Given xyz = 1 for x, y, z being positive numbers.
Prove that
x + y + z >= (z/x)^(1/3)+(x/y)^(1/3)+(y/z)^(1/3).

I think this might work:

[hide="Solution?"]

To make it look nicer let $ x\equal{}a^{3}, y\equal{}b^{3}, z\equal{}c^{3}$ where $ a,b,c \in \mathbb{R}^{\plus{}}$ such that $ abc\equal{}1$. Hence:

$ a^{3} \plus{} b^{3} \plus{}c^{3} \geq \frac{c}{a} \plus{} \frac{a}{b} \plus{} \frac{b}{c}$

is the equivalent form.

$ a^{3} \plus{} b^{3} \plus{}c^{3} \geq \frac{c}{a} \plus{} \frac{a}{b} \plus{} \frac{b}{c} \iff a^{3} \plus{} b^{3} \plus{}c^{3} \geq bc^{2} \plus{} ca^{2} \plus{} ab^{2}$
[/hide]

[quote="BanishedTraitor"]I think this might work:

[hide="Solution?"]

To make it look nicer let $ x \equal{} a^{3}, y \equal{} b^{3}, z \equal{} c^{3}$ where $ a,b,c \in \mathbb{R}^{ \plus{} }$ such that $ abc \equal{} 1$. Hence:

$ a^{3} \plus{} b^{3} \plus{} c^{3} \geq \frac {c}{a} \plus{} \frac {a}{b} \plus{} \frac {b}{c}$

is the equivalent form.

$ a^{3} \plus{} b^{3} \plus{} c^{3} \geq \frac {c}{a} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \iff a^{3} \plus{} b^{3} \plus{} c^{3} \geq bc^{2} \plus{} ca^{2} \plus{} ab^{2}$
[/hide][/quote]
 all right :
 use  ineq  Cauchy's :
$ b^3\plus{}c^3\plus{}c^3  \geq 3bc^2$
 similar we have $ 3(a^3\plus{}b^3\plus{}c^3 ) \geq 3 ( bc^2 \plus{} ca^2 \plus{} ab^2 )$

[quote="kritikos"]Cauchy ineq is AM-GM.[/quote]
"The Cauchy's inequality is AM-GM" it's true for $ \frac{a\plus{}b}{2}\geq\sqrt{ab}$ ( where  $ a\geq0,$ $ b\geq0$ ) only. :wink:

[b]BanishedTraitor wrote:[/b]
[quote]I think this might work:
Solution?[/quote]
i think the last inequality is true by using rearangement.


geometry
perimeter
trigonometry
circumcircle
trig identities
Law of Sines
area of a triangle

If you've figured these out already, or know the solution please let someone who has not seen it already attempt it.

Prove that for a triangle with side lengths a,b, and c,  r= a/sinA=b/sinB=c/sinC, where r is the radius of the inscribed circle of the triangle.

Prove that for a triangle with side lengths a,b, and c, R= sqrt (((s-a)(s-b)(s-c))/(s)), where s is the semi-perimeter of the triangle and R is equal to the radius of the circumscribed circle.

Are you sure about the first one?  Try a 3-4-5 rt. triangle for example.  That doesn't work.  You probably wanted the Extended Law of Sines:

\[2R=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.\]

[hide="solution1"]
let's say triangle ABC is a right triangle, with C=90.
draw an imaginary circumcircle (AB will be diameter)

so then, $\displaystyle \Sin A = \frac{a}{D}$
$\displaystyle = \frac{a}{2R}$
$\displaystyle \text { thus, if we simplify it, } 2R = \frac{a}{\sin A}$

similarly, we can apply the same thing with sinb and sinC, and since they are all equal, QED. or AWD, or $W^5$.
[/hide]

Yea, thanks I did have my r's mixed up.  Anyway the second one's a little harder but still very doable.  For a big hint see below

[hide]Think Heron[/hide]

i know i thought of Heron's too, but CAN we use that as a lemma to this proof? i mean i think that since this is too similar to Heron, we might need to prove HEron's formula before we go on and prove this one. If we dont have to prove HEron first, then this is a piece of cake.

I thought if it's something notable you don't have to prove it.  By something notable I mean something with a name like AM-GM.

the area of a triangle is also rs so with heron you get the answer, but that's not much of a proof (like xxreddevilzxx said)


abstract algebra
group theory
superior algebra
superior algebra unsolved

Is there an infinite non-abelian group with finite center and finite commutator subgroup? Give a proof, state a counterexample or provide us with the appropriate references. ;)

- By the way, you can find the August proposal [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293302][color=orange]here[/color][/url].

[quote="coquitao"]Is there an infinite non-abelian group with finite center and finite commutator subgroup? Give a proof, state a counterexample or provide us with the appropriate references.[/quote]

Extraspecial groups are like this.  The groups $ \langle x_i, y_i, z : x_i^p \equal{} y_i^p \equal{} z^p \equal{} 1, y_ix_i \equal{} x_i y_i z, y_i x_j \equal{} x_j y_i, y_i y_j \equal{} y_j y_i, x_i x_j \equal{} x_j x_i \rangle$ where i varies over some index set and j ≠ i have center and derived subgroup of order p.  These posts would be more interesting if they described a question in context.  Are you asking about adjustments of Schur's lemma or are you asking about the structure of groups whose conjugacy classes are all of bounded size?  P. Hall has some nice variations on Schur's lemma, and Robinson's textbook and some of his and Newman's "just-non" papers describe interesting properties and uses of groups with nice conjugacy classes.  In particular, the extraspecial groups play a very interesting role in Newman's "On a class of nilpotent groups".


C1 and C2 are two non - intersecting circles such that no point of C1 lies inside C2 and vice versa. The external common tangents to both circles intersect at S. A third circle C3 touches C1 in A and C2 in B. Prove that A, B and S are collinear.

By "touches," you mean "is tangent to," right?  Assuming that is the case:

[hide]Let the center of our third circle be O.  Draw in triangle C1C2O and points S, A, and B.  Also, draw the tangent from S to circles C1 and C2.  Let r1 be the radius of circle C1 and r2 the radius of circle C2.

Now, we know that line C1C2 passes through S.  Draw radii from C1 and C2 to the points of tangency with the tangent through S.  This gives us two similar triangles by a shared angle at S and right angles.  Thus, we have C1S : C2S = r1 : r2.  Now, examine the expression

C1S/SC2 * C2B/BO * OA/AC1.

If the radius of circle O is r, this expression equals

r1/r2 * r2/r * r/r1 = 1.  Therefore, by Menelaus, the three points S, A, B are colinear.[/hide]

Let an external common tangent to two circles touch them at $M, N$ respectively; an inversion of pole $S$ and power $SM\cdot SN$ sends $C_1$ to $C_2$ and reverse, so image of $C_3$ will be tangent to $C_1$ and $C_2$ as well, hence $C_3$ remains unchanged under the said inversion, meaning $A$ goes to $B$ and $B$ to $A$, so they are collinear with $S$.

Best regards,
sunken rock


Two mothers have two children each.  The first woman's first child is a boy and at least one of the second woman's children is a boy.  What is the probability for both of the first woman's children to be boys? The second woman?

the first woman's child is a boy
therefore, the prob that the second chiled is a boy is 1/2.

For the second woman, there are these possibilities:

BG
BB
GB

so it is 1/3.

i am assuming of course that there are no inbetweens.

tell me if im wrong

Good job. That's right.

After thinking about the question for a while there is something that bothers me.  For the second woman there are three choices BB BG or GB as starrynight stated.  This is how I went about solving the problem.  But what if you go about it this way; since one of her children has to be a boy it has to be either the first child or the second child.  If it's the first child then there are two choices for the other child BB or BG.  If it's the second child then there are two choices for the first one BB or BG. So this gives us four overall choices two of which are the wanted outcome.  This gives us 1/2 instead of 1/3.  Can someone tell me why this happens and why the way I just explained is not acceptable?

Yes -- because you've just counted the BB case in two different places, even though it isn't twice as likely to occur.


calculus
integration
logarithms
trigonometry
calculus computations

[hide="Click to enter the dungeon"]
Using the relation

$\ln (r^{2}-2 r \cos \theta+1) =-2 \sum_{n=1}^{\infty}\frac{r^{n}\cos n\theta}{n}$,

we have

$\ln{(17-8\cos{x})}= 4 \ln 2+\ln \left( \frac{1}{4^{2}}-\frac{2}{4}\cos{x}+1\right) = 4 \ln 2-2 \sum_{n=1}^{\infty}\frac{\cos nx}{n4^{n}}$.

Since $\int_{0}^{\pi}\cos nx \; dx = 0$ for each $n$, this integral is reduced to

$\int_{0}^{\pi}\ln(17-8\cos x) \; dx = \int_{0}^{\pi}4 \ln 2 \;dx = 4\pi \ln 2$
[/hide]

Generalizing this integral, we have

[hide="Click to exit the dungeon"]
$\int_{0}^{\pi}\ln(r^{2}-2r \cos\theta+1) \; d\theta = \begin{cases}0 & |r| \leq 1 \\ 2\pi \ln |r| & |r|>1 \end{cases}$
[/hide]


geometry
geometric transformation
homothety
ratio
angle bisector
geometry unsolved

Let $ABC$ be an acute triangle, $w$ be the circumcircle. Let $M_{1},M_{2},M_{3}$ be the midpoints of $BC,AB,AC$ respectively. The altitudes from $A,C$ to $BC,AB$ intersect $w$ at $L_{1},L_{2}$. $P_{3}$ is the orthocenter of triangle $BM_{1}M_{2}$. Prove that $M_{3}P_{3}$ is perpendicular to $L_{1}L_{2}$

Let $H_{1}, H_{2}, H_{3}$ be the feet of the altitude from $A, C, B$.
$L_{1}, L_{2}$ are the symmetrics of the orthocentre $H$ w.r.t. $BC, AB$. Hence $H_{1}H_{2}\| L_{1}L_{2}$.
Consider the homothety with centre $B$ and ratio $-\frac{1}{2}$. It sends $\Delta ACB$ in $\Delta M_{2}M_{1}B$ and $H$ in $P_{3}$. Also $BP_{3}= P_{3}H$. Then $M_{3}P_{3}$ is a diameter of the Feuerbach circle of $\Delta ABC$. The line $HP_{3}$ is the angle bisector of $\measuredangle H_{1}H_{3}H_{2}$ and so $P_{3}$ is the midpoint of arc $H_{1}H_{2}$. Then $M_{3}P_{3}\bot H_{1}H_{2}$.


Let $ a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$. Then using $ xyz=1$, our inequality is equivalent to
\[ \frac{9}{2}\le \sum\frac{x}{y+z}\cdot \sum x\le 3\sum \frac{x^{2}}{y+z}\]
for the left part, use $ \sum\frac{x}{y+z}\ge \frac{3}{2}\ge \frac{9}{2(x+y+z)}$, which is true since $ x+y+z\ge 3$.

now for the rigth side, we need to prove that
\[ \sum\frac{x^{2}}{y+z}\ge \frac{x+y+z}{3}\sum\frac{x}{y+z}\]
denote $ p=x+y,q=y+z,r=z+x$, then this is equivalent to
\[ \frac{x^{2}}{q}+\frac{y^{2}}{r}+\frac{z^{2}}{p}\ge \frac{x+y+z}{3}\left(\frac{x}{q}+\frac{y}{r}+\frac{z}{p}\right) \]
which in turn is equivalent to
\[ 2\sum x^{2}rp\ge \sum xyrp+\sum zxrp \]
which is true by rearrangement.

P.S. I think we don't need $ a>b>c$. :wink:

[quote="nayel"]Let $ a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$. Then using $ xyz=1$, our inequality is equivalent to
\[ \frac{9}{2}\le \sum\frac{x}{y+z}\cdot \sum x\le 3\sum \frac{x^{2}}{y+z}\]
[/quote]

$ \sum\frac{x}{y+z}\cdot \sum x\le 3\sum \frac{x^{2}}{y+z}$
It's obvious true by Chebyshev inequality :wink:

[quote="10maths_tp0609"]$ \sum\frac{x}{y+z}\cdot \sum x\le 3\sum \frac{x^{2}}{y+z}$
It's obvious true by Chebyshev inequality :wink:[/quote]

I already thought of that before posting my solution, and I think it's not true.

WLOG let $ x\le y\le z$. then $ \frac{x}{y+z}\le \frac{y}{z+x}\iff (x-y)(x+y-z)\le 0$, which is not always true. :P

EDIT: Thank you hungkhtn. I get it now. :blush:

Not that, nayel. Actually,

\[ x(x+z) \le y(y+z) \Leftrightarrow (x-y)(x+y+z) \ge 0,\]


Let $x = a+b \sqrt{2}, y = c+d \sqrt{2}$.  We are given

$xy = (ac+2bd)+(ad+bc) \sqrt{2}= 1+\sqrt{2}$

Which is an element with norm $-1$, hence its two constituent factors must have norms $1,-1$.  The conclusion follows.

[quote="Beat"]Show that if integers $a, b, c$ and $d$ satisfy $bc+ad = ac+2bd = 1$, then they
also satisfy $a^{2}+c^{2}= 2b^{2}+2d^{2}$.[/quote]
[hide="real solution (no norms, just standard)"]
We want to factor $a^{2}+c^{2}-2b^{2}-2d^{2}=(a+c)^{2}-2(b-d)^{2}-2$
We have $(b-a)(c-d)=db$ etc.
i'll finish up tomorrow...
[/hide]

[quote="me@home"]
real solution (no norms, just standard)
i'll finish up tomorrow...[/quote]
There is no way you can simply calculate this algebraically.
Take $a=2,b=1,c=0,d={1\over 2}$. Then $bc+ad = ac+2bd = 1$, but $a^{2}+c^{2}\not = 2b^{2}+2d^{2}$.
So you'll need $a,b,c,d\in\mathbb{Z}$ and use the divisibility properties of $\mathbb{Z}$ somehow.

Of course you could simply write down
$(a^{2}-2b^{2})(c^{2}-2d^{2})=(ac+2bd)^{2}-2(bc+ad)^{2}=-1$
but it would be very hard to see how you got this formula.
It's much more natural to calculate in $\mathbb{Z}[\sqrt{2}]$. (with all that standard norm stuff  :wink: )


geometry
perimeter
inequalities
function
area of a triangle
Heron\u0027s formula
inequalities unsolved

Perimeter and area of an arbitrary triangle are given. Sought is the triangle such that the difference between the length of the longest side and the length of the shortest side is a maximum.

It seems it is a bloody problem at the first sight. Maybe I am wrong.
Let's move it to the Inequalities Section.

Applying Heron's formula: $xyz(x+y+z)=A^2$, $x+y+z=\frac P2$. Also, the differences between $x,y,z$ are equal to the differences between the sides.

WLOG, $x\ge y\ge z>0$. We want to maximize $x-z$ subject to the constraints $xyz=\frac{2A^2}{P}$ and $x+y+z=\frac P2$.

Use Lagrange multipliers: $(1,0,-1)=\lambda(yz,xz,xy)+\mu(1,1,1)$
Then $\lambda xz+\mu=0=\lambda(xy+yz+zx)+3\mu=\lambda y(x+z)+2\mu.$
This gives us $y(x+z)=2xz$, so $y$ is the harmonic mean of $x$ and $z$.
Therefore $x+z+\frac{2xz}{x+z}=\frac P2$ and $\frac{2x^2z^2}{x+z}=\frac{2A^2}{P}.$

Let $u=x+z,v=x-z$. Then $4xz=u^2-v^2$. Substituting, our system becomes
$u+\frac{u^2-v^2}{2u}=\frac P2$ and
$\frac{(u^2-v^2)^2}{8u}=\frac{2A^2}{P}$.
The first equation gives us $u^2-v^2=Pu-2u^2$. Substituting in the second, we have $\frac u8\cdot(P-u)^2=\frac{2A^2}{P}$.
This is a cubic equation, and there's no reason for the solution to be nice.
The steps after this are merely substitution and quadratics.

The general solution to this problem involves the solution of cubic equations, and I'm not going there.

The appearance of cubics is not necessarily fatal, as it could mean trisection of some angle that appears naturally in the problem.  (I don't have such an interpretation, at the moment.)

For maximization of $|x - z|$, note that $(x-z)^2$ is a known function of $y$, via $xz$ and $x+z$.

[quote="fleeting_guest"]For maximization of $|x - z|,$ note that $(x-z)^2$ is a known function of $y,$ via $xz$ and $x+z.$[/quote]
This approach leads to the following: Find $\frac{P^2}{4}-2Py+3y^2$ given that $y^2(P-2y)=\frac{8A^2}{P}.$
It is an angle trisection equation, but I have no idea whether that's a natural angle.


function
IMO 1987
IMO
algebra
functional equation
algebra solved
fe

One Nice Solution:

Let $N$ be the set of non-negative integers. Put $A = N - f(N)$
(the set of all $n$ such that we cannot find $m$ with $f(m) = n$).
Put $B = f(A)$.

Note that $f$ is injective because if $f(n) = f(m)$, then $f(f(n)) = f(f(m))$ so $m = n$.
We claim that $B = f(N) - f(f(N))$.
Obviously $B$ is a subset of $f(N)$ and if $k$ belongs to $B$,
then it does not belong to $f(f(N))$ since $f$ is injective.
Similarly, a member of $f(f(N))$ cannot belong to $B$.

Clearly $A$ and $B$ are disjoint.
They have union $N - f(f(N))$ which is ${0, 1, 2, ... , 1986}$.
But since $f$ is injective they have the same number of elements,
which is impossible since ${0, 1, ... , 1986}$ has an odd number of elements.

 :)

Solved for the more general case:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1987&t=2749

what if you:

   Take the derivative of $ f(f(n))$ which is one and therefore derivative of $ f(n)$ is also 1, which means $ f$ is a linear function, and since f must have integer coefficients in the form of $ an\plus{}b$, plug it in $ f(f(n))$ and see that b cannot be integer, a contradiction.

You can't differentiate this function. That would be $ \lim_{h \rightarrow 0} \frac{f(x\plus{}h) \minus{} f(x)}{h}$, and the concept of a limit doesn't make sense on the natural numbers or any other discrete set.

Define $a(n)=f(n)-n$. Then, from $f(f(n))=n+1987$ we get $(f(f(n))-f(n))+(f(n)-n)=1987$, or $a(f(n))+a(n)=1987$ for every $n$ natural.

If we replace $n$ by $f(n)$ in $a(f(n))+a(n)=1987$, we get $a(f(f(n)))+a(f(n))=1987=a(f(n))+a(n)$, and then $a(n+1987)=a(n)$ for every natural $n$. Thus, $a$ is an 1987-periodic funcion.

Now, let $a_1=a(1), a_2=a(2),...,a_{1987}=a(1987)$ be the 1987 possible values of $a(n)$ (Since $a$ is periodic, this is possible). We must pair the 1987 $a_i$ in the following mannner: $a_i$ is paired with $a_j$ if $a_i +a_j =1987$ (we can always do this, because $a(f(n))+a(n)=1987$ for every $n$ natural). If there's two or more $j$ satisfying the pairment, anyone serve. So, we can divide 1987 numbers in pairs, and this is a contradiction, because 1987 is odd. Finally, the problem is solved.

$f(x)$must be one-to-one.$f(f(n))=n+1987$so $f(f(n))$ will miss 1987 values.
suppose that $f(n)$ miss exactly $k$ distinct values $c_1,c_2,...,c_k$so
$f(f(n))$will miss $c_1,c_2,...,c_k,f(c_1),f(c_2),....,f(c_k)$so if $f(m)=w$so $w\ne f(c_1),f(c_2),....,f(c_k),m\ne c_1,c_2,...,c_k$so there is $n$ such that $f(n)=m$ hence $f(f(n))=w$so $f(f(n))$ misses $2k$ values.
so it's contradiction.

[quote="varun.tinkle"]Assuming that a function exsists....   suppouse.... the function is
$ ax+b $ ...
Similarly we can check for higher degrees of equation........[/quote]

But .... you cant check ALL the possible functions to achieve the result!

And what's the meaning of "degrees of equation" ? What is the degree of the function $f(n)=1+\left\lfloor e^{175\sin(n)}\right\rfloor$ ?

Dont forget that there exists infinitely many non polynomial functions

i can't find pco's solution who can help me?[quote=varun.tinkle]Wrong solution thanx pco for ur help[/quote]

[quote=darkeagle]I'm interested to know what is the number of function which verified $f(f(x))=x+k$ where k is even.[/quote]
I suppose you mean "functions from $\mathbb N\to\mathbb N$"

If so, just look at the general solution and count.

You'll find $\frac{k!}{\frac k2!}$

[quote=orl]Prove that there is no function $f$ from the set of non-negative integers into itself such that $f(f(n))=n+1987$ for all $n$.[/quote]
Observe if $i\geq 1987\implies f(i)\geq 1987$ and $f$ is injective. Now assume for some $i<1987$ we have $f(i)\geq 1987$ Then $f(f(f(i)-1987))=f(i)-1987+1987\implies f(f(i)-1987)=i$ Hence if for $i<1987$ we have $f(i)>1987\implies$ there exist $j<1987$ such that $f(j)=i$ and as $f(f(i))\geq 1987 \forall i\in N$ So if we define $S=\{i\in N, 0\leq i<1987| f(i)<1987\}$  and
$T=\{i\in N, 0\leq i<1987| f(i)\geq 1987\}$ then $|T|\geq |S|$ But as $f(f(i))\geq 1987$ So $|S|\geq |T|$ Hence $|T|=|S|$ But there are odd number of $i$ such that $0\leq i<1987$ So Such Function can't exists.

We apply the "$fff$ trick" to obtain
$$f(n)+1987=f(f(f(n)))=f(n+1987)\qquad (1).$$
This means that $f \pmod{1987}$ is has period $1987$, so we can work in $\mathbb{Z}_{1987}$. In this case, $f$ is an involution, but since $1987$ is odd there exists some $0 \leq a<1987$ with $f(a) \equiv a$, so $f(a)=a+1987c$ for some $c \geq 0$. Then, by $(1)$ we have $f(a+1987k)=a+1987(c+k)$, so
$$a+1987=f(f(a))=f(a+1987c)=a+1987\times 2c \implies c=\frac{1}{2},$$
absurd. $\blacksquare$

It can be easily seen that the function $f$ is injective

Our problem function produces natural values from $1988$ onwards

Suppose the function $f$ alone does not generate $N$ values. So the function $f(f(n))$ does not produce $2N$ value.

So, if a function wants to be true in the problem, the following equation must have a solution

$2N=1978$
And this issue has no natural answer!

So the function does not work in the problem $\blacksquare$

[quote=Pluto04]$f(n + k) = f(f(f(n))) = f(n) + k$
and it follows by induction on m that
$f(n + km) = f(n) + km$
for all $n, m$ $\in$ $N$.
Now take an arbitrary integer p, 0$\leq$p$\leq$ k−1, and let $f(p) = kq+r$,
where $q$  $\in$ N and 0$\leq$r$\leq$ k − 1. Then
$p + k = f(f(p)) = f(kq + r) = f(r) + kq.$
Hence either $q = 0$ or $q = 1$ and therefore
either $f(p) = r, f(r) = p + k or f(p) = r + k, f(r) = p.$
In both cases we have p not equal to  r which shows that f defines a pairing of
the set $A = {0, 1, . . . , k}$. Note that different functions define different
pairings of $A.$
Conversely, any pairing of $A$ defines a function $f : N \rightarrow N$ with
the given property in the following way. We define f on A by setting
$f(p) = r, f(r) = p + k$ for any pair $(p, r)$ of the given pairing and
$f(n) = f(q) + ks$ for $n \geq k + 1,$ where $q$ and $s$ are respectively the
quotient and the remainder of $n$ in the division by $k$.
So,it is clear that if $k$ is an odd integer then,there are no such functions and we are done.[/quote]
The best solution so far

[hide=SOLUTION]$\textbf{Solution.}$ Given, $f(f(n))=n+1987$.
Notice that if $f(n_1)=f(n_2)\Rightarrow f(f(n_1))=f(f(n_2))=n_1+1987=n_2+1987\Rightarrow n_1=n_2$. This means that $f$ is one-one.
Taking $f$ on both sides, we get
$f(f(f(n)))=f(n+1987)=f(n)+1987\ (\star)$.
So, for any $i\ge1987$, we get $f(i)=f(i-1987)+1987\ge1987$ because $f(i-1987)$ is always a non-negative integer. So, if $f(i)<1987\Rightarrow i<1987\ (*)$.
Let $S=\{0,1,\cdots,1986\}$.
Let $A,B\subset S$ be defined as $A=\{i\ :\ f(i)\ge1987\}$ and $B=\{i\ :\ f(i)<1987\}$. So, $A\cup B=S$ and $A\cap B=\phi$.
Let $k\in A\Rightarrow f(k)\ge1987$. Put $n=f(k)-1987$ in $(\star)$ to get $f(f(k))=k+1987=f(f(k)-1987)+1987\Rightarrow f(f(k)-1987)=k<1987$. So, by $(*)$, we conclude that $f(k)-1987<1987$. This means that $f(k)-1987\in S$. However, $f(f(k)-1987)<1987\Rightarrow f(k)-1987\in B$.
Since $f$ is one-one, we conclude that every element in $A$ corresponds to a distinct element in $B$ (call this correspondence $(1)$).

Let $m\in B\Rightarrow f(m)<1987\Rightarrow f(m)\in S$. Also, $f(f(m))=m+1987\ge1987\Rightarrow f(m)\in A$.
Since $f$ is one-one, we conclude that every element in $B$ corresponds to a distinct element in $A$ (call this correspondence $(2)$).

Combining $(1)$ and $(2)$, we get a bijection between $A$ and $B$. So, $|A|=|B|$, where $|.|$ represents cardinality of set. Also, since $A\cup B=S$ and $A\cap B=\phi$, we use Principle of Inclusion-Exclusion to get
$2|A|=|A|+|B|=|A\cup B|+|A\cap B|=|S|+|\phi|=1987\Rightarrow |A|=\frac{1987}{2}$, which is an absurd conclusion, giving us the required contradiction.

This proves that $\nexists$ any such function $f$. $\blacksquare$[/hide]
[hide=REMARK]$\textbf{Remark.}$ We can replace $1987$ with any other odd number and still get no such function. However, if $1987$ is replaced by an even number (say $2k$), then we can safely have $f(x)=x+k$. [/hide]

\[\rightarrow \rightarrow \rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow \rightarrow \rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow \rightarrow \rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow\]

Suppose there existed such a function. We have $f(n + 1987) = f^3(n) = f^2(f(n)) = f(n ) + 1987$, so $f(n + 1987) = f(n) + 1987$ for all nonnegative integers $n$.

If $f(a) = f(b)$, then $f(f(a)) = f(f(b))$, so $a + 1987 = b + 1987$, which means $a = b$, so $f$ is injective. Additionally, every integer greater than or equal to $1987$ is in the image of $f$.

We have $a\equiv b\pmod{1987}$ implies $f(a)\equiv f(b)\pmod{1987}$. Now we prove a helpful claim.

Claim: $f(n)\not\equiv n\pmod{1987}$ for all $n\ge 0$.
Proof: Suppose some $n$ satisfied $f(n) = n + 1987k$, where $k$ is an integer. Note that \[n + 1987 = f(f(n)) = f(n + 1987k) = f(n) + 1987k = n + 1987 \cdot 2k,\] so $k = \frac{1}{2}$, absurd. $\square$

Now we define $g\colon \{0,1,2\ldots, 1986\} \to \{0,1,2\ldots, 1986\}$ such that $g(n)\equiv f(n)\pmod{1987}$ for any $0\le n\le 1986$. We already know $g$ has no fixed points by our previous claim. Additionally, we have $g(g(n))\equiv f(g(n)) \equiv f(f(n)) \equiv n\pmod{1987}$, so $g(g(n)) = n$, which means $g$ is bijective.

Due to $g$ being an involution, it only has cycles of length $1$ or $2$, but since $g$ has no fixed points, it has only cycles of length $2$. This is impossible since $1987$ is odd. Therefore we have no such functions $f$.

Note that $f$ is injective, and thus the arrows graph on $\mathbb{Z}_{\ge 0}$ induced by $f$ is compromised of multiple disjoint chains. Realize that $f(n+1987) = f(f^2(n)) = f^2(f(n)) = f(n)+1987$. Consider the chain \[ n \to f(n) \to f^2(n) \to \ldots.\] Note  that it must contain at most two distinct residues modulo $1987$ since $f^k(n) \equiv n \pmod{1987}$ for even $k$. If $f(n) \equiv n \pmod{1987}$, then the chain is a cycle, a contradiction. Otherwise, each disjoint chain corresponds to a unique [I]codeset[/i] $\{a, b\}$ of residues, and all codesets are disjoint. However, by parity reasons, at least one residue is not covered by the codesets, which is impossible.

Consider any integer $n$ and look at the values of $f^k(n)$ modulo $1987$; they cover exactly two distinct residues. On the other hand, every set of residues must be disjoint, but as $1987$ is odd, one set can never be covered, contradiction.

Did blackpink exist in 1987?

Let $k = 1987$.
FTSOC suppose that such an $f$ exists. Note that $f$ is injective. Consider the chains of the form $a, b, a + k, b + k, \dots$ formed by $f$ when interpreted as a graph. Each chain contains two residues $\pmod{k}$, however there are $1987$ residues, contradiction.

We'll show the following claim, which clearly implies the problem.
[color=#f00][b]Claim:[/b][/color] There exists a function $f \colon \mathbb{N}_0 \to \mathbb{N}_0$ such that $f(f(x)) = x + k$ iff $k$ is even.
[i]Proof.[/i] Note that for $k$ even, we have $f(x) = x + \frac{k}{2}$ clearly satisfies the equation, so it suffices to show that there's no such $f$ for $k$ odd. Assume for sake of contradiction that there exists such $f$ and $k$ is odd. Note that $f$ must be injective, since $f(a) = f(b) \implies f(f(a)) = f(f(b)) \implies a + k = b + k \implies a = b$. Also, we have $f(f(f(x))) = f(x + k)$, but on the other hand, $f(f(f(x))) = f(x) + k$, so $f(x + k) = f(x) + k$. Now, assume there exist two $a, b$ with $a \not \equiv b \pmod{k},$ and $f(a) \equiv f(b) \pmod{k}$. We have $f(a) + kx = f(b) + ky$ for some $x, y$ which gives $f(a + kx) = f(b + ky)$, which gives $a + kx = b + ky,$ so $a \equiv b \pmod{k}$, a contradiction, so $f(a) \equiv f(b) \iff a \equiv b \pmod{k}$. So, the values $\{f(0), f(1), \cdots, f(k - 1) \}$ modulo $k$ must be a permutation, and we also have $f(x \pmod{k}) \equiv f(x) \pmod{k}$ But, since $f(f(x)) = x + k \equiv x \pmod{k},$ the length of each cycle in the permutation must divide $2$; that is, each cycle is either of length $1$ or length $2$. Note that we must have at least one cycle of length $1$, since $k$ is odd, since the sum of each cycle length is $k$. So, $f(e) \equiv e \pmod{k}$, for some element $0 \le e < k$. We can't have $f(e) = e$, as then $f(f(e)) = e$, a contradiction, so $f(e) \ge e + 1987$, so we have $f(f(e)) \ge f(e) + 1987 \ge e + 1987 \cdot 2$, a contradiction. So, there's no such function, and we're done.

Okay, hard problem. The generalization is quite normal though, but well I think someone did count number of functions when $N$ is even, right? Here's a solution... Would really love to learn arrows, alas! I don't get them a lot of time, how do people do this witch-craft?

[i]Solution:[/i] Label the original functional equation as $(\clubsuit)$ and let $N = 1987$. It is obvious that function $f$ is injective. By the triple involution trick, we have that
\[f(f(f(x))) = f(x + N) = f(x) + N.\]
This means it is sufficient to show that $f$ is not well-defined on $S \coloneqq \{0,1,2, \dotsc, N-1\}$.

[b][color=red]Claim:[/b][/color] For $r \in S$, $f(r) < 2\cdot N$.

[i]Proof:[/i] For an arbitrary $r \in S$, it easy to show by induction that $f(Nk + r) = Nk + f(r)$ for any $k \in \mathbb{Z}_{\ge 0}$. Put $x = Nk + r$ in $(\clubsuit)$. We get
\[f\left(Nk + f(r)\right) = r + N(k+1).\]
If $f(r) > 2\cdot N$, then upon writing $f(r) = 2\cdot N + \epsilon$ for $\epsilon \in \mathbb{Z}_{\ge 0}$, we get the desired contradiction. $\square$
-----------------------------------
 Let $r_i \in S$. From the above claim, if $f(r_1) = r_2$, then $f(r_2) = r_1 + N$. This creates a chain and pins down all $f(r_1 \pmod{N}), f(r_2 \pmod{N})$. Similarly, another possibility is $f(r_3) = r_4 + N$. Here, $r_3,r_4$ will create a [i]similar[/i] chain as $r_1,r_2$ did. Therefore, we can naturally pair $r_i$ which create these chains. But, since $N$ is odd. One of the elements will get paired to themselves. So, we get that there is at least one $m$ such that either $f(m) = m$ or $f(m) = m + N$ for $m \in S$.

If $f(m) = m$, then we put it in $(\clubsuit)$ to get a contradiction. This means $f$ can't have a fixed point. If $f(m) = m + N$, then again putting in $(\clubsuit)$ we would get $f(m + N) = m + N$ which gives a fixed point, impossible. Therefore there exists no such $f$ as desired. $\blacksquare$

[b]Remark.[/b] An obvious generalization follows. Clearly odd $N$ won't work and is well-illustrated in the above solution. If $N = 2k$, then $f(x) = x + k$ works.

[quote=torch]This problem should seem very familiar to MathPath applicants this year......[/quote]

Well, it isn't too uncommon for math camps to have unoriginal problems on their qualifying tests, and you can't really blame MathPath for that (although I agree, it is wrong to have a problem on your qualifying test intentionally directly based on an older problem).

Assume $f^k(n) = n$ for some $k \ge 1$ and $n \in \mathbb{Z}_{\ge 0}$. Then $n + 1987k = f^{2k}(n) = n$, a contradiction. Thus $f$ doesn't contain a cycle. Since $f$ is injective and $\text{Im}(f)$ contains all numbers greater or equal to 1987, hence the graph of $f$ is a finite number of chains. Call a positive integer startpos if it isn't in $\text{Im}(f)$. Then all chain has exactly one startpos. Assume there were $k$ startpos and $a_1, a_2, \dots, a_k$ be the startposes. Then in order to cover $\mathbb{Z}_{\ge 0}$, $\{a_1, a_2, \dots, a_k, f(a_1), f(a_2), \dots, f(a_k)\}$ must form complete residue class module $1987$. But $2k \neq 1987$, a contradiction. Hence there are no such function. $\blacksquare$


logarithms
function
complex analysis
complex analysis unsolved

I know of two ways to show that:
$1-\frac{1}{2}+\frac{1}{3}-... = \ln 2$.

1)Using Taylor's Theorem on $f(x) = \ln (1+x)$.

2)Using Abel's Theorem for the endpoints of $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$.

But is there a way to do this with Complex Analysis?
Because I have a way to show that,
$\ln (1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-...$ for $|x|<1$.

[b]Proof:[/b] Consider the complex logarithm function $f(z) = \ln (1+z)$, work in the neighborhood $|z+1|<1$. This function is holomorphic because we stay away from the branch on $(-\infty,-1]$. So $f(z)$ is in fact an analytic function, with $f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\cdot z^{n}$. Now if we chose $y=0$ in $z=x+iy$ we end up with:
$\ln (1+x) = x-\frac{x^{2}}{2}+...$ for all $|x|<1$.

1)But can we modify this somehow to include more points?

2)Does Abel's theorem generalize to complex series?

[quote="Kent Merryfield"]Abel's theorem does generalize to complex series with no problems. But at the end of all of this, you're not producing an argument that was any different than what you already had.[/quote]
Yes, I understand that, but that was a seperate question.

More specifically my question is:
Can we somehow use the remarkable property that holomorphic functions on open disks are analytic functions to show that it does indeed converge on that point.

Of course you can get convergence at $ z=1$ for the function $ \ln(1+z)$ provided you define the branch cut in the usual way.  (or so that the cut joins the branch points at $ -1$ and $ \infty$ in such a way that it avoids the positive real axis, but that would be non-standard).

The idea is just to use "analytic continuation".  You have established the fact that $ \ln(1+z)$ coincides with a particular Taylor series on a disk.  By the "Uniqueness theorem", you can carry that correspondence to anywhere $ \ln(1+z)$ is analytic which would include the point $ z=1$, thus giving you the complex proof you desire.

Edit:  After posting, I realized that the last part of what I wrote was misleading and probably not too helpful.  Yes, $ \ln(1+z)$ is the unique analytic continuation of the power series beyond the disk and is valid wherever it is analytic.  That does not mean that the original power series will similarly converge at all of those points.  In fact, here the radius of convergence is not increased.  But $ \ln(1+z)$ can be used to calculate the sum of the power series at a particular point provided it converges there.  It so happens that it does converge at $ z=1$ which means we can just evaluate it as $ \ln 2$, but that does not prove it converges there.  That must be established separately.  Sorry, I will be more careful before posting.


number theory
greatest common divisor
relatively prime

let (a,b,c) denote the GCD of a,b,c
let [a,b,c] denote the lowest common multiple of a,b,c

prove: xyz=(xy,yz,xz)[x,y,z]

Not a pretty proof, but a proof nonetheless:

[hide]Assume x,y,z are pairwise relatively prime. Then (xy,yz,xz) is 1 and [x,y,z] is xyz.

Assume that x,y are divisible by n, x/n,y/n are relatively prime, and z is relatively prime to both of x and y. Then (xy,yz,xz) is n and [x,y,z] is xyz/n (because n appears twice in the product).

Assume that x,y are divisible by a (maximum possible), y,z are divisible by b (maximum), and x,z divisible by c (maximum). Then (xy,yz,xz)=abc and [x,y,z]=xyz/abc because each of a,b,c appears twice in the product.[/hide]

This does not work for three numbers.
[hide="Counter-Example"]Let $x = abc$, $y = cde$, and $z = efg$. $LCM(x,y,z)$ is then $abcdefg$. The $GCF$ is 1,  and the product of $xyz$ is $abc^2de^2fg$ so this does not work. It does, however, work always with two numbers.[/hide]

Tarquin: I suggest you reread the problem. It asks to prove that (xy,yz,xz)[x,y,z]=xyz, not that (x,y,z)[x,y,z]=xyz. Considering this, work out your "counter-example" again. You will find that it does, in fact, work.

xyz=(x, y, z)[xy, yz, zx]
is true also.

jb05, what case does 2, 2, 2 fall under? There are quite a few more trivial cases. :)


function
real analysis
real analysis unsolved

Let A,B be two finit and disjoint sets(throng, "multimi" in romanian)
and the functions f:R->R with the following properties:
a) f is continuous on R-A and discontinuous on A
b) g is continuous on R-B and discontinuous on B
Prove that if f and g have PD (Darboux's property) then f+g has also PD.

It's in fact Intermediate Value Property (which is also in Romanian, rather than PD) we can use IVP :).


MATHCOUNTS
LaTeX
algebra
polynomial
complex numbers

Solve for $ x$:

$ \sqrt{10x^{2}}\minus{}2\equal{}\sqrt{5x\minus{}25}$

Be sure to check for extraneous solutions. And put your answer in simplest radical form. No decimals!

I think the answer is a complex number... :maybe:

Yeah, I was intending for you to find the complex solutions.

The relevant quartic is $ 100x^4\minus{}100x^3\plus{}445x^2\minus{}290x\plus{}841\equal{}0$, which doesn't factor nicely.

This is where quartic formula should be used.

Are you serious? Who would be crazy enough to do that? :huh:  :o  :wallbash_red:

[quote="12markkram34"]I did today in [b]Algebra II class[/b]. The teacher made some typos and gave us this problem instead of a simpler one.[/quote]

Put this in HSB then, MATHCOUNTS is obviously only for MATHCOUNTS problems and topics.

[quote="12markkram34"]I did today in Algebra II class. The teacher made some typos and gave us this problem instead of a simpler one.[/quote]

Oh really? And the solutions are?

[quote="12markkram34"]I did today in Algebra II class. The teacher made some typos and gave us this problem instead of a simpler one.[/quote]

Can you solve this with RRT, Descartes', etc.?

I'm too lazy to try, but can at least two of the roots be found?

Nope, the roots are all complex. :wink: I'm still waiting for 12markkram34 to post his solutions.

There are no solutions in the reals, and I don't think radicals are well defined for complex numbers (at least, complex numbers are not MC level).

Actually radicals are well defined...

just take the root of the magnitude of the complex and halve the argument of the complex and you got it.

Isn't the formula for quartics like several pages long and requires an extremely deep understanding of advanced topics to generate?

 :o

[b]james4l:[/b] That's the usual convention, but it's not totally standard.
[b]mathemagician1729:[/b] [url=http://planetmath.org/encyclopedia/QuarticFormula.html]The quartic formula[/url]: real pretty, right? :P

Who has enough time to type that out and check for latex syntax errors?  I tried to program the cubic formula in my calculator and gave up in 5 minutes...

...and what are the other conventions at taking exponentials of complex numbers?

[quote="12markkram34"]Solve for $ x$:

$ \sqrt {10x^{2}} \minus{} 2 \equal{} \sqrt {5x \minus{} 25}$

Be sure to check for extraneous solutions. [b]And put your answer in simplest radical form. No decimals![/b][/quote]

[quote="12markkram34"]The four roots are approximately $ 0.8824555320336759\pm1.4564587992741715i$ and $ \minus{} 0.382455532033676\pm1.65943597828203i$

Just plug and play to get rid of the extraneous solutions.[/quote]

Good job! I'm glad you can punch in numbers to polynomial root finders and follow your own instructions!


Let $a,b,x\in \mathbb{N}$ with $b>1$ and such that $b^{n}-1$ divides $a$. Show that in base $b$, the number $a$ has at least $n$ non-zero digits.

This seems not from the ISL 1996...

*note to self: $ x$ should be $ n$*

[edit: ok, I think it is from ISL 1993 instead.]

[edit 2: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=128468]here[/url] is a special case of this problem.]

[edit 3: Now all we need is a solution...]


Let the (x, y)-plane be divided into regions by n lines, any two of which may or may not intersect. Describe a procedure whereby these regions may be colored using only two colors so that regions with a common line segment
as part of their boundaries have different colors.

"Show that the plane divided by $ n$ lines can be colored in exactly two colors".

This is an induction proof. It is certainly true for $ n=1$. Now say it is true for $ n=k$. So draw $ k$ lines and we know we can color them all by hypothesis. Now place the $ k+1$ line and [b]invert[/b] all the colors to the right of this line. So now you have successfully colored a $ k+1$ line region.


1- In the equations

$ f(x)\equal{}|3x\minus{}a|$,and $ f(f(x))\equal{}x$ ,and
$ x$and $ a$ represent real numbers. If the sum of the x-values satisfying the pair of equations is 27, find the value of $ a$.

2- solve

 $ \arcsin(\frac{1}{2})\equal{}x$

For the second problem, take the sin of both sides: $ \frac{1}{2}\equal{}\sin{x}\Rightarrow x\equal{}\frac{\pi}{6}, \frac{5\pi}{6}$


algebra
polynomial
trigonometry
complex numbers
geometry unsolved
geometry

We can use complex numbers.  Let $P_{1}$, ..., $P_{n}$ be the $n$th roots of unity, with $P_{1}= 1$.  Then the problem translates to showing that \[\prod_{\omega \neq 1}|1-\omega| = n \, ,\] where $\omega$ ranges over all the $n$th roots of unity.  We will actually prove the more precise identity \[\prod_{\omega \neq 1}(1-\omega) = n \, .\] Let $f(z) = z^{n-1}+z^{n-2}+\cdots+1$.  The roots of the monic polynomial $f$ are the $n$th roots of unity besides $1$.  Thus \[f(z) = \prod_{\omega \ne 1}(z-\omega) \, .\] Let's plug in $z = 1$.  We get \[f(1) = \prod_{\omega \ne 1}(1-\omega) \, .\] That's exactly the product we are interested.  Plugging into the definition of $f$, we have $f(1) = n$.  Thus the product is $n$.

Is this right??  :P

Let O be the center of the unit circle. The n equal angles formed at the center joining consecutive vertex with O measure 360/n. Let x=180/n, then those angles measure 2x. The line P1Pk (k=2,3,..,n) using cosines law is [using the fact that the angle P1OPK measures 2(k-1)x]:

P1Pk^2=|P1O|^2+|PkO|^2-2|P1O||PkO|cos[2(k-1)x]
          =1+1-2cos[2(k-1)x]
          =2(1-cos[2(k-1)x])
          =2*2*{sin[(k-1)x]}^2 (using half-angle formula)

then

P1Pk=2*sin[(k-1)x].

Then we have that:

|P1P2|*|P1P3|*...*|P1Pn|=2^(n-1)*A

Where A = sin x * sin 2x *...*sin[(n-1)x]
            =sin[1*180/n]*sin[2*180/n]*...*sin[(n-1)*180/n]
            =2^(1-n)*n

The  we have that:

|P1P2|*|P1P3|*...+|P1Pn|=2^(n-1)*A=2^(n-1)*2^(1-n)*n = n

[quote]=sin[1*180/n]*sin[2*180/n]*...*sin[(n-1)*180/n]
=2^(1-n)*n
[/quote]
Im a little unclear how you got from the former to the latter equality, probably just a brain fart.

I got that equality form here:

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

thats a really cool page, and an equally cool identity.

[quote="jamiemorgan88"]Label the the vertices of a regular n-gon inscribed in a unit circle cyclically ($P_{1},P_{2},...,P_{n}$).Show that \[\left|P_{1}P_{2}\right|\left|P_{1}P_{3}\right|...\left|P_{1}P_{n}\right|=n\] [/quote]

You must mean "a problem from Prof. Vakil's [i]book[/i]".  The product of diagonals in a regular n-gon, and the amazing solution using complex numbers, is an ancient and famous problem.


yek soale gashang

Agar $ x,y,z\ge 0$ va $ x^2\plus{}y^2\plus{}z^2\equal{}1$  ,sabet konid

\[ x\plus{}y\plus{}z\plus{}\frac{1}{xyz}\ge 4\sqrt{3}\]

ba makhraj moshtark giri darim:
$ x^2yz\plus{}xy^2z\plus{}xyz^2\plus{}1\geq 4\sqrt{3}xyz$
hala age 1 hesaB hendeC beyne 3 jomle aval taraf chap begirim darim:
$ 3xyz\sqrt[3]{xyz}\plus{}1\geq4\sqrt{3}xyz$
hala bejaye 1 moadelesho miarim darim:
$ 3xyz\sqrt[3]{xyz}\plus{}x^2\plus{}y^2\plus{}z^2\geq4\sqrt{3}xyz$
hala age dobare 1 hesabi hendesi beyne jomalat samte chap bezanim darim:
 $ 3xyz\sqrt[3]{xyz}\plus{}x^2\plus{}y^2\plus{}z^2\geq4\sqrt[4]{3x^3y^3z^3\sqrt[3]{xyz}}$ ke in bayad
az $ 4\sqrt{3}xyz$ bozorgtar bashe ke in moadele ine ke sabet konim:
$ \frac{1}{\sqrt{3}}\geq\sqrt[3]{xyz}$ ke esbate inam baye morabae hendec va inke $ x^2\plus{}y^2\plus{}z^2\equal{}1$ hale :)

sb-555 rahe hallet kheily gashang bood

Hala yek soale jadid

Agar $ a,b,c>0$ va $ a^2\plus{}b^2\plus{}c^2\equal{}1$ sabet konid
\[ a\plus{}b\plus{}c\plus{}\frac{1}{abc}\ge \frac{4\sqrt{3}}{9}(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\]

[quote="khashi70"]Man ye rahe hal vase soali ke gozashti daram vali benazare khodam kheyli ghashang nist :
az oonjayi ke $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$ pas $ a \plus{} b \plus{} c\leq \sqrt 3$
pas kafie neshoon bedim :
$ a \plus{} b \plus{} c \plus{} \frac {1}{abc}\geq \frac {12}{9}(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$
ba zarb do taraf too $ abc$ hokm moadele ine ke neshoon bedim:
$ 3(a \plus{} b \plus{} c)abc \plus{} 3\geq 4(ab \plus{} bc \plus{} ca)$
ke inam age hamgenesh konin bayad sabet konim :
$ 3(a \plus{} b \plus{} c)abc \plus{} 3(a^{2} \plus{} b^{2} \plus{} c^{2})^{2}\geq 4(ab \plus{} bc \plus{} ca)(a^{2} \plus{} b^{2} \plus{} c^{2})$
ke inam moadele ine ke :
$ (a \minus{} b)^{4} \plus{} (b \minus{} c)^{4} \plus{} (c \minus{} a)^{4} \plus{} (a^{4} \plus{} b^{4} \plus{} c^{4} \minus{} a^{2}bc \minus{} b^{2}ac \minus{} c^{2}ab)\geq 0$ ke inam ba $ AM \minus{} GM$ ya muirhead badihie .
pas masale halle .[/quote]

Rahe halle man ham hamine :wink:

Agar $ a,b,c>0$ va $ a^2\plus{}b^2\plus{}c^2\equal{}3$ sabet konid
\[ 5(a\plus{}b\plus{}c)\plus{}\frac{3}{abc}\ge 18\]


Let $\ell$ be a line through $G$, the centroid of $\triangle ABC$.  Let $a,b,c$ denote the distances from $A,B,C$ to $\ell$.  Prove that
\[ a^3+b^3+c^3+2abc = ab(a+b)+bc(b+c)+ca(c+a). \]

Factor that as $(a+b-c)(a-b+c)(-a+b+c) = 0$. Now, note that the line $\ell$ necessarily separates the triangle in two as it passes through the interior; WLOG (as the identity is symmetric) let $B, C$ be the two vertices who lie in the same halfplane. Then, if $M$ is the midpoint of $BC$ and the foot from it to $\ell$ is $P$, we have $b + c = 2MP = a$ where the first equality follows from obviousity (ok, trapezoid) and the second from similarity as $\frac{AG}{GM} = 2$.


symmetry
absolute value
combinatorics proposed
combinatorics

At the vertices of a regular octagon are written eight nonnegative integer whole sum is odd.Peter is allowed to make moves of the following form;he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices.Prove that Peter can make a sequence of moves,after which the number $0$ appears at all eight vertices.

The condition that the sum of all integers is odd in unnecessary. I won't use it in the proof below.

Make a move at a largest integer among all vertices (if there several such integers make move at each of them in consecutive manner). Such move cannot increase the original integer. And if it decrease the number, it also decreases the sum of all numbers at the vertices.
Therefore, after a finite number of moves we end up with a stable configuration where a move applied to any largest integer does not change it.

Let $a$ be a largest integer in a stable configuration. Then one of its neighbors must equal $a$ while the other must be $0$. Similarly, yet another neighbor of the latter integer $a$ must be $0$ as well. Hence, in four consecutive vertices of the octagon we have integers $0,a,a,0$. Forget about them for a while and focus on the other 4 vertices of the octagon. After a finite number of moves at the largest integers there, we end up with one of the following two configurations (up to symmetry):
$0,a,a,0,0,b,b,0$
$0,a,a,0,b,b,0,c$
where $b$ is the largest integer besides $a$.

Now both configurations can be trivially reduced to all-zeros:
$0,a,a,0,0,b,b,0\to 0,a,a,0,0,b,b,b\to 0,a,a,0,0,b,0,b\to 0,a,a,0,0,b,0,0$
$\to 0,a,a,0,0,0,0,0\to 0,a,a,a,0,0,0,0\to 0,a,0,a,0,0,0,0\to 0,a,0,0,0,0,0,0$
$\to 0,0,0,0,0,0,0,0$
and
$0,a,a,0,b,b,0,c \to 0,a,a,0,b,b,0,0 \to 0,a,a,0,b,b,b,0 \to 0,a,a,0,b,0,b,0$
$\to \dots$ as above $\to 0,0,0,0,0,0,0,0.$


do you think it would help? at all...
give opinions for this...

or just get a ti89 and it's really easy

jorian

:what?:  You can find factors with a calculator?

A [i]graphing[/i] calculator that has  programming capabilities, yes...

@Jorian:  yup... I wish you could bring TI89s to MATHCOUNTS.

Well like you can so...your comment becomes pointless?
But if you want to buy a TI-89 do it because you are too lazy to learn to use mathematica because it is a HUGE bargain as it's thosands of times more powerful than a TI-89, and it's price isn't drastically higher

If you want a TI-89 to cheat on tests (although that's going away now, no more punching away on AMC), don't do it only for Mathcounts, buy it because your clueless AP Calc teacher worships it and forces you to learn to do calculus on it, which actually turns out to be somewhat useful, in DESPERATE situations. Like OMG I need to calculate this integral NOW and it's really complicated and all I have is my TI-89.

just a quick tip:
i, also, have about 50 programs on my TI to do all this weird stuff (just in case i need them), but i found that its better just to memorize or derive them.
one, you'll understand them better, and two, many competitions (even the AMC now) do not allow calculators.
just my two cents. =)

[quote="footballrocks41237"][quote="mgao"]if this will help:

[hide="how to do a factor program on TI calc"]
prgm FACTOR
:Lbl A
:Input "NUMBER TO FACTOR?",D
:1->F
:Lbl B
:D/F->E
:round(E,0)->G
:If E=G and F$\le$D
:Disp F
:F+1->F
:If F>D
:Goto A
:Goto B
[/hide]

The program will not pause and wait for pressing enter between factors, I couldn't work that out.

[hide="example"]
_____________________
|NUMBER TO FACTOR?18|
|:::::::::::::::::::::::::1|
|:::::::::::::::::::::::::2|
|:::::::::::::::::::::::::3|
|:::::::::::::::::::::::::6|
|:::::::::::::::::::::::::9|
|::::::::::::::::::::::::18|
_____________________

The :::::'s are blank space...
[/hide][/quote]

If you have your link cable...and your calculator isn't being stupid like mine...then you can download this nifty little program from Ticalc.org

http://www.ticalc.org/archives/files/fileinfo/67/6783.html[/quote]

Or, if you want it to pause, change the [b]disp[/b] to a [b]pause[/b] in the line: Disp F

So, it should read:

:Pause F

[quote="erdogankerem123"][quote="footballrocks41237"][quote="mgao"]if this will help:

[hide="how to do a factor program on TI calc"]
prgm FACTOR
:Lbl A
:Input "NUMBER TO FACTOR?",D
:1->F
:Lbl B
:D/F->E
:round(E,0)->G
:If E=G and F$\le$D
:Disp F
:F+1->F
:If F>D
:Goto A
:Goto B
[/hide]

The program will not pause and wait for pressing enter between factors, I couldn't work that out.

[hide="example"]
_____________________
|NUMBER TO FACTOR?18|
|:::::::::::::::::::::::::1|
|:::::::::::::::::::::::::2|
|:::::::::::::::::::::::::3|
|:::::::::::::::::::::::::6|
|:::::::::::::::::::::::::9|
|::::::::::::::::::::::::18|
_____________________

The :::::'s are blank space...
[/hide][/quote]

If you have your link cable...and your calculator isn't being stupid like mine...then you can download this nifty little program from Ticalc.org

http://www.ticalc.org/archives/files/fileinfo/67/6783.html[/quote]
I tried the link, but it won't work... anyone else tried it yet?
It is probably easier to make the program yourself rather than downloading it. It's a fairly simple program.[/quote]

I have not tried the link, but just curious; this program isn't especially helpful in Mathcounts, is it. It is much faster to use formulas than it is to run this programs.


Given positive integers $ a,b,c$ such that $ \gcd(a,b,c)\equal{}1$ and $ \gcd(a,b), \gcd(b,c),\gcd(c,a)$ are coprime and greater than 1, what is the largest number that cannot be expressed in the form $ xa\plus{}yb\plus{}zc$ for nonnegative integers $ x,y,z$.

[quote="SplashD"]Given positive integers $ a,b,c$ such that $ \gcd(a,b,c) \equal{} 1$ and $ \gcd(a,b), \gcd(b,c),\gcd(c,a)$ are coprime and greater than 1, what is the largest number that cannot be expressed in the form $ xa \plus{} yb \plus{} zc$ for nonnegative integers $ x,y,z$.[/quote]
Although this isn't the same, perhaps it can help?
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366618#366618


inequalities
geometry
analytic geometry
geometry proposed

I'm sure this is well known, I however couldn't find it.

Let $ABC$ be a triangle and let $X,Y,Z$ be points on the segments $BC,CA,AB$ respectively so that $AX, BY, CZ$ are concurrent.
Show that $[XYZ]\leq [ABC]/4$ where $[\ldots]$ denotes the area of $\ldots$.

I can help you a little to remember it by the well know formula of barycentric coordinate.
If concurrent point has barycentric coordinate $(x,y,z)$ then
$S_{XYZ}=\frac{2xyz}{(x+y)(y+z)(z+x)}S_{ABC}$

Denote $\frac{ZA}{ZB}=z,\frac{XB}{XC}=x,\frac{YC}{YA}=y.$ We have $xyz=1$ by Ceva's theorem.

Now,  $\frac{[XYZ]}{[ABC]}=1-\frac{[AYZ]}{[ABC]}-\frac{[BZX]}{[ABC]}-\frac{[CXY]}{[ABC]}.$

And $\frac{[AZY]}{[ABC]}=\frac{AZ.AY}{AB.AC}=\frac{z}{(z+1)(1+y)}$, $\frac{[BZX]}{[ABC]}=\frac{x}{(1+z)(1+x)}$, $\frac{[CXY]}{[ABC]}=\frac{y}{(1+x)(1+y)}$. Therefore we need prove

$\frac{z}{(z+1)(1+y)}+\frac{x}{(1+z)((1+x)}+\frac{y}{(1+x)(1+y)}\geq \frac{3}{4}$. But it is easy :D


$x-y\sqrt{x^2-y^2}=a\sqrt{1-x^2+y^2}=I$

$y-x\sqrt{x^2-y^2}=b\sqrt{1-x^2+y^2}=II$
[b]____________________________[/b]
$I^2\Rightarrow x^2+y^2(x^2-y^2)-a^2(1-x^2+y^2)=2xy\sqrt{x^2-y^2}$
$II^2\Rightarrow y^2+x^2(x^2-y^2)-b^2(1-x^2+y^2)=2xy\sqrt{x^2-y^2}$

- - - - - - - - - - - - - - - - - - - $\Downarrow$ - - - - - - - - - - - - - - - - - - - - - - -

$\boxed{x^2+y^2(x^2-y^2)-a^2(1-x^2+y^2)=x^2(x^2-y^2)+y^2-b^2(1-x^2+y^2)}$
$\Updownarrow$
$\boxed{(x^2-y^2)^2+(b^2-a^2-1)(x^2-y^2)+a^2-b^2=0}$

Solving the last equation as quadratic in unknown $z=x^2-y^2$ yields:
$z_{1}=x^2-y^2=0,z_{2}=x^2-y^2=a^2-b^2+1$

Substitution of these into the inital system finally gives:

a)$z_{1}$ corresponding to the solution [b]x=y[/b]=a=b

b)$z_{2}$ corresponding with linearized system:
$x-y\sqrt{a^2-b^2+1}=a\sqrt{b^2-a^2}$
$y-x\sqrt{a^2-b^2+1}=b\sqrt{b^2-a^2}$

and its solutions :
[b]x=[/b]$\frac{\sqrt{b^2-a^2}}{b^2-a^2}(a+b\sqrt{a^2-b^2+1)}$
[b]y=[/b]$\frac{\sqrt{b^2-a^2}}{b^2-a^2}(b+a\sqrt{a^2-b^2+1)}$

(Discussion regarded:consider the definition in $\mathbb{R}$ with respect to the condition $y^2\leq x^2\leq y^2+1$)

[color=indigo]I admit I'm surprised that such a  cutie  remained without a single reply here!
It was given to 16 year old students (10th grade) on national competion in former Yugoslavia held nearly 17 year ago![/color]


MATHCOUNTS
geometry
geometric transformation
reflection
algebra
polynomial

Yes, although normally one would post them in the Getting Started, Intermediate, or Advanced forum (depending on the difficulty) or possibly in the Mathcounts or AMC forum (if it were similar to competition problems).

[quote="mathnovice"]Am I allowed to post exercises from a math book?

Several math books that I have have exercises/ problems but no solutions, and I was wondering if I could post them here to see if anyone had the solution.[/quote]

Most mathematics discussion forums, for example the sci.math forum on Usenet, have participants who live in fear of their forums becoming homework-help forums, and yet questions like this are tolerated. Many participants give full solutions to questions that are clearly homework problems, as long as they think the original poster made an honest effort to solve the problem before asking. It appears that you are asking about posting questions from books that you are using for your self-study, which should make the questions very welcome here.

As for copyright concerns, if you quote a LIMITED number of problems from a math book, credit them to their original source, and don't take money for the answers you get, you should be in the clear (under the "fair use" doctrine) to post copyrighted material in this manner.

As for academic honor code concerns, IF you are thinking about posting exercises from class homework, then if your class teacher says that you are permitted to form a study group, or to ask for hints from other persons, you should be okay with asking questions here. But if your teacher says that all your solutions must reflect your own individual work ALONE, with no outside aids permitted, then I think you shouldn't ask here about school homework. (Although, in that case, you would still be in the clear to ask about similiar questions you see in other, supplemental books.)

As for netiquette issues, I would post each question in its appropriate level on one of the three problem-solving forums here: the Getting Stared Forum (probably below the level you are working at), the Intermediate forum (probably right where you are these days), or the Advanced forum (which wouldn't include much high school homework from a typical United States high school, but might include some of what you are working of for self-study). Label each question a "Question:" in the subject line (so that we all know that the posts are not challenge problems, and are thus fair game for anyone to answer) and include enough of what the question is about to alert people to what the question is about. (As an example, you could post something labeled "Question: Third-Degree Polynomials" or "Question: Cramer's Rule" or whatever.) It would be extremely helpful to other participants to cite the problem exactly enough (e.g., "exercise 7 from chapter 5 on page 214") so that those of us who have the same book can LOOK IT UP.

I would be glad to see more questions of the kind you are proposing here on the AoPS Web site as long as we respect the legalities and academic honor code issues involved.

Best wishes on your studies.


第二届东南数学奥林匹克将于7月上旬(5-13日)在福州一中举行.闽浙赣,MO协作体和国外50多所中学的高一学生参加.

[quote="liyi"]are you 林常?[/quote]
   YES.

   在MATHLINKS久仰您的大名,我们似乎还在同一城市,只是无缘相见.

其实是我久仰您.
我只是一个普通的学生而已, 以前我参加数学夏令营什么的都见过您很多次.
因为我不搞数学竞赛, 成绩不好, 所以您不认识我.

林常教授，真荣幸能在这里遇到您。我在去年福州一中的培训中听过您的课。

林老师：
我把您在橙子奥数工作室上发的第二届东南数学奥林匹克试题翻成了英文，发在

http://www.mathlinks.ro/Forum/viewtopic.php?t=44857

请您多指正错误！

谢谢ZHAOLY.您的译文比命题组的要好.

请问,在那里能找到 东南MO 的结果和官方解法?

[quote="lin"]

     http://www.aoshoo.com/bbs1[/quote]
Thanks you very much, lin.


Prove that among (ab+1) mice, there is a sequence (a+1) mice of which one of is descended from the preceeding, or there are (b+1) mice of which none descends from the other.

I guess you guys don't like mice. This is well known.


Given that $ a, b, c,$ and $ d$ are four distinct digits and $ \frac{a}{b}$ and $ \frac{c}{d}$ are proper fractions, what is the greatest value for $ \frac{a}{b}\plus{}\frac{c}{d}$? Express your answer as a mixed number.

Isn't it infinite? \[ \frac{n\minus{}2}{n\minus{}1}\plus{}\frac{n\minus{}1}{n}\]

We are given the restriction that $ a,b,c,d$ are digits

$ \frac{8}{9}\plus{}\frac{6}{7}\equal{}\boxed{1\frac{47}{63}}$

Wait is there a "legit" way of doing this?


Choose two multidigit numbers.  Announce the numbers and use your calculator to get their product, but do not announce it.  Instead, pick one of the nonzero digits in the product and then read off all of its other digits in any order.  Explain mathematically how it is always possible to predict the omitted digit.

I've been trying to get this one for nearly two months now.  Any help would be appreciated.

[hide="hint"]take everything mod $ 9$[/hide]


inequalities
induction
integration
calculus
inequalities unsolved

[quote="arqady"][quote="cezar lupu"]Arqady, your inequality can be proved without induction very easy. ;)[/quote]
Yes! For instence, use $\int^n_1\frac{1}{\sqrt x}dx$ :)[/quote]

If possible , no calculus please ,an elemantary solution is prefered :)

[quote="bomb"]Use pairwise summation

Bomb[/quote]

At least elaborate abit instead of just stating " pairwise summation " ..... well I only manage to get the lower boundary using " pairwise summation " but still cant figure out how it works on the upper boundary  :?

Use AMGM to come up with something, I can't remember waht and then you add all up.
Done for RHS.

Bomb

[quote="shyong"]
If possible , no calculus please ,an elemantary solution is prefered :)[/quote]
$2(\sqrt{n+1}-\sqrt n)=\frac{2}{\sqrt{n+1}+\sqrt n}<\frac{2}{\sqrt n+\sqrt n}=\frac{1}{\sqrt n}.$
Hence, $1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt100} >2(\sqrt2-\sqrt1+\sqrt3-\sqrt2+...+\sqrt{101}-\sqrt100)=2(\sqrt101-1)>2(10-1)=18.$
$2(\sqrt n-\sqrt{n-1})=\frac{2}{\sqrt n+\sqrt{n-1}}>\frac{2}{\sqrt n+\sqrt n}=\frac{1}{\sqrt n}.$
Hence, $\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+...+\frac{1}{\sqrt100}<2(\sqrt2-\sqrt1+\sqrt3-\sqrt2+...+\sqrt100-\sqrt99)=2(\sqrt100-\sqrt1)=18.$
Hence, $18<1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt100}<19.$ :roll:

Yah , thats what I want arqady , thank you  :D


I seem not to be able to use [img id=] anymore in the classroom...

Probably, my computer is incredibly old with only 256mb of memory, seeing as my dad took my 2gb memory/updated one.  :(

During the MIT Math Jam, Dr. Patrick said custom avatars had been removed.

we should gather everyone and vote on this. i dunno. just a suggestion.

[quote="Valentin Vornicu"][quote="coolts"]we should all have a huge voting rebellion to have the avatars back...[/quote]We would just punish the starter/leader of the rebellion :)[/quote]

:)

i personally don't care cuz i never used them anyway.


greatest common divisor
number theory
relatively prime

Prove that $2^{2^{m}}+1$ and $2^{2^{n}}+1$ are relatively prime for any $n, m \in \mathbb{Z^{+}}$.

[hide]
(We are assuming $m \neq n, m > n$)

Assume that $\mu = 2^{2^{m}}+1$ and $\nu = 2^{2^{n}}+1$ are not relatively prime, i.e. $gcf(\mu,\nu) = k$ and $k \neq 1$.

Let $\phi = 2^{2^{m}-2^{n}}$.  It is clear that $(\nu-1) \phi = \mu-1$, and $\phi \nu = \mu-1+\phi$.

Hence $\phi \nu = k \nu-1+\phi$.  Thus, $\nu (\phi-k) = \phi-1$ so $\nu = \frac{\phi-k}{\phi-1}$.

If $k > 1$, $\nu$ is not an integer.  Thus, $k = 1$.  We have a contradiction, thus $gcf(\mu,\nu) = 1$.

[/hide]

Proofs are not generally allowed in the Getting Started Forum, so I will move it to Intermediate.

Are you kidding me? All of the problems in Getting Started are proofs  :lol:

[quote="Teki-Teki"]
Hence $\phi \nu = k \nu-1+\phi$.
[/quote]

Where is this coming from? Just because $v$ and $u$ aren't relatively prime doesn't mean that one is a multiple of the other.

Of course, the problem needs the condition $m \neq n$.

[hide="One line"] WLOG $m < n$.  Then $2^{2^{m}}+1 | 2^{2^{n}}-1$. [/hide]


[hide]There are $4!=24$ different four-digit integers that can be made from 1, 2, 3, and 4.

Each digit appears in each place $6$ times. $6(10000+1000+100+10)=\boxed{66660}$.[/hide]

[quote="DUKEFAN"]What is the sum of all four-digit positive integers that can be written with the digits 1,2,3 and 4 if each digit must be used exacty once in each four-digit integer?
Is there a quick way to do this problem? It takes me forever![/quote]
[hide]
there are $4!$ four-digit numbers. there are $6(1+2+3+4)$ sums for each digit, adding to $60$ for each digit. so it is $60(1111)$, which is $66660$
[/hide]


Find all triplets of positive integrers such that $\frac{1}{a+2}+\frac{1}{b+2}=\frac{1}{2}+\frac{1}{c+2}$.

For $a \geq 2$ and $b \geq 2$, $L.H.S. \leq \frac{1}{2}$, which gives no solutions.
Therefore, both $a$ and $b$ cannot be greater than 1. So, without loss of generality, let $a = 1$. For convenience, let $b+2 = y$, $c+2 = z$. This gives
\[ \frac{1}{3}+\frac{1}{y} = \frac{1}{2} + \frac{1}{z} \]
\[ \frac{z-y}{yz} = \frac{1}{6} \]
\[ z = \frac{6y}{6-y} = \frac{36}{6-y}-6 \]
For $z$ to be an integer, $y$ can only be $3,4,5$ and correspondingly $z$ can be $6,12,30$. So the solutions $(a,b,c)$ are $(1,1,4)$ $(1,2,10)$ $(1,3,28)$.


Three non-zero real nos. a,b,c are said to be in harmonic progression if 1/a +1/c=2/b.find the three term harmonic progressions a,b,c of strictly increasing positive integers in which a=20 and b divides c.

We have $ \frac{1}{a} \plus{} \frac{1}{c} \equal{} \frac{2}{b}$ with $ a \equal{} 20$, $ b$ dividing $ c$ and $ a < b < c$.
Substituting $ a \equal{} 20$ in the equation, we get

$ b(c\plus{}20) \equal{} 40c$

Since $ b$ divides $ c$, we let $ c \equal{} bk$ where $ k$ is a positive integer. This gives us

$ b(bk\plus{}20) \equal{} 40bk \Rightarrow bk \equal{} 40k \minus{} 20 \Rightarrow b \equal{} 40 \minus{} \frac{20 }{k}$.

Hence, $ k$ divides $ 20$ and $ k > 1$ since $ b > a$.
For $ k \equal{} 2$, we get $ a \equal{} 20, b \equal{} 30, c \equal{} 60$.
For $ k \equal{} 4$, we get $ a \equal{} 20, b \equal{} 35, c \equal{} 140$.
For $ k \equal{} 5$, we get $ a \equal{} 20, b \equal{} 36, c \equal{} 180$.
For $ k \equal{} 10$, we get $ a \equal{} 20, b \equal{} 38, c \equal{} 380$.
For $ k \equal{} 20$, we get $ a \equal{} 20, b \equal{} 39, c \equal{} 780$.


Let $a,$ $b$ and $c$ are non-negative numbers such that $(a+b)(a+c)(b+c)=8.$
Prove that $\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^{3}+b^{3}+c^{3}}{3}}.$

anything to do with $\frac{a+b+c}{3}\geq 1$ ?? ...hmm...so how do u do it?? :blush:

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=945097#p945097


Found a pretty fun game from the xkcd forums.

Basically, you gain as much time that separates the last post and your post.

So e.g.

A: 4:00 (first)
B: 4:23
C: 4:26
B: 4:41
A: 5:50

A has 1:09, B has 23+15=0:38, and C has a mere 3 min.

Additional rules:
1) You win if you satisfy both of the following: a) you have OVER NINE THOUSAND minutes = 150 hours = 6.25 days, and b) your accumulated time is at least twice as much as the runner-up.
2) No doubleposting ever. (you can however delete your previous post and repost but only if you were the last to post)
3) No moderator abusements.
4) No editing.
5) Keep a running total scoreboard in hide tags.

[hide=SCORES][/hide]

[hide="SCORES!!!"]
[b]math_explorer: 4:04[/b]
[/hide]

This should be interesting. Reasonably. Suggestion: bold the people with most time?
(Excuse to bold my entry  :P )

1:58
[hide="SCORES!!!"][b]math_explorer: 4:04 [/b]
bluecarneal: 1:58[/hide]

+0:18  :rotfl:
[hide="SCORES!!!"]
math_explorer: 4:22
bluecarneal: 1:58
[/hide]

+2:14

[hide="scores"]math_explorer: 4:22
bluecarneal: 4:12[/hide]

+0:25

[hide="scores"]math_explorer: 4:22
bluecarneal: 4:12
cf249:  0:25[/hide]

+3:03
[hide="SCORES"][b]bluecarneal: 7:15[/b]math_explorer: 4:22
cf249: 0:25[/hide]

+0:23

[hide="scores"]bluecarneal: 7:45
math_explorer: 4:22
15!!!: 1:08
cf249: 0:25
ernie: 0:23
Richard_Min: 0:20 [/hide]

+0:53

[hide="scores"]bluecarneal: 7:45
math_explorer: 4:22
cf249: 1:18
15!!!: 1:08
ernie: 0:23
Richard_Min: 0:20 [/hide][/quote]

argh +0:15

[hide="scores"]bluecarneal: 7:45
math_explorer: 4:22
cf249: 1:18
15!!!: 1:08
ernie: 0:23
Richard_Min: 0:20
james4l: 0:15 [/hide]

+0:34

[hide="scores"]bluecarneal: 7:45
math_explorer: 4:22
15!!!: 1:42
cf249: 1:18
ernie: 0:23
Richard_Min: 0:20
james4l: 0:15 [/hide]

SIGH

I'll make necropost II sometime today.

It took 21:06:03 for you to win.


algebra
polynomial
modular arithmetic
number theory

A_n, n>=1 is a sequence of positive integers such that a_n=2a_(n-1)+n^2 for all integers n>1. Compute the remainder obtained when a_2004 is divided by 1000 if a_1=1

[hide="Method One? (Without characteristic polynomials)"]
Well, we can see that, if I didn't mess up the dummy variable,
\[a_{2004}= \sum_{k=1}^{2004}\left( k^{2}\cdot 2^{2004-k}\right) = 2^{2004 }\sum_{k=1}^{2004}\frac{k^{2}}{2^{k}}\]
which doesn't lead me anywhere, since the nice sum $\sum_{m=1}^\infty \frac{m^{2}}{2^{m}}= 6$ isn't applicable, and we can't reduce the power of the $2^{2004}$ without messing alot of stuff up in mods, since $2$ doesn't have an inverse in $\bmod 1000$. Can anyone salvage this?  :P [/hide]
[hide="Hint (Method Two)"] What is $b_{n}= a_{n+1}-a_{n}$?
[hide="Further Hint"] What is $c_{n}= b_{n+1}-a_{n}$?
[hide="Once you get a recursive equation, you can use"] [b] characteristic polynomials[/b]. The rest should be easy from there. [/hide]
[/hide]
 [/hide]

(EDITED: Method One)

You don't need all those steps...you just have to add a $an^{2}+bn+c$ to the equation that you get from your characteristic polynomial

Characteristic equations never need to come into it.

[quote="mdk"]$a_{n}, n\geq1$ is a sequence of positive integers such that $a_{n}=2a_{n-1}+n^{2}$ for all integers $n>1$. Compute the remainder obtained when $a_{2004}$ is divided by 1000 if $a_{1}=1$.[/quote]

[hide="Solution"]Let $a_{n}= b_{n}+an^{2}+bn+c$ for constants $a, b, c$.  Then for $n > 1$ we have \[b_{n}+an^{2}+bn+c = 2b_{n-1}+(2a+1)n^{2}+(2b-4a)n+(2c-2b+2a).\] Then choose $a =-1$, $b =-4$ and $c =-6$ so that we have $b_{n}-n^{2}-4n-6 = 2b_{n-1}-n^{2}-4n-6$ and thus $b_{n}= 2b_{n-1}$.  In addition, $1 = b_{1}-1-4-6$ so $b_{1}= 12$ and $b_{n}= 12\cdot 2^{n-1}$.  Thus $a_{2004}= 12\cdot 2^{2003}-2004^{2}-4\cdot2004-6$.

Now, there's still a little bit of number theory left to solve the question.  Fermat's Little Theorem and the Chinese Remainder Theorem give that $2^{2003}\equiv 2^{8}= 16 \pmod{1000}$, so our remainder is $12\cdot 8-16-16-6= 58$.[/hide]


HI
I'm a 14-year-old Chinese Senior highschool student and i'm going to take part in the AMC in 2009.
I want to know how could I prepare for the conpetition, now i have some knowledge of fuctions, geology, analytic geology, calculous, permutation, combination and equations. What should I learn next?
I'm now learning about vector quantity.

Problem solving, I guess.  It's what the AMC tests are all about.

If you haven't already, take some old years' tests, and assess from there.  Also, AoPS Volume 1 and AoPS Volume 2 can help you a whole bunch preparing for the AMC's.

Geology isn't gonna help you much on the AMC's either.

[quote="AIME15"]
Geology isn't gonna help you much on the AMC's either.[/quote]
I think (s)he means geometry.


Fred picked four numbers out of a hat. The average of the four numbers is 9. If three of the numbers are 5, 9 and 12, then what is the fourth number?

[hide]

The sum of the numbers is 9*4=36,
36-9-12-5=10
the fourth number is 10[/hide]


[quote="Jenia"]Could you help me to buy the thing in your Internet shop. I have
found the thing that is not sold in Russia. But there is a problem, the shop
does not send the goods to Russia. Compensation for your help is guaranteed.
Pls, write to e-mail afinans@rambler.ru[/quote]

To what are you referring?

The only internet shop I'm aware of would be the bookstore. Jenia, could you specify which items you would like and where the site details that it can't ship to Russia?


function
inequalities
inequalities unsolved

[quote="Shishkin"]$(\frac{1}{-x^{2}+2x+3})''=2(-x^{2}+2x+3)^{-3}*(7x^{2}-14x+11)>0$
for $1 \ge x \ge 0$. So by Karamatas inequality max get when x=1,y=0, z=0.
As $\frac{2}{3}+\frac{1}{4}=\frac{11}{12}$ we get what we need.The second is true.[/quote]
That's right it is only the Karamata inequality and Cauchy-Swart! and Good solution Gackiem!


geometry
rectangle
perpendicular bisector

3.Tony the Tortoise walked 10 inches during the first hour of his journey. He walked one and one-half times that distance during the second hour, and in gerneral during the nth hour he walked (1+1/n) times the distance walked during the n-1st hour. How many inches did Tony walk during the first three hours of his journey?

7.Terrell wondered how many blades of grass were in his 60-foot by 90-foot rectangular backyard. He picked a square region three inches on a side, which contained 520 blades of grass. If the grass was uniformly distributed throughout the backyard, how many blades were in the entire backyard?

8.Each circular pulley shown has a radius of 12 inches. The
shortest distance between the pulleys is 20 inches. What
is the length of the continuous belt that ioops around both
pulleys? Express your answer to the nearest whole number.

9.Two ladders, both 6 meters in length, are leaned up against opposite vertical walls in a 3-meter-wide corridor, as shown. How far above the ground do the two ladders cross? Express your answer as a decimal to the nearest tenth.[/hide][/url]

[quote="darit28"]Tony the Tortoise walked 10 inches during the first hour of his journey. He walked one and one-half times that distance during the second hour, and in gerneral during the nth hour he walked (1+1/n) times the distance walked during the n-1st hour. How many inches did Tony walk during the first three hours of his journey?[/quote][/quote]
1st hour: 10 inches
2nd hour: 15 inches
3rd hour: 22.5 inches
Total distance: 47.5 inches.
EDIT: Oh I just realized, he said $ 1\frac{1}{n}$, not $ 1\frac{1}{2}$
....
I feel ashamed.
So yeah, it would be 45 because the 3rd hour would be 20 inches, which gets you 45 inches total.

Nope, it's a whole number.

Reread the question.

1st hour: 10 inches
2nd hour: 15 inches
3rd hour: 20 inches

Total Distance: 45 inches

This was for #3, and nope, I believe 45 was the exact answer.

#7: I think you just find the area of a garden in square inches and divide it by 9 and multiply it by 520 to get 44928000.

#8,9: I think the question has the pictures to them.

I couldn't get the picture on? If you know how please tell me

Ok since I already did #6, I'll do 7 because 8 and 9 are impossible without the corresponding pictures.
In a 60 feet x 90 feet region, there are 5400 square feet, (obviously.) or 777600 square inches.
And since a 9 inch region contains 520 blades of grass,
So in that 60' x 90' region, there would be $ 777600\div9 \equal{} 84600$ of those 9 square inch regions.
And in each 9 square inch region, there are 520 blades of grass.
So there would be $ 84600\times520 \equal{} 449280000$ blades of grass.
Oh wait, you got the picture for #9.
Ok...
So you can prove that the middle triangle is equilateral...
Which you can work out the height from there.
$ \frac{3\sqrt{3}}{2}$

For #9, prove that if you drop a perpendicular from the intersection to the floor, you get the perpendicular bisector, then use similar triangles.

I estimate the answer to be
[hide="answer"]
[u]2.6[/u]
[/hide]

For 8, you can make the lines connecting the circles the opposite sides of a rectangle, since they have to be tangent to the circle. You know that the part in the middle of those lines is 20 inches, and the sides have a length equal to the radius (so two sides of 12). The part of the pulley that goes around the circle is half of the circle's circumference, or 12 pi. Since the pulley goes around half of both circles, the total around circle part is 24 pi, and so the total length is 24 pi +88 or 163 inches.


vector
linear algebra
matrix
algebra
function
domain
abstract algebra

Find the general form of all 2x2 matrices $A$ such that $A^{2}=0$.

[quote="mlok"]Think of $A$ as a linear transformation whose range coincides with its kernel. Pick a vector $\vec b$ in the range, and let $\vec a$ be some of its preimages. In the basis $\langle a,b\rangle$ this transformation is represented by the matrix $\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$. Going back to the original basis, find that $A=B\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}B^{-1}$ for some invertible matrix $B$. Unless $A$ is itself zero, of course.[/quote]

how does the range coincide with the kernal. isn't the kernal a subset of the domain?

You can think of $A$ as a map from $\mathbb R^{2}$ to $\mathbb R^{2}$. In this interpretation $A^{2}$ represents the same map applied twice. Then $\ker A=\{x\in\mathbb R^{2}\colon Ax=0\}$ and $\mathrm{range }A=\{Ax\colon x\in\mathbb R^{2}\}$ are both subspaces of $\mathbb R^{2}$, and the condition $A^{2}=0$ is equivalent to $\mathrm{range }A\subseteq \ker A$.

I wrote that the range and kernel coincide, because the proper inclusion $\mathrm{range }A\subset \ker A$ occurs only in the trivial case $A=0$.


How do you find the integral divisors of a number?

[quote="DaReaper"]Can you tell me the number of postive integral divisors of $440$?

and about that link... :huh:[/quote]
In order to find the number of positive integral factors a number has you must first prime factorize it. When you do so you get $2^{3}\cdot5\cdot11$. Then you add all the exponents by one so now you have 4,2, and 2. Then you multiply them $4\cdot2\cdot2=16$ factors.

[quote="ckck"]
In order to find the number of positive integral factors a number has you must first prime factorize it. When you do so you get $2^{3}\cdot5\cdot11$. Then you add all the exponents by one so now you have 4,2, and 2. Then you multiply them $4\cdot2\cdot2=16$ factors.[/quote]

But are the "integral divirsors" and "integral factors" same thing..

and $16$ was one of the options ont the test..but so was $8,12,14,16$ and $NOTA$ (meaning none of the above)

[quote="DaReaper"]But are the "integral divirsors" and "integral factors" same thing..[/quote]

Generally, yes.

[quote="t0rajir0u"]Generally, yes.[/quote]

you say generally...what do you mean generally...whats the catch? :(  ...:mad:

There are several qualifying words.

[b]Positive:[/b]  If this word is not here, then the problem [i]might[/i] be asking you for the negative divisors/factors as well.  (This is more likely if the word is "divisor.")

[b]Proper:[/b]  "Proper factors" of a number are all the divisors/factors besides itself.  Additionally, if a question says "proper factor" then it probably means "positive proper factor."  I don't think you'll ever see the phrase "proper divisor," so that's one difference.  Sometimes "factor" means "proper factor," but this is a horrible thing to do to people and most well-designed tests won't do it.


Given [b]x[/b] = (1,1,1,1)^T and [b]y[/b] = (8,2,2,0)^T:

i) determine the angle theta between [b]x[/b] and [b]y[/b]

If [b]p[/b] is defined the vector projection[b] p[/b] of[b] x[/b] onto [b]y[/b] : (4/3, 1/3, 1/3, 0)^T

ii) Verify that [b]x[/b]-[b]p[/b] is orthogonal to [b]p[/b]

iii) Compute ll[b]x[/b]-[b]p[/b]ll2,  ll[b]p[/b]ll2,  ll[b]x[/b]ll2  and verify that the Pythagorean Law is satisfied.

I have the end result for two of these but I don't understand the systematic steps that need to be taken to get there.

The reason no one has answered you is that these are just straightforward calculations. There doesn't seem to be anything to explain. What about it aren't you getting?


abstract algebra
induction
superior algebra
superior algebra solved

Prove that Q cannot be a Z[i] module.

[i]Proposed by Voica Cristian, my algebra teacher.[/i] I'm not sure it is his problem but he gave it ot me.

there is something wrong with this problem. If you are reffering to (Q,+) that is a commutative group, and it is well known that all commutative groups have a structure of Z module ...

Yes there is smth wrong with the problem. It is not my falut, it is html's.
You may notice that the module is written in italic. Well that is because the wonderful html compiled the "["i"]" when I italicised the problem source.

So the problem is:
Prove that (Q,+) is not an Z[i] module.

in back there is no html code [code][i][/code] that is a BB code :D. Have you solved it Mihai? (wouldn't it be fun to have this problem on the exam tomorrow? :D:D) ...

Yes I solved it. This is why I am posting it here.

It would be nice to have it in exam tomorrow because I'd have one less problem to worry about :D

just look at the number i.1=k which is a rational (suppose that Q is a modulus over Z[i]). From the modulus definition properties it is easy to check that i.m=km where m is an integer. Also from the fact that i.(1/2+1/2)=i.1=k => i.1/2=k/2 and so by induction i.q=kq for all rationals q. Consider now i.(i.q)=i.(kq)=k^2q but in the same time from the definition of the modulus we again have i.(i.q)=(i^2).q=(-1).q=-q from the definition of modulus. So in fact we have obtained that k^2=-1, and k is a rational. Contradiction, so the assumption is false.

Same proof can be adapted from any ring of the form Z[ir] where r is a real number.

What is your solution Mihai?


This is how it works: the first person says a letter of the alphabet. The next person says another letter of the alphabet, and so on. ALL OF THESE PEOPLE MUST BE INTENDING TO SPELL A WORD. If the letter that the last person says doesn't seem to spell a word, then someone may "challenge" him/her. The challenged person must be able to come up with a word that they are trying to spell, or else they are out of the game. If, however, you are thinking of a valid word, then the person who challenged you is out of the game. Oh, by the way, the word you say must be MORE THAN 3 LETTERS.

At the same time, you are trying NOT to get stuck with saying the last letter of the word, or else you are out. But you don't get out if the "word" is 3 letters or less.

Here is a sample play:

qwertythecucumber: <starts> G
Wickedestjr: E
ernie: O
nick42: Q
Wickedstjr: Challenge.
nick42: um...uh...dang!
-nick42 is out-

Once someone is out, we start again from the last person who had a letter. So, ernie would start.

Obviously, we would need a certain number of people in a certain order. So...sign up!!! I'm putting up 6 slots + me.

1. qwertythecucumber
2.
3.
4.
5.
6.
7.

[quote="zapi2007"]can i join even if i'm #8?

:([/quote]
The problem with having a lot of players is that it becomes too many to keep track of, the last few people many not get to play since the word would be too short, and just plain inconvenience. What we'll do is let others play in the next games. This also means that the people who play in this game should be good sports and let others try it out in the next game.

You probably won't end up waiting a long time; this game is usually pretty fast-paced, but I haven't tried holding one online before, so for that reason, it may be slower.

1. qwertythecucumber
2. Brut3 Forc3
3. cf249
4. eggylv999
5. Arvind_sn
6. Isabella2296
7. Discrete_Math

OK let's start now:

Q

Lol nice, brut3 forc3 says u by default.

unless we're using proper nouns (Qatar) or non-English words....O.O

Amendment: if you don't post the next letter within 3 days, then you will be kicked out automatically. (To prevent from a potential freeze).

Fortunately, ex post facto does not apply here (yay, I get to apply Social Studies stuff!)...that means as of now, Brut3 Forc3 has 3 days.

[quote="Arvind_sn"][quote="qwertythecucumber"]

Fortunately, ex post facto does not apply here (yay, I get to apply Social Studies stuff!)...that means as of now, Brut3 Forc3 has 3 days.[/quote]

Sorry, just wanted to say

I KNOW WHAT EX POST FACTO IS TOO, I JUST LEARNED IT LAST WEEK :D[/quote]
ooooh that's so cool!

anyways...eggy has to go now. I'll pm him (or her, for all I know).

Eggy's not responding in the Elimination Game either...he's getting pwned. :-/

T! I lose. :(

Brut3 Forc3  now must begin a new word.

Wow I was like "QUO" is a word until I reread the rules...

Sorry guys! You oughta PM me next time...
If I'm not out, then... [b]J[/b]

cmon eggy. this is like the second time >.<

JUR :mad:  :mad:  :mad: WHAT IS WRONG???

PEOPLE. WAKE UP.

Discrete_Math is out, Brut3 Forc3 needs to start new word.

Discrete_Math is out because he failed to submit within 3 days, not because there was no possible word to be made.

Continuing Brut3Forc3's word:
Z[b]U[/b]

No need to revive when we already have a revival
:wink:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=325231]Here.[/url]
I'd lock this is I could...
*cough*

[size=84][color=red]Moderator PhireKaLk6781 says: This game is actually quite a different game, so I'll let both threads run. [/color][/size]

So what do we do now? ZUC[b]H[/b]? Evidently the list idea didn't work out so well, sadly.

Psh.
Phire.

Anyways,
I won't mod the game,
yet I will start sign-ups.

1.
2.
3
4.
5.
6.
7.
8.
9.
10.

Sweet and simple.

my game in the games section didnt get off its feet...

:(

o well

join

Oh yeah the word spells Zuca for this: [url]http://www.zuca.com/[/url]

This is going off topic from the original rules. Make a new thread.


Let p prime number,n and m nonnegative integers.If  $p^2=2^n.3^m+1$ then find the maximum value of p.

Max $p$ is $17$.

Rewrite the given equation as $(p-1)(p+1) = 2^n \cdot 3^m$. Since $\gcd(p-1,p+1) = 2$ or $1$ (if $p=2$), it follows that :
$p-1 = 2^a \cdot 3^m,p+1 = 2^{n-a}$ or $p+1 = 2^a \cdot 3^m,p-1 = 2^{n-a}$, where $a =1$ or $a=n-1$ if they are non-zero.

In the first case :
$2^{n-a} = 2^a \cdot 3^m + 2$.
This forces $n-a \geq 2$ so that $a=1$ and the equation reduces to $2^{n-2} - 3^m = 1.$
This is the well-known equation $2^x - 3^y = 1$ which leads to $n-2=1,m=0$ or $n-2 = 2, m=1.$ This gives $p=3$ or $p=7.$

In the second case :
$2^a \cdot 3^m = 2^{n-a} + 2$.
If $n-a = 0$ then $a=n=0$ and $m=1$ so that $p=2.$
If $n-a=1$ then $2^a \cdot 3^m = 4$ so that $m=0,a=2,n=3$ and $p=3.$
If $n-a \geq 2$ then $a=1$ and the equation reduces to $3^m - 2^{n-2} = 1.$
This is the well-known equation $3^x-2^y = 1$ which forces $m=1,n-2=1$ or $m=2,n-2=3$. The first case leads to $p=5$ and the latter to $p=17.$

Pierre.


the sequence $U_{n}=2-(\frac{2}{3})^{n}$ converges to L. Find L
plz xplain how u get the answer.

[hide]Wouldn't the solution be 2, because (2/3)^infinity is 0?[/hide]

[hide]Not quite.... 2/3 ^ (an arbitrarily large number) isn't 0... it's close, just not quite 0. Beware of that distinction.[/hide]


$ \phi(k)\equal{}det(A\plus{}kB)\equal{}ak^3\plus{}bk^2\plus{}ck\plus{}d$ is a polynomial of degree $ 3$ in $ k$ with $ d\equal{}det(A)$.
$ \frac {\sum_{k\equal{}1}^{n}{[det(A\plus{}kB)\plus{}det(A\minus{}kB)\minus{}2det(A)]}}{n^3}\equal{}\frac {\sum_{k\equal{}1}^{n}{\phi(k)\plus{}\phi(\minus{}k)\minus{}2d}}{n^3}\equal{}\frac {\sum_{k\equal{}1}^{n}{2bk^2}}{n^3}$ which converges to $ \dfrac{2b}{3}$.
$ \psi(k)\equal{}det(kA\plus{}B)\equal{}dk^3\plus{}ck^2\plus{}bk\plus{}a$ and $ b\equal{}\psi'(0)$. If $ (A_i),(B_i)$ are the columns of $ A,B$ then $ \psi(k)\equal{}det(kA_1\plus{}B_1,kA_2\plus{}B_2,kA_3\plus{}B_3)$ and $ \psi'(k)\equal{}det(A_1,kA_2\plus{}B_2,kA_3\plus{}B_3)\plus{}det(kA_1\plus{}B_1,A_2,kA_3\plus{}B_3)\plus{}det(kA_1\plus{}B_1,kA_2\plus{}B_2,A_3)$. QED

We can do better: ($ \minus{}2det(A)$ is useless and the result is valid in any dimension)
Let $ A,B\in\mathcal{M}_m(\mathbb{C})$ where $ m$ is odd. Then $ lim_{n \to \infty} \frac {\sum_{k \equal{} 1}^{n}{[det(A \plus{} kB) \plus{} det(A \minus{} kB) ]}}{n^{m}} \equal{} \frac {2b}{m}$ with $ b\equal{}det(A_1,B_2.\cdots)\plus{}det(B_1,A_2,B_3,\cdots)\plus{}\cdots$.
Proof: use similar method and $ \sum_{k\equal{}1}^n{k^p}\sim\dfrac{n^{p\plus{}1}}{p\plus{}1}$ when $ n$ tends to infinity and $ p>\minus{}1$.


Prove that:$ \sqrt {1 \plus{} \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} ...}}}} \equal{} 2$

Isn't it interesting :D

[quote="mavropnevma"]I think this is attributed to Ramanujan. Quite nice, not trivial, and with a nice solution by Fine in the A.M.M.[/quote]

Can you post the solution you know?
I spent a lot of time in this problem,I think I had a two-page proof(rather long) for it :) (wish it was true)

Yes, in a couple of days or so ... unless someone else posts it before  :)

using $ x=\sqrt{1+(x-1)(x+1)}$
$ 2=\sqrt{1+1\times3}=\sqrt{1+\sqrt{1+2\times4}}=\sqrt{1+\sqrt{1+2\sqrt{1+\cdots}}}$
but I heard that it isn't complete solution..

$ 0 \equal{} 1 \minus{} 1$

then

$ 0 \equal{} \displaystyle\sum_{i \equal{} 0}^{\infty}( \minus{} 1)^i$

or

$ n \equal{} n \plus{} n \plus{} 1 \minus{} n \minus{} 1...$

Don Equis, are you sure that sum converges to 0? :S

I'm sure that it doesn't converge. I'm showing why things like

[quote="jedaihan"]using $ x = \sqrt {1 + (x - 1)(x + 1)}$
$ 2 = \sqrt {1 + 1\times3} = \sqrt {1 + \sqrt {1 + 2\times4}} = \sqrt {1 + \sqrt {1 + 2\sqrt {1 + \cdots}}}$
but I heard that it isn't complete solution..[/quote]

prove nothing.

He assumes that because it works for all n it has to work for infinity, but things are not like that.

sorry :blush:
I am not convinced that expression is right, but I see that in my book(korean popular educational books).. I think this book is full of error :blush:

This is my own proof (not Fine's). Denote for $ x > 0$
\[ f_n(x) \equal{} \sqrt{1 \plus{} x\sqrt{1 \plus{} (x\plus{}1)\sqrt{1 \plus{} \cdots \plus{} (x \plus{} n \minus{} 1)\sqrt{1 \plus{} (x\plus{}n)}}}},\]
\[ g_n(x) \equal{} \sqrt{1 \plus{} x\sqrt{1 \plus{} (x\plus{}1)\sqrt{1 \plus{} \cdots \plus{} (x \plus{} n \minus{} 1)\sqrt{1 \plus{} (x\plus{}n)(x\plus{}n\plus{}2)}}}}.\]
It is easily seen that $ g_n(x) \equal{} x\plus{}1$, since $ 1 \plus{} (x\plus{}n)(x\plus{}n\plus{}2) \equal{} (x\plus{}n\plus{}1)^2$, and it telescopes.
On the other hand, $ g_{n}(x) > f_{n}(x) > f_{n\minus{}1}(x) > \cdots > f_0(x) \equal{} \sqrt{1\plus{}x}$, so the sequence $ (f_n(x))_{n \geq 1}$ is increasing and upper bounded by $ x\plus{}1$, hence convergent to some limit $ \ell(x) \leq x\plus{}1$.

But $ f_n^2(x) \equal{} 1 \plus{} xf_{n\minus{}1}(x\plus{}1)$, and of course also $ g_n^2(x) \equal{} 1 \plus{} xg_{n\minus{}1}(x\plus{}1)$, therefore $ g_n(x) \minus{} f_n(x) \equal{} x \frac {g_{n\minus{}1}(x\plus{}1) \minus{} f_{n\minus{}1}(x\plus{}1)} {g_n(x) \plus{} f_n(x)} <$ $ \frac {x} {(x\plus{}1) \plus{} \sqrt{x\plus{}1}} (g_{n\minus{}1}(x\plus{}1) \minus{} f_{n\minus{}1}(x\plus{}1))$.

By iterating we get $ g_n(x) \minus{} f_n(x) < \prod_{k\equal{}0}^{n\minus{}1}\frac {x\plus{}k} {(x\plus{}k\plus{}1) \plus{} \sqrt{x\plus{}k\plus{}1}} (g_{0}(x\plus{}n) \minus{} f_{0}(x\plus{}n)) \equal{}$ $ \prod_{k\equal{}1}^{n\minus{}1}\frac {x\plus{}k} {(x\plus{}k) \plus{} \sqrt{x\plus{}k}} \left ( \frac {x} {(x\plus{}n) \plus{} \sqrt{x\plus{}n}} ((x\plus{}n\plus{}1) \minus{} \sqrt{x\plus{}n\plus{}1}) \right )$.

Now, $ \frac {x} {(x\plus{}n) \plus{} \sqrt{x\plus{}n}} ((x\plus{}n\plus{}1) \minus{} \sqrt{x\plus{}n\plus{}1})) < x$, while $ \prod_{k\equal{}1}^{n\minus{}1}\frac {(x\plus{}k) \plus{} \sqrt{x\plus{}k}} {x\plus{}k} >$ $ 1 \plus{} \sum_{k\equal{}1}^{n\minus{}1} \frac {1} {\sqrt{x\plus{}k}} >$ $ 1 \plus{} (n\minus{}1)\frac {1} {\sqrt{x\plus{}n\minus{}1}} >$ $ 1 \plus{} \sqrt{(n\minus{}1)/2}$ for $ n > x\plus{}1$.

This means $ 0 \leq (x\plus{}1) \minus{} \ell(x) < g_n(x) \minus{} f_n(x) < \frac {x} {1 \plus{} \sqrt{(n\minus{}1)/2}} \to 0$ for $ n \to \infty$, so $ \ell(x) \equal{} x\plus{}1$.
For $ x\equal{}1$ we get therefore the value to be $ 2$.

Very nice solution mavropnevma :lol:
My idea is almost same as yours,but my proof is much longer
In your solution,you denoted
\[ f_{n}(x) \equal{} \sqrt {1 \plus{} x\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} \cdots \plus{} (x \plus{} n \minus{} 1)\sqrt {1 \plus{} (x \plus{} n)}}}},\]
Different from yours,I denoted $ f(x) \equal{} \sqrt {1 \plus{} x\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} \cdots}}}$ which is an infinite algebraic expression
My thought:
Step 1;try to prove $ f(x)$ is bounded
Step 2:try to prove $ f(x)\le n \plus{} 1$
Step 3:try to prove $ f(x) \equal{} x \plus{} 1$
Here is my specific solution
Proof for step 1

Proof for the rest

P.S Can you post Fine's solution,mavropnevma :lol:

Oh, it will take me a few days to locate Nathan Fine's proof. I know where to look for it (I hope...) - it's basically an analysis of the behavior of $ f(x)$ that leads to the convergence to $ 1\plus{}x$.

Historical Note. In fact Ramanujan's problem was for $ x\equal{}2$, so the nested square roots' value was $ 3$.

I understand this thread is old (I think), I feel compelled to give the general formula since no one has mentioned it.

[b]Question:[/b] Prove $$\sqrt{1+\color{blue}{{\sqrt{1+2\sqrt{1+3\sqrt{1+\&\text{c}.}}}}}}=2$$

Let's look at the blue nested radical. We have to prove that$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\&\text{c}.}}}}=3$$This isn't took difficult by using a general formula derived from the expansion of the binomial of degree two.

Start with$$(x+a)^2=x^2+2ax+a^2$$Replacing $a$ with $n+a$ gives$$x+n+a=\sqrt{ax+(a+n)^2+x(x+2n+a)}$$

Note that we only considered the positive case for the square root. Noticing that $x+2n+a$ is eerily similar to $x+n+a$, we replace $x$ with $x+n$ to get$$x+2n+a=\sqrt{a(x+n)+(a+n)^2+(x+n)(x+3n+a)}$$And setting $x:=x+n$ into $x+2n+a$ gives$$x+3n+a=\sqrt{a(x+2n)+(a+n)^2+(x+2n)(x+4n+a)}$$Eventually, we notice the pattern for the $k^{\text{th}}$ value$$x+kn+a=\sqrt{a[x+(k-1)n]+(a+n)^2+[x+n(k-1)][x+(k+1)n+a]}$$And putting these nests back, we get the general formula$$x+n+a=\sqrt{ax+(a+n)^2+x\sqrt{a(x+n)+(a+n)^2+(x+n)\cdots}}$$Setting $a=0,n=1,x=2$ gives$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\&\text{c}.}}}}=3$$And we are done.


trigonometry
logarithms
function
algebra proposed
algebra

solve
$ log_2 \sqrt{tan x \plus{} 3} \equal{} log_3 (3.tan y)$
$ log_2 \sqrt{tan y \plus{} 3} \equal{} log_3 (3.tan x)$

[quote="nguyensilva"]solve
$ log_2 \sqrt {tan x \plus{} 3} \equal{} log_3 (3.tan y)$
$ log_2 \sqrt {tan y \plus{} 3} \equal{} log_3 (3.tan x)$[/quote]

Let $ u\equal{}\tan(x)$ and $ v\equal{}\tan y$. The system is equivalent to :

$ \log_2(\sqrt{u\plus{}3})\equal{}\log_3(3v)$ and $ \log_2(\sqrt{v\plus{}3})\equal{}\log_3(3u)$

$ \iff$ $ v\equal{}f(u)$ and $ u\equal{}f(v)$ where $ f(x)\equal{}\frac 13(x\plus{}3)^{\frac{\log(3)}{2\log(2)}}$

So $ f(f(u))\equal{}u$ but $ f(x)$ is an increasing function so :
$ f(x)>x$ $ \implies$ $ f(f(x))>f(x)>x$
$ f(x)<x$ $ \implies$ $ f(f(x))<f(x)<x$
And so $ f(f(x))\equal{}x$ $ \iff$ $ f(x)\equal{}x$

$ f(x)$ is concave and so $ f(x)\minus{}x$ is concave and since $ f(0)\plus{}0>0$, we have at most one root to $ f(x)\minus{}x\equal{}0$ for $ x>0$

And since we have the trivial root $ f(1)\equal{}1$, we have a unique $ (u,v)$ solution which is $ (1,1)$

Hence the roots of the original equation : $ (x,y)\equal{}(\frac{\pi}4\plus{}m\pi,\frac{\pi}4\plus{}n\pi)$


It's really strange.
We need to prove:
$ (x \plus{} y \plus{} z)^2(xy \plus{} yz \plus{} xz)^2 \plus{} 4(xy \plus{} yz \plus{} xz)^2*xyz \plus{} (xy \plus{} yz \plus{} xz)^2\ge 36x^2y^2z^2 \plus{} (xy \plus{} yz \plus{} xz)^2(x \plus{} z \plus{} y).$
And
$ \frac {4(x \plus{} y \plus{} z)^2(xy \plus{} yz \plus{} xz)^2}{9} \ge 36x^2y^2z^2$
$ \frac {(x \plus{} y \plus{} z)^2(xy \plus{} xz \plus{} yz)^2}{4} \plus{} (xy \plus{} yz \plus{} xz)^2\ge(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} xz)$

[quote="behdad.math.math"]if $ x,y,z\geq0$, prove that

$ \sqrt{(x\plus{}y\plus{}z)^2\plus{}4xyz\minus{}(x\plus{}y\plus{}z)\plus{}1}(xy\plus{}yz\plus{}zx)\geq 6xyz.$[/quote]If $ x,y,z>0$, then

$ \left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)\sqrt{4xyz\plus{}(x\plus{}y\plus{}z)^2\minus{}x\minus{}y\minus{}z\plus{}1}\geq 9,$

with equality if and only if $ x\equal{}y\equal{}z\equal{}\frac{1}{2}.$

[quote="Ji Chen"][quote="behdad.math.math"]if $ x,y,z\geq0$, prove that

$ \sqrt {(x \plus{} y \plus{} z)^2 \plus{} 4xyz \minus{} (x \plus{} y \plus{} z) \plus{} 1}(xy \plus{} yz \plus{} zx)\geq 6xyz.$[/quote]If $ x,y,z > 0$, then

$ \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)\sqrt {4xyz \plus{} (x \plus{} y \plus{} z)^2 \minus{} x \minus{} y \minus{} z \plus{} 1}\geq 9,$

with equality if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{2}.$[/quote]
Let $ x\plus{}y\plus{}z\equal{}p,xy\plus{}yz\plus{}zx\equal{}\frac{p^2\minus{}q^2}{3},xyz\equal{}r$
so it's equalvent to $ \frac{p^2\minus{}q^2}{3r}\sqrt{4r\plus{}p^2\minus{}p\plus{}1} \geq 9$
use $ r \leq \frac{(p\plus{}2q)(p\minus{}q)^2}{27}$
<=$ \frac{p\plus{}q}{(p\plus{}2q)(p\minus{}q)}\sqrt{\frac{4(p\plus{}2q)(p\minus{}q)^2}{27}\plus{}p^2\minus{}p\plus{}1} \geq 9$
$ \iff f(q)\equal{}\frac{4}{27}p^3\minus{}\frac{4}{9}pq^2\plus{}\frac{8}{27}q^3\minus{}p\plus{}1\plus{}2pq\minus{}q^2 \geq 0$
$ f'(q)\equal{}(p\minus{}q)(2\minus{}\frac{8}{9}q)$
if $ q>\frac{9}{4}$
$ f(q) \geq f(p)\equal{}p^2\minus{}p\plus{}1>0$
if$ q\leq \frac{9}{4}$
=>$ f(q) \geq f(0)\equal{}\frac{4}{27}p^3\minus{}p\plus{}1\equal{}\frac{(p\plus{}3)(2p\minus{}3)^2}{27} \geq 0$


For every natural number n, prove that

$1 =\sum_{k=0}^{\infty }\frac{n+k}{\left( n+1\right) \left( n+2\right)
...\left( n+k+1\right) }$
$=\frac{n}{n+1}+\frac{n+1}{\left( n+1\right) \left( n+2\right) }+\frac{n+2}{\left( n+1\right) \left( n+2\right) \left( n+3\right) }+...$.

  Darij

Oh yes...

$\frac{n+k}{\left( n+1\right) \left( n+2\right) ...\left( n+k+1\right) }=\frac{1}{\left( n+1\right) \left( n+2\right) ...\left( n+k\right) }-\frac{1}{\left( n+1\right) \left( n+2\right) ...\left( n+k+1\right) }$

I was so stupid not to see this myself... my proof was somewhat more involved.  :blush:

  Darij


advanced fields
advanced fields unsolved

Suppose $A$ and $B$ are sets of finite Lebesque outer measure (on $\mathbb{R}$). Show that
\[ m^*(A\cup B)=m^*(A)+m^*(B)  \]
if and only if there are measurable sets $A_1,B_1$ such $A\subset A_1,\,B\subset B_1$ and $m(A_1\cap B_1)=0$.

Examine the proof in http://www.mathlinks.ro/Forum/viewtopic.php?p=359544#p359544.


What is the smallest positive integer that is a perfect square and
is also a sum of six consecutive positive integers?

lets say 6 consecutive integers are $ a\minus{}3, a\minus{}2, a\minus{}1, a, a\plus{}1,$ and $ a\plus{}2$.

the sum is 6a-3, which is multiple of 3.

therefore, their sum must be a square of multiple of 3.

however, if you square even multiple of 3, 2a-1 becomes also even, which is not true.

so its square of odd multiple of 3.

least odd multiple of 3 is 3 and if you square it you get 9.

but not all of them are positive.

next odd multiple of 3 is 9. and if you square it, you get 81.

and all of 6 integers are positive.

answer : 81


O is the centre of a circle.at any point T of the circle tangent tp is drawn and PBA is secant. if <apt=90 then prove PA^2+PB^2=4(OT)^2

What is ''PBA is secant''? If you want help, why you not use LaTex?

Not everybody knows LaTex. But you could have written the vertices in capitals. It tends to turn people off. PBA is secant means that P, A, B are collinear and A and B lie on the circle.

simply draw OC perpendicular to AP (the feet is C) then
$OB^{2}=OT^{2}=OC^{2}+BC^{2}=(\frac{PA-PB}{2})^{2}+PA\cdot PB$
 :) Forgive me, I don't know why I can't type dot product in latex :blush:

EDIT: fixed it! Thanks, N.T.Tuan :)

[quote="andy_tok"]simply draw OC perpendicular to AP (the feet is C) then
$OB^{2}=OT^{2}=OC^{2}+BC^{2}=(\frac{PA-PB}{2})^{2}+PA\cdot PB$
 [/quote]
Type \cdot :D


floor function
calculus
calculus computations

is it  possible to sum  :maybe:

$ \sum_{n\equal{}0}^\infty \; \frac{1}{(2n\plus{}1)^{\frac{3}{2}}}$

While I'm not sure if that I can be done, I found out that it is possible to sum
$ \sum_{i\equal{}1}^{\infty}\frac{1}{(2n)^{2}}$ and $ \sum_{i\equal{}1}^{\infty}\frac{1}{(2n\plus{}1)^{2}}$.

At least I think this is right. :blush:

$ \sum _{i\equal{}1}^n f(i)\equal{}\sum _{i\equal{}1}^{\left\lfloor \frac{n}{2}\right\rfloor } f(2 i\plus{}0)\plus{}\sum _{i\equal{}0}^{\left\lfloor \frac{n\minus{}1}{2}\right\rfloor } f(2 i\plus{}1)$

$ \sum _{n\equal{}0}^{\infty } \frac{1}{(2 n\plus{}1)^{3/2}}\equal{}\frac{1}{4} \left(4\minus{}\sqrt{2}\right) \sum _{n\equal{}1}^{\infty } \frac{1}{n^{3/2}}\equal{}\frac{1}{4} \left(4\minus{}\sqrt{2}\right) \zeta \left(\frac{3}{2}\right)$

If you can give a analitic form to this zeta you are Euler.

How do you get the $ 1/4*(4 \minus{} \sqrt{2})$ ?? I can't follow :(

That might look clearer as $ 1\minus{}\frac1{2^{3/2}}$.


I am looking for a good trigonometry textbook. Please post your recommendations here.

The standard book(at least in India) is..

[b]S.L. Looney's Trignometry I and II[/b](2 books)

It's easily available in India but don't know about else where. It's a VERY old book ( first wriiten in 1880s)

I can recommend two:

1. Trigonometry by Gelfand and Saul (229 pages) published by Birkhauser on Amazon for 22.95 dollars.
    Very clear and systematic exposition.

2. 103 Trigonometry Problems: From the Training of the USA IMO Team by Andreescu and Feng also published by Birkhauser on Amazon for 49.95 dollars.
    First 62 pages give brief review of all trigonometry with examples of how to employ trigonometric techniques to problem solving. Following that are 52 introductory problems (Intermediate Level) and 51 advanced problems (Olympiad level). Complete solutions to all of the problems complete the book with a glossary and a list of 52 other books for further reading.

What on earth does Trigonometry include?

tricycle, I think the second book is little bit too advanced for 4everwise.

Trigonometry is basically study of triangles. You learn about $\sin, \cos, \tan$ and etc with their graph. Also, you learn basic laws like Law of Cosine and Law of Sine (I don't think Law of Tangent is covered much).

Iono about the first one either, only times I've heard of Gelfand are authoring graduate texts...

EDIT: NM, flipped through the contents at amazon, it looks great :D


[quote="Ji Chen"]If $ a,b,c,d$ be positive numbers, then

$ \frac {a}{b \plus{} kc} \plus{} \frac {b}{c \plus{} kd} \plus{} \frac {c}{d \plus{} ka} \plus{} \frac {d}{a \plus{} kb}\geq \frac {4}{1 \plus{} k}$

holds for $ 0\leq k\leq 1$.[/quote]
Cauchy works. :)

Really!!!
Post it please, shfdfzhjj!
 :)

[quote="onlylove_math"][quote="Ji Chen"]If $ a,b,c,d$ be positive numbers, then

$ \frac {a}{b \plus{} kc} \plus{} \frac {b}{c \plus{} kd} \plus{} \frac {c}{d \plus{} ka} \plus{} \frac {d}{a \plus{} kb}\geq \frac {4}{1 \plus{} k}$

holds for $ 0\leq k\leq 1$.[/quote]

By C-S and AM-GM we need $ (a \plus{} b \plus{} c \plus{} d)^2\ge\ 8(ac \plus{} bd)$

Is it true? :)[/quote]
Wrong $ a\equal{}c\equal{}1,1;b\equal{}d\equal{}1$