I need some help with the following problems. Some hints on how to solve them will suffice, of course.

Let $V=M_{nn}(\mathbb{R})$ be the vector space of square matrix with real coefficients and $E \in V$ a fixed matrix.

1.
a) Prove that the subset $W=\{B\in V|BE=EB\}$ is a subspace of $V$.

I'm not sure how to prove the closure under addition.

b) Let $n=2$ and $E=[(0,1),(0,0)]$. What should be the form of a matrix $\in V$ in order to appartain to $W$.

Do you confirm this form $[(0,a),(a,0)]$ ?

c) Find a basis of $W$ when $n=2$ and $E=[(0,1),(0,0)]$.

d) What if $E=I$, the identity matrix?

2.
a) Determine for which values of $k$ the following vectors are linearly independent: $v_{1}=[1,-1,3], v_{2}=[2,k,6],v_{3}=[k-1,k,6]$.

Is is true that for no $k$?

b) Extract a basis $\{v_{1},v_{2},v_{3}\}$ for the subspace $W=Span\{v_{1},v_{2},v_{2}\}$.

Thanks to anybody who will help me.

[quote="Stewie"]
I'm not sure how to prove the closure under addition.
[/quote]
If $X\in W$ and $Y\in W$ then $(X+Y)E=XE+YE=EX+EY=E(X+Y)$ therefore $X+Y\in W$.

I've solved all of them, except for 2. b).

Problem 2 b) seems kind of weird.

[quote]b) Extract a basis $\{v_{1},v_{2},v_{3}\}$ for the subspace $W=Span(v_{1},v_{2},v_{2})$.[/quote]

Is it right that $v_{2}$ is repeated?  Or is that a typo.

If vectors $v$ and $w$ are linearly independent, then $\{v,w\}$ is a basis for $span(v,w)$.  If they aren't, and $v$ is non-zero, then $v$ alone is a basis for $span(v,w)$.  There are some other possibilities to consider too.

Do we have enough information to answer this question?  Are the $v$'s supposed to be from earlier?

That was a typo, sorry. The three vectors are the ones from 2.a).

Is it true that those vector are lin. ind. for $k\ne0,1$ ?

Well, it's easy enough to check -- just slap them into a matrix and do Gaussian elimination (being careful not to divide by 0).  (The answer to your question, incidentally, is no: the vectors are independent at $k = 0, 1$, but there are two other values at which they are dependent.)


[hide]Volume of a Cylinder= $\pi r^2h$

Thus, when you substitute the given values, you have:
$785 \ in^3=10\pi r^2$

Divde both sides by $10\pi$:
$\frac{785}{10\pi}=r^2$

$r=\sqrt{\frac{785}{10\pi}}$[/hide]

[quote="LalaMcla"][size=150][i]A cylinder is 10 inches tall. if the volume of the cylinder is 785 cubic inches[/i], [b]what is the radius of the cylinder???[/b][/size] :?:  :?:[/quote]

[size=150]whats with the super size font?[/size] :?

[hide]$V=\pi r^2\cdot h$

$785=\pi r^2 \cdot 10$

then solve for r[/hide]

[quote="SplashD"][quote="mathgeniuse^ln(x)"][quote="LalaMcla"][size=150][i]A cylinder is 10 inches tall. if the volume of the cylinder is 785 cubic inches[/i], [b]what is the radius of the cylinder???[/b][/size] :?:  :?:[/quote]

[hide]We know that the volume of a cylinder with radius r and height h, the volume is $V=\pi r^{2} h$.

So, we equate this to be $785=\pi r^{2} *10$

So $r^{2}=\frac{785}{10\pi}$

Therefore $r=\frac{\sqrt{314\pi}}{2\pi}$

So that is the answer

$\frac{\sqrt{314\pi}}{2\pi}$

We weren't supposed to approximate anything right.[/hide][/quote]

IF you were to approximate it, it would be about $4.9987$[/quote]

[hide]Yeah, which would give 5, if you approximate $\pi=3.14$.[/hide]

[hide]
Formula for the area of a cylinder:
$V = \pi r^2 h$

So we have
$\pi r^2 \times 10 = 785$

$\pi r^2 = \frac{785}{10} = \frac{157}{2}$

$r^2 = \frac{157}{2\pi}$

$r = \sqrt{\frac{157}{2\pi}} \approx 4.999$
[/hide]

Wow, that was my first time solving a problem with LaTeX!  :D


For real numbers $ x_{i}>1,1\leq i\leq n,n\geq 2,$ such that:

$ \frac{x_{i}^2}{x_{i}\minus{}1}\geq S\equal{}\displaystyle\sum^n_{j\equal{}1}x_{j},$ for all $ i\equal{}1,2\dots, n$

find, with proof, $ \sup S.$

Just my thoughts:

First:
$ \frac{x_{i}^2}{x_i\minus{}1}\geq x_i\plus{}\sum_{j\ne i}{x_j}>x_i \plus{} n\minus{}1$ then $ x_{i}^2>x_{i}^2\plus{}(n\minus{}1)x_i\minus{}x_i\minus{}(n\minus{}1)$
We get: $ x_{i}<\frac{n\minus{}1}{n\minus{}2}$ or $ 1<x_i<1\plus{}\frac{1}{n\minus{}2}$ [b](*)[/b].
This gives for our sum: $ S<n\plus{}\frac{n}{n\minus{}2}\equal{}n\plus{}1\plus{}\frac{2}{n\minus{}2}$.

Second:
$ x_{i}^2\minus{}Sx_i\plus{}S\geq 0$ so $ 1<x_i\leq\frac{S\minus{}\sqrt{S^2\minus{}4S}}{2}$ and $ x_i\geq\frac{S\plus{}\sqrt{S^2\minus{}4S}}{2}$
But as we see from [b](*)[/b] for $ n\geq 3: \; x_i <2$. This means that $ 1<x_i\leq\frac{S\minus{}\sqrt{S^2\minus{}4S}}{2}$ are only valid interval.
And this means that for maximized $ S$ should be: $ \;\frac{S_{\text{max}}\minus{}\sqrt{S_{\text{max}}^2\minus{}4S_{\text{max}}}}{2}\cdot n \equal{} S_{\text{max}}$.
Solving this equation gives: $ S_{\text{max}}\equal{}\frac{4}{1\minus{}\left(1\minus{}\frac{2}{n}\right)^2}\equal{}\frac{n^2}{n\minus{}1}$.
We can check: if all $ x_i\equal{}\frac{n}{n\minus{}1}$ then $ S\equal{}\frac{n^2}{n\minus{}1}$.


Hello! I would like this thread to be a collection of the problems and solutions in the book "The All-Time Greatest MATHCOUNTS Problems."

Rules:

PM me with your solution to the last problem I post. I will then post the best solution to the problem.

DO NOT LOOK IN THE BACK OF THE BOOK FOR SOLUTIONS.  :)

OK:

[hide="Problem 1"]

Find the measure of AB in the diagram:
 [geogebra]4250cd5c07a19fc53fa17a7285448faa94d5103c[/geogebra][/hide]

If you already have the book, why do you need solutions?  If you don't have the book, you can trade for it in the "Trading MATHCOUNTS Resources" thread, or you can buy it.

That's a good point. So should i just delete this topic?

Yes i think you should. This is not spam as this topic is to be deleted anyway

What about the Mathcounts Marathon then?

What do you mean "What about it." We're not saying anything's wrong with it.

[quote="RoFlLoLcOpT"]This could be a marathon-ish thing for people who don't have the book.[/quote]

Either way, this could be a copyright violation...

My book says, "Reproduction of any portion of this book without the expressed permission of the
MATHCOUNTS Foundation is strictly prohibited."

So, no; we cannot have this thread.

Wait... will I be penalized for quoting [i]that[/i] part of the book?  :P

It is a copyright violation if you post all the problems, but only a few you don't understand is okay.

10 spam posts have been made in less than 2 hours in this thread.

Lol, sorry PoP, I thought it would be a good idea. Anyways, someone wanna tell me how to delete this whole thread?

Contact an active mod.

EDIT: If that fails, contact an admin.
2nd EDIT: Or you could edit your first post and let the subject say something like "Delete this" or "ignore this".


Let $ U$ be the circumcenter of triangle $ ABC$ and $ P$ be a point on the extension of $ UA$ beyond $ A$. Lines $ g,h$ are symmetric to $ PB,PC$ with respect to $ BA,CA$ respectively. Let $ Q$ be the intersection of $ g,h$. Find the locus of points $ Q$ as $ P$ takes all possible locations.

Using the sin form of Menelaus Theorem, it is easy to prove that $ g, h$ and $ AU$ are concurrent.

I think that the locus of $ Q$ is the segment $ AU$ except the point $ U$. But I can't prove that.

[size=117][color=darkblue][b]Nice and easily problem, [u]Cancer[/u] ![/b][/color][/size]

[quote="cancer"] [color=darkred]Let $ w = C(O,R)$ be the circumcenter of triangle $ ABC$ . For a mobile point $ M\in OA$ , $ A\in (OM)$ define the lines $ g,h$ which are symmetric lines of $ MB,MC$ w.r.t. $ BA$ , $ CA$ respectively. Find the geometrical locus of the point $ L\in g\cap h$ .[/color] [/quote]
[color=darkblue][b]Proof.[/b] Denote the diameter $ [AA']$ of the circle $ w$ and $ \{\begin{array}{c} L_1\in OA\cap g \\
 \\
L_2\in OA \cap h\end{array}$ .

Observe easily that $ \{\begin{array}{c} A'B\perp BA \\
 \\
A'C\perp CA\end{array}$ . Therefore, $ \{\begin{array}{c} \frac {AL_1}{AM} = \frac {A'L_1}{AM} = \frac {BL_1}{BM} \\
 \\
\frac {AL_2}{AM} = \frac {A'L_2}{AM} = \frac {BL_2}{BM}\end{array}$ ,

i.e. $ A'$ is conjugate with $ A$ w.r.t. both $ \{M,L_1\}$ and $ \{M,L_2\}$ .

Reciprocally, $ L_1$ , $ L_2$ are the harmonical conjugate of $ M$ w.r.t. $ \{A,A'\}$ .

In conclusion, $ L_1\equiv L_2\equiv L$ and [b]the geometrical locus of $ L$ is the open segment $ (AA')$[/b] .[/color]


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry unsolved

In $ \Delta ABC$, $ AB\equal{}AC$ & $ \angle BAC \equal{} 80^o$.$ M$ is a point within the triangle such that $ \angle MBC \equal{} 30^o ,\angle MCB \equal{} 10^o$. Find $ \angle BAM$

Take D the symmetrical of M about BC; the triangles ACD and BDM are equilateral, ABD is isosceles with <BAD = 20 degs.
Since BDM is equilateral, M is on the perpendicular bisector of BD, hence  <BAM = 10 degs.

Best regards,
sunken rock

Or, we could let P be the reflection of M over the perpendicular bisector of BC. Then $ AM \equal{} AP$, but also $ \angle PCA \equal{} \angle MBA \equal{} 20 \equal{} \angle PCM \implies PM \equal{} PA$. So $ \triangle PMA$ is equilateral, so $ \angle MAP \equal{} 60$ and $ \angle BAM \equal{} \angle PAC \equal{} 10$.

In sunken rock's solution, how do we know ACD is equilateral?

To: dgreenb801

You were right, I was wrong; however it seems your solution takes the same assumption AC = CM...

I shall try to work on this issue.
Best regards,
S. Fulger

[b]sunken rock[/b]'s claim is correct.

If we take $ D$ as the reflection of $ M$ about $ BC$, $ A$ is the circumcenter of $ \triangle BCD$ since $ A$ lies on the perpendicular bisector of $ BC$ and $ \angle BAC \equal{} 80^{\circ} \equal{} 2(180 \minus{} \angle BDC)$.

So, $ AB \equal{} AD \equal{} AC \implies \angle BAD \equal{} 20^{\circ}$.

Now, $ \angle MBD \equal{} \angle MDB$, so $ M$ lies on the bisector of $ \angle BAD$.

Hence $ \boxed{\angle BAM \equal{} 10^{\circ}}$


Let $ a,b,c$ be positive reals. Prove that

$ \prod(a^2\plus{}1)\geq\frac{a\plus{}b\plus{}c}{2}\sqrt{\frac{3}{2}(a\plus{}1)(b\plus{}1)(c\plus{}1)(abc\plus{}1)}$

[quote="Ligouras"]Let $ a,b,c$ be positive reals. Prove that

$ \prod(a^2 \plus{} 1)\geq\frac {a \plus{} b \plus{} c}{2}\sqrt {\frac {3}{2}(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)(abc \plus{} 1)}$[/quote]

Hard and nice  because   the strongest one is

$ \prod(a^2 \plus{} 1)\geq\frac {a \plus{} b \plus{} c}{2}\sqrt {\frac {3 \plus{} k}{2}(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)(abc \plus{} 1)}$

where $ k \equal{} 0.00687556710....$


linear algebra
matrix
vector
linear algebra unsolved

An invertible real matrix is a product of positive invertible matrices if and only if it has positive determinant.

Does "positive matrix" mean that for all non-zero vectors $x$ we have $\langle Ax,x\rangle>0$?

Sorry for the stupid question.. :blush:

The first variant. Anyway, if the entries are positive, the entries of the product are also positive and there are many counterexamples.  :)

I realized how stupid my question was and edited my post before getting your response :). Anyway, thank you for clarifying it.

As for the problem, looks like the idea used [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43798]in this topic[/url] (the one about the transvections) works here as well:

One of the implications is trivial, so we'll assume that $A$ has positive determinant. $A$ is the product of several transvections $T_{ij}(\alpha)$ (the matrix having $1$ on the main diagonal, $\alpha$ in position $(i,j),\ i\ne j$, and $0$ everywhere else) and $D=\mbox{diag}(1,1,\ldots,1,\det A)$, which is positive definite.

The above means that it suffices to prove that a transvection is a product of positive definite matrices. It's easy to see that if $n$ is large enough, then $T_{ij}\left(\frac\alpha n\right)$ is positive definite, and since $T_{ij}(\alpha)=T^n_{ij}\left(\frac\alpha n\right)$ for all $n$, the desired result follows.


[quote="Rushil"]If $a,b,c,x$ are real numbers such that $abc \not= 0$ and \[ \frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c},  \] then prove that $a = b = c$.[/quote]

The problem is wrong as stated: Take any three reals a, b, c such that a + b + c = 0, and take $x=\frac12$. Then,

$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}=-\frac12$.

Here is a possible corrected version of the problem:

[color=blue][b]Problem.[/b] If four real numbers a, b, c, x satisfy $abc\neq 0$ and

$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}$,

then either a = b = c, or a + b + c = 0 and $x=\frac12$.[/color]

[i]Solution.[/i] Since $abc\neq 0$, the numbers a, b, c are all nonzero, so we can divide by them without remorses.

Denote

$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}=k$.

Then,

$xb+\left(1-x\right)c=ka$;
$xc+\left(1-x\right)a=kb$;
$xa+\left(1-x\right)b=kc$;

adding these three equations together, we arrive at

$\left(xb+\left(1-x\right)c\right)+\left(xc+\left(1-x\right)a\right)+\left(xa+\left(1-x\right)b\right)=ka+kb+kc$,

what simplifies to $a+b+c=k\left(a+b+c\right)$. Hence, either k = 1, or a + b + c = 0.

[b]First consider the case when a + b + c = 0.[/b]

In this case, a = - (b + c), so that $xb+\left(1-x\right)c=ka$ becomes $xb+\left(1-x\right)c=k\left(-\left(b+c\right)\right)=-kb-kc$. Addition of $kb-\left(1-x\right)c$ yields $xb+kb=-kc-\left(1-x\right)c$; in other words, $\left(x+k\right)b=\left(x-k-1\right)c$.

Now, if $x+k\neq 0$, then this equation becomes $\frac{b}{c}=\frac{x-k-1}{x+k}$. Similarly, $\frac{c}{a}=\frac{x-k-1}{x+k}$ and $\frac{a}{b}=\frac{x-k-1}{x+k}$. Hence,

$\left(\frac{x-k-1}{x+k}\right)^3=\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}=1$,

and thus $\frac{x-k-1}{x+k}=1$. Thus, $\frac{b}{c}=\frac{x-k-1}{x+k}=1$, so that b = c. Similarly, c = a. Thus, a = b = c, and we are done.

If x + k = 0, then the equation $\left(x+k\right)b=\left(x-k-1\right)c$ yields x - k - 1 = 0 (since c is nonzero). Hence, x + k = x - k - 1 = 0, so that k = - k - 1, and thus 2k = -1; hence, $k=-\frac12$, so that x + k = 0 yields $x=-k=-\left(-\frac12\right)=\frac12$. Hence, we have a + b + c = 0 and $x=\frac12$. So in this case we are done as well.

[b]Now, we consider the case when k = 1.[/b]

In this case, the equation $xb+\left(1-x\right)c=ka$ becomes $xb+\left(1-x\right)c=a$, what is equivalent to $xb+c-xc=a$, i. e. to $xb-xc=a-c$, i. e. to $x\left(b-c\right)=-\left(c-a\right)$. Similarly, $x\left(c-a\right)=-\left(a-b\right)$ and $x\left(a-b\right)=-\left(b-c\right)$. Mulitplying these three equations, we get

$x\left(b-c\right)\cdot x\left(c-a\right)\cdot x\left(a-b\right)=\left(-\left(c-a\right)\right)\cdot\left(-\left(a-b\right)\right)\cdot\left(-\left(b-c\right)\right)$,

i. e.

$x^3\left(b-c\right)\left(c-a\right)\left(a-b\right)=-\left(b-c\right)\left(c-a\right)\left(a-b\right)$.

Now, if one of the terms b - c, c - a, a - b equals 0, then we are quickly done (in fact, assuming that b - c = 0, we see that $x\left(b-c\right)=-\left(c-a\right)$ yields c - a = 0, so that a = b = c). If not, then we can divide this equation by $\left(b-c\right)\left(c-a\right)\left(a-b\right)$, and thus we get $x^3=-1$, so that x = -1. Hence, the equation $xb+\left(1-x\right)c=a$ becomes $-1b+\left(1-\left(-1\right)\right)c=a$, what simplifies to -b + 2c = a. In other words, c - (b - c) = a, so that b - c = c - a. Similarly, c - a = a - b. So the numbers b - c, c - a, a - b are all equal to each other; but their sum is 0, and thus they are all equal to $\frac03=0$. Hence, a = b = c. Thus, the problem is solved in all cases.

  darij

[quote="Rushil"]If $a,b,c,x$ are real numbers such that $abc \not= 0$ and \[ \frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c},  \] then prove that $a = b = c$.[/quote]
A very simple solution:
First assume $a<b<c$. Then:
$\frac{1}{a} > \frac{1}{c}$
$xb > xa$
$(1-x)c > (1-x)b$, so:
$\frac{xb + (1-x)c}{a} > \frac{xa + (1-x) b }{c}$, contradiction.
So two of $a,b,c$ are equal. wolog $a=b$. Have
$\frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b}$
$xb+(1-x)c=xc+(1-x)a$. Subst $b$ with $a$:
$xa+c-xc=xc+a-xa$
$(c-a)+2x(a-c)=0$
$(a-c)(2x-1)=0$
Have $a=c$ or $x=1/2$ which also yields $a=c$.
Hence $a=b=c$, qed

[quote="rem"][quote="Rushil"]If $a,b,c,x$ are real numbers such that $abc \not= 0$ and \[\frac{xb+(1-x)c}{a}= \frac{xc+(1-x)a}{b}= \frac{xa+(1-x) b }{c},\] then prove that $a = b = c$.[/quote]
A very simple solution:
First assume $a<b<c$. Then:
$\frac{1}{a}> \frac{1}{c}$
[/quote]

This isn't necessarily true. Take $a=-1$ and $c=1.$

[quote="Rushil"]If $a,b,c,x$ are real numbers such that $abc \not= 0$ and
\[\frac{xb+(1-x)c}{a}= \frac{xc+(1-x)a}{b}= \frac{xa+(1-x) b }{c}, \]
then prove that $a = b = c$.[/quote]

Cross-multiplying the first and third terms we get
$xbc+(1-x)c^{2}= xa^{2}+(1-x)ab$
$\Longrightarrow x(c^{2}+a^{2}-ab-ca) = c^{2}-ab$
And similarly,
$x(a^{2}+b^{2}-bc-ab) = a^{2}-bc$
$x(b^{2}+c^{2}-ca-bc) = b^{2}-ca$

Adding the three equations we get
$2x(a^{2}+b^{2}+c^{2}-bc-ab-ca) = (a^{2}+b^{2}+c^{2}-bc-ab-ca)$
$(2x-1)(a^{2}+b^{2}+c^{2}-bc-ab-ca) = 0$

Case 1:
$(a^{2}+b^{2}+c^{2}-bc-ab-ca) = 0$
$\Longrightarrow \frac{1}{2}((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) = 0$
$\Longrightarrow a=b=c$

Case 2:
$x=\frac{1}{2}$
In the three equations we get
$\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}$
Add 1 to all three equations
$\frac{a+b+c}{c}=\frac{b+c+a}{a}=\frac{c+a+b}{b}$
So we again get $a=b=c$ or $a+b+c=0$

[quote="Rushil"]If $ a,b,c,x$ are real numbers such that $ abc \not \equal{} 0$ and
\[ \frac {xb \plus{} (1 \minus{} x)c}{a} \equal{} \frac {xc \plus{} (1 \minus{} x)a}{b} \equal{} \frac {xa \plus{} (1 \minus{} x) b }{c},
\]
then prove that $ a \equal{} b \equal{} c$.[/quote]
um...
i thought if we added the top and the bottom together respectively we could obtain
xb+c(1-x)=a
xc+a(1-x)=b
xa+b(1-x)=c

OR a+b+c=0

Since the three fractions are equal, they equal the sum of the numerators over the sum of the denominators, which is 1, unless a+b+c=0. If two of a,b,c are equal, then all three are. So we find $ x \equal{} \frac {a \minus{} c}{b \minus{} c} \equal{} \frac {b \minus{} a}{c \minus{} a} \equal{} \frac {c \minus{} b}{a \minus{} b}$ if no two are equal and a+b+c is not 0. But if we have let's say a>b>c or a>c>b we get a contradiction because one of the fractions will be positive and the other negative, thus they are all equal.
If $ a\plus{}b\plus{}c\equal{}0$ we can just substitute, we have
$ \frac{xb\plus{}(1\minus{}x)c}{\minus{}b\minus{}c}\equal{}\frac{xc\plus{}(1\minus{}x)(\minus{}b\minus{}c)}{b}$, which is equivalent to
$ 2x(b^2\plus{}bc\plus{}c^2)\equal{}b^2\plus{}bc\plus{}c^2$, in which case $ x\equal{}1/2$ or $ a\equal{}b\equal{}c\equal{}0$.

its too simple a problem to be hyped up so much..!

Put x=1 and solve taking RHS as k or else take two sets of eqns and solve for x separately and then compare them

[quote="murgi"]INMO in India is rare in mistake............... :o
Probably they wanted to make a,b,c non-negative real no.
I think in that case the sum holds good. :)[/quote]

not rare but i have seen it getting pointed as wrong quite a good number of times

they really should check these problems more thoroughly

Assuming, $a+b+c \neq 0$, let \[ \lambda = \frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c} \]
Then, \[ (a+b+c)\lambda = xb + (1-x)c + xc + (1-x)a + xa + (1-x)b = a+b+c\]
Therefore, $\lambda = 1$, which means that
$xb+(1-x)c = a ..... (1)$
$xc+(1-x)a = b ..... (2) $
$xa+(1-x)b = c ..... (3) $
From adding $(1)$ and $(2)$, we get
\[x(b-a) = b - c \Rightarrow x = \frac{b-c}{b-a}\]
Similarly, by adding $(2)$ and $(3)$, as well as $(3)$ and $(1)$, we get
\[x = \frac{b-c}{b-a} = \frac{c-a}{c-b} = \frac{a-b}{a-c}\]
Note that here, we are assuming so far that $ a \neq b \neq c $. But by cross-multiplying the three relations, we will get
\[ a^2 + b^2 + c^2 = ab + bc - ca = ab + ca - bc = ca + bc - ab \]
This leads us to $ ab=bc=ca $
Note that according to question, $ a, b, c \neq 0$ (as otherwise $abc = 0$) therefore $ab = bc \Rightarrow a  = c$
Hence, $ab = ca = a^2 \Rightarrow a = b = c$
Hence $a=b=c$ or $a+b+c=0$.

$$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c} \implies \begin{cases} xb^2+bc-xbc=xac+a^2-xa^2 \\\ xc^2+ac-xac=xab+b^2-xb^2 \\\ xa^2+ab-xab=xbc+c^2-xc^2 \end{cases}$$ $$ \implies (2x-1) (a^2+b^2+c^2)=(2x-1)(ab+bc+ca) \implies \begin{cases} x=\frac{1}{2} \implies a+b+c=0 \\\ a^2+b^2+c^2=ab+bc+ca \implies a=b=c \end{cases}$$

I think that darij grinburg is correct.
His anticipation that the problem was possibly wrong ,I think,is true.
Cause I am arriving at similar results taking a+b+c=0.

broski the question is incomplete
its either a =  b = c or a+b+c = 0


LaTeX
function
algebra
polynomial
binomial theorem
combinatorics unsolved
combinatorics

The following file contains a question of recurrence relation and my solution. Could anyone check if my answer is correct? Thanks.

Basically i think the combinatorial argument was right but i have trouble with this part:
[quote="eleanorwang"]
and the coefficients in the Taylor series for this rational function count the number of combinations of the type considered.
[/quote]
What was that supposed to mean?

My alternative for that would be to use the binomial theorem approximation to solve that  :)

If you would take the time to learn a little bit of [[LaTeX]] (see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194030]here[/url]) you would be much more likely to get comments from forum-members as compared with what happens when you attach a file to your post.  For example, I would have told you 2 days ago that your solution was fine had you written it up here.  (It's not clear to me that the latter part of your solution, when you re-write the function $ g$, is necessary to answer the question.  Also, I think you'll find that in general your life will be simpler if you leave polynomials ($ 1 \plus{} x \plus{} x^2$, for example) as polynomials instead of writing them as rational functions ($ \frac {1 \minus{} x^3}{1 \minus{} x}$).)


Does the series $\sum_{n=1}^{\infty}\frac{1}{p_{n}}$ converge?
p_n are all the prime numbers.

rustam bu qatorning uzoqlashuvchiligi osongina ko'rsatiladi.Ayrlend,Rouzenning Sovremenniye teori chisel kitobida eng zur isboti bor!


Lately I've been getting alot of strange e-mails from AOPS users, but I know that it isn't them because they would not send me junk mail with viruses attached, so I don't know who it is guys, but someone knows your e-mail accounts.

I don't want to mention any names, but I know I've gotten about 20 of these so far...

It's not from them, and it certainly hasn't got anything to do with the site. It's a mail worm. You should delete those messages without opening any attachements or such.

is it the netsky D@mm virus?

more info here: http://securityresponse.symantec.com/avcenter/venc/data/w32.netsky.d@mm.html

i like to put them all in a folder for yahoo mail cuz it does no harm cuz yahoo won't let you download it

i got about 100 in my folder and some of them are kinda funny like people from neopets, or netscape, or amd, or microsoft... but you always know it's a virus if it's exactly 24K and is a reply

But is the sender random?

Or does it copy someone else's email address?

[quote="theone853"]once the virus is in your inbox, it'll send the virus to everyone in your address book or whoever you've sent a email to recently...[/quote]

One reason I never use the address book...


Find all integer solutions of the equation \[\left\lfloor \frac{x}{1!}\right\rfloor+\left\lfloor \frac{x}{2!}\right\rfloor+\cdots+\left\lfloor \frac{x}{10!}\right\rfloor =1001.\]

We need $ x>\frac{1001}{1\plus{}\frac{1}2\plus{}\ldots\plus{}\frac{1}{10!}}>\frac{1001}{e\minus{}1}>582$. So, trying $ x\equal{}583,584,585$ respectively, we get $ 999,1001,1002$. Since the LHS is obviously nondecreasing when $ x$ increases, $ x\equal{}584$ is the only solution.


Let f(x) = ax^2+bx+c be such that |f(x)|<=1 for 0<=x<=1. Prove that |a|+|b|+|c|<=17


Eum... may I ask how do you "note that" without using software?

It was not said that using math software is forbidden.

Ah ok. :) Of course it is not forbidden, but I think there also is a nice solution to find this factorisation handwise.

let $ x \equal{} 2^{5*2 \plus{} 1}, y \equal{} \minus{} 2^{5*1 \plus{} 1}, z \equal{} \minus{} 1$, then $ 10^3 < x \plus{} y \plus{} z \equal{} 1983 < 5*10^3$ and $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} 2^{33} \minus{} 2^{19} \minus{} 2^{17} \minus{} 1$.
it is well known that $ x \plus{} y \plus{} z | x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$

bye
paolo :wink:

[quote="bboypa"]it is well known that $ x \plus{} y \plus{} z | x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$[/quote]
Hi bboypa (aka Jordan), how do you demonstrate this?

Bye
GioMott [hide="..."]aka Oblomov[/hide]

[quote="GioMott"][quote="bboypa"]it is well known that $ x \plus{} y \plus{} z | x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$[/quote]
Hi bboypa (aka Jordan), how do you demonstrate this?[/quote]$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz\equal{}(y\plus{}x\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}xz)$

Alternately, the polynomial $ x^3 \minus{} (3yz)x \plus{} (y^3 \plus{} z^3)$ has the root $ x \equal{} \minus{}y\minus{}z$.


superior algebra
superior algebra theorems

1, R is a ring in which a^4=a for any a in R,then R is commutative.
2, R is a ring in which a^3=a for any a in R,then R is commutative.

anyone know the proof??

First a general result: given a ring $R$ in which the implication $a^2=0\Rightarrow a=0$ holds, every idempotent element $e$ belongs to the center $Z(R)$ of $R$. Take any $a\in R$. We have $(ea-eae)^2=(ae-eae)^2=0$, as can easily be checked, so $ea=eae=ae$. In particular, this implies that in a ring $R$ in which $a^n=a,\ \forall a$, $a^{n-1}\in Z(R),\ \forall a$.

Another result which applies to both situations (and to other situations, more general than this): If we solve the problem ($a^n=a,\forall a\in R\Rightarrow R$ is commutative) for rings with a unit, then we can solve it in the general case. Let $R$ be a ring in which $a^n=a,\ \forall a$. Then for all $a,\ a^{n-1}$ is a non-zero idempotent element which we will denote by $e_a$. By the above result, the ring $e_aR$ is commutative, because it's easy to check that it's a ring with the unit $e_a$ in which $x^n=x,\ \forall x$. This means that for all $x,y,a\in R$, we have $e_axb=e_abx=bxe_a$. If we take $x=a$, then $e_ax=xe_a=a$, and we get $ab=ba,\ \forall a,b\in R$, so we're done.

All that shows that it suffices to prove our two problems in the case when $R$ has a unit $1$. Now we prove them one by one.

[b](1)[/b] Suppose that $a^3=a,\ \forall a\in R$. Then $R$ is commutative.

By the first lemma above, $a^2\in Z(R),\ \forall R$. From $a^3=a,(1+a)^3=1+a=1+a^3$ we also get $3a+3a^2=0\in Z(R),\ \forall a\in R$. Since $a^2\in Z(R)$, this implies that $3a\in Z(R),\ \forall a\ (*)$. We also have $2a+2a^2=(a+a^2)^2\in Z(R),\ \forall a$, and, since $a^2\in Z(R)$, we get $2a\in Z(R),\ \forall a$. Subtracting this from $(*)$, we get $a\in Z(A),\ \forall a$, or, in other words, $R$ is commutative, as desired.

[b](2)[/b] If $a^4=a,\ \forall a\in R$, then $R$ is commutative.

Again, from the first lemma, $a^3\in Z(R),\ \forall a$. We also have $a+a^2=(1+a)^3-1-a^3\in Z(R),\ \forall a\ (*)$ (notice that $2=0$ in $R$). This means that $ab+ba=[(a+b)+(a+b)^2]-(a+a^2)-(b+b^2)\in Z(R),\ \forall a,b$. In particular, $ab+ba$ commutes with $a$ for all $a,b$, so $a^2b+aba=aba+ba^2,\ \forall a,b\Rightarrow a^2b=ba^2,\ \forall a,b$. We have thus obtained $a^2\in Z(R),\ \forall a$. This, together with $(*)$, implies the conclusion: $a=a+a^2-a^2\in Z(R),\ \forall a$.


1) Let a_n = 6^n + 8^n. Determine the remainder on dividing a_83 by 49.

2) What is the largest 2-digit prime factor of the integer n = 200C100.

[hide]1) Writing a_n (7-1)^n + (7+1)^n and developping with binomial formula :

a_n = (-1)^(n-1)7n + (-1)^n + 7n+1  modulo 49.

2) Every prime factor < 100 in the numerator are cancelled at least once by the denominator.

61 appears once in the denominator and twice in the numerator (2*61 and 3*61).

Next prime is 67 with appears only once in the numerator (only 2*67, 3*67 > 200).

So biggest 2 digits prime is 61.[/hide]


[quote="mychrom"]Show that $ 10^{10}$ cannot be written as a product of two natural numbers which do not contain $ 0$ in their decimal representation.[/quote]

$ 10^{10} \equal{} 2^{10} \cdot 5^{10}$

Since $ 2 \cdot 5 \equal{} 10$ any product which contains a $ 2$ and $ 5$ will have a 0 in its decimal representation.

Therefore it could only feasibly be $ 2^{10} \times 5^{10}$, But $ 2^{10} \equal{} 1024$ which contains a 0,

Hence there is no possible way of doing it.


[hide="answer"]$\text{Add both equations together. Get } 2t+l-l+w-w=60 \text{. The }l\text{s and the }w\text{s cancel out leaving the equation as: }2t=60 \text{. Divide both sides by 2 and get }t=30\text{ answer: }\boxed{\text{c}}$[/hide]

[hide]:rotfl:  :rotfl: t - w + l + t + w - l = 32 + 28 .
2t=60 => t=30[/hide]

[color=darkred][size=75][Hide your answers. What's so funny?][/size][/color]

[quote="KaSmIR"][hide]:rotfl:  :rotfl: t - w + l + t + w - l = 32 + 28 .
2t=60 => t=30[/hide][/quote]What's so funny?

[quote="apusey"]Why are all the problems I have seen of the old AMC 12s so easy?[/quote]Simple: Because back then there was no AOPS and people weren't as smart do they had to make the tests easier.

It's because only the easier AMC problems would be posted in middle school forums.

Yea that is probably true.
Plus the hard ones probably would be posted in the AMC section.

[hide]
Using elimination, add the 2 lines together.
$2t+l-l+w-w=60$
$2t=60$
$t=30$
$C$[/hide]

Wait isn't the question (1986 AHSME #6) about a table and two identical blocks of wood??  Maybe the picture couldn't have been put in.

EDIT: The equations are what is used to solve for the problem but that isn't the actual question lol.

Wait WHOAAAA you're right.  I don't have moderatorship of the Middle School math forum, so can someone copy and paste the following code into the original post?  Thanks!

[quote]
Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1.  Length $r$ is found to be $32$ inches.  After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches.  How high is the table?

[asy]
size(300);
defaultpen(linewidth(0.8)+fontsize(13pt));
path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle;
path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle;
path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle;
draw(table^^shift((14,0))*table);
filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray);
draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5));
draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5));
unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle);
unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle);
label("$r$",(8.5,5.75));
label("$s$",(22,5.75));
[/asy]

$\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}$[/quote]

@above

Thx for working on it! I was really confused lol

Fixed.  Thanks for spotting that problem!

Also, it would be great if you could fix it here as well:
https://www.artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_6
https://www.artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems

Thx again!

\bump

Could a moderator fix the above pages?  Thx!

The wiki is publicly editable, so anyone can make those changes.

They've been fixed now.

I represented the height of the table as a different variable. I solved a system of linear equations such that x+y (the dimensions of the rectangles) is 2.
You get 30.


Triangles $ ABC$ and $ ADC$ are isosceles with $ AB \equal{} BC$ and $ AD \equal{} DC$. Point $ D$ is inside $ \triangle ABC$, $ \angle ABC \equal{} 40^\circ$, and $ \angle ADC \equal{} 140^\circ$. What is the degree measure of $ \angle BAD$?

$ \textbf{(A)}\ 20 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 60$

[hide] Since $\angle ABC=40$, we know that \[\angle BAC=\angle BCA=\frac{180-40}{2}=70\]
We also know that since $\angle ADC=140$, that \[\angle DAC=\angle DCA=\frac{180-140}{2}=20\]  Hence, $\angle BAD=\angle BAC-\angle DAC=70-20=50$.  Therefore, the answer is $D$. [/hide]

i remember making a stupid subtraction mistake on this one, for some reason i subtracted from 200 instead of 180??  :mad:

Basically remember that the base angles of an isoceles triangle are the same.
I hope my picture does the trick.  :P
So 70-20=50.


AMC
AMC 8
AMC 10
AIME
AMC 10 A
AMC 12
geometry

ehh... You want opinions? In my opinion, they... are. average. AMC 8 score... is pretty good in comparison to most people. AMC 10 is  :wink: ... and err... your goals. seem astronomical... a 63 points improvement in one year? O.O.

If I set high goals I'll work harder. Hopefully. Maybe. Not really.

I think i just got really nervous because when i took the test packet home, the problems seemed a lot easier, and people who got the same or lower score than me on the amc8 usually did much better

[quote="miller4math"][quote="jmerry"]A goal of 150, ever, is unrealistic. If your nervousness really affected you, you'll probably do substantially better next time- but that's still just a 108 or something. Once you learn the material, you'll be able to come pretty close to your AMC 10 scores on the 12, so you will likely qualify for the AIME via AMC 12 some year. Just don't bet on next year.[/quote]
yes i agree. I came pretty close to a 150 (not really though 138)  :rotfl:
but still. It's pretty hard to do all the harder problems and make NO careless errors.[/quote]
On the 10 not REALLY, but when you get good, your 12 scores should look like your 10 scores, like the above poster said. Next year I'll have to somehow sneak into the AMC10A room and have a shot at a perfect.

[quote="jmerry"]A goal of 150, ever, is unrealistic. If your nervousness really affected you, you'll probably do substantially better next time- but that's still just a 108 or something. Once you learn the material, you'll be able to come pretty close to your AMC 10 scores on the 12, so you will likely qualify for the AIME via AMC 12 some year. Just don't bet on next year.[/quote]

uh... how can your amc12 score approach your 10 score? I took the amc10 three times and got 138, 145.5, and 144. But never have I gotten anything over 120 on my amc12.

bluelobster, which math class are you enrolled in? if you are in geometry or lower, that may be a cause. alot of the tougher AMC 10  problems may require higher level geometry and a little bit of trigonometry/maybe alg. 2. and if you haven't even encountered topics such as trig./higher level geometry and tricky counting, AMC 10 maybe quite challenging. the reason i say this is you have performed very well on the AMC 8, which shows a good knowledge of algebra/basic geometry and counting principals. i think AIME is well within reach for you. in 7th grade, i had barely seen an AMC 10 and i managed to make AIME. so nothing is impossible if you just work hard.

[quote="matt276eagles"][quote="jmerry"]A goal of 150, ever, is unrealistic. If your nervousness really affected you, you'll probably do substantially better next time- but that's still just a 108 or something. Once you learn the material, you'll be able to come pretty close to your AMC 10 scores on the 12, so you will likely qualify for the AIME via AMC 12 some year. Just don't bet on next year.[/quote]

True, you should never shoot for 150, as it's near impossible to solve all the problems in 75 minutes and not make a single mistake.  But making the AIME next year is not out of the question.  When I was in 8th grade, I got 107.5 on the AMC 10, and I didn't take the AMC 12.  Then in my 9th grade year I got 150 on the AMC 10 and 130 on the AMC 12.  Granted, I improved a lot during that year, but still, this shows that you can improve enough to qualify for the AIME next year if you work at it.[/quote]

Aim for an 150 on AMC 10 when you can get 7 - 10 on AIME, By that time all the problems will be pretty easy and you will get a perfect score if there are no stupid mistakes. Also, AMC 12 is easier to pass but harder to qualify to USAMO. Study Aops Volume 1.

Setting a goal of 150 can only hurt you. Try to set a more realistic, reachable goal and really work toward it. Take some online classes (like the ones offered by AoPS), study from good contest prep books, and do a practice AMC 10 every 2-3 weeks. That way, you can see your score improve, and then you can modify your goal when necessary.

No, setting a goal of 150 is NOT unrealistic. I knew someone who felt sick on the day of the AMC 10 and got like an 80. The following year they got a 144.

He/she wasn't sick though... they were merely nervous...

Are those seriously your Ohio results? Did you get in =D

Has everyone gotten their official scores for the a tests?

i just got mine today and i didn't make any bubbling errors.  :D

I got my score, but I marked my answers in the bookles so I knew it for a couple of weeks.

It really depends on how you set your goals. I set one for 150 AMC 10 this year, and I got a 132. I know setting a goal any lower now would be foolish, but setting a goal high early led me to overconfidence.

then i guess i'll ask my teacher tomorrow.

[quote="bluelobster"]2007-2008 Season: AMC8-23 AMC10-87   :rotfl:[/quote]

i think it really depends on your grade. if you were in grade 7~8, i would say it's pretty good, but if you're in gr 10, i would say about average

and i think you set too high goals for next year.

seeing how i managed to go up from 122.5 in gr 8 to 144 in gr 9 :), i should say it isn't impossible, though it might be very tough

and yes, it is virtually impossible to get the same in AMC 12 as you did in AMC 10 the year before. i dropped from 144 in AMC 10 to 130.5 in AMC 12 this year :(

Yeah, and if you can get 150 on the AMC 10, then you can definitely get a lot higher than 100 on the AMC 12.

mathcrazed-posting how well you did in AMC 10 in your sig last year as a 7th grader makes me feel better about my score.

blobster-post more on AoPS.

why should he????when i started out abacadaea i posted like 2 posts in my first month!!!!!

ehh...his name is bluelobster...LOl BLOBSTER!!!(not that different)

Lol...I've posted 28+ and the first month's not over!

Edit: Bragging about the AIME already, abacadaea?

[quote="zolojetto"]wow!!!so you know chineese, we were talking about chineese and then he just said a bunch of stuff in chineese that i didnt understand and then for some reason he put that in his sig...i think he'll get 5(which is a complement abacadaea!!!)[/quote]

Sorta. I really can't understand characters, though (only some spoken Chinese). He says he will get 15 problems right on the AIME.  I would agree that 5 is a compliment, although I'm shooting for 10 as an eighth grader.

wow...well i wont disclose what grade he is in...but we were having that conversation because i am just starting to learn chineese(well i just started today actually cause i was really really really... bored)

and good luck on AIME hope you get a 10...i probably wont make AIME depending on how everyone else did...but i have a feeling that i didnt...definetely next year...high school W00T W00T!!!!!

wow people here have never seen such scores before as an 87 ^_^;

honestly guys, when i was a freshman I got 60's on the AMCs...and now i can do ~10 problems on the AIME (hopefully...)

[quote="illusiat"][quote="bluelobster"]2007-2008 Season: AMC8-23 AMC10-87   :rotfl:[/quote]
uh...ur AMC 10 scores are pretty bad
I'm in 8th grade, and I got a ~120 on a practice test 4 months ago without studying. I'm not even smart at math.
(Unless ur in 7th grade or lower, then ur allright. Still time to improve)[/quote]
Wow... thats very nice words of encouragement.
I have a friend that scored a 93 last year on the amc10b. But this year, he got a 106.5 on his amc12a and 126 on his 10b, and those weren't his best scores either.
But I think if you got a 23 on the amc8 and an 87 on the amc10, I think you just had a bad day on the amc10. If I were you, I wouldn't worry about it too much. Just practice for next year.

You beat me in the AMC 8 :lol:
There's definitely time to improve; I jumped up 46.5 if you compare my 10B last year and 10A this year...which I'm only doing since I only scored decently on the 10A last year by guessing.

[quote="matt276eagles"][quote="jmerry"]A goal of 150, ever, is unrealistic. If your nervousness really affected you, you'll probably do substantially better next time- but that's still just a 108 or something. Once you learn the material, you'll be able to come pretty close to your AMC 10 scores on the 12, so you will likely qualify for the AIME via AMC 12 some year. Just don't bet on next year.[/quote]

True, you should never shoot for 150, as it's near impossible to solve all the problems in 75 minutes and not make a single mistake.  But making the AIME next year is not out of the question.  When I was in 8th grade, I got 107.5 on the AMC 10, and I didn't take the AMC 12.  Then in my 9th grade year I got 150 on the AMC 10 and 130 on the AMC 12.  Granted, I improved a lot during that year, but still, this shows that you can improve enough to qualify for the AIME next year if you work at it.[/quote]

1. Getting a 150 is by no means easy, but it's not impossible either. Plenty of people do it every year.

2. Even if its unrealistic, why shouldn't you shoot for it? matt276eagles got a 145.5 this year - do you think his goal should be anything but 150 next year?

If you have already had a score of 145.5, 150 is an obvious and reasonable goal. If your past score is 87, it isn't. "Plenty of people" is about one in every 2500 takers, using last year's numbers- and those numbers feel higher than normal to me.


I will be posting the 2006 Mock AIME 5 later today. It's been in production for about 3 months now since I've been really lazy and all. With any luck, the difficulty will be approximately equal to the difficulty of this year's AIME, but I guess we'll find out if that's true or not soon. I tried to make it harder than the Mock AIME 1, but didn't pound as many geometry problems in as the other Mock AIME writers, so this test is probably a bit short in that category.

In any case, I'm thinking of just letting people take it and then letting loose a thread for discussion instead of having me grade them since I don't have time. Discussion will probably start sometime next weekend before the real AIME happens; this way people can still take the test even as discussion is going on and then check their answers and get solutions, too. And anyone who doesn't have a three-hour block on such short notice can just work out a few problems whenever they have time. It's a win-win situation!

Anyway, go ahead and post your thoughts here (if there are any...) and I'll probably have the AIME up in this thread around 9 PM EST (I hope).

EDIT: Here's the the link to the test:

[url=http://wangsblog.com/jeffrey/]2006 Mock AIME 5[/url]

Great news! Not sure if I can take it but good job on doing this. We appreciate this.

Therefore, this thread is stickied.

Yeah, thanks for taking the time to make it!

Hmm... few geometry problems? I might try to squeeze in 1 1/2 to 2 hours to work on some of the problems, then. :)

It's my first time, I hope I can do good :)

Fun stuff, fun stuff. I'll work out a couple problems, more likely than not.

OK, it's posted now. Someone set up a discussion forum when they feel like it.

8 seems interestingly-weird to me already. I wonder. . .

[quote="paladin8"][quote="chess64"]Eh. . . why is #5 so easy? :P[/quote]

It's not that easy, is it? :?

EDIT: On second though, I should've made it bigger so people couldn't just multiply it out and stuff.[/quote]

No, it is easy. You can do it in your head. :)

I greatly appreciate the time you took to actually make this...thanks!

I definitely will not have time to do this today.  I may sometime this week though..if not, then this weekend.

[quote="Silverfalcon"]paladin, do you mind if I set up discussion today? I think since this is AIME problems, the chances are that you will not be able to get the full solutions just by the weekend.[/quote]

Yeah, that's fine. I'm not planning on writing up solutions anyway.

Do you have the answers? I want to check mine.

Can someone PLEASE tell me the answers so I can check mine?

There's a discussion thread in this forum...

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3745701_2006_jeffrey_wang_2nd_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS. Answer key is [url=https://artofproblemsolving.com/community/c5h77102p446788]here[/url]


Let $ n\geq 3$ be an integer and $ x_{1}\leq x_{2}\leq ...\leq x_{n}$ positive real numbers.

Prove that

$ \frac{x_{n}\cdot x_{1}}{x_{2}}\plus{}\frac{x_{1}\cdot x_{2}}{x_{3}}\plus{}\frac{x_{2}\cdot x_{3}}{x_{4}}\plus{}...\plus{}\frac{x_{n\minus{}1}\cdot x_{n}}{x_{1}}\geq x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}$.

I don't know why I believe this problem has a very short solution. I tried a lot to solve it, but I didn't manage. It's very tricky.

This inequality is very interesting and I have spent a lot of time on it.
I am sure somebody will find a beautiful solution.

well actually the source is Sankt Petersburg Olympiad ( LMO 2000 ). I have a solution but it is rather unexpected (it is not mine). Please let me know when you want me to post it, ok?

I would like to know it as soon as possible.
Thank you very much.

suppose that 0 \leq x \leq y and 0<a \leq 1. we then have 1 \geq a and y \geq x \geq ax which means that(1-a)(y-ax) \geq 0 that is ax+ay \leq a^2x+y or

x+y \leq ax + y/a (1)

Setting (x,y,a) = (x_n,x_n.x_{n-1}/x_2, x_1/x_2 ) in (1) we find that

x_n + x_{n-1}x_n / x_2 \leq x_nx_1/x_2 + x_{n-1}x_n /x_1 (2)

Now for i=1,n-2 we set (x,y,a) = ( x_i, x_{n-1}x_{i+1}/x_2, x_{i+1}/x_{i+2} ) and summing the obtained inequalities we get

(x_1+x_2+...+x_{n-2})+x_{n-1} \leq ( x_1x_2/x_3 + ... + x_{n-2}x_{n-1}/x_n ) + x_{n-1}x_n / x_2 (3)

From (2) and (3) we obtain the required inequality.

A beautiful solution for a beautiful problem. But really it is very difficult and almost impossible to find.
Russian inequalities are very original and  ingenious.

Thank you very much, Valentin.

yes manlio your observation is keen. I have modified the post.


Who's all going to participate in the math counts competition on February 13, 2010?  :)

This would more appropriately belong in the Middle School forum.

[quote="Maybach"]I think that might be the date of the chapter in my state, oddly enough.  ARE YOU  FROM WILBUR WRIGHT MIDDLE SCHOOL?  If so, PM me and tell me who you are.[/quote]

I'm pretty sure that most chapters are held on the same day.  :) Ours' is on the same day...

Edit: Oh, you're right AIME (just checked the Mathcount's website)

The chapter coordinator decides when the chapter competition is...it must be in February though (chapters in snow-prone areas usually have their competitions earlier in case of a snow-out).

Ours is on the 6th, but we sometimes have lots of snow so I guess that's why. Although it hasn't snowed a lot for a while.

I guess I'm not participating in the one on the 13th because ours is on the 6th too. And it just snowed recently.

noo idea when our chapter is---snow canceled our school round :(

2/13/10 is Fort Worth Texas Chapter. There probably a ton of other chapters on that date.

Ours in Feb 6th (at 8:00 in the morning UGH!). I have to wake up and drive 45 minutes to north charlotte. Not fun  :(

I have to go to a different chapter this year because the rest of my team members had scheduling issues.  It's probably one of the last chapters to go, as it's on 2/27.


any help would be much appreciated.

consider an inviscid fluid with the velocity field:

1. u(x,y,x,t)=(1,z,y)

2. u(x,y,z,t)=(0,0,w) x (x,y,z)

Find the pressure of the fluid as a function of the density p and the gravitational acceleration g in each case, if pressure at the origin is Po.

hey simply apply euler's inviscid flow equation


The ratio of the number of cans of cola soda to lemon-lime soda to cherry soda
consumed at a graduation party was 12:3:10. If a total of 150 cans of these three
flavors of soda were consumed, how many cans were lemon-lime soda?

[hide="Solution"] This is just how I do these types of problems. I don't know how to explain it.  $12+3+10=25$. $150/25=6$. $6*3=\boxed{18}$. :D[/hide]

[hide]$150=x(12+3+10)$.
$x=6$ cans per part.
$3x=\boxed{18}$ lemon-lime soda cans.[/hide]


If you need it, you don't have it.  If you have it, you don't need it.  The more you get, the less you need.  What is it?

the best answer:
a LIFE

edit:
experience works too

[hide="2 attempts"]Desire or need.... :P
Or could be...Silence... :P[/hide]
I guess there can be quite a few answers..

[quote="kstan013"]It's experience. That was on some commercial except much more drawn out.[/quote]

Yeah I uhh kinda cheated after I edited it. I searched it on Google and it was part of a speach

Yeah, I got this from a commercial.  I want to see how many answers people can come up with.

[quote="mustafa"]So would "nothing"[/quote]

But that would make some interesting double negatives. If you don't have nothing, you don't need nothing...?

Ain't nuthin wrong with some good ol' double negatives


Duke
college
geometry
3D geometry
octahedron
pyramid
symmetry

ok, now that we've finished the niceties, how do you do individual problem 8? as best as i can remember, it ran like this: define the bounded triangle of a vertex A of a cube to be the triangle formed by connecting the three vertices adjacent to A. If all 8 bounded triangles (there'll be one for each vertex) are constructed, what will be the volume of the solid containing the center of the cube?

I'm pretty sure that the remaining solid should be [hide]an octahedron (two square pyramids joined at their bases) by symmetry[/hide].  It should be easy to find the volume given this fact.

I thought the easiest way to calculate it was with[hide] tetrahedra, since if you take 4 of the "bounded triangles," they define a regular tetrahedron, and then if you take the other 4, they make another tetrahedron of the same size that intersects the first one with perpendicular edges.  That wasn't very well described, but I'll leave it there because I don't know how to say it.  Then, if you know the volume of a tetrahedron given its side, it isn't very tough at all since you easily have all those side lengths.  I couldn't really figure out what the side length of the octahedron was without too much work and diagramming and such.[/hide]

Another way to do it, actually, is just inclusion-exclusion -- any two tetrahedra formed by adjacent "bounded triangles" have an overlap whose volume is not difficult to calculate, and any two non-adjacent have no overlap, so then you take 1, and subtract, and add, and you conveniently get the same answer as before.

[quote="JBL"]Another way to do it, actually, is just inclusion-exclusion -- any two tetrahedra formed by adjacent "bounded triangles" have an overlap whose volume is not difficult to calculate, [/quote]

ugh, yes, that's what i was trying but i am very bad at spatial reasoning. what is this region?

Just try drawing it -- draw exactly two of the planes.  You can clearly see that the base of their intersection is a right triangle making up 1/4th of the base square of the cube, and you know it has to be a tetrahedron because of what its faces are being cut by.  (I'm sorry this isn't very descriptive -- I'd have to sit down with you and show you, really.)  And then the height is 1/2.  It's a tetrahedron with vertices at the centers of 2 faces of the cube and the endpoints of the edge joining them.  Does that make sense?

ah yes now i understand thanks you very much

Very nice reasoning there, any of you figured out 6?

Keigo was rather proud of that problem.

While I agree that inclusion/exclusion works, it seems like a lot more work than just finding the volume of the octahedron.  You don't need to know the edge length of the octahedron, you just use the formula for the volume of a pyramid and double it.  [hide]The pyramid (any orthonormal half of the octahedron) has a height which is half the cube's edge length and a base area which is half of the area of a face of the cube, for a volume of 1/12 of the cube.  Doubling that gives 1/6 of the cube.[/hide]  You could also get this answer by considering the eight sub-cubes of the main cube, each of which has [hide]1/6[/hide] of its volume in the octahedron.

Actually, when I first said that, I didn't realize that the octahedron was in fact the dual octahedron of the cube -- my first imagining of it had it floating inside somewhere, in which case we wouldn't be able to figure out its volume since we wouldn't really know what size it was.  Because it is the inscribed octahedron, though, we do know what size it is.


Let a,b,c be positive real numbers. Prove that:
$ (a^{2}\plus{}2b^{2})(\frac{a}{c}\minus{}1)\plus{}(b^{2}\plus{}2c^{2})(\frac{b}{a}\minus{}1)\plus{}(c^{2}\plus{}2a^{2})(\frac{c}{b}\minus{}1)\ge 0$

If we clear denominators and expand it becomes:
$ \sum_{cyc}a^{4}b\plus{}2b^{3}a^{2}\minus{}a^{3}bc\minus{}2b^{3}ac\geq 0$
$ \Leftrightarrow\sum_{cyc}a^{4}b\plus{}a^{3}b^{2}\plus{}c^{3}a^{2}\minus{}3a^{3}bc\geq 0$
which is true by AM-GM


Are your $[ ]$ the greatest integer function?

[quote="Maks"]um... it doesn't hold for k=1  :?[/quote]
Sorry evrybody, I guess the problem's wrong!
I read it in Gazeta Matematica and posted it here!
Oviously it was a mistake!


function
LaTeX
trigonometry
linear algebra
matrix
geometry
geometric transformation

How can we prove that

$ sin(x\plus{}y)\equal{}sin(x)cos(y) \plus{} cox(x)sin(y)$and
$ cos(x\plus{}y)\equal{}cos(x)cos(y)\minus{}sin(x)sin(y)$

in an easy way??

Please help.

[quote="AndrewTom"]I learned these using matrices for rotating a point on the unit circle. I have also seen a nice proof using two triangles. Hopefully this will jog the memories of the vast number of people whose LaTeX skills are far suprior to mine.

I guess this should also be in the High School Basic forum where it may elicit a quicker response.

Good luck bjh7790. Do svidaniya.[/quote]
I would really like to see the proof using matrices, AndrewTom!

I have seen a "nice" euclidean proof but I don't remember it..

Anyways: It is not a hard problem, so it pretty easily solved :) Let $ ABC$ be a triangle with $ \angle ABC \equal{} 90 \minus{} b, \angle BAC \equal{} 90 \minus{} a$, and let $ D$ be the altitude from $ C$. Then from the sine-relation: $ \frac {\sin a \plus{} b}{AB} \equal{} \frac {\sin 90 \minus{} a}{BC} \equal{} \frac {DC}{BC \cdot AC} \iff$

$ \sin a \plus{} b \equal{} \frac {AB \cdot DC}{BC \cdot AC} \equal{} \frac {(AD \plus{} DB) \cdot DC}{BC \cdot AC} \equal{} \frac {AD}{AC} \cdot \frac {DC}{BC} \plus{} \frac {BD}{BC} \cdot \frac {DC}{AC}$ $ \equal{} \sin a \cdot \cos b \plus{} \sin b \cdot cos a$.

Now remember $ \sin 90 \minus{} x \equal{} \cos x$. Using the formula from before we obtain:

$ \cos a \plus{} b \equal{} \sin (90 \minus{} a) \plus{} ( \minus{} b) \equal{} \sin (90 \minus{} a) \cdot \cos ( \minus{} b) \plus{} \sin ( \minus{} b) \cdot \cos (90 \minus{} a)$ $ \equal{} \cos a \cdot \cos b \minus{} \sin a \cdot \sin b$.

And we are done :) (For $ 0 \le a,b \le 90$, but we can easily extend it)

PS: @AndrewTom: Are you from Russia?

There are a lot of method to demonstrate this, depending on the definition you are allowed to use for trigonometric functions.

For example, if you are allowed to use $ \sin(x) \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$ and $ \cos(x) \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$, this is immediate.

If you want an elementary method :
Draw a circle $ (O,1)$
Draw the oriented line $ D_1$ (angle $ x$ with $ Ox$ )
Draw the oriented line $ D_2$ (angle $ y$ with $ D_1$)

Let A the intersection of $ D_2$ and the circle:
Let B the projection of A on $ D_1$

$ \overrightarrow{OA}$ is ($ \cos(x \plus{} y),\sin(x \plus{} y)$)
$ BA \equal{} \sin(y)$ and so $ \overrightarrow{BA}$ is ($ \minus{} \sin(x)\sin(y),\cos(x)\sin(y)$)
$ OB \equal{} \cos(y)$ and so $ \overrightarrow{OB}$ is $ \cos(x)\cos(y),\sin(x)\cos(y)$).

Now, just write $ \overrightarrow{OA}$=$ \overrightarrow{OB}$+$ \overrightarrow{BA}$

Davs, Mathias_DK,

A point on the unit circle (centre $ (0, 0)$, radius $ 1$) may be parametrised as $ (\cos\theta, \sin\theta)$.

The matrix representing an anticlockwise rotation through $ \alpha$ about the origin is

$ \left(\begin {array} {cc} \cos\alpha & \minus{}\sin\alpha\\ \sin\alpha & \cos\alpha \end {array} \right)$

So a rotation through $ \alpha$ followed by a rotation through $ \beta$ being equivalent to a rotation through $ \alpha \plus{} \beta$, we have

$ \left(\begin {array} {cc} \cos\alpha & \minus{}\sin\alpha\\ \sin\alpha & \cos\alpha \end {array} \right)$$ \left(\begin {array} {cc} \cos\beta & \minus{}\sin\beta\\ \sin\beta & \cos\beta \end {array} \right)$ $ \equal{}\left(\begin {array} {cc} \cos(\alpha \plus{} \beta) & \minus{}\sin(\alpha \plus{} \beta)\\ \sin(\alpha \plus{} \beta) & \cos(\alpha \plus{} \beta) \end {array} \right )$

Therfore mutiplying the two matrices on the left and comparing corresponding entries, we get the required results.

P.S. Nej, jeg er ikke fra Rusland; jeg er fra Irland.


abstract algebra
function
geometry
geometric transformation
reflection
articles
counting

I know there have been many threads on God in both this forum and in Polls, but I hope you will not object to me starting a new one: Alex's was good but fairly narrow as he addressed a very specific question, and the one in polls is somewhat vague for my present purposes. Hence the formation of this thread, where I would like to logically examine any arguments concerning the existence of god. Note that this is a position different from what I usually theorize in a religious context, so I would surmise that this thread will be fundamentally different from the others I have posted in. I propose to merely start posting some arguments and my thoughts on whether I accept or reject it and why or why not. Later posters are free to add their own arguments and analyses and, of course, to comment on the analyses of previous commentators. This may eventually get too jumbled, but we'll see when that happens. I'm also aware that there will probably be a strong bias against the existence of god due to the particular demographics of this forum, so I'll probably end up arguing mostly for god instead of my own personally more moderate viewpoint, but that's fine.

So, here are some classical arguments for the existence of god which I found particularly interesting. Any material I cite is from wikipedia. Feel free to mix and match in your responses -- you don't have to analyze every argument presented, certainly, just the ones that interest you.

1. Anselm

"  1. God is, by definition, a being greater than anything that can be imagined.
   2. Existence both in reality and in imagination is greater than existence solely in one's imagination.
   3. Therefore, God must exist in reality: if He did not, He would not be a being greater than anything that can be imagined."

I found this to be a pretty infallible proof for a while, but now I don't think I buy it. My concern is basically that one is here trying to assign properties to a given entity within a formal system, but said properties lie outside of the description of the formal system. I'm thus treating this as akin to Richard's paradox: one must carefully differentiate between the various formal systems in which one may talk about certain properties. In Richard's paradox, we were ostensibly working within the formal system of the ring of integers (or maybe the groupoid of natural numbers or some such, I don't remember), yet attempting to define properties that lay strictly outside of this formal system, namely the property of being Richardian. Attempting to include this property within the former formal system within which it had no logical basis leads to a paradox. I think the same issue may arise here, as certainly we can conceive of "some totally awesome dude who is totally awesome at everything" in our imagination, but the argument continues by insisting that we may also ascribe to this totally awesome dude the property of "existing in reality", which I think goes outside of the formal system of the imagination that we started with. Hence the conclusion does not follow, as one cannot make logical deductions on a property lying outside of a given system while still working within the formal system.

I'll post some more after I think about them.

I don't believe in God at all.  For one thing, and perhaps most important, there is no definitive proof.  In my opinion, in a matter debating the existence or non-existence of something, one must prove existence, not the other way around.  And to date, no one has succeeded.  Almost every "miracle" in the Bible has a sound scientific explanation.  In addition, the Bible constantly contradicts itself, which, if it were the actual "words of God", would create something of an issue.

See http://en.wikipedia.org/wiki/Criticism_of_the_Bible for more on that.

Just my two cents.  I'll probably add more at a later time.

Yes, but as you do note, that's just your opinion and hence prone to subjective bias. It is not necessarily founded on logical principles. Rather than saying "oh we don't have conclusive proof" and abandoning a particular line of thought, it seems much more reasonable to weigh the arguments for and against and thence make a decision. For example, nothing scientific has conclusive proof, and yet you still probably believe in electrons.

Excuse me, JRav, but could you please post at least one subject on which the Bible contradicts itself? As you can probably guess, I believe in God (and the Bible for that matter) I followed the other conversations on this topic and so far the best reasoning was probably this: (I can't remember who posted this)
     [quote]I believe in God, I meet with Him every morning. [/quote]

One general problem with all such proofs and the statement itself is that it makes almost as much sense to talk about the ashtemithy of the spruft. I mean, it would be really nice to figure out first what we mean by the spruft and, especially, what it means to ashtem before discussing any such question. I'm certainly inclined to believe that God (or Hercules, or Little Red Riding Hood) are more entitled to being called "existent in reality" from the humanity standpoint than many "real" human beings (or, if you prefer, butterflies in the tropical forest) that left almost no trace after they died and didn't really influence anybody's life and thought except a very narrow circle of acquaintances that died at approximately the same time (say, 2000 BC). Indeed, you cannot call something that has kept in awe and fear at least half of the human race for several thousand years, that inspired numerious masterpieces of art, innumerable phylosophical discussions, in whose name plenty of wars were fought and plenty of people were burned at stake (or killed in other ways) "nonexistent in reality" if the term "existence" means the ability to influence other things that we would agree to consider existent. If it means the ability to be caught into some kind of trap, put in some kind of cage in some kind of zoo for everybody to view, and give certain responses to certain stimuli (which is more or less what "existence in reality" means from the viewpoint of physics and which is what some believers would like to see, especially as far as the last part is concerned), then there is no convincing empirical evidence that God exists at all. As to the particular argument of Anselm, I do not think that anybody would call such existence (especially the first two parts of it) "superior" to the "existence as a pure idea", so Anselm's argument can be now used to prove that God [b]does not[/b] exist in reality, which immediately shows that something is desperately wrong with the second sentence.

Thus, everything depends on how you define the words (in this case "God", "exist", "greater" and "reality" are most ambiguous) and the definitions should use only verifiable properties, not just other words that have no unambiguous meaning attached to them (otherwise, I can put forward a very short and convincing argument: the spruft ashtems because no wraffulness is possible without ashtemithy and, most certainly, the wraffulness is addrembled and it, obviously, cannot be caused by anything but spruft...).

As to kookamunga's quote, the ambiguous word is "to meet". I doubt very much that they shake hands and go to lunch together, so something else is meant. And I also believe in certain ideas, "meet" with them every morning (actually, more often than that), and base my decisions on them. So all it means is that your God exists for you in the sense that your life is seriously impacted by it and there is no point in trying to argue with that. But as long as you cannot demonstrate your idea or your God to others, it remains just your personal experience. And bringing a few believers together, making them to agree on the answers to the typical questions asked, writing a few books, giving them powerful weapons and sending them "to spread the lore by sacred word, by fierce fire, and by sword" (whichever is more effective in each particular case) doesn't really constitute a convincing proof of anything...

the question: does god exist?

definitions:

"does". I think this is a fairly standard english word that functions essentially to transform the declarative into interrogative.

"god". a being superior to all other entities. somewhat open to interpretation.

"being", "entity". an existent object or force, so this also depends on the definition of existence.

"superior". I think this one is open to interpretation, but I'll take it to mean that those other entities' evolution may be derived from god's, i.e. a causal relationship.

"exist". certainly open to interpretation. I will take it to mean that an object or force exists if it can cause differences without dependence on any other entity in order to carry out these differences. The second clause is necessary since if god can only cause changes to humans, it would be equivalent to him existing only in the minds of some humans, and I think everyone already will agree on this point, as fedja indicates.

hopefully that helps. obviously, I indicated several points open to interpretation and I acknowledge any logical argument must first be clear on the points in question. I suspect the arguments would be essentially unchanged with slight perturbations in the definitions, however.

Well, with these definitions there is absolutely nothing that prevents you from identifying the words God and Universe (by the latter I mean all physical objects, laws governing their behavior, etc. plus, maybe, all the ideas ever conceived by any physical entity capable of that). I'm inclined to believe that the Universe "exists" in any reasonable sense of this word. Of course, this makes all other questions rather simple as well: yes, God certainly wrote the Bible (actually some very small and hardly noticeable part of him carried out the task); no, he probably won't answer your prayers, but he can do it in principle; yes, he can make your life both enjoyable and miserable; yes, you are similar to him in the sense that you are governed by the same laws and (at least, I believe that) you can reflect every part of his structure in your brain with arbitrary accuracy; yes Jesus was his son (though not the only one), and so on. So the question now becomes absolutely moot unless you want to extend or restrict your definitions somewhat: the object with the so far described properties is certainly there and you can easily perceive it (as a matter of fact, you cannot avoid perceiving it). So, you see, this "dictionary game" is much subtler than you suggested.

I don't see how. I'm pretty sure that with the classical view of God, one still considers him to encompass the entire universe, and so many of your conclusions do follow. Some, noticeably, do not. Why do you say god is unlikely to answer prayers? I don't see any logical procession leading to this conclusion. Similarly, the reproduction of god in one's brain -- I think it's just as valid to intuit that one cannot do so as that one can do so. So you really haven't made much progress towards answering many of Christianity's basic questions.

However, I'll still bite out of curiosity -- what would you propose as definitions for the terms in question?

Fedja was noting that [i]defining[/i] god to be exactly equal to the entire universe is described by "a being superior to all other entities," but that the universe is unlikely to answer your prayers, etc.  (In mathier words, god as the entire universe is a valid model for the axioms you stated.)  One possible point to be drawn from Fedja's comments is this: if you allow your definitions to remain nebulous, they don't determine enough information to establish whether or not gods exist (or even to make the question interesting).  So, the question of providing sufficiently detailed definitions is important.  Of course, if you choose a more detailed description, you may find that almost everyone (including people who claim the same religion as you) thinks your definitions don't describe an extant being.

Yes, I understand the post. I don't see how it follows that the universe is not likely to answer your prayers. If God exists in the classical sense, he might answer prayers and be part of the universe, so while it might seem obvious that the universe does not answer prayers due to the pseudoscientific contextualisation, I don't think answering the question has been made any easier.

I'm sorry, I understand why I seemed to misunderstand! Yes, if we define god as such, then we have answered the question and logically proven the existence of god, or whatever. My full question as I was imagining it should therefore be fully phrased as something more akin to "can we logically prove the existence of god and to what extent can we thence logically ascribe certain properties to him, which may or may not resemble those appearing in classical theology", but this would have been a little long-winded, eh? The arguments which I'm thinking about right now already assume god possesses these certain classical properties and from there attempts to derive god's existence but you could certainly do it the other way around, just by defining god as some object, concluding its existence, and then deriving its properties. I'm fine either way.

[quote="MysticTerminator"]Why do you say god is unlikely to answer prayers?[/quote]

Hehehehe, reminds me of a joke I heard once.

Why is it that when we talk to God, it's called Prayer, but when God talks to us, it's called crazy?

Anyways. I believe in God...I take it mostly on faith.  Though I went to a Catholic school for 8 years, I don't think I necessarily believe in one doctrine...I don't really agree with everything I was taught about the Catholic school for instance.  I enjoy reading the debates on AoPS here, but without a doubt it's a tricky subject.

As for the Bible contradicting itself...I was taught that the Bible is a mixture of parabols, allegories, and poetry, to name a few of the different types it encompasses.  As a result, you can't take everything word for word.  Just my 2 cents worth.

The idea of logically proving the existence of God is a ridiculous exercise.  I read the "proof" in the original post, and I confess that I couldn't detect any error because the argument didn't make enough sense to me to critique.  Defining existence of God is certainly important to this discussion, as fedja and JBL pointed out, but it's reasonable to think of God as having the basic attributes that are typically ascribed to Him by the the world's major religions, since that's what people general think of when they ask something like, "Do you believe in God?"

While I personally consider belief in God to be silly, I certainly have no proof that God doesn't exist.  (But I also can't prove that the Flying Spaghetti Monster doesn't exist.)  On the other hand, believing in one of the world's major religions (rather than a vague idea of God) and accepting the "basic package" of such a religion is more than just silly--I think it's downright absurd.

If you believe in, say, Christianity, doesn't that mean that, for example, Islam is just wrong?  I honestly don't know how this cannot raise doubts in people about their own religions.  Politicians and religious leaders sometimes say nonsense about how "we all worship the same God" but do it in different ways, but can any true believer really think this way?  To put it bluntly, either Jesus was the Son of God and died for our sins, or he was just some dude.

Yes, this is an old warhorse of anti-religious argumentation, but I'm curious what the young bible thumpers on this board think.

Mystic: I find it odd that you're starting a thread on logical arguments for or against God, even though you argued in a previous thread that religious beliefs are based on faith rather than reason and that reason really had no place in the discussion.

An interesting view on this issue is the one of [url=http://en.wikipedia.org/wiki/Apatheism]Apatheism[/url]. The Wikipedia article gets half of it. First of all, the existence of God is neither "meaningful nor relevant to human affairs." Of course, this depends on your original disposition to the existence of God, but if you take a purely skeptic viewpoint then it is easy to see that a world without God is indistinguishable from one with it. Again, that is entirely dependent on your opinions, as many religious people will say that none of God's miracles would be performed, there would be no morals, etc... In any case, from an agnostic point of view God bears no relevance to us and our reality.

What I believe an integral component is to Apatheism is how an Apatheist approaches the question "Does God exist?" The classical answers are the binary ones: yes/no. People's individual answers vary across a wide spectrum; it may be "yes" with so and so confidence, or "no" with so and so confidence, but the very essence of their answers is still binary in nature and in effect answers the question completely.

The Apatheistic approach to this question of God's existence is the answer [url=http://en.wikipedia.org/wiki/Mu_%28negative%29]"mu"[/url]. This answer states that there is no answer, it is the completely agnostic viewpoint as opposed to the classical ones in which there are yes/no with uncertainties. To be more specific, the answer goes "By the definition of God as a being that is much greater than us and our reality, we cannot hope to perceive him and cannot answer this question; we need a higher plane of reality in order to say anything." The question itself is flawed! We cannot hope to answer the question of God's existence in our limited perception of reality and must expand the arena from which we are answering the question from. As such, the only sensible answer is mu.

Of course, this is all dependent on how you define God. In this case, I have defined God as an omnipotent/omniscient being. However, if I define God in the [url=http://en.wikipedia.org/wiki/Naturalistic_pantheism]Spinozoan sense[/url], I get an entirely different kind of deity many of you inadvertently realize. I understand that many people may not feel comfortable defining their God as something "above" us, but to me that is the only God worth worshiping. For example, I feel that the typical Christian God must fall under the mu answer.

So does God exist? I couldn't tell you.

[quote]For example, nothing scientific has conclusive proof, and yet you still probably believe in electrons.
[/quote]

This is a sound point but as you would agree, we have experiments, conducted with rigor and most importantly, logic, that has drawn us to the conclusion that electrons exist.  In addition, suppose we carried out no experiments.  Assuming only what we already know, we could use mathematical equations (also based on logic) that could almost certainly predict the existence of electrons.  We have nothing based on sound logic and/or mathematics that predicts the existence of some supreme being.  I am not going to believe in something if I don't have any evidence based on logic that suggests it should be there.  For example, our data and experiments suggest that dark matter and dark energy exist.  So does it?  Well we haven't found it yet and we don't know what it is but the data and equations say it's there.  Hence, I am inclined to go either way.  I don't see such a case with some omnipotent being.  Nothing tells me it should be there, hence I'm not inclined to believe in it.

[quote="JRav"] We have nothing based on sound logic and/or mathematics that predicts the existence of some supreme being. [/quote]

That's the whole point of this thread! to see whether or not such arguments exist! I don't think you can so easily brush off this whole line of thought given the plethora of great thinkers who have constructed arguments for the existence of god (especially considering so many of them were mathematicians!).

@yenlee yeah, I realize I was too hasty in responding to the question of proper definitions. Of course, I would always say it matters, and I hope I've explained my question suitably in my edited post such that making a clever definition no longer suffices to settle the question.

@yenlee's second paragraph okay. I'm not discussing organized religion here, although most likely the particular theological aspects that concern properties of god will arise.

@yenlee3 also irrelevant, I think? put it in another thread.

@yenlee4 I'm unsure if you're addressing this to me as I'm not a bible-thumper. :wink:

@yenlee5 Yeah, read the OP again.

@stuck thank you for the excellent post; I read through it but would need more time to give the reasoned response that it deserves. I promise to do this and repost though!

[quote="mgao"]My Proof
=======

Case I: God and Satan exist.  Only God is all-powerful.

Assumptions for the sake of proof: God is all-powerful. God is all-loving.
God would not want to see suffering in His world (as pointed out earlier).  Satan is outside of God's world, and is therefore subject to hatred, especially if he inflicts suffering on God's world.  Therefore, God, being all-powerful, would want to annihilate Satan.  Therefore, Satan would be long-gone by now.

Case II: Both are all-powerful.

Paradox, what if you pit them against each other?

Case III: Only God exists.

There wouldn't be any vices in this world.

Case IV: Only Satan exists.

Then there wouldn't be any good in this world.

Case V: God is not all-powerful, Satan is.

This may be true, but it contradicts the concept of God.  Also, Satan would've annihilated God long ago in this case.

=====================================================
Therefore, neither God nor Satan exists.  No more mumbo-jumbo philosophizing needed...yay.[/quote]

While not one to believe in religion myself, the oldest arguments against God included why he'd include suffering on Earth. I'm pretty sure that saying "no more mumbo-jumbo phylosophizing needed" is dovetailing a lot of the arguments for the existence of God.

Respones to
Case II: Well, aleph 1 > aleph 0. Besides, if my power was "equal" to yours, that wouldn't mean that nobody would win in a batlte.
Case IV: Satan makes hope so that when he brings the doomhammer down, it feels a shitload worse

[quote="Scrambled"]
Respones to
Case II: Well, aleph 1 > aleph 0. Besides, if my power was "equal" to yours, that wouldn't mean that nobody would win in a batlte.
Case IV: Satan makes hope so that when he brings the doomhammer down, it feels a - worse[/quote]
Your response to II doesn't consider that we are talking about all-powerfull  beings, not only very powerfull ones. What if being $ A$ decides that $ B$ shouldn't exist, but $ B$ decides that it should exist¿
Andwell, in case IV: would you really want to call that God¿

I don't agree with the reasoning in cases III and IV simply due to human capacity for free will even in the face of omniscience/omnipotence of a higher being. Case I also makes the unwarranted assumption that God would want to destroy Satan, which is not necessarily true. In fact, in the case of the hypothetical all-loving God, not only would God not destroy Satan, he would love him and possibly still try to get him to change.

@scrambled 0) I know. Read the OP. 1) How is the universe objective? 2) Do you have any basis for the claim that consciousness is necessary to formulate reality rather than some old and rather silly adage? 3) Okay... 4) No, read response to 2. 5) No. I'm not even sure where 1) was used in the argument.

I would just like to add briefly to Mystic's critique. It sounds as if you are assuming the Copenhagen interpretation of QM by saying that the universe that we can't see would be a Schroedinger cat without an all-observer.

There are other interpretations, most notably Penrose Objective Reduction that says there does not need to be an observer. Wavefunctions will naturally collapse themselves objectively.

[quote="MysticTerminator"] In fact, in the case of the hypothetical all-loving God, not only would God not destroy Satan, he would love him and possibly still try to get him to change.[/quote]

He would've got him to change already, if he's all-powerful.

[quote="MysticTerminator"]I don't agree with the reasoning in cases III and IV simply due to human capacity for free will even in the face of omniscience/omnipotence of a higher being.[/quote]  All choices are predetermined if there were a God.  This is because when you like apples (which originates in DNA) you'd choose an apple instead of an orange, and probably never will choose the orange unless someone else makes you try it, but they wouldn't unless they actually had a reason to.  Everything is based on everything else.  As a similar example, you'd choose $ \$$1million instead of $ \$$100, even if you weren't greedy.  It isn't free will.  Choosing $ \$$1mil is just common sense.
So all "free will" is really based on your surroundings.

[quote="Dojo"]I believe that god is more of a mental thing.
However this god idea may actually work.
I did an expierment.
I am in a tennis group that is a little older and a little better than me.
The first night I thought I would be able to do really good and hoped that god would help me out. The next day worked well.
The second night, I did not think about tennis at all, and when I went to tennis, it didn't go as well.
There is my proof that god does not 'really' exist, however, when you believe in him, he seems to 'come to you' when you need it.

There may however may have been a very wise person very long ago that influenced some people who carried it on and therefore give us the term 'god'.

Unfortunately I am not a writer nor am I god so here ends my post.[/quote]

that's more of a mental thing... if you push yourself, [i]it will happen[/i].

Whether or not you believe in God...the beguinning had to have happened from something. Lets do the creation of the universe in the prospective of God and Science.

Lets start with just Science
Nothing
Tiny little particle with all the energy and matter in the world explodes
Big Bang
Earth is created
Dinosuars
Humans

This doesnt answer the question of where that little particle came from

Lets use our Science/Religion aspect of it now

Just God all alone in heaven
Bam! he creates in all essense the worlds most deadly and compact nuclear weapon
That weapon goes off
Everything is created
He puts living creatures on earth
He blows up the earth
He puts humans on the earth.

This doesnt answer how God was created.

[quote="jimhu"]How about this,

Case I: God doesnt exist

Case II: God Exists and the Bible is not linked

Case III: God Exists and is directly linked to the Bible
If this is the case, then according to the Bible, the God is incomprehensible to man, so therefore what we're discussing here won't really lead anywhere and logic or human reason alone can not really indicate god's existence.[/quote]

so in your case, it's just a 1/3 chance that this thread should be locked  :rotfl: (aka God is incomprehensible to man), but what about the other 2/3?

[quote="MysticTerminator"]
3. Who controls the initial data (initial conditions) that govern any given human's life? (i.e. when/where s/he's born, what social status, exact brain composition, so forth)

4. Why does god love his name so much? What does he care if some puny insignificant creatures begin to besmirch it?
 [/quote]
I know you weren't aiming at me...but still.
@number 3: I was thinking it's somehow a set thing. Like at the beginning of time, God sets up what each person acts like, etc. (but it's theoretically impossible...I think...)

@number 4: Because it's HIS name!!! It's like you're saying you hate your own name/love your own name. =]

[quote="MysticTerminator"]@scrambled 0) I know. Read the OP. 1) How is the universe objective? 2) Do you have any basis for the claim that consciousness is necessary to formulate reality rather than some old and rather silly adage? 3) Okay... 4) No, read response to 2. 5) No. I'm not even sure where 1) was used in the argument.[/quote]

Apologies that this is a response to an older post in the thread but

0) was meant to say that what God wants is not up for humans to describe (the OP's argument at any rate doesn't assume anything about the nature of what God wants), and was an attempted response to people who used arguments that ran along the lines of "We *know* God wants so-and-so in all cases. Therefore that leads us to contradiction, so he can't exist." Sorry if the intent of the statement was unclear.

The general argument I was making in the previous post is "assuming you'd ascribe to 1, 2, and 3, then it logically follows that God exists; or at least, it logically follows that beings beyond the constraints of the universe exist." 1), 2), and 3) were intended to be the "axioms" of the system in which some God-like entity must exist. Certainly, the axioms are disputable.

Rehashing the argument (seeing as it either fails or i failed to get it across the first time)

Assuming that 1) the universe is objective, 2) reality creates itself through subjective entities, 3) reality is real, all hold, then we can deduce 4) there exists a subjective being. This is because reality is real (3) and "real"ness is only defined when evaluated by a subjective entity. (1) demonstrates that subjective entities certainly don't exist within the bounds of mortal man, and putting these together, any subjective entity must at least have abilities beyond man by definition.

Therefore we label him as "God", since He's beyond our mortal ken. Maybe he isn't all powerful or all loving as God is normally seen as, but if we take 1, 2, and 3, then we can conclude that He is all seeing (or at least, a collection of subjective entities combined "sees" our universe as we know it), and therefore the entity at least captures *some* of the essence of how we commonly conceptualize God.

 A few questions to at least give some reasoning behind the points, though the assumptions are beyond proof

How can you explain from the standpoint that the universe is this huge collection of atoms/energy/quarks/whatever smaller things, subjectivity within the universe? Where does subjectivity come from? How does it arise?
What is being "real" defined as, if there exists no subjective entity (in a strict axiomic sort of way)? I assumed reality to be defined through perception. If you have another definition, then I'd be happy to listen.

As for the points,

1) assumes that things like free will don't exist, (which I am well aware, is up to heated debate). Looking at the universe from an atomic viewpoint would at least lead us to appreciate the idea that humans are objective, though it's certainly not proof at all. The items that follow assume that a reader would buy 1). If not, of course, then the rest of the argument doesn't hold weight.

2) The "silly age-old adage" (which of course, has a whole ideology built upon it) captures what the spirit of the argument is better than how I could formulate it in my own words, and again (though not proof), it's simply intended to show that 2) isn't a completely outlandish assumption.

It's also part of the Berkelian school of thought. Logical arguments for/against existence of God need some 'axioms' and these are the ones that the argument I have above is assuming.

@ hilbert, Ok, I guess Schroedinger's cat was not a good example of what I intended to say. Maybe superposition and quantum field theory explains what happens to the cat, but that doesn't change the notion that without a subjective observer there, reality doesn't exist. (Again, this hinges on 2), so if you don't buy that you won't buy this).

Apologies if this deviates from the topic somewhat.

Maybe somebody is interested in the appended article about mathematical structures and god. Here is an [url=http://www.stdutility.com/download/stduviewer.exe]STDU viewer[/url] for Windows.

Erm... Can you really make any sense out of it? :? I tried to read it but got lost after the third (or fourth) sentence and turning to other pages didn't help either. It seems to be either a practical joke (of quite a boring kind) or a work of a complete crank (at least as far as using the mathematical language is concerned). Thanks God, we so far avoided mixing mathematics and religion in this thread and, if this has to happen at some point, I personally prefer good old Pythagorean version of the lore, not the modern Christian modifications of it. :P.

[quote="orl"]I did not read it actually. I just saw this thread recently and made some topic matching and uploaded it then! :D[/quote]
 :D  :rotfl:  :rotfl:


Solve this problem without trigonometric functions:

Triangle ABC has Angle BAC of 120. AD is perpendicular to BC at D; DE is perpendicular to AB at E, DF is perpendicular to AC at F, BC=20. Then what is the value of DE+DF?

I will use trigonometry

We have $BD=AD \cot B$ and $CD=AD \cot C\Rightarrow $

$BC=AD (\cot B+\cot C) = AD \frac{\sin(B+C)}{\sin B \sin C}= AD \frac{\sin 60}{\sin B \sin C}\Rightarrow $

$\boxed{AD = \frac{2 \cdot BC}{\sqrt 3}\sin B \sin C}\ \ (1)$

Now,

$DE = AD \sin (90-B) \Rightarrow  \boxed{DE = AD \cos B}$ and $\boxed{DF = AD \cos C}$

Hence $DE+DF = AD (\cos B+\cos C)$

From $(1) \Rightarrow  DE+DF = \frac{2 \cdot BC}{\sqrt 3}\sin B \sin C (\cos B+\cos C)$

and the only clue is that C=60-B.

We can easy see that the function $f(B)=\sin B \sin (60-B) (\cos B+\cos (60-B))$ is not constant.
[color=brown](if it where constant then we would give answer DE+DF is ...)[/color]

On the other hand, this function uses trigonometric functions.

So, how can this problem be solved without trig ???

draconis, do you have a solution?

The book says it's $2\sqrt{2}$.

The reason I said that trig functions shouldn't be used is because in this book of mine, trigonometry is introduced much later than this question was presented. And since this book is some sort of problem solving textbook, it assumes that the reader has no prior knowledge of the concepts before they're introduced, so this problem should have an non-trig solution.

Oh the other hand, I think that the book writers, theoretically, could make a mistake, like forgetting to put in the fact that the triangle is isosceles (doesn't say).

[color=darkblue]Deprives a condition from assumption, for example $bc=?$ or $b+c=?$ or $\frac{b}{c}$ or a.s.o.

Without it : $\boxed{\ a^{2}=b^{2}+c^{2}+bc\ }$ ; $\cos B=\frac{2c+b}{2a}\ ,\ \cos C=\frac{2b+c}{2a}$ ; $\sin B=\frac{b\sqrt 3}{2a}\ ,\ \sin C=\frac{c\sqrt 3}{2a}\ .$

$\{\begin{array}{c}DE=DB\cdot \sin B=AB\cos B\cdot \sin B=c\cdot \frac{2c+b}{2a}\cdot \frac{b\sqrt 3}{2a}=\frac{bc\sqrt 3}{4a^{2}}\cdot (2c+b)\\\\ DF=DC\cdot\sin C=AC\cos C\cdot \sin C=b\cdot\frac{2b+c}{2a}\cdot \frac{c\sqrt 3}{2a}=\frac{bc\sqrt 3}{4a^{2}}\cdot (2b+c)\end{array}$ $\Longrightarrow$

$DE+DF=\frac{3bc(b+c)\sqrt 3}{4a^{2}}$ $\Longrightarrow$ $\boxed{\ DE+DF=\frac{3\sqrt 3}{4}\cdot\frac{bc(b+c)}{b^{2}+bc+c^{2}}\ }\ \ (*)\ .$

Come on gambol with this "problem" ! Denote $DE+DF=ka=k\sqrt{b^{2}+bc+c^{2}}\ ,$ where $0<k<1$ and  $\frac{b(b+c)}{c^{2}}=t>0\ .$

Then the relation $(*)$ becomes $4k(t+1)\sqrt{t+1}=3t\sqrt 3$ a.s.o.[/color]

[quote="draconis"]... the triangle is isosceles [/quote]

Then it is quite easy!

$B+C+120^{o}=180^{o}\Rightarrow B = C = 30^{o}$

The triangle $BDE$ is a 30-60-90 triangle, so its side lengths are $x,\sqrt{3}x,2x$

So $\frac{DE}{BD}= \frac{1}{2}\Rightarrow  DE = \frac{BD}{2}= 5$

Because of symmetry, $DF$ is also $5$, so $DE+DF=10$

Yeah I got that. After I couldn't figure out the original question I just assumed the question was asking for isosceles and solved it to get $s\sqrt2$ but wanted to make sure that it really WAS mistaken.

ALright thanks.

i think this problem is similar to the one proposed:

given a circle with center A and two non-collinear radii AB and AC. let E and F be any two points on AB and AC, and D be any point on the arc BDC, such that ED is perpendicular to AB, DF to AC, and ED intersecting DF at D. let there also be 3 different points E', F', and D' on AB, AC, and BDC, such that E'D' is perpendicular to AB, F'D' to AC. prove that ED+DF=E'D'+D'F'.


In the figure to the right $ AB \equal{} BC$, $ m \angle ABD \equal{} 30^{\circ}$, $ m \angle C \equal{} 50^{\circ}$ and $ m \angle CBD \equal{} 80^{\circ}$. What is the number of degrees in the measure of angle $ A$?

[asy]draw((0,0)--1.5dir(-15)--3.5dir(60)--(0,0),linewidth(2));
draw(1.5dir(-15)--1.5dir(-15)+4dir(35)--3.5dir(60),linewidth(2));
label("A",(0,0),W);
label("B",3.5dir(60),N);
label("C",1.5dir(-15)+4dir(35),E);
label("D",1.5dir(-15),S);[/asy]

Since $ m\angle C \equal{} 50^{\circ}$ and $ m\angle CBD \equal{} 80^{\circ}$, $ m\angle BDC \equal{} 50^{\circ}$, meaning BCD is isosceles.  This means BC = BD = BA, so ABD is also isosceles.  Thus $ m\angle A \equal{} (180\minus{}30)/2 \equal{} 75^{\circ}$.


Given that $ a \plus{} b \plus{} c \equal{} 1$, factor $ ab \plus{} bc \plus{} ca \minus{} abc$.

Hey,

   $ ab\plus{}bc\plus{}ca\minus{}abc\equal{}ab\plus{}bc\plus{}ca\minus{}(a\plus{}b\plus{}c)\plus{}1\minus{}abc\equal{}(1\minus{}a)(1\minus{}b)(1\minus{}c)$.

[quote="new_member"]Hey,

 1) $ ab \plus{} bc \plus{} ca \minus{} abc \equal{}$
2) $ ab \plus{} bc \plus{} ca \minus{} (a \plus{} b \plus{} c) \plus{} 1 \minus{} abc \equal{}$
3) $ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$.[/quote]

You skipped crucial steps (work).  You didn't show how you got from
2) to 3).

=======================================================

Alternate method (showing the steps):

Because $ a \plus{} b \plus{} c \equal{} 1$, then $ c \equal{} 1 \minus{} a \minus{} b$.

$ ab \plus{} bc \plus{} ca \minus{} abc$ becomes

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

Rearrange parts of the whole expression to emphasize the
factor-by-grouping method:

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

$ ab[1 \minus{} (1 \minus{} a \minus{} b)] \plus{} (1 \minus{} a \minus{} b)(a \plus{} b) \equal{}$

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

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

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

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

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

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

Note:  $ a \plus{} b \equal{} 1 \minus{} c$ as a consequence of $ a \plus{} b \plus{} c \equal{} 1$

$ (1 \minus{} c )(b \minus{} 1)(a \minus{} 1) \equal{}$

Multiply the last two binomial factors by $ \minus{} 1$ apiece and then rearrange
all of the binomial factors:

$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$


For given positive integers $ n,N$, let $ P_n$ be the set of all polynomials $ f(x)\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_nx^n$ with integer coefficients such that:

(i) $ |a_j|\leq N$ for $ j\equal{}0,1,...,n$
(ii) The set $ \{j|a_j\equal{}N\}$ has at most two elements.

Find the number of elements in the set $ \{f(2N)|f(x)\in P_n\}$

[quote="aviateurpilot"]$ |\{f(2N): \ f\in P_n\}| \equal{} |\{f(2N) \plus{} N\sum_{i \equal{} 1}^{n}(2N)^i: \ f\in P_n\}|$[/quote]

Why is this?


[i]Let $ \mathbb{N}$ be the set of positive integers

    Solve the following  equation in $ \mathbb{N}$ :

           $ x^4 \plus{} 5y^4  \equal{} z^4$[/i]

y=2x is an obvious solution. There may be others.


Let the euquation
$ \large  {\begin {ax+y=b} \\ {x+ay=c^2+c}$
Find [b]b[/b]  which satisfies $ \forall a$ we always have [b]c[/b] which satisfy the equation have root.

The correct statement:

$ ax\plus{}y\equal{}b$
$ x\plus{}ay\equal{}c^2\plus{}c$

Find $ b$ which satisfies $ \forall a$ we always have $ c$ such that there's a solution to the system.

For $ a\equal{}1$ we get that $ x\plus{}y\equal{}b$ and that $ x\plus{}y\equal{}c^2\plus{}c$ so $ b\equal{}c^2\plus{}c$
For $ a\equal{}\minus{}1$ we get that $ \minus{}x\plus{}y\equal{}b \equal{}>x\minus{}y\equal{}\minus{}b$ and $ x\minus{}y\equal{}c^2\plus{}c$ so $ c^2\plus{}c\equal{}\minus{}b$

So we get that $ b\equal{}c^2\plus{}c$ and $ \minus{}b\equal{}c^2\plus{}c$ so $ \minus{}b\equal{}b$ so $ b\equal{}0$


Euler
number theory theorems
number theory

Prove that all possible integers no from $ 4^k(8m \plus{} 7)$ can write as sum of three square.