[quote="Mersenne"]Twenty bored students take turns walking down a hall that contains a row of closed lockers, numbered 1 to 20. The first student opens all the lockers; the second student closes all the lockers numbered 2, 4, 6, 8, 10, 12, 14, 16, 18, 20; the third student operates on the lockers numbered 3, 6, 9, 12, 15, 18: if a locker was closed, he opens it, and if a locker was open, he closes it; and so on. For the ith student, he works on the lockers numbered by multiples of i: if a locker was closed, he opens it, and if a locker was open, he closes it. What is the number of lockers that remain open after all the students finish their walks?[/quote]
This has been posted very recently, and several times before that also.

PLEASE SEARCH.

Actually, I did. I saw a topic just like this before I ever posted my first post, and use search all the time now to check that I won't be annoying anybody with a re-post of a question.

[quote="seamusoboyle"]Did YOU start searching 3 posts into your aops career!?! I know i didnt![/quote]
No, but then someone yelled at me in capitals so I started using search  :P .


[quote="math10"]Let $ a_1,a_2,..,a_n\geq 0$.Prove that:
  $ (n \minus{} 1)\sum_{i \equal{} 1}^{n}a_i^2 \plus{} n\sqrt [n]{\prod_{i \equal{} 1}^{n}a_i^2}\geq (\sum_{i \equal{} 1}^{n}a_i)^ 2$[/quote]
This problem,I only have a solution (using Mixing Variables)
Can anyone post solution other?

it's Turkevici inequality
but I don't think it satisfies you  :(

[quote="math10"]Let $ a_1,a_2,..,a_n\geq 0$.Prove that:
  $ (n \minus{} 1)\sum_{i \equal{} 1}^{n}a_i^2 \plus{} n\sqrt [n]{\prod_{i \equal{} 1}^{n}a_i^2}\geq (\sum_{i \equal{} 1}^{n}a_i)^ 2$[/quote]
Let $ a_1 \leq a_2 \leq \cdots \leq a_n$
let $ \prod_{i \equal{} 1}^{n}a_i \equal{} 1$,and  $ a_n\geq a_{n \minus{} 1} \geq \cdots \geq a_1$
$ f(a_1,a_2,\cdots,a_n) \equal{} (n \minus{} 1)\sum_{i \equal{} 1}^{n}a_i^2 \plus{} n \minus{} (\sum_{i \equal{} 1}^{n}a_i)^ 2$
$ f(a_1,a_2,\cdots,a_{n \minus{} 1},a_n) \minus{} f(a_1,a_2,\cdots,\sqrt {a_{n \minus{} 1}a_n},\sqrt {a_{n \minus{} 1}a_{n}})$
$ \equal{} ((n \minus{} 2)(\sqrt {a_{n \minus{} 1}} \plus{} \sqrt {a_{n}})^2 \minus{} 2\sum_{i\equal{}1}^{n \minus{} 2}a_i)(\sqrt {a_{n \minus{} 1}} \minus{} \sqrt {a_{n}})^2\geq 0$
so $ f(a_1,\cdots,a_n)\geq f(a_1,\cdots,\sqrt {a_{n \minus{} 1}a_{n}}$,$ \sqrt {a_{n \minus{} 1}a_{n}})\geq \cdots \geq$
$ f(a_1,\sqrt [n \minus{} 1]{a_{2}a_{3}\cdots a_{n}},\cdots,\sqrt [n \minus{} 1]{a_{2}a_{3}\cdots a_{n}})\geq 0$

[quote="ye109"][quote="math10"]Let $ a_1,a_2,..,a_n\geq 0$.Prove that:
  $ (n \minus{} 1)\sum_{i \equal{} 1}^{n}a_i^2 \plus{} n\sqrt [n]{\prod_{i \equal{} 1}^{n}a_i^2}\geq (\sum_{i \equal{} 1}^{n}a_i)^ 2$[/quote]
Let $ a_1 \leq a_2 \leq \cdots \leq a_n$
let $ \prod_{i \equal{} 1}^{n}a_i \equal{} 1$,and  $ a_n\geq a_{n \minus{} 1} \geq \cdots \geq a_1$
$ f(a_1,a_2,\cdots,a_n) \equal{} (n \minus{} 1)\sum_{i \equal{} 1}^{n}a_i^2 \plus{} n \minus{} (\sum_{i \equal{} 1}^{n}a_i)^ 2$
$ f(a_1,a_2,\cdots,a_{n \minus{} 1},a_n) \minus{} f(a_1,a_2,\cdots,\sqrt {a_{n \minus{} 1}a_n},\sqrt {a_{n \minus{} 1}a_{n}})$
$ \equal{} ((n \minus{} 2)(\sqrt {a_{n \minus{} 1}} \plus{} \sqrt {a_{n}})^2 \minus{} 2\sum_{i \equal{} 1}^{n \minus{} 2}a_i)(\sqrt {a_{n \minus{} 1}} \minus{} \sqrt {a_{n}})^2\geq 0$
so $ f(a_1,\cdots,a_n)\geq f(a_1,\cdots,\sqrt {a_{n \minus{} 1}a_{n}}$,$ \sqrt {a_{n \minus{} 1}a_{n}})\geq \cdots \geq$
$ f(a_1,\sqrt [n \minus{} 1]{a_{2}a_{3}\cdots a_{n}},\cdots,\sqrt [n \minus{} 1]{a_{2}a_{3}\cdots a_{n}})\geq 0$[/quote]
NO,I need a solution other


Does there exist an ordered pythagorean triple where the two legs of the triangle are odd?  If so, state which one.

[hide]No there is not. If we let u>=v, all sets of pythagorean triples are defined by (2uv, u^2-v^2, u^2+v^2), where 2uv and u^2-v^2 are the legs.[/hide]

geez
Why do all the experts post on the beginners forum?
put it in a spoiler at least...

[hide]Or, if you don't know the formula for pythagorean triples, you can think of it this way: perfect squares are either 0 or 1 mod 4. Even perfect squares are 0 mod 4, and odd perfect squares are 1 mod 4. Two odd perfect squares add up to 2 mod 4, which can't be a perfect square. Thus there are no pythagorean triples with the smaller number odd.

(0 and 1 are quadratic residues mod 4; 2 is a quadratic non-residue mod 4.)[/hide]

Oh, sorry, I thought you needed help with the problem (and were not issueing it as a challenge).


The product of two numbers is 24 and their sum is 10. What is the larger number?

Trial and error yields 6 and 4, so the answer is $ \boxed{6}$.

We have:
x*y=24
x+y=10 Thus, y=10-x
Plug it in the equation:
x(10-x)=24
10x-x^2=24
x^2-10x+24=o
(x-4)(x-6)=0
x= {4,6}
Thus, the greater number we are looking for is 6. :D

Both ways work, but with numbers this small, I think that trial and error is faster.

[quote="quattayangel"]
x^2-10x+24=o
(x-4)(x-6)=0[/quote]

this=guess&check :)


[quote="CircleSquared"]I found this problem in a Chinese problem book, and I had tons of trouble with it :huh: :

There are 100 children numbered 1 to 100 standing in a circle. Starting from the 1st one, the count off 1,2,1,2,...  Any child who counts the number 1 sits down.   What is the number of the last child standing?

Does anyone have a good solution :?:

Thanks!

EDIT: Mods, feel free to move this if it is above MC level.  Although I did once see a MC Nat problem similar to this one...[/quote]

[hide]In the first stage, the odd numbered children are to sit down; let's divide the rest of the numbers by 2 to obtain the new list of 1 through 50.

In the next stage, the odd numbered chilldren sit down.

We repeat this forever until there is one child standing.

Therefore, we can simply count the number that includes the highest power of $2$ - which in this case is $64$, if I am not mistaken. :roll: [/hide]

really, i got

[hide]$72$

I found out that in the first go-around, the odd-numbered kids sit.  then, without restarting the number, the multiples of $4$ stay standing.  after that roung, the multiples of $8$ stay up.  since $100$ was the last number to sit down, 8 is safe, so the last six numbers are $8$ $24$ $40$ $56$ $72$ $88$.  once again, $8$ is safe, so the last $3$ numbers are $8$ $40$ and $72$.  $8$ gets lucky again, so $8$ and $72$ are the last two numbers.  $8$ isn't as lucky because 72 was a $2$, so $72$ is the answer.[/hide]

I think that there was a Masters Round similar to this problem, i am trying to find the link and the formula though.  i will edit this post if i find it.

Here's the answer in the book, and my answer:

[hide]
IDK how they got this, but:

They subtracted the highest power of 2 possible from 100 such that the result is an integer.  Then they multiplied the result by 2 to get 72.

When I did the problem myself, I got 64.

 :huh:
[/hide]

That would be from the master's round in the all-time greatest problems book.

[quote="mchoi815"]is there a way to get master's rounds?
does the all-time greatest problems book have them all?[/quote]

All-time Greatest just has one set of masters, the lightbulb one.  I think Mathcounts gave the Master's Round with their tests before 2002, so if you buy a test from 2001 or earlier it should have the Master's Round, or if your coach has old tests, they should contain the Master's Rounds.

[quote="rd5493"]so i was right, but the formula is

[hide]let $x$=the highest power of $2$

$(100-2^{x})2=y$

where $y$ is the answer (72 in this case)[/hide][/quote]

so if it was every three children, would all the 2's just change to 3's?

[i]Solving[/i] this kind of problem (with small numbers) is MATHCOUNTS-level, and such problems have appeared in contests and in handbooks.

[i]Analyzing[/i] general cases of this problem is above MATHCOUNTS-level.  This is a variant of the "Josephus Problem".  You can search MathWorld or ask in a higher-level Forum.  (I suggest High School Intermediate.)

Please do not attempt to analyze this problem in the MATHCOUNTS Forum.


Wow nice solution, while mine is so long thank you  :)

the solution
$T\leq\frac{3-\sqrt{5}}{2}$
can come faster by $T\leq 1$ and
$T\leq(1-S)^{2}\leq(1-T)^{2}$


geometry
incenter
circumcircle
ratio
angle bisector
geometry solved

Let a triangle $ ABC $ with $ AB = AC $ ,the incenter $  I $ ,  the altitude $ AD  $ and a point  $ M $  on  the circumcircle of triangle $  IBC$   , which is not in the interior of triangle $  ABC$ . Prove that $ MI $  bisects angle $ \angle  AMD $.

   Babis

[i]( I think that it deserves to have this problem here . The autor , Ioan Patrascu ,  has  written a good geometry book in Romanian and this seems to be a good problem. )[/i]

The lines $AB,\ AC$ are tangent to the circumcircle $w$ of the $\triangle BIC$. I note the second intersection $N$ between the line $AI$ and the circle $w$. Because the division $(A,D,I,N)$ is harmonically and $MI\perp MN$ results that  $\angle AMI\equiv \angle DMI$.
See LEMMA from http://www.mathlinks.ro/Forum/topic-46146.html

[quote="stergiu"]Let a triangle $ ABC $ with $ AB = AC $ ,the incenter $  I $ ,  the altitude $ AD  $ and a point  $ M $  on  the circumcircle of triangle $  IBC$   , which is not in the interior of triangle $  ABC$ . Prove that $ MI $  bisects angle $ \angle  AMD $.[/quote]

This generalizes to an arbitrary triangle ABC; however, of course, the point D becomes the foot of the angle bisector rather than of the altitude from the vertex A (for an isosceles triangle ABC with AB = AC, that's the same thing, but for a non-isosceles triangle ABC, it isn't). So we have to prove:

[color=blue][b]Theorem.[/b] Let I be the incenter of an arbitrary triangle ABC, and let D be the point of intersection of the line AI with the line BC. Let M be a point on the circumcircle of triangle IBC. Then, the line MI bisects the angle AMD.[/color]

[i]Proof.[/i] Since the point I is the incenter of triangle ABC, the line BI is the angle bisector of the angle ABC. Equivalently, the line BI is the angle bisector of the angle ABD. Hence, since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, in the triangle ABD we have $\frac{AI}{DI}=\frac{AB}{DB}$. Similarly, $\frac{AI}{DI}=\frac{AC}{DC}$. Thus, denoting $k=\frac{AI}{DI}$, we have $\frac{AI}{DI}=\frac{AB}{DB}=\frac{AC}{DC}=k$. Now, it is well-known that the locus of all points X such that $\frac{AX}{DX}=k$ is a circle, a so-called Apollonius circle of the segment AD; since $\frac{AI}{DI}=\frac{AB}{DB}=\frac{AC}{DC}=k$, this circle passes through the points I, B and C, i. e. this circle must be the circumcircle of triangle IBC. Hence, since the point M lies on this circumcircle of triangle IBC, it must satisfy $\frac{AM}{DM}=k$. Since $k=\frac{AI}{DI}$, this becomes $\frac{AM}{DM}=\frac{AI}{DI}$. Now, the angle bisector of an angle of a triangle - and only the angle bisector - divides the opposite side in the ratio of the adjacent sides; hence, this equation yields that the line MI is the angle bisector of the angle AMD of triangle AMD. And the theorem is proven.

  darij

Dear :  Levi and Darij , thank you ! Very nice solutions and generalization!

 Babis


probability
number theory
least common multiple

[quote="kidwithshirt"]A bowl contains fewer than 50 marbles, and each is either red green or blue. The probability of drawing a red marble is 2/5 and the probability of drawing a greeb marble is 3/7. If two marbles are drawn without replacement, what is the probability that both are blue? Express your answer as a common fraction.

The sum of the digits of a three-digit number is 26. The number is multiplied by 7, then by 11, then by 13. How many times will the digit 9 occur in the final product?[/quote]
[hide="1"]
$1-(\frac{2}{5}+\frac{3}{7})=\frac{6}{35}, \frac{6}{35}\cdot\frac{5}{34}=\boxed{\frac{3}{119}}$
[/hide]

[hide="2"]$7\times11\times13=1001$ so three extra zeroes are added to the number. Since the number is either $899, 989,$ or $998$, there are two nines when multiplied by $1000$ and when adding itself two more nines are added, so $\boxed{4}$ nines. [/hide]

Hah I just did this sprint

[hide="the first one"]
Since the denominators has 5ths and 7ths, the total amount of marbles must be a mulitple of both. The lcm is 35.
Thus there are $1-\frac 25 - \frac 37=\frac{6}{35}=6$ marbles that are blue.
The probability of choosing two without replacement is thus
$\frac{6}{35}\cdot\frac{5}{34}=\frac{3}{7\cdot17}=\frac{3}{119}$[/hide]
[hide="the second one"]
The key to this one is that $11\cdot13=143$ and $143\cdot7=1001$
Now looking at the 3 digit number, it must be 899,989,998.
Mulitply any of those by 1001 and get
the first 3 digits are the number and the last 3 are it again
(899899,989989,998998)
All of them have 4 nines[/hide]

[hide="1"]We see that the probablity of drawing a blue marble is 6/35. Just to make things simple, make the number of marbles in the bowl 35. Therefore, the probability of drawing two blue marbles is 6/35*5/34, or 3/119.[/hide]
[/hide]

[hide="1"]3/119. there are 35 marbles, 14 are red, 15 are green, so 6 are blue. 6/35*5/34 reduces to 3/119. [/hide]


geometry
3D geometry
tetrahedron
ratio
USAMTS

Given a tetrahedron, where would the parallel plane to the base be if it divides the tetrahedron in half by volume?

[hide]$\frac{1}{2^\frac{1}{3}}*height$ from the tip? since the tetrahedrons must be similar, the ratio of the volume is the linear ratio^3...[/hide]

you arent supposed to talk about USAMTS problems until after the deadline I think...


Some time ago, it was posted a hard inequality:
$a,b,c>0, a+b+c=3$. Prove that $\sum_{cyc}\frac{a}{b^{2}+c}\geq \frac{3}{2}$
see here: http://www.mathlinks.ro/viewtopic.php?t=145137
So, I have a qeustion related to this one. Is the following true?
$\sum_{cyc}\frac{a}{b^{2}+c^{2}}\geq \frac{3}{2}$
Of course, with the same conditions: $a,b,c>0$ and $a+b+c=3$

[quote="Zamfirmihai"] Is the following true?
$\sum_{cyc}\frac{a}{b^{2}+c^{2}}\geq \frac{3}{2}$
Of course, with the same conditions: $a,b,c>0$ and $a+b+c=3$[/quote]
It's not true. Try $c\rightarrow0^{+}$ and $a=b\rightarrow1.5^{-}.$ :wink:
But if $a^{2}+b^{2}+c^{2}=3$ ( $a,$ $b$ and $c$ are still positive numbers ) then it's true. :)


Let $ \overline{MN}$ be a diameter of a circle with diameter 1. Let $ A$ and $ B$ be points on one of the semicircular arcs determined by $ \overline{MN}$ such that $ A$ is the midpoint of the semicircle and $ MB\equal{}\frac{3}5$. Point $ C$ lies on the other semicircular arc. Let $ d$ be the length of the line segment whose endpoints are the intersections of diameter $ \overline{MN}$ with chords $ \overline{AC}$ and $ \overline{BC}$. The largest possible value of $ d$ can be written in the form $ r\minus{}s\sqrt{t}$, where $ r, s$ and $ t$ are positive integers and $ t$ is not divisible by the square of any prime. Find $ r\plus{}s\plus{}t$.

I think this will help: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1455383#1455383

 :)


[quote="MellowMelon"]There's a bajillion such functions. Pick any subset $ S$ of the real numbers including $ 36, 47, \pi/3$. For $ x \in S$ let $ f(x) \equal{} x$ and for $ x \in \mathbb{R} \minus{} S$ let $ f(x)$ be any element of $ S$. This works and happens to be all of them.[/quote]
Close, but you missed one. There's actually a bajillion and one.

The wording of the problem actually said find [b]a[/b] function, sorry. I found the solution that you have but I tryed to find a more interesting solution and couldn't. I don't know if a more interesting solution exists but this problem was marked with a star (that means difficult) in Spivak's Calculus, so I have reason to believe that there is. It doesn't require anything about precalculus to solve because it was in the a chapter before limits are even introduced.


geometry
induction
combinatorics proposed
combinatorics

Given a convex polyhedron with an even number of edges.

Prove that we can attach an arrow to each edge, such that for every vertex of the polyhedron, the number of the arrows ending in this vertex is even.

We prove by induction on $n$, the number of vertices of the polyhedron, that we can orient the edges such that at most one vertex has odd in-degree, and this vertex can be chosen arbitrarily. The base cases (for polyhedra with few vertices) are clear, and assume we've shown this for polyhedra with $n-1$ vertices.

Pick the vertex $x$ which we allow to have an odd in-degree. After its elimination, what's left is still a combinatorial polyhedron. According to the induction hypothesis, we can orient its edges such that at most one vertex has odd in-degree, and this vertex is among those adjacent to $x$ in the original polyhedron. Now add $x$ back, and orient the edges incident with it such that the in-degrees of all the neighbours of $x$ become even. Only $x$ can have an odd in-degree, and we're done.

Now, for our purposes, all this is good enough, since if in a directed graph with an even number of edges all the vertices except maybe for one have even in-degrees, then all the vertices must have even in-degrees, because the sum of all the in-degrees is the number of arcs.

My solution original was the following:
Orientate the edges by random.
By simple summation over the number of incoming edges we see that the number of vertices with odd incoming degree is even.
So when there is one ($A$), there's another one ($B$). Now choose two of them and 'connect' them by taking a path of (nondirected) edges from $A$ to $B$ and invert all arrows on that path:
the parity of $A$ and $B$ will change (thus get even), the others will not change. So the number of 'odd vertices' reduces by two, and reaches zero after finitely many such steps.


Zic problema din minte, dar in mare o sa sune la fel :

Fie un dreptunghi de dimensiuni $2m$ si $2n+1$ care este impartit in patrate unitate. Se cere numarul $K$ maxim pentru care : oricum am "scoate" $K$ patrate unitate din dreptunghi in regiunea ramasa sa se poata inscrie cel putin un disc de diametru $1,1$.

[i]Dan Schwarz, Bucuresti[/i]

nu cumva trebuia zis "numarul $K$ minim"? pt ca daca scoatem *toate* patratele unitate, atunci sigur ca putem inscrie un disc de diametru 1.1 in regiunea ramasa (decat intr-un singur caz: in care dreptunghiul e 2 pe 1.)

nu, deci se cere numarul $K$ maxim, pentru care oricum am elimina $K$ patrate unitate sa se poata inscrie un disc de diametru $1,1$ in regiunea ramasa [i]formata de patrate unitate[/i], sry ca nu m-am exprimat clar initial

un disc de diametru 1,1 (adica 11/10)... un "disc" e un cerc impreuna cu interiorul lui...

se observa ca un disc de diametru 1,1 se poate inscrie intr-o asemenea regiune daca si numai daca se poate inscrie un "L" (format din trei patrate unitate) in regiunea aia. deci o regiune in care *nu* se poate inscrie un disc de diametru 1,1 e una formata din "dungi" (adica regiuni $1\times n$, $n\in \mathbb{N}$, de patrate unitate), in care patratele unitate din doua dungi distincte n-au laturi comune. aria maxima unei regiuni formate numai din dungi am impresia ca e $2m(n+1)=2mn+2m$. (de exemplu, in cazul in care se iau dungi paralele cu latura de lungime $2m$. daca luam dungi paralele cu latura $2n+1$, obtinem o arie mai mica. asta inca nu am putut sa demonstrez cu precizie, dar cred ca asa e.) numarul $K$ maxim ar rezulta atunci ca e $2m(2n+1)-[2mn+2m+1]=2mn-1$.

pohoatza, stii cumva raspunsul?

asta e raspunsul pleurestique . O sa atasez eu subiectele si baremele .

Raspunsul corect e mn-1.Ideaa de baza este sa aratam ca pt k maxim pe tabla va ramane un patrat 2X2.Artam ca k<mn. Impartim tabla in patrate 2X2 astfel incat ultima linie sa fie de forma 1X2m.Notam coltul de sus (de unde se incepe impartirea) cu cifra 1, cifra de langa cu 2 ,cifra de sub 2 cu 3 iar cifra de sub 1 cu 4.Pe acelasi principiu notam celelalte patrate 2X2.Daca indepartam patratele ce au inscrise cifra 3 vom indeparta in total mn patrate.Se observa cu usurinta ca in regiune ramasa nu se poate inscrie un cerc de diametru 1,1.Deci k<mn.Aratam ca K=mn-1 convine.Din Principiul Cutiei pastrand aceeasi idee de eliminare a patratelelor ce sunt notate cu 3, va ramane un patrat 2X2.De aici concluzia problemei.

ba nu! nu e adevarat ca nu se poate inscrie un disc de diametru 1.1 in regiunea ramasa care ati descris-o....nu e nevoie de un patrat 2x2 ca sa putem inscrie discul. e suficient (si de fapt necesar) sa avem un "L" format de trei patrate unitate, si in regiunea descrisa de dv. sunt de fapt multe L-uri.

:huh:  Cred ca este putin probabil sa se poata inscrie un cerc de diametru 1,1 intr-o configuratie in forma de L alcatuita din 3 patratele unitate.

Am discutat azi problema la un seminar pentru valori generale m si n ale dimensiunilor dreptunghiului. Problema se complica un pic dar devine mult mai interesanta. Astept solutiile voastre pentru varianta generala.


Let $ a,b,c>0,m,n>0,k\geq{1},$prove that

         \[ \frac{a^m\plus{}b^m\minus{}2c^m}{a^n\plus{}b^n\plus{}kc^n}\plus{}
           \frac{b^m\plus{}c^m\minus{}2a^m}{b^n\plus{}c^n\plus{}ka^n}
           \plus{}\frac{c^m\plus{}a^m\minus{}2b^m}{c^n\plus{}a^n\plus{}kb^n}\ge0.\]

[quote="fjwxcsl"]Let $ a,b,c > 0,m,n > 0,k\geq{1},$prove that
\[ \frac {a^m \plus{} b^m \minus{} 2c^m}{a^n \plus{} b^n \plus{} kc^n} \plus{} \frac {b^m \plus{} c^m \minus{} 2a^m}{b^n \plus{} c^n \plus{} ka^n} \plus{} \frac {c^m \plus{} a^m \minus{} 2b^m}{c^n \plus{} a^n \plus{} kb^n}\ge0.
\]
[/quote]

It is equivalent :

$ \sum\frac{(k\minus{}1)(a^n\minus{}b^n)(a^m\minus{}b^m)}{(c^n\plus{}a^n\plus{}kb^n)(b^n\plus{}c^n\plus{}ka^n)} \geq 0$

Done ! :)

Suppose $ a\geq b\geq c\Rightarrow a^m\plus{}b^m\minus{}2c^m\geq c^m\plus{}a^m\minus{}2b^m\geq b^m\plus{}c^m\minus{}2a^m$
                                                   $ a^n\plus{}b^n\plus{}kc^n\leq c^n\plus{}a^n\plus{}kb^n\leq b^n\plus{}c^n\plus{}ka^n$
$ \Rightarrow LHS\geq\frac{1}{3}\sum(a^m\plus{}b^m\minus{}2c^m)\sum\frac{1}{a^n\plus{}b^n\plus{}kc^n}\equal{}0$


Can someone explain me what "Sturm's method" to prove inequalities is,
or give an appropriate link ?

It is a variational method. For example if a+b=const then its product ab as bigger as a and b closer

to give a simple example say you want to prove that a^2 + b^2 \geq 1/2 if a+b=1.

Then you notice that if a+b=1, and a<1/2<b, then by replacing the pair (a,b) with (a',b') with a<a'<1/2<b'<b, a'+b'=1 we observe that a^2+b^2 \geq a'^2+b'^2 (computation) and thus by replacing (a',b') with a pair of smaller difference between the terms you can shirnk the inequality.

So you can consider that a=b=1/2 and you obtain the inequality.
Usually it is hard to apply Sturm, and where is Sturm, you can apply an easier inequality to solve the problem.

Also, in a contest, it is very likely not to receive any points for a incomplete Sturm approach of a inequality, so it's better to leave this for the general knowledge (and not apply it in an olympiad :D - well that's my opinion at least)

So IMO 1999/2 can be done with Sturm's method (a sort of at least) ?
(Check the second solution at http://www.kalva.demon.co.uk/imo/isoln/isoln992.html )

I found something about it in one of the romanian selection tests.
(The third one of 2002.) That's why I was asking for it.

imo 1999/2 also has a very nice proof using hardy-littlewood-polya majorization inequality.it's a very cute problem

guess where did it get proposed .... yes you are right - in Romania :D :D


Just wondering... what rating do you start with?

What do you mean, it might change? You start at 1200 everytime you make a new account

Hello,
please do not revive a dead thread :wink:. This hasn't been posted in for several years. It's okay if you revive a thread if you have a good solution, answer, etc.

@Mrdavid445,
I think BOGTRO means it will change due to you winning, loosing games.

At that time (Almost 3.5 years ago!), FTW was just beginning (sigh nostalgia). Many things about the future of FTW, including the starting rating, were unclear at the time.

Are you sure this is in the right forum?
Moderators, please move this topic.

[quote="BOGTRO"]Again, at the time, IIRC, there was only one FTW forum. It was later divided up into the 3 sub-forums we have now. It's best to just ignore this thread, it's 3.5 years old.[/quote]

Oh sorry my bad.
:blush:


Prove that, for all positive integers $ n$,

$ ((n)(n \plus{} 1)(\sqrt {\sin^3 1^\circ} \plus{} \sqrt {\sin^3 2^\circ} \plus{} ... \plus{} \sqrt {\sin^3 n^\circ)}/2)^{\frac {2}{3}} \ge n \sqrt [n]{n! \sin 1^\circ \sin 2^\circ \sin 3^\circ ... \sin n^\circ)}$

By the way, I haven't tried solving this in an easy way, so I'm not sure if it's triv.

[hide="Step 1"]

$ n(n + 1)/2 = 1 + 2 + ... + n$

[hide="Step 2"]

$ (1 + 2 + ... + n)^2 = (1^3 + 2^3 + ... + n^3)$

[hide="Step 3"]

By Holder's Inequality:

$ (1^3 + 2^3 + ... + n^3)^{1/3}((\sin 1^\circ)^{3/2} + (\sin 2^\circ)^{3/2} + ... + (\sin n^\circ)^{3/2})^{2/3} \ge (\sin 1^\circ + 2 \sin 2^\circ + ... + n \sin n^\circ)$

By AM-GM:

$ (\sin 1^\circ + 2 \sin 2^\circ + ... + n \sin n^\circ)/n \ge \sqrt [n]{n! \sin 1^\circ \sin 2^\circ ... \sin n^\circ}$

[/hide]

[/hide]

[/hide]

[hide=" :) "]

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[/hide]

Very nice solution!

If this is your own, made-up inequality, I would suggest replacing the sines with something else.  First of all, sin(n) could be negative, which messes up the problem completely.

The solution given works if sin(n) is replaced by any positive function of n.  You could either replace sin(n) by arbitrary $ a_1,a_2,...,a_n > 0$, showcasing the problem's generality, or you could try to trick the problem solver into thinking the exact nature of the $ a_n$ is important.  In the later case probably you should try to make $ \frac {\sqrt{a_n}}{n}$ nearly equal to $ 1$, because this makes the inequality sharp.


I'm not sure if I'm able to post here my question, but I don't know where else.

I've searched in the net about a pdf or djvu copy of the Titu book - "101 Problems in Algebra from the Training of the US IMO Team", but couldn't find. If someone know there is such a thing, can he give a link, please?  :blush:

It's a violation of copyrights. So, even if you do want the book, keep your request private. Remember the recent furor about Pirate Bay?


Let $ ABC$ be a triangle,$ D$ is midpoint of $ AB$.The excircle $ (O)$ opposite the angle $ A$ touch $ AB$ at $ M$.Let$ I$ is midpoint of$ MC$.Prove that $ D,I,O$ are collinear

Dear mathlinkers,
this problem is an extraversion of this one whixh is well knowed:
let ABC be a triangle, A' the midpoint of BC, I the incenter and D the point of contact of the incircle and BC.
The line A'I pass through the midpoint of AD.
Sincerely
Jean-Louis


In the convex quadrilateral $ABCD$, $AB$ and $CD$ meet at $P$,$BC$ and $AD$ meet at $Q$.The point $O$ is in the interior of $ABCD$ such that $\angle BOP=\angle DOQ$.Prove that $\angle BOA+\angle COD=180^0$

Using directed angles modulo 180°, this is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=15103 part (b).

  Darij


[quote="hanschinaman"]Prove $((a+b)(b+c)(c+d)(d+a))^{3}\geq 16(abcd)^{2}(a+b+c+d)^{4}$, where $a,b,c,d$ are positive real numbers.[/quote]
$(a+b)(b+c)(c+d)(d+a)\geq(a+b+c+d)(abc+abd+acd+bcd)\Leftrightarrow(ac-bd)^{2}\geq0.$
Hence, remain to prove that $(abc+abd+acd+bcd)^{3}\geq16a^{2}b^{2}c^{2}d^{2}(a+b+c+d).$
Let $abc=x,$ $abd=y,$ $acd=z$ and $bcd=t.$
Then  $(abc+abd+acd+bcd)^{3}\geq16a^{2}b^{2}c^{2}d^{2}(a+b+c+d)\Leftrightarrow$
$\Leftrightarrow(x+y+z+t)^{3}\geq16(xyz+xyt+xzt+yzt),$which true by Muirhead.


function
number theory unsolved
number theory

Show that a^phi(n) = 1 (mod n) for any integers a and n with (a,n) = 1
Is this a well known theorem?

Very well known (Euler's theorem, or Euler's Totient theorem)

Let $\{a_{1},a_{2},a_{\phi(n)}\}$ be the invertible elements of $\overline{\{1,n\}}$.  Say that $a$ is invertible.

We claim that $\{a_{1},a_{2},a_{\phi(n)}\}=\{aa_{1},aa_{2},aa_{\phi(n)}\}$

The invertibility of all elements in the second set is trivial - to prove uniqueness, if we have $aa_i=aa_j$, then $a(a_i-a_j)=0\iff a_i-a_j=0\iff a_i=a_j$

Now multiply everything together.  We get $a^{\phi(n)}\prod_i a_i=\prod_i a_i$ and since the product of invertible elements is invertible, we may multiply through by the inverse to get $a^{\phi(n)}=1$.

There is also a group theorethic proof if you want...


(i) (easy): AB is the diameter of a circle S with centre O. We construct C(0) to be tangent to AB at O and also internally tangent to S. Circle C1 is constructed to be tangent to AB, internelly tangent to S and externally tangent to C0. If AB=4, find the radius of C(1).
(ii) Construct an infinite sequence C(2), C(3), ... such that C(n) is tangent to AB, is internally tangent to S and externally tangent to C(n-1). For each n, find a closed expression for the radius of C(n).
for (i) the answer is 1/2 (easy!).
for (ii) i have the following recursion (which i cant solve):
C(n+1) = (3*C(n) - 2*C(n)^2 - 2*C(n)*:sqrt: (2-4*C(n)) )/ (2*C(n) + 1)^2.

Im getting \[ C(n) = \frac{2}{2+(1+\sqrt 2)^{2(n-1)}+(1-\sqrt 2)^{2(n-1)}} \]
with the recurrence you gave. Induction should work.


[url=http://www.mathlinks.ro/Forum/download.php?id=1123]PDF Version[/url]

1. Let $n$ be a positive integer and let $a_n$ be the number of
ways to write $n$ as a sum of positive integers, such that any two
summands differ by at least 2. Also, let $b_n$ be the number of
ways to write $n$ as a sum of positive integers of the form $5k\pm
1$, $k \in \mathbb{Z}$.

Prove that $\displaystyle \frac {a_n}{b_n}$ is a constant for all
positive integers $n$.

2. Let $ABCD$ be a convex quadrilateral, and let $K$ be a point
on side $AB$ such that $\angle KDA=\angle BCD$. Let $L$ be a point
on the diagonal $AC$ such that $KL \parallel BC$.

Prove that $\angle KDB=\angle LDC$.

3. The sequence $\{x_n\}_n$ is defined as follows: $x_1=0$, and
for all $n\geq 1$
 \[(n+1)^3x_{n+1}=2n^2(2n+1)x_n+2(3n+1). \]
Prove that $\{x_n\}_n$ contains infinitely many integer numbers.

[quote="siuhochung"][quote="Valentin Vornicu"]such that any two
summands differ by at least 2.[/quote]

Sorry, what does this mean?[/quote]It means that any terms in the sum are different by at least 2. For example $10=1+3+6$ could contain such summands, however $10=1+4+5$ could not, because the difference between 4 and 5 is less than 2, so only the first decomposition is counted in $a_n$.

[quote="Armo"]Can I post my solution here?  :?[/quote]NO!   :o
It's strictly against regulations and good spirit.  Only submit solutions to valentin@mathlinks.ro in an editable format (for example .tex, .doc).

Do solutions have to be in by the 19th (as stated on the PDF version) or by the 24th (2 weeks after posing), or by some other date?

Solutions are due this Sunday, namely 19th.

I want solution this as following
Let [tex] y_{n}=n^2x_{n}+2[/tex]
We have [tex]y_{1}=2 [/tex]and [tex]y_{n+1}=2y_{n}(2n+1)/(n+1)[/tex]
from that  [tex]y_{n}=2^n .(2n+1)!!/n![/tex ]
and now using theory of Willson with[tex] n=p\in P[/tex]


LaTeX
symmetry
induction
algebra
binomial theorem
Pascal\u0027s Triangle

The following problems are a bit troubling for me (They are all from ch.16 end of chapter AoPs Volume 2):

problem 271 and 272. I am sorry for not typing these up because I don't have access to LaTeX at this time. I have read the solutions, but they make no sense to me :huh: . It would be helpful if any of you can post your original solutions to these problems or even [b]explain in detail[/b] the solutions provided the solutions manual.

272. Find the coefficient of the fourth term of $(1-2x)^\frac{1}{3}$.

[hide]Apply the binomial theorem:

$\displaystyle{\frac{1}{3} \choose 3}(1)^\frac{-26}{3}(-2x)^3$

To evaluate $\displaystyle{\frac{1}{3} \choose 3}$ we multiply:

$\frac{(\frac{1}{3})(\frac{-2}{3})(\frac{-5}{3})}{(3)(2)(1)}$

After multiplying everything together, we are left with:

$\boxed{\frac{-40}{81}}$[/hide]

:?
How did you go from \displaystyle{\frac{1}{3} \choose 3}
to \frac{(\frac{1}{3})(\frac{-2}{3})(\frac{-5}{3})}{(3)(2)(1)}?

I tried the pascal's triangle but it didn't work.

You have to put $\$$ signs around it to make it display, haha.

Anyway, a different definition of combinations is needed to evaluate those containing non-integer or negative terms.

I'll try to explain it with just words:

1)  Start with $\frac{1}{3}$ and keep multiplying by the number that is exactly one less than the number you started with.  Do this three times (becausing you are choosing 3)

2)  For the denominator, simply do $3!$

Then evaluate.

Also, I think it would be easier to understand if you looked at the explaination and the solution at the same time to make sure you follow all the steps.

[quote="Inspired By Nature"]:?
How did you go from $\displaystyle{\frac{1}{3} \choose 3}$
to $\frac{(\frac{1}{3})(\frac{-2}{3})(\frac{-5}{3})}{(3)(2)(1)}$?

I tried the pascal's triangle but it didn't work.[/quote]

Recall that ${n\choose{m}}=\frac{n(n-1)(n-2)...(n-k+1)}{m!}$ if m is a whole number.

271. Define two sequences of rational numbers as folows: let $a_0 = 2$ and $b_0 = 3$, and recursively define $a_n = \frac{a_{n-1}^2}{b_{n-1}}$ and $b_n=\frac{b_{n-1}^2}{a_{n-1}}$. Find $b_8$, leaving your answer in the exponential form $m^n/p^q$.

It can be shown that all $a_n$ can be expressed as $\frac{2^x}{3^y}$.  Because of the symmetry they share, we can expressed $b_n$ as $\frac{3^x}{2^y}$.  Let $x_n$ and $y_n$ be the series of $x$ and $y$ in $a_n$ and $b_n$.  We see that $x_{n+1} = 2x_n + y_n$ and $y_{n+1} = 2y_n + x_n$.  Since $x_0 = y_0 + 1$, we use induction to show that $x_n = y_n + 1$.  Then $x_{n+1} = 3x_n - 1 \implies x_n = \frac{3^n + 1}{2}$.  So $x_8 = \frac{3^8 + 1}{2} = 3281 \implies y_8 = 3280$.  Then $b_8 = \frac{3^{3281}}{2^{3280}}$

:lol:
Thanks a lot you guys, that really helped. Feel free to post your sequences questions here.


Let $ 0\le x\le 1$ .Find the maximum of

   \[ 13\sqrt{x^2\minus{}x^4}\plus{}9\sqrt{x^2\plus{}x^4}\]

hello, it is $ 13\sqrt{x^2\minus{}x^4}\plus{}9\sqrt{x^2\plus{}x^4} \leq 16$ and will be obtained
at $ x\equal{}\frac{2}{5}\sqrt{5}$.
Sonnhard.


algebra
polynomial
number theory unsolved
number theory

Determine all polynomial $P(x)\in R[x]$ fulfill the following property:
For all the polynomial $Q(x)\in R[x]$ ,degQ≤degP and [P(x)]=[Q(x)] implying P(x)≡Q(x)...

Still nothing!!!  ;) .I read it in a notebook without solution and I have no any idea.And you??

Id try, but im not fimilar with the notation!


Let $\{a_{n},\, n\ge1\}$ be any sequence of positive numbers. Denote by $b_{n}$ the number of terms $a_{k}$ such that $a_{k}\ge\frac{1}{n}.$ Prove that at least one of the series $\mathop{\sum}_{n=1}^{\infty}a_{n}$ and $\mathop{\sum}_{n=1}^{\infty}\frac{1}{b_{n}}$ is divergent.
(G. Shevchenko)

If $\sum_{n=1}^\infty a_{n}$ is convergent.\
assume $\sum_{n=1}^\infty a_{n}<M$
Then,$b_{n}\le Mn$
So $\sum_{n=1}^\infty b_{n}\ge \sum_{n=1}^\infty \frac 1{Mn}$
is divergent.

Assume convergence for a_n.
Since this sequence is made of positive numbers and assuming cauchy criteria (a_n->0), i will considere the "worst case", which is that if a_n is a sequence with some a_(n+k) > a_n. The numbers satisfying this condition mut be finite, if not, we would have a sum arbitrarly large of a_(n+k)>a_n and this would break down the cauchy criteria ( lim a_n-->0) so k is a finite number. If this property follow some period, we will take k'={max(k_1,K-2...,K_n}
Now we just use comparison:
$\sum \frac{1}{b_{n}}\geq \sum \frac{1}{n+k'}$which is the translated harmonic series.


a point P is chosen at random in the coordinate plane.  What is the probability that the unit circle with center P contains exactly two lattice points in its interior?

The book's solution i couldn't understand, but from my understanding the probability should be zero.

[quote="kysuke"]It's not zero.  You can easily translate the unit circle so that it includes 2 lattice points.[/quote]

my bad, i mean it should be 100%

The circle may also contain three or four lattice points. I believe the answer is $ 4 \minus{} \frac {2\pi}{3} \minus{} \sqrt {3}$.

[hide]
Consider the unit lattice point square containing $ P$. Then the only lattice points the unit circle about $ P$ can contain are the four vertices of that square. We do not want $ P$ to be within distance of $ 1$ from any $ 3$ of the vertices.

[asy]pathpen=linewidth(0.7);size(100);pen g=rgb(0.6,0.6,0.6),d=linetype("4 4");path p=arc((1,1),1,180,270)--arc((0,0),1,0,90)--cycle,q=arc((0,1),1,270,360)--arc((1,0),1,90,180)--cycle;D(unitsquare);fill(p,g);fill(q,g);D(p,d);D(q,d);[/asy]

So it must be in one of those white regions above. It's now just standard sectors.
[/hide]

But still how did you calculate the unshaded area of the graph?

start by finding the area of 2 "footballs" (the complete shaded area * 2)
and subtract the middle section (of which 1 fourth of it is shaded faintly gray bordered by points 1, 2, and 3)
find this by finding the area of the sector minus the darkest gray area ( the quadrilateral bordered by points 1, 2, 3, and 4 of which the lower right angle is 15 degrees)

Let's focus on a single region. edit- beaten;
[asy]pathpen=linewidth(0.7);size(100);pen g=rgb(0.6,0.6,0.6),r=rgb(1,0.2,0.2)+linewidth(0.7),d=linetype("4 4");pair P=(1-3^.5/2,1/2);path p=arc((1,1),1,180,270)--arc((0,0),1,0,90)--cycle,q=arc((0,1),1,270,360)--arc((1,0),1,90,180)--cycle;D(unitsquare);fill((0,0)--(.5,.5)--(0,1)--(1,1)--(1,0)--cycle,rgb(0.8,0.8,0.8));fill(p,g);fill(q,g);D(p,d);D(q,d);D((1,0)--P);D(P--(1,1/2)--(1,0)--(0,0),r);D(arc((1,0),1,150,180),r);D(arc((1,0),1,90,150));MP("1/2",(1,1/4),E);MP("1",(P+(1,0))/2,S);[/asy]

The area of [the bottom] half of the shaded region is the sum of the areas of the $ 30 \minus{} 60 \minus{} 90$ triangle and the $ 30^{\circ}$ sector, which is $ A \equal{} \frac {1}{2} \cdot \frac 12 \cdot \frac {\sqrt {3}}2 \plus{} \frac 1{12} \pi$. The area of the unshaded region is $ 1 \minus{} 2A \equal{} 1 \minus{} \frac {\sqrt {3}}{4} \minus{} \frac {\pi}{6}$; multiplying by four gives the answer (since the probability is over the area of each unit square, which is $ 1$).


[quote="mysteryfigure"]ABCDE is a patogon, connect A with B,C,D,E     B with C,D,E    C with D,E  & D with E, so that a "star" is made.The area of the five protruding triangles of the star are all 1. please give the total area of the patogon.

PS: is there anyone who can tell me how to upload a graph that was drawn by myself onto Mathlinks Forum?[/quote]

A patogon?? :) Sorry, I couldn't avoid imagining some kind of long time extinct paleocenic mammal :)

Convert your graph to jpg, don't know what are the other extensions allowed, and just click the Add Attachment button below your message.

Daniel

sorry for my poor english.the "patogon" should be "pentagon".so the problem should be put in this way:

ABCDE is a pentagon, connect A with B,C,D,E B with C,D,E C with D,E & D with E, so that a "star" is made.The area of the five protruding triangles of the star are all 1. please give the total area of the pentagon.

hint: 我想应该用相似三角形+4点共圆来解决。

You wrote in the figure that all the sides of the pentagonon $FGHIJ$  are  1 ? :?

I mean the area of the triangles AFG,BFJ,CIJ,IDH,GHE are 1. :)


probability
geometry
inequalities
analytic geometry

Two real numbers are chosen at random between 0 and 10, inclusive. What is the probability that they are within one unit of each other? (Round to the nearest hundredth)
I know the answer, but I need a solution.

[hide="Answer"].19[/hide]

My solution

[hide]
Since the reals are continuous, we can use geometry to solve this one.

Consider the area $S$ of the Cartesian plane bounded by the points $(0,0)(10,0)(10,10)(0,10)$.  Then, let $x$ be the first real we choose, and let $y$ be the second real.  From the problem, we have two inequalities:

$y \le x + 1$
$y \ge x - 1$

We graph these inequalities.  Let the region of the plane that satisfies these inequalities be denoted $A$.

The probability is thus $\frac{S \cap A}{S}$, where $S \cap A$ is the region of the plane that satisfies the inequalities and is within $A$.

Numerically, this comes to [b]19/100[/b]

[/hide]

[hide="note"]If you have the second volume of the AoPS book, you can use geometry to solve this problem. I do not want to illustrate, but the answer does come out to 19/100, [/hide]

That's essentially where I learned to do this sort of stuff as well.  Really useful books!


trigonometry
geometry
circumcircle
incenter
geometric transformation
reflection
homothety

Given triangle ABC with AB>BC>CA. CD is a heigth, M and N are the middle points of AB, AC. P and Q are tangent points of inscribed circle with AB, AC respectively. Prove that the center of the inscribed circle into triangle DMN lies on line PQ.

 :alien:

[img]http://img263.imageshack.us/img263/8977/30936818ou6.png[/img]
The problem is easy,but i wasn't able to find a beautiful proof :(
Solution:
Let $ \angle BAC \equal{} 2\alpha,\angle ABC \equal{} 2\beta$,$ BC \equal{} a,AC \equal{} b,AB \equal{} c,s \equal{} \frac {a \plus{} b \plus{} c}{2}$.
Then $ \boxed{MP} \equal{} MA \minus{} AP \equal{} \frac {c}{2} \minus{} \frac {b \plus{} c \minus{} a}{2} \equal{} \boxed{\frac {a \minus{} b}{2}}$ and $ \angle{MDN} \equal{} \pi \minus{} 2\alpha$.
Since $ AP \equal{} AN$,then $ \angle DPQ \equal{} \frac {\pi}{2} \minus{} \alpha$,hence if $ J$ is the point of intersection of the line $ PQ$ and the angle bisector of $ \angle MDN$,then
$ \boxed{DJ \equal{} JP}$.
It is obvious that $ AD \equal{} b\cos{2\alpha}$,hence $ PD \equal{} s \minus{} a \minus{} b\cos{2\alpha}$.Let $ K$ be the midpoint of $ PD$,then $ PK \equal{} \frac {s \minus{} a \minus{} b\cos{2\alpha}}{2}$,and
$ \sin{\alpha} \equal{} \frac {PK}{PJ}$,hence $ \boxed{PJ \equal{} \frac {s \minus{} a \minus{} b\cos{2\alpha}}{2\sin{\alpha}}}$.
Let $ J'$ be the intersection point of the line $ PQ$ and the angle bisector of $ \angle DMN$.
It is obvious,that $ \angle DMJ' \equal{} \beta$ and $ \angle PJ'M \equal{} \frac {\pi}{2} \minus{} \alpha \minus{} \beta$.
From the sine law:
$ \frac {MP}{\cos{\alpha \plus{} \beta}} \equal{} \frac {PJ'}{\sin{\beta}}$.So
$ \boxed{PJ' \equal{} \frac {(a \minus{} b)\sin{\beta}}{2(\cos{(\alpha \plus{} \beta)})}}$.
Now if we prove that $ PJ' \equal{} PJ$,we will be done,or
$ \boxed{\frac {(a \minus{} b)\sin{\beta}}{2(\cos{(\alpha \plus{} \beta)})} \equal{} \frac {s \minus{} a \minus{} b\cos{2\alpha}}{2\sin{\alpha}}}$
The last equation can be proved for example,by substitution
$ a \equal{} 2R\sin{2\alpha},b \equal{} 2R\sin{2\beta},c \equal{} 2R\sin{(\pi \minus{} 2\alpha \minus{} 2\beta)}$,where is the radius of circumcircle of the $ \triangle ABC$.
Then we will obtain not hard trigonometrical equation.
Q.E.D :)
P.S:
If someone wants i could explain the last step in details.
P.S.S:
If anybody knows a nice proof,without trigonometry,please post it.Thank you.

Denote $ a \equal{} BC, b \equal{} CA, a \equal{} BC$ and $ s$ the semiperimeter of a $ \triangle ABC.$ Let $ (I)$ be the incircle and $ (I_b)$ the excircle in the angle $ \angle B.$ Let $ P, P_b$ the tangency points of $ (I), (I_b)$ with $ AB.$ Let perpendicular to the angle bisector $ AI$ through the incenter $ I$ cut $ AB$ at $ Z.$ Perpendicular to $ AI$ through the excenter $ I_b$ is the external bisector of the angle $ \angle A$ and it cuts $ AB$ at $ A.$ Since $ BA \equal{} c, BP \equal{} s \minus{} b, BP_b \equal{} s,$ it follows that

$ \frac {BZ}{s \minus{} b} \equal{} \frac {BZ}{BP} \equal{} \frac {BA}{BP_b} \equal{} \frac {c}{s}.$

Let $ D$ be the foot of the altitude from $ C$ and $ A'$ a reflection of $ A$ in $ CD$. Then $ A'A \equal{} 2b \cos A \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{c}$ and $ BA' \equal{} BA \minus{} A'A \equal{} \frac {a^2 \minus{} b^2}{c}.$ Semiperimeter of the $ \triangle A'BC$ is then $ s' \equal{} \frac {1}{2} \left(a \plus{} b \plus{} \frac {a^2 \minus{} b^2}{c}\right) \equal{} \frac{(a \plus{} b)(s \minus{} b)}{c}.$ Incircles $ (I'), (I)$ of the $ \triangle A'BC, \triangle ABC$ are centrally similar with center $ B$ and coefficient

$ \frac {r'}{r} \equal{} \frac {S(A'BC)}{S(ABC)} \cdot \frac {s}{s'} \equal{} \frac {a^2 \minus{} b^2}{c^2} \cdot \frac {cs}{(a \plus{} b)(s \minus{} b)} \equal{} \frac {s(a \minus{} b)}{c(s \minus{} b)}.$

Let perpendicular to the angle bisector $ AI$ through the incenter $ I'$ of the $ \triangle A'BC$ cut $ A'B \equiv AB$ at $ Z'$. Then

$ \frac {BZ'}{BZ} \equal{} \frac {r'}{r},\ \ BZ' \equal{} \frac {c(s \minus{} b)}{s} \cdot \frac {s(a \minus{} b)}{c(s \minus{} b)} \equal{} a \minus{} b,$

$ Z'A \equal{} BA \minus{} BZ' \equal{} c \minus{} (a \minus{} b) \equal{} 2(s \minus{} a) \equal{} 2\ PA.$

The $ \triangle AMN \sim \triangle ABC$ are centrally similar with center $ A$ and coefficient $ \frac {_1}{^2}$ and so are their incircles $ (J), (I).$ If $ D'$ is the foot altitude of the $ \triangle AMN$ through $ N,$ then $ D$ is a reflection of $ A$ in $ ND'.$  The $ \triangle DMN \sim \triangle A'BC$ are are also centrally similar with center $ A$ and coefficient $ \frac {_1}{^2}$ and so are their incircles $ (J'), (I').$ Applying the above result to the $ \triangle AMN, \triangle DMN$ then solves the problem.

[quote="Klaas"]Given triangle ABC with AB>BC>CA. CD is a heigth, M and N are the middle points of AB, AC. P and Q are tangent points of inscribed circle with AB, AC respectively. Prove that the center of the inscribed circle into triangle DMN lies on line PQ.[/quote]

I'm going to be brief here: Let E be the point on the line AB such that $ CE\equal{}CA$ (but $ E\neq A$, apart from the case when $ A\equal{}90^{\circ}$). According to [url=http://www.mathlinks.ro/viewtopic.php?t=149173]IMO Shortlist 2006 problem G4[/url], the line PQ bisects the segment AJ, where J is the incenter of triangle EBC. Now, the triangle EBC is the image of the triangle DMN under the homothety with center A and ratio 2. Thus, the midpoint of the segment AJ is the incenter of triangle DMN, and we are done.

  darij


$ f$ is a real-valued function on the reals such that $ f(x \plus{} 19)\leq f(x) \plus{} 19$ and $ f(x \plus{} 94)\geq f(x) \plus{} 94$ for all $ x$. Show that $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$ for all $ x$.

Repeatedly using

$ f(x \plus{} 19) \le f(x) \plus{} 19, f(x \plus{} 94) \ge f(x) \plus{} 94$

yields

$ f(x \plus{} 19\cdot 94) \le f(x) \plus{} 19 \cdot 94, f(x \plus{} 19\cdot 94) \ge f(x) \plus{} 19\cdot 94$

The equality occurs, and there must be

$ f(x \plus{} 19) \equal{} f(x), f(x \plus{} 94) \equal{} f(x)$

And so

$ f(x \plus{} 1) \equal{} f(x \plus{} 1 \plus{} 94) \equal{} f(x \plus{} 19\cdot 5) \equal{} f(x \plus{} 19\cdot 4) \equal{} \cdots \equal{} f(x)$

Anyway, the claim is also true if $ 19, 94$ is replaced by $ a,b$ which are coprime.


I am not quite sure how to solve this equation because of the use of fractions. I know how to solve them without fractions.

y =  1x/2  -	52

-18y = 910 - 8x

Like my topic title says: "System of Linear Equations (Elimination Method)".

You must be quidding me.... multiply the first equation by $18$.


While doing a calculus word problem (yes, i know calculus is not allowed in this forum but it's the algebra giving me trouble in this particular problem.), i came up with this following equation i had to solve after doing some calculus mumbo jumbo.

e^(t) - t = 100 (solve for t)

*e = euler's number. Sorry i can't get latex to work*

So anyways, that equation looks simple enough but is there anyway to solve it without using a graphing calculator? I used my TI-83 to plot the graphs of f(x) = e^x and g(x) = 100 + x and found the intersection point. Also, giving a more accurate answer, i used my TI-89 Titanium to numerically solve it.

Hi,
I think that there's NO other solution. I explain why:
It's a "transcendental" equation, not a polynomial one, like \[x^{2}-3x=7\] wich has solutions with radicals. :huh:

You could rearrange the terms to get:

$e^{t}=100+t$
$t=\ln(100+t)$
$t=\ln(100)+\ln(1+\frac{t}{100})$
$t=\ln(100)+\ln(1+\frac{\ln(100+t)}{100})$
$t=\ln(100)+\ln(1+\frac{\ln(100)+\ln(1+\frac{t}{100})}{100})$
$\vdots$

You can substitute and expand this forever, and use some kind of sums to actually find the solution, but that does not belong in High School Basics.

Moved to Intermediate Topics

Generally most arbitrary algebraic expressions don't have a closed-form solution.  Just use your calculator.


floor function
trigonometry
logarithms
LaTeX
function
number theory

[quote="black.phoenix"]I'd like to know if it's difficult to get a formula that gives the number of divisors a positive integer has.

Usually one writes the given number as a product of powers of primes and the number of divisors is the product of every exponent of every prime plus 1.

For example, the number 2004:

2004 | 2
1002 | 2
501   | 3
167   | 167
1

2004 = 2^2 * 3^1 * 167^1

This way, 2004 has (2+1)*(1+1)*(1+1) = 3*2*2 = 12 divisors.

d(2004) = 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002, 2004

Usually, if the representation of a number N in primes is N = (p1^e1)*(p2^e2)*...*(pn^en) so every exponent of every prime can assume the values ranging from 0 to ei, i = 1, 2, ..., n generating a divisor of N. The total number is (e1+1)*(e2+1)*...*(en+1)

For example:

3 = 2^0 * 3^1 * 167^0
12 = 2^2 * 3^1 * 167^0
668 = 2^2 * 3^0 * 167^1

My doubt is if there's a formula that given a number N it gives the number of divisors the number has.[/quote]

i don't really see the difference between the formula that is already established and the one that you are having trouble finding?

I think (s)he wants to be able to find the number of divisors without having to prime factorize it.

For all elementary number theory is concerned, you've pretty much figured most of it out, with your formula based on prime factorizations. For more information about how difficult factorizing (and thus getting divisors) can be for large numbers, look up the RSA cryptosystem and why it's successful.

[quote="black.phoenix"]My doubt is if there's a formula that given a number N it gives the number of divisors the number has.[/quote]

Clarify what you mean by a "formula."  The operation you've already described,

$ \sigma_0 \left( \prod_{k\equal{}1}^n p_k^{e_k} \right) \equal{} \prod_{k\equal{}1}^n (e_k \plus{} 1),$

is a formula; it's just not directly in terms of $ N$.  We could also say, for example,

$ \sigma_0(N) \equal{} \sum_{k\equal{}1}^{N} \left\lfloor \left| \cos \frac{N \pi}{k} \right| \right\rfloor$

which is a formula directly in terms of $ N$ that is simply difficult to evaluate.  (Evaluating it comes down to checking each number from $ 1$ to $ N$ to determine whether it is a factor of $ N$, which is worse than the previous approach.)

What you seem to want is a [b]closed[/b] formula.  A closed formula for $ \sigma_0(N)$ would be a fantastically difficult result, and there are plenty of reasons to suspect that such a formula does not exist, as MellowMelon mentions.  Among other things, it would imply a constant-time [url=http://en.wikipedia.org/wiki/Primality_test#Fast_deterministic_tests]primality test[/url].

[quote="black.phoenix"]Do you have the demonstration for this formula that uses cosine to find the number of divisors?

One more question. How do you generate these "formula images" inside the message?[/quote]

Just look carefully at the formula given by t0rajir0u : $ \sigma_0(N) \equal{} \sum_{k \equal{} 1}^{N} \left\lfloor \left| \cos \frac {N \pi}{k} \right| \right\rfloor$

Since $ \cos(x)\in[\minus{}1,\plus{}1]$, $ |\cos(x)|\in[0,1]$ and $ \lfloor|\cos(x)|\rfloor$ is $ 0$ except when $ \cos(x)\equal{}1$ or $ \cos(x)\equal{}\minus{}1$, hence when $ x\equal{}k\pi$

So the sum $ \sum_{k \equal{} 1}^{N} \left\lfloor \left| \cos \frac {N \pi}{k} \right| \right\rfloor$ just counts the cases where $ \frac {N \pi}{k}\equal{}n\pi$, so counts the cases where $ \frac{N}{k}\in\mathbb{N}$ and so just counts the divisors of $ N$.

Magic !

And useless ... !
 :lol:

T0r: why constant-time?  The calculation would still involve $ N$ and so would still grow as $ N$ grows.  (Addition and multiplication are not constant-time operations.)

[quote="Identity"]If there was it would be too easy to detect primes, yeah?[/quote]

"Detecting primes is in P" is a true (although not very good, in this case) formalization of the statement that detecting primes is easy.

It's clear that "factoring $ N$ can be done in time $ t(N)$" implies that "computing $ d$ can be done in time $ t(N) \plus{} O(\log N)$ (or something very close to that)" -- it's not immediately obvious that some sort of converse is true.  In particular, knowing how many divisors something has doesn't necessarily help you factor it at all, e.g. in the case of the product of two large primes.  That said, it does seem rather unlikely that the computation of $ d$ can be done any faster than factoring: heuristically, it's not at all clear how you would count the factors of a number without knowing what any of them are.

[quote="JBL"]It's clear that "factoring $ N$ can be done in time $ t(N)$" implies that "computing $ d$ can be done in time $ t(N) \plus{} O(\log N)$ (or something very close to that)" -- it's not immediately obvious that some sort of converse is true.  In particular, knowing how many divisors something has doesn't necessarily help you factor it at all, e.g. in the case of the product of two large primes.  That said, it does seem rather unlikely that the computation of $ d$ can be done any faster than factoring: heuristically, it's not at all clear how you would count the factors of a number without knowing what any of them are.[/quote]

Why $ t(N) \plus{} O(\log N)$ ?
$ t(N)$ is for factoring the number but what $ O(\log N)$ means?

[quote="black.phoenix"]One more question. How do you generate these "formula images" inside the message?[/quote]

Click on the "LaTeX" link on the left sidebar, and the tutorial will tell you how. ;)

$ O(f)$ (pronounced "big oh of $ f$") means, "some function that grows at most as quickly as a constant multiple of $ f$."  In this case, it just includes all of the arithmetic associated with calculating $ (a_1 \plus{} 1)\cdots(a_k \plus{} 1)$ once you've factored $ n \equal{} p_1^{a_1} \cdots p_k^{a_k}$.  (Each of the numbers $ a_i$ is at most $ \log_2 n$.)


John ordered $4$ pairs of black socks and some additional pairs of blue socks. The price of the black socks per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of the two colors had been interchanged. This increased the bill by $ 50\%$. The ratio of the number of pairs of black socks to the number of pairs of blue socks in the original order was:

$\textbf{(A)}\ 4:1 \qquad
\textbf{(B)}\ 2:1 \qquad
\textbf{(C)}\ 1:4 \qquad
\textbf{(D)}\ 1:2 \qquad
\textbf{(E)}\ 1:8$

I believe it is C, I would write the solution tomorrow

[hide]
set K as the cost of 1 pair of black socks
therefore, $ K/2$ is the cost of 1 pair of blue socks
set x as the # of blue sock pairs
then, set the equation as stated in the problem using the above variables
$ 1.5(4K\plus{}\frac x2K))\equal{}2K\plus{}Kx$
Simplifying, we get $ 6K\plus{}3K \frac x4\equal{}2K\plus{}Kx$
Solve for x (cancel the K's) to get $ x\equal{}16$
x is the number of blue sock pairs so $ 4: 16\equal{}1: 4$
So the answer is C[/hide]

Sorry, my fractions are screwed up though  :(


modular arithmetic
number theory proposed
number theory

Find all ordered triples of primes (p,q,r) such that p|(q^r)+1 , q|(r^p)+1 , r|(p^q)+1.

:(
Please,give me a solution in this topic or give me a link for the other topic  :)

I´m think that to solve use order and primitive roots.. but one case is hard to solve  :(

[quote="Pascual2005"]i am sure this one already appeared[/quote]

So am I. Actually, I remember posting sort of a solution, but it wasn't exactly complete, disregarding some cases (:?; I'm not entirely sure about that).

Assume none of the numbers is $2$. We then find the order of $q\pmod p$ to be either $2$ or $2r$.

[b]a)[/b] Assume first that it's $2$. We can't have $p|q-1$, so $p|q+1$. The order of $r\pmod q$ is either $2$ or $2p$, but it can't be $2p$, because then $p|2p|q-1$, and $p|q+1$, false, so the order is $2$. As above, we get $q|r+1$. In exactly the same way we get $r|p+1$. From here we get a contradiction, because the ordering of the numbers is absurd.

[b]b)[/b] We can now assume that the order of $q\pmod p$ is $2r$, the order of $r\pmod q$ is $2p$, and the order of $p\pmod r$ is $2q$. From here we get $2r|p-1,2p|q-1,2q|r-1$, and, again, we reach a contradiction.

We have thus shown that at least one of the numbers is $2$. Let $p=2$.

We have $q|r^2+1,r|2^q+1$. The order of $2\pmod r$ is either $2$ or $2q$. If it's $2$, then we get $r=3,q=5$. If, on the other hand, it's $2q$, we get $2q|r-1\Rightarrow q|r-1$, meaning that $q|r^2+1\equiv 2\pmod q$, so $q=2$, and $r=5$.

We have two solutions, up to cyclic permutations: $(p,q,r)$ is either $(2,5,3)$ or $(2,2,5)$.


calculus
integration
trigonometry
absolute value
calculus computations

Calculate the exact value of:

$\int^\pi_{0}{|\sin{x}+\cos{x}|}\,dx$

If you need a hint, it is here:
[hide="Hint"]
Express $\sin{x}+\cos{x}$ as $r\sin{(x+\alpha)}$ where $r, \alpha \in\Re$[/hide]

I've been a TA for second-quarter calculus several times now, and absolute values always throw students off. My (general) advice:
It's a piecewise defined function. Make that explicit, and split the interval where you switch formulas.

No solutions submitted yet? Follow the above post's advice: split the interval into sections, and make appropriate sign changes to get the true value. Also, a sketch is [b]very[/b] useful, in order to locate precisely where sign changes are required.  :ninja:

Well, I gues I'll post the solution (although it's pretty much been answered):

$\int_{0}^\pi |\sin x+\cos x| \ dx = \sqrt{2}\left(\int_{0}^{3\pi/4}\sin\left(x+\frac{\pi}{4}\right) \ dx-\int_{3\pi/4}^\pi \sin\left(x+\frac{\pi}{4}\right) \ dx \right) =$$2\sqrt{2}$

[quote="pkothari13"]Why did you split the integral up like that?[/quote]

[quote="jmerry"]It's a piecewise defined function. Make that explicit, and split the interval where you switch formulas.[/quote]

I apologize, I should have been more specific. Because there is an absolute value sign, shouldn't the two integrals be [b]added [/b]together, and not subtracted?

Yes, but for $\frac{3\pi}4\le x\le \pi,$ $\left|\sin\left(x+\frac{\pi}4\right)\right|=-\sin\left(x+\frac{\pi}4\right)$ and Jimmy simply factored that minus sign out of the integral.

[color=darkblue][size=134][b]You can gambol "seriously"[/b][/size] if you will use the following relations (you can prove easily it) :
$\blacktriangleright\ 1.\ \int_{a}^{b}f(x)\ dx=\int_{a}^{b}f(a+b-x)\ dx\ .$
$\blacktriangleright\ 2.\ \int_{a}^{b}f(x)\ dx=\int_{-\frac{b-a}{2}}^{\frac{b-a}{2}}f\left(\frac{a+b}{2}-x\right)\ dx\ .$
$\blacktriangleright\ 3.\ \int_{-c}^{c}f(x)\ dx=\int_{0}^{c}[f(x)+f(-x)]\ dx\ .$
Two particular cases :
$\blacktriangleright\ 3.3.\ \ \ \ f(-x)=f(x)\Longrightarrow$ $\int_{-c}^{c}f(x)\ dx=2\cdot\int_{0}^{c}f(x)\ dx\ .$
$\blacktriangleright\ 3.2.\ \ \ \ f(-x)=-f(x)\Longrightarrow$ $\int_{-c}^{c}f(x)\ dx=0\ .$
===============================================================================================================
$\int_{0}^{\pi}\left|\sin x+\cos x\right|\ dx=$ $\int_{0}^{\pi}\left|\sin x-\cos x\right|\ dx=$ $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left|\sin x-\cos x\right|\ dx=$ $\int_{0}^{\frac{\pi}{2}}\left[ \left| \sin x-\cos x\right|+\left|\sin x+\cos x\right|\right]\ dx=$ $2+\int_{0}^{\frac{\pi}{2}}\left|\sin x-\cos x\right|\ dx=$ $2+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left|\frac{\sqrt 2}{2}(\cos x-\sin x)-\frac{\sqrt 2}{2}(\cos x+\sin x)\right|\ dx=$ $2+\sqrt 2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left|\sin x\right|\ dx=$ $2+2\sqrt 2\int_{0}^{\frac{\pi}{4}}\sin x\ dx=$ $2+2\sqrt 2\cdot\left(-\frac{\sqrt 2}{2}+1\right)=2\sqrt 2\ .$[/color]


calculus
integration
derivative
trigonometry
calculus computations

how would i do this?

Find the derivative of the function.

g(x) =     integral sign    3x^2 (top)                      1/((7+t^4)^1/2)   dt
              tan(x) (bottum)

i am not sure what i supposed to do here, how do i get the anti D of this?

Do you mean the derivative wrt $ x$ of $ g(x) \equal{} \int^{3x^2}_{\tan x} \frac{1}{7 \plus{} t^4} \, dt$? The antiderivative is the integral

sorry but i edited the question after a bit , the 7+t^4 is in a root, so its 1/((7+t^4)^(1/2))

and for this i am pretty sure, i find the anti D, of it and plug in the 3x^2 in x, then tan x in x, and then subtract, but i cant get the anti D

Then $ g(x) \equal{} \int^{3x^2}_{\tan x} \frac {1}{\sqrt{7 \plus{} t^4}} \, dt$ and we are looking for $ \frac{dg}{dx}$.

Here I think you can just apply the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part]fundamental theorem of calculus[/url] and the answer is $ \frac{1}{\sqrt{7 \plus{} (3x^2)^4}} \minus{} \frac{1}{\sqrt{7 \plus{} \tan^4 x}}$.

Why do you need the antiderivative?

I think that you have to consider the inner derivatives as well, that is, multiply the first term in Aidan's response by $ 6x$ and the second one with $ 1/\cos(x)^2$.

thanks adrian, and then thanks homander, i got it now


function
real analysis
real analysis unsolved

Hello!
Let $V \in B(\mathbb{R}), V \not \approx \emptyset$ Is the function $f: R\to R$ $B(\mathbb{R})$-measurable if:

a)  $\{\begin{array}{cc}x \mbox ;\ \ x \in V \\ -x \mbox;\ \ x \not \in V \end{array}$

b) $\{\begin{array}{cc}1+x \mbox;\ \ x \in V \\ 3 \mbox;\ \ x \not \in V \end{array}$
?

Thanks for help!

${\bf Theorem}$ Let $A\cap B = \emptyset$ , $A,B \subset X$  and $f_1: A\to R$ , $f_2: B\to R$ and
$f(x)=\{\begin{array}{cc}f_1(x) \mbox ;\ \ x \in A \\ f_2(x) \mbox;\ \ x \in B \end{array}$
If $A,B,f_1,f_2$ - measurable then $f: A\cup B \to R$ - measurable.

It follows that the booth function in problem is measurable.

So the secend function is not $B(R)$ measurable?

[quote="anik_andrew"]I think it is measurable.
$1+x$ is measurable on $V$
$3$ is measurable on $CV$
and $V,CV$ is measurable...
So $f$ is measurable...[/quote]

I think that it it doesn't mean that $f$ is measurable.
I think that theorem is :
$f,g$ - measurable $\Rightarrow \ f+g$ - measurable
not
$f,g$ - measurable $<\ =\ >\ f+g$ - measurable

i dont understand what you talk about...
i writed another theorem

Ok, i understand your theorem. I wrote wrong. So function $a)$ is measurable to?


Hi,

When does the index come out? Is it different for A or B AMC tests? When do they send emails?

Thanks

The cutoffs certainly won't come out until after AIME-II.  After that, I seem to recall it being fairly quick (a week, maybe?), since there's not a lot of turnaround time between AIME-II and USAMO.

The cutoff may or may not be different from the A to the B tests, or from AIME I to II.  It depends on what sort of score distribution the committee sees.


[quote="Jericho kid"]When is the challnge set due? On my account it says that it is due on Mar 25, 2009 (Wed.) and thats not possible.[/quote]

belongs in your class message board


Diophantine equation
number theory proposed
number theory

Suppose that $x$ and $y$ are two natural numbers such that $3x^2 + x = 4y^2 +y$ . Prove that $x - y$ is the square of a whole number.

We can say that [tex]x>y[/tex] and so [tex]x-y=k[/tex] with [tex]k>0[/tex]. We rewrite the equation:
[tex]3(y+k)^2+(y+k)=4y^2+y[/tex]
[tex]y^2-6ky-(3k^2+k)=0[/tex]
[tex]y=3k+\sqrt {k(12k+1)} [/tex]

if y is a natural number then k(12k+1) must be a square, but GCD(k,12k+1)=1 and then k=x-y is a square.
If we want to find all the solutions we have to solve the pell equation [tex]a^2-12b^2=1[/tex] and the first couple is [tex]a=7[/tex] and [tex]b=2[/tex] then
[tex](x,y)=(4b^2+ab,3b^2+ab)=(30,26)[/tex]


Okay, this should be trivial for you guys...

Lets say that I have the recursive relation $ f(m \plus{} 1) \equal{} f(m) \plus{} m \plus{} 1$.
How do I yield $ f(m) \equal{} m(m \plus{} 1)/2$?

You need to be given an initial value, like $ f(0) \equal{} 0$. [hide="Hidden by mod"]Then you can verify the result by induction.[/hide]

[quote="MellowMelon"]You need to be given an initial value, like $ f(0) \equal{} 0$. Then you can verify the result by induction.[/quote]

sorry, :blush: i meant to have f(1)=1

[hide="Hidden by mod"]This is probably a little stretched but...

$ f(1) \equal{} 1$
so

$ f(2) \equal{} f(1) \plus{} 1 \plus{} 1$
$ f(3) \equal{} f(2) \plus{} 2 \plus{} 1$
$ f(4) \equal{} f(3) \plus{} 3 \plus{} 1$
...
$ f(1 \plus{} k \minus{} 1) \equal{} f(1 \plus{} k \minus{} 2) \plus{} 1 \plus{} k \minus{} 2 \plus{} 1$
$ f(1 \plus{} k) \equal{} f(1 \plus{} k \minus{} 1) \plus{} 1 \plus{} k \minus{} 1 \plus{} 1$

If you add them up

$ f(2) \plus{} f(3) \plus{} ... \plus{} f(1 \plus{} k) \equal{} f(1) \plus{} f(2) \plus{} ... \plus{} f(1 \plus{} k \minus{} 1) \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} ... \plus{} k \plus{} (1)(k)$
canceling

$ f(1 \plus{} k) \equal{} f(1) \plus{} \frac {(k \plus{} 1)(k)}{2} \plus{} k$
$ f(1 \plus{} k) \equal{} k \plus{} 1 \plus{} \frac {(k \plus{} 1)(k)}{2}$
substitute like $ m \equal{} k \plus{} 1$
$ f(m) \equal{} m \plus{} \frac {(m)(m \minus{} 1)}{2}$
$ f(m) \equal{} \frac {m^2 \plus{} m}{2}$
$ f(m) \equal{} \frac {(m \plus{} 1)(m)}{2}$

I know theres something wrong with this proof but its the basic idea.[/hide]


[quote="L_Euler"]We have a,b,c are positive real numbers such that $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \equal{} 1$.

Prove that: $ \frac {a^{2} \plus{} bc}{a\sqrt {b \plus{} c}} \plus{} \frac {b^{2} \plus{} ca}{b\sqrt {c \plus{} a}} \plus{} \frac {c^{2} \plus{} ab}{c\sqrt {a \plus{} b}}\geq \sqrt {2}$.[/quote]
Let $ \sqrt a\equal{}x,$ $ \sqrt b\equal{}y$ and $ \sqrt c\equal{}z.$
Hence, we need to prove that
$ \sum_{cyc}\frac{x^4\plus{}y^2z^2}{x^2\sqrt{y^2\plus{}z^2}}\geq\sqrt2(x\plus{}y\plus{}z).$
But $ \sqrt2x\sqrt{x^2\plus{}y^2}\leq\frac{2x^2\plus{}y^2\plus{}z^2}{2}.$
Thus, it remains to prove that $ \sum_{cyc}\frac{2(x^4\plus{}y^2z^2)}{x(2x^2\plus{}y^2\plus{}z^2)}\geq x\plus{}y\plus{}z,$ which is equivalent to
$ \sum_{sym}(4x^7y^3\minus{}2x^7y^2z\minus{}x^6y^3z\plus{}5x^5y^5\minus{}3x^5y^4z\minus{}x^5y^3z^2\minus{}2x^4y^3z^3)\geq0,$ which is obvious.

We prove the stronger
\[ \sum\frac {a^2 + bc}{a\sqrt {b + c}}\geq \sum \sqrt {b + c} \geq 2
\]
[hide="Solution"]
\[ \iff \sum\frac {(a - b)(a - c)}{a\sqrt {b + c}}\geq 0
\]
make $ a\geq b\geq c$. then reqroup terms and we need only show:
\[ \frac {1}{a\sqrt {b + c}} + \frac {1}{c\sqrt {a + b}}\geq\frac {1}{b\sqrt {a + c}}
\]
Anyway we can use crude approximations here as equality $ (a = b = c)$ is already taken care of in the rest of the regrouping:
\[ \frac {b\sqrt {a + c}}{a\sqrt {b + c}} + \frac {b\sqrt {a + c}}{c\sqrt {a + b}}\geq \frac {b}{a} + \sqrt {\frac {a + c}{a + b}}\geq \frac {b}{a + b} + \sqrt {\frac {a}{a + b}}\geq \frac {a}{a + b}+{\frac {b}{a + b} = 1
}\]
which is what we wanted to prove.
(The regrouping is straightforward, I'll post but only if people really care)
[/hide]


x,y -integers and $ x,y \neq 0$
Prove taht
$ (x^2 \minus{} y^2)^2 < 4xy(x^3 \plus{} y x^2 \plus{} y)$

Already out of time  :!:

hello, do you mean
$ (x^2\minus{}y^2)^2<4xy(x^3\plus{}yx^2\plus{}y^3)$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean
$ (x^2 \minus{} y^2)^2 < 4xy(x^3 \plus{} yx^2 \plus{} y^3)$?
Sonnhard.[/quote]
It's also wrong. :wink:


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a finite non-trivial group so that any non-trivial subgroup is solvable. Show that $ G$ is abelian.

nonsense (and you can choose a counterexample by yourself)

This is not true: $ A_{5}$ is the smallest non-solvable group and hence all its proper subgroups are solvable.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14205[/img]

 :(

isn't the same that the other?  :maybe:
EDIT: it's different, sorry, i dont read well  :(

I change a little the notation
[color=brown]Given a circle $ (I)$ and a point $ O$, draw another circle centered at $ O$ so that the first circle can be inscribed and the second circumscribed to the same triangle[/color]

http://mathworld.wolfram.com/EulerTriangleFormula.html

$ OI^2 \equal{} R(R \minus{} 2r)$

$ OI^2 \equal{} R^2 \minus{} 2rR$

$ R^2 \minus{} 2rR \plus{} r^2 \equal{} r^2 \plus{} OI^2$

$ (R \minus{} r)^2 \equal{} r^2 \plus{} OI^2$

The positive sulution is $ R \equal{} r \plus{} \sqrt {r^2 \plus{} OI^2}$, which is constructible

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14285[/img]

 :thumbup:


Functional Analysis
real analysis
real analysis unsolved

Let operator $ A: l^2\rightarrow l^2$ be difined with $ x\equal{}(x_1,x_2,...)\in l^2$,  $ Ax\equal{}(0,x_1,x_2,...)$. A is an operator compact, or not?

This is an isometry. (That is, $ d(Ax,Ay)\equal{}d(x,y)$ for all $ x$ and $ y.$)

Now tell me: can an isometry on an infinite dimensional Hilbert space [i]ever[/i] be compact?

No this operator is not a compact operator. because If we take the sequence $ (e_n)$ ($ e_1 \equal{} (1,0,0,...),e_2 \equal{} (0,1,0,0,...)...$) which is bounded. The image of $ (e_n)$ under $ A$ is the sequence $ A(e_1) \equal{} (0,1,0,...),A(e_2) \equal{} (0,0,1,0,...)$  which has no convergent subsequence. so $ A$ is not a compact operator.


For an arbitrary point $ M$ inside triangle $ ABC$, let $ AM$ intersect the circumcircle of triangle $ ABC$ again at $ A'$. Find $ M$ that minimizes $ \frac{MB\cdot MC}{MA'}$

It is posted in this forum but I can't find the link, the main idea is prove that
$ MAMBMC\ge r(R^{2}-OM^{2})$ by the result $ R'\ge r$ with $ R'$ is circumradius of pedal triangle of $ M$ and $ r$ is inradius. Equality when $ M\equiv I$ incenter.


geometry
geometric transformation
reflection
circumcircle
trigonometry
trig identities
Law of Sines

Let $ CL$ be a bisector of triangle $ ABC$. Points $ A_1$ and $ B_1$ are the reflections of $ A$ and $ B$ in $ CL$, points $ A_2$ and $ B_2$ are the reflections of $ A$ and $ B$ in $ L$. Let $ O_1$ and $ O_2$ be the circumcenters of triangles $ AB_1B_2$ and $ BA_1A_2$ respectively. Prove that angles $ O_1CA$ and $ O_2CB$ are equal.

AC/BC=AL/BL=AL/LB2=A2L/BL=CA1/BC=CA/CB1
=>  A1A2//CL//B1B2   =>  ∠BA1A2=∠AB1B2
BA2/sin∠BA1A2=AB2/sin∠AB1B2  =>  BO2=B1O1
∠O2BA1=90-(180-∠BA2A1)=90-∠CLA=90-∠AB2B1=∠O1B1A =>
⊿O2BC≌⊿O1B1C

This solution more or less is an expansion of the one above.  Note that it suffices to prove that $\angle O_2CA=\angle O_1CB$.

WLOG let $AC>AB$, and let $A_L=AA_1\cap CL$, $B_L=BB_1\cap CL$.  It is well known that $A_1$ lies on $BC$ and $B_1$ lies on $AC$ and that $\triangle ACA_1$, $\triangle BCB_1$ are both isosceles.  (The proof of this basically comes down to showing that $\triangle AA_LC=\triangle A_1A_LC$ and similarly for the triangles associated with $B_L$.)  Thus $AA_1\parallel BB_1\implies AB_1=A_1B$.  Next, we know that by the definition of reflection $AA_L=A_LA_1$ and $AL=LA_2$, so $CL\parallel A_1A_2$.  Similarly, $CL\parallel B_1B_2$, so $A_1A_2\parallel B_1B_2$ and $\angle A_1A_2B+\angle B_1B_2A=180^\circ$.  Therefore by the Extended Law of Sines, we have \[\begin{cases}\dfrac{AB_1}{\sin\angle B_1B_2A}&=2R_{(B_1B_2A)},\\ \dfrac{A_1B}{\sin\angle A_1A_2B}&=2R_{(A_1A_2B)}\end{cases}\implies R_{(B_1B_2A)}=R_{(A_1A_2B)},\] where $R_{(ABC)}$ denotes the circumradius of $\triangle ABC$.

Finally, let $H_1$ and $H_2$ be the feet of the projections from $O_1$ and $O_2$ to $BC$ and $AC$ respectively.  We know that $O_1A_1=O_2A$ by our work above, and in addition we know that $H_1A_1=\tfrac12(A_1B)=\tfrac12(AB_1)=H_2A$.  Therefore by Pythag or HL or whatever we have $O_2H_2=O_1H_1$.  Thus since $H_2A+AC=H_1A_1+A_1C\implies H_1C=H_2C$, we have that $\triangle O_2H_2C=\triangle O_1H_1C\implies \angle O_2CA=\angle O_1CB$, as desired.  $\blacksquare$

What a great problem!

$AB_1=AB_2$
$B_2L=BL$
$AL=A_2L$
$B_2L-AL=BL-A_2L$
OR $B_2A=A_2B $
$m(B_2AB_1)=m(CAB)=m(A_2BA_1)$
Comparing we  get $triangleB_1AB_2=triangleA_1BA_2$
                                             $O_1A=O_2A_1$
                                             $m(O_1AB_1)=m(O_2A_1B)$
                                             $m(CAO_1)=m(CA_1)$
                                             $AC=A_2C$
                                             $triangleCAO_1=triangleCA_1O_2 [S-A-S]$
                                             $m(O_1CA)=m(O_2CB) $                                Q.E.D.

[quote="TripteshBiswas"]AB’=AB”…….(1)
B”L=BL
AL=A”L
B”L-AL=BL-A”L
OR B”A=A”B ……….(2)
m(B”AB’)=m(CAB)=m(A”BA’)……….(3)
Comparing  1,2,3 we  get ∆B’AB”=∆A’BA”
                                             →O’A=O”A’
                                             →m(O’AB’)=m(O”A’B)
                                             →m(CAO’)=m(CA’O”)
                                              AC=A’C
                                              ∆CAO’=∆CA’O” [S-A-S]
                                              →m(O’CA)=m(O”CB)                                 Q.E.D.[/quote]

Your No. 3 equality is wrong (2nd part!)!

Bets regards,
sunken rock

$O_1$ is circumcentre  of $trianleAB_1B_2$ SO $m(O_1AB_1)=90-m(B_1B_2A)$
$O_2$ is the circumcentre of $trianleA_2A_1B$ so $m(O_2A_1B)=90-m(A_1A_2B)$
$m(B_1B_2A)= m(A_1A_2B)$ gives $m(O_1AB_1)= m(O_2A_1B)$
So Please kindly tell me the detail of wrong of my solution sunken rock.

[hide=Solution]
Drop perpendiculars from $O_1$ and $O_2$ onto $AB_1$ and $BA_1$ respectively; call them $M_1$ and $M_2$. Since $CA=CA_1$ and $CB=CB_1$, $CM_1=CM_2$. Also, $\angle{O_1M_1C}=\angle{O_2M_2C}=90^{\circ}$,so it suffices to show that $O_1M_1=O_2M_2$. Since $AB_1=BA_1$, we just need to show $\angle{B_1O_1A}=\angle{A_1O_2B}$, to show this, by basic arcs and angles we need to show $\angle{B_1B_2A}+\angle{BA_1A_1}=180^{\circ}$. So we need to show $B_1B_2 \parallel A_1A_2$. Since $LB_1=LB_2$ and $LA_1=LA_2$, to show $\angle{LB_2B_1}=\angle{LA_2A_1}$, we just need to show $\angle{B_2LB_1}=\angle{A_2LA_1}$. Since $\angle{B_2LB_1}+2\angle{B_2LC}=180^{\circ}$ and $\angle{A_2LA_1}+2(180^{\circ}-\angle{A_2LC})=180^{\circ}$, the problem comes down to showing $\angle{B_2LC}=180^{\circ}-\angle{A_2LC}$, which is true as the sum of the angles make up line $B_2A_2$.
[/hide]

I do not understand how djmathman got this:

Finally, let $H_1$ and $H_2$ be the feet of the projections from $O_1$ and $O_2$ to $BC$ and $AC$ respectively. We know that $O_1A_1=O_2A$ by our work above, and in addition we know that $H_1A_1=\tfrac12(A_1B)=\tfrac12(AB_1)=H_2A$

Can someone explain.

The equality $O_1A = O_2A$ is a result of the fact that the radii of the two circles are equal, while $H_1A_1 = \tfrac12(A_1B)$ comes from the fact that $H_1$ is the midpoint of $\overline{A_1B}$ (and similarly for $H_2A = \tfrac12(AB_1)$).

I'm surprised this solution isn't already on the thread, hopefully it isn't a fakesolve.

In fact, we claim $(AB_1B_2)$ and $(BA_1A_2)$ are reflections over $CL$. It suffices to show that the reflection of $A_2$ over $CL$ lies on $(AB_1B_2)$.  By homotheties at $A$ and $B$ we know $A_1A_2\parallel B_1B_2\parallel CL$. Now let $P\neq A$ be on $(AB_1B_2)$ such that $AP\parallel CL$. If we can show $AA_1A_2P$ is a rectangle, we'll be done. Since $AP\parallel B_1B_2$, $APB_1B_2$ is an isosceles trapezoid and hence the perpendicular bisector of $AP$ is also the perpendicular bisector of $B_1B_2$. But from right triangle $BB_1B_2$ this line passes through $L$, so from right triangle $AA_1A_2$ it is also the perpendicular bisector of $AA_1A_2$ and we're done.


floor function
logarithms
algebra unsolved
algebra

Let $ \{x\}: \equal{}x\minus{}\lfloor x\rfloor$. Prove that the set $ \{\{\log{(n!)}\} | n\in\mathbb{N}^*\}$ is dense in $ (0;1)$.

Somebody plese help me with some idea to solve this problem. My approach is by proving $ \{\{\log{n}\} | n\in\mathbb{N}^*\}$ is dense in $ (0,1)$. This is because of the fact that if we take some fixed $ n\in\mathbb{N}*$ then $ \{\{k\log{n}\} | k\in\mathbb{N}^*\}\subset\{\{\log{n}\} | n\in\mathbb{N}^*\}$, but by Kronecker's theorem, we deduce that $ \{\{k\log{n}\} | k\in\mathbb{N}^*\}$ is dense in $ (0,1)$.


ratio logarithms quadratics LaTeX trigonometry algebra

The equation  \[ x^{logx^4}\equal{}\frac{x^8}{1000}\] has two positive solutions. What are the solutions?

[quote="DaReaper"]The equation
\[ x^{logx^4} \equal{} \frac {x^8}{1000}
\]
has two positive solutions. What are the solutions?[/quote]
We have
$ logx^{logx^4}\equal{}log\frac{x^8}{1000}
\rightarrow logx^4 logx\equal{}logx^8\minus{}log10^3
\rightarrow 4(logx)^2\minus{}8logx\plus{}3\equal{}0$
So let $ logx\equal{}a$
We have $ 4a^2\minus{}8a\plus{}3\equal{}0$........

hmm... i dont think that is correct since the problem also asks for the ratio of the larger to the smaller, which is 10 (according to the answer key). and 3/2 and 1/2 are not in a ratio of 10.

is the log in the original problem supposed to be $ \log_x 4$?  or a different base than $ 10$?  because the way it is written, there is nothing wrong with Dimitris' solution.

hello, i assume that $ \log x^4\equal{}\ln(x^4)$ the logarithm with the base $ e$, the Euler's constant, taking the logarithm of both sides we get $ \ln(x^4)\cdot\ln(x)\equal{}8\ln(x)\minus{}\ln(1000)$, from here we get
$ 4\ln(x)^2\minus{}8\ln(x)\plus{}\ln(1000)\equal{}0$ and this is a quadratic equation in $ \ln(x)$.
Sonnhard.

no the way it is written in the original problem is correct. and yes, it is base 10. However, one of my teammates said the roots are 1 and 1/10, thus, making the ratio 10. Which is supposedly correct.

Go back and read the very first response again, thinking about it more carefully this time.  Dimitris gave you a quadratic in the variable $ a \equal{} \log x$.  (By the way, please note that \log is a command in LaTeX, just as \sin, \cos, etc.  But don't write \logx :))  You found the solutions for $ a$ -- now you must go back and find the associated $ x$ values!  (The difference of 1 between those two solutions is the same as the ratio 10 between the associated values of $ x$ -- and it's quite easy to check that your friend is wrong about what the two roots are.)

hello, if the given logarithm is to the base $ 10$ we can taking the logarithm of both sides and we have $ \lg(x^4)\lg(x)\equal{}\lg(x^8)\minus{}\lg(1000)$ and from here we get
$ 4\lg(x)^2\minus{}8\lg(x)\plus{}3\equal{}0$
Sonnhard.

ohhhh.....yes, you're right. i did just find $ a$, i didnt go back and plug those variables for the $ \log$ s  (hah, it worked  :lol: )


In class today we were learning series and I started toying around to find a shortcut and I came up with this forumula that sums up all the terms of an arithmetic series no matter what it is.

Example:   $3+7+11+.....+479$
$3$ being $t_{1}$ and $t_{n}$ being the last number in the series.

The formula I came up with is

$\frac{(t_{n}-t_{0})(t_{n}+t_{1})}{2(t_{2}-t_{1})}$

[b]Why does this work????[/b]

Wait.  If $3 = t_{1}$ then what's $t_{0}$?

Anyway, I'm going to assume that $t_{0}=-1$ is the "zeroth" term of the arithmetic sequence if it had any.

In that case, notice that since it's an arithmetic sequence, $t_{2}-t_{1}= t_{1}-t_{0}$ and so forth.

Now, $\frac{t_{n}-t_{0}}{t_{1}-t_{0}}= n$ (can you see why?)  This just gives the number of terms in the sequence, since it's one more than the number of times you added your common difference.  In this case, that's $\frac{479-(-1)}{7-3}= 120$.

$\frac{t_{n}+t_{1}}{2}$ is the average of the first and last terms, which is the average of [i]all[/i] of the terms (again; can you see why?).  In this case, that's $\frac{3+479}{2}= 241$.

The key here is that [i]if you replace every member of a group of numbers by the average of the numbers, they have the same sum.[/i]  (This follows from the definition of the average.)  So

$3+7+11+...+479 = 241+241+241+...+241 = 120 \cdot 241 = 28920$

A better way to phrase your formula would be

$\left( \frac{t_{n}-t_{0}}{t_{1}-t_{0}}\right) \cdot \frac{t_{n}+t_{1}}{2}$

Or, even better, just

$\frac{n( t_{n}+t_{1}) }{2}$

My book cares to differ. The answer they claim is $28920$

$t_{0}=3$ or $t_{0}=(-1)$ ???

Because $n$ should be $120$ if I'm not wrong.

Probably just a careless mistake but Thanks for explaining this to me! I appreciate it.

[quote="Jeremic"]$t_{0}=3$ or $t_{0}=(-1)$ ???

Because $n$ should be $120$ if I'm not wrong.[/quote]

Yes, sorry, I confused myself there  :oops:  If we let $t_{0}=-1$ be the "zeroth" member of the sequence, and [b]don't[/b] count it in the summation, then $\frac{t_{n}-t_{0}}{t_{1}-t_{0}}= n = 120$.


calculus
integration
trigonometry
derivative
abstract algebra
vector
linear algebra

Given DE is D^2(y)+5Dy+4y=cos 2x ........where D=>(dy/dx)

We are to find the general solution.(I have done it already)
Is the particular integral unique?If not,find another PI and show that the general solution is the same for all PI.

I did the first part and got the general solution as

y=C1 exp(-x)+C2 exp(-4x)+(1/10)cos 2x

I cannot understand why the PI would be non-unique?Can anyone please check if I am correct in having that answer?

$\frac 1{10}\cos 2x$ is, clearly, incorrect because its derivative gives you a term with $\sin 2x$ that doesn't cancel out with anything. The particular solution is not unique, of course: you can add any solution of the homogeneous equation to it and get another particular solution. The point is that you cannot do anything else, so the general solution stays the same regardless of which particular solution you started with.

Oh!!!I was wrong to write that.Using method of undetermined co-efficients,I assumed the form of solution: a sin 2x+b cos 2x
and got a=(1/10) and b=0.

Does this make sense now?

I did not want to have a [u]particular solution[/u].The question says that the [u]particular intergral[/u] is non-unique...

OK,I got it:

Think of the equation as Ly=f where L is a linear operator. To find a general solution you probably solved the homogeneous problem Ly=0 first then found a particular solution by trying it with a probable form. Let us write it formally: let h be a solution of Lh=0, and let p be some particular solutioin of Lp=f. Then you can see that p+h is also a particular solution: L(p+h)=Lp+Lh=f+0=f. A nice way of saying it is that the particular solution is not unique because L has a nontrivial kernel. You are writing the general solution for Ly=0 means you finding the kernel of L, and adding a particular solution to it means you are parallel shifting this kernel. The kernel of L is a linear space so the soluton set of the nonhomogeneous equation is an affine space, and this can be found by shifting the kernel of L by any vector (part.sol.) of this affine space. Perhaps you can see the situation clearly if you think L as a matrix and y,h,p,f etc as vectors.


Let  $r,s\in R$ with  $0<r<s$  and let  $a,b,c\in \left(r,s\right)$  . Prove that

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac{3}{2}+\frac{(r-s)^2}{2r(r+s)}$

Very nice your solution pavel . I think the constructor is the mathlinker   pvthuan.
Cogratulations also to him for the nice creation.


real analysis
topology
real analysis unsolved

Show that for any $ \epsilon >0$, there exists a dense open subset $ G$ of $ \mathbb{R}$ such that $ m^*(G)<\epsilon$ where $ m^*$ denotes the Lebesgue Outer Measure.

Just choose an open set containing $ \mathbb{Q}$, approximating the measure of $ \mathbb{Q}$, as in the definition of the outer measure


if $ a,b,c$ are positive, and $ abc\leq{1}$, prove that
         $ \dfrac{a}{b}\plus{}\dfrac{b}{c}\plus{}\dfrac{c}{a}\geq{a\plus{}b\plus{}c}$

It's an old problem..
We have:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}\ge a\plus{}b\plus{}c$ because $ abc \le 1$

Stronger!
$ 2(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\plus{}1 \ge \frac{21(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}$
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{9(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}$

[quote="v235711"]if $ a,b,c$ are positive, and $ abc\leq{1}$, prove that
         $ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{a}\geq{a \plus{} b \plus{} c}$[/quote]
this problem is very nice, this is my solution,
let put that, $ a\equal{}\frac{x}{y}  , b\equal{}\frac{y}{z}   , c\equal{}\frac{z}{x}\minus{}1$
we have $ abc\equal{}1\minus{}\frac{z}{x}\leq 1$   and the inequality becomes,
$ \frac{x^2y^2}{z}\plus{}\frac{y^2z^2}{x}\plus{}\frac{x^2z^2}{y} \geq x^2z\plus{}y^2x\plus{}z^2x$
$ \Leftrightarrow x^3y^3\plus{}y^3z^3\plus{}z^3x^3\geq x^3z^2y\plus{}y^3x^2z\plus{}z^3x^2y$
wish is obviously true!


What is the tens digit in the sum $ 7! \plus{} 8! \plus{} 9! \plus{} \cdots \plus{} 2006!$?

Since all are divisible by 7!, you just take the tens digit of 7!, or 5040, which turns out to be 4.  :D

7! + 8! would not have a tens digit of 4; what makes this sum any different from that one?

EDIT : I think I figured I may have figured it out. Since 5040 times 81 has a 4 in the tens digit,  the sum of this sequence would never change the tens digit since the next numbers (10, 11,....) will just add 0's.Can someone confirm ?

Hmmm... could someone elaborate? I'm still not sure I understand the above posts. Thanks in advance!

The answer is the tens digit of $7!+8!+9!$ because for $10!+$, the tens digit is $0$

[hide=solution]the tens digit of 7! is 4 and if you multiply this by 8 you get 32 which if you take away the 3 is the tens digit for 8!. If you repeat this process 1 more time you get another tens digit of 8. 4+2+8=14 so the tens digit is $\boxed{4}$[/hide]

what
how did you get that equation
not to be rude i just don't understand


inequalities
logarithms
calculus
derivative
algebra
polynomial
inequalities proposed

[quote="can_hang2007"]There exists a solution using Cauchy Schwarz. :)[/quote]
Great! :lol:
My solution is S.O.S,and it is very ugly. :(

[quote="shfdfzhjj"]Waiting for nice proof. :)[/quote]

Any other idea? :maybe:

The following is clearly not a nice solution  :oops: , but at least is simple enough.

Well, if $ a\equal{}\frac xy,b\equal{}\frac yz,c\equal{}\frac zx$ and $ g(u)\equal{}\frac{u(4u\plus{}41)}{8u\plus{}1}$ then the inequality is equivalent to $ g(a)\plus{}g(b)\plus{}g(c)\ge15$ for all positive $ a,b,c$ with $ abc\equal{}1$. We set $ u\equal{}\ln a,v\equal{}\ln b,w\equal{}\ln c$ and $ f(u)\equal{}g(e^u)$ so the ineq is also equivalent to $ f(u)\plus{}f(v)\plus{}f(w)\ge 15$ for all $ u,v,w\in\mathbb R$ with $ u\plus{}v\plus{}w\equal{}0$. We have $ f''(x)\equal{}\frac{e^x(41\minus{}312 e^x\plus{}96e^{2x}\plus{}256 e^{3x})}{(1\plus{}8e^x)^3}$ which is positive if $ x\ge 0$ (look at the derivative of the numerator). So we can apply Vasc's RCF Theorem reducing it to the case when $ a\equal{}b\ge 1$, $ c\equal{}a^{\minus{}2}$, which gives the following equivalent polynomial: \[ 8a^6\minus{}38a^5\plus{}49a^4\plus{}24a^3\minus{}79a^2\plus{}32a\plus{}4\ge 0.\] It is \[ (a\minus{}1)^2(8a^4\minus{}22a^3\minus{}3a^2\plus{}40a\plus{}4)\ge 0.\] But if $ x\equal{}a\minus{}1\ge0$ we have to prove $ 8x^4\plus{}10x^3\minus{}21x^2\plus{}27\ge 0.$ It's derivative is $ 2x(16x^2\plus{}15x\minus{}21)$, which is positive for $ x\ge 1$. We check that $ 8x^4\plus{}10x^3\minus{}21x^2\plus{}27\ge 0$ for $ x\equal{}1$ and so for $ x\ge 1$, so we only have to check it when $ x\leq 1$. In that case we have \[ 8x^4\plus{}10x^3\minus{}21x^2\plus{}27\equal{}8\left(x\plus{}\frac58\right)^2x^2\plus{}\frac{193}8(1\minus{}x^2)\plus{}\frac{23}8\ge 0\] and we're done.

[quote="shfdfzhjj"]Waiting for nice proof. :)[/quote]
The full my proof :)
See also here: http://canhang2007.wordpress.com


For positive integers $a_1,a_2 ,\ldots,a_{2006}$ such that $\frac{a_1}{a_2},\frac{a_2}{a_3},\ldots,\frac{a_{2005}}{a_{2006}}$ are pairwise distinct, find the minimum possible amount of distinct positive integers in the set$\{a_1,a_2,...,a_{2006}\}$.

I think this is a combinatorial problem... :P

[hide]The minimum possible amount of distinct integers of the sequence $[a_n]_1^{2006}$ can be given when we repeat in a maximum possible way,the numerator or denominator of the distinct elements of the set $\{\frac{a_1}{a_2},\frac{a_2}{a_3},\ldots,\frac{a_{2005}}{a_{2006}}\}$ and its happens keeping constant  a member of the sequence, for example:

If $a_1=x$, $a_2=y$ and $\{x,y \in Z|x$ ≠ $y\}$, $x$ is the constant and $y$ is fixed,a example of a set in the condition described above is:

$\{\frac{x}{y},\frac{y}{x},\frac{x}{2y},\frac{2y}{x},\frac{x}{3y},\frac{3y}{x},...,\frac{x}{1003y}\}$ such that $a_{2k-1}=x$, $a_{2k}=ky$ for $1 \le k \le 1003$.

note that the set follows the division postulate,since two fractions with equal numerator or denominator to be distinct must have different denominator or numerator,consequently the distinct members of the sequence are:
$x,y,2y,3y,...,1003y$ and the amount is $1004$.[/hide]