If $x , y , z >0$ and $x+y+z = 1$ , find the maximum value of the expression
   $A = xy+yz+zx-3xyz$

 [u]Babis[/u]

[hide="NOT 2/9"]
... but $\boxed{\frac{1}{4}}$, attained when $a=b=\frac{1}{2}$ and $c=0$. To prove the inequality, WLOG let $a\geq b\geq c$ and rewrite as
\[ab(1-3c)+c(a+b)\leq \frac{1}{4}\]
As $a\geq b\geq c$, $c\leq \frac{1}{3}$ so $1-3c$ is nonnegative. Also, $a+b=1-c$ and $ab\leq \frac{(a+b)^{2}}{4}=\frac{(1-c)^{2}}{4}$. This gives us
\[ab(1-3c)+c(a+b)\leq\frac{(1-c)^{2}(1-3c)}{4}+c(1-c)\]
so it remains to prove
\[(1-c)^{2}(1-3c)+4c(1-c)\leq 1\]
But the left hand side is $(1-c)((1-c)(1-3c)+4c)=(1-c)(1+3c^{2})=1-c+3c^{2}-3c^{3}$. The above inequality then follows from
\[1-(1-c)^{2}(1-3c)-4c(1-c)=c(3c^{2}-3c+1)=c\left( \frac{3}{4}(2c-1)^{2}+\frac{1}{4}\right) \geq 0\]
[/hide]

[hide="I dont know if this can go anywhere, but with some manipulation I get that..."]
$(x+y+z)^{3}=1 \implies 3A-1= \sum x^{3}-4 \sum x^{2}-3 \sum xy$
and
$(x+y+z+1)^{3}=8 \implies 6A-7= 2 \sum x^{3}-6 \sum x^{2}-3 \sum x-12xyz$
(All sums here are symmetric.)[/hide]

[quote="scorpius119"][quote="bigboy82892"][quote="scorpius119"][hide="NOT 2/9"]
... but $\boxed{\frac{1}{4}}$, attained when $a=b=\frac{1}{2}$ and $c=0$. To prove the inequality, WLOG let $a\geq b\geq c$ and rewrite as
\[ab(1-3c)+c(a+b)\leq \frac{1}{4}\]
As $a\geq b\geq c$, $c\leq \frac{1}{3}$ so $1-3c$ is nonnegative. Also, $a+b=1-c$ and $ab\leq \frac{(a+b)^{2}}{4}=\frac{(1-c)^{2}}{4}$. This gives us
\[ab(1-3c)+c(a+b)\leq\frac{(1-c)^{2}(1-3c)}{4}+c(1-c) \]
so it remains to prove
\[(1-c)^{2}(1-3c)+4c(1-c)\leq 1 \]
But the left hand side is $(1-c)((1-c)(1-3c)+4c)=(1-c)(1+3c^{2})=1-c+3c^{2}-3c^{3}$. The above inequality then follows from
\[1-(1-c)^{2}(1-3c)-4c(1-c)=c(3c^{2}-3c+1)=c\left( \frac{3}{4}(2c-1)^{2}+\frac{1}{4}\right) \geq 0 \]
[/hide][/quote]
Umm... didn't the question say $c>0$?[/quote]

[hide="Well..."]in that case we can consider $x=y=\frac{1-\epsilon }{2}$ and $z=\epsilon$ (for some reason i used $a,b,c$ in my solution...) as $\epsilon$ approaches 0, the expression approaches 1/4. So there would technically be [b]no[/b] maximum, but 1/4 would still be the best bound.
[/hide]

(of course, that wouldn't be necessary if it just said that the variables were nonnegative rather than positive.)[/quote]

 [b]YES ! [/b]
I must change the conditiion to $x , y , z \geq 0$.  Really sorry , but I saw the problem under these conditions and I had not solved the problem , before posting it.Thank you !


\equiv $ \equiv$

\neq $ \neq$

[url]http://www.artofproblemsolving.com/Forum/posting.php?f=123&mode=smilies[/url]
(These are the available emoticons for use in messages, signatures, etc. Click the "more emoticons" link on the left, underneath the visible emoticons, to see this list. Then, click on the emoticon you with to insert. )

EDITS: sorry, grammar fail  :oops:

[quote="topofmath"]Hi. Just wondering how to type:

1) The modular congruence sign.
2) The not-equal sign
3) More emotions in my signature.

Thanks you.[/quote]

Note: For other symbols in $ \text{\LaTeX}$ that you don't know you can find them [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]here[/url].


Let $ P$ be a 30-sided polygon inscribed in a circle. Find the number of triangles whose vertices are the vertices of $ P$ such that any two vertices of each triangle are separated by at least three other vertices of $ P$.

Official Solution:
Let $ A$ be a vertex of $ P$. First we shall count the number of such triangles having $ A$ as a vertex. After taking away $ A$ and 3 consecutive vertices of $ P$ on each side of $ A$, we are left with 23 vertices from which we can choose two vertices in such a way that, together with $ A$, a desired triangle can be formed. There are $ \binom{2\plus{}(23\minus{}3\minus{}2)}{2}\equal{}\binom{20}{2}$ ways to do so. Hence there are $ 30\binom{20}{2}\times\frac{1}{3}$ (because each triangle is counted thrice) = 1900 such triangles.

What I would like to know is how they get $ \binom{2\plus{}(23\minus{}3\minus{}2)}{2}\equal{}\binom{20}{2}$ ways to choose the remaining two vertices after one is fixed..

I don't get it either. Also, what's wrong with this method: Choose 3 out of 21 vertices (12C3) and "add" in 3 more between each pair.

Ubemaya, I think your method fails because the points are distinct so you can't arbitrarily take 9 out if you don't know where they are supposed to be. But I'm not sure, since I probably would have believed you if we didn't know your answer was wrong.

As for the author's question, I'm not sure exactly how they got that, but here is a different calculation that gets the same result (assuming you haven't already found one):

Let the $ 23$ points be OOOO...OOOO. Pick the "left" point first. If it is the first point, there are $ 19$ choices for the "right" point (just count). If it is the second, there are $ 18$. Third has $ 17$, and so on. So there are $ 19\plus{}18\plus{}17\plus{}\cdots\plus{}1 \equal{}\frac{19\cdot 20}{2}\equal{}\binom{20}{2}$.

I used the same method as yours initially, but then i double counted because i didn't assign "left" and "right" points. Thanks =)


anybody know if there are other taiwan amc exams besides the 2002 one?  If not, why did they only do that for one year?

Answer:  There are no other contests prepared specfically for Taiwan other than the one given in January 2002.  After that contest was used by students in Taiwan, we released it on the Web for practice purposes and we designated it as the 2002 AMC 10 P and 2002 AMC 12 P, P for Practice.

The reason is that when the Taiwan Secondary Education Association originally contracted to participate in the AMC contests, they did not believe that it was possible to fit the February administrations of the regular A and B dates into their academic calendar.  This was partly because of their academic calendar and partly because of where the Chinese New Year fell that year.

Since then, both sides have agreed that it is easier to administer the AMC 10 and 12 contests at the same time.

Steve Dunbar
AMC Director


Estos son los problemas de la final de la olimpiada de Uruguay en mi nivel
(no están ordenados por dificultad)

1. Demostrar que en cualquier conjunto de 11 enteros positivos siempre es posible seleccionar 6 de sus elementos tales que su suma sea múltiplo de 6.

2. Sea $\triangle ABC$ triángulo isósceles en $A$. Llamamods $D$ al punto medio de $AC$, $E$ a la proyección ortogonal de $D$ sobre $BC$ y $F$ al punto medio de $DE$.
Demostrar que $BF$ y $AE$ son perpendiculares si y solo sí $\triangle ABC$ es equilátero.

3. Encontrar todos los enteros positivos $a, b, c$ tales que $MCM(a,b,c)=a+b+c$

4. Sean los polinomios $P(x)=6x^2-24x-4a$ y $Q(x)=x^3+ax^2+bx-8$
Encontrar todos los valores reales de $a$ y $b$ para que $P(x)$ y $Q(x)$ tengan toas sus raíces reales no
negativas.

[i]Editado por el Moderador: Le puse Latex![/i]

ok

1. entre 5 enteros positivos siempre hay 3 que suman multiplo de 3. Entre 3 enteros positivos siempre hay 2 que suman multiplo de 2.

Con los 11 numeros puedo armar tres grupos de tres numeros cada uno la suma de cada uno multiplo de 3 (usando la parte 1)

Entre estos tres grupos puedo escoger 2 con suma multiplo de 2 (usando parte 2)

Estos dos grupos tienen en total 6 numeros y su suma es multiplo de 6.

3. Sea $a$ el mayor de los numeros, los 3 numeros iguales no es solucion. entonces $a$ divide a $b+c$ de donde $a=b+c$. Asi pues se obtiene que $a+b+c=2a$  sea $b$ el segundo mayor. entonces $b$ divide a $2a$ o $2(b+c)$ entonces $b$ divide a $2c$ se sigue que $b=2$c o $b=c$. Asi pues las potenciales triplas a servir son: $c,c,2c$ o $c,2c,3c$ se ve que solo la segunda sirve.

Generalización del problema 1:
Para cada $n$ natural encuentra el menor número $k$ tal que es posible garantizar que entre cualesquiera $k$ enteros, es posible escoger $n$ cuya suma es múltiplo de $n$.
Ese problema apareció en los selectivos de México para la IMO o la Ibero (no recuerdo cual). Nadie lo pudo terminar. Seguro ya lo postearon por allí, pero sería interesante intentarlo un rato acá.

Y la respuesta es $2n-1$ este es un celebre teorema de Erdos.

calramente $2n-2$ no son suficientes, $n-1$ $1's$ y $n-1$ $0's$.

la prueba es un corolario de lo siguiente: Sea $p$ un numero primo, entonces entre $2p-1$ enteros puedo seleccionar $p$ tales que su suma es multiplo de $p$.

La prueba es: sea $A=(a_1,a_2...,a_{2p+1})$ el conjunto de numeros, para un conjunto $T$ sea $S(T)$ la suma de los elementos de $T$ y sea $B$ el conjunto de subconjuntos de tamaño $p$ de $A$.

Consideremos la siguiente suma, modulo $p$

$\sum_{T\in B} (S(T)^{p-1}-1)$

si $S(T)$ es diferente de $0$ para todo $T$ en $B$, entonces la suma es $0$ por Fermat.

Pero por otro lado, podemos ver esta suma tambien de la siguiente manera: cada producto asi: $a_1^{t_1}a_2^{t_2}...a_k^{t_k}$ Con $t_1+t_2+...t_k=p-1$ y $k<p$ sale $\binom{2p-1-k}{p-k}$ veces en la suma. pero es facil ver que este coeficinte es multiplo de $p$ para $0<k<p$. De donde  $\sum_{T\in B} S(T)^{p-1}=0$ modulo $p$.

esto implicaria que $\binom{2p-1}{p}$ sea multiplo de $p$, pero sin embargo esto no es cierto.

Para ver que si para $m$ funciona y $n$ funciona entonces para $mn$ funciona, un argumento identico al que presente en mi pri mer post de este topic funciona.


inequalities
algebra
polynomial
inequalities proposed

$\bigstar$
If $a,b,c,d$ are non-negative with sum $4$ then
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+4abcd \le 16. \]
$\bigstar$
If $a,b,c,d$ are non-negative with sum $4$ then
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 8. \]

I think the following stronger holds
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+8abcd \le 16.\]
Notice that in the this inequality, equality can holds for $a=b=c=d=1$ or $(a,b,c,d)=(2,0,2,0)$.

For two initial inequalities, I have a few-line solutions. So I am waiting a nice solution from you.

You are right, VASC. I am stupid not to realize it before, because to prove it is really easier than prove the inequality
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+8abcd \le 16\]
(since it is fourth-degree). Anyway, all of them are killed by my "few-line" solutions.

[quote="hungkhtn"][quote="Vasc"][quote="hungkhtn"]I think the following stronger holds
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+8abcd \le 16. \]
[/quote]
Actually, the following stronger inequality holds
\[\sum a^{2}(b+c)+2\sum abc \le 16. \]
[/quote]
Just want to ask, do you solve the second part, VASC?[/quote]
The second inequality is equivalent to

$\sum a(a+d)(b+c)\le 16$.

[quote="Vasc"][/quote][quote="hungkhtn"][quote="Vasc"][quote="hungkhtn"]I think the following stronger holds
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+8abcd \le 16. \]
[/quote]
Actually, the following stronger inequality holds
\[\sum a^{2}(b+c)+2\sum abc \le 16. \]
[/quote]
Just want to ask, do you solve the second part, VASC?[/quote][quote="Vasc"]
The second inequality is equivalent to

$\sum a(a+d)(b+c)\le 16$.[/quote]

It is so so ingenious solution, VASC. But I mean do you solve the second part
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 8?\]

[quote="Vasc"][/quote][quote="hungkhtn"][quote="Vasc"][quote="hungkhtn"]I think the following stronger holds
\[a^{2}(b+c)+b^{2}(c+d)+c^{2}(d+a)+d^{2}(a+b)+8abcd \le 16. \]
[/quote]
Actually, the following stronger inequality holds
\[\sum a^{2}(b+c)+2\sum abc \le 16. \]
[/quote]
Just want to ask, do you solve the second part, VASC?[/quote][quote="Vasc"]
The second inequality is equivalent to

$\sum a(a+d)(b+c)\le 16$.[/quote]
Is it because $(a+d)(b+c)\leq \frac{(a+b+c+d)^{2}}{4}=4$ then $\sum a(a+d)(b+c)\le 4\sum a=16$

[quote="hungkhtn"] It is so so ingenious solution, VASC. But I mean do you solve the second part
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 8? \]
[/quote]
I think you mean
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 16, \]with equality for $a=b=2$ and $c=d=0$, or any cyclic permutation.
We have
$\sum ab(a+b+c) \le \sum ab(a+b+c+d)=4\sum ab =4(a+c)(b+d) \le (a+c+b+d)^{2}=16.$

[quote="Vasc"][quote="hungkhtn"] It is so so ingenious solution, VASC. But I mean do you solve the second part
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 8? \]
[/quote]
I think you mean
\[ab(a+b+c)+bc(b+c+d)+cd(c+d+a)+da(d+a+b) \le 16, \]
with equality for $a=b=2$ and $c=d=0$, or any cyclic permutation.
We have
$\sum ab(a+b+c) \le \sum ab(a+b+c+d)=4\sum ab =4(a+c)(b+d) \le (a+c+b+d)^{2}=16.$[/quote]

:(( why it is so simple as that. I am sorry.

And this is the hardest and nicest one

Let $a,b,c,d$ be non-negative real numbers with sum $4$. Prove that
\[ab(a+b+2c)+bc(b+c+2d)+cd(c+d+2a)+da(d+a+2b) \le 16.\]

The equality holds for $a=b=c=d=1$ or $a=c=1,b=2,d=0$. This is very special.

I am waiting for a nice solution for it. My solution is also very short and nice.

[quote="hungkhtn"]And this is the hardest and nicest one

Let $a,b,c,d$ be non-negative real numbers with sum $4$. Prove that
\[ab(a+b+2c)+bc(b+c+2d)+cd(c+d+2a)+da(d+a+2b) \le 16. \]The equality holds for $a=b=c=d=1$ or $a=c=1,b=2,d=0$. This is very special.
I am waiting for a nice solution for it. My solution is also very short and nice.[/quote]
We have
$\sum ab(a+b+2c)=\sum ab(a+b+c+d)=4(a+c)(b+d)\le (a+c+b+d)^{2}=16$.

[quote="Vasc"][quote="hungkhtn"]And this is the hardest and nicest one

Let $a,b,c,d$ be non-negative real numbers with sum $4$. Prove that
\[ab(a+b+2c)+bc(b+c+2d)+cd(c+d+2a)+da(d+a+2b) \le 16. \]
The equality holds for $a=b=c=d=1$ or $a=c=1,b=2,d=0$. This is very special.
I am waiting for a nice solution for it. My solution is also very short and nice.[/quote]
We have
$\sum ab(a+b+2c)=\sum ab(a+b+c+d)=4(a+c)(b+d)\le (a+c+b+d)^{2}=16$.[/quote]

It is out of everything I expected. Thank you, VASC, since my solution is much harder. I will think about another.

From the solution it appears that equality holds whenever $a+c=2. b+d=2$. Am I missing something?

Ok. This one is left for you, VASC. It is not as easy as before.

[b]Problem. (hungkhtn dedicates to VASC)[/b]
If $a,b,c,d$ are non-negative and $a+b+c+d=3$ then
\[ab(a+2b+3c)+bc(b+2c+3d)+cd(c+2d+3a)+da(d+2a+3b) \le 6\sqrt{3}. \]
It must be very hard, I think.

Sorry VASC. It is $a+b+c+d=3$, not $a+b+c+d=4$.

[quote="hungkhtn"]If $ a,b,c,d$ are non-negative with sum $ 4$ then

\[ a^{2}(b \plus{} c) \plus{} b^{2}(c \plus{} d) \plus{} c^{2}(d \plus{} a) \plus{} d^{2}(a \plus{} b) \plus{} 8abcd \le 16.\]
Notice that in the this inequality, equality can holds for $ a \equal{} b \equal{} c \equal{} d \equal{} 1$ or $ (a,b,c,d) \equal{} (2,0,2,0)$.[/quote]
[quote="Vasc"]Actually, the following stronger inequality holds

\[ \sum a^{2}(b \plus{} c) \plus{} 2\sum abc \le 16.\]
It is equivalent to

$ \sum a(a \plus{} d)(b \plus{} c)\le 16$.[/quote]
[quote="me@home"]Is it because $ (a \plus{} d)(b \plus{} c)\leq \frac {(a \plus{} b \plus{} c \plus{} d)^{2}}{4} \equal{} 4$ then $ \sum a(a \plus{} d)(b \plus{} c)\le 4\sum a \equal{} 16$[/quote]
$ \frac {(a \plus{} b \plus{} c \plus{} d)^3}{4} \minus{} \sum_{cyc}{a^{2}(b \plus{} c)} \minus{} 2\sum_{cyc}{abc} \equal{} \sum_{cyc}{\frac {a}{4}(a \minus{} b \minus{} c \plus{} d)^2}\geq 0$.

[hide="Another proof."]$ (a \plus{} b \plus{} c \plus{} d)^3 \minus{} 4\sum_{cyc}{a^{2}(b \plus{} c)} \minus{} 8\sum_{cyc}{abc}$

$ \equal{} 4a[(a \minus{} c)^2 \plus{} (b \minus{} d)^2] \plus{} (c \minus{} a)(a \minus{} b \minus{} c \plus{} d)^2 \plus{} (b \plus{} d \minus{} 2a)(a \plus{} b \minus{} c \minus{} d)^2\geq 0$,

which is clearly true for $ a \equal{} \min\{a,b,c,d\}$. :blush: [/hide]

Thank you very much, it is very nice.

Did you solve the last one? problem with $ 6\sqrt{3}$?

[quote="hungkhtn"]Let $ a,b,c,d$ be non-negative real numbers with sum $ 4$. Prove that

$ ab(a \plus{} b \plus{} 2c) \plus{} bc(b \plus{} c \plus{} 2d) \plus{} cd(c \plus{} d \plus{} 2a) \plus{} da(d \plus{} a \plus{} 2b) \le 16.$

My solution is very short and nice.[/quote]
[b]Proof[/b] $ (a \plus{} b \plus{} c \plus{} d)^3 \minus{} 8(bcd \plus{} cda \plus{} dab \plus{} abc) \minus{} 4[a^2(b \plus{} d) \plus{} b^2(c \plus{} a) \plus{} c^2(d \plus{} b) \plus{} d^2(a \plus{} c)]$

$ \equal{} (a \plus{} b \plus{} c \plus{} d)(a \minus{} b \plus{} c \minus{} d)^2\geq 0$.

[b]Generalization[/b] For any $ a,b,c,d\geq 0$, the following inequality

$ 4[a^2(b \plus{} pc \plus{} d) \plus{} b^2(c \plus{} pd \plus{} a) \plus{} c^2(d \plus{} pa \plus{} b) \plus{} d^2(a \plus{} pb \plus{} c)]$

$ \leq(a \plus{} b \plus{} c \plus{} d)^3 \minus{} q(bcd \plus{} cda \plus{} dab \plus{} abc) \minus{} \frac {16(8 \minus{} 4p \minus{} q)abcd}{a \plus{} b \plus{} c \plus{} d}$

holds if and only if $ \left(q\geq3 \wedge p\leq \frac {136 \minus{} 9q \minus{} (3q \minus{} 8)^{\frac {3}{2}}}{108}\right)\vee\left(q\leq3 \wedge p\leq 1\right)$.

The special case $ q \equal{} 0,p \equal{} 1$ see: http://www.mathlinks.ro/Forum/viewtopic.php?t=148354

[quote="hungkhtn"]If $ a,b,c,d$ are non-negative and $ a \plus{} b \plus{} c \plus{} d \equal{} 3$, then

$ ab(a \plus{} 2b \plus{} 3c) \plus{} bc(b \plus{} 2c \plus{} 3d) \plus{} cd(c \plus{} 2d \plus{} 3a) \plus{} da(d \plus{} 2a \plus{} 3b) \le 6\sqrt {3}$.[/quote]
Actually, the following stronger inequality holds

$ a^2(b \plus{} 2d) \plus{} b^2(c \plus{} 2a) \plus{} c^2(d \plus{} 2b) \plus{} d^2(a \plus{} 2c)$

$ \leq \frac {2}{3\sqrt {3}}(a \plus{} b \plus{} c \plus{} d)^3 \plus{} \left(3 \minus{} \frac {32}{3\sqrt {3}}\right)(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, c \equal{} (\sqrt {3} \minus{} 1)d$. :yinyang:

[b]Generalization[/b] If $ a,b,c,d\geq 0$ and $ r > 0$, then

$ a^2(b \plus{} rd) \plus{} b^2(c \plus{} ra) \plus{} c^2(d \plus{} rb) \plus{} d^2(a \plus{} rc)$

$ \leq\frac {r^2}{2(r^2 \minus{} r \plus{} 1)^{\frac {3}{2}} \minus{} (r \plus{} 1)(r \minus{} 2)(2r \minus{} 1)}(a \plus{} b \plus{} c \plus{} d)^3$

$ \minus{} \frac {37r^4 \minus{} 32r^3 \plus{} 118r^2 \minus{} 32r \plus{} 37}{32(r^2 \minus{} r \plus{} 1)^{\frac {3}{2}} \minus{} (r \plus{} 1)(r \minus{} 5)(5r \minus{} 1)}(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} c \equal{} d$.:yinyang:

For $ r\to 0$, we obtain

$ 27(a^2b \plus{} b^2c \plus{} c^2d \plus{} d^2a)\leq4(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 37(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, c \equal{} 2d$.

[hide="Other special cases."]$ r \equal{} \frac {8}{3}$:

$ 25[a^2(3b \plus{} 8d) \plus{} b^2(3c \plus{} 8a) \plus{} c^2(3d \plus{} 8b) \plus{} d^2(3a \plus{} 8c)]$

$ \leq36(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 301(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, 3c \equal{} 2d$;

$ r \equal{} \frac {8}{5}$:

$ 243[a^2(5b \plus{} 8d) \plus{} b^2(5c \plus{} 8a) \plus{} c^2(5d \plus{} 8b) \plus{} d^2(5a \plus{} 8c)]$

$ \leq 400(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 3241(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, 5c \equal{} 4d$;

$ r \equal{} \frac {15}{7}$:

$ 432[a^2(7b \plus{} 15d) \plus{} b^2(7c \plus{} 15a) \plus{} c^2(7d \plus{} 15b) \plus{} d^2(7a \plus{} 15c)]$

$ \leq 1225(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 10096(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, 7c \equal{} 5d$;

$ r \equal{} \frac {15}{8}$:

$ 49[a^2(8b \plus{} 15d) \plus{} b^2(8c \plus{} 15a) \plus{} c^2(8d \plus{} 15b) \plus{} d^2(8a \plus{} 15c)]$

$ \leq 144(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 1177(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, 4c \equal{} 3d$;

$ r \equal{} \frac {21}{5}$:

$ 64[a^2(5b \plus{} 21d) \plus{} b^2(5c \plus{} 21a) \plus{} c^2(5d \plus{} 21b) \plus{} d^2(5a \plus{} 21c)]$

$ \leq 225(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 1936(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} 0, 5c \equal{} 3d$.[/hide]

[quote="Ji Chen"][quote="Ji Chen"]$ \frac {(a \plus{} b \plus{} c \plus{} d)^3}{4} \minus{} \sum_{cyc}{a^{2}(b \plus{} c)} \minus{} 2\sum_{cyc}{abc} \equal{} \sum_{cyc}{\frac {a}{4}(a \minus{} b \minus{} c \plus{} d)^2}\geq 0$.[/quote]
$ \frac {m(a \plus{} b \plus{} c \plus{} d)^3}{4} \plus{} (1 \minus{} 3m)\sum_{sym}{bcd} \minus{} \sum_{cyc}{a^{2}(b \plus{} mc)}$

$ \equal{} \frac {a \plus{} c}{4m}[(ma \minus{} b \minus{} mc \plus{} d)^2 \plus{} (m \minus{} 1)(3m \plus{} 1)(b \minus{} d)^2]$

$ \plus{} \frac {b \plus{} d}{4m}[(a \plus{} mb \minus{} c \plus{} md)^2 \plus{} (m \minus{} 1)(3m \plus{} 1)(a \minus{} c)^2]\geq 0$

for $ m\geq 1$.[/quote]
It is really very very nice  :o How could you find it?  :o

[quote="hungkhtn"]If $ a,b,c,d$ are non-negative with sum $ 4$, then

$ a^{2}(b \plus{} c) \plus{} b^{2}(c \plus{} d) \plus{} c^{2}(d \plus{} a) \plus{} d^{2}(a \plus{} b) \plus{} 8abcd \le 16$.[/quote]
[quote="Vasc"]Actually, the following stronger inequality holds

$ \sum a^{2}(b \plus{} c) \plus{} 2\sum abc \le 16$.[/quote]
For $ a,b,c,d\geq 0$ and $ a \plus{} b \plus{} c \plus{} d > 0$, the following inequality

$ a^2(b \plus{} mc) \plus{} b^2(c \plus{} md) \plus{} c^2(d \plus{} ma) \plus{} d^2(a \plus{} mb)$

$ \leq\frac {m}{4}(a \plus{} b \plus{} c \plus{} d)^3 \minus{} n(bcd \plus{} cda \plus{} dab \plus{} abc) \minus{} \frac {16(3m \minus{} 1 \minus{} n)abcd}{a \plus{} b \plus{} c \plus{} d}$

holds if and only if $ \left(\frac {16}{27}\leq m\leq1 \wedge f(m,n)\leq0\right)\vee(m\geq1 \wedge n\leq 3m \minus{} 1)$,

where the irreducible polynomial

$ f(m,n) \equal{} 4mn^7 \minus{} m(27m \plus{} 16)n^6 \plus{} 2(39m^3 \plus{} 55m^2 \plus{} 24m \minus{} 8)n^5$

$ \minus{} (125m^4 \plus{} 306m^3 \plus{} 277m^2 \minus{} 56m \minus{} 16)n^4$

$ \plus{} 4m(30m^4 \plus{} 111m^3 \plus{} 155m^2 \plus{} 9m \minus{} 33)n^3$

$ \minus{} m(69m^5 \plus{} 356m^4 \plus{} 870m^3 \plus{} 28m^2 \minus{} 403m \plus{} 104)n^2$

$ \plus{} 2m(11m^6 \plus{} 75m^5 \plus{} 318m^4 \plus{} 278m^3 \minus{} 777m^2 \plus{} 415m \minus{} 64)n$

$ \minus{} m(3m^7 \plus{} 26m^6 \plus{} 173m^5 \plus{} 396m^4 \plus{} 669m^3 \minus{} 2086m^2 \plus{} 1331m \minus{} 256)$.

[b]Remark[/b] $ f\left(\frac{16}{27},\frac{17}{54}\right)\equal{}f(1,2)\equal{}0$.

[hide="Special case."]$ m \equal{} \frac {16}{27},n \equal{} \frac {17}{54}$:

$ a^2(27b \plus{} 16c) \plus{} b^2(27c \plus{} 16d) \plus{} c^2(27d \plus{} 16a) \plus{} d^2(27a \plus{} 16b)$

$ \leq 4(a \plus{} b \plus{} c \plus{} d)^3 \minus{} \frac {17}{2}(bcd \plus{} cda \plus{} dab \plus{} abc) \minus{} \frac {200abcd}{a \plus{} b \plus{} c \plus{} d}$,

with equality if $ a \equal{} b \equal{} 0, c \equal{} 2d$.[/hide]

[quote="Ji Chen"][quote="hungkhtn"]If $ a,b,c,d$ are non-negative and $ a \plus{} b \plus{} c \plus{} d \equal{} 3$, then

$ ab(a \plus{} 2b \plus{} 3c) \plus{} bc(b \plus{} 2c \plus{} 3d) \plus{} cd(c \plus{} 2d \plus{} 3a) \plus{} da(d \plus{} 2a \plus{} 3b) \le 6\sqrt {3}$.[/quote]Actually, the following stronger inequality holds

$ a^2(b \plus{} 2d) \plus{} b^2(c \plus{} 2a) \plus{} c^2(d \plus{} 2b) \plus{} d^2(a \plus{} 2c)$

$ \leq \frac {2}{3\sqrt {3}}(a \plus{} b \plus{} c \plus{} d)^3 \plus{} \left(3 \minus{} \frac {32}{3\sqrt {3}}\right)(bcd \plus{} cda \plus{} dab \plus{} abc)$,

with equality if $ a \equal{} b \equal{} c \equal{} d$. [/quote]
$ 2(a \plus{} b \plus{} c \plus{} d)^3 \minus{} \left(32 \minus{} 9\sqrt {3}\right)(bcd \plus{} cda \plus{} dab \plus{} abc)$

$ \minus{} 3\sqrt {3}\left[a^2(b \plus{} 2d) \plus{} b^2(c \plus{} 2a) \plus{} c^2(d \plus{} 2b) \plus{} d^2(a \plus{} 2c)\right]$

$ \equal{} 8a(a \minus{} b \plus{} c \minus{} d)^2 \plus{} (16 \minus{} 9\sqrt {3})a\left[(a \minus{} c)^2 \plus{} (b \minus{} d)^2\right]$

$ \plus{} \left(113 \minus{} 65\sqrt {3}\right)(b \minus{} a)(a \minus{} d)^2$

$ \plus{} \frac {13 \plus{} 4\sqrt {3}}{44}(d \minus{} a)\left[b \plus{} \left(17 \minus{} 11\sqrt {3}\right)c \plus{} 2\left(25 \minus{} 14\sqrt {3}\right)d \plus{} \left(39\sqrt {3} \minus{} 68\right)a\right]^2$

$ \plus{} \left[2b \plus{} \left(2 \plus{} \sqrt {3}\right)c \plus{} \left(\frac {87}{11}\sqrt {3} \minus{} \frac {453}{44}\right)d \minus{} \left(\frac {98}{11}\sqrt {3} \minus{} \frac {277}{44}\right)a\right]\left[b \plus{} \left(1 \minus{} \sqrt {3}\right)c \plus{} 2\left(2 \minus{} \sqrt {3}\right)d \plus{} 3\left(\sqrt {3} \minus{} 2\right)a\right]^2$

$ \geq0$.

[hide="Remark."]$ 3 \minus{} \frac {32}{3\sqrt {3}} \equal{} \minus{} 3.15\cdots$;

$ 16 \minus{} 9\sqrt {3} \equal{} 0.411\cdots$;

$ 113 \minus{} 65\sqrt {3} \equal{} 0.416\cdots$;

$ \frac {87}{11}\sqrt {3} \minus{} \frac {453}{44} \equal{} 3.40\cdots$.[/hide]

[quote="hungkhtn"]Ok. This one is left for you, VASC. It is not as easy as before.

[b]Problem. (hungkhtn dedicates to VASC)[/b]
If $ a,b,c,d$ are non-negative and $ a \plus{} b \plus{} c \plus{} d \equal{} 3$ then
\[ ab(a \plus{} 2b \plus{} 3c) \plus{} bc(b \plus{} 2c \plus{} 3d) \plus{} cd(c \plus{} 2d \plus{} 3a) \plus{} da(d \plus{} 2a \plus{} 3b) \le 6\sqrt {3}.
\]
It must be very hard, I think.[/quote]
It is not very hard, I think. Denote $ f(a,b,c,d)\equal{}LHS$, we have
\[ f(a,b,c,d)\minus{}f(a\plus{}c,b,0,d)\equal{}c(b\minus{}d)(a\plus{}c\minus{}b\minus{}d)\]
and
\[ f(a,b,c,d)\minus{}f(0,b,a\plus{}c,d)\equal{}\minus{}a(b\minus{}d)(a\plus{}c\minus{}b\minus{}d)\]
From now, we have
\[ f(a,b,c,d) \le \max\{f(a\plus{}c,b,0,d),f(0,b,a\plus{}c,d)\}\]
Similarly,
\[ f(a,b,c,d) \le \max\{f(a\plus{}c,b\plus{}d,0,0),f(a\plus{}c,0,0,b\plus{}d),f(0,b\plus{}d,a\plus{}c,0),f(0,0,a\plus{}c,b\plus{}d)\}\]
Moreover, it is easy to check that
\[ \max\{f(a\plus{}c,b\plus{}d,0,0),f(a\plus{}c,0,0,b\plus{}d),f(0,b\plus{}d,a\plus{}c,0),f(0,0,a\plus{}c,b\plus{}d)\} \le 6\sqrt{3}\]
Equality holds for $ a\equal{}d\equal{}0,b\equal{}3\minus{}\sqrt{3},c\equal{}\sqrt{3}.$ :)

[quote=Ji Chen]
Another proof. $ (a \plus{} b \plus{} c \plus{} d)^3 \minus{} 4\sum_{cyc}{a^{2}(b \plus{} c)} \minus{} 8\sum_{cyc}{abc}$
$ \equal{} 4a[(a \minus{} c)^2 \plus{} (b \minus{} d)^2] \plus{} (c \minus{} a)(a \minus{} b \minus{} c \plus{} d)^2 \plus{} (b \plus{} d \minus{} 2a)(a \plus{} b \minus{} c \minus{} d)^2\geq 0$,
which is clearly true for $ a \equal{} \min\{a,b,c,d\}$. :blush: [/quote]
Shorter
\[(a \plus{} b \plus{} c \plus{} d)^3 \minus{} 4\sum_{cyc}{a^{2}(b \plus{} c)} \minus{} 8\sum_{cyc}{abc} = (c+a)(d+a-b-c)^2+(b+d)(a+b-c-d)^2 \geqslant 0.\]


In a diagram, DE is parallel to AC.
BD = 4, DA = 6
and EC = 8.  Find BC to the nearest tenth.

If i assume that ABC is a triangle nd D is on AB nd E is on BC.Then ABC nd DBE are similar triangles.
So BD:DA=BE:EC then calculating BE=16/3 So,BC=BE+EC=40/3
>tarik adnan

[quote="ckck"]Assuming that B is the vertex of triangle BAC, and that D is on side Ba and E is on side BC, I get the followong answer.

[hide]Using similiar triangles we have the following equation $\frac{2}{5}=\frac{x}{x+8}$. Solving for x we get $\frac{8}{3}$. Then add that to 8 to get $\frac{32}{3}$. This to the nearest tenth is $10.67$.[/hide][/quote]

That's hundredth.  He said tenth.

To i_like_pie,

I am not seeking angle B.  The question is asking for side BC, which the other two guys answered.

Thanks!

If you ever post a problem that came with a diagram, be sure to attach it.

Good idea BUT HOW do I attach a diagram here?


trigonometry
geometry
AMC
AIME
analytic geometry
number theory
\/closed

How fast would one expect to finish these two books (thoroughly and able to comprehend/grasp the most out of the book), assuming that the person is in Precalculus, and has experience with trigonometry and geometry? When I say this, I mean spending one hour solid a day.

It depends a lot on that person's ability.

I think one big aspect of "finishing" the books is doing most if not every problem provided. Only by doing the problems will you reinforce the chapter's contents.

So, the time it takes depends on your own speed.

Definitely make sure you go all the way through it at least once, though. I stopped at Chapter 12 and haven't started back since, and it kills me on some topics..

Yeah, I really got stuck on chapter 12 specifically.  I got to it.. hit a road block, so I tried it for a few days, and then moved on when I realized I still didn't get it.  Analytic Geometry is rough lol.

Its not about working an hour a day and trying to finish the books.  You can read through the entire book in a few days, and not really gain anything.  You learn more by reading through the book very slowly.  If you get stuck on a concept, leave it for a few days.. move on or just let it simmer for a while in your head.  Come back to the chapter a little bit later and see if it looks any better then.  Worst case, you can just ask someone on the forum.

And another thing... SOLVE THE PROBLEMS.  Especially if you're not in a crunch for time, you should solve the problems.

I personally would spend at least a few months finishing the books.  Volume 1 should probably only take like a month and a half if you work constantly but Volume 2 should take at least a solid three months because it involves math most people would find... impossible.

Would you probably get Art and Craft of Problem Solving while you do AoPS or after?

[quote="Pakman2012"]Art and Craft is only useful if you're trying to learning how to solve proofs and AIME questions 11-15.  I would only get it if you're perfectly comfortable with all but the last few chapters of Art of Problem Solving Volume 2.[/quote]

When you say perfectly comfortable, do you mean that you can solve all the problems quite rapidly and efficiently?  :huh:

As in, you can use most of the concepts fairly efficiently.. on most of the problems.   You're not gonna be able to get all of the problems, without some help from the solutions manual or what not.

Practice tests, practice tests, practice tests.

Practice problems, practice problems, practice problems.

Like thats basically all of what got me to do well on AIME, along with learning quite a few important concepts in AoPS classes... specifically Intermediate Counting.  Tough course... especially if you slack like me, but it paid dividends when I figured out how to solve questions I normally wouldn't be able to on AMC and AIME, largely because I'd put in the time with the transcripts.

Number theory classes on AoPS helped me quite a bit... at least it helped my confidence when approaching number theoretical problems.  Even if I didn't have to use any high powered tools.. it helped knowing that I had them haha.

Problem series classes might be the best way to learn for a minimalist haha.  From what I heard, the AIME Problem Series was really good.  AMC 12 problem series DEFINITELY raised my score by like.. 15 points purely on the merit that I'd seen more of the difficult questions.  I'm not saying that would happen to everyone.. but it was well worth it.

[quote="cooljoe"]Where should someone with an AIME score of 2 start?[/quote]

Somebody scoring a 2 on the AIME probably still has a lot to learn from AoPS Volume 1 and the Introductory subject books.


linear algebra
matrix
analytic geometry
vector
function
algebra
polynomial

Let $ M$ be a square matrix with exactly one $ 1$ in each column and all other entries $ 0$.  Show that the only possible eigenvalues of $ M$ are $ 0$ and roots of unity.

Let's work with $ A \equal{} M^T$, so each row has exactly one $ 1$ (they have the same eigenvalues). Then supposing $ \alpha\neq 0$ an equation of the form $ Ax \equal{} \alpha x$ , $ x\neq 0$ can be translated to: $ \forall i$ ,there is a $ j$ such that $ x_i \equal{} x_j/\alpha$, where $ i,j\in\{1,\dots n\}$. Take $ x_i\neq 0$, denote it with $ y_0$. So we can build an infinite sequence $ y_0, y_1\dots$ from the coordinates of $ x$ (a finite set), satisfying $ y_i \equal{} y_{i \plus{} 1}/\alpha$, and they all are nonzero, for $ y_0$ is nonzero. This means there will be $ i < j$, such as $ y_i \equal{} y_j$, which means $ \alpha^{j \minus{} i} \equal{} 1$.

Let $ MX\equal{}\lambda X$ for some nonzero $ \lambda$ and nonzero $ X$; Let $ X^T\equal{}(x_1 x_2....x_n)$.
Let there be $ r$ nonzero rows in $ M$. It is clear that, if the $ k^{th}$ row is zero, then, $ x_k\equal{}0$.
Remove all the zero rows from $ M$ and the corresponding zero entries in $ \lambda X$. Now, the $ n\minus{}r$ columns, multiplied with the $ n\minus{}r$ zero elements of $ X$ in the product $ MX$ can also be removed. This leaves us with an $ r*r$ matrix, say, $ B$, and, the column vector $ Y$ comtaining the left over elements, say $ (x_1....x_r)$ in $ X$., satisfying $ BY\equal{}\lambda Y$

Now, this $ B$ has the property that, every columnn has exhactly one 1, and rest all zeroes, and, every row has exhactly one 1 and rest all zeroes (that's because, there are no zero rows, and there are exhatly $ r$ 1's in the matrix).  Hence, $ BY$ is a permutation of the elements, $ (x_1...x_r)$. Now, since, not all of $ x_1,...x_r$ are zero, there exists atleast one nonzero expression among: the sum of $ x_1....x_r$; the sum of products taken two at a time;...........; the product of $ x_1......x_r$. That will give us, some integer power of $ \lambda$ equal to 1.

Let $ f$ be the linear function associated to $ M$. If $ e_i$ are the vectors of the basis then $ f(e_i)=e_j$. Considering the orbit of the $ e_i$ under the action $ e_i\rightarrow{e_j}$ or $ i\rightarrow{j}$, we change the order of these vectors. $ e_1=1\rightarrow{f(e_1)=e_j=2}$ if $ j>1$, else $ =1$ and STOP. If $ j>1$ then $ 1\rightarrow{2}\rightarrow{f(e_j)=e_k=3}$ if $ e_k$ has not been used, else $ =$ the new value of $ e_k$ and STOP; and so on.
When the  STOP occurs we have $ 1\rightarrow{2}\cdots\rightarrow{k}\rightarrow{j}$ with $ j\leq{k}$.
We consider one $ e_i$ that has not been used. Then $ e_i=k+1\rightarrow\cdots$ and when the STOP occurs we obtain:$ k+1\rightarrow{k+2}\cdots\rightarrow{l}\rightarrow{j}$ with $ j\leq{l}$.
When the process finishes then we obtain a matrix similar to $ M$ that is a blockmatrix in a triangular form (Darij will not be happy). The elements of the diagonal are companion matrices s.t. their last column is zero or has exactly one term equal to $ 1$ (and the other are $ 0$).
The charac polynomial of $ M$ is the product of the charac polynomials of these matrices; these polynomials are in the form $ x^k$ or $ x^k-x^q$ with $ q<k$.
QUESTION: is $ M$ a semi simple matrix ?

$ M\equal{}\begin{pmatrix}1&1&0\\0&0&1\\0&0&0\end{pmatrix}$ has minimal polynomial $ x^2(x\minus{}1)$, so it's not diagonalizable.

The problem, as stated, allows for a shorter argument:  a product of $ n \times n$ matrices of the given form is again a matrix of the given form, and there are $ n^n$ of them, so among the sequence $ I, M, ... M^{n^n}$ two are identical.  But loup's is the argument I was originally thinking of.


algebra
polynomial
number theory unsolved
number theory

Prove that $x^n-x^{n-1}-x^{n-2}-...-x-1$ is  Irreducibile in $Z[x]$

I am kind of BADLY stuck in this problem .
I think there is a heavy theorem (Rouche) ,whichs solves this problem but I prefer a high school solution.

[quote="silouan"]$x$ is real number ? Thanks[/quote]
$x$ is a variable ;)

PS: the theorem you probably meant is Rouche's theorem.

Briefly:
Suppose $\displaystyle f(x)=u(x)v(x)$ then $g(x)=-x^{n} f(\frac{1}{x})=u_{1}(x)v_{1}(x)=x^n+x^{n-1}+...+x-1$

Note $g(0)=-1, g(1)= n-1$ , hence there exists $\alpha \in (0,1)$ solution for $g(x)$

$\displaystyle g(x)=(x-\alpha)(x^{n-1}+q_{1}x^{n-2}+...+q_{n-1})$ and it is easy to observe that $1=q_1\leq q_2\leq...\leq q_{n-1}$

Now prove lemma that if $P(x)=q_{1}x^{n-1}+q_{2}x^{n-2}+...+q_{n}$ with $q_1\leq q_2\leq...\leq q_{n-1}$

then any solution  $\epsilon$ satisfies  $|\epsilon| > 1$.

Proof: Suppose $|\epsilon| \leq 1$ for $\displaystyle q_{1}\epsilon^{n-1}+q_{2}\epsilon^{n-2}+...+q_{n-1}\epsilon=-q_n$

$\displaystyle q_{1}\epsilon^{n}+q_{2}\epsilon^{n-1}+...+q_{n-1}\epsilon^{2}-q_{1}\epsilon^{n-1}-q_{2}\epsilon^{n-2}+...-q_{n-1}\epsilon=q_n-q_n\epsilon$

$\displaystyle q_{1}\epsilon^{n}+(q_{2}-q_{1})\epsilon^{n-1}+...+(q_{n}-q_{n-1})\epsilon=q_n$

$\displaystyle |q_n|\leq |q_{1}\epsilon^{n}|+|(q_{2}-q_{1})\epsilon^{n-1}|+...+|(q_{n}-q_{n-1})\epsilon|\leq$

$\displaystyle |q_{1}|+|(q_{2}-q_{1})|+...+|(q_{n}-q_{n-1}|=q_n$, which holds iff $|\epsilon|=1$ and $q_{i}=q_{j}$ - contradiction.

Then $g(x)=u_{1}(x)v_{1}(x)$ and one from them WLOG $u_{1}(x)$ has roots with modulo greater than 1.

But $g(0)=u_{1}(0)v_{1}(0)=-1$ and $|u_{1}(0)|=|\epsilon_{i_{1}}\cdot \epsilon_{i_{2}}\cdot ...\epsilon_{i_{l}}|> 1$ - contradiction.


How many positive integers have exactly three proper divisors, each of which is less than 50?

[hide="Solution"]To have 3 proper divisors, we need 1 and two other divisors. The ways in which this can occur is if we have a number which is the product of two primes, or the square of a prime number.

Case 1: There are 15 primes less than 50, and we must choose two at a time. The number of these is thus ${15 \choose 2}=105$

Case 2: There are 4 prime squares less than 50, which are the squares of 2, 3, 5, and 7. These will be our other proper divisors.

Therefore, our final answer is 105+4=$109$[/hide]

Ah, this surprisingly was a problem many people got wrong last year because they forgot to consider a few cases... :roll:

Did I get it wrong?  :?

Edit: Okay.  :)

Oh no! I was just saying for everyone doing this before looking at the spoilers! Good solution.

[hide="Answer"]Since $n$ is not a proper divisor of $n$, we must have $(a+1)(b+1)...-1=3\Rightarrow (a+1)(b+1)...=4$ for $n=p_1^ap_2^b...$. Therefore, we can have either two primes multiplied together or one perfect cube where the base is a prime. The largest perfect cube with the other divisors less than $50$ is $343$, which has $1$, $7$, $49$ as divisors, so our options are $2$, $3$, $5$, $7$. With primes, we have $1$, $a$, $b$, as divisors of $ab$, and the first three must be less than $50$. There are $15$ primes less than $50$, so we have $\binom{15}{2}=105$, and our answer is $105+4=\boxed{109}$.[/hide]

Actually, it's prime cubes rather than prime squares. But the prime still must be less than or equal to 7, so the answer doesn't change.

wat r proper divisors?

i know tat for the 15 choose 2 thing.. for a, b being primes, the factors r 1 * ab and a * b...and also for a*a, the factors are 1 * a^2 and a * a

how is that 3 for each?
one has...
1 ab
a b
the other...
1 a^ 2
a a

do u not count the 1's? and count the a's two times?

A proper divisor is any integer that evenly divides a number integer $n$, excluding $n$ itself. For example, the proper divisors of $18$ are $1,2,3,6,9$.

So, in this case, you have a multiple $ab$ of two primes $a$ and $b$. $ab$ is not a proper divisor of $ab$, but $1,a,b$ are, so all numbers of that form will have three proper divisors. Also, for any perfect cube $m^3$, you have $1,m,m^2$ as proper divisors, so those also have three; however, you can only use prime values of $m$. If you used composite, say $m=4$, you would have $1,2,4,8,16,32$ as proper divisors of $64$. ;)

[quote="LawOfSigns"]How are 4, 9, 25, and 49 counted?

4 has only 1 and 2 as proper divisors

9 has only 1 and 3

25 has only 1 and 5

49 has only 1 and 7[/quote]
They're not; as I pointed above, it's the cubes of primes that work (that is, 8,27,125,343). Strangely enough, the primes being cubed are still 2,3,5,7, so the answer is still the same.

Can we change the official solution? I got really confused after reading it.

Call the positive integer $n$.

$n$ must have $4$ positive integer divisors.

Case 1: $n=p^3$ for some prime $p$.
Then $p$ and $p^2$ are less than $50$. This gives $4$ ways.

Case 2: $n=pq$ for primes $p$ and $q$.
Then both primes $p$ and $q$ are less than $50$. This gives $\binom{15}{2}=105$ ways as there are $15$ primes less than $50$.

Thus, the answer is $4+105=\boxed{109}$.


Download các đề thi TOEFL các năm

[url=http://luyenthianhvan.blogspot.com/]http://luyenthianhvan.blogspot.com/[/url]

ummm no

untill you explain what you want us to download, i stronly urge reveryone to not click on the link.

Looking at the title, I'm pretty sure it's online TOEFL Test. (I'm only guessing so do not think that I know Vietnamese)

[quote="Zmastr"]What does a vietnamese keyboard look like?  hmm....[/quote]
It totally likes any other standard keyboards,if somesone want to type in Vietnamese s(he) has to set up a software.for example if I type"aa"(after turning off this software) ,it will change to the symbol" â"
I wondered that  the title of topic  was written in Vietnamese  :D .It means that" download some problems of TOEFL test in years ".I am sure that this links he post is very safe(no virus,spyware,..etc),If you care about TOEFL test  you can try to visit this links.
I am a beginer to learn ENglish so :
[b]If anyone find any grammatical mistakes in upper sentences ,plz correct if possible :) thanks so much  :D [/b]

It should have been more like this:
----------------------------

It [i]looks like[/i] any other standard [i]keyboard[/i][i];[/i] if [i]someone[/i] [i]wanted[/i] to type in Vietnamese [i]s/he[/i] [i]would have[/i] to set up software. [i]For[/i] example, if I type "aa" (after turning off [shouldn't this be on?] this software), it will change [i]into[/i] the symbol "â".
I wondered [i]why[/i] the title of [i]the[/i] topic was written in Vietnamese  :D . [i]It means[/i] "download some problems of TOEFL test in years" [Not sure what you meant here, but I think you mean "download some TOEFL test problems from past years"]. I am sure that [i]the link[/i] he [i]posted[/i] is very safe (no [i]viruses[/i], spyware,... etc). If you care about TOEFL tests you can try to visit [i]the link[/i].
I am a beginer to learn English so [There are a variety of ways to say this in a more fluent way; "I have just begun to learn English" is probably the best.]:
[b]If anyone [i]finds[/i] any grammatical mistakes in [i]the sentences above[/i] , plz correct if possible :) thanks so much  :D [/b]

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

Most of the problems are with fluency, and not with grammar. A test grader probably wouldn't have made so many corrections. There is no space before a comma, but there is one after a comma. The same goes for periods.

Your English is very good, in my opinion. You are not a beginner.  :)

I probably made a little dialectical mistake myself somewhere in there  :oops: .

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

About the link - I clicked it and there is not really a "download", it's just a little quiz thingy.

600 bài test mới nhất của http://luyenthianhvan.org

http://luyenthianhvan.org/2008/10/toeic-test-vocabulary.html

Trên trang này có đầy đủ các tài liệu toeic, toefl, anh văn trình độ A,B,C, ôn tập ngữ pháp, kiểm tra trực tuyến, và còn nhiều, nhiều điều hấp dẫn trên đấy nữa. Các bạn từ từ khám phá nha. Chúc các bạn thành công.


function
topology
real analysis
real analysis theorems

Definition of metric space says:

A metric space is a set $ X$ together with a real valued function $ d: X \times X \rightarrow R$ with property ...

My question is, since $ X$ can be any set (no restriction specified), assume $ X \equal{} (1,3)$

FYI, $ (1,3)$ means all the value between 1 and 3 excluding 1 and 3.

But I found that X must be open and closed...(since empty set is open and closed)

Does that mean (1,3) is open and closed too in this metric space? I'm confused, since in usual sence, (1,3) is open and not closed.

I got confused when I was working on definition of dense set. (A is dense in X if  cl(A)=X. But we only need to prove X is in cl(A), since the fact that cl(A) is in X is obvious. But why is it obvious? Because X is closed(?).)

Oh, the definition of openness depends on metric d...is this where I made mistake I think?

Does this mean that metric space itself always should be open and closed?

Yes and yes.  Openness and closedness of a set are not intrinsic properties -- they depend on whether you are considering the set as part of some larger metric space, and if so, what this larger metric space looks like.  Every metric space is both open and closed in itself.

To the last question of the first proof, it's obvious because you're taking the closure of A [i]in X[/i], i.e., your definition of the closure is something like, "a point x in X is in the closure of A if ...."


function
number theory
Diophantine equation

I'm looking for a good math software that will help me do those hard problems. They say that the best are Mathematica and Maple. There are also Matlab, Mathcad, etc. What would you recommend?

Maple's Trademark is "Command the Brillance of a thousand mathematicians. Goto for a review: [url] so I guess having a thousand mathematicians would help you do those hard problems.
[url]http://www.deskeng.com/articles/03/july/cover/index.htm[/url].
If you are setting for low standards. Download the FREE Virtual TI by Rusty Wagner from [url]ticalc.org[/url]. It does not have any TI-OS-ROMS. Download a TI-89 one at [url]http://ti89games.free.fr/Html/emu.htm#2[/url]. There are some more on [url]http://electronika2.tripod.com/indice_tir.htm[/url]. Cool Fact! THe Virtual TI simulates everything on your caluclator FLASH APPS, games, everything.

In theory you have to own a calculator to use the ROMS since TI has a copyright on them. I didn't download them because I own a TI-89 and TI-83+ and I used the VTI to transfer to my computer.

I have no idea if your last two posts had anything to do with my question.  They confused me more, not less.

What type of problems are you talking about?  All of those softwares are very limited in their ability to, say, prove a geometric theorem.  However, they are all probably quite good about, say, calculating some sum of many many terms to a great degree of accuracy.  What exactly are you expecting your program to do for you?

THese software have the capabilities to do those function problems, and those find the values of solutions to equation, and those geometric problems.
I am mainly refering to "hard" problems as in college level classes such as Number THeory, Linear Algebra.
AND the Virtual TI is just something useful for people who want to use their calculator on the computer just like "mathematical software as in Maple"

I'm pretty sure all those programs can do linear algebra and number theory problems without too much trouble.

Eh, what types of number theory problems are we talking about here? I mean, perhaps something like a diophantine equation (brute-forcing it), but no proofs I don't think.

Sometimes computers can prove things. Their proofs are a bit strange though. http://www.math.upenn.edu/~wilf/AeqB.pdf is a nice example of things computers can prove.


Well now you can solve for $ a$ in terms of $ b$ or $ b$ in terms of $ a$ from that quadratic.  Then, once you plug it in, you get a general expression for $ a/b$.

You needn't get $ a^2\minus{}2ab\plus{}b^2\equal{}0$.
We have $ \frac{a\plus{}b}{2}\equal{}2\sqrt{ab}\Rightarrow 4a\minus{}4\sqrt{ab}\plus{}b\equal{}3a$
$ \Rightarrow 2\sqrt{a}\minus{}\sqrt{b}\equal{}\sqrt{3a}$ (because $ a>b>0$)
$ \Rightarrow \sqrt{a}(2\minus{}\sqrt{3})\equal{}\sqrt{b}$
$ \Rightarrow \frac{\sqrt{a}}{\sqrt{b}}\equal{}\frac{1}{2\minus{}\sqrt{3}}$
$ \Rightarrow \frac{a}{b}\equal{}9\equal{}(2\plus{}\sqrt{3})^2\equal{}7\plus{}4\sqrt{3}$
 :wink:[/u]

[quote="ThetaPi"]If $ \frac {a \plus{} b}{2} \equal{} 2\sqrt {ab}$, where $ a > b > 0$, find a possible value of $ \frac {a}{b}$.

I managed to get $ a^2 \minus{} 2ab \plus{} b^2 \equal{} 0$, but that doesn't seem very useful.

What if we wanted to find all possible values of $ \frac {a}{b}$?[/quote]

If you have $ a^2 \minus{} 2ab \plus{} b^2 \equal{} 0$, then $ (a\minus{}b)^2\equal{}0 \implies a\minus{}b\equal{}0 \implies a\equal{}b \implies \frac{a}{b}\equal{}1$

I don't see how you got that though, since $ \frac {a \plus{} b}{2} \equal{} 2\sqrt {ab}\implies a\plus{}b\equal{}4\sqrt{ab} \implies a^2\plus{}2ab\plus{}b^2\equal{}16ab \implies a^2\minus{}14ab\plus{}b^2\equal{}0$

[quote="metafor"]Kyoshiro's answer is correct, except for the nine. Where did that nine come from? O.o[/quote]
Oh! Thanks.
Well, It's just an incidental idea! :lol:
When I saw $ \minus{}4\sqrt{ab}$, I thought of equality: $ 4a\minus{}4\sqrt{ab}\plus{}b\equal{}(2\sqrt{a}\minus{}\sqrt{b})^2$

[quote="21 7 15"][i]now a^2-2ab+b^2=0
divide through by b^2
then let a/b=x
you'll arive at x^2-2x+1=0
x=1,so a/b=1 :D [/i][/quote]ThetaPi incorrectly got $ a^2\minus{}2ab\plus{}b^2\equal{}0$. The rest is just bogus. Kyoshiro's answer is correct.

But then thetapi's equation implies a = b because if it were then a + b /2 = sqrt{ab} by AM-GM not 2sqrt{ab}
Sorry im feeling 2 lazy 4 LaTeX
EDIT: Didnt read 1000th users comment

[quote="lgd0612"]if 7+4sqr rt(3) is one solution

does 7-4sqr rt(3) work as well?
but a and b have to be imaginary #s[/quote]Hmm $ 7\minus{}4\sqrt3$ is not a possible value because it is less than 1 because we have the condition $ a>b>0\Rightarrow\frac{a}{b}>1$.

ah  i was careless

i thought 7-4sqr rt(3) was going to be a negitive #
  :oops:

Yepp, I did some algebra error then. What's wrong with me lol.

[hide="Another solution"]Let $ \frac{a}{b}\equal{}k\ (k>1)$
$ \Rightarrow a\equal{}kb \Rightarrow b\plus{}kb\equal{}4b\sqrt{k}$
$ \Rightarrow k\minus{}4\sqrt{k}\plus{}1\equal{}0 \Rightarrow k\equal{}7\plus{}4\sqrt{3}$[/hide]


If AD and BF are bisectors,AB=4 ,BC=3 and CA=2 then find $ S_{AFD}: S_{ABC}$.

hello, i have got
$ \frac{S_{AFD}}{S_{ABC}}\equal{}\frac{4}{21}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have got
$ \frac {S_{AFD}}{S_{ABC}} \equal{} \frac {4}{21}$
Sonnhard.[/quote]
Yes,that's the answer.How did you got it?

If I have provided an outline to the solution because the solution itself is somewhat long.

[hide="Outline"]
Note: $ [ABC]$ denotes the area of triangle $ \triangle ABC$.

Let the intersection of the cevians be $ E$. Let $ a \equal{} [CFD]$, $ b \equal{} [EFD]$, $ c \equal{} [AFE]$, $ d \equal{} [BED]$, $ e \equal{} [AEB]$.

1. Find the ratios $ AF: FC$ and $ BD: DC$ from the Angle Bisector Theorem. (If the given cevians aren't angle bisectors, then I assume they are medians, in which case the two ratios are just $ 1: 1$.)

2. Use the fact that the ratio of the areas of two triangles that share the same height is equal to the ratio of the corresponding bases of the triangles. (Hint: Take a look at $ \triangle ADB$ and $ \triangle ADC$ considering the bases to lie along $ \overline{BC}$. There is only more pair of such triangles, see if you can find them.)

3. Compute $ [ABC]$ using Heron's Formula.

4. Use the ratios from Step 2 to come up with 4 equations. The fifth equation will be that the sum of the areas of all the triangles mentioned before is whatever you got in Step 3.

5. Finally, find $ b \plus{} c$ and divide that by the number you got in Step 3 to get the answer.
[/hide]

If you still need help, let me know and I'll add some more detail.


I don't think there is a way.  When the administrators enabled dollar signs in addition to <math> tags, the $ \$$ thing got messed up.

[quote="#H34N1"]How do you make the $ \$$ signs for the wiki? I have a wiki that I would like to use the $ \$$ signs instead of the <math> tags. Thanks.[/quote]

@Archimedes1, solafidefarms, I think he's asking about how to let $ \$$ turn into <math> tags on a different wiki, not how to type $ \$$ signs on this wiki. That would probably require some programming. I imagine it's possible to just do it by adding some JavaScript to the personal [i]monobook.js[/i] file. For example, add an onSubmit() or onClick() handler to the "Save page" button on the wiki, or add a separate tab that would convert all $ \$$ into <math> tags (see Wikipedia > User:Omegatron/monobook.js/addlink.js). Then if there are an even number of $ \$$ (using regular expressions or something), convert alternatively each to <math> or </math>; otherwise alert an error.

Or if you have "special" abilities on the wiki (like administrator), I guess just ask our admins how they managed to do it (I'm assuming Valentin did it?)

The client-side version would not be a good solution, as any subsequent edits would have the <math> tags instead of the dollar signs.

A simple thing to do would be to interpret all the dollar signs as math tags with a pre-save-in-db function in the MediaWiki file that recieves the editing form.


algebra
polynomial
Support
function
real analysis
real analysis unsolved

Given $A,B \subset \mathbb{R}^{n}$, define the Minkowski sum $A+B= \{a+b \mid a \in A, b \in B \}$. Suppose that $A$ is compact and $B$ is a closed ball of radius $r$.
(1) Prove that if $A$ is convex, then the volume of $A+B$ is a polynomial in $r$.
(2) Is the converse to (1) true?

it has already been posted but received no answer : http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=90537
Any new ideas ?

Your idea of approximation with polytopes works (at least for (1)), but I hope that there is an approximation-free approach, maybe using the support function $h_{A}(x)=\sup\{\langle x,a\rangle \colon a\in A\}$. This function determines a convex body, and it converts Minkowski sums into ordinary sums. Is there a formula for volume of $A$ in terms of $h_{A}$?

i posted a generalized version here:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=13874
and you can find a complete solution :
Rolf Schneider,convex bodies ,The Brunn Minkowsky theory,cambridge university press (1993).


calculus
derivative
geometry
trapezoid
3D geometry
sphere
function

[quote="keta"]Hi guys, I have been practicing on some maximization-minimization problems and I have been stuck on solving those problems in which only variables are introduced (i.e.: How can you use the local extreme theorems when there are only variables? Here are a few questions that have boggled me:

1. Find the dimensions of the trapezoid of greatest area that can be inscribed in a semicircle of radius r. (I can't figure out the general position of the trapezoid so that I can assign variables, how do I work with unknown r?).

2. Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius r.

3. A solid is formed by adjoining a hemisphere to each end of a right circular cylinder. The total volume of the figure is 12 cubic inches. Find the radius of the cylinder that produces the minimum surface area.

Can someone please explain all the steps needed in these problems, especially the first two? Any help is GREATLY APPRECIATED. Thanks.[/quote]

The hard part of most differentiation problems lies in finding the quantity to be minimized or maximized.

1.  A point (rcost,rsint) 'generates' a trapezoid with larger base 2r and smaller base 2rcost and height rsint.

Thus A(t) = 1/2(2r+2rcost)*(rsint)

The rest is trivial.

2.  A sphere is identical with respect to all dimensions.  Thus let (rcost,rsint) be an arbitrary point on the sphere and see what the volume of such a CIRCULAR cylinder will be.  Now minimize that.

3.  You have a quantity in two variables and a constraint.  This constraint gives you either height at a function of radius or vice versa.  Now you can reduce to one variable.


calculus
integration
special factorizations

Find the largest possible integer $ n$ such that $ n^2 \plus{} 2008n$ is a perfect square.

Ugh Im so used to there being no solution that I never actually learned how to solve it if there is one.

So by completing the square you get:
$ (n \plus{} 1004)^2 \equal{} q^2 \plus{} 1004^2$
$ n\equal{}\sqrt{q^2\plus{}1004^2} \minus{}1004$
and Im pretty sure you can ignore the 1004, so basically what your trying to find is the largest value of $ q$ for which $ \sqrt{q^2\plus{}1004^2}$ is an integer. How would you do that?

$ q^2 \plus{} 1004^2 \equal{} k^2$
$ (k\plus{}q)(k\minus{}q)\equal{}1004^2$
$ k\minus{}q\equal{}2, k\plus{}q\equal{}502\cdot 1004 \Rightarrow q\equal{}502^2\minus{}1$

[hide="What I got"]I'm getting $ 251$:

Let $ n^2+2008n=q^2$. We have the following:
\begin{eqnarray*} n^2+2008n & = & q^2 \\
2008n & = & q^2-n^2 \\
2008n & = & (q+n)(q-n) \\
2^3 \cdot 251 \cdot n & = & (q+n)(q-n) \end{eqnarray*}
So we have two (integral) factors of $ 2008n$ that differ by $ 2n$. The possibilities are the following (and what $ n$ would be if the two factors differed by $ 2n$, and there are two possibilities for each because either one could be bigger):
\[ 1 \times 2008n \longrightarrow n=\frac{1}{2006} \text{ or } \frac{1}{2010} \\ \\
2 \times 1004n \longrightarrow n=\frac{1}{501} \text{ or } \frac{1}{503} \\ \\
4 \times 502n \longrightarrow n=\frac{1}{125} \text{ or } \frac{1}{126} \\ \\
8 \times 251n \longrightarrow n=\frac{8}{249} \text{ or } \frac{8}{253} \\ \\
251 \times 8n \longrightarrow n=\frac{251}{10} \text{ or } \frac{251}{6} \\ \\
502 \times 4n \longrightarrow n=251 \text{ or } \frac{251}{3} \\ \\
1004 \times 2n \longrightarrow n=251 \\ \\
2008 \times n \longrightarrow n=-2008 \text{ or } \frac{2008}{3}\]
We see that the largest integral value of $ n$ (out of $ 251$ and $ -2008$) is $ \boxed{251}$. If we plug that into the original expression, we get $ 251^2+2008 \cdot 251 = 567009=753^2$.

I probably did something wrong, though[/hide]

[quote="PowerOfPi"]
So we have two (integral) factors of $ 2008n$ that differ by $ 2n$. The possibilities are the following (and what $ n$ would be if the two factors differed by $ 2n$, and there are two possibilities for each because either one could be bigger):
[/quote]

I dont understand this part and how you got the list of values of n. :oops:

[quote="athunder"][hide="Solution #1"] Therefore, $ z^2 \plus{} 1004^2$=$ (n \plus{} 1004)^2$

So basically we have a number $ z$ when added to $ 1004^2$ becomes a perfect square.[/quote]
I think you lost a "squared" between these two steps...

[quote="Thunder365"]I dont understand this part and how you got the list of values of n. :oops:[/quote]
[hide="Another explanation"]
Think of it like this problem, where $ a$ and $ b$ are positive integers:
$ a^2\equal{}24\plus{}b^2$
Where we get $ (a\minus{}b)(a\plus{}b)\equal{}24$. We find factors of $ 24$ that have a product of $ 24$, and set the two expressions equal to the factors. Of course, this takes some case work, but we can find all the possible pairs $ (a,b)$.

Now back to the problem, which is essentially the same as above, except that we have a variable on the RHS.

We had two expressions, $ k\plus{}n$ and $ k\minus{}n$, that had a product of $ 2008n$. We know $ k$ is an integer because its square is a perfect square. We know $ n$ is an integer because we are given that. Thus, the two expressions have integer values and a product of $ 2008n$. The difference between the expressions is $ 2n$.

The above paragraph just concludes stuff about the two expressions.

Now, we find the factors of $ 2008n$ (and treat $ n$ as a number, not a variable), so we can set them equal to the two expressions. These pairs of factors are what I listed before the arrows, in the list.

(Note that these things I am about to say were not said in the post, but this is how I got the list.) Now, since the expressions differ by $ 2n$, we can solve for $ n$. For example, for the first case, I let $ 1 \pm 2n\equal{}2008n$ and solved, getting two values of $ n$, which I listed beside the factors. Note that for the second last case, one value was omitted because you get $ 1004\equal{}0$, so there was no solution for that case.

I hope that explains things.
[/hide][/hide]


I don't have a clean solution started yet but I think it's obvious that one such x is [hide="answer"]2000[/hide]

And if my preliminary work is correct, there is a way to show that it is the only one.  Show that this is the only intersection of the two functions (LHS and RHS).

[color=blue]Substract 3 from each side and put them in factor, I get the solution of this equation, but i am not sure whether it has other solutions or not. However I can get one: 2000. :blush:  :D [/color]

Set $y=2000+x$ and $g(y,t)=\sqrt{1+x/t}$.  It's clear that we need to show that $x=0$ is the only solution to

$g(y,21)+g(y,22)+g(y,23)=g(y,1977)+g(y,1978)+g(y,1979)$

It is not difficult to show that $g(y,21)>g(y,1978)$ for $x>0$ and the inequality is reversed for $x<0$.  The answer follows easily from this.


[quote="Elio (n)"]The class has 32 pupils , 7 of them play football , 16 of them play volleyball , 16 of them play basketball . 4 of them play football and volleyball , 6 of them play football and basketball , 3 of play volleyball and basketball , 1 pupil plays the three sports . How many pupils in that class don't play any sport?[/quote]

I think there is something wrong with the problem,because we can find the number of those who only play football is -2

[quote="sunehra"]Let $ n$ be the pupils that don't play any sport. By the Principle of Inclusion-Exclusion, we have

$ 7 \plus{} 16 \plus{} 16 \minus{} 4 \minus{} 6 \minus{} 3 \plus{} 1 \plus{} n \equal{} 32 \implies 27 \plus{} n \equal{} 32 \implies n \equal{} \boxed{5}$.[/quote]
I think we have:
$ 7\plus{}16\plus{}16\minus{}4\minus{}6\minus{}3\minus{}1\plus{}n \equal{} 32\implies 27\plus{}n \equal{} 32\implies n \equal{}\boxed{7}$

cgcgbcbc is right. Problem's impossible.


Hi Derek,

[quote="Derek"]Can somebody show me how to solve a continued radical.

For example,  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/870d2aa1f3a203f1eac62458c5e2af27.gif[/img][/quote]
[hide]Let $S=\sqrt{3+\sqrt{3+\sqrt{3+\dotsb}}}$

Then $S^2-3=S,$
and using the quadratic formula, we can find the exact value of S.[/hide]

[quote="Derek"]Can somebody show me how to solve a continued radical.

For example,  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/870d2aa1f3a203f1eac62458c5e2af27.gif[/img][/quote]

Right.

Set that equation to S, and you have $S = \sqrt{3 + S}$, which is how Mathew Got his equation.

Oh ok, I see now.

Thank you for your help.

[hide]so the asnwer would be \displaystyle $S=\frac{1+\sqrt{13}}{2}=2.302775638...$[/hide]

[quote="jli"][hide]so the asnwer would be 2.302775638[/hide][/quote]The answer is given by the positive solution to $S^2-3=S,$ as given above. This number is $S=\frac{1+\sqrt{13}}{2a}$, which is not rational because 13 is not a perfect square. The number you gave is rational. Therefore your answer isn't right.


In maple i have defined a string:
A:="gfhgfnsdklqsemhpjsdmoqezhjfjjfdjgfdjkhqhqsdy"

I want to see wich words of 3 consecutive letters apears more than one time in the string.
For example the first word is "gfh", the second "fhg" ...

What i tried to do is defining a row with those 3-characters string :
T:=n->[substring(A,n..n+2)]:

But if i do T[1]; I onle get T_1 so i think that is not correct.

please ask questions if I am not clear.

Thanks in advance

Hello Amazigh!

Try T(1) instead of T[1]...

Just in case, the answer to your original question is "fdj". Try the following Maple input:

A:="gfhgfnsdklqsemhpjsdmoqezhjfjjfdjgfdjkhqhqsdy";
ListTools[FindRepetitions]([substring(A,i..i+2)$i=1..length(A)-2]);

Have a nice day!
Sergey Markelov
markelov@mccme.ru


A and B are two M2(R) matrices. we know $A^{2}=B^{2}=I_{2}$ and $AB \neq BA$. is this enough to show that $(BA)^{n}\neq I_{2}$ , no matter what $n\in\mathbb{N}$

The answer is "No". Just take $A=\left[\begin{matrix}a&1-a^{2}\\1&-a\end{matrix}\right]$ and $B=\left[\begin{matrix}b&1-b^{2}\\1&-b\end{matrix}\right]$. Then the element in the left upper corner of $AB$ is $ab+(1-a^{2})$, which is different from the corresponding element in $BA$ if $a^{2}\ne b^{2}$. Now, the trace of $BA$ is $2-(a-b)^{2}$, which can be made any number less than $2$ you wish. In particular, you can make it $2\cos \frac{2\pi}n$ if $n\ge 3$, which will result in the matrix $BA$ being similar to $\left[\begin{matrix}e^{2\pi i/n}&0\\0&e^{-2\pi i/n}\end{matrix}\right]$, so $(BA)^{n}=I_{2}$. Of course, $(BA)^{1}\ne I_{2}$ because otherwise $B$ and $A$ would commute. I leave the remaining question of whether one can achieve $(BA)^{2}=I_{2}$ to somebody else ;).


trigonometry
function
calculus
calculus computations

Is there another way to solve this limit without using L'hopital's Rule????

I've tried everything I can think of to manipulate this into function in order to be able to plug in the limit without getting something indeterminate.

Lim x->pi/4  (1-Tanx)/(Sinx-Cosx)

Please help....... :(

How about some algebra?

$\frac{1-\tan x}{\sin x-\cos x}=\frac1{\cos x}\cdot\frac{\cos x-\sin x}{\sin x-\cos x}=-\frac1{\cos x}$.

Thanks!!!

Funny, I got all the way to (1/cos)(cos-sin)/(sin-cos) but just didn't see that (cos-sin)/(sin-cos) was just -1


geometry
projective geometry
combinatorics unsolved
combinatorics

S={1;...15}.Assume that there are exist n subset A_1...A_n satisfying:
$|A_i|=7$
$|A_i\cap A_j|\leq\ 3$
For every subset M of S;|M|=3.there are exist A_k such that $M\in A_k$
Find the least value of n

This is the number of covering designs C(15,7,3)=15 which can be obtained as a projective geometry of dimension 3 over GF(2).
See Section 3 from the paper [url=http://www.ccrwest.org/cover/cover.pdf]"New Constructions for Covering Designs"[/url] at [url=http://www.ccrwest.org/cover.html]La Jolla Covering Repository[/url].

[quote="Bluesea"]Do you have solution for this particular case?My english is so poor,i can't understand this paper[/quote]

[url=http://www.ccrwest.org/cover/t_pages/t3/k7/C_15_7_3.html.gz]This is it[/url].


calculus
integration
calculus computations

Change variable $y=-x$ gives $\int_{-1}^{1}\frac{e^x}{(e^x+1)(x^2+1)}dx$

Twice of the first integral gives $2\int_{0}^{1}\frac{dx}{x^2+1}$

For the second integral sme technic

perhaps the handbook of Alan Jeffery helps

Don't need this book the change variable x=1/y

$\int_{0}^{+\infty}\frac{y^\alpha}{(y^2+1)(y^\alpha +1)}dy$

Twice of the given integral is $\int_{0}^{+\infty}\frac{1}{(y^2+1)}dy$

[quote="Moubinool"][quote="liyi"]The same trick works for the following problem, which was a competition problem of USSR in 1977.

Prove that the value of $\int_0^\infty \frac{dx}{(x^2+1)(x^\alpha+1)}$ is independent of $\alpha$[/quote]

For this one the change variable is $x=1/y$[/quote]
Yes, you are right. It was my mistake... :blush: I didn't read your solution to the previous two integrals carefully...


geometry
3D geometry
sphere
percent
ratio
search

An infinite number of congruent spheres are packed in space hexagonally*. To the nearest tenth, what percent of the space is in the interior of one of these spheres?

*say you have 4 balls, you can stack them into a pyramid. Pack the rest of the space similiarly.

The ratio you are talking about is $\frac{\pi}{\sqrt{18}} = 0.74048\ldots$. Calculating the ratio is easy. Getting bounds on this ratio is hard. In fact, it is interesting to note that while it seems that this is the best possible packing of spheres for $3$ dimensions, its proof still remains as an open problem.

Are you sure it's still open?

This problem is Kepler's conjecture, and as far as I know, Hales has proved it.
But, yes, I think that no "formal" proof has been given yet. (Hales used computers.)

Hmmm.. A search on [url=http://en.wikipedia.org/wiki/Kepler%27s_conjecture]wikipedia[/url] reveals that indeed the problem was proven.  :blush:

Unfortunately, with 250 pages and 3 gigabytes of programs, data and results, it's kinda hard to accept it as a proof. In my opinion, Hales' proof is more of a verification, an example of an enormous amount of brutal whacking on a seemingly 'innocent' question.

This problem can actually be generalised to $n$ dimensions, where we want to find the maximum 'density' of spheres that can be packed in an $n$ dimensional space. The last I read, the exact maximum bound is an open problem for all dimensions $n \geq 3$ (definitely not up-to-date). If we don't even have a 'nice' proof for $3$ dimensions, then the proof for higher dimensions seem hopeless without depending on linear programming methods.  :(   :(

I have read (somewhere...) that 10,11 and 13 dimensions seem to be vaguely special cases in that the densest (known)
packing for most dimensions are very similar (cross-sections of each other) except these dimensions- though how much of this
proof and how much computer I dont know

The so-called special cases are dimensions $8$ and $24$. They are known as the $E_8$ lattice and the Leech lattice respectively, and they are highly symmetric in their respective dimensions.


In my opinion i think we can find a formula for predicting chances and randomness it is impossible, what do you think ?

impossible? why ?
possible? how ?

The questions of whether randomness exists in the real world and whether it exists in an abstract mathematical sense are not the same. The first question is debatable (and in fact it is still hotly debated, especially by philosophers and physicists). However, I don't see how you could say that randomness and probability theory don't exist in an abstract mathematical sense. The theory is there.

Also, that theory of randomness has a great many practical uses. Random models describe the motion of large numbers of particles very well- that's the core of thermodynamics. The motion of any particular particle is determined by its initial state and its collisions, but there's no way to get enough of that information to build a useful nonrandom model of a container of gas.

Random models extract useful information out of extreme complexity. It doesn't really matter whether the process being modeled is random or not.

It's an open question in computational theory whether pseudorandom generators (deterministic functions whose output is "random" in the sense that there is no efficient algorithm that can predict the outcomes more than epsilonically better than random guessing) exist.  The notion of randomness that they are trying to mimic, of course, is of a function whose outputs are independent and uniformly distributed.  Calling something random in this sense is [i]equivalent[/i] to saying that you can't predict it.

i disagree, u lot are putting restrictions, we have to be far-sighted. you think you know everything.in my opinion humans brains are too small+ feeble for finding that complexity in randomness if we think negative, i say its possible by time.
there is a definite formula for any random thing. no formula is complete. if you think a formula is complete then we can add something else to it like +1 or ^2 or do many things which would be true for another thing.we cant stop doing this can we ? that formula wouldnt be wrong.
"useful information out of extreme complexity" how can you judge that easy and be so sure that its impossible, without being able to know that complexity completly.

sorry ppl dont get it bad , i am just saying my opinion.

Statistical mechanics and thermodynamics work. Trying to model a volume of air as about $10^{24}$ particles interacting through various forces... doesn't, because we can't measure that much data, store that much data, or run computations of that size, much less solve the equations. I'm not talking about human limits here- I'm talking about computer limits.

Also, embracing random models is not "thinking negative". Random models do have a great deal of order and predictability to them, often much more than deterministic but chaotic models.


[quote="vittasko"][b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given with circucircle $ (O)$ and let $ (K)$ be, an arbitrary circle taken as cord its side-segment $ BC.$ Let be the point $ D\equiv BC\cap AK,$ where $ K$ is the center of $ (K)$ and we denote as $ E,\ F,$ the points of intersection of $ (K),$ from the circles $ (O_{1}),\ (O_{2})$ with diameters $ BD,\ CD,$ respectively. Prove that $ PQ\parallel BC,$ where $ P\equiv AB\cap DE$ and $ Q\equiv AC\cap DF.$[/color][/size][/b]

Kostas Vittas.[/quote]
Let $ J\equal{}DP\cap AO_1, R\equal{}DQ\cap AO_2$
Because $ KO_1\perp BE$ and $ \angle BED\equal{}90^o$ then $ KO_1//DE$, similarly $ KO_2//DF$
Using Thales's theorem we have $ \frac{AJ}{AO_1}\equal{}\frac{AD}{AK}\equal{}\frac{AR}{AO_2}$
Then $ JR//O_1O_2$
Applying Menelaus's theorem for triangles $ JO_1D$ and $ RDO_2$ we get $ \frac{AJ}{AO_1}.\frac{BO_1}{BD}.\frac{PD}{PJ}\equal{}\frac{AR}{AO_2}.\frac{CO_2}{CD}.\frac{QR}{QD}$
$ \Rightarrow \frac{PD}{PJ}\equal{}\frac{QR}{QD}\Rightarrow PQ//JR\rightarrow$ QED

Since $ BE$ and $ CF$ are radical axes of $ (K),(O_1)$ and $ (K),(O_2),$ then  $ KO_1$ and $ KO_2$ are respectively parallel to $ DP$ and $ DQ$ and go through the midpoints $ M,N$ of $ PB$ and $ QC.$ Hence, by Menelaus' theorem we obtain:

$ \frac {KD}{KA} \equal{} \frac {BM}{MA} \equal{} \frac {CN}{NA} \ \Longrightarrow \ MN \parallel BC$

Therefore, $ MN$ becomes the midline of the trapezoid $ PQCB$ $\Longrightarrow$ $ PQ \parallel BC$


I started this thread because i tend to make a LOT of careless mistakes, especially on sprints and other tests where speed is an issue (well it is for me so don't brag). I just wanted to know if there are any tips people can give me to improve my accuracy. I've tried almost everything and i'm getting desperate :(

I used to be just like you. I averaged like 20 on any sprint i took, no matter the difficulty. I practiced doing problems quickly, and I can now do them pretty well.

If you don't know how to do a problem or want to do it faster, post the question on AoPS and someone's going to help you.

It's not that i don't know how to do them. I wouldn't be as disappointed if i didn't know how to do it but got it wrong than if i knew how to do it but didn't read the question or something like that...this totally isn't making sense.

I think time is a problem for me because i worry too much about getting things wrong so i slow down on others. Is there a way to prepare or focus yourself or something so that you won't have to worry?? Maybe i'm psyching myself out... :maybe:

the way i check to see if i made a careless mistake is to just see if my answer makes sense and if you are given an algebra problem try plugging your answer into the equation/expression.
i should have used these tips on the school round

i got a 36/46, when i could easily could have gotten a 40 or 41.
 :mad:

I should've had a 46 but because of carless mistakes my score dropped to 22 :mad: :mad:

he wasnt criticizing anyboday :maybe:

wow...that mus be A LOT of careless mistakes .  i used to make about 5 careless ones on each test in the biggining of the year but i got better just beforre the real thing.  i dunn, try underlining the question and all the important stuff.

what kind of careles mistakes do you make?

you should look at my
mathcounts preparation guide for state thread
for help in reducing stupid mistakes

[quote="jhredsox"][quote="jhredsox"]bpms, don't become ubemaya. please.

espeacially when you criticize others about it. please.

:lol:

jorian[/quote]

are you trying to make a point that mistakes aren't careless or something? you're usually not the one to spam.

jorian[/quote]

you're spamming  jhredsox

a careless mistake still fits under the description of a MISTAKE . Everyone makes mistakes.

to make a mistake = to understand, interpret, or evaluate wrongly; misunderstand; misinterpret.

I think careless mistakes are just ways to make an excuse to increase your self esteem, especially ones with low self-esteem.

Well there's one way to cure it. when you do it, find out what you did wrong. don't just say, ok, i just did this wrong. actually do it out. don't be afraid to go into algebra and stuff, and check your work. when you check your work, there's not much sense in just looking over your work. actually do it out again, and when possible, try to find an alternate way to do the problem.

I missed AIME 2 times this year because of careless mistakes (at least for the first one). The second one, I didn't mark a correct answer--lame i know.

A way to aviod mistakes is to check your answers, and sometimes to do them twice

Sry to get off topic, but... wait missed that #1 on AIME last year?  That was like a mathcounts problem..

Checking your work is a good idea on a previous problem if you think you cant do anymore on teh sprint.  Instead of wasting like 3 mins trying to figure out a problem, use those three minutes to check your work on another problem that you werent 100 percent sure  about.  But in the end, its how many problems you can do, nto the accuracy(especially true on sprint).  Ill take a score of getting 25 right, 5 wrong over a score of gettin 22 right, 0 wrong anyday...

But I wouldnt totally recommend doing problems 2x because on mathcounts, there isnt enough time..but check over the ones that you  think you might have made a computational or even reading mistake on when you cant figuer out the remaining ones on the test, as its pointless to waste time looking at a rpoblem you cant do

do every problem in one minute, and then you have 10minutes, left. and then start checking from the front, not from the back...this is how I got a perfect score on the school, (but not on the chapter) :blush:

[quote="Ubemaya"]I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad:  :mad:  :mad: .

BTW, accuracy counts just as much as how many you can do IMSAMO.[/quote]

Oh boy! Mr. Sarcasm.

 :roll:

@undefined117: ya, my math teacher said she cried when she looked at ur pamphlet!

and this thread is getting [i]really[/i] off topic... It'll probably get locked, but at least it served it's purpose.

Thanks!

[quote="undefined117"]Wow.. it also sucks a lot when you type the number wrong in the calculator.. like I did on #1 on an AMC8 one year.. i was enraged..[/quote]

lucky for you amc is banning calculators

back on topic:

I took a practice 2006 State Sprint and my score dropped 8 points because for some of them

1) arithmetic error (simple)
2) counted incorrectly the number of tiems so and so
3) being dumb and forgot how to do a simple problem

i believe the best way to avoid careless mistakes is to double check...usually I only double check ones that I might have doubts on...if it is a simple answer try not to check it until the end, when you have completed the test

[quote="cognos599"]if nobody made careless mistakes then there would be like 20 46's.[/quote]

Haha if nobody made careless mistakes the US wouldn't be in war with Iraq now

[quote="7h3.D3m0n.117"]back on topic:

I took a practice 2006 State Sprint and my score dropped 8 points because for some of them

1) arithmetic error (simple)
2) counted incorrectly the number of tiems so and so
3) being dumb and forgot how to do a simple problem

[/quote]

that sounds exactly like something i would do! except not on the practices, but the real thing :( .

hehe I usually don't get my MATHCOUNTS paper back so i never know which ones i got wrong due to error

[quote="Ubemaya"]I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad:  :mad:  :mad: .

BTW, accuracy counts just as much as how many you can do IMSAMO.[/quote]

who are you ubemaya

[quote="Ubemaya"]I should have had a 36 at Chapters (reflecting my skill level), but due to careless mistakes I got a 44 :mad:  :mad:  :mad: .

BTW, accuracy counts just as much as how many you can do IMSAMO.[/quote]

Are you so mad that you reversed the 44 and the 36?

I had a lot of time left after I did the chapter, so I did the whole sprint again. I got 25 out of 30.

The targets were really fun problems. REALLY fun problems!!!

[quote="Arvind_sn"]personally i thought the targets were really easy. i mean most of us in the hall were done by thirty seconds or something...[/quote]

I agree. Except most people in the hall also didn't use that extra time to [u]check.[/u]


Is the number $2^{10}+5^{12}$ prime or composite?


You should read the Introduction to Counting and Probability. It explains everything you want to know.

alright well lets start with the 3's instead of dividing by threes lets see if we can speed it up by dividing by nines.
Seeing that the sum of the digits of the original number is a a multiple of nine, we divide by nine to get 3087, seeing how the sum of its digits is 18 we divide by nine again to yield 343 note that 343 is just $ 7^3$ So basically we divided by nine twice or instead we can say we divided by 3 4 times and 7 three times so 3*4 = 12
dang beaten to it.

yes well you actually explained it :P

gj though, and sorry


probability
combinatorics unsolved
combinatorics

16 players take part in a tennis tournament. The order of the matches is chosen at random. There's always a player better than another one. In a match, the better wins.
Calculate:
a) The probability that all the 4 best players reach the semi-final

b) The probability that the sixth player reach the semifinal.

Note: The order of the matches doesn't mind players'skill. The winner of the first match plays with the winner of the second, etc.

Enjoy!

In other words, we have a single-elimination bracket-style tournament.

[hide="(a)"] can only happen if each of the 4 best players lies in a different quadrant of the original setup, the probability of which is $ 1\cdot \frac{12}{15}\cdot \frac{8}{14}\cdot \frac{4}{13}$.[/hide]

[hide="(b)"] can only happen if the three other players in the same quarter of the bracket with player 6 are ranked lower, the probability of which is $ \frac{10}{15}\cdot \frac{9}{14}\cdot \frac{8}{13}$.  This is nearly twice as likely as in part (a)[/hide]

The b) is ok...can you explain better the part a)?
Another question..shouldn't I count the probability 4! times?

The answer I gave is larger than $ \frac{1}{24}$, so you certainly shouldn't multiply it by 4!.  In fact, the factor of 4! is included in my answer: write $ 1 = \frac{16}{16}$ and the numerator of the fraction becomes $ 4! \cdot 4^{4}$.

I found the condition a) "they lie in different quadrants" but i can't understand how to express it.
In particular where you multiply 12/15 by 8/14 and 4/13...
Then, call the 4 best players A B C D, how can i express that all these ways are right:
1A                   2B                    3C               4D
1B                   2A                    3C               4D
1D                   2A                    3B               4C
etc...
Forgive me, i'm a beginner :P

The first player has 16 spaces to go, so $ \frac{16}{16}$.  The second player has 12 spaces remaining in other quadrants out of 15 total, so $ \frac{12}{15}$.  The third player has 8 spaces remaining in other quadrants out of 14 total, so $ \frac{8}{14}$.  The fourth player has 4 spaces remaining in other quadrants out of 13 total, so $ \frac{4}{13}$.

Doing it your way, there are $ 4!$ ways to assign each player to a different quadrant.  In addition, there are $ 4^{4}$ ways to assign each player to a position within the quadrant.  This gives us $ 4! \cdot 4^{4}$ arrangements, each of which occurs with probability $ \frac{1}{16\cdot15\cdot14\cdot13}$.


inequalities
function
inequalities proposed
n-variable inequality

Prove that if $x_1, x_2 \cdots ,x_n \ge 0$ add up to $1$, then
\[\sqrt{1-x_1}+\sqrt{1-x_2}+\cdot+\sqrt{1-x_n} \ge n-1\]

I assumed you meant x_i >= 0. That expression is a concave function in each variable, so it reaches its minimum when the variables take the ends of the interval [0,1] as values, so its minimum is reached when n-1 variables are 0 and one of them is 1. It's not very rigorous, but I think this argument works.

It is really the easiest.

Ofcourse we have to assume that x_i>=0 (If not so, we can take x1=x2=...=x_n-1=1, xn=2-n and the inequality not holds).
Then we have 1-xi <=1 => sqrt(1-xi) >= 1-xi. Add up we have the result!

Sometimes, the result can be received in such way!

Namdung

Well, harazi did name it 'the EASIEST inequality'... :D


trigonometry
limit
geometry proposed
geometry

Determine the extreme values of

$ \sin^2 A \plus{} \sin^2 B \minus{} \sin^2 C$

[quote="Ligouras"][color=darkred]In any triangle $ ABC$ exists the full-relation $ \sin^2 A \plus{} \sin^2 B \minus{} \sin^2 C\in\left[ \minus{} \frac 14\ ,\ 2\right)$ .[/color] [/quote]

$ \mathrm{\underline{Proof}.}\ f(A,B,C)\equiv\sin^2 A \plus{}$ $ \sin^2 B \minus{} \sin^2 C\equal{}\sin ^2(B\plus{}C) \plus{} \sin (B \plus{} C)\sin (B \minus{} C)\equal{}$

$ \equal{} \sin A\left[\sin (B \plus{} C) \plus{} \sin (B \minus{} C)\right] \equal{}$ $ 2\sin A\sin B\cos C\implies \boxed {f(A,B,C) \equal{} 2\sin A\sin B\cos C}$ .

$ \blacktriangleright\ f(A,B,C) < 2\ \ ;\ \ \left(\forall \right)x\in\left(0,\frac {\pi}{2}\right)\ ,\ g(x)\equiv f\left(\frac {\pi}{2},\frac {\pi}{2} \minus{} x,x\right) \equal{} 1 \plus{} \cos 2x$ and $ \lim_{x\searrow 0}\ g(x) \equal{} 2$ .

$ \blacktriangleright\ f(A,B,C) \equal{} \cos C[\cos (A \minus{} B) \plus{} \cos A] \equal{}\left[\cos C \plus{} \frac 12\cdot \cos (A \minus{} B)\right]^2 \plus{}$ $ \frac 14\cdot\sin^2(A \minus{} B) \minus{} \frac 14\ge \minus{} \frac 14$

$ \mathrm{\ with\ equality\ iff\ }A \equal{} B \equal{} 30^{\circ}$ . $ \mathrm{In\ conclusion\ ,}\ \min_{\triangle ABC}\ f(A,B,C) \equal{} \minus{} \frac 14\ \ \ \mathrm{and}\ \ \sup_{\triangle ABC}\ f(A,B,C) \equal{} 2$ .

$sin^2 A + \sin^2 B - \sin^2 C\le |sin^2 A + \sin^2 B - \sin^2 C|$
${{= 2|\sin A\sin B\cos C}|\le 2|\sin A||\sin B||\cos C}|\le 2$
$\implies sin^2 A + \sin^2 B - \sin^2 C<2.$
$sin^2 A + \sin^2 B - \sin^2 C\ge -\frac 14 :$
1995, I prove
$a^2 - b^2 - c^2 \leq R^2 \iff sin^2 B+sin^2 C-\sin^2 A\ge -\frac 14$
$\iff cosAsinBsinC\ge-\frac 18.$
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=480082]《Hunan mathematics communication》(China Changsha)No.3(1995)[/url]
1996, I prove
$cos^2 A+\lambda(sin^2 B+\sin^2 C)\ge \lambda-\frac {\lambda^2}{4}\implies  sin^2 B+sin^2 C-\sin^2 A\ge -\frac 14.$
《Middle school Mathematical Monthly》(china Suzhou)No.9(1996)


probability
geometry
rectangle
analytic geometry
similar triangles
AMC

1. The number of 5-digit numbers in which every two neighbouring digits differ by 3 is
(A) 40
(B) 41
(C) 43
(D) 45
(E) 50

2. A $3 \times 3$ square is divided into nine $1 \times 1$ unit squares. Different integers from 1 to 9 are written into these nine unit squares. Consider the pairs of numbers in the squares sharing a common edge. What is the largest number of pairs where one number is a factor of the other number?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 12

3. Each point on the four sides of a 1m $\times$ 1m square is coloured one of $n$ colours so that no two points that are exactly 1m apart are coloured the same. What is the smallest $n$ for which such a colouring can be made?

4. A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?

For number one, i made a little tree and got 47, must've made a silly mistake.  :(

For number two, I get something like 9, with this arrangement

7 5 4
9 1 8
3 6 2

It obviously can't be 12 because that's how many edges there are.

On 3: I would say that the answer is 5. Am I correct?

oops,duh, I was thinking of a cube...

[quote="solafidefarms"]On 3: I would say that the answer is 5. Am I correct?[/quote]

Hmmm.... Do the points have to be in the center of a side? Otherwise the answer is 1.

For #1, I got (D) 45...
I drew a tree diagram, just like 236factorial did. Is that correct?

Okay, I thought 8 - 2 was 3 for a second, so I also got 45 as the answer (D). For number one, that is.

[quote="Iversonfan2005"][quote="math92"]for number 1 i get D
for number 2 i get B

I am not sure what i did though.[/quote]

please don't post answers without solutions, especially if someone's already given the same answer ;)[/quote]

It would be nice if you gave some answers to begin with  :roll:

Well number one is definitely (D) and I got that one wrong... I miscounted and got 47 too, but I ran out of time and just put 50. Dang! I don't know any other solution, other than brute counting and the tree diagram though, with which you get 45.

Anyway, 236Factorial, you're my lifesaver! I answered 9 for question 2 and I thought I got it wrong all along, but now it turns out there [i]is[/i] a way to get 9 (I hope there's no way to get 10).

Numbers 3 and 4 are among the last 5 questions, in which competitors nominate an integer answer between 0 and 999 (and they're really difficult), so I don't really know the answers. If I have the time, I'll scan the question booklet and post it in AoPS.  :) There's even an interesting question on how to cut the cake in the best possible way to avoid your brother or sister getting the larger piece.

[quote="aidan"]I hope there's no way to get 10. [/quote]

I can't see anything that will involve 10. 5 and 7 are prime, and the only number from  1 and 9 they can share a factor with is 1. The 5 and seven will be sharing sides with at least 4 different numbers (put 5 on the corner and 7 right next to it. We counted the one that 5 and 7 share). Say one of these numbers is 1. Only one of the numbers can be touching it adjacently, so there must be at least three out of the twelve possible meetings where one is not a factor of the other. So the maximum is 9.

[quote]Only one of the numbers can be touching it adjacently, so there must be at least three out of the twelve possible meetings where one is not a factor of the other. So the maximum is 9.[/quote]

Excellent observation... didn't think of that. Here's one more question I need someone's opinion on:

14. Two dice are thrown at random. The probability that the two numbers obtained are the two digits of a perfect square is

(A) $\frac {1}{9}$

(B) $\frac {2}{9}$

(C) $\frac {7}{36}$

(D) $\frac {1}{4}$

(E) $\frac {1}{3}$

So the total number of options are
$6\cdot6 = 36$
We know our perfect square must be (inclusive) 11-16,21-26,31-36,41-46,51-56,61-66

We see 16, 25, 36, and 64 work.

We might jump the gun and go! "Ooh! 1/9" And get it wrong

But we have to realize if we roll 61, 52, 63, or 46 we also get the correct answer

So we have $4\cdot2 = 8 $ways to get 2 digits of perfect square so

$\frac{8}{36} = \boxed{\frac{2}{9}}$

[hide]There are 36 total possibilities. The possible ones are (1,6), (6,1), (2,5), (5,2), (3,6), (6,3), (6,4), (4,6). That's 8/36=2/9,

Remember that the dice only goes up to 6. [/hide]

[quote]We might jump the gun and go! "Ooh! 1/9" And get it wrong [/quote]

 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

I think I should use that on my friends.... I said 2/9 and they kept saying 1/9.

Hey everybody, I'm new to this and it is my first post!

I also did the intermediate test on Thursday over here in New Zealand and found some questions hard.

For #30, does anyone have an answer for it:

[i]A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer.[/i]

And for #23, the area of the shaded rectangle - I wrote (D) between 7/16 and 1/2, but I'm not sure about it. A friend of mine wrote (E) more than 1/2 and we all thought he was crazy, but he might end up with the last laugh!

And, finally, could someone fill me in on #17. The 12cm^2 square was folded and you needed to find the length of the fold line...

Thanks,
Jordan.

Welcome... it's a nice community over here.  :) I have no idea at all about #30.

The answer for #23 is definitely (D) as the area is found to be 15/32 square units. There are 4 similar triangles in the diagram, if you have your question paper.

For #17, the answer is 4. It is said that the total visible area of the paper is half shaded and half white. So you can form the equation: $12 - x^2 = \frac {x^2}{2}$, where $x = US$. So $x^2 = 8$ and $x = 2 \sqrt {2}$. The the fold line is $\sqrt {2} \times 2 \sqrt {2} = 4$.

[quote="nat mc"][quote="aidan"]My name is Louis and my father has cooked me an L-shaped cake for my birthday. He says that I must cut it into three pieces with a single cut, so that my brother and sister can have a piece too. So, I have to cut it

He says that I have to be polite and let them have the first choice of the pieces, but I just know that they’ll be greedy and leave the smallest possible piece for me. So I want to cut the cake so that my little piece will be as big as possible. If I do this, how big, in square centimeters, will my piece be?[/quote]
I got the answer to be 80, is that right? I used some little coordinate type things.[/quote]

well, this is my solution:
[hide]
from the diagram(in the middle), we know that in order to cut three pieces in one slice, the line must touch the turning pt. therefore, it will make a right triangle with sides $30 - x$ and $20 - y$, and two trapezoids.  therefore the total area is:

$\displaystyle \frac{(30-x)(20-y)}{2} + (y + 10)*5 + (x + 20) * 5 = 10*30 + 10*10 = 400$

simplifying get:

$-20y -10x +xy = -100$

adding 200 and factor yields:

$(20-x)(10-y) = 100$

since we want $y$ to be as large as possible, $x = 0$, and $y = 5$. plug in and we get $(y+10)*5 = 75$
[/hide]

[quote="Maths_Crazy"]
[i]A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer.[/i]
[/quote]

I think it's 13. $1^2+2^2+5^2+10^2=130$. That's guess and check...

[quote="nat mc"]I got 80, by making the top most portion being a triangle with base 8 and height 20, area 80. the cut goes through the corner, and the right most portion is a 10 by 10 square minus a triangle base 4 height 10, to get area 100-20=80[/quote]
Here is the solution for that. I tried it your way, then I tried it this way.  You want the two smallest areas to be equal in this case, so that his isn't really the smallest. so we have
$\frac{20*2a}2=100-\frac{10a}2$
$20a=100-5a$
$25a=100$
$a=4$
$100-5*4=\boxed{80}$

Ok... good. I answered 80 after solving some equation but some friends claimed it was 75. Made me worried for a moment.

I got question 30 wrong too... I guessed 5! Even worse...

I think 130 is the only integer that equals to the sum of the squares of its four smallest divisors. Here is my solution for #4, but i m not sure if it good. I'd be good to see someone else give a solution.
[hide="edited solution"]Let the positive integer that equals to the sum of the squares of its four smallest positive divisors be “n”. n must be divisible by 1. Also it must be divisible by 2, since the sum of four odd numbers squared is still even. The sum of the squares of the two smallest divisors is 5, therefore the next two are an odd and an even divisor. This means that if the odd divisor is ‘x’, and the even divisor is ‘2x’, unless the even divisor is a power of 2.
Therefore n=1^2+2^2+x^2+(2x)^2=5+5x^2=5(x^2+1). Hence 5 is a divisor of n thus 3 can’t equal ‘x’ since 5 would be the next smallest divisor. Therefore x=5 and 2x=10, and n= 1^2+2^2+5^2+10^2=130.
Thus the largest prime that divides n is 13.
Examining the case in which the even divisor is a power of 2, then n must be divisible by 4 and 8. Therefore the odd divisor of n must be 3,5,7. However none of these satisfy the conditions. [/hide]

[quote="236factorial"]Couldn't you use a calculator in the AMC? Wait, are you in australia?[/quote]

No you can't.

Yes, I was doing the Australian MC, however, I live in New Zealand, slightly south-east of Australia (if you didn't know already)!

[quote="p.c"]and n= 12+22+52+102=230.
Thus the largest prime that divides n is 13.[/quote]
I assume you mean $1^2+2^2+5^2+10^2=130$

[quote="aidan"]The code probably got messed up. Anyway:

[quote]Also it must be divisible by 2, since the sum of four odd numbers squared is still even.[/quote]

Why is this so? Someone explain... don't really get this.[/quote]

An odd number squared is odd. four odds added together is even.

Anyone got an answer for #12...the question with the triangle inside a 6x6 square.

[quote]An odd number squared is odd. four odds added together is even.[/quote]

Get this now... thanks. Anyway the answer for #12 is $10.5cm^2$. You minus off the area of the right-angles triangles from the rectangle around triangle we need.

OK thanks, what would #15 be?

P.S. Sorry for asking for the answers - I'm trying to figure out my mark (currently 69 out of 120).

No problem... #15 is $(E)$, $\sqrt {2} : 1$ We set up equations as follows: $\frac {2x}{y} = \frac {y}{x}$, as the two rectangles are similar. Cross multiply to get $y^2 = 2x^2$. Then ${\frac {y}{x}}^2 = 2$, so $\frac {y}{x} = \sqrt {2}$. We can substitute back to get $x = 1$.

OK, thanks.

Final question, #25.

What is the maximum n of reflex angles in a polygon?

The answer is $n - 3$. In a polygon with $n$ sides, the total number of degrees in the polygon is $180(n - 2)$. Now a reflex angle is an angle $\geq 180$ and $\leq 360$. So there can at most be $n - 3$ angles greater than 180 degrees in any polygon (if there were $n - 2$ reflex angles, then every angle would have to be 180, and the shape wouldn't be a polygon).

[quote="aidan"]The answer is $n - 3$.[/quote]

Yyyyeeeessss!!

74 out of 120 then for me...

[quote]74 out of 120 then for me...[/quote]

But you have to remember that how well you do depends on how well everyone does in New Zealand...

Ok guys, it's been a while (because it takes ages for the test to get marked and returned), but guess what I got?

(Highlight below)

[color=white][size=117][b]HIGH DISTINCTION[/b][/size]

I actually ended up with 75/120, enough to be in the top 1% of candidates in New Zealand.[/color]

I would like to thank all of you guys for answering the many questions I had after the test.

Cheers,
Jordan.


Let $ n \ge 3$ be a prime number and $ a_{1} < a_{2} < \cdots < a_{n}$ be integers. Prove that $ a_{1}, \cdots,a_{n}$ is an arithmetic progression if and only if there exists a partition of $ \{0, 1, 2, \cdots \}$ into sets $ A_{1},A_{2},\cdots,A_{n}$ such that
\[ a_{1} \plus{} A_{1} \equal{} a_{2} \plus{} A_{2} \equal{} \cdots \equal{} a_{n} \plus{} A_{n},\]
where $ x \plus{} A$ denotes the set $ \{x \plus{} a \vert a \in A \}$.

The problem is from Romanian TST 1998 and was proposed by Vasile Pop.

You can find it [url=http://projectpen.wordpress.com/2008/04/17/08s-an-arithmetic-partition-of-the-positive-integers/]here[/url]. I remember I spend so much time solving and writing the second solution in [url=http://projectpen.files.wordpress.com/2008/04/pen08s-draft.pdf]the pdf[/url]. Good old days when I had time to write for PEN :). Actually, not even now am I sure if that solution is fully corect. I really wonder if anyone has solved it during the TST. Valentin, any information? :)


[quote="alex2008"]If $ a,b,c$ are positive numbers, then :

$ \frac {a^{2} \minus{} bc}{4(a^{2} \plus{} b^{2}) \plus{} c^{2}} \plus{} \frac {b^{2} \minus{} ca}{4(b^{2} \plus{} c^{2}) \plus{} a^{2}} \plus{} \frac {c^{2} \minus{} ab}{4(c^{2} \plus{} a^{2}) \plus{} b^{2}}\geq 0$

[/quote]
It's true for all real $ a,$ $ b$ and $ c$ such that $ a^2\plus{}b^2\plus{}c^2\neq0.$


superior algebra
superior algebra solved

A collection $C$ if subsets of a given set is called a monotone class if it is closed under increasing countable unions and decreasing countable intersections.