Last month, I sent a long email to Bryan Sullivan of ARML Committee about ARML books. I was trying to see if ARML would let AoPS to have their problems up in the Resource page. I didn't receive a reply until now. It was a short but a good news.

[quote="From email"]We are printing the past 2 books?83-88(June) and 89-03 for ARML[/quote]

I know it would be better if we can simply add them (so not paying :roll: ) but again, ARML books are good books that you should buy. Only issue was that out of 4 books, only 2 books were easy to be accessed (89-94 and 95-03) and still, they were pretty hard to get (well, get both of them I guess). So, if this email works out, it's a good news for everybody ------ to have ARML contest problems from 83 to 03 in two books!  :)

There's one practical consideration, from a coach's point of view. A significant portion of ARML involves group work - shared problem solving. It's very hard to practice the Team Round, the Power Round, or the Relay round without actually getting everyone into the same room and have them experience the same time pressure and the same interactions as in the real thing. You have to practice leadership, you have to practice coordination.

So at practice sessions, I'd like to be able to use a Team Round or a Power Round that [i]isn't[/i] something that a third of the people in the room have already seen. And for that reason, I don't want to see it all posted out here in the open.


geometry
perimeter
linear algebra
matrix

Write 3 equations in 3 variable for each problem

The sum of the digits of a three digit number is 11.  The hundreds digt exceeds the sum of the tens digit and the units digit by 1.  When the digits are reversed, the new number is 396 less than the original number.  Find the number.  Write the three equations

One side of a triangle is 3 inches longer than another side of the riangle.  The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle.  If the perimeter is 29 inches, find the length of each side.  Writge the three equations

#1
[hide]we can have the original #'s hundreds digit be x, tens digit be y and ones digit be z. our first equation is $x+y+z=11$ our second equation would be $x=y+z+1$ our 3rd equation deals with the value of the 3 digit numbers so in order to have the value of the digits xyz, we need to write the value as $100x+10y+z$ we then get our equation as $100x+10y+z=100z+10y+x=396$ we simplify it to $x-z=4$ we have 3 nice equations now. we can solve with matrices or whatever and we get x=6, y=3, z=2. Therefore, our original number is $632$[/hide]

[hide="Solution for #1"]
$a+b+c=11 \\b+c+1=a \\100c+10b+a=100a+10b+c-396$

Elementary matrix operations or basic algebra yields:

$a=6\\b=3\\c=2$
[/hide]

EDIT: Urrrrrrg, I was 10 seconds too late :(.

[hide="Solution to #2"]
Let $a, b, c$ be the sides of a triangle.

$a=b+3 \\b+c-1=a \\a+b+c=29$

Elementary row reduction leads to $a=14, b=11, c=4$.
[/hide]

the broken light:

do not double post please

Solution 1:

[hide]Let a be the hundreds digit, b be the tens digit, and c be the unit digit of the original number.

(1) a + b + c = 11

We can write abc = 100a + 10b + c
                    cba = 100c + 10b + a
=> abc - cba = 99a - 99c = 396
Then we simply to have: (2) a - c = 4

(3) a = b + c + 1

Then we solve three equations. From (2) we have a = c + 4. Substitue in (3), b = 3.

Then a + c = 11 - 3 = 8

Solve: a + c = 8, a - c = 4, we will have c = 2, and a = 6.

So the orginial number will be 632.

Sorry for not typing by latex. [/hide]

- TO brokenlight: Your solution for number 2, the first equation doesn't consist of three variables.

[quote="yvette484848"]thankyou so much :D
can u figure out this problem
The sum of Adam, Betsy and Carol's ages is 47.  Half the sum of Adam's and Betsy ages less one is Carol's age.  If Adam is 6 years older than Betsy, find each person's age.

I can figure out each person's age, but i cant find the three equations, so can someone help please[/quote]

is the answer 19, 13,15?? this is what i did  :D

x+y+z= 47

1/2 (x+y) -1= z

x=y+6

just solve these  easy equations and get the answer ...

if there 's any problem then i 'll post how to solve the equations too.... :D


A large bag of coins contains pennies, dimes, and quarters.  There are twice as many dimes as pennies and three times as many quarters as dimes.  An amount of money which could be in the bag is

$ \textbf{(A)}\ \$306 \qquad \textbf{(B)}\ \$333 \qquad \textbf{(C)}\ \$342 \qquad \textbf{(D)}\ \$348 \qquad \textbf{(E)}\ \$360$

Let $ p$ be the number of pennies in the bag, $ d$ be the number of dimes in the bag and $ q$ be the number of quarters in the bag. From the given we have $ 2p \equal{} d$, and $ 3d \equal{} q$. The total number of cents in the bag is $ p \plus{} 10d \plus{} 25q \equal{} \frac {d}{2} \plus{} 10d \plus{} 75d \equal{} \frac {171d}{2}$. Thus, the total number of cents in the bag must be a multiple of $ \frac{171}{2}$. $ \boxed{(C)}$ is the only answer that fits.

We have that $ 2p\equal{}d$ and $ 3d\equal{}6p\equal{}q$.  Letting the value of a penny be $ 1$, then our total is $ p\plus{}20p\plus{}150p\equal{}171p$.  The only value in our answers divisible by $ 171$ is $ \boxed{C}$.


vector
calculus
LaTeX
analytic geometry
graphing lines
slope
derivative

Sorry, I've never used $\LaTeX$ before.

Find the slope of the tangent line at $(1,2,322)$ to the curve of intersection of the surface

$z=2x^{3}+5y^{6}$

a) the plane $x=1$

b) the plane $y=2$

I'm really not sure how to go about this one. Should I take the derivative of $z$? Take the partials of $z$?? Let $x=1$ for the derivative???

THanks for any help.

First of all use $\LaTeX$. See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php]here[/url] for more.
$\line(1,0){200}$

Now the problem could be solved in different ways, but one way is to consider parameterizations.
Because we are investigating a curve in $\mathbb{R}^{3}$ and it can be described by
a vector $\mathbf{r}(t)$ which sweeps out the curve.
The derivative of this vector $\mathbf{r}'(t)$ is the tangent vector to the curve, which is more or less the same
as the slope.

a) For the first one, $\mathbf{r}(y)=(x(y),y,z(y))=(2,y,2+5y^{6})$.
   Then $\mathbf{r}'(y)=(0,1,30y^{5})$ and this is evaluated at $(1,\mathbf{2},322)$
  and you have the direction of the tangent line there.

Observe that the curve lie in the plane $x=2$ so the tangent line could not have a direction out of that plane, which is why we have a zero in $\mathbf{r}'(y)=(\mathbf{0},1,30y^{5})$

b) This is very similar to a), so perhaps you could solve it yourself.

I recommend looking over the geometric representations of partial derivatives.

Yes I agree and it is probably what is expected here.

@[b]pilly151[/b] Recall what the partial derivative with respect to say $y$ means and you are
almost done.

Actually you could also touch [url=http://en.wikipedia.org/wiki/Directional_derivative]this subject[/url], and just choose direction in the plane $x=2$,
so that the directional derivative coincides with the usual.


sketch the curve in the argand diagram determined by $ arg(z\minus{}1)\equal{}arg(z\plus{}1)\plus{}\frac{\pi}{4}$

[hide]how would you know that this shape is represented by a circle?[/hide]

Answer

Let $ z \equal{} a\plus{}ib$

$ \Rightarrow\arg{((a\minus{}1)\plus{}ib)}\equal{}\arg{((a\plus{}1)\plus{}ib)}\plus{}\frac{\pi}{4}$

$ \Rightarrow\tan{(\arg{((a\minus{}1)\plus{}ib)})}\equal{}\tan{\left(\arg{((a\plus{}1)\plus{}ib)}\plus{}\frac{\pi}{4}\right)}$

$ \Rightarrow\frac{b}{a\minus{}1}\equal{}\frac{\tan{(\arg{((a\plus{}1)\plus{}ib)})}\plus{}\tan{\frac{\pi}{4}}}{1\minus{}\tan{(\arg{((a\plus{}1)\plus{}ib)})}\tan{\frac{\pi}{4}}}$

$ \Rightarrow\frac{b}{a\minus{}1}\equal{}\frac{\frac{b}{a\plus{}1}\plus{}1}{1\minus{}\frac{b}{a\plus{}1}}$

$ \Rightarrow\frac{b}{a\minus{}1}\equal{}\frac{a\plus{}b\plus{}1}{a\minus{}b\plus{}1}$

$ \Rightarrow b(a\minus{}b\plus{}1) \equal{} (a\minus{}1)(a\plus{}b\plus{}1)$

$ \Rightarrow ab\minus{}b^{2}\plus{}b \equal{} a^{2}\plus{}ab\plus{}a\minus{}a\minus{}b\minus{}1$

$ \Rightarrow a^{2}\plus{}b^{2}\minus{}2b\minus{}1 \equal{} 0$

This is the locus of a circle in the complex plane with centre $ (0,\, 1)$ and radius $ \sqrt{2}$

Hmm. I didn't think of it like that. THanks for your detailed proof
----------------------------------

[b](a)[/b]On an Argand diagram shade in the region containing all the points representing complex numbers $ z$ such that both $ |z|\le2$ and $ \frac{\pi}{4}%Error. "learg" is a bad command.
(z+2)\le\frac{\pi}{2}$

[b](b)[/b]Find the possible values of $ |z|$ and arg(z) for such complex numbers $ z$

Why is the modulus and argument radiating from the origin. Why not the chord the modulus and the angle between the positve x-axis and chord the argument??

[hide="diagram of (a)"][url=http://img132.imageshack.us/my.php?image=87504685wu7.jpg][img]http://img132.imageshack.us/img132/5849/87504685wu7.th.jpg[/img][/url][/hide]

Hello sludgethrower!

Be careful : you used many "$ \implies$" so you know the locus is necessarilty on the circle you've found. But you are not sure the locus is the [u]whole [/u]circle.

For example $ z=(1-\sqrt{2})i$ is on the circle but ${{ \arg(z-1)-arg(z+1)=-\frac{3\pi}{4}}\neq\frac{\pi}{4}}$

In fact, the required locus is only the part of the circle above x-axis.

Yes, I see where the problem lies:

${ \tan{x}=\tan{y+\frac{\pi}{2}}}\Longleftrightarrow x = y+\frac{\pi}{2}+n\pi$ where $ n\in Z$

Is there any way that I can remedy my proof so that it becomes more specific?


Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2\geq 3$, prove that
\[ \frac{1+xy}{z^2+xy}+\frac{1+yz}{x^2+yz}+\frac{1+zx}{y^2+zx}\geq3.  \]

anyway,it is a nice inequality.
and here is a solution for $x^2+y^2+z^2 \le 3$
by Cauchy-Schwarz,we get:
$\left(\sum_{cyc}(z^2+xy)(1+xy) \right)\left(\frac{1+xy}{z^2+xy}+\frac{1+yz}{x^2+yz}+\frac{1+zx}{y^2+zx} \right) \ge \left(3+xy+yz+zx \right)^2$
so we should only to prove:
$\left(3+xy+yz+zx \right)^2 \ge 3\left(\sum_{cyc}(z^2+xy)(1+xy) \right)$
$\iff 9+6\sum_{cyc}xy+(\sum_{cyc} xy)^2 \ge 3\sum_{cyc}x^2y^2+3\sum_{cyc}xyz^2+9+3\sum_{cyc}xy$
$\iff 3\sum_{cyc}xy \ge 2\sum_{cyc}x^2y^2+\sum_{cyc}xyz^2$
by $x^2+y^2+z^2 \le 3$
we should only to prove:
$(\sum_{cyc}xy)(\sum_{cyc}x^2) \ge 2\sum_{cyc}x^2y^2+\sum_{cyc}xyz^2$
$\iff \sum_{sym}x^3y \ge 2\sum_{cyc}x^2y^2$
obvious true :)

[quote="zhaobin"]hi,I think there must be something wrong,
take $x=y=z=2$,then the inequality is obvious false.
and I guess the condition may be $x^2+y^2+z^2 \le 3$ or $x^2+y^2+z^2 = 3$
right :) ?[/quote]

I am sorry, the condition must me $x^2+y^2+z^2\leq3$.

Nice solution [b]zhaobin[/b], but I have two questions:

[quote="zhaobin"]$9+6\sum_{cyc}xy+\boxed{(\sum_{cyc}xy)^2} \ge 3\sum_{cyc}x^2y^2+3\sum_{cyc}xyz^2+9+3\sum_{cyc}xy$[/quote]
Shouldn't it be $\sum_{cyc}x^2y^2$ :?:

[quote="zhaobin"]$(\sum_{cyc}xy)(\sum_{cyc}x^2) \ge 2\sum_{cyc}x^2y^2+\sum_{cyc}xyz^2$
$\iff \sum_{sym}x^3y \ge 2\sum_{cyc}x^2y^2$[/quote]
How do you get that final inequality? Why it is true?

Thanks :)

[quote="Ezbakhe Yassin"]Nice solution [b]zhaobin[/b], but I have two questions:

[quote="zhaobin"]$9+6\sum_{cyc}xy+\boxed{(\sum_{cyc}xy)^2} \ge 3\sum_{cyc}x^2y^2+3\sum_{cyc}xyz^2+9+3\sum_{cyc}xy$[/quote]
Shouldn't it be $\sum_{cyc}x^2y^2$ :?:
[/quote]

NO,It should be $\boxed{(\sum_{cyc}xy)^2}$
because:$\left(3+xy+yz+zx \right)^2=\left(3+xy+yz+zx \right)^2$ :)
[quote="Ezbakhe Yassin"]
[quote="zhaobin"]$(\sum_{cyc}xy)(\sum_{cyc}x^2) \ge 2\sum_{cyc}x^2y^2+\sum_{cyc}xyz^2$
$\iff \sum_{sym}x^3y \ge 2\sum_{cyc}x^2y^2$[/quote]
How do you get that final inequality? Why it is true?

Thanks :)[/quote]
for the final inequality,I think it is just muirhead's.
or you can prove it by:
$x^3y+y^3x \ge 2x^2y^2$
and similar another two.

[quote="wudud"]sorry ! from which book did you got the problem ?  :huh:[/quote]

My book!  :P  :blush:

It has not been published. I hope this year.

There is a shorter proof but it uses a stronger result
$\sum_{cycl}\frac{3+3xy}{z^2+xy} \ge \sum_{cycl}\frac{x^2+y^2+z^2+3xy}{z^2+xy} = 3+\sum_{cycl}\frac{(x+y)^2}{z^2+xy} \ge 9$
where the last inequality follows from
[url]http://www.artofproblemsolving.com/Forum/topic-60128.html[/url]


geometry
geometric transformation
reflection
trigonometry
AMC
USA(J)MO
USAMO

jeje, en vista de q la participacion foro no aparenta ser muy constante y tiene pinta de desertico, se nos ocurrio a gabriel (digo, jutaro, oh no! revele su identidad secreta jeje) y a mi hacer una especie como de topic nuevo con problemas y tratar de que fueramos lo mas constantes... se me ocurrio que tal vez c/ persona propusiera tres problemas a la semana (todos los voluntarios q quieran) y al cabo de la semana cada quien presenta las soluciones a los problemas propuestos, ojala incluidos los de uno, y asi se obtiene una especie de "retroalimentacion" entre los participantes... jejeje, asi seamos solo jutaro y yo...

los dejo con la inquietud latente... espero q se apunte bastante gente para asi entre todos poder mover un poquito mas este foro un poco oxidado... obviamente cualquier sugerencia u opinion al respecto la cantidad de problemas, el tiempo o lo q sea, sera bienvenida... sin mas por el momento propongo otros problemas que habia propuesto en otros topics q parecen estar muertos...

1. (TST EEUU 2002) Sea p un primo mayor que 5. Para cada entero x, se define
$f_{p}(x)=\sum_{k=1}^{p-1}\frac{1}{(px+k)^{2}}.$
Probar que para todos los enteros positivos x,y, el numerador de $f_{p}(x)-f_{p}(y)$, escrita como fracción irreducible ("when written in lowest terms"), es divisible por $p^{3}$.

2. (TST EEUU 2003) Hallar todas las ternas de primos (p,q,r) tal que
$p|q^{r}+1, \ q|r^{p}+1, \ r|p^{q}+1.$

3. (Daniel Campos  :D , Mathematical Reflections 1/2007) Sean $a,b,c$ reales positivos tales que $a+b+c+1=4abc$. Probar que
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\geq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}.$

bno campos yo estoy de acuerdo con ud en q este foro es ligeramente menos activo q el foro ingles :maybe: probablemente los demas compañeros q suelen participar aqui esten muy ocupados por el momento y no tengan tiempo, o noc :o pero bno pienso q aunq seamos poquitas personas podemos ir reanimando poco a poco el foro hasta q volvamos a los tiempos dorados del foro hispano :D

en cuanto a los problemas q ud acaba de proponer, estan bastante guapos jajajajajaja me gusta especialmente su desigualdad, asi q cuando sea tiempo vere si muestro mi solucion :D

[quote="campos"]
3. (Daniel Campos  :D , Mathematical Reflections 1/2007) Sean $a,b,c$ reales positivos tales que $a+b+c+1=4abc$. Probar que
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\geq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}.$[/quote]

[hide="Mi solución al lado izquierdo"]Sean $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$ entonces la condición se convierte en $xy+yz+xz+xyz=4$. Sea $S=x+y+z$, nos piden demostrar que $S\geq 3$, pero es conocido que
                        $\frac{S^{2}}{3}\geq xy+xz+xz$
                        $\frac{S^{3}}{27}\geq xyz$
entonces $\frac{S^{2}}{3}+\frac{S^{3}}{27}\geq 4$; si $S<3$ entonces $\frac{3^{2}}{3}+\frac{3^{3}}{27}> 4$ lo cual no es cierto, luego $S\geq 3$. [/hide]

Por ahora propongo dos problemas:
4) (Belarús 2005) Halle todos los enteros positivos $n$ para los cuales existen números primos $p$ y $q$ con $p+2=q$ y tales que los números $2^{n}+p$, $2^{n}+q$ también son números primos.

5) Sea $I$ el incentro de un triángulo $ABC$ ( $AB<BC$ ). $M$ es el punto medio del segmento $AC$, $N$ es el punto medio del arco $ABC$ de la circunferencia circunscrita al triángulo $ABC$. Pruebe que $\angle IMA=\angle INB$.

espero contribuir a reactivar el foro  :)

$Tipe$

jeje, que bueno, parece q hay un poco de interes! posteo mi sol al prob 5 de tipe.

[hide]
Sea $D$ el punto de tangencia del incírculo con $AC$. Luego, $\tan IMA=\frac{r}{DM}=\frac{r}{b/2-(s-a)}=\frac{2r}{a-c}$.

También tenemos que $\tan INB=\frac{BI}{BN}=\frac{r}{2R\sin (B/2)\sin((A-C)/2)}$, pues $BI=\frac{r}{\sin (B/2)}$ y $BN=2R\sin((A-C)/2)$.

Luego, hay que probar que $a-c=4R\sin (B/2)\sin ((A-C)/2)$, pero

$a-c=2R(\sin A-\sin C)=4R\sin((A-C)/2)\cos((A+C)/2)$
$=4R\sin (B/2)\sin ((A-C)/2)$, y con esto concluye la prueba.[/hide]

[quote="tipe"]4) (Belarús 2005) Halle todos los enteros positivos $n$ para los cuales existen números primos $p$ y $q$ con $p+2=q$ y tales que los números $2^{n}+p$, $2^{n}+q$ también son números primos.[/quote]
[color=blue]Por falta de tiempo, aún no enviaré la demostración de lo que he logrado hasta ahora. Lo haré más tarde.
Sólo necesito descartar este caso: $p=3;q=5;3<n\equiv 1\left({\bmod 2}\right)$.
¿Alguien puede ayudarme, por favor?[/color]

creoq con p= 3 q=5 y n=1 , hay solucion, tb si n=3, solo he avanzado hasta esos valores , faltaria demostrar q para valores mas grandes no hay solcucion

[color=blue]En las mismas estoy.
Ya descarté todos los otros casos para $p$ y $q$. Sólo me falta lo que acaba de mencionar.[/color]

[color=green]aca les dejo mis sols a 2 de los problemas q propuso campos[/color]

[hide="Sol Prob 03"]
[quote="campos"]
3. (Daniel Campos  :D , Mathematical Reflections 1/2007) Sean $a,b,c$ reales positivos tales que $a+b+c+1=4abc$. Probar que
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\geq \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}.$[/quote]

La idea de la solución es efectuar sustituciones adecuadas que trivialicen cada desigualdad. Primero observemos que si escribimos la condición $a+b+c+1=4abc$ de la siguiente manera
\[4=\frac{a+b+c+1}{abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\]
podemos hacer $x=1/a$, $y=1/b$, $z=1/c$. Obtenemos una nueva condición $xy+yz+zx+xyz=4$, y ahora tenemos que probar que $x+y+z \geq 3 \geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$.

Veamos primero que $x+y+z \geq 3$. La condición $xy+yz+zx+xyz=4$ puede reescribirse como
\[\frac{x}{2}\cdot\frac{y}{2}+\frac{y}{2}\cdot\frac{z}{2}+\frac{z}{2}\cdot\frac{x}{2}+2\cdot\frac{x}{2}\cdot\frac{y}{2}\cdot \frac{z}{2}=1 \]
Dado que esta expresión tiene la forma $pq+qr+rp+2pqr=1$, podemos efectuar la sustitución
\[x=\frac{2p}{q+r},\ y=\frac{2q}{r+p},\ z=\frac{2r}{p+q}\]
Así que nuestra desigualdad es equivalente a
\[2\left(\frac{p}{q+r}+\frac{q}{r+p}+\frac{r}{p+q}\right) \geq 3 \]
o bien
\[\frac{p}{q+r}+\frac{q}{r+p}+\frac{r}{p+q}\geq \frac{3}{2}\]
que no es más que la desigualdad de Nesbitt.

Para probar la otra desigualdad $\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\leq 3$, usamos otra sustitución conocida. Observemos que la condición es equivalente a tener
\[(\sqrt{xy})^{2}+(\sqrt{yz})^{2}+(\sqrt{zx})^{2}+\sqrt{xy}\cdot \sqrt{yz}\cdot \sqrt{zx}=4 \]
Luego podemos hacer
\[\sqrt{yz}=2\cos A,\ \sqrt{zx}=2\cos B,\ \sqrt{xy}=2\cos C \]
donde $A$, $B$, $C$ son los ángulos de un triángulo acutángulo. Despejando $x$, $y$, $z$ obtenemos
\[x=\frac{2\cos B \cos C}{\cos A}, \frac{2\cos C \cos A}{\cos B},\ \frac{2\cos A \cos B}{\cos C}\]
Luego la desigualdad $\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\leq 3$ es equivalente
\[\cos A+\cos B+\cos C \leq \frac{3}{2}\]
que es un resultado conocido. Luego tenemos el resultado.

[/hide]

[color=green]bno mi solucion al otro problema esta medio barroca jajajajaja xq ahorita no he pulido bn el argumento... pero = aqui les va[/color]
[hide="sol a prob 02"]
[quote="campos"]
2. (TST EEUU 2003) Hallar todas las ternas de primos (p,q,r) tal que
$p|q^{r}+1, \ q|r^{p}+1, \ r|p^{q}+1.$
[/quote]

Sean $a=\mathrm{ord}_{p}\ q$, $b=\mathrm{ord}_{q}r$, $a=\mathrm{ord}_{r}p$. Como $q^{r}\equiv-1 \pmod{p}$, resulta que $q^{2r}\equiv 1\pmod{p}$, de donde $a \in \{1,\ 2,\ r,\ 2r \}$. Descartamos fácilmente los valores $1$ y $r$, que contradirían la hipótesis $q^{r}\equiv-1\pmod{p}$. Obtenemos resultados análogos para $b$ y $c$. Ahora lo que nos queda por hacer es un ligeramente tedioso agotamiento de casos... Sin perder generalidad podemos reducir el problema a analizar los siguientes casos:

:arrow: $a=2r,\ b=2p,\ c=2q$. Como $2r=\mathrm{ord}_{p}\ q|p-1$, resulta que $r|p-1$. Así que tenemos en este caso que $r|p-1$, $p|q-1$, $q|r-1$, y esto implica que $r \leq p-1$, $p \leq q-1$, $q \leq r-1$. Sumando las dos primeras desigualdades obtenemos $r \leq q-2$, lo cual contradice $q \leq r-1$.

:arrow: $a=2, b=2p, c=2q$. Entonces tenemos que $p|q-1$, $q|r-1$ y además $q^{2}\equiv 1 \pmod{p}$. Entonces $-1 \equiv q^{r}\equiv 1 \pmod{p}$, de donde $p=2$. De la misma manera obtenemos $q=2$, lo cual es absurdo, pues es claro que no pueden haber dos primos iguales, dadas las condiciones del problema.

:arrow: $a=2r$, $b=c=2$. Tenemos que $p \equiv 1 \pmod{r}$, $r^{2}\equiv 1 \pmod{q}$. Entonces, como en el caso anterior $-1 \equiv p^{q}\equiv 1 \pmod{r}$, de donde $r=2$. Entonces $q|2^{2}-1=3$, esto es, $q=3$, y fácilmente verificamos que debe cumplirse que $p=5$.

:arrow: $a=b=c=2$. Por la definición de orden debemos tener $q \equiv-1 \pmod{p}$, $r \equiv-1 \pmod{q}$, $p \equiv-1 \pmod{r}$, esto es, $p|q+1$, $q|r+1$, $r|p+1$, de donde $p \leq q+1$, $q \leq r+1$, $r \leq p+1$. Esto nos lleva a $-1 \leq p-r \leq 2$, y por simple inspección de los posibles valores de $p-r$ vemos que se obtienen los mismos valores para $p$, $q$, $r$ que en el caso anterior.

Concluimos que $\{p,\ q,\ r\}=\{2,\ 3,\ 5\}$.  :)
[/hide]

jeje q bueno, me alegra q este topic este tomando cada vez mas fuerza... les pediria q colaboraran, si fueran tan amables, en dos aspectos. 1) si van a proponer un problema tratar de seguir con la numeracion. 2) a la hora de subir las soluciones tratar de "esconderlas", porq sino a la hora de cargar nos va a tomar mucho tiempo, ademas q no los q no hayan resuelto los problemas no vean las soluciones si no desean... jeje y si no fuera mucha molestia, alguien q me diga como poderle cambiar el "click to reveal hidden content" por un nombre un poquito mas bonito jejeje... tuanis... los dejo con mis soluciones... el otro de tipe todavia no lo he intentado...

[hide]
1. la idea es extender el concepto de congruencias a los racionales, considerando $1/x$ como el inverso de $x$. todas las congruencias seran tomadas modulo $p^{3}$ para evitar tanta notacion. por el teorema de euler-fermat tenemos que $x^{p^{2}(p-1)-2}$ es "congruente" con $x^{-2}$ luego resulta q

$f_{p}(x)\equiv \sum_{k=1}^{p-1}(px+k)^{p^{2}(p-1)-2}\equiv \sum_{k=1}^{p-1}p^{2}x^{2}k^{p^{2}(p-1)-4}{{p^{2}(p-1)-2}\choose 2}+pxk^{p^{2}(p-1)-3}{{p^{2}(p-1)-2}\choose 1}+k^{p^{2}(p-1)-2}$

luego, $f_{p}(x)-f_{p}(y)\equiv \sum_{k=1}^{p-1}3p^{2}(x^{2}-y^{2})k^{p^{2}(p-1)-4}-2p(x-y)k^{p^{2}(p-1)-3}$

probaremos que $\sum_{k=1}^{p-1}k^{p^{2}(p-1)-4}\equiv 0$ (mod $p$) y que $\sum_{k=1}^{p-1}k^{p^{2}(p-1)-3}\equiv 0$ (mod $p^{2}$).

ahora consideramos modulo $p^{2}$. por el teorema de euler fermat tenemos que

$\sum_{k=1}^{p-1}k^{p^{2}(p-1)-3}\equiv \frac{1}{2}\sum_{k=1}^{p-1}k^{p^{2}(p-1)-3}+(p-k)^{p^{2}(p-1)-3}\equiv \frac{1}{2}\sum_{k=1}^{p-1}pk^{p^{2}(p-1)-4}{{p^{2}(p-1)-3}\choose 1},$

entonces para probar esto habria que probar que

$\sum_{k=1}^{p-1}k^{p^{2}(p-1)-4}\equiv 0$ (mod $p$),

lo cual por el pequeño teorema de fermat es equivalente a

$\sum_{k=1}^{p-1}k^{-4}\equiv 0$ (mod $p$).

tenemos tambien que probar que $\sum_{k=1}^{p-1}k^{p^{2}(p-1)-4}\equiv 0$ (mod $p$),

lo cual es equivalente a $\sum_{k=1}^{p-1}k^{-4}\equiv 0$ (mod $p$)

ahora, notese que en el intervalo $[1,p-1]$ los unicos numeros q son sus propios inversos son $1$ y $p-1$ y luego es claro que todo numero en $[2,p-2]$ tiene un inverso unico en el intervalo $[2,p-2]$. luego,

$\sum_{k=1}^{p-1}k^{-4}\equiv \sum_{k=1}^{p-1}k^{4}$ (mod $p$)

para hallar la formula $S4_{n}=\sum_{i=1}^{n}i^{4}$ podemos suponer que $S4_{n}=an^{5}+bn^{4}+cn^{3}+dn^{2}+en+f$, y luego usar que
$S4_{n+1}=S4_{n}+(n+1)^{4}$ y comparar coeficientes, para llegar a que

$S4_{n}=n^{5}/5+n^{4}/2+n^{3}/3-n/30=\frac{n(n+1)(6n^{3}+9n^{2}+n-1)}{30}$

*otra forma de llegar a esta formula es resolver el sistema con 6 incognitas (a,b,c,d,e,f) y usar matrices  :D jeje, q fue mi primer approach cuando resolvi este problema, jeje, pero luego encontre esta manera mil veces mas corta jeje*

luego, $\sum_{k=1}^{p-1}k^{4}=\frac{p(p-1)(6(p-1)^{3}+9(p-1)^{2}+(p-1)-1)}{30}$, lo cual es divisible por $p$ porq $(p,30)=1$, con lo cual concluye la prueba!  :D
[/hide]

[hide]
3. la idea de la solucion es usar q condicion es equivalente a
$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}=4$.

para la ecuacion $x^{2}+y^{2}+z^{2}+xyz=4$ en los reales positivos es bien sabido que existe un triangulo acutangulo tal que $(x,y,z)=(2\cos A,2\cos B, 2\cos C)$.

luego, $(\sqrt{1/ab},\sqrt{1/bc},\sqrt{1/ca})=(2\cos A, 2\cos B, 2\cos C)$. entonces la desigualdad es equivalente a las desigualdades

$\sum_{cic}\frac{\cos A\cos B}{\cos C}\geq \frac{3}{2}\geq \sum_{cic}\cos A$, las cuales son ambas bien conocidas  :D y estamos listos![/hide]

Para que reemplazes el "Click  to reveal hidden content" tienes que escribir:
 {hide=Titulo que quieres que aparezca} Lo que quieres ocultar  {/hide}

donde las llaves debes reemplazarlas por corchetes (no encontre otra forma de explicarlo, espero que sea clara  :lol: )

$Tipe$

En el problema 3 yo hice el reemplazo trigonometrico, uno salio con jensen, pero todavia no se como probar que:
$ \sum_{cyc}\frac{\cos A \cos B}{\cos C}\ge \frac{3}{2}$
¿Como se puede probar eso?

[color=green]otra solucion al problema de geometria de tipe... x supuesto no podiamos conformarnos con una solucion trigonometrica :)

bonito problema tipe, podrias indicarme la fuente? (olimpiada y año, libro, etc...) [/color]

[hide]Sea $P$ el punto medio del menor arco $AC$ del circuncírculo. Es claro que $B$, $I$, $P$ están alineados, y además $N$ y $P$ son puntos diametralmente opuestos en el circuncírculo. Además abemos que $P$ es el circuncentro del triángulo $ACI$, esto es, $PI=PA$. Pero observemos que $\angle NAP=\pi/2$, dado que $NP$ es un diámetro del circuncírculo del triángulo $ABC$. Entonces $PA^{2}=PM \cdot PN$, de donde $PI^{2}=PM \cdot PN$, y $PI$ es tangente al circuncírculo de $IMN$. Luego $\angle PIM=\angle INM=\theta$. Pero observemos que, como $AB<AC$, resulta que $\angle IPN=\pi/2-B/2-C$, por lo que $\angle IMN=\theta+\angle PIM=\pi/2+\theta-B/2-C$, y en consecuencia $\angle AMI=B/2+C-\theta$. Por otra parte, $\angle NBP=\pi/2$, puesto que $MN$ es diámetro del circuncírculo del triángulo $ABC$; así que $\angle BNP=\pi/2-\angle IPM=B/2+C$. Por tanto $\angle BNI=B/2+C-\theta$, esto es, $\angle BNI=\angle AMI$, y hemos terminado.[/hide]

[color=green]para samper: esa desigualdad es una consecuencia de una desigualdad mas fuerte q aparece en las memorias del semi de la ibero de colombia, me imagino q ud debe tener ese material a la mano, puede encontrarla en la pagina 176 del folleto[/color]

6 ) en una tabla de 10x10 estan escritos los enteros del 1 al 100 . desde cada fila selecionamos el tercer mayor numero. Mostrar que la suma de estos numeros no es menor que la suma de los numeros en alguna fila

[hide="{para samper}"]bueno, una es hacer $x,y,z$ a cada esos terminos y quedaria probar que

$xy+xz+yz+2xyz=1$ implica que $s=x+y+z\geq \frac{3}{2}$.

sin embargo, esto se sigue (como dijo tipe en su post) por ma-mg y la condicion, porq tenemos que $\frac{s^{2}}{3}+\frac{2s^{3}}{27}\geq 1$... si $\frac{3}{2}>s$ entonces $1>1$, lo cual es una contradiccion y ya...

o si no, factorizar el polinomio $\frac{s^{2}}{3}+\frac{2s^{3}}{27}-1=\frac{2}{27}(s-\frac{3}{2})(s-3)^{2}$, lo cual implica que $s\geq \frac{3}{2}$[/hide]

[quote="Jutaro"][color=green]otra solucion al problema de geometria de tipe... x supuesto no podiamos conformarnos con una solucion trigonometrica :)

bonito problema tipe, podrias indicarme la fuente? (olimpiada y año, libro, etc...) [/color]
[/quote]
Estaba esperando que alguien ponga una solución geométrica, bueno la fuente es la siguiente:
Olimpiada Rusa 2005, ronda 4, grados 9 y 10.

PD: Mi solución es muy parecida a la tuya  :lol:, bonito problema por cierto.

$Tipe$

[color=green]hola gente, les dejo un problema q me gusto bastante, es de nuestro selectivo para esta imo[/color]

18. Sean $x_{1},\ x_{2},\ldots,\ x_{6}$ las raíces del polinomio $P(x)=x^{6}+ax^{5}+bx^{4}+cx^{3}+bx^{2}+ax+1$. Mostrar que \[\prod_{k=1}^{6}(x_{k}^{2}+1)=(2a-c)^{2}\]

[hide="Mi solución al 18"]La ecuación es equivalente a $(x^{3}+x^{-3})+a(x^{2}+x^{-2})+b(x+x^{-1})+c=0$, haciendo el cambio de variable: $y=x+x^{-1}$ tenemos
\[(y^{3}-3y)+a(y^{2}-2)+by+c=0 \\ \Longleftrightarrow y^{3}+ay^{2}+(b-3)y+(c-2a)=0 \]
Sean $m, n, p$ las raíces de esta ecuación, entonces $mnp=2a-c \ldots (1)$.
Por otro lado, las raíces de la ecuación original son las raíces de la ecuaciones:
\[x+x^{-1}=y=m \\ x+x^{-1}=y=n \\ x+x^{-1}=y=p \]
Sin pérdida de generalidad podemos suponer que $x_{1}, x_{2}$ son raíces de la primera ecuación, $x_{3}, x_{4}$ de la segunda y $x_{5}, x_{6}$ de la tercera, luego $x_{1}+x_{1}^{-1}=m$, de donde $x_{1}^{2}+1=mx_{1}$ y de manera análoga con las otras raíces, con esto:
\[\prod_{k=1}^{6}(x_{k}^{2}+1)=mx_{1}.mx_{2}.nx_{3}.nx_{4}.px_{5}.px_{6}=(mnp)^{2}\]
pues $\prod_{k=1}^{6}x_{k}= 1$, usando $(1)$ concluimos la demostración.[/hide]

P.D. Sería interesante que postees los otros problemas de su selectivo para la IMO.
$Tipe$

[color=green]hey tipe, sabes q yo hice exactamente lo mismo q vos en ese problema :lol: el chiste del problema era observar q el polinomio es reciproco de grado par, y entonces es posible hacer ese cambio de variable, etc...

bueno ya q tipe me lo sugeria, les dejo otro problema del selectivo q acabamos de hacer... :D esta es una version equivalente de un problema de la ISL 1988, creo[/color]

19.  Sea $f: \mathbb{N}\rightarrow \mathbb{N}$ una función q satisface $f(f(m)+f(n))=m+n$ para cualesquiera $m,\ n \in \mathbb{N}$. Determinar todos los posibles valores de $f(2007)$.

[quote="Jutaro"]

19.  Sea $f: \mathbb{N}\rightarrow \mathbb{N}$ una función q satisface $f(f(m)+f(n))=m+n$ para cualesquiera $m,\ n \in \mathbb{N}$. Determinar todos los posibles valores de $f(2007)$.[/quote]

loa naturales incluyen 0 o no?

para nosotros los enteros no negativos incluyen el cero, y los naturales no, yo resolvi el problema sin haber incluido el cero :)

Recien he intentado el problema 1, mi solucion es algo diferente a la de campos; pero coincido en tener que demostrar que $S_{4}=1^{4}+2^{4}+....+(p-1)^{4}$ en congruente a $0(mod p)$, $p>5$, para resolver mas rapida la situacion se puede considerar:$S_{n}=1^{n}+2^{n}+...+(p-1)^{n}$, con $p-1$ no divide a $n$, sea $g$ la raiz primitiva en modulo $p$, como $(g;p)=1$, entonces: $g,2g,3g,..,(p-1)g$ es el sistema reducido de residuos, luego: $S_{n}\equiv g^{n}S_{n}(modp)$, y $g^{n}$ no es congruente a 1 modp, ya que $p-1$ no divide a $n$, luego: $S_{n}\equiv0(modp)$, en este caso: $p-1$ no divide a 4, ya que p>5

[quote="M4RI0"]Interesantes las dos demostraciones, la mia se basa en el hecho de que $X$ es el circuncentro de $PQC$[/quote]

[color=green]casualmente estaba revisando unas notas de yiu y encontre este problema, y aca saque dos soluciones, una siguiendo la sugerencia de mario y otra siguiendo las ideas de cuenca, realmente me parecio interesante este problema xq es posible dar tanto una solucion basica como tambien una mas conceptual[/color]

[hide]
Sea $\angle PAQ=\alpha$; entonces $\angle ACQ=\pi/2-\alpha$, dado que $\angle AQB=\pi/2$. Pero como $PX$ y $PY$ son tangentes a la circunferencia tenemos que $\angle XPQ=\angle XQP=\alpha$. Como $XP=XQ$ y $\angle PXQ=\pi-2\alpha$, se sigue que $X$ es el circuncentro del triángulo $CPQ$. Ahora bien, si $\angle BAQ=2\theta$ entonces $\angle BPQ=\theta$, y en consecuencia $\angle CPX=\pi/2-\alpha-\theta$, y como $X$ es el circuncentro de $CPQ$ resulta que $\angle PCX=\angle CPX=\pi/2-\alpha$. Por tanto $\angle PAB+\angle PCX=\pi/2$, es decir, $AB$ y $CX$ son perpendiculares, como queríamos demostrar.
[/hide]

[color=green]aqui esta una solucion mas "tecnica"[/color]

[hide]
Sean $R=AQ \cap BP$, $S=AB \cap PQ$. Vemos que $PQ$ es la polar de $X$; además $RS$ es la polar de $C$. Luego $PQ \cap RS= S$ es el polo de $CX$, de donde $AS$ es perpendicular a $CX$, como queríamos.
[/hide]

[hide="respuesta"]
El único valor que puede tomar $ f(2007)$ es 2007
[/hide]
[hide="Prueba"]
Primero hecho que nos sera útil:
$ \forall x\equal{}a\plus{}b\plus{}c\plus{}d\ge4, x\in\mathbb{Z}$ tenemos que $ f(x)\equal{}f(a)\plus{}f(b)\plus{}f(c)\plus{}f(d)$
En efecto:
$ f[f(a)\plus{}f(b)]\plus{}f[f(c)\plus{}f(d)]\equal{}a\plus{}b\plus{}c\plus{}d \Rightarrow$
$ f(a\plus{}b\plus{}c\plus{}d)\equal{}f\{f[f(a)\plus{}f(b)]\plus{}f[f(c)\plus{}f(d)]\}\equal{}f(a)\plus{}f(b)\plus{}f(c)\plus{}f(d)$
 por la condición incial.(use [ ]y { } para que se mire más claro pero todos signinfican ( ) )

De aqui vemos que $ f(6)\equal{}f(3)\plus{}3f(1)\equal{}2f(2)\plus{}2f(1)$ de donde que $ f(3)\equal{}2f(2)\minus{}f(1)$
$ f(4)\equal{}4f(1)$
También $ f(8)\equal{}4f(2)\equal{}f(4)\plus{}f(2)\plus{}2f(1)\equal{}6f(1)\plus{}2f(2)$ entonces $ f(2)\equal{}2f(1)$ y $ f(3)\equal{}3f(1)$
En resumen:
$ f(1)\equal{}f(1)$
$ f(2)\equal{}2f(1)$
$ f(3)\equal{}3f(1)$

Ahora probaremos por inducción que $ \forall \ n\in\mathbb{Z}^\plus{} \ f(n)\equal{}nf(1)$
○Supongamos que es cierto para todo $ 1\le n\le 4k\minus{}1$, con $ k\ge1$.
○Probaremos que es cierto para $ 4k$, $ 4k\plus{}1$, $ 4k\plus{}2$ y $ 4k\plus{}3$:
$ f(4k)\equal{}4f(k)\equal{}4kf(1)$
$ f(4k\plus{}1)\equal{}3f(k)\plus{}f(k\plus{}1)\equal{}3kf(1)\plus{}(k\plus{}1)f(1)\equal{}(4k\plus{}1)f(1)$
$ f(4k\plus{}2)\equal{}2f(k)\plus{}2f(k\plus{}1)\equal{}2kf(1)\plus{}2(k\plus{}1)f(1)\equal{}(4k\plus{}2)f(1)$
$ f(4k\plus{}3)\equal{}f(k)\plus{}3f(k\plus{}1)\equal{}kf(1)\plus{}3(k\plus{}1)f(1)\equal{}(4k\plus{}3)f(1)$
Todas estas por el hecho que mencionamos al inicio y la hipótesis de inducción. Esto termina la inducción.

Ahora como $ f$ toma solo valores enteros postivos tenemos que $ f(n)\equal{}nf(1)\ge n$, con igualdad sí y solo sí $ f(1)\equal{}1$.
Note que $ 2\equal{}f(2f(1)))\ge 2f(1)\ge 2$, lo cual implica que $ f(1)\equal{}1$ y que $ f(n)\equal{}n\forall n \ \in\mathbb{Z}^\plus{}$

En particular $ f(2007)\equal{}2007$
[/hide]

 :)

encontré un problema muy interesante, de la ronda 44 de una olimpiada que se llama la IMTS (International Mathematics Talent Search), que lo veo muuuuuy difícil, pero me gusta (el problema original no incluye la parte A, esa la inventé yo , pero no creo que sea demasiado difícil :D )

[hide="Problema 20"]
El lenguaje de una extraña isla del Pacífico sur tiene sólo dos letras: la A y la B.
Las "palabras" se forman según los sgtes. postulados:

1) A es una "palabra"
2) Si tenemos una "palabra" dada, el añadir una B al final forma una nueva "palabra"
3) Si una "palabra" tiene en su escritura una secuencia "AAA", el cambiar la secuencia por una B forma otra "palabra"
4) Si aparece una secuencia "BBB" en la "palabra", el quitar la secuencia forma una nueva "palabra"
5) Si tenemos una "palabra" cualquiera, el duplicar la secuencia forma una nueva "palabra"

a) ¿Será B una "palabra"? :|
b) Demostrar que BAABAA NO es una "palabra"
[/hide]

[quote="Rubén D."]
El lenguaje de una extraña isla del Pacífico sur tiene sólo dos letras: la A y la B.
Las "palabras" se forman según los sgtes. postulados:

1) A es una "palabra"
2) Si tenemos una "palabra" dada, el añadir una B al final forma una nueva "palabra"
3) Si una "palabra" tiene en su escritura una secuencia "AAA", el cambiar la secuencia por una B forma otra "palabra"
4) Si aparece una secuencia "BBB" en la "palabra", el quitar la secuencia forma una nueva "palabra"
5) Si tenemos una "palabra" cualquiera, el duplicar la secuencia forma una nueva "palabra"

b) Demostrar que [b]BAABAA [/b] NO es una "palabra"
[/quote]

El problema está mal escrito ya que la palabra BAABAA se puede construir facilmente:
A
AA
AAAA
BA
BAB
BABB
BABBBABB
BAABB
BAABBB
BAA
BAABAA

La pregunta original es:
Demostrar que [b]BAABAABAA [/b] NO es una "palabra" (notese la diferencia con BAABAA)
La respuesta es muy fácil, tanto para a) como para b):
[hide="Pista"]Mire módulo 3.[/hide]

Problema 21, otro problema del IMTS:

$ f(x) = x\cdot\left\lfloor x\cdot\left\lfloor x\cdot\left\lfloor x\right\rfloor\right\rfloor\right\rfloor \ \forall x \in \mathbb{R}$
a)Determinar $ x$ tal que $ f(x) = 2001$
b)Demostrar que $ f(x) = 2002$ no tiene solución

Yo soy brasileño, entonces perdone por mi español horrible.

Una otra y buena soluccion para el problema 19:

Suponga f(m)>m. Entonces f(m)+f(n)>m+n=f(f(m)+f(n)).Aca, temos x>f(x), contradicione.Se hiciermos la suposicione para f(m)<m, novamente obtuvemos contradicione.Entonces, f(n)=n y f(2007)=2007.

  :D


function
induction
algebra unsolved
algebra

Find all functions f: R - R satisfying for all x and y
[b][f(x+y)]^2 = [f(x)]^2 + [f(x)]^2[/b]

[hide="hint"]Let $g(x)=[f(x)]^2$ and then $g(x+y)=g(x)+g(y)$[/hide]

so now we have to find functions satisfying
g(x+y) = g(x) + g(y).

I've only have g(0) = 0
and g(2x) = 2g(x)   what next?

I've never done such tasks. I don't know from to begin
Please help

this example is very well-known and very easy also... let $g(1)=t$. then we have by induction $g(n)=nt$, $g\left( \frac1n \right)=\frac{t}{n}$ and $g \left( \frac{m}{n} \right) = \frac{mt}{n}$. so we have $g(x)=tx$ for all $x \in \mathbb{Q}$. if $g$ is continouos, the result holds also for $x \in \mathbb{R} - \mathbb{Q}$. if not, there is an interesting result about the linearly independent irrationals. i read about it somewhere, but i forgot it. can anyone tell me what was it?

The result I think you're talking about is that either $g$ is everywhere linear, or the graph of $g$ is dense in $\mathbb{R}^2$ (if $g$ satisfies $g(x+y)=g(x)+g(y)$).


algebra
polynomial
Rational Root Theorem

Okey dokey, edited this to hopefully a full solution.

Suppose x=y. Then 2xn = 0, thus x=y=0.

Suppose x<y. Then to keep the right hand side positive, we must have n odd. Then we will get exactly the same solutions as in the x>y case but with x and y flipped around.

So now we have x>y. Let x=y+k, k>0, and let c = y/k where c is rational > 0.

Then simplifying the equation we get (c+1)n + cn = k.
Thus we get x = (c+1)n+1 + (c+1)cn and y = c(c+1)n + cn+1.
All we have to do now is choose rational c which makes these both whole numbers, or to make things easier to make y-x (which is k) and y (which is ck) whole numbers.

Obviously c a whole number is a solution. We also eliminate the solutions where n=0 and x-y = 2. Now we look for other solutions.

k = (c+1)n + cn
By the rational root theorem thingy, all rational solutions must have denominator 1 or 2. So the denominator must be 2 for a non whole number. But k is obviously odd. Thus if c is of the form a/2 for an odd integer a, then ck cannot be a whole number.

So overall we have the following solutions:
(0,0)
((c+1)n+1 + (c+1)cn, c(c+1)n + cn+1) for all whole numbers c and all whole numbers n>=1.
(c(c+1)n + cn+1, (c+1)n+1 + (c+1)cn) for all whole numbers c and all whole odd numbers n>=1.

And (c+2, c) for all whole numbers c.

Did I miss any?

Sorry I didn't answer sooner :oops:

I think you might be missing a few sol'ns.  For example, what about (43425,39951,3) = (x,y,n)?

Ah. I got around to looking at this and having paper and pen at the same time, and everything is correct until I said: But k is obviously odd.
So can anyone find all solutions to k = (a/2+1)^n + (a/2)^n, with a,k,n integers? (n positive)

I'll try! :D I don't know how far I'll get though.

Okay, this is what I got solving for (a,k,n)

[hide]In order for k to be an integer, the two terms on the right must either both be integerson or both be half of an odd integer.  Clearly, they will always agree in that aspect, but we must find when they fall into one of those two categories.  If a is even, then it automatically works.  If a is odd, however, the two terms on the right will be decimals.  The only way for these to both be .5 is to have n=1.  Thus we have the two solutions for (a,k,n) of:

(2m,(m+1)^n+m^n,n) where m is any integer and n is an integer n :ge: 1

and

(2m-1,2m,1) where m is any integer.

I have to go, so I can't convert this into solutions for x,y right now.[/hide]

This challenge problem is still open, right?

Yeah, it still hasn't been solved yet.

By the way - they don't both have to end in 0.5. Using the counterexample that was given to my proof, let a=23 and n=3.

Okay, I think that I've got it now

[hide]I was wrong before.  If a is odd, they don't have to end in .5

Substituting 2x-1 for a, what we need instead is for (2x+1)^n + (2x-1)^n  :equiv: 0 mod 2^n.

Multiplying out the left side shows us that for n even, this can never happen, because the 1's don't cancel and thus create an extra 2.  However, for n odd, they cancel, and if x is even, then it works.  Thus, my new set of solutions for a odd (keeping the old set, because it has more than this includes) we have this solution for (a, k, n):

(4x-1,((4x-1)/2+1)^n+((4x-1)/2)^n,2n-1) where x is an integer and n is a natural number.

I have to leave again, so I can't convert into solutions of x,y right now.[/hide]

I'm not sure how you have managed to cancel anything out mod 2^n. Try x=4 and n=5, say. Then (2x-1)^n + (2x+1)^n has only 4 factors of 2.

sorry, I confused myself.  Finally, I think that this is the answer.

[hide]Instead of it having 2^n-1 as the least guaranteed factor of two for the sum, it is instead 4.  Therefore, our new solution must have x with a factor of 2^n-2.  Therefore, the final answer (hopefully) is

(x*2^(n-1)-1,((x*2^(n-1)-1)/2+1)^(2n-1)+((x*2^(n-1)-1)/2)^(2n-1),2n-1) where x is an integer, and n is a natural number.[/hide]

Now, are there any problems with that?  I want to be sure before I get the x,y answers.

Okay, I checked my work, and so this is the final answer. (I'm not going to rewrite everything that TripleM and I have already posted).

[hide]Because of the solution that I gave, the possible values of (a,k,n) are:

(2m,(m+1)^s+m^s,s)

(2m-1,4m,1) and

(m*2^(r-1)-1,((((m*2^(r-1)+1)/2)^(2s-1)+(((m*2^(r-1)-1)/2)^(2s-1)),2s-1)

Working backwards, we can find that the value of y is 2*a*k, and the value of x is (2a+1)*k.  Therefore, are solutions are:

(0,0,n), where n is a whole number.

(16 :^2: -4m,16m :^2: -8m,1) where m is a natural number.

((2m+1)(m+1)^s+(2m+1)m^s,2m(m+1)^s+2m^(s+1),s) where m and s are natural numbers.

and (not fully simplified because I have to leave)

((m*2^r-1)*((((m*2^(r-1)+1)/2)^(2s-1)+(((m*2^(r-1)-1)/2)^(2s-1)),(m*2^r-2)*((((m*2^(r-1)+1)/2)^(2s-1)+(((m*2^(r-1)-1)/2)^(2s-1)),2s-1) where m, r, and s are natural numbers.

Additional solutions are also findable by switching the x and y in the second, switching x and y and substituting 2s-1 for s in the third, and switching x and y in the third.[/hide]

Please tell me if I have missed anything or made any false assumptions.


calculus
integration
trigonometry
geometry
calculus computations

I think that is interesting people give problems.

Because the objective of AoPS is not make a Table of Integrals.

You can easy check the answer in some programs. MAthematica, Maple. etc...

I am interested to see an analytical proof if this is possible...
Mathematica Integrate cannot solve it, but NIntegrate successfully evaluate it... this is the purpose of this post.

even without outer square root, you'll have extremely nasty result, which is of course non-elementary.

$\int_{0}^{1}\left(\frac{1}{2}+\sqrt{\frac{1}{2}-\left(x^{2}-\frac{1}{2}\right)^{2}}\right) \, dx$

$ \begin{array}[t]{cl}= & \frac{i \left(2+\sqrt{2}\right)}{6 \sqrt{1+\sqrt{2}}}E\left(i \sinh^{-1}\sqrt{2+2 \sqrt{2}}|-3+2 \sqrt{2}\right)\\ & \quad-\frac{i\sqrt{1+\sqrt{2}}}{3}F\left(i \sinh^{-1}\sqrt{2+2 \sqrt{2}}|-3+2 \sqrt{2}\right)+\frac{2}{3}\end{array}$

(this result is given by [i]Mathematica[/i] 5.1)

[quote="heartwork"]How much is:

$\int_{0}^{1}\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+x^{2}-x^{4}}}dx$[/quote]

Well, I haven't yet an answer, but a start: complete the square

$J: =\int_{0}^{1}\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+x^{2}-x^{4}}}dx=\int_{0}^{1}\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}-\left(x^{2}-\frac{1}{2}\right)^{2}}}dx$

apply the substitution

\[x^{2}-\frac{1}{2}=\frac{1}{\sqrt{2}}\sin\theta \Rightarrow x=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}\sin\theta}\Rightarrow dx=\frac{\frac{1}{2\sqrt{2}}\cos\theta d\theta }{\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}\sin\theta}}\]

and note that $0\leq x\leq 1\Rightarrow-\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4}$ to get

\[J=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}-\frac{1}{2}\sin^{2}\theta}}\cdot\frac{\frac{1}{2\sqrt{2}}\cos\theta d\theta }{\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}\sin\theta}}\]
\[{{=\frac{1}{2\sqrt{2}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos\theta \sqrt{\frac{1+\sqrt{2}\cos\theta}{1+\sqrt{2}\sin\theta}}}\, d\theta }\]

I'm wondering if there could be a transcription error from the original source. Because

$\int_{0}^{1}\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}-x^{2}+x^{4}}}dx$

is, shall we say, a different case.

I can guarantee that I want:

$\int_{0}^{1}\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+x^{2}-x^{4}}}dx$

[quote="Zarark"]Sorry for the last post. but is usefull write someting in the post. explain some things.

because no one what to solve problems like that, without text.

If mathematica did not solve this, there are 2 reasons

1 - It's impossible to give a analytical solution
2 - You will wast 10 years of your life trying to solve this.[/quote]

I have seen much easier (simpler) integrals that Mathematica failed to solve so....  :D

[quote]I have seen much easier (simpler) integrals that Mathematica failed to solve so....[/quote]

MAthematica is not God

the thing is. Why I will try to solve this!? whats the purpose?

make a book of list of integrals?

If you set up a interesting problem.

but things apart like that is not interesting to solve.

[quote="Zarark"]

Why I will try to solve this!?
If you set up a interesting problem.
but things apart like that is not interesting to solve.[/quote]

Well the original problem asks for the area determined by:

$x^{4}+y^{4}= x^{2}+y^{2}$

I've tried this and then I got the problem I've posted...

Try polar form; the equation becomes $r^{2}=\frac1{1-\frac12\sin^{2}2\theta}$, and the area is $\int_{0}^{2\pi}\frac{r^{2}}{2}\,d\theta$.
It's not pretty, but it's routine.


Let $ABC$ be a triangle. The excircle of $ABC$ with respect to $A$ touches $BC$ at $A'$. Points $B',C'$ are defined similarly. A line $a$ is drawn through $A'$ paralel with $AI$ ($I$ is the incenter). Lines $b,c$ are defined similarly. Prove that $a,b,c$ are concurrent.


Find all $ x\in\mathbb{R}$ such that
\[ \frac{(10-\sqrt{3x})^{2}}{\left(3+\sqrt{\sqrt{3x}-1}\right)^{2}}>3x-15. \]


Find the value of:

$(1+i\sqrt{3})^8$

Good luck!

[hide="Hint"] Use De'Moivres..., or Binomial it, but I doubt you'll want to calculate all those terms.[/hide]


A mixture containing 2.24L $ H_2$ and 1.12L $ D_2$ at NTP is taken inside a bulb and connected to another similar bulb by a stop cork with a tiny opening.The 2nd bulb is fully evacuated..The stop cork is opened for certain time and then closed. The first bulb is now found to contain 0.1g of $ D_2$. Calculate percentage of gases in 2nd bulb.

[b]Doubt:[/b]

OK I know that $ \frac {w1}{w2} \equal{} \sqrt{\frac {M2}{M1}} \frac {P1}{P2}$ . Now what should I take as P1 and P2. The initial partial pressure of the different gases? But it is changing as time goes isnt it?

uhhh... ithink ur thinking too hard.

Grahams law states that the rate of effusion of two gasses is inversely proportional to the sqrt of the molar masses of the gasses. So the ratio of the H2 and D2 is sqrt of their molar mass. Further, they tell u that after some effusion time, .1g D2 is left. Which must mean that X-.1 must have effused, where X is total mass of D2 initially. convert that to volume and use the ratio to h2 to find how much of the H2 is in the tube... then just find the percents...

As a side, you dont need the IGL as its at stp (which i assume is same as ntp (normal temp, pressure?)) so 22.4l/mol is a handy conversion factor to use

But why dont we consider the pressure thing?

[quote="Conjurer"]
Now what should I take as P1 and P2. The initial partial pressure of the different gases? But it is changing as time goes isnt it?[/quote]

It is the pressures at which the gases are allowed to diffuse, not the pressure at which the entire mixture of hydrogen and deutirium is being collected.

Oh so P1/P2 =1 here. Thnx that solves my doubt.


modular arithmetic
prime numbers
Perfect Powers
number theory
IMO Shortlist

Let $p$ be a prime number and let $A$ be a set of positive integers that satisfies the following conditions:

(i) the set of prime divisors of the elements in $A$ consists of $p-1$ elements;

(ii) for any nonempty subset of $A$, the product of its elements is not a perfect $p$-th power.

What is the largest possible number of elements in $A$ ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[hide]Let our primes be $q_1, q_2, \ldots, q_{p-1}$. By considering the exponents of each prime, our problem reduces to finding the largest number of (not necessarily distinct) $p-1$-tuples of elements in $\mathbb{Z}/p\mathbb{Z}$ such that the sum of any set of elements in the set of them is not $(0,0,\ldots,0)$ mod $p$.

We have that $k=(p-1)^2$. $(p-1)^2$ is attainable, as the set $\{ \underbrace{(1, 0, 0, \ldots, 0)}_{p-1 \mbox{ times }}, \underbrace{(0,1, 0, \ldots, 0)}_{p-1 \mbox{ times }}, \ldots, \underbrace{(0, 0, 0, \ldots, 1)}_{p-1 \mbox{ times }} \}$ works.

We will now show that with $(p-1)^2 + 1$ tuples, we can find a non-empty set of them that sum to 0. Let the elements of our set be $T_1 = (e_{1,1}, e_{2,1}, \ldots, e_{p-1,1}), T_2 = (e_{1,2}, e_{2,2}, \ldots, e_{p-1,2}), \ldots, T_{(p-1)^2 + 1} = (e_{1,(p-1)^2 + 1}, e_{2, (p-1)^2 + 1}, \ldots, e_{p-1,(p-1)^2 + 1})$.

Consider the following system of equations in the $(p-1)^2 + 1$ variables $a_1, a_2, \ldots, a_{(p-1)^2 + 1}$:

\begin{align*}
a_1^{p-1} e_{1,1} + a_2^{p-1} e_{2, 1} + \cdots + a_{p-1}^{p-1} e_{p-1,1} &\equiv 0 \pmod{p} \\
a_1^{p-1} e_{1,2} + a_2^{p-1} e_{2, 2} + \cdots + a_{p-1}^{p-1} e_{p-1,2} &\equiv 0 \pmod{p} \\
& \vdots \\
a_1^{p-1} e_{1,(p-1)^2 + 1} + a_2^{p-1} e_{2, (p-1)^2 + 1} + \cdots + a_{p-1}^{p-1} e_{p-1,(p-1)^2 + 1} &\equiv 0 \pmod{p}.
\end{align*}.

The sum of the degrees of this equation is $(p-1)^2 < (p-1)^2 + 1$. This system has the solution $(a_1, a_2, \ldots, a_{p-1}) = (0, 0, \cdots, 0)$, so we may apply Chevalley's theorem to find a solution $(a_1, a_2, \ldots, a_{p-1})$ in which not each entry of the tuple is 0 modulo $p$.

Let $a_{b_1}, a_{b_2}, \ldots, a_{b_n}$ be the nonzero entries. Note that $a_i^{p-1} \equiv 0 \pmod{p}$ iff $i \neq b_j$ for any $j$, and it is 1 iff $i = b_j$ for some $j$. Therefore, we have that

\begin{align*}
e_{b_1, 1} + e_{b_2, 1} + \cdots + e_{b_n,1} &\equiv 0 \pmod{p} \\
e_{b_1, 2} + e_{b_2, 2} + \cdots + e_{b_n, 1} &\equiv 0 \pmod{p} \\
& \vdots \\
e_{b_1, (p-1)^2 + 1} + e_{b_2, (p-1)^2 + 1} + \cdots + e_{b_n, (p-1)^2 + 1} &\equiv 0 \pmod{p}
\end{align*}

It follows that the sum of the $T_{b_1}, T_{b_2}, \ldots, T_{b_n}$ is $(0,0,\ldots,0)$ mod $p$, so our proof is complete.[/hide]

Solution from [i]Twitch Solves ISL[/i]:

Let $D = p-1$. Then the question (thinking in terms of the exponents) can be phrased as follows:

[i]What's the largest multiset of vectors in ${\mathbb F}_p^{D}$ 	such that no nonempty subset has zero sum?[/i]

We claim the answer is $D \cdot (p-1)$. A construction is given by taking [list] [*]$p-1$ copies of the vector $\left< 1, 0, \dots, 0\right>$; [*]$p-1$ copies of the vector $\left< 0, 1, \dots, 0\right>$; [*]\dots; [*]$p-1$ copies of the vector $\left< 0, 0, \dots, 1 \right>$. [/list] To show it's best possible, suppose we have vectors $v_1$, \dots, $v_N$ with coordinates given as \[ v_k = \left< v_{k,1}, v_{k,2}, \dots, v_{k,D} \right> 	\qquad k = 1, 2, \dots, N.  \] Then, we construct the polynomial \[ 	F(X_1, \dots, X_N) 	= \prod_{i=1}^D 	\left[ 1 - \left( \sum_{k=1}^N X_k v_{k,j} \right)^{p-1} \right] 	- \prod_{i=1}^N (1-X_i) \] If we imagine the $X_i \in \{0,1\}$ as Bernoulli random variables indicating whether $v_k$ is used in a set or not, then $F(X_1, \dots, X_N) \neq 0$ exactly when the $X_i$'s equal to $1$ correspond to a nonempty subset of the vectors which have vanishing sum.
Now assume for contradiction $N < D \cdot (p-1)$. Then the largest degree term is given in the latter product; it is exactly $(-1)^{N+1} X_1 X_2 \dots X_N$. So if we quote Alon's combinatorial nullstellensatz, it now follows that there is a choice of $X_i \in \{0,1\}$ such that $F(X_1, \dots, X_N) \neq 0$ which is a contradiction.

Solution with derfu37 and a hint to use Chevalley's from Al3jandro0000.

We claim the answer is $(p-1)^2$. This is achievable by taking the set
\[ A = \{ q_i^{pj+1}\vert(i,j)\in\{1,\ldots,p-1\}^2 \} \]
where $\{q_i\}_{i=1}^{\infty}$ is the sequence of primes. Now we show $|A|>(p-1)^2$ is impossible. Let $n=|A|$. Note that by condition (i) we can represent the elements of $A$ as vectors in $\mathbb F_p^{p-1}$. Let these be
\[ v_i = (e_{i1},\ldots,e_{i(p-1)})\text{ for } 1\le i\le n. \]
Now, we define $f_j\colon\mathbb F_p^n\rightarrow\mathbb F_p^n$ as
\[ f_j(x_1,\ldots,x_n) = \sum_{i=1}^n x_i^{p-1}e_{ji}. \]
Consider the system of equations $f_j(x_1,\ldots,x_n)=0$ for $1\le j\le p-1$. Note that $(0,\ldots,0)$ is a trivial solution. Suppose for the sake of contradiction that $n>(p-1)^2$. Then, we have
\[ n>(p-1)^2=(p-1)\cdot(p-1)=\sum_{j=1}^{p-1}\deg(f_j). \]
Thus, quoting [i]Chevalley's Theorem[/i] there must exist a nontrivial solution to the system. However, since $x_i^{p-1}\in{0,1}$, this nontrivial solution specifies a nonempty subset of $A$ for which the product is a perfect $p-$th power, yielding the desired contradiction. $\blacksquare$

One can easily rephrase this problem using the prime factorizations of the elements of $A$; this asks for the size of the largest multiset of vectors of $\mathbb{F}_p^{p-1}$ where no nonempty subset has a sum of zero.

The answer is $(p-1)^2$; this can be obtained by taking\[\bigcup_{\textbf{x}\in \mathbb{F}_p^{p-1} \\ |\textbf{x}|=1}\{\textbf{x} \ \text{repeated} \ p-1 \ \text{times}\}.\]
Assume for contradiction that there is a solution of vectors $v_1, \ldots, v_N$ with $N>p-1$ that works, where we write\[ v_i = (e_{i1},\ldots,e_{i(p-1)}) \]for each $i$. Now define\[ f_k(x_1,\ldots,x_N) = \sum_{i=1}^N x_i^{p-1}e_{ki} \]for all $1 \le k \le p-1$. Since $(0, \ldots, 0)$ is a common root of all $f_k$, the Chevalley-Warning theorem guarantees a nontrivial root common to all $f_k$. By Fermat's Little Theorem, each $x_i^{p-1}$ is either $0$ or $1$, so there is a nonempty subset of $\{v_1, \ldots, v_N\}$ with a zero sum by using this nontrivial root, a contradiction. Hence $(p-1)^2$ is the size of the largest multiset of vectors of $\mathbb{F}_p^{p-1}$ where no nonempty subset has a sum of zero, and we are done.

Chevalley-Warning essentially nukes this.

The answer is $(p-1)^2$. For construction, pick primes $q_1, q_2, \cdots, q_{p-1}$, and use $q_1, q_1^{p+1}, \cdots, q_1^{1+p(p-2)}$ and similarly for the other $q_i$'s. This obviously satisfies both conditions.

For the proof of maximality, set the $q_i$ to be the same and choose $k = (p-1)^2 + 1$ distinct elements from $A$, which we will label $x_1$ through $ x_k$. For each $j$, let $e_{ij}$ be the exponent of $q_i$ in $x_j$.

Now, for the polynomials $$f_i(X_1, \cdots, X_k) = X_1^{p-1}e_{i1} + X_2^{p-1}e_{i2} + \cdots + X_k^{p-1} e_{ik},$$ by Chevalley-Warning there exists a nontrivial solution to the system $$f_1(x_1, x_2, \cdots, x_k) \equiv f_2(x_1, x_2, \cdots, x_k) \equiv \cdots \equiv 0 \pmod p.$$ Pick a subset $B$ of all nonzero $x_i$ in this set that satisfy the condition. Then by definition and Fermat's Little Theorem it follows that $\sum_{j \in B} e_{ij} \equiv 0 \pmod p$ for each $i$, thus the product of the elements in $B$ is a perfect $p$th power, contradiction.

Claim: If $x_j = (x_{j,1},\cdots, x_{j,n}) \in (\mathbb{Z}/p\mathbb{Z})^n$ ($j\in \{1,\cdots,K\}$) such that for all $e_1, \cdots, e_K \in \{0,1\}$ (not all zero), then $\sum e_jx_j \ne 0$. Then $K\le n(p-1)$ (this bound is optimal because I can have $p-1$ copies of $(\underbrace{0, \cdots, 0}_{j}, 1, \underbrace{0, \cdots, 0}_{n-1-j})$ for $j=0,\cdots,n-1$)

Proof: Assume $K\ge n(p-1)+1$. Consider the polynomial

$$P(f_1, \cdots, f_K) = \prod\limits_{j=1}^n (1-(\sum_{k=1}^K f_kx_{k,j})^{p-1})$$

If $\sum e_jx_j \ne 0 \iff$ there exists $t$ such that $\sum_k e_k x_{k,t} \ne 0$. Then $1-(\sum_k e_k x_{k,t})^{p-1}=0$ in $\mathbb{Z}/p\mathbb{Z}$. Hence $P(f_1,\cdots,f_K)=0$ for all $f_1,\cdots,f_k \in \{0,1\}$ that are not all zero, and $P(0,\cdots,0)=1$.

We first flatten the polynomial i.e. replace $f_k^j$ with $f_k$ when $j\ge 2$ (i.e. we take this polynomial and take its remainder mod $f_1^2-f_1, f_2^2-f_2, \cdots, f_K^2-f_K$). Call this polynomial $\tilde P$. Now note $$\tilde P + (1-f_1)(1-f_2)\cdots (1-f_K)$$ vanishes in $\{0,1\}^K$, but one can show that $$v\colon [K] \to (\mathbb{Z}/p\mathbb{Z})^{2^K}, v_S = ( \prod\limits_{j\in S} f_j )_{(f_1,\cdots,f_k)\in \{0,1\}^K}$$ are linearly independent in $\{0,1\}^K$ (or we induct by seeing this as a linear poly in $f_1$ namely $R(f_2,\cdots,f_K) f_1 + Q(f_2,\cdots,f_K)$ and since this poly is 0 when $f_1\in \{0,1\}$, $R(f_2,\cdots,f_K), Q(f_2,\cdots,f_K)$ satisfy the problem conditions for $K-1$) so $$\tilde P + (1-f_1)(1-f_2)\cdots (1-f_K) \equiv 0$$

This is absurd since $\deg \tilde P < K$

Yet another C-W nuke solution

[b][color=pink]Claim:[/color][/b] $|A| = (p-1)^2$ is achieveable.
[i]Proof.[/i] Let $p_n$ denote the $n$th prime for $1 \le n \le p - 1$. Then $p-1$ occurences of each $p_n$ suffices. $\blacksquare$

[b][color=pink]Claim:[/color][/b] $(p-1)^2$ is maximal.
[i]Proof.[/i] FTSOC assume there exists some $A$ of size $n = p^2 - 2p$.  Enumerate $A$ as $a_{1}, a_{2}, \dots, a_n$ and the prime divisors as $p_1, p_2, \dots, p_n$.  For each $i$, let $a_i = p_1^{c_{i1}}p_2^{c_{i2}}\dots p_n^{c_{in}}$.
Construct the $p-1$ polynomials $f_1, f_2, \dots f_{p-1}$ in ${\mathbb F}_p$ such that \[ f_i(x_1, x_2, \dots, x_n) = \sum_{j=1}^{n} c_{ij}x_j^{p-1}. \]
Note that $\deg f_i = p-1$ and that $f_i$ is $0$ iff the elements of $A$ at its nonzero inputs have a product with $\nu_{p_i}$ divisible by $p$.
Since $n > \deg f_i \cdot (p-1) = (p-1)^2$, and $(0, 0, \dots, 0)$ is one of the common roots of every polynomial, by Chevalley-Warnining there exists another root, contradiction. $\blacksquare$

Let $D = p-1$. Then the question (thinking in terms of the exponents) can be phrased as follows:

[i]What's the largest multiset of vectors in ${\mathbb F}_p^{D}$ 	such that no nonempty subset has zero sum?[/i]

We claim the answer is $D \cdot (p-1)$. A construction is given by taking [list] [*]$p-1$ copies of the vector $\left< 1, 0, \dots, 0\right>$; [*]$p-1$ copies of the vector $\left< 0, 1, \dots, 0\right>$; [*]$......$; [*]$p-1$ copies of the vector $\left< 0, 0, \dots, 1 \right>$. [/list] To show it's best possible, suppose we have vectors $v_1$, \dots, $v_N$ with coordinates given as \[ v_k = \left< v_{k,1}, v_{k,2}, \dots, v_{k,D} \right> 	\qquad k = 1, 2, \dots, N.  \] Then, we construct the polynomial \[ 	F(X_1, \dots, X_N) 	= \prod_{i=1}^D 	\left[ 1 - \left( \sum_{k=1}^N X_k v_{k,j} \right)^{p-1} \right] 	- \prod_{i=1}^N (1-X_i) \] If we imagine the $X_i \in \{0,1\}$ as Bernoulli random variables indicating whether $v_k$ is used in a set or not, then $F(X_1, \dots, X_N) \neq 0$ exactly when the $X_i$'s equal to $1$ correspond to a nonempty subset of the vectors which have vanishing sum.
Now assume for contradiction $N < D \cdot (p-1)$. Then the largest degree term is given in the latter product; it is exactly $(-1)^{N+1} X_1 X_2 \dots X_N$. So if we quote Alon's combinatorial nullstellensatz, it now follows that there is a choice of $X_i \in \{0,1\}$ such that $F(X_1, \dots, X_N) \neq 0$ which is a contradiction.                                                                                $\square$

The answer is $(p-1)^2$

Interpret each number as a $\mathbb{F}_p^{p-1}$ vector with entries corresponding to their $\nu_q$ modulo $p$: we are asked to find the largest multiset of vectors such that no nonempty subset of them sum to the $0$ vector. For a construction, we may take $p-1$ copies of each of the $p-1$ vectors with one entry equal to $1$ and the rest $0$.

I will now prove that $k:=|A|>(p-1)^2$ is impossible. To that end, suppose we had $k$ vectors $v_1,\ldots,v_k$ and the $j$-th element of $v_i$ was $a_{ij}$. Consider the following polynomial in $\mathbb{F}_p[x_1,\ldots,x_k]$:
$$P(x):=\prod_{i=1}^{k} (1-x_i)-\prod_{j=1}^{p-1} \left(1-\left(\sum_{i=1}^k a_{ij}x_i\right)^{p-1}\right).$$
If we restrict $x_i \in \{0,1\}$ for all $i$, then choosing some subset of the vectors and summing them is equivalent to setting some subset of the $x_i$ to $1$ (and the rest $0$). If we choose the empty set, i.e. $x_i=0$ always, then the first and second products are both $1$. Otherwise, the first product clearly vanishes, and by hypothesis, since some entry of their sum is nonzero, Fermat's little theorem guarantees that the second product vanishes as well. Thus $P$ vanishes for any choice of $(x_1,\ldots,x_k) \in \{0,1\}^k$, but since $k>(p-1)^2$ it contains the maximal-degree term $(-1)^kx_1\ldots x_k$, which yields a contradiction by combinatorial nullstellensatz. $\blacksquare$


Is there an enumeration $a_1,a_2,...$ of the positive integers such that $a_1/a_2,a_2/a_3,...$ is an enumeration of ${\mathbb Q} \cap (0, \infty)- \{1\}$?

Yes. Let's enumerate the rationals $r_1,r_2,\ldots,r_k$, then perform the following steps:

Step $k$
a) If the rational $r_k$ is enumerated in $\frac{a_1}{a_2}$ etc. until now, do nothing. Otherwise choose two positive integers $a,b$ with $\frac ab=r_k$, $a,b$ not alredy enumerated and $\frac ma$ not already enumerated, and add them. (Clearly we can choose such $a,b$, for example taking them large enough).By $m$ we denote the last number written in the sequence.

b)If $k$ is enumerated, do nothing. Otherwise pick up an $a$ with $\frac ma, \frac ka, a$ not already enumerated, and add $a$ with $k$. (Again we can choose such an $a$ , for example by taking it large enough)

Such a sequence also appeared on the 2002 Putnam, A5:
[url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2002.pdf]www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2002.pdf[/url]
(The problem doesn't state it, but every positive rational does appear exactly once.)

This sequence is related to the Stern-Brocot tree:
[url=http://www.cut-the-knot.org/blue/Stern.shtml]www.cut-the-knot.org/blue/Stern.shtml[/url]


thanks to indianamath alot more people are coming on the indiana forum. :gathering:

there's only like 5 ppl posting. out of which, some aren't even from indiana. i pity this forum...


geometry
perimeter
Support
3D geometry
prism
trigonometry
area of a triangle

1.  How do you find the area of a regular pentagon when all they give you is a perimeter of 9090 feet (no apothem)?

2.  A 90 lb. and a 84 lb. person sit at opposite sides of a 15-foot seesaw.  How do you figure out where to put the support in order to balance the seesaw?

3.  700 mile trip takes 8 hours flying against the wind.  The return trip is 5 hours with the wind.  How do you find the rate of the wind in miles per hour?

4.  How do you find the area of a scalene triangle when they only give you the length of the sides?  sides equal 37, 46, and 54

5.  How do you find the volume of a hexagonal prism when they only give you the length of one side of the hexagon (10) and the height (35)?  Again no apothem - I could do it if I had the apothem and everything I look up says you need the apothem!

[hide="#4"]Heron's formula[/hide]
[hide="1,5"]there is a way to find the height of a regular pentagon in terms of its sides.  I have to go right now and its pretty long so I won't write it out but it includes the pythagorean theorem, similar triangles, and extending two adjacent sides[/hide]
i know how to find 3 so ill post it if it hasn't been answered when I get back

[quote="13375P34K43V312"]Pentagons are not uniquely determined by their perimeter.[/quote]
Note that it is a regular pentagon.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38807

[quote="13375P34K43V312"]Wow, I didn't even see regular. Anyway, I think the formula is like $\frac{s^{2}}{4}\sqrt{25+10\sqrt{5}}$, I don't remember exactly. Proven easiest with trig, naturally, but I won't get into it now.[/quote]
Did you completely ignore the link I posted?!?!?!?
Here is the proof with trig
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=240837#p240837

Here's #5 (If you don't want to use that link...)

[hide]For the hexagonal prism, you can split the hexagon into 6 equilateral triangles and then you can find the area by finding the height of each triangle by the pythagorean theorem $10^{2}-5^{2}=75$ so then the area of each triangle is $(5\sqrt3*10)/2=25\sqrt3$ multiply by six to get the area of the hexagon and get $150\sqrt3$. Finally, multiply by 35 to get $5250\sqrt3$ for the final volume.[/hide]

Correct me if I'm wrong!  :P


ok im kinda a noobie with calclulus

If an object moves along a straight line according to the position function s(t) = 0.5t^3 -2t + 1 in the interval 0<= t <=5

find s(0) and s(5) and explain its meaning

well, we have: $\acute{s}(t) = 1.5t^{2}-2$
$\acute{s}(t) > 0 \rightarrow t > \frac{2}{\sqrt{3}}$
$\acute{s}(t) = 0 \rightarrow t = \frac{2}{\sqrt{3}}$
$\acute{s}(t) < 0 \rightarrow t < \frac{2}{\sqrt{3}}$
As a consequence $min\left( s(t) \right) = s\left( \frac{2}{\sqrt{3}}\right) = 1-\frac{8}{3\sqrt{3}}$ .
and $s(0) = 1 > s\left( \frac{2}{\sqrt{3}}\right)\;,\;s(5) = 53.5 > s\left( \frac{2}{\sqrt{3}}\right)$
thus in this interval $s\left( \frac{2}{\sqrt{3}}\right)$ is the minimum and $s(5)$ is the maximum. This is all I know, was this what you wanted???

[quote="Cvert"]SinaJackson,

I am new to this sight an would like to know how to get the symbols that your are using in your equations?[/quote]
Go to the LaTeX link: [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

[quote="sinajackson"]was this what you wanted???[/quote]

I believe he's asking something simpler.  That is, we want to calculate $s(0)$ and calculate $s(5)$, then explain what they mean physically.

Of course, $s(0)$ gives the position at time zero and $s(5)$ gives the position at time $5$.


And a 5 is not at all bad for a middle-schooler.

I think the best way is to work through old AIMEs. Find problems that challenge you significantly, but that aren't impossible, and work on those. Eventually the level of difficulty of problems that challenge you will increase. I think that with sufficient and disciplined practice in that manner, an 8 at the very least is certainly attainable.

[quote="Boy Soprano II"][quote="Ubemaya"]What do you do after you work through all the AIME problems?[/quote]
By that time you should probably have started doing olympiad problems.[/quote]
Or arml problems
http://www.arml.com/

Every single AIME is in the resources section.

Yeah, AoPS Vol. II and previous AIMEs, pretty much.
5 is not bad at all. I took it in 8th grade and got 1.  :roll:

[quote="1337h4x"][quote="Boy Soprano II"]ARML problems are generally easier than AIME problems (compare :  5 mins/question vs. 16 mins/question).  If you've done anywhere near that many AIME problems, you should probably start working on olympiad problems.[/quote]

Wait, has there been a change in the time limit on AIME? I thought it was 12 minutes per question. Also, do you know where I can get some of the older AIME's? Most of the AIME's from the 90's are not on the Wiki.[/quote]
No, I think that was a typo.

Oh, I was just testing how long it would take you people to notice, of course :P

[quote=aberde]Do the AIME problem series, and know all topics from intermediate counting and probability. For Number Theory, you should know CRT, Euler’s theorem, FLT, Euclidean algorithm, Wilson’s theorem, orders, and exponent lifting. When I did these things and knew all the topics I mentioned, I was able to consistently receive 13s, 14s, and 15s on AIME tests.[/quote]

Yes, that would be very helpful for a 24-25 year old person :-D

In case you haven't noticed, this is from 2007.

This thread is older than I am.  Why the massive bump  :play_ball:

What's the point of necroposting on threads this old...?

This thread's age is 3 times my age

EDIT: I'm good at math but make the dumbest mistakes on planet earth: I'm 9

[quote=revak]This thread's age is 3 times my age[/quote]

You're five yet you have better grammar than me

[quote=1=2]I'm a 7th grader(11 year old) who qualified for the AIME, but the best I can do on a previous AIME is 5 points. How do I prepare????

P.S. I tried the AIME problem series, but it got too hard too fast. :([/quote]

[s]take the course i just sent in c&p lawlll [/s]

this thread is older than me, either im young, or aops is old

Man, if you're younger than 14, then I am dumb. I am 12 and only started comp math half a year ago.

[quote=1=2]I'm a 7th grader(11 year old) who qualified for the AIME, but the best I can do on a previous AIME is 5 points. How do I prepare????

P.S. I tried the AIME problem series, but it got too hard too fast. :([/quote]

Are you a rising 7th grader? If you're 11, you must've skipped a grade. Also a score of 5 on the AIME is very good for an 11 year old, you still have years to prepare lol

[quote=pi271828][quote=Bedwarspro]woah something is older than u! never expected it!!11!1!1![/quote]

:o ikr!1!11!!1!

in all seriousness, for ppl looking for advice, use search function, do intermediate series, and vol. 2[/quote]

yep
People ask these questions sooooooo many times


geometry
rectangle
algebra
polynomial
real analysis
binomial theorem
binomial coefficients

$(x)_{i}=\frac{(x+i-1)!}{(x-1)!}$, so $\frac{(3n-p+3)_{i}}{i!}=\frac{(3n-p+i+2)!}{(3n-p+2)!i!}={3n-p+i+2\choose i}$, and the sum becomes \[\sum_{i=0}^{p}{3n-p+i+2\choose i}{n+p-i\choose p-i}.\] I'm sure that $\sum_{i=0}^{p}{a\choose i}{b\choose p-i}={a+b+1\choose p}$ is a reasonably well known identity, but I can't think of a simple proof of it.

Edit: The identity should have said $\sum_{i=0}^{p}{a+i\choose i}{b-i\choose p-i}={a+b+1\choose p}$

[quote="Diarmuid"]I'm sure that $ \sum_{i \equal{} 0}^{p}{a\choose i}{b\choose p \minus{} i} \equal{} {a \plus{} b \plus{} 1\choose p}$ is a reasonably well known identity, but I can't think of a simple proof of it.[/quote]
In fact, Vardermonde's identity is
$ {a \plus{} b\choose p}\equal{}\sum_{i \equal{} 0}^{p}{a\choose i}{b\choose p \minus{} i}$.
I know there's a better proof, but usually I remember it from
$ (1 \plus{} x)^{a \plus{} b} \equal{} (1 \plus{} x)^{a}(1 \plus{} x)^{b}$,
then, using Newton's binomial theorem at both sides, and finally looking at the coefficient of $ x^{p}$.

Oops - I mistyped the original identity... I'll fix it now.

Well corrected!
Unfortunately, your new identity is not "reasonably well known" by me :maybe:

the following poof of Vandermonde's Identidy is very beautiful! :

Claim: $\sum_{i=0}^{p}{a\choose i}{b\choose p-i}={a+b\choose p}$
Proof:
In the first quadrant of the xy-plane consider the rectangular grid defined by the points $(0,0),(a+b-p,0),(a+b-p,p),(0,p)$, where the segements of length one that join the lattices inside this rectangle are drawn.

It is easy to see that the number of shortest routes from $(0,0)$ to $(a+b-p,p)$ is queal to: ${{a+b-p+p}\choose{p}}={{a+b}\choose{p}}$.
(i can explain later if needed)

Now lets count in a different way.
Consider the line segment joining: $P_{0}(a,0),P_{1}(a-1),...,P_{p}(a-p,p)$. Note that each shortest route from $(0,0)$ to $(a+b-p,p)$ passes through exactly one point $P_{i}$.  So the number of shortest routes from $(0,0)$ to $P_{i}$ is ${{a-p+p}\choose{p}}={{a}\choose{p}}=A$.  And the number of shortest routes from $P_{i}$ to $(a+b-p,p)$ is ${{p-i+a+b-p-(a-i)}\choose{p-i}}={{b}\choose{p-i}}=B$. So there are $AB$ shortest routes from $(0,0)$ to $(a+b-p,p)$ passing through each $P_{i}$. Summing up for each $P_{i}$ we get $\sum_{i=0}^{p}{AB}=\sum_{i=0}^{p}{a\choose i}{b\choose p-i}={a+b\choose p}$. and we are done.

I guess the problem is still open then...

I'm sure the identity is correct, and I've been trying to think of a nice combinatorial proof, but so far I've got nothing.

[quote="Diarmuid"]$\sum_{i=0}^{p}{a+i\choose i}{b-i\choose p-i}={a+b+1\choose p}$[/quote]
I've got a proof of your identity, but you're not going to like it.  :P It uses an identity from [url=http://mathworld.wolfram.com/BinomialCoefficient.html]MathWorld[/url]:
${n\choose k}=(-1)^{k}{k-n-1\choose k}$.
Don't be frightened by the negative numbers that appear. You'll see that the identity holds, if you consider the similarities of
$\frac{q!}{r!}$ and $\frac{(-r-1)!}{(-q-1)!}$

Now,
$\sum_{i=0}^{p}{a+i\choose i}{b-i\choose p-i}= \sum_{i=0}^{p}(-1)^{i}{-a-1\choose i}(-1)^{p-i}{p-b-1\choose p-i}=$
$=(-1)^{p}\sum_{i=0}^{p}{-a-1\choose i}{p-b-1\choose p-i}=(-1)^{p}{p-a-b-2\choose p}={a+b+1\choose p}$.

Please correct me if any step is wrong.
I'll try to find some other proof, more elegant than this one.

On the contrary, I like it very much -  a very neat proof.

[quote="lordWings"]I had my doubts because it's not evident that Vandermonde's identity holds for the "negative binomial coefficients" too.[/quote]
Let Vandermonde's identity \[(\pounds)\; \; \; \; \; \;{a+b \choose n}=\sum_{k=0}^{n}{a\choose k}{b\choose n-k}\; .\] Consider $a$ as variable. In this case we have an identity between two polynomials of degree $\le n$ in variable $a \; .$ It's sufficient to prove it only for $n+1$  distinct values of $a .$ Therefore if $(\pounds)$  is proved anly for $a \in{\mathbb Z}\; ,$ then $(\pounds)$ is true for all real (or complex) values of $a$ .

Thanks a lot! Your answer was very illuminating.

By the way, my usual reference with this kind of identities was at this [url=http://binomial.csuhayward.edu/Identities.html]link[/url], but it doesn't seem to work anymore. The cache at google still shows the text, but not the images with the identities. Does anyone have a substitute online reference? There's a copy of "Concrete mathematics" at my library, but a quick online reference is always handy.

[Edit 1]
I've been checking the "Concrete mathematics" book, and Diarmuid's identity appears as number $5.26$ at table $169$, but without any proof.
Looking at the context, many proofs use Vandermonde's identity (called Vandermonde's convolution in the book, the identity $5.22$) and what I called  the "negative binomial coefficients" (called the "upper negation" in the book, the identity $5.14$), so the proof I gave is probably the one intended by the author.

[Edit 2]
Yes. I just checked again, and proving the identity is the basic problem number $5.14$, and those are exactly the hints.

Somehow, I still feel there may be a more elegant solution, with a combinatorial meaning (lost here with the negative values), even if it gets more complicated.

If you want to dodge negative numbers in your binomial coefficients, we'll have to assume $b \geq p$.  Assuming this is true, how many ways are there to take a line of $a+b+1$ squares and color $p$ of them black?

On one hand, there are $\binom{a+b+1}{p}$ ways.

On the other hand, since $b \geq p$ there are at least $a+1$ nonblack squares.  Some number of black squares (call it $i$) appear before the $(a+1)^{st}$ nonblack square.  For each $i$ there are $\binom{a+i}{i}$ choices for how to color the squares before that $(a+1)^{st}$ nonblack square, and $\binom{b-i}{p-i}$ choices for which squares after that square are black.

This gives the other side of your identity.

$\sum_{s+t=n}{\binom{i+s-1}{s}\binom{j+t-1}{t}}=\binom{n+m-1}{n}, m=i+j$

Consider the number of the natural number solution of the equation $x_{1}+x_{2}+\cdot+x_{m}=n$, which is $\binom{n+m-1}{n}$ virtually.

On the other side, $x_{1}+x_{2}+x_{i}= s, x_{i+1}+x_{i+2}+\cdot+x_{i+j}=t$, the number of the solution should be $\binom{i+s-1}{s}\binom{j+t-1}{t}$.

Thus we get the identity.

I think this is the best way to prove this identity.  :blush:

Our solutions are actually very closely related.  In fact, the way I originally learned to prove the formula you cited "virtually" is the block argument I gave in my proof.   Your $a_{i}$ would in my construction count the number of nonblack squares between each pair of black squares.


Who do you think will win teh 2009 nationals. i predict that bobby shen will not choke this year.

I'm not in Wisconsin, but I think that it'll be Bobby Shen and Darryl Wu in the semifinals, and at least one of them in the finals.  The thing is, there's about 10 people that have a decent chance at winning, and it sorta all depends on who has a good day...

Um why does this thread exist? Predictions are basically wrong $ 90\%$ of the time for MATHCOUNTS...

I don't live in Wisconsin either, but this an interesting question. Last year, there were so many 6th and 7th graders who made countdown (the majority were). All of them, I think, have a great shot at being champion. This is a great year for MathCounts (not talking about finnancial/economic stuff), as everyone will be working harder than ever, especially at the top. And, we never know, there could be a surprise out there, too.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="N.T.TUAN"]In the questions below: $G$ is a ﬁnite group; $H \leq G$ a subgroup of $G; |G : H |$ the index of $H$ in $G; |X |$ the number of elements of $X \subseteq G; Z (G)$ the center of $G; G'$ the commutator subgroup of $G; N_{G}(H )$ the normalizer of $H$ in $G; C_{G}(H )$ the centralizer of $H$ in $G$; and $S_{n}$ the $n$-th symmetric group.

Suppose that $n \geq 1$ is such that the mapping $x \mapsto x^{n}$ from $G$ to itself is an isomorphism. Prove that for each $a \in G, a^{n-1}\in Z (G).$[/quote]
Define $\phi: G\to G$ as $\phi(x)=x^{n}$, which is an automorphism.

The case $n=1$ is trivial. Because $a^{n-1}=a^{0}=1$.

The case $n=2$ is also. Note $\phi(xy)=\phi(x)\phi(y)$ hence $(xy)^{2}= x^{2}y^{2}$ hence $xyxy=xxyy$ thus $xy=yx$ so $G$ is abelian. And hence $a^{n-1}=a^{2-1}$ is certainly in $Z(G)$ because $Z(G)=G$.

Now let us consider a more general case $n\geq 3$.
Consider the inner product $xyx^{-1}$.
Then $\phi(xyx^{-1}) = \phi(x)\phi(y)\phi(x^{-1})$
Thus, $(xyx^{-1})^{n}= x^{n}y^{n}x^{-n}$
So, $xy^{n}x^{-1}= x^{n}y^{n}x^{-n}$
Then, $y^{n}x^{n-1}= x^{n-1}y^{n}$.
But $\phi$ is an automorphism so for any $a\in G$ we can write $a=\phi(y)$ so some $y\in G$. That means $a=y^{n}$ for some $y\in G$.
Meaning, $ax^{n-1}= x^{n-1}a$ for all $a\in G$.
So we see that the $n-1$ degree of any element is in $Z(G)$.
Q.E.D.

[quote="N.T.TUAN"][quote="sylow_theory"]
 $y^{n}x^{n-1}= y^{n}x^{n-1}$.
[/quote]
You are true! But here must is $y^{n}x^{n-1}=x^{n-1}y^{n}$  :wink:[/quote]

Thanks, I fixed it.


Let $ A\equal{}\left(\begin{array}{cc}1 & 3\\ \ 2 & 4\end{array}\right).$ Prove that: $ A^{n\plus{}2}\equal{}5A^{n\plus{}1}\plus{}2A^n$ for all $ n \in \mathbb{N}^*.$

Take the Cayley-Hamilton identity $ A^2\minus{}(\text{tr}(A))\cdot A\plus{}(\det A)\cdot I\equal{}0$, and multiply by $ A^n$.


What is the units digit of the product of the first three positive composite numbers?

The first three positive composite numbers are $ 4$, $ 6$, and $ 8$. The product of these three numbers is $ 4\times6\times8\equal{}4\times48\equal{}2\times96\equal{}192$. The units digit of $ 192$ is $ \fbox{2}$.


hi! Can anyone teach me a great way to learn logs( No hard stuff; I'm going to study Geometry.)

$log_{a}{b}$ is the solution to $a^{x}= b$.

5 main log principles.
$1) \log_{a}{a^{x}}= x$
$2) a^{\log_{a}{x}}= x$
$3) \log_{c}{ab}= \log_{c}{a}+\log_{c}{b}$
$4) \log_{c}{\frac{a}{b}}= \log_{c}{a}-\log_{c}{b}$
$5) \frac{\log_{c}{b}}{\log_{c}{a}}= \log_{a}{b}$

In general,
$\log a$ means $\log_{10}a$
$\ln a$ means $\log_{e}a$
$\lg a$ means $\log_{2}a$

Practice with log equations until you get used to using them.  Prove all the formulas to know why they make sense.  People here have said this billions of times, but you should prove things or find out why they work instead of memorizing.

[quote="archimedes1"]Prove all the formulas to know why they make sense.[/quote]@tridecalogism_lover: As archimedes1 says, the only was to understand logs is to prove the properties I gave you.  You can assume $1)$ by definition, but try to use it to prove $2)-4)$.

Why assume 1) by definition? The proof is kind of redundant but it's possible. I guess more importantly is to work 1) out conceptually in your mind without written proof, don't just take it for granted.

How else would you prove the others if you have no way of knowing how to deal with $\log$? It's a definition; you kind of have to take it for granted.

An additional property that follows easily from the others in the above list but that you should probably be able to recognize quickly is $x \log{a}= \log{a^{x}}$

Remember, the [url=http://www.artofproblemsolving.com/Wiki/index.php/Log]AoPSWiki[/url] is your friend  :wink:

[quote]How else would you prove the others if you have no way of knowing how to deal with \log? It's a definition; you kind of have to take it for granted. [/quote]

His identity labeled "1)" is not the definition of the logarithm...

The most simple definition for that is that it's the inverse function of an exponential function.


number theory
least common multiple
floor function
logarithms
modular arithmetic
function
inequalities

Show that ${2n \choose n} \; \vert \; \text{lcm}(1,2, \cdots, 2n)$ for all positive integers $n$.

Since $ n! \equal{} \prod_{p_{i}\text{ prime}}p_{i}^{\sum^{\lfloor\log_{p_{i}}(n)\rfloor}_{k \equal{} 1}\left\lfloor\frac {n}{p_{i}^k}\right\rfloor}$ and $ \text{lcm}(1,\ldots,n) \equal{} \prod_{p_{i}\text{ prime}}p_{i}^{\lfloor\log_{p_{i}}(n)\rfloor}$, there is to prove that
\[ \binom{2n}{n} \equal{} \prod_{p_{i}\text{ prime}}p_{i}^{\sum^{\lfloor\log_{p_{i}}(2n)\rfloor}_{k \equal{} 1}\left\lfloor\frac {2n}{p_{i}^k}\right\rfloor \minus{} 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor}\left|\prod_{p_{i}\text{ is prime}}p_{i}^{\lfloor\log_{p_{i}}(2n)\rfloor}\right. \equal{} \text{lcm}(1,\ldots,2n),\]
so it suffices to show that $ \sum^{\lfloor\log_{p_{i}}(2n)\rfloor}_{k \equal{} 1}\left\lfloor\frac {2n}{p_{i}^k}\right\rfloor \minus{} 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor\le\lfloor\log_{p_{i}}(2n)\rfloor$, which is obvious since $ \left\lfloor\frac {2n}{p_{i}^k}\right\rfloor \minus{} 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor\in\{0,1\}$.

[Moderator edit: fixed some typos. - darij]

[quote="Peter"]Since $ n! \equal{} \prod_{p_{i}\text{ prime}}p_{i}^{\sum^{\lfloor\log_{p_{i}}(n)\rfloor}_{k \equal{} 1}\left\lfloor\frac {n}{p_{i}^k}\right\rfloor}$[/quote]
There's no need to know this...  :maybe:
a more general statement holds:
for every $ k < 2n$:
$ 2n{2n \minus{} 1\choose k}\mid \text{lcm} (2n,2n \minus{} 1,\cdots ,2n \minus{} k)$
(note that by letting $ k \equal{} n$ we will derive the first problem)
Let $ p^{\alpha}$ be the greatest power of $ p$ which divides $ \text{lcm} (2n,2n \minus{} 1,\cdots ,2n \minus{} k)$.hence there exists a natural number $ t$,in which $ p^{\alpha}\mid 2n \minus{} t$,therefore for every $ i\leq\alpha$ we would have $ p^i\mid 2n \minus{} t$ or in other words $ 2n \plus{} t \minus{} l\equiv l\pmod{p^i}$,($ 1\leq l\leq t$) thus exactly $ \lfloor\frac {t}{p^i}\rfloor$ numbers from the set $ \{ 2n \minus{} t \minus{} 1,\cdots ,2n\}$ and $ \lfloor\frac {k \minus{} t}{p^i}\rfloor$ numbers from the set $ \{2n \minus{} t \plus{} 1,\cdots , 2n \minus{} k\}$ are divisible by $ p^i$.on the other hand,we know that $ \lfloor\frac {t}{p^i}\rfloor \plus{} \lfloor\frac {k \minus{} t}{p^i}\rfloor\leq\lfloor\frac {k}{p^i}\rfloor$ which means that the number of multiples of $ p^i$ in the two sets mentioned above are less than the number of multiples of $ p^i$ in the set $ \{1,2,\cdots ,k\}$.

thus the greatest power of $ p$ that divides $ 2n {2n \minus{} 1\choose k} \equal{} \frac {2n(2n \minus{} 1)\cdots (2n \minus{} t \plus{} 1)(2n \minus{} t \minus{} 1)\cdots (2n \minus{} k)(2n \minus{} t)}{k!}$ is less than or equal to $ \alpha$,hence:

$ 2n{2n \minus{} 1\choose k}\mid\text{lcm} (2n,2n \minus{} 1,\cdots ,2n \minus{} k)$

Just wondering, but does this solution work? :

[b]Solution (My-don't-know-if-its-correct try) :[/b]

[b]Lemma:[/b]
If $x\ge0$ and is real number, then we have
\[\ 2[x]\le [2x] \le 2[x]+1 \Rightarrow 0 \le [2x]-2[x] \le 1\]

[b]Proof:[/b]
If $x$ is an integer, then clearly, we have $2[x]=[2x]$. When $x$ is real, then let $x=[x]+\{x\}$, assume that $\{x\} < 0.5$ then, $[2x]=[2[x]+2{x}]=2[x]$. If $\{x\} \ge 0.5$ then we have $[2x]=[2[x]+2{x}]=2[x]+1>2[x]$ $\Box$.

Let $p$ be a prime, such that $p|\dbinom{2n}{n}$ hence, $p|lcm(1,...,2n)$. Let $k$ be the greatest integer such that $p^k \le 2n <p^{k+1}$. Now we apply Legenrde's Function, we have (as the exponent of $p$)

\[\ \sum_{i=0}^{\infty} \lfloor \frac{2n}{p^i} \rfloor -2 \sum_{i=0}^{\infty} \lfloor \frac{n}{p^i} \rfloor \le k \]

Our aim now is to prove that the above inequality is true. And note that the RHS is equal to $k$ is because of our definition.

Now we note that when $i>k$ the LHS will not produce any more positive values, for all $i \le k$, we can apply the [b]lemma[/b], which gives us,

\[\ 0 \le \sum_{i=0}^{\infty} \lfloor \frac{2n}{p^i} \rfloor -2 \sum_{i=0}^{\infty} \lfloor \frac{n}{p^i} \rfloor \le k(1) \]
as desired $\Box$

EDIT: some typos

[quote="Peter"]... which is obvious since $ \left\lfloor\frac {2n}{p_{i}^k}\right\rfloor \minus{} 2\left\lfloor\frac {n}{p_{i}^k}\right\rfloor\in\{0,1\}$.[/quote]
Is this not exactly the crux of your proof? What new do you bring to Peter's ?

OMG :P Sorry, I thought we had completely different solutions due to his "complex notation" (i just typed up my solution right after i thought of it :P )


Are there two 2003 digit numbers A and B with the following properties: the digits of B are a rearrangement of those of A, and the digits of A+B are all 9's? Is there such a pair of numbers with 2004 digits?

Those with $2004$ digits exist: just take numbers $A=33\ldots 366\ldots 6,\ B=66\ldots 633\ldots 3$, where $xx\ldots x$ means the digit $x$ taken $1002$ times.

Those with $2003$ digits don'e exist. First of all, notice that when we add the last two digits, we can get at most $18$, so if we get a number ending in $9$ then that number must be exactly $9$. The same happens with the next digit and so on. This means that addition is unusually simple for these two numbers: we just do it for each pair of corresponding digits. In turn, this means that the number of $0$s equals the number of $1$s, the number of $2$s equals the number of $7$s, and so on, meaning that $A,B$ must have an even number of digits (even, unlike $2003$ :)).


Let $ a,b,c\geq 0$.Prove that:
 $ (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2\geq abc(a \plus{} b \plus{} 2c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)$

my solution is rather ugly ^^

by multiplying out we have:

$ \sum_{sym}a^4b^2\plus{}\sum_{sym}a^3b^3\plus{}6\sum_{sym}a^3b^2c\plus{}\sum_{sym}a^4bc\plus{}10a^2b^2c^2\geq$
$ 7\sum_{sym}a^3b^2c\plus{}\sum_{sym}a^4bc\plus{}16a^2b^2c^2$
then subtract terms, which appear on both sides of the inequaltiy
$ \sum_{sym}a^4b^2\plus{}\sum_{sym}a^3b^3\geq\sum_{sym}a^3b^2c\plus{}\sum_{sym}a^2b^2c^2$ (*)

And because of Muirhead Inequality is
I $ \sum_{sym}a^4b^2\geq\sum_{sym}a^3b^2c$
([4;2;0]$ \geq$[3;2;1])
and II $ \sum_{sym}a^3b^3\geq\sum_{sym}a^2b^2c^2$
([3;3;0]$ \geq$[2;2;2])

By adding I and II we have (*)

[quote="math10"]Let $ a,b,c\geq 0$.Prove that:
 $ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\geq abc(a \plus{} b \plus{} 2c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)$[/quote]

My solution :)
Setiing:$ p \equal{} a \plus{} b \plus{} c \equal{} 3;q \equal{} ab \plus{} bc \plus{} ca;r \equal{} abc$
Inq
$ < \equal{} > (3q \minus{} r)^2 \ge r(r \plus{} 3q \plus{} 54)$
$ < \equal{} > 9q^2 \minus{} 9qr \minus{} 54r \ge 0$
$ < \equal{} > 3q(q \minus{} 3r) \plus{} 6(q^2 \minus{} 9r) \ge 0$
It's always true  :P

[quote="math10"]Let $ a,b,c\geq 0$.Prove that:
 $ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\geq abc(a \plus{} b \plus{} 2c)(b \plus{} c \plus{} 2a)(a \plus{} c \plus{} 2b)$[/quote]
By Am-Gm: $ LHS \geq\ \frac{64}{81}(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^2$$ \ge\ \frac{64}{27}abc(a\plus{}b\plus{}c)^3 \geq\ RHS$
 :)


I have a machine to shuffle 53 cards, and this machine would permute the cards in the same way (same method) every time. After a finite number of shuffles, I eventually get back to the original order. What is the largest possible number of shuffles it could take?

It's the maximum value the term $\mathrm{lcd}(a_1,a_2,...,a_n)$ with $a_1+a_2+...+a_n=53$ can reach (the order of a permutation build up by $a_i$-cycles).
From this point the brutal fork would do, but there should be a better approach.


inequalities
function
algebra unsolved
algebra

Prove that there is no function $ f: R^\plus{}\minus{}>R^\plus{}$ such that

$ f^2(x) \ge f(x\plus{}y)(f(x) \plus{} y)$

for any x, y > 0.

Let's denote $ g(x) \equal{} \frac 1{f(x)}$ , and that leeds to $ g(x \plus{} y)\geq g(x)(1 \plus{} yg(x))$ , already solved on this site . See http://www.mathlinks.ro/viewtopic.php?t=67094
 :cool:


superior algebra
superior algebra unsolved

Let $ A \equal{} \bigoplus_{n\ge 0} A_{n}$ be a commutative (unitary) $ \mathbb{N} \minus{}$graded ring.
Assume that $ I$ is an ideal in $ A,$ which is generated by homogeneous elements of $ A.$
Suppose that for every homogeneous elements $ a, b \in A:$
$ ab\in I$ implies $ a \in I$ or $ b^{n}\in I$ for some $ n\in \mathbb{N}$

Prove that $ I$ is primary ideal in $ A.$

[hide="Note"]
I've already invented an argument, but I found this problem somewhat interesting, so I want to share it, and watch other approaches:)[/hide]

The definition of a primary ideal that I know is this:
A primary ideal of a ring $ A$ is an ideal $ I$ such that : $ ab \in I$ and $ a\not \in I \Rightarrow \exists n \in \mathbb{N}$ such that $ b^n \in I$.
So your $ I$ is indeed primary ideal in $ A$.

There is a subtle logical twist here: What you need to show is that if the condition is fulfilled for [b]homogeneous[/b] elements, then it is satisfied for [b]arbitrary[/b] elements (which are finite sums of homogeneous), too.


The incircle of a triangle $ABC$ is tangent to $BC$ at $D$. Let $E\in[CA]$. Prove that the incircles of triangles $EDC$, $EDB$, $EAB$ have a common tangent line $(l)$.

Let ABCD be a convex tangential quadrilateral with an incircle (I). Let $(I_1)$ be an arbitrary circle inside the quadrilateral touching the sides DA, AB, sufficiently small for a tangent to $(I_1)$ from the vertex D different from DA to meet the line AB at a point E inside the segment (AB). Let $(I_2)$ be the incircle of the triangle $\triangle EBC$ and let the common external tangent of the circles $(I_1), (I_2)$ different from AB meet the lines DA, DE, CE, CB at points K, M, N, L, respectively. (The quadrilaterals AEMK, BENL are tangential.) Then the quadrilateral CDMN is also tangential.

The posted problem is a limiting case of this generalization, when the vertex C and the tangency points of the quadrilateral incircle (I) with the sides BC, CD coincide into the tangency point C of the incircle (I) of the triangle $\triangle ABD$ with the side BD.

Let $D_1, B_1$ be tangency points of the circle $(I_1)$ with the sides DA, AB, respectively $A_2, C_2$ tangency points of the circle $(I_2)$ with the sides AB, BC, respectively, $E_1, E_2$ tangency points of the circles $(I_1), (I_2)$ with the lines ED, EC, respectively, and $K_1, L_2$ tangency points of the circles $(I_1), (I_2)$ with the line KL, respectively. Since the quadrilateral ABCD is tangential,

$AB + CD = BC + DA$

Substracting segments $AB_1, BA_2$ from the left side and equal segments $AD_1 = AB_1, BC_2 = BA_2$ from the right side,

$B_1A_2 + CD = CC_2 + DD_1$

Substituting equal segment $K_1L_2 = B_1A_2$ on the left and equal segments $DE_1 = DD_1, CE_2 = CC_2$ on the right,

$K_1L_2 + CD = CE_2 + DE_1$

Substracting segments $MK_1, NL_2$ from the left side and equal segments $ME_1 = MK_1, NE_2 = NL_2$ from the right side,

$MN + CD = CN + DM$

which means that the quadrilateral CDMN is tangential, as required.


In some of the $2n$ cells of a $2$x$n$ table there are one or more coins. In each step we choose a cell with at least two coins , remove two coins and put one either on the upper cell , or on the cell to the right. If we start with at least $2^{n}$ coins on the table , prove that we can play so that we bring at least one coin to the upper right cell.

[quote="paladin8"][hide="hint"] Invariance. Label each cell with an appropriate power of $2$. [/hide][/quote]

OK. I understand that you label it by $2^{x}$, where $x$ is the "distance" from the bottom left corner. Then the total sum must be at least $2^{n}$. That still doesn't imply what we want, I think...


[quote="ac1072"]For triangle $ABC$with sides $a,b,c$and area $S$,
proof that $\frac{2ab+2bc+2ca-(a^{2}+b^{2}+c^{2})}{4S}=\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}$[/quote]
We have
$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}=\frac{4+4sinA/2.sinB/2.sinC/2)R}{4RcosA/2.cosB/2.cosC/2}$
$r=4RsinA/2.sinB/2.sinC/2$
$p=4RcosA/2.cosB/2.cosC/2$
$\to RHS=\frac{4R+r}{p}$

[quote="kcbu"]
We have
$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}=\frac{4+4sinA/2.sinB/2.sinC/2)R}{4RcosA/2.cosB/2.cosC/2}$
$r=4RsinA/2.sinB/2.sinC/2$
$p=4RcosA/2.cosB/2.cosC/2$
$\to RHS=\frac{4R+r}{p}$[/quote]
Can you explain it in detail? :(

$\left\{\begin{array}{c}\tan \frac{A}{2}=\frac{r_{a}}{p}\ ,\ \mathrm{a.s.o.}\\\\ r_{a}+r_{b}+r_{c}=4R+r\\\\ \end{array}\right\|$ $\Longrightarrow$ $\boxed{\sum\tan\frac{A}{2}=\frac{4R+r}{p}}\ge \sqrt 3\ .$

$\left\{\begin{array}{c}S=pr\ ,\ \sum a=2p\\\\ \sum bc=p^{2}+r^{2}+4Rr\\\\ \sum a^{2}=2\left(p^{2}-r^{2}-4Rr\right)\end{array}\right\|$ $\Longrightarrow$ $\boxed{\frac{2\sum bc-\sum a^{2}}{4S}=\frac{4R+r}{p}}\ .$

[color=darkblue][b][u]Details.[/u][/b][/color]

$1.\blacktriangleright pr^{2}=(p-a)(p-b)(p-c)=$ $p^{3}-p^{2}(a+b+c)+p(ab+bc+ca)-abc$,

$abc=4Rpr$ $\Longrightarrow$ $r^{2}=-p^{2}+\sum bc-4Rr$ $\Longrightarrow$ $\boxed{\sum bc=p^{2}+r^{2}+4Rr}$.

$2.\blacktriangleright\sum (p-b)(p-c)=3p^{2}-p\sum (b+c)+\sum bc=-p^{2}+\sum bc$ $\Longrightarrow$

$\boxed{(p-b)(p-c)+(p-c)(p-a)+(p-a)(p-b)=r(4R+r)}$.

$3.\blacktriangleright\sum r_{a}=\sum\frac{S}{p-a}=$ $\frac{S}{(p-a)(p-b)(p-c)}\cdot \sum (p-b)(p-c)$ $\Longrightarrow$

$\sum r_{a}=\frac{pr}{pr^{2}}\cdot r(4R+r)$ $\Longrightarrow$ $\boxed{r_{a}+r_{b}+r_{c}=4R+r}$.

[quote="ac1072"]Thank a lot.But why $\sum r_{a}=\sum\frac{S}{p-a}$?[/quote]
because we have $S=(p-a).r_{a}$
It's easy to prove.


Source: 360 PMC
On side $BC$ of a triangle $ABC$ consider points $M$ and $N$ such that $\angle BAM=\angle CAN$. Prove that
$\frac{MB}{MC}+\frac{NB}{NC}\geq 2\frac{AB}{AC}$.

first, on the book, it has some symbol above $BAM$ and $CAN$, it looks like a long hat...does this mean angle $BAM$ and angle $CAN$? I assumed it means angle on the above problem..

[quote="shobber"]Let $\angle{BAM}=\angle{CAN}=\alpha$ and $\angle{BAN}=\angle{CAM}= \beta$. Now use the fact that: $\frac{BM}{CM}=\frac{AB \sin{\alpha}}{AC \sin{\beta}}$ and $\frac{BN}{CN}=\frac{AB\sin{\beta}}{AC \sin{\alpha}}$, we get: \[\frac{BM}{CM}+\frac{BN}{CN}=\frac{AB}{AC}\left( \frac{\sin{\alpha}}{\sin{\beta}}+\frac{\sin{\beta}}{\sin{\alpha}}\right) \geq 2 \cdot \frac{AB}{AC}\] [/quote]
Wait, i still don't understand how u got
$\frac{BM}{CM}=\frac{AB \sin{\alpha}}{AC \sin{\beta}}$ and $\frac{BN}{CN}=\frac{AB\sin{\beta}}{AC \sin{\alpha}}$..
Is it Steiner's theorem? (that's what the solution in my book stated, but i don't know what Steiner's theorem is....)
Thanks,

$\frac{BM}{CM}=\frac{\triangle{ABM}}{\triangle{ACM}}=\frac{AB \cdot AM \cdot \sin{\alpha}}{AC \cdot AM \cdot \sin{\beta}}=\frac{AB \cdot \sin{\alpha}}{AC \cdot \sin{\beta}}$


Let $A\in XY\cap UW, B\in XZ\cap VW$; since $\Delta XUY\sim\Delta ZVX\implies\frac{AY}{AX}=\frac{BX}{BZ}\iff\frac{AY}{XY}+\frac{BZ}{XZ}=1$, so easily, $W\in YZ$.

Best regards,
sunken rock

[color=darkred][b][u]Lemma.[/u][/b] Let $XYZ$ be a right triangle for which $XY\perp XZ$. Denote the projections $U$, $V$ of the points $Y$, $Z$ respectively to a mobile

line $d$ for which $X\in d$. Define the point $W$ for which $WU\perp XY$ and $WV\perp XZ$. Ascertain the geometrical locus of the point $W$.[/color]

[b][u]Proof[/u].[/b] Denote the projections $A$ , $B$ of $U$ , $V$ on $XY$ , $XZ$ respectively and $M\in UA\cap YZ$ , $N\in VB\cap YZ$ . Thus,

$\triangle UXY\sim\triangle VZX\implies$ $\frac{MY}{YZ}=$ $\frac {AY}{XY}=$ $\frac {BZ}{XZ}=$ $\frac {NZ}{YZ}\implies$ $MY=NZ\implies$ $M\equiv N\equiv W\implies$ $W\in BC$ .


Sandy typed a positive six-digit integer, but the two 9s she typed didn’t show. If 2010 appeared, how many different positive six-digit integers could she have typed? One such integer to include is 920910

Just count the total number of ways to choose 2 place values for the 9's.

$ C(6,2)\equal{}\frac{6*5}{2}\equal{}\boxed{15}$ possible integers.