foundation has put a one-million dollar price tag on.  There's an elaborate set of[url=http://www.claymath.org/millennium/Rules_etc]rules[/url] that Perelman should follow to collect his million dollars.  This is supposed to begin with the submission of a detailed proof to a well-known refereed journal, something Perelman hasn't done and shows no signs of doing.  As far as anyone can tell, his attitude is that he's not interested in the million dollars.  If you look closely at the rules, it doesn't necessarily have to be Perelman who writes up the proof.  Someone else may do it, with Perelman still getting the money.   Ultimately the question of the million dollars is to be decided by the Scientific Advisory Board of the Clay Mathematics Institute, and one question they will have to face is  whether to split the award between Perelman and Hamilton.

Another interesting question concerns the Fields medal, the most prestigious award in mathematics. These are awarded every four years at the International Congress of Mathematicians, the next one of which will take place in Madrid in the summer of 2006. One stipulation for the award of the Fields medal is that a recipient must be under the age of 40.  Seeing Perelman speak, I had assumed he was already at least forty, but this is not so clear.  No one seems to be sure exactly what his age is and whether he will be under 40 in 2006.  Some news reports from spring 2003 referred to him as being in his late 30s or even 40, some recent ones claim that he is now 37.  His first scientific paper was published in 1985, so he would have had to have been 19 or younger at the time to be under 40 in 2006. If Perelman really is under 38 now, he's a sure thing for a 2006 Fields medal.

For a really dumb news article about this, go [url=http://www.guardian.co.uk/uk_news/story/0,3604,1298728,00.html]here[/url] (no, proving the Riemann hypothesis won't bring down the internet, and Perelman's Poincare proof won't explain the nature of the universe).
[/hide]

He refused a prize : [url]http://news.bbc.co.uk/2/hi/science/nature/5274040.stm[/url] :maybe:

He was awarded Fields Medal this year, and declined.

I don't get it?  Some articles say the Poincare conjecture implies that a banana is the same as a football?  But isn't this something all topologists have always known? :maybe:
Would someone please, in correct mathematical formulations, explain once and for all what this conjecture says? I am familiar with homotopy, homology and (topological)n-varieties.

Is this the correct formulation  :
[quote]If a compact three-dimensional manifold M3 has the property that every
simple closed curve within the manifold can be deformed continuously to a point,
does it follow that M^3 is homeomorphic to the sphere S^3?[/quote]

I assume "can be deformed continuously to a point" is just another way of saying "simple connected"?
And what is his definition of threedimensional manifold?  The usual one?
[hide]- each point has open neighbourhood homemorphic to $\mathbb{R}^{3}$
- Haussdorf space
- existence of a countable basis[/hide]

Please scroll up to my post, you seem to know quite a bit about it, it would be a shame if you missed it because we simultaneously posted.

[quote="Gyan"][quote]Up to today (21/8/96), is there any rigoruos proof accepted in the math community about Poincaré Conjecture, made by Perelman, those two chinese people or any other?. I ask this because I saw in a lot of places that Perelman proved it, others said that it's not true, others that the chinese proved it, others not, so I'm a bit confused (actually, big confused)[/quote]

Almost everyone believes that Perelman proved it (or has a full proof), it was just that no body can make him write a paper (or do it formally).  Around 2002 or so he was in US (Stony Brook)and did gave a few lectures etc ... No body has  found any flaws  but some steps are not that easy to see.

Chinese (and some other noted mathematicians too) who have filled in the gaps etc (in completing  the proof etc)  give full credit to Perelman.

For example, to claim Millenium prize, one has to formally write and submit the paper and it seems it is unlikely because he just doesn't care enough to do it.

Funny thing is, no one even knows, how old (exactly) is he (He is near 40 years old - Fields Medal require an age under 40 ) .[/quote]

His attitude towards the problem is very interesting. He is so interested in the problem, and so dis-interested in everything else. Doesn't seem as flat out crazy as some famous mathmeticians in the past, but still a bit of a kooky guy. I'd be currious to hear his views on other things in the world besieds math ( and math too  :lol:  ).

[quote="Gyan"][quote]Up to today (21/8/96), is there any rigoruos proof accepted in the math community about Poincaré Conjecture, made by Perelman, those two chinese people or any other?. I ask this because I saw in a lot of places that Perelman proved it, others said that it's not true, others that the chinese proved it, others not, so I'm a bit confused (actually, big confused)[/quote]

Almost everyone believes that Perelman proved it (or has a full proof), it was just that no body can make him write a paper (or do it formally).  Around 2002 or so he was in US (Stony Brook)and did gave a few lectures etc ... No body has  found any flaws  but some steps are not that easy to see.

Chinese (and some other noted mathematicians too) who have filled in the gaps etc (in completing  the proof etc)  give full credit to Perelman.

For example, to claim Millenium prize, one has to formally write and submit the paper and it seems it is unlikely because he just doesn't care enough to do it.

Funny thing is, no one even knows, how old (exactly) is he (He is near 40 years old - Fields Medal require an age under 40 ) .[/quote]

But he knows that having a complete proof would be huge important to math, so why doesn't he just lost some time and write a complete proof? He's "hurting" math without doing it.

Another thing: He was born on the 13/6/66, so he's already 40

[url=http://www.icm2006.org/dailynews/fields_perelman_info_en.pdf]Here[/url] I found the real situation up-to-date about Poincaré and its generalization. (Read the 2º paragraph)

How do you know that having solved a particular mathematics problem by one mathematician is a "CERTIFIED GENIUS?" Would you consider an individual a genius(Regardles of
subject he contributed.)if such intellectual,individual solved a
given problem.Genius is not just solving problem:Its more on
originality.creativity,focus attention,and perseverance!To this include,ability to withstand the pressures.Of course,'QUALITY'
ones contributed ideas are important.Even great "Criminals" in history had their share of so -called genius.Don't just simply used,or believe overworn semantics.

                                                      Respectfully,
                                                     A.Eddington

more about him: [url]http://en.wikipedia.org/wiki/Grigory_Perelman[/url]

He hates any attention, but everyone do against what he want.

He want to be low profile. No attention whatsoever.

He don't want to appear in any news media but his story turns up in CNN,BBC,every major newspaper. Every major media say "Perelman get perfect 42 on IMO 1982. Perelman is strange. Perelman lives with mother. Perelman declined Fields medal, Clay award".

He don't want to appear on the internet but every math discussion forum talk about him. Every math discussion forum say "Perelman get perfect 42 on IMO 1982. Perelman is strange. Perelman lives with mother. Perelman declined Fields medal,clay award".

Perelman hates all this attention.

Perelman want to be treated like the average joe who work an average office job walking on the street where everyone do not pay attention to. No newspaper talk about average joe, no math forum talk about average joe. That's what perelman want, be an average joe that nobody pay attention to.

Here's a great article by The New Yorker on the issue [url]http://www.newyorker.com/fact/content/articles/060828fa_fact2[/url]


trigonometry
inequalities
geometry unsolved
geometry

A convex polygon $ A_1A_2 . . .A_n$ is inscribed in a circle with center $ O$ and radius $ R$ so that $ O$ lies inside the polygon. Let the inradii of triangles $ A_1A_2A_3, A_1A_3A_4, . . . , A_1A_{n\minus{}1}A_n$ be denoted by $ r_1, r_2, . . . , r_{n\minus{}2}$. Show that:
$ r_1 \plus{} r_2 \plus{}...\plus{} r_{n\minus{}2} \le R\bigg( n \cos{\frac {\pi}{n}} \minus{} n \plus{} 2\bigg)$.

By Mikami and Kobayashi theorem for the cyclic polygon $ A_1A_2A_3A_4 \ldots A_n,$ we have

$ r_1 \plus{} r_2 \plus{} r_3 \plus{} r_4 \plus{} \ldots \plus{} r_{n \minus{} 2} \equal{} \delta_1 \plus{} \delta_2 \plus{} \delta_3 \plus{} \delta_4 \plus{} \ldots \plus{} \delta_n \minus{} R \cdot (n \minus{} 2) \ , (*)$

Where $ \delta_1, \delta_2 , \delta_3, ... ,\delta_n$ denote the  distances from $ O$ to $A_1A_2,A_2A_3,A_3A_4 \ldots A_nA_1.$ Let $ \phi_1,\phi_2,\phi_3 , \ldots , \phi_n$ denote the corresponding central angles of the chords $ A_1A_2,A_2A_3,$ etc $\Longrightarrow$ $ \delta_1 \equal{} R \cdot \cos \frac {\phi_1}{2}$ and similarly for $ \delta_2 , \delta_3,\ldots, \delta_n$

Substituting $\delta_n$ into $ (*),$ the subsequent sum of the radii equals

$R \cdot \left ( \cos \frac {\phi_1}{2} \plus{} \cos \frac {\phi_2}{2} \plus{} \cos \frac {\phi_3}{2} \plus{} ... \plus{} \cos \frac {\phi_n}{2} \right) \plus{} R \cdot (2 \minus{} n)$

Since $ O$ is inside the polygon, then $\frac {\phi_n}{2}$ is not greater than $ \frac {\pi}{2}.$ Thus, by Jensen's inequality

$ \cos \frac {\phi_1}{2} \plus{} \cos \frac {\phi_2}{2} \plus{} \ldots \plus{} \cos \frac {\phi_n}{2} \le n \cdot \cos \frac {\pi}{n}$

$ \Longrightarrow \ r_1 \plus{} r_2 \plus{} r_3 \plus{} r_4 \plus{} \ldots \plus{} r_{n \minus{} 2} \le R \left(n \cdot \cos \frac {\pi}{n} \plus{} 2 \minus{} n \right)$


If $ a,b,c>0$, then

$ \sum\frac{a^2\plus{}b^2}{(2c\plus{}b)^2}\plus{}2\sum\frac{ab}{(2a\plus{}c)(2c\plus{}b)}\geq\frac{4}{3}$

[quote="lipako"]If $ a,b,c > 0$, then

$ \sum\frac {a^2 \plus{} b^2}{(2c \plus{} b)^2} \plus{} 2\sum\frac {ab}{(2a \plus{} c)(2c \plus{} b)}\geq\frac {4}{3}$[/quote]

only prove

$ \sum\frac {a^2 \plus{} b^2}{(2c \plus{} b)^2} \geq \frac {2}{3}$

$ \sum\frac {ab}{(2a \plus{} c)(2c \plus{} b)}\geq\frac {1}{3}$

and

$ \sum\frac {a^2 \plus{} b^2}{(2c \plus{} b)^2} \geq \frac {(\sum a^2)^2 \plus{} (\sum ab)^2}{\sum (a^2(2c \plus{} b)^2)}$

only prove

$ \frac {(\sum a^2)^2 \plus{} (\sum ab)^2}{\sum (a^2(2c \plus{} b)^2)}\geq \frac {2}{3}$

 $ \Longleftrightarrow \minus{} 2a^2cb \minus{} a^2c^2 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} 2ab^2c \minus{} 2abc^2 \plus{} 3a^4 \plus{} 3b^4 \plus{} 3c^4\geq 0..........(*)$

assume $ a \equal{} max{(a,b,c)}$

 $ (*)\Longleftrightarrow (3a^2 \plus{} (3c \plus{} 3b)a \plus{} bc \plus{} 2c^2 \plus{} 2b^2)(a \minus{} b)(a \minus{} c) \plus{} ((2b \plus{} 2c)a$ $ \plus{} 3b^2 \plus{} 4bc \plus{} 3c^2)(b \minus{} c)^2\geq 0$

and

$ \sum\frac {ab}{(2a \plus{} c)(2c \plus{} b)}\geq\frac {1}{3}\Longleftrightarrow 2a^2c \plus{} 2b^2a \plus{} a^2b \plus{} 2bc^2 \plus{} c^2a \plus{} b^2c \minus{} 9abc\geq 0$

from AM-GM the inequality holds.


Σ=(n+1/n+2)(1/2^n)

Σ=(a^n.n^n/n!)  for a>0

n=1---> Infininty
Which criteria are we going to use?

use the comparison test for the first one...with 1/2^n

these kind of questions belong in the "calculus computations" section.

for the second, there is the ratio test. The ratio of the $(n+1)$st term to the $n$th is equal to
\[a\left( 1+\frac{1}{n}\right)^{n}\]
so when $a<\frac{1}{e}$, it converges, and when $a>\frac{1}{e}$, it diverges. That leaves the mysterious case $a=\frac{1}{e}$...

[quote="scorpius119"] That leaves the mysterious case $a=\frac{1}{e}$...[/quote]

By Stirling's approximation, this is asymptotically

$\sum{ \frac{1}{ \sqrt{2 \pi n}}}$

Which diverges by p-series.


In a triangle ABC, prove that
a)a cosA + b cosB = c cos(A-B)
b)(a+b-c)/(a+b+c)= tan(A/2)tan(B/2)
c)If A = pi/3, prove that
   1/(a+b) + 1/(a+c) = 3/(a+b+c)

[color=red][b]1.[/b][/color] Divide the equation by the rigth side:

$(?)\ \frac{a \cos \alpha + b \cos \beta}{c \cos{(\alpha - \beta)}} = 1$

Substitute $a = 2R \sin \alpha$, $b = 2R \sin \beta$, $c = 2R \sin \gamma$, where $R$ is the triangle circumradius, and use $\sin \gamma = \sin{(\alpha + \beta)}$. Then use the following identity, evident if you express $x, y$ as $x = \frac{x + y}{2} + \frac{x - y}{2}$, $y = \frac{x + y}{2} - \frac{x - y}{2}$:

$\sin x + \sin y = 2 \sin{\frac{x + y}{2}} \cos{\frac{x - y}{2}}$

$\frac{a \cos \alpha + b \cos \beta}{c \cos{(\alpha - \beta)}} = \frac{\sin \alpha \cos \alpha + \sin \beta \cos \beta}{\sin \gamma \cos{(\alpha - \beta)}} =$

$= \frac{\sin 2 \alpha + \sin 2 \beta}{2\sin{(\alpha + \beta)} \cos{(\alpha - \beta)}} = \frac{\sin 2 \alpha + \sin 2 \beta}{ \sin{2 \alpha} + \sin{2 \beta}} = 1$

[color=red][b]B.[/b][/color] Substitute $a = 2R \sin \alpha$, $b = 2R \sin \beta$, $c = 2R \sin \gamma$, where $R$ is the triangle circumradius, and use $\sin \gamma = \sin{(\alpha + \beta)}$:

$\frac{a + b - c}{a + b + c} = \frac{\sin \alpha + \sin \beta - \sin \gamma}{\sin \alpha + \sin \beta + \sin \gamma} = \frac{\sin \alpha + \sin \beta - \sin{(\alpha + \beta)}}{\sin \alpha + \sin \beta + \sin{(\alpha + \beta)}} = $

$= \frac{2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}} - 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha + \beta}{2}}}{2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}} + 2 \sin{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha + \beta}{2}}} = \frac{\cos{\frac{\alpha - \beta}{2}} - \cos{\frac{\alpha + \beta}{2}}}{\cos{\frac{\alpha - \beta}{2}} + \cos{\frac{\alpha + \beta}{2}}} =$

$= \frac{-2 \sin{\frac{2 \alpha}{2}} \sin{\frac{-2 \beta}{2}}}{2 \cos{\frac{2 \alpha}{2}} \cos{\frac{2 \beta}{2}}} = \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = \tan \alpha \tan \beta$

[color=red][b]C.[/b][/color] Multiply both sides by $a + b + c$ and rearrange:

$(?)\ \frac{a + b + c}{a + b} + \frac{a + b + c}{a + c} = 3$

$(?)\ \frac{c}{a + b} + \frac{b}{a + c} = 1$

Substitute $a = 2R \sin \alpha$, $b = 2R \sin \beta$, $c = 2R \sin \gamma$. Use the facts that $\alpha = \frac \pi 3$, $\gamma = \pi - (\alpha + \beta) = \frac{2 \pi}{3} - \beta$,

$\sin{\frac \pi 3} = \sin{\frac{2 \pi}{3} = \frac{\sqrt 3}{2}}$, $\cos{\frac \pi 3} = -\cos{\frac{2\pi}{3} = \frac 1 2}$:

$\frac{c}{a + b} + \frac{b}{a + c} = \frac{\sin \gamma}{\sin \alpha + \sin \beta} + \frac{\sin \beta}{\sin \alpha + \sin \gamma} = \frac{\sin^2{\beta} + \sin^2{\gamma} + (\sin \beta + \sin \gamma ) \sin \alpha}{\sin^2{\alpha} + \sin \beta \sin \gamma + (\sin \beta + \sin \gamma) \sin \alpha }$

$\sin^2{\beta} + \sin^2{\gamma} = \sin^2{\beta} + \sin^2{( \frac{2 \pi}{3} - \beta )} = \sin^2{\beta} + ( \frac{\sqrt 3}{2} \cos \beta + \frac 1 2 \sin \beta )^2 =$

${= \frac 5 4 \sin^2{\beta} + \frac 3 4 \cos^2{\beta} + \frac{\sqrt 3}{2} \sin \beta \cos \beta = \frac 3 4 + \sin \beta ( \frac 1 2 \sin \beta + \frac{\sqrt 3}{2} \cos \beta} ) = $

${= \frac 3 4 + \sin \beta ( -\cos{\frac{2 \pi}{3}} \sin \beta + \sin{\frac{2\pi}{3}} \cos \beta} ) = \frac 3 4 + \sin \beta \sin{( \frac{2 \pi}{3} - \beta )} =$

$= \frac 3 4 + \sin \beta \sin \gamma$

In addition, $\sin^2{\alpha} = \sin^2{\frac{\2 \pi}{3}} = \frac 3 4$. As a result,

$\frac{c}{a + b} + \frac{b}{a + c} = \frac{\frac 3 4 + \sin \beta \sin \gamma + (\sin \beta + \sin \gamma ) \sin \alpha}{\frac 3 4 + \sin \beta \sin \gamma + (\sin \beta + \sin \gamma) \sin \alpha } = 1$


function
combinatorics unsolved
combinatorics

For $r,n\in \mathbb {N}$ ,let $f(r,n)$ denote the number of partitions of $n$ of the form $n=n_1+n_2+...+n_s$ where ,for $i=1,2,...,s-1$ , $n_i\geq rn_{i+1}$, and let $g(r,n)$ denote the number of partitions of $n$ ,where each part is of the form $1+r+r^2+...+r^k$ for some $k\in \mathbb {N^*}$. Show that \[ f(r,n)=g(r,n) \]

There's a simple bijective proof.

To each $s$-uple of non-negative integers $(n_1,n_2,\ldots,n_s)$ such that $n_19+\ldots+n_s=n$ and $n_i\ge rn_{i+1},\ \forall 1\le i\le s-1$ we associate the $s$-uple $(a_1,\ldots,a_s)$ of non-negative integers such that $n=a_1+a_2(1+r)+\ldots+a_s(1+\ldots+r^{s-1})$ as follows: take $a_s=n_s,\ a_{s-1}=n_{s-1}-rn_s,\ldots, a_1=n_1-rn_2$.

It's easy to check that the application $(n_i,\ i=\overline{1,s})\mapsto(a_i,\ i=\overline{1,s})$ is a bijection.

thanks for your nice solution, grobber and i think it isn't easy for me to make a bijective function  :D


quadratics
Vieta
analytic geometry
conics
parabola
graphing lines
slope

$ \boxed{1}$ Let $ a,\ b,\ c$ be postive integers such that $ a \plus{} b \plus{} c,\ a^2 \plus{} b^2 \plus{} c^2,\ a^3 \plus{} b^3 \plus{} c^3$ are the multiple of 7.

(1) Prove that $ ab \plus{} bc \plus{} ca$ is the multiple of 7.

(2) Prove that $ abc$ is the multiple of 7.

(3) Prove that all of $ a,\ b,\ c$ are the multiple of 7.

$ \boxed{2}$ Let $ N$ be the number of the real root of the equation with respect to $ x: (x^2 \plus{} a|x| \plus{} a \plus{} 1)(x^2 \plus{} (a \plus{} 1)|x| \plus{} a) \equal{} 0$.
Note that you count the double root as one solution.

(1) Find the condition of $ a$ such that $ N \equal{} 0$.

(2) Find the condition of $ a$ such that $ N \equal{} 4$.

(3) Find the maximum value of $ N$ and the condition of $ a$ for which $ N$ is maximal.

For #1 we can start with part 3 and just use residues, make a chart mod 7
$ n^1$ 0 1 2 3 4 5 6
$ n^2$ 0 1 4 2 2 4 1
$ n^3$ 0 1 1 6 1 6 6

From the residues of $ n^3$ we find one must be 0 mod 7, the other 1, and the last 6 mod 7, or all three are 0 mod 7. Assume the first case is true.  Then the residues of a,b,c are either (0,1,6), (0,2,5), (0,3,4), but none of these work from the $ n^2$ chart. So they are all divisible by 7.

$ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2\left( ab\plus{}bc\plus{}ca \right)$

$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)\left( a^2\plus{}b^2\plus{}c^2 \minus{}ab\minus{}ac\minus{}bc \right)$

I'm not sure if this is right, but:

[hide="2"]
(1) We need the discriminant of both factors to be less than zero.

Two cases:

Case 1. For $ x \ge 0$, we must have

$ a^2 - 4(a + 1) < 0 \implies (a - 2)^2 < 5 \implies a \in (-\sqrt{5} + 2,\sqrt{5} + 2)$

and

$ (a + 1)^2 - 4a < 0 \implies (a - 1)^2 < 0 \implies a \in (-\infty,1)$

Finding the intersection of the solution sets, we have $ \boxed{a \in (-\sqrt{5} + 2, 1)}$.

Case 2. For $ x < 0$, we have the exact same conditions as Case 1 because the negative is removed when the coefficient of $ x$ is squared.

(2) We must have both of the discriminants be equal to 0.

Case 1. For $ x \ge 0$, we have $ \boxed{a \in \{\}}$. (The two discriminants can never both be 0.)
Case 2. For $ x < 0$, same thing occurs.

(3) We notice that if $ x_1$ is a root of the quartic, then so will $ -x_1$. Therefore, $ N \le 8$. This condition occurs when both roots and real and distinct. Note that when we have quadratics of the form $ ax^2 - c$ have two roots which are negations of each other, so we must have $ b \not= 0$. We must have the discriminant be strictly greater than 0.

Case 1. For $ x \ge 0$, $ a \in (-\infty, -\sqrt{5} + 2) \cup (\sqrt{5} + 2), \infty)$.
Case 2. For $ x < 0$, ditto.

The interval contains a value that will make $ a$ $ 0$ in one of the factors, specifically, $ -1$. We must exclude that from our set. Therefore, ${ \boxed{a \in (-\infty, -1) \cup (-1,-\sqrt{5} + 2) \cup (\sqrt{5} + 2), \infty)}}$.

[/hide]

[hide="Question 2"]
By solving for $ |x|$, we get

$ |x|_{1,2} \equal{} { \minus{} a\pm\sqrt {a^2 \minus{} 4a \minus{} 4}\over 2}, |x|_3 \equal{} \minus{} 1,|x|_4 \equal{} \minus{} a$

(1) From the expression for $ |x|_4$, obviously $ a > 0$, and in that case $ |x|_{1,2}$ will yield no solution too - either the discriminant will be negative, or $ \minus{} a \plus{} \sqrt {a^2 \minus{} 4a \minus{} 4} < 0$.

Hence $ a > 0$.

(2) For more than zero solutions, we necessarily must have $ a < 0$, but in that case $ |x|_4$ yields two solutions. Hence the discriminant for $ |x|_{1,2}$ must be zero, since otherwise we'd get four simple roots from that branch, hence six in total: $ a^2 \minus{} 4a \minus{} 4 \equal{} 0\implies a \equal{} 2 \minus{} 2\sqrt {2}$ (since we discard the positive branch). For that value, $ |x|_{1,2}$ will be positive, so we'll have four solutions.

Hence $ a \equal{} 2 \minus{} 2\sqrt {2}$

(3) Maximum number of solutions is six, when all of $ |x|_{1,2}$ and $ |x|_4$ are positive. As in (2), we must have $ a < 0$. Let's put $ b \equal{} \minus{} a$, so $ b > 0$. Then $ |x|_{1,2} \equal{} {b\pm\sqrt {b^2 \plus{} 4b \minus{} 4}\over 2}$, so we must examine $ b \minus{} \sqrt {b^2 \plus{} 4b \minus{} 4}>0\iff 4b \minus{} 4<0\iff b<1\iff a>\minus{} 1$. Also, the discriminant must be positive, and from (2) we get that this means $ a < 2 \minus{} 2\sqrt {2}$.

Hence $ \minus{} 1<a < 2 \minus{} 2\sqrt {2}$[/hide]

[hide="Q1, Part 3, alternative"]
From using the results from part 1 and 2

$ (a\plus{}k)(b\plus{}k)(c\plus{}k) \equiv abc\plus{}k(ab\plus{}bc\plus{}ca) \plus{} k^2(a\plus{}b\plus{}c) \plus{} k^3 \equiv k^3 \mod 7$ for some integer $ k$.

Now suppose one of $ a,b,c\not\equiv 0 \mod 7$, wlog let it be a. Now we can chose some $ k\not\equiv 0 \mod 7$ such that $ a\plus{}k\equiv 0 \mod 7$

$ \therefore (a\plus{}k)(b\plus{}k)(c\plus{}k) \equiv 0 \mod 7$ but $ k^3\not\equiv 0 \mod 7$ contradiction!

Hence $ a\equiv b\equiv c\equiv 0 \mod 7$[/hide]

Nice, ocha! Can we generalize this problem to all primes except for 2?

[quote="MellowMelon"]Vieta's relations work in any field (in particular, the finite one with $ p$ elements). However, the proofs of a) and b) require that the field is not characteristic $ 2$ or $ 3$ (that is, $ 2 \not\equiv 0$ and $ 3 \not\equiv 0$), so you might need $ p > 3$.[/quote]

Sorry, I don't really know what a field is. Are you saying that;

If we have positive integers $ a,b,c$ and some prime $ p$, such that $ p|(a\plus{}b\plus{}c), p|(a^2\plus{}b^2\plus{}c^2)$ and $ p|(a^3\plus{}b^3\plus{}c^3)$
Then
1. $ (a\plus{}b\plus{}c)^2 \equal{} a^2\plus{}b^2\plus{}c^2 \plus{} 2(ab\plus{}bc\plus{}ca)$
  which implies $ 2(ab\plus{}bc\plus{}ca) \equiv 0 \mod p \Longrightarrow (ab\plus{}bc\plus{}ca) \equiv 0 \mod p$ if $ p\not \equal{} 2$

2. $ a^3\plus{}b^3\plus{}c^3\minus{}3abc \equal{} (a\plus{}b\plus{}c)(a\minus{}b)(b\minus{}c)(c\minus{}a)$
   Which implies $ 3abc \equiv 0 \mod p \Longrightarrow abc \equiv 0 \mod p$ if $ p\not \equal{} 3$

Therefore we can generalise for $ p>3$

[quote="ocha"][quote="MellowMelon"]Vieta's relations work in any field (in particular, the finite one with $ p$ elements). However, the proofs of a) and b) require that the field is not characteristic $ 2$ or $ 3$ (that is, $ 2 \not\equiv 0$ and $ 3 \not\equiv 0$), so you might need $ p > 3$.[/quote]

Sorry, I don't really know what a field is. Are you saying that;

If we have positive integers $ a,b,c$ and some prime $ p$, such that $ p|(a \plus{} b \plus{} c), p|(a^2 \plus{} b^2 \plus{} c^2)$ and $ p|(a^3 \plus{} b^3 \plus{} c^3)$
Then
1. $ (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)$
  which implies $ 2(ab \plus{} bc \plus{} ca) \equiv 0 \mod p \Longrightarrow (ab \plus{} bc \plus{} ca) \equiv 0 \mod p$ if $ p\not \equal{} 2$

2. $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a \minus{} b)(b \minus{} c)(c \minus{} a)$
   Which implies $ 3abc \equiv 0 \mod p \Longrightarrow abc \equiv 0 \mod p$ if $ p\not \equal{} 3$

Therefore we can generalise for $ p > 3$[/quote]

You are right.

I have not received correct answer to Problem 2, (2), (3).

[hide="Appendix to Question 2"](2) For $ a<\minus{}1$, $ |x|_{1,2}$ will give one positive and one negative number, hence there would be two solutions. Combined with two solutions coming from $ |x|_4$, that gives us four.

Hence $ a\equal{}2\minus{}2\sqrt{2}\lor a<\minus{}1$

(3) With $ a<0$, let's put $ b\equal{}\minus{}a$, so $ b>0$. Then for six solutions, all of the following must be satisfied:

$ b>0$

$ b^2\plus{}4b\minus{}4>0$

$ {b\plus{}\sqrt{b^2\plus{}4b\minus{}4}\over 2}>0$

$ {b\minus{}\sqrt{b^2\plus{}4b\minus{}4}\over 2}>0$

The second condition (combined with the first) gives $ b>2\sqrt{2}\minus{}2$. The third is then automatically satisfied. The fourth gives $ b^2>b^2\plus{}4b\minus{}4\iff b<1$. Hence $ b\in(2\sqrt{2}\minus{}2,1)\iff a\in(\minus{}1,2\minus{}2\sqrt{2})$, so the previous answer to (3) was correct.[/hide]

Now all of problem have been solved.

@Farenhajt Your answer to Problem (3) was correct. :blush:

Comment for Problem 2

Let $ |x|\equal{}t\ (t\geq 0)$, we are to consider the equation $ (t^2\plus{}at\plus{}a\plus{}1)(t\plus{}1)(t\plus{}a)\equal{}0$, especially
the number of real roots of the quadratic equation : $ t^2\plus{}at\plus{}a\plus{}1\equal{}0$ can be the $ t$ coordinate of the intersection point of the parabola $ y\equal{}\minus{}t^2\minus{}1\ (t\geq 0)$
and the line $ y\equal{}a(t\plus{}1)$ which passes through the point $ (\minus{}1,\ 0)$ in regardless of the slope $ a$.

[color=blue] 1st place: mnjm   :winner_first:  :first:
  2nd place: iijm  :winner_second:
3rd place: tucky :winner_third: [/color]

$ \displaystyle \displaystyle \displaystyle \begin{tabular}{|c|c|c|c|c|c\parallel{}c|}\hline Name & Problem 1 & Problem 2 & Total Score \\
\hline mnjm & 48 & 39 & 89 \\
\hline iijm & 36 & 29 & 65 \\
\hline tucky & 50 & 6 & 56 \\
\hline Average & 23.2 & 4.1 & 27.3 \\
\hline\end{tabular}$

Entry 55


wat is the arml?

and wen is teh ct arml?

ARML=American Regional Mathematics League

http://www.arml.com

just the forum to get questions answered.. :)


Prove that in an acute triangle \[ \frac {\pi}3\le\frac {\alpha a^2\plus{}\beta b^2\plus{}\gamma c^2}{a^2\plus{}b^2\plus{}c^2}<\frac {\pi}2\]
where $ a$, $ b$, $ c$ are the sides, $ \alpha$, $ \beta$, $ \gamma$ are the angles of the triangle respectively.

(LHS) if $ a \geq b \geq c$, $ \alpha \geq \beta \geq \gamma$, and Chebyshev Inequality kills.

(RHS) Obviously $ \alpha < \frac {\pi}{2}$ and analogously.

In a  triangle  , We have
 \[ (\frac {\pi}3)^n\le\frac {\alpha a^n\plus{}\beta b^n\plus{}\gamma c^n}{a\plus{}b\plus{}c}<\frac {\pi^n}2\]
where $ a$, $ b$, $ c$ are the sides, $ \alpha$, $ \beta$, $ \gamma$ are the angles of the triangle respectively, $ n$ is positive integer.


I have one question regarding the proof of triangle inequality. Consider the proof below. I have read this in a book.

$|x+y| \leq  |x|+|y|$
${|x+y|}^{2}\leq {(|x|+|y|)}^{2}$

$|x+y|^{2}= x^{2}+2xy+y^{2}= |x|^{2}+2|xy|+|y|^{2}= (|x|+|y|)^{2}$

Since ${|x+y|}^{2}\leq {(|x|+|y|)}^{2}$, therefore
$|x+y| \leq  |x|+|y|$

Why did $2xy$ became $2|xy|$? By the way the reason in the book was that $x \leq |x|$. I just don’t get it.

[quote="guile"][quote="ArcticMonkey"]Probably it should be \[|x+y|^{2}= x^{2}+2xy+y^{2}\leq |x|^{2}+2|xy|+|y|^{2}= (|x|+|y|)^{2}\] [/quote]

But how did $2xy$ became $2|xy|$

Can you state the reason?[/quote]

the value of a number is always less than or equal to it's absolute value.

basically, you can view |x| and |y| to be two variables, say a and b, then when you square their sum, the middle term is 2ab and you can substitute back in to get 2|x||y|=2|xy|.

Also, you have

$\left|a+b\right|^{2}=\left(a+b\right)\cdot\left(a+b\right)=\left|a\right|^{2}+2a\cdot b+\left|b\right|^{2}$

By the Cauchy-Schwartz Inequality, $a\cdot b\leq \left|a\right|\left|b\right|$ so you have

$\left|a+b\right|^{2}\leq \left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}=\left(\left|a\right|+\left|b\right|\right)^{2}$

which completes the proof.


[quote="vittasko"][b][size=100][color=DarkBlue]An equilateral triangle $ \bigtriangleup ABC$ is given with circumcircle $ (O)$ and let $ \bigtriangleup A'BC,\ \bigtriangleup B'AB,\ \bigtriangleup C'AB$ be, three triangles erected on its side-segments $ BC,\ AC,\ AB$ respectively, outwardly ( or inwardly ) to it, such that $ \angle A'BC \equal{} \angle B'AC \equal{} \angle C'AB \equal{} \angle \omega < 120^{o}.$ The line segments $ A'B,\ A'C,\ B'C,\ B'A,\ C'A,\ C'B,$ intersect the circumcircle $ (O)$ of the given triangle at the points so be it $ K,\ L,\ M,\ N,\ P,\ Q,$ respectively. Prove that the lines connecting the midpoints of the opposite side-segments of the cyclic hexagon $ KLMNPQ$ ( convex or non-convex ), are concurrent at one point.[/color][/size][/b]

Kostas Vittas.[/quote]
Dear [b]Kostas Vittas[/b], the problem seems nice but it is not right according to my figure?! Can you check the problem again. Thanks in advance


geometry
parameterization
integration
trigonometry
calculus
calculus computations

is there a simple way to find the area under the curve when you have a parametric curve, without having to eliminate the parameters

Sure. Just integrate $\int y\,dx.$ Just make sure you draw plenty of pictures so you can see what you're doing.

Example: what is the area under one arch of a cycloid? The cycloid is given by

$\begin{array}{c}x=at-a\sin t\\ y=a-a\cos t\end{array}$

Then $dx=(a-a\cos t)\,dt.$ The area under the curve is

$\int y\,dx=\int_{0}^{2\pi}(a-a\cos t)^{2}\,dt$

$=a^{2}\int_{0}^{2\pi}(1-2\cos t+\cos^{2}t)\,dt=a^{2}(2\pi+0+\pi)=3\pi a^{2}.$


The first 20 numbers of an arrangement are shown below. What would be the value of the $ 40^{\mathrm{th}}$ number if the arrangement were continued?\\
Row 1: 2, 2\\
Row 2: 4, 4, 4, 4\\
Row 3: 6, 6, 6, 6, 6, 6\\
Row 4: 8, 8, 8, 8, 8, 8, 8, 8\\

there are total 2+4+6+8+10=30 numbers until Row 6 begins.

40th number should be in the Row 6, and Row 6 is made of 12s.

answer : 12


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Please help me -  :lol:
Let R be a commutative integral domain integral closed. Prove that if I is the maximal reflexive ideal then I is a prime ideal and the height of I is one.


[quote="dead222"][quote="Mamat"]Hello guys. I need to find Kyrgyzstan IMO team :). If you see them could you tell them to visit this page and write about the olympiad. All of the except one know me so it won't be so difficult. Thx.[/quote]

sorry its russian
hi Mamat.  u nas 3 gramoti i 1 bronza ( po moemy uge okonchatelno)ermeku ne hvatilo 1 bala hotia govorit chto v 1 zadache vse bilo pravilno interesno uvidet6 v chem on oshibalcia)   no kak vidish > chem v proishlom gody) v prinsipe 3 zadachi bili dostypnimi dla kagdogo mb ne smogli sobratcia mb ploho gotovilic6 (ia sovsem ne gotovilcia po etomu manshion vpolne normalno:) posle togo kak ne hvatilo 1 bala na gaotikovskoi hoyia nekotorii sudii ego dali a kazahi zabrali propalo vse gelanie naverno vse zavisit skoree ne ot togo skolko ti gotovishcia)mb tyt prosto ochen6 garko mb ne sud6ba )bronza u 10 kl tak chto nadeys6 na sled god bydyt eshe lutshii resultati)
                                                                                balai.[/quote]

It says: We have 3 mansions and 1 bronze. Ermek needed 1 point to get bronze  :( , we hope next year it will be better.

Sorry for russian guys.

[quote="Mamat"]Hello guys. I need to find Kyrgyzstan IMO team :). If you see them could you tell them to visit this page and write about the olympiad. All of the except one know me so it won't be so difficult. Thx.[/quote]
Privet Mamat. Kak uje znaesh, u nas 1 bronza i 3 gramoty. U men'a na 1 ball men'she bronzy. Dumal 1 zadachu reshil na
vse 100. Ostal'noe ty uje znaesh. prosti chto ne opravdali nadejdy. Jalko chto eto posledn'aa olimpiada dl'a nas. Kak sam. Znau ty seichas v Bremene. Molodets, chto ne zabyl nas. Esli chto - pishi. Do vstrechi.
Ermek  (ermek_01@mail.ru)

sorry again :D

Ermek glavnoe chtoby ty byl dovolen! Vash rezultat odin iz luchshih, ved' vas bylo vsego 5 chelovek. Nu davay mne itdi nado :), do vstrechi.

[quote="ducpbc"][quote="Mamat"]Hello guys. I need to find Kyrgyzstan IMO team :). If you see them could you tell them to visit this page and write about the olympiad. All of the except one know me so it won't be so difficult. Thx.[/quote]

You can go to:[url]http://imo2007.edu.vn[/url][/quote]

 Thx i found them!


Prove that in any triangle $ABC$ the following inequality holds:

          $2(ab+bc+ca)-(a^2 + b^2 + c^2)\leq\frac{9abc}{a+b+c}$ ;)

are you sure it's not supposed to be
\[\displaystyle 2(ab+bc+ca)-(a^2 + b^2 + c^2)\leq\frac{9abc}{(a+b+c)} \]

[quote="amd2357"]\[\displaystyle 2(ab+bc+ca)-(a^2 + b^2 + c^2)\leq\frac{9abc}{(a+b+c)} \][/quote]

This is inequality 3 in http://www.mathlinks.ro/Forum/viewtopic.php?t=22367 .

  darij

It's exactly Schur's ineq after expanding all  ;)


geometry
Euler
projective geometry
geometry proposed

The segment lines $ B'Q,$ $ C'Q,$ as the medians of the triangle $ \bigtriangleup AB'C',$ intersect its sidesegments $ AC',$ $ AB',$ at their midpoints $ M,$ $ N,$ respectively.

It is easy to show that the segment line $ B''P,$ $ ($ connecting the midpoints of $ AC,$ $ AH$ $ ),$ passes though the point $ M$ and similarly the segment line $ C''P,$ passes through the point $ N.$

We denote as $ D,$ the intersection point of $ B'C',$ from the segment line $ PQ$ $ ($ [b][size=100]Euler line[/size][/b] of the triangle $ \bigtriangleup AB'C'$ $ ).$

Apply the [b][size=100]Pappus theorem[/size][/b], for the triads of collinear points $ B',$ $ B'',$ $ N$ and $ C',$ $ C'',$ $ M,$ we conclude that the point $ R\equiv B'C''\cap B''C',$ lies on $ PQ.$

So, we have the concurrency of the segment lines $ B'C'',$ $ B''C',$ $ DP,$ at point $ R$ and based on [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup DB'B'',$ $ \bigtriangleup PC''C',$ are perspective and hence, the points $ X\equiv DB'\cap PC'',$ $ A\equiv B'B''\cap C''C'$ and $ Z\equiv DB''\cap PC',$ are collinear.

The triangles $ \bigtriangleup DB'B'',$ $ \bigtriangleup PA_{1}A_{2},$ are also perspective with the point $ Q$ as their perspector and so, we conclude that the points $ X,$ as the inbtersection point of $ DB',$ $ PA_{1}$ and the point $ Z,$ as the one of $ DB'',$ $ PA_{2}$ and the intersection point of $ B'B'',$ $ A_{1}A_{2},$ are collinear.

That is the segment line $ A_{1}A_{2},$ passes through the point $ A,$ as the intersection point of $ XZ,$ $ B'B''.$

Hence, the points $ A,$ $ A_{1},$ $ A_{2},$ are collinear and the proof is completed.

Kostas Vittas.

Thank you dear vittasko for your very nice solution. My solution is base on [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997]your very nice problem[/url], really your problem is true for four points on a cricle, we don't need a convex quadrilateral.
Now we consider four point $ B,C',B',C$ on a circle we can consider $ B'C''$ as Euler line of triangle $ ABB'$ and similarly $ C'B''$ is as Euler line of $ ACC'$, and of course $ PQ$ is Euler line of $ AB'C'$ by vittasko's problem then three lines is concurent and more this point lie on Euler of $ ABC$, by the concurrency of three line we can see two triangle $ C'B''P$ and $ B'C''Q$ are perspective thus by Desarques theorem the intersection of three side are colinear.
Thank you vittasko again about nice concurrent problem of four Euler lines  :) .

[quote="gemath"]Given triangle $ ABC$ with orthocenter $ H$ and $ B',C'$ are feet of altitude form $ B,C$ and $ B'',C''$ are midpoint of $ AC,AB$ resp.
Let $ P$ is mid point of $ HA$ and $ Q$ is centroid of triangle $ AB'C'$, let $ C''P\cap B'Q=A_{1}$ and $ C'P\cap B''Q=A_{2}$ prove that $ A,A_{1},A_{2}$ are colinear.[/quote]
Nice problem and yet the draw is intricately. I am an idea. The problem is equivalently with the following :  the rays $ [AA_{2}$ and $ [AA_{3}$ are isogonally in the angle $ \widehat{BAC}$, where the point $ A_{3}$ is the intersection between the line $ BB''$ and the bisector of the side $ [AB]$, i.e. the point $ A_{3}$ is the homologously to the point $ A_{1}$ in the similarity $ ABC\sim AB'C'$ because, in the triangle $ AB'C'$, the line $ C''P$ is the bisector of the side $ [AB']$ and the ray $ [B'Q$ is the median from the vertex $ B'$.

Good Idea Virgil Nicula! thank you for your interest  :)
Sorry the above I have a typo the perspectrive triangle must be $ B'B''P$ and $ C'C''Q$ thanks vittasko for your remark. :)

[quote="Virgil Nicula"] .....
Nice problem and yet the draw is intricately. I am an idea. The problem is equivalently with the following :  the rays $ [AA_{2}$ and $ [AA_{3}$ are isogonally in the angle $ \widehat{BAC}$, where the point $ A_{3}$ is the intersection between the line $ BB''$ and the bisector of the side $ [AB]$, i.e. the point $ A_{3}$ is the homologously to the point $ A_{1}$ in the similarity $ ABC\sim AB'C'$ because, in the triangle $ AB'C'$, the line $ C''P$ is the bisector of the side $ [AB']$ and the ray $ [B'Q$ is the median from the vertex $ B'$.[/quote]
Indeed it is a good idea dear [b][size=100]Virgil[/size][/b].

I still have not any proof of the indispensable result that the segment line $ AA_{2},$ is also isogonal conjugate of $ AA_{3},$ with respect to the angle $ \angle BAC,$ the same as the segment line $ AA_{1}$ you mentioned and I am looking forward in a solution of this equivalent problem.

Best regards, Kostas vittas.


[quote="bowei"]YOU TRICKSTER![/quote]

lol Agreed.

Interesting, but not surprising. Possibly because of the previous posts...

So funny lol

Actually the file size betrays you :D

LOL, I can't believe I fell for that. Nice one.

All I see is a sock puppet  :(

Ohhhh! Now I see it! I wasn't staring at it very long...Haha..

[quote=silverdimension]that really funny lol  :rotfl:[/quote]

hahaha

[quote=Valentin Vornicu]Best optical illusion. Click on the picture so that it opens up in its own page. Stare at the picture for a while. You should be able to see a giraffe.[/quote]

Wow. Just wow. UHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHGGGGGGGGGGH

[quote=AnimalLover111]cool. look at this :P[/quote]

the peices accually expand a bit while moving, to compensate for the extra piece

[quote=Bob_Smart][quote=jgao2022]theres four people with illegal usernames

in a single thread[/quote]

what u mean[/quote]

spaces and symbols

[quote=Bob_Smart][quote=jgao2022][quote=AnimalLover111]cool. look at this :P[/quote]

The animation secretly adds chocolate that has the same amount as the piece left over. Watch this video if you don't know what I mean.[url=https://www.youtube.com/shorts/6aDB9VLnyZQ]Click here for explanation.[/url][/quote]

i said that already[/quote]

ok but why can't I expand on it?

[s]bro really wants to start a flamewar[/s]

thanks for the giraffe thing. my whole family got a huge laugh today :)

15 year old thread nice
op could have waited like a few more days until april fools tho

[url]https://cdn.artofproblemsolving.com/attachments/b/1/0abe71996b66a8d473e693927283d4081a6feb.gif[/url] Its a gif

[quote=RAYRAY7512][quote=Valentin Vornicu]Best optical illusion. Click on the picture so that it opens up in its own page. Stare at the picture for a while. You should be able to see a giraffe.[/quote]

IT'S THE FIRST USER!!![/quote]

wow i didnt notice
ik who the first was but i didnt know they made this thread lol


What is the greatest number of whole 4-inch by 6-inch notecards
that can be cut from a 25-inch by 27-inch sheet of poster board if
every cut must be made parallel to a side of the poster board? No
taping or gluing is allowed.

Is there a way to generalize this problem so we don't have to guess and check all the time?


The number $ 2^{1993} \plus{} 3^{1993}$ is a multiple of 5. What is the units digit of the quotient $ \frac{2^{1993} \plus{} 3^{1993}}{5}$?

How do you get that 3mod4 and 23mod25 get 23mod100?

If x is 23 mod 25, then it's 23, 48, 73 or 98 mod 100, but only 23 of these 4 is 3 mod 4.

Yeah, it works because of the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]Chinese Remainder Theorem[/url].


isn't
$\left(r^{\frac{x}{y}}cis \left(\frac{x\theta}{y} + \frac{2\pi kx}{y}\right)\right)^{\frac{y}{x}}=\left(r^{\frac{x}{y}}cis \left(\frac{x}{y}(\theta+2\pi k)\right)\right)^{\frac{y}{x}}=r \cdot cis \left(\theta+2\pi k\right)$
 :?

[quote="breez"]$(r^{\frac{x}{y}}cis \frac{x\theta}{y} + \frac{2\pi kx}{y})^{\frac{y}{x}}$
$= r cis \theta + 2\pi k + \frac{2\pi k}{x}$ if raising to y power first, then taking x root.
$= r cis \theta + 2\pi k + \frac{2\pi ky}{x}$ if taking x root then raising to y.

These two perspectives seem to indicate that $\frac{2\pi ky}{x}$ and $\frac{2\pi k}{x}$ yield the same values of $\theta$. How is this possible?[/quote]
$\left[r^{\frac{x}{y}}\textrm{cis}\left(\frac{x\theta}{y}+\frac{2\pi kx}{y}\right)\right]^{\frac{y}{x}}$

$=\left[r^{\frac{x}{y}}\textrm{cis}\left(\theta +2\pi k\right)^{\frac{x}{y}}\right]^{\frac{y}{x}}$

$=r\textrm{cis}\left(\theta +2\pi k\right)$

$=r\textrm{cis}\left(\theta\right)$

But your taking it to the x root, so shouldn't you add $\frac{2\pi k}{x}$, $k = 0$ to $(x-1)$?


Dear All,

I know that a few ARML teams currently use airlines for their transportation needs.  I
am wondering if it's possible to setup some sort of special arrangements with one
airline so that we can all get much better rates and decrease the financial cost
involved.  In order to properly negotiate such an agreement, can all the teams that
currently use airlines send me the following information (math@jackpo.org) :

1)  How many people fly to ARML
2)  Where are they flying from?
3)  Where are they flying to?
4)  Current rates obtained
5)  Current airlines used
6)  Whether any special negotiated rate is already being used.  (and if so, any
contact(s) that you can share would be greatly appreciated).

Thanks for everyone's cooperation,
Ming Jack Po

As far as I know, most of the Washington team has already booked tickets so it might be too late for that... we're flying with Alaska Airlines (from WA to Las Vegas) but we're not doing it as a team (though most people are on the same flight).

Dear Paladin,

It's definitely true that it is too late.  I am currently attempting to negotiate a better rate for future ARMLs.  So if you know the numbers for Washington, can you send them to me?

Thanks,
Jack

Well since im on Paladin's team, ill give u the info:

1) 30 people although right now the coachs got only 28 (at least according to the emails that i got)
2) seattle
3) Las Vegas
4) Around mid 200s (at least mine, like paladin said most of us are on the same flight anyways)
5) Alaska
6) no


geometry Asymptote circumcircle MATHCOUNTS

Given $ \triangle ABC$ with $ AB=AC=7$ and $ BC=2$:
Let $ M$ and $ N$ be the midpoints of $ AB$ and $ AC$ respectively.  For some point $ O$ in the interior of the triangle, lines $ MO$ and $ NO$ are perpendicular to $ AB$ and $ AC$, respectively.  Now let $ O$ be the midpoint of side $ DE$ of equilateral triangle $ \triangle ADE$.  Find the side length of the equilateral triangle.

This problem seems too easy...

[hide]
[asy]pair B,A,C,O,M,N; B=(0,0); A=7*dir(81.7867892983); C=2*dir(0); O=circumcenter(A,B,C); M=midpoint(A--B); N=midpoint(A--C); draw(A--B); draw(B--C); draw(C--A); draw(M--O); draw(O--N); dot(O); label("O",O,S); dot(midpoint(A--B)); label("M",midpoint(A--B),W); dot(midpoint(A--C)); label("N",midpoint(A--C),E); label("A",A,N); label("B",B,SW); label("C",C,SE);[/asy]

(Sorry, drawing D and E is too hard for me on Asymptote)

Note that O is the circumcenter of triangle ABC. Hence, the circumradius AO has length:
$ AO = R = \frac{abc}{4[ABC]}= \frac{98}{16\sqrt{3}}= \frac{49\sqrt{3}}{24}$

This is the length of the altitude of triangle ADE, since it is equilateral. Hence,
$ AD = \frac{2}{\sqrt{3}}\cdot AO = \frac{2}{\sqrt{3}}\cdot \frac{49\sqrt{3}}{24}= \boxed{\frac{49}{12}}$[/hide]

[quote="vishalarul"]This problem seems too easy...[/quote]

I didn't find it too easy.  What level of MATHCOUNTS competition is this from?

is the formula for circumradius product of the sides divided by 4 times the area of the triangle?

[quote="ninadthanawala"]Excuse me. Could you please give a simpler description.

                                                                                Thank You[/quote]

[hide]youdonotknowme said the formula
$ [ABC] = abc/4R$,

where $ a,b,c$ are the sides, $ R$ is the circumradius (i.e. the radius of the unique circle that passes through all three vertices of some triangle), and $ [ABC]$ is the area of the triangle.

In the problem, $ O$ is the point at which the perpendicular bisectors meet. This, by compass-and-ruler constructions, makes $ O$ the circumcenter of the triangle (i.e. the center of the unique circle that passes through all three vertices of some triangle).

I used this formula to find $ R = AO$. $ AO$ is the radius since $ O$ is the center and $ A$ is on the circle.

Note that $ AO$ is the altitude of the equilateral triangle, from which we can easily determine the side of the triangle.[/hide]

EDIT: since when did triangles have 4 vertices?

[quote="vishalarul"] (i.e. the radius of the unique circle that passes through all [b]four[/b] vertices of some triangle[/quote]
Just pointing out a typo...


calculus
integration
logarithms
trigonometry
calculus computations

I am stuck with the following problems:

1. $ \displaystyle\int \cosh (2w \plus{} 1) dw$
2. $ \displaystyle\int \sinh(3t) dt$
What are $ \sinh$ and $ \cosh$?
3. $ \displaystyle\int \frac {x\cos(x^2)}{\sqrt {\sin(x^2)}}$
Setting $ w \equal{} \sin(x^2)$ as well as the $ w \equal{} \cos(x^2)$: doesn't work.
4. $ \displaystyle\int \frac {x \plus{} 1}{x^2 \plus{} 2x \plus{} 19}$
I let $ w\equal{}x\plus{}1$ and the denom becomes $ w^2 \plus{} 18$.
I got $ \frac {w^2}{2} \cdot \ln |w^2 \plus{} 18|$ but I don't think the $ w^2 \plus{} 18$ works here.

[quote="jeez123"]What are $ \sinh$ and $ \cosh$?
[/quote]
They are hyberbolic fuctions jeez:
$ \sinh x \equal{} \frac {e^{x} \minus{} e^{ \minus{} x}}{2}$

$ \cosh x \equal{} \frac {e^{x} \plus{} e^{ \minus{} x}}{2}$
This will help u to solve the first two or if u are already an expert in this(unlike me) u have direct formulas for those integrals.

For the third first substitute $ y \equal{} \sin {x^{2}}$ u will get it :huh:
$ dy \equal{} \cos {x^{2}}.2x.dx$
and then $ \int \frac {1}{2\sqrt {y}}\,dy \equal{} \sqrt {y} \plus{} C \equal{} \sqrt {\sin{x^{2}}} \plus{} C$

For the fourth multiply and divide by 2:
$ \frac {1}{2} \times \int \frac {2.x \plus{} 2}{x^{2} \plus{} 2x \plus{} 19}\,dx$
Now substitute $ y \equal{} x^{2} \plus{} 2x \plus{} 19$ then $ dy \equal{} (2x \plus{} 2).dx$ so the integral becomes:
$ \int \frac {1}{2y}\,dy \equal{} \frac {1}{2}\ln|y| \plus{} C\equal{}\frac{1}{2}\ln|x^{2}\plus{}2x\plus{}19| \plus{} C$

4. $ \boxed{\int \frac{f'(x)}{f(x)}\ dx\equal{}\ln |f(x)|\plus{}C}$


number theory
greatest common divisor
number theory proposed

Determine the gcd of all numbers of the form $p^8-1$, with p a prime above 5.

I've found $480$.

We have $p^8 - 1 = (p-1)(p+1)(p^2 +1 )(p^4+1)$.
Clearly, the desired gcd, say $d$, cannot have a prime factor $q$ greater than 5 since for $p=q$ it will not divide $q^8-1$.

Now, if $p=7$ it leads to $7^8-1 = 2^6 \cdot 3 \cdot 5^2 \cdot 1201$
Thus $d$ divides $2^6 \cdot 3 \cdot 5^2$. (1)

Since $p>5$ we have $p^2 = 1$ mod[3] and $p^2 = 1$ mod[8].
It follows that $p^8-1 = 0$ mod[3] $(p^2+1)(p^4+1)=4 $ mod[8] and that $p^8-1 = 0$mod[2^5].
Thus $d = 0$ mod[2^5].
In another hand for $p=11$ it is easy to see from the above decomposition that $11^8 - 1 \neq 0$ mod[2^6], thus $d \neq 0$ mod[2^6]. (2)
In the same way, it is also easy to verify that $11^8 - 1 \neq 0$ mod[25], thus $d \neq 0$ mod[25]. (3)

But, since $p > 5$ we deduce that $p^2 = 1$ or $-1$ mod[5], so that $p^2-1 = 0$ mod[5], thus $d= 0 $ mod[5].

Then, $d= 0$ mod[$2^5 \cdot 3 \cdot 5$] and from (1),(2),(3) we deduce that $d=480$ as claimed.

Pierre (yeah! 1000)

yep, correct

btw congrats on your 1000th post! :D


The diagonals of a convex quadrilateral $ABCD$ intersect in the point $E$. Let $U$ be the circumcenter of the triangle $ABE$ and $H$ be its orthocenter. Similarly, let $V$ be the circumcenter of the triangle $CDE$ and $K$ be its orthocenter. Prove that $E$ lies on the line $UK$ if and only if it lies on the line $VH$.

easy problem.
 I just Answered this one, to remove it from MATHLINKS Unsolved problems( those problem are VERY rare) , Even ashegh and lomos_lupin are able to solve it.

let $U,E,K$ be collinear hence $UK\perp CD\Longrightarrow \ \ 90^{\circ}-\angle A=\angle UEB=\angle DEK=90^{\circ}-\angle D\Longrightarrow \ \ \angle A=\angle D$ hence quadrilateral $A,B,C,D$ is cyclic.
and again with easy angle chasing you can prove that $H,E,V$ are collinear.

i only give u the idea!

u know$U,G,H$,are colinear,and $\frac{UG}{GH}$, is constant.which $G$,

is the centroid.

u should know that projection keep it fix,and if u project the shape and

change triangle$EUH$,to an equalirateral triangle,u can firmly find the

answer.

sorry because i really dont have time to post the complete solution.

to amir.s:

i will kill u at school :lol:


Let $P$ be a Poset with countable elements. $P$ has all its antichains of some bounded size, then $P$ is the union of finitely many chains.

is it true?

Isn't it pretty obvious? :? I mean, we just apply Dilworth's Theorem for finite posets and use the kind of argument that arises in all compactness results:

Let $\{x_n\}_{n\ge 1}$ be our poset, and let $P_n$ be the poset with elements $x_1,\ldots,x_n$, with the order induced by that in $P$. If $\ell$ is the height (length of the largest chain) of $P$, then there will be some $n$ for which $P_n$ has height $\ell$. We can assume WLOG that $x_1<\ldots<x_\ell$ is such a chain of maximal length. Then each of the $x_i,\ i=\overline{1,\ell}$ belongs to exactly one of the antichains used to partition $P_n$ for every $n\ge\ell$, which we'll denote by $A_n^i$.

There are infinitely many $n$ for which $x_{\ell+1}$ belongs to $A_n^{i_1}$ for some fixed $i_1$. Among these, there are infinitely many for which $x_{\ell+2}\in A_n^{i_2}$ for some $i_2$, and so on. For each $k\ge 1$, we place $x_{\ell+k}$ in $A_n^{i_k}$ for the appropriate $n$.

Sorry if I misunderstood something..


graph theory
combinatorics unsolved
combinatorics

Ok i promise this will be my last post regarding hamiltonian cycles   ;)
Prove that every 4-connected panar graph is hamiltonian!  :D

[quote="manuel"] Prove that every 4-connected panar graph is hamiltonian!  :D[/quote]
what is panar graph :?:

planar graph is a graph that can be drawn without crossings!!
 seshadri, is the proof complicated?

well, i really don't know..i think it might be rather hard!
i had tried to solve this some time ago and i didn't get it then.i can give a reference to the original paper of Tutte:
 'A Theorem on Planar graphs' in Transactions of the American Mathematical Society,vol 82(1956), 99-116.i don't know if there are any simplications or even if the proof appears in any graph theory book.

it's not in Diestel's book!
thanks for the reference!