Putnam
quadratics
geometry
3D geometry
function
Stanford
college

Show that the curve $x^{3}+3xy+y^{3}=1$ contains only one set of three distinct points, $A,B,$ and $C,$ which are the vertices of an equilateral triangle.

This curve is Folium of Decarte. This problem is interesting to me. :)

[quote="Kent Merryfield"]But this particular curve is degenerate. Factor it![/quote]

I know, $x^{3}+y^{3}+3xyz=1\Longleftrightarrow x^{3}+y^{3}+(-1)^{3}-3xy(-1)=0$
$\Longleftrightarrow (x+y-1)\{x^{2}+y^{2}+(-1)^{2}-xy-x(-1)-y(-1)\}=0$$\Longleftrightarrow (x+y-1)(x^{2}+y^{2}-xy+x+y+1)=0.$

[quote="Bictor717"][quote="Kent Merryfield"]And what is the set $x^{2}+y^{2}-xy+x+y+1=0?$[/quote]
It's the point (-1,-1). Consider the equation as a quadratic in x. The discriminant is $-3(y+1)^{2}$, so x is real-valued when y=-1. Then x=-1.[/quote]

Alternatively, note $(x-y)^{2}+(x+1)^{2}+(y+1)^{2}= 2(x^{2}+y^{2}-xy+x+y+1$.  This one I found very annoying -- I don't think it was transparent enough to be B1.

[quote="Kent Merryfield"]And what is the set $x^{2}+y^{2}-xy+x+y+1=0?$[/quote]

Completing square yields what you want.

$x^{2}+y^{2}-xy+x+y+1=0$

$\Longleftrightarrow x^{2}-(y-1)x+y^{2}+y+1=0.$

$\Longleftrightarrow \left(x-\frac{y-1}{2}\right)^{2}+\frac{3}{4}(y+1)^{2}=0.$

$\Longleftrightarrow (x,y)=(-1,-1).$

I wound up making headway on it and figuring out what was going on by making the substitutions $u=x+y,v=x-y.$

Start by moving the $3xy$ over to the right, then adding in the middle terms needed to complete the cube:

$x^{3}+3x^{2}y+3xy^{2}+y^{3}=3x^{2}y+3xy^{2}-3xy+1$

$(x+y)^{3}-1=3xy(x+y-1)$

Now make the substitution I was talking about, and note that $u^{2}-v^{2}=4xy:$

$u^{3}-1=\frac34(u^{2}-v^{2})(u-1).$

There's still more to go through from here, but I found that much to be useful.

I noticed very quickly that the equation looked a lot like $(x+y)^{3}$ with $x+y=1$.  In fact, I should have factored $x+y-1$ immediately when I saw this!  But instead (I wasn't able to focus much yesterday anyway) I messed around with it and managed to factor it.  Then I used the discriminant on the second factor as did Bictor.

I really like the problem.

This should make things apparent:

$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$
$= \frac12(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$

The second factor on the RHS above directly  yields the point $(-1,-1)$ when we substitute $a = x,\ b = y$ and $c =-1$.

The other factor of course gives $x+y = 1$ from which it is easy to see that there is only one set of three points, which form an equilateral triangle, whose area is $\sqrt6$.

Sadly, this is the only problem I got correct.  :(

I thought it actually looked like a folium of descartes (you know, with the loop), and had no idea as to how to proceed on such a problem.

I moved on to B2 in about 2 minutes.

[quote="FieryHydra"] which form an equilateral triangle, whose area is $\sqrt6$.

 this is the only problem I got correct.  :([/quote]

UM.....
if i recall correctly,
No, the side length of the triangle is $\sqrt{6}$ the area is $\frac{3\sqrt{3}}{2}$

i just cant believe i missed the easy factoring solution, but i did it long way with (-1.-1) and y=1-x and got it.
hopefully a 10 on this one.

[quote="amirhtlusa"][quote="FieryHydra"] which form an equilateral triangle, whose area is $\sqrt6$.

 this is the only problem I got correct.  :([/quote]

UM.....
if i recall correctly,
No, the side length of the triangle is $\sqrt{6}$ the area is $\frac{3\sqrt{3}}{2}$[/quote]

You are correct amirhtlusa. My bad. I do remember getting $\sqrt6$ as the side length and $\frac{3\sqrt3}{2}$ as the area. Thanks!

[quote="APSStudent"]I had a question about the cubic formula in general, is there a closed form for the cubic formula--in the same way there is a closed form of the quadratic formula?[/quote]
1) There was no need to quote that very long post before asking your question (you could have just quoted the first couple of lines or something).

2) Yes. But you'll rarely see the whole general solution written anywhere explicitly because it's way too unwieldy. There's also a general solution for quartic equations, which is similarly enormous. There's nothing in general for higher order equations, although certain special cases may be solvable. You can see the method for solving cubics in that post you quoted. It works for any equation.

Generally, in contests, if you need to solve an equation of degree 3 or higher, you'll be able to use the rational root theorem to find one of the roots by hand, or find some other clever way to guess at one of the factors (as in this problem).

[quote="Xevarion"]

1) There was no need to quote that very long post before asking your question (you could have just quoted the first couple of lines or something).
[/quote]

Agreed, I quoted a bit too much, I've edited it--so it should look more compact.

[quote="Xevarion"]

2) Yes. But you'll rarely see the whole general solution written anywhere explicitly because it's way too unwieldy. There's also a general solution for quartic equations, which is similarly enormous. There's nothing in general for higher order equations, although certain special cases may be solvable. You can see the method for solving cubics in that post you quoted. It works for any equation.

Generally, in contests, if you need to solve an equation of degree 3 or higher, you'll be able to use the rational root theorem to find one of the roots by hand, or find some other clever way to guess at one of the factors (as in this problem).[/quote]

Thanks for the reply, while I am familiar with the methods of solving cubics, such as the Cardano-Tartaglia method and Tschirnhaus transformation.

I have not even seen a closed form solution ever written down in a textbook or online. By closed form--I mean a solution to,

$ {{ax^3} \plus{} {bx^2} \plus{} {cx} \plus{} {d}} \equal{} {0}$

in terms of,

 $ a, b, c, d$

That is why (since you have mentioned it exists)--what it is?

Thanks,

-APSStudent

I don't know. I've never seen it, I don't want to work it out myself, and I don't really want to search around online either (you can try this if you want). You can get it by taking the Cardano solution (in kunny's link) and doing the reverse of that transformation. It will be a huge expression. The reason no one ever writes it is that it's easier to do the transformation and then use the Cardano solution.

The key is to notice the convenient factorization. The equation can be written as
$$x^3 + 3xy + y^3 -1 = (x+y)^3 - 3xy(x+y-1) -1 = (x+y-1)((x+y)^2+(x+y)+1-3xy)=(x+y-1)(x^2+y^2-xy+1+x+y) = 0$$
If $x+y=1$, we get a line. If $x^2 + y^2 -xy+1+x+y=0$, then $$2x^2+2y^2-2xy+2+2x+2y = 0 \Longleftrightarrow (x+1)^2+(y+1)^2 + (x-y)^2 = 0$$
Clearly, this only happens when $x=y=-1$. Therefore, we have a line and a point. Clearly, only one equilateral triangle can be drawn.

Good Laugh.

We use the key identity that $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$. Let $a = x$, $b = y$, $c = -1$, which gives us our problem statement. Hence, it factors to\[(x + y - 1)(x^2 + y^2 + 1 - xy + x + y) = 0\]so either $x + y = 1$ or $(x+1)^2 + (y+1)^2 + (x - y)^2 = 0$. Hence the locus of all points satisfying the problem is the line $x + y = 1$ and the point $(-1, -1)$. The other two vertices of this triangle are clearly unique so the triangle is unique.

In fact, the distance from $(-1, -1)$ to $x + y = 1$ can be calculated to be $\tfrac{3}{\sqrt{2}}$, which also happens to be the height of the unique equilateral triangle. Hence, the equilateral triangle has area $\tfrac{h^2}{\sqrt{3}} = \tfrac{3\sqrt{3}}{2}$.

[i]Solved with andyxpandy99.[/i]

[hide=Solution]
Note that \begin{align*}
x^3+3xy+y^3&=1\\
\leftrightarrow x^3+3xy(x+y)+y^3&=1+3xy(x+y)-3xy\\
\leftrightarrow (x+y)^3-1&=3xy(x+y-1)\\
\leftrightarrow (x+y-1)(x^2+xy+y^2+x+y+1)&=0\\
\leftrightarrow (x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)&=0\\
\end{align*}

Thus, either $x+y=1$, or $(x,y)=(-1,-1)$. So, the curve is one line and one other point not on the line. The only equilateral triangle with all vertices on the curve must have one vertex be $(-1,-1)$, and the two other vertices be on the line. The point $(-1,-1)$ has distance $\frac{3\sqrt{2}}{2}$ from the line $x+y=1$. The side length of the triangle is $\frac{2}{\sqrt{3}}$ times that, which is $\sqrt{6}$. Thus the area of the equilateral triangle is $\frac{(\sqrt{6})^2\cdot\sqrt{3}}{4}=\boxed{\frac{3\sqrt{3}}{2}}$.
[/hide]

oh titu...

We have
\begin{align*}
x^3+3xy+y^3&=1\\
x^3+y^3+(-1)^3-3xy(-1)&=0\\
(x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)&=0,
\end{align*}
so the graph of the curve is the union of $x+y=1$ and $(x,y)=(-1,-1)$. From here the result is immediate. $\blacksquare$

Consider the behavior of the curves $C_b$: $f(x,y)=x^3+y^3+bxy-1=0$, when $b\in (-3-\epsilon,3+\epsilon)$ in a neighborhood of the point $A=(-1,-1)$.
The system $\dfrac{\partial f}{\partial x}=0,\dfrac{\partial f}{\partial y}=0,f(x,y)=0$ give the singular points of $C_b$.
For every $b\not= 3$, there are no singular points; for $b=3$, we obtain the sole point $A$ which is an isolated point of $C_3$.
When $b=3-h,h>0$ small , $C_b$ has no points close to $A$.
When $b=3+h$, the curve $C_b$ presents a kind of small ellipse close to $A$.

In fact, when $b<3$ the complex set $C_b$ contains two conjugate curves that don't intersect; yet, they intersect -when $b=3$- in the isolated singular  real point $A$.
A simpler example is as follows:
Consider the curves $C_a$: $x^2+y^2-a=0$, where $a\leq0$. If $a<0$, then  $C_a$ contains the conjugate complex curves
 $\{(x,y);x\in\mathbb{R},y=i\sqrt{x^2-a}\}\},\{(x,y);x\in\mathbb{R},y=-i\sqrt{x^2-a}\}$ that don't intersect.
When $a=0$, these $2$ curves intersect in the isolated point $(0,0)$.
Of course, when $a>0$ small, we obtain a little circle close to $(0,0)$.

The curve can be factored to $.5((x+1)^2+(y+1)^2+(x-y)^2).$ It follows that the curve consists of a point $(-1,-1)$ and the line $x+y=1.$ We need one of the vertices to be $(-1,-1)$ for the triangle with three points on this curve to be formed. Moreover, the distance from $(-1,-1)$ to $(.5,.5)$ is $3\sqrt{2}/2$, in fact this is the altitude from $(-1,-1).$ So the area of the triangle is $\boxed{3\sqrt{3}/2}.$

[quote=loup blanc]Zeta, read my post.
I don't want to solve the problem; moreover the same solution has already been written at least 4 times above (5 times with yours).
I explain how an isolated point of an algebraic curve can be dynamically created.[/quote]

I see. I guess my solution happens to be "the most obvious one."

We see $x^3+3xy+y^3-1$ factors as $(x+y-1)((x+1)^2+(y+1)^2-(x+1)(y+1))=0$ and we may notice that the second factor is zero only at $x=y=-1$ since it is the average of $(x+1)^2+(y+1)^2$ and $(x-y)^2.$ Since the curve is a point and a line we see that one of $A,B,C$ must be $(-1,-1)$ and the other two are on $x+y=1.$ Their midpoint must be the foot from $(-1,-1)$ to $x+y=1,$ and now it is easy to finish. The triangle has height $\frac{3\sqrt2}2$ so it has area $\frac{3\sqrt3}2.$

Notice that $$x^3+y^3+3xy-1=(x+y)^3-3x^2y-3xy^2+3xy-1=(x+y)^3-3xy((x+y)-1)-1$$ so the given equation is equivalent to $(x+y)^3-1=3xy((x+y)-1)$. Notice that this implies that $x+y=1$ is a solution. Since this is a line, an equilateral triangle can be made from any point such that $x+y\ne 1$ and $2$ points on $x+y=1$. Therefore, it suffices to prove that there is only one solution to $(x+y)^3-1=3xy((x+y)-1)$ such that $x+y\ne1$. Since we've already considered $x+y=1$, we can divide by $x+y-1$ to get $(x+y)^2+(x+y)+1=3xy$ or $x^2-xy+y^2+x+y+1=0$. We can rewrite this as $(x+1)^2+(y+1)^2-(x+1)(y+1)=0$. Clearly, $(-1, -1)$ is a solution to this. Plugging in $x=-1+a$ and $y=-1-a$ into the original equation gives $a=0$, and we have already considered $x=y=-1$, so we multiply both sides by $x+y+2=(x+1)+(y+1)$ to get that $(x+1)^3+(y+1)^3=0$ so $x+y+2=0$, which we've already shown only has solution $(-1, -1)$. Now, we calculate the area of this unique equilateral triangle to be $\frac{3\sqrt2}{2}\cdot\frac{3\sqrt\frac23}{2}=\frac{3\sqrt3}{2}$.


Let be $ n\in \mathbb{N}\ ,\ n\ge 2$ and $ a_1,a_2,...,a_{2n}\in \mathbb{R}_\plus{}$ such that $ s\equal{}a_1\plus{}a_2\plus{}...\plus{}a_{2n}$ . Show that :

$ \frac{a_1}{s\plus{}a_{n\plus{}1}\minus{}a_1}\plus{}...\plus{}\frac{a_n}{s\plus{}a_{2n}\minus{}a_n}\plus{}\frac{a_{n\plus{}1}}{s\plus{}a_1\minus{}a_{n\plus{}1}}\plus{}...\plus{}\frac{a_{2n}}{s\plus{}a_n\minus{}a_{2n}}\ge 1$

When does the equality occurs ?

nice, By Cauchy-Scwarz,
$ \left(\frac{a_1}{s\plus{}a_{n\plus{}1}\minus{}a_1}\plus{}...\plus{}\frac{a_{2n}}{s\plus{}a_n\minus{}a_{2n}}\right)\left(a_1s\plus{}a_1a_{n\plus{}1}\minus{}a_1^2\plus{}...\plus{}a_{2n}s\plus{}a_{2n}a_{n}\minus{}a_{2n}^2\right) \geq s^2$
and we have that,
$ a_1s\plus{}a_1a_{n\plus{}1}\minus{}a_1^2\plus{}...\plus{}a_{2n}s\plus{}a_{2n}a_{n}\minus{}a_{2n}^2\equal{}2(a_1a_{n\plus{}1}\plus{}...\plus{}a_{2n}a_n)\plus{}s^2\minus{}\sum_{i\equal{}1}^{2n}a_i^2$
and it's easy to probe that, $ 2(a_1a_{n\plus{}1}\plus{}...\plus{}a_{2n}a_n) \leq \sum_{i\equal{}1}^{2n}a_i^2$
therefore $ a_1s\plus{}a_1a_{n\plus{}1}\minus{}a_1^2\plus{}...\plus{}a_{2n}s\plus{}a_{2n}a_{n}\minus{}a_{2n}^2 \leq s^2$
then $ \frac{a_1}{s\plus{}a_{n\plus{}1}\minus{}a_1}\plus{}...\plus{}\frac{a_n}{s\plus{}a_{2n}\minus{}a_n}\plus{}\frac{a_{n\plus{}1}}{s\plus{}a_1\minus{}a_{n\plus{}1}}\plus{}...\plus{}\frac{a_{2n}}{s\plus{}a_n\minus{}a_{2n}} \geq 1$
equality occurs when, $ a_1\equal{}a_2\equal{}...\equal{}a_{2n}$

[quote="peine"]nice, By Cauchy-Scwarz,
$ \left(\frac {a_1}{s \plus{} a_{n \plus{} 1} \minus{} a_1} \plus{} ... \plus{} \frac {a_{2n}}{s \plus{} a_n \minus{} a_{2n}}\right)\left(a_1s \plus{} a_1a_{n \plus{} 1} \minus{} a_1^2 \plus{} ... \plus{} a_{2n}s \plus{} a_{2n}a_{n} \minus{} a_{2n}^2\right) \geq s^2$
and we have that,
$ a_1s \plus{} a_1a_{n \plus{} 1} \minus{} a_1^2 \plus{} ... \plus{} a_{2n}s \plus{} a_{2n}a_{n} \minus{} a_{2n}^2 \equal{} 2(a_1a_{n \plus{} 1} \plus{} ... \plus{} a_{2n}a_n) \plus{} s^2 \minus{} \sum_{i \equal{} 1}^{2n}a_i^2$
and it's easy to probe that, $ 2(a_1a_{n \plus{} 1} \plus{} ... \plus{} a_{2n}a_n) \leq \sum_{i \equal{} 1}^{2n}a_i^2$
therefore $ a_1s \plus{} a_1a_{n \plus{} 1} \minus{} a_1^2 \plus{} ... \plus{} a_{2n}s \plus{} a_{2n}a_{n} \minus{} a_{2n}^2 \leq s^2$
then $ \frac {a_1}{s \plus{} a_{n \plus{} 1} \minus{} a_1} \plus{} ... \plus{} \frac {a_n}{s \plus{} a_{2n} \minus{} a_n} \plus{} \frac {a_{n \plus{} 1}}{s \plus{} a_1 \minus{} a_{n \plus{} 1}} \plus{} ... \plus{} \frac {a_{2n}}{s \plus{} a_n \minus{} a_{2n}} \geq 1$
equality occurs when, $ a_1 \equal{} a_2 \equal{} ... \equal{} a_{2n}$[/quote]

Can you explain it clearly,

$ \boxed{2(a_1a_{n \plus{} 1} \plus{} ... \plus{} a_{2n}a_n) \leq \sum_{i \equal{} 1}^{2n}a_i^2}$

Thanks.

[quote="PKMathew"]

Can you explain it clearly,

$ \boxed{2(a_1a_{n \plus{} 1} \plus{} ... \plus{} a_{2n}a_n) \leq \sum_{i \equal{} 1}^{2n}a_i^2}$

Thanks.[/quote]

$ \Leftrightarrow (a_1\minus{}a_{n\plus{}1})^2\plus{}...\plus{}(a_{2n}\minus{}a_n)^2\ge 0$

This seems like a special case of http://www.mathlinks.ro/viewtopic.php?t=168801. Which I don't remember how to prove... I think I was in my "discovering cauchy" time.... I'll try to remember and post it.

:)


LaTeX
algebra
polynomial
Rational Root Theorem

:)  :D
Show that the number x given by
$x=1+\sqrt{2+\sqrt{1+\sqrt{2+...}}}$
is irrational.

Hrmm...I wouldn't have expected that to turn up in Pre-Olympiad.

On a similar note:

Prove $\pi$ is transcendential.

@Farenhajt
wouldn't it be $((x-1)^2-2)^2=x$ instead

$x=1+\sqrt{2+\sqrt{x}}$

square, and rearrange, square, and use rrt to show x is not rational

ah.. thanks man... i gotta look it up on the net to find out what it is..  :blush:

Continuing from above, you'll get $x^4-4x^3+2x^2+3x+1=0$.. Suppose a rational root $x_0=p/q$ exists, where $p,q$ are integers. From rational root theorem, $p|1$ and $q|1$, which implies $p=+-1$, $q=+-1$, and thus $x_0=+-1$.
Subbing $x_0=1$ to $((x-1)^2-2)^2=x$, you'll get a contradiction.
Subbing $x_0=-1$ to $((x-1)^2-2)^2=x$, you'll get another contradiction.
Thus, x can't be a rational number ( x is irrational )..

is my working correct?

That looks right :) Also, you can use \pm in $\LaTeX$ to get $\pm$.


Find all f:R->R such that
f(f(x))=x and f(x+y)=f(x)+f(y)
for all x,y∈R

Last is Cauchy's equation with solution $ f(x)\equal{}ax$. Put this into first we get $ a^2\equal{}1$ and two solutions $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$.

No, Cauchy equation usually requires additional assumption on continuity, monotonicity, measurability or smth. else. Given just $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$, we can have a lot of "weird" solutions. This problem seems to be closer to real analisys(i think i have a solution), but i would like to see something algebraic.

In fact, there are a lot of functions satisfying $ f(x \plus{} y) \equal{} f(x) \plus{} f(y), \ f(f(x)) \equal{} x$. Let $ \{a_{\alpha}, \alpha\in \mathbb{A} \}$ be a basis of $ \mathbb{R}$ over $ \mathbb{Q}$, let $ b\in \{ \plus{} 1, \minus{} 1\}^{\mathbb{A}}$ and define $ f(\sum_{\alpha}q_{\alpha}a_{\alpha}) \equal{} \sum_{\alpha} q_{\alpha}a_{\alpha}\cdot b_{\alpha}$. It is clear that $ f$ satisfies both conditions.  But, we say something about the functions $ \mathbb{R}^{ \plus{} }\to \mathbb{R}^{ \plus{} }$, there will be only one solution $ f(x) \equal{} x$.
The following problem is somewhat similar: $ f: \mathbb{R}^{ \plus{} }\to\mathbb{R}^{ \plus{} }, \ f(x \plus{} y) \equal{} f(x) \plus{} f(y), \ f(z)f(\frac 1z) \equal{} 1$.


Let (X,d)be a compact metric space and f_n be a sequence of countinuous functions from X to R.We assume that for all ε>o ,for all x in X,there is a positive δ_x  such that for all n in N and for all y in X, |f_n (x)- f_n (y) |< ε,if |x-y|< δ_x.
a.If for all x in X,there is a positive M_x  such that |f_n (x) |≤M_x for all n in N,show that there is a positive  M such that |f_n (x) |≤M, for all x in X and n in N
b.  Show that there is a positive δ,such that for all x and y in X and all n in N, |f_n (x)- f_n (y) |<ε,if |x-y|< δ

That property is called [url=http://en.wikipedia.org/wiki/Equicontinuity]equicontinuity[/url], and the point of this problem is to show that an equicontinuous family on a compact space is uniformly equicontinuous.*

The basic idea: pick an $ \epsilon$, take all of those $ \delta_x$ balls, and find a finite subcover of the space. There is some small $ \delta$ such that every point has a $ \delta$-ball around it contained entirely in one of that finite collection of $ \delta_x$ balls. That's (b).
Now, what about (a)? Cover the space with finitely many $ \delta$-balls, and use the bound at the center together with uniform equicontinuity to extend to a bound on the whole ball. The maximum of those finitely many bounds is still finite.

* A side note on terminology: you presumably know what "uniform" or "uniformly" means. "Equi" is uniform in a different variable, when we already have a direction to reasonably call uniform. This appears most often in dealing with sequences of functions $ f_n(x)$, where variation over $ x$ can be "uniform" and variation over $ n$ can be "equi".

[color=green]Moderator edit: redundant quote deleted[/color]

Thanks. I'm just not quite sure how to do b though, as I am having difficulty choosing delta.

The maximum size ball you can fit around $ x$ in one of those open sets is a continuous function of $ x$. It has a minimum.

http://en.wikipedia.org/wiki/Lebesgue's_number_lemma

It's in Munkres, and should be in any intro topology book, because it is very useful when dealing with compact metric spaces.


function
limit
inequalities
algebra
functional equation
algebra solved

Find all continuous functions $f: R \rightarrow R$ satisfying:
$f(f(x))=f(x)+2x$; $\forall x \in R$.

There are only two solutions, namely $f(x)=-x$ and $f(x)=2x$.

Suppose $f$ is a solution.
Clearly $f$ is injective. Thus, $f(0) = 0$ and since $f$ is continuous, $f$ is monotonic. Therefore it has limits in $+ \infty$ and $-infty$. Since, the range of $fof - f$ is $R$, we deduce that the range of $f$ is $R$ too.
Now, $f$f is a bijective function from $R$ onto $R$, and we denote $f^{-1}$ its inverse.

Let $f^0 = f$, and $f^{n+1} = fof^n$ for all integers $n$.
Let $g(x) = 2x-f(x)$ and $h(x) = f(x)+x$.
Let $x$ be fixed.
Then, using the given functional equation, we easily deduce that $f^{n+2}(x) = f^{n+1}(x) + 2f^{n}(x)$.
Solving the recurrence relation, we deduce that $f^n(x) = \frac {g(x)} 3 (-1)^n + \frac {h(x)} 3 2^n$.

- First case : if $f$ is increasing.
Then, from $f(0)=0$ we deduce that $f^n(x) >0$ for all $x > 0$ and all integers $n$.
Let $x_0 > 0$ be given.
If $g(x_0) \neq 0$ then lim $2^n \frac {h(x)} x = 0$ when $n$ tends to $- \infty$ thus according to the sign of $g(x_0)$ we may found $n$ sufficientely large such that $f^n(x_0) < 0$. A contradiction.
Thus $g(x_0) = 0$ and $f(x) = 2x$ for all $x \geq 0$.

We deduce that $f(x)= 2x$ for all $x < 0$ in the same manner.
Thus $f(x) = 2x$ for all $x$.

- Second case : $f$ is decreasing.
In a similar manner, we deduce that $f(x)=-x$.

Pierre.

Alternative solution. We can show as pbornsztein did that [tex]f[/tex] is injective, monotonic, and the image [tex]f(\mathbb{R})=\mathbb{R}[/tex]. Let [tex]S=\sup_{x\in \mathbb{R}}(\frac{f(x)}{x})[/tex] and [tex]I=\inf_{x\in \mathbb{R}}(\frac{f(x)}{x})[/tex]. Take two sequences [tex]\{x_i\}^{\infty}_{i=0}[/tex] and [tex]\{y_i\}^{\infty}_{i=0}[/tex] such that [tex]\lim_{i\to \infty}\frac{f(x_i)}{x_i}=S[/tex] and, similarly, [tex]\lim_{i\to \infty}\frac{f(y_i)}{y_i}=I[/tex]. If [tex]S=0[/tex], then [tex]\lim_{i \to \infty}\frac{f(f(x_i))}{f(x_i)}=\lim_{i \to \infty}\left(1+\frac{2x_i}{f(x_i)}\right)[/tex] implies that [tex]I=-\infty[/tex], but [tex]\lim_{i \to \infty}\frac{f(f(y_i))}{f(y_i)}=\lim_{i \to \infty}\left(1+\frac{2y_i}{f(y_i)}\right)=1[/tex], so [tex]S\geq1[/tex]. Contradiction. Similarly, the following equalities and inequalities take place: [tex]-\infty < S,I < +\infty[/tex] and .[tex]SI \neq 0[/tex]. Using [tex]\lim_{i \to \infty}\frac{f(f(x_i))}{f(x_i)}=\lim_{i \to \infty}\left(1+\frac{2x_i}{f(x_i)}\right)[/tex], we arrive at [tex]I=1+\frac{2}{S}[/tex], since [tex]S=\sup_{x\in \mathbb{R}}(\frac{f(x)}{x})=\sup_{x\in \mathbb{R}}(\frac{f(f(x))}{f(x)})[/tex] cause of [tex]f(\mathbb{R})=\mathbb{R}[/tex]. Similarly [tex]S=1+\frac{2}{S}[/tex], so [tex]S+2=SI=2+I[/tex], so [tex]S=I[/tex] and [tex]f(x)=kx[/tex] for all [tex]x\in \mathbb{R}[/tex]. It is easy to find [tex]k[/tex] which is equal to [tex]2[/tex] or [tex]-1[/tex].


geometry
3D geometry
tetrahedron
circumcircle
analytic geometry

In $ xyz$ space, given 3 points $ A(1,\ 0,\ 0),\ B(0, \minus{} 2,\ 0), C(0,\ 0,\ 4)$. Let $ P$ be the cenre of circumcircle of $ \triangle{ABC}$.

Draw a line passing through the pont $ P$ and perpendicular to the plane $ ABC$. Take the point $ Q$ on the line.

(1) Find the $ x$ coordinate of $ P$.

(2) Let $ R$ be one of the point of the circumcircle of $ \triangle{ABC}$ such that $ \angle{PRQ} \equal{} 60^\circ$. Find the $ x$ coordinate of $ Q$.

(3) Under the condition of question (2), find the volume of tetrahedron $ QABC$.

No, I'm pretty sure he means [url=http://en.wikipedia.org/wiki/Incenter]excircle[/url].

Okay, I lied.

[quote="brightzhu"]Do you mean the circumcircle of the triangle?[/quote]

Yes. :blush:

I'm very sorry that I made a typo. I have just edited it.

kunny


combinatorics
algorithm
game strategy
search algorithms
IMO Shortlist

[quote="orl"]The lock of a safe consists of 3 wheels, each of which may be set in 8 different ways positions. Due to a defect in the safe mechanism the door will open if any two of the three wheels are in the correct position. What is the smallest number of combinations which must be tried if one is to guarantee being able to open the safe  (assuming the "right combination" is not known)?[/quote]

Someone who knows the solution of this?

The problem belongs to the chapter of Combinatorics called "Packings and Coverings".
A typical question would be: given a set $X$ of $n$ elements, what is the minimal number of triplets in a family $\mathcal{F}$ of subsets of $X$ of cardinality $3$, such that any pair of elements of $X$ is contained in at least one triplet of $\mathcal{F}$ (triplets covering doubletons)? This question has been fully answered, sometimes towards the end of the XX century - but I lack a reference.
In the problem at hand, the situation is complicated by needing also that the pair be in the right position (in both doubletons and triplets, position matters).

The answer is $32$

[b]Lower Bound[/b]
Assume there exists a set $S$ of $31$ triplets that cover all combinations. Let $S_i$ be the triplets in $S$ with first component equal to $i$, for $i=1,2,...,8$. There must exist some $i$ such that $|S_i| \le \textstyle\lfloor\frac{31}{8} \rfloor = 3$. wlog $i=1$

Consider the second and third components in the triplets of $S_1$. Suppose $k_1\le 3$ numbers appear in some triplet as a second component and $k_2\le 3$ as a third component. Then there are $(8-k_1)(8-k_2)\ge 25$ triplets $(1,y,z)$ that don't share two components with any triplet in $S_1$. Let the set of triplets be $K$. So among the other sets $S_i$ there exists a triplet of the form $(i,y,z)$ that meets $(1,y,z)$ in the last two components for each $(1,y,z)\in K$.

If each $S_i$ for $i\ge 2$ contained a triplet $(i,j,k)$ that doesn't meet a triplet in $K$ in two components then $|S| \ge |S_1| + |K| + 7 \ge 0 + 25 + 7 = 32$ contradiction. So there exists a set, say $S_2$, such that every triplet in $S_2$ meets an element of $K$ in two components (i.e. the last two). But now there are at least $k_1$ elelemts are that aren't a second component of a triplet in $S_2$ and at least $k_2$ elements that aren't a third component. So there are at least $k_1k_2$ triplets $(2,y,z)$ that don't meet any element of $S_2$ in two components. Call these triplets $K'$. No triplet can meet a triplet in both $K$ and $K'$ because triplets in $K$ and $K'$ have distinct components. Meaning that $|S| \ge |K| + |K'| = (8-k_1)(8-k_2) + k_1k_2 = 32 + 2(4-k_1)(4-k_2)$ where $k_1,k_2 \le 3$. This gives $|S| \ge 34$ and we have our contradiction.

[b]Upper Bound[/b]
For all $0\le i,j \le 3$ take the triplets $(i,j,i+j \mod 4)$ and $(4+i,4+j,4 +(i+j \mod 4))$ (which is $2\cdot 4^2=32$ triplets in total). Now for any triplet $(a,b,c)$ one of the pairs $(a,b),(b,c)$ or $(c,a)$ will be in the same interval $\{1,2,3,4\}$ or $\{5,6,7,8\}$. So in one of $(i,j,i+j \mod 4)$ or $(4+i,4+j,4 + (i+j\mod 4))$ we will be able to select $i,j$ to get the desired result.


algebra
polynomial
factorial
abstract algebra
function
domain
calculus

Let G be the additive group of all polynomails in x with integer coefficients. Show that G is isomorphic to the group Q+ of all positive rationals (under multiplication). (Hint: Use the Fundamental Theorem of Arithmetic to construct an isomorphism).

Attempted solution:

I wanted to do something like $ \phi \left(\frac{p_1^{t_1}\cdots p_n^{t_n}}{q_1^{s_1}\cdots q_m^{s_m}} \right) \equal{} \sum_{k\equal{}0}^n x^{p_k} t_k \minus{} \sum_{k\equal{}0}^m x^{q_k} s_k$ but of course then you only get polynomials with prime powers.

Then why don't you take $ p_k$ to $ k \minus{} 1$?

since $ \mathbb{Z}$ is factorial, $ \mathbb{Q}_{>0}$ is free abelian with basis consisting of all prime numbers. and $ \mathbb{Z}[X]$ is also free abelian with an infinite countable basis.

-oo-, I don't know exactly what is by "Z is factorial", but I see how that idea works.

A unique factorization domain is any integral domain in which every nonzero noninvertible element has  essentially unique decomposition as the product of prime elements or irreducible elements.


Find the last two digits of a sum of eightth powers of $100$ consecutive integers.


I can't come this year due to other commitments, but how do I join in the future? Do I just email Mr. Yang? What pre-requisite tests do I need to take?

Any help in this matter would be appreciated.

I don't know how your region does it but at least where I live, there is a high school math olympiad that extends for a few weeks.  The top scorers are then invited to training sessions and the ARML team is built from those people.  I would imagine your region has a similar process.

From the FAQ at [url]http://www.cscc.edu/docs/math/faculty/ayang2.htm[/url]

Q: How can I join Ohio ARML?
A: Mostly, if you score high on OCTM, AMC, or Ohio Math League, you will be invited to join Ohio ARML. Among these competitions, only OCTM provides Ohio ARML with participants’ personal contact information. Therefore, if your school does not participate in OCTM, it is suggested that you contact Prof. Alan Yang at Columbus State Community College.

Some students get in through their school's or district's recommendations. Teachers, Gifted Coordinators, or Gifted Services teachers, especially from districts (or schools) that do not actively participate in math competitions, may also recommend students. Those wishing to nominate students should submit the appropriate information to Prof. Alan Yang at: Math Department, Columbus State Community College, 550 E. Spring Street, Columbus, Ohio 43215.

Ohio ARML members who still are eligible will get the first round of invitations in February. Sometime in mid March, the next round of invitations will be sent out. If needed, another round of invitations will be sent out in early April.

Cool. So I just send him AMC scores next year? I'm not sure if we do OCTM.

If you are interested in joining ARML '09 shoot me an e-mail at
ayang@cscc.edu

Alan

If you are interested in joining ARML '09, shoot me an e-mail at
ayang@cscc.edu

Alan

darn, I continue to fail at math while under time pressure

Darn, so do I That's why I phailed nats...must do better this year.


geometry
angle bisector
geometry unsolved

this problem is equivalent to this.we have $ a$ and $ h_a$ and $ l\minus{}a$ and i think its open.

"this problem is equivalent to this."   ?   :D

Let $ V,D,M$ denote the unknown feet of the altitude, angle bisector and median on $ BC,$ respectively. The angle between the altitude $ h_a$ and the angle bisector $ AV$ is presicely half of the given difference $ |\beta - \gamma|.$ Thus, choose $ D$ on a fixed line $ a$ (the sideline BC), construct the A-vertex on the perpendicular to $ a$ at $ D$ with the given distance $ h_a$ and draw the angle bisector of $ \angle BAC,$ which is a A-ray $ v$ such that $ \angle(v,h_a) = \frac {_1}{^2}|\beta - \gamma|$ $ \Longrightarrow$ $ V \equiv v \cap a.$ Perpendicular to $ AV$ through $ A$ is the external bisector of $ \angle BAC$ meeting $ a$ at $ V'.$ The circle with diameter $\overline{ VV'}$ is the A-Apollonius circle, which is orthogonal to the circle $ (M)$ with diameter $\overline{ BC},$ since $ (B,C,V,V') = - 1.$ Hence, if $ N$ is the midpoint of $ VV'$ and $ P$ is one of the intersections of $ (N)$ and $(M)$ $ \Longrightarrow$ $ NP \perp MP.$ The right $ \triangle MPN$ is constructible since its catheti $ NP$ and $ MP$ are known as half of the distances $ VV'$ and $ BC$ $\Longrightarrow$ length of the hypotenuse gives the position of $ M$ on the sideline $a.$ Circle centered at $ M$ with radius $ \frac {_1}{^2}BC$ cuts $ a$ at $ B,C$ and $\triangle ABC$ is completed.

"construct the A-vertex on the perpendicular to $ a$  at $ D$  with the given distance $ h_{a}$ and draw the angle bisector of $ \angle BAC$  which is a A-ray $ v$ such that  $ \angle(v,h_{a}) \equal{}\frac{_{1}}{^{2}}(\beta\minus{}\gamma)$ $ \Longrightarrow$ $ V\equiv v\cap a$ "

There's two feet of altitude onto BC:  $ D$ and $ V$ ?

 :D

[quote="Luis González"] choose $ D$ on a fixed line $ a$ (the sideline BC), construct the A-vertex on the perpendicular to $ a$ at $ D$ with the given distance $ h_a$ and draw the angle bisector of $ \angle BAC,$ which is a A-ray $ v$ such that $ \angle(v,h_a) \equal{} \frac {_1}{^2}|\beta \minus{} \gamma|$ $ \Longrightarrow$ $ V \equiv v \cap a.$ [/quote]
Read carefully, we have to fix a sideline $ a$ and then choose a point $ D$ as the foot of the A-altitude. But if you mean that vertices $ B$ and $ C$ are previously fixed, then there's no problem either, because we can construct $ \triangle A'B'C' \cong \triangle ABC.$ Then $AB=A'B'$ and $AC=A'C'.$


For the threads containing problems in all my class message boards, the solutions are shown and I recall that all of them were hidden before. It appears the hidden function was tinkered with and needs to be fixed.

Yes, this also happens to me.  It's not just in classes, either.

For example:

[hide]
It's not hidden, is this?
[/hide]

Another note:  It doesn't show up in preview function either.

[quote="AIME15"]Now it's hidden, but when you click it, nothing happens.[/quote]Guys I'm working on it, I will let you know when it's fixed :)

OK main functionality is fixed. Please report any other bugs you find (I've made a major update to the javascript on the site so there are bound to be some things broken.)

It appears that, when you click on hide tags in blogs, it redirects to the top of the page instead of revealing the hidden content.

Also, when I'm posting this message, under "Topic Review", when I click my hide tag, it doesn't open and automatically scrolls up to the top, like cf249 said.

It doesn't work @ http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237933


$ a_{1},a_{2},...$ is a sequence of real numbers where $ a_{n}$ is the arithmetic mean of the previous $ n-1$ terms for $ n>3$ and              $ a_{2004}=7$. $ b_{1},b_{2},...$is a sequence of real numbers in which b_n is the geometric mean of the previous n-1 terms for $ n>3$ and $ b_{2005}=6$. If $ a_{i}=b_{i}$ for $ i=1,2,3$ and $ a_{1}=3$, then compute the value of $ a_{2}^{2}+a_{3}^{2}$


proof that for $ a,b,c>0$ we have this inequality:
$ \forall t ,2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ac)\plus{}\sum_{cyc} a^{2t}(a\minus{}b)^2 \ge \sum a^t(a\minus{}b)(a\minus{}c)$ with equality if and only if $ a\equal{}b\equal{}c$

no one found a solution ?! am still waiting  :(

I just analyze it as SOS:
ineq$ \Leftrightarrow\sum(a^{2t}\plus{}1\minus{}\frac{a^t\plus{}b^t\minus{}c^t}{2})(a\minus{}b)^2\geq 0$
who can evaluate it

did you continue this SOS analysis ? coz actually i have a very simple solution, i want just to see a simpler one

[quote="othman"]proof that for $ a,b,c > 0$ we have this inequality:
$ \forall t ,2(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ac) \plus{} \sum_{cyc} a^{2t}(a \minus{} b)^2 \ge \sum a^t(a \minus{} b)(a \minus{} c)$ with equality if and only if $ a \equal{} b \equal{} c$[/quote]
We have
\[ LHS \equal{}\sum (a\minus{}b)^2\plus{}\sum a^{2t} (a\minus{}b)^2 \equal{}\sum [(a\minus{}c)^2\plus{}a^{2t}(a\minus{}b)^2] \ge 2\sum a^t \left| (a\minus{}b)(a\minus{}c)\right| \ge \sum a^t (a\minus{}b)(a\minus{}c)\]
:)


A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more ``bubble passes''.  A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$, with its current predecessor and exchanging them if and only if the last term is smaller.

The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass.  The numbers compared at each step are underlined.

\[ \begin{array}{c} \underline{1 \quad 9} \quad 8 \quad 7 \\ 1 \quad \underline{9 \quad 8} \quad 7 \\ 1 \quad 8 \quad \underline{9 \quad 7} \\ 1 \quad 8 \quad 7 \quad 9 \end{array}  \]

Suppose that $n = 40$, and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order.  Let $p/q$, in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\text{th}}$ place.  Find $p + q$.

[hide]Notice that in order for $r_{20}$ to reach $r_{30}$, it must be true that $r_{20}$ is the greatest of all the terms $r_1,r_2,\ldots,r_{30}$. This is because if $r_{20}$ is less than one of those terms, it won't be passed to the next position. This can happen with $\frac{1}{30}$ probability.

We also see that if $r_{30}>r_{31}$, it will be passed to the $31^{\text{st}}$ position. Therefore, we must have $r_1,r_2,\ldots,r_{30}<r_{30}<r_{31}$, which can happen with $\frac{1}{31}$ probability. Therefore, our answer is $\frac{1}{3}\cdot \frac{1}{31}=\frac{1}{930}$ and $p+q=931$.   [/hide]

actually r20 doesn't have to be the greatest of all the terms r1,r2,....r30  r20 can be the smallest number in the sequence but being settled at the 31 position at the beginning of the game

If the bubble pass starts at $ a_{20}$ and continues all the way to $ a_{40}$, then the probability would be $ 1/(11*10)$. However, there's a $ 1/2$ probability that $ a_{20}>a_{21}$ in comparing those two, and then another $ 1/2$ probability that $ a_{20}>a_{22}$ and so on... so shouldn't the probability be $ (1/2)^{11}$? (I'm pretty sure there's an equal $ 1/2$ probability for either element being greater in each comparison.) What's wrong with this reasoning?

Note that the above is equivalent to saying that we if pick a random element $ a_i$ out of $ \{1,2,3,4,5\}$, find the probability that that element is the biggest, which is clearly $ 1/5$. However, if we consider this in another way of first picking $ a_i$ and another element and compare them, and then eliminate the element that's smaller. So there's $ 1/2$ probability of $ a_i$ being greater than the first element from the rest four, and then another $ 1/2$ probability of $ a_i$ being greater than the element from the rest three, and so on; so continuing we should get $ 1/2^4$, but those two answers aren't the same...  :huh:

EDIT: Just realized that my solution was wrong :). 4everwise is right.

@cdymdcool Indexes determine starting place of the numbers.

[quote=me for the past month]How do I manage to solve the hardest problems and yet fail on the easiest ones?[/quote]

[hide=Solution]We only care about the first $31$ terms of the sequence. The condition rewrites itself as $r_{31}$ and $r_{20}$ being the largest and second-largest elements of the sequence, respectively. There's $\frac{1}{31}$ probability $r_{31}$ is the largest, and given that, there's $\frac{1}{30}$ probability $r_{20}$ is the second largest, so the answer is $\frac{1}{31\cdot30}=\frac{1}{930}\implies\boxed{931}$.

Only the first $31$ number matter; It is necessary and sufficient to have
\[ r_{31} > r_{20} > r_{30}, \dots, r_{21}, r_{19},\dots, r_1.\]
For any $31$ element subset of $\{1,2,\dots, 40\}$, there are $31!$ ways to assign the values to the $r_i$, and $29!$ of these will have the above property ($r_{31}, r_{20}$ have to be largest, second largest resp. and the remaining 29 can be permuted freely). The answer is
\[ \frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}\to\boxed{931}.\]


calculus
integration
function
Ring Theory
superior algebra
superior algebra theorems

Show that ring $\mathbb{Z}[\sqrt{2}]$ is euclidean.Show furthermore that its units are given by $(1+\sqrt{2})^n;-(1+\sqrt{2})^n(n\in\mathbb{Z})$ and determine its prime elements.

A norm you can use is $N: a+\sqrt 2 b \mapsto a^2-2b^2$, and then you have only to show that for every $z \in \mathbb{Q}[\sqrt 2]$, there are $a,b \in \mathbb{Z}$ with $|N(z-a-b\sqrt 2)| < 1$ (this commes from dividing $N(c-qd)=N(r)<N(d)$ through $N(d)$ when $c=qd+r$).

Now the idea to bring it into some lattice:
Consider the lattice generated by $x+\sqrt 2 y \mapsto (x+\sqrt 2 y , x-\sqrt 2 y)$ and you are (since $|x+y|,|x-y| < 2 \implies |xy|<1$) left to prove that for all $(x,y) \in \mathbb{Q}^2$ or even $\in \mathbb{R}^2$ there exists such $a,b \in \mathbb{Z}$ with $|(x-a)+(y-b)|,|(x-a)-(y-b)|<2$.
But the last one means that squares with diagonals of length $4$ and parallel to the axes cover, when placed at the integral points $(a+\sqrt 2 b , a-\sqrt 2 b)$, the whole $\mathbb{R}^2$.
This holds here since that lattice is generated by $(1,1)$ and $(\sqrt 2, -\sqrt 2)$.

Take a unit $u=x+\sqrt 2 y$ such that $x,y>0$. Then take the greatest $n$ such that $(1- \sqrt 2)^n (x+\sqrt 2 y) \geq 1$ and we want to prove that equality occurs. Well, assume the contraposition, then we have a unit $u=x+\sqrt 2 y$ with $1<u<1+ \sqrt 2$. Simple checking gives that not both, $x$ and $y$, can be positive, but can't also both be negative (since that means $1<u<0$), thus either one of them is $0$ (which will be $y$ then) or they have different signs. But in the second case just take conjugates=inverses to get both of them positive again,  giving $u^{-1}>1 \implies u<1$, contradiction.
In general, you can reduce every oterh case to this one by taking inverses and multiplying with $-1$.

Well, every prime either remains prime or splits into a product of two primes since $N(p)=p^2$.
So when there are no $a,b$ with $a^2-2b^2=p$, the prime $p$ remains prime. But that is at least unsolvable when $c^2 \equiv 2 \mod p$ is unsolvable.
This are also all primes that remain prime:
For all others, take such an $c$ (as an integer) and $p$ divides $(c+ \sqrt 2)(c- \sqrt 2)$, but none of the factors, thus can't be prime anymore.

[quote="ZetaX"]A norm you can use is $N: a+\sqrt 2 b \mapsto a^2-2b^2$
[/quote]

But i think.Norm to be function nonnegative? :roll:

Take $|N|: a+b\sqrt 2 \mapsto |a^2-2b^2|$ instead ;)

You are right!This is a problem other:Find all free square positive integer number $d>1$ such that $\mathbb{Z}[d]$ is  euclidean.

Do you mean $\mathbb{Z}[\sqrt{d}]$¿ Then the problem should be very hard...

[quote="ZetaX"]Do you mean $\mathbb{Z}[ \sqrt{d}]$¿ Then the problem should be very hard...[/quote]

Yes,i think so.

I think that is an open problem :) However one can prove that if $d<0$ the only euclidean rings of integers $\mathbb{Z}[\sqrt d]$ are for $d=-1,-2,-3,-7,-11$.

Actually you can look up Sloane's [url=http://www.research.att.com/~njas/sequences/A048981]A048981[/url].

[quote="Omid Hatami"]I think my teacher said if $\mathbb Z[\sqrt d]$ was PID for each $d$ then Fermat's last theorem would be proved easily[/quote]
Are you sure about that¿ I think you need every number field to be PID for that (than UFD follows and the standard arguments work)...


For what real values of $ x$ is

$ \sqrt {x \plus{} \sqrt {2x \minus{} 1}} \plus{} \sqrt {x \minus{} \sqrt {2x \minus{} 1}} \equal{} A,$

given (a) $ A \equal{} \sqrt {2}$, (b) $ A \equal{} 1$, (c) $ A \equal{} 2$, where only non-negative real numbers are admitted for square roots?

wow....

old IMO's are SO trivial...

it's pretty sad i guess

this is WAY too trivial for IMO

its  not  too trivial,  its  kinda  according to the people  in those times.  anyways  the  imo`s have become tougher and tougher  since  the 90`s .

this is not the case  only with  imo`s  . we also  can see that  the old admisiion or enterance tests of  many  famous insittes  which  have world wide reputation  are pretty easy. :blush:


有什么线性代数的习题集比较好的啊,介绍一下吧.

You certainly mean quatre-vingt-treize ...

quatre-vingt-treize???
It is 93 in French...  :|

这是雨果写的一部小说叫做《九三年》……


[size=150]Welcome to the very first, and hopefully not the last, [b]The Mole: AoPS Deception[/b]![/size]

[u]PLAYER LIST[/u]:
2. xpmath
5. miyomiyo
6. BOGTRO [size=59]replacing kimmystar94 [/size]
7. hm29168 [size=59]replacing woser456[/size]
8. CatalystOfNostalgia

Eliminated:
Temperal (Fraud)
Hamster1800 (Test 1)
erabel (Test 2)
mnmath (Test 3)
zapi2007 [size=59]replacing Cyrazeno[/size] (Test 4)
Klebian (Test 5)

Sorry to those that applied and didn't make the cut; I tried to make the game as diverse, interesting, and sure to last as possible. You're now making up the replacement list, though!

First, let’s get some logistics out of the way. Exactly 9 of you are playing the game, trying (not really) to win hundreds of thousands of dollars. However, one of you is THE MOLE, who is trying to thwart your efforts and stop you from winning money. Each round, you will participate in several tasks, in which you shall work together to earn money for the pot. The mole, detesting all of you, will sabotage these tasks, while evading all of your watchful eyes. At the end of each round, there will be a test, questions about the mole. Whoever receives the lowest score will be executed, and lose all chances of winning the game. The final two players, along with the mole, will take part in a final test, and whichever player scores the highest will win the game! Well, that is, if the mole doesn’t defeat all of you…

You see, the mole can win too. If (s)he prevents you from reaching your target pot(known only to myself and the mole), and prevents both of the final two players from getting a perfect score on the last test, then the mole will steal the cash and be the victor! But that certainly won’t happen…

THE RULES:
1. There are ten of you, and some more temporary players will not be here long at all. Get over it.
2. There will be outside communication, of any form, allowed at all times in the game UNLESS STATED OTHERWISE.
3. All deadlines are final, and will not for any reason be changed.
4. To be clear, lying and deception are a part of the game, and are encouraged.
5. Don’t disappear: replacements aren’t fun.
6. All players, including the mole, can be replaced.
7. Tasks will test your abilities, and are not meant to be hurtful, nor easy. Don’t take anything personally.
8. I reserve the right to(but don’t plan to) change the tasks or rules at any times.
9. Tests will proceed as follows: I will post the questions, with a three day time limit. The lowest scorer will be executed. In the case of a tie, whoever turned it in later will be executed.
10. There will be several possible exemptions offered during the game. If a player wins one, they will not be required to fill out the round’s test, and will  automatically advance to the next round.
11. Expect surprises along the way: I, like The Mole, am unpredictable.
12. This is a game. Have fun, and don’t ruin it for yourself or others.

Well, here we go. Good luck, make a lot of money, and don’t let the mole take advantage of you.

----------------

QUID PRO QUO

[color=blue][size=150][b][u]TASK 1: The Invader[/u][/b][/size][/color]

Hey, didn’t I say that there were only 10 players? I think that I did anyway…Well, yes I did. That means that one of you eleven is a fraud! Your first task will be to discover this culprit, and reveal him/her for the scoundrel they truly are. For clues, well, see this and the first post of the topic…To submit, one of you must post, in bold and word for word, [b]"The fraud is _______"[/b]. Communication is allowed during this task.

Reward: $ \$20,000$

Deadline: 2/06/08 8:00 EST(About 24 hours from this edit-this is a quick one!)

----------------

QUID PRO QUO

To clarify - we only get one submission, yes?

[quote="perfect628"]2. There will be outside communication, of any form, allowed at all times in the game UNLESS STATED OTHERWISE.[/quote]Is this what you meant?

EDIT: Do we posts clues we find in the thread?

I suppose this wasn't clear...You can say whatever you want in thread, and out of the thread, at any time, unless otherwise stated. It will be clear at times when you can not talk to each other.

Well then,
[quote="perfect628"]Sorry to those that applied and didn't make the cut; I tried to make the game as diverse, interesting, and sure to last as possible. You're now making up the replacement list, though! [/quote]The fraud would be somebody who "didn't make the cut" and would thus be making up the "replacement list".  This may be referring to the mafia replacement list....

Just a thought.

Not possible - unless the answer is Klebian. There's currently no mafia replacement list so the only person who that could possibly apply to is Klebian. I'd rather look for some other clues.

True.  If you compare the "The Mole" thread with signups with the player list, a bunch of people didn't post in that thread but are on the list.

I'M THE FRAUD!

(since jokes like this have gotten me erroneously lynched in mafia games *cough*Peter game 31*cough*, this is a joke for the record)

I'm suspicious of the first few posters in this thread; they seem to have noticed the topic awfully quickly...

well...
okay...
do we know if the fraud knows that they are a fraud?

oh. and read perfect's signature/last statement of first post..

QUID PRO QUO..

any guesses?

EDIT: I read wiki's version, and it means "something for something"...
it might be a clue..

[quote="miyomiyo"][quote="Temperal"]I'm suspicious of the first few posters in this thread; they seem to have noticed the topic awfully quickly...[/quote]
(a) I have no life
(b) I'm currently watching Super Tuesday with cnnpoliticis.com and AoPS open
(c) I like posting[/quote]

eh... same here...
well.. not the first part of b..

21 hours until deadline.

----------------

QUID PRO QUO

Well, it was inevitable that I'd be knocked out.  It was very fun nonetheless, great job perfect!

I might join the second one but you might want to give us a little break between games.

Yay. Moleish behavior that I noticed:

1. In the challenge where the mole had to pick the truth out of three statements, the only one the mole got correct was miyomiyo's. It was about miyomiyo's age and grade. Well of course woser456 was going to get it right, they know each other!

2. In the challenge where you all (final 5) picked me to survive, I claimed (yes, I was lying, and I do not have a brother), that we would win 10,000 if I survived. hm29168 said that he should die, because we would win more if I survived. But that was a lie, as he knew we would win 40,000 if I survived.

3. Three Questions task: hm29168 put me down as all three answers. That makes absolutely no sense, as it would mean I was both meaner and nicer than miyomiyo. In addition, when guessing my answers, he put himself as all three, that also makes no sense and would never be right.

4. Final task: he was uberstalling us.

Also perfect: hm29168 wins 337,000, not 377,000.

I think I was trying to act like the mole too much. Eh.

[quote="miyomiyo"]I might join the second one but you might want to give us a little break between games.[/quote]
Hmm...Well the few weeks between now and USAMO should suffice...Any longer and it would probably fail. As it is, I'm going to have to go really fast.

[quote="BOGTRO"]Psst... If I AM the mole in the next game I'll probably drop clues about that[/quote]

GOOD PLAN. Now maybe if you're clever you'll use REVERSE PSYCHOLOGY. oh man i cannot WAIT to see what happens.

Just as a tip of advice for the next game. SamL told me that most of you didn't split answers (miyomiyo doesn't count), and that was your downfall. I split answers on every single test other than the last two. If you don't understand what "splitting answers" means, even if you think person X is the mole, you shouldn't necessarily answer the questions as if person X is actually the mole. Because you can never be sure (I knew I had nothing to lose on the last test because I knew miyomiyo wasn't the mole, and probably wasn't going to beat me).

It just so happened that lots of miyomiyo and hm29168 fell in lots of the same categories on execution quiz 7 (the one I got 10/10), so I would have still gotten like 8 had I been wrong.

[quote="BOGTRO"]
Thanks to [b]kimmystar94[/b], for backing down, and allowing me

to step in, effectively crushing kimmy  :P jk
[/quote]

thanks so much...  :mad:  hehe..
don't worry. i'm gonna apply for the next mole... teehee!

I don't think the Mole should win if only one person figured out who he is. I mean, hypothetically (obviously this didn't happen), miyomiyo could have gotten like 10 on the test because my play was so amazing that he was thoroughly convinced that I was the mole.

The Mole should win if both players score <27 on the test and the pot is not reached. I mean, isn't that what the original rules (first post) said? Except instead of <27, non-perfect scores.

So according to the original rules, don't I win the game?

[quote="CatalystOfNostalgia"]Yay. Moleish behavior that I noticed:

3. Three Questions task: hm29168 put me down as all three answers. That makes absolutely no sense, as it would mean I was both meaner and nicer than miyomiyo. In addition, when guessing my answers, he put himself as all three, that also makes no sense and would never be right.

4. Final task: he was uberstalling us.[/quote]

Comments: For point 3, I don't know ANYBODY here, and I didn't even pay attention to the questions, so I just put down the first player name that popped into my head for that.

For point 4, I was extremely busy on the day when all of the Final Task information got posted. I wasn't even intentionally trying to stall you guys.

But, whatever, I'm still happy I won.  :P

Can I still play in mole game 2?[/quote]

Just as my evidence:

[quote="SamL"]You see, the mole can win too. If (s)he prevents you from reaching your target pot(known only to myself and the mole), and prevents both of the final two players from getting a perfect score on the last test, then the mole will steal the cash and be the victor! But that certainly won’t happen… [/quote]

Defense:

[quote="perfect628"] 11. Expect surprises along the way: I, like The Mole, am unpredictable

8. I reserve the right to(but don’t plan to) change the tasks or rules at any times[/quote]

Well, as BOGTRO pointed out, I was allowed to change them, and I did. The point of that rule is to make The Mole sabotage(pot total), while remaining unknown. I wasn't expecting a player to not try on the quizzes...This was an unpredictable flaw--And one that shouldn't have happened. Look at it this way: Catalyst, you won, but the players lost.

[quote="SamL"]...This was an unpredictable flaw--And one that shouldn't have happened.[/quote]

 :P  :D

Anyway, does this occur in the actual Mole? Like the one on TV?

No, but it's different online...In real life, they're playing for money, and ratings. Online, everyone needs a win condition. So it has to be implemented, as otherwise The Mole is just there, and doesn't have to try very hard. He can just blatantly sabotage everything, so everyone knows who it is, or sabotage nothing, so the ot skyrockets, but no one knows who it is.

[quote="perfect628"]I wasn't expecting a player to not try on the quizzes...This was an unpredictable flaw--And one that shouldn't have happened.[/quote] :lol:

The really unpredictable thing that messed up the game was that my random quizzes got me this far.

Yes...I suppose...If you play in The Mole 2, let's hope it doesn't happen again.


Is there a good way to encourage many other people to post on this forum? IT is a good way to discuss with other people.  :)

Get your yellowbook and call all 1 million numbers in there.

 :roll:

Sigh* I was just kidding.

The best way is by word of mouth. Tell your friends, and even better, tell your teachers.

Tell them that it increases brainpower :D

Don't forget to mention that you get to meet cool people like me, the greatest loser on the face of the earth!

[quote="b-flat"]There is already quite a large number of members - over 23000!

However, telling people you know, especially at math club meetings, etc, will greatly help spreading popularity.[/quote]
Only about 1200 of them have 50+ posts. I myself had <20 posts until about 6 months ago.

And only 8685 of them have any posts, 6390 at least two.

But do we really want more spamming people here...  :roll:

[quote="anirudh"]So what is the point of 20000+ people, when only like 1000 of them are active.

Secondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.[/quote]

Hmm. Well I got the feeling that all of the forums are going and creeping a bit higher than the difficulty level. For example, I find it strange when a pre-olympiad forum has olympiad+ level questions.

[quote="anirudh"]So what is the point of 20000+ people, when only like 1000 of them are active.

Secondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.[/quote]
yea, but some people decide that they need to improve and start spamming furiously for some unknown reasons(*cough anirudh cough moogra cough*)

[quote="anirudh"]So what is the point of 20000+ people, when only like 1000 of them are active. [/quote]What do you understand by active? Some people register accounts and subscribe to topics so that they receive email notifications of when the topic has been posted into. Others check the forum regularly and use the unread messages feature.

[quote="anirudh"]Secondly, some people just sign up, then just don't come back again from sheer disappointment of not being able to solve any problems. I know a couple of kids like that from my school.[/quote]
I was one of those people until about 6 months ago when I needed practice for state and I saw that I was at the level of the mathcounts forum, so I started posting like crazy.

Most forums have a very low percentage of people who post actively; I think AoPS actually has a higher percentage.

With every so many users that sign up, you will get a new user who posts, so you want lots of people to sign up.  Like VV said, not everyone wants to post, but still visit, etc.


[color=darkblue]Sa se detemine numerele naturale $n, p$ unde $p$ este prim si
$n^{3}+np=p^{2}+1.$[/color]

solutia mea se incadreaza la categoria brute-force
imi e destul de lene sa o postez, dar e foarte straight-forward
exista vreo solutie interesanta?
[hide="raspuns"]
37,10
[/hide]

[color=darkblue]Hm... nu stiu daca solutia mea e gen brute-force. Iat-o:
[hide]Avem $(n-1)(n^{2}+n+1)=p(p-n)$ de unde $p>n$
Cum $p$ este prim rezulta ca din $p|(n-1)(n^{2}+n+1)$ avem
$p|n-1$ sau
$p|n^{2}+n+1$
Prima varianta este exclusa, dat fiind ca ar rezulta ca $p\leq n-1<p-1$
$\implies p|n^{2}+n+1$
Rezulta ca exista $k$ natural astfel incat
$pk=n^{2}+n+1$
Inlocuim acum in relatia initiala:
$(n-1)pk=p(p-n)\implies (n-1)k+n=p$
Inmultim cu $k$ si vom obtine:
$pk=nk+k^{2}n-k^{2}=n^{2}+n+1$
De aici vom considera ca o ecuatie de gradul 2 in $n$, unde discriminantul $d$ trebuie sa fie patrat perfect.
$(k^{2}+k-3)^{2}<d<(k^{2}+k-2)^{2}$ pentru $k>3$
Verificandu-se valorile lui $k=$1,2,3, vom obtine pentru $k=3$ solutiile $(n,p)=(10,37)$
[/hide]
Ok, asa te-ai gandit, sau altfel?[/color]

tot geva in genu
[hide="click"]
din cauza ca $p|n^{2}+n+1$, rezulta ca $n-1 | p-n$, deci $p-n=a(n-1)$, si de aici il inlocuiesc pe $p$ si apoi pun conditia ca $p|n^{2}+n+1$ si tot umblu la divizibilitate pe acolo pana cand ajung la $a=3$ sau ceva, ca da la sfarsit un polinom de gradu 3 in $a$.
[/hide]


function
induction
calculus
floor function
algebra proposed
algebra

Find all functions $ h: \mathbb{N}\to\mathbb{N}$ satisfying the functional relation $ h(h(n))+h(n+1)=n+2$ for all $ n\in\mathbb{N}$.

[quote="Rust"]By induction $ h(n)=[\phi n+c]$, were $ \phi=\frac{\sqrt 5-1}{2},c=\frac{3(3-\sqrt 5)}{2}=3-3\phi$.[/quote]

Can you post your solution .Thanks  :)

[quote="pco"][quote="Rust"]By induction $ h(n)=[\phi n+c]$, were $ \phi=\frac{\sqrt 5-1}{2},c=\frac{3(3-\sqrt 5)}{2}=3-3\phi$.[/quote]

I'm sorry, Rust, but your result is wrong again (calculus error, maybe).
With your formula, :
$ h(2)=[2\phi+c]=[\sqrt{5}-1+\frac{9-3\sqrt 5)}{2}]=[\frac{7-\sqrt 5)}{2}]=2$
$ h(3)=[3\phi+c]=3$

$ h(h(2))+h(3)=5\neq 2+2$[/quote]

$ h(n)=[n\alpha]-n+1$

[quote="quangpbc"]Find all functions $ h: \mathbb{N}\to\mathbb{N}$ satisfying the functional relation :

$ h(h(n))+h(n+1)=n+2$ for all $ n\in\mathbb{N}$[/quote]

The unique solution is $ h(n)=1+\lfloor(\phi-1)n\rfloor$ where $ \phi=\frac{1+\sqrt{5}}{2}$

$ 1).$ $ h(n)=1+\lfloor(\phi-1)n\rfloor$ is a solution.
Let $ a=\phi-1$. We have $ a^{2}=1-a$
$ h(h(n))=1+\lfloor ah(n)\rfloor=1+\lfloor a+a\lfloor an\rfloor\rfloor$
$ h(h(n))=1+\lfloor a+a(an-\{an\})\rfloor$
$ h(h(n))=1+\lfloor a+a^{2}n-a\{an\})\rfloor$
$ h(h(n))=1+\lfloor a+n-an-a\{an\})\rfloor$
$ h(h(n))=n+1+\lfloor a-an-a\{an\})\rfloor$

Then, we must notice that $ a-an-a\{an\}<0$ and that $ -a+an+a\{an\}$ can't be an integer :  $ -a+an+a\{an\}=$ $ -a+an+a(an-\lfloor an\rfloor)$ $ =-a+n-a\lfloor an\rfloor$ and this number can't be integer, else $ a$ would be a rational number.

Since $ a-an-a\{an\}<0$ and is not an integer, then  $ \lfloor a-an-a\{an\})\rfloor=-1-\lfloor-a+an+a\{an\})\rfloor$ and so :
$ h(h(n))=n-\lfloor-a+an+a\{an\})\rfloor$
$ h(h(n))=n+2-\lfloor 2-a+an+a\{an\})\rfloor$
$ h(h(n))=n+2-(1+\lfloor an\rfloor+\lfloor 1-a+(a+1)\{an\})\rfloor)$

On the other hand, we have :
$ h(n+1)=1+\lfloor an+a\rfloor$
$ h(n+1)=1+\lfloor an\rfloor+\lfloor \{an\}+a\rfloor$

So it remains to show that $ \lfloor 1-a+(a+1)\{an\})\rfloor=\lfloor \{an\}+a\rfloor$ :
If $ \{an\}<1-a$, RHS is $ 0$ and $ 0<1-a+(a+1)\{an\}<1-a+1-a^{2}=1$. So LHS is $ 0$ too.
If $ \{an\}\geq 1-a$, RHS is $ 1$ (since  $ 1\leq\{an\}+a<2$) and $ 2>1-a+(a+1)\{an\}\geq 1-a+1-a^{2}=1$. So LHS is $ 1$ too.

And so $ h(h(n)=n+2-h(n+1)$
Q.E.D.

$ 2).$ $ h(n)=1+\lfloor(\phi-1)n\rfloor$ is the [u][b]unique [/b][/u]solution
Let $ h(1)=u$
Then $ h(u)+h(2)=3$ which allow two cases : $ h(2)=1$ or $ h(2)=2$

Case 1 : $ h(2)=1$ and $ h(u)=2$
Then $ h(h(2))+h(3)=4$ gives $ h(3)=4-u$ and $ u$ can only be $ 1,2$ or $ 3$ but :
$ u=1$ implies $ h(1)=u=1$ but we already have $ h(u)=h(1)=2$
$ u=2$ implies $ h(u)=h(2)=2$ but we already have $ h(2)=1$
$ u=3$ implies $ h(u)=h(3)=2$ but we have $ h(3)=4-u=1$
So no solution exist in case 1.

Case 2 : $ h(2)=2$ and $ h(u)=1$
Then $ h(h(2))+h(3)=4$ gives $ h(3)=2$
Then $ h(h(3))+h(4)=5$ gives $ h(4)=3$
Its then easy to show that we must have $ u=1$ : consider $ m$ as the first positive integer for which $ h(m)=1$
Then, if $ m>1$, $ h(h(m-1))=m+1-h(m)=m$
But, from $ h(h(n))+h(n+1)=n+2$ we have $ h(n+1)<n+2$ and so $ h(n)\leq n$ $ \forall$ $ n>1$
$ h(m-1)>1$ ($ m$ is the lowest) and so $ h(h(m-1))\leq h(m-1)$. But $ h(2)=2$ and so $ m-1\neq 2$ and $ m-1>1$ and so $ h(m-1)\leq m-1$ and so $ h(h(m-1))\leq m-1$, in contradiction with $ h(h(m-1))=m$

So $ m=u=1$
So $ h(1)=1$
So $ h(n)\leq n$ $ \forall n$
So $ h(n+1)=n+2-h(h(n))$ and $ h(n)$ is totally defined as soon as $ h(1)$ is defined
So there is a unique solution to the required equation.

And so this unique solution is $ h(n)=1+\lfloor(\phi-1)n\rfloor$


How many non-similar triangle have angles whose degree measures are distinct positive integers in arithmetic progression?

$ \textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 59 \qquad \textbf{(D) } 89 \qquad \textbf{(E) } 178$

[hide]Call the three angles $a-d, a, a+d$ with $d>0$.  Then $(a-d)+a+(a+d)=3a=180^\circ$ so $a=60^\circ$.  Thus $d$ is any integer from $1$ to $59$ so $C$[/hide]

[hide="Answer"]Since we have three angles, the middle angle must be $60$ so that the three sum to $180$. The other angles can range from $(59,61)$ to $(1,119)$, so the answer is $\boxed{C}$.[/hide]

I basically did the same thing, but I did it in a longer way.  I made the three angles angles in an arithmetic progression and used the sum rule.  The sum rule states that the second and third angle have to equal 120, so the second angle is 60.  If we let x be the first angle, then 0<x<60, so that leaves 59 values for x.

PS: How do you get the mathematical symbols?[/i][/code]


A sequence of n positive integers is full if for eack k > 1, k only occurs if k-1 occurs before the last occurrence of k. How many full sequences are there for each n?

The answer is $ n!$. This is surprising since just looking at the "full" sequences, they don't look much like permutations. For example when $ n\equal{}3$ we have the following sequences:

111
112
121
122
123
212

A very nice and clear exposition of a bijection between the set of "full" sequences and permutations can be found [url=http://www.rose-hulman.edu/mathjournal/archives/2005/vol6-n2/paper6/v6n2-6pd.pdf]here[/url].


function algebra polynomial geometry perimeter

Let $ f$ be a polynomial function such that, for all real $ x$, $ f(x^2 \plus{} 1) \equal{} x^4 \plus{} 5x^2 \plus{} 3$.

For all real $ x$, $ f(x^2 \minus{} 1)$ is:

(1983 AHSME #18)

How many non-congruent [b]obtuse/acute/right[/b] triangles such that the perimeter in cm and the area in square centimetres are numerically equal?

(Solve for each case, i.e. obtuse, acute and right)

I know the answer is quite clearly [b]infinitely many,[/b] but could someone supply a proof?

This is similar to question 24 on the 1983 AHSME[/b]

Just plug in $ \sqrt{x^2\minus{}2}$ for x. Then note that if you let $ g(x)\equal{}f(x^2\plus{}1)$ then $ g(x)\equal{}x^4\plus{}5x^2\plus{}3$ for infinitely many x, so it is identically equal to it. This means that the relation holds for all x, real or complex, so whether or not $ \sqrt{x^2\minus{}2}$ is real does not effect the final answer. Then just multiply it out.

[quote="herefishyfishy1"][quote="jli"][quote="ThetaPi"]Let $ f$ be a polynomial function such that, for all real $ x$, $ f(x^2 \plus{} 1) \equal{} x^4 \plus{} 5x^2 \plus{} 3$.

For all real $ x$, $ f(x^2 \minus{} 1)$ is:

(1983 AHSME #18)
[/quote]

$ x^4 \plus{} 5x^2 \plus{} 3 \equal{} (x^2 \plus{} 1)^2 \plus{} 3(x^2 \plus{} 1) \minus{} 1$ so $ f(y) \equal{} y^2 \plus{} 3y \minus{} 1$

Then $ f(x^2 \minus{} 1) \equal{} (x^2 \minus{} 1)^2 \plus{} 3(x^2 \minus{} 1) \minus{} 1 \equal{} x^4 \plus{} x^2 \minus{} 3$

and for the second question...
Right Triangles:
There is a general form for pythagorean triples so we are done.

ummm for the other cases, we only have to find one triangle and we can just scale it.  13-14-15 triangle is acute and has area 84.

take a look at this link:
http://mathworld.wolfram.com/IntegerTriangle.html[/quote]

Umm... You can't scale it. Then it would NOT be numerically equal anymore.[/quote]

right i read it wrong...again


if n is natural and 3n+1 and 4n+1 are perfect squares then prove that 56 divides n

[quote="Albanian Eagle"][hide="2 facts that could be useful"]squares can only be $ 0,1\pmod{8}$ and $ 0,1,2,4\pmod{7}$  :wink: [/hide][/quote]

$ 0,1,4\pmod{8}$ :)

yes  :lol: , but we won't need the $ 0$ nor the $ 4$


When I compile the PDF on my USAMTS solutions, the first page overruns and I can't see most of my solution. I don't know what happened. Could someone give me some pointers on how to fix this problem?

Does it run off the side of the page or the bottom of the page?

Because if it is the side, add some "\\"s to bring the LaTeX down to the next line.


algebra
polynomial
number theory
greatest common divisor
complex numbers
Galois Theory
complex analysis

Prove that it is impossible for the root of unity to have a real part in the form $ \sqrt{k\plus{}1}\minus{}\sqrt{k}$ for some positive integers $ k$.

Note that either $ k \plus{} 1$ or $ k$ is not a perfect square. Let us say $ k$ is not a perfect square. Then $ \sqrt {k \plus{} 1} \minus{} \sqrt {k}$ is conjugate to $ \sqrt {k \plus{} 1} \plus{} \sqrt {k}$ over $ \mathbb Q$ (since $ k$ and $ k \plus{} 1$ are coprime). Therefore, there exists an automorphism $ \phi: \bar {\mathbb Q} \to \bar {\mathbb Q}$ that maps $ \sqrt {k \plus{} 1} \minus{} \sqrt {k}$ to $ \sqrt {k \plus{} 1} \plus{} \sqrt {k}$.

Now assume that $ \xi$ is a root of unity and $ \mathrm{Re}(\xi) \equal{} \sqrt {k \plus{} 1} \minus{} \sqrt {k}$. Then
\[ \frac {\xi \plus{} \xi^{ \minus{} 1}}{2} \equal{} \mathrm{Re}(\xi) \equal{} \sqrt {k \plus{} 1} \minus{} \sqrt {k},
\]
and
\[ \frac {\phi(\xi) \plus{} \phi(\xi)^{ \minus{} 1}}{2} \equal{} \phi(\sqrt {k \plus{} 1} \minus{} \sqrt {k}) \equal{} \sqrt {k \plus{} 1} \plus{} \sqrt {k}.
\]
On the other hand, since $ \xi$ is a root of unity, $ \phi(x)$ is also a root of unity and $ |\phi(x)| \equal{} 1$. Therefore,
\[ \left|\frac {\phi(\xi) \plus{} \phi(\xi)^{ \minus{} 1}}{2}\right| \leq 1.
\]
We get a contradiction.

[quote="Yury"]Note that either $ k \plus{} 1$ or $ k$ is not a perfect square. Let us say $ k$ is not a perfect square. Then $ \sqrt {k \plus{} 1} \minus{} \sqrt {k}$ is conjugate to $ \sqrt {k \plus{} 1} \plus{} \sqrt {k}$ over $ \mathbb Q$ (since $ k$ and $ k \plus{} 1$ are coprime). Therefore, there exists an automorphism $ \phi: \bar {\mathbb Q} \to \bar {\mathbb Q}$ that maps $ \sqrt {k \plus{} 1} \minus{} \sqrt {k}$ to $ \sqrt {k \plus{} 1} \plus{} \sqrt {k}$.
[/quote]

hi yury, thanks for replying.. i have questions, what do you mean conjugate [b]over[/b] $ \mathbb{Q}$. and also why there is an automorphism.. what theorem did you use?

thanks.

Two numbers are conjugate over $ \mathbb Q$ if they are roots of the same irreducible polynomial (over $ \mathbb Q$) with rational coefficients (see http://en.wikipedia.org/wiki/Conjugate_element ).

The theorem I am using is a basic theorem from the Galois theory ( http://en.wikipedia.org/wiki/Galois_theory ):

[b]Theorem.[/b] For every two conjugate numbers $ \alpha$ and $ \beta$ there exists an automorphism of $ \mathbb C$ that maps $ \alpha$ to $ \beta$.[size=75] (The statement applies to other fields as well.)[/size]

We can prove the statement "explicitly" without using the Galois theory (but essentially this will be the same proof).

For simplicity, suppose that $ k \plus{} 1$ is a perfect square and $ k$ is not. Denote $ \alpha \equal{} \sqrt {k \plus{} 1} \minus{} \sqrt {k}$ and $ \beta \equal{} \sqrt {k \plus{} 1} \plus{} \sqrt {k}$. Observe that $ \alpha, \beta \notin \mathbb Q$ but $ \alpha \plus{} \beta$ and $ \alpha\beta$ are in $ \mathbb Q$. Let $ \xi$ be a root of unity as in my post above; say, $ \xi^d \equal{} 1$. We have $ \xi^2 \plus{} 1 \minus{} 2\xi \alpha \equal{} 0$ (why?). Thus $ (\xi^2 \plus{} 1 \minus{} 2\xi \alpha)(\xi^2 \plus{} 1 \minus{} 2\xi \beta) \equal{} 0$. That is, $ \xi$ is a root of the polynomial
\[ f(x) \equal{} (x^2 \plus{} 1 \minus{} 2\alpha x) (x^2 \plus{} 1 \minus{} 2\beta x).
\]
Observe that $ f(x)$ has rational coefficients (use the fact that $ \alpha \plus{} \beta$ and $ \alpha\beta$ are in $ \mathbb Q$). We get that $ \xi$ is a root of $ x^d \minus{} 1$ and a root of $ f(x)$. Therefore, $ \xi$ is a root of $ g(x) \equiv \mathrm{gcd}(f(x), x^d \minus{} 1)$ (see http://en.wikipedia.org/wiki/Greatest_common_divisor_of_two_polynomials ). Note that $ g(x)$ is a polynomial with rational coefficients since both $ x^d \minus{} 1$ and $ f(x)$ have rational coefficients; also every root $ r$ of $ g(x)$ is a root of unity (why?) and a root of $ f(x)$. Therefore, $ |r| \equal{} 1$ and $ \mathrm{Re} (r) \in \{\alpha, \beta\}$. Since $ \beta > 1$, $ \mathrm{Re}(r) \equal{} \alpha$. Thus $ r\in \{\xi, \bar\xi \}$; and $ g(x) \equal{} x \minus{} \xi$ or $ g(x) \equal{} (x \minus{} \xi)(x \minus{} \bar \xi)$. But neither of these two polynomials have rational coefficients! We get a contradiction.

In general, if $ k \plus{} 1$ is not a perfect square we need to consider the polynomial $ f(x) \equal{} \prod(x^2 \plus{} 1 \minus{} 2 (\pm\sqrt {k \plus{} 1} \pm\sqrt {k}))$. The rest of the argument is very similar.

Thank you very much yury. I understand your nice proof  :)


geometry
rectangle
ARML
analytic geometry
vector

I think there might be a way to do this analytically but I can't see a nice way.
Problem: In rectangle ABCD, AB=6 and BC=8.  Equilateral triangles ADE and DCF are drawn on the exterior of ABCD.  Find the area of triangle BEF.

Source: 1996 Chicago ARML tryouts #24

You could easily find BE, EF and BF.  Then, use Herons formula.
Or I guess you could partition BEF into three sections so that the center point where they're partitionedis 3 from BC and 4 from AB.  Then you could find the areas.

[quote="K81o7"]Or I guess you could partition BEF into three sections so that the center point where they're partitioned is 3 from BC and 4 from AB. Then you could find the areas.[/quote]

Another way to approach this is to introduce coordinates; put $B$ at the origin, $A$ and $C$ on the axes, and find $E=(x_1,y_1)$ and $F=(x_2,y_2)$. The area is then $\frac12|x_1y_2-x_2y_1|.$ This formula can be derived in general by dissections like the one quoted above.

The area will have a $\sqrt{3}$ in it; don't expect a rational answer with the irrational coordinates.

Thanks-that works very well.  And there was a rad3 in the area. Is that the same as one of those proof without words formulas running in the boxes on the left part of the screen?

That's what I got, by setting up the picture in coordinates and finding the area the way jmerry described it.  My coordinates had A at (0, 0) and the verticies of the triangle at (0, 6), (4, -4sqrt(3)), (8+3sqrt(3), 3)

[quote="WarpedKlown1335"]Is the answer by any chance 36 + 25 :rt3: ?  I just found the area of ABCFE and then subtracted the areas of triangles ABE and BCF.[/quote]

That is what i got  :D
[hide]I drew the rectangle like:
A-------D
B-------C

Then I assigned (0,0) to point B, the point E in relation to B becomes (4, 6 + 4 :rt3:) and F becomes (8 + 3 :rt3:) Then using the shoelace method the area of the triangle is 36 + 25 :rt3:[/hide]

edit - Wumbate beat me to it  :blush:

[ABCFE] = [ABCD] + [ADE] + [CDF] + [DEF] = 48 + 9 :rt3: + 16 :rt3: + 12 = 60 + 25 :rt3: .

[ABE] + [BCF] = 24.

[BEF] = [ABCFE] - [ABE] - [BCF] = 60 + 25 :rt3: - 24 = 36 + 25 :rt3: .

Using analytic geometry for this problem is kind of overkill don't you think?   ;)

Yeah, you are all right that is the official answer.  I don't think analytical geometry is really that much overkill...most of the time i'm more likely to figure out a problem like this with coordinates and vectors, etc. than with "pure" geometrical approaches.  But either way, you get the same answer. :)


find max of \[ \frac{15}{2}a^2 + 4ab  \]
if \[ a^2 + b^2=2  \]

what i did was \[ \frac{15}{2}a^2 + 4ab\leq k  \]
\[ k=(a^2+b^2)\frac{k}{2}  \]
\[ \frac{k-15}{2}a^2 - 4ab + \frac{k}{2}b^2 \geq 0  \]
I assumed that it should be in this form \[ (1) (\sqrt{\frac{k-15}{2}}a - \sqrt{\frac{k}{2}}b)^2\geq 0  \]
and thus \[ 2\sqrt{\frac{k-15}{2}}\sqrt{\frac{k}{2}}=4  \]
so the answer is\[ k= 16  \]

I would like to know if i can assume (1) or is there another way to prove it...
thanks

yes you can. In fact when you write your solution, you only need to write
$\frac{15}{2}a^2+4ab \leq \frac{15}{2}a^2+\frac{a^2+16b^2}{2} = 8(a^2+b^2) =16$


vector
algebra
system of equations
linear algebra

Prove that every linear system of equations falls into one of the three categories:
no solution, one solution, or infinitely many solutions.

Given a linear system $ \mathbf{Ax} \equal{} \mathbf{b}$, if $ \mathbf{b}$ is not in the column space of $ \mathbf{A}$ then no solutions are possible.  Otherwise, the set of solutions is a translate of the nullspace of $ \mathbf{A}$ (which is a much stronger statement than the one desired) and in particular is a vector space.  Dimension zero is one solution; dimensions one and higher are infinite (in a vector space over an infinite field).


Here is my conjecture.

Let $ b\in\mathbb{Z}$ such that $ b > 1$.  Define, for each $ n\in\mathbb{N}$, $ S_{b}(n)$ and $ P_{b}(n)$ to be the sum and the product of digits of $ n$, when expressed in the base-$ b$ representation.
For each positive integers $ u$ and $ v$, there exists $ n$ for which
\[ uS_{b}(n)\plus{}vP_{b}(n) \equal{} n\,.\]

It's false.
Doesn't exist n such that:
$ S_{10}\plus{}2P_{10}\equal{}n$
In this case you can prove  $ n$ must have 4 or 5 digit.

[quote="Vinxenz"]It's false.
Doesn't exist n such that:
$ S_{10}\plus{}2P_{10}\equal{} n$
In this case you can prove  $ n$ must have 4 or 5 digit.[/quote]

$ n\equal{}669$
$ S_{10}(669)\equal{}21$
$ P_{10}(669)\equal{}324$

$ 21\plus{}2*324\equal{}669$


[quote="romano"]Let $ a_n $ be a sequence satisfying : $ a_n = 1 + \frac{n+1}{2n}.a_{n-1} $ for n >=2 . Find $ \lim_{n\to \infty} a_n $ ?[/quote]

The answer seems to be 2, no matter what a_1 is. But I don't know why.

[quote="pleurestique"]$\lim_{n \to \infty} \frac{n+1}{2n} = \frac 12$, so if $\lim_{n \to \infty} = a$ then $a = 1+\frac a2$, from which $a = 2$.[/quote]

This isn't a complete proof, is it? All it shows is that if the limit exist, the limit is 2.

and it seems to be true that it always exists.  i'll post with something rigorous tomorrow, unless someone else posts in the meantime.  gotta sleep..


Let $ x_1,x_2,\cdots,x_n$ such that $ x_1x_2\cdots x_n \equal{} 1$,prove that
\[ \frac {1}{1 \plus{} (n \minus{} 1)x_1} \plus{} \frac {1}{1 \plus{} (n \minus{} 1)x_2} \plus{} \cdots \plus{} \frac {1}{1 \plus{} (n \minus{} 1)x_n} \ge 1\]

\[ \ge \frac {1}{1 \plus{} x_2 \plus{} \cdots \plus{} x_n} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_3 \plus{} \cdots \plus{} x_n} \plus{} \cdots \plus{} \frac {1}{1 \plus{} x_1 \plus{} \cdots \plus{} x_{n \minus{} 1}}\]

\[ \ge \frac {1}{n \minus{} 1 \plus{} x_1} \plus{} \frac {1}{n \minus{} 1 \plus{} x_2} \plus{} \cdots \plus{} \frac {1}{n \minus{} 1 \plus{} x_n}\]

for \[ 1 \ge\frac{1}{1\plus{}x_{2}\plus{}\cdots\plus{}x_{n}}\plus{}\frac{1}{1\plus{}x_{1}\plus{}x_{3}\plus{}\cdots\plus{}x_{n}}\plus{}\cdots\plus{}\frac{1}{1\plus{}x_{1}\plus{}\cdots\plus{}x_{n\minus{}1}}\]

we have by AM-GM $ x_{2}\plus{}\cdots\plus{}x_{n} \ge n\minus{}1$

by the similary way we are going to find $ \frac{1}{1\plus{}x_{2}\plus{}\cdots\plus{}x_{n}}\plus{}\frac{1}{1\plus{}x_{1}\plus{}x_{3}\plus{}\cdots\plus{}x_{n}}\plus{}\cdots\plus{}\frac{1}{1\plus{}x_{1}\plus{}\cdots\plus{}x_{n\minus{}1}} \leq n.\frac{1}{n} \equal{} 1$

[quote="zhaobin"]why?$ x_{2} \plus{} \cdots \plus{} x_{n}\ge n \minus{} 1$[/quote]
He is wrong because by AM-GM and the condition, we would have $ x_{2} \plus{} \ldots \plus{} x_{n}\geqslant (n \minus{} 1)\sqrt[n\minus{}1]{\frac{1}{x_1}}$...

[quote="Martin N."][quote="zhaobin"]why?$ x_{2} \plus{} \cdots \plus{} x_{n}\ge n \minus{} 1$[/quote]
He is wrong because by AM-GM and the condition, we would have $ x_{2} \plus{} \ldots \plus{} x_{n}\geqslant (n \minus{} 1)\sqrt [n \minus{} 1]{\frac {1}{x_1}}$...[/quote]
But why $ (n \minus{} 1)\sqrt [n \minus{} 1]{\frac {1}{x_1}} \ge n\minus{}1$...


Find all functions $ f: \mathbb{N}\rightarrow\mathbb{N}$ such that for all $ m,n\in\mathbb{N}$,

$ (2^m\plus{}1)f(n)f(2^mn)\equal{}2^mf(n)^2\plus{}f(2^mn)^2\plus{}(2^m\minus{}1)^2n$

The equation equivalent:
$ (2^mf(n) \minus{} f(2^mn))(f(2^mn) \minus{} f(n)) \equal{} n(2^m \minus{} 1)^2$
Because $ 2^mf(n) \minus{} f(2^{m}n)\equiv f(n) \minus{} f(2^{m}n) (\mod 2^m \minus{} 1)$ so easy to check that :
$ 2^m \minus{} 1|f(2^{m}n) \minus{} f(n)$
Exist $ (g_n,h_n)\in N$ such that:
$ 2^mf(n) \minus{} f(2^{m}n) \equal{} g_n(2^m \minus{} 1)$ and $ f(2^{m}n) \minus{} f(n) \equal{} h_n(2^m\minus{}1)$ where $ g(n).h(n) \equal{} n$
$ \Rightarrow f(n) \equal{} g_n \plus{} h_n,f(2^{m}n) \equal{} 2^{m}h_n \plus{} g_n,f(n).g(n) \equal{} n$
We can find $ f(m)$ if we know all $ f(n)$ where n is an odd number.
Easy to check that this function satisfy condition . For example $ g_n \equal{} 1,h_n \equal{} n$ then $ f(n) \equal{} n \plus{} 1$


algebra
polynomial
calculus
integration
algebra proposed

Let $f(X) = a_{0}+a_{1}X+\hdots+a_{n-1}X^{n-1}+X^{n}\in\mathbb{Q}[X]$ be a monic polynomial with integral coefficients. Show that if $\overline{a_{0}}+\overline{a_{1}}X+\hdots+\overline{a_{n-1}}X^{n-1}+X^{n}\in\mathbb{Z}_{p}[X]$ is irreducible (for some prime $p$), then $f(X)$ is irreducible over $\mathbb{Q}$.

EDIT: Added something.

Isn't it obvious? If $f(x)$ would be reducible over $\mathbb{Z}$, let's say $f(x) = p(x)q(x)$ then of course also $f(x) \equiv p(x)q(x) \mod{p}$. So $f(x)$ is irreducible over $\mathbb{Z}$ and the irreducibility over $\mathbb{Q}$ follows from Gauss's Lemma.


A telephone call costs $ \$1.19$ for the first three minutes and $ \$0.28$ for each additional minute. If a call started at 10:55:00 AM and ended at 11:06:00 AM, how much did the call cost?

the elapsed time is 9 min. so the price is $ \$1.19\plus{}\$0.28*(9\minus{}3)\equal{}\$2.87$

I'm pretty sure the elapsed time is $ 11$ minutes so $ 1.19\plus{}0.28(11\minus{}3)\equal{}\boxed{3.43}$?


Let $ P(x)$ be a polynomial of degree n with real coefficients and let $ a \geq 3$. Prove that $ \max_{0 \leq j \leq n\plus{}1}|a^j \minus{} P(j)| \geq 1$

We can write $ P(x) \equal{} \sum_{i \equal{} 0}^n (x_i\plus{}(a\minus{}1)^i)C_x^i.$ Thent $ b_j\equal{}P(j)\minus{} a^j \equal{}\sum_{i\equal{}0}^jC_j^ix_i, j<n\plus{}1$, and
$ b_{n\plus{}1}\equal{}P(n\plus{}1)\minus{}a^{n\plus{}1}\equal{}\minus{}(a\minus{}1)^{n\plus{}1}\plus{}\sum_{i\equal{}0}^{n}C_{n\plus{}1}^ix_i.$
It give $ x_i\equal{}(\minus{}1)^i\sum_{j\equal{}0}^i(\minus{}1)^jC_i^jb_j.$
Let $ |b_i| < b$. Then we calculate $ b_{n \plus{} 1} \equal{} P(n \plus{} 1) \minus{} a^{n \plus{} 1}$.
It give
\[ b_{n \plus{} 1} \equal{} \sum_{i \equal{} 0}^n((a\minus{}1)^i \plus{} x_i)C_{n \plus{} 1}^i \minus{} a^{n \plus{} 1} \equal{}\minus{} (a\minus{} 1)^{n \plus{} 1} \plus{} \sum_{i \equal{} 0}^nC_{n \plus{} 1}^ix_i\] or
\[ b_{n\plus{}1}\equal{}\minus{}(a\minus{}1)^{n\plus{}1}\plus{}\sum_{j\equal{}0}^nb_j\sum_{i\equal{}j}^n(\minus{}1)^{i\minus{}j}C_i^jC_{n\plus{}1}^i.\] It give
\[ (a\minus{}1)^{n\plus{}1}\equal{}(\minus{}1)^n\sum_{j\equal{}0}^{n\plus{}1}(\minus{}1)^jC_{n\plus{}1}^jb_j.\]
Obviosly $ \sum_{i \equal{} 0}^{n\plus{}1}(\minus{}1)^iC_{n \plus{} 1}^ib_i \equal{} b2^{n \plus{} 1}\theta$, were $ \ \minus{} 1\le theta\le 1$.
It give $ b\ge (\frac{a\minus{} 1}{2})^{n \plus{} 1} \equal{} c_n$. Exist minimal polinom, suth that $ b \equal{} c_n$.

Your proof is not right :). My result is extremely true. Besides, $ a > 3$


function
real analysis
real analysis unsolved

f is continuous on [a,b]
∀x∈[a,b] x is local minimum.
show that f is a constant function
thanks~

Hardly an unexpected result, but, it turns out, not as trivial as one may think. :D (I'm not saying it's hard... it's just not *that* trivial.)

Anyway, since $ f$ is continuous, the intermediate value theorem applies and thus there exist $ x_m\in [a,b]$ and $ x_M\in [a,b]$ such that $ m \equal{} f(x_m)$ is a global minimum of $ f$ and $ M \equal{} f(x_M)$  is a global maximum of $ f$. (So, $ m\leq M$. We want to show $ m \equal{} M$ - then we are obviously done.)

Without loss of generality, we assume $ x_m\leq x_M$. If $ x_m \equal{} x_M$ we have $ m \equal{} f(x_m) \equal{} f(x_M) \equal{} M$ and we're done. So, $ x_m < x_M$. Now we'll take a look at $ S$ defined with $ S \equal{} \{x\in [a,b]: x > x_m \wedge f(x) \equal{} M\}$. Since $ x_M\in S$, $ S$ is not empty and since $ S\subseteq [a,b]$, there exists $ t \equal{} \inf S$.

Obviously, $ t\geq x_m$. If $ t \equal{} x_m$, then we have a sequence $ (r_k)_{k\in\mathbb{N}}\subseteq S$ such that $ r_k\rightarrow x_m$. Since $ f$ is continuous, $ f(r_k)\rightarrow f(x_m)$ and therefore, $ M \equal{} m$.

So, the only case left is $ t > x_m$. We'll prove that isn't possible. Since $ f$ is continuous, $ t\in S$ (again, the same argument with the sequence as in the previous paragraph). So, $ f(t) \equal{} M$. We know that $ t$ is a local minimum, so there exists $ \delta > 0$ such that $ f(x)\geq f(t) \equal{} M$ for all $ x\in\langle t \minus{} \delta,t]$. However, since $ M$ is a global maximum, $ f(x)\leq M$ for all $ x$. So, $ x\in\langle t \minus{} \delta,t]\rightarrow f(x) \equal{} M$. Now, since $ t > x_m$, $ t > \max\{x_m,t \minus{} \delta\}$. So, there exists $ p$ such that $ t > p > \max\{x_m,t \minus{} \delta\}$. (For instance, you can take $ p \equal{} \frac {t \plus{} \max\{x_m,t \minus{} \delta\}}{2}$.) In that case, then, $ p > x_m$ and $ f(p) \equal{} M$. So, $ p\in S$. However, since $ p < t \equal{} \inf S$, that is a contradiction.

Hmmm to complicated. But the approach is right. Take $ M \equal{} \max_{[a,b]} f$ and $ x' \equal{} \inf\{x\in[a,b]: f(x) \equal{} M\}$, $ x'' \equal{} \sup\{x\in[a,b]: f(x) \equal{} M\}$. Obviously, $ f(x') \equal{} f(x'') \equal{} M$ and we have $ f(x)<M$ for $ x\in[a,x')\cup(b,x'']$. Since $ x',x''$ are points of local minimum, we have that the latter set is empty, i.e. $ x' \equal{} a$, $ x'' \equal{} b$.

I would submit another proof although
much less elementary. I employ the result: the ordinates of the extremum points of any function,
continuous or not, determine a  set at most countable (A.Shoenflies: Die
Entwickelung der Lenhre von dem Punktmannigflatigkeiten, Jahrebsbericht Deutschen Mathematiker--Vereinigung 8, Leipzig,
1900). First of all we take away from $ [0,1]$ all the open intervals where
$ f$ is constant and define by $ I$ the set obtained which is still closed. Moreover $ f\colon I\to {\bf R}$
is continuous. Define $ A_k \equal{} \{x\in I\colon f(x) \equal{} y_k\}$ and observe that each $ A_k$ is
closed by the continuity of $ f.$ Since any point of $ I$ is a minimum point, it results that $ I \equal{} \bigcup_{k \equal{} 1}^\infty A_k.$ Now $ I$ is complete (as a closed subset of the complete space $ [0,1]$) and Baire's theorem assures that
at least one of the $ A_k's,$ say $ A_{k_0},$ is dense at least at one point. In other words $ {\buildrel {\rm o} \over {\overline A}}_{k_0} \equal{} {\buildrel {\rm o} \over A}_{k_0} \ne\emptyset.$ This means that there exists an
interval, say $ (c,d),$ such that $ (c,d)\subset A_{k_0}.$ But this is impossible because any point of $ (c,d)$ is a
strict minimum having deleted all open intervals where $ f$ is constant.


assume P1 ,..Pr are prime number,and k,c1,..,cr are natural numbers prove that there  exist infinite n such that C(n,k)/(Pi^ci)is integer but C(n,k)/(Pi^(ci+1))is not an integer.

Assume that k is in each interval of the form[piai, piai+1)

We choose n s.t. n+piai is divisible with piai+ci but not with piai+ci+1. It seems to me right now that this works, but there might, of course, be a mistake. Can someone please check it?


For each $n\ge 1$, we call a sequence of $n$ (s and $n$ )s "legal" if the parentheses match up, somehow. For example, if $n=4$, the sequence (()()()) is legal, but ()())(() is not. Let $l_{n}$ denote the number of legal arrangements for the $2n$ parentheses. Find a recurrence relation for $l_{n}$.


$f$:$R^n\rightarrow R^n$  and $n\geq 2$
$f$ is $C^{1}$ and $C>0$ lipschitzienne

If exist a solution $x$ of $x'=f(x)$, $T$-periodic

Prove $T\geq \frac{2\pi}{C}$


calculus
integration
analytic geometry
trigonometry
geometry
calculus computations

Find the volume of the region between surfaces $ x^2\plus{}z^2\equal{}b^2$, $ x^2\minus{}y^2\minus{}z^2\equal{}0$, where $ x>0$.

[quote="tmrfea"]Find the volume of the region between surfaces $ x^2 \plus{} z^2 \equal{} b^2$, $ x^2 \minus{} y^2 \minus{} z^2 \equal{} 0$, where $ x > 0$.[/quote]

Cylinder/Cone intersection Volume: is not nice... The volume integral in modified cylindrical coordinates (swap x and z in regular cyl coords) is

\[ V\equal{}\int_{0}^{2\pi}\int_{0}^{{\scriptstyle \frac{b}{\sqrt{1\plus{}\cos^2\theta}}}}\int_{r}^{{\scriptstyle \sqrt{b^2\minus{}r^2\cos^2\theta} }}r\, dx\, dr\, d\theta \equal{} {\scriptstyle \frac{1}{3}}b^3\int_{0}^{2\pi} \frac{1\minus{}\left( 1\plus{}\cos^2\theta \right) ^{{\scriptstyle \minus{}\frac{1}{2}}}}{\cos^2\theta}d\theta\]

Maple give this in terms of Elliptic integrals as

\[ V\equal{} 4b^3 \left( \mbox{EllipticE}( \i ) \minus{} \mbox{EllipticK}( \i )\right)\]

It is no wonder that this is the case when you study the geometry at hand...

[img]http://i306.photobucket.com/albums/nn249/arinkovsky/intersectionCylandcone.jpg[/img] [img]http://i306.photobucket.com/albums/nn249/arinkovsky/intersectionCylandcone_back.jpg[/img]

[img]http://i306.photobucket.com/albums/nn249/arinkovsky/intersectionCylandcone_elipse.jpg[/img]


Give $ a,b,c$ are non-negative real number.Prove that
$ \frac3{4}\le \frac{a^2}{(a\plus{}b)(a\plus{}c)}\plus{}\frac{b^2}{(b\plus{}a)(b\plus{}c)}\plus{}\frac{c^2}{(c\plus{}a)(c\plus{}b)}\le 1$

[quote="dduclam"]Give $ a,b,c$ are non-negative real number.Prove that
$ \frac3{4}\le \frac {a^2}{(a \plus{} b)(a \plus{} c)} \plus{} \frac {b^2}{(b \plus{} a)(b \plus{} c)} \plus{} \frac {c^2}{(c \plus{} a)(c \plus{} b)}\le 1$[/quote]
The inequality equaivalent to $ \sum_{sym}a^2b \ge 6abc$
and $ 2abc \ge 0$.Which is obviously true.

Yes,very easy. A similar problem:
Find the min and max value of $ P\equal{}\sum\frac{a^3}{(a\plus{}b)(a\plus{}c)(a\plus{}d)}$  for $ a,b,c,d\ge0$

it was appeared in Mathlink Forum, I remember that pvthuan purposed above problem and Japanese (Naoki Sato) solved it by AM-GM  :)
The inequality is equivalent to:
\[ \sum \left(\frac{a}{a\plus{}2}\right)^3\ge \frac{4}{27}\]
Using the following result:
 \[ \left(\frac{x}{x\plus{}2}\right)^3\ge \frac{2x\minus{}1}{27}\]
It is really nice but not hard too  :)

You are right,Hoang. Find max problem already in pvthuan'book. Now,can we find the min value of P?
Mean find the max value of k such that:
 $ \sum_{cyc}a^3(b\plus{}c)\ge k(a\plus{}b)(b\plus{}c)(c\plus{}a)(a\plus{}b)$

[quote="dduclam"]
Mean find the max value of k such that:
 $ \sum_{cyc}a^3(b \plus{} c)\ge k(a \plus{} b)(b \plus{} c)(c \plus{} a)(a \plus{} b)$[/quote]
Are you sure? Maybe find the maximum value of $ k$ for which en inequality
\[ \sum_{cyc}a^3(b \plus{} c)\ge k(a \plus{} b)(b \plus{} c)(c \plus{} a)(a \plus{} b\plus{}c)\]
holds for all non-negative $ a,$ $ b$ and $ c.$ :wink:

Sory,it's better is:  find the maximum value of k such that:
 $ \sum_{cyc}^{a,b,c,d}a^3(b \plus{} c)(b\plus{}d)(c\plus{}d)\ge k (a\plus{}b)(a\plus{}c)(a\plus{}d)(b\plus{}c)(b\plus{}d)(c\plus{}d)$
for all non-negative real numbers $ a,b,c$.


Given 187 integers, both the sum and product of these integers are 188.

What's the sum of the absolute value of these integers?

236.

[hide="rationale"]
for the product to be 288, there must be a few factors of 288, and the rest are either 1 or -1.

using the prime factors 2, 2, and 47 doesn't work because an even amount of remaining 1's and -1's (284 of them) can't make the extra 188-2-2-27=odd number that we need

combining factors into 4 and 47 gives the odd amount needed.

the numbers, then, are:

4, 47, 24 -1's, and 161 1's
[/hide]

I think you accidentally said 288 at the beginning of your solution when you meant 188 ;)  .  Other than that it looks good to me.


Does anyone know the specific sources (or inspirations) for this year's Putnam and USAMO?

The solutions on the AMC page state the problem authors for the USAMO.

Oops, I didn't realize they had posted solutions. Sorry.


function
logarithms
number theory
totient function

find the first 3 and the last 3 digits of 2^5127. I was able to find the last three digits using Euler's totient function and mods, but I have no idea how to find the first three digits.  :?:

I think you can do the first three digits this way.
[hide]Take the log of 2 and multiply it by 5127. Then subtract the whole number part of your answer and let the result be $x$. Then $10^x$. The first three digits should be your answer... The reason this should work is because when you take the log of two and multiply it by 5127, you get the value of $x$ such that $10^x = 2^{5127}$ But the whole number part is just a bunch of zeros, so subtracting that part and then doing the $10^x$ of that should result in the $\frac {2^{5127}}{10^{1543}}$ because 1543 is the whole number part of $5127 * \log 2$.[/hide]

[quote="probability1.01"]Can you use a calculator?[/quote]

Well, I learned to use one back almost twenty years ago. I suppose you can if you know how to use it, if you don't, you can just learn  :D  :D  :D .

Just kidding. Really, I don't have an idea about how to compute the first three digits without a calculator that can handle logarithms!


hi im a new member and i want know some moroccain friend
because i'm it :roll:

Ain't Moroccan but your country is very good :D


I read posts in this forum and it makes me go "WOW! :lol: " and there are lots of good problems (hard!!!) especially proof ones.

So, here is my question:

How do you solve or at least practice your skill on proof? I know there is an article about it in this site but I want to hear from this forum since there are lots of good problems in here.

I'm hoping to take some proof writing competitions but I'm not sure where to start.

Any suggestion or your idea is GREAT! :)

Do you mean proof by actual writing-out part, or solving problems that require you to prove something?

When I first started on math at Canadian National camp, the organizers made participants write out many messy proofs, and gave feedbacks. Usually logical ones and geometry problems are hard to write out.

If you mean the latter, I guess you have to solve lots of problems..  :lol:

Mathematical circles differs from the AoPs books in that it has tons of clever proof exercises instead of "finding this" problems. Maybe that's what your looking for?


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]

The last three digits of $ N$ are $ x25$. For how many values of $ x$ can $ N$ be the square of an integer?

[hide][quote="sunehra"]We can rewrite it as $ 100x \plus{} 25 \equal{} k^2$ for some integer $ k$. Factoring, we have $ 5^2(4x \plus{} 1) \equal{} k^2$. So $ 4x \plus{} 1$ must also be a perfect square.

$ 4x \plus{} 1 \equal{} 9 \implies x \equal{} 2 \\
4x \plus{} 1 \equal{} 25\implies x \equal{} 6$

Therefore, $ N$ is a perfect square for $ \boxed{x \in \{2,6\}}$.[/quote][/hide]

:)

Note that $ N$ doesn't necessarily have to equal $ \overline{x25}$; only the last two digits of $ N$ are $ x25$. Thus, $ 0$ would also be a solution, and the actual answer is $ x\in \{0,2,6\}$. :wink:

wow, what an ez question, 25squared= 625. 15squared=225, 45sqared=2025. :rotfl:

Don't revive old threads, especially if you're not going to contribute to the discussion.  Anyway, you provided no proof at all of your answer.

AIME15, $ \frac{1}{6}^\text{th}$ of a month is certainly not a revival, though the post was indeed totally unnecessary. Also, I don't see how the $ 2025$ answer can count.

Are the last three digits in the form $ \overline{x25}$ where $ x$ is a digit? Note that $ N$ doesn't need to be only three digits. :wink:


abstract algebra
group theory
superior algebra
superior algebra unsolved

How do i show that the non-abelian groups of order 8 are either $D_{4}$ or $Q_{8}$? Similarly, for order 12, either $D_{6}$, $A_{4}$ or $\mathbb{Z}/3\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$.

Use Sylow theorems to determine the number and the structure of groups of order 8 and 12. Obviuosly the direct product of cyclic groups is an abelian group, so it excludes from non-abelianity C(8), C(4)xC(2), C(2)xC(2)xC(2). Show that D(4) and Q(8) are non-abelian giving a pair of elements which do not commute (this is simple, one is a dihedral group, the other is the quaternions' group). For the other case, in a silmilar way, C(12)  and C(4)xC(3) are abelian. To determine the non-abelianity of the others use the same procedure used before, a little bit more complex is to find the incriminated elements in the semidirect product.

For the first part you know that the order of an element in G is a divisor of the order of group, for 8 we have 2, 4, 8 (if it's 8 you have done) and for 12 we have 2, 3, 4, 12. Now use the third Sylow theorem to determine the number of Sylow p-groups, and so on... it's a little bit tedious, i know  :|


Let $ a\equal{}\sqrt[1992]{1992}$. Which number is greater
\[ \underbrace{a^{a^{a^{\ldots^{a}}}}}_{1992}\quad\text{or}\quad 1992?
\]

$ \underbrace{a^{a^{a^{\ldots^{a}}}}}_{1992}>1992\equal{}(\sqrt[1992]{1992})^{1992}\equal{}a^{1992} \Leftrightarrow \underbrace{a^{a^{a^{\ldots^{a}}}}}_{1991}>1992 \Leftrightarrow \dots \dots \Leftrightarrow a>1992$, which is impossible, so $ 1992$ is larger.

Fix integer $m>1$, let $A_0=1<m,\; a=m^{1/m},\; A_n = a^{A_{n-1}}$.
Then $1\le A_0 < A_1 < m,\; A_n = \underbrace{a^{a^{a^{\ldots^a}}}}_{n}$ and
$(1\le A_{n-1}<A_n<m)\implies (1\le A_n = a^{A_{n-1}}<a^{A_n}=A_{n+1}<a^m=m)$.
Thus $1\le A_{n-1} < A_n < m$ for each positive integer $n$.

In particular, $A_m < m$ with $m=1992$.


Show that $ \forall x,y,z>0$ we have:
$ \sum (x+y)^{2}z \leq \frac{3}{2} \prod(x+y)$.

made by Mnich W.

cheers! :D :D

Let 2a = x+y, 2b = y+z and 2c = z+x.
Then, the inequality reduces to
 \sum a 3  - 2 \sum a 2 b +  \sum abc   \geq 0,  (where the sums are symetrics) which is Schur.

Pierre.

Or we observe that $(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz$ and after multiplications our inequality becomes $(x+y+z)(xy+yz+zx)+3xyz \le \frac{3}{2}(x+y)(y+z)(z+x)$.If we substitute the identity in the inequality we have to prove that $(x+y)(y+z)(z+x) \ge 8xyz$ which is obvious by AM-GM.


analytic geometry
graphing lines
slope
calculus
calculus computations

A curve in the plane is defined parametrically by the equations $ x\equal{}t^3\plus{}t$ and $ y\equal{}t^4\plus{}2t^2$. Find the equation of the line tangent to the curve at $ t\equal{}1$.

I figured out that at $ t\equal{}1$, it is $ (2,3)$, but then I don't know how to find the derivative...

[hide="Hint"]
$ \frac {dy}{dt}\equal{}4t^3\plus{}4t$ and $ \frac {dx}{dt}\equal{}3t^2\plus{}1$. For $ t\equal{}1$, the values, respectively, are $ 8$ and $ 4$. Note also that $ \frac {dy}{dx}\equal{}\frac {dy}{dt}.\frac {dt}{dx}\equal{}8.\frac {1}{4}\equal{}2$. Then the gradient of the tangent line is $ 2$. The equation will be $ y\equal{}2x\minus{}1$. I am sorry if I am wrong... :(
[/hide]

[quote="Qruni"][hide="Hint"]
$ \frac {dy}{dt} \equal{} 4t^3 \plus{} 4t$ and $ \frac {dx}{dt} \equal{} 3t^2 \plus{} 1$. For $ t \equal{} 1$, the values, respectively, are $ 8$ and $ 4$. Note also that $ \frac {dy}{dx} \equal{} \frac {dy}{dt}.\frac {dt}{dx} \equal{} 8.\frac {1}{4} \equal{} 2$. Then the gradient of the tangent line is $ 2$. The equation will be $ y \equal{} 2x \minus{} 1$. I am sorry if I am wrong... :(
[/hide][/quote]
Using that, I think I solved it.

[hide="Am I right?"]took my point (2,3) and my slope 2 to get the equation $ y\minus{}3\equal{}2(x\minus{}2)$
$ y\minus{}3\equal{}2x\minus{}4$
$ y\equal{}2x\minus{}1$[/hide]