geometry
circumcircle
incenter
cyclic quadrilateral
power of a point
radical axis
geometry proposed

[quote="Virgil Nicula"]Given are $ \triangle ABC$ and a point $ P$ . Denote the projections $ D$ , $ E$ , $ F$ of the point  $ P$ to $ BC$ , $ CA$ , $ AB$ respectively

and the point $ R\in EF\cap BC$ . The circumcircle of $ \triangle DEF$ cut again the line  $ BC$ in $ M$. Prove that $ RP\perp AM$.

[b]Particular cases.[/b]

$ 1\ \odot\ P: \equal{} H$ ([u]ortocenter[/u] of $ \triangle ABC$). In this case the point $ M$ is the midpoint of $ [BC]$  $ \Longrightarrow\ RH\perp AM$ .

$ 2\ \odot\ \ P: \equal{} I$ ([u]incenter[/u] of $ \triangle ABC$). In this case the point $ M$ is $ D\ \Longrightarrow\ RI\perp AD$.[/quote]

[img]http://img97.imageshack.us/img97/9950/48732798.png[/img]

The line through point $ A$ pendicular to $ RP$ at $ K$ and cut $ BC$ at $ M'$

It's easy to see that $ FPKE$ and $ DPKM'$ are concyclic .

$ \Longrightarrow RF.RE \equal{} RD.RM'$  $ ( \equal{} RP.RK)$

$ \Longrightarrow EFDM'$ is concyclic

It means $ M' \equiv M$

Let be the point $ M\equiv BC\cap (O),$ where $ (O)$ is the circumcircle of the triangle $ \bigtriangleup DEF$ and we denote as $ K,$ the point of intersection of the line segment $ AM,$ from the circumcircle $ (O_{1}),$ of the cyclic quadrilateral $ AEPF.$

So, we have $ \angle AKP \equal{} 180^{o} \minus{} \angle AFP \equal{} 90^{o}$ $ ,(1)$

Hence, from $ \angle PDM \plus{} \angle PKM \equal{} 180^{o}$ $ ,(2)$ we conclude that the quadrilateral $ PKMD$ is cyclic, with circucircle so be it $ (O_{2}).$

We conclude now that the line segment $ PK,$ as the radical axis of the circles $ (O_{1}),\ (O_{2}),$ passes through the point $ R\equiv BC\cap EF,$ as the radical center of the circles $ (O),\ (O_{1}),\ (O_{2})$ and the proof is completed.

Kostas Vittas.


may be you can help me.i want to send my messages as stikies,what should i do ?please tell me if you know how... :?:

Only moderators can turn a post into a sticky or announcement.

In general we want to keep the forums clear of announcements and sticky posts except those that are essential to the forums.  Otherwise they just become clutter.

Yes, MCrawford is right.

I'm sort of confused of announcements and stickies: sometimes I see forum guidelines as stickies and some as announcements. Which one is more correct?

It doesn't matter all that much.  Announcements are higher priority as they remain at the tops of the forums.

I'll eventually standardize the forum guidelines, but that will likely be after I'm done writing the AoPS Intro Number Theory book, so perhaps in October.


quadratics
algebra
polynomial
Vieta
geometry
3D geometry
system of equations

[quote="boxedexe"]Express $x^{5}+y^{5}=11$ in terms of elementary symmetric polynomials.  In our case, we'll only need to use $\sigma_{1}=x+y$ and $\sigma_{2}=xy$.  Since you know $x+y=1$, you have a quadratic in $xy$.  Find all possible values of $xy$. You can then use Vieta or normal substitution to solve the possible systems for $x,y\in\mathbb{C}$.[/quote]
What exactly is a vieta substitution?

let x^5+y^5= (x+y)(x^4-x³y+x²y²-xy³+y^4) in order to simplify

Now we know x+y is one, so  x^5+y^5= (x^4-x³y+x²y²-xy³+y^4)

Lets add x²y² and subtract x²y² from the equation to pair up some quadruples. Generally, do the same for cubes and squares
Now its x^4-x³y+[b]2x²y²[/b]-xy³+y^4[b]-x²y²[/b]
When we factor, we get (x²+y²)²-x³y-xy³-x²y²
Combine and manipulate= (x²+y²)²-xy(x²+y²+2xy-xy)
=(x²+y²)²-xy(x+y)²+2x²y²
Now change (x²+y²)² to (x²+y²+2xy-2xy)² and then to [(x+y)²-2xy]².

The RHS expression now reads: [(x+y)²-2xy]²-xy+x²y²
We know that x+y=1 so after some plugging and chugging, we get:
(1-2xy)²-xy+2x²y²

Multiply out the first variable and get: 1-4xy+4x²y²-xy+x²y²
Simplifying, we get: 1-5xy+5x²y²=x^5+y^5
To make it even easier, make xy=z
Now it reads 5z²-5z+1=11
5z²-5z-9=0
now solve the system of equations
x+y=1
xy=(5+root5)/10

Wait a minute... the answer is imaginary or am I wrong. Plz someone check my work.


This is - as far as I know - the only Belgian problem that ever made the IMO paper. Her it goes :

Let ABC be an equilateral triangle and E the set of all points contained in the three segments AB, BC and CA (including A, B and C). Determine whether, for every partition of E into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.

And I really like it !

Nice problem

I've been working the IMOs and just encountered it a few days ago, so I won't post a solution...

Here's a hint tho: [hide] Choose the vertices of a regular hexagon such that each each side of the triangle is trisected by them. Then what can you say about the colors of the points at the vertices?[/hide]


function
parameterization
algebra unsolved
algebra

Let a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfy \[ f(x^n\minus{}y^n)\equal{}(x\minus{}y)\left[ f(x)^{n\minus{}1}\plus{}f(x)^{n\minus{}2}f(y)\plus{}\cdots\plus{}f(x)f(y)^{n\minus{}2}\plus{}f(y)^{n\minus{}1} \right].\] Prove that $ f(rx)\equal{}rf(x)$ for all rational $ r$ and all real $ x$.

[quote="james digol"]Let a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfy
\[ f(x^n \minus{} y^n) \equal{} (x \minus{} y)\left[ f(x)^{n \minus{} 1} \plus{} f(x)^{n \minus{} 2}f(y) \plus{} \cdots \plus{} f(x)f(y)^{n \minus{} 2} \plus{} f(y)^{n \minus{} 1} \right].\]
Prove that $ f(rx) \equal{} rf(x)$ for all rational $ r$ and all real $ x$.[/quote]

I suppose $ n$ is a parameter $ \in\mathbb N$

Let $ P(x,y)$ be the assertion $ f(x^n\minus{}y^n)\equal{}(x\minus{}y)\sum_{k\equal{}0}^{n\minus{}1}f(y)^kf(x)^{n\minus{}1\minus{}k}$
Let $ Q(z)$ be the assertion $ f(zx)\equal{}zf(x)$ $ \forall x\in\mathbb R$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,0)$ $ \implies$ $ f(x^n)\equal{}xf(x)^{n\minus{}1}$
$ P(0,x)$ $ \implies$ $ f(\minus{}x^n)\equal{}\minus{}xf(x)^{n\minus{}1}$

So (comparing the two last lines) : $ f(\minus{}x)\equal{}\minus{}f(x)$

1) $ n$ even
===========
Multiplying both sides of $ P(x,y)$ by $ f(x)\minus{}f(y)$, we get $ R(x,y)$ : $ (f(x)\minus{}f(y))f(x^n\minus{}y^n)\equal{}(x\minus{}y)(f(x)^n\minus{}f(y)^n)$
Consider $ x,y\ne 0$ such that $ |x|\ne|y|$ and $ |f(x)|\ne |f(y)|$
Then, dividing $ R(x,y)$ by $ R(x,\minus{}y)$, we get $ \frac {f(x)\minus{}f(y)}{f(x)\plus{}f(y)}\equal{}\frac{x\minus{}y}{x\plus{}y}$ and so $ \frac{f(x)}x\equal{}\frac{f(y)}y$
And it is easy to get rid of the restrictions $ |f(x)|\ne |f(y)|$ and to conclude $ f(x)\equal{}xf(1)$ $ \forall x$
And so $ Q(z)$ true $ \forall z$
Q.E.D.

2) $ n$ odd
========
$ f(\minus{}x)\equal{}\minus{}f(x)$ $ \implies$ $ Q(z)\implies Q(\minus{}z)$

Suppose now $ Q(z)$ :
$ P(x,zx)$ $ \implies$ $ f((1\minus{}z^n)x^n)\equal{}(x\minus{}zx)\sum_{k\equal{}0}^{n\minus{}1}z^kf(x)^{n\minus{}1}$ $ \equal{}xf(x)^{n\minus{}1}(1\minus{}k^n)$ $ \equal{}(1\minus{}k^n)f(x^n)$
So $ f((1\minus{}z^n)x)\equal{}(1\minus{}z^n)f(x)$ $ \forall x\ge 0$ and, since $ f(\minus{}x)\equal{}\minus{}f(x)$ : $ f((1\minus{}z^n)x)\equal{}(1\minus{}z^n)f(x)$ $ \forall x$
So $ Q(z)$ $ \implies$ $ Q(1\minus{}z^n)$

Suppose now $ Q(z)$ :
$ P(|z|^{\frac 1n}x,0)$ $ \implies$ $ f(|z|x^n)\equal{}|z|^{\frac 1n}xf(|z|^{\frac 1n}x)^{n\minus{}1}$
Since $ Q(z)$ $ \implies$ $ Q(|z|)$, this last equlity becomes $ |z|f(x^n)\equal{}|z|^{\frac 1n}xf(|z|^{\frac 1n}x)^{n\minus{}1}$
And so $ |z|xf(x)^{n\minus{}1}\equal{}|z|^{\frac 1n}xf(|z|^{\frac 1n}x)^{n\minus{}1}$
And so $ f(|z|^{\frac 1n}x)^{n\minus{}1}\equal{}(|z|^{\frac 1n}f(x))^{n\minus{}1}$
Since $ n$ odd, $ f(x^n)\equal{}xf(x)^{n\minus{}1}$ shows that $ f(x)$ and $ x$ has same signs and so $ f(|z|^{\frac 1n}x)\equal{}|z|^{\frac 1n}f(x))$ and so $ Q(|z|^{\frac 1n})$

So we know :
T1 : $ Q(z)\implies Q(\minus{}z)$
T2 : $ Q(z)\implies Q(|z|^{\frac 1n})$
T3 : $ Q(z)\implies Q(1\minus{}z^n)$

From this, it is easy to conclude :
For any $ z>0$ : $ Q(z)$ $ \implies$ $ Q(z^{\frac 1n})$ $ \implies$ $ Q(\minus{}z^{\frac 1n})$ $ \implies$ $ Q(1\minus{}(\minus{}z^{\frac 1n})^n)$ and so $ Q(z\plus{}1)$
And, since $ Q(1)$ is true, $ Q(n)$ is true $ \forall n\in\mathbb N$ and so $ Q(n)$ is true $ \forall n\in\mathbb Z$

And it's immediate to conclude $ Q(z)$ true $ \forall z\in\mathbb Q$
Q.E.D.


This sum was given to me by a friend. He had an answer but wanted to be sure. Can you try this so that we can verify the answer?

Suppose that f is a function defined on the real numbers for which f(a+b) = f(a) + f(b) + ab
for all a,b. What can f be?

[hide="Solution"]
Note that $ f(n \plus{} 1) \equal{} f(n) \plus{} f(1) \plus{} n \implies f(n \plus{} 1) \minus{} f(n) \equal{} f(1) \plus{} n$

With the finite difference formula, we can determine by summing consecutive differences that $ f(n \plus{} m) \minus{} f(n) \equal{} mf(1) \plus{} mn \plus{} \frac {m(m \minus{} 1)}{2}$

However, from the original formula, we know that $ f(n \plus{} m) \minus{} f(n) \equal{} f(m) \plus{} mn$

Equating the two, we have $ f(m) \plus{} mn \equal{} mf(1) \plus{} mn \plus{} \frac {m(m \minus{} 1)}{2} \implies f(m) \equal{} \frac {m^2}{2} \plus{} \left( f(1) \minus{} \frac {1}{2} \right) m$

$ f(1)$ is a free variable here, so we replace it with $ C \plus{} \frac {1}{2}$. This gives the necessary condition $ f(x) \equal{} \frac {x^2}{2} \plus{} Cx$.

We return test this form in the original formula, and find that this form of $ f$ is also a sufficient condition: ie, there are no restrictions on $ C$. Thus, $ \boxed{f(x) \equal{} \frac {x^2}{2} \plus{} Cx \; , \; \; C \in \mathbb{R}}$.
[/hide]

[quote="smokinjoe"][hide]my ans is f(x)=f'(0)*x + x^2/2
method...its not std but here goes
the imp point is tht the 2 variables a and b are independent of each other...i.e with respect to 1 the other is a constant

so d(a)/db=0
diff the given eqn wrt b
v get:
f'(a+b)=f'(b)+a
now put b=0
v get f'(a)=f'(0) + a
integrate
f(a)=a*f'(0) + (a^2)/2 + c(c=0 easy to show)
f(a)=a*f'(0) + (a^2)/2
the function cud be known exactly if f'(0) is known but i dun think it is[/hide][/quote]The problem with this solution is that you assumed f is differentiable, which is not given in the problem. And no calculus in Intermediate Topics ;)


I've gone over and thoroughly done problems in chapters 1-8 only about 10000 times, but I just don't have a knack for algebra and number theory. Usually I won't remember concepts for over a week...How do you guys work on stuff like this?

i guess i have a knack for algebra :rotfl:

what you should do is study more practice problems on saab's website

then you will get a knack for algebra

Try switching chapters every once in a while. i.e. Do some algebra then geometry then number theory then back to algebra then maybe some exponents and then some more stuff.


[color=blue] [size=150]$6^{th}$  Iberoamerican Olympiad[/size]
Córdoba, ARGENTINA. [1991]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1. [/b] A cada vértice de un cubo se asigna el valor de +1 o -1, y a cada cara el producto de los valores asignados a cada vértice. ¿Qué valores puede tomar la suma de los 14 números así obtenidos?.[/quote]
Al cambiar el número de un vértice se modifican 4 valores, el vértice que se cambió y las tres caras que comparten dicho vértice, $\lambda_1,\lambda_2,\lambda_3,\lambda_4.$ Si $\lambda_1=\lambda_2=\lambda_3=\lambda_4,$ la nueva suma $S'$ es $S \pm8.$ Si $\lambda_1=\lambda_2=\lambda_3$ y $\lambda_1=-\lambda_4,$ la nueva suma $S'$ es $S \pm 4.$ Si $\lambda_1=\lambda_2$ y $\lambda_3=\lambda_4=-\lambda_1,$ la nueva suma $S'$ es $S.$

Como toda asignación puede obtenerse a partir de una dada mediante sucesivos cambios en un vértice y $-14 \le S \le 14,$ entonces los valores posibles de $S$ son $14, 10, 6, 2, - 2, - 6, - 10$ y $-14.$

No se puede obtener -14 pues no se puede lograr que los 14 números sean negativos, ya que si los 8 vértices son negativos las caras son positivas. No se puede obtener 10 pues para ello debe haber 12 positivos y 2 negativos. Con un vértice negativo, S = 6 y con dos vértices negativos hay además dos caras (por lo menos) negativas, que son las que tienen a uno de los vértices pero no al otro.

[quote][b]Problema 2.[/b] Dos rectas perpendiculares dividen un cuadrado en cuatro partes, tres de las cuales tienen cada una área igual a 1. Demostrar que el área del cuadrado es cuatro.[/quote]
Sea $P_1P_2P_3P_4$ el cuadrado dado. $M$ es un punto interior a éste por el cual pasa una recta que corta a $\overline{P_1P_2}$ y $\overline{P_3P_4}$ en $P,T$ y otra recta perpendicular a la anterior que corta a $\overline{P_2P_3}$ y $\overline{P_4P_1}$ en $S,U.$ $O$ es el centro de $P_1P_2P_3P_4.$ La paralela por $O$ a $PT$ corta a $\overline{P_1P_2}$ y $\overline{P_3P_4}$  en $E,G$ y la paralela a $SU$ por $O$ corta a $\overline{P_2P_3}$ y $\overline{P_4P_1}$ en $F,H.$ Sean $X \equiv EG \cap SU$ y $Y \equiv FH \cap TP.$  Denotemos $S,$ $S_A,$ $S_B,$ $S_C,$ $S_D,$ $S_R$ las áreas de $P_1P_2P_3P_4,$ $UHYM,$ $PEOY,$ $SFOX,$ $TGXM,$ $OXMY.$

Asumimos sin perdida de generalidad que $O$ yace en el interior de $MPP_2S.$ Es claro que las rectas $EG$ y $FH$ dividen $P_1P_2P_3P_4$ en cuatro cuadriláteros con igual área. Por lo tanto, se tendrá

$\frac{_1}{^4}S-S_B+S_A=\frac{_1}{^4}S-S_A-S_C-S_R=\frac{_1}{^4}S-S_D+S_C=1.$

De éstas, se desprende que $S_A+S_D=S_B+S_C$ $\Longrightarrow$ los paralelogramos $PEGT$ y $SFHU$ tiene igual áreas. Como además, éstos tiene alturas iguales, entonces se sigue que $PE=FS$ $\Longrightarrow$ $OXMY$ es un cuadrado. Por tanto $S_A=S_C$ y $S_B=S_D.$ Luego, usando las expresiones conseguidas anteriormente, se llega a que $S_A=S_B=S_C=S_D=0$ $\Longrightarrow$ $O \equiv M$ y la conclusión se sigue.

[quote]Problema 4. Encontrar un número N de cinco cifras diferentes y no nulas, que sea igual a la suma de todos los números de tres cifras distintas que se pueden formar con cinco cifras de N.[/quote]
El número buscado $N=abcde$ habrá de ser igual a la suma $abc + abd + ... + cde.$ Esta suma tendrá $3 \cdot 4 \cdot 5 = 60$ sumandos; como primera cifra de los sumandos aparecerá igual número de veces las cinco cifras $a, b, c, d$ y $e,$ es decir habrá 60/12 sumandos cuya primera cifra es a, 12 sumandos cuya primera cifra es b, etc. Igual ocurrirá con las segundas y terceras cifras de los sumandos, de modo que se tendrá

$N = (100a + 10b + c) + (100a + 10b + d) + ... + (100c + 10d + e) =$

$= (1200a + 120a + 12a) + (1200b + 120b+ 12b) + ... + (1200e + 120e + 12e) =$

$= (1200 + 120 + 12)(a + b + c + d + e) = 12 \cdot 111 (a + b + c + d + e)$

Llamando $S = a + b + c + d + e$ resultará $N = 12 \cdot 111 \cdot S.$

$S$ será máximo cuando las cifras de $N$ sean $9, 8, 7, 6, 5.$ Así, $S = 35.$ $S$ será mínimo cuando las cifras de $N$ sean $1, 2, 3, 4, 5.$ Así, $S = 15.$

Luego, S debe estar comprendido entre 15 y 35; pero la primera igualdad indica que N debe ser múltiplo de 9, ya que tanto 12 como 111 son divisibles por 3. Por lo tanto, también debe ser múltiplo de 9 la suma S de las cifras de N. Ahora bien, los únicos múltiplos de 9 comprendidos entre 15 y 35 son 18 y 27. Para S = 18 resulta N = 23976, solución no válida pues tiene S = 27≠18. Para S = 27 resulta N = 35961, que es la solución buscada pues S = 27.

[quote][b]Problema 5.[/b] Sea $P(x,y)=2x^2-6xy+5y^2.$ Diremos que un número entero $a$ es un valor de $P$ si existen números enteros $b$ y $c$ tales que $a = P(b, c).$

I) Determinar cuántos elementos entre 1 y 100 son valores de $P.$

II) Probar que el producto de valores de $P$ es un valor de $P.$[/quote]
Los valores de $P$ son los enteros que pueden expresarse como suma de los cuadrados de dos enteros. Observemos en primer lugar que $P(x, y)$ puede expresarse en la forma $P(x,y)=(x-y)^2+(x-2y)^2,$ con lo cual los valores de $P$ son sumas de dos cuadrados. Por otro lado si $A=B^2+C^2,$ resolviendo $x-y=B$ $\cup$ $x-2y=C,$ resulta $P (2B - C, B - C) = A,$ lo cual prueba nuestra afirmación. La parte I) se resuelve pues contando los elementos entre 1 y 100 que son sumas de dos cuadrados.

II) Suponemos que $A$ y $A'$ son valores de $P.$ Con lo establecido previamente, $A$ y $A'$ pueden expresarse como $A=B^2+C^2$ y $A'={B'}^2+{C'}^2.$ Multiplicando éstos se llega a la conclusión.

[quote][b]Problema 6.[/b] Se dan tres puntos nos alineados $M,N,H.$ Si $M,N$ son los puntos medios  de los lados $AC,AB$ de un triángulo $\triangle ABC$ cuyo ortocentro es $H,$ probar que tal triángulo es construible con regla y compas.[/quote]
Sea $D$ el pie de la A-altura (desconocido) y $P \equiv AH \cap MN,$ es decir la proyección ortogonal de $H$ en $MN.$ Consideremos el $\triangle ABC,$ en el cual $A$ y $H$ yacen en lados distintos de la recta $MN,$ la otra solución posible se tratará por analogía. Es bien conocida la relación métrica $AD \cdot HD=DB \cdot DC=4 PN \cdot PM,$ por consiguiente

$HP=PD-HD=\frac{AD}{2}-HD=\frac{AD}{2}- \frac{4 PN \cdot PM}{AD} \Longrightarrow$

$\frac{AD^2}{2}-AD \cdot HP= 4PN \cdot PM \Longrightarrow  AD=HP+ \sqrt{HP^2+8 PN \cdot PM}$

Se podrá pues construir un segmento de igual medida que la A-altura, consiguientemente determinar el vértice $A$ y completar facilmente el $\triangle ABC$ reflejando $A$ en $M$ y $N.$


Let $ a \equal{} \frac {\pi}{7}$. Prove that:

                        $ \frac {1}{sina} \equal{} \frac {1}{sin2a} \plus{} \frac {1}{sin3a}$

Nobody?
[hide="Hint"]
Ptolemy's Theorem
[/hide]

It's very simple:

$ \frac{1}{\sin a}\equal{}\frac{1}{\sin{2a}} \plus{} \frac{1}{\sin{3a}} \ \Longleftrightarrow \ \sin a \ ( \sin 2a \plus{} \sin 4a) \equal{} \sin2a \cdot \sin 4a \ \Longleftrightarrow$

$ \Longleftrightarrow \ \sin a \ (2 \sin3a \cdot \cos a) \equal{}  2 \sin a \cdot \cos a \cdot \sin 4a$

That's a trigonometric solution. Here's a geometric from the book (for those who don't have it):
[hide]
Let regular hexagon $ A_1A_2...A_6$ inscribed in a circle. By applying Ptolemy's theoremto $ A_1A_3A_4A_5$ we get:
$ A_1A_3 * A_5A_4 \plus{} A_3A_4 * A_1A_5 \equal{} A_1A_4 * A_3A_5$
which means:
$ sin2a$ $ sina$ $ \plus{}$ $ sina$ $ sin3a$ $ \equal{}$ $ sin3a$ $ sin2a$
[/hide]

I too thought in complete trigonometric approach:
$ \sin{4a} \equal{} \sin{3a}$
$ \Rightarrow4\sin{a}\cos{a}\cos{2a} \equal{} 3\sin{a} \minus{} 4\sin^3{a}$
$ \Rightarrow4\cos{a}(1 \minus{} 2\sin^2{a}) \equal{} 3 \minus{} 4\sin^2{a}$
$ \Rightarrow4\cos{a} \minus{} 8\sin^2{a}\cos{a} \equal{} 3 \minus{} 4\sin^2{a}$
$ \Rightarrow6\cos{a} \minus{} 8\sin^2{a} \equal{} 3 \minus{} 4\sin^2{a} \plus{} 2\cos{a}$
$ \Rightarrow2\cos{a}(3 \minus{} 4\sin^2{a}) \equal{} 3 \minus{} 4\sin^2{a} \plus{} 2\cos{a}$
$ \Rightarrow\sin{2a}(3 \minus{} 4\sin^2{a}) \equal{} \sin{3x} \plus{} \sin{2a}$
$ \Rightarrow\frac {\sin{2a}\sin{3a}}{\sin{a}} \equal{} \sin{3a} \plus{} \sin{2a}$
$ \Rightarrow\frac {1}{\sin{a}} \equal{} \frac {1}{\sin{2a}} \plus{} \frac {1}{\sin{3a}}$.......Done! :)


Find all triangles satisfying:

$ \sum{cyc}\frac{1}{a^2}\equal{}\frac{9}{1\plus{}2abc}$

Nobody?

A huge hint:

[hide]Obviously enough try to prove that LHS>=RHS and equality will hold only for equilateral triangles of side 1.
To prove the inequality use two well-known results.

One is an inequality that has appeared on a competition a little more than a decade ago and has caused much discussion... :wink:

Two is an inequality where equality holds only when all variables equal 1. (What does the denominator of RHS remind you of???) :wink:

I think now the inequality will be immediately solved.[/hide]

Solved by Silouan in the Greek forum.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204781


geometry
perimeter
combinatorics unsolved
combinatorics

$w$ is a given circle that its perimeter is colored with 2 colors (White and black).
It is known that ,at least one side of each diameter of this circle is White.

Can we say that half of the perimeter of $W$ is white?

Can you please specify your question¿ I don't understand what it's asked for...  :?

$W$ is a given circle in a plane .

The points on the Perimeter of this circle are colored with 2 colors(white and black)

For each diameter of this circle like $AB(A$ and $B$ belong to the perimeter of $W)$.

At least one of the $A$ and $B$ is white.

Can we say that half of the perimeter of $W$ is White?(I think the answer is no)

What do you mean by "half of the perimeter", exactly? The sets involved might not be measurable. You need to define stuff like this rigorously, man :). No matter how intuitive such notions appear, people might not know what you're trying to say.

Hmm, I think you are right .It doesnt make sense to ask this kind of question.
I am gonna read some stuff about these measurable sets to get a little Academic view about this stuffes .Sorry for the inapproporiate Question.


Joan can swim $ \frac34$ of the race in 18 minutes. At this rate, how many more minutes will it take her to finish the race?

$ \frac34x\equal{}18$   where $ x$ is total time needed.
solving for x...
$ x\equal{}18*\frac43\equal{}24$

[quote="GameBot"]Joan can swim $ \frac34$ of the race in 18 minutes. At this rate, how many [i][b]more [/b][/i]minutes will it take her to finish the race?[/quote]
Finishing off...
$ 24 \minus{} 18 \equal{} \boxed{6}$.
Alternately, you could have seen that she has $ 1 \minus{} \frac {3}{4} \equal{} \frac {1}{4}$ of the race left.  So
\[ \frac {1}{4}x \equal{} \frac {\frac {3}{4}x}{3} \equal{} \frac {18}{3} \equal{} \boxed{6}\]

agh  more
didn't see :) but ftw would've marked me right since i'm so pro


Last year, we just wrote our AoPS names on our nametags.

Thats kind of boring... You should have an exclusive AoPS club and all the other people will be jealous...

thats kind of weird. plus, like half the states don't know who is going to nats...

[quote="mathluver2.718"]Here's an idea...

Get all the AoPSers together, and get everybody to write down their user name on slips of paper, without looking at anybody else's. Hand all the slips to one person, and then look at them, and try to guess who's who.

I wish I was going... :([/quote]

Thats the point. We have to find some way of getting them together...

BTW- the entire MO team is putting their AoPS name as the name being printed onto their nametag, so they wont be hard to find. :P

Yes, you can request that a "different" name from your real name be put on your nametag on the registration form. :D

[quote="tinytim"]The user has 4 numbers in his/her username...  :ninja:[/quote]

What? :huh:

[quote="Meta_Knight"][quote="xpmath"]This is too annoying to do and it won't work anyway.

Just write your username on your nametag and if you want, you can hang around the AoPS table (I assume there will be one again this year).

Also, I'm not saying exactly who, but there are some AoPSers I'd rather not meet...[/quote]

who? :P[/quote]

The user xpmath is referring to...

[quote="tinytim"]The user has 4 numbers in his/her username...  :ninja:[/quote]

hmm that could be me??   :?: :ninja:
or maybe someone else...

meh..heres an idea...all go to tinytim's room. His nametag will say tinytim, so that wont be too difficult. :P

[quote="Temperal"][quote="xpmath"]some AoPSers I'd rather not meet...[/quote]

one especially.

also aops meetups in general are stupid ideas.[/quote]
WHAT??? *gasp* is it me??!?!?!?! :(  :o

nooooo, it would never be you...  :wink:

Yeah, i guess my idea wouldn't really work... :P
w/e :blush:
Have fun @ nats! :wink:

Make AoPS nametags a seperate color from each other.

As CountdownKing said, in the thread from last yr, someone suggested that all AoPSers write their usernames on their nametags, so, when pintrading or socializing during down time, we were able to identify each other.  I think there's plenty of time to meet fellow AoPSers, and I wish we had thought to do something the prev two years, I could've met more people  :D

[quote="connaissance"]a meetup? that'd be kinda hard, considering that we're going to be with our teams for most of the time. Unless breakfast would do for a meetup, I don't see enough time for that. Also, there are going to be 228 students there... unless you spend like an hour looking at everyones' nametag, I don't see how anyone could distinguish AoPS people from their nametags can single each other out...[/quote]

I probably spent more time with the Michigan team than with Ohio, excluding the competition and the nights.

There is quite a bit of free time. Last year, since it was Denver, there wasn't something "special" to do. This year it's at Disney World, so people will probably be going to more places than last year.


geometry
incenter
inequalities unsolved
inequalities

Let $ABC$ be a triangle with circumcentre $O$ and incentre $I$, and let $DEF$ be the pedal triangle of interior point $M$ of triangle $ABC$ (with $D$ on $BC$, etc.). Then

\[ OM\geq OI\Rightarrow r^{'}\leq \frac{r}{2} \]

where $r$ and $r^{'}$ are the inradii of triangles $ABC$ and $DEF$, respectively.

Sorry to bring up such an old problem.We have that $\frac{S_{DEF}}{S_{ABC}}=\frac{R^{2}-OM^{2}}{4R^{2}}\leq\frac{r}{2R}$,from $OM\geq OI$.Now we have $S_{DEF}\geq 3r'^{2}\sqrt{3}$,$rS_{ABC}=r^{2}p$ and $2p\leq 3\sqrt{3}R$.Put these toghether and we solved the inequality. :lol:


Find all functions \[ f: \Re \to  \Re \]that satisfy:\[ x^2f(x)+f(1-x)=2x-x^4 \]

Make $x=y$ and then $x=1-y$. You will obtain a system and now it´s just solve it.


[quote="Interval"]Graph each rational function using transformation (shifting, compressing, stretching, and reflection).

(1) R(x) = (3/x)

(2) R(x) = (x - 4)/x

What are the steps to graph rational functions?

What exactly is a rational function?[/quote]
A rational function is a funtion that stays rational.
For the second one use the distributive property.

How do I use the distributive law here?

Distributive law to graph rational functions?

[quote="Interval"]Graph each rational function using transformation (shifting, compressing, stretching, and reflection).

(1) R(x) = (3/x)

(2) R(x) = (x - 4)/x

What are the steps to graph rational functions?

What exactly is a rational function?[/quote]

Answer is attached.


I'm soo stupid to get this :oops:

find the inverse of the matrix [u]USING THE GAUSS JORDAN[/u] method

1  -2  -1
2  3   1
3  5   0

anyone out there can help?

why did this thread get moved? gauss-jordan is like algebra 2

Adjoin the identity matrix to the right of the given matrix.
$ \left(
\begin{array}{rrr|rrr}
1&\minus{}2&\minus{}1&1&0&0\\
2&3&1&0&1&0\\
3&5&0&0&0&1\\
\end{array}\right)$
Add multiples of the first row to the second and third rows so that the first entries in those rows become 0.
$ \left(
\begin{array}{rrrrrr}
1&\minus{}2&\minus{}1&1&0&0\\
0&*&*&*&*&*\\
0&*&*&*&*&*\\
\end{array}\right)$
Add multiples of the new second row to the first and third rows so that the second entries in those rows become 0.
$ \left(
\begin{array}{rrrrrr}
1&0&*&*&*&*\\
0&*&*&*&*&*\\
0&0&*&*&*&*\\
\end{array}\right)$
Repeat for the third entry of the third row. Divide each row by the appropriate constants so that the leftmost columns are the identity matrix. The inverse is the rightmost columns.
$ \left(
\begin{array}{rrrrrr}
1&0&0&*&*&*\\
0&1&0&*&*&*\\
0&0&1&*&*&*\\
\end{array}\right)$

Welcome to the forums. Please try to work it out yourself first, but feel free to ask for clarifications on any steps; be specific. Don't ask me to do the actual calculation for you because this is your homework and doing your homework for you is not the purpose of this forum.

so i'd add -2 times the first row to the second and -3 times the first to the third, and then add -2 times the second- OH!

thx. i tried it, but i couldnt get the correct pattern cuz i didnt know it, teacher wuz just like think of all t3h combinations and find the right one

Does anybody want to prove why the gauss jordan method works? (I think the argument is quite nice!).

[hide="Intuitive Explanation"]We have $ (A|I)$. Row operations of the gauss jordan method, taken collectively, can be expressed as a linear transform $ T$ with a nonzero determinant. Then, we have from the left side that $ TA \equal{} I \implies T \equal{} A^{ \minus{} 1}$. Then, the right side becomes $ TI \equal{} A^{ \minus{} 1}$. Thus, applying $ T$ to the left and right halves of $ (A|I)$ gives $ (I|A^{\minus{}1})$[/hide]

Well thats correct except you have to explain why the row operations can be represented as a matrix.


group theory
abstract algebra
function
topology
advanced fields
advanced fields unsolved

Let $X$ be a topological group. Prove that $\pi_{1}(X)$ is abelian.

Do you perhaps need "connected" as a hypothesis?

uhm.. i think it's not even necessary: (path-)connected components are "lateral classes" of the (path-)connected component containing identity of the group, so you can restrict your attention to this single component, which turns out to be a subgroup (so every component is actually a "lateral class").

as regarding the problem, you can consider $\pi_{1}(X,e)$, and, given $\alpha, \beta$ paths, the function $(s,t)\mapsto \alpha(s)\beta(t)$, which is continuous, and by a homeomorphism gives a homotopy, since you can see it as a function $S(I)\to G$, where $S$ is suspension, and the two "endpoints" are $(0,0)$ and $(1,1)$.

ok, i know i gave no "real proof", but i think the idea is pretty clear...

You can give any space any group structure together with discrete topology -> you could show that all homotopy groups are abelian....

Of course, the homotopy groups are trivial with discrete topology.


Out of 22 students in class, 9 play football and tennis, 10 play football and baseball, and 11 play
baseball and tennis. How many students play all 3 sports if it is known that each student plays at
least one sport, and each sport is played by a total of 15 students ?

Please post in the appropriate section :wink:

The answer is $ 7$, and this is a direct application of inclusion/exclusion formula.

I move it into the combinatorics forum, even it should be more in an beginner forum...

Pierre.


modular arithmetic
invariant
induction
algebra
polynomial
vector

Assume $n$ is a positive integer. Considers sequences $a_0, a_1, \ldots, a_n$ for which $a_i \in \{1, 2, \ldots , n\}$ for all $i$ and $a_n = a_0$.

(a) Suppose $n$ is odd. Find the number of such sequences if $a_i - a_{i-1} \not \equiv i \pmod{n}$ for all $i = 1, 2, \ldots, n$.

(b) Suppose $n$ is an odd prime. Find the number of such sequences if $a_i - a_{i-1} \not \equiv i, 2i \pmod{n}$ for all $i = 1, 2, \ldots, n$.

Is is just something like $n!(1-\frac{1}{n^n})$
because at each of the $a_i$ there is a $\frac{1}{n}$ chance it isn't satisfied

[hide=" (a) Use inclusion-exclusion:"]

Let $0\le k <n$. There are $\binom nk$ ways to choose $k$ of the conditions to be disregarded.
Each condition "synchronizes" two neighboring entries in the sequence, so we get $n-k$ lumps of entries who "move in unison". There are $n^{n-k}$ possibilities for this.

For $k=n$, we have to have $1+2+\dots+n = \frac{n(n+1)} 2 \equiv 0 \mod n$ which is true for $n$ odd. There are $n$ possibilites in this case (instead of 1 as given by the term $n^{n-k}$).

Thus, we get $\sum_{k=0}^n (-1)^k \binom nk n^{n-k} -(n-1)$.
The binomial theorem gives $(n-1)^n-(n-1)$.

(For even $n$, we would get $(n-1)^n-1$.) [/hide]

[hide="(b) Expand on the previous method"]
Now we choose the $k$ places, where we disregard the condition, but we have two possibilities to do so. Unfortunately, the condition for $i=n$ is actually just one condition, so we have to do two terms for each $k$ to distinguish whether $i=n$ is involved or not.

Therefore, the sum for $k<n$ is
\[ \sum_{k=0}^{n-1}((-1)^k \binom {n-1}k 2^k n^{n-k}+ (-1)^k \binom {n-1}{k-1} 2^{k-1} n^{n-k}\\ =n(n-2)^{n-1}-(n-2)^{n-1}\\= (n-1)(n-2)^{n-1}.  \]

As before, we have to treat the case $k=n$ separately.

We have to find the number of ways to choose $\epsilon_i \in \{1,2\}$ such that $\sum_{i=1}^{n} \epsilon_i i \equiv 0 \mod n$ (and then divide by two, since $\epsilon_n$ is "the same" in both cases).

Since $n$ is still odd, this boils down to finding subsets $S$ of $\{1,2,\dots, n\}$ with $\sum_{x\in S} x \equiv 0 \mod n$. This is certainly true if $S$ contains all or none of the elements. For all other sets, we can consider the shifts of $S$ by adding $i$ to each entry of $S$. Since $n$ is prime and we excluded the sets that are invariant under the shift by 1, the sequence of shifted sets must have period $n$. The sum of the elements of $S$ with $m$ elements is shifted like $s, s+m, s+2m, \dots$. Since $m$ is not divisible by $n$, we get each residue mod $n$ exactly once, so there are
$2+\frac{2^n-2}n$ such sets.

In summary, we get for the number of sequences:

\[ (n-1)(n-2)^{n-1}- (2+\frac{2^n-2}n)/2\\ = (n-1)(n-2)^{n-1}-\frac{2^{n-1}-1}n-1.  \]

[/hide]

so is the answer some silly formula like that approximately equal for large n to mine?

kunterbunt,

I checked your solution for b) and I think you messed up your calculations, more precisely in the second sum and the last part of the reasoning.

I got [hide="this."]$(n-1)[(n-2)^{n-1}-1]$[/hide]

Could you either point out an error in my argument or give your reasoning?

In the first part, the second sum isn't $-(n-2)^{n-1}$ you are missing one of the terms of the sum, $2^{n-1}$.

On the last part of your solution, you are only considering the possible diferences between the terms of the sequence but you need to multiply by $n$ because there are $n$ diferent $a_{0}$. My doubt about your solution arrised because the final formula should always be divisible by $n$ and at least for n=7 your formula doesn't give a multiple of 7.

You are absolutely right, thank you for your correction. I don't know why I didn't check the result. For completeness, I include the fixed version below.

[hide=" (a) Use inclusion-exclusion:"]

Let $0\le k <n$. There are $\binom nk$ ways to choose $k$ of the conditions to be disregarded.
Each condition "synchronizes" two neighboring entries in the sequence, so we get $n-k$ lumps of entries who "move in unison". There are $n^{n-k}$ possibilities for this.

For $k=n$, we have to have $1+2+\dots+n = \frac{n(n+1)}2 \equiv 0 \mod n$ which is true for $n$ odd. There are $n$ possibilites in this case (instead of 1 as given by the term $n^{n-k}$).

Thus, we get $\sum_{k=0}^{n}(-1)^{k}\binom nk n^{n-k}-(n-1)$.
The binomial theorem gives $(n-1)^{n}-(n-1)$.

(For even $n$, we would get $(n-1)^{n}-1$.) [/hide]

[hide="(b) Expand on the previous method"]
Now we choose the $k$ places, where we disregard the condition, but we have two possibilities to do so. Unfortunately, the condition for $i=n$ is actually just one condition, so we have to do two terms for each $k$ to distinguish whether $i=n$ is involved or not.

Therefore, the sum for $k<n$ is
\[\sum_{k=0}^{n-1}((-1)^{k}\binom{n-1}k 2^{k}n^{n-k}+(-1)^{k}\binom{n-1}{k-1}2^{k-1}n^{n-k})\\ =n(n-2)^{n-1}-(n-2)^{n-1}+2^{n-1}\\ = (n-1)(n-2)^{n-1}+2^{n-1}. \]
As before, we have to treat the case $k=n$ separately.

We have to find the number of ways to choose $\epsilon_{i}\in \{1,2\}$ such that $\sum_{i=1}^{n}\epsilon_{i}i \equiv 0 \mod n$ (and then divide by two, since $\epsilon_{n}$ is "the same" in both cases). Afterwards, we have to multiply by $n$, because we have fixed the differences between the $a_{i}$ and can choose $a_{0}$ freely.

Since $n$ is still odd, this boils down to finding subsets $S$ of $\{1,2,\dots, n\}$ with $\sum_{x\in S}x \equiv 0 \mod n$. This is certainly true if $S$ contains all or none of the elements. For all other sets, we can consider the shifts of $S$ by adding $i$ to each entry of $S$. Since $n$ is prime and we excluded the sets that are invariant under the shift by 1, the sequence of shifted sets must have period $n$. The sum of the elements of $S$ with $m$ elements is shifted like $s, s+m, s+2m, \dots$. Since $m$ is not divisible by $n$, we get each residue mod $n$ exactly once, so there are
$2+\frac{2^{n}-2}n$ such sets.

In summary, we get for the number of sequences:
\[(n-1)(n-2)^{n-1}+2^{n-1}-n(2+\frac{2^{n}-2}n)/2\\ = (n-1)((n-2)^{n-1}-1). \]
[/hide]

This can certainly be proved directly, too. I specifically didn't look for a simpler proof because my result didn't look promising...

Here's a different approach:
(a)
Let $ c_{i} = a_{i+1} - a_{i}$ for $ i=1,2 ... n$ (where $ a_{n+1} = a_{1}$). We have $ c_{1} + ... + c_{n} \equiv 0 (mod n)$, and $ c_{i} \neq i (mod n)$. Consider the sequence $ b_{i} = c_{i} - i$: we require $ b_{1} + ... + b_{n} \equiv c_{1} + ... + c_{n} - [1+2+ ... +n] \equiv 0 (mod n)$, since $ n$ is odd so $ 1+2+ ... + n \equiv n* \frac{n+1}{2} \equiv 0 (mod n)$. Also, $ b_{i} \neq 0 (mod n)$, since $ c_{i} \neq i (mod n)$. We can see easily see that each such set $ b_{i}$ corresponds to exactly $ n$ choices for a set $ a_{i}$ [since there are $ n$ choices for $ a_{1}$, and after all $ a_{i}$ are determined by the $ b_{i}$ sequence]. Thus we need to count the number of sets $ b_{i}$, with $ b_{i} \neq 0 (mod n)$ and $ b_{1} + ... + b_{n} \equiv 0 (mod n)$, and multiply this number by $ n$.
Using recurrences, we can quickly do this: Define $ c_{k,l}$ to be the number of sets $ b_{1}, b_{2} ... b_{k}$ such that $ b_{1} + ... + b_{k} \equiv l (mod n)$, and $ b_{j} \neq 0 (mod n)$ for $ 1 \leq j \leq k$. We seek to find $ c_{n,0}$. We have $ c_{1,0} = 0, c_{1,t}=1$ for $ t=2,3...n$. Since $ b_{k} = 1$ or $ 2$ or ... $ n$. If $ b_{k} = p$, the number of such sets is equal to the number sets where $ b_{1} + ... + b_{k-1} = l-p$, which is $ c_{k-1, l-p}$. Thus we have $ c_{k,l} = \sum_{t: 0 \leq t \leq n-1, t \neq l} c_{k-1,t} = [\sum_{t} c_{k-1,t}] - c_{k-1,l} = (n-1)^{k-1} - c_{k-1,l}$, since in our summation $ t \neq l (mod n)$ since $ b_{l} = 0$ is not possible, and $ \sum_{t} c_{k-1,t} = (n-1)^{k-1}$, since the number of sequences $ b_{1} ... b_{k-1}$ together is equal to $ (n-1)^{k-1}$ ($ n-1$ choices for each of the $ k-1$ possible variables). Using this recurrence, we can prove by induction the following: $ c_{k,0} = \frac {(n-1)^k + 1}{n} - 1$ when $ k$ is odd, and $ c_{k,0} = \frac {(n-1)^k - 1}{n} + 1$ when $ k$ is even.
To finish off, since $ n$ is odd, we have $ c_{n,0} = \frac {(n-1)^n+1}{n} - 1$. Recalling that our answer is this number multiplied by $ n$, the number of satisfactory sets $ a_{i}$ is: $ (n-1)^n + 1-n$.
(b) This is harder since our recurrence method will not work to completion.
As in (a), define the sequence $ c_{i}$, and define the sequence $ b_{i} \equiv c_{i} - i (mod n)$. This time we require $ b_{i} \neq 0, i (mod n)$, and we must count how many sequences $ b_{i}$ such that $ b_{1} + ... + b_{n} \equiv 0 (mod n)$. Since $ n$ is prime, denote $ n=p$. As in (a), define the sequence $ c_{k,l}$.
Using the same method we used to obtain the recurrence for $ c_{k,l}$ in part (a), we shall do so again now. We obtain, that for $ k \leq p-1$, that $ c_{k,l} = (p-2)^{k-1} - c_{k-1,l} - c_{k-1,l-k}$. Note that we require $ k \leq p-1$, since for $ k=p$, $ c_{k} \neq 0, k (mod n)$, but $ 0 \equiv p (mod p)$, so instead we have that $ c_{p,l} = (p-2)^{p-1} - c_{p-1, l}$. We seek to compute $ c_{p,0}$. Thus we have: $ c_{p,0} = (p-2)^{p-1} - c_{p-1, 0}$. But we have $ c_{p-1, 0} = (p-2)^{p-2} - c_{p-2, 0} - c_{p-2, 1}$, so substituting back in: ${ c_{p, 0} = (p-2)^{p-1} - (p-2}^{p-2} + c_{p-2, 0} + c_{p-2, 1}$. But we can break down  $ c_{p-2, 0}$ and $ c_{p-2, 1}$ further, and doing so we obtain $ c_{p, 0} = (p-2)^{p-1} - (p-2)^{p-2} + 2^1* (p-2)^(p-3) - c_{p-3, 0} - c_{p-3, 1} - c_{p-3, 2} - c_{p-3, 3}$. Now, a pattern emerges, and we can utilise it to break down $ c_{p, 0}$ in terms of $ c_{1, 0}, c_{1,1} ... c_{1, p-1}$, all of which we know (since $ c_{1,0} = c_{1,1} = 0$, and the rest are equal to $ 1$). We define the following set recursively: $ S_{1} = {0}$, and if $ S_{k} = {a_{1} ... a_{t}}$, then $ S_{k+1} = {a_{1}, ... a_{t}, a_{1} + k, a_{2} + k, ... a_{t} + k}$ (with repeated elements included. From the pattern that emerges when we keep breaking down $ c_{p,0}$ using the recurrence, we obtain the following: $ c_{p,0} = (p-2)^{p-1} - (p-2)^{p-2} + 2^1 * (p-2)^{p-3} - 2^2* (p-2)^{p-4} ... - 2^{p-3}*(p-2)^{1} + \sum_{t \in \{S_{p-1}\}} c_{1,t}$.
Now we note that in this sum, by a trivial induction, $ {S_{p-1}}$ consists of $ 2^{p-2}$ elements; and thus since $ c_{1,t} = 1$ for all $ t$, except when $ t \equiv 0, 1 (mod p)$, $ \sum_{t \in \{S_{p-1}\}} c_{1,t} = 2^{p-2} - X$, where $ X$ is equal to how many elements in $ S_{p-1}$ are equivalant to $ 0, 1 (mod p)$. Next, we can easily see, by induction, that $ S_{n}$ consists of the sum of all subsets of $ {1,2 ... n-1}$ (including the empty subset); thus we need to find how many subsets of $ (1,2 ... p-2)$ have sum equivalant to $ 0$ or $ 1$ in $ Z_{p}$. However, we can see that a subset of $ (1,2 ... p-1)$ has sum equal to $ 0$ in $ Z_{p}$ either when: [1] it doesn't have $ p-1$, and the remaining elements sum to $ 0 (mod p)$ or [2] it does have $ p-1$, and the remaining elements, a subset of $ (1,2 .. p-2)$ sum to $ 1 (mod p)$. Thus, effectively, our number $ X$ is equal to the number of subsets of $ (1,2 ... p-1)$ which have sum divisible by $ p$. But we can see, that this number is equal to half the number of subsets of $ (1,2 ... p)$ which have sum divisible by $ p$ since, a subset of $ (1,2 ... p)$ has sum divisible by $ p$ [1] when it doesn't contain $ p$, and the remaining elements have sum divisible by $ p$ [2] it does have $ p$, and the remaining elements have sum divisible by $ p$. So now we must count how many subsets of $ {1,2 .. p}$ have sum divisible by $ p$, half that to find $ X$, and then compute $ c_{p,0}$ from the above formula.
My way of counting such subsets of $ (1,2 .. p)$ is complete overkill: it's a slight variant of the method used in Special Prize solution to IMO 95/6. For $ k \neq 1, p$, we shall prove that the number of subsets of $ (1,2 ... p)$, of size $ k$, whose sum is divisible by $ p$, is equal to $ \frac {\binom pk}{p}$. Let $ a_{m}$ be the number of subsets, size $ k$, whose sum is equivalant to $ k (mod p)$.
Consider a $ p$-th root of unity, $ w$. Consider polynomial $ P(x) = x^{p} - 1 = \prod_{1 \leq j \leq p} (x - w^j)$. Consider the sum of products of roots, taken $ k$ at a time. Since coefficient of $ x^{p-k}$ in $ P(x)$ is $ 0$, we have:
$ 0 = \sum_{1 \leq j_{1} < j_{2} ... <j_{k} \leq p} w^{j_{1}}*w^{j_{2}} ... w^{j_{k}}$
But the RHS is equal to $ w^{s}$, whenever $ j_{1} + ... + j_{k} \equiv s (mod p)$. There are $ a_{s}$ such sets. So we have the following:
$ 0 = a_{0} + a_{1} * w^{1} + a_{2} * w^{2} ... + a_{p-1}*w^{p-1}$.
Since the minimal polynomial of $ w$ in $ Q[x]$ is $ 1+x+x^2... + x^{p-1} = 0$, we thus have $ a_{0} = a_{1} ... = a_{p-1}$. But since $ a_{0} + .. + a_{p-1} = \binom pk$, we have $ a_{0} = \frac {\binom pk}{p}$.
Thus the number of subsets, of size $ k$ which have sum $ 0 (mod p)$ is $ \frac {\binom pk}{p}$: so the total number of such subsets is $ 2 + \frac {\sum_{1 \leq k \leq p-1} \binom pk }{p} = \frac {2^p + 2p - 2}{p}$ (using Binomial Theorem, and the fact that we have two extra sets, $ 1$ for each $ k=1$, $ k=p$). Thus $ X = \frac {2^{p-1}-1}{p} + 1$.
Finishing off now is just computations. We have: $ \sum_{t \in \{S_{p-1}\}} c_{1,t} = 2^{p-2} - X = 2^{p-2} - [\frac {2^{p-1}-1}{p} + 1]$, so then, from above,
$ c_{p,0} = (p-2)^{p-1} - (p-2)^{p-2} + 2^1 * (p-2)^{p-3} - 2^2* (p-2)^{p-4} ... - 2^{p-3}*(p-2)^{1} + \sum_{t \in \{S_{p-1}\}} c_{1,t} = (p-2)^{p-1} - (p-2)^{p-2} [ 1 + Y + Y^2 ... + Y^{p-3} ] + 2^{p-2} - [\frac {2^{p-1}-1}{p} + 1] = \frac {(p-1)((p-2)^{p-1}-1)}{p}$,
 where $ Y = \frac {-2}{p-2}$, since we can use sum of geometric series. Here I've skipped all of the computations, they're very simple.
The answer is $ p$ times $ c_{p,0}$ (going back to the very beginning), i.e. $ (p-1)[ (p-2)^{p-1} - 1]$.

[hide]
We work in $\mathbb{Z}/n\mathbb{Z}$ (including with indices). Let's first shift each $a_i$ by subtracting $1 + 2 + \ldots + i = \binom{i + 1}{2}$. This just amounts to changing the condition for (a) to needing all $a_{i+1} \neq a_i$ and in (b) needing all $a_{i + 1} - a_i \neq 0, i + 1$.

Next define $x_i = a_i - a_{i-1}$, $1 \leq i \leq n-1$.

(a) We are counting the number of vectors of $x_i$ such that no $x_i = 0$ and $\sum{x_i} \neq 0$. By a standard generating function argument, this number is just:

$P(1) - \frac{P(1) + P(\omega) + \ldots + P(\omega^{n-1})}{n}$

where $P(t) = (t + t^2 + \ldots + t^{n-1})^{n-1}$, and $\omega$ is a primitive $n$-th root of unity. Since $P(\omega^i) = (-1)^{n-1}$ for $i \neq 0$, we get the answer to be $\frac{1}{n}((n-1)^{n-1} - (n-1))$. Remembering that we only counted the number of $x_i$, we multiply by $n$, corresponding the the choice of $a_0$ arbitrarily. The answer is $(n-1)^{n-1} - (n-1)$.

(b) The number of $x_i$ with no $x_i = 0, i$ and $\sum{x_i} \neq 0$ is:

$P(1) - \frac{P(1) + P(\omega) + \ldots + P(\omega^{n-1})}{n}$

where $P(t) = \prod_{i = 1}^{n-1}{((1 + t + t^2 + \ldots + t^{n-1}) - 1 - t^i)}$. Then $P(\omega^i) = Q(-1)$ where $Q(x) = \prod{(x-\omega^i)}$, hence $Q(-1) = 1$, and the answer is $\frac{1}{n}((n-1)(n-2)^{n-2} - (n-1))$. Multiplying by $n$ we get the answer $(n-1)(n-2)^{n-2} - (n-1)$.

[/hide]

Part a.) It is equivalent to find the number of tuples $(x_1, x_2, \dots , x_n)$ where $n \nmid x_i-i$ and $n \mid \sum x_i$. Let $T(X)=X+X^2+\dots+X^n$ and $\epsilon$ be a primitive $n$ th root of unity. Define the polynomial $$h(X) \overset{\text{def}}{:=} \prod^n_{j=1} \left(T(X)-X^j\right)=\sum_{k \ge 0} a_kX^k.$$ Note that the desired answer is $n \cdot \left(\sum_{j \ge 0} a_{jn}\right)$. This evaluates to $\sum^n_{j=1} h(\epsilon^j)$. Note that $h(1)=(n-1)^n$ and $h(\epsilon^j)=-1$ for all other indices, so $\boxed{(n-1)^{n}-(n-1)}$ is the answer.

Part b.) Following the notations of Part a.) define $$h(X) \overset{\text{def}}{:=} \prod^n_{j=1} \left(T(X)-X^j-X^{2j \bmod n}\right),$$ and note that we require $\sum^n_{j=1} h(\epsilon^j)$ as the answer. Clearly $h(1)=(n-1)(n-2)^{n-2}$ and all other summands are $-1$ so we obtain the $\boxed{(n-1)(n-2)^{n-1}-(n-1)}$ as the answer.


If $ x_1$, $ x_2$, $ x_3$, $ x_4$ are four reals such that $ 0 < x_i \leq 1/2$ for every $ i\in\left\{1,2,3,4\right\}$, then prove that

$ \frac {\sum_{i\equal{}1}^4 x_i^4}{\sum_{i\equal{}1}^4 \left(1 \minus{} x_i\right)^4} \geq \frac {\prod_{i\equal{}1}^4 x_i}{\prod_{i\equal{}1}^4 \left(1 \minus{} x_i\right)}$.

This was posted at least at two places (what doesn't mean solved at any of them):

http://www.mathlinks.ro/Forum/viewtopic.php?t=27254
http://www.mathlinks.ro/Forum/viewtopic.php?t=1834

  Darij


Susan ran 100 complete laps around the track in 5 days. Each day, she ran 6 more laps than the previous day. How many laps did she run on the last day?

On the last day, say she ran $ x$ laps. Then we have $ x \plus{} (x \minus{} 6) \plus{} (x \minus{} 12) \plus{} (x \minus{} 18) \plus{} (x \minus{} 24) \equal{} 5x \minus{} 2\cdot30 \equal{} 5x \minus{} 60 \equal{} 100$

$ \implies5x \equal{} 160\implies x \equal{} \boxed{32}$.


Hi all,

It seems that this forum needs a good moderator to keep it going. ;)
Please submit all your preferences for moderatorship in this topic.

Thanks,
Valentin Vornicu

I suggest Mr. Namdung is the most approppriate one could moderate the Vietnam forum.

[quote="behemoth"]I dont think so .
I think we need a person who have many expericence in math like me (:D).So I think the moderator for this forum is nguyenkims .He knows Vietnamese (because he's Vietnamese) , he has many years in university and he's very good in math.[/quote]
how about  linkgreencold  .I know him .he is new member but he have many expericence . he is a moderator of website hkt-music.net.  he is very good in maths . his english is verry well.  and he speak vietnamese well of course because he live in vietnam

Who's calling me? [url=http://diendantoanhoc.net/][color=blue]fast access to diendan[/color][/url]. :lol:
[url=http://mathnfriend.net/][color=blue]fast access to diendan[/color][/url]

it is right .I think the MOD must be linkgreencold,I know him,he has experience in doing the work of a MOD

sorry because I learn E badly.   but   I think "linkgreencold"  is a  avaiable person to become moderator of this section Asia , especially VietNam country .
          because i know  linkgreencold ,he is a good  student at maths  and he is so funny .
 when I make friend with him , he helped me in many fields , ex: maths , informations ....
                          THANKS , LINKGREENCOLD .
 I support  him become moderator of this section.
       he is  one candidate who is  capable of winning in List( members:nguyenkims, namdung...) of candidates.

[quote="N.T.TUAN"]Hi all,
Now pvthuan and me are Mods of this box, we will try :D[/quote]

Em bầu thầy Nam Dũng hay thầy Thuận_I think Mr Nam Dung or Mr Thuan can be able to do this(Hic ko biết nói tiếng anh có đúng hông nữa,gà tiếng nước ngoài quá :blush: )

[quote="N.T.TUAN"]Hi Ly,
Now, N.T.TUAN and pvthuan are Mods of this box. And pvthuan is my private secretary. :P[/quote]
gr8 is it  :D

[quote="pardesi"]
gr8 is it  :D[/quote]
What is ''gr8''?  :lol:

[quote="N.T.TUAN"]
What is ''gr8''?  :lol:[/quote]
'gr8' $ \rightarrow$ gr'eight' $ \rightarrow$ great  :wink:

[quote="N.T.TUAN"]Thanks for help! :P[/quote]
u r most welcome  :wink:

I think none of the people that you guys nominated shouldn't run for the election. Don't you guys think this forum is soooooooooooooo boring. New post appears one in a while and no one responds it. If this keeps going on, in the near future no one would come to this website to post or talk.


inequalities
Cauchy Inequality
IMO
IMO 2005

[quote="math92"]cool, i want one.  What will the design look like?[/quote]Finally they are ready to be purchased :P
You can purchase the t-shirts following this link

http://www.zazzle.com/contributors/products/gallery/browse_results.asp?cid=238898388108976697

Anyway I am attaching some of the pictures (all shirts have the same front - the back is different).

Nice!  I'll have to get one of those soon.

"You have read at least one post from Grobber"
 :rotfl:  :rotfl:  :rotfl:
the shirt's nice and "you know if you are a mathlinker" is very funny!

Sweet!!!!!!
I want one!!!!
the "you know you are a mathlinker" was very funny :rotfl:

[quote="DiaDi"]neat indeed!...but isn't there a method to buy them from Romania too?   :roll:   :blush:[/quote]You can buy them from anywhere. Just need a valid internet Credit Card (all BCR cards are ok I think, and almost all other cards issued by banks in Romania will probably work) and have it shipped in Romania.

well...yeah...i already knew that....just thought they could be bought in some stores here or i dunno :? ....I'd like one too anyway :D

[quote="Xixas"]And it's impossible in Lithuania, I bet...[/quote]You can buy them online and have them shipped to Lithuania.

Valentin,how can get a mathinks t-shirt?

You can buy them online from here: http://www.zazzle.com/contributors/products/gallery/browse_results.asp?cid=238898388108976697

I think those 3 teams who give the best performance on stage at the IMO Opening Ceromony should get the ML T-Shirts Nr. 1-6 per team as prize each.  :)

[quote="Dave Vather"]Maybe host team should have T-shirts as a prize too, don´t you think?
However, are they going to sell them in IMO?[/quote]We are not  selling them. Zazzle is (which is a specialized site in doing promotion stuff). You can buy them online from there.

I've never bought something from internet by a credit card. Why can't we buy t-shirts from Romania :(?

I believe in Cauchy inequality, the name Schwarz is spellled "Schwarz", not "Schwartz", as it is the common spelling error that everyone makes. The inequality is named after Hermann Amandus Schwarz.

[quote="lewissac"]May I know is there any other way I can buy that shirt? cuz my parents and I do not own any credit card nor american express.[/quote]You have to find someone willing to buy it for you in your country and/or someone to buy it and ship it there ...  :?


I have already submitted my solution, but I realized that I made a mistake in my proofs for one of the questions... I know i did the correct procedures to obtain the solutions... what is the max do you think I could get for that one question? 3 or 4?

wow.. you submitted quite early.

Anyway, it depends on the nature of the mistake.  You could get a 5 if you are just talking about poor wording (then again, depending how bad the wording is, you could get much worse).  If it is just a small error, you will probably only lose a point - the larger the error, the more points you'll lose.

This is why I would not recommend submitting solutions so far in advanced.  With more than a month to work, I like to make sure my solutions aren't just correct but also well-written.

Your case is one of the reasons I have not submitted my solutions yet.

ok... next time will just make sure and be careful... i checked many times, and made sure things are correct, but didn't see that mistake until after i submit my solutions ;.; oh well...

( i dont know if its rude or something to post something of ur own in some elses topic so plz tell me if it is becuasse this topic seems to fit wut im about to ask)
is grahamer counted in ur solution grade cause i am pretty sure that if anyone read my solution it would be very clear except for the fact that theres no periods, comas, etc ( i[b] never[/b] notice puncuation when i read) so as long as its understandible i have a chance at five points right

From the [url=http://www.usamts.org/Rules/U_RulesComplete.php]rules[/url] (#17):
[quote]Each solution will be given a score of 0, 1, 2, 3, 4, or 5. Solutions will be graded both on mathematical correctness and on writing style.[/quote]

When I finish my solutions that early, I wait a week or so, print them out, and read them out loud, then correct, lather, rinse, repeat. That helps immensely. I think at this point I've caught all my errors (there were a few).

is it okay if i write mine by hand?? I mean that doesn't make me a total n00b does it??

It makes you a n00b at $ \text{\LaTeX}$, but not at mathematical problem solving and solution writing (which is more important). Just make sure your solutions are legible.

Im actually not a noob at $ \text{\LaTeX}$ im just too lazy to put it on my computer. (is that how it works- u download it for  microsoft word)?

You may definitely write your solutions by hand, and it will not affect the grading one bit (provided your handwriting is readable).


abstract algebra
linear algebra
linear algebra unsolved

Find $ Hom_{\mathbb Z}(\mathbb Q,\frac {\mathbb Q}{\mathbb Z})$.


[quote="Peter VDD"]I think the source file is not searched for... could that be?

regardless of what I put in the source field, it cannot be found. :( (and the text explicitly says also look in source)[/quote]No, that could not be. There might be a problem with your search string, or with [i]some[/i] source fields (especially old ones). I will look into this (much) later.

I've changed the source of all flanders problems to have flanders, and not in the rest of the post. Flanders gives no problem matches any more. This is something that keeps coming back... can you show me one example of a new post where you can find on source?

I had the same problem with searching for "South Africa".

Didn't find any posts with South Africa in the source!

Hmm, this is strange. I will try to fix (if I find the problem :D) it this upcoming weekend.

Ok, problem has been identified and fixed :) Old source fields should also pop up now ;)


inequalities
function
Cauchy Inequality
inequalities proposed

For $ a,b,c$ positive reals prove that the following inequality holds -
$ \frac{a}{b\plus{}c} \plus{}\frac{b}{a\plus{}c} \plus{}\frac{c}{a\plus{}b} \plus{}\frac{3 (abc)^{\frac{1}{3}}}{2(a\plus{}b\plus{}c)} \ge 2$
i know a solution using pqr but thats messy so wishing someone could post simpler solutions.

[quote="arqady"]
The following inequality is stronger.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c)} \ge 2\]
[/quote]

doesn't it just follow from Schur's?  :lol:

[quote="arqady"][quote="geniusbliss"]For $ a,b,c$ positive reals prove that the following inequality holds -
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3 (abc)^{\frac {1}{3}}}{2(a \plus{} b \plus{} c)} \ge 2$
[/quote]
The following inequality is stronger.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c)} \ge 2\]
[/quote]

The following seems to be true as well with the same condition:

\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {12abc(ab\plus{}bc\plus{}ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a\plus{}b\plus{}c)^2} \ge 2\]

[quote="shaaam"]

The following seems to be true as well with the same condition:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {12abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a \plus{} b \plus{} c)^2} \ge 2\]
[/quote]
Yes, it's true!  :)

[quote="geniusbliss"]so many stronger ineqs but not even one solution has been posted! someone plz post a solutions either for my sum or arqady's or shaam's stronger ineq.(which i hope avoids terrible computations!)[/quote]

Hi. Arqady's inequality follows from Schur. You can do it by expanding, which is not terrible at all.

Another way of proving Arqady's inequality is to prove following stronger inequality:
\[ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2\]
whcih may not look ugly.

For my inequality, it is [b]not[/b] possible to use the following estimation:
\[ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {12abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2\]
Because the above statement is [b]false.[/b]

My solution uses expanding and sum of the squares.

geniusbliss, I'll prove the shaaam's inequality:
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {12abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a \plus{} b \plus{} c)^2} \ge2\]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, the shaaam's inequality is equivalent to $ f\left(w^3\right)\geq0,$ where
$ f\left(w^3\right) \equal{} 27u^5 \minus{} 36u^3v^2 \plus{} (5u^2 \plus{} 4v^2)w^3.$
But $ f$ is increasing function. Thus, $ f$ gets a minimal value, when $ w^3$ gets a minimal value,
which happens when $ w^3 \equal{} 0$ or two numbers from $ \{a,b,c\}$ are equal.
Id est, for the proof we need to check only two cases:
1)$ c \equal{} 0,$ $ b \equal{} 1,$ which is obviously true;
2)$ b \equal{} c \equal{} 1,$ which gives $ a^2(a \plus{} 4)(a \minus{} 1)^2\geq0.$ Done! :)

[quote="arqady"]geniusbliss, I'll prove the shaaam's inequality:
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {12abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a \plus{} b \plus{} c)^2} \ge2\]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, the shaaam's inequality is equivalent to $ f\left(w^3\right)\geq0,$ where
$ f\left(w^3\right) \equal{} 27u^5 \minus{} 36u^3v^2 \plus{} (5u^2 \plus{} 4v^2)w^3.$
But $ f$ is increasing function. Thus, $ f$ gets a minimal value, when $ w^3$ gets a minimal value,
which happens when $ w^3 \equal{} 0$ or two numbers from $ \{a,b,c\}$ are equal.
Id est, we need to check two cases:
1)$ c \equal{} 0,$ $ b \equal{} 1,$ which is obviously true;
2)$ b \equal{} c \equal{} 1,$ which gives $ a^2(a \plus{} 4)(a \minus{} 1)^2\geq0.$ Done! :)[/quote]

Thanks Arqady!. This method may look confusing in some ways, but I see, it so [b]effective[/b].  :)
Using this method, it should be possible to sharpen my inequality.
My question may sound silly.  :oops:  Why $ abc$ gets minimal value when two numbers from $ \{a,b,c\}$ are equal?

[quote="shaaam"] Why $ abc$ gets minimal value when two numbers from $ \{a,b,c\}$ are equal?[/quote]
Because we need to keep  $ a,$ $ b$ and $ c$ as real numbers.
Indeed, $ a,$ $ b$ and $ c$ are roots of the equation $ x^3 \minus{} 3ux^2 \plus{} 3v^2x \minus{} w^3 \equal{} 0.$
If $ u \equal{}$constant and $ v^2 \equal{}$ constant then, $ w^3$ can move between  the maximal point ordinate and  the minimum point ordinate of the function $ g(x) \equal{} x^3 \minus{} 3ux^2 \plus{} 3v^2x$ or maybe $ w^3 \equal{} 0.$
But for extremal point of $ f$ two numbers from $ \{a,b,c\}$ are equal.  :wink:

[quote="geniusbliss"]For $ a,b,c$ positive reals prove that the following inequality holds -
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3 (abc)^{\frac {1}{3}}}{2(a \plus{} b \plus{} c)} \ge 2$
i know a solution using pqr but thats messy so wishing someone could post simpler solutions.[/quote]

It's old result. By Cauchy-Schwarz Inequality, we must to prove
\[ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)}\plus{}\frac{3\sqrt[3]{abc}}{2(a\plus{}b\plus{}c)}\ge2\]
\[ \Leftrightarrow\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{3\sqrt[3]{abc}}{a\plus{}b\plus{}c}\ge2\]
\[ \Leftrightarrow\sum a^3\plus{}\sum bc(b\plus{}c)\plus{}3\sqrt[3]{abc}(ab\plus{}bc\plus{}ca)\ge2\sum bc(b\plus{}c)\plus{}6abc\]
By Cauchy Inequality
\[ 3\sqrt[3]{abc}(ab\plus{}bc\plus{}ca)\ge9abc\]
We only need to prove
\[ \sum a^3\plus{}3abc\ge\sum bc(b\plus{}c)\]
which is Shur Inequality, we are done.

Of course, we also have a similar one
\[ \left(\frac {a}{b\plus{}c}\right)^{2}\plus{}\left(\frac {b}{c\plus{}a}\right)^{2}\plus{}\left(\frac {c}{a\plus{}b}\right)^{2}\plus{}\frac{15\sqrt[3]{abc}}{4(a\plus{}b\plus{}c)}\ge2\]

[quote="shaaam"]
The following seems to be true as well with the same condition:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {12abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a \plus{} b \plus{} c)^2} \ge 2\]
[/quote]
It's equivalent to
\[ \sum a(a \plus{} b)(a \plus{} c) \plus{} \frac {12abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\ge 2\sum bc(b \plus{} c) \plus{} 4abc\]

\[ \Leftrightarrow\sum a^3 \plus{} 3abc \minus{} \sum bc(b \plus{} c)\ge4abc\left(1 \minus{} \frac {3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}\right)\]

\[ \Leftrightarrow\sum a(a \minus{} b)(a \minus{} c)\ge4abc\frac {\sum (a \minus{} b)(a \minus{} c)}{(a \plus{} b \plus{} c)^2}\]

\[ \Leftrightarrow X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} a)(b \minus{} c) \plus{} Z(c \minus{} a)(c \minus{} b)\ge0\]
with $ X \equal{} a \minus{} \frac {4abc}{(a \plus{} b \plus{} c)^2}$ and $ Y,Z$ similar.
Without loss of generality, assume $ a\ge b\ge c$, we have $ X\ge Y\ge Z\ge0$

So by Vornicu-Schur Inequality, we are done.


I have a geometry software called Geometer's Sketchpad that I would like to use to construct figures. Would this be allowed or should I use traditional paper and pencil methods to construct my figures?

I wouldn't expect you to need this.  Plus, you'd have to drag a printer along since Geomter's Sketchpad isn't installed on the lab computers and in general, schools don't like you hooking your computer up to their printer.  Paper and pencil will do fine.

Sorry, I don't believe I was clear. I meant for the Qualifying Test.

I'm pretty sure you can use anything you want, as long as the work and thought process is your own.

If you meant using Geomter's Sketchpad (or any other software) on the quiz, that's allowed, as long as you don't use brute force. MathPath graders do not look at the presentation, as long as it is legible, so there is no difference.

(or at least that's how it was last year.)


A secretary works in a buiding located 9 blocks east and 8 blocks north of his home. Every day he walks 17 blocks to work.
(a) how many different routes are possible for him?
(b) how many different routes if the street in the easterly direction, which begins 4 blocks east and 3 blocks north of his home, is under water (and he can't swim)?

Please see my comments on your other thread.  Question (a) is a standard exercise in elementary combinatorics, and the answer is $ \binom{9 \plus{} 8}{8}$.  For (b), just count by breaking the thing your counting into the 4 or 5 cases (depending on the meaning of the restriction) for when he takes his fourth north block.


Given $a,b,c>0$, prove that
\[ \frac1{a\sqrt{8a+b}}+\frac1{b\sqrt{8b+c}}+\frac1{c\sqrt{8c+a}}\geq\frac1{\sqrt{abc}}. \]
This is suggested by an inequality by Phan Thanh Nam.

[quote="Together"]It likes VMEO2 in this month.... :mad:

So I think Phan Thanh Nam is the proposer of the inequality in VMEO2 _ month 11 :D[/quote]

The problem proposed by Phan Thanh Nam is
\[ \frac1{a\sqrt{a+b}}+\frac1{b\sqrt{b+c}}+\frac1{c\sqrt{c+a}}\geq \frac3{\sqrt{2abc}}. \]

[quote="Tiks"]$\frac1{a\sqrt{8a+b}}+\frac1{b\sqrt{8b+c}}+\frac1{c\sqrt{8c+a}}\geq\frac1{\sqrt{abc}}$. <=> $\sum(bc/(8a^2bc+ab^2c))^{1/2}\geq 1$

But $(8a^2bc+ab^2c)^{1/2}\leq (1/6)(bc+8ac+9ab)$=>

$\sum(bc/(8a^2bc+ab^2c))^{1/2}\geq \sum(6bc/(bc+8ac+9ab))$

Let $ab=x,bc=z$ and $ac=y$=>

$\sum(6bc/(bc+8ac+9ab))=6(\sum(z/(z+8y+9x)))= 6(\sum(z^{2}/(z^{2}+8zy+9xz)))$
I think you  now that $a_1^{2}/b_1+ a_2^{2}/b_2+ a_3^{2}/b_3\geq (a_1+a_2+a_3) ^2/(b_1+b_2+b_3)$=>

$6(\sum(z^{2}/(z^{2}+8zy+9xz)))\geq 6(x+y+z)^2/(x^2+y^2+z^2+17(xy+xz+yz))\geq 1$
becous $x^2+y^2+z^2\geq xy+xz+yz$.[/quote]

Multiplying the two sides of the inequality by $\sqrt{abc}$, we get
\[ \sqrt{\frac{bc}{8a^2+ab}}+\sqrt{\frac{ca}{8b^2+bc}}+\sqrt{\frac{ab}{8c^2+ac}}\geq1. \]
If we substitute $x=bc, y=ca, z=ab$ then we get
\[ \frac{x}{\sqrt{8yz+zx}}+\frac{y}{\sqrt{8yz+zx}}+\frac{z}{\sqrt{8yz+zx}}\geq1. \]
Now we use the inequality
\[ \frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\geq\frac{(x+y+z)^2}{a+b+c} \]
so that it is sufficient to show that
\[ (x+y+z)^2\geq x\sqrt{8yz+zx}+y\sqrt{8xz+xy}+z\sqrt{8yx+zx}. \]

By Cauchy-Swharz inequality, we have
\[ (x\sqrt{8yz+zx}+y\sqrt{8yz+xz}+z\sqrt{8yz+zx})^2= (\sqrt x\sqrt{8xyz+x^2z}+\sqrt y\sqrt{8xyz+y^2z}+\sqrt z\sqrt{8xyz+z^2x})^2\leq (x+y+z)(24xyz+x^2z+y^2x+z^2x). \]
Note that we have $x^3+y^3+z^3\geq x^2z+y^2x+z^2x$ and $8xyz\leq (x+y)(y+z)(z+x)$, we have
\[ 24xyz+x^3+y^3+z^3\leq 3(x+y)(y+z)(z+x)+x^3+y^3+z^3=(x+y+z)^3. \]
Therefore, we get

\[ (x\sqrt{yz+zx}+ y\sqrt{8xz+yx}+z\sqrt{8yx+xz})^2\leq(x+y+z)^4. \]

And the result follows. This completes the proof. :)

[quote="pvthuan"][quote="Together"]It likes VMEO2 in this month.... :mad:

So I think Phan Thanh Nam is the proposer of the inequality in VMEO2 _ month 11 :D[/quote]

The problem proposed by Phan Thanh Nam is
\[ \frac1{a\sqrt {a \plus{} b}} \plus{} \frac1{b\sqrt {b \plus{} c}} \plus{} \frac1{c\sqrt {c \plus{} a}}\geq \frac3{\sqrt {2abc}}.\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-60-unknown-author.html


At the Doggie Dash, dogs and their owners ran a two-mile race. There were 330 heads and 1040 legs in the race. How many dogs were in the race?

Let $ d$ be the number of dogs and $ p$ be the number of owners.

We then get

$ d\plus{}p\equal{}330$
$ 4d\plus{}2p\equal{}1040$

$ 2d\plus{}p\equal{}520$
$ d\equal{}\boxed{190}$ dogs were in the race.


Given a convex polygon of k veritces $V_1, V_2, ... V_k$ with interior angles be $T_1, T_2, ... T_k $ respectively. Also given side length of edge $(V_i,V_{i+1})= L_i$. Let $M_1, M_2, ... M_k$ be  arbitrary  points such that $M_i $ lies on  side $(V_i, V_{i+1})$ and length of $ (V_i, M_i) = L_{i1} $ and $(M_i, V_{i+1}) = L_{i2} $ where i = 1 to k.
So $ L_i = L_{i1} + L_{i2}. $

Now prove that
$
\sum _{i = 1}^ k 2* L_{(i-1)2}*L_{i1} - \sum_{i = 1 }^ k  [ L_{(i-1)2}^2 + L_{i1}^2 ]* cos(T_i) >= 0.
$

where  $L_{02} = L_{k2}.$

whoa.. dude you need to learn some latex

Sorry dude now i edited it check it out. :)

Awesome.  I just started learning Latex myself a few days ago... of course, I still haven't figured out this problem, but at least I can read it. ;)


geometry
combinatorics proposed
combinatorics

Two square sheets have areas equal to $ 2003$. Each of the sheets is arbitrarily divided into $ 2003$ nonoverlapping polygons, besides, each of the polygons has an unitary area. Afterward, one overlays two sheets, and it is asked to prove that the obtained double layer can be punctured $ 2003$ times, so that each of the $ 4006$ polygons gets punctured precisely once.

Assume that the polygons are vertices of a bipartite graph. (First set of vertices are the polygons on the first sheet, second set of the vertices are the polygons on the second sheet) Draw an edge between two vertices iff their intersection isn't empty after overlaying. We can easily prove that any k polygons from the first sheet must have intersection with at least k polygons from the second sheet. (First set of polygons have an area of k but if they have intersection with less than k polygons they can have an area of at most k-1.) Hence this graph satisfies the conditions of Hall's Marriage Theorem, which tells us that there is a one to one correspondance between the first set of polygons and second set of polygons. Than puncture one point from every 2003 intersection. Done.

Congratulations, Umut, you've found precisely the same solution as I did. Actually, I hoped this problem could've forced the person, who is trying to solve it, to struggle a bit, but the problem turned out to be not that hard for mathlinkers.

nevertheless the problem is quite fascinating, isn't it?

Yes, this is a nice question. It's a useful method to look for the Hall's conditions in this type of finding a one to one correspondance questions.


Did anyone here get into Harvard?

If you did, what should I do to prepare for getting accepted there?

The first question, I don't have any idea.

With respect to the second question, I would recomend you to read (carefully) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30872]this thread[/url], because probably you are going to obtain the same answers (of course, with the few modification applicable).

Best,

I did!!!!
See the other post about Harvard.

me too!!! Are you definitely going or are you deciding between Harvard and another college? I'm currently trying to deciding between Harvard and Yale...and maybe Amherst too (when I get its finaid package). Has anyone visited these schools and what are you opinions?


inequalities
vector
function
limit
advanced fields
advanced fields unsolved

Let T be a linear mapping from a dense vector V subspace of L^1 into the space of measurable complex valued functions satisfying the weak type inequality

m({x:|Tf(x)|> :k}) <= C||f||/k (||  || is the L1 norm) for all f  in  v and all k>0 then T can be extended to L1 as a linear mapping satisfying the same inequality.[/code]

Here's the all-important first step:

[b]Lemma[/b]: Suppose $\{f_n\}$ is a Cauchy sequence in $V.$ Then there exists a subsequence $\{f_{n_k}\}$ such that $\lim_{k\to\infty}T(f_{n_k})(x)$ converges almost everywhere.

Proof: We can choose a subsequence such that $||f_{n_k}-f_{n_{k-1}}||_1\le2^{-2k}.$ Then we write $f_{n_m}=g_0+\sum_{k=1}^mg_k$ where $||g_k||\le2^{-2k}.$

From the linearity of $T,\, T(f_{n_m})=T(g_0)+\sum_{k=1}^mT(g_k)$

From the weak-type (1,1) inequality, $m(\{T(g_k)>2^{-k}\})\le\frac{C2^{-2k}}{2^{-k}}=C2^{-k}.$

Thus in $\sum_{k=m+1}^{\infty}T(g_k),$ the terms of the series are pointwise less than $2^{-k}$ [b]except[/b] on a set of measure at most $\sum_{k=m+1}^{\infty} C2^{-k}.$ This can be made arbitrarily small by making $m$ large enough.

Hence, the series is absolutely convergent except on a set of arbtirarily small measure - which is to say the series converges almost everywhere.

[The word "except" entered on edit to fix typo.]

So far we have this: given $f\in L^1$ we can find a sequence $\{f_n\}$ of functions in $V$ that converge to $f$ in $L^1$ and such that $T(f_n)$ converges both pointwise and in measure to a function which we will call $F$.

We would like to set $T(f)=F.$

There are two things left to prove. The first is that $T(f)$ is well-defined. (That is, the definition does not depend on the sequence.) The second is that $T(f)$ satisfies the appropriate weak-$L^1$ size estimate. I'll leave the last as an exercise, but I will address the first issue here.

Suppose that we have two sequences $\{f_n\}$ and $\{g_m\}$ in $V$ that both tend to $f$ in $L^1.$ Let $F=\lim T(f_n)$ and $G=\lim T(g_m).$ We can find $N$ such that if $n>N$, then $m(\{|T(f_n)-F|>\frac{\epsilon}3\})<\frac{\epsilon}3.$ Similarly, if $m>M,$ then $m(\{|T(g_m)-G|>\frac{\epsilon}3\})<\frac{\epsilon}3.$

Now choose $n$ and $m$ so that $||f_n-g_m||_1\le||f_n-f||_1+||g_m-f||_1<\frac{\epsilon^2}{3C}.$

It follows that $m({|T(f_n)-T(g_m)|>\frac{\epsilon}3\})<\frac{\epsilon}3.}$

Put these estimates together to find that $m(\{|F-G|>\epsilon\})<\epsilon.$ Since we can say this for all $\epsilon>0$, it follows that $F=G\,\text{a.e.}$ and the operator $T$ is well defined.

Thanks for the reply but i dont see how the operator should be defined in L^1.

It's defined as a (pointwise) limit. You said that $V$ is dense in $L^1$, so for any $f\in L^1,\,\exists\,f_n\in V$ such that $\lim f_n=f$ in the $L^1$ sense. We define $T(f)(x)=\lim T(f_n)(x).$ Post #2 is devoted to showing that this limit exists, and post #3 to showing that it is well defined.

But dont you need convergence in measure for ´We can find $N$ such that if $n>N,$ then $m(\{|T(f_n)-F|>\frac{\epsilon}3\})<\frac{\epsilon}3$'?

Yes, we do need convergence in measure. If you examine the proof in #2 carefully, (oops, must fix typo) you'll see that I also directly showed convergence in measure as well as pointwise convergence.


A game of chance is based on spinning a 1-10 spinner two times in succession. The player wins if the larger of the two numbers is greater than 5.

How many possible outcomes are there for win and loose?

This was a school qualifier to the district olmypiad for the state olympiad.)

[hide="Hint"] Count how many ways you can lose first. [/hide]

[hide="hint"]If you know the number of ways to lose, and you know the total number of combinations, you know the number of ways to win.[/hide]

[hide="bigger hint"]To lose, you must have both spins less than or equal to 5.[/hide]

[hide]
Probability that each spin is $\le$ five:
$\frac{1}{2}$ We need to use dependent probability here, because we're making two spins which each count.

You therefore have a $\frac{1}{4}$ chance of losing.
The amount of distinct combinations where no group is repeated (i.e. 21 = 12) is equal to $(9+8+7+6+5+4+3+2+1)$ The one is in there because the digits in 99 do not show up at all until 99.

This is equal to 45.

What did I do wrong? This number doesn't divide by four evenly. I think I'm on the right track though...

[/hide]

[quote="DiscreetFourierTransform"][hide]We need to use dependent probability here, because we're making two spins which each count.[/quote]

Not sure where you're getting this.

And in the second part of your argument you confuse probability with permutations.  You can't use the number of distinct combinations as the basis for a probability argument because they don't have the same probability.

E.g.:  The probability of getting two $2$s is lower than the probability of getting a $1$ and a $3$.

[quote="8parks11"]A game of chance is based on spinning a 1-10 spinner two times in succession. The player wins if the larger of the two numbers is greater than 5.

How many possible outcomes are there for win and loose?

[hide]
This was a school qualifier to the district olmypiad for the state olympiad.)[/quote]
So, there are 100 possible spins (First can be 1-10, second can be 1-10).
To lose, you need 1-5 each time. 5*5 = 25.
That means 100-25 = 75 ways to win, 25 to lose[/hide]


function
limit
topology
complex analysis
complex analysis unsolved

Suppose that $\{f_n\}$ is a sequence of functions on connected and open subset of $\mathbb C$ and $|f_n|<M$. suppose that $f_n(z)$ is convergent for each $z\in S$ which $S$ is a subset of $U$ which has an accumulation point in $U$. Prove that $f_n$ is uniformly convergent on each compact subset of $U$.

Those had better be analytic functions on $U$; do specify these things.

The family is uniformly bounded, so it is a normal family. Every subsequence of the $f_n$ has a further subsequence which converges normally (uniformly on compact sets). The limit $f$ of any such subsequence must be equal to $\lim f_n(z)$ for each $z\in S$. By the identity principle, every normally convergent subsequence converges to the same function $f$. Since normal convergence is given by a metrizable topology, we can apply Urysohn's principle:
If every subsequence of $x_n$ has a further subsequence converging to $x$, the original sequence $x_n$ converges to $x$.
Using this, $f_n$ converges to $f$ normally and we are done.

I don't think that would have been much trouble on my class's exam.


advanced fields
advanced fields unsolved

In the Presence of $ T_1$ space $ X$ ,

[b]X[/b] has  [b]B.W.P [/b] $ \Longleftrightarrow X$ has Countably compact property.

B.W.S (Bolzano Weirestrass Property) : every infinite set subset of $ X$ has a limit point .

Countably Compact : every countable open cover has a finite Subcover.

I need only to prove this side $ \Leftarrow$ i.e. sufficient part.

Waiting your replay.

Thx

[quote="Kalle"]Suppose $ (a_n)$ be a sequence with no convergent subsequence.
Let $ B_n \equal{} X \minus{} \{a_k : k \geq n\}$.  Each $ B_n$ is open, and they form a cover of $ X$, without a finite subcover[/quote]

Thank you  :D


ratio
calculus
trigonometry
geometry proposed
geometry

Let $ ABC$ a triangle where $ \overline{BC} \equal{} 16$ and $ h_a \equal{} 8$, find $ \max\left\{\dfrac{\overline{AB}}{\overline{AC}}\right\}$.

Note: $ h_a$ denote the relative altitude to the side $ BC$.

[hide="Answer"]$ 2$[/hide]

Say that $ BC\equal{}2$, $ h_a\equal{}1$. Call the ratio $ r$. It is clear that $ r\equal{}1$ occurs. Consider when $ r>1$. Let $ B$, $ C$ are fixed on a plane. $ A$ lies on the intersection of a line distance one from $ BC$, and on the circle of apollonius with ratio $ r$. As $ r$ increases, the radius of the corresponding circle of apollonius decreases. This continues until the circle is tangent to the line. In otherwords, $ \angle BCA\equal{}90$. The maximal ratio is then $ \sqrt{5}$.

Again $ BC\equal{}2,\;h_a\equal{}1$. Let $ H$ be the foot of the altitude from $ A$, and $ M$ - the midpoint of $ BC$. If $ MH\equal{}x$
 then $ AB^2\equal{}(1\plus{}x)^2\plus{}1, \; AC^2\equal{}(1\minus{}x)^2\plus{}1$. From this with calculus I got $ \max \frac{AB}{AC}\equal{}\sqrt 2\plus{}1$ and is obtained when $ MH\equal{}\sqrt 2$

[quote]Say that $ BC \equal{} 2$, $ h_a \equal{} 1$. Call the ratio $ r$. It is clear that $ r \equal{} 1$ occurs.[/quote]

Fine.

[quote]
Consider when $ r > 1$. Let $ B$, $ C$ are fixed on a plane. $ A$ lies on the intersection of a line distance one from $ BC$, and on the circle of apollonius with ratio $ r$. As $ r$ increases, the radius of the corresponding circle of apollonius decreases. This continues until the circle is tangent to the line.[/quote]

It's not clear for me. The relation between the radius of circle of Apollonius and the ratio $ \dfrac{\overline{AB}}{\overline{AC}}$ is given for $ R\equal{}\dfrac{r\cdot \overline{BC}}{|r^2\minus{}1|}$, where $ R$ is the radius of circle of Apollonius and $ r$ is the ratio between $ AC$ and $ BC$. And i'm stuck  :oops:.

[quote="delta"]Again $ BC \equal{} 2,\;h_a \equal{} 1$. Let $ H$ be the foot of the altitude from $ A$, and $ M$ - the midpoint of $ BC$. If $ MH \equal{} x$
 then $ AB^2 \equal{} (1 \plus{} x)^2 \plus{} 1, \; AC^2 \equal{} (1 \minus{} x)^2 \plus{} 1$. From this with calculus I got $ \max \frac {AB}{AC} \equal{} \sqrt 2 \plus{} 1$ and is obtained when $ MH \equal{} \sqrt 2$[/quote]

If $ MH \equal{} \sqrt 2$ then $ HC\equal{}1\minus{}x\equal{}1\minus{}\sqrt{2}<0$...

[quote="LoboSolitario"]
If $ MH \equal{} \sqrt 2$ then $ HC \equal{} 1 \minus{} x \equal{} 1 \minus{} \sqrt {2} < 0$...[/quote]
Meaning $ C$ is between $ M$ and $ H$ and $ \angle BCA > 90^\circ$

I would like to see an entire geometric solution. Somebody can help? I'm almost sure that the Apollonius circle is the kee.

The appolonius' circle is still tangent to that line, but It is shifted over a bit. This should give the same answer.

For sure Altheman, but i didn't find any mistakes in delta approach, did you?

About time, after delta solution, if it's not clear yet, the answer given in my first message is wrong.

This problem was moved 2 times from different foruns, any idea for a full geometric solution? I'm sure this is not that hard.

Thanks in advance.

P.S. Trig is allowed too.


I have some questions about Math Zoom

first: what do we do in the study session?

second: how do you determine which group and level we're in.

third: can we bring laptops, etc. if we need it for other work (like programming :D)

Thanks and post other questions here!

[quote="helagha"]I have some questions about Math Zoom

first: what do we do in the study session?[/quote]
For the self-study time in the evenings, you can do homework, review, study, ask questions.  Sometimes we have activities led by  counselors.

[quote]second: how do you determine which group and level we're in.[/quote]
It is determined by your math contest experience, assessment test, etc.  It can be adjusted.

[quote]third: can we bring laptops, etc. if we need it for other work (like programming :D)
Thanks and post other questions here![/quote]
Laptops are not required for the math classes, but you may bring it for other work.

 :)


Find all positive integers x, y so that 7^x + x^4 + 47 = y^2
I don't remember the source.

I have already given a solution for this problem on the forum:

if n is odd then consider the ecuation modulo 4 and you get that  7^x + x^4 + 47==3(mod 4) and a perfect square can't be 3(mod 4)

for x=2p you get that for p>=4

7^(2p)+(2p)^4+47 is between two squares: (7^p)^2 and (7^p+1)^2 and this is easy to prove!

take now p=1,2,3 and get only sol for p=2 which is y=52 and x=4

cheers!! :D :D

what's your solution J?

yes, I got exactly the same solution ;)
sorry, that I didn't check the forum for the problem before I posted it ;)

I think the author is Cristinel Mortici.


Can u help me?

A student scored 85 on an exam containing 100 problems.  She tried to solve all the problems.  Her score was figured by subtracting two times the number of wrong answers from the number of correct answers.  How many problems did she solve correctly.

[hide]She solved 95 problems correctly. Since she missed 5, there would be 10 points taken off, giving her a score of 85. [/hide]

huh?/

the question doesn't make any sense

how could a girl be a he later on? :D

Oops...... my bad.  :blush: I guess on most problems I just assume it's a he and not a she if I don't look at it that carefully.

This makes me think of my planets project I'm doing in science.... Instead of our creature being an it or whatever, since we didn't decide what gender it would be, I noticed that I was automatically writing that it was a he.

Weird... :blush:

I'll edit that.  :)


trigonometry
function
complex numbers
algebra unsolved
algebra

Is there an exact result?

If not, how accurate do you wish the approximation to be?

[quote="tranthanhnam"]Caculate  value :$cos7^o$ ( Don't use caculator )[/quote]

 Allows using Maclaurin Series for $cos(x)$ ?

I was taught a way to calculate some exact values:

1) Angle $15^\circ$:
$15^\circ=45^\circ-30^\circ$. Using trigonometry, we get
$\sin 15^\circ=\frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos 15^\circ=\frac{\sqrt{6}+\sqrt{2}}{4}$.

2) Angle $18^\circ$:
$5\cdot 18^\circ=90^\circ$, so $2\cdot 18^\circ+3\cdot 18^\circ=90^\circ$ and then $\cos(2\cdot 18^\circ)=\sin(3 \cdot 18^\circ)$.
After some trigonometry equalities to put it all in terms of $x=\sin 18^\circ$, we get $4x^2+2x-1=0$. The positive solution is $\sin 18^\circ=\frac{\sqrt{5}-1}{4}$.

3) Angle $3^\circ$:
$3^\circ=18^\circ-15^\circ$, so we act like in 1) again.
$\cos 3^\circ=\frac{\sqrt{10-5\sqrt{3}}-\sqrt{2-\sqrt{3}}+\sqrt{20+4\sqrt{5}+10\sqrt{3}+2\sqrt{15}}}{8}$. I hope you haven't given up yet.

4) Angle $1^\circ$:
$3\cdot 1^\circ=3^\circ$.
After some trigonometry equalities to put it all in terms of $x=\cos 1^\circ$, we get $4x^3-3x-c=0$, with $c=\cos 3^\circ$.
Using Cardano's solution to the simplified third-degree equation $x^3+px+q=0$, being $x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$, we should get the [b]exact[/b] answer. The result will be a really long and hard expression, but that didn't stop mathematicians a century ago (without calculators), and shouldn't stop us either.

5) Angle $7^\circ$.
This should be an easy calculation using previous results. For instance, we could get our answer from $7^\circ=2\cdot 3^\circ+1^\circ$.

I certainly won't try, but it could be done!

[Edit: I corrected Cardano's formula from another book. I'll give another formula later].

I have a simpler expression for $cos(3^o)$,

$\frac{\sqrt{8+\sqrt{3}+\sqrt{3}\sqrt{5}+\sqrt{10-2\sqrt{5}}}}{4}$

I used the fact that $2sin(126^o)=\phi$ and the values for $120^o$ and then the half-angle formula.

oh! how about $\cos \sqrt{2}$  it becomes more and more complex...

$\cos(\sqrt 2 )$ is transcedental, thus in what terms do you want to express it¿

[quote="lordWings"]I was taught a way to calculate some exact values:

1) Angle $15^\circ$:
$15^\circ=45^\circ-30^\circ$. Using trigonometry, we get
$\sin 15^\circ=\frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos 15^\circ=\frac{\sqrt{6}+\sqrt{2}}{4}$.

2) Angle $18^\circ$:
$5\cdot 18^\circ=90^\circ$, so $2\cdot 18^\circ+3\cdot 18^\circ=90^\circ$ and then $\cos(2\cdot 18^\circ)=\sin(3 \cdot 18^\circ)$.
After some trigonometry equalities to put it all in terms of $x=\sin 18^\circ$, we get $4x^2+2x-1=0$. The positive solution is $\sin 18^\circ=\frac{\sqrt{5}-1}{4}$.

3) Angle $3^\circ$:
$3^\circ=18^\circ-15^\circ$, so we act like in 1) again.
$\cos 3^\circ=\frac{\sqrt{10-5\sqrt{3}}-\sqrt{2-\sqrt{3}}+\sqrt{20+4\sqrt{5}+10\sqrt{3}+2\sqrt{15}}}{8}$. I hope you haven't given up yet.

4) Angle $1^\circ$:
$3\cdot 1^\circ=3^\circ$.
After some trigonometry equalities to put it all in terms of $x=\cos 1^\circ$, we get $4x^3-3x-c=0$, with $c=\cos 3^\circ$.
Using Cardano's solution to the simplified third-degree equation $x^3+px+q=0$, being $x=\sqrt{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt{\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$, we should get the [b]exact[/b] answer. The result will be a really long and hard expression, but that didn't stop mathematicians a century ago (without calculators), and shouldn't stop us either.

5) Angle $7^\circ$.
This should be an easy calculation using previous results. For instance, we could get our answer from $7^\circ=2\cdot 3^\circ+1^\circ$.

I certainly won't try, but it could be done![/quote]
The Cardano's formula you use is not true,because,$(\frac{q}{2})^2 + (\frac{p}{3})^3 <0$

We can use a correct version  :P  of Cardano's formula to get the three complex solutions to the cubic equation, using the complex numbers carefully. We might have to deal with complex numbers to finally reach a real value (Cardano himself was astonished, complex numbers were created precisely to deal with these kind of things, in a time when even negative numbers were not very popular). More details:[url=http://en.wikipedia.org/wiki/Cubic_equations]wikipedia[/url] (look for the paragraph that begins with "Note that there are six possibilities...").

But the point was that $\cos 1^\circ$ can be calculated exactly with only additions, substractions, multiplications, divisions, square roots and cubic roots.

but finally,where is the formula $\cos 1^o$,using Cardano's I cannot find the exact answer,It depends a function of $\arccos$

I'll quote the text from Wikipedia (I haven't fully checked it, but it seems trustworthy):
[hide="text"]Equation: $x^3+px+q=0$, $p\neq 0$, $q\neq 0$.
Solution: $x=\frac{p}{3u}-u$, where $u=\sqrt[3]{\frac{q}{2}\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$.
Comment: Note that there are six possibilities in computing $u$, since there are two solutions to the square root ($\pm$), and three complex solutions to the cubic root. However, which sign of the square root is chosen does not affect the final resulting  $x$.[/hide]

I suppose that there's no problem if $\frac{q^2}{4}+\frac{p^3}{27}<0$, since you can work with complex numbers. I'll check it later.

[Edit: correcting errors]

[quote="epitomy01"]how would you prove that $cos(\sqrt 2)$ is transcendental?[/quote]
It follows from the theorem of Lindemann-Weierstrass.


limit
absolute value
calculus
calculus computations

Determine whether each the following sequences converges or diverges. If it converges, find the limit; otherwise
show the sequence diverges.

These look like homework problems, and the best way to help you is to first look at what you have done and then point out what you have done wrong [Since you haven't answered JBL's question, I will just add a few hints for your problems]:

i. For large n, $ \frac {2n \plus{} 1}{3n \minus{} 1}$ looks like $ \frac {2n}{3n}$.

iii. $ e^x \equal{} \lim_{n \to \infty}\left({1 \plus{} \frac {x}{n}}\right)^{n}$

iv. Compare the absolute value of the given sequence with the absolute value of $ b_{n} \equal{} n$. Look up a theorem that compares two sequences.

v. Do you know how a sequence is related to its subsequences?

vi. Compare the size of the top and that of the bottom.

Or to rephrase that hint for vi: in moving from $ a_n$ to $ a_{n\plus{}1},$ you multiply by (what?) which is approximately the same as (what?)  - and if you keep multiplying by that, what happens in the long run?


Let $f,g$ be polynomials with rational coefficients. Let $F,G$ denote the sets of values of $f,g$ at rational numbers. Prove that $F=G$ id and only if there exist $a\in\mathbb Q\backslash\{0\}$ and $b\in\mathbb Q$ which $f(x)=g(ax+b)$

The problem is so exciting, Hey guys try it.


Alex needs to catch a train. The train arrives randomly some time between 1:00 and 2:00, waits for 10 minutes, and then leaves. If Alex also arrives randomly between 1:00 and 2:00, what is the probability that the train will be there when Alex arrives?

The solution provided is a 2D diagram with 60 base 60 height (sample space 3600). Probability of Alex and train are both present is 550/3600. why...

Thanks in advance!

EDITED: sorry. the probability is 550/3600.

550? That can't be a probability, even if it's a percent. Are you sure that's what it showed?

This is an example of geometric probability. (You can read more about it [url=http://www.artofproblemsolving.com/Wiki/index.php/Geometric_probability]here[/url].)
Basically, it involves making a 2x2 square from 1:00 to 2:00 on each "axis."
Finding the times that Alex catches the train is basically the area on the 2d plane that satisfies the conditions.
The 60 * 60 sample space is basically the time in minutes between 1:00 and 2:00


Aqui van los problemas de la seman 2, soluciones hasta el 8 de agosto!

Suerte!!!

1.

Sean $abc$ números reales positivos tales que $abc=1$, pruebe que:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c \]
2.

Sea $a>2$ un entero, pruebe que existe un número $n$ con exactamente $2006$ divisores primos tal que $n|a^{n}-1$. Que pasa si $a$ es $2$?.

3.

Sean $ABC$ un triangulo y sea $P$ un punto en su interior y $M$ un punto en el segmento $PH$, donde $H$ es el ortocentro del triangulo.  denotemos por $f(X)$ perimetro del triangulo pedal de un punto $X$.

Pruebe que $f(P)\geq f(M)$.

4.

En el 2050 habra 2050 ciudades en el mundo atendidas por 11 aerolineas, de modo que una ruta entre dos ciudades es cubierta por exactamente una aerolinea. Pruebe que una aerolinea puede ofrecer un viaje que pase por un número impard e ciudades y vuelva de nuevo a la ciudad inicial. ¿Que pasa con 2006 ciudades?

[quote="Pascual2005"]

4.En el 2050 habra 2050 ciudades en el mundo ... ¿Que pasa con 2006 ciudades?[/quote]

Pascual, es correcto  los numeros 2050??
o son 2005... ??

Gracias ...

[quote="José"]El dato del año 2050 no tiene importancia, hay que tratar de demostrar el problema para 2050 ciudades, y para 2006 ciudades[/quote]

claro ...logicamente me refiero al numero y no al año...
y mi pregunta es si es correctamente 2050 y no 2005, ..
puesto que existe un indicio de extender la idea de la solucion
a un numero, que es 2006, osea su consecutivo ...
es esto.. lo que me indujo a hacer la consulta ..... ;)

Carlos ..

[quote="Pascual2005"]El dato es 2050![/quote]

Claro, pero capaz, como algunos problemas en Argentina, que la pregunta es para $n$, y después preguntan también qué pasa para $n+1$. Bueno, suerte en la resolución!

En el problema 4, si la ruta de $A\rightarrow B$ esta cubierta por una aerolinea tambien  la ruta de $B\rightarrow A$ esta cubierta por la misma aerolinea?

si!

Aqui cuelgo los trabajos que recibi...

Felicidades a quienes participaron, colgare las soluciones restantes mañana...

amigo que paso con el resto de soluciones?

3. Prueben el siguiente lema: Sea $l$ una linea y $P$ un punto que varia en la linea, entonces $f(P)$ es convexa. Para esto basta probar que $f(P)+f(Q)\geq f(M)$ donde $M$ es el punto medio de $PQ$, para este fin use el siguiente lema:

Lema: Sea $ABCD$ un cuadrilatero y $M,N$ los puntos medios de $AB$ y $CD$, entonces $AC+BD\geq 2MN$.

Con este lema probar el hecho mencionado anteriormente es trivial.

El lema implica que si una linea pasa por el ortocentro, esta función es convexa con minimo en $H$ (pruebe que H es el minimo $f(H)$).

El resultado se sigue por convexidad.

4. Pista: pruebe por induccion en $n$ que en $2^{n}+1$ ciudades hay ciclo impar usando que un grafo no tiene ciclos impares unicamente si es bipartito. Y pruebe por induccion tambien que en $2^{n}$ ciudades esto no ocurre necesariamente.

[quote="Pascual2005"]3. Prueben el siguiente lema: Sea $l$ una linea y $P$ un punto que varia en la linea, entonces $f(P)$ es convexa. Para esto basta probar que $f(P)+f(Q)\geq f(M)$ donde $M$ es el punto medio de $PQ$, para este fin use el siguiente lema:

Lema: Sea $ABCD$ un cuadrilatero y $M,N$ los puntos medios de $AB$ y $CD$, entonces $AC+BD\geq 2MN$.

Con este lema probar el hecho mencionado anteriormente es trivial.

El lema implica que si una linea pasa por el ortocentro, esta función es convexa con minimo en $H$ (pruebe que H es el minimo $f(H)$).

El resultado se sigue por convexidad.
[/quote]

Mi solucion es asi, siento no abertela mandado jejeje

Solución Chen:

Reto 2
1.-Usando la desigualdad: $MA %Error. "geMG" is a bad command.
$
$\frac{\frac{a}{b}+\frac{a}{b}+\frac{b}{c}}{3}\ge{\frac{a^{2}}{bc}}^\frac{1}{3}=a$
$\frac{\frac{b}{c}+\frac{b}{c}+\frac{c}{a}}{3}\ge{\frac{b^{2}}{ac}}^\frac{1}{3}=b$
$\frac{\frac{c}{a}+\frac{c}{a}+\frac{a}{b}}{3}\ge{\frac{c^{2}}{ab}}^\frac{1}{3}=c$
Sumando, obtenemos el resultado.

Yo mande esta solucion a tiempo, no se si es que no le llego a pascual o que:

Problema 2
Lema 1: Lifting the exponent
Si p es un primo impar y a,b son enteros y n un entero positivo que cumplen que:
$p\not| a$
$p^{\alpha}\parallel a-b$
$p^{\beta}\parallel n$
Entonces tambien se cumple que:
${p^{\alpha+\beta}\parallel a^{n}-b^{n}}$
Este lemma es conocido
Lema 2: Caso $p=2$ del lema 1(tambien es conocido)
Si $2\not| a$
$2^{\alpha}\parallel \frac{a^{2}-b^{2}}{2}$
$2^{\beta}\parallel n$
$\implies 2^{\alpha+\beta}\parallel a^{n}-b^{n}$

(a)Probaremos que para todos los entero $t$ existe un $n$ producto de $t$ primos que cumple que:
$n|a^{n}-1$ si $a>2$.
Usaremos induccion fuerte.
Caso Base:
Como $a > 2$ se tiene que existe un primo $p_{1}$ tal que $p_{1}|a-1$ y como $a-1|a^{p}_{1}-1$ ya esta el caso base.
Paso inductivo:
Supongamos que para todo entero $k$ menor que $t$ funciona y que siempre existe un primo que no divide a $n$ y que si divide a $a^{n}-1$(en todos los casos anteriores). Mostraremos que para $t=k+1$ tambien funciona.
Entonces sean  $n_{i}=\prod_{j=1}^{i}p_{j}$ que cumplen que $n_{i}|a^{n_{i}}-1$. Lo unico que tenemos que probar para acabar es que existe un primo $p_{k+1}$, que no divide a $n_{k}$ y que divide a $a^{n_{k}}-1$, porque entonces $n_{k}\cdot p_{k+1}| a^{n_{k}}-1 | a^{n_{k}\cdot p_{k+1}}-1$.
Por el Lema 1 tenemos que $p_{i}^{\alpha_{i}}\parallel a^{n_{i}}-1$ entoces $p_{i}^{\alpha_{i}}\parallel a^{n_{k}-1}-1$ y $p_{i}^{\alpha_{i}}\parallel a^{n_{k}}-1$ para todos los $i<k$ $(*)$ (en el caso que $2|n_{i}$ podemos escoger siempre a $2$ como $p_{1}$ y por el lema 2 la potencia no sube despues de $n_{2}$. Eso significa que con 2 pasa lo mismo que con los otros primos)

Ahora miremos que pasa con la potencia de $p_{k}$. Para eso sabemos que $n_{k}= n_{k-1}\cdot p_{k}$.
Ahora si:
$p_{k}^{\alpha_{k}}\parallel a^{n_{k-1}}-1$ y $\alpha_{k}$ es positivo por la hipotesis de induccion.
y tenemos que:
$p_{k}\parallel p_{k}$
entonces por el Lema 1 tenemos que:
$p_{k}^{\alpha_{k}+1}\parallel a^{n_{k}}-1= \left(a^{n_{k-1}}-1\right)\left(1+a^{n_{k-1}}+...+a^{n_{k-1}(p_{k}-1)}\right)$
Ahora usando $(*)$ tenemos que ningun primo $p_{i}$ para $i <k$ divide el lado derecho del ultimo producto. Y ademas $p_{k}\parallel \left(1+a^{n_{k-1}}+...+a^{n_{k-1}(p_{k}-1)}\right)$. Entonces si no hubiera un primo $p_{k+1}$ se tendria que tener que:
$p_{k}= \left(1+a^{n_{k-1}}+...+a^{n_{k-1}(p_{k}-1)}\right)$ que es imposible pues $p_{k}< p_{k}^{\alpha_{k}}<a^{n_{k-1}}$.  Con eso queda demostrado el problema.  Pues en particular se cumple para $t=2006$.
(b) Si $a=2$ la unica solucion para $n|2^{n}-1$ es $n=1$.
Demostracion:
Supongamos que existe un $n$ que cumple las condiciones del problema y es mayor que $1$.
Lo primero es que $n$ tiene que ser impar porque $2^{n}-1$ es impar para $n$ mayor que $1$.
Sea $p$ el menor primo que divide a $n$. Sabemos que:
$2^{n}\equiv 1$ (mod $p$)
Sea $d$ el orden de $2$ modulo $p$. Tenemos que $d|n$ pero $d<p$ entonces $d$ tiene que ser $1$, pero si eso pasa significa que $p| 2-1=1$ que es una contradiccion.
Con esto queda demostrado el problema.