ratio
geometry
geometric transformation
rotation
power of a point
radical axis
geometry proposed

Given the triangle $ABC$ we consider the points $X,Y,Z$ such that the triangles $ABZ,BCX,CAZ$ are equilateral, and they don't have intersection with $ABC$. Let $B'$ be the midpoint of $BC$, $N'$ the midpoint of $CY$, and $M,N$ the midpoints of $AZ,CX$, respectively. Prove that $B'N' \bot MN$.

Cute one :).

Consider the transformation we get by composing the spiral similarity cetnered at $B$ of angle $\frac\pi 6$ and ratio $\frac{\sqrt 3}2$ with the rotation of angle $\frac \pi 3$ centered at $C$ (the rotation comes first, then the spiral similarity). This transformation is again a spiral similarity of angle $\frac \pi 2$. The image of $Y$ is $M$, and the image of $B$ is $N$, which means that the image of the line $BY$ is $MN$, which shows that $BY\perp MN$. Now, since $BY\|B'N'$, we get the conclusion.

P.S. the above even gives the ratio of the lengths of the two segments.

wow! your solutions is Great! Another nice idea is to use the fact that $MN$ is parallel to the line between the centers of $ABZ,BCX$. Hence by radical axis, and since the Fermat point of $ABC$ belongs to $BY$ and to the intersection of the circles $ABZ,BCX$, we get $MN\bot BY$ and the conclusion follows.

jojo, excelent, es una muy interesante solución la de grobber.

Una mínima generalización del problema es:

Given the triangle ABC we consider the points A', B', C' such that the triangles BCA', CAB', ABC' are equilateral an externals to ABC. Let Ca, Ba, Ab, Cb, Bc y Ac be the midpoints of BA', A'C, CB', B'A, AC' y C'B. The intersection of BaBc and CaCb is X, the intersection of AbAc and CaCb is Y, and the intersection of  BaBc and AbAc is Z. Prove that XYZ is homotetic with the triangle NaNbNc (centers of BCA', CAB', ABC'), with center F and reason 1/2. (perdón por mi inglés, is terrible).

pretty easy I think, the rotation (CCA  :pi:/3 2) of BY from C is AX, and the rotation (CA :pi:/6 2/ :sqrt: 3) of MN from B is AX, too. So we get it.

3500th post  :thumbup:  :coolspeak:  :clap2:

Complex number bash ftw.

[hide="Solution"]
[hide="Hints from AoPS"]
127)  Let $A$ be the origin.  What must be true about $(m-n)/(n'-b')$ if and only if $\overline{MN} \perp \overline{B'N'}$?
85)  If you're stuck with a ratio that you'd like to show is inamginary, here's a hint.  If $\frac{z_1}{z_2}=\frac{z_3}{z_4}=k$, then what does $\frac{z_1b+z_3c}{z_2b+z_4c}$ equal?  [/hide]
Let $A$ be the origin

We desire:  $\frac{m-n}{b'-n'}$ being imaginary.

We know:  $b-a=\zeta(z-a)\implies b=\zeta*z$

Also, $x-c=\zeta(b-c)$ and $y=\zeta(c)$.

Now we have a HIGHLY computationally intense problem so let's try it.

$m=\frac{z+a}{2}=\frac{\frac{b}{\zeta}+a}{2}$.  Next, $n=\frac{x+c}{2}=\frac{2c+\zeta(b-c)}{2}$.  Also, $b'=\frac{b+c}{2}$ and $n'=\frac{c+y}{2}=\frac{c+\zeta(c)}{2}$.

$\frac{m-n}{b'-n'}=\frac{\frac{b}{\zeta}-2c-\zeta(b-c)}{b-\zeta*c}$ which we need to be imaginary.

This is the same thing as $\frac{\frac{b}{\zeta}-2c-\zeta(b)+\zeta*c}{b-\zeta*c}$

This is also the same as $\frac{b-\zeta^2(b)-2\zeta*c+\zeta^2*c}{b\zeta-\zeta^2*c}$

Therefore, we get $\frac{b(1-\zeta^2)+c(\zeta^2-2\zeta)}{b\zeta+c(-\zeta^2)}$

We now prove $\frac{1-\zeta^2}{\zeta}=\frac{\zeta^2-2\zeta}{-\zeta^2}$.

This is crucial in the next part of the problem.

$\frac{1-\zeta^2}{\zeta}=\frac{\zeta-2}{-\zeta}$ or $1-\zeta^2=-\zeta+2$  Therefore, we get $\zeta^2-\zeta+1=0$ which is true by the definition of $\zeta$.

Yay, lemma proven.

Find $\frac{bz_1+cz_3}{bz_2+cz_4}$ where $\frac{z_1}{z_2}=\frac{z_3}{z_4}$.

We note that letting this value be $k$, we get $z_1=kz_2$.

Therefore, we have $\frac{bkz_2+ckz_4}{bz_2+cz_4}=k$.

Therefore, we must now prove that $\frac{1-\zeta^2}{\zeta}$ is imaginary.

Since $z$ is a primitive 6th root of unity, we get $z=e^{\frac{\pi}{3}i}$ or $z=e^{\frac{5\pi}{3}i}$.

Therefore, $z=\frac{\sqrt3}{2}+\frac{i}{2}$ or $z=\frac{1}{2}-\frac{i\sqrt3}{2}$.

In the first case, we get $z^2=e^{\frac{2\pi}{3}i}$ or therefore it is the same as $\frac{-1}{2}+\frac{\sqrt3}{2}i$

We then get $\frac{\frac32-\frac{\sqrt3}{2}i}{\frac{\sqrt3}{2}+\frac{i}{2}}$
$= \frac{3-\sqrt3*i}{\sqrt3+i}=\frac{(3-\sqrt3*i)(\sqrt3+i)}{4}=\frac{3-4\sqrt{3}i-3}{4}=-\sqrt{3}i$.

In the second case, we get $z^2=e^{\frac{4\pi}{3}i}=-\frac{1}{2}-\frac{\sqrt3}{2}i$.
Also, we get $\frac{1-z^2}{z}=\frac{3+\sqrt3i}{1-\sqrt3i}=\frac{(3+\sqrt3i)(1+\sqrt3i)}{4}=\frac{4\sqrt3i}{4}=\sqrt3i$.

We are finally done.  $\Box$.  [/hide]

We denote the affix of each uppercase letter by a lowercase letter.Then by rotation formula we have,

$x=c+(b-c)\epsilon$
$y=a+(c-a)\epsilon$
$z=b+(a-b)\epsilon$

Thus $n=c+\frac{(b-c)\epsilon}{2}$
         $n'=\frac{(c+a)+(c-a)\epsilon}{2}$
         $m=\frac{(a+b)+(a-b)\epsilon}{2}$
         $b'=\frac{b+c}{2}$
It is just calculation to see that $\frac{m-n}{b'-n'}=i\mathbb{R}$ for some nonzero real $\mathbb{R}$.

[u][b]Note[/b][/u]:Here $\epsilon$ represents $cos60^{\circ}+isin60^{\circ}$.

Bye.

Maths is the doctor of science...

Sayantan..

Dear Mathlinkers,
for a more general proof, you can see
http://perso.orange.fr/jl.ayme  vol. 16   p. 43-44

Sincerely
Jean-Louis


Find all prime divisors $ p$(p<1500) of $ 2^{37}\minus{}1$
I tried to find  greater prime divisors,but failed

You can chek only $ p\equal{}1\mod 37$ and $ p\equal{}\pm 1\mod 8$.
It mean $ p\equal{}1\mod 296$ or $ p\equal{}223\mod 296$.

hello, i have $ 2^{37}\minus{}1\equal{}223\cdot616318177$.
Sonnhard.


This must have been posted before.

[hide="Hint"]Show that the product of four consecutive integers is one less than a perfect square.[/hide]

[hide]$ x(x\plus{}1)(x\plus{}2)(x\plus{}3) \plus{} 1 \equal{} x^4 \plus{} 6x^3 \plus{} 11x^2 \plus{} 6x \plus{} 1 \equal{} (x^2 \plus{} 3x \plus{} 1)^2$[/hide]

We use the fact that if $ m$ is a perfect square then $ m\minus{}1$ can not be a perfect square.  :)

[quote="B_Gasperov"]We use the fact that if $ m$ is a perfect square then $ m \minus{} 1$ can not be a perfect square.  :)[/quote]

What about $ m \equal{} 1$?

The actual property you're using is that between two consecutive perfect squares there isn't another perfect square.  This is true of any monotonic sequence of integers and can be used to solve several other related problems.

This is part a) of a problem from the Spanish olympiad. It can be a perfect square only if one of the four terms equals zero.

now part b):

Prove that the product of 4 consecutive positive integers is never a perfect cube.

[quote="Pul de Algodoncito"]Prove that the product of 4 consecutive positive integers is never a perfect cube.[/quote]

Let $ n(n\plus{}3)\equal{}2k$. Then $ (n\plus{}1)(n\plus{}2)\equal{}2k\plus{}2$. Thus \[ P\equal{}n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}4k(k\plus{}1)\]
Let $ P\equal{}(2t)^3$. Then $ 4k(k\plus{}1)\equal{}8t^3\implies k(k\plus{}1)\equal{}2t^3$. So the only possible solution to this equation is $ k\equal{}1,t\equal{}1$, as $ (k,k\plus{}1)\equal{}1$. But $ k\equal{}1$ implies $ n(n\plus{}3)\equal{}2$, which is impossible. Hence the conclusion.

Let the 4 consecutive positive integers be x,x+1,x+2 and x+3
Their product = {x*(x+3})*{(x+1)*(x+2)} = (x^2+3x+1)^2 -1.
So, the product of them can't be a perfect square.


trigonometry
geometry
area of a triangle
Heron\u0027s formula

I can't do this problem...  :(

Two circles have radii a,b. The distance between their centers is d. They intersect at P and Q. The distance from P to Q is?

Express your answering using a,b,and/or d.

this seems an easy problem. let O1, O2 be the centers of the circles.  draw the O1a, where it's tangent to the second circle (that would be point P), and draw O2b were it's tangent to the first one(point P). then draw the line, d. it is obvious that the angle O1PO2 is 90 :^o: and all you need to do now is to figure out the altitue of the line PH, that is orthagonal to D, and then multiply it by two. and that would be the length of PQ.

[hide]Oops messed up the first time.
$2a \sin \left( \cos^{-1} \left( \frac {b^2 - a^2 - d^2}{-2ad} \right ) \right)$[/hide]

Here were the choices.. Since no answer doesn't seem to give me "variable" answer..

A) $\frac{[d^2 - (a-b)^2]^{\frac{1}{2}} [(a+b)^2 - d^2]^{\frac{1}{2}}}{d}$

B) $(a^2 + b^2 - d^2)^{\frac{1}{2}}$

C) $[(a-b)(a+b)]^{\frac{1}{2}}$

D) $\frac{a+b+d}{3}$

E) None of the above

[hide]Answer: [b]A [/b]

 Using Heron's formula, $\triangle PO_1O_2 =\frac{1}{2}dh=\frac{1}{2}\sqrt{\frac{a+b+d}{2}\cdot \frac{a+b-d}{2}\cdot \frac{b+d-a}{2}\cdot \frac{d+a-b}{2}}$[/hide]

think of the triangle with vertices of the centers and $P$.  The sides are $a,b,d$.  The distance from $P$ to the base is $\displaystyle \frac{1}{2}$ of $PQ$.
To find the height, use Heron's formula.
$\displaystyle \frac{1}{2}dh=\sqrt{\frac{a+b+d}{2}\cdot \frac{a+b-d}{2}\cdot \frac{b+d-a}{2}\cdot \frac{d+a-b}{2}}$

$h={2}\displaystyle \frac{\sqrt{\frac{a+b+d}{2}\cdot \frac{a+b-d}{2}\cdot \frac{b+d-a}{2}\cdot \frac{d+a-b}{2}}}{d}$

$PQ={4}\displaystyle \frac{\sqrt{\frac{a+b+d}{2}\cdot \frac{a+b-d}{2}\cdot \frac{b+d-a}{2}\cdot \frac{d+a-b}{2}}}{d}$

$PQ=\frac{\sqrt{d^2 - (a-b)^2} \sqrt{(a+b)^2 - d^2}}{d}$

Answer is $\text {A}$


[quote="#H34N1"]This is the second time I'm posting this, but its back again!

Every time I post in the MC forum, I get this:

[code]
SQL requests not achieved

DEBUG MODE

SQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 19209) GROUP BY ug.user_id' at line 7

SQL Request: SELECT MAX(a.auth_value) as allowed, ug.user_id FROM phpbb_auths a, phpbb_user_group ug WHERE a.obj_type = 'f' AND a.obj_id = 132 AND a.auth_name = 'auth_watch_forum' AND a.group_id = ug.group_id AND ug.user_id IN (, 19209) GROUP BY ug.user_id
Line : 1189
File : functions_post.php
[/code]

Can someone please see what's wrong?[/quote]
The post still gets made, right?

[quote="#H34N1"]Yes, but the messages is quite annoying.[/quote]

yes that has happened to me many times and i recall someone else posting this bug somewhere...

That was me :roll:

I couldn't find it so I posted it again.

Bug reproduced, and finally fixed, thanks to Darij :)


Find all triples of positive integers $ (p,q,n)$, with $ p$ and $ q$ primes, satisfying:

$ p(p\plus{}3)\plus{}q(q\plus{}3)\equal{}n(n\plus{}3)$.

Here is my solution.
By $ mod \ 3$, we have $ p^2 \plus{} q^2 \equiv \ n^2$. Hence, $ p$ or $ q$ is equal to $ 3$, otherwise $ n^2 \equiv \ 2$. Now, w.l.o.g. suppose that $ p \equal{} 3$, so the equation becomes $ 18 \plus{} q(q \plus{} 3) \equal{} n(n \plus{} 3)$. Solving like a 2-nd degree equation in $ n$, we have the discriminant equals to $ \Delta \equal{} \sqrt {4q(q \plus{} 3) \plus{} 81}$. Since $ n$ is a integer and $ \Delta$ is odd, $ 4q(q \plus{} 3) \plus{} 81 \equal{} (2k \plus{} 1)^2$, where $ k$ is a nonnegative integer. Hence $ q(q \plus{} 3) \equal{} (k \minus{} 4)(k \plus{} 5)$.
Now suppose that $ k \equal{} qt \minus{} 5$, where $ t$ is a nonnegative integer. So $ q \plus{} 3 \equal{} (qt \minus{} 9)t$. So $ q \plus{} 3\geq qt \minus{} 9$, so $ t\leq \frac{q\plus{}12}{q}$, and it's easy to see that $ t\leq 7$. By brute force, we can observe that $ t \equal{} 5$ and $ p \equal{} 2$. Substituting in the equation, the only positive integer $ n$ is $ 4$.
Analogously, the case $ k \equal{} qt \plus{} 4$ don't generate solutions.
The unique triple $ (p,q,n)$ is $ (2,3,4)$.

The above solution is incorrect, as $3(3 + 3) + 7(7 + 3) = 88 = 8(8 + 3)$. Here's my attempt.

Considering the expression modulo 3, we note that the only quadratic residues modulo 3 are 0 and 1, thus
$$
p(p + 3) + q(q + 3) = n(n + 3) \iff p^2 + q^2 \equiv n^2 \pmod{3},
$$
will have solutions only if $p^2 + q^2 \equiv 0$ or $1 \pmod{3}$.

If $p^2 + q^2 \equiv 0$, then $p = q = 3$, but then there is no solutions for $n$.

This implies that one of $p$ or $q$ must be 3. WLOG let $p = 3$, then
$$
18 + q(q + 3) = n(n + 3) \implies 18 = (n-q)(n + q + 3).
$$
Checking factor pairs of 18, we find that the only solutions are $(p, q, n) = (3, 2,4)$ and $(3, 7, 8)$,
so all of the solutions to the equation are
$$
(p, q, n) \in \{(2, 3, 4), \; (3, 2, 4),\; (3, 7, 8), \; (7, 3, 8)\}.
$$

As people above me have proven, one of the primes p or q equals 3. WLOG q=3
Then we expand and get 18+q²-n²=3(n-q) using diffrence of squares and changing sides we get 18=(n-q)(3+q+n) so we have n>q and because q≥2 we have 3+q+n≥8 and 3+q+n divides 18 so either 3+q+n=9 and n-q=2 or 3+q+n=18 and n-q=1


I had a simple conceptual doubt once and forgot all about that. Now that I recall it could someone help me out...

There is a block placed on a rough horizontal surface such that it just touches a wall perpendicular to the floor. Now if a slight increasing external force is applied pushing it against the wall then what are the normal forces (from the wall) and the frictional forces.

I mean suppose the limiting friction is 10N. And a force of 5N is applied then is the entire 5N opposed by friction or the normal force of the wall. If not then how does the 5N divide among the two opposing forces ?

I am sorry if my doubt is really stupid :-) .But please help..

The 5N countering the force pushing against the wall is the normal force, but I don't see where you're getting the frictional force from.

Constructing the free body diagram for this system, if we suppose the walls are below and to the right of the block, you should get four forces as follows:
-Pushing force to the right
-Weight acting downwards
-Normal force acting upwards
-Normal force acting left

The left normal force counters the right force and the upwards normal force counters the downwards weight.

Thanks its clear now...
There is no friction when there is no tendancy to move


How can I solve $\sqrt{164}$ without using a calculator?

[quote="blueshark"]How can I solve sqrt{164} without using a calculator?[/quote]To make that appear as $\sqrt{164}$, just type \sqrt{164} and type dollar signs before and after it. If you're going to be around here for a while you should learn $\LaTeX$. It's very easy.

Anyway, I'm not sure what you mean when you say you want to "solve $\sqrt{164}$." If you want to find it's decimal expansion, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48990]here[/url]. It's an irrational number, so it will have an infinite nonrepeating decimal expansion, so you won't be able to actually write $\sqrt{164}$ completely as a decimal.

By solve,do u mean like till infinite digits??cause we can write sqroot164 as 2*sqroot41 and then find sqroot 41 by the normal method(as taught in school in around 8th grade).We can find the appprox value by this method.

Thanks for your reply.  What I meant to ask is: What is
$\sqrt{164}$ correct to the nearest integer?  How can I find the answer without using a calculator?

For example, I know how to solve problems like $\sqrt{169}$ which equals 13 because 13 times 13 = 169, right?  The same for applies for the sqaure root of 36, 25, 81, 9, 4, 16, 100, etc.   I recall my teacher saying that numbers like 4, 9, 16, 25, 36, 81, and 100 are called perfect squares, right?

How do I answer the question above without using a calculator?

something like this ?

$\sqrt{144}<\sqrt{164}<\sqrt{169}$ so $12<\sqrt{164}<13$  :huh:

This is basically through guessing the approximate like what is the nearest integer for

$\sqrt{3724}$ ? So you know $\sqrt{3600}=60$ so the integer should be starting with $6$ . Then just work out some number $61^2=3721$ , $62^2=3844$ ,$63^2=3969$ and you will get the nearest integer to be $61$  :)

i'm sorry for the horrifying diagram,but thats the method i follow for findin square roots.if u multiply 6.4 by 2,u get 12.8 and sqroot 164 is 12.806.....

[hide]
since 164 is between $12^2$ and $13^2$ the answer is betwen 12 and 13.  164 is closer to 169 than 144 so the answer is definetly bigger than 12.5.  Seeing that 164 is only 5 off from 169 and 20 off from 144 we can estimate that the answer is 12.8 or 12.9.  Trying 12.8 i get 163.84 so 12.8 is close.  Trying 12.9 i get 166.41 so the answer is approximatley 12.8.[/hide]

If you want to find the value of $\sqrt{164}$..

[hide]Find the nearest perfect square, 169, and take the square root of that, which is 13. Divide 164 by 13, to get (I like to cheat with calculators) 12.62. Take the average of that and 13 to get 12.81. Now divide 164 by 12.81 to get 12.805, take the average of 12.81 and 12.805 to get about 12.806. Continue until you get to the desired place. [/hide]

[hide]
164 is in between 12^2=144 and 13^2=169,
164 is closer to 169 then 144, so it will be more than 12.5.
Try 12.8, you get 163.84
Try 12.9, you get 166.41
163.84 is closer to 164 than 166.41
$\sqrt{164} \approx \framebox{12.8}$[/hide]

I hope you mean 12.8 in boxes  :)

That's a method too. There are many.

I want to thank all the tutors who took time to explain my question.

well it is my first post in this sub-forum
$\sqrt{164}  \approx \frac{333}{26}$  :D  :D
I use this formula \[ \sqrt{x} \approx \frac{2x + p}{2 \sqrt{x+p}} \]
i choose $p=5$.
want a proof??? :)

I don't know if this was posted earlier, but you could always use the binomial theorem.  ;)


algebra unsolved
algebra
combinatorics
Combinatorial sum

Find The Closed form of:

$\sum_{k=0,n}(-1)^k(n-k)!(n+k)!$

[quote="Megus"]It is: $(2n)!(1-\frac{1}{2(n+1)})$ if I didn't miscalculated.

I used a little changed Australian problem: $\sum _0 ^{2n} \frac{(-1)^k}{C_{2n} ^k}=2-\frac{1}{n+1}$[/quote]
how to prove the australian problem?

you can use induction.
i'll try to write a detailed solution in a moment

To prove australian problem first prove that:

$\frac{1}{C _{2n}^k}=\frac{2n+1}{2n+2}(\frac{1}{C_{2n+1}^{k}}+\frac{1}{C_{2n+1}^{k+1}})$

i just checked.. it's from the 2001 TST.  you can find them at http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html, though there aren't any solutions posted.


Heloo everyone.
I will be very pleased if s.o. of you can write me links or problems,which have proofed with Stewart's theorem.
Thank you.

[quote="Andreas"]Use the Search function to get some problems involving Stewart's theorem. [url]http://www.mathlinks.ro/Forum/search.php[/url]
And also it not regards Middle School, of course.  :!:[/quote]
Thank you


Λοιπον να μια λιγο περιεργη ασκηση.Καλοκαιρινη πειτε....
Την αφιερωνω σε ολους .

Αν $a,b,c,d \in \mathbb{R}$ με

$a+b+c+d<-1$
$a-b+c-d<-1$
$\mid 4a+2c \mid=\mid 3b+d \mid$
$\mid b \mid \leq 1$

νδο $ac\leq \frac{3}{8}$

 :)

Ωραία η άσκηση Socrates   :) . Πού την βρήκες ??
Να μια προτεινόμενη λύση :

[hide="Λύση"]Θεωρούμε το πολυώνυμο $p(x) = a x^{4} +b x^{3} + c x^{2} +d x +1$.

Έχουμε $p(1)=a+b+c+d +1<0$ και $p(-1)=a-b+c-d +1<0$ .

Αφού $p(0)=1>0$ το πολυώνυμο έχει δύο ρίζες $\rho_{1} \in (-1,0)$ και $\rho_{2} \in (0,1)$.

[u][b]Περίπτωση 1[/b][/u]

Άν $\boxed{a>0}$ αφού  $\lim_{x \to + \infty}p(x)=\lim_{x \to -\infty}p(x)=+ \infty$ και $p(1), \, p(-1) \, <0$

το πολυώνυμο έχει άλλες δύο ρίζες $\rho_{3} \in (-\infty,-1)$ και $\rho_{4} \in (1, \infty)$ .

[u][b]Περίπτωση 2[/b][/u]

$\boxed{a<0}$

Έπίσης $p'(x)=4ax^{3}+3bx^{2}+2cx+d$ άρα από την συνθήκη

$\\ \mid 4a+2c \mid=\mid 3b+d \mid$ παίρνουμε $p'(1)=0$ ή $p'(-1)=0$ .

Υποθέτουμε ότι $p'(1)=4a+3b+2c+d=0$.

Αυτό μπορεί να συμβεί όταν $(4a+2c)(3b+d) \leq 0$ .

$\bullet$ Άν  $(4a+2c)(3b+d) =0$ τότε $p'(-1)$ συνεπώς το $p'(x)$ έχει τρεις ρίζες.

$\bullet$ Αν $4a+2c <0$ και $3b+d>0$ τότε $p'(-1)=(3b+d)-(4a+2c) >0$.
Σε αυτή την περίπτωση αν υποθέσουμε ότι $p'(x)>0$ για $x \in (- \infty ,1)$ καταλήγουμε σε
άτοπο γιατί το πολυώνυμο έχει ρίζα στο $(0,1)$ .
Άρα κάπου αλλάζει πρόσημο και έτσι παίρνουμε πάλι 3 ρίζες .

$\bullet$ Αν $4a+2c >0$ και $3b+d<0$ τότε $p'(-1)<0$ .
Τώρα αφού $\lim_{x \to - \infty}p'(x)=+ \infty$ το $p'(x)$ έχει ρίζα $\xi \in (- \infty ,-1)$ άρα έχει 3 ρίζες .
[u][b]
Περίπτωση 3[/b][/u]

$\\ a=0$  :D

Από όλα τα παραπάνω το $p''(x)=12ax^{2}+6bx+2c$ έχει 2 ρίζες
άρα $D = 36b^{2}-96ac \geq 0 \Rightarrow \frac{3}{8} \cdot b^{2} \geq ac$

και αφού $|b| \leq 1$ παίρνουμε $\boxed{ac \leq \frac{3}{8}}$ .[/hide]

Πήρα πολλές περιπτώσεις και αυτό δεν μου αρέσει.
Θα προσπαθήσω για κάτι πιο κομψό αύριο .  :)


algebra
polynomial
vector
complex numbers
superior algebra
superior algebra unsolved

1. Let $\alpha, \beta$ be complex numbers of degree 3 over $\mathbb{Q}$. Determine the possibilities for $[\mathbb{Q}(\alpha, \beta) : \mathbb{Q}]$.

2. Let $\alpha, \beta$ be complex roots of irreducible polynomials $f(x), g(x) \in \mathbb{Q}[x]$. Prove that $f(x)$ is irreducible in $\mathbb{Q}[\alpha]$ iff $g(x)$ is irreducible in $\mathbb{Q}[\beta]$.

1. Suppose $[\mathbb{Q}(\alpha): \mathbb{Q}]=3$ and $[\mathbb{Q}(\beta): \mathbb{Q}]=3$. Then $3~\leq[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]\leq~9$.
If $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$ then $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]=3$. If no common basis elements of $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ as vector spaces over $\mathbb{Q}$, then $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]=9$.

[quote="abdul"]If no common basis elements of $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ as vector spaces over $\mathbb{Q}$, then $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]=9$.[/quote]
This is not true, you can still get degree $6$.

2. is a matter of degree calculation, too.

When does this equality $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]=9$ hold?

This what i think too when i made $9$ as upper bound.
But how can still get degree $6$ if no common basis elements of $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$?

I just read about field extensions for first time some hours ago..

You can even have $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$ but disjoint bases, simply as bases can be choosen quite random (you mean something else than bases I think).
One example for degree $6$: $\alpha = \sqrt[3]{2}$ and $\beta = \zeta_{3}\sqrt[3]{2}$ with $\zeta_{3}^{3}=1 , \zeta_{3}\neq 1$.


[quote="toanhocmuonmau"]This is MIC round 3 and its solution... We are looking for your ideas. :)

Regards
Vo Quoc Ba Can[/quote]
One of it was from here:http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=48&Id=14604&page=2 :)


A recursive sequence has terms in the form $ a_n/b_n$ and is defined such that all even numbered terms are the reciprocal of the immediately preceding term and in all odd numbered terms, $ a_n$ and $ b_n$ are equal to $ a_{n \minus{} 2} \plus{} 1$ and $ b_{n \minus{} 2} \plus{} 1$ respectively. If $ t_1 \equal{} 2/3$ what is:

$ (a_{2005} \plus{} a_{2006} \plus{} a_{2007} \plus{} a_{2008})/(b_{2005} \plus{} b_{2006} \plus{} b_{2007} \plus{} b_{2008})$

[hide="Solution"]

Notice that

$ a_{2k} \equal{} a_{2k \minus{} 1} \plus{} 1$ and
$ a_{2k \plus{} 1} \equal{} a_{2k}$
for any integer k.

Also notice that

$ b_{2j} \equal{} b_{2j \minus{} 1} \minus{} 1$ and
$ b_{2j \plus{} 1} \equal{} b_{2j} \plus{} 2$

We can turn the recursive definition of sequence $ \{t\}$ into a... a... (the word is slipping from my mind) definition where you can find the term with the index.

$ a_x \equal{} (x \plus{} 3)/2$ if x is odd
$ a_x \equal{} (x \plus{} 4)/2$ if x is even
$ b_x \equal{} (x \plus{} 5)/2$ if x is odd
$ b_x \equal{} (x \plus{} 2)/2$ if x is even

Therefore,

$ (a_{2005} \plus{} a_{2006} \plus{} a_{2007} \plus{} a_{2008})/(b_{2005} \plus{} b_{2006} \plus{} b_{2007} \plus{} b_{2008})$

$ \equal{} (1004 \plus{} 1005 \plus{} 1005 \plus{} 1006)/(1005 \plus{} 1004 \plus{} 1006 \plus{} 1005)$

$ \equal{} 1$

We can generalize this. For indices $ 4n \plus{} 1, 4n \plus{} 2, 4n \plus{} 3, 4n \plus{} 4$ (a is an integer),

$ (a_{4n \plus{} 1} \plus{} a_{4n \plus{} 2} \plus{} a_{4n \plus{} 3} \plus{} a_{4n \plus{} 4})/(b_{4n \plus{} 1} \plus{} b_{4n \plus{} 2} \plus{} b_{4n \plus{} 3} \plus{} b_{4n \plus{} 4})$

$ \equal{} [(2n \plus{} 2) \plus{} (2n \plus{} 3) \plus{} (2n \plus{} 3) \plus{} (2n \plus{} 4)]/[(2n \plus{} 3) \plus{} (2n \plus{} 2) \plus{} (2n \plus{} 4) \plus{} (2n \plus{} 3)]$

$ \equal{} 1$

[/hide]

[quote="Aryth"][...] (the word is slipping from my mind) [...][/quote]

Explicit?  :wink:


function
inequalities proposed
inequalities

Let $ n \geq 3$ be an odd integer. Determine the maximum value of
\[ \sqrt{|x_{1}\minus{}x_{2}|}\plus{}\sqrt{|x_{2}\minus{}x_{3}|}\plus{}\ldots\plus{}\sqrt{|x_{n\minus{}1}\minus{}x_{n}|}\plus{}\sqrt{|x_{n}\minus{}x_{1}|},\]
where $ x_{i}$ are positive real numbers from the interval $ [0,1]$.

[quote="JoeBlow"][hide="no proof"]
$ n \minus{} 2 \plus{} \sqrt {2}$. My proof is inelegant.
[/hide][/quote]
After solving it I understand what you mean! I solved it making extensive use of mixing variables, to reduce it to maximizing a function defined for each partition of n.

We have a continuos function on a compact set $ [0,1]^n$, hence there is an optimal point $ (x_1,...,x_n)$.
Note now that
0) impossible to have $ x_{i \minus{} 1} \equal{} x_{i} \equal{} x_{i \plus{} 1}$;
1) if $ x_i\leq x_{i \minus{} 1}$ and $ x_i\leq x_{i \plus{} 1}$, then $ x_i \equal{} 0$;
2) if $ x_i\geq x_{i \minus{} 1}$ and $ x_i\geq x_{i \plus{} 1}$, then $ x_i \equal{} 1$;
3) if $ x_{i \plus{} 1}\leq x_i\leq x_{i \minus{} 1}$ or $ x_{i \minus{} 1}\leq x_i\leq x_{i \plus{} 1}$, then $ x_i \equal{} \frac {x_{i \minus{} 1} \plus{} x_{i \plus{} 1}}{2}$.
It follows that $ (x_1,...,x_n)$ looks like
\[ (0,\frac {1}{k_1},\frac2{k_1},...,1,\frac {k_2 \minus{} 1}{k_2},...,\frac2{k_2},\frac1{k_2},0,\frac1{k_3},...,\frac1{k_l}),
\]
where $ k_1$, $ k_2$, ..., $ k_l$ are natural numbers, $ k_1 \plus{} k_2 \plus{} ... \plus{} k_l \equal{} n$, $ l$ is even clearly. Then the function is this point equals
\[ S \equal{} \sqrt {k_1} \plus{} \sqrt {k_2} \plus{} ...\sqrt {k_l}.
\]
Using the fact that $ l$ is even and $ \sqrt {k} < \sqrt {k \minus{} 1} \plus{} 1$ we conclude that maximal possible value of $ S$ is $ n \minus{} 2 \plus{} \sqrt 2$ ($ l \equal{} n \minus{} 1$, $ k_1 \equal{} k_2 \equal{} ... \equal{} k_{l \minus{} 1} \equal{} 1$, $ k_l \equal{} 2$ in this case).

It is not a real olympiad problem in sense of inner beauty.

since $ n$ is odd, there must be an $ i$ such that both $ x_i$ and $ x_{i \plus{} 1}$ are both belong to $ [0,\frac {1}{2}]$ or $ [\frac {1}{2},1]$. without loss of generality let $ x_1\leq x_2$ and $ x_1$, $ x_2$ belong to $ [0,\frac {1}{2}]$. We can prove that $ \sqrt {x_2 \minus{} x_1} \plus{} \sqrt {Ix_3 \minus{} x_2I}\leq \sqrt {2}$. If $ x_3 > x_2$, $ \sqrt {x_2 \minus{} x_1} \plus{} \sqrt {x_3 \minus{} x_2}\leq 2\cdot \sqrt {\frac {x_3 \minus{} x_1}{2}}\leq \sqrt {2}$; else $ x_1,x_2,x_3$ are all belong to $ [0,\frac {1}{2}]$. Hence, $ \sqrt {x_2 \minus{} x_1} \plus{} \sqrt {Ix_3 \minus{} x_2I}\leq \sqrt \frac {1}{2} \plus{} \sqrt \frac {1}{2}$. Also all of the other terms of the sum are less then or equal to $ 1$. summing them gives the desired result.
example is $ (0,\frac {1}{2},1,0,1,\ldots ,1)$
note: all the indices are considered in modulo $ n$

We can assume there exists some $ i$ for which $ x_i \leq x_{i\plus{}1} \leq x_{i\plus{}2}$ because if it didnt exist: we can WLOG suppose $ x_1\equal{}max{x_i}$. Now $ x_1 \geq x_2$. It follows that $ x_2 \leq x_3$ because if not our assumption would be faulty. Now we continue, it would be: $ x_1 \geq x_2 \leq x_3 \geq x_4 .... \leq x_n \geq x_1$, because n is odd. It's a contradiction that $ x_1\equal{}max{x_i}$.  So there exists $ i$ such that $ x_i \leq x_{i\plus{}1} \leq x_{i\plus{}2}$. Now we can use AM-QM: $ \sqrt{_x{i\plus{}2}\minus{}x_{i\plus{}1}}\plus{}\sqrt{x_{i\plus{}1}\minus{}x_i} \leq \sqrt{2}\sqrt{x_{i\plus{}2}\minus{}x_i}$. Now, maximum is attained if all roots are equal to $ 1$. Now it easy to check this is attainable, if $ i$ is even, then $ (x_1,x_2,...,x_i,x_{i\plus{}1},x_{i\plus{}2},...,x_n)\equal{}(0,1,...,1,\frac{1}{2},0,...,1)$. Similarly if $ i$ is odd. The maximum is $ n\minus{}2\plus{}\sqrt{2}$.

I think that this is pretty similar to IMO 2021/2. Here we can use shifting to let the biggest number be $1$ or the smallest to be $0$, and then induction will probably finish.


$ A*\begin{pmatrix}
    2&4\\
        1&3
        \end{pmatrix}\equal{}\begin{pmatrix}
    2&4\\
        1&3
        \end{pmatrix}*\begin{pmatrix}
    1&1\\
        11&5
        \end{pmatrix}$
$ A\equal{}\begin{pmatrix}
    1&1\\
        11&5
        \end{pmatrix}$where A is 2x2 matrix


Find all functions $ f : \mathbb{R}\to\mathbb{R}$ such that for any $ x,y\in \mathbb R$,
\[ f(x^2 - y^2) = (x - y)\cdot [f(x) + f(y)].\]

[quote="cnyd"]$ f: R\rightarrow R$ $ ,$     $ \forall x,y\in\ R$            $ f(x^2 \minus{} y^2) \equal{} (x \minus{} y).[f(x) \plus{} f(y)]$[/quote]

[hide="My solution"]
Let $ P(x,y)$ the assertion $ f(x^2\minus{}y^2)\equal{}(x\minus{}y)(f(x)\plus{}f(y))$

$ P(x,x)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,0)$ $ \implies$ $ f(x^2)\equal{}xf(x)$
$ P(\minus{}x,0)$ $ \implies$ $ f(x^2)\equal{}\minus{}xf(\minus{}x)$ $ \implies$ $ f(\minus{}x)\equal{}\minus{}f(x)$
$ P(x,\minus{}y)$ $ \implies$ $ f(x^2\minus{}y^2)\equal{}(x\plus{}y)(f(x)\minus{}f(y))$

So $ (x\minus{}y)(f(x)\plus{}f(y))\equal{}(x\plus{}y)(f(x)\minus{}f(y))$ and so $ xf(y)\equal{}yf(x)$ and so $ \frac{f(x)}{x}\equal{}\frac{f(y)}{y}$ $ \forall x,y\neq 0$

So $ f(x)\equal{}ax$ $ \forall x$ and it is easy to check back that this solution fits the initial equation.
[/hide]

$ \frac{f(x)}{x}\equal{}\frac{f(y)}{y}$ $ \implies$  $ f(x) \equal{} ax$ Could you explain better?

[quote="cnyd"]$ \frac {f(x)}{x} \equal{} \frac {f(y)}{y}$ $ \implies$  $ f(x) \equal{} ax$ Could you explain better?[/quote]

$ \forall x\neq 0$ $ \frac{f(x)}{x}\equal{}\frac{f(1)}{1}$ and so $ f(x)\equal{}f(1)x$ $ \forall x$


A particle moves through the first quadrant of the figure as follows: during the first minute it moves from the origin to (1,0). Then, it continues to follow the pattern indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in one minute. Which point will the particle reach after exactly 2 hours?

[img]http://www.kangurusa.com/clark/pdb/kangur_2005_0708/output/kangur_2005_0708_30.png[/img]

Um I'm sorry I can't really understand how the pattern goes, can you clarify that?

I think I'm misreading the diagram or the question somehow, but I'll try anyway.

[hide="Solution"]2 hours is 120 minutes, so 120 moves are made. The pattern tells us that the particle essentially moves from $ (a,b)$ to $ (a\plus{}4,b\plus{}2)$ every 22 minutes (if I read your diagram correctly...)

$ 120\equal{}5\cdot 22\plus{}10$, so we will make 5 of these translations from $ (0,0)$. After 110 minutes, the particle is at $ (4\cdot 5,2\cdot 5)\equal{}(20,10)$.

Now we have 10 moves left. Checking the diagram, if we start at $ (a,b)$, we move to $ (a\plus{}3,b\plus{}1)$ in 10 steps, so we end up at $ \boxed{(23,11)}$.[/hide]


Prove that a continuous function $ f: M \rightarrow \mathbb{R}$, all of whose values are integers, is constant provided that $ M$ is connected.

In trying to start this problem as I started looking at $ f^{ \minus{} 1}(i)$ where $ i \in \mathbb{Z}$, is $ f^{ \minus{} 1}(i)$ open and why?

$ f^{\minus{}1}(\{i\})\equal{}f^{\minus{}1}(i\minus{}\tfrac12,i\plus{}\tfrac12).$

That's why it's open. And it's also closed, since $ \{i\}$ is a closed set.

So basically you're saying that for whatever size ball we try to set up around $ i$, that ball still contains $ i$ and nothing else right?

There's a theorem in my book that states:

If $ M$ is connected, $ f: M \rightarrow N$ is continuous, and $ f$ is onto, then $ N$ is connected. The continuous image of a connected is a connected.

In writing my proof, can I invoke the theorem above. Then say that because the image of $ f$ has to be connected, and that $ \mathbb{Z}$ is a disjoint union of clopen proper singleton sets (defn. of disconnected), then $ f$ has to have a singleton set as an image, and in our set $ \mathbb{Z}$ - a constant.

Do I have to be more specific, or can this suffice?


function
Asymptote
calculus
calculus computations

For any real $a$ find the number of real roots for the equation $x=\tanh ax$

Write the equation as $\frac{t}{a}=\tanh t$. Since $t=0$ is always a root, it suffices to look at $t>0$. Two important oservations are: (1) $\tanh$ is a concave function with a horizontal asymptote; and (2) $(\tanh t)'=1$ at $t=0$.


Ben left at 10:30 from Toronto and arrived in Sudbury at 2:04. He drove from Toronto to Parry Sound on a 4 lane highway at 100km/hr. Then from Parry Sound to Sudbury he drove on a 2 lane highway at 90km/hr. If the distance from Toronto to Parry Sound is 335 km, at what time did he switch from the 4 lane to 2 lane highway?

Total time is 3 hours, 34 minutes.  $ t\equal{}\frac{d}{v}\Rightarrow 90D\plus{}100d\equal{}3.5\bar{6}$ and $ D\plus{}d\equal{}335$  Now we have two equations and two unknowns.  Once we solve for $ d$, we have $ 100d\equal{}t$ so we can get the amount of time to get to the 2-lane highway and just add it to 10:30.


is this a planar graph? If yes, how does its dual graph look like?
[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=266[/img]

It is planar and the dual graph has vertices A, B, C and edges AB, AB, BC, BC, AC, AC. If you want to know how to construct it, look at http://mathworld.wolfram.com/DualGraph.html


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

hello, for $ (i)$ it is 4Catalan's constant -$ \frac{1}{4}\pi^2$.
Sonnhard.

After substitution $ x\equal{}\sin t$ the integral becomes $ \int_0^{\frac{\pi}{2}} \left(\frac{t}{\sin t}\right)^2 \, \cos t \; dt$

And this is $ \int_0^{\frac{\pi}{2}} t^2 \; \left(\frac{\minus{}1}{\sin t}\right)' \, dt\equal{}\left[\frac{\minus{}t^2}{\sin t}\right]_0^{\frac{\pi}{2}}\plus{}\int_0^{\frac{\pi}{2}} \frac{2t}{\sin t} \, dt\equal{}\minus{}\frac{\pi^2}{4}\plus{}4\, \underbrace{\int_0^{\frac{\pi}{4}} \frac{2t}{\sin 2t} \, dt}_{\equal{}J}$

Whereas $ J\equal{}\int_0^{\frac{\pi}{4}} \frac{t}{\sin t \; \cos t} \, dt\equal{}\minus{}\int_0^{\frac{\pi}{4}}  t \, \frac{1}{\cot t} \, \frac{\minus{}1}{\sin^2 t} \, dt
\equal{}\underbrace{\Big[\minus{}t \, \ln\cot t \Big]_0^{\frac{\pi}{4}}}_{\equal{}0}\plus{}\int_0^{\frac{\pi}{4}} \ln\cot t \; dt$

$ \equal{}\int_0^1 \frac{\minus{}\ln u}{1\plus{}u^2} \, du\equal{}\underbrace{\Big[\minus{}\ln u \; \arctan u \Big]_0^1}_{\equal{}0}\plus{}\int_0^1 \frac{\arctan u}{u} \, du\equal{}G$ , hence $ I\equal{}4\, G\minus{}\frac{\pi^2}{4}$ .

After substitution $ x\equal{}\sin t$ the second integral $ I$ becomes $ \int_0^{\frac{\pi}{2}} \frac{t}{\sqrt{1\plus{}\sin^2 t}}\, \cos t \, dt$ .

And this is $ \int_0^{\frac{\pi}{2}} t \, \left[\text{arsinh}(\sin t)\right]' \, dt\equal{}\underbrace{\left[t\, \text{arsinh}(\sin t)\right]_0^{\frac{\pi}{2}}}_{\equal{}\frac{\pi}{2} \, \text{arsinh} 1}
\minus{}\underbrace{\int_0^{\frac{\pi}{2}} \text{arsinh}(\sin t)\, dt}_{\equal{}J}$ .

Because of $ \text{arsinh} z\equal{}\sum_{n\equal{}0}^\infty {\minus{}\frac12 \choose n} \frac{z^{2n\plus{}1}}{2n\plus{}1}\equal{}\sum_{n\equal{}0}^\infty (\minus{}1)^n \, \frac{1}{2^{2n}} \, {2n\choose n} \, \frac{z^{2n\plus{}1}}{2n\plus{}1}$

$ J\equal{}\sum_{n\equal{}0}^\infty (\minus{}1)^n \frac{1}{2^{2n}} {2n\choose n} \frac{1}{2n\plus{}1} \int_0^{\frac{\pi}{2}} \sin^{2n\plus{}1} t \; dt$, whereas $ \int_0^{\frac{\pi}{2}} \sin^{2n\plus{}1} t\; dt\equal{}\frac{1}{2n\plus{}1} \left[\frac{1}{2^{2n}} {2n\choose n}\right]^{\minus{}1}$ .

Therefore $ J\equal{}\sum_{n\equal{}0}^\infty (\minus{}1)^n \frac{1}{(2n\plus{}1)^2}\equal{}G$ , and hence $ I\equal{}\frac{\pi}{2} \, \text{arsinh}\, 1\minus{}G$.


inequalities
function
logarithms
trigonometry
geometry solved
geometry

How to prove that: In any triangle $\triangle ABC$, \[8 \omega ^3 \le ABC\], where $\omega$ is the Brocard angle?

Nice problem, in $Prasolov's$ book you can find a nice solution using Jensen(and a couple of other inequalities with Brocard's angle)!

Do you mind to post the solution here? I don't have this book... Thanks so much.  :)

[quote="yptsoi"]How to prove that: In any triangle $\triangle ABC$, \[8 \omega ^3 < ABC\], where $\omega$ is the Brocard angle?[/quote]

Yptsoi, I guess you meant the inequality to be $8\omega^3\leq ABC$ (the sign is $\leq$, and equality is obtained for equilateral triangles ABC).

In fact, this is the famous [url=http://mathworld.wolfram.com/YffConjecture.html][i]Yff inequality[/i][/url]. Here is the proof as given in Prasolov's book:

Consider the function $f\left(x\right)=\ln\frac{x}{\sin x}=\ln x-\ln\sin x$. The functions

$f^{\prime}\left(x\right)=\frac{1}{x}-\cot x$ and $f^{\prime\prime}\left(x\right)=\frac{1}{\sin^2 x}-\frac{1}{x^2}$

are both easily seen to be positive for $0<x<\pi$. Hence, the function f(x) is monotonically increasing on the interval $\left]0;\;\pi\right[$ and convex on this interval. Therefore, because of the monotonicity, the well-known inequality $\omega\leq 30^{\circ}$ yields $f\left(\omega\right)\leq f\left(30^{\circ}\right)$, and because of the convexity, we can apply the Jensen inequality to the angles $\omega$, $\omega$, $\omega$, $A-\omega$, $B-\omega$, $C-\omega$ (these angles all lie in the interval $\left]0;\;\pi\right[$, since $\omega<A$, $\omega<B$, $\omega<C$), and obtain

$\frac{f\left(\omega\right)+f\left(\omega\right)+f\left(\omega\right)+f\left(A-\omega\right)+f\left(B-\omega\right)+f\left(C-\omega\right)}{6}\geq f\left(\frac{\omega+\omega+\omega+\left(A-\omega\right)+f\left(B-\omega\right)+f\left(C-\omega\right)}{6}\right)$
$=f\left(\frac{A+B+C}{6}\right)=f\left(\frac{180^{\circ}}{6}\right)$       (since A + B + C = 180° by the sum of angles in a triangle)
$=f\left(30^{\circ}\right)$.

Together with $f\left(\omega\right)\leq f\left(30^{\circ}\right)$, this yields

$\frac{f\left(\omega\right)+f\left(\omega\right)+f\left(\omega\right)+f\left(A-\omega\right)+f\left(B-\omega\right)+f\left(C-\omega\right)}{6}\geq f\left(\omega\right)$,

so that

$f\left(\omega\right)+f\left(\omega\right)+f\left(\omega\right)+f\left(A-\omega\right)+f\left(B-\omega\right)+f\left(C-\omega\right)\geq 6f\left(\omega\right)$,

and thus

$f\left(A-\omega\right)+f\left(B-\omega\right)+f\left(C-\omega\right)\geq 3f\left(\omega\right)$

Now, since $f\left(x\right)=\ln\frac{x}{\sin x}$, this becomes

$\ln\frac{A-\omega}{\sin\left(A-\omega\right)}+\ln\frac{B-\omega}{\sin\left(B-\omega\right)}+\ln\frac{C-\omega}{\sin\left(C-\omega\right)}\geq 3\ln\frac{\omega}{\sin\omega}$.

Since the logarithm function is monotonically increasing, we can rewrite this as

$\frac{A-\omega}{\sin\left(A-\omega\right)}\cdot\frac{B-\omega}{\sin\left(B-\omega\right)}\cdot\frac{C-\omega}{\sin\left(C-\omega\right)}\geq \left(\frac{\omega}{\sin\omega}\right)^3$.

In other words,

$\frac{\left(A-\omega\right)\left(B-\omega\right)\left(C-\omega\right)}{\sin\left(A-\omega\right)\sin\left(B-\omega\right)\sin\left(C-\omega\right)}\geq\frac{\omega^3}{\sin^3\omega}$.

But a well-known formula for the Brocard angle $\omega$ of triangle ABC yields

$\sin\left(A-\omega\right)\sin\left(B-\omega\right)\sin\left(C-\omega\right)=\sin^3\omega$.

Thus, our inequality above becomes

$\left(A-\omega\right)\left(B-\omega\right)\left(C-\omega\right)\geq \omega^3$.

Now, by the AM-GM inequality for the positive numbers $\omega$ and $A-\omega$, we have $\omega+\left(A-\omega\right)\geq 2\sqrt{\omega\left(A-\omega\right)}$; in other words, $A\geq 2\sqrt{\omega\left(A-\omega\right)}$. Similarly, $B\geq 2\sqrt{\omega\left(B-\omega\right)}$ and $C\geq 2\sqrt{\omega\left(C-\omega\right)}$. Thus,

$ABC\geq 2\sqrt{\omega\left(A-\omega\right)}\cdot 2\sqrt{\omega\left(B-\omega\right)}\cdot 2\sqrt{\omega\left(C-\omega\right)}$
$=8\sqrt{\omega^3\cdot\left(A-\omega\right)\left(B-\omega\right)\left(C-\omega\right)}$
$\geq 8\sqrt{\omega^3\cdot\omega^3}$      (since $\left(A-\omega\right)\left(B-\omega\right)\left(C-\omega\right)\geq \omega^3$)
$=8\omega^3$,

and the Yff inequality is proven.

In the above, I used two well-known facts: $\omega\leq 30^{\circ}$ and $\sin\left(A-\omega\right)\sin\left(B-\omega\right)\sin\left(C-\omega\right)=\sin^3\omega$. Let me sketch the proofs of these two facts:

In order to prove the inequality $\omega\leq 30^{\circ}$, one should note that the cotangent function is monotonically decreasing on the interval $\left]0;\;\pi\right[$, so that proving $\omega\leq 30^{\circ}$ is equivalent to showing $\cot\omega\geq \cot 30^{\circ}$; but we know that $\cot\omega=\cot A+\cot B+\cot C$, and we have $\cot 30^{\circ}=\sqrt3$, so that it is enough to show $\cot A+\cot B+\cot C\geq\sqrt3$, what is problem 4 in http://www.mathlinks.ro/Forum/viewtopic.php?t=3607 .

As for establishing the identity $\sin\left(A-\omega\right)\sin\left(B-\omega\right)\sin\left(C-\omega\right)=\sin^3\omega$, let's denote by $\Omega$ the Brocard point of triangle ABC such that $\measuredangle \Omega BC=\measuredangle \Omega CA=\measuredangle \Omega AB=\omega$. Then, $\measuredangle \Omega AC=\measuredangle BAC-\measuredangle \Omega AB=A-\omega$ and similarly $\measuredangle \Omega BA=B-\omega$ and $\measuredangle \Omega CB=C-\omega$. But since the lines $A\Omega$, $B\Omega$ and $C\Omega$ are concurrent (namely, at the point $\Omega$ ;) ), the Ceva theorem in the trigonometric form yields

$\frac{\sin\measuredangle\Omega AC}{\sin\measuredangle\Omega AB}\cdot\frac{\sin\measuredangle\Omega BA}{\sin\measuredangle\Omega BC}\cdot\frac{\sin\measuredangle\Omega CB}{\sin\measuredangle\Omega CA}=1$.

In other words,

$\frac{\sin\left(A-\omega\right)}{\sin\omega}\cdot\frac{\sin\left(B-\omega\right)}{\sin\omega}\cdot\frac{\sin\left(C-\omega\right)}{\sin\omega}=1$.

Hence,

$\sin\left(A-\omega\right)\sin\left(B-\omega\right)\sin\left(C-\omega\right)=\sin^3\omega$.

And the proof is complete.

  Darij

Thanks so much for your clear proof, Darij!  :lol: And I have edited the inequality.


function quadratics calculus integration algebra domain

[hide]Looking at mod 4, all squares are equal to 0 mod 4 or 1 mod 4, so we just need a function that is always equal 2 mod 4 or 3 mod 4. Some samples might be $4x+3, 8x+2, or 4x+2$[/hide]

[hide](I'm assuming you mean functions with integer coefficients that never have square values when you plug in any integer.)

There are many functions that satisfy this condition. For example, you look just look at quadratic residues (mods of squares) in modulo $n$, and if $k$ is not one of them, then $nx+k$ satisfies the condition. Also, functions like $x^{2}+2$ and $x^{2}-2$ even work too. And depending on whether or not you consider constants as functions, things like $f(x)=2$ and $f(x)=3$ obviously work too.  :P [/hide]

[hide="one of the answers"]
$\boxed{3x+2}$
[/hide]
[hide="because"]

If a perfect square will be congruent to 2 mod 3, then let us test the possible roots in mod 3.

If root is 0 mod 3, then the perfect square is 0 mod 3.

If root is 1 mod 3, then the perfect square is 1 mod 3.

If root is 2 mod 3, then the perfect square is 1 mod 3.

We have explored all the possible roots in mod 3, and none of them yield a 2 in mod 3 when squared.  Thus $3x+2$, because any such number when x is an integer would be congrent to 2 mod 3, making it not a perfect square.[/hide]

This is a rather trivial problem, since all you have to do is ensure that there exists an integer $p$ such that $f(x) \bmod p$ is always a quadratic non-residue.

Another trivial example for positive integers $k$:  $f(k) = k^{2}+1$.

If you wanted a function on the reals instead, you could just get something that is always negative, such as $-(x^{2}+1)$.

[quote="Farenhajt"][quote="bpms"]Or how about $x^{2}+1$[/quote]

It's a square for $x=0$.[/quote]
How about $x^{2}+1.1$ then :D. THanks for pointing that out...

Well, we can easily put $f(x)=\pi$ :) but I'm sure that the intention of the problem was "the function which returns integer values for integer arguments".

[quote="lotrgreengrapes7926"]Since the question was trivial, can we find a cubic function $f(x)$ ($\mathbb{Z}\to\mathbb{Z}$) such that $f$ is never a perfect square for integral $x$?[/quote]

I already gave a concise argument why this is trivial.  Consider

$f(x) = 4x^{3}+3$

[hide]
$5x+2$ i think works. you can't get a sqrt for it...
$3x+2$ i think also works, because you cannot get a square root for it.

a general forumla would be like that u need a function that = $2\mod4$ or $3\mod4$[/hide]

[hide]$f(x)=x^{2}+2$ is another function, right?[/hide]

You would have to specify the domain.  For example, $4x+2$ is never a perfect square along the domain of natural numbers (easily proven by working in mod 4).  But it could be a perfect square under the domain of rational ones.  For example, if $x=\frac12$, then $4x+2=2^{2}$.

But here's an example that is NEVER a perfect square, even under the real domain:
[hide]$\sin(x)+2.5$
Its mininum is at $1.5$, while its maximum is at $3.5$.  There are no perfect squares between those values, so it can never be a perfect square, no matter what the value of $x$ is.[/hide]

[quote="lingomaniac88"]But here's an example that is NEVER a perfect square, even under the real domain:
[hide]$\sin(x)+2.5$
Its mininum is at $1.5$, while its maximum is at $3.5$.  There are no perfect squares between those values, so it can never be a perfect square, no matter what the value of $x$ is.[/hide][/quote]

The problem is even more trivial in the reals, since you can just use $f(x) = c$ a constant where $c$ is not a square, or you can specify the range of $f(x)$ to be the reals minus the squares.  (Actually, you could do a similar thing for the integers; $f(x) = 2$ is never square.)

anirudh:  This has already been discussed.  One way to ensure that a function is never a square on the [b]integers[/b] is to ensure that it is never a quadratic residue $\bmod p$ for some $p$, not necessarily $3, 4, 5$.  (For example, you could use the function $17x+3$.)


Can anyone tell me how can i draw something with pstricks
i made a very simple example but nothing appears

\documentclass[a4paper]{article}
\usepackage{pstricks}
\usepackage{pst-plot}
\begin{document}
\begin{center}
  \begin{pspicture}(0,0)(100,100)
    \psline[linecolor=blue, linewidth=1](0,0)(50,50)
  \end{pspicture}
\end{center}
\end{document}

i use the latex package that comes in suse 9.1 and pdflatex to obtain the pdf file

thanks

Have a look in this topic:
http://www.mathlinks.ro/Forum/viewtopic.php?t=22791


Let $ a, b, c$ be non-negative real numbers. Prove or disprove that

$ \sum\frac{a^3 \minus{} 1}{b^2 \minus{} bc \plus{} c^2}\ge \frac{(a \plus{} b \plus{} c)^3 \minus{} 12}{(a \plus{} b \plus{} c)^2}$ .

________
 :oops:

[quote="Ligouras"]Let $ a, b, c$ be non-negative real numbers. Prove or disprove that

$ \sum\frac {a^3 \minus{} 1}{b^2 \minus{} bc \plus{} c^2}\ge \frac {(a \plus{} b \plus{} c)^3 \minus{} 12}{(a \plus{} b \plus{} c)^2}$ .

________
 :oops:[/quote]

You can try $ a \equal{} b \equal{} c \equal{} 1$.  :)
With the condition $ a^3\plus{}b^3\plus{}c^3\equal{}1$ then the follow is obvious true. :)
\[ \sum {\frac{{1 \minus{} a^3 }}{{b^2  \minus{} bc \plus{} c^2 }}}  \ge \frac{{3\left( {a \plus{} b \plus{} c} \right)^3  \minus{} 9}}{{\left( {a \plus{} b \plus{} c} \right)}}
\]


Euler
number theory
Diophantine equation

[quote="sagar_astakala"]Can we prove that x = 2 is the only one solution of the equation

3^x + 4^x = 5^x ?[/quote]

You could claim this is due to Fermat's Last Theorem. Still...

Consider mod $ 3$

$ 0^x \plus{} 1^x\equal{}2^x$
Therefore, $ x \equal{} 2m$

Consider mod $ 5$

$ 3^x\plus{}4^x\equal{}0^x$
Therefore, $ x\equal{}2m\plus{}2$

Hence $ 5^{2m\plus{}2}\equal{}3^{2m\plus{}2}\plus{}4^{2m\plus{}2}$
Therefore $ 5^{2m\plus{}2}\minus{}3^{2m\plus{}2}\equal{}2^{4m\plus{}4}$

$ (5^{m\plus{}1}\plus{}3^{m\plus{}1})(5^{m\plus{}1}\minus{}3^{m\plus{}1})\equal{}2^{4m\plus{}4}$

So $ 5^{m\plus{}1}\plus{}3^{m\plus{}1}\equal{}2^a$ and $ 5^{m\plus{}1}\minus{}3^{m\plus{}1}\equal{}2^b$

Adding $ 2 \cdot 5^{m\plus{}1} \equal{} 2^a \plus{} 2^b$
Hence $ 5^{m\plus{}1}\equal{}2^{a\minus{}1}\plus{}2^{b\minus{}1}$

Therefore $ b\minus{}1\equal{}0 \Rightarrow b\equal{}1$

Therefore $ 5^{m\plus{}1}\minus{}3^{m\plus{}1}\equal{}2 \Rightarrow m\equal{}0$

Hence $ x\equal{}2$ is the sole solution

[quote="sagar_astakala"]Can we prove that x = 2 is the only one solution of the equation

$ 3^x \plus{} 4^x \equal{} 5^x$ ?[/quote]
$ 3^x \plus{} 4^x \equal{} 5^x \Leftrightarrow (\frac{3}{5})^x\plus{}(\frac{4}{5})^x\equal{}1$
 + If $ x>2 \Rightarrow (\frac{3}{5})^x\plus{}(\frac{4}{5})^x <(\frac{3}{5})^2\plus{}(\frac{4}{5})^2\equal{}1$
 +If $ x<2 \Rightarrow (\frac{3}{5})^x\plus{}(\frac{4}{5})^x >(\frac{3}{5})^2\plus{}(\frac{4}{5})^2\equal{}1$
So $ x\equal{}2$ is the only one solution of the equation.

I think that this trivial solution is also the only one to $ 3^x\plus{}4^y\equal{}5^z$ in integers $ x,y,z$.

[quote="t0rajir0u"]Integer $ x$, and solutions such that $ abc \neq 0$.  Wiles' proof is beyond undergraduate study.[/quote]

And some postgraduates or lecturers too!

[quote="t0rajir0u"][quote="SimonM"][quote="sagar_astakala"]Can we prove that x = 2 is the only one solution of the equation

3^x + 4^x = 5^x ?[/quote]

You could claim this is due to Fermat's Last Theorem. [/quote]

The question is for real $ x$.  NguyenDungTN has the correct solution.

[quote="Riders on the Storm"]fermat's last theorem says that for any diophantine equation a^x + b^x = c^x, no x for which x>2 produces integer solutions for a, b, and c. the proof is very interesting, but it delves into very deep number theory. check it out.[/quote]

Integer $ x$, and solutions such that $ abc \neq 0$.  Wiles' proof is beyond undergraduate study.[/quote]

sorry i was confused. i thought you just wanted to prove that it doesnt work for x=3 in general. i think euler had an excellent proof of this.


function
calculus
derivative
calculus computations

My question is how to find the criticals numbers for this function (solve for x for this particular function) and find the intervals of increasing and decreasing?

x^2 - 1/x

1st step: find the derivative ($2x+1/x^{2}$)
2nd step: identify the values of $x$ for which the derivative does not exist
3rd step: identify the values of $x$ for which the derivative is equal to $0$.

The numbers found in steps 2 and 3 are critical numbers. When you plot then on the real line, you will get a partition of the line into some collection of intervals. The derivative has constant sign within each interval. You can determine the sign by testing some point of each interval. (+) indicates increasing function, (-) decreasing.

So in this case how do i solve for x when 2x + 1/x^2. Im not sure what method to use after i recuced it down to 2x^3 + 1 then 2x^3 = -1, im just wondering how do you solve it now?

Just one more question. When you said plug in values im just wondering which values and which equation, the derivative of the function or the original equation?

You want to know the sign of the derivative in each interval such as $(0,\infty)$, so you pick a point, e.g. $x=1$ and plug it into the derivative. Got a positive number? The derivative is positive, and the original function is increasing.


inequalities
geometry
circumcircle
inradius
geometry open

Let $ \triangle ABC$. Find the maximal constant $ k$ such that  the following inequality holds :
\[ \frac {r}{R} \plus{} k\frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{R^2}\le \frac {1}{2}\]
[hide] I knew $ k \ge \frac {1}{16}$ :( [/hide]

1993,I got
In $ \triangle ABC$. the following inequality holds :
\[ R\ge \22r \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{8R}.\]

In $ \triangle ABC$. the following inequality holds :\[2r+\frac {1}{8R}\sum(b-c)^2\le R\le 2r+\frac {3}{16r}\sum(b-c)^2.\]
\[2r+\frac {1}{16R}\left(\sum|b-c|\right)^2\le R\le 2r+\frac {1}{16r}\left(\sum|b-c|\right)^2.\]

[quote="Allnames"]Thanks a lot for keeping this problem tuned up over three years.

I have had a solution for $k = 20$ and any constant backwards to 16.

That still remains hidden to me of how this solution performs after that?

Interesting question then I think? Hopefully some fellows could raise a certain for this and I could close up this 3 year old post.

Thanks a lot sqing and have a nice 12/12/12 date.[/quote]
a nice 12/12/12 date.
Good luck.
Thank Allnames.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2347259#p2347259:
 that in any $\triangle\, ABC$ with side-lengths $a,b,c$ , circumradius $R$ and inradius $r$ the following inequality holds:
$\boxed{\ \frac 1{16R}\cdot\left(\sum\, |b-c|\right)^2+2r\, \le\, R\, \le\, 2r+\frac 1{16r}\cdot\left(\sum\, |b-c|\right)^2\ }$ .


function
logarithms
complex analysis
complex analysis unsolved

It is know that $ e^{f(z)}$ is an entire function provided $ f(z)$ is also an entire function. My question is as follows: Is $ e^{e^z}$ an entire function? Since $ e^{e^z}$ never equals to $ 0$ and $ 1$ then it contradicts to the Little Picard Theorem, am I wrong?  :(

It is true that $ e^{e^z}\ne 0$ but you have to reconsider your idea about $ 1$ (Hint: $ 0$ is not the only possible value of $ w$ for which $ e^w\equal{}1$ ;) ).

Oh thanks fedja  :oops: , I did figure out the problem, indeed, $ e^{\minus{}i\pi} \equal{} 1$ and $ e^{\ln(\minus{}i\pi)} \equal{} \minus{}i \pi$.


O is a point inside △ABC,OA=OB+OC.R,S are inside △ABO,ACO.  A,R,O,C are concyclic and AR=RC   A,B,O,S are concyclic and AS=BS
Prove: the circumcircles of △ROB △SOC have only one common inner tangent

here is my solution:
[url]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=15715[/url]

Thank you,dear mr.danh :lol:
I know nothing about invertion,but I will try to understand


Suppose that any $ n\in \mathbb{N}$ and $ n\ge 2$ ,

prove that  $ 2!4!6!\cdots(2n)! >\left((n+1)!\right)^n$

Can this problem be done with induction??

For $ n\equal{}2$, $ 2!.4!\equal{}48>(3!)^2\equal{}36$
Now assume the statement is true for $ n\equal{}k$
$ 2!.4!.\dots.(2k)!>((k\plus{}1)!)^k$

Then it follows that
$ 2!.4!.\dots (2k)!.(2k\plus{}2)!>(2k\plus{}2)!((k\plus{}1)!)^k$ (by assumption)
$ RHS>((k\plus{}1)!)^{k\plus{}1}$
This follows from $ (2k\plus{}2)!>(k\plus{}1)!$, so $ (2k\plus{}2)!((k\plus{}1)!)^k>((k\plus{}1)!)^k.(k\plus{}1)!\equal{}((k\plus{}1)!)^{k\plus{}1}$
Which means the statement is true for $ n\equal{}k\plus{}1$
Then, it's proved by induction

I'm not sure that  my solution is correct. I just learnt how to use induction :P

[quote="IW@IT"]Which means the statement is true for $ n \equal{} k \plus{} 1$[/quote]  Nope -- you need the RHS of the final expression to be $ ((k \plus{} {\bf 2})!)^{k \plus{} 1}$, so the comparison you did between $ (2k \plus{} 2)!$ and $ (k \plus{} 1)!$ needs to be between $ (2k \plus{} 2)!$ and $ (k \plus{} 2)^{k \plus{} 1} (k \plus{} 1)!$.  This is a bit trickier to argue than what you did, but not terribly much :)


calculus
induction
geometric series
real analysis
real analysis unsolved

I need to show the collection of positive integers without 0 in their decimal expansion sum to a number less than 90.  Is this a variation of the Cantor set and how do I start this problem?

Thanks.

$91$ has no $0$ in its decimal expansion and $91>90$ ? I think I misunderstood the problem ...

No. The problem is this classic (which I heard in my calculus class 35 years ago):

Let $S=\{n\in \mathbb{N}:$ the decimal repesentation of $n$ does not contain the digit $0\}.$

Prove that $\sum_{n\in S}\frac1n\le 90.$

This is

$1+\frac12+\cdots \frac19+\frac1{11}+\cdots+\frac1{99}+\frac1{111}+\cdots$

[toadsmack7 left out the word "reciprocal." Without that word, the problem is nonsense.]

I suppose you could find a relationship to a Cantor set if you tried, but you don't have to and that may have little to do with solving the problem.

Yes, sorry about that.  Kent has the formula right.  Any ideas on how to solve this??

It seemed like that the most reasonable approach would be to consider numbers with the same number of digits at a time.

So for one digit we get: \[\sum_{k=1}^{9}\frac{1}{k}\] .

The reason we think about number of digits at a time is if we think about the "chance" that a number has no zeroes, it is something like $.9^{n-1}$ where $n$ is the number of digits (the first digit cannot be zero).  So it seems like we might be able to get a nice geometric series that we can actually compute for.  The only question is which $.9^{n-1}$ of the total sum do we keep and which do we throw out?  Well out of the numbers $\frac{1}{10}, \dots, \frac{1}{19}$ we throw out $\frac{1}{10}$.  This is the biggest one of these $10$.  This is promising, we throw out $20^{-1}$ out of the next ten, etc. etc.  In other words, we throw the "biggest" tenth in some sense.  So we expect that we will be able to bound the entire sum by: \[\sum_{n=0}^{\infty}(\frac{9}{10})^{n}\sum_{k=10^{n}}^{10^{n+1}-1}\frac{1}{k}\] So now we actually establish this inductively.

Actually, we don't since I have a headache.  But the induction is not very tricky.   Notice also that the bound should be not bad since the approximation gets pretty much arbitrarily good as we go along.  So now to bound this sum, we pull out the handy dandy bound of $\log(x) < \sum_{k=1}^{x}\frac{1}{k}< \log(x)+1$ as usual (just look at what happens when we integrate).  So our bound is going to be about $(\log(10)+1) \sum_{n=0}^{\infty}(\frac{9}{10})^{n}= 30,40$ ish. Hmm, why the 90 then?  Did I make some mistake?

The problem has been posted and solved before.  Here is one link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104330[/url].

Haha, there are so many easier ways.  I guess I should have realized that $90 = 9+9\frac{9}{10}+9(\frac{9}{10})^{2}+\cdots$.  Yay for cryptic comment.


I found a fun video forum. Videos are update regularly. See http://www.funnywebvideos.com/

The page looks pretty good.....only thing, it seems to take quite some time to load...

I was quite surprised to see the man who changed the history of skateboarding can stiil laugh after that accident......


Fie un cerc de centru $O$ si raza $r$,  $AB$ diametru. $P$ un punct fixat pe $AB$.
Fie $M$ si $N$ doua puncte pe cerc, $P$ apartine lui $MN$.
Sa se afle pozitia punctelor $M$ si $N$ pe cerc astfel incat aria lui $AMBN$ sa fie maxima.

[quote="dina013"][color=darkred][b]Fie un punct fix $P$ care apartine unui diametru fix $AB$ al cercului $w$. Notam doua puncte mobile $\{M,N\}\subset w$ pentru care $P\in MN$.
Sa se stabileasca  pozitiile punctelor $M$ si $N$ pentru care aria patrulaterului $AMBN$ este maxima.[/b][/color][/quote]
[color=darkblue][b][u]Demonstratie.[/u][/b] Se observa ca $[AMBN]=[ABM]+[ABN]=\frac{AB}{OP}\cdot [OPM]+\frac{AB}{OP}\cdot [OPN]=$ $\frac{AB}{OP}\cdot ([OPM]+[OPN])=$ $\frac{AB}{OP}\cdot [OMN]$.
Asadar, aceasta problema se reduce la urmatoarea problema cunoscuta (vezi o carte mai veche, [b]"Maxime si minime"[/b] de [u]Ion Ionescu[/u]) :[/color]

[quote][color=darkred][b]Fie un punct fix $P$ din interiorul cercului dat $w=C(O)$. Notam doua puncte mobile $\{M,N\}\subset w$ pentru care $P\in MN$.
Sa se stabileasca  pozitiile punctelor $M$ si $N$ pentru care aria triunghiului $MON$ este maxima.[/b][/color][/quote]

[color=darkblue][b]P.S.[/b] Voi reveni curand tot in acest mesaj (sunt chemat de sus la masa de pranz !).[/color]


The problem marathon posted by Silverfalcon has not had enough visitors recently. If you have an interesting problem that you think would help others, please answer the previous question and post it. Thanks.  :)

hm... not sure what that is... can you post a link?

mabye if some of us link the basics marathon to our signature, we'd get some more hits...... I'll consider doing it......


superior algebra
superior algebra unsolved

Serre says that for a finite number of elements $ q_1,...,q_n \in \bf{Q}$ and $ \epsilon_1,...,\epsilon_n \in \{ \minus{} 1, \plus{} 1\}$ one can find an element $ x \in \bf{Q}$ such that the Hilbert symbol satisfies $ (x,q_i) \equal{} \epsilon_i$. What goes wrong if we want to do this for infinitely many $ q_i$?

Principielly it's the same as for the Chinese remainder theorem. You construct it in some way, and this is a purely algebraic construction, so no limit-like things can be done (in both theorems).


Including the endpoints, how many ordered pairs on the line segment with endpoints $ (0,0)$ and $ (16,8)$ have both integer coordinates?

the first thing we should see is that this line has slope $ \frac{1}{2}$

from there, we can see that for every $ y$, there will be an $ x$ that has an integer value

so, we just find how many $ y$'s there are, and we will have our answer

we have the integers $ 0\minus{}8$ for $ y$ values, so there are $ 9$ ordered pairs that have integer coordinates


Gauss
quadratics
number theory unsolved
number theory

second comes straight from the Gauss lemma for the Jacobi symbol

[quote="Mathx"]Let$ n $ be natural number and $n=p_1p_2...p_k$ and $p_i$ are odd prime numbers.prove that:

1-  $\frac{p_1^2-1}{8}+...+\frac{p_k^2-1}{8}\equiv \frac{(p_1p_2...P_k)^2-1}{8}$    $(\mod 2)$

2-  $\left(\frac{2}{n}\right)=(-1)^{\frac{n^2-1}{8}}$[/quote]

From the Gauss lemma we now that $\left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{2}}$ for odd $p$ prime.

Now let's divide primes on these of the form $8k+1,8k+3,8k+5,8k+7$.
Thus for $p=8k+1$ : $\frac{p^2-1}{8}= 0$   $(\mod 2)$
             $p=8k+3$ : $\frac{p^2-1}{8}= 1$   $ (\mod 2)$
             $p=8k+5$ : $\frac{p^2-1}{8}= 1$   $(\mod 2)$
             $p=8k+7$ : $\frac{p^2-1}{8}= 0 $   $(\mod 2)$
On the right side we have $\frac{(p_1p_2...p_k)^2-1}{8} = \frac{7^{2l_1}5^{2l_2}3^{2l_3} -1}{8} = \frac{9^{l_2+l_3}-1}{8}$ $(\mod 2)$, where
$l_1,l_2,l_3$ are exponents of respectively - $7,5,3$ and where $l_2+l_3$ is the number of prime divisors of the form $8k+3$ and $8k+5$.

If the sum of numbers of primes of the form $p=8k+3$,$p=8k+5$ is even then the left side is $0$ $(\mod 2)$.
The right side is $\frac{81^n-1}{8} = 0$ $(\mod 8)$ (cause $\frac{80}{8}= 0$ $(\mod 2)$, where $n$ is a positive integer.
If the sum of primes of the form $p=8k+3,p=8k+5$ is odd then the left side is $1$ $(\mod 2)$.
The right side is $\frac{9*81^n-1}{8}=\frac{9(81^n-1)}{8}+1=1$ $(\mod 2)$.
Thus we always have
$\frac{p_1^2-1}{8}+...+\frac{p_k^2-1}{8}\equiv \frac{(p_1p_2...P_k)^2-1}{8}$    $(\mod 2)$, so
$\left(\frac{2}{n}\right)=\prod_{1 \leq i \leq k}\left(\frac{2}{p_i}\right)=(-1)^{\frac{p_1^2-1}{8}+...+\frac{p_k^2-1}{8}}=(-1)^{\frac{(p_1p_2...P_k)^2-1}{8}}$ (applying the quadratic reciprocity theorem - $\left(\frac{p_i}{2}\right)=1$, q.e.d.


How can I use (x,y,z) i.e. 3-dimensional points in asymptote?

[asy]draw((0,0,0)--(1,1,1));[/asy]

That doesn't work but that is the sort of hting I want to do. How would I do it?

I believe this would work:

[code]
[asy]
import three;
triple stuffgoeshere=(0,0,0);
dot(stuffgoeshere, red);
draw(O--(1,1,1), blue);
[/asy]
[/code]

[asy]import three;
triple stuffgoeshere=(0,0,0);
dot(stuffgoeshere, red);
draw(O--(1,1,1), blue);[/asy]

Of course, you could make much more complicated stuff than that, but its just to get you started. I hope you understand the code.


can we use a calculator?
[hide]77.3%[/hide]

[hide]541/700*100=77.3% (rounded to the nearest tenth)[/hide]

[quote="colts18"]The BigTown Outdoors Club recently held an election for new officers. The turnout for the special election meeting was good. Of the 700 members, 541 showed up to vote. To the nearest tenth of a percent, what percent of the members voted in the special election?[/quote]

[hide]541/700=.7728=77.28%[/hide]

[hide="to jli"]it says to the nearest tenth percent.  your answer is to the hundreth percent ;) [/hide]


Let $ a, b, c$ be positive numbers.Prove:
$ \frac{a^3}{a^2\plus{}2bc}\plus{}\frac{b^3}{b^2\plus{}2ac}\plus{}\frac{c^3}{c^2\plus{}2ab}\ge\frac{a^3\plus{}b^3\plus{}c^3}{a^2\plus{}b^2\plus{}c^2}$

$ (b\minus{}c)^2\ge 0 \Leftrightarrow a^2\plus{}b^2\plus{}c^2\ge a^2\plus{}2bc\Leftrightarrow \dfrac{a^3}{a^2\plus{}b^2\plus{}c^2}\le \dfrac{a^3}{a^2\plus{}2bc}$.
Aplying cyclic sum, we get the result

It's very easy :(.
to be continue
$ Let: a,b,c \ge 0$
chứng minh:
$ \sum\frac{a}{b\plus{}c} \plus{}\frac{3abc}{2(a^2b\plus{}b^2c\plus{}c^2a)} \ge 2$

P/s:From maths.vn


Given two triangles and such that the lengths of the sides of the first triangle are the lengths of the medians of the second triangle. Determine the ratio of the areas of these triangles.

The ratio is 3/4; we just construct the triangle having the medians as sides by formung some parallelograms. This is possible because the sum of the vectors AM, BN, CP (where M, N, P are midpts of BC, CA, AB) is 0, so there is a triangle having sides equal to AM, BN, CP and parallel to these.


Prove that among sixteen consecutive integers it is always possible to find one that is relatively prime to all the rest.

I don't think this belongs in the MATHCOUNTS forum

Hmm....this looks familiar. Exactly the same as number 21b from the midterm.

I have no clue how to do it.


Each side of hexagon ABCDEF has a length of at least 5 cm and AB = 7 cm. How many centimeters are in the least possible perimeter of hexagon ABCDEF?

[hide]We set $ \overline{AB}\equal{}7$ and let all the other sides have length $ 5$. Then, \[ 5\cdot5\plus{}7\equal{}\boxed{32}\]
$ \Box$[/hide]


Hi community ,
 i'm paining with this little problem.

[quote]let ABC an isocele triangle of apex(vertex) B and M be an arbitrary point of the circumcircle $ \Gamma$ of ABC .
let H be the orthogonal projection of C on the line (BM) et I the point of intersection of the two lines (CH) and (AM).
What is the locus (geometric place) of I when M describes $ \Gamma$	?[/quote]

i can see that this is gonna be :
[hide]the circle of center B and whose radius is the common measure of BA and BC [/hide]

i've drawn this figure with GeoGebra ; i hope it helps :

[img]http://safaelmarwa.googlepages.com/olymp5fig2.PNG[/img]

but i didn't succeed in proving it .

i need your help.

thanks.

[hide="note"]By angle chasing, it's easy to see that $ \angle AIC \equal{} 1/2\angle ABC$[/hide]

Indeed man , on this arch it's true but i don't see this holds when M lies on one of the two other archs ?
Besides :maybe: i'm still unsure about the case of M = B , is the line (BM) then reduced to the point B ? or is it the tangent to the circle at B ?
With Geogebra software , dragging M will drag I and show you a bit what 'll happen , but the result is a bit weird ..
Any more suggestions to completely solve this problem ?

[quote="plane geometry"]Denote N is midpoint of AC  then we have NH is the mid-line of AI which means NH=1/2AI  and MH//AI[/quote]

I can't see what you're talking about  :oops:

Hey community , F1 is still pressed here  :huh: ?


Ring Theory
superior algebra
superior algebra unsolved

probably it's standard, but i found it just nice (and i could be interested in "more elementary" solutions)..

prove that $ \prod_{k\equal{}0}^\infty \mathbb{Z}$ is not projective (as a $ \mathbb{Z}$-module).

$ \mathbb{Z}$ PID => (free = projective). and it is not free.

i didn't find it obvious to prove that it's not.
at least, i don't actually see it, i just needed to show it.

by the way, isn't there way to prove it straight by definition, without involving the theorem? i used that too, but professor gave us this exercise before stating the theorem.


I am a visual basic fanatic. Here we discuss Visual Basic 5.0/6.0 Programming.

Even though I dont like visual basic I took it last term in college. With hard work, I managed to get an A. What do you find so interesting about it?

Its interesting because think about some probability problems. You can program to simulate the condition and do lets say 100000 trials and you can find the probability by brute forcing.

Heh, You know what Dijkstra said about BASIC, don't you?
[hide]It is practically impossible to teach good programming to students that have had a prior exposure to BASIC: as potential programmers they are mentally mutilated beyond hope of regeneration
[/hide]

I agree...

But visual basic is a programming language that really doesnt have a finite structure to it.  Kind of like HTML!  You can get away with doing almost anything and the code will run fine.

You should get into programming a bit more advanced that VB, C++.NET is a lot more capable and cleaner that VB.

The only problem is that you develop a crutch after you program in it...just cuz of the nature of the language.

[quote="henryyangrui"]Who's  Dijkstra?[/quote]
Dijkstra was a famous Computer Scientist (He died in 2002, I believe). His single source shortest path finding algorithm (Known as Dijkstra's algorithm) is why he is so famous (Though he may have also probably found many other famous algorithms which I may not be knowing of). See his wikipedia page here: [url]http://en.wikipedia.org/wiki/Edsger_Dijkstra[/url]


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I'm not realy a math type person, But if i know the steps in solving a problem generaly it isn't a problem.  Now the reason im posting is because i dont know the steps in solving this one, Nor how to search for or describe the problem. I can however. represent it visualy as follows. I know it's not the best drawn but hopefully someone with a kind heart can lend me a hand. I feel kind if like i should know this, and feel a bit silly needing to get help  :oops: .  I figured this was the appropriate section based on my own expirence with this type of problem.

[img]http://img132.imageshack.us/img132/746/untitledxw5.png[/img]

The green angle looks like 90 degrees.

If you were talking about the triangles angle it's simple,

$\tan \theta = \frac{5}{4}$
$\theta = \tan^{-1}(\frac{5}{4})$

Also, you can use the law of cosines.

$c^{2}=a^{2}+b^{2}-2ab\cos{C}$

The blue object rotates, And im looking for the angle between its front and the bottom of the triangle. It looks near 90 degrees in this example but it changes as the object is rotated...

Also the green was drawn incorrectly, sorry, I need the equation for finding this as the blue rotates around its point.

[img]http://img140.imageshack.us/img140/8656/untitled2rn0.png[/img]

How fast is the blue object rotating? (I'm guessing it's rotating in the clockwise direction and you want the angle between the horizontal [from the triangles leg] and the blue line).

Is this for a game by any chance?

It isnt rotating at any particular speed, and i don't think we need to incoperate speed, i need only calculate the angle at the time of calculation. I't may be rotating in either direction, but i don't think it's relevent if im only calculating it as a current position.

Why do you ask if it is for a game?

Let $\angle G$ be that green angle you drew. We know that complementary angle to the left of it is $tan\frac{5}{4}$. If that blue cross has rotated $x^\circ$ clockwise then we know that $x$ has the domain $[0,tan\frac{5}{4}]$. So we have
\[\angle G=90-(tan\frac{5}{4}-x) \]
as your formula.

Can you provide the information you know (such as coordinate of the top of the blue line with respect to the center), because I'm unsure as to what I can use to base calculations upon. (if that makes sense..)

I was just curious  :D

Say the center of the object is 0 (likley wont be but does this matter?)

And the point at the top of the triangle is at -4,5 or so, These are just random... Mabey i should attempt to find away to express it more clearly.

As for it having to do with a game, Not exactly..

Edit: To be more broad, I need to make the object face a givin point, They can both be anywhere nearby one another on the plane (within 20 or so units)


Evaluate
\[ \left\lfloor \sum_{n = 1}^{10^9} n^{- 2 / 3} \right\rfloor  \]


convert cyclo heptane---------> methyl benzene

Well, plz tell me if dis is fine:

cycloheptane ---------> Br2 (free radical path) bromocyclo heptane
then Dehydrohalogenation using sm alc KOH.
Then Ozonolyse it.
Then Wolff Kishner Reduction.
Lastly, Aromatization with Mo2O3 or V2O5 supported by Al2O3 at 10-20 atm and 773K.

wont the last step form a seven membered aromatic system.....

also...
i think this conversion works out in just 1 step..

This conversion is kind of ridiculous...

Anyway, just use a suitable metal catalyst, or a transition metal salt like $ \ce{RhCl3}$.

1. No one would ever think of making toluene from cycloheptane, since toluene is readily available in every laboratory.

2. Conversion of saturated carbon rings to benzene rings is seldom used in synthesis. Perhaps the rare examples start with some ring system that have already some insaturation.

I dont think it gives a seven member aromatic ring. I think toluene is formed. i ve seen it in sm book. <Grade 11 NCERT book  :P >

What seven membered aromatic ring could form?

u are right Euclidean Geometer....

It forms the required compound only...

But this reaction can be directly done by aromatization i.e treating cycloheptane in a red hot iron tune....(not actually mine--chemrock's method)....

My method is trating cycloheptane with Se and we get toluene as the product

exactly the topic i was gonna start  :D
Carcul u once asked me for a referrence to convert cyclohexane into benzene using Pd/Se/C but i cannot find it. But that was what my prof told me. Do u know the mechanism for that? Is it by adsorption or sth related to surface chemistry?  :maybe: . Also are there other reagents (other than using allylic chlorination?)  :)

Advice: don't waste your time with that. Such processes have at best only an academic interest.

i know this was to come ....i didn't want to disturb the discussion...but having been here for more than 1.5 year this was tentative :D


algebra
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1,1,2,3,5,8,13,21,34,55,89,144,233,....

Is there a formula for this so as called the "Fibonnaci Sequence"

 :rotfl:  :blush:

The question's been discussed (and locked, too...)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130146[/url]

[quote="learner1729"]okay guys i was using the above formulas to make a program to find the nth term in the fibonnacci sequence. they didn't work and they were off in terms of the signs. so i went to wolframs math world and i got binets formula and it really does work

here is  the binets formula that i obtained from math world

{(1+sqrt50^N)-(1-sqrt5)^N} divided by(2^N times sqrt 5)

N represents the term in the sequence, e.g. first term, third term[/quote]

Sorry to dissappoint you but that was [u]just[/u] posted. Please do not post things that have already been posted.

all right  but mine is a whole lot easier to understand, just one variable to plug in

[quote="DiscreetFourierTransform"]There's a few formulas.

Binet's is the well known one.
Another is the general recursive relationship:
Let $F_{n}$ be defined as the Nth Fibonacci number.
Define $F_{0}= 1$ and $F_{1}= 1$

Then the general recursive relation is $F_{n}= F_{n-2}+F_{n-2}$
Or, $F_{n+2}=F_{n+1}+F_{n}$

Here's how we'd go about solving this for Binet's Formula (beware that this is well above MC Level, but I'll put it in for anyone who is interested:

For all linear recursions, we can generate a characteristic polynomial that defines the recursion. For example, the polynomial for the Fibonacci numbers is
$x^{2}= x+1$
(This comes from the fact that the Nth term becomes $x^{0}$, and the N+1th term becomes $x^{1}$)

Rearranging this equation, we get
$x^{2}-x-1 = 0$

Now we can set up a general recurrence formula, in the form
$F_{n}= A\cdot(R_{1})^{n}+B\cdot(R_{2})^{n}$, where A and B are constants, and $R_{1}$ and $R_{2}$ are the roots of the characteristic polynomial.

The roots of the above eqn, by the Quadratic Formula, are
$\frac{1 \pm \sqrt{5}}{2}$ Call this $\phi$

Thus, we can sub these in, and set up a system of equations:
$F_{0}= A(\phi)^{0}+B(\phi)^{0}$
$F_{1}= A(\phi)^{1}+B(\phi)^{1}$

$1 = A+B$
$1 = A\phi+B\phi$

Solving this yields values for A and B, which leads us to the explicit formula for the Fibonacci Series (Binet's Formula).

Here's a practice problem to try.

There is a recurrence relation defined as $F_{n}= 3F_{n-1}-2F_{n}$ with initial values $F_{0}= 2$, $F_{1}= 3$

[hide="Step One: Define the characteristic polynomial"]
$F_{n+2}= 3F_{n+1}-2F_{n}$
$x^{2}= 3x-2$
$x^{2}-3x+2 = 0$[/hide]

[hide="Step Two: Find the Roots"]
$(x-2)(x-1) = 0$
$x = 2; 1$[/hide]

[hide="Step Three: Set up the recurrence relation"]
$F_{n}= A(r_{1})^{n}+B(r_{2})^{n}$
$F_{n}= A(2)^{n}+B(1)^{n}$
$F_{n}= A(2)^{n}+B$[/hide]

[hide="Step Four: Set up system of equation to find A and B"]
$2 = A(2)^{0}+B$
$3 = A(2)^{1}+B$[/hide]

[hide="Step Five: Solve system, getting values of constants A and B"]
$2 = A+B$
$3 = 2A+B$
$A = 1; B = 1$[/hide]

[hide="Step Six: Finish the explicit formula"]
$F_{N}= 2^{n}+1$[/hide][/quote]

i'm confused

what is the $F_{N}= 2^{n}+1$ for?

not for fibonacci i think. i didn't work.

jorian

it was too complex so i had a hard time trying to understand it

-jorian


trigonometry
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Is there anything wrong with the formula $F = G\frac{m_{1}m_{2}}{r^{2}}$ ?

I used it to develop a long equation about the speed of the planet orbiting the object at any given moment and tested it with Earth, Neptune and Jupiter and in all cases I got about 20% error.

The formula I developed, by the way, is

$v = \sqrt{GM\frac{(a-c\cos \theta)^{2}}{b^{4}}\sqrt{\frac{b^{8}\cos^{2}\theta}{a^{4}(a-c\cos \theta)^{2}}+b^{2}\left( 1-\frac{b^{2}\cos \theta}{a^{2}(a-c\cos \theta)}\right)}}$,

where v is the speed of the planet, G is the gravitational constant, M is the mass of the object being orbited, a is the semimajor axis of the orbit, b is the semi-minor axis, c is $\sqrt{a^{2}-b^{2}}$ and $\theta$ is the angle made between the orbiting object, the object being orbited and the point where the orbiting object is the farthest away.

$F = G\frac{m_{1}m_{2}}{r^{2}}$

we cannot proove that this formula is true or false. but for pratical purpose this formula gives the rigth solution.

your problem can be:

in the set up of your problem you did not considerate that in the universe have more than 3 objects :)

or a mistake in simplification

if you had said how you derive this formula, because there is best formulas for the velocity of planets. try use polar coordinates

it's impossible to help you. you just said "this formula is rigth" but it's impossible for we deduce what problem you made just with the answer

what's the question for 42? dump

What I did was really, really long, and I did it three times and I got the same equation, so it can't be the simplifying. I did use polar coordinates, hence the theta. I'm not sure what you mean by universe having more than 3 objects. My derivation of the formula was completely mathematical. I only applied it to the planets to see if I was right.

What I was asking was, is there a more accurate form of the formula $F = G\frac{m_{1}m_{2}}{r^{2}}$?

Yes, but it involves general relativity.

you don't need to use Einstein on the problem dude.

Kepler did a lot of things without calculus and another usefull things.

what I mean for more than 3 planets is

In the movement of the earth on the sun, for our calculation we need to think that. MArs, pluto, etc. all those planets have a interference on the movement of the earth. it's not a perfect ellipse, but the equation says it is a ellipse

and what i mean for using Polar coordinates is the following.

use the equation of an ellipse, in pollar coordinates, with some algebra an calculus you can have the "velocity" that is equal to the derivative

you can see sites about derivative in polar coordinates
and the First Law of Kepler, the deduction and proof.

the solution is in Kepler

Of course, for everyday purposes you don't need GR. I just pointed out that there [i]is[/i] a more precise law...

Yes! That is exactly what I did! The polar coordinates for an ellipse!

Kepler doesn't tell me anything. He just says T^3 = a^2 for all planets and that the areas are the same when the planet moves the same amount of time. How am I supposed to get the speed of the object using Kepler's laws? I doubt the planets have that much amount of an effect on earth that it decreases its velocity by 20%. There has to be a better explanation.

I have not been able to make sense out of that formula - it has an awful lot of moving parts, with roots inside roots and exponents up to 8.

There should be a much, much simpler formula for velocity based on simple conservation of energy:

$-\frac{GMm}r+\frac12mv^{2}=C,$ where $r$ is the (variable) distance and $C$ is a constant.

There is one slight (classical) improvement on the accuracy. Instead of the smaller object revolving around the larger, they both revolve around their common center of mass. Take that into account and you wind up multiplying or dividing various things by a factor of $\frac{M}{M+m}.$ Of course, even for Jupiter and the sun, that's pretty close to 1 - no 20% lurking in that correction.

err emilgouliev. what you said is the third law of kepler.

the first is what we want.
With it we have the equation of the movement of a planet using the mass of the sum, the constant G, the actual position of the plannet and the velocity

see http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

I think a better solution is

${x=r \cos (\theta )}$   (Parametric equations of the movement. r is not constant)
${y=r \sin (\theta )}$

${r=\frac{p}{\epsilon \cos (\theta )+1}}$  (In PolarCoordinates)

${v=\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{r \cos (\theta )+\frac{dr}{d\theta }\sin (\theta )}{\cos (\theta ) \frac{dr}{d\theta }-r \sin (\theta )}}$

After a few calculations
${\frac{dy}{dx}=-(\epsilon+\cos (\theta )) \csc (\theta )}$

the next problem is just to express $P, \epsilon$ in terms of the others constants.

yagaron: $v$ does not equal $\frac{dy}{dx}.$ As a scalar, $v$ equals $\frac{ds}{dt},$ where $s$ measures arclength.

$v=\sqrt{\left(\frac{dr}{dt}\right)^{2}+\left(r\frac{d\theta}{dt}\right)^{2}}.$

To which we can add two more conditions.

The conservation of momentum condition:

$r\frac{d\theta}{dt}=\text{constant}$

and the conservation of energy condition, which I gave above.

What I did was to use $a_{c}= \frac{v^{2}}{r_{c}}$. I substituted in $\frac{GM}{r^{2}}$ for $a_{c}$, and $r\cos \theta-\frac{b^{2}r\cos \theta}{a}$ for $r_{c}$, and then substitute in $\frac{b^{2}}{a-c\cos \theta}$ for $r$, which gave me that whole thing.

Is there a way to find $\frac{d\theta}{dt}$ using Kepler's laws?

ohh ya...

thanks Kent. i forgot!!! we are using vectors.

sorry for the mistake

[quote="gauss202"][quote="emilgouliev"]What I did was to use $a_{c}= \frac{v^{2}}{r_{c}}$. I substituted in $\frac{GM}{r^{2}}$ for $a_{c}$.[/quote]

This is not accurate. You're using a formula for centripetal acceleration that is only accurate for objects moving in a circle.

[quote="emilgouliev"]
Is there a way to find $\frac{d\theta}{dt}$ using Kepler's laws?[/quote]

Angular Momentum, $\omega = \frac{d\theta}{dt}$.[/quote]

No, it is. You were probably taught that $a = \frac{v^{2}}{r}$ works in a circle, but in fact it works for any curved path. The $r_{c}$ is the radius of the circle that's tangent to the path at that point.

I found my mistake, by the way. I forgot to subtract the $c$ when doing the radius. The new equation is:

$v = \sqrt{GM \frac{(a-c\cos \theta)^{2}}{b^{3}}\sqrt{1-\left(\frac{b^{2}\cos \theta}{a(a-c\cos \theta)}-\frac{c}{a}\right)^{2}+\left( \frac{b^{2}}{a^{2}}\right) \left( \frac{b^{2}\cos \theta}{a-c\cos \theta}-c \right)^{2}}}$, if $0 < \theta < \frac{\pi}{2}$ or $\frac{3\pi}{2}< \theta < 2\pi$

$v = \sqrt{GM \frac{(a-c\cos \theta)^{2}}{b^{3}}\sqrt{1-\left( \frac{c}{a}-\frac{b^{2}\cos \theta}{a(a-c\cos \theta)}\right)^{2}+\left( \frac{b^{2}}{a^{2}}\right) \left( c-\frac{b^{2}\cos \theta}{a-c\cos \theta}\right)^{2}}}$, if $\frac{\pi}{2}< \theta < \frac{3\pi}{2}$

This new equation gave an amazing .0004% error for Earth.

[quote="emilgouliev"]
No, it is. You were probably taught that $a = \frac{v^{2}}{r}$ works in a circle, but in fact it works for any curved path. The $r_{c}$ is the radius of the circle that's tangent to the path at that point.[/quote]

My objection isn't to the "r" it is to the "v". The v in the formula for centripetal acceleration is only the tangential component of the velocity. In a circle, of course, the tangential velocity is equal to the total velocity - but in an elliptical orbit they are not [i]exactly[/i] the same.

In an orbit with relatively low eccentricity this error should be very small. Or it may also be possible that this error has a small net effect over the course of an entire orbit.

[quote="emilgouliev"]
No, it is. You were probably taught that $a = \frac{v^{2}}{r}$ works in a circle, but in fact it works for any curved path. The $r_{c}$ is the radius of the circle that's tangent to the path at that point.[/quote]

Well, there are several problems with that.

That is the formula you get if you decompose the acceleration into two components, one normal to the path and one tangent to the path.

The acceleration tangent to the path is $\frac{d}{dt}|v|.$ That is, it is the rate of change of the speed (and hence dependent on the rate of change of the kinetic energy). Note that for the body in elliptical orbit, the speed is not constant - it is greatest when the body is closest to the sun and least when the body is furthest from the sun.

The acceleration normal to the path is indeed equal to $\frac{v^{2}}{r_{c}}$ where $r_{c}$ is the radius of curvature (the distance to the center of the best-approximating circle). However, please note two issues. The first is that $r_{c}\ne r.$ At both aphelion and perihelion, $r_{c}<r,$ while at some of the points in between, $r_{c}>r.$ Note also that the center of that approximating circle is not where the sun is - that is, the normal to the curve does not generally point toward the sum. The total acceleration - the vector sum of the two components - does point toward the sun, but in general, neither the tangent nor the normal components of the acceleration point in that direction.

In retrospect, that decomposition into normal and tangential accelerations is [b]not[/b] useful in this instance, and I would not recommend pursuing this line any further.

It is clear to me now that emilgouliev's formula has to be incorrect, because of the concerns expressed above. I would strongly suggest scrapping this approach and starting over. Don't directly use the acceleration at all, rather derive the velocity from conservation of energy.

[quote="gauss202"]
In a circle, of course, the tangential velocity is equal to the total velocity - but in an elliptical orbit they are not [i]exactly[/i] the same.[/quote]

OK.

[quote="Kent Merryfield"]Note that for the body in elliptical orbit, the speed is not constant - it is greatest when the body is closest to the sun and least when the body is furthest from the sun. [/quote]

Which is why I have the theta in the equation. When you put in $\pi$ for $\theta$ you get the maximum speed, and if you put $0$ for $\theta$, you get the minimum speed.

[quote="Kent Merryfield"]The acceleration normal to the path is indeed equal to \frac{v^{2}}{r_{c}} where r_{c} is the radius of curvature (the distance to the center of the best-approximating circle). However, please note two issues. The first is that r_{c}\ne r. At both aphelion and perihelion, r_{c}<r, while at some of the points in between, r_{c}>r. Note also that the center of that approximating circle is not where the sun is - that is, the normal to the curve does not generally point toward the sum. The total acceleration - the vector sum of the two components - does point toward the sun, but in general, neither the tangent nor the normal components of the acceleration point in that direction. [/quote]

I already took those into consideration when creating my equation.

[quote="Kent Merryfield"]It is clear to me now that emilgouliev's formula has to be incorrect, because of the concerns expressed above.[/quote]

The only reason it is incorrect is because of what gauss202 pointed out, so besides that you should not have any problems with it.

By the way, where did you find data about the velocities of the planets in terms of $\theta$?

I didn't, actually. I only used the maximum and the minimum velocties, although I should the othe ones, too. My physics teacher found another mistake in it, which I have to fix.

[quote]

${r=\left\{\frac{p \cos (\theta )}{\epsilon \cos (\theta )+1},\frac{p \sin (\theta )}{\epsilon \cos (\theta )+1},0\right\}}$
${v=\frac{dr}{dt}}$ , ${a=\frac{dv}{dt}}$

${V^{2}=v.v=\frac{p^{2}\left(\epsilon^{2}+2 \cos (\theta ) \epsilon+1\right) \left(\frac{d\theta }{dt}\right)^{2}}{(\epsilon \cos (\theta )+1)^{4}}}$

${\text{An}=a.\frac{(v\times a)\times v}{\sqrt{(v\times a).(v\times a) (v.v)}}=\frac{G M}{r.r}}$

${\left(\frac{d\theta }{dt}\right)^{2}=\frac{G M (\epsilon \cos (\theta )+1)^{3}\sqrt{\epsilon^{2}+2 \cos (\theta ) \epsilon+1}}{p^{3}}}$

${V^{2}=\frac{G M \left(\epsilon^{2}+2 \cos (\theta ) \epsilon+1\right)^{3/2}}{p (\epsilon \cos (\theta )+1)}}$
[/quote]

I think this is rigth, or not?

[quote="Fabiomr1"]err....

it should be an elipse, not a circle
and, since u used "r", it's a considerable error

that's the error[/quote]

What are you talking about?

Is the planet that is orbited fixed, or are the planets orbiting each other around the centre of mass?

yagaron, I'm actually confused on what stands for what.

Thaakisfox, the planet being orbited is fixed.

Then why all the fuss with that long formula?
Here is a much simpler one:
The orbiting component of the velicity of the orbiting planet:
$v_{\phi}=\frac{\Psi}{R}= \frac{\Psi}{p}(1+e\cos \phi)$
Here:
$\Psi =\frac{2\pi}{T}\sqrt{a^{3}p}=\sqrt{GMp}$ where $p$ is the parameter of the ellipse, $e$ is the excentricity, $a$ is the semi-major axis, $T$ is the period of orbit of the planet.
The radial component of the velocity:
$v_{R}= \dot R =\frac{dR}{d\phi}\frac{d\phi}{dt}=\frac{pe \dot \phi\sin\phi}{(1+e\cos\phi)^{2}}=\frac{R^{2}\dot\phi}{p}e \sin\phi$
So
$v_{R}=\frac{\Psi}{p}e\sin\phi$
So the square of the velocity:
$v^{2}=v_{R}^{2}+v_{\phi}^{2}=\frac{{\Psi}^{2}}{p^{2}}(e^{2}\sin^{2}\phi+(1+e\cos\phi)^{2})$
So from here:
$v^{2}=\frac{{\Psi}^{2}}{p}\left ( \frac{e^{2}-1}{p}+\frac{2}{R}\right )$
But we also know that:
$\frac{1-e^{2}}{p}=\frac{1}{a}$
So the velocity of the orbiting planet:
$v=\Psi \sqrt{\frac{1}{p}\left (\frac{2}{R}-\frac{1}{a}\right )}=\sqrt{GM\left (\frac{2}{R}-\frac{1}{a}\right )}$

Now this formula looks much simpler to me  :D


Can you help me prove this inequalities by using Jensen inequality?
Thanks.


well the thing is to take the square root

$\sqrt{2400}\approx 48$

that means u divide 2400 by all primes less than 48...

when you have a quotient you divide until you reach a prime

I know it might  be confusing  :lol:

nonononono
Just start dividing.
2400/2=1200   2
1200/2=600    2
600/2=300     2
300/2=150     2
150/2=75      2
75/3=25        3
25/5=5          5
5/5=1          5

Looking at the crooked right column :D we see that we have 5 2s, a 3, and 2 5s, which is a total of $8$

yeah kstan013's way is much much simpler in this case as the number 2400 is convieniant but my way works for any number

[quote="epatjn"]The question is this:

the number 24 can be made by multiplying 4 prime numbers, 2, 3, 2, 2.

how many primes must be multipled to make 2400? is there a formula to help me determine that?[/quote]
2400/2=1200/2=600/2=300/2=150/2=75
SO just 2, and the prime factors of 75 which are 3 and 5

[quote="epatjn"]The question is this:

the number 24 can be made by multiplying 4 prime numbers, 2, 3, 2, 2.

how many primes must be multipled to make 2400? is there a formula to help me determine that?[/quote]
[hide]Knowing that $24=2^{3}\times3$, and $100=2^{2}\times5^{2}$, you multiply them together to get $2400=2^{5}\times3\times5^{2}$[/hide]

[quote="lifesapain"]Yes, all you have to do is make a factor tree, simple as that.
[hide]You and up with 2 to the power of 5, 5 squared, and 3, so 8 in all. [/hide]Right?[/quote]

...as a future warning please do not revive topics more than 5 days old


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra solved

How can I find the splitting field of $x^{p}-x-a \in \mathbb{F}_{p}[x]$ where $a \neq 0$, and why should the Galois group of this splitting field be cyclic?

Elementary solutions are preferred since google gave me only advanced answers :)

first obverse that $x^{p}-x-a$ is separable and has no root in $\mathbb{F}_{p}$, and with every root $\alpha$ (we fix one in a splitting field), $\alpha+i$ is also a root. so $\alpha+i$, where $i$ run's through $\mathbb{F}_{p}$, are all roots. thus, the splitting field is $\mathbb{F}_{p}(\alpha)$, and we have an injective homomorphism $G \to \mathbb{F}_{p}, f \mapsto f(\alpha)-\alpha$, where $G$ denotes the galois group. but $\mathbb{F}_{p}$ has only the two obvious subgroups,  so that $G \cong \mathbb{F}_{p}$.

Actually, every extension of finite fields is Galois with cyclic Galois group, canonically generated by the Frobenius automorphism. owk

@owk: indeed :-). but I think the concrete situation here is also interesting ;-).


function
superior algebra
superior algebra unsolved

Wasn't sure if I should post this here, but...

If f: A -> B let us define f_image: P(A) -> P(B) by
f_image(E) = {b in B | b = f(x) for some x in E}.
Prove that the function f_image is onto if and only if f is onto.

P(A) means the power set of A.

So far, all I have is the following:
Prove that: If f is onto, then f_image is onto.
Assume f is onto. We want f_image is onto, or, for all Y in P(B), there exists an X in P(A) such that f_image(X) = Y.

I know it isn't much, but I really have no idea what to do now. Can anyone help? Tell me if you need something clarified.

[Edit] I forgot - please reply ASAP if you can. I need this by Friday.

I hope you are not asking anybody here to do your homework. Just use the definitions, and common sense. You'll be able to do it.

No, I'm not trying to get people to do my homework. I'm just asking someone if they could point me in the right direction...I really have no idea how to continue.


it should be [hide](2x:^2:+5)(2x-3)[/hide]

Yes, but I think that it would be nice to know how you managed to conjure the right expressions to multiply.

Q: 4x:^3:-6x:^2:+10x-15

A: [hide](2x:^2:+5)(2x-3)[/hide]

Solution:[hide]well, hmmmmmmm...whered my pencil go...aha...*pats himself on the back*...well, the constants equal -15. and on a hunch i would use 3 and 5. nevermind the signs for now. i knew that a polynomial times  :pm: 3 or :pm: 5 equaled -6x:^2:. 3 would divide that. so then we know that 2x:^2: and -3  are in the answer. they have to be on different factors though, if you want to multiply them. also, -3 times a number equals -15. so the other constant is 5. then its easy from there. just plug in what you know. and you get: [hide](2x:^2:+5)(2x-3)[/hide][/hide]

if this didnt make sense to you, cause i didnt really explain it nicely, just ask.

Well, a nice little trick that works with a couple (not many, but some) factorising questions:

[hide]

4x:^3:-6x:^2:+10x-15 = x:^2:(4x-6) + 5(2x-3) = 2x:^2(2x-3)+5(2x-3) = (2x:^2:+5)(2x-3).

[/hide]

thats a cool trick sphink. my solution may have been longer, but it only took about 10 seconds for me, but that may not occur all the time. your way would be better for cases one is unsure of

Rep,

Do you know the rational roots theorem?

Any rational root of a polynomial has a numerator that is a divisor of the constant of the polynomial and a denominator that is a divisor of the lead coefficient of the polynomial.  Can you see why this is true?

thats good  to know

P.S.
wow, i didnt realize that i passed the 300 mark. yeha


What is the smallest $ n$ for which $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\cdots\plus{}\frac{1}{n}>3?$

Hi Everybody,

I don't quite understand how you did the graph. Could you elaborate? Thanks!

--mathdriven

sorry to revive this, but can anyone tell me a way to do it without brute forcing?
Can you pm me if you know?


Does $ a^4 \plus{} b^4 \plus{} c^4 \leqslant 2(a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 a^2 )$  hold for every $ a,b,c$  if it is known that $ a^2 \plus{} b^2 \plus{} c^2 \leqslant 2(ab \plus{} bc \plus{} ca)$.

why do you posted this problem two time,
see her, http://www.mathlinks.ro/viewtopic.php?t=271098


If  $a,b,c$ are positive reals and $ab+bc+ca=3$ prove that

$a+b+c\geq abc+2$

Firstly we prove that $a+b+c\geq 3.$
Indeed, $a+b+c\geq 3$ $\Leftrightarrow$ $(a+b+c)^2\geq 3(ab+bc+ca)$, obvious.
From $ab+bc+ca=3$, we obtain \[ a^2 b^2+b^2 c^2 +c^2 a^2+2abc(a+b+c)=9\geq 3abc(a+b+c)\Longrightarrow  \]
\[ \Longrightarrow abc(a+b+c)\leq 3 \Longrightarrow abc \geq\dfrac{3}{a+b+c}.  \]
It follows \[ a+b+c\geq abc+2 \Leftrightarrow abc\geq a+b+c-2.  \]
Then its enough to prove that $\dfrac{3}{a+b+c}\leq a+b+c-2.$
We have $\dfrac{3}{a+b+c}\leq a+b+c-2\Leftrightarrow (a+b+c+1)(a+b+c-3)\geq 0$ and hence the conclusion.

Actually, this problem appeared on JBMO shortlist in 2003. I gave for this inequality two solutions. One of them is exactly like Marius, but the other one is much more easier. Here it goes:

By AM-GM we get $3=ab+bc+ca\geq3\sqrt [3] {abc}$.From here it results that $abc\leq 1.$
So if we prove that $a+b+c\geq 3$ we are done here. But this follows from this well-known inequality $(a+b+c)^{2}\geq 3(ab+bc+ca).$  :D  ;)

If$a,b,c\in R: a^2+b^2+c^2=2$ we have:
    $\mid a+b+c-abc\mid\le 2$