Λοιπόν θα κάνω μια υπόθεση.Είμαι κατά 99% σίγουρος ότι αληθεύει αλλά δεν μπορώ να το αποδείξω . To Conjecture εχει ως  εξής :

Έστω πολυώνυμο  $P(x)=a_nx^n+a_{n-1}x^{n-1}+.....+a_0$   με θετικούς ακεραίους συντελεστές ώστε

$a_n+n=a_{n-1}+n-1=.....=a_0$ .Για ένα τέτοιο πολυώνυμο κάνω την υπόθεση ότι

είναι  <<irreducible>> δηλαδή δεν γράφεται ως γινόμενο δύο πολυωνύμων με ακεραίους θετικούς συντελεστές και βαθμού > 1.

Πράγματι αυτό ισχύει για τα πολυώνυμα για $n=2,3,4$  αφού

αυτά είναι τα $x^2+2x+3$  ,   $x^3+2x^2+3x+4$,  ..........

Βοηθηστε με να αποδείξουμε αυτό το πολύ χρήσιμο λήμμα
Ευχαριστώ
Σιλουανός

ΛΟΙΠΟΝ ΜΑΓΚΕΣ ΝΟΜΙΖΩ ΟΤΙ ΤΟ ΕΛΥΣΑ ΣΙΛΟΥΑΝΕ ΚΑΙ ΟΙ ΛΟΙΠΟΙ ΜΑΘΛΙΝΚΕΡΣ ΛΟΙΠΟΝ ΠΡΟΣΕΞΕΤΕ
ΕΣΤΩ ΟΤΙ ΥΠΑΡΧΕΙ ΠΟΛΥΩΝΥΜΟ ΠΟΥ ΙΚΑΝΟΠΟΙΕΙ ΤΙΣ ΣΥΝΘΗΚΕΣ ΤΟΥ ΠΡΟΒΛΗΜΑΤΟΣ ΜΕ Ν>=3
ΤΟΤΕ
Α1=Α0-1
Α2=Α0-2
…
ΑΝ=ΑΟ-Ν
ΑΝ ΑΘΡΟΙΣΟΥΜΕ ΚΑΤΑ ΜΕΛΗ
Α1+Α2+…+ΑΝ=Ν*Α0-Ν*(Ν+1)/2 (1)
ΑΠΟ ΤΗΝ ΑΛΛΗ ΕΧΟΥΜΕ OTI AN=1 ΔΙΟΤΙ
ΑΝ ΑΝ>=2 ΤΟΤΕ ΑΙ>= 2 ΓΙΑ Ι=1,2,3,,Ν
ΑΝ=Α0-Ν>=2
ΑΝ>=5 (2)
ΧΡΗΣΙΜΟΠΟΙΩΝΤΑΣ ΤΗΝ (2) ΣΤΗΝ (1) ΕΧΟΥΜΕ ΟΤΙ
5>=Ν ΑΤΟΠΟ ΜΕ ΔΟΚΙΜΕΣ
ΑΡΑ ΑΝ=1 ΑΝ-1=2 ,,,,,. Α0=Ν+1
ΚΑΙ
Π(Χ)=ΧΝ+2ΧΝ-1+3ΧΝ-2+…+Ν+1
ΕΔΩ ΕΧΟΥΜΕ
Α0=Ν+1
Α0+Α1+Α2+..+ΑΝ=1+2+3+…+Ν=(Ν+2)*(Ν+1)/2 (3)
ΑΝΤΙΚΑΘΙΣΤΩΝΤΑΣ ΤΗΝ (3) ΣΤΗΝ (1) ΕΧΟΥΜΕ
 (Ν+2)*(Ν+1)/2=Ν*(Ν+1)- Ν*(Ν+1)/2
ΟΠΟΥ ΜΕΤΑ ΑΠΟ ΠΡΑΞΕΙΣ
(Ν+1)*2=0
Ν=-1 ΑΤΟΠΟ
ΑΡΑ ΔΕΝ ΥΠΑΡΧΕΙ ΠΟΛΥΩΝΥΜΟ ΜΕ ΤΙΣ ΣΥΝΘΗΚΕΣ ΤΟΥ ΠΡΟΒΛΗΜΑΤΟΣ
ΠΑΡΑΚΑΛΩ ΟΛΟΥΣ ΤΟΥΣ ΜΑΘΛΙΝΚΕΡΣ ΝΑ ΕΛΕΓΞΟΥΝ ΑΜΕΣΑ ΤΗΝ ΟΡΘΟΤΗΤΑ ΤΗΣ ΛΥΣΗΣ ΚΑΙ ΝΑ ΜΕ ΕΝΗΜΕΡΩΣΟΥΝ ΓΙΑ ΤΟ ΑΝ ΕΙΝΑΙ ΣΩΣΤΗ Η ΛΑΘΟΣ ΚΑΙ ΠΟΥ ΕΙΝΑΙ ΛΑΘΟΣ ΕΤΣΙ ΩΣΤΕ ΝΑ ΔΙΟΡΘΩΘΕΙ ΑΜΕΣΑ ΟΚ?

[quote="ambros"]ΑΡΑ ΔΕΝ ΥΠΑΡΧΕΙ ΠΟΛΥΩΝΥΜΟ ΜΕ ΤΙΣ ΣΥΝΘΗΚΕΣ ΤΟΥ ΠΡΟΒΛΗΜΑΤΟΣ
[/quote]

Νομίζω ότι δεν έχεις καταλάβει το πρόβλημα . :(

ΤΟ ΧΩ ΚΑΤΑΛΑΒΕΙ  :mad: ΑΠΛΩΣ ΕΧΩ ΥΠΟΘΕΣΕΙ ΟΤΙ  ΥΠΑΡΧΕΙ ΚΑΠΟΙΟ ΠΟΛΥΩΥΜΟ REDUCIBLE OK?  :)


$\mathbb R^\ast = \mathbb R \setminus \left\{ 0 \right\}$.

[quote="perfect_radio"]Let $a,b \in \mathbb R^\ast$. Find all functions $f : \mathbb R \to \mathbb R$ such that \[f \left( x-\frac{b}{a}\right)+2x \leq \frac{a}{b}\cdot x^{2}+\frac{2b}{a}\leq f \left( x+\frac{b}{a}\right)-2x , \, \forall x \in \mathbb R .\] [i]Dorin Mărgidanu[/i][/quote]
Replacing $x$ by $x+\frac{b}{a}$ in the first inequality yields
$f(x)+2(x+\frac{b}{a}) \leq \frac{a}{b}{(x+\frac{b}{a})}^{2}+\frac{2b}{a}$

$\Rightarrow f(x) \leq \frac{a}{b}x^{2}+\frac{b}{a}$ ... (I)

Again, replacing $x$ by $x-\frac{b}{a}$ in the second inequality yields
$\frac{a}{b}{(x-\frac{b}{a})}^{2}+\frac{2b}{a}\leq f(x)-2(x-\frac{b}{a})$

$\Rightarrow \frac{a}{b}x^{2}+\frac{b}{a}\leq f(x)$ ... (II)

Combining (I) and (II) gives $f(x) = \frac{a}{b}x^{2}+\frac{b}{a}, \forall x \in \mathbb R$


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

a)Show that the alternating group A4 is a group of order 12 which contains the
identity, three elements of order 2, and eight elements of order 3.
This part I managed allright

b)Show that the three elements of order 2 form a single conjugacy class, while the
eight elements of order 3 fall into two conjugacy classes each of size 4.

This part had me a bit puzzled.
I proved that {(1 2)(3 4), (1 3)(2 4),(1 4),(2 3)} form a conjugacy class.

The second part I got 4 conjugacy classes instead of 2.. I think I must have misundertodd the concept..
{(1 2 3),(1 3 2)}
{(1 2 4)(1 4 2)}
{(1 3 4)(1 4 3)}

Also I do not have a clue how to go about the third part:
(c) Suppose that H is a subgroup of A4 of order 6. Show that H is a normal subgroup
of H. Hence show that no such subgroup H can exist.

The 3-cycle $(1,2,3)$ is not in the same conjugacy class as its square $(1,3,2)$; it's in the same class as $(1,3,4)$, $(2,4,3)$, and $(4,2,1)$. I got those by conjugating with other 3-cycles.

For part (c): Show that in any group $G$, a subgroup $H$ of index 2 is normal by looking at its cosets.
Now, what would a normal subgroup of order 6 look like? It would contain complete conjugacy classes of the original group. Since a group of order 6 has only two elements of order 3, this can't happen.

[quote="sunshiney_smile"]I am sorry stil not clear on how you get conjugacy classes..
I tried a
g=(1,2,3)
h=(1,3,2)

x=(2,3)
and this (x)-1.(g)(x)=h
So how are they not conjugate?

How do you get the conjugacy class? as in how do you determine what elements are conjugate to each other?

I have recently started studying this.. so apologise for asking questions that will be trivial to someone experienced.
Thank you[/quote]
(2,3) is not in A_4

g=(1,2,3)
h=(1,3,2)

x=(2,3,4) .. is this right. From what we were told write h below g and that is the two-line notation for x.

and (2,3,4) is an element of A4
so (1,23) conjugate to (1,3,2)

Is this the right way?

[quote="jmerry"]You don't get $(1,3,2)$; you get $(1,3,4)$ or $(1,4,2)$ depending on which side you write things.[/quote]

For what do i get ( 1 3 4) or (1 4 2)
I'm so confused.

That's the result of conjugating $(1,2,3)$ with $(2,3,4)$. Work it out in detail; function notation helps.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ R$ be a ring with 1 and $ I$ be a right ideal of $ R$.
Is $ RI$ a right ideal of $ R$?

if $ RI$ means the subgroup generated by products $ r*i$ with $ r \in R, i \in I$: trivially yes.


I've been doing cube puzzles recently and heard that there were other twisty-like puzzles (tetrahedron, megaminx, etc.)

I was able to do the 2x2 and 3x3 by myself without looking at any guides, and when I got the pyraminx, I could do the same. I don't really want to use a guide, so could you guys list the twisty puzzles from easiest to hardest in your opinion?

Thanks.

2x2 cube
3x3 rubik's cube
4x4 rubik's revenge
pyraminx
5x5 rubik's professor
v-cube 6
v-cube 7
pyramorphix
skewb diamond
alexander's star
rubik's magic
square-1
rubik's master magic

not in this order...just some I know :wink:

Rubik's magic is so easy... I've gotten sub-second (ok only 2 times). Now that I'm out of practice, I average 1.3 :( It should be at the top of your list.

I used to be interested in such puzzles, but then I realized there were too many [url]http://www.geocities.com/jaapsch/puzzles/[/url] that it was hopeless.

pyraminx is also easy, it was the only rubiks like puzzle i solved without using a guide
all others i needed 1
idk what rubiks magic is but it sounds easy
square 1 is uber-- i could barely get it to a cube


Which 3D graphing programs do you use/recommand?


inequalities
geometry
perimeter
function
inequalities proposed

I need help with the following inequalities:

Given $ a, b, c$ the sides of a triangle with perimeter $ 2p$.
Prove the following:
1) $ \frac{1}{p\minus{}a}\plus{}\frac{1}{p\minus{}b}\plus{}\frac{1}{p\minus{}c}\ge 2\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)$

2) $ a^2\plus{}b^2\plus{}c^2\ge \frac{36}{35}\left(p^2\plus{}\frac{abc}{p}\right)$

3) $ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{3abc}{(a\plus{}b)(b\plus{}c)(a\plus{}c)} < 2$

4) $ \frac{a}{\sqrt{b\plus{}c\minus{}a}}\plus{}\frac{b}{\sqrt{c\plus{}a\minus{}b}}\plus{}\frac{c}{\sqrt{a\plus{}b\minus{}c}}\ge \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

5) $ \frac{a}{2b\plus{}2c\minus{}a}\plus{}\frac{b}{2a\plus{}2c\minus{}b}\plus{}\frac{c}{2a\plus{}2b\minus{}c}\ge 1$

[hide="1"]
By AM-HM, or Cauchy:
\[ \sum{\frac {1}{p - a}} = \sum{\frac {1}{2}\left(\frac {1}{p - b} + \frac {1}{p - c}\right)}\ge\sum{\frac {2}{p - b + p - c} = \sum\frac {2}{a}}
\]
[/hide]
[hide="2"]
Let $ s = 2p$ and $ t = ab + bc + ca$.

It is true that $ s^2\ge 3t$ since $ 2(s^2 - 3t) = (a - b)^2 + (b - c)^2 + (c - a)^2\ge 0$

On the other side, by AM-GM, $ s^3\ge 27abc$.

Consequently:
\[ 35s(a^2 + b^2 + c^2) = 35s^3 - 70st\ge\frac {35s^3}{3} = 9s^3 + \frac {8s^3}{3}\ge 9s^2 + 72abc \Rightarrow
\]

\[ a^2 + b^2 + c^2\ge\frac {36}{35}\left(\frac {s^2}{4} + \frac {2abc}{s}\right)
\]
[/hide]
[hide="3"]
\[ 0 < (a + b - c)(b + c - a)(c + a - b) =
\]

\[ (a + b)(b + c)(c + a) - \sum{a(a + b)(a + c)} - abc + \sum{bc(b + c)} =
\]

\[ 2(a + b)(b + c)(c + a) - 3abc - \sum{a(a + b)(a + c)}\Rightarrow
\]

\[ 2(a + b)(b + c)(c + a) > 3abc + \sum{a(a + b)(b + c)}\Rightarrow
\]

\[ 2 > \frac {3abc}{(a + b)(b + c)(c + a)} + \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b}
\]
[/hide]
[hide="4"]
This was nice.
Let $ x=\sum{\frac{a}{\sqrt{b+c-a}}}$. Note that $ 0<\sum{a(b+c-a)=4t-s^2}$

By Holder with three sequences,
\[ x^2(4t-s^2)=\left(\sum{\frac{a}{\sqrt{b+c-a}}}\right)^2\left(\sum{a(b+c-a)}\right)\ge (a+b+c)^3=s^3\Rightarrow\]
\[ x\ge\sqrt{\frac{s^3}{4t-s^2}}\ge\sqrt{\frac{s^3}{t}}\ge\sqrt{3s}=\sqrt{(1+1+1)(a+b+c)}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\]
The last inequality holds by Cauchy and the others because $ s^2\ge 3t$.
[/hide]
[hide="5"]
By Cauchy,
\[ (6t-s^2)\sum{\frac{a}{2b+2c-a}=\left({\sum{a(2b+2c-a)}\right)\left({\sum{\frac{a}{2b+2c-a}\right)\ge (a+b+c)^2=s^2\Rightarrow}}}}\]
\[ \sum{\frac{a}{2b+2c-a}\ge\frac{s^2}{6t-s^2}\ge\frac{s^2}{3t}\ge 1}\]
[/hide]

:)

4)
Ravi substitution and $ \frac{x}{\sqrt{y}}\plus{}\frac{y}{\sqrt{x}}\geq \sqrt{2(x\plus{}y)}$
5)
Cauchy

Super thanks to ElChapin & Albanian Eagle.

Other one for number four

First we prove that $ \sqrt{a}+\sqrt{b}+\sqrt{c}\ge\sqrt{b+c-a}+\sqrt{c+a-b}+\sqrt{a+b-c}$
[hide]
Since the inequality is symmetric we may suppose that $ a\ge b\ge c$.
$ (a,b,c)$ majorizes $ (b+c-a,c+a-b,a+b-c)$ because
$ a\ge b+c-a$ and
$ a+b\ge b+c-a+c+a-b=2c$ and
$ a+b+c=2c+a+b-c$

Then, as $ f(x)=\sqrt{x}$ is a concave function, by Karamata inequality, it follows what we wanted.
[/hide]
By Cauchy,
\[ \sum{\frac{a}{\sqrt{b+c-a}}\ge\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{b+c-a}+\sqrt{c+a-b}+\sqrt{a+b-c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}}\]

:)


Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.

I wasn't able to follow [b]al.M.V.[/b]'s sketch. I'm sorry if my solution is very similar to his one.
For $ n\equal{}6$ take a regular hexagon divided into $ 6$ equilateral triangles.
Check the image. The first hexagon is made up of $ 4k$ congruent right triangles. The second one is made up of $ 4k\plus{}2$ congruent right triangles, where $ k\geq2$ is equal to $ 1\plus{}$length of horizontal side of the hexagon.
For odd $ n$ we use right rectangles with sides $ 1,3$ and $ \sqrt{10}$ respectively, arranged like in the third picture. If $ k$ is the length of the right vertical side, then we have a hexagon made up of $ 4\plus{}2k\plus{}1\equal{}2k\plus{}5\geq7$ congruent triangles.

Of course, it is obvious that each small rectangle consists of 2 right triangles.
[img]http://www.fileden.com/files/2007/1/24/687621/Mathematics/apmo90_5.PNG[/img]

Just in case the above image doesn't show up, the attached image is also available:

[quote=al.M.V.]Just use right-angled triangles with side lengths 1, 2, and root(3). Show that if it's possible for n, it's possible for n+2 (just add 2 triangles which form a rectangle in the middle). Then we are only left to show n=6 and n=7 are possible. Easy problem. :)[/quote]

I think your solution can't work when n is odd.Can you explain it?

Look at the third diagram. It is possible to write triangles with such lengths.


geometry
inradius
inequalities
circumcircle
trigonometry
geometric transformation
reflection

Show that for any triangle 2r ≤ R (where r is the inradius and R is the circumradius).

Hm, this is Euler's Triangle Inequality, shown easily using Euler's Triangle Formula, stating $d=\sqrt{R(R-2r)}$ where $d$ is the distance between the inradius and the circumradius, although that isn't really a proof.

[hide="Second proof from Engel"]
I love this proof.

Consider the triangle formed by the midpoints of the sides.  The circumradius of this triangle is $\frac{R}{2}$ and this circumcircle must be larger than the incircle of the original triangle.  Hence, $\frac{R}{2} \ge r$.
[/hide]

[hide="First proof from Engel"]
$A = rs = \frac{abc}{4R} = \sqrt{ s(s-a)(s-b)(s-c)}$
$\frac{R}{r} = \frac{sabc}{4A^2} = \frac{sabc}{4s(s-a)(s-b)(s-c)} = \frac{2abc}{(a+b-c)(a-b+c)(-a+b+c)} \ge \frac{2abc}{ \sqrt{a^2b^2c^2} } = 2$.

(It was previously proven that $(a+b-c)(a-b+c)(-a+b+c) \le abc$ using 'symmetric functions' - personally, the other proof is better.)
[/hide]

[hide="Here are fifth methods"] $\triangle ABC$; its incircle $C(I,r)$, its circumcircle $C(O,R)$; the second intersection $A'$

of the circumcircle with the line $AI$, the reflection $A''$ of the point $A'$ w.r.t. the

circumcentre $O$, the projection $D$ of the incentre $I$ on the side $[AC]$

$(IA'=A'B=A'C,\ A'A''=2R,\ ID=r)$.

[u]The first method.[/u] $\triangle AID\sim \triangle A''A'C\Longrightarrow\frac{AI}{A''A'}=\frac{ID}{A'C}\Longrightarrow IA\cdot IA'=2Rr$

is the power of the incentre $I$ w.r.t the circumcircle $\Longrightarrow$ $R^2-OI^2=2Rr$

$\Longrightarrow\boxed {OI^2=R(R-2r)}\Longrightarrow$ $R\ge 2r.$

[u]The second method.[/u] $\triangle AA'A''$ $\Longrightarrow$ $\cos \frac{B-C}{2}=\frac{AA'}{A'A''}=$

$\frac{IA+IA'}{2}\cdot \frac 1R\ge\frac 1R\sqrt{IA\cdot IA'}=\frac 1R\sqrt{2Rr}=\sqrt{\frac{2r}{R}}\Longrightarrow$

$\boxed {\sqrt{\frac{2r}{R}}\le \cos \frac{B-C}{2}}\le 1$ $\Longrightarrow$ $R\ge 2r.$

[u]The third method.[/u] $a^2\ge a^2-(b-c)^2\Longrightarrow a^2\ge 4(s-b)(s-c)$ a.s.o. $\Longrightarrow$

$a^2b^2c^2\ge 64(s-a)^2(s-b)^2(s-c)^2\Longrightarrow\boxed {abc\ge 8(s-a)(s-b)(s-c)}\Longrightarrow$

$4Rrs\ge 8sr^2\Longrightarrow R\ge 2r$.

$(abc=4RS=4Rrs,\ \frac{S^2}{s}=(s-a)(s-b)(s-c)=sr^2$ are wellknown$).$

[u]The fourth method [/u](Shobber's idea).

$\cos A+\cos B+\cos C=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C=$

$2\sin \frac C2\cos \frac{A-B}{2}+1-2\sin^2\frac C2=1+2\sin \frac C2\left(\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right)=$

$1+4\sin \frac A2\sin \frac B2\sin \frac C2=1+\frac{4(s-a)(s-b)(s-c)}{abc}=1+\frac{4sr^2}{4Rrs}=1+\frac rR$

$\Longrightarrow\boxed {\cos A+\cos B+\cos C=1+\frac rR}$. From the third method results

$\frac{(s-a)(s-b)(s-c)}{abc}\le \frac 18$. Therefore, $1+\frac rR=\cos A+\cos B+\cos C=$

$1+\frac{4(s-a)(s-b)(s-c)}{abc}\le 1+\frac 12\Longrightarrow R\ge 2r.$

[u]The fifth method[/u] ([b]the best[/b]).

$(s-b)(s-c)+s(s-a)=bc$; $\sum bc=s^2+r^2+4Rr$;

$\sum (s-b)(s-c)=r^2+4Rr$. From the inequality

$9xyz\le (x+y+z)(xy+yz+zx),\ (x: =s-a,\ y: =s-b,\ z: =s-c)$

we obtain $9sr^2\le sr(4R+r)$ $\Longrightarrow 2r\le R.$

[b]Remark.[/b]

$3(r_ar_b+r_br_c+r_cr_a)\le (r_a+r_b+r_c)^2\Longrightarrow \boxed {s\sqrt 3\le 4R+r}.$

$(s-a)+(s-b)+(s-c)\ge \sqrt [3] {(s-a)(s-b)(s-c)}\Longrightarrow \boxed {3r\sqrt 3\le s}.$

$s\sqrt 3\le 4R+r,\ r\le \frac{s\sqrt 3}{9}\Longrightarrow s\sqrt 3\le 4R+\frac{s\sqrt 3}{9}\Longrightarrow \boxed {s\le \frac{3\sqrt 3}{2} R}.$

Therefore, $\boxed {3\sqrt 3 r\le s\le \frac{3\sqrt 3}{2}R}$. In conclusion, $R\ge 2r.$

$3xyz(x+y+z)\le (xy+yz+zx)^2,\ 3(xy+yz+zx)\le (x+y+z)^2\Longrightarrow$

$(x: =s-a,\ y: =s-b,\ z: =s-c)\ \ s\sqrt 3\le 4R+r;\ \boxed {3r(4R+r)\le s^2}.$ [/hide]

[quote="t0rajir0u"][hide="Second proof from Engel"]
I love this proof.

Consider the triangle formed by the midpoints of the sides.  The circumradius of this triangle is $\frac{R}{2}$ and this circumcircle must be larger than the incircle of the original triangle.  Hence, $\frac{R}{2} \ge r$.
[/hide]
[/quote]
This is actually from the Hungarian mathematician Adam, as Engel has quoted. :)

[quote="t0rajir0u"][hide="Second proof from Engel"]
I love this proof.

Consider the triangle formed by the midpoints of the sides.  The circumradius of this triangle is $\frac{R}{2}$ and this circumcircle must be larger than the incircle of the original triangle.  Hence, $\frac{R}{2} \ge r$.
[/hide]

This is true because ([b]Feuerbach's theorem)[/b] the incircle is interior tangent to the Euler circle ! It is not evidently!!!

Virgil, you can use \boxed to put a box around your answers.  For example, $\boxed{x^2+\frac 1x}.$

By the way, what do you mean by

[quote] the reflection $A'$ of the point $A$ w.r.t. the incentre $I$, the reflection $A''$ of the point $A'$ w.r.t. the circumcentre $O$[/quote]

?

I've tried what I thought you meant, but by no means do I get $A'A''=2R$.

Thanks, Joml88. You are true ! I corrected "the foolishness" from the text. I like the [b]"boxed {...}".[/b] Again, thank you ! I added and the solution of the Shobber's idea.


I got a 6 on the AIME II. Two stupid mistakes, one adding, one subtracting.

Grrr.

um..what is the point of this thread? there is already a thread on this forum where we brag about our aime scores ;) and blame our score on careless mistakes

this is AIME II, shobhit.

let's be respectful toward other members, please.

Yes, Shobhit.  You need to be kind and considerate!  Did anyone else take the AIME II?  I know Phillip was considering it...


geometry
parallelogram
geometry unsolved

Hi, does anyone know how to prove this using parallelogram proof?

[color=green]BE is a median of triangle ABC.  THe line through C and the midpoint H of BE meets AB at F.  Prove that AF/FB = 2.
You must use a //ogram proof.  (Hint: Extend BE to 3/2 of its lenght)[/color]

Thanks so much guys. :lol:

Using the hint it becomes trivial

[hide="Here is another proof"]

Take the symmetric point $M$ of $C$ with respect to $B$
Then $B$ is the midpoint of $MC \Rightarrow  AB$ is a median of the triangle $AMC$
We will show that $F$ is the centroid of  $\triangle AMC$

Since $B,E$ are the midpoints of $MC,AC,$ hence we have $BE \parallel MA$

So $\triangle CMA \sim \triangle CBE$

The two triangles have a common angle at $C$.
Hence the median $CH$ of $\triangle CBE$ is also a median of $\triangle CAM$

But this line intersects the other median $AB$ at the point $F$, so $F$ is the centroid of $\triangle AMC\Rightarrow  AF = 2FB$[/hide]

$L\in HC\ ,\ EL\parallel AB\Longrightarrow FB=EL\ ,\ AF=2\cdot EL\Longrightarrow \frac{AF}{FB}=2\ .$

[b]Remark.[/b] The quadrilateral $BFEL$ is a parallelogram and the segment $EL$ is the middle-line w.r.t. the side $[AF]$ in the triangle $CAF$.

[b]Remark.[/b] You can choose differently the point $L$: the point $H$ be the middlepoint of the segment $FL$ ! Continue alone.


Hello.

If for positive real numbers $ a$,$ b$,$ c$,   $ 2b^2\equal{}a^2 \plus{}c^2$ holds, prove that then:

$ \frac{1}{a\plus{}b} \plus{} \frac{1}{b\plus{}c}\equal{}\frac{2}{a\plus{}c}$

[hide="Proof"]
We have $ 2b^{2}\equal{}a^{2}\plus{}c^{2}$.

Then,
$ 2b^{2}\plus{}(2ab\plus{}2ac\plus{}2bc)\equal{}a^{2}\plus{}c^{2}\plus{}(2ab\plus{}2ac\plus{}2bc)$
$ 2(b^{2}\plus{}ab\plus{}ac\plus{}bc)\equal{}(a^{2}\plus{}ab\plus{}ac\plus{}bc)\plus{}(c^{2}\plus{}ab\plus{}ac\plus{}bc)$
$ 2(a\plus{}b)(b\plus{}c)\equal{}(a\plus{}b)(a\plus{}c)\plus{}(a\plus{}c)(b\plus{}c)$

Dividing both sides by $ (a\plus{}b)(b\plus{}c)(a\plus{}c)$, we get:
$ \frac{2}{a\plus{}c}\equal{}\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}b}$[/hide]


I'm a noob in geometry right now  :D so I missing a point on this

Why $ \angle J \equal{} \angle X$ on this?
 [geogebra]70ed4a12fa19dc38cbd88dfeed91a1e4fe8af5b2[/geogebra]

And why here it's $ \angle J \equal{} 180^{\circ} \minus{} \angle X$
 [geogebra]5ab7fbee1e6ae30893140a6bcbce0c7b86087201[/geogebra]

Note: $ \angle ZYJ \equal{} 90^{\circ}$ in both figures. And this is my first time on GeoGebra

Thanking you

http://en.wikipedia.org/wiki/Inscribed_angle_theorem

For the below, there's no A, if you meant X, again check the wiki above.

Thank you "Bugi"  :)  and yes I meant $ X$


Two dice are thrown.  What is the probability that the product
of the two numbers is a multiple of 5?  Express your answer as a common fraction.

One of the dice must be 5. The other dice can have 6 choices.

So there are 2*6 - 1 combinations since (5,5) is counted twice.

So the probability is 11/36.

Another solution could be by subtracting the probability of the product not being a multiple of 5, or 25/36, subtracting, we get 11/36


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

Let triangle $ ABC$ has circumcircle $ (O)$ and altitudes $ AM,BN$. Let $ D$ be a variable point on the circle such that $ D$ is distinct from $ A,B$ and $ DA,BN$ has a common point $ Q$. Let $ BD$ cut $ AM$ at $ P$. Prove that the midpoint of $ PQ$ lies on a fixed line as $ D$ varies on the circle $ (O)$

[color=darkblue]Denote by $ N_1$ and $ M_1$ intersections of the altitudes $ BN$ and $ AM$ with the circumcircle of triangle $ ABC$. Let $ H$ be orthocenter of the triangle and $ D$ the intersection of the lines $ PQ$ and $ MN$.
It is well known that $ N_1$ is reflection of $ H$ w.r.t. point $ N$, so $ HN \equal{} NN_1$ and $ \angle{N_1HA} \equal{} \angle{HN_1A}$. So

 $ \triangle{NN_1A}\equiv\triangle{NHA}\sim \triangle{MHB}$

 then

  $ \frac {MH}{NN_1} \equal{} \frac {BM}{NA}$    [b] (1)[/b].

    But $ \angle{CBD} \equal{} \angle{CAD}$, so

    $ \triangle{MBQ}\sim \triangle{NAP}$ $ \Rightarrow$ $ \frac {MQ}{NP} \equal{} \frac {BM}{NA}$   [b]  (2)[/b]

     From (1) and (2) we obtain that

  	 $ \frac {MQ}{NP} \equal{} \frac {MH}{NN_1} \equal{} \frac {MH}{NH}$ $ \Leftrightarrow$ $ \frac {MQ}{MH} \equal{} \frac {NP}{NH}$

  Then apply Menelaus Theorem to the triangle $ QHP$ and transversal $ \overline{MDN}$:

   $ \frac {MQ}{MH}\cdot\frac {HN}{NP}\cdot \frac {PD}{DQ} \equal{} 1 \Leftrightarrow \frac {PD}{DQ} \equal{} 1$

 So midpoint of $ PQ$ always lies on the line $ MN$.[/color]


$ ABC$ is a triangles such that $ \angle A\equal{}3x,\angle C\equal{}4x$. $ D\in\overline{AC}$ and $ E\in\overline{BD}$. If
1 - $ \overline{CE}\perp\overline{BD}$.
2 - $ \overline{CE}$ bisect $ \angle C$.
3 - $ AC\equal{}2CE$.
Find $ x$.

$ \frac{\sin 7x}{\sin 3x} \equal{} \frac{\sin B}{\sin A} \equal{} \frac{CA}{BC} \equal{} \frac{CA}{EC} \cos 2x \equal{} 2 \cos 2x$ $ \Longrightarrow$

$ \sin 7x \equal{} \sin 5x \plus{} \sin x$ $ \Longrightarrow$ $ 2 \sin x \cos 6x \equal{} \sin x$ $ \Longrightarrow$
$ \sin x \equal{} 0$ or $ \cos 6x \equal{} \frac{_1}{^2}$ $ \Longrightarrow$ $ x\equal{} k \cdot 180^\circ$ or $ x \in \{10^\circ, 50^\circ\} \plus{} k \cdot 60^\circ.$
$ 0^\circ < \angle A \equal{} 3x < 90^\circ$ $ \Longrightarrow$ $ x \equal{} 10^\circ.$

Thanks, yetti.
Is there any solution without trigonomatry?

[quote="mrmohamed"][quote="mathson"]$ ABC$ is a triangles such that $ \angle A \equal{} 3x,\angle C \equal{} 4x$. $ D\in\overline{AC}$ and $ E\in\overline{BD}$. If
1 - $ \overline{CE}\perp\overline{BD}$.
2 - $ \overline{CE}$ bisect $ \angle C$.
3 - $ AC \equal{} 2CE$.
Find $ x$.[/quote]
hi mathson
could u put acondition for A ,C
for example   A < 90  or  A > 90  similiary   C
and so on
thanks[/quote]

No, I think there is no conditions for $ A,C$.


Let $ ABC$ be an acute-angled triangle. The circle $ k$ with diameter $ AB$ intersects $ AC$ and $ BC$ again at $ P$ and $ Q$, respectively. The tangents to $ k$ at $ A$ and $ Q$ meet at $ R$, and the tangents at $ B$ and $ P$ meet at $ S$. Show that $ C$ lies on the line $ RS$.

[quote="moldovan"] [color=darkred]Let $ ABC$ be a triangle. The circle $ k$ with diameter $ AB$ intersects $ AC$ and $ BC$ again at $ P$ and $ Q$ respectively.
 The tangents to $ k$ at $ A$ and $ Q$ meet at $ R$ and the tangents at $ B$ and $ P$ meet at $ S$ . Show that $ C$ lies on the line $ RS$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $ PQ\cap AB\ne\emptyset$ and denote $ T\in PQ\cap AB$ .

Apply [u]the [b]Pascal[/b]'s theorem[/u] $ (p)$ to the cyclical quadrilaterals :

$ \left\|\begin{array}{cccc} AAPQQB\ : & \left|\begin{array}{c} R\in AA\cap QQ \\
\ C\in AP\cap QB \\
\ T\in PQ\cap BA\end{array}\right| & \stackrel{(p)}{\Longrightarrow} & R\in CT \\
 \\
BBQPPA\ : & \left|\begin{array}{c} S\in BB\cap PP \\
\ C\in BQ\cap PA \\
\ T\in QP\cap AB\end{array}\right| & \stackrel{(p)}{\Longrightarrow} & S\in CT\end{array}\right\|\ \Longrightarrow\ \overline {CRST}$ .

[b][u]Remark[/u].[/b] If $ PQ\parallel AB$ , then $ C\in RS$ and $ \overline {CRS}\parallel AB$ .[/color]


Find a if a and b are integers such that x^2 - x - 1 is a factor of ax^17 + bx^16 + 1. (1988/13)

I know that the solution has been presented in a previous math jam and in another thread, but I was just wondering how someone would come up with either of the two most popular solutions to this problem. Frankly, for one of the solutions, I would never think to note that the expressions (involving the roots) pq and p+q have nice numerical values and to multiply ap^17+bp^16+1 and aq^17+bq^16+1 by q^16 and p^16, respectively, and then make the appropriate substitutions. Its just not something I would probably do in a polynomial problem like this. Would someone care to enlighten me on the thought process associated with solving this problem? Thanks.

I don't imagine I would think of that either, although I sometimes come up with things in contests that amaze me (for better or for worse). What I would do would be to multiply x2-x-1 by x15 and then doing various other operations to it until I got ax17+bx16+1, so I would simply solve it ``by construction.'' To me, that's the easiest way of solving polynomial problems.


[quote="santosguzella"]Let $ a_{1}\ge ... \ge a_{n}\ge a_{n+1}=0$ be real numbers. Show that:
\[ \sqrt{\sum_{k=1}^{n}}\le \sum_{k=1}^{n}\sqrt{k}(\sqrt{a_{k}}-\sqrt{a_{k+1}}) \]
[/quote]
Please edit your problem.

If the right version is:
$ \sqrt{\sum_{k=1}^{n}a_{k}}\leq\sum_{k=1}^{n}\sqrt{k}(\sqrt{a_{k}}-\sqrt{a_{k}+1})$
I'm sure that this problem from ISL 1997.If someone want i could post solution(you can also find it at http://www.kalva.demon.co.uk/short/sh97.html ).


modular arithmetic
floor function
logarithms
ceiling function
function
geometry
3D geometry

Wouldn't it be a good idea to list answers so people can check and make sure they didn't make any mistakes?

Well here's what I got:

1. 12

2. $2^\frac{1}{2} + 2^\frac{1}{6} + 2^\frac{-1}{6} + 2^\frac{1}{3} + 2^0 + 2^\frac{-1}{3}$ (there might be other ways to express this)

3. $\frac{(n-2)(n-1)}{2 \times 3^6}$

4. Any number of the form $2 \times 3^m - 2$ or $3^m - 2$, where m is an integer $\ge 1$. Ie 4, 7, 16, ...

5. I didn't solve this but I've heard 140...

Those look right.  The only thing is that #4 states in the problem that $n>2$ so $n=1$ shouldn't be part of the answer.

[quote="Philip_Leszczynski"]Um.... I put $\frac{(n-1)(n-2)}{2}\cdot(\frac{1}{3})^6$ for #3. Will they take off points for that?[/quote]
I highly doubt it....your answer is mathematically correct, and changing between the two forms is trivial.

At first I wrote it as $(2-(n\mod 2))$, then realized that it could be written differently, and probably more acceptably. :)

I wrote my answer to #4 as

$3^rx - 2, x,r\in \mathbb{N}, x < 3$

You know, there IS no shame in giving two equations that generate all results  :P (for those of us who didn't bother to find some fancy way to give one equation).

My equation followed logically from my solution method, though, and it allowed me to find a clever extension of the solution.

For number 2, Instead of doing difference of cubes followed by difference of squares, I just used difference of 6ths ... that is

a^6 - b^6 = (a - b)(a^5 + (a^4)b +(a^3)(b^2) + (a^2)(b^3) + a(b^4) + b^5)

They wont take off for that, will they?

JB

well...i go for the difference of 6th (if that's the technical term), but i prove it first before i use it :roll:

How would you go about "proving" a simple application of the distributive property? Surely you don't have to prove so simple a statement as

$(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5) = a^6-b^6$

hmm, i should have done this usamts round, i got the expected value and the powers of 2 problems in less than an hour, but i thought that my answers were too easy, so i just forgot about it and concentrated on school...oh well, and for the last one, what did you guys get, wasn't the point just the orthocenter? i am not sure how to prove that though

In #4, I made a small careless mistake.  I wrote $3^x-2$, x greater than 1, integer.  I hope they don't go crazy on that one...

[quote="JesusFreak197"]I think he might just mean stating that (in terms of $a$ and $b$), then apply it to the equation, so that they know what he's doing. That's what I did with the cubic step, since that's not [i]quite[/i] as well known as the square step.[/quote]

Here's what I said ...

"For any real a and b:

a^6 - b^6 = (a - b)(a^5 + (a^4)b +(a^3)(b^2) + (a^2)(b^3) + a(b^4) + b^5)

Therefore, we multiply both the numerator and denominator of

1/(2^(1/2) - 2^(1/3))

by

[2^(1/2)]^5 + [2^(1/2)]^4*[2^(1/3)] + [2^(1/2)]^3*[2^(1/3)]^2 + [2^(1/2)]^2*[2^(1/3)]^3 + [2^(1/2)]*[2^(1/3)]^4 + [2^(1/3)]^5

= 2^(5/2) + 2^(7/3) +2^(13/6) + 2^2 + 2^(11/6) + 2^(5/3)

Yielding

[2^(5/2) + 2^(7/3) +2^(13/6) + 2^2 + 2^(11/6) + 2^(5/3)]/[(2^(1/2))^6 - (2^(1/3)^6)]

..."

*Sorry about the poor typesetting ....*

I think that should be okay.  The difference of two powers theorem is fairly well known, and I stated what I was using before I actually went into it.

What do you think?

JB

I don't see anything wrong with that. Also, that formula should also work for non-real $a$ and $b$.

You should learn to use Latex, though. It's very simple for problems like that, and makes it much easier to read.

For convenience:

"$a^6-b^6=(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$

Therefore, we multiply both the numerator and the denominator of

$\frac{1}{2^\frac{1}{2}-2^\frac{1}{3}}$

by

$(2^\frac{1}{2})^5+(2^\frac{1}{2})^4(2^\frac{1}{3})+(2^\frac{1}{2})^3(2^\frac{1}{3})^2+ (2^\frac{1}{2})^2(2^\frac{1}{3})^3+(2^\frac{1}{2})(2^\frac{1}{3})^4+(2^\frac{1}{3})^5$
$=\frac{2^\frac{5}{2}+2^\frac{7}{3}+2^2+2^\frac{11}{6}+2^\frac{5}{3}+2^\frac{3}{2}}{(2^\frac{1}{2})^6-(2^\frac{1}{3})^6}$"

[quote="banblink182"]For Question 4, I gave a recursive formula instead of an empirical one. That is, I said that 4 worked, and f(4), f(f(4)), f(f(f(4))), etc. I said that 7 worked, and f(7), f(f(7)), etc. worked as well. f(x)=3x+4. No one seemed to have done it this way, and I was just wondering if there was something wrong with it.[/quote]
Basically you set up a recursive equation - there's nothing wrong with that (that's how I came up with my solution, which I then converted into a closed-form solution).  However, in that case you have to treat the two possible starting values.  4 is one such starting value, but 1 is another.  If you left out one of these values, then you've left out half of the solutions, which would leave your solution incomplete.

JesusFreak --

Thanks for TeX-ing my thing ... I really appreciate it.

JB

[quote="d343seven"][quote="ZennyK"]
I think you can safely bet that you're wrong pretty much any time you end up with a floor or ceiling function as part of your answer. ;)[/quote]

The first answer I got was something like $n \equiv -2 \pmod {3^{\lfloor \log_3 {n+2} \rfloor}}$ and it still gave the correct values.  (It's just hard to find the values)  So floors don't mean that it won't give the right values.   I changed it to the 2 formulas form later though.[/quote]

I'm pretty sure that you woudn't have gotten full credit for that answer.  It gives you the right answer, but it's not at all simplified.

[quote="ZennyK"]I think you can safely bet that you're wrong pretty much any time you end up with a floor or ceiling function as part of your answer. ;)[/quote]

My final answer had a floor in the exponent. It was something like $\left(2-\frac{1-(-1)^n}{2}\right)3^{\lfloor \frac{n+1}{2}\rfloor}-2$. (If that's wrong, then it wasn't that, because I tested the first few values of $n$ and they all gave the correct answers in order.) I think you actually needed to use the floor if you only had one equation, since the power of $3$ would stay constant for two values, then move up to the next power.

[b]EDIT:[/b] Equation corrected.

[quote="ZennyK"][quote="dangvh"][quote="JesusFreak197"]#4 can also be written with $\left(2-\frac{1-(-1)^n}{2}\right)$ at the beginning so that it simplifies to one equation.[/quote]

ah that's smart!!... argh i should have thought of that... i simplified it into one equation but it made it really weird looking... i got $\displaystyle n=2^{a\mod 2}\cdot 3^{\left\lceil\frac{a+1}{2}\right\rceil}-2$. where $a\in %Error. "bbmath" is a bad command.
{N}$ it still gives all the right numbers though.. but it's unnessarily complicated  :blush:[/quote]
I think you can safely bet that you're wrong pretty much any time you end up with a floor or ceiling function as part of your answer. ;)[/quote]

well i put the equation into my graphing calculator and all the numbers it gave worked... i'm pretty sure it's right though

[quote="JesusFreak197"][quote="ZennyK"]I think you can safely bet that you're wrong pretty much any time you end up with a floor or ceiling function as part of your answer. ;)[/quote]

My final answer had a floor in the exponent. It was something like $\left(2-\frac{1-(-1)^n}{2}\right)3^{\lfloor \frac{n+1}{2}\rfloor}$. (If that's wrong, then it wasn't that, because I tested the first few values of $n$ and they all gave the correct answers in order.) I think you actually needed to use the floor if you only had one equation, since the power of $3$ would stay constant for two values, then move up to the next power.[/quote]
umm you need to subtract two after that, but i'm sure you did that since i read some of your previous posts.. but in order for you equation to work you have to says that $n\ge 2$

[quote="zanttrang"]Basically you set up a recursive equation - there's nothing wrong with that (that's how I came up with my solution, which I then converted into a closed-form solution).  However, in that case you have to treat the two possible starting values.  4 is one such starting value, but 1 is another.  If you left out one of these values, then you've left out half of the solutions, which would leave your solution incomplete.[/quote]

Got it. Thank you  :)

[quote="dangvh"]umm you need to subtract two after that, but i'm sure you did that since i read some of your previous posts.. but in order for you equation to work you have to says that $n\ge 2$[/quote]

Oh, right, I forgot to mention that (in the post). :oops:  :roll:

Why would you have to mention that $n>2$ if the problem itself already mentioned that for you?

he'd have to mention that $n\ge 2$ not $n>2$. This is irrelevant though, because his formula gives values of $n$ that are greater than $2$. He just chose a weird variable to express it as. In his equation, i guess he should have chosen a different variable for this reason, you'll see that other posts use $x$ or $a$. In JesusFreak's case, he seemed to have just chosen a variable $n$. This is kind of confusing, but he just needed to state that his variable had to be $\ge 2$. Then $n>2$ you refer to is the value of $n$ (the one given by the problem) in which the $n^{th}$ person will win... heh that's kinda confusing isnt it...

I meant that I forgot to put in the $-2$ at the end.

As for the values, they have no reason to take off points because my answer includes $1$.


algebra
polynomial
function
combinatorics unsolved
combinatorics

[quote="fedja"]Let $M_n$ be the maximal coefficient of the polynomial $P_n(z)=(z-1)(z^2-1)(z^3-1)\ldots (z^n-1)$. If I do not mistake, there exists a nice combinatorial proof showing that the numbers $M_n$ grow quite slowly (say, slower than exponentially in $n$). Does anyone know such a proof (or any other simple proof of this statement)?[/quote]

I don't...  :(  ... but would like to know one! ;)

This is a consequence of [url=http://planetmath.org/encyclopedia/PentagonalNumbersTheorem.html]Franklin's proof[/url] of Euler's [url=http://mathworld.wolfram.com/PentagonalNumberTheorem.html]Pentagonal Number Theorem[/url]:

$\prod_{k=1}^{\infty}(1-x^k)=\sum_{k=-\infty}^{\infty}(-1)^k x^{k(3k+1)/2}$

The bijective proof desribed in the first link corresponds to the following picture:

The coefficient of $x^N$ on the left hand side is a generating function for partitions of $N$ into parts of different length, e.g. 24=10+9+5, represented by the diagram
xxxxxxxxxx
xxxxxxxxx
xxxxx

We have to count this partition with $(-1)^{number of parts=lines of the diagram}x^{number of x's}$.

Set up a bijection between partitions like this:

1)
xxxxxxxxx[color=red]x[/color]
xxxxxxxx[color=red]x[/color]
xxxxx

and

2)

xxxxxxxxx
xxxxxxxx
xxxxx
[color=red]xx[/color]

Very few truncated staircases remain and give the terms on the right.

Now, if the product on the left stops at $k=n$, we can set up the same bijection, but the size of the first line of the diagram is bounded, so we can't go from 2) to 1) if the first line has already length $n$. Obviously, still most of the terms cancel, although it might be a bit annoying to write down an exact statement.


logarithms
function
calculus
calculus computations

[quote="AndrewTom"]Why do graph-plotters only plot (x-1)^(2/5) for values of x >=1?[/quote]
Couldn't understand the question, please rephrase. :(  :(
Ah. i get it, for x<1, they get the complex 5 th root of a negative which is 'nearest ' to +ve real axis, so negative real root is not considered.

Because the fractional powers are usually computed using the $ \log$ function and it is undefined for negative arguments. Some clever math software tools recognize fractions in powers for what they are and give you the entire thing but most of them compute $ 2/5$ in binary first (which results in some ugly approximate number) and then proceed via $ \log$.

Because if $ x<1$ then $ (x \minus{} 1)^{\frac {2}{5}}$ is not real. :)

Nope, it is real (you can take the fifth root of a negative number in $ \mathbb R$ and then square it, or square the negative number first and take the fifth root afterwards: the result is the same). ;)

While those have other complex roots (or branches), they have one root $ \in \mathbb{R}$. With the equation $ x^5 = - 1$ or $ x^3 = - 1$, $ x = - 1$ is a real solution.

Although with $ ( - 1)^\frac {2}{6}$, I see $ ( - 1)^\frac {2}{6} = ( - 1)^\frac {1}{3} = - 1$, but I also see ${ ( - 1)^\frac {2}{6} = \left[( - 1)^2\right]^\frac {1}{6}} = 1$. :huh: (in terms of real roots).


geometry
limit
projective geometry
geometry theorems

i came across a proof of menelaus using projective geometry ...in this the author uses the idea that if $ N$ is projected to a point $ N^{'}$ belonging to the line at infinity then $ \frac{A'N^{'}}{B^{'}N^{'}}\equal{}1$
it seems obvious enuf but is there a proof?

Well...$ \lim_{x\rightarrow \infty} \frac{x\plus{}k}{x}\equal{}1$...

What more than this are you looking for?


geometry
angle bisector
geometry proposed

$ AE,CF,BD$ are concur at $ Q$ in triangle $ ABC$.$ AQ$ is perpendicular to $ DF$ at $ P$. $ BPDC$ is cyclic.prove that $ BD\equal{}CD$

$ R \equiv FD \cap BC$ and  let $ M$ and $ M'$ be the midpoint of $ BC$ an the orthogonal projection of $ D$ onto $ BC.$ We shall prove that $ M\equiv M',$ then $ \triangle BDC$ is isosceles with base $ BC.$ It's enough to see that $ (B,C,E,R) \equal{} \minus{} 1$ and that the quadrilateral $ PDM'E$ is cyclic on account of the right angles at $ P,M'.$ Then we have

$ \overline{RB} \cdot \overline{RC} \equal{} \overline{RE} \cdot \overline{RM}$ and $ \overline{RP} \cdot \overline{RD} \equal{} \overline{RE} \cdot \overline{RM'} \equal{} \overline{RB} \cdot \overline{RC}$

$\Longrightarrow  \overline{RM}$ $ \equal{} \overline{RM'}$ $ \Longrightarrow  M \equiv M'.$

Nice! my solution is like your! quite simple:)

we use the following well-known lemma:
let $ AD$ be the altitude of $ \Delta ABC$ and let the lines $ CF,BE$ intersect each other on $ AD$ then $ AD$ is the angle bisector of $ EDF$.
by the above lemma it follows that $ AP$ is the angle bisector of $ BPC$ .now let this line intersect $ (BPDC)$ at $ S$ it implies that $ \angle BDS\equal{} \angle SDC$.but we know that $ \angle SPD \equal{} 90^\circ$ or in other words $ DS$ is also the diameter.
so $ DS$ is the diameter and also the angle bisector of $ \angle BDC$ so we get $ BD\equal{}CD$.


modular arithmetic
number theory unsolved
number theory

If $v\equiv 1 (mod 3)$ and $v^{2}-v\equiv 0 (mod 12)$ , then
$v\equiv k_{i}(mod 12)$ .  Find $k_{i}, i\in\mathbb{N}_{0}$ Please explain in details because I have just started doing number theory.
Thanks

We have $12$ cases occur $v=12k+0,12k+1,...,12k+11$. Now, checking! :D

Ok.I understand (i didnt consider the first condition)


Find the maximum number $ c$ such that for all $n \in \mathbb{N}$ to have \[ \{n \cdot \sqrt{2}\} \geq \frac{c}{n}\] where $ \{n \cdot \sqrt{2}\} \equal{} n \cdot \sqrt{2} \minus{} [n \cdot \sqrt{2}]$ and $ [x]$ is the integer part of $ x.$ Determine for this number $ c,$ all $ n \in \mathbb{N}$ for which $ \{n \cdot \sqrt{2}\} \equal{} \frac{c}{n}.$

Let $ [n\sqrt 2] \equal{} m$, then $ 2n^2 \minus{} m^2 \equal{} a\ge 1$ There are infinitely many solutions $ m^2 \minus{} 2n^2 \equal{} \minus{} 1$.
Therefore for such $n$ we get $ \{n\sqrt 2\} \equal{} \sqrt {m^2 \plus{} 1} \minus{} m \equal{} n\sqrt {2} \minus{} \sqrt {2n^2 \minus{} 1} \equal{} \frac {\sqrt {2}}{4n}(1 \plus{} \frac {1}{4n^2} \plus{} \frac {3}{16n^4} \plus{} \cdots)$. Therefore the best coefficient is $ c \equal{} \frac {\sqrt 2}{4}$, and $ \{n\sqrt 2\} > \frac {\sqrt 2}{4n} \ \forall n\in N$.


Determine the best constant $ k$ for which the iequality below holds

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\leq\frac{k}{\sqrt{3}}(\frac{9}{8R}\plus{}\frac{15}{16r})$

[quote="Ligouras"]Determine the best constant $ k$ for which the iequality below holds

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\leq\frac {k}{\sqrt {3}}(\frac {9}{8R} \plus{} \frac {15}{16r})$[/quote]

$ k_{min} \equal{} 1$

$ k\equal{}1$

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\leq\frac {1}{\sqrt {3}}(\frac {9}{8R} \plus{} \frac {15}{16r})........(*)$

substitution $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$

$ (*)\Longleftrightarrow \sum_{sym} (75x^5y^2\minus{}53x^5yz\plus{}225x^4y^3\minus{}141x^4y^2z\minus{}163x^3y^3z\plus{}57x^3y^2z^2)\geq 0$

from AM-GM and Muirhead theorem the inequality holds.

 :D


In this topic, we talk about our Middle School NEML scores and past NEML scores. Well, I created this topic because I'm kind of tired of seeing the same topics every time I go to this board. For those of you who doesn't know what the Middle School New England Mathematics League is, its a 30 minute 40 question test designed to train your accuracy and speed, although the questions are very easy (about as easy as school or easy chapter countdown questions).

6th Grade: 39
7th Grade: 38
8th Grade: 40

hmmm i got like

39
40
40

my advice is to uh use your calculator at any point whether necessary or unnecessary; the last thing you want to do is be mad because you got a 39 and said 26+54=70.

Err...

40
38
39

I added fractions incorrectly or something in 7th, and last year, thought there were 30 days in July.

for high school depressingly i have 6/4/6/6/6 so far because on the second meet i solved for the wrong thing and didn't see the obvious resemblance of the last problem to a problem which I had written up the solution to before...twice

Lol I'm 6/5/6/6/6 or 6/6/5/6/6 or something. When did I miss one? It was the thing with the finding a so that it had two positive solutions. I didn't notice the positive thing, found an a that gave two solutions, and just stopped.

[quote="shentang"]And this is NEML?

6th grade- 38 stunk because I can't count how many pictures in a pattern
7th grade- 39 just because I misread a 3 for a 1, ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!
8th grade- goal is 41

Yay I beat alan! First time ever in my life or second......[/quote]

dude, I suck at NEML...

Err...
38
something
something
Because I don't have my score yet.

Congrats to my school, Jonas Clarke Middle School, for having the highest team score on the NEML in all 3 grades(7th was tied). The 8th grade score was beat by only one school in the country(Miller in CA)!!!

Great job Clarke!!!

Results are posted here :

[url]http://www.mathleague.com/reports/leagues.htm[/url]

I go to a pretty poor school in Georgia, but we've had 199 before too. Back when we had smart people...
We still won this year though, so yay.
(Why am I posting this here?!)


sure that the statement doesn't need to be cyclic in some way? :)

Cuz now I guess it's simply to prove that:
$\dsp\frac{a^2}{a+1}\geq \frac89a - \frac43$
which is trivial...

Is it $\dsp\sum_{cyc} \frac{a^2}{b+1}$ perhaps?

Anyway manlio, these are all pretty nice questions :) what does their source "GM_x" mean? :)

[quote="Peter VDD"]

Is it $\dsp\sum_{cyc} \frac{a^2}{b+1}$ perhaps?

[/quote]

By rearrangement, that is stronger than $\dsp\sum_{cyc} \frac{a^2}{a+1}$, so there's not much of a difference.

GMA means Gazeta Matematika A. It is a nice Romanian journal about mathematical problems.

I went on a long trip to go and copy them.

Sorry, but I don't remember the years and the numbers of the problems as I have had little time to look at  them and when  I copied them I was in a great hurry.

However I have again a lot of inequalities of this kind of difficulty to post for the forum but before I prefer to see the solutions of these ones.


Does every monotone increasing function $f : \mathbb[0,1] \rightarrow \mathbb[0,1]$ have a fixed point?
What about every monotone decreasing function?

[quote="cezar lupu"]Yes, every monotone increasing function $f$ defined as $f: [a,b]\to [a,b]$ has at least a fix point.
This is called Knaster theorem. ;)[/quote]

can you prove it  :roll:
and what about the monotone decreasing function ? the ansewr is[color=red] no [/color].give a counter example  ;)

take the set $E=\{x | f(x) \geq x\}$. This set is non empty.
Take $a=sup(E)$.
If $f(a)>a$ then it is also true for some $x>a$, because $f$ is increasing. Impossible
If $f(a)<a$ then it is also for all $a- \epsilon<y<a$. impossible.

conclusion follows ..

I have a similar problem from the Romanian olympiad:

[color=blue]Let $f: [0,1] \to [0,1]$ be function continuous in $0, \, 1$, with lateral limits in every point and which satisfies \[ f(x-0) \leq f(x) \leq f(x+0) , \, \forall x \in \left( 0,1 \right) . \]

Prove that there is $x_0 \in \left[ 0,1 \right]$ such that $f(x_0) = x_0$.[/color]

As for Fermat-Euler's question:

[quote="Fermat -Euler"]and what about the monotone decreasing function ? the ansewr is[color=red] no [/color].give a counter example  ;)[/quote]

This part is a lot easier: On the interval $\left[ 0,\frac12 \right)$ let function decrease linearly from $1$ to $\frac34$. On the other interval, let it is decrease linearly from $\frac14$ to $0$. Of course we can find infinitely many examples.

The other part (proved by alekk) is much more interesting.

I draw a graph and looks like this function can become a counterexample.
let $f(x)=1 - \frac{x}{k}$ for $x \leq \frac{1}{k}$, $f(x)=\frac{1}{k} - \frac{x}{k}$ for $\frac{1}{k} < x < 1$.

Let $M=${$x\in [a,b]/$x<= f(x)}. $M\subseteq[a,b]$  $\exists u=supM$ $u\geq x$  $\forall x \in M$ $\Rightarrow f(u) \geq f(x) \geq x \forall x \in M$

My solution:

Let M={$x\in [a,b]/ f(x)\geq x$}.$a\in M$ ,$M\subseteq[a,b]$  $\exists u=supM$ $u\geq x$  $\forall x \in M$ $\Rightarrow f(u) \geq f(x) \geq x \forall x \in M$ $\Rightarrow f(u) \geq x \forall x \in M$ $\Rightarrow f(u) \geq u \Rightarrow f(f(u)) \geq f(u) \Rightarrow f(u) \in M \Rightarrow u \geq f(u) \Rightarrow f(u)=u (q.e.d)$ ;)

$(i)$. Yes, if $f$ is increasing, $f$ has a fixed point. Assume the contrary that $f(x)\neq x$ for all $x\in [0,1]$, then $0<f(0)$ and $f(1)<1$. Let $p=\sup\{x | f(x)>x \}$, so $f(p)>p$. Consequently,
$$ f(x) < x, \quad \forall x>p \; .........(1) $$
Setting $x:= f(p)$ to $(1)$ gives us $f(f(p)) < f(p)$ which implies $f(p)<p$ since $f$ is increasing. Contradiction.

$(ii)$. Not always, for example:
$$ f(x) = 1-x^2  \; ,\forall  x\in [0, \frac{1}{2}), $$
$$ f(x) = (1-x)^2  \; ,\forall  x\in [ \frac{1}{2}, 1]. $$


function
algebra
functional equation
algebra solved

I have this functional equation, for which I have a 4 pages solution....  :?

here goes:

[b]find out all functions f:Z-->R for which we have:

f(h+k-hk)=f(h)+f(k)-f(hk),for any h,k integers. [/b]

g'luck with it!!! :D
cheers!

you should try it... it is not that hard .. it is just that I have found a very long solution... perhaps you may find one more direct!!!

 \sqrt :D  2

It looks to me like the functions f(x)=ax+b check the relation. I think this is the answer, but I haven't really given it any thought. I'll try to solve it. I don't know where to start, though.

the answer is more complicated than you would think! :D :D

(don't worry I am not trying to scare you! ) :)

cheers!

Okay.. guess it's kinda hard?! :?

here is a simmilar problem that is more simple and simmilar to the first one, that I am sure it will help for the 1st one! :D

[b] find out all functions f:Z-->R for which we have:
f(h+k+hk)=f(h+k)+f(hk),for any h,k integers. [/b]

cheers! :D :D

The autor of f(h+k+hk)=f(h+k)+f(hk) is M.O. Drimbe, isn't he?

I don't know!  Perhaps you know better! :D :D
It doesn't say who proposed it! But it is kind of an  old problem..

Cheers! :D :D

Where is it from? I found it in "200 de ecuatii functionale pe N,Z,Q"
http://www.gil.ro

Could you please post a soln? I've tried to solve it for some time and then I got scared :D  :D  :D

The problems are old.. Italian problem actually.. Giuseppe Ancona was the one who made them :)

the solutions are very long, especcially for the other one!!!

ciao!


[quote=" in Hungkhtn's book he"] It easy to prove that :
$\frac{(a^{2}+b^{2}+c^{2})(a+b+2c)}{(a+b)(b+c)(c+a)}\geq \frac{(2t^{2}+c^{2})(2(t+c)}{2t(t+c)^{2}}$
with $t=\frac{a+b}{2}$
[/quote]
But i think it isn't easy . I need a proof

Just use $a^{2}+b^{2}\geq 2t^{2}$ by RMS-AM and $(a+c)(b+c)\leq (t+c)^{2}$ by AM-GM on $a+c$ and $b+c$.


Easy one :lol:

In base 8 (Oct) 31is equal to 25 in base 10 (Dec)

"To Be or  not to be. That is the question"  what is the answer?
(Why it is negative one?")

Once I asked my son to buy some apples, how may he asked? I said "6 or 7" he brought home 7.

Next time, in similar case, I said "7 or 8" and he brought home 15.

Why?

Halloween owns Christmas, just like the Great Pumpkin owns Santa.

um...no its the other way CHRISTMAS owns halloween and SANTA owns the great pumpkin in charlie brown

Gyan wrote:Once I asked my son to buy some apples, how may he asked? I said "6 or 7" he brought home 7.

Next time, in similar case, I said "7 or 8" and he brought home 15.

Why?

Spoiler: [hide]Because he still carried the 7 from before?[/hide]

R. -  Hint: We are applying same kind of logic  as OCT28=DEC31...
(Ti- calculator can help)

what does that mean? i still don't get it.

Spoiler: [hide]31 in base 8 (OCTal)=25 in base 10 (DECimal)[/hide]

Hey no solutions yet! Ok here  is the answer to the  second question: (Forst than will become almost, but not quite, giveaway)

[hide]In Binary representation:

6 = '0101'

7 = '0111'

8 = '1000'

Now just find out  6 'o'r 7   (You can use TI calcualtor , if you want, just use boolian "or")

'0101' | '0111'  = '0111' = 7

and '0111' | '1000' = '1111' = 15

  [/hide]

[quote="uglyman621"]wow gyan.....

thats a wierd problem

and what does octal mean?????[/quote]
BASE 8

[quote="gameworld7"]how are you so sure that your answer is correct?[/quote]

Because he made the problem...  :maybe:

how D0 :varepsilon: s H :varepsilon: kn0w th :varepsilon: c0 :gamma:  :gamma:  :varepsilon: ct answ :varepsilon:  :gamma: ( :varepsilon: v :varepsilon: n iF hE IVIADE IT)

first of all dont ever post ur responses like that again
second if u have a better answer to that question feel free to post it

i I)on't hav :varepsilon:   a choic :varepsilon: .  IVIy sist :varepsilon:  :gamma:  IVI :varepsilon: ss :varepsilon: I)  up IVIy  coIVIput :varepsilon:  :gamma: . :(

then dont post that day! (cuz no SANE person can read it... unless they look at it for a long time)

and i lyk christmas better (course thats not what the topics about, but christmas is SO much better than halloween!)

halloween is dull...christmas is just when you get and get and get and get and oh yeah don't forget to give :lol:

yup!  givings important! (that way OTHERS can get... and give to u :lol: )

there's a saying:the more you give, the more you get.

that's not true all the time...

I disagree; the act of gift-giving can be very rewarding.

Especially if it involves gag gifts.

the thing is....

u can give things ppl, and they dont give it to u THAT year....

BUT they usually feel REAL guilty and get u something special the NEXT year(thinking ur giving them something again)

thats how it works w/ me...

um...wouldn't that make people even more angrier at you?


function
limit
real analysis
real analysis unsolved

function $ {f(x)$ at  $ x \equal{} 0$ continuious   ， and
 $ \lim_{x \rightarrow 0}\frac {f(2x) \minus{} f(x)}{x} \equal{} A.$
prove： $ f'(0) \equal{} A$.

proof
\[ \lim_{x \rightarrow 0}\frac {f(2x) \minus{} f(x)}{x} \equal{} A.\]

\[ f(2x) \minus{} f(x) \equal{} Ax \plus{} o(x), f(\frac {x}{2^{k \minus{} 1}}) \minus{} f(\frac {x}{2^{k}}) \equal{} A\frac {x}{2^{k}} \plus{} o(\frac {x}{2^{k}}),\]

\[ \sum_{k \equal{} 1}^n (f(\frac {x}{2^{k \minus{} 1}}) \minus{} f(\frac {x}{2^{k}}) ) \equal{} A \sum_{k \equal{} 1}^n \frac {x}{2^{k}} \plus{} o(\sum_{k \equal{} 1}^n\frac {x}{2^{k}}),\]

\[ f(x) \minus{} f(\frac {x}{2^n}) \equal{} Ax(1 \minus{} 2^{ \minus{} n}) \plus{} o((1 \minus{} 2^{ \minus{} n})x).\]
$ { n \rightarrow \infty },$
\[ f(x) \minus{} f(0) \equal{} Ax \plus{} o(x),\]

\[ \lim_{x \rightarrow 0}\frac {f(x) \minus{} f(0)}{x} \equal{} A\]

[quote="yzymat"]

\[ \lim_{x \rightarrow 0}\frac {f(x) \minus{} f(0)}{x} \equal{} A\]
[/quote]

In my opinion, to get the avove result you must prove that $ \frac{o(x)}{x}$ is infinitesimal when $ x \rightarrow 0$. Anyway the result proves that $ f'(0)\equal{}A$ and not $ \forall x$ as you write at the beginning of proof.

That's the definition of $ o(x)$. It means "asymptotically strictly smaller than $ x$", and here it's understood that is "as $ x \to 0$".

Of course, the second objection is still valid ;).

[quote="prester"][quote="yzymat"]
\[ \lim_{x \rightarrow 0}\frac {f(x) \minus{} f(0)}{x} \equal{} A\]
[/quote]

In my opinion, to get the avove result you must prove that $ \frac {o(x)}{x}$ is infinitesimal when $ x \rightarrow 0$. Anyway the result proves that $ f'(0) \equal{} A$ and not $ \forall x$ as you write at the beginning of proof.[/quote]

Thanks!  I have  edited

Couldn't you just say that, by L'Hopital's rule, $ \lim_{x\to 0}\frac {f(2x) \minus{} f(x)}{x} \equal{} \lim_{x\to 0}2f'(2x) \minus{} f'(x) \equal{} f'(0) \equal{} A$?

Edit: Oh, of course. I thought for some reason that we were given that $ f$ is differentiable, but now I see that we're only given continuity. Thanks

1. There was another topic on this exact question just a few months ago. I haven't read yzymat's proof all that carefully, but it seems to somewhat resemble the proof I posted on that other topic.

2. No, grn_trtle, you can't do that. You can't do that because there is no hypothesis that says $ f$ is differentiable anywhere at all. That it is differentiable at zero is something that must be proved here.

Probably the proof is right but let me pose just a question:  are we sure that
$ \lim_{n\to\infty} o((1 \minus{} 2^{ \minus{} n})x) \equal{} o( \lim_{n\to\infty} (1 \minus{} 2^{ \minus{} n}) x)$
or we need the continuity of $ o(x)$ ?  From the statement $ f(x)$ seems continuous at $ x\equal{}0$ and at no other points.

Hi.  I am new to the forum so hello everyone.  I have a naive solution to the problem.

Let  $ L\equal{}\lim_{x\to0}\frac{f(x)\minus{}f(0)}{x}$

Notice that by letting $ y\equal{}2x$ we get

$ \lim_{x\to 0}\frac{f(2x)\minus{}f(0)}{x}\equal{}2\lim_{x\to0}\frac{f(x)\minus{}f(0)}{x}$

Write,

$ \frac{f(x)\minus{}f(0)}{x}\equal{}\frac{f(2x)\minus{}f(0)}{x}\minus{}\left(\frac{f(2x)\minus{}f(x)}{x}\right)$

let $ x\to 0$ to get

$ L\equal{}2L\minus{}A$

which proves the result.

I guess there is no guarantee that $ L$ exists.  Can we add $ f$ differentiable at $ x\equal{}0$ to the hypothesis ?  :)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299969]This[/url] is the topic I was looking for - I finally found it.

Go to my post #7 on that topic for a complete proof. To answer mathador: while his naive proof does require an assumption of differentiability at $ 0,$ the original problem does not require differentiability. Continuity at zero does suffice, which is what my proof shows.


algebra
function
domain
modular arithmetic
quadratics
Ring Theory
number theory unsolved

Prove that we can show every prime $p=8k \pm 1$ with this form $x^{2}-2y^{2}$

$\mathbb Z[\sqrt 2]$ is an Euclidean domain, so is a number in $\mathbb Z[\sqrt 2]$ is not prime, then it's decomposable.

Since $p\equiv\pm 1\pmod 8$, it means that $2$ is a quadratic residue $\pmod p$, so there are $a,b,\ p\not|\ a,b$ s.t. $p|a^2-2b^2=(a-\sqrt 2b)(a+\sqrt 2b)$. Since $p\not|\ a-\sqrt 2b,a+\sqrt 2b$, it means that $p$ is not a prime in $\mathbb Z[\sqrt 2]$, so it must be decomposable, meaning that we can find $x,y$ s.t. $x-\sqrt 2y$ is a prime in $\mathbb Z[\sqrt 2]$ and $x-\sqrt 2y|p$. However, this means that $x+\sqrt 2y|p$, so $x^2-2y^2|p\ (*)$. By applying the norm function to $(*)$, we will find that either $p=x^2-2y^2$ or $p=-(x^2-2y^2)$. In the first case we're done, and in the second case we have $p=(x+2y)^2-2(x+y)^2$, and we are, again, done.

alternatively, if $p$ is prime and is one the form $8k+1,8k+7$ then $2$ is a quadratic residuum $mod$ $p$.
Applying Thue's theorem ([url=http://mathworld.wolfram.com/ThuesTheorem.html]mathworld.wolfram.com/ThuesTheorem.html[/url]) for $a$ such that $a^2=2$ $(\mod p)$ we receive the answer
(however after few additional steps).


Suppose the first term of a sequence $ a_1\equal{}1$,

$ a_{n\plus{}1} \equal{} \frac{1}{16} \left[1\plus{}4a_n\plus{}(1\plus{}24a_n)^\frac{1}{2}\right]$

Find the general formula of $ a_n$

[quote="Bovy11"][hide]
Let $ b_n \equal{} \sqrt {1 \plus{} 24a_n}$,then $ a_n \equal{} \frac {b_n^2 \minus{} 1}{24}$,and we have

$ 4b_{n \plus{} 1}^2 \equal{} (b_n \plus{} 3)^2$
$ \equal{} > 2b_{n \plus{} 1} \equal{} b_n \plus{} 3$
$ \equal{} > b_{n \plus{} 1} \minus{} 3 \equal{} \frac {1}{2}(b_n \minus{} 3)$ and $ b_1 \equal{} 5$,thus we have

$ b_n \equal{} \left(\frac {1}{2}\right)^{n \minus{} 2} \plus{} 3$.

$ a_n \equal{} \frac {b_n^2 \minus{} 1}{24} \equal{} \frac {1}{24}\left[\left(\frac {1}{2^{n \minus{} 2}} \plus{} 3\right)^2 \minus{} 1\right]$[/hide][/quote]

Great :blush:  just the same as my draft before. :roll:


Joe has one book each for algebra, geometry, history, psychology, Spanish, English and physics in his locker. If he chooses three books from his locker, how many different sets of three books could he choose?

There are a total of $ 7$ books in his locker.  Because we are just looking for the number of sets of three books he can choose, the order we choose the books does not matter.  So the number of different sets of three books could he choose is simply \[ \binom{7}{3}\equal{}\frac{7!}{3!4!}\equal{}\frac{7\cdot \cancel{6}\cdot 5}{\cancel{3\cdot 2}}\equal{}\boxed{35}\]


modular arithmetic
floor function
quadratics
induction
algebra
functional equation
number theory proposed

Let $n\geq 0$ be an integer and let $p \equiv 7 \pmod 8$ be a prime number. Prove that
\[ \sum^{p-1}_{k=1} \left \{ \frac {k^{2^n}}p - \frac 12 \right\} = \frac {p-1}2 . \]

[i]Călin Popescu[/i]

This is one nice problem (in fact, all of them seem to be very nice, except maybe for the functional equation ;)).

Let's solve it for $n=1$, and then we'll see why this is enough.

We must compute $\sum_{k=1}^{p-1}\left(\frac{k^2}p-\frac 12-\left\lfloor\frac{k^2}p-\frac 12\right\rfloor\right)$, which is equal to $\sum_{k=1}^{p-1}\left(\frac{2k^2}p-\frac 12-\left\lfloor\frac{2k^2}p\right\rfloor+1\right)-\sum_{k=1}^{p-1}\left(\frac{k^2}p-\left\lfloor\frac{k^2}p\right\rfloor\right)\ (*)$, because of the identity $\left\lfloor x\right\rfloor+\left\lfloor x+\frac 12\right\rfloor=\left\lfloor 2x\right\rfloor$ applied to $x=\frac{k^2}p-\frac 12$.

Now, $(*)$ is equal to $\sum_{k=1}^{p-1}\left(\left\{\frac{2k^2}p\right\}-\left\{\frac{k^2}p\right\}\right)+\frac{p-1}2$, which equals $\frac{p-1}2$ because $2$ is a quadratic residue $\pmod p$, so the residues $2k^2\pmod p$ are the same as the residues $k^2\pmod p$.

The identity is proven for $n=1$. This is enough, because the residues $k^{2^n}\pmod p$ are the same as the residues $k^2\pmod p$, since $-1$ is not a quadratic residue $\pmod p$, meaning that the equation $x^4=1$ has $2$ solutions $\pmod p$, so there are $\frac{p-1}2$ fourth powers $\pmod p$, and we can apply induction: there are also $\frac{p-1}2$ $8$'th powers, and so on, and these are precisely the quadratic residues.

It says $n\ge 0$ in the text, but if it's also true for $n=0$, it's certainly not interesting enough to work out :).

Edit: edited in response to prowler's post below :)

My solution is almost the same :)

@ grobber I think you meant $[x]+[x+1/2] = [2x]$

:huh: $\left\lfloor x\right\rfloor+\left\lfloor\frac x2\right\rfloor=\left\lfloor 2x\right\rfloor$. What do I miss here?

[quote="Myth"]:huh: $\left\lfloor x\right\rfloor+\left\lfloor\frac x2\right\rfloor=\left\lfloor 2x\right\rfloor$. [b]What do I miss here?[/b][/quote]Grobbers edit of his post due to prowler's post :D

I know, but for some uknown reasons I receive topic reply notifications with 40 minutes delay this evening!

[quote="grobber"]This is one nice problem (in fact, all of them seem to be very nice, except maybe for the functional equation ;)).

Let's solve it for $n=1$, and then we'll see why this is enough.

We must compute $\sum_{k=1}^{p-1}\left(\frac{k^{2}}p-\frac{1}{2}-\left\lfloor\frac{k^{2}}p-\frac{1}{2}\right\rfloor\right)$, which is equal to $\sum_{k=1}^{p-1}\left(\frac{2k^{2}}p-\frac{1}{2}-\left\lfloor\frac{2k^{2}}p\right\rfloor+1\right)-\sum_{k=1}^{p-1}\left(\frac{k^{2}}p-\left\lfloor\frac{k^{2}}p\right\rfloor\right)\ (*)$, because of the identity $\left\lfloor x\right\rfloor+\left\lfloor x+\frac{1}{2}\right\rfloor=\left\lfloor 2x\right\rfloor$ applied to $x=\frac{k^{2}}p-\frac{1}{2}$.

Now, $(*)$ is equal to $\sum_{k=1}^{p-1}\left(\left\{\frac{2k^{2}}p\right\}-\left\{\frac{k^{2}}p\right\}\right)+\frac{p-1}2$, which equals $\frac{p-1}2$ because $2$ is a quadratic residue $\pmod p$, so the residues $2k^{2}\pmod p$ are the same as the residues $k^{2}\pmod p$.

The identity is proven for $n=1$. This is enough, because the residues $k^{2^{n}}\pmod p$ are the same as the residues $k^{2}\pmod p$, since $-1$ is not a quadratic residue $\pmod p$, meaning that the equation $x^{4}=1$ has $2$ solutions $\pmod p$, so there are $\frac{p-1}2$ fourth powers $\pmod p$, and we can apply induction: there are also $\frac{p-1}2$ $8$'th powers, and so on, and these are precisely the quadratic residues.

It says $n\ge 0$ in the text, but if it's also true for $n=0$, it's certainly not interesting enough to work out :).

Edit: edited in response to prowler's post below :)[/quote]

I wanna ask two questions :
1) Why $2$ is a quadratic residue $(mod p)$ ?
2) Why if $x^{4}\equiv 1 (mod p)$ has exactly $2$ solutions, then there are $\frac{p-1}{2}$ distinct fourth power in $mod p$ ? Isn't it possible that if $a$ and $b$ are distinct quadratic residues $mod p$, then $a^{2}\equiv b^{2}(mod p)$ ?
(Of course $a \equiv b (mod p)$ is impossible, but $a+b \equiv 0 (mod p)$ is possible, for example $a = 1, b = 4, p = 5$)
Maybe $p \equiv 7 (mod 8)$ has a role in both parts. Thanks..

1) Because $p\equiv 7(\mod p)$
and $\left( \frac{2}{p}\right) =(-1)^{\frac{p^{2}-1}{8}}$

2) Suppose $a$ and $b$ are distinct quadratic residue $\mod p$ such that $a\equiv-b(\mod p)$
so $ab=-a^{2}(\mod p)$

but $ab$ and $a^{2}$ are quadratic residue $\mod p$ and $-1$ is not quadratic residue $\mod p$

Contadiction..

Just be legendre symbol $(\frac {k^{2^n}}{p})=(\frac {k}{p})^{2^n}=1$. Now consider the set $\{k^{2^n}\}$ mod ($p$). Suppose there are two same elements into that set, the $p|{a^{2^n}-{b^{2^n}}}$ also $p|a^{p-1}-b^{p-1}$ and that implies $p|a^2-b^2$ since $gcd(p-1,2^n)=2$. Now obviously $\{k^{2^n}\}$ mod ($p$)=$\{k^2\}$ mod ($p$).
 So it's enough do for $n=1$. Now we've to show $\sum_{k=1}^{p-1}\{\frac {k^2}{p}-\frac {1}{2}\}=\sum_{k=1}^{p-1}\{\frac {(ak)^2}{p}-\frac {1}{2}\}$. Since we've $\{k^2\}$ mod ($p$)=$\{(ak)^2\}$ mod ($p$) for all $a\in\mathbb N$. Now take such $a$ for which $\frac {(ak)^2}{p}>\frac {1}{2}$ for all $k<p$.Now just using $[x]+[x-\frac{1}{2}]=[2x-1]=[2x]-1$ (as we're taking $x=\frac {(ak)^2}{p}>\frac {1}{2}$) we're done.


Each point in the plane is coloured with one of two distinct colours. Prove that there is an equilateral triangle with side 1 or [tex]\sqrt{3}[/tex] whose vertices have the same color.

I think this is from some Chinese olympiad. Do the same thing as in the other similar problem you have recently posted. You get the same thing: all points distance $2$ apart have the same color. Derive from here that all points of the plane have the same color.

I think, maybe it was on ARO in third round (ARO has 5 rounds).  :?


i need some help with my calculus homework.
I have this question which one of the following integrals diverges. I found out which one, but I don't know how to prove it.
It is Int(1/(sqrt(x)+ln(x)),x=1..infinity) or see bottom photo how do you prove it is divergent

0 < 1/(sqrt(x)+ln(x)) ~ 1/sqrt(x) on +oo

Does int (1..+oo, 1/ :sqrt:x dx) converge ??

[quote="belenos"]0 < 1/(sqrt(x)+ln(x)) ~ 1/sqrt(x) on +oo[/quote]

This is true, but I think it requires some justification. Proving it isn't very hard though once you know what ~ means.

Another way would be to show that ln (x) < sqrt (x) for x > 1. So 1/(2*sqrt (x)) < 1/(sqrt (x) + ln (x)). Then just show that the integral of this smaller integrand on the left diverges.


function
number theory unsolved
number theory

Show that there exists a function $ f: N\to N$ such that
$ f^{(2008)}(n) \equal{} 2n$

We divide the set of the odd positive integers into infinite disjoint $ 2008$-uples, like this:
${ \{1, 3, 5, ..., 4015}\}$
$ \{4017, 4019, 4021, ..., 8031\}$ etc.

Now for each of these sets, say $ \{a_1, a_2, ..., a_{2008}\}$, we define the infinite sequence $ \{b_i\}$ with $ b_i = a_i$ for all $ i \leq 2008$, and $ b_i = 2{a_{i - 2008}}$ for $ i > 2008$.

We'll show that every positive integer occurs exactly once in exactly one sequence $ \{b_i\}$.

This is trivial for the odd integers. Now, every even integer can be expressed as $ 2^k \times P$, where $ P$ is an odd integer. It's clear that $ 2^k \times P$ belongs to the sequence $ \{b_i\}$ where $ P$ belongs. Until now, we know that every positive integer occurs at least once in at least one sequence $ \{b_i\}$.

Now, let's suppose some even integer occurs more than once in a sequence $ \{b_i\}$ or occurs in two or more different sequences $ \{b_i\}$. In that case, we have $ 2^{k_1} \times P_1 = 2^{k_2} \times P_2$. But then it must be $ k_1 = k_2$, and it follows that $ P_1 = P_2$. So every positive integer occurs exactly once in exactly one sequence $ \{b_i\}$, as wanted.

So now we define the function $ f: \mathbb{N} \to \mathbb{N}$ such that $ f(b_i) = b_{i + 1}$, and we are done.


algebra
system of equations
algebra proposed

Determine all triplets of real numbers $ x,y,z$ satisfying \[1\plus{}x^4\leq 2(y\minus{}z)^2,\quad  1\plus{}y^4\leq 2(x\minus{}z)^2,\quad  1\plus{}z^4\leq 2(x\minus{}y)^2.\]

note that $ 1\plus{}x^4\geq 2x^2$, then we have $ (y\minus{}z)^2\geq x^2$ and similar inequalities... suppose wlog that $ x\geq y\geq z$.

we have that $ x\minus{}y\geq |z|, x\minus{}z\geq |y|, y\minus{}z\geq |x|$...

adding the first and the last one it follows that $ x\minus{}z\geq |x|\plus{}|z|$. so $ x\geq 0, z\leq 0$ and we must have equality at $ x\minus{}y\geq\minus{}z$ and $ y\minus{}z\geq x$. then $ y\equal{}x\plus{}z$.

substituting in the original system of equations it follows that $ x^2\equal{}z^2\equal{}1$. we conclude that $ (x,y,z)\equal{}(1,0,\minus{}1)$ (and the other five symmetric permutations)..


Hi, I am a rising junior in HS and in the linear algebra textbook I am using it says "If v is a scalar multiple of u OR if u is a scalar multiple of v." But isn't that the same thing?

They're almost the same thing (over a field), with an exception when one is zero.


Let $ f_0\equal{}f_1\equal{}1$ and $ f_{i\plus{}2}\equal{}f_{i\plus{}1}\plus{}f_i$  $ (i\ge0)$

Find all real solutions of the equation

$ x^{2010}\equal{}f_{2009} \cdot x\plus{}f_{2008}$

[quote="binaj"]Let $ f_0 \equal{} f_1 \equal{} 1$ and $ f_{i \plus{} 2} \equal{} f_{i \plus{} 1} \plus{} f_i$  $ (i\ge0)$

Find all real solutions of the equation

$ x^{2010} \equal{} f_{2009} \cdot x \plus{} f_{2008}$[/quote]

Just search "baltic way 2009" and .... http://www.mathlinks.ro/Forum/viewtopic.php?t=311538  just 10 days ago :(


Like other some other people, I'm having a very hard time deciding which college to go to. I got into MIT, Princeton, etc, but those schools are very expensive and i've ruled those out.

In terms of affordable schools, GA Tech would be about 8000 a year and University of Arizona would be free. I'm not sure whether I'd want to do science or engineering, and I've heard that which school you go to for undergrad doesn't matter as long as you make the most of it.

Any suggestions/advice?

Thanks in advance...any help is greatly appreciated.

why not apply for financial aid at schools like MIT and Princeton?

Do you feel that there are resources you are likely to need that will be unavailable to you at University of Arizona? If you're a self-motivated person and you don't think there are any specific resources lacking at University of Arizona, you'll probably do just as well there as you would at MIT or Princeton, and you'll save yourself a lot of money. So if you can't think of any specific resources that will not be available at Arizona that you're likely to need, I'd recommend going there.

I, on the other hand, would recommend going to MIT or to Princeton as long as your family won't go bankrupt if you go there, and the financial aid offers from each of those schools should be calculated to prevent that. What do your parents think you should do, and why?

I agree with tokenadult that you should go to MIT or Princeton if it's at all possible. Even if you aren't able to think of what additional resources you will be missing out on that doesn't mean that you won't necessarily have the same experience at either place.

And I disagree, because I don't think it's fair to put your parents in any type of financial distress for your education, and a lot of big name schools are more name brand than anything. Also, the amount a college expects you to pay is disgustingly high. I have a friend who, after applying for financial aid, still has to pay so much money for college that her mother has stopped putting money into her own 401k. That's not a good thing. However, my cultural background advocates taking care of your own finances, to a certain point. My parents will not be going into debt to pay for my college, and I am not responsible for taking care of them financially when they are retired.
I actually recommend GT. A lot of our graduates go there, and you can get a really good education. I'm assuming $\$$8k a year won't be a particular financial hardship?The only thing I would warn you about is the male-female ratio. Maybe this won't bother you much, but I believe it's about 70% male. It bothers some of my friends that attend the school. Anyways, if you want to be an engineer, that's a wonderful school to go to. I really don't think you're going to get 5x the educational experience at MIT or Princeton, even though it will cost 5x as much. Do a cost-benefits analysis, and visit them all if you possibly can. You may find that you hate one of the schools.

Another thing about colleges, though, is how you fit with your peers. It's not as simple as a cost-benefit analysis or whatever, and the prestige alone can possibly help you later in life, as sad as that sounds. In my opinion, if MIT is not crippling, that is, you won't actually go into *debt* or risk debt in the future, then you should go there, especially with financial aid and such.

I am also in a similar situation.Financial problems have  blocked my way to college.
I have been admitted to UT Austin(economics) but I don't have enough money
to attend the college. :( Can anyone help me?(although it is alittle late).

[hide]I have searched  a lot for scholarships,aids at many sites(fastweb,etc) but I haven't found anything.At first me and my family thought that we will be able to pay,but then
we found it hard to pay such a great money.
[/hide]

[quote]And I disagree, because I don't think it's fair to put your parents in any type of financial distress for your[/quote]
FWIW,  speaking as a parent, can assure you that not all the parents think their kids education is  a  financial burden.  For many parents, what  their son/daughter can do to  make them  really happy  is to do get in a school like MIT/Princeton... they would happily pay the tuition. :)

In any case GT is pretty nice, and if you like to go there, you won't be making a wrong choice but, financial reason alone ( or even a major criteria)  is not,  (in my opinion)  the right way to make that choice.

Please keep in mind:

If your parents can't afford it, MIT/Princeton would provide all the  financial aid you need.

(For example Harvard's tuition is $\$0$ for those whose parents make less than $\$60K$.)

 MIT has a very good reputation of providing good financial aid and in many case (in fact  [i] every[/i]  recent case,  where people shared  the  information),   when all the financial aid was figured in, was  less expensive  than "other" schools which were though to be cheaper.
Obviously, not knowing you situation,  I could easily be wrong, but if you have not gotten the financial figures form MIT/GT/Princeton etc.. you may not be comparing the right things. ..  Yes people complain about filling up FAFSA and CSS profiles etc.. but I think you must fill it out.

Anyway these are just some thoughts.

[quote="Gyan"](For example Harvard's tuition is $\$0$ for those whose parents make less than $\$60K$.)[/quote]

This is not quite true -- there is still a (relatively small) expected student contribution for families of that income.  Princeton didn't introduce a similar system, but they did replace all loans with grants, which is a pretty darn good deal.  You should definitely look at the financial aid offers from Princeton and MIT -- and call up their financial aid officers, if the offers aren't high enough, letting them know you'll be going to GTech (or Arizona) because of the cost.

JBL -   Thanks for the correction - but for practical purpose it still almost nil.  right?   -( BTW this  picked up from the  recent  news  reported in most newspapers in India  as  Summers was visiting India to " deepen engagement between India and the US" ( to open, I believe,   a Harvard research center there, .. some  chairs  in Cambridge etc..) and was trying to attract good students by mentioning that Harvard education is affordable to  all -(Many  Didn't know   Mumbai/India  has a large number of Harvard Alumni!)

[quote="Gyan"]but for practical purpose it still almost nil.  right?[/quote]

Yes -- it's supposed to be within the range that the student can reasonably earn holding a term-time job, and maybe over the summer.  I don't know the exact numbers, but it can't be too much more than a couple of thousand dollars a year.

[quote="froufrou4189"]
I've heard that which school you go to for undergrad doesn't matter as long as you make the most of it.[/quote]

This really depends on what you want to do after undergrad.  I would guess this is more likely true if you are headed to grad school afterwards (and work hard/excel in undergrad, etc), but is not as clearly true if you want to head directly from undergraduate school into a highly competitive field.  (In fact, if that field were the financial world, it's probably not true at all.)

In general, I'll note that much of your education comes from your peers, as will many of the opportunities you have later in life.  This is one of the main advantages of a place like MIT/Princeton over a Georgia Tech/Arizona.

To spider_boy: Talk to the college, and start hunting for a job.  It will be tough, but plenty of people juggle work and school.

FWIW: A news item and some perspective  that Ivy leagues     may be less expensive than many think.  To attract  the best and brightest, they tell that    you need not be very rich to go there: A story from Times of India:
[url=http://timesofindia.indiatimes.com/articleshow/1490065.cms]Enroll at Yale for free[/url][quote] Are you smart, intelligent and 18? Then stop worrying about quotas and think Ivy League — that too with full financial aid. That was Yale professor and director of its world fellows programme Daniel Esty making his pitch for India’s best and brightest.

[b]Yale[/b], which recently expanded its under-graduate financial aid programme along with [b]Harvard, Stanford and University of Pennsylvania, is now emphasising its affordability as well as accessibility[/b]. Parents of undergrads whose annual income is less than $\$$ 45,000   (or about Rs 20 lakh) [b]no longer have to pay any portion of the cost of education. Full financial aid includes tuition, room and board[/b]. What’s more, Esty says there’s no cap on the number of foreign students. Currently, Yale has 125 Indian students out of 11,000 students and would be happy to hike the number. ‘‘We want to attract the best talent and lack of funds should not come in the way of a student applying,’’ he said. Yale offers aid — full or part —[b] to 60% of undergrads and its financial aid budget for 2006-07 is approximately $\$$59 million,[/b] an 8% increase over last year. I[b]n addition, all international students will get tickets for one return trip home every year.[/b]

‘‘We noticed that many foreign students were not going home during the vacation because they could not afford the journey. The tickets will help make their Yale experience as positive as possible,’’ said George Joseph, assistant secretary of international affairs at Yale.[/quote]

Interesting that Harvard President was pitching similar theme  recently.


geometry
trigonometry
parallelogram
geometric transformation
homothety
reflection
geometry proposed

Denote the sides and angles of the triangle $\triangle ABC$ as usual: $a = BC, b = CA, c = AB$, $\alpha = \measuredangle A, \beta = \measuredangle B, \gamma = \measuredangle C$ The triangles $\triangle AB'C' \sim \triangle ABC$ are similar with similarity coefficient $\cos \alpha$, because the angle $\alpha$ is common and $AB' = AB \cos \alpha$, $AC' = AC \cos \alpha$. Likewise, the triangles pairs $\triangle A'BC' \sim \triangle ABC$, $\triangle A'B'C \sim \triangle ABC$ are similar with similarity coefficients $\cos \beta, \cos \gamma$, respectively. Let $R, R_A, R_B, R_C$ be the circumradii of the triangles $\triangle ABC, \triangle AB'C', \triangle A'BC', \triangle A'B'C$, respectively:

$R = \frac{R_A}{\cos \alpha} =\frac{R_B}{\cos \beta} =\frac{R_C}{\cos \gamma}$

The distances from the vertices $B', A'$ to the orthocenters $A'', B''$ of the triangles $\triangle AB'C', \triangle A'BC'$ are equal:

$B'A'' = R_A \cos{\measuredangle AB'C'} = R \cos \alpha \cos \beta$

$A'B'' = R_B \cos{\measuredangle BA'C'} = R \cos \beta \cos \alpha$

The lines $B'A'' \parallel A'B''$ are parallel, because they are both perpendicular to the line $AC' \equiv AB \equiv C'B$. Thus the quadrilateral $A'B'A''B''$ is a parallelogram and the corresponding sides $A'B', A''B''$ of the orthic triangle $\triangle A'B'C'$ and the orthocenter triangle $\triangle A''B''C''$ are (oppositely) parallel and equal: $A'B' \parallel B''A''$ and $A'B' = A''B''$. In exactly the same way (by a cyclic exchange of the vertices $A, B, C$, altitude feet $A', B', C'$ and orthocenters $A'', B'', C''$), it can be shown that the remaining pairs of corresponding sides $B'C', B''C''$ and $C'A', C''A''$ of the triangles $\triangle A'B'C', \triangle A''B''C''$ are also (oppositely) parallel and equal: $B'C' \parallel C''B'', C'A' \parallel A''C''$ and $B'C' = B''C'', C'A' = C''A''$. Thus these 2 triangles are homothetic with homothety coefficient -1.

[quote="grobber"]Let $A'B'C'$ be the orthic triangle of $ABC$, and let $A'',B'',C''$ be the orthocenters of $AB'C',A'BC',A'B'C$ respectively. Show that $A'B'C',A''B''C''$ are homothetic.[/quote]

This problem is actually very easy:

The altitudes AA', BB' and CC' of triangle ABC intersect at the orthocenter H of this triangle. Then, $HB^{\prime}\perp CA$, what rewrites as $HB^{\prime}\perp AB^{\prime}$. On the other hand, since the point A" is the orthocenter of the triangle AB'C', we have $C^{\prime}A^{\prime\prime}\perp AB^{\prime}$. Thus, HB' || C'A". Similarly, HC' || B'A". Thus, the quadrilateral B'HC'A" is a parallelogram. Consequently, we have the vectorial identity $\overrightarrow{C^{\prime}A^{\prime\prime}}=\overrightarrow{HB^{\prime}}$. Similarly, $\overrightarrow{A^{\prime}C^{\prime\prime}}=\overrightarrow{HB^{\prime}}$. Thus, it follows that $\overrightarrow{C^{\prime}A^{\prime\prime}}=\overrightarrow{A^{\prime}C^{\prime\prime}}$; hence, the quadrilateral C'A'C"A" is a parallelogram. Since the diagonals of a parallelogram bisect each other, it follows that the segments C'C" and A'A" bisect each other. In other words, the midpoint of the segment C'C" is also the midpoint of the segment A'A". Similarly, the midpoint of the segment C'C" is also the midpoint of the segment B'B". Hence, the segments A'A", B'B" and C'C" have the same midpoint. Then, the points A", B", C" must be the reflections of the points A', B', C' in this midpoint. Hence, the triangles A'B'C' and A"B"C" are symmetric to each other with respect to this midpoint. Of course, this yields the homothety of these triangles.

By the way, one can easily show that the common midpoint of the segments A'A", B'B" and C'C" is the center of the Taylor circle of triangle ABC; if the triangle ABC is acute, then this midpoint is also the Spieker point of the orthic triangle A'B'C'.

  Darij


$ A\equal{}\sqrt{\frac{b\plus{}c}{1\plus{}a}}\plus{}\sqrt{\frac{a\plus{}c}{1\plus{}b}}\plus{}\sqrt{\frac{a\plus{}b}{1\plus{}c}}$
since $ \sqrt{x}$ is concave : by jensen inequaliy ;
$ A\leq3\sqrt{\frac{1}{3}(\frac{b\plus{}c}{1\plus{}a}\plus{}\frac{a\plus{}c}{1\plus{}b}\plus{}\frac{a\plus{}b}{1\plus{}c}})$

so :    $ A\leq3\sqrt{\frac{2(a\plus{}b\plus{}c)}{3(a\plus{}b\plus{}c)}}\equal{}\sqrt{6}$

 :)

[quote="mehdi cherif"]$ A \equal{} \sqrt {\frac {b \plus{} c}{1 \plus{} a}} \plus{} \sqrt {\frac {a \plus{} c}{1 \plus{} b}} \plus{} \sqrt {\frac {a \plus{} b}{1 \plus{} c}}$
since $ \sqrt {x}$ is concave : by jensen inequaliy ;
$ A\leq3\sqrt {\frac {1}{3}(\frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {a \plus{} c}{1 \plus{} b} \plus{} \frac {a \plus{} b}{1 \plus{} c}})$

so :    $ A\leq3\sqrt {\frac {2(a \plus{} b \plus{} c)}{3(a \plus{} b \plus{} c)}} \equal{} \sqrt {6}$

 :)[/quote]
You have mistake, mehdi cherif. Try $ a\equal{}b\equal{}\frac{1}{2},c\equal{}0$. :)

PS: This is a problem on crux magazine (I don't remember exactly the year. :()

[quote="manlio"][quote="can_hang2007"][quote="nkht-tk14"][u] [i]Problem[/i][/u]

                       [i] Let $ a , b , c$ be nonnegative reals numbers such that

                                       $ a \plus{} b \plus{} c \equal{} 1$

                 Find the greatest value  of this expression

    $ A \equal{} \sqrt {\frac {1 \minus{} a}{ 1 \plus{} a}} \plus{} \sqrt {\frac {1 \minus{} b}{ 1 \plus{} b}} \plus{} \sqrt {\frac {1 \minus{} c}{ 1 \plus{} c}}$

  Scorpius119 , Argady  . Let  perform :lol:

  Love An forever

  Huỳnh Võ Phương An[/i][/quote]
:)[/quote]

Sorry, I cannot open this file.  :(  It looks BLACK.

Thank you very much.[/quote]
I still open it as normal. Maybe your adode reader program have some mistakes?


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If 11.63 and -2.41 are the roots for x of

x^2 + kx + w =0,

find the value of k-w.
Express your answer as a decimal rounded to 4 significant digits.

[hide]
answer: 18.81
[/hide]

How to solve this problem.. Please help me!

[hide]Okay, if we have a quadratic with roots a and b, then

$(x-a)(x-b)=x^{2}-(a+b)x+ab$

If that doesn't make sense, just multiply it out.

We know that a and b are 11.63 and -2.41, so we can solve for k and w.

$(x-11.63)(x-(-2.41))=(x-11.63)(x+2.41)=x^{2}-(11.63-2.41)+11.63(-2.41)$

So, $k=-(11.63-2.41)=-9.22$ and $w=11.62(-2.41)=-28.0283$

Now, we subtract.

$-9.22-(-28.0283)=18.8083$  Just round to $18.81$, which is four significant figures.  There you go![/hide]


1)Let be $ n\in \mathbb{N}\ , \ x\in \mathbb{R}$ . Solve the next equation :

$ \{nx\} \minus{} \{(n \plus{} 1)x\} \equal{} x$

2)Solve in the set of real numbers this equation :

$ \left\lfloor x^2 \minus{} x \plus{} 1\right\rfloor \equal{} x$

3)Solve in the set of natural numbers this equation :

$ \lfloor \sqrt {1}\rfloor \plus{} \lfloor \sqrt {2}\rfloor \plus{} \lfloor \sqrt {3}\rfloor \plus{} ... \plus{} \lfloor \sqrt {n^2 \plus{} 2n}\rfloor \equal{} 3723$

1. Sorry if this sounds stupid, but what do those {} brackets indicate? >.>

2. Obviously x must be an integer. The intuitive answer is x = 1, but I cannot say for certain if there are others. (I'm 99% sure that's the only one)

3. There should only be a single solution for n to that... I'm too lazy to solve it now, so I'll rewrite it in summation notation...

$ \sum^{n^2 \plus{} 2n}_{j \equal{} 1}\lfloor\sqrt {j}\rfloor$

Set j = k^2.

Clearly the value of $ \lfloor\sqrt {j}\rfloor$ must equal $ \lfloor k\rfloor$.

EDIT>
Okay, basically:

$ 2^2 \minus{} 1^2$ 1's in the sum;
$ 3^2 \minus{} 2^2$ 2's in the sum;
$ 4^2 \minus{} 3^2$ 3's in the sum;
...

Knowing that we get $ 323 \equal{} n^2 \plus{} 2n$. Factoring gives you $ 0 \equal{} (n \plus{} 19)(n \minus{} 17)$, so n = 17?

[quote="yanglunj"]1. Sorry if this sounds stupid, but what do those {} brackets indicate? >.>[\quote]

For a real $ x$, $ x\equal{}[x]\plus{}{x}$, where $ [x]$ is an integer and $ 0 \le {x}<1$

[b]1)[/b] Case 1:$ \{nx\} \ge \{(n\plus{}1)x\}$, then $ \{\minus{}x\}\equal{}x \Leftrightarrow \{\minus{}x\}\equal{}\{x\} \Leftrightarrow \{2x\}\equal{}0$, so $ \{x\}\equal{}0$ or $ \frac{1}{2}$, $ x\equal{}0$ or $ \frac{1}{2}$

Case 2: $ \{nx\} < \{(n\plus{}1)x\}$, then $ \{\minus{}x\}\minus{}1\equal{}x$, similarly we have $ \{2x\}\equal{}0 \Leftrightarrow x\equal{}\minus{}\frac{1}{2}$

So $ x\equal{}0$ is solution for all $ n$, $ x\equal{}\frac{1}{2}$ is the solution for all odd $ n$, $ x\equal{}\minus{}\frac{1}{2}$ is the solution for all even $ n$.

[b]2)[/b] $ x \le x^2\minus{}x\plus{}1<x\plus{}1$
Left inequality is $ (x\minus{}1)^2 \ge 0$, which is true for all $ x$
Right inequality is $ x(x\minus{}2)<0$, so $ 0<x<2$ is the solution.

[b]3)[/b] $ 1 \times 3\plus{} 2 \times 5\plus{} ...\plus{} n \times (2n\plus{}1)\equal{}3723$
$ 2(1^2\plus{}2^2\plus{}...\plus{}n^2)\plus{}(1\plus{}2\plus{}...\plus{}n)\equal{}3723$
$ \frac{n(n\plus{}1)(2n\plus{}1)}{3}\plus{}\frac{n(n\plus{}1)}{2}\equal{}3723 ......$

An estimation: $ \frac{2n^3}{3}<3723$, so $ n \le 17$. After checking , $ n\equal{}17$ is the answer.

[quote="yanglunj"][quote="alex2008"]2)Solve in the set of real numbers this equation :

$ \left\lfloor x^2 \minus{} x \plus{} 1\right\rfloor \equal{} x$[/quote]

Wouldn't x be limited to just an integer since the LHS is an integer?[/quote]

You are correct. $ x$ should be $ 1$.


geometry
3D geometry
geometric transformation
rotation
superior algebra
superior algebra solved

Keeping it very simple and very concrete ...

Let $G$ be the group of permitted motions of a $3\times 3\times 3$ Rubik cube. (Mod out by the rigid rotations of the whole cube.)

Let $F$ have the action of rotating the front face (facing you) of the cube by 90 degrees clockwise. Let $R$ have the action of rotating the right hand face of the cube by 90 degrees clockwise.

Clearly, $o(F)=o(R)=4.$

1. What is $o(FR)$ ?

2. What is $o(FRF^{-1}R^{-1})$ ?

My cube has the colors Blue, Red, White, Orange, Yellow and Green, in the arrangement shown in the cube net below (White being the front and Orange the right hand side):

[code]
    BBB
    BBB
    BBB

RRR WWW OOO YYY
RRR WWW OOO YYY
RRR WWW OOO YYY

    GGG
    GGG
    GGG
[/code]

The action $(FR)^{7}$ is the smallest power of $FR$ that leaves the edges fixed (and unflipped) and only permutes and rotates the vertices, so the order of $FR$ is 7 times the order of $(FR)^{7}$.

I name the vertices in the following (probably nonstandard) way: First the color of the front or back face of the vertice cube, and then the other two colors counter-clockwise. So the vertices that are moved by F and R are these:
[code]
WBR WOB YBO
WRG WGO YOG
[/code]
I number the vertices and their rotation:
1(id) 2(id) 3(id)
4(id) 5(id) 6(id)

The action $(FR)^{7}$ yields this arrangement of the cube:
[code]
    BBG
    BBB
    WBW

RRG OWO BOW RYY
RRR WWW OOO YYY
RRG OWO BOW RYY

    YGY
    GGG
    GGB
[/code]

The vertices now have this form:
[code]
OWG OBW RGW
OGY OYB RWB
[/code]

Comparing with the original state gives this permutation of the vertices and  (in cycle notation) the permutation of each vertices faces.

5(123) 2(132) 4(132)
6(132) 3(123) 1(123)

We now work with the permutations of 6 objects, each of which has 3 rotational states (3-cycles of their face colors), and want to find the order of the permutation given above.

The permutation $\tau=(123456 \mapsto 524631)$ is the 5-cycle (15346), so the action $(FR)^{7 \cdot 5}$ fixes the vertice's position and only rotates them. Since vertice 2 is always rotated 1/3 into the same direction, the action $(FR)^{35}$ does not fix the orientation of that vertice, so the order of $FR$ is a multiple of 105 (maybe 105 itself).

I have now calculated that $(FR)^{35}$ rotates all vertices 1/3 into the same direction, so $(FR)^{105}= id$.

LiMa

Thank you, LiMa. Yes, the product has order 105. Now, what about the commutator?


Yep here are the solutions. You can discuss the problems here too if you feel like it.

Haha, what's interesting about number 5 is that when I did it, I read it as "none can be increased or decreased by more than one place..."

Hmmm... that's an interesting problem too. I think the answer to that is like $F_{10}$ or something.

SO that's the solution to 12, huh?? Pretty tricky :( Oh well.

ARRGH I'M STILL PISSED THAT I MISSED #12...

IT IS A VERY TRICKY PROBLEM

$20\cdot 29$ is NOT equal to $870$... $20\cdot 29$ is NOT equal to $870$... :wallbash:

Made a dumb mistake on #8, as well. I knew it gave 30-60-90 triangles, but for some reason I think I didn't find the sides of the smaller equilateral triangle properly. :huh:

Also, you thought that I answered #11, but I answered #13. According to your answers, I'm not correct anyway, but I wanted to check something. Why can't it be solved in the following way:

$(x+y)(y+z)(x+z)=(y^2+xy+xz+yz)(x+z)=(y^2+96)(x+z)\Rightarrow$
$xy^2+y^2z+96(x+z)=y(xy+yz)+96(x+z)=y(96-xz)+96(x+z)\Rightarrow$
$96(x+y+z)-xyz=1680$

And then solve for that? That gives $x+y+z=2+8+9=19$, $xyz=2(8)(9)=144$, $96(17)-144=1680$, so we have $100(2)+10(8)+9=289$.

huh why to construct this point in 15 I do not see it is very nice equationally but what is motivation

I dunno if you feel like being all right triangley maybe.

[quote="JesusFreak197"][quote="probability1.01"]Errr $96 \cdot 17 = 1824$?[/quote]

Err, typo. $2+8+9=19$, and $96(19)=1824$.[/quote]

Err.. well I didn't bother to check but I think you must have lost a condition somewhere--$2 \cdot 8 + 2 \cdot 9 + 8 \cdot 9 \equiv 2 \not\equiv 0 \equiv 96 \mod 8$.

That method does not work because it is only necessary and not sufficient. And there is no easy way (at least yet) to narrow down the possible solutions. Therefore, you would have to do a lot of guessing and checking and hope that eventually something will turn up. And you still don't know if it's the minimum or not.

Yeah, I know, that's the problem. What's wrong with the solution, though?

Misread #12 - make that 3 trivial mistakes for me. I counted how many ROLLS Al made, not the number of turns, and I was off by exactly a factor of 6.

Half of the problems I missed were due to stupid mistakes: Forgetting that 3|87 and 60, using x+y, y+z, and z+x instead of x, y, and z in the final step, and misreading what a turn was in the problem statement. I missed another because I didn't read the question right and solved a much simpler problem...that's why on the real AIME, I bring sugar and caffeine. It really helps - I know from experience. Doing an AIME in the morning versus after dark also helps, but that might just be me.

Interesting, I got #5 very quickly:Nine gamblers are ranked in accordance with their wealth. When they are ranked again after a few rounds of poker, none of their rankings increased by more than one place. How many different possible rankings are there this time?
Here's how I thought about it: It's the same as no player decreases a slot. And in Mathcounts coundown round, you can't decrease more than a slot!! So there would be 8 matches, so the answer is $ 2^8$. I know it's the same as Alex's solution, but I think it's cool that I used the Matchounts competition format to help me with a problem.


Prove that the eigenvalues of A* are the complex conjugates of the eigenvalues of A.

The conjugate conjugates both eigenvectors and eigenvalues. The transpose scrambles eigenvectors but leaves the same eigenvalues- the rank of $ \lambda I\minus{}A$ and its transpose are equal.


I'm a rising 7th grader looking for the answer to the following neurology question:

To start my post, please take the examples of acetylcholine.

The neurotransmitter,aceytlcholine, is used in the PNS(peripheral nervous system) and CNS (central nervous system); for the purpose of activating muscles and is responsible for certain excitatory reactions.

Now just as acetylcholine, serves a function along those lines, is [i]there a neurotransmitter that is secreted during memory-oriented tasks.[/i]

My first guess is that aforesaid neurotransmitter, would be secreted by the hippocampus or possibly hypothalamus, but I was wondering if there was an exact name.

Thanks for the help!

-arb95

Could the neurotransmitter you're looking for be glutamic acid (glutamate)? It's the main excitatory neurotransmitter in the brain and plays a role in memory formation and synaptic plasticity.

It basically works in memory by stimulating AMPA receptors and NMDA receptors. AMPA receptors contribute to learning and memory, while NMDA receptors can promote long-term beneficial changes in the brain.

I haven't been able to find a specific place in the brain where glutamate is synthesized. Astrocytes--a type of glial cell--located in the brain and spinal cord transport and release it, but I'm not sure to where or if they are the only cells that do this.

Anyway, I hope this helps a little. :)

I did some research to my question and unearthed some more questions as is science:

PART I
I searched neurotransmitters secreted by the hippocampus, to look deeper into the subject.

It seemed as if norephinephrine came up in many scientific studies as being secreted from the hippocampus.

[b]Due to the fact that the hippocampus has a major role in memory, would you concur that norephinephrine has a similar role?[/b]

PART II
I also found a neurotransmitter, of the name dopamine, with a function of memory, learning, and many others (sleep, mood, attention, reward, motivation, motor functions, etc.)

Now, if I was correct with the role of norepinephrine than let us analyze the atomic structures:
[img]http://www.biochemistry.bham.ac.uk/osmart/msctox/dopamine.jpg[/img][img]http://images.cambridgesoft.com/chemstore/CBN/norepinephrine_IUPAC.jpg[/img]

The atomic structure and function are eerily similar. They both have the Benzene ring of carbons, and possess the HO, OH, and NH(sub 2) molecules.

[b]Now I lead to the question, is function deeply rooted in atomic structure?[/b]

-arb95

[quote][b]Due to the fact that the hippocampus has a major role in memory, would you concur that norephinephrine has a similar role?[/b][/quote]

Actually, norepinephrine is a stress hormone that acts in the brain and body during a "fight or flight" reaction. Also, norepinephrine is mainly created in and released from the locus ceruleus ("blue spot") in the brain stem. That area's synthesis of norepinephrine is what gives it its blue color. The locus ceruleus does connect to the hippocampus, but that connection is part of the brain's reward system.

[quote][b]Now I lead to the question, is function deeply rooted in atomic structure?[/b][/quote]

Yes. Atomic structure is everything when it comes to neurotransmitter function. That's why drugs such as morphine work. Morphine's structure--especially the part of the molecule that bonds to a neuron--is very similar to the brain's own endorphins', so it works in a similar way to depress pain receptors and other parts of the nervous system.

The reason that dopamine and norepinephrine look so similar is that they are both catecholamines, hormones/neurotransmitters released as part of stress or reward. Another catecholamine you may be familiar with is epinephrine or adrenaline. Again, similar shapes lead to similar functions.

Excellent questions! :D

[b]How do neurologists discover these neurotransmitters?[/b] I can only imagine the gruelling process it must undertake.

Thanks in advance.

-arb95

Postscript: I'm quite interested in this subject. Are there any books, courses, or resources, anyone would reccommend?

[quote]How do neurologists discover these neurotransmitters? I can only imagine the gruelling process it must undertake.[/quote]

I'm really not sure. I think to find neurotransmitters scientists might look at the brains of other animals, such as rats, to see what pathways in the brain are activated during certain activities. They might also cut these brains into pieces so they can look at the structures that were activated and find concentrations of the different chemicals used. That's just a guess on my part, though.

As far as books and resources go, there is a very good neuroscience book available online from the British Neuroscience Association. It's actually used in the United States National Brain Bee.

[url=http://www.bna.org.uk/publications/brain_sci.html]Neuroscience: Science of the Brain[/url]

 :)