quadratics
algebra
quadratic formula
algebra unsolved

Can someone explain step-by-step how to get the solution to this equation?

[b]z^2-(3+2i)z+(1+3i)=0[/b]

I looked in the back of the book, and the answer is : [b]2+i[/b], [b]1+i[/b]I just need to know how to get there.

--I know it has to the with the quadratic formula but things got very complicated. Maybe there's an easier way?

Thanks (:

[quote="kunny"]You can use the formula $ \boxed{ax^{2}+bx+c=0\ (a\neq 0)\Longrightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$[/quote]

I know that but it's really complicated with the multiplying of the imaginary numbers and stuff. i would be able to do it if there weren't any imaginary numbers or if it was less complex.

can anyone show me how to do it step by step?

[quote="kunny"]All right, bz0inks.
$ z^{2}-(3+2i)z+1+3i=0\Longrightarrow z=\frac{3+2i\pm \sqrt{D}}{2}$.

$ D=(3+2i)^{2}-4(1+3i)=9+12i+4i^{2}-4-12i=1$, yielding $ z=\frac{3+2i\pm 1}{2}=\boxed{1+i,\ 2+i}$[/quote]

yay (: I get it now it made it easier using the D
thanks.


Let A be an nxn mx with A^4=I and let P be a projection mx. Show that B=(1/4)[P + A^-1PA + A^-2PA^2 + A^-3PA^3] is a projection mx, that AB=BA and that C(B) = C(P), assume C(P) is an A-subspace that is if X is in C(P) so is AX.

Thanks!

Did anyone understand what I was talking about? Because I didn't how to use the code to post the pretty format, I am sorry about it!


abstract algebra
vector
induction
Ring Theory
superior algebra
superior algebra theorems

Let $R=k[v]/(v^2)$ for some commutative field with $1$, $k$.
Prove that every $R$ module is a direct sum of modules isomorphic to $R$ or $R/vR$.

Let's just write $R$ as $k[v]$ for $v^2=0$, to make things simpler.

Let $M$ be an $R$-module, and let $N$ be the submodule consisting of those elements o $R$ annihilated by $v$. In particular, $N$ is a $k$-vector subspace of $M$, so it has a complement $P$, i.e. $_kM=_kN\oplus_kP$, where the $_k$ means that we consider the sets as vector spaces over $k$ (as opposed to $_RM$, for instance, which will denote the $R$-module $M$).

Let $\{e_\alpha\}_{\alpha\in A}$ be a basis for $_kP$, and let $\{f_\beta\}_{\beta\in B}$ be a basis for a complement of the $k$-vector subspace of $_kN$ spanned by $\{ve_\alpha\}_{\alpha\in A}$. It should be easy to check now that $_RM$ is the direct sum of the modules generated by $\{e_\alpha,ve_\alpha\},\ \alpha\in A$ (these will be the summands isomorphic to $_RR$ in our direct sum; the choice is not unique, of course) and the modules generated by $f_\beta,\ \beta\in B$ (the summands isomorphic to $_RR/vR$).

I hope I didn’t miss anything.

The outline of my proof was:
-Consider $L=vM$. Then $L$ is an $R/vR$ vector space and choose a basis $\{vm_i|\ i\in I\}$ for it.
-Let $M'=\sum_iRm_i$. I proved that $M'=\oplus_iRm_i$ and each direct summand is isomorphic to $R$
-$vM\subset M'$, so $M/M'$ is an $R/vR$ vector space.
-Let $\{\bar n_j|\ j\in J\}$ be a family of element of $M$ that yields a basis as an $R/vR$ module modulo $M'$ in $M/M'$.
-Get a family $\{n_j|\ j\in J\}$ of elements of $M$ obtained from $\{\bar n_j\}$ by adding some elements of $M'$ such that $vn_j=0$ for all $j$. By the way we obtain the $n_j$s, they are still a basis for $M/M'$ as an $R/vR$ module when reduced modulo $M'$.
- Let $N=\sum_jRn_j$. I proved that the sum is direct and that each summand is a copy of $R/vR$.
- Finaly, I proved $M=M'\oplus N$

Writting all these in a clear form, took me about $2$ $A4$ pages. I don't know if grobber's proof is shorter, but the ideas are similar

Does this result generalize to $R=k[v]$ with $v^n=0$?
I mean: is every $R$ module a direct sum of $R/v^mR$ with $m\in\overline{1,n}$?

Let us consider a more general situation. Let $R$ be a commutative local Artinian principal ideal ring. The lattice of its ideals is the chain $R>\pi R>\ldots\pi^{n-1}R>\pi^nR=0$, where $n$ is the index of nilpotency of the Jacobson radical $rad(R)$ and $\pi$ is an element from $R$ such that $rad(R)=\pi R$. (That's why such rings are also called finite-chain rings.) Note that any element $r$ from $R$ can be represented as $r=\alpha\pi^a$, where $\alpha$ is a unit in $R$ and $a=0,1,\ldots,n$. The described class of rings is rather wide. For instance, it contains fields, $\mathbb Z_{p^n}$, $k[v]/(v^n)$.

[b]Theorem.[/b][i] Any $R$-module $M$ can be represented as a direct sum of cyclic modules (which are isomorphic to $R/\pi^aR$, $a=1,2,\ldots,n$).
[/i]
[i]Proof.[/i] Let us use induction on $n$. If $n=1$ then $R$ is a field and the result follows immediately. Len $n>1$. Consider a homomorphism $\phi\colon M\to M\pi^{n-1}$ defined by $\phi(m)=m\pi^{n-1}$. $M\pi^{n-1}$ can be viewed as a $R/(\pi)$-module. $R/(\pi)$ is a field, so $M\pi^{n-1}$ is a semisimple $R$-module. That is $M\pi^{n-1}=\oplus_{i\in I}E_i$, where $E_i$ is a simple $R$-module. For every $i\in I$ there exists $x_i\in M$ such that $E_i=x_i\pi^{n-1}R$. It is not difficult to verify that the elements $x_i$, $i\in I$ are linearly independent over $R$ (use the fact that $M\pi^{n-1}$ is a [i]direct[/i] sum of $E_i$). Let $F=\oplus_{i\in I}x_iR$, $F$ is a free submodule of $M$. Since $R_R$ is injective (Baer's criterion) and $R$ is Noetherian we see that $F$ is injective. Thus $M=F\oplus N$ for some submodule $N\le M_R$. We have $M\pi^{n-1}=F\pi^{n-1}\oplus N\pi^{n-1}=M\pi^{n-1}\oplus N\pi^{n-1}$, hence $N\pi^{n-1}=0$. Thus $N$ can be considered as a $R/(\pi^{n-1})$-module. By the induction $N$ is a direct sum of cyclic $R/(\pi^{n-1})$-modules. But every $R/(\pi^{n-1})$-submodule is an $R$-submodule (and vice versa).  The result follows.

It is very interesting to describe the class of rings (not necessary commutative) such that every right $R$-module is a direct sum of cyclic submodules. I cannot do it and i don't know if it is already done.


quadratics
algebra
quadratic formula
special factorizations
factoring quadratics
polynomial
Rational Root Theorem

Hello all. I'm having somewhat of a crisis. I only learned quadratics (correct name?) last year. I'm talking about binomials/trinomials. I know that there are methods used to solve these trinomials etc, one of them involves FOIL, but I didn't learn that. I learned something that my teacher called "Cross Method".

My teacher was an asian man, it was his first year teaching, and he wasn't [b]completely[/b] comfortable speaking english, but he was definitely a math wiz.

Anyway, I'm hoping someone here knows of this "cross method" and can provide an example or two. The way I remember it last year, it could be used for any binomial/trinomial, and I remember being happy that it was so easy, but...I can't remember, for the life of me, how to do it.

If I don't get some kind of memory refresher, I'll have to learn everything from scratch, as I go, and I'm afraid the course may be overwhelming.

Anyway, in short: If anyone knows of a method for solving quadratics called "cross method" please post an example, or some steps. It'd really save me, I can't begin to explain that. ><

Sorry, I didn't catch all of that.  Which one of these problems are you trying to solve:

Expand $(3x+4y+z)(4x+2y+4z)$

OR

Factor $3a^{2}+7b^{2}+4c^{2}+3a+4b+c$

OR

Factor $x^{3}+4x^{2}+3x+4$

OR

post a sample problem to solve if it was not like any of these

There are three methods I know, factoring, completing the square, and the quadratic formula. I have neitheer heard of either of those methods, but maybe they're just different names.

[b]Factoring[/b]
$0=x^{2}+2x-15$
$0=(x-3)(x+5)$
Notice how this was factored. If u multiply the two constants, you get -15. When I added them, I got 2.
Now $x-3$ or $x+5$ must equal 0. That means $x=3$ or $x=-5$

[b]Completing the square[/b]
$0=x^{2}+2x-15$
$16=x^{+}2x+1$
$16=(x+1)^{2}$
$\pm4=x+1$
$\pm4-1=x$
$x=3$ or $x=-5$

[b]Quadratic Formula[/b]
The quadratic formula says that if $ax^{2}+bx+c=0$ then $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Since $0=x^{2}+2x-15$, we plug these values into the the quadratic formula
$x=\frac{-2\pm\sqrt{2^{2}-4(a)(-15)}}{2(1)}$
We evaluate this to get $x=3$ or $x=-5$

Do any of these methods sound familiar? Do you need any of them explained more?[/b]

It's used for factoring (and can be used to expand too) in quadratic formulas that are written in the format
0= [i]a[/i]x^2 + [i]b[/i]x - [i]c[/i]
or
0= x^2 + 2x - 15

It's the same end result as your factoring method, but it's a different process. It involves crossing the numbers somehow, I just can't remember how it's done, and my searches aren't coming up with anything.

Sorry if I'm vague, I'm struggling to remember this "cross method" and I've never learned the other way until two days ago.

If there's more info needed, I'll do my best to supply, just ask :/

i think your talking about, because there kind of is some "crossing" involved.

$(x+a)(x+b)=x^{2}+ax+bx+ab=x^{2}+(a+b)x+ab$
So like I said, you need to find two numb that sum to the middle term and have a product that is the second term.

$(cx+a)(dx+b)=cdx^{2}+adx+bcx+ab=cdx^{2}+(ad+bc)x+ab$
So the x^2 term is the product of the x-coefficients. The constant is the product of the two constants. And the middle term is the sum of the product of the outer coefficients and the product of the inner coefficients.

When you picture it all, I can kind of see the numbers "crossing" over each other, or something like that...

does this seem to be what you learned?

The "cross method" is probably the method of factoring quadratics that is essentially equivalent to use of the Rational Root Theorem; you guess at possible factorizations until one works out.

For example, suppose we wanted to factor $x^{2}-8x+15 = 0$.  The "cross method" is equivalent to guessing that the factorization is either

$(x \pm 3)(x \pm 5)$

Or

$(x \pm 1)(x \pm 15)$

You expand them (the content of the "cross method" is a shorthand way of performing this expansion; it's not really necessary) to see if they work.  In fact,

$x^{2}-8x+15 = (x-3)(x-5)$

Because $8 = 3+5$.

The idea is that given the quadratic $x^{2}+bx+c$ we find the factors of $c$.

Specifically, look at all the ways of writing $c$ as a product of integers $mn$.  Then test the factorization

$(x+m)(x+n) = x^{2}+(m+n)x+c$

To see if $b = m+n$.

[quote="Zaryef13"][quote="t0rajir0u"]
For example, suppose we wanted to factor $x^{2}-8x+15 = 0$.  The "cross method" is equivalent to guessing that the factorization is either
[/quote]

this sounds very familiar, but I can't remember how to do it on paper and it's killing me.

I'm not very good at math to begin with, but the cross method, I understood. "Guessing until you find something that works" makes me picture what I was doing to solve the problems, but I can't remember the steps.

Everything else is over my head. Thanks for all the replies, and replying so quickly, but I'm getting kind of lost, aside from the above quote ><[/quote]
Maybe this is it?
$(x+a)$
$(x+b)$
xx+xb+ax+ab?


vector
geometry
3D geometry
tetrahedron
algebra
linear equation
linear algebra

[img]http://img232.imageshack.us/img232/4980/vector1e.jpg[/img]

This problem comsists of a solution
But I cannot get the idea :(

[img]http://img245.imageshack.us/img245/2930/vector2.jpg[/img]

How can I get the answer underlined by red line... :o

Well...I am quite confusing about some concepts

If two vectors in R3 are linearly dependent they must be collinear (both vectors are lying along the same straight line) . If three vectors in R3 are linearly dependent they must either be collinear (all three vectors are lying along the same straight line) or coplanar (two of them are linear independent and the third one is lying in the plane expanded by the two linearly independent vectors). Hence three vectors are linearly independent iff their triple scalar product is not zero.

Four or more vectors in R3  will always be linearly dependent.

[b]If two vectors in R3 are linearly dependent they must be collinear[/b]
WHat is the difference between collinear and coplanar?
The former means on the same line..The latter means on the same plane?
If two vectors can be in coplanar in this case?

[b]Four or more vectors in R3  will always be linearly dependent.[/b]

It means one vector must be on the same plane with another vector since it is in 3-dimensional space?

On your first post: by "projection", it means orthogonal projection; we want to drop a perpendicular, and find the shortest distance to that plane. What do we know about planes in $ \mathbb{R}^3$? There's only one perpendicular direction; if $ \mathbf{v}$ is a normal to the plane, $ PR$ (where $ R$ is the point $ (\minus{}1,2,3)$ we're projecting) is parallel to $ \mathbf{v}$. The plane has equation $ \mathbf{v}\cdot\mathbf{x}\equal{}c$, and $ P\equal{}R\minus{}t\mathbf{v}$, where $ \mathbf{v}\cdot(R\minus{}t\mathbf{v})\equal{}c$. That's a linear equation for $ t$.
$ Q$ is just the intersection of a line $ s(2,1,0)\plus{}(1\minus{}s)(\minus{}1,2,3)$ with the plane; again, that's an easy linear equation.

Where do the messy square roots come from? Normalization, and I'd rather do it in a way that avoids the square roots. Note that that linear equation I wrote has $ |v|^2$ as the coefficient of $ t$, so you have to divide by it in solving. The method shown divides by $ |v|$ and then again by $ |v|$, by scaling to a unit normal.

On your second post- yes, "coplanar" means "in the same plane. This has a different meaning for points than for vectors; any two vectors are automatically coplanar since they span a space of dimension at most two, while any three points are automatically coplanar.
On the other hand, four or more vectors in $ \mathbb{R}^3$ can be linearly dependent without any three of them being linearly dependent and coplanar; this is completely normal. As a typical example, take the four vectors from the center of a tetrahedron to its vertices; any three span the whole space.


Support
function
real analysis
real analysis unsolved

Let $ \varphi_j$ be a sequence of $ C^{\infty}_c(\mathbb{R}^n)$ functions with the property that $ \varphi_j(x)\not\equal{}0$ for each $ j$.  If $ \text{supp }\varphi_j \to \{x\}$, can we find $ \varepsilon>0$ such that $ B(x,\varepsilon)\supset \text{supp }\varphi_n$ for all sufficiently large $ n$?

No. An example of support sets that could do it on $ \mathbb{R}$:

$ S_n \equal{} [ \minus{} \frac1n,\frac1n]\cup [n,n \plus{} 1]$

There are olther ways to do it- we don't need to go to $ \infty$. If you want the result, you need something stronger like the diameters going to zero.


inequalities
logarithms
number theory proposed
number theory

Hi,

I am searching for a good inequality for
$2^{\omega(n)}\mbox{ with }\omega(n)=\sum_{p|n}{1}.$
I know
$2^{\omega(n)}\leq n^{\frac{1}{\ln \ln(n)}}$
but I want a better one.

Thanks for help,
Isabel

It is easy to show, that
$e^{\omega(n)}\leq n^{\frac{1+o(1)}{\ln \ln(n)}}.$

$\omega (n)=\omega (rad(n)),$ were $rad(n)=\prod_{p|n}p$.
\[n\ge rad(n)\ge \prod_{k=1}^{m}p_{k}, \ m=\omega (n). \]
were $p_{k}$ - k-th prime. We have $\prod_{k=1}^{m}p_{k}=exp(\sum_{k}ln p_{k})=exp(p_{m}(1+o(1))).$
Therefore $\omega (n)\le \pi(ln(rad(n)))(1+o(1))$. It give $2^{\omega (n)}<e^{\omega(n)}\le n^{\frac{1+o(1)}{ln ln(n)}}$.

but this isn't a big advancement, or is it ?
I'm seeking for some better (at least asymtotical, $n\to \infty$) estimates...

Let $p_{k}$ be kth prime. Take $n=p_{1}p_{2}\ldots p_{k}$. Then
$n^{\frac1{\ln\ln n}}=e^{k(1+o(1))}$.


limit
function
calculus
integration
real analysis
real analysis unsolved

In the following  let $\{\; \cdot \; \} $ denotes the fractional part . If  $ F(x):= \frac{6x^3+3x^2-3x-1}{2} , $  [b]find   with proof [/b]
\[ \lim\limits_{n\to \infty}\frac{1}{n}\sum\limits_{k=1}^{n}F\left(\left\{\frac{n}{k}\right\}\right)  \; . \]

The function $g(x)=F(\{\frac1x\})$ is piecewise continuous on $[0,1]$, hence Riemann integrable. The sum is a Riemann sum for this integral, so it clearly converges to $\int_0^1g(x)\, dx$.

The only thing that remains is the tedious task of calculating this integral.

[quote="jmerry"]...$g(x)=F(\{\frac1x\})$ i... The sum is a Riemann sum for this integral, so it clear  converges to $\int_0^1g(x)\, dx $. The only thing that remains is the tedious task of calculating this integral.[/quote] Of course, the existence of the limit (:=L )  it's clear.   But what about   L    ? Here it's seems to be very difficult.


geometry
circumcircle
trigonometry
geometry proposed

Given a regular heptagon $ ABCDEFG$ such that $ GA \equal{} \frac {3}{5},\ GB \equal{} b,\ GC \equal{} c$. Let $ P$ be the intersection point of the lines $ CD$ and $ EF$.

(1) Prove that $ CP \equal{} CE$.

(2) Find the value of $ \frac {1}{b} \plus{} \frac {1}{c}$.

O is the heptagon circumcenter, r circumradius. From isosceles $ \triangle OCD,\ \triangle OCE$ with OC = OD = OE = r,
$ \angle PCE \equiv \angle DCE = \angle DCO - \angle ECO = (\frac{\pi}{2} - \frac{\pi}{7}) - (\frac{\pi}{2} - \frac{2\pi}{7}) = \frac{\pi}{7}$
From isosceles $ \triangle CDE$ with CD = DE

$ \angle CEP = \angle CED + \angle DEP = \angle DCE + \pi - \angle DEF =$
$ \frac{\pi}{7} + \pi - (\pi - \frac{2\pi}{7}) = \frac{3 \pi}{7}$
$ \angle CPE = \pi - (\angle PCE + \angle CEP) = \frac{3\pi}{7}$
$ \triangle CPE$ is iscocseles with CP = CE.

Let $ O = 0$ and $ D = -r$ negative real number. Heptagon vertices are complex roots of $ z^7 + r^7 = 0,$ $ \{A, B, C, D, E, F, G\} = r \exp (\frac{\text{i}(2k + 1)\pi}{7}),\ k = 0, 1, ..., 6.$ Factoring out the D-root and dividing by $ \frac{z^3}{r^3} \neq 0,$
$ \frac{z^3}{r^3} - \frac{z^2}{r^2} +\frac{z}{r} - 1 + \frac{r}{z} - \frac{r^2}{z^2} + \frac{r^3}{z^3} = 0$

If $ z = r\exp (\text{i} \phi),$ then $ x = \frac{z}{r} + \frac{r}{z} = 2 \cos \phi$ and $ \frac{z^2}{r^2} + \frac{r^2}{z^2} = x^2 - 2,$ $ \frac{z^3}{r^3} + \frac{r^3}{z^3} = x^3 - 3 (\frac{z}{r} + \frac{r}{z}) = x^3 - 3x.$ Substituting for z, $ x_k = 2 \cos \frac{(2k + 1) \pi}{7},\ k = 0, 1, 2$ are roots of

$ x^3 - 3x - (x^2 - 2) + x - 1 = x^3 - x^2 - 2x + 1 = 0$

Denoting GA = BC = a, by the sine theorem for the $ \triangle GBC \cong \triangle CEF,$
$ \frac{a}{b} + \frac{a}{c} = \frac{\sin \frac{\pi}{7}}{\sin\frac{2\pi}{7}} +\frac{\sin \frac{\pi}{7}}{\sin\frac{4\pi}{7}}$
Using $ \sin \frac{4\pi}{7} = \sin \frac{3\pi}{7}$ and formulas for half and third of an angle,
$ \frac{a}{b} + \frac{a}{c} = \frac{1}{2\cos\frac{\pi}{7}} +\frac{1}{3 - 4\sin^2\frac{\pi}{7}} = \frac{1}{2\cos\frac{\pi}{7}} +\frac{1}{4\cos^2\frac{\pi}{7} - 1} =$
$ = \frac{1}{x_0} + \frac{1}{x_0^2 - 1} = \frac{x_0^2 - 1 + x_0}{x_0^3 - x_0} = 1$

Yes, that's right. But Your solution is very difficult for Junior high school students . We can use a famous theorem.

Here is the generalized old problem for which you have already offered the solution from Japan.

[color=blue]Kobe University entrance exam 1958

Let $ a$ be the side length of a regular heptagon and $ b,\ c$ be the distinct two diagonal lengths.
Prove that $ \frac {1}{a} \equal{} \frac {1}{b} \plus{} \frac {1}{c}.$[/color]

We can use Polemy theorem for quadrilateral ABCF, $ ab \plus{} ac \equal{} bc$ and divide the equation by by abc.


limit
real analysis
real analysis unsolved

[b][u]Problem[/u][/b]:
Let $(a_{n})_{n=1}^{\infty}$ is a sequence of positive numbers such that $\varlimsup_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}=1$ and $\varliminf_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}<1$.Prove that there is a subsequence $(a_{n_{k}})_{k=1}^{\infty}$ of $(a_{n})_{n=1}^{\infty}$ such that :
(i)$\lim_{k\rightarrow \infty}a_{n_{k}}^{\frac{1}{n_{k}}}=1$ and
(ii)$\lim_{k\rightarrow \infty}|a_{n_{k}}^{2}-a_{n_{k}+1}.a_{n_{k}-1}|^{\frac{1}{n_{k}}}=1$

I tried to prove it many times,but I haven't had any success! :(

(i)
Let $b_{n}=a_{n}^{\frac{1}{n}},\varlimsup_{n\rightarrow \infty}b_{n}\neq \varliminf_{n\rightarrow \infty}b_{n}$, so $b_{n}$ is not constant, but bounded
 1 is an accumulation point for $b_{n}$ from Bolzano-Weierstrass theorem $\implies$ there is a $b_{n_{k}}$ subsequence of
$b_{n}$ wich converge to 1

(ii)$b_{n_{k}}\to 1\implies \forall \epsilon >0~ \exists N_{\epsilon}\in N~ so~ that~ |b_{n_{k}}-1|<\epsilon~ if~ n_{k}>N_{\epsilon}$
So $(1-\epsilon)^{n_{k}}\leq a_{n_{k}}\leq (1+\epsilon)^{n_{k}}\implies$
$|a_{n_{k}}^{2}-a_{n_{k-1}}a_{n_{k+1}}|^\frac{1}{n_{k}}<((1+\epsilon)^{2n_{k}})-((1-\epsilon)^{2n_{k}})^{\frac{1}{n_{k}}}<(1+\epsilon)^{2}$ So $|a_{n_{k}}^{2}-a_{n_{k-1}}a_{n_{k+1}}|^\frac{1}{n_{k}}\to 1$

[quote="cckek"]
(ii)$b_{n_{k}}\to 1\implies \forall \epsilon >0~ \exists N_{\epsilon}\in N~ so~ that~ |b_{n_{k}}-1|<\epsilon~ if~ n_{k}>N_{\epsilon}$
[/quote]
I think it's not true.In fact $|b_{n_{k}}-1|<\epsilon~ if~ k>N_{\epsilon}$
not with $n_{k}$


geometry
perimeter
algebra solved
algebra

We have a 4x4 board which has + in every square except for the second square on the first row, where we have -. We are allowed to make the following moves:

(1) change the signs of all the squares in a row;
(2) change the signs of all the squares in a column;
(3) change the signs of all the squares in any diagonal (you can, for example, change the sign of only a corner square because itm can be seen as a diagonal; when I say "diagonal" I'm not referring just to the 2 large diagonals of the square);

Can we reach the situation when all the squares have the + sign by these 3 moves and startibg from the position I've mentioned?

A move changes the signs of 2 squares on the perimeter of the big square. Therefore the final configuration cannot be reached.

Nice and easy, isn't it? :D That's why I like Russian problems (many of them are MUCH tougher, but they are all based one some nice ideas).

Am I wrong or Iandrei's argument doesn't work (because of the diagonal with only a corner square)?
But, it may be used for the set of the 8 squares which are on the perimeter of the big square and which are not corners. So....

Pierre.

That's true. In fact, this was my original idea, but I thought iandrei's idea worked, and since it could be explained in fewer words, I didn't mention my solution :D

Errr I'm very sorry, I knew the problem beforehand (I solved it some time ago) and I didn't read the whole text :D

No problem, I was just checking before moving it into the solved section...

Pierre.


dude karthic (sorry if ms) i was looking through a cross country forum and i saw someone mention that he is really good. i also remember that you said he got into mit.

do you know anything about how fast he runs in cross country or track and field?

Ooh... Do tell!  I'm a runner too!

By the way, drunner2007, are you in high school still, and if so, what team do you run for?

i'm a senior in hs. well i ran (but not this year) for columbus high school and northside high school. my pr's are,

5k: 16:05
1mile: 4:49
2mile: 10:32

i didn't run this year because i had to do several other stressful things, and spend more time at the local college, which obviously conflicts with after school activities sometimes. i am looking at running in college though (caltech or mit probably).

That is a pretty freaking good mile time???

[quote="drunner2007"]
i didn't run this year because i had to do several other stressful things[/quote]

That's too bad, particularly senior year  :o.  We had one kid at our school that was really fast, but he had to quit his senior year (the year I joined).

I run for Etowah.  Right now, my PRs are:

5K: 17:06
2 mile: 10:13
1 mile: 4:44
800m: 2:07

Yea, he's the same Vibin. I'll ask him tomorrow  :P But yea he got into MIT, etc. and pretty much pwnz at life (2400 on SAT, etc.)

[quote="Potato Theory"]I found him at GATFXC.
1600m - [url]http://mirror.milesplit.com/meet/results/43805[/url]

By the way, did you know Ben Hubers ran a 4:07 mile today?![/quote]

sweet. ben hubers placed 3rd at nike outdoor nationals last year in the 1mile. and arguably, he could have placed top 5 in the 2mile.

since you're in atlanta, have you ever ran against him?

your times are very fast  :) . you run the 800 too?

yeah i've heard lots about vidim since i found out that he got into mit. 2400 is impressive, even though a lot of people here like to talk bad about the sat. i also saw that he was the only person in ga to be a semifinalist in the team selection thing for the u.s. physics team.

[quote="Potato Theory"]Yeah, in cross-country at the regional meet.  I started out fast, so I was in first for about a half mile before he breezed past me, and I literally mean breezed.  It looked like he was jogging, and I was trying an all out-sprint to stay with him...  :roll:

I just started running the 800, but the 2 mile is the main event I focus on.

And yes, I think 2400 is incredibly impressive.  It's higher than anyone I know...[/quote]

that's cool. the only time i saw ben hubers up close was at the state cross country meet during my sophomore year (2 years ago). our team was arriving late, and the championship race was just finishing. he was running neck-to-neck with this guy named sam that was a senior, but sam edged him out and he placed 2nd.
they were gunning it though (like low to mid 15:00s). and it was like the first time he was really noted as an elite runner. i think i remember hearing that he was in the 17:00s at the beginning of that season.

so what happened to that guy on your team that quite senior year? i don't know of many other people that have done that, so i'm curious.

It was weird.  I think he quit because he didn't like the training methods (I think he was pretty far ahead of the rest of the team), but at that time, there were different coaches than I have now.  His name was Vinny something; not sure if you've heard of him.  I think he ran mid to low 16s.  Currently, he's a professional wakeboarder, so I guess he is still good at sports...

And also with Ben, at the McEachern meet the other day, he was there, but having a practice instead of running.  He was doing 1000s at 68 pace, followed by a 30 second break, then 600s at a 62 pace, and he kept getting faster each time!

By the way, is there a fast freshman named Charlie Sparks at your school?

so he was just practicing while everyone else was in the meet lol?
yeah charlie goes to my school but i hardly know him. literally, i have never been in a same class or been around him after school or anything, so i havn't talked to him yet. he's pretty fast, but really fast if you consider that he's a freshman. he won state in 3a during cross country, but the guy from westminster that was expected to win tripped near the beginning of the race. most people tell me that he's pretty cocky, but i don't know for sure.

how did you hear about him? from state or something?

[quote="Potato Theory"]Well, there was this post on GATFXC by him asking who all the fast freshmen were...

Also, I saw his name on the 3200m run for under 10:00 this year at a 9:57, and I noticed that he was the only freshman.  A freshman at our school who joined this year caught up to me and actually beat me by .8 seconds for a 10:13.16, which is why I was interested in freshman times.[/quote]

yeah those are the type of freshman that have the potential to be in contention for footlocker nationals.


geometry
probability
AMC
AMC 10
number theory

I have not yet taken the AMC 10, and I am wondering what will be important for the test. What formulas, or concepts are very important to score relatively high? What areas are most important (combinatorics, algebra, geometry, or number theory)? And how well should I know each of these areas, so that I can study (also any recommended books or other resources).

All four areas that you listed are important for the AMC 10, but given what you have told me, I think that you should study Combinatorics and Number Theory the most - try [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]Introduction to Counting and Probability[/url] and [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206]Introduction to Number Theory[/url] textbooks.

Additionally, try the [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=1]Intro. CP[/url] and [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=2]Intro. NT[/url] classes.

Also, this is a very common question you are asking. Just look through this forum and you'll find plenty of help with AMC10 preparation.  :)


Find the biggest natural number $n$ with the propriety

$\frac{1}{\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{4+\sqrt{15}}}+...+\frac{1}{\sqrt{2n+\sqrt{4n^2-1}}} < 3\sqrt{2}$

I'm lost at how you get this:

${\sqrt{4k+2\sqrt{(2k+1)(2k-1)}}=\sqrt{2k+1}+\sqrt{2k-1}}$


Find all continuous functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x\plus{}f(y\plus{}f(z)))\equal{}f(x)\plus{}f(f(y))\plus{}f(f(f(z)))\]
for all reals $ x,y$, and $z$.

[quote="discredit"]Find all continuous functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that

$ f(x \plus{} f(y \plus{} f(z))) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(f(z)))$

for all reals $ x,y,z$[/quote]

[hide="My solution"]
Let $ P(x,y,z)$ be the assertion $ f(x \plus{} f(y \plus{} f(z))) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(f(z)))$

$ f(x) \equal{} 0$ is obviously a solution. So we'll now look for non-zero solutions.

1) non-zero solutions are such that $ f(f(a))\neq 0$ for some real $ a$
Suppose $ f(f(x)) \equal{} 0$ $ \forall x$ then $ P(x,y,f(u))$ $ \implies$ $ f(x \plus{} f(y)) \equal{} f(x)$ $ \forall x,y$
So, we have a periodic continuous fonction, with period as little as we want, and so a constant, and so $ f(x) \equal{} 0$ $ \forall x$
Q.E.D.

2) non-zero solutions are surjective.
If $ f(f(f(0)))\neq 0$, then $ P(x,f(0),0)$ $ \implies$ $ f(x \plus{} u) \equal{} f(x) \plus{} v$, with $ u \equal{} f(2f(0))$ and $ v \equal{} 2f(f(f(0)))\neq 0$
If $ f(f(f(0))) \equal{} 0$, then, using 1) above, $ \exists a$ such that $ f(f(a))\neq 0$. Then $ P(x,a,0)$ $ \implies$ $ f(x \plus{} u) \equal{} f(x) \plus{} v$ with $ u \equal{} f(a \plus{} f(0))$ and $ v \equal{} f(f(a))\neq 0$

From $ f(x \plus{} u) \equal{} f(x) \plus{} v$ and $ v\neq 0$, we get $ f(nu) \equal{} f(0) \plus{} nv$ $ \forall n\in \mathbb Z$ and so, since $ f(x)$ is continuous, $ f(x)$ is surjective.
Q.E.D.

3) The only non-zero solutions are $ f(x) \equal{} ux$ and $ f(x) \equal{} \minus{} 2x \plus{} b$
Let $ b \equal{} f(0)$
Since $ f(x)$ is surjective, the initial assertion may be written $ f(x \plus{} f(y \plus{} z)) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(z))$
Setting $ x \equal{} 0$, we get $ f(f(y \plus{} z)) \equal{} f(0) \plus{} f(f(y)) \plus{} f(f(z))$ and so, since $ f(f(x))$ is continuous, a classical Cauchy equation and : $ f(f(x)) \equal{} ax \minus{} b$
So $ f(b) \equal{} \minus{} b$

Putting this in $ f(x \plus{} f(y \plus{} z)) \equal{} f(x) \plus{} f(f(y)) \plus{} f(f(z))$, we get $ f(x \plus{} f(y \plus{} z)) \equal{} f(x) \plus{} a(y \plus{} z) \minus{} 2b$ and so :
$ Q(x,y)$ : $ f(x \plus{} f(y)) \equal{} f(x) \plus{} ay \minus{} 2b$
Notice here that $ a\neq 0$. Else $ f(x \plus{} f(y)) \equal{} f(x) \minus{} 2b$ and so (taking $ y \equal{} f^{[ \minus{} 1]}(0)$) $ b \equal{} 0$ and $ f(f(x)) \equal{} 0$ $ \forall x$ and so (using point 1) $ f(x) \equal{} 0$

$ Q(ax,b)$ $ \implies$ $ f(ax \minus{} b) \equal{} f(ax) \plus{} ab \minus{} 2b$
$ Q(0,f(x))$ $ \implies$ $ f(ax \minus{} b) \equal{} af(x) \minus{} b$
Equating the two RHS implies $ af(x) \equal{} f(ax) \plus{} (a \minus{} 1)b$

$ Q(ax,f(y))$ $ \implies$ $ f(ax \plus{} ay \minus{} b) \equal{} f(ax) \plus{} af(y) \minus{} 2b$ $ \equal{} f(ax) \plus{} f(ay) \plus{} (a \minus{} 3)b$
$ Q(ax \plus{} ay,b)$ $ \implies$ $ f(ax \plus{} ay \minus{} b)) \equal{} f(ax \plus{} by) \plus{} (a \minus{} 2)b$
Equating the two RHS implies $ f(ax \plus{} ay) \equal{} f(ax) \plus{} f(ay) \minus{} b$ and so, since $ a\neq 0$, $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ with $ g(x) \equal{} f(x) \minus{} b$
Hence $ f(x) \equal{} ux \plus{} b$

Pluging this back in $ P(x,y,z)$, we get $ (u \plus{} 2)b \equal{} 0$
Q.E.D.

So the solutions are :
=====================
$ f(x) \equal{} 0$
$ f(x) \equal{} ux$
$ f(x) \equal{} \minus{} 2x \plus{} b$

[/hide]

[color=#f00]Also posted [url=https://artofproblemsolving.com/community/c6h219275p1216243]here[/url]. -Amir[/color]


A ball thrown from an 96 foot high building with a vertical
velocity of 80 feet per second will take how long to hit the ground?

[hide]
gravity: approx. $-32 ft/s^{2}$

height of an object at time $t$ thrown at angle $a$ at velocity $v$ at height $h$ with gravity $g$: $y=\frac{1}{2}gt^{2}+v_{0}\sin{(a)}t+h_{0}$

$a=90$
$v_{0}=80$
$h_{0}=96$

and we want $y$ to be zero (when it hits the ground)

$0=-16t^{2}+80t+96$
$0=t^{2}-5t-6$
$0=(t-6)(t+1)$
$t=6$ or $t=-1$

$\boxed{t=6\text{ seconds}}$ because the time cannot be negative.
[/hide]


AMC
USA(J)MO
USAMO
MATHCOUNTS
USAMTS
AMC 12
AMC 10

Well folks, I'm here eating a bowl of life cereal and browsing the forums when I came upon an interesting topic.

When in a math competition which is better to attain, speed in solving a math problem or accuracy? That is, doing all the problems very fast and then checking them over, or working slowly and cautiously on each and every problem?

What do you guys think? What are your strategies?

Peace,
IW.OAV

I voted "Mix and Match" because it depends on a lot of factors.

First, if the question is which do i LIKE to do,  it would have to be speed.  Doing problems quickly is more fun and more interesting, because you go through more problems (although some may argue that studying a problem for hours is more interesting because you learn more about specific things.  I have to agree with that one, too.  But speed is FUN.)

But I do better when I am slow and careful.  In fact, I have to say I'm pretty accurate when I want to be.

But which is better?  That's a tough one.  On USAMO, obviously going slowly but thoroughly is important.  In MATHCOUNTS, you have to be ridiculously fast (and I know someone will quote and contradict that last statement).  In AMC, I'd say the first 16 or so problems are easy and should be gone through quickly; the last 9 should go slowly (this is an estimate... maybe the 24th one will be easy while the 3rd is hard...but in general).  On USAMTS, you can take as long as you want (to a certain extent - but I don't think anyone works on a USAMTS problem from its release date to its deadline, nonstop).

But like I said somewhere, I'm gonna race through AMC next year and then take my time checking the answers when I'm done.  How's that for a mix and match?

Going through the AMC's really fast and then checking it may work for some, but probably not me.  I'm a slow problem solver (which is why my symbol in math jams is a turtle  8-) ).  Also, the speed must be based on accuracy.  You can have all the speed you want but will score a zero if you don't have accuracy (or even knowing how to do the problem).  I also voted "mix and match" because when you have a problem that you've seen many times before and you're confident that you can solve it with ease, use speed on it!  However, if you had, say, 10 minutes to do the last 5 problems of an AMC, then you should probably tackle about 3 of them unless you're really good....

-interesting_move

[quote]Going through the AMC's really fast and then checking it may work for some, but probably not me. I'm a slow problem solver. [/quote]

Same deal here. And, although rushing through the first 16 problems is not too bad, I tried that this year and made a couple of silly mistakes (note that I did not have time in the end to check back, but that's just me).

Fierytycoon

See, but I've always been really slow but really accurate, and so I haven't tried rushing through the first 16.  I don't realize how badly it'll go for me.  And don't bother telling me it's a bad idea, because I'm too fixed on this idea (or too stupid) to change my mind.

Just kidding!  But my definition of "rushing through the first 16" may be different from yours.  Rushing for me may be 3 mins for the first problem!

For me the AMC12 boiled down to flipping coins on the last dozen questions... My brain shuts down after long continuous thinking sessions...

Thanks to my lucky penny I was one away from a perfect score...

Thank you Mr. Penny.

Hey, I could sell those on Ebay.... I'd become a mogul.
IW.OAV

no questions now on how you thought of your username

*laughs* just kidding mr. man your one cool cat

i'm used to taking each problem and working them nice and slow, unless its those ridiculusly easy ones. it worked for me at mathcounts, where my definition of slow is 1 min per problem. for the amc 10 i did it slow but finished early but i was 2 lazy 2 check and thus got plenty wrong from careless mistakes, so that method failed me. overall, ithink the best way is to work slow but make sure ( or at least try to make sure) that ur correct.

If your definition of slow is 1 min per problem, then what is your definition of fast?

-interesting_move

probably like 30 seconds or less. some sprints are ridiculosly easy, but its also easy to make silly mistakes, which a lot of people (like me) do.

of course for harder competitions like the AMC series, fast would be like 3 minutes and slow would be 5, by my standards.

I don't think your times for the AMC make much sense. There are 75 minutes for 25 problems, so you should spend an average of just under 3 minutes on each problem. (From past experience, you don't want to work until the last second, as finishing early gives you a little leeway for relaxing and going over what you think you might have done wrong.) I'd say fast for an AMC problem is <= 2 minutes, and slow shouldn't be more than 4.

Yeah, that makes sense.  I wish I could work at an average rate of 2 min/problem on the AMC.  But some problems take like 5 minutes for me.

-interesting_move

eh. no wonder i don't do so well... it might just be that i miscalculated the times, cause i don't bother to look at the clock after every problem.


[quote="zhaobin"]I find it is not hard.
Is there something with me?
I can prove:$a^4c^3+b^4a^3+c^4b^3 \ge abc(a^2b^2+b^2c^2+c^2a^2)$
Am I wrong?[/quote]
Dear $zhaobin$ you are right,here is another way to prove this ineq.
\[ a^4c^3+b^4a^3+c^4b^3 \ge abc(a^2b^2+b^2c^2+c^2a^2) \]
<=>
\[ \frac{a^4c^4}{abc^2}+\frac{b^4a^4}{bca^2}+\frac{c^4b^4}{acb^2}\geq a^2b^2+b^2c^2+c^2a^2 \]
but  we have that
 \[ \frac{a^4c^4}{abc^2}+\frac{b^4a^4}{bca^2}+\frac{c^4b^4}{acb^2}\geq \frac{(a^2b^2+b^2c^2+c^2a^2)^2}{abc(a+b+c)}\geq a^2b^2+b^2c^2+c^2a^2 \]
 ;)

[quote="toetoe"]
It doesn't seems hard the inequality .I think it should posted in

an other forum.
[/quote]
It doesn't seem hard the inequality .I think it should be posted in

an other forum  :D .


What is the smallest possible product of a two-digit number and a three-digit number obtained from five distinct digits?

A three digit number times a two digit number is algebraically represented by \[ abc \cdot de\] Since $ a$ is the hundreds digit is worth the most, it must be $ 1$. It may not be $ 0$ because it said  [i]three digit number[/i]. Next, $ b$ must be $ 0$ and $ d$ must be $ 2$ because these are least numbers possible.

However, for the last one, we have casework for when $ c,e \equal{} (4,3),(3,4)$. \[ 103 \cdot 24 \equal{} 2472\]\[ 104 \cdot 23 \equal{} 2392\]

Thus, our answer is $ \boxed{2392}$.

Can you tell us how to do that algebraically?

Um, I was just looking for a counter-example because your logic just didn't seem right, and I just stumbled across that.  :blush:

I think the logic behind this is that the two-digit number should be made as small as possible. That is why we made it 10 and set the three-digit number equal to 234, giving us 2340. Unfortunately, I don't know why this works.


If 2^5.a^b=25ab, where 25ab is a 4-digit integer, find a and b.

Thank you in advance for any help you can provide.

Techin

THis isn't the correct way, but I think it works.

You know that $ a^b\equal{}\frac{25ab}{32}$. From here, guess and check. Guessing $ a^b\equal{}79,80,81$. Only $ 81$ is a power of a number, so $ a\equal{}9$, and $ b\equal{}2$.

abcd=cd^2 where abcd and cd are 4-digit and 2-digit integers respectively.  Find abcd.

Thank you,
Techin

Do you mean the product abcd or are a,b,c,d digits?

Are you sure that you are not forgetting some restriction? If not, there are several answers.

a, b, c, d are digits of a 4-digit interger.  c, d are digits of a 2-digit integer.  The question does not have any other restrictions.

Techin

Thank you, fish.

How do you figure out teh answer?

All I did was to limit my search of ab from 31 to 99 (because the square of the number has to be a 4-digit number).  Then I further limited my search to a 2-digit number ending with a  1, 5, 6.  Using this strategy, I hadto try 15 numbers (31, 35, 36,... 71, 75, and 76) before I got the answer.

Is there a better way?

Techin

The square of any number ending in 5 will end in 25, so you can rule that out.

Expanding $ (10a\plus{}1)^2$, you get $ 100a^2\plus{}20a\plus{}1$, so the last digit is 1, and the 3rd digit is (the last digit of) 2a, which does not equal a, unless a is 0, which is impossible. So the number doesn't end in 1.

Expand $ (10a\plus{}6)^2$ to get $ 100a^2\plus{}120a\plus{}36$. So that last digit is 6, and the 3rd digit is (3 + the last digit of 2a (which is the second to last digit of 120a)) mod 10.
Solve. (You can just check the values of a from 1 to 9). You also know that a is odd, because 3+2a is odd. Only $ a\equal{}7$ works. You can probably narrow it down further, but it would be easier to use brute force at this point.

Therefore, the number is $ \boxed{76}$


quadratics
Vieta
algebra
number theory proposed
number theory

thats that why i asked. seanie don't said nothing, so first time i believed that it must be solved in $R$.

[hide="hint "]$x^{2}+y^{2}\geq 2xy \Rightarrow z\geq 2+\frac{1}{xy}$ [/hide]
the hint is for N, i guess we use the same ideea for Z, too.

posted many and many times!
this equation have infinite solutions, but z = 3.

Hint:
1) if x>y and (x,y) is one solution, $(\frac{y^{2}+1}{x},y)$ also is.

[quote="Seanie_G"]Solve:

$x^{2}+y^{2}+1 = xyz$[/quote]
Must be $z\ge 3$ (If x,y,z are natural).
For all z>2 we have Pell's equation $t^{2}-ay^{2}=-4$, were $t=2x-yz,a=z^{2}-4$. Solution is $t+y\sqrt a =(t_{0}+y_{0}\sqrt a )(c+d\sqrt a )^{n}$.
Because $-4=-2^{2}$ and equation $t^{2}-ay^{2}=-1$ had solution if $(\frac{-1}{a})=1$. Therefore we had solutions for all z, suth that $z^{2}-4$ had not prime divisor $p=3(mod 4)$.

uhh i dont think thats right. give me a solution triple where z isn't equal to 3.

If z>3, and $3\not |z$, then $3|z^{2}-4$ don't work. Therefore $z=3m$. For example work $z=15$ (give $z^{2}-4=13*17$).

so what is one solution triple where z not equal to 3? if z can equal 15 as u claim, what solutions (x,y) are there

Yes z=15 don't work (x,y - rational). Exactly one solution is $x_{0}=\frac{2884}{5}, y_{0}=\frac{194}{5}$, oter solutions are $x+y\sqrt{221}=(x_{0}+y_{0}\sqrt{221})(c+d\sqrt{221})^{n}$.

hallo,
cant we use the fact that the equation $x^{2}+y^{2}+z^{2}=kxyz$ has solutions only for $k=1$ or $k=3$ (infinite). From here we have $z=3$ or $z=1$. So we have two cases:
1. $z=1$ we have: $x^{2}-xy+y^{2}+1=0$ a quadratic equation in $x$, witch has no solutions.
2. $z=3$ we denote the discriminant, which has to be a square, by $q^{2}=5y^{2}-4$. so we have to solve $q^{2}-5y^{2}=-4$.

Actually the equation $x^{2}+y^{2}+z^{2}=kxyz$ is called Markov's equation and it can be easily done by Fermat's Method of Infinite Descend.
For the equation $x^{2}-3xy+y^{2}+1=0$ we have that if $x=y$ it follows that $x=y=1.$ So if $x>y$ then by Viete we see that the equation has other positive integer root $x'<y$ and so all the solutions are to cosecutive terms of the sequence $a_{n}$ for which $a_{1}=a_{2}=1,a_{n+1}=3a_{n}-a_{n-1}.$
Also very interesting is the equation $q^{2}-5q^{2}=-4$ in positive integers. It's easily seen that $p\geq 2q$ and so with the substitution $p=x-2y,q=y$ it can be reduced to the equation $x^{2}-xy-y^{2}=-1.$ The last equation is a part of IMO 81, problem 3. The solutions can be expressed by Fibonacci's sequence.

yeah infinite descent was the way i went about it. and then u could just solve the recurrence u get from vieta's to get a closed form expression. @Rust: x,y,z are positive integers


Find the x-and y-intercepts of each polynomial function below.

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

(2) y = -(x^2 + (1/2)) (x - 1)^2 times (x + 1) ( x - 2)

the x intercepts are in the form of $(a,0)$ where $a$ is a factor $(x-a)$ of the equation. convert each linear factor to the form $(x-a)$, and you'd have every x intercept.
the y intercept is just what you get when u substitute 0 for $x$.

This reply I understand.

Please, do not use textbook terminology when answering questions.  Try to make it as easy for students as you possibly can.

Okay?

If I want math jargon, I simply read my math books, which usualy takes me no where.

Tell me, which two equations did you equate?

why did you set them equal to each other? theyre two separate graphs
[quote="Interval"]Find the x-and y-intercepts of [b]each[/b] polynomial function below.

[/quote]
for (1) it's simply $(1,0),(-1,0),(2,0),(0,-1)$
for (2) it's just $(1,0),(-1,0),(2,0),(0,1)$

[quote="Interval"]Find the [b]x-and y-intercepts[/b][/quote]
May be Interval didn't write correctly the problem.

what do you mean? it's just asking for the 3 x-intercepts and the y-intercept for each function.

My error...

But the secret of the problem are doing the things that it dont say to you....

er....

sorry

[quote="ProtestanT"]for (1) it's simply $(1,0),(-1,0),(2,0),(0,-1)$
for (2) it's just $(1,0),(-1,0),(2,0),(0,1)$[/quote]

If it was for that, then I don't think Interval have posted an easy problem like this  :ninja:

I posted two different equations.  The books tells me to find the x-and y-intercept for each equation separately.


Given the above figure, with a square and equillateral triangle of sides 2 inside the circle, find the radius of the circle.

My method: I used the circle equation, assuming h=0:
$x^{2}+(y-k)^{2}=r^{2}$
with the points $(1,0);(-1,0);$ and $(0, 2+\sqrt{3})$. The first two points are the bottom corners of the square and the third is the top of the triangle. Using elimination and substitution I found
$r=2$.

Are there more elegent geometric strategies?

See the attached diagram.

$\triangle OAE\cong\triangle ODE$ (the corresponding sides equal), hence $\angle OAE=\angle ODE$, but because of the symmetry, $\angle ODE={1\over 2}\angle CDE=30^\circ$, which gives $\angle OAB=90^\circ-\angle OAE=60^\circ$. Therefore, $\triangle OAB$ is an isosceles with the base angle of $60^\circ$, i.e. it's equilateral, giving $r=a$.


[hide]Draw an altitude to find two 3-4-5 triangles, so the area is 12.[/hide]

[hide="solution"]The other leg must have a length of 5. By the pythagorean theorum, the height of the triangle is 4. Therefore,the area is 24/2=12[/hide]

[hide]Well, if one leg is 5, then the other has to be as well. Using Pythagoras or Heron's we obtain the answer [b]12[/b].[/hide]

Like mtms said, you can do the problem in two ways:

Using Phthagoras, or Heron's Formula, we get the answer to be 12

[quote="Arvind_sn"]Find the area of an isosceles triangle where the base is 6 and one of the legs is 5. Explain your answer :!:  :!:  :!:[/quote]

[hide]

As other posters have demonstrated, the area is $\boxed{12}$.

I'll explain the solution with more detail:

The triangle has sides $5,5$ and $6$. Hence, we can find the area applying the Heron's Formula.

If you don't know it, you can find the altitude of the triangle corresponding to the base, which has length $6$. We use pithagoras: $h= \sqrt{5^{2}-3^{2}}=4$. Now, the area is $\frac{1}{2}*b*h=\frac{1}{2}*6*4=\boxed{12}$.

Please, remember that there are young posters in this section, so remember to explain your solution with detail. [/hide]

[quote="Totally Zealous"][hide]Find that it's two 3-4-5 triangles. The area is 12.[/hide]

The name isoceles triangle, or so I've heard, comes from a story.
A young madien called Celes (SA-leece) was kidnapped by an evil villian and imprisoned on top of a tower. Celes knows about a secret and a young prince needs to know it. Celes didn't get a chance to tell him before she got kidnapped. The tower is 1000 high and bulit to top of mountains and jagged rocks. The ground is a big wide forest in a circle of a radius of 1000 feet, with the center being the tower. The forest is filled with traps and feral beasts and poisonous plants. The young prince stands at the edge of the forest and holds up a telescope pointed right at Celes. Celes wrote the secret on a sheet of paper and held it facing the young prince. The young prince saw the message and Celes. The triangle formed, from the young prince, Celes, and the base of the tower formed and isocles triangle. The distance from the base of the triangle to Celes and from the base of the triangle to the young prince was the same. When the young prince quickly told his friend about the secret, he quickly said "I-saw-Celes" but it was so fast it sounded like isoceles. The name Isoceles is applied from then on to every triangle that has 2 equal sides.[/quote]
 :) Nice story.

wait...
didnt it say isoceles? not right?  :maybe:

Well, you're right, the triangle is right ans isosceles. But well, only the matter that the triangle formed was isosceles was the one that reamined in time.

[quote="Arvind_sn"]Find the area of an isosceles triangle where the base is 6 and one of the legs is 5. Explain your answer :!:  :!:  :!:[/quote]

[hide]5,5,6

so the semiperimeter is 8
(8*3*3*2)^1/2
(16*9)^1/2
12
[/hide]


linear algebra
matrix
calculus
calculus computations

a)   Find the Wronskian of 1,x,x^2,..., x^(n-1)

b)  Show that

w(g(x)f1(x), g(x) f2(x),...,g(x)fn(x))= [g(x)^n * w(f1(x) , f2(x),..., fn(x))

c) Show that e^rx , xe^(rx), x^2 * e^(rx),...,x^(n-1) * e^(rx) are linearly independent on (-infinite,infinite)

[hide="Part a)"]$ W(1,x,\cdots,x^{n-1})=\begin{\vmatrix}
1&x&x^{2}&\cdots&x^{n-1}\\
0&1&2x&\cdots&(n-1)x^{n-2}\\
0&0&2&\cdots&(n-1)(n-2)x^{n-3}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&(n-1)!
\end{vmatrix}$
Notice anything about that matrix?[/hide]

So you just multiply the diagonals and you should be able to find the answer for the a).  How about for b) and c)?

The Wronskian of $ g(x) f_1(x), g(x) f_2(x), ...$ can be written as the product of the Wronskian of $ f_1(x), f_2(x), ...$ times a matrix whose determinant you should be able to find.


induction
number theory unsolved
number theory

Let the sum of the first $ n$ primes be denoted by $ S_n$. Prove that for any positive integer $ n$, there exists a perfect square between $ S_n$ and $ S_{n\plus{}1}$.

The link provided above points to a problem that states that $S_n$ and $S_{n+1}$ are never simultaneously perfect squares. The same poster links here from [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=433250&p=2448232#p2448232[/url], where the same statement that $S_n$ and $S_{n+1}$ are never simultaneously perfect squares is made. So we are in the presence of a circular argument, where in fact the very statement that there exists a perfect square (strictly) between $S_n$ and $S_{n+1}$ has never been proved.

Take the largest square $a^2 \leq S_n$. Assume $a^2 \leq S_n < S_{n+1} \leq (a+1)^2$. This means $p_{n+1} = S_{n+1} - S_n \leq (a+1)^2 - a^2 = 2a+1$. This leads to $(p_{n+1} - 1)^2 \leq 4a^2 \leq 4S_n$.
But we can prove by induction that $(p_{n+1} - 1)^2 > 4S_n$ for $n\geq 4$; this shows the above assumption is incorrect, hence $S_n < (a+1)^2 < S_{n+1}$. For the small values $n=1,2,3$ we can easily check that $2 = S_1 < 2^2 < S_2 = 5$, $5 = S_2 < 3^2 < S_3 = 10$, $10 = S_3 < 4^2 < S_4 = 17$.

Indeed, for $n=4$ we have $(p_{n+1} - 1)^2 = (11 - 1)^2 = 100 > 68 = 4(2+3+5+7)= 4S_n$. Assume this holds up to some $n\geq 4$. We have $(p_{n+2} - 1)^2 \geq  ((p_{n+1} + 2) - 1)^2 = ((p_{n+1} - 1) + 2)^2 = (p_{n+1}-1)^2 + 4(p_{n+1}-1) + 4$. By the induction hypothesis we have $(p_{n+1}-1)^2 + 4(p_{n+1}-1) + 4 > 4S_n + 4p_{n+1} = 4S_{n+1}$, thus $(p_{n+2} - 1)^2 > 4S_{n+1}$, and we are done.


calculus function logarithms induction inequalities trigonometry geometry

In 1980s, just after the Cultural Revolution, math, physics, chemistry are taken seriously. During that period, exams are hard. Some math olympiad problems from Russia or other countries appeared in National Matriculation. But many people complained that mathematics is too hard for them. So after 1990s, the difficulty is reduced. Linear Algebra and pre-calculus are removed from the textbooks and the exam problems became easier. But now it changed a little. pre-calculus is back, but no linear algebra. After I know the math courses in Romania from Vornicu and the courses in Germany from orl, I was sore and disappointed -- Too little math is taught in high school! "National Matriculation" is very important to every high school student in China. What university a student is able to enter depends on his score of the examination. So nearly everyone spend a large time on revising the schoolwork -- this nearly takes a whole year! All the courses have finished, and what they do is just "doing problems"! A student in the last grade of high school are often required to do over thousands of problems on each subject. The homework is so heavy that many of them don't have enough time to finish it. This is absolutely a waste of time! If we have some new courses in the last grade(not have courses of revision), I'm sure single-variable calculus can be completely taught in high school, also some linear algebra.

the following are math courses in high school (textbook in 1980s)

Book I
  Sets and Simple Logic
  Mapping and Functions
  Power Functions, Exponential Functions, Logarithms
  Sequences and Math Induction
  Inequalities Proofs

Book II
  Plane Vectors(mainly definitions and dot product)
  Trignometry (including inverse trignometric functions)

Book III
  Solid Geometry (lines, planes, polyhedrons, revolving solids,

Book IV
  Analytical Geometry (lines, circles, quadratic curves, translation and rotation of axes)
  Polar Coordinates
  Parametric Equations
  Determinants and System of Linear Equations
  Complex Numbers

Book V
  Permutations, Combinations, Binomial Theorem
  Probability
  Limits
  Derivatives
  Integrals

Well, I always think that Chinese students are very hardworking
I don't know if they're doing too much, since a IMO-1990 team member told us that each IMO Chinese team member had to do 2000 questions each year at that time
Then again Chinese students got great results in IMO... Being an IMO-2002 gold medalist I barely got a silver in CMO 2003  :(

Yes, they always need to do a lot of problems. Before they go to the training camp, they themselves may do over thousand a bit easier problems each year. I've heard a contestant who did five thousand problems in a month! You need to do many to gain experiences, CMO problems often don't have some important ideas, but careful discussions and various of skills.

[quote="liyi"]Before they go to the training camp, they themselves may do over thousand a bit easier problems each year. [/quote]

Well its not that bad if you think about it. That's about 3 problems per day, and I'm sure many of us do more than that per day on average.

ok. it was my typo mistake...
it may be one month ...

[quote="Fierytycoon"]Well its not that bad if you think about it. That's about 3 problems per day, and I'm sure many of us do more than that per day on average.[/quote]

So USA also scores high in IMO. :)

Yeah, 1 month sounds better. 36 problems a day every day would be tough if you're not serious about training.

[quote="PiDeltaPhi"]Depends what problems you are doing  :) .3 IMO questions/day is quite a lot[/quote]
it also depends what kind of IMO questions / day you are doing ... if you are doing only problem 1 or 4 type of problems, that is not to hard. On the other hand solving a couple of 3-s and 6-s from IMO-s is a really hard job ... :)

i don't what kind of problems they did. but i think they are not very difficult, at least easier than CMO, maybe just like the additional problem of the first stage contest in China.
They didn't think write down the whole solution or considered every details of the problem, once they found the key to a problem, they turned to another.

Chinese students don't study Number theory, which is an essential for many of you.
but they study Solid geo, which is said to have 'died out' in another thread here.

i think instead of sticking to the book, chinese education system should be a little more open and free.. some of those old traditions are not good.. i think china could be improving alot faster if it focused more on practical issues.. not really related to this thread but i just needed to voice my concerns

There is another thing:
During the 1950s to the 1980s, Chinese students should only pay 10 years to learn primary school and high school(5 years for primary school, and 5 years for high school), but now, it has changed to 12 years, 6 years for primary school, and 6 years for high school. And the students attending primary school's ages is much older, most of them attend primary school when 6 or 7 years old, but my father, who learn primary school during the time 1960s, went to primary school when 4 years old(maybe he was too young, but most of the other student also attended primary school when 5 years old).

[quote="liyi"]Yes, they always need to do a lot of problems. Before they go to the training camp, they themselves may do over thousand a bit easier problems each year. I've heard a contestant who did five [b]thousand[/b] problems in a month! You need to do many to gain experiences, CMO problems often don't have some important ideas, but careful discussions and various of skills.[/quote]
it's like posting on forums
someone can post 5000 posts in a month, and some can only post 5. ;)

P.S.Math Induction isn't taught in Chinese high school now, IMO.


function
limit
real analysis
real analysis unsolved

Problem:
For a function $ f: \mathbb{R}^n \rightarrow \mathbb{R}$ and a natural number, show that

$ \lim_{x \to 0} \frac {f(x)}{\left\| x \right\|^{m \plus{} 1}} \equal{} 0$ implies $ \lim_{x \to 0} \frac {f(x)}{\left\| x \right\|^{m}} \equal{} 0$

but the converse is not true
========================
I know that the norm of x denoted by $ \left\| x \right\|$ can be written as $ \sqrt {\sum_{i \equal{} 1}^{n}x_{i}^2} \equal{} \sqrt {x_{1}^2 \plus{} ... \plus{} x_{n}^2}$.  Then we can rewrite $ \left\| x \right\|^{m \plus{} 1} \equal{} \left\| x \right\|^m \left\| x \right\|$

Then the limit can be rewritten as

$ \lim_{x \to 0} \frac {f(x)}{\left\| x \right\|^{m \plus{} 1}} \equal{} \lim_{x \to 0} \frac {f(x)}{\left\| x \right\|^{m}} \frac {1}{\left\| x \right\|}$ (1)

By the multiplication of limits,

Let $ g(x) \equal{} \frac {f(x)}{\left\| x \right\|^m}$ and $ h(x) \equal{} \frac {1}{\left\| x \right\|}$ from (1)

 then

$ \lim_{x \to 0} [g(x)h(x)] \equal{} AB \equal{} 0$

So, for $ \lim_{x \to 0} g(x) \equal{} A$ and $ \lim_{x \to 0} h(x) \equal{} B$ , A or B must be equal to 0. I am lost from here.

I believe I say that $ \lim_{x \to 0} \frac {f(x)}{\left\| x \right\|^{m}} \equal{} 0$

Also, I can not think of an example for the converse.

Much appreciated for your time and help.

$ f(x)\equal{}\|x\|^m$ satisfies $ \lim_{x\to 0} f(x)\|x\|^{\minus{}m\minus{}1}\equal{}0$ but not $ \lim_{x\to 0} f(x)\|x\|^{\minus{}m}\equal{}0$.

To prove the forward direction, you need to write

$ \lim_{\|x\|\to0}\,\frac{f(x)}{\|x\|^m}\equal{} \lim_{\|x\|\to 0}\,\|x\|\cdot\frac{f(x)}{\|x\|^{m\plus{}1}}$

You got a little tangled on the direction of the logic there.

Both this and a previous question are questions about size. To think like an analyst, you pretty much have to constantly think about size - what's big, what's small, what's bigger than what else. The only thing that matters to this question is that as $ \|x\|\to0,$ $ \|x\|^{m\plus{}1}$ tends to zero faster than $ \|x\|^m$ does. And you can find rates of decay that lie in between these two, if you want to.

[quote="blahblahblah"]$ f(x) \equal{} \|x\|^m$ satisfies $ \lim_{x\to 0} f(x)\|x\|^{ \minus{} m \minus{} 1} \equal{} 0$ but not $ \lim_{x\to 0} f(x)\|x\|^{ \minus{} m} \equal{} 0$.[/quote]

Hello Blah, I understand that if $ f(x) \equal{} \|x\|^m$ then $ \lim_{x\to 0} f(x)\|x\|^{ \minus{} m} \equal{} 1 \neq 0$, but I do not understand how this is a counterexample.

Since $ \lim_{x \to 0} \frac {f(x)}{\| x \|^{m \plus{} 1}} \equal{} 0 \implies \lim_{x \to 0} \frac {f(x)}{\| x \|^m} \equal{} 0$

then wouldn't the converse be

 $ \lim_{x \to 0} \frac {f(x)}{\| x \|^{m}} \equal{} 0 \implies \lim_{x \to 0} \frac {f(x)}{\| x \|^{m \plus{} 1}} \equal{} 0$

So, to find a counterexample the function, then shouldn't $ \lim_{x \to 0} \frac {f(x)}{\| x \|^{m}} \equal{} 0$ but $ \lim_{x \to 0} \frac {f(x)}{\| x \|^{m \plus{} 1}} \neq 0$?

Thank you Kent for the hint. I was able to prove it.

You're right - blahblahblah got what he said a little tangled. Try $ f(x) \equal{} \|x\|^{m \plus{} 1}$ as your counterexample to the converse.


Can you make all the numbers from $ 1$ to $ 20$ using only $ 4$s (for example, $ 8\equal{}4\plus{}4$)?

yes,
for one, 4 divided by 4=1
for two, (4+4)/4
for three, (4+4+4)/4
for four, just four
for five, (4+4+4+4+4)/4
and so on

You can make all integers with only 4s since you didn't specify a number of 4s:
1=4/4
2=4/4+4/4
3=4/4+4/4+4/4
4=4/4+4/4+4/4+4/4
and so on...
Even though there are easier ways.

I know that 1-100 can be made using everything taught in Algebra 1 plus cotangent, natural log, and floor value. See [url=http://www.wheels.org/math/44s.html]here[/url] for a list of 1-116, or just Google "Four fours".


This is a particular case (p=5/2) of the conjecture from page 447:
If $x_{1},x_{2},...,x_{n}$ are positive numbers such that $x_{1}+x_{2}+...+x_{n}\geq n$, then for any $p>1$ the inequality holds
$\frac 1{x_{1}^{p}+x_{2}+...+x_{n}}+\frac 1{x_{1}+x_{2}^{p}+...+x_{n}}+...+\frac 1{x_{1}+x_{2}+...+x_{n}^{p}}\leq 1$.


Are you sure you don't have a typo in the second term there?

you have as your second term x*(x - 2)/(x - 2).  Is one of those minus signs supposed to be a plus sign?

[hide]

if you factor the top and bottom of the left side all the terms cancel out and the right side also cancels out to just x.  This leaves x=1 but this doesn't work because it is a restricted value because if put back into the equation it produces 0 in the denominator of the left side.  No solution[/hide]

Ah.. I had 1 too, but I wasn't sure so....
Okay, thanks..


function
Alcumus
videos
arithmetic sequence

[hide="Hint"]If you subtract 1 from every number, the size of the list hasn't changed.  This new list is easier to work with than the original.[/hide]

Or if it is just that type of problem in general than the formula for the number of numbers in a sequence is :

$ \frac{highest\minus{}lowest}{interval}\plus{}1$

Heres another way to do this:
[hide="hint"]Find a function for this set such that f(1)=1, then solve for when f(x)=2008[/hide]

[hide="answer"] In the function $ f(x)\equal{}mx\plus{}b$,m can be found by
$ m\equal{} \frac{x_2\minus{}x_1}{y_2\minus{}y_1}$
$ \equal{}(4\minus{}1)/1$
$ \equal{}3$
and $ b$ can be found by
$ f(x)\equal{}3x\plus{}b$
$ f(1)\equal{}3\plus{}b$
$ 1\equal{}3\plus{}b$
 $ b\equal{}\minus{}2$.
Now substitute 2008 for f(x).
$ 2008\equal{}3x\minus{}2$
$ x\equal{}670$

so the sequence starts at $ f(1)$ and ends at $ f(670)$, meaning there are 670 numbers.
This process looks long and tedious, but its actually really fast and easy.[/hide]

I certainly don't want to figure out the equation of a line if I don't have to.  My method is quick and easy.  The number of elements in $ \{0,3,6,\ldots,2007\}$ is simply the number of time 3 divides 2007, plus $ \{0\}$, which is 669+1=670.

You can add 2 to each element, then divide each element by 3. So

\[ \{1, 4, 7, \dots, 2005, 2008 \}\]

becomes

\[ \{3, 6, 9, \dots, 2007, 2010 \}\]

and then, after dividing by 3, you get:

\[ \{1, 2, 3, \dots, 669, 670 \}\]

Note how the number of elements in the set did not change. Now, this is very easy to count. There are $ \boxed{670}$ elements in the set.

$ U_n\equal{}a\plus{}(n\minus{}1)b$
$ 2008\equal{}1\plus{}(n\minus{}1)3$
$ 3(n\minus{}1)\equal{}2008\minus{}1$
$ 3(n\minus{}1)\equal{}2007$
$ n\minus{}1\equal{}\frac{2007}{3}$
$ n\minus{}1\equal{}669$
$ n\equal{}670$

I am sorry, if my answer's wrong

Er, the above posts say 670, so why not?

It's correct, he just didn't check that all of the previous posts came up with the same answer.

There is a formula to solve this. That formula is...

Last number minus First Number over the interval of the numbers...and then you add 1.

For example...
Lets say you were asked to find how many numbers were in the set of 1, 5, 9......., 65, 69.
You would take the Last Number (69) and the First Number (1) and then subtract...yielding 69-1 = 68.
Okay. Then you would divide by the interval, which in this case is 4.
68/4 = 17.
Now we have 17, but we are not done yet! We still have to add 1!
17 + 1 = 18. So there are 18 numbers in the set.

Proof that this works?
Just list all the numbers in the set until you reach the last number!
1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69.
Now count the numbers and see for yourself.

Hope this helped!  :lol:

[quote="Lowgear1"]Can somebody tell me the easiest way to do this?
How many numbers are in the list.
1,4,7................2008
reply[/quote]

Obviously, this is an arithmetic sequence.
To find the number of terms in an arithmetic sequence, with first term a1, and last term a(n), the formula is...
((a(n)-a1)/increment size)+1 ( the a(n) and a(1) are subscripts.)
In this question, you get ((2008-1)/3)+1=(2007/3)+1=669+1=[b]670[/b]

[quote="Thunder365"]Heres another way to do this:
[hide="hint"]Find a function for this set such that f(1)=1, then solve for when f(x)=2008[/hide]

[hide="answer"] In the function $ f(x) \equal{} mx \plus{} b$,m can be found by
$ m \equal{} \frac {x_2 \minus{} x_1}{y_2 \minus{} y_1}$
$ \equal{} (4 \minus{} 1)/1$
$ \equal{} 3$
and $ b$ can be found by
$ f(x) \equal{} 3x \plus{} b$
$ f(1) \equal{} 3 \plus{} b$
$ 1 \equal{} 3 \plus{} b$
 $ b \equal{} \minus{} 2$.
Now substitute 2008 for f(x).
$ 2008 \equal{} 3x \minus{} 2$
$ x \equal{} 670$

so the sequence starts at $ f(1)$ and ends at $ f(670)$, meaning there are 670 numbers.
This process looks long and tedious, but its actually really fast and easy.[/hide][/quote]

i've never seen this used this way before. thanks for showing.

[quote="Lowgear1"]Can somebody tell me the easiest way to do this?
How many numbers are in the list.
1,4,7................2008
reply[/quote]

Add 2 to each element of the list. Then the list is $ 3,6,9,...,2010$ and divide the list by 3 to get $ 2010\div3\equal{}670$ elements.

Here is a really good Alcumus video that has [b]very[/b] similar questions.

[url]http://www.artofproblemsolving.com/Alcumus/Videos/ShowVideo.php?video_id=33&m=cfb44dec9aeca81036175a5ea412e73b[/url]

Look at the first post, it's at [b]July[/b].  :(

Look. I didn't check. I just saw "New Posts made" and it wasn't locked.

Also, this is one of the basic C&P things so you may want to study that in more detail in Intro to C&P. Very good book, helped a lot.


The number $n$ has $2002$ digits, all of which are 2. What is the greatest common divisor of $n$ and $1111$? Please show your work.

[hide="partial solution"]For a number to be divisible by 1111, the sum of the 1, 5, 9th, etc. digits, 2, 6, 10th, etc. digits, 3, 7, 11th, etc. digits, and 4, 8, 12th, etc. digits have to be the same. For a number with all the same digits to be divisible by 1111, it must have a number of digits that is a multiple of 4. 2002 isn't divisible by 4, so that numbe isn't divisible by 1111.
Using alternating digit divisors, that number is divisible by 11, but I'm not sure whether it is divisible by 101 (since 1111 = 11*101). My answer is currently 11, but it may be 101. [/hide]

[hide]$1111=11\cdot 101$. Therefore, $11$ and $101$ are the only options, since we can easily show that $n$ is not divisible by $1111$, since the number of digits in $n$ is not a multiple of four, which is required to be divisible by $1111$. $11$ will obviously work, but $101$ is larger, so we must next see if that is a viable option. If we take the first few digits of $n$ and start dividing by $101$, we come up with the pattern $22002200...$. This pattern goes in pairs, and $n$ has $1111$ pairs of numbers, meaning that, if $101$ is a factor of $n$, then the last two numbers have to be the $22$ pair, which, after some thought, we see is correct.

Therefore, the answer is $101$.[/hide]

Let's think about this...

[hide="Comments for 236 and 197"]Didn't 236! assert that the number was divisible by 11?

Didn't JF197 assert that the number was not divisible by 1111?

Didn't JF197 assert that the number was divisible by 101?

But $1111=11\times 101$.  If the number is divisible by both $11$ and $101$, then the number must also be divisible by $1111$.  Therefore, at least one of the above assertions must be false.[/hide]

well its obviously divisible by 11, and not 1111, so it is 11

i found if we take the first few digits of n such that n is divisible by 4, only then will n be divisible by 101. (for example 2222 is divisible by 101 but 222222 is not) since 2002 is not divisble by 4, i got 11. strange pattern...


geometry
circumcircle
incenter
analytic geometry
geometric transformation
reflection
ratio

Recently, I have been trying to find something nice about cevian triangles to prove the conjecture 3 (see below) which I posted in ML a few months ago; and luckily I did find some and hopefully they are helpful. Although they are conjectures, I checked them thoroughly with computer and I believe they are true.

Given $ABC$ a triangle and $P$ a point in its plane. Let $A'B'C'$ be the Cevian triangle and $A_{1}B_{1}C_{1}$ be the pedal triangle of $P$ with respect to $ABC$.

Then the four pedal circles of $A, B, C, P$ with respect to $BCP$, $CPA$, $ABP$, $ABC$ are concurrent at a point $F$. This point $F$ is the nine-point center of the quadrilateral $ABCP$ (i.e. the common point of the four nine-point circles of $ABC$, $BCP$, $CAP$, $ABP$ ( :) This part is easy :) )

[b] Conjecture 1 [/b] Moreover, $F$ lies on the circumcircle of $A'B'C'$

Actually this conjecture is pretty easy if $ABCP$ is cyclic. An immediate and interesting corollary is that the Feurbach point $F$ lies on the circumcircle passing through the feet of the internal bisectors of $ABC$  (A proof of this fact can be found at http://forumgeom.fau.edu/FG2001volume1/FG200117index.html )

[b] Conjecture 2 [/b] Let $H$ be the orthocenter of $A'B'C'$. Then the anti-Steiner point $S$ of the line $HP$ with respect to $A'B'C'$ lies on the circumcircle of $A_{1}B_{1}C_{1}$. Note that this point is not $F$.

Finally, the last conjecture is the problem I posted here a few months ago

[b] Conjecture 3 [/b] Let $F, F_{a}, F_{b}, F_{c}$ be the Feurbach points of $ABC$. Then the Steiner line of $F$ with respect to $F_{a}F_{b}F_{c}$ passes through the incenter of similitude of the incircle and the nine-point circle of $ABC$.

If you know about any of these conjectures (which I think the chance is very high because you can't imagine how surprised you will be when you draw them out), I would be appreciative if you could post the solutions or the sources here.

Hi dear Treegoner!

First I would like to say "Thank you!" for your very nice conjectures; I hope they all will prove to be true  :) .

I tried to prove your first conjecture and actually I got some results :blush: .

So, I am using the complex numbers theory in the Gemoetry.

I seted $P$ to be the center of complex plane,that is the point with coordinate $0$, and after some (a bit boring :oops: ) work with liner,and not only liner,equations I found that
\[f=\frac{n}{2m}\]
where
\[n=\sum{\overline{c^{2}}(\overline{a}b-\overline{b}a)}\]
\[m=\sum{\overline{c^{2}}(\overline{a}b^{2}-\overline{b}a^{2})}.\]

And it is easy to show that

\[a'=\frac{a(\overline{b}c-\overline{c}b)}{(\overline{b}a-\overline{a}b)+(\overline{a}c-\overline{c}a)}\]
\[b'=\frac{b(\overline{c}a-\overline{a}c)}{(\overline{c}b-\overline{b}c)+(\overline{b}a-\overline{a}b)}\]
\[c'=\frac{c(\overline{a}b-\overline{b}a)}{(\overline{a}c-\overline{c}a)+(\overline{c}b-\overline{b}c)}\]

So :) ,the only thing that is left to prove the cojecture is to check(but I havn't checked it :blush: ) that
\[t=\overline{t}\]
where
\[t=\frac{(a'-c')(b'-f)}{(a'-f)(b'-c')}.\]

I havn't tried the other two conjectures because I am not sure what are  "anti-Steiner point " or "Steiner line" or "similitude". If you could tell me what mean those expretions, I might find some interesting results for you :) .

It seems that the coordinate of $F$ is quite nice, even though I don't think it is nice enough for your final computation. Anyway, a proof by using complex number is pretty cool.  :)

Recently, Quang Tuan Bui gave me a reference to Hyacinthos message 12403 in which you will find some more interesting concurrent circles at that same point. According to Pierre Ehrmann in his reply, this theorem was proved by Poncelet during his time in prison.

About the anti-Steiner point, let me tell you briefly. It is well known that given a point $K$ in the circumcircle of $ABC$. Then the Steiner line which connects the reflections of $K$ through the sides of the triangle will pass through the orthocenter $H$ of $ABC$. Conversely, given each line $l$ passing through $H$, one can find a unique point $K$ such that $l$ is the Steiner line of $K$. This point $K$ is called the anti-Steiner point of $l$.

Given two circles (O) and (O') with radii $R$ and $r$ respectively. Then the incenter of similitude of these two circles is the point defined uniquely by the algebraic ratio: $\frac{PO}{PO'}=-\frac{R}{r}$.


Given $a,b,c,d$ be non-negative real numbers satisfy the condition $a+b+c+d=4$. Find the minimum of the expression
\[ P=(1+a^2)(1+b^2)(1+c^2)(1+d^2). \]

Very nice inequality. I think it may have been suggested from APMO inequality
\[ (2+a^2)(2+b^2)(2+c^2)\geq 9(ab+bc+ca). \]
For this one, my friend, Nghi, has a generalization.

Great, Vasc.

I see that you and I have many similar idea.  :D

I deduce in from the general result, replace $1$ by $k \ge 1$. It's not from APMO. Althought, it's very nice and hard problem. The result is $\frac{10^4}{9^3}$.

I will think this question. I think it's very hard.

Now I hope someone can be solved my first problem?

[quote="Vasc"]Hung, I solved some years ago the following problem:
If $a_1,a_2,...,a_n$ are non-negative numbers such that $a_1+a_2+...+a_n=s$ and $s \geq n$, then
$P(a_1,a_2,...,a_n) \geq P(r,r,...,r,\frac 1{r})$ with $(n-1)r+\frac 1{r}=s$,
where
$P=(1+a_1^2)(1+a_2^2)...(1+a_n^2)$.  ;)[/quote]

Dear Vasc,

According to your result then min = 16. But by Hung then the min = 10000/729 = 13.717...

Can you explain me the situation?

I think for first problem, Mixing variables works.

Namdung

I forgot to specify $r<1$. For $n=4$ we have $r=\frac 1{3}$.

OK, after VASC or Namdung can post a solution for this one, I hope any other can solve it. The hints is mixing variable method, but more difficult.

Hung,  does SMV Method work in the case $s<n$? Have you a result ?

Sorry, VASC. I didn't try to solve it. But I will think about it this night. (Oh, Champion League  :( ). SMV only get the best effect for 4-variable problems, but may be I.G.I or U.M.V work.  :)

yes UMV (or AC) method works!
$(1+a^2)(1+b^2)\geq (1+(\frac{a+b}{2})^2)^2$ if $a+b\le2$
so three variables should be equal.
for the case s<n I don't think we can use this method...

No, UMV or AC only work for one numbers or more is equal $0$, but in my problem, it's not true. We need to use a stronger method than classical mixing varibales.

hey $ln 1+x^2$ has only one inflection point ( x=1 ) so to reach the minimum n-1 variables should be equal and smaller than 1
when $s\geq n$ the minimum happens as Vasc said at $P(r,r,...,s-(n-1)r=\frac{1}{r})$ but when $s<n$ $(n-1)r+\frac{1}{r}=s$ has no solution <1 so the minimum happens at $P(a,a,.....a=\frac{s}{n})$ or$P(0,0,0,....,s)$
Right?

I think if $s\le n$ is harder thank expect. If $n=4$ then the minium attack when $a=b=c=1/3, d=3$ so in this case, if we let $s\to 4$ then the minimum will attack when $a,b,c\to 1/3$ and $d\to 3$.

I only solve this one in case $n=4$.

well after n-1 variables are equal to each case it's not difficult to find the minimum right? :)

Of course, but how to proof that the minimum attack when $n-1$ variables equal. Notice that for some problem it's not true.

of course we can say that n-1 variables should be equal :
read the first sentence of my post above $\uparrow$

Actually, I don't know thoughtly about RCF and LCF. But I think your solution is not clear enough to understand, why to tranform it to the case some variables equal or $n-1$ variables equal?

Thank to VASC, too if you help me this work?

I think Albanian Eangle used the IPT theorem of Myth
[quote="Albanian Eagle"]hey $ln 1+x^2$ has only one inflection point ( x=1 ) so to reach the minimum n-1 variables should be equal [/quote]  and it is correct . :D
I think the problem is solved

[quote="Albanian Eagle"] but when $s<n$, $(n-1)r+\frac{1}{r}=s$ has no solution <1 so the minimum happens at $P(a,a,.....a=\frac{s}{n})$ or$P(0,0,0,....,s)$
Right?[/quote] It is not very right, because the equation $(n-1)r+\frac{1}{r}=s$ can have a solution <1.
The general answer is:
$P(a_1,a_2,...,a_n) \geq \min\{P(s/n,s/n,...,s/n), P(r,r,...,r,\frac 1{r})\}$
with $(n-1)r+\frac 1{r}=s$ and ${r<1}$.

Althouht I dont like this solution, but I must accept it. My solution is only vaild for $n=4$ and for the problem to find the best constant for which
\[ f(a_1,a_2,...,a_n) \ge f(s/n,s/n,...,s/n) \]
is true for all $a_1,a_2,...,a_n$. But indeed, it only used primary mathematics.

sorry
the equation actually can have no solution >1 :D

the equation has no solution when $s^2<4(n-1)$ so we will look the extremities
that's why I said that we can work this case by case...

[quote="Namdung"][quote="hungkhtn"]For -variable problem, mixing variables is very good and usful method. But for 4 more more variables problem, this work seem to be no good, because we only make two variables equal. It's the reason why we need to improve this method for problem which involve more than 3 variables. I often use a method call SMV. (Stronger Mixing Variables).[/quote]

Mixing variables doesnt not means mixing two varibales, but n-1 variables into 1 also.

SMV I think we should call: Strong Mixing Variables.[/quote]

Yes, sure. I wil think about it again,


calculus
integration
modular arithmetic
induction
algebra
binomial theorem

Prove that $ \dfrac {7^n\minus{}1} {6}$ is an integer for positive integral values of $ n$.

[hide]We must prove that $ 6$ divides $ 7^n\minus{}1$. Thus, we examine the numerator in modulo $ 6$. We see that:
\[ 7^n\minus{}1 \equiv \left(6\plus{}1\right)^n\minus{}1 \pmod{6}\]If we expand the base using the binomial theorem, we'll see that all the terms are equivalent to $ 0$ modulo $ 6$ except the last term, which is $ 1^n\equal{}1$. Then, $ 1\minus{}1 \equiv 0 \pmod{6}$, and we are done.[/hide]

[hide]Factor $ 7^n\minus{}1$. $ (7\minus{}1)(7^{n\minus{}1}\plus{}7^{n\minus{}2}\plus{}\hdots\plus{}1)$.
This is obviously divisible by 6.[/hide]

[hide="Sure"] $ \frac{7 \minus{} 1}{6} \equal{} 1$ and $ \frac{7^{n\plus{}1} \minus{} 1}{6} \minus{} \frac{7^n \minus{} 1}{6} \equal{} 7^n$.  All this argument really does, in my opinion, is motivate mathcrazed's argument, which is direct. [/hide]

[hide="Induction for the lazy"]Base Case: For $ n \equal{} 1$, duh. :P
Assume $ 6|7^k \minus{} 1$.
Then $ 7^{k \plus{} 1} \minus{} 1 \equal{} 7(7^k \minus{} 1) \plus{} 6$ is divisible by $ 6$.
Yay.[/hide]
edit: Grr, never mind.


Let a quadrilateral $ ABCD $ . A point $ O $ satisfies : $ <AOK = <COP $ . $  K = AB $   :cap: $  CD ,  P = AD $  :cap: $  BC $ . Prove that : the bisector of $ <AOC $ and the bisector of $ <BOD $ are perpendicular .

Your problem follows from another one posted by Peter Scholze in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15103]another thread[/url]. In fact, if one uses directed angles, your angle relation < AOK = < COP becomes < AOK + < COP = 180 (mod 180), and then, using Peter Scholze's problem b), one gets < AOB + < COD = 180 (mod 180). Now, using this equation, it's a very easy angle chase to prove that the angle bisectors of the angles < AOC and < BOD are perpendicular (I will not show it here since the proof is very situation dependent).

  Darij


quadratics
number theory unsolved
number theory

Consider the form Q = (1, 0, 13). Show that if p is a prime not equal to 13 represented by Q, then $\frac{-13}p = +1$ and $p \equiv 1 mod4$

We have
 $p\equiv x^2+13y^2\equiv x^2+y^2\equiv 0, 1 or 2 (/mod4)$,
but p can't be even (2 obviously doesn't satisfy the given condition so $p\equiv 1 (/mod4)$
We have $x^2\equiv -13y^2 (\mod p)$, so
$(x/y)^2\equiv -13(\mod p)$, so $\left(\frac{-13}p\right) =1$.


algebra polynomial calculus derivative inequalities

let f(x) be a polynomial of degree n with real coefficients and such that f(x) :ge: 0 for every real number x.  Show that f(x) + f'(x) + f''(x) + ... + f(n)(x) :ge: 0 for all real x.  (f(k)(x) denotes the kth derivative of f(x).)

i didn't create the problem, and that is how it was worded, but i took n :ge: 0 or f(x) = 0

Yes, I believe I've seen this somewhere before and that it's supposed to be for any n.

Let g(x) = f(x) + f'(x) + f''(x) + ... + f^(n)(x)

A few facts about f and g :
. g'(x) = g(x) - f(x).
. g as same degree as f.
. g(x) -> +oo near -+oo so there exists x0 such as g(x) >= g(x0) for all x  (classical calculus result).

But then g'(x0) = 0 (Fermat's theorem) or g(x0) = f(x0) >= 0.

[quote="Ravi B"]Belenos, you are supposed to prove the inequality for all x, not just for one x.[/quote]

I don't understand ????

Since x0 was chosen as the minimum for g, g(x) >= g(x0) >= 0.

Oops.  Sorry, I missed that last line.  Nice proof.


Find the least value of the expression:
[b]H[/b]=$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
where x,y,z are positive number satisfying
$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}=2006$

lots of mistakes on the solution I attempted.

my main goal was to show that h had a lower bound when assuming $x\leqq\leqq$ but the other inequalities I came up with did not help me at all. tough problem

There are a bunch of problems:
Since you showed that H is less than or equal to that number, you could only have demonstrated that it is a maximum, not a minumum.  In fact, though, you have problems at the step where you substituted for x and z -- some switched inequalities, and it's not clear how you could substitute in any value for x (since x can be as close to 0 as you desire).  Also, for any such problem, there are two halves to showing an extremum: (1) show that more extreme values can't be reached and (2) show the value that you are claiming can be reached.  You haven't done (2) yet.

[hide]
WLOG let $x\geq y\geq z$. Then $y+z\leq z+x\leq x+y$ and $\frac{x^2}{y+z}\geq \frac{y^2}{x+z}\geq \frac{z^2}{x+y}$. Applying Chebyshev to these two sequences ordered in opposite directions gives

\[ 3(x^2+y^2+z^2)\leq \left( \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y} \right) (y+z+z+x+x+y) \]
\[ \Rightarrow \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\geq \frac{3(x^2+y^2+z^2)}{2(x+y+z)} \]

By Cauchy, \[ \left( \sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\right)^2 \] \[ \leq (1^2+1^2+1^2)(x^2+y^2+y^2+z^2+z^2+x^2) \] which gives $x^2+y^2+z^2\geq \frac{1003\cdot 2006}{3}$.

Also, by RMS-AM, $\sqrt{x^2+y^2}\geq \frac{x+y}{\sqrt{2}}$ and similar inequalities hold for the other two cases. Adding the three cases gives $x+y+z\leq 1003\sqrt{2}$.

From the inequalities on $x^2+y^2+z^2$ and $x+y+z$, it follows that $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\geq \frac{1003}{\sqrt{2}}$. There is equality when $x=y=z=\frac{1003\sqrt{2}}{3}$, so the minimum is $\boxed{\frac{1003}{\sqrt{2}}}$.
[/hide]


Let AD, BE, and CF be the bisectors of the interior angles of triangle ABC, with D, E, and F lying on the perimeter. If angle EDF is 90 degrees, determine all possible values of angle BAC.

It has been posted many times. For example:
http://www.mathlinks.ro/Forum/viewtopic.php?t=138151
http://www.mathlinks.ro/Forum/viewtopic.php?t=114998


Post your score if you got above 135 on the AMC 10A 2007 and you are from Indiana.

Sorry if it seemed like bragging, but I'm in 8th grade and I just want to know where I stand compared to other people. This is my first year making it to AIME and I'm pretty excited.

Well, most of the really smart people in like 9th and 10th grade in Indiana took the 12 instead of the 10, but 135 is a pretty good score on the amc's.  I think Richard Ni, the guy who won Indiana mathcoutns got a 138, as well as Jeffrey Shen (5th)-for other 8th grade comparison..I cant name a 9th or 10th grader that beat you though..

When the stats come out, you can always look at them..Im betting that the 8th and 7th graders did better than the 9ith and 10th graders..cause I know like atleast 5 that got 130+

No, Jeffery got 114 on AMC 10a, he got 138 on 10b (I looked up the scores on the sheet our teacher gave back) and Richard  got 138 on 10B

I got 132 on 10a and 114 on 12b (after misbubbling one (#6), misreading another one (#17), and losing my mind on the one with the particle..1/9 would never even be a logical choice...)

ya, i made 1 careless mistake on the 10a (forgot i was using complimentary counting), but i'm not really complaining. It was waaaay better than my last year's score! :)

[quote="randomdragoon"]I know an 8th grader who got a 150 on the AMC 10A ...
[hide=" :ninja: "]His name is Carl.[/hide]

But he's not from Indiana, so I guess this makes this post spam[/quote]

I think this whole thread is more or less spam, so no worries.

I'm a 1st grader from Indiana; I got 150 on the AMC 10A.

I had a 150 on the 10B when I was in preschool, yo. And then I got abducted by aliens and they sucked all the math out of my brain so I had to start over learning everything.  :roll:

Yo I got a 42 at HMMT but I guess I'm from North Carolina and I'm not talking about the AMC 10A, so darn.

I got #7 wrong in Target Round of MATHCOUNTS (states) in 2003.

Please refrain from this type of posting in the future (both the original poster and the people who responded)


For the short answers on the Caltech Application, if it asks you to list 3 books, should you then explain them? I just want to be on the safe side. Thanks

Do you mean the Caltech app? I know of no such question on the Stanford app (I earlied to Stanford, by the way).

Sorry, I corrected it. I was talking about CALTECh, so thanx for correcting me.

I would also like an opinion about this... :blush:


Solve --

$ 3x^{log_{5}2}\plus{}2^{log_{5}x}\equal{} 64$

The second log is log base 5 of x.

[hide="hint"]
Use the equality $  a^{\log_{b}c}\equal{}c^{\log_{b}a}$
[/hide]

[quote="karthik_k"]Solve --

$ 3x^{log_{5}2}+2^{log_{5}x}= 64$

The second log is log base 5 of x.[/quote]

Try \log, instead of log.

[hide="Solution"]Let $ y = 2^{\log_{5}x}$. Then, taking $ \log_{x}$ of both sides,

$ \log_{x}y =\log_{x}2^{\log_{5}x}$

$ \log_{x}y =\log_{5}x\cdot\log_{x}2$

By the Chain Rules, that simplifies to:

$ \log_{x}y =\log_{5}{2}$

Raising both sides to the x-th power, we get:

$ y = x^{\log_{5}{2}}$

From our definition of $ y$,

$ 2^{\log_{5}{x}}= x^{\log_{5}{2}}$

Finally, we can solve the problem. Note that:

\begin{eqnarray*}64&=&3 x^{\log_{5}{2}}+2^{\log_{5}{x}}\\ &=& 3 x^{\log_{5}{2}}+x^{\log_{5}{2}}\\ &=& 4x^{\log_{5}{2}}\end{eqnarray*}

So:

$ x^{\log_{5}{2}}= 16$

Since $ 2^{\log_{5}{x}}= x^{\log_{5}{2}}$,

$ 2^{\log_{5}{x}}= 16$

$ \log_{5}{x}= 4$

$ x = 5^{4}=\boxed{625}$[/hide]

[quote="erdogankerem123"]What are the Chain Rules?[/quote]
[[Logarithm]]

I think that the solution is much easier if we use that x^{\log_{5}2}=2^{\log_{5}x} (using the hint of red_dog)

This is easy to prove. We take \log_{5} on both sides and we have
\log_{5}(x^{\log_{5}2})=\log_{5}(2^{\log_{5}x})  which becomes
(\log_{5}x)*(\log_{5}2)=(\log_{5}2)*(\log_{5}x) which is obvious

So in the exercise:
3x^{log_{5}2}+2^{log_{5}x}= 3*(2^{\log_{5}x})+2^{log_{5}x}=4*(2^{\log_{5}x}) which is equal to 64, so we get that 2^{\log_{5}x}=16 or 2^{\log_{5}x}=2^4
Finally we get \log_{5}x=4 and from here we take that x=5^4

Gregg, in order to use Latex you must write the mathematical symbology between dollar signs (the symbol above the 4 on your keyboard).

[quote="gregg"]I think that the solution is much easier if we use that $ x^{\log_{5}2}\equal{} 2^{\log_{5}x}$ (using the hint of red_dog)

This is easy to prove. We take $ \log_{5}$ on both sides and we have
$ \log_{5}(x^{\log_{5}2}) \equal{}\log_{5}(2^{\log_{5}x})$  which becomes
$ (\log_{5}x)\cdot(\log_{5}2) \equal{} (\log_{5}2)\cdot(\log_{5}x)$ which is obvious

So in the exercise:
$ 3x^{\log_{5}2}\plus{}2^{\log_{5}x}\equal{} 3\cdot(2^{\log_{5}x})\plus{}2^{\log_{5}x}\equal{} 4\cdot(2^{\log_{5}x})$which is equal to $ 64$, so we get that $ 2^{\log_{5}x}\equal{} 16$ or $ 2^{\log_{5}x}\equal{} 2^{4}$
Finally we get $ \log_{5}x \equal{} 4$ and from here we take that $ x \equal{} 5^{4}$[/quote]

Submit to the power of the dollar signs  :ninja:  :D

Thanks for the advice!!!!!!!!

$ I think that the solution is much easier if we use that x^{\log_{5}2}\equal{}2^{\log_{5}x}(using the hint of red_{d}og) This is easy to prove. We take\log_{5}on both sides and we have\log_{5}(x^{\log_{5}2})\equal{}\log_{5}(2^{\log_{5}x}) which becomes (\log_{5}x)*(\log_{5}2)\equal{}(\log_{5}2)*(\log_{5}x) which is obvious So in the exercise: 3x^{log_{5}2}\plus{}2^{log_{5}x}\equal{} 3*(2^{\log_{5}x})\plus{}2^{log_{5}x}\equal{}4*(2^{\log_{5}x}) which is equal to 64, so we get that 2^{\log_{5}x}\equal{}16 or 2^{\log_{5}x}\equal{}2^{4}Finally we get\log_{5}x\equal{}4 and from here we take that x\equal{}5^{4}$

I think that the solution is much easier if we use that $ x^{\log_{5}2}\equal{}2^{\log_{5}x}$ (using the hint of red_dog)

This is easy to prove. We take $ \log_{5}$ on both sides and we have
$ \log_{5}(x^{\log_{5}2})\equal{}\log_{5}(2^{\log_{5}x})$which becomes
$ (\log_{5}x)*(\log_{5}2)\equal{}(\log_{5}2)*(\log_{5}x)$ which is obvious

So in the exercise:
$ 3x^{log_{5}2}\plus{}2^{log_{5}x}\equal{} 3*(2^{\log_{5}x})\plus{}2^{log_{5}x}\equal{}4*(2^{\log_{5}x})$ which is equal to 64, so we get that $ 2^{\log_{5}x}\equal{}16$ or $ 2^{\log_{5}x}\equal{}2^{4}$
Finally we get $ \log_{5}x\equal{}4$ and from here we take that $ x\equal{}5^{4}$


Let $a,b,c$ be positive reals. Find the biggest real $C$, such that the following inequality holds:

$\frac{(2a+b+c)(a+2b+c)(a+b+2c)}{abc}\geq C$

Unless I missed a subtle aspect of it, this question is too easy for Pre-Olympiad.
$a+a+b+c \geq 4 \sqrt[4]{a^{2}bc}$ and multiply.
So $C \geq 64$.
Fill in $a=b=c$, and get that $64 \geq C$.
So $C=64$

I think Jan's proof is valid, but just worded a little awkwardly. Basically, from the AM-GM we know that $C$ is at least $64$ (since we know that $C = 64$ works). Then, given an equality case, we know that $C$ is at most $64$ (since we know that if $C > 64$ then the case $a = b = c$ fails). It follows that $C = 64$ is the best constant.

Come on. I think it's clear what I mean.
rem, you are clever enough to understand my solution...

We know that $C$ is the largest bound, so when i prove that $64$ is a bound, it follows that $C \geq 64$, yes?

This question belongs in Intermediate or Getting Started.

OMG I was being stupid, sorry. I did not see that you skipped the reasoning that $C$ is the largest possible constant, so I thought you had a mistake. Sorry. :blush:

So, I'm sorry that this problem was that easy. I had a proof which was quite more difficult and I didn't think of AM-GM.


In a game of bridge, 4 players are dealt with 13 cards each

What is the probability of one of the players has all same suit?

There are $\frac{52!}{13!^{4}}$ ways to deal the cards. Now lets use inlusion and exclusion to count the positive cases:

$P_{1}$:a given player has same suit: 4 ways to choose the suit,
$\frac{39!}{13!^{3}}$ ways to deliver the other cards

$P_{2}$:two given players have same suit: 4*3=12 ways to chose the suits $\frac{26!}{13!^{2}}$ ways to deliver the other cards.

$P_{3}$:3 given players have same suit:  4*3*2=24 ways to chose the suits, 1 way to deliver the rest

$P_{4}$: all the players have same suit: $P_{4}=P_{3}$.

The number of positive cases is ${4 \choose 1}P_{1}-{4 \choose 2}P_{2}+{4 \choose 3}P_{3}-{4 \choose 4}P_{4}=16\frac{39!}{13!^{3}}-72\frac{26!}{13!^{2}}+4\cdot 24-24$, the probability is ths number divided by $\frac{52!}{13!^{4}}$. I'm afraid the result won't be nice.

It can be approximately computed as $\frac{16}{{52\choose 13}}=2.5\cdot 10^{-11}$ (which is an upper approximation).