[quote="bigknick"]Find the lengths of all sides of a triangle if they form an arithmetic sequence with the difference r/4, where r is length of inradius.[/quote]

WLOG $ a\le b\le c$ and $ s = \frac {a + b + c}{2}.$ This implies $ b = \frac {2}{3}s,a = \frac {2}{3}s - \frac {r}{4},c = \frac {2}{3}r + \frac {r}{4}.$ From $ s(s - a)(s - b)(s - c) = \frac {A^2}{16} = \frac {r^2s^2}{16}$ one gets
\begin{align*} r^2 & = 16\frac {(s - a)(s - b)(s - c)}{s} \\
r^2 & = 16\frac {\left(s - \left(\frac {2}{3}s - \frac {r}{4}\right)\right)\left(s - \frac {2}{3}s\right)\left(s - \left(s + \frac {r}{4}\right)\right)}{s} \\
r^2 & = \frac {16}{3}\left(\frac {s}{3} - \frac {r}{4}\right)\left(\frac {s}{3} + \frac {r}{4}\right) \\
\left(1 + \frac {1}{3}\right)r^2 & = \frac {16}{27}s^2 \\
r & = \sqrt {\frac {48}{108}}s \\
r &= \frac{2s}{3} \end{align*}

This gives the results $ a=\frac{s}{2},b=\frac{2s}{3},c=\frac{5s}{6}$ for any value of s.


$K_{1}$ and $K_{2}$ ( which are circles with centers $O_{1},O_{2}$, and radii $r$ and $R$ respectively ) intersect in two points $M,N$. $O_{1}O_{2}$ meets $K_{1},K_{2}$ at $Q,P$ respectively. $MP$ and $MQ$ intersect $K_{1}$ and $K_{2}$ in $A,B$ respectively and $O_{1}A \cap O_{2}B = C$. $O_{1}M$ and $O_{2}M$ intersect $K_{1}$ and $K_{2}$ in $S,T$ respectively.
Is it True that:
$CM$ passes through the midpoint of $ST$ if and only if $r=R$?
( If yes or no, why? )

salam

yes,it's true.

let $\angle MO_{1}O_{2}=\alpha\ ,\ \angle MO_{2}O_{1}=\beta$
notice that $\angle O_{2}TM=\angle TMO_{2}=\angle SMO_{1}=\angle MSO_{1}$ hence quadrilateral $STO_{1}O_{2}$ is cyclic and $\angle O_{1}NO_{2}=\angle O_{1}MO_{2}=180^{\circ}-\angle O_{2}MT=180^{\circ}-\angle O_{2}TM$ hence 5 points $S,T,O_{1},O_{2},N$ lie on a circle$(\mathcal C)$.

$\angle O_{2}MB=\angle O_{1}MO_{2}-\angle O_{1}MP=180^{\circ}-\alpha-\beta-(90^{\circ}-\frac{\alpha}{2})=90^{\circ}-\frac{\alpha}{2}-\beta\Longrightarrow \angle MO_{2}B=\alpha+2\beta\Longrightarrow \angle BO_{2}O_{1}=\alpha+\beta$
and $\text{Arc}\ O_{1}NO_{2}$ in circle $\mathcal C$ is equal to $2\alpha+2\beta$  hence $CO_{2}$ is tangent to $\mathcal C$ similarity $CO_{1}$ is tangent to $\mathcal C$.

now let $ST\cap O_{1}O_{2}=X\ ,\ CM\cap ST=Y$ notice that $CM$ is polar of $X$ wrt $\mathcal C$ therefore $(XY,ST)=-1$ so $Y$ is midpoint of $ST$ iff $ST\parallel O_{1}O_{2}$

whcih means $O_{1}M=O_{2}M$

[b]I like very much your proof, Amir.S ![/b] In the our situation, must be  remembered (for another problems with two secant circles) that $STO_{2}NO_{1}$ is cyclically (in the circle $w$) and $CO_{1}$, $CO_{2}$ are tangents to the circle $w$ !


trigonometry
inequalities unsolved
inequalities

Let $ \Delta ABC$. Prove that :
$ cos (\frac {A \minus{} B}{2}) \plus{} cos (\frac {B \minus{} C}{2}) \plus{} cos (\frac {C \minus{} A}{2}) \leq cos (\frac {3A \minus{} \pi}{9}) \plus{} cos (\frac {3B \minus{} \pi}{9})\plus{}cos (\frac {3C \minus{} \pi}{9})$

hello, do you mean
$ \cos \left(\frac{A\minus{}B}{2}\right)\plus{}\cos \left(\frac{B\minus{}C}{2}\right)\plus{}\cos \left(\frac{C\minus{}A}{2} \right) \leq\\
 \cos\left (\frac{3A\minus{}\pi}{9}\right)\cos \left(\frac{3B\minus{}\pi}{9} \right)\cos\left (\frac{3C\minus{}\pi}{9}\right)$?
Sonnhard.

[quote="conan_naruto236"]Let $ \Delta ABC$. Prove that :
$ cos (\frac {A \minus{} B}{2}) \plus{} cos (\frac {B \minus{} C}{2}) \plus{} cos (\frac {C \minus{} A}{2}) \leq cos (\frac {3A \minus{} \pi}{9}) \plus{} cos (\frac {3B \minus{} \pi}{9}) \plus{} cos (\frac {3C \minus{} \pi}{9})$[/quote]
I think Karamata ineq works here with inconvex $ cos x$


The value of $ x \minus{} y^{x \minus{} y}$ when $ x \equal{} 2$ and $ y \equal{} \minus{}2$ is:

$ \textbf{(A)}\ \minus{}18 \qquad
\textbf{(B)}\ \minus{}14\qquad
\textbf{(C)}\ 14\qquad
\textbf{(D)}\ 18\qquad
\textbf{(E)}\ 256$

[hide]
Plug in and calculate.
$ 2 \minus{} (\minus{}2)^{2\minus{}(\minus{}2)}\equal{}2\minus{}(\minus{}2)^4\equal{}2\minus{}(16)\equal{}\minus{}14$

$ \boxed{\text{B}}$[/hide]


[quote="pvthuan"]Prove or disprove that for $a,b,c>0$

\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac34+\Bigl(\frac{a}{a+b}\Bigl)^2+\Bigl(\frac{b}{b+c}\Bigl)^2+\Bigl(\frac{c}{c+a}\Bigl)^2\][/quote]

This is wrong. Counterexample: a = 1, b = 34, c = 30.

  Darij


$\iff 4(a+b+c)^{3}\ge 27(a^{2}b+b^{2}c+c^{2}a+abc)$
This is well known inequality.

[quote="The soul of rock"]Ah, what's a big mistake! I'm sorry!
The correct form
Let $a,b,c\ge 0$. Prove that
$4(a^{3}+b^{3}+c^{3})+12(ab^{2}+bc^{2}+ca^{2})\ge 15(a^{2}b+b^{2}c+c^{2}a)+3abc$[/quote]

We may assume that c=min{a,b,c} and let us write a=c+p,b=c+q ($p,q \geq 0$ )
Te inequality becomes $4p^{3}+4q^{3}+7cp^{2}+7cq^{2}+12pq^{2}\geq 15p^{2}q+9pcq$
But $7c(p^{2}+q^{2}) \geq 9cpq$
and we will prove that $4p^{3}+4q^{3}+12pq^{2}\geq 15p^{2}q$
If $q \geq p$ then inequalities easy
If $q \leq p$ then $p=q+x (x \geq 0)$


man these questions are hard..

given forumlae are these:
-should be familiar

[code]Terms forumlae

tn= ar^n-1

sum sequence formulae

sn = a ( r^n  -  1 )
     ----------------
           r   -  1[/code]

a = first term         r = ratio        n = term number

pleases show work, thanks

1. Given the expression of Sn[size=100]n[/size]= n^2 +2n for the sum of the first n terms of a series, determine an expresssion, in simplified form, for the n^th term
t[size=100]n[/size] in terms of n

2. x-9, x-1, 3x-3 are the second, third, and fourth terms of a geometric sequence of non-zero terms. determine the value of the first term

3. The sum of the first three terms of a geometric sequence is -14, while the seecond term is 12. Find all such sequences

4. The sum of the first two terms of a geometric sequence is 16. The sum of the reciprocals of the first two terms is 1/3. Find the first two terms of the sequence

[hide="Solution to 2"] Since the ratio of a geometric sequence is constant, $\frac{x-1} {x-9}=\frac{3x-3} {x-1}=3$ Therefore the ratio is 3. Now we need to solve using this fact. Since the ratio is 3, $3(x-9)=x-1$. Distribute: $3x-27=x-1$. Add 27 to both sides and subtract x from both sides. $2x=26$. Therefore, $x=13$. Since the ratio is $3$ and the 2nd term is$x-9$ (which we can now determine to be $4$), we can simply divide 4 by 3 to obtain $\boxed {\frac{4} {3}}$[/hide]

[hide="Solution to 4"] The problem states $a+ar=16$ and $\frac{1} {a}+\frac{1} {ar}=\frac{1} {3}$. Multiply the second equation by $(a)(ar)$ to get $a+ar=\frac{a^2r} {3}=16$ (You know that it equals 16 because $a+ar=16$ is stated in the original problem). So you mess with $\frac{a^2r} {3}=16$ by multiplying both sides by 3 to get $a^2r=48$ Then divide both sides by $a^2$ to get $r=\frac{48} {a^2}$.

Substitute that back into $a+ar=16$ to get $a+\frac{48a} {a^2}=16$ which cancels out into $a+\frac{48} {a}=16$ Multiply the entire equation by $a$ to get $a^2+48=16a$ Then subtract 16a to make a quadratic $a^2-16a+48=0$ which factors into $(a-12)(a-4)=0$.

Therefore the first term is 12 or 4. If you try them out in combination with $r=\frac{48} {a^2}$, they actually both work.

If the first term is 12, then the ratio is $\frac{1} {3}$ and the second term is 4.
If the first term is 4, then the ratio is $3$ and the second term is 12.[/hide]

Too tired to do either of the other ones.


Suppose $(u,v)$ is an inner product on $\mathbb R^{n}$ and $f: \mathbb R^{n}\longrightarrow\mathbb R^{n}$ is an isometry, that $f(0)=0$.
1) Prove that for each $u,v$ we have $(u,v)=(f(u),f(v)$
2) Prove that $f$ is linear.

This is not difficult. Write that $\parallel f(x)+f(y)\parallel =\parallel x+y\parallel $ and square it in order to obtain a). Then observe that because $f$ is an isometry we have $\parallel f(ax+by)-af(x)-bf(y)\parallel ^{2}=0$ (just develop the square of the norm and use a)).

there is a stronger result:
http://en.wikipedia.org/wiki/Mazur-Ulam_theorem


What is the largest six-digit number whose digits are all different such that the sum of the squares of the digits in the even-numbered positions equals the sum of the squares of the digits in the odd-numbered positions?

Please post the [b]whole[/b] solution. Do not post only hints, beginning steps, etc.

I basically have a solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245790]here[/url]. The problem is really just tedious. :wink:

Yes, but the problem is that I don't understand your solution.

How did you get $ 9867$ from $ 9876$? And a whole bunch of other stuff?

I'm not as smart as you; even MATHCOUNTS level problems prove challenging for me  :( .

Oh! That's because if we let the number be $ \overline{9876ab}$, then we have to have $ 9^2 \plus{} 7^2 \plus{} a^2 \equal{} 130 \plus{} a^2 \equal{} 8^2 \plus{} 6^2 \plus{} b^2 \equal{} 100 \plus{} b^2$. But we've already used $ 9,8,7,6$, so $ b\le5\implies b^2\le25$, so $ 100 \plus{} b^2$ is at most $ 125$, and thus it's impossible for $ 100 \plus{} b^2 \equal{} 130 \plus{} a^2$.

Then we go down from $ \overline{9876ab}$ and see that $ \overline{9867ab}$ is the highest we can go.


[quote="math beginner"]A flexible chain of length n  is hanging from one end x=0 but oscillates horizontally. Let the x axis point downward and the u axis point to the right. ASsume that the force of gravity at each point of the chain equals the weight of the part of the chain below the point and is directed tangentially along the chain. Assume that the oscillations are small. Find the PDE satisfied by the chain....[/quote]The PDE of a dynamic system in generalized coordinates is given by $\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\partial L}{\partial \dot{q}}\right) -\dfrac{\partial L}{\partial q} = 0$, where $L\left(\dot{q},q,t\right) = T\left(\dot{q},q,t\right)-V\left(\dot{q},q,t\right)$.

This yields a system of PDEs, one for each coordinate in $\vec{q}$.

That said, in the given case: $T =\frac12\mu\left(r\right)\left(\dot{r}^2 +r^2\dot{\theta}^2\right)$ and $V = \mu\left(r\right)\cdot g\cdot r\left(1-\cos\theta\right)$, where $r$ is the distance up from the bottom of the chain.

Since the links are discrete $\mu\left(r\right) = m_0\left(r-1\right)$, where $m_0$ is the mass of a single link.

At this point, to actually write the PDEs is a simple but time consuming exercise that I shall leave to others.


integration
calculus
real analysis
real analysis unsolved

For what $ p\ge 1$ is the following true?

Let $ A$ be a measure space of total mass $ 1,$ and $ f\in L^p(A).$ Let $ m\equal{}\int_Af.$ Is $ \|f\minus{}m\|_p\le\|f\|_p?$

Alternative formulation, same problem: suppose $ X$ is a random variable such that $ E(|X|^p)<\infty.$ Is $ E(|X\minus{}E(X)|^p)\le E(|X|^p)?$

oh..my stupid mistake .  sorry I was wrong in my example  . thanks Ravi B. :blush:
[EDIT]

Where did $ 2c$ come from in your example?  The result is definitely true for $ p \equal{} 2$, right?  My guess, based on zero knowledge, is that it's false for other $ p$.

[quote="Ravi B"]The result is definitely true for $ p \equal{} 2$, right?[/quote]
Yes, of course. $ E((X\minus{}E(X))^2)\equal{}E(X^2)\minus{}E(X)^2\le E(X^2).$

[quote]My guess, based on zero knowledge, is that it's false for other $ p$.[/quote]
At least take a look at $ p\equal{}1$ before you make that leap.

It's false for other $ p$, unless $ A$ is excessively granular. A constructive example: choose some $ q\in (0,\frac12)$. For each $ r\in (0,1)$, let $ f_r$ be equal to $ r$ on a set of measure $ q$, and equal to $ r \minus{} 1$ on the complement. We have $ m \equal{} \int f_r \equal{} q \plus{} r \minus{} 1$, so $ f_r \minus{} m$ is $ 1 \minus{} q$ on a set of measure $ q$, and $ q$ on the complement of measure $ 1 \minus{} q$.

Now, $ \int |f_r^p| \equal{} qr^p \plus{} (1 \minus{} q)(1 \minus{} r)^p$, and $ \int |f_r \minus{} m|^p \equal{} q(1 \minus{} q)^p \plus{} (1 \minus{} q)q^p$. We want to show that there is some value $ x \equal{} r$ for which $ qx^p \plus{} (1 \minus{} q)(1 \minus{} x)^p$ is smaller than its value at $ x \equal{} 1 \minus{} q$. A bit of calculus shows that the minimum occurs where $ \left(\frac {x}{1 \minus{} x}\right)^{p \minus{} 1} \equal{} \frac {1 \minus{} q}{q}$, which is a different place as long as $ p\neq 2$.
At $ p \equal{} 1$, the above calculus is different and the minimum occurs at the endpoint $ x \equal{} 1$. At $ p \equal{} \infty$, it's not calculus at all and the minimum occurs at $ x \equal{} \frac12$. In both cases, the minimum value is strictly smaller than the value at $ 1 \minus{} q$, and the conclusion is the same.

This came out of a side issue of a lecture I was preparing for tomorrow. Fortunately for the sake of the lecture, $ \|f\minus{}m\|_p\le2\|f\|_p$ will be good enough for the purpose I'll be using it for, so I don't really need what I asked here.


[color=green]Are there integers a, b and c such that a divides bc, but a doesn't divide b and a doesn't divide c?[/color]

I think the answer is NO, but I don't konw how to prove it?  Can anyone give me some clue?  Thanks alot  :lol:

please post in the appropriate forum henceforth, suedenation. as for your question take a=4. b=6, c=2.

What's wrong with this problem being posted here?

The smallest example; $a = 4, b = c = 2$.

This is only impossible for $a$ a prime.


Let x,y,z be real numbers such that x+y+z=0.Show that
6(x^3+y^3+z^3)^2=<(x^2+y^2+z^2)^3

We will use that when $ x\plus{}y\plus{}z\equal{}0, x^{3}\plus{}y^{3}\plus{}z^{3} \equal{} 3xyz$.

Then, we have to prove that:

$ 54(xyz)^{2} \leq (x^{2}\plus{}y^{2}\plus{}z^{2})^{3}$

But this follows from AM-GM applyed to $ x^{2}, y^{2}, z^{2}$

[quote="msecco"]We will use that when $ x \plus{} y \plus{} z \equal{} 0, x^{3} \plus{} y^{3} \plus{} z^{3} \equal{} 3xyz$.

Then, we have to prove that:

$ 54(xyz)^{2} \leq (x^{2} \plus{} y^{2} \plus{} z^{2})^{3}$

But this follows from AM-GM applyed to $ x^{2}, y^{2}, z^{2}$[/quote]
No it doesn't... $ (x^2 \plus{} y^2 \plus{} z^2)^3 \ge 27(xyz)^2$ by AM-GM. Actually it is equivalent to $ (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$...

(It has equality e.g. when $ x \equal{} y \equal{} 1, z \equal{} \minus{} 2$, so it can not follow by AM-GM!)

Replacing $ z \equal{} \minus{} x \minus{} y$ in the original inequality one gets:
$ 8x^6 \plus{} 24x^5y \minus{} 6x^4y^2 \minus{} 52x^3y^3 \minus{} 6x^2y^4 \plus{} 24xy^5 \plus{} 8y^6\geq0$.
Observing that this is cyclic for $ x$ and $ y$ one may easily see the factorisation:
$ 2(x \minus{} y)^2(2x \plus{} y)^2(x \plus{} 2y)^2\geq0$ which is obviously true.


Find all positive integer $ m$ such that $ 1444...44$ (have $ m$ 4s) is perfect square.


algebra
polynomial
function
integration
real analysis
real analysis unsolved

It is given a continuous function $f: [0,1 ]\rightarrow \mathbb{R}$ such that $\int_0^1x^{2n}f(x)dx=0,\forall n\in \mathbb{N}^*$. Prove that $f(x)=0,\forall x \in [0,1]$.

I think it can be done by using the fact that any continuous function on a closed interval can be uniformly aproximated by a sequence of polynomials. But I don't know exactly how. Can anybody help me?

You need approximation by polynomials of the form $P(x)=Q(x^2)$. If you have this, you get, for every such $P,\ \int Pf=0\ (*)$. Now if you take $P_n\to f$ uniformly, $(*)$ applied to $P_n$ instead of $P$ turns into $\int f^2=0$, and you're done.

In order to prove that polynomials in $x^2$ approximate every continuous function on $I=[0,1]$ arbitrarily well, just apply the [url=http://en.wikipedia.org/wiki/Weierstrass_approximation_theorem]Stone-Weierstrass Theorem[/url]. The set of such polynomials is an algebra $\mathcal A$ in $\mathcal C=C(I,\mathbb R)$ which separates points and contains a non-zero constant function, so the hypotheses of the theorem are satisfied, and the conclusion tells you that the closure of $\mathcal A$ is $\mathcal C$ itself.

Edit:

You actually need $\int x^{2n}f=0$ for all $n\ge 0$, not just $n\ge 1$ for the proof above to work.

A slight variation on the idea: let $g(x)=\sqrt{x}f(\sqrt{x}).$

$g$ is continuous on $[0,1],$ and for $m=0,1,2,\dots,$

$\int_0^1x^mg(x)\,dx=2\int_0^1u^{2m+2}f(u)\,du=0.$

Now use the ordinary Weierstrass theorem: the approximation of continuous functions by polynomials. We should be able to prove that $g$ is identically zero, which will suffice to prove that $f$ is zero.

For the benefit of our original poster, I'll flesh out the details a little. (grobber is correct, of course - I'm just explaining a little.) By Weierstrass, given $\epsilon>0,$ we can write $g(x)=P_m(x)+h(x),$ where $P_m$ is a polynomial of degree $m$ ($m$ is allowed to depend on $\epsilon$) and $|h(x)|<\epsilon$ for all $x\in[0,1].$

Now, $\int_0^1(g(x))^2\,dx=\int_0^1g(x)P_m(x)\,dx+\int_0^1g(x)h(x)\,dx$

$\le 0+\epsilon\int_0^1|g(x)|\,dx\le\epsilon\left(\int_0^1(g(x))^2\mdx\right)^{\frac12}.$

(That last bit is by the Cauchy-Schwarz inequality.)

If $\int_0^1(g(x))^2\,dx=0$ then we are done. If it's not zero, divide by its square root to get

$\left(\int_0^1(g(x))^2\mdx\right)^{\frac12}<\epsilon.$

With $\epsilon$ arbitrary, this will force us to have $\int_0^1(g(x))^2\,dx=0.$ With $g$ continuous, we have that $g$ is identically zero.

Here is another beautiful conequence of the Weierstrass aproximation by polynmials:
Find all continuous functions $f: [-1,1] \rightarrow \mathbb{R}$ such that $\int_{-1}^1x^{2n}f(x)dx=0,\forall n\ge 0$.

This problem is strongly related with the first one but none of the two methods can be applied directly.

Hint:[hide="clicktext"] use the function $g(x)=f(x)+f(-x)$[/hide]


inequalities function calculus derivative

Let $ a,b,c\in\left[\frac{1}{3},3\,\right]$.Prove that
\[ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}a}\geq\frac{7}{5}\]

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge \frac 32 > \frac 75\;\; \forall x,y,z\in\mathbb R$  (Nesbitt's inequality) :huh:

[quote="bertram"]$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac 32 > \frac 75\;\; \forall x,y,z\in\mathbb R$  (Nesbitt's inequality) :huh:[/quote]
the problem is to prove $ \frac {a}{ a \plus{} b} \plus{} \frac {b}{c \plus{} b} \plus{} \frac {c}{c \plus{} a} \ge \frac 75$.
the fuction$ f: x\rightarrow \frac{x}{x\plus{}A}$ is concave so the the minimum occur when $ a,b,c \in {3 , \frac13}$.

[quote="tchebytchev"][quote="bertram"]$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac 32 > \frac 75\;\; \forall x,y,z\in\mathbb R$  (Nesbitt's inequality) :huh:[/quote]
the problem is to prove $ \frac {a}{ a \plus{} b} \plus{} \frac {b}{c \plus{} b} \plus{} \frac {c}{c \plus{} a} \ge \frac 75$.[/quote]

:( I should really buy glasses.

[quote="bertram"]$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\ge \frac 32 \forall x,y,z\in\mathbb R$ [/quote]
I think, you are wrong. :wink:

[quote="tchebytchev"] so the the minimum occur when $ a,b,c \in {3 , \frac13}$.[/quote]

I think that you are wrong because the equality holds when (a,b,c)=(3,1/3 ,1)
And the hind to the solution is to prove that $ f(a,b,c)\geq f(a,b,\sqrt{ab})$

If you take the derivative of $ f(a,b,c)\equal{}\frac a{a\plus{}b}\plus{}\frac b{b\plus{}c}\plus{}\frac c{c\plus{}a}$ with respect to c it will vanish for $ c\equal{}\sqrt{ab}$.

[quote="apollo"]Let $ a,b,c\in\left[\frac {1}{3},3\,\right]$.Prove that
\[ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}\geq\frac {7}{5}
\]
[/quote]
My ugly proof:
Let $ a\equal{}\max\{a,b,c\}.$
We obtain: $ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}\geq\frac {7}{5}\Leftrightarrow$
$ \Leftrightarrow(3a\minus{}2b)c^2\minus{}(2a^2\minus{}ab\minus{}3b^2)c\plus{}3a^2b\minus{}2ab^2\geq0.$
Thus, it remains to prove that $ (2a^2\minus{}ab\minus{}3b^2)^2\minus{}4ab(3a\minus{}2b)^2\leq0,$ which equivalent to
$ (a\minus{}b)(a\minus{}9b)(4a^2\plus{}b^2)\leq0,$ which obviously true.
What your proof, apollo? Thank you.


Alcumus FTW blogs articles MATHCOUNTS \/closed

What in the world is it? All I can find is [url=http://www.artofproblemsolving.com/Alcumus/invaliduser.php]this,[/url] which doesn't exactly explain it.

I don't think it's something I can give away

EDIT: I just realized something very strange if the above is true...

I thin I know what Alcumus is now (not from this thread, from somewhere else). Thanks anyways.

Everyone seems a bit shady in regards to this.  :maybe:

[quote="Nerd_of_the_Ages"]Alcumus isn't released to the public yet, but eventually will be. I would know this because... nvm. People weren't supposed to tell, but it obviously leaked out. AoPS wasn't extremely strict on keeping it super top secret, although they didn't want it to be told about.[/quote]

Check your blog, page 20 or 21.  That might explain part of it.

Um, has anyone actually read the last paragraph of this?
[url]http://www.artofproblemsolving.com/Focus_article.pdf[/url]
The article says:
[quote]AoPS is beginning development on a new project, [i]Alcumus[/i], which will combine a database of problems, wideo lessons, interactive applets, and references to textbooks with machine learning software to provide an automated interactive educational experience.[/quote]
This much information wasn't so much leaked as proclaimed.  AoPS has a prominent link to this Focus article on the Art of Problem Solving home page.  It's not quite "under wraps," evidently.

Apparently, only people taking the AoPS MATHCOUNTS class have access to it.

[quote="FrogottenLegends"]Nerd of the Ages is right. It's open to a few select schools in the nation. Can't say who...[/quote]

Dope. I was talking about one of the other online classes, not actual "physical" teaching.

Why would you not want to tell it to the public?

Why keep it all such a secret?

[quote="Nerd_of_the_Ages"]Not true. People who took it have access to it, but other people have it, too.[/quote]

i took the mathcounts problem seires before...its not open for me though...


inequalities
search
inequalities unsolved

Show that for all positive real numbers $x_1,x_2,\ldots,x_n$ with product 1, the following inequality holds
\[ \frac 1{n-1+x_1 } +\frac 1{n-1+x_2} + \cdots + \frac 1{n-1+x_n} \leq 1. \]

Romania 1999 TST and it has been posted at least 6 times! Please, search it in solved section.

Oops...thank you anyway!

(the worksheet I got this from said it was Vietnam 98!)

[quote="colorfuldreams"][quote="arqady"][quote="chinacai"]who can solve the question？[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1965995#p1965995[/quote]

can it solve by Jensen ?[/quote]
Certainly, by the similar way:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1967083#p1967083

http://www.artofproblemsolving.com/community/c6h1101773p4968742

Let $x_i$ be positive real numbers such that  $\sum _{i=1}^n \frac{1}{x_i} =\sum _{i=1}^n x_i $ , [url=http://www.artofproblemsolving.com/community/c6h1353550p7400601]show that:[/url]
$$\sum _{i=1}^n \frac{1}{n-1+x_i} \le 1$$
[url=http://www.artofproblemsolving.com/community/c6h534460p3063186]here[/url]

[quote=chinacai]who can solve the question？[/quote]

See also [url=https://artofproblemsolving.com/community/c1068820h2050333]https://artofproblemsolving.com/community/h2050333[/url].


algebra
polynomial
calculus
derivative
real analysis
real analysis unsolved

Let P(x) = (an)X^n + (a(n-1))X^(n-1) + ... + a0, with real coefficients. Suppose P(x) has only real zeros. Prove that
Q(x) = ((an)/n!)X^n + ((a(n-1))/(n-1)!)X^(n-1) + ... + a0 has only real zeros.

[quote="joao1"]Let $ P(X) \equal{} a_{n}X^{n}\plus{}a_{n\minus{}1}X^{n\minus{}1}\plus{}\ldots\plus{}a_{0},$ with real coefficients. Suppose $ P(X)$ has only real zeros. Prove that
$ Q(X) \equal{}\frac{a_{n}}{n!}X^{n}\plus{}\frac{a_{n\minus{}1}}{(n\minus{}1)!}X^{n\minus{}1}\plus{}\ldots\plus{}a_{0}$ has only real zeros.[/quote]

This problem is indeed nice. :)

We're going to prove the following
[b]Claim :[/b] let $ f, g\in\mathbb{R}[X]$ be two polynomials having only real zeros, then the same goes for $ g(D)(f),$ where $ D$ denotes the differentiation operator over $ \mathbb{R}[X].$

If $ g$ is constant, it's obvious, so, WLOG, let $ g$ be a monic polynomial with degree $ m\geq 1.$
It can then be written as $ g(X) \equal{} (X\minus{}r_{1})\cdots (X\minus{}r_{m}).$

So it can easily be seen that, to prove our claim, it's enough to prove the following
[b]Lemma :[/b] for any $ f\in\mathbb{R}[X]$ having only real zeros and for any $ c\in\mathbb{R},$ the same goes for $ f'\minus{}cf.$

Again, it is clear if $ f$ is constant, so let $ n\geq 1$ be its degree and let $ x_{1}<\ldots < x_{k}$ be the zeros of $ f,$ with respective multiplicities $ \alpha_{1},\ldots,\alpha_{k}.$

[u]1st case :[/u] $ c \equal{} 0$
the $ x_{i}$ are zeros of $ f'$ of order $ \alpha_{i}\minus{}1,$ which yields $ n\minus{}k$ zeros of $ f'$ counted with multiplicity. Now if $ k > 1,$ we get $ k\minus{}1$ more zeros of $ f',$ which are pairwise distinct and distinct from the previous ones (thanks to Rolle). Done.
[u]2nd case :[/u] $ c$ [u]arbitrary[/u]
here again, the multiplicity of $ x_{i}$ as a root of $ f'\minus{}cf$ is $ \alpha_{i}\minus{}1.$
Now let $ h(x) \equal{} e^{\minus{}cx}f(x),$ so that $ h'(x) \equal{} e^{\minus{}cx}(f'(x)\minus{}cf(x))$ and we get $ k\minus{}1$ more zeros (again, thanks to Rolle). Summing it up, this makes $ n\minus{}1$ zeros counted with multiplicity for $ f'\minus{}cf.$
But, as its degree is $ n,$ this implies it has only real zeros. Done.

Now that our claim is proved, it's enough to use it with :
$ f(X) \equal{} a_{n}X^{N\minus{}n}\plus{}a_{n\minus{}1}X^{N\minus{}n\plus{}1}\plus{}\ldots\plus{}a_{0}X^{N}$ and $ g(X) \equal{}\frac{X^{N}}{N!}$ (both of them have only real zeros) to solve the problem.

[quote="mathmanman"]Now that our claim is proved, it's enough to use it with :
$ f(X) \equal{} a_{n}X^{N\minus{}n}\plus{}a_{n\minus{}1}X^{N\minus{}n\plus{}1}\plus{}\ldots\plus{}a_{0}X^{N}$ and $ g(X) \equal{}\frac{X^{N}}{N!}$ (both of them have only real zeros) to solve the problem.[/quote]That's a very interesting approach, but it seems to me that $ g(D)f(x) \equal{} a_{0}$, a constant.

[quote="perfect_radio"]No, $ g(D)(f) \equal{}\sum_{k \equal{} 0}^{n}\frac{a_{k}}{k!}x^{k}$ really holds, since $ \left(\frac{x^{N}}{N!}\right)^{(k)}\equal{}\frac{x^{N\minus{}k}}{(N\minus{}k)!}$.

By the way, mathmanman, your solution owns. I was like: WOW :wow: .[/quote]

 But should that be $ f(D)(g) \equal{}\sum_{k \equal{} 0}^{n}\frac{a_{k}}{k!}x^{k}$  :?:

Anyway,nice solution.


calculus
derivative
function
real analysis
real analysis unsolved

hi , my question is very short we can claculate the nieme diif of (f.g) with the formula of leibnitz but can we calculate the nieme diff of (fog)=f(g) according to some math professors such formula do not exist but can we calculate it ???

nieme? I've never heard or seen that word before.

:rotfl:   That's the french word for $n$-th. He probably wants some formula for the $n$-th derivative of the composition of two functions $f,g$. In that case, yes, such a formula exists and it is called Faa di Bruno's formula. Try some google with it, it's so long that I really don't have the patience to write it here.

sorry for the nieme thing , i looked for the formula huge thx

Oh, wouldn't it be n-ième?  It looks like one word as nieme, which   "nth" is, but French was totally unexpected I believe...


Consider the ODE given by $ \frac{dy}{dt} \equal{} \frac{at \plus{} by \plus{} k}{ct \plus{} dy \plus{} l} \textup{.........................................(A)}$.

(i) For $ a \equal{} \minus{} 1, b\equal{}2, c\equal{}2, d\equal{}\minus{}1, k\equal{}l\equal{}0$, solve the ODE.

(ii) For $ a \equal{} \minus{} 1, b\equal{}2, c\equal{}2, d\equal{}\minus{}1, k\equal{}5, l\equal{}\minus{}4$, the ODE can be solved by making use of the substitutions $ t \equal{} T \plus{} m$ and $ y \equal{} Y \plus{} n$ (where $ m$ and $ n$ are suitably chosen real numbers) to convert the ODE to the following form:
\[ \frac{dY}{dT} \equal{} \frac{2Y \minus{} T}{2T \minus{} Y}\]Determine appropriate values for $ m$ and $ n$.

(iii) For any given non-zero values of $ a, b, k, l$, determine all non-zero values of $ c$ and $ d$ for which the equation $ (A)$ cannot be solved by the method given in part (ii).

Is there a solution with a nicer form? o.0


A radio tower is 100 feet high. How long does a guy wire need to be if it is attached to the top of the tower?  The guy wire is also attached to a point 30 feet from the base of the tower.

Good but how did yo get the answer?

I want the steps needed to find the answer.

[hide]I got a different answer, probably did something wrong though. But its a right triangle with legs 100 and 30. using Pythag. the length of the guy wire will be $10\sqrt{109}$.[/hide]

i think mgao subtracted 900 from 10000 instead of adding it
i also get
[hide]
10(sqrt(109))[/hide]

Let's assume that the wiire will be exactly straight.

[hide="solution"]

X=length of guy wire

$X=\sqrt{10000+900}$ by pythagorean theorem

$X=\sqrt{10900}=10\sqrt{109}$[/hide]

I thank all of you for your great tips and answer.


[quote="white_horse_king88"]Your are given 3 distinct bookshelves and n different books. Given that the arrangements of the books matter, if the n books are placed randomly on those 3 bookshelves - each bookshelf equally likely to be placed upon - what is the probability that no bookshelf is empty?[/quote]

This is really an interesting problem  :)

Well , the total number of ways can be counted in this way :

If we dont consider the order , then the question is equivalent to find the number of combination of non-negative integer $a,b,c$ such that

$a+b+c=n$

From this we will get the combination is $\binom{n+3-1}{n}=\binom{n+2}{n}$

But since the order matter , we just need to times $n!$ back to the combination making the total number of ways = $\binom{n+2}{n}n!$

Now using same method we have to compute the number of ways such that no shelf is empty , and it is actually equivalent to find non-negative integer $a,b,c$ such that

$a+b+c=n-3$ .  ( this is because every shelf must have at least one book so three books are placed in the three shelfs leaving only $n-3$ books left .)

Now the number of ways is $\binom{n-1}{n-3}\times n!$

Hence the probability is
$\frac{\binom{n-1}{n-3}}{\binom{n+2}{n}}$
= $\frac{(n-1)(n-2)}{(n+1)(n+2)}$

Well , this question is beautiful because it can also actually be generalised to find the total number of ways to arrange $m$ books into $n$ shelfs and using same method it can be calculated as $\binom{n+m-1}{n}n!$

 :)

I dont know if i got that answer, but i had this:

[hide]
$\frac{1}{n(n+1)}$
[/hide]

i am lazy and will put up my solution later. I dont think this is right though (must have been some kinda mistake somewhere... i still feel wrong)


modular arithmetic
number theory
relatively prime

Find the least n such that 1+4+9+.....+n^2  is divisible by 200.

Sorry for writing n instead of 200. It was just a mistake. :(

I think you should specify that $n>1$. The next sum divisible by $n$ would be $140\implies 7| 140$.

Masoud Zargar

EDIT: I guess eth problem changed.

Note- $a\mid{b}$ means $a$ divides $b$

$\sum_{i=1}^{n}{i^2}=\frac{n(n+1)(2n+1)}{6}$

$200\mid\frac{n(n+1)(2n+1)}{6}\Rightarrow1200\mid\n(n+1)(2n+1)$

$1200=2^4*3*5^2\Rightarrow2^4\mid{n(n+1)(2n+1)}$ and $3\mid{n(n+1)(2n+1)}$ and $5^2\mid{n(n+1)(2n+1)}$

$n\equiv0\pmod{3}\Rightarrow3\mid{n}\Rightarrow3\mid{n(n+1)(2n+1)}$
$n\equiv1\pmod{3}\Rightarrow3\mid{2n+1}\Rightarrow3\mid{n(n+1)(2n+1)}$
$n\equiv2\pmod{3}\Rightarrow3\mid{n+1}\Rightarrow3\mid{n(n+1)(2n+1)}$

So $3\mid{n(n+1)(2n+1)}$ for all $n$.

Now $2n+1$ is odd, and only one of $n$ and $n+1$ can be even, so either $2^4\mid{n}$ or $2^4\mid{n+1}$, so $n\equiv0$ or $15\pmod{16}$

If $5\mid{n}$,
$n\equiv0\pmod{5}\Rightarrow{n+1}\equiv1\pmod{5}$
$n\equiv0\pmod{5}\Rightarrow{2n+1}\equiv3\pmod{5}$,
so neither $n+1$ nor $2n+1$ is divisible by $5$.  If $5\mid{n+1}$,
$n+1\equiv0\pmod{5}\Rightarrow{n}\equiv4\pmod{5}$
$n+1\equiv0\pmod{5}\Rightarrow{2n+1}\equiv4\pmod{5}$,
so neither $n$ nor $2n+1$ is divisible by $5$.  And if $5\mid{2n+1}$,
$2n+1\equiv0\pmod{5}\Rightarrow{n}\equiv2\pmod{5}$
$2n+1\equiv0\pmod{5}\Rightarrow{n+1}\equiv3\pmod{5}$, so neither $n$ nor $n+1$ is divisible by $5$.  Since no two of $n$, $n+1$, and $2n+1$ can be divisible by $5$ simultaneously, one must be divisible by $25$ for $25\mid{n(n+1)(2n+1)}$.  So $n\equiv0$ or $12$ or $24\pmod{25}$.

The solutions for the two congruences modeled are:
$n\equiv0$, $112$, $224$, $175$, $275$, or $399\pmod{400}$,
So the lowest $n$ for which $200$ divides $1^2+2^2+3^2+4^2+...+n^2$ is $112$.

[quote="jrshoch"]Also, if the problem had read $n$ instead of $200$, wouldn't $5$ be the lowest number?[/quote]

Yes.  Bonus question: what are [i]all[/i] positive integers $n$ such that $1^2 + \ldots + n^2$ is divisible by $n$?

[hide]
$1+4+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}$.

So we want $4$ powers of two and $2$ powers of five on the top. Since $\gcd(n, n+1) = \gcd(n, 2n+1) = \gcd(n+1,2n+1) = 1$ (linear combinations), the $2$ powers of five and the $4$ powers of two must each come from a single term. If we just look to minimize $n$ and get the powers of five, we see $n = 12$ is the smallest because $2(12)+1 = 25$, but that doesn't work. Then we can try $n = 24$ but that doesn't work, either. Same goes for $n = 25$. It's easy to see that none can equal $50$ because then there won't be enough powers of $2$.

But anyway, we want to look for a multiple of $25$ that's either close to something divisible by $16$ or about double of one. We look around and we're like $225$ is close to $2(112)$ so let's pick $112$ and then it's divisible by $200$. If you really wanted to you could try solving stuff like

$25k \equiv 0, \pm 1 \pmod{16}$

or

$\frac{25k-1}{2} \equiv 0, \pm 1 \pmod{16}$,

whichever one is cooler and gives a solution first. [/hide]

EDIT: For JBL's question, I said they're all relatively prime, so if $n$ is divisible by $2$ or $3$ the denominator will go like woosh and then the sum won't be divisible by $n$. Otherwise, we're ok, though, since the $2$ and $3$ will appear in the other terms.


Decode the following quote from the famous science fiction writer Isaac Asimov.

“Ol ok irujhq, imjlojnojh irujhq, ojqfoluyeq irujhq, lrul ok lrq wmdojujl tuilma oj kmioqlc lmwuc. Jm kqjkoyeq wqiokomj iuj yq duwq ujc emjhqa golrmnl luzojh ojlm uiimnjl jml mjec lrq gmaew uk ol ok, ynl lrq gmaew uk ol goee yq.”

"It is change, continuing change, inevitable change, that is the dominant factor in society today. No sensible decision can be made any longer without taking into account not only the world as it is, but the world as it will be."  :lol:


geometry
circumcircle
vector
analytic geometry
search
arithmetic sequence
algebra

[b]Problem 1[/b]

(1) Let $ a,\ b$ be rational numbers such that $ \displaystyle \frac {6 \plus{} \sqrt {2}}{\sqrt {2} \minus{} 1} \equal{} a \plus{} b\sqrt {2}.$ Find the value of $ a.$

(2) Let $ a,\ b$ be real numbers such that $ \displaystyle \frac {12 \minus{} 21i}{2 \minus{} i} \equal{} a \minus{} bi.$ Find the value of $ b.$

(3) Find the minimum value satisfying the following inequality.

$ \left(\frac {1}{2}\right)^{x \minus{} 1} \plus{} \left(\frac {1}{2}\right)^{x \plus{} 3} \minus{} 2^{ \minus{} 2x}\geqq \frac {1}{4}.$

(4) Find the range of $ p$ for which the following cubic equation has 3 distinct roots.

$ \displaystyle \frac {1}{4}x^3 \plus{} \frac {3}{4}x^2 \minus{} 6x \plus{} p \equal{} 0.$

(5)Let $ n$ be a positive number. Find the other solutions in such way that the cubic equation :$ x^3 \plus{} nx^2 \plus{} (n \minus{} 6)x \plus{} 1 \equal{} 0$ has a postive integer root.

[b]Problem 2[/b]

(1) Find the 10 th term of the sequence:$ 3,\ 10,\ 23,\ 42,\ 67,\ 98,\ \cdots .$

(2) Let $ S_k$ be the sum from the first term to the $ k$ th term of the arithmetic sequence with the first term 1, the common difference 1.Find $ \displaystyle \sum_{k \equal{} 1}^9 \frac {1}{S_k}.$

[b]Problem 3[/b]

Given that a triangle such that $ AB \equal{} 1,\ \angle{ACB} \equal{} 30^\circ.$

(1) Find the diameter of the circumcircle of $ \triangle{ABC}.$

(2) The points $ A,\ B,\ C$ and $ D$ whch is different from other points are on the circle. If $ BD \equal{} \sqrt {2},$ find $ \angle{ABD}\geq 30^\circ.$

(3) In the last question, if $ DC \equal{} \sqrt {3},$  find the area of the quadrilateral$ ABDC.$

[b]Problem 4[/b]

Given that two vectors $ \overrightarrow{a} \equal{} (1,\ 2),\ \overrightarrow{b} \equal{} (4,\ 1).$ For the origin $ O,\ \overrightarrow {OP}$ is expressed as $ \overrightarrow{OP} \equal{} s\overrightarrow{a} \plus{} t\overrightarrow{b}$ for real numbers $ s,\ t\ (s\geq 0,\ t\geq 0).$

(1) Let the coordinate of $ P$ as $ (x,\ y).$ Express $ s$ interms of $ x,\ y.$

(2) If $ s \plus{} t\leq 3,$ find the area of the region in which the point $ P$ can be moved.

(3) If an area in which the point $ P$ can be moved is three times as that of the result of (2), find the range for which $ s \plus{} t$ can be valued.

[hide="4"]
1)
$ <x,y> \equal{} s<1,2> \plus{}t<4,1>$

$ x \equal{} s \plus{} 4t$
$ y \equal{} 2s \plus{} t$

solving for t and resubstituting,

$ x \equal{} s \plus{} 4(y\minus{}2s)$

$ s \equal{} \frac{4y \minus{} x}{7}$

2)
solving for t in terms of y and x similarly to above, we get $ t \equal{} \frac{2x\minus{}y}{7}$

Substituting into $ s\plus{}t \leq 3$

$ \frac{4y\minus{}x}{7} \plus{} \frac{2x\minus{}y}{7} \leq 3$

$ x \plus{} 3y \leq 21$

This is a triangle with base 21 and height 7. The area is $ \frac{21*7}{2} \equal{} \frac{147}{2}$

3) If the area is three times the original, then the lengths of the bases is $ \sqrt{3}$ times the original

$ x \plus{} 3y \leq 21\sqrt{3}$
Working backwards
$ \frac{4y\minus{}x}{7} \plus{} \frac{2x\minus{}y}{7} \leq 3\sqrt{3}$

$ s \plus{} t \leq 3\sqrt{3}$
[/hide]

For 2), by observing the sequence $ (a_n\minus{}a_{n\minus{}1})$, we obtain $ a_{n\plus{}1}\equal{}a_n\plus{}6n\plus{}1$.

1)

[hide]$ \frac{6 \plus{} \sqrt2}{\sqrt2 \minus{} 1} \equal{} \frac{6 \plus{} \sqrt2}{\sqrt2 \minus{}1} \times \frac{\sqrt2 \plus{} 1}{\sqrt2 \plus{} 1}$

multiplying fractions you get

$ 8 \plus{} 7\sqrt2$

Therefore $ a \equal{} 8$[/hide]

2) Works the same way

[hide]$ \frac {12 \minus{} 2i}{2 \minus{} i} \equal{} \frac {12 \minus{} 2i}{2 \minus{} i} \times \frac {2 \plus{} i}{2 \plus{} i}$

multiplying the fractions we get

$ \frac {26 \plus{} 8i}{5}$

Therefore $ b \equal{} \frac {8}{5}$[/hide]

1)  :)

2)  :(

Take a look the expression carefully. :)

[quote="kunny"][b]Problem 1[/b]
(3) Find the minimum value satisfying the following inequality.

$ \left(\frac {1}{2}\right)^{x - 1} + \left(\frac {1}{2}\right)^{x + 3} - 2^{ - 2x}\geqq \frac {1}{4}.$
[/quote]
I like this problem :lol:
[hide="solution"]

It is equivalen to:

$ \frac {2}{2^x} + \frac {1}{8}\cdot\frac {1}{2^x} - (\frac {1}{2^x})^2 - \frac {1}{4}\ge 0$

So lets sustitute $ \frac {1}{2^x} = a$, it becomes:

$ 2a + \frac {a}{8} - a^2 - \frac {1}{4}\ge 0$ or $ 8a^2 - 17a + 2\le 0$

Wich factors as $ (8a - 1)(a - 2)\le 0$

We must have  $ 8a - 1\ge 0$ and $ a - 2\le 0$, or $ 8a - 1\le0$ and $ a - 2\ge 0$

It is easy to verify that the second case doesn't work. And the first one yields $ a\in[\frac {1}{8},2]$
thus $ x\in [ - 2,3]$ I
f haven't commited any mistakes the anwser should be $ \min\{x\}=-2$.
[/hide]
[quote="kunny"]
(4) Find the range of p for which the following cubic equation has 3 distinct roots.

$ \displaystyle \frac {1}{4}x^3 + \frac {3}{4}x^2 - 6x + p = 0$
[/quote]
Nice problem too.
[hide="solution"]
First we note that the equation can't have: trhee equal roots, two roots equal to $ 0$, two equal non-real roots; and remember that can't have and ood number of non-real roots. Now we search for the values of $ p$ for wich the equation has two equal real roots.
So it can be written as $ (x^2+2ax+a^2)(x-r)=0$ for some $ a,r \in \mathbb{R}$. Equating coeficients:
$ 2a-r=3$
$ a^2-2ar=-6$
$ -a^2r=4p$
Sustituting $ r=2a-3$ in the second equation yields $ a^2-4a^2+6a+6=0\Leftrightarrow a^2-2a-2=0$  so $ a=1+\sqrt{3}, 1+-\sqrt{3}$; $ r=2\sqrt{3}-1, -2\sqrt{3}-1$ for wich $ p=\frac{7\sqrt{3}-2}{4}, -\frac{7\sqrt{3}+-2}{4}$.
The answer I think is $ p\in\{{\mathbb{R}-\{\frac{7\sqrt{3}-2}{4}, -\frac{7\sqrt{3}+-2}{4}\}\}}$
[/hide]

[quote="kunny"]
5)Let n be a positive number. Find the other solutions in such way that the cubic equation $ x^3 + nx^2 + (n - 6)x + 1 = 0$ has a postive integer root.
[/quote]
[hide="solution"]
By the rational root theorem, if the equation has a rational root it must be $ -1$ or $ 1$, we search for  it to have $ 1$ as root. So $ (1)^3+n(1)^2+(n-6)(1)+1=1+n+n-6+1=2n-4=$. From this we see that $ n=2$ is the only value wich works.
[/hide]
 :)

so is my solution wrong or was there a quicker way?

Your solution is right, but you made a mistake to look the expression in error. :)


Can someone tell me what arbelos journal is? And is it helpful to prepare for USAMO?

The AMC says it is very helpful to prepare for the USAMO.  I plan to get them entirely for that purpose.  Here's the description...Check out what it says about it in the USAMO teachers' manual on http://www.unl.edu/amc

Can you give me the direct link to the description?

[url=http://www.unl.edu/amc/d-publication/d6-books/testsoln.html#4045]The Arbelos[/url]

btw, are there ones after 1988??
And how many problems does each booklet have???
are there anything other than the problems???

sorry for lots of questions... thanks in advance.

I have the Arbelos journals and they're pretty nice. Some decent discussion of problems as well as actual problems, which will help beginners with olympiad especially.

Which would you say is better, the journals or Engel's book? Also, how long does it take to ship the journals from the AMC?

[quote="fanaticsm"][quote="paladin8"]It's a journal, like it says. I have 2 of them; they are basically an exploration of some topic followed by some problems related to the topic (e.g. a recursions section that I just read).[/quote]

Which volumes do you have?[/quote]

I have the 1982-1983 book, which is No. 1 through No. 5 (dunno about volume #...), and Volume 5 (Second Edition).

Does it matter which volume you being with? Also, how do you recommend using the Arbelos Journals (studying directly from them, just using them as problem books, etc.)?


modular arithmetic
arithmetic sequence
arithmetic series

please help me  :(

if  A= 1^5 + 2^5 + 3^5 + .......+ (101)^5

find the remains after we divide A by 7

[b]@ Ahmad [/b] : Is the remainder 6? I got the remainder as 6, tell me whether this is correct or not?

This basicaly cycles since it's remainders mod 7. We have that

$ A \equiv 1^5\plus{}2^5\plus{}3^5\plus{}4^5\plus{}5^5\plus{}6^5\plus{}0^5\plus{}.... \mod7$

$ A \equiv (1^5\plus{}2^5\plus{}3^5\plus{}4^5\plus{}5^5\plus{}6^5\plus{}0^5)14 \plus{} 1^5\plus{}2^5\plus{}3^5 \mod7$

$ A \equiv 1^5\plus{}2^5\plus{}3^5 \mod7$

$ A \equiv 1\plus{}32\plus{}243 \mod7$

$ A \equiv 276 \mod7$

$ A \equiv 3 \mod7$

The remainder is $ \boxed{3}$.

Please move this to the High School Basics.

What's the answer to this problem? Can anybody clarify whether my answer is correct or not?

Sorry to say but your answer is wrong, I see no error in modularmarc101's solution.

I tried it myself just to be sure and I got the same thing he did, would you mind explaining how you solved yours.

As Agr_94_Math said, please mods move this to High School Basics.

[b] @ tony pr [/b] : Well, I  solved the problem in the following manner:

[b] Step 1 [/b]

There exists a formula for calculating the sum of an Arithmetic progression like the one given by [b] ahmad [/b], since the number of terms in the sequence was 101, so, firstly, I decided to find the common difference.

    S= n/2 [2a+ (n-1)d] - Formula for calculating the sum of the arithmetic series

where n= no of terms in the sequence (In this case, it was 101)

a- first term in the series

d- common difference

After much thinking, I managed to figure out the common difference 'd', which happened to be (n^5- 5n^4+10n^3-10n^2+5n-1) -> (This actually happens to be of the form: (n^5) - ((n-1)^5)

[b] Step 2 [/b]

I substituted this in the sum formula, to get the sum of the series.

After calculating the sum, I managed to divide it by 7 to find the remainder.

I do know that this is a highly inefficient technique, my knowledge of Math is somewhat restrained, I need to refresh my knowledge of Math.

P.S.: I think, I have just found where I committed the mistake.  :lol:

Yes, The answer is:[b] 3 [/b], I made a small error while calculating the common difference.

I don't see where you find a common difference?

For example the difference of the first two terms..: $ 2^5\minus{}1^5 \equal{} 31$ from what you say the difference between the 3rd term and the 2nd term must equal $ 31$ so let's try it: $ 3^5\minus{}2^5 \equal{} 211$ (I think) which definately does not equal $ 31$ which means it does not have a common difference and is not in a arithmetic progression.

Okay, can you tell me how you managed to solve this problem?

I started by writting out the sum of $ 1^5\plus{}2^5\plus{}3^5\plus{}4^5\plus{}5^5 \plus{} 6^5$ in mod 7, then I noticed that this was the same as $ 1^5\plus{}2^5\plus{}3^5\plus{}(\minus{}3)^5\plus{}(\minus{}2)^5\plus{}(\minus{}1)^5\equal{}0$ in mod 7 and with this it's easy to see that the sum is the same as:

$ 1^5\plus{}2^5\plus{}3^5\plus{}\cdots\plus{}98^5\plus{}99^5\plus{}100^5\plus{}101^5\equiv 0\cdot 14 \plus{} 99^5\plus{}100^5\plus{}101^5\equiv 1^5\plus{}2^5\plus{}3^5\equiv 3 \pmod{7}$

Basicly it's the same method that modularmarc101 used only with the difference of $ 1^5\plus{}2^5\plus{}3^5\plus{}(\minus{}3)^5\plus{}(\minus{}2)^5\plus{}(\minus{}1)^5\equal{}0$ in the solution which doesn't change the answer.

[quote="modularmarc101"]This basicaly cycles since it's remainders mod 7. We have that

$ A \equiv 1^5 \plus{} 2^5 \plus{} 3^5 \plus{} 4^5 \plus{} 5^5 \plus{} 6^5 \plus{} 0^5 \plus{} .... \mod7$

$ A \equiv (1^5 \plus{} 2^5 \plus{} 3^5 \plus{} 4^5 \plus{} 5^5 \plus{} 6^5 \plus{} 0^5)14 \plus{} 1^5 \plus{} 2^5 \plus{} 3^5 \mod7$

$ A \equiv 1^5 \plus{} 2^5 \plus{} 3^5 \mod7$

$ A \equiv 1 \plus{} 32 \plus{} 243 \mod7$

$ A \equiv 276 \mod7$

$ A \equiv 3 \mod7$

The remainder is $ \boxed{3}$.[/quote]

thanks modularmarc101  :clap2:


how do you type  in a latex piece of [b]text[/b]

So if I need to make a pdf doc and write "Belgi is..." then how do I do? I always write Belgi$\ddot{e}$ is... in my docs, but that's kind of ugly. Is there a better way to write that in pdf docs? Or is it just WinEdt not allowing to type those symbols? :?

\"e

doesn't

Belgi\"e

work?

you don't need to put \"e in dollar signs

yes, incredible. where did you learn that? I didn't find it in any latex guide :(

thanks! :)

Play around with \', \`, \", \H, \c, etc. I found those ones in Knuth's The $\text{\TeX}$book. There are a bunch more that I need to look up every time.

[quote="ComplexZeta"]Play around with \', \`, ", \H, \c, etc. I found those ones in Knuth's The $\text{\TeX}$book. There are a bunch more that I need to look up every time.[/quote]

$\'a, \`a, "a, \H a, \c a$

$\'e, \`e, "e, \H e, \c e$

$\'i, \`i, "i, \H i, \c i$

$\'o, \`o, "o, \H o, \c o$

$\'u, \`u, "u, \H u, \c u$

the  will be most used though

Am I wrong or don't I see a ^ here like ? How should I do that?

hey, why aren't the e and the o rows displaying??? :? :? :?


function AMC USA(J)MO USAMO logarithms limit algebra

differentiating both sides, f'(x/y)=(f'(x)f(y)-f'(y)f(x))/f(y)^2 let x=y
the terms in the numberator are the same when we substitute, so the fraction=0 if f(y)^2 does not equal 0.

then we would get 0=f(1), or f(y)=0. if f(Y)=0, this is the same as f(x)=0 but then f'(1) would not equal 2. so f(1)=0

i dont have any paper with me so thats all i got for now. please check to see if i made any mistakes; my calc is very rusty-i never used it for mathcounts...

You can't just differentiate $ \frac {f(x)}{f(y)}$ with respect to $ x$ and $ y$ simultaneously - you have to temporarily treat $ f(y)$ as a constant if you're differentiating with respect to $ x$. For the left side, don't forget the $ \frac 1y$ that gets kicked out because of the chain rule.

Edit: Using my guess and check powers, one possible function is $ f(x) \equal{} x^2$.

$ f\left(\frac xy\right) \equal{} \frac {x^2}{y^2}$
$ f'(1) \equal{} 2(1) \equal{} 2$

 :D

[hide="Sketch of solution"]Set $ x\equal{}y\equal{}1$ and obtain $ f(1)\equal{}1$. Then $ f\left(\frac{1}{y}\right) \equal{} \frac{1}{f(y)}$ and $ f(xy) \equal{} \frac{f(x)}{f\left(\frac{1}{y}\right)} \equal{} f(x)f(y)$. It is well-known that all functions that satisfy this are of the form $ |x|^a$ or $ \operatorname{sgn}(x)|x|^a$. Differentiating, we have $ a\equal{}2$, so either $ f(x)\equal{}x^2$ or $ f(x)\equal{}x^2\operatorname{sgn}(x)$.[/hide]

It's also not true.  The condition that $ f(x)$ is differentiable at a point is enough to make that conclusion, but without it there are many pathological solutions.  In particular, I believe we can take $ f(x) \equal{} e^{g(\ln |x| )}$ where $ g(x)$ is a solution to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url].

[quote="Yongyi781"]In the USAMO, "well-known" doesn't cut it.[/quote]

That's why I said sketch. :P

[hide="An attempt at a complete solution"]As above, we have $ f(xy)=f(x)f(y)$. Now by the condition $ f'(1) = 2$ the limit $ \lim_{\epsilon\to 0 \frac{1+\epsilon}{\epsilon}}$ exists and is equal to 2. For $ x>0$ we then have
\begin{align*}
2\frac{f(x)}{x} &= \frac{f(x)}{x}\lim_{\epsilon\to 0} \frac{f(1+\epsilon)-f(1)}{\epsilon} \\
&= \lim_{\epsilon\to 0}\frac{f(x+x\epsilon)-f(x)}{x\epsilon} \\
&= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\
&= f'(x).
\end{align*}
So the function is differentiable for $ x>0$ with $ f'(x) = 2\frac{f(x)}{x}$. Note, however, that $ (f(x)x^{-2})' = f'(x)x^{-2}-2f(x)x^{-3} = x^{-2}\left(f'(x)-2\frac{f(x)}{x}\right) = 0$ so $ f(x) = cx^2$. From $ f'(1) = 2c = 2$ we then conclude $ f(x) = x^2$ for $ x>0$.

Then, we have $ f(0) = f(2)f(0) = 4f(0)\implies f(0) = 0$ and $ 1 = f(1) = (f(-1))^2\implies f(1) = \pm 1$. This gives the solutions $ f(x)=x^2$ and $ f(x)=\operatorname{sgn}(x)x^2$, respectively.[/hide]


Circles $ S_1$ and $ S_2$ with centers $ O_1$ and $ O_2$, respectively, pass through the centers of each other. Let $ A$ be one of their intersection points. Two points $ M_1$ and $ M_2$ begin to move simultaneously starting from $ A$. Point $ M_1$ moves along $ S_1$ and point $ M_2$ moves along $ S_2$. Both points move in clockwise direction and have the same linear velocity $ v$.

$ (a)$ Prove that all triangles $ AM_1M_2$ are equilateral.

$ (b)$ Determine the trajectory of the movement of the center of the triangle $ AM_1M_2$ and find its linear velocity.

[color=darkred][b][u]Easy and very nice problem ![/b][/u][/color][hide="Hint."][color=darkblue] Prove easily that $ \triangle\ AO_1M_1\sim\triangle\ AO_2M_2\implies$  $ \triangle\ M_1AM_2$ is equilateral. The geometrical locus of the center $ G$ of $ \triangle\ M_1AM_2$ is the circumcircle of $ \triangle\ O_1AO_2$. The velocity $ v_g$ of the point $ G$ is $ v_g \equal{} \frac {v}{\sqrt 3}\ .$[/hide] [b]Remark.[/b] Here is an extension.[/color][quote="Virgil Nicula"][color=darkred]Let $ C_1 \equal{} C(O_1,r_1)$ , $ C_2 \equal{} C(O_2,r_2)$ be two secant circles and let $ A\in C_1\cap C_2$. Two points $ M_1$ and $ M_2$ begin to move simultaneously starting from $ A$. The point $ M_1$ moves along $ C_1$ with the velocity $ v_1$ and the point $ M_2$ moves along $ C_2$ with the velocity $ v_2$. Both points move in clockwise direction and $ \frac {v_1}{v_2} \equal{} \frac {r_1}{r_2}$.
$ 1\blacktriangleright$ Prove that all triangles $ AM_1M_2\sim O_1AO_2$.
$ 2.\blacktriangleright$ Find the trajectory of the movement of the circumcenter of $ \triangle AM_1M_2$ and the its velocity.[/color][/quote]


Prove that for each integer $ m, 0\leq m<n$, the function $ F_m(p)\equal{}\sum_{j\equal{}0}^m\binom{n}{j}p^j(1\minus{}p)^{n\minus{}j}$ is strictly decreasing on $ (0,1)$.

$ F_m'(p) \equal{} \minus{} (n\minus{}m)C_n^mp^m(1 \minus{} p)^{n \minus{} 1 \minus{} m} < 0.$


geometry
parallelogram
ratio
trapezoid
angle bisector
geometry solved

Let a triangle $ABC$  , the median $AM$ and the bisector $AD$  of angle $A$ . The perpendicular from $B$ to $AD$  meets $AM$  at  $E$. Prove that $DE // AB$

[u]Babis[/u]

[b]NOTE[/b]
 [i]The problem in Moskow -Olympiad  1995  was the following.[/i]

Let a triangle $ABC$  , the median $AM$ and the bisector $AD$  of angle $A$ . The parallel from  $D$ to $AB$   meets $AM$  at  $E$. Prove that $BE \perp AD$.

   [i]Please ,  send  your nice  solutions to this problem too , if they are essencially different from the solutions to the above problem , which have been allready sent .[/i]

Let $AB=c$, $AC=b$ and $BC=a$. We can find that  $BD=\frac{ac}{b+c}$ and $MD=\frac{a(b-c)}{2(b+c)}$. Then $\frac{MD}{BD}=\frac{b-c}{2c}$. Let the line $BE$ meet line $AC$ at $F$. Then $AB=AF=c$ and $FC=b-c$. Let the line $CE$ meet $AB$ at $H$. Then by Ceva we have $\frac{BM \cdot CF \cdot AH}{CM \cdot AF \cdot BH}=1$. Then we have $(b-c)(c-BH)=c \cdot BH$. Now $BH=\frac{c(b-c)}{b}$ and $c-x=\frac{c^2}{b}$. Then by Van Obel`s theorem we have $\frac{AE}{EM}=\frac{c-BH}{BH} \cdot \frac{c}{b-c}=\frac{2c}{b-c}$ and $\frac{EM}{AE}=\frac{b-c}{2c}$. Now we can see that $\frac{EM}{AE}=\frac{DM}{BD}$ from which we conclude that $DE||AB$.

I will prove this stronger problem :.

[b]Let $ABC$ be a triangle .Let $F$ be any point on side $AC$.let $M,H,G$ be midpoints of $BC,BF,BA$ respectively.
Let $AH$ intersects $BC$ at $D$ .and $BF$ intersects $AM$ at $E$ .Prove that $DE \parallel AB$.[/b]
[b]Proof[/b].we know $M,H,G$ are collinear.now consider complete quadrilateral formed by $AB,BM,AM,DE$ .Let $DE$ intersects AB at $X$ . we know $MB,MH,MA,MX$ are $harmonical rays$.so $(BA,GX)=-1$ .and since $G$ is midpoint of $AB$ , so $X$ is in $\infty$ and so $DE\parallel AB$.
Note that this problem trivially implies our problem. So the proof is complete.

Just in order to contribute to the diversity of solutions to this problem:

[color=blue][b]Problem.[/b] In a triangle ABC, let M be the midpoint of the side BC, and let the angle bisector of the angle CAB intersect the side BC at a point D. The perpendicular from the point B to the line AD meets the line AM at a point E. Prove that DE || AB.[/color]

[i]Solution.[/i] Let the parallel to the line AB through the point C meet the parallel to the line CA through the point B at L. Then, CL || AB and BL || CA. Hence, the quadrilateral ABLC is a parallelogram; thus, its diagonals AL and BC bisect each other. In other words, the midpoint M of the segment BC is also the midpoint of the segment AL.

Let the two perpendicular lines AD and BE intersect each other at a point R. Then, < ARB = 90°, so in the right-angled triangle ARB, we have < ABR = 90° - < BAR. On the other hand, since the line AD is the angle bisector of the angle CAB, we have $\measuredangle BAR=\frac{\measuredangle CAB}{2}$; also, since BL || CA, we have < ABL = 180° - < CAB. Thus, $\measuredangle ABR=90^{\circ}-\measuredangle BAR=90^{\circ}-\frac{\measuredangle CAB}{2}=\frac{180^{\circ}-\measuredangle CAB}{2}=\frac{\measuredangle ABL}{2}$. Thus, the line BE bisects the angle ABL. Now, recall the following well-known fact:

[b](*)[/b] The angle bisector of an angle of a triangle divides the opposite side in the ratio of the two adjacent sides.

Applying this statement [b](*)[/b] to the triangle ABL with its angle bisector BE, we get $\frac{LE}{AE}=\frac{BL}{AB}$. Since ABLC is a parallelogram, BL = AC, so this becomes $\frac{LE}{AE}=\frac{AC}{AB}$.

On the other hand, applying the statement [b](*)[/b] to the triangle ABC with its angle bisector AD, we get $\frac{CD}{BD}=\frac{AC}{AB}$.

Thus, $\frac{LE}{AE}=\frac{CD}{BD}$. Hence, $\frac{AL}{AE}=\frac{AE+LE}{AE}=1+\frac{LE}{AE}=1+\frac{CD}{BD}=\frac{BD+CD}{BD}=\frac{BC}{BD}$. Now, since the point M is the midpoint of both segments BC and AL, we have $AL=2\cdot AM$ and $BC=2\cdot BM$, so that this becomes $\frac{2\cdot AM}{AE}=\frac{2\cdot BM}{BD}$, so that $\frac{AM}{AE}=\frac{BM}{BD}$. Thus, by Thales, DE || AB, and we are done.

  Darij

I like behzad's solution for a special reason,  but here is my solution:

   Let line BE intersect AD in Y.

   Drop the other shoe, so to speak, in the form of the perpendicular from vertex C

   onto line AD at point X. Draw circle k centered at M through points Y and X. We know

   that line XM is the mid-line of ABC parallel to AB. Line XM intersects circle k on line

   BE.  A circle p tangent to circle k at Y and passing D will also go through E

   (as BE is parallel to CX).  Now we have two tangent at Y circles k and p  and

   a common inscribed angle  DYE.   Two corresponding lines through the feet of

   the inscribed angle are parallel.

   Thank you.

   M.T.

  Darij, I want to ask you what is X in your posts?

We denote as $K,$ the midpoint of the side segment $AB$ and as $L,$ the intersection point of $AD,$ $BE.$

From the right angled triangle $\bigtriangleup LAB,$ we have that $\angle ALK = \angle KAL = \angle LAC$ $,(1)$

From $(1),$ we conclude that $KL\parallel AC$ and so, the points $K,$ $L,$ $M,$ are collinear.

Hence, the quadrilateral $ABDE$ is a trapezium, with $AB\parallel DE$ and the proof is completed.

Kostas vittas.


How many people can enter the same Math Jam? :|

I don't recall, but it's a relatively high number.  We've had Math Jams with over 100 people before.


linear algebra
matrix
algebra open
algebra

Consider a sequence of matrices $ A_n\equal{}[a_n (i,j)]_{n*n}$ for every $ n \geq 1$.
Given $ a_1 (1,1)\equal{}1$.
Define :
$ S_n (r)\equal{}\sum_{i\equal{}1}^{n} \frac{a_n (r,i)}{n\plus{}i}$for every$ 1\leq r \leq n.$
$ T_n\equal{}\frac{1}{2n\plus{}1} \minus{}\sum_{i\equal{}1}^{n} \frac{S_n (i)}{n\plus{}i}$ for every $ n \geq 1.$
Use the following to get $ A_{n\plus{}1}$ from $ A_n$ :
$ a_{n\plus{}1}(i,j)\equal{}a_n (i,j) \plus{} \frac{S_n (i) S_n (j)}{T_n}$for$ 1\leq i,j \leq n$
$ a_{n\plus{}1}(n\plus{}1,j)\equal{}a_{n\plus{}1}(j,n\plus{}1)\equal{}\frac{\minus{}S_n (j)}{T_n}$for$ 1\leq j \leq n$
$ a_{n\plus{}1}(n\plus{}1,n\plus{}1)\equal{}\frac{1}{T_n}$
Prove that $ a_n (i,j)$ is an integer for every $ 1\leq i,j \leq n$ and for every $ n \geq 1$.
Is it true that $ a_n (1,1)\equal{}n^2$?

Can You Prove that
$ a_n (i,j) \equal{} (\minus{}1)^{i\plus{}j} (i\plus{}j\minus{}1)$ $ {{n\plus{}i\minus{}1}\choose {n\minus{}j}}$ $ {{n\plus{}j\minus{}1}\choose {n\minus{}i}}$ $ {{i\plus{}j\minus{}2}\choose {i\minus{}1}}^2$

??


vector
advanced fields
advanced fields unsolved

Prove that a normed space X is separable iff A a subspace of X and X/A a quotient space are separable.

Quite urgently need a proof for this ... in particular (<=) this implication.

$ A$ has to be closed. review the definition of the quotient norm. if $ \{a_{n}\}$ is dense in $ A$ and $ \{x_{n}\text{~mod~}A\}$ is dense in $ X/A$, then $ \{a_{n}\plus{}x_{m}\}$ is dense in $ X$.

for the other direction: since the quotient map is continous, a dense subset of $ X$ is mapped to a dense subset of $ X/A$. besides, separable metric spaces are exactly those with a countable base of the induced topology. clearly, subspaces inherit this property.


[quote="Ravi B"]I believe $z = e^{2\pi i / n}$ works.[/quote]

can you explain why this works and other numbers don't?

We have that [tex]z^n[/tex] is real iff [tex]z=re^{2ki \frac \pi n}=r(\cos(2k \frac \pi n)+i\sin (2k \frac \pi n))[/tex]. But we have [tex](z+1)^n[/tex] so it must be:
[tex]z+1=r(1/r+\cos(2k \frac \pi n)+i\sin (2k \frac \pi n))=s(\cos(2j \frac \pi n)+i\sin (2j \frac \pi n))[/tex]
[tex]\left\{ \begin{array} {cc} 1+r\cos(2k \frac \pi n)=s\cos(2j \frac \pi n)  \\ r\sin (2k \frac \pi n)=s\sin (2j \frac \pi n)\end{array} \right[/tex]
that works for example if [tex]k=2j, r=1, s=2\cos (2j \frac \pi n)[/tex]

[quote="perfect_radio"]let $n \in \mathbb{N}, \geq 3$. Determine $z \in \mathbb{C}-\mathbb{R}$ such that $z^n$ and $(1+z)^n$ are real numbers.[/quote]

Well. That is to say that if A=(0,0), B=(-1,0). Find all P on the plane but not the x-axis such that $cos(n(\pi-\angle PAB))=cos(n\angle PBA)=0$
To find all such P, you just draw n lines through a and n lines through B such that the angles between the lines and the x-axis are $\frac{k\pi}{n}$.
Then all the intersections are all the point you seek.


OK, a lot of you probably already know this series of problems, and I can't guarantee that this hasn't been posted before. But I searched for these problems and couldn't find one. So let's see if you can solve these  :D

You are in an [i] almost [/i] unescapable room. You know you are going to run out of air in an hour, you'll die of either thirst or hunger in 1/2 a hour, and you're too young to die. "Useless" materials have been carelessly put in the room, such as --

-a calendar
-a baseball bat
-a matress
-a grand piano
-a mirror
-a table

Assuming that all of you are smart and not bored to death by this point, and also assuming that you know the one that is just like this one, but with a table and saw, you must--

1. Get water to drink.

2. Get something to eat

3. Find the [b]three[/b] ways to escape!

Good luck!

3. crush the door with piano, bat, table? 3 ways? :P

[hide="3"]
Look in the mirror, see what you saw, take the saw and cut the table in half, two halves make a whole, climb through the hole.

Other two solutions: Cut the baseball bat or calendar instead :P
[/hide]

AIME15, nice job using the mirror, and finding a way other than the three I intended(cutting the baseball bat or calendar). However, you still need to eat, drink and find two more ways to escape! By the way, I doubt you can eat the bat(maybe grow it into a tree? not enough time), and to clarify, there is no door, and the room's material is [i]way[/i] too hard (diamond?) to crush with any of your materials.

Here's a hint:  [hide] For Number 2, use the calendar. [/hide]

Good luck!  :P

[hide="1"]Put the grand piano and mattress on the table, the table goes pop, and drink the pop.

The mattress has springs, springs give off water, drink the water.[/hide]

[hide="2"]Calenders have dates. Eat the dates (they are good for you too).[/hide]

[hide="3"]You could use the keys on the piano, but since there is no door, this doesn't work.[/hide]

You wait until you [b]run out[/b] of air... or food...or water.

Or, alternatively, run around until you run out of energy.

You hit a home run with the baseball bat and run home.

[hide="1"]
Smash the mirror into pieces of glass, then drink the glass of water.

[/hide]

Having a glass (container) doesn't mean it has water.

Good idea though.

Take the springs of the mattress, drink from them.
A calendar has dates, eat the dates.
Look in the mirror, see what you saw. Take the saw and cut the table in half. Two halves make a whole. Climb through the hole, you're free.
Or, alternatively, hit a home run with the bat, and run home.  :P


geometry
angle bisector
Pythagorean Theorem
similar triangles

Consider $ \triangle ABC$ where $ AB\equal{}5$, $ BC\equal{}6$, and $ AC\equal{}7$. Circle $ \omega$ is drawn tangent to $ \overline{AB}$ and $ \overline{AC}$ with center $ O$ on $ \overline{BC}$. Draw line $ \Gamma$ tangent to $ \omega$ and parallel to $ \overline{BC}$. Choose point $ D$ on $ \Gamma$ such that $ \overline{AD}$ passes through $ O$. Compute the length of $ AD$.

made this problem, just wanna see how hard it is...

i have a feeling i should change the side lengths to make it less ugly

[hide]

Extend sides $ AB$ and $ AC$ to meet line $ \Gamma$ at $ E$ and $ F$.

Circle $ O$ is an incircle of $ \bigtriangleup AEF$. Since $ AD$ passes through $ O$, $ AD$ is an angle bisector of $ \angle A$.

By the Angle Bisector Theorem, $ BO \equal{} \frac {5}{2}$.

The altitude of $ \bigtriangleup ABC$ from $ A$ is $ 2\sqrt6$. Let it intersect $ BC$ at $ G$.

$ AO$=$ \frac {\sqrt105}{2}$, so $ GO \equal{} \frac {3}{2}$.

Now, draw the radius to intersect $ AB$ at $ H$.
Let $ x \equal{} BH$. $ HA \equal{} 5 \minus{} x$.

Use Pythagorean Theorem to set up two equations:
$ (AH)^2 \plus{} (OH)^2 \equal{} (AD)^2$
$ (BH)^2 \plus{} (OH)^2 \equal{} (BO)^2$

$ x^2 \minus{} 10x \plus{} 25 \plus{} r^2 \equal{} \frac {105}{4}$.
$ x^2 \plus{} r^2 \equal{} \frac {25}{4}$.

Solving for $ r$ gives $ r \equal{} \sqrt6$.

Now, extend $ AG$ to meet line $ \Gamma$ at $ K$.
$ AK \equal{} AG \plus{} r$
$ \equal{} 3\sqrt6$

Drop a perpendicular down from $ O$ to line $ \Gamma$ which it intersects at $ L$.
Let $ LD \equal{} y$.
Now set up a similar triangles proportion to solve for $ y$.

$ \frac {3\sqrt6}{\frac {3}{2} \plus{} x} \equal{} \frac {\sqrt6}{y}$.
$ y \equal{} \frac {3}{4}$.

Now, use Pythagorean on $ \bigtriangleup OLD$ to find $ OD$.

$ OD \equal{} \frac {\sqrt105}{4}$.

$ AD \equal{} AO \plus{} OD$

$ \equal{} \boxed{\frac {3\sqrt105}{4}}$.

(Dang, I used a lot of letters for points)
[/hide]


You can memorize it. Other than that, I don't know.

Find the answer to this problem. Then try it with a 4 by four grid. Try different types and you can probably come up with a formula

[quote="The Pink Panther"]The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different
sets of three points of this grid can be the three
vertices of an isosceles triangle?

Is there a faster way to do this?  I need a fast and efficient way to do this..[/quote]

Hmm..well I guess you can split it into cases..idk...I think thats the easiest way.
Perhaps one of our Top 10 National Cders can help you out :wink:

Yes, do casework based on the length of the two equal sides.

[quote="Math Champion"]There is no other way than just splitting it into cases and counting. You could also use symmetry. This would speed it up.[/quote]

Yeah true....Symmetry is a good way to solve it. Find the number of triangles you can make with one side, then multiply by 4!

That was an exclamation point, SuperNerd, not a factorial sign.


The sequence 2; 3; 5; 6; 7; 10; 11….. consists of all positive integers that are not squares or cubes. What is the 500th term in the sequence?

(A)527 (B) 528 (C) 529 (D) 530 (E) 531

[hide] Consider where the number $ 500$ is in the sequence. It has $ 22$ squares behind it ($ 1^2$ to $ 22^2$) and $ 7$ cubes behind it. However, two of these numbers, $ 1$ and $ 64$, are in both sets. So $ 500$ is the $ 500\minus{}22\minus{}7\plus{}2\equal{}473rd$ term in the sequence. Thus, we need $ 27$ more terms. We can go up to $ 527$. However, we must go up one more (since $ 8^3\equal{}512$). Thus our number is $ \boxed{528}$, or B.[/hide]


function
trigonometry
integration
calculus
vector
complex numbers
calculus computations

If $ F(t)$ is defined only for $ 0\leq t\leq\pi$, show that by defining $ F(\minus{}t): \equal{}\minus{}F(t)$, $ 0<t<\pi$ and constructing the Fourier series for this function over the interval, one arrives at a Fourier sine series
$ \sum_{n\equal{}1}^{\infty}\beta_{n}\sin{nt}$
for $ F$, with coefficients given by
$ \beta_{n}\equal{}\frac{2}{\pi}\int_{0}^{\pi}F(t)\sin{nt}\ dt$.

First, $ F(t)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}c_{n}e^{int}$.  $ F(t)$ is odd, so this series turns into
$ \sum_{n\equal{}1}^{\infty}c_{n}e^{int}\minus{}e^{\minus{}int}\sum_{n\equal{}1}^{\infty}2ic_{n}\sin{nt}\equal{}\sum_{n\equal{}1}^{\infty}\beta_{n}\sin{nt}$.
I can't get the coefficients in the form desired, though.

If $ F$ is odd, then $ c_{ \minus{} n} \equal{} \minus{} c_n,$ which includes that $ c_0 \equal{} 0$

Break up the sum $ \sum_{n \equal{} \minus{} \infty}^{\infty}c_ne^{int}$ into pairs of terms:

The $ n$th term is $ c_ne^{int}$ and the $ \minus{} n$ tern is $ c_{ \minus{} n}e^{ \minus{} int}.$

$ c_ne^{int} \plus{} c_{ \minus{} n}e^{ \minus{} int} \equal{} c_ne^{int} \minus{} c_ne^{ \minus{} int}$

$ \equal{} c_n(e^{int} \minus{} e^{ \minus{} int}) \equal{} (2ic_n)\cdot\frac {e^{int} \minus{} e^{ \minus{} int}}{2i} \equal{} 2ic_n\sin nt.$

So, assuming $ F$ is odd, we have $ \beta_n \equal{} 2ic_n$ or $ c_n \equal{} \minus{} \frac {i}2\beta_n.$

As for the integral form: $ \beta_n \equal{} 2ic_n \equal{} \frac {2i}{2\pi}\int_{ \minus{} \pi}^{\pi}F(t)e^{ \minus{} int}\,dt$

$ \equal{} \frac {i}{\pi}\left(\int_0^{\pi}F(t)e^{ \minus{} int}\,dt \plus{} \int_0^{\pi}F( \minus{} t)e^{int}\,dt\right)$

$ \equal{} \frac {i}{\pi}\int_0^{\pi}F(t)(e^{ \minus{} int} \minus{} e^{int})\,dt$ since $ F( \minus{} t) \equal{} \minus{} F(t)$

$ \equal{} \minus{} \frac {i}{\pi}\int_0^{\pi}F(t)(e^{int} \minus{} e^{ \minus{} int})\,dt$

$ \equal{} \frac {2}{\pi}\int_0^{\pi}F(t)\frac {e^{int} \minus{} e^{ \minus{} int}}{2i}\,dt \equal{} \frac {2}{\pi}\int_0^{\pi}F(t)\sin nt\,dt.$

If $ F$ is real-valued, then $ c_{ \minus{} n} \equal{} \overline{c_n}.$ That makes the real part of $ c_n$ an even function in $ n$ and the imaginary part of $ c_n$ an odd function in $ n.$

Therefore, if $ F$ is both real and odd, then $ c_n$ is odd in $ n$ and pure imaginary, and we can write $ \beta_n \equal{} \minus{} 2\operatorname{Re}(c_n).$

I'll leave the parallel discussion of even functions for JRav to do on his own. One way to look at this: the set of all simple harmonic motions of frequency $ n$ (also known as the set of solutions of the ODE $ y'' \plus{} n^2y \equal{} 0$) is, for $ n\ne 0,$ a two-dimensional vector space. As is true for any two-dimensional vector space, we have a choice of possible bases. One possible basis is $ \{e^{int},e^{ \minus{} int}\}.$ Another possible basis is $ \{\cos nt,\sin nt\}.$ All of the algebra above is really just a change of basis and can be re-cast as something involving invertible $ 2\times 2$ matrices. Here's a third basis that gets used sometimes: the "cas" or Hartley form: $ \{\cos (nt) \plus{} \sin (nt),\cos( \minus{} nt) \plus{} \sin( \minus{} nt)\}.$

Thanks.  I know how to re-arrange the series to yield a sine, that wasn't hard n(i.e. the first part of what you wrote).  It was just coming up with the integral expression for $ \beta_{n}$ that I needed.  Doing the same for even functions shouldn't be much different then.

I just noticed that Kent has a $ \text{\TeX}$ error.  In the second-to-last integral, you have $ e - int$ instead of $ e^{ - int}$.

[color=green][That's easy to fix.][/color]


1)If the real numbers $x,y,z$ have the properties: $x+y+z=2$ and $xy+yz+zx=1 $ then:
$x,y,z \in [0, \frac{4}{3}]$.

2)If the real numbers $x,y,z$ have the properties: $x+y+z=5$ and $xy+yz+zx=3$ then:
$x,y,z \in [-1, \frac{13}{3}]$.

I have solved them both using Rolle's sequence. What is your solution?

cheers! :D :D :cool:

your problem can solve by viet.! You can try solve it .it is very interested  :D

I had solution of your problem ! Do you want see it?  :(

a) x+y+z=2 ,xy+xz+yz=1
b) x+y+z=5, xy+xz+yz=3
  then x+y=2-z
         xy  =1-z(x+y)=1-z(2-z)
   => x,y is real numbers so that X 2 -(2-z)X+1-z(2-z)(1)
   we find z sothat (1) have x,y is  real
  similary  x+z=2-y and y+z=2-x and give (b)
 we have  problem :lol:


A square is inscribed in a circle, which is inscribed in an equilateral triangle. The side length of the triangle is $ 60$ cm. What is the area of the square?

It is. Did you use a $ 30\minus{}60\minus{}90$ triangle to get the radius and then a $ 45\minus{}45\minus{}90$ to  get the side of the square?

This is how I did it:

Using the formula $ a\equal{}rs$ where $ a \equal{} \text{area}$, $ r \equal{} \text{radius of inscribed circle}$ and $ s \equal{} \text{semi perimeter}$, we get $ 900\sqrt3 \equal{} 90r$ (I'm assuming you can find the area and semi perimeter yourself).  $ r \equal{} 10\sqrt3$.  That is the radius of the circle.  The diameter of the circle is then $ 20\sqrt3$, which is also the diagonal of the square.  Since we have the diagonal, we can figure out the sides of the square by the $ 45\minus{}45\minus{}90$ triangle properties, which we get to be $ 20\sqrt3 \times \frac{\sqrt2}{2} \equal{} 10\sqrt6$.  Squaring gives us an area of $ \boxed{600}$ of the square.

This is my way:
1.Equilateral triangle area=l^2*3^.5/4
2.inradius circle=area/s where s is semiperimeter
3.inradius=l*3^.5/6=10*3^.5
4.(square side)^2*2=4(inradius)^2
5.square area=4(inradius)^2/2


hello,
id like to ask how is it possible to draw a 3d outline of the text
like in this picture:  http://i35.tinypic.com/a15rae.jpg
im new to asymptote, pardon if its anything obvious
but have problems where to start,
will be grateful for any tips
best
zb

A start:
[code]
settings.render=0;
import three;
currentprojection=perspective(0,3,10,up=Y);

size(15cm);
path[] text = texpath(Label("reward",fontsize(50pt)));
draw(extrude(text,10*Z),white,black,nolight);
[/code]

thanks a lot!

if i can - 2 related questions: what is happening with some of the letters, they seem to be missing a part (vide 'e' in your example) is there anything there can be done about it?

how can i extrude (cut through) the text into another object, e.g. a cube with holes shaped as letters?

once again thanks for your help

zb

[code]
import three;
currentprojection=perspective(0,3,10,up=Y);

size(20cm);
path[] text = texpath(Label("reward",fontsize(50pt)));

draw(extrude(text,10*Z),white,black+1bp,nolight);

draw(surface(text),white,nolight);
draw(shift(10*Z)*surface(text),white,nolight);

[/code]

[asy]
settings.render=0;
import three;
currentprojection=perspective(0,3,10,up=Y);

size(15cm);
path[] text = texpath(Label("Don't forget [asy] at the start and end!",fontsize(50pt)));
draw(extrude(text,10*Z),white,black,nolight);
[/asy]


[quote="ernie"]Six bags of marbles contain $ 18$, $ 19$, $ 21$, $ 23$, $ 25$, and $ 34$ marbles, respectively.  One bag contains chipped marbles only.  The other $ 5$ bags contain no chipped marbles.  Jane takes three of the bags and George takes two of the others.  Only the bag of chipped marbles remains.  If Jane gets twice as many marbles as George, how many chipped marbles are there?

$ \textbf{(A)}\ 18 \qquad \textbf{(B)}\ 19 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 25$[/quote]
[hide]I can't think of any other way of doing this other than brute-forcing it.
With brute-forcing: Jane gets the marble bags: 34, 25, and 23, whereas George gets the marble bags: 21 and 18. Thus there are 19 chipped marbles.[/hide]

$ 18\plus{}19\plus{}21\plus{}23\plus{}25\plus{}34 \equal{} 140$. When one of these is taken away, what remains must be divisible by 3. 23 satisfies this condition.


Anyone help with this one?

(2+sqrt(5))^1/3 + (2-sqrt(5))^1/3 = x

Use $(a+b)^3=a^3+b^3+3ab(a+b)$, we have $x^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})}\ x $

$\Longleftrightarrow x^3+3x-4=0\Longleftrightarrow (x-1)(x^2+x+4)=0$.since $x^2+x+4=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0$, yielding the non zero real number is $x=1$.

[hide]
Let $p=\sqrt[3]{2+\sqrt{5}}$ and $q=\sqrt[3]{2-\sqrt{5}}$.

We have
$p^3+q^3=(2+\sqrt{5})+(2-\sqrt{5})=4$,
$p^3q^3=(2+\sqrt{5})(2-\sqrt{5})=-1\Longleftrightarrow pq=-1$

Then,
$(p+q)^3=p^3+q^3+3pq(p+q)$
$\longrightarrow x^3=4-3x$
and the rest is the same as kunny's solution.
[/hide]


counting
distinguishability
geometry
factorial
geometric transformation
rotation

You have twelve one-foot tiles to tile a three foot by four foot stoop. If nine tiles are white, two are black, and one is red, how many distinguishable tilings are possible?

how would you do this?

is there a fast method to do these kinds of problems? :ninja:

ehh... err... I hope I did this right. I am taking geometry this year and am currently working EXTRA hard on geometry, so my algebra is not as "ripe"

   12!/9!2!1! which is factorial

 12x11x10x9x8x7x6x5x4x3x2x1/9x8x7x6x5x4x2x1x1

cross out some numbers and cancel out

then get (12x11x10) / 2 which equals 660

[quote="lotrgreengrapes7926"]If rotations are different, that is the correct answer. Think of it as a long $1\times12$ grid, and each tile has a unique corresponding tile in the $3\times4$ grid. This becomes the same question as "how many ways are there to arrange 9 As, 2 Bs, and 1 C?" The answer is, of course, $\frac{12!}{9!2!1!}$. :D[/quote]
That's if you can't rotate them. However, I'm pretty sure you can.

Hmm...if you rotate it and it is the same, that means it must be colored in this fashion:
[code]
1  2  3  4
5  6  6  5
4  3  2  1
[/code]
With matching numbers being the same color. Of course, there is one red, so it cannot possibly be rotated and stay the same.

Is this reasoning correct?? :maybe:

Oh, they're square tiles! I was thinking of bricks. :blush:

In, that case, it's much easier. You both got it right.


Prove that $\sqrt{2}^{\sqrt{2}}$  is an irrational number .

What is your solution? Post a solution, please!  :wink:

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


What is the area for a Regular Hexagon with a side which length is x?
How can you prove it?

Its area is 6 times the area of the triangle.
[hide="Proof"]
The Hexagon can be split up into 6 equilateral triangles. Find the area of 1 of those triangles, and multiply by six to get the area.
[/hide]

[hide]use equilateral triangles to find $ a\equal{}\dfrac{1}{2}s\sqrt{3}$ where $ a$ is the apothem. now use the equation $ Area\equal{}\dfrac{1}{2}ap$ to find $ Area\equal{}\dfrac{1}{4}s\sqrt{3}\times6s$ or $ Area\equal{}\dfrac{3}{2}s^2\sqrt{3}$[/hide]

[hide]We can draw a line from the center to each vertex, thus creating 6 triangles. We can prove these are equilateral:Each angle at the center is $ \frac {360}{6} = 60^{\dot}$. Since each triangle is isosceles, the other angles are also $ 60^{\dot}$, so they are equilateral. Now, the side length of each triangle is $ x$;the area is then $ \frac {x^2\sqrt {3}}{4}$ (see if you can prove this by drawing a perpendicular), so the total are is 6 times that or $ \frac {3x^2\sqrt {3}}{2}$.[/hide]

It would just be 6*(sqrt(3))x^2/4, since (sqrt(3))x^2/4 is the area of each of the six equilateral triangles. So it would be (3sqrt(3)/2)x^2

any regular polygon with perimeter [i]P[/i] and an apothem [i]A[/i] has an area of 1/2*[i]PA[/i]..... an apothem is the distance from the center of the polygon to the midpoint of a side.... I think...


z|xy where x,y,z are all positive integers. x and y are both less than z.

I have found that either x|z or y|z.  Why is this?


Show dat there is no real constant c>0 such dat cos√(x+c)=cos√x for real no.s  x ≥ 0.

hello, what is with $ c\equal{}4\pi^2$ and $ x\equal{}0$?
Sonnhard.

It has to be true for all real $ x \ge 0$ though; the claim is that there exists no positive real constant $ c$ such that $ \cos \sqrt {x \plus{} c} \equal{} \cos \sqrt {x}$ for all $ x \ge 0$.

[hide="hint"]It's not a difficult indirect proof. You just need to plug in some nice values for x: 0 and c.[/hide]


[quote="thanhak5"]1. For $ a,b,c > \frac {1}{\sqrt {6}}$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
. Prove that
        $ \frac {a^2 \plus{} 1}{\sqrt {2a^{2} \plus{} 3ab \minus{} c^2}} \plus{} \frac {b^2 \plus{} 1}{\sqrt {2b^{2} \plus{} 3ab \minus{} a^2}} \plus{} \frac {c^2 \plus{} 1}{\sqrt {2c^{2} \plus{} 3ac \minus{} b^2}}\geq 2(a \plus{} b \plus{} c)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155354


In isosceles triangle $ABC$, $A$ is located at the origin and $B$ is located at $(20, 0)$.  Point $C$ is in the first quadrant with $AC = BC$ and $\angle BAC = 75^\circ$.  If $\triangle ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive y-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2}+q\sqrt{3}+r\sqrt{6}+s$ where $p$, $q$, $r$, $s$ are integers.  Find $(p-q+r-s)/2$.

Yeah, I know, I was just posting them for the resources section.  I need to add a 2007 year to the AIME, and I don't feel like experimenting with it right now, so if another RManager wants to add the problems, go ahead, or else I will do it some time this week.


This year in Iowa City I couldn't find many good places to eat.  We went to a nice italian place, but it was waaay too expensive (plus splitting a tip 8 ways became confusing, even with graphing calculators).

Next year, I don't want to repeat this.  The yellow pages are useless because I don't know where anything is, and they don't tell me how good the food is.  And I think the locals all speak Iowan too.

So what are some good places to eat?  If you were in Vegas or Penn State you can post too, just say specifically that's where it is :).

there was a food festival thats really good. i dont know if they have it every year since it was my first year.

at vegas we ate at in n out and at the campus food services. Campus food is alright and in n out is as lame as it always is.

The Minnesota team always eats together at a place called the Brown Bottle several blocks from Burge.  It kind of sounds like that might be the place you tried, though.  Aside from that, I have no idea what's good, since that's the only place I've ever eaten besides the university dining hall.  In fact, the whole "going off on your own for dinner" concept is a bit foreign to me, as neither of the teams I've traveled with have done any of that while on-site.

I can't really say what's good in Vegas, either, since as ShoeFactory said, all our meals there were at the dining hall.

At the Penn State site, go into town and you'll see about 999 great restaurants...PHILLY CHEESESTEAK.  Also, don't miss the Creamery....ask anyone about it.

I'll second the creamery- though recently the creamery changed location.  Also, you should go to Mio Zio- they have the best plastic forks I have ever seen.  There's also a pretty good sub shop whose name I forget.

And, FYI, Penn State in not in Philadelphia- it's several hours away.

yeah, the plastic forks at that italian place were amazing.  :D and the calzones there were good.

aargh i hate calzones

if i were u jus go to the closest restraunt wit a lot of cars in the parking lot
if it has a lot of cars parked that means that there r a lot of customers
if a lot of customers come there, its good

QED

Sorry someone told me it was in philadelphia  :oops: .   I'll fix that :)


How would you solve ;
$\sum_{m}\sum_{n}n^{-m}$
 but without changing the order of sumation
(i hope you understand what i mean) :lol:

The sum equals $\infty$, since

$S=\sum_{m=2}^{\infty}\sum_{n=1}^{\infty}n^{-m}=\sum_{m=2}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n^{m}}=\sum_{m=2}^{\infty}\zeta(m)=\infty$
since $\zeta(m)\rightarrow 1$ as $m \rightarrow \infty$.

[quote="Kent Merryfield"]If the sum on $n$ were to start at $n=2,$ then it would be a different problem. (Mr. Doe should have been more explicit.)[/quote]

Yes ,that is right!!
Ok,if we have
$\sum_{m,n\geq 2}n^{-m}$.Then first we do $\frac{1}{n^{2}}(1+\frac{1}{n}+\frac{1}{n^{2}}+\dots )=\frac{1}{n(n-1)}$
After that, we have ,after changing the order of sumation :
$\sum_{k=0}^{\infty}\frac{1}{(k+1)(k+2)}=1$
(please correct me ,if Im wrong)
But I interested how would we solve it if we didnt change the order of sumation.If anyone has any bright ideas please share it. :lol:

If we don't change the order of summation, we can only calculate the inner sum at all half the time. The exact value of $\zeta(n)$ for odd $n$ is an open question.
Interchanging the order of summation is the [b]only[/b] way to solve this problem.

Ok,lets leave this problem behind .

Let $S: =\{a^{b}~|~ a,b\in \mathbf{N}\backslash \{1\}\}$ Find :
$\sum_{n\in S}\frac{1}{n-1}$

First I did:
$\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{n^{k}-1}=$
$-\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{1-n}\frac{1}{1+n+n^{2}+\dots+n^{k-1}}=$
$-\sum_{n=2}^{\infty}\frac{1}{1-n}\sum_{k=2}^{\infty}\frac{1}{1+n+n^{2}+\dots+n^{k-1}}$ .But I am stuck here and I have no idea how to solve it (although it was a pretty good start :lol: ).

You've missed a subtlety- your iterated sum is not the same as the original. We only count each perfect power once, but your version counts terms such as $\frac1{64-1}$ multiple times.
This problem was previously discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=36404]here[/url].


[quote="frt"]If $a$, $b$, $m$, $n$ are positive numbers and $m+n=1$, prove that
\[\sqrt{ma+nb}\ge m\sqrt{a}+n\sqrt{b}\]
[/quote]

[hide="uhh..."]jensen?[/hide]

[quote="frt"]If $a$, $b$, $m$, $n$ are positive numbers and $m+n=1$, prove that
\[\sqrt{ma+nb}\ge m\sqrt{a}+n\sqrt{b}\]
[/quote]

I kind of bruteforced it but it's not that bad...

[hide]$a+b\ge 2\sqrt{ab}$

$man+nbm\ge 2mn\sqrt{ab}$

$ma(1-m)+nb(1-n)\ge 2mn\sqrt{ab}$

$ma-m^{2}a+nb-n^{2}b\ge 2mn\sqrt{ab}$

$ma+nb\ge m^{2}a+n^{2}b+2mn\sqrt{ab}$

$ma+nb\ge (m\sqrt{a}+n\sqrt{b})^{2}$

$\sqrt{ma+nb}\ge m\sqrt{a}+n\sqrt{b}$

...yeah.[/hide]

[quote="scorpius119"][hide="uberhint"]
the LHS is $\sqrt{(ma+nb)(m+n)}$
[/hide][/quote]
wow nice
[hide="to finish off, even though there isnt much to finish off"]
By cauchy,
$(ma+nb)(m+n)\ge(m\sqrt a+n\sqrt b)^{2}$
$\sqrt{(ma+nb)(m+n)}\ge m\sqrt a+n\sqrt b$
$\sqrt{(ma+nb)}\ge m\sqrt a+n\sqrt b$
[/hide]


So I know I haven't been visiting AoPS much lately, but I just have to ask about this.  Besides Edward and Dickerson, who won? :P

gj to all who participated, and good luck at nats!

Edward of course won first place, but surprisingly, Dickerson was 5th place. First place team was Blackmill Road or something like that.

My team got 6th and I got second at CD. We don't know the scores or places. All in all, good fun was had by all!

Question, did anyone from RTMS place? o.o

[quote="weird888"]what was the winning score??[/quote]
The general policy on AoPS is that you may not discuss scores until all states have had their competition. However, you may talk about what places certain individuals or teams got.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261916]No score discussion![/url]

But Georgia tends to have winning scores a few points less than every other state, since we have state at the zoo in a tent.

Who made Nats for the Georgia team? Just curious.... :D

@Arvind- David Xing was from RTMS and he placed 5th.

@hm29168- I sat next to you during CD.  :P

Nationals is coming up soon enough; good luck to everyone who made it :D

It's Blackmon Road. It's a school in Columbus.

How'd the Georgia team do at nats?  I know that Bobby won cd and David Yang (that very rude brat) won written, but how did Georgia's people place?

bobby shen won. w00t. XD

what happened to darryl wu? the one with the chewy thing? o.o

hmmm....I was sitting between Mike and Edward

5th!!!!!!!!!!!!!!! That's really messed up  :wallbash_red:

EDIT: Wow I'm retarded, this post was from a year ago....no wonder you guys were talking about different people rofl


[quote="hien"]Let $a,b,c$ be positive integer numbers and assume $a=max\{a,b,c\}$. Find minimum of ${\frac{a}{b}+2\sqrt{1+\frac{b}{c}}}+3\sqrt[3]{1+\frac{c}{a}}$[/quote]
The fact that $\lbrace a,b,c\rbrace \in \mathbb{Z_{+}}\implies a = 3,\,\, b = 2,\,\, \text{and }c = 1 \text{ or }a = 3,\,\, b = 1, \text{ and }c=2.$

$\frac{a}{b}+2\sqrt{1+\frac{b}{c}}+3\sqrt[3]{1+\frac{c}{a}}=\begin{cases}\frac{3}{2}+2\sqrt{1+\frac{2}{1}}+3\sqrt[3]{1+\frac{1}{3}}\text{ or}\\ \frac{3}{1}+2\sqrt{1+\frac{1}{2}}+3\sqrt[3]{1+\frac{2}{3}}\end{cases}$

Evaluating these expressions with a calculator reveals that the former is minimum.

For your information, dear "teacher", if $a=b=c$ we obtain even smaller value.

And that 1-2-3 business is particularly ingenious...

[quote="Farenhajt"]For your information, dear "teacher", if $a=b=c$ we obtain even smaller value.[/quote]
True, but finding the minimum of the given function when $\lbrace a,b,c \rbrace = \lbrace a\rbrace$ is rather trivial, don't you think?

[quote="Farenhajt"]And that 1-2-3 business is particularly ingenious...[/quote]

Thank you.


[quote="José"][quote="most-wanted"]i got a simple equation
so here it is.

8 computers have to  work for 20  hours and have download about 150 GB of data.

after 2 hours one computer would be damage.

how long would it takes for the rest.
and how many new computer should be add on, that the same amount of data should be download on exact time.[/quote]

Does all of them download the same bytes in the same time???[/quote]

yeah there are

[hide]Is that 133.125?[/hide] :roll:

Math is my life :!:

[quote="most-wanted"]i got a simple equation
so here it is.

8 computers have to  work for 20  hours and have download about 150 GB of data.

after 2 hours one computer would be damage.

how long would it takes for the rest.
and how many new computer should be add on, that the same amount of data should be download on exact time.[/quote]

1)Each computer download $\frac{15}{16}$ GB per hour. Then, after $2$ hours, all computers have download $15$ GB. Then, we have $135$ GB for $7$ PC's. Then, you need $\frac{144}{7}$ hours to download all.

2)You need $8$ PC's... (think why)


Show that $(k^3)!$ is divisible by $(k!)^{k^2+k+1}$.

First let's prove a lemma :

[b]Lemma[/b] : For positive integers $m,n$ $\dfrac{(mn)!}{(m!)^nn!}$ is a positive integer.

Proof:

$\dfrac{(mn)!}{(m!)^n}={2m \choose m}{3m \choose m}...{mn \choose m}$

In addition we know that  for positive integers $m,k$ ${mk \choose m}=k{mk-1 \choose m-1}$ thus
$\dfrac{(mn)!}{(m!)^n}=n!{2m-1 \choose m-1}{3m-1 \choose m-1}...{mn-1 \choose m-1}$ so simply
$\dfrac{(mn)!}{(m!)^nn!}={2m-1 \choose m-1}{3m-1 \choose m-1}...{mn-1 \choose m-1}$, q.e.d.

So using the lemma twice we know that $\dfrac{(k^2*k)!}{(k^2!)^kk!}\dfrac{(k^2!)^k}{(k!)^{k^2}(k!)^k}=\dfrac{(k^3)!}{(k!)^{k^2+k+1}}$ is a positive integer...


Prove that for any set of real numbers x_i (i=1,..,n)

  sum(i,k=1,..,n)cos(x_i-x_k) \geq 0

sum(i,k=1,..,n)cos(x_i-x_k) = sum (i < j) (2 + 2*cos (xi-xj))

And 2 + 2*cos (xi-xj) >= 0. Is that it ?


$ \sqrt{x\plus{}5}\equal{}x^2\minus{}5$
(Solve the equation step by step for real numbers)

Putting $ t\equal{}\sqrt{x\plus{}5}$, we get $ (t^2\minus{}5)^2\minus{}5\equal{}t$. If $ f(t)\equal{}t^2\minus{}5$, then $ f(f(t))\equal{}t$. Since $ t\geqslant 0$, the function is strictly increasing, hence $ f(f(t))\equal{}t\iff f(t)\equal{}t\iff t^2\minus{}t\minus{}5\equal{}0\iff t\equal{}{1\plus{}\sqrt{21}\over 2}$ (negative solution is discarded). That gives $ x\equal{}t^2\minus{}5\equal{}t\equal{}{1\plus{}\sqrt{21}\over 2}$

Let $ t\equal{}\sqrt{x\plus{}5}$. So $ x\equal{}t^2\minus{}5$ and $ t \ge 0$. We have
$ t\equal{}x^2\minus{}5$
$ x\equal{}t^2\minus{}5$
Substract the second equation from the first to get $ t\minus{}x\equal{}(x\minus{}t)(x\plus{}t)$ or $ (x\minus{}t)(x\plus{}t\plus{}1)\equal{}0$.
When $ t\equal{}x$ we obtain $ x^2\minus{}x\minus{}5\equal{}0$ or $ x\equal{}\frac{1\plus{}\sqrt{21}}{2}$.
When $ t\equal{}\minus{}(x\plus{}1)$ we obtain $ x^2\plus{}x\minus{}4\equal{}0$ or $ x\equal{}\minus{}\frac{1\plus{}\sqrt{17}}{2}$.


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let be given a cyclic quadrilateral $ ABCD.$ Let  $ H_{1},\,H_{2},\,H_{3},\,H_{4}$ be the orthocentre of triangles $ BCD,\, CAD,\, ABD$ and $ CAB,$ respectively.

$ 1.$ Prove that $ H_{1},\,H_{2},\,H_{3},\,H_{4}$ are concyclic.
$ 2.$ Prove that $ AH_{1},\,BH_{2},\,CH_{3},\,DH_{4}$ are concurrent.

1. O is a circumcenter and M is a midpoint of BC. $ AH_{4}=2OM=DH_{1}$ and$ AH_{4}$ is parrallel to $ DH_{1}$. So AD is parallel to $ H_{1}H_{4}$. Therefore each side of$ H_{1}H_{2}H_{3}H_{4}$ is parrallel to each side of ABCD. so proven

2.midpoint of $ AH_{1}$=minpoint of$ BH_{2}$=minpoint of$ CH_{3}$=midpoint of $ DH_{4}$. So proven.


logarithms
real analysis
real analysis unsolved

hi all,

let a>0 a real. Does the series Sum( 1/(ln(n)+2*cos(a*n)),k=7..infinity) converge ?

hi all,

Let $a>0$ be a real. Does the series $\sum_{n=7}^{\infty}\frac{(-1)^n}{\ln n+2\cos(an)}$ converge?

[corrected, see alekk's clarification]

:D  sorry .. it is:
Sum( (-1)^n/(ln(n)+2*cos(a*n)),k=7..infinity)

For example, for $a=\pi/2$ series diverges.

my teacher tell me it was quite hard to find those "a" such that the series converges.

I have already felt it. Do you think your teacher solved this problem?

P.S. Take a look to http://www.mathlinks.ro/Forum/viewtopic.php?p=130102 for grobber's problem.


inequalities
quadratics
algebra
polynomial
inequalities unsolved

Prove that for all x,y,z>=0 and u,v>=1 we have
        sum_{cyc_(x,y,z)}((x-u*y)*(x-u*z)*(x-v*y)*(x-v*z))>=0.

[quote="fjwxcsl"]Prove that for all $x,y,z\geq 0$ and $u,v\geq 1$ we have
        $\sum_{cyc_{(}x,y,z)}\((x-u*y)*(x-u*z)*(x-v*y)*(x-v*z))\geq 0$.[/quote]

Putting $p=\frac{u+v}2$ and $q=\frac{u-v}2$, we obtain a quadratic polynomial in $q^{2}$. It's determinant must be negative, so it is sufficient (in fact equivalent) to prove $\forall\ p\ge 1$
$\ \left(2(1+p^{2})\left(\sum x^{2}y^{2}\right)-4pxyz(x+y+z)\right)^{2}\le \left(\sum x^{2}y^{2}\right)\left(\sum (z-px)^{2}(z-py)^{2}\right)$. But what now? :wallbash: CS isn't strong enough... Is there a nicer way than bunching? :oops:


Y DID I DO TAT

WHO STARTED IT

AAARGGGHHH I DONT KNO

Do you even read the freaking definition of spam? Do you remember this forum being unable to access for 24 hours and all post count inside removed? People like you being stupid when creating this topics are the cause of all this.
[size=9][color=white]And to all the idiots who can't read sarcasm, this is an example of it.[/color][/size]


Given that $ x,y,z,t$≥0.Prove the ineq:
$ 3(x^2\plus{}y^2\plus{}z^2\plus{}t^2)\plus{}4\sqrt{xyzt}$≥$ (x\plus{}y\plus{}z\plus{}t)^2$

[quote="SUPERMAN2"][quote="quykhtn-qa1"][quote="SUPERMAN2"]Given that $ x,y,z,t$≥0.Prove the ineq:
$ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt}$≥$ (x \plus{} y \plus{} z \plus{} t)^2$[/quote]
This inequality by  Suranyi :)[/quote]
Could you post the solution please?[/quote]
I will post solution but now I'm very busy,sorry :)

Anyone help me?I really need a solution.Thanks.

[quote="SUPERMAN2"]Given that $ x,y,z,t$≥0.Prove the ineq:
$ 3(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \plus{} 4\sqrt {xyzt}$≥$ (x \plus{} y \plus{} z \plus{} t)^2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6397
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204175


Find the greatest prime on the smallest aritmethic progression of seven primes.


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Of the numbers $\displaystyle 3^{\frac{1}{3}}, 4^{\frac{1}{4}}, 5^{\frac{1}{5}}, 6^{\frac{1}{6}}$ which is greatest?

The biggest one is[hide] 3^(1/3).  If you take 2^(1/2) and 4^(1/4), they are equal.

4=2^2

4^(1/4)=(2^2)^(1/4)=2^(2/4)=2^(1/2)

If you look at 1^(1/1) and 2(1/2), 2^(1/2) is bigger.  2^(1/2) is about 1.414, and 1^1=1.  This is a rough curve, which means 5^(1/5) is smaller than 3^(1/3), and, likewise, 6^1/6 is also smaller.  Therefore, 3^(1/3) is the biggest number.[/hide]


As an application of the aldol reaction, devise a method for the following synthesis:

Methylcyclohexene $\longrightarrow$ 1-cyclopentenyl methyl ketone.

yes ozonolysis and aldol should do.
And btw Carcul could you tell me what aromatic thallation is?  :D

Yes, that's right, although a detailed explanation of the synthesis should be provided in a reasonable answer.

What do you think it is?

[quote="Carcul"]Very good explanation. Your lucky is that I am not your teacher.[/quote]

Was that sarcastic like SSSNVR (our english teacher  :D  S for sarcastic) ?
Do you want me to explain in detail with the mechanism fully?

first perform Ozonolysis.We get a product containing an aldehyde and a ketone..
The Base extracts the B-Hydrogen from the ketone rather than the aldehyde.
This is because if it were to extraxt from the aldehyde ,ketone being unreactive, the reaction would have no process. So B-hydrogen of the ketone is abstacted ad ntramolecular aldol condensation takes place..

That explanation is much better. However, you cannot say "ketone being unreactive".

ok ok i get it  :D  Yes ketone is less reactive  :)

[quote="Carcul"]Very good explanation. Your lucky is that I am not your teacher.[/quote]
what would have been the consequences had u been our teacher :)

u mean a Nazi Concentration Camp  :D
Actually the best part is the inverse happens here in our school atleast in most cases...the teacher is not allowed to go out until he explains in detail what happens the mechanism....actually it's due to our sylabbi which focusses more on how much u know than on what and how well u know :rotfl:

[quote="Carcul"]You would not leave classroom until you explain to me in detail what reactions happen an why (of course, explaining that to me in the board by writting the mechanisms and speaking). Depending on my mood, extra exercises in the classroom, extra homework, or a surprise exame in the next few days could also be considered.[/quote]

I wouldnt mind a surprise test. But surely if u set it i would probably not even scrape through  :oops:
Yes as pardesi said our system looks @ more knowledge than details  :rotfl:  And btw are you, Carcul sir :D a teacher in a college or a school?

that's true pardesi.......... :D  :D  :D

[quote="pardesi"]i would like calling him Carcul Sir ....i have learnt more organic from him than i have at school :)[/quote]

Absolutely spot on  :D

[quote="Pardesi"]u mean a Nazi Concentration Camp[/quote]

Yes, something like that, but without the wired fence.  :lol:

Yeah I would also prefer calling him carcul sir


Let $ \triangle ABC$ with circumcircle $ \odot O$, $ DB$ and $ DC$ is the tangent line.

$ AE \perp AD$ and $ AE \cap BC \equal{} E$.

$ B'$ and $ C'$ are on $ ED$ satisfying that $ |B'D|\equal{}|BD|$, $ |C'D|\equal{}|CD|$.

Prove that:

$ \triangle ABC \sim \triangle AB'C'$

Whoa, this problem is a nice one!
I have solved it just a minute before and I can say that I like it :)