invariant
induction
combinatorics proposed
combinatorics

Let $G = (V,E)$ be a tree and $f: V \mapsto V$ be a mapping with the property that for each vertex $x,y \in V$ and $y \in U(x)$, it is implied that $f(y) \in U(f(x))$. This statement says that the environment is an invariant of the mapping  $f$: the image of the environment of $x$ is part of the environment of $f(x)$. The environment $U(x)$ of a vertex $x \in V$ is defined as $U(x) = \{y \in V|d_g(x,y) \leq 1\}$ where $d_g(x,y)$ is the distance between the two vertices $x$ and $y$. In other words the environment $U(x)$ contains $x$ and and all vertices that are directly connected with $x$ by an edge. Prove for which graphs the following statement is true: There is vertex $x \in V$ with $f(x) \in U(x)$. This statement says that every mapping $f$ has a fixed point $x$ or an "almost" fixed point $x$: $f(x)$ can be mapped on $x$ or is a neighbor of $x$. One can easily checks that it fails for quadrilaterals. A little bit more work is required to find a permuation that fails in the case of a icosahedron. But of course that are just trivial cases that you get  a feeling for the problem. You shall find a general answer.

Ok, it is easier than I thought...

We will prove the result by induction on the number $n$ of vertices of the tree.
Clearly for $n=1,2$ the result holds. Now assume it holds for any tree with $n$ and consider a tree $T$ with $n+1$ vertices.
Let $A$ be a leaf and $B$ be the only neighbour of $A$.

Let $M \in U(B)$ and assume that $f(M)=A$. Then $f(M) \in U(f(B))$ so that $A \in U(f(B))$. It follows that $f(B)=A$ or $f(B)=B$ and we are done.
Thus, we may assume that if $M \in U(B)$ then $f(M) \neq A$. (1)

Now, consider the mapping $g$ defined on $T' = T \setminus \{A \}$ by
$g(x) = B$ if $f(x)=A$, and $g(x) = f(x)$ otherwise.

It is straightforward to verify that $g$ satisfies the statement of the problem.
By the induction hypothesis, there is a vertex $M$ from $T'$ such that $g(M) \in U(M)$.
If $g(M) = f(M)$ we are done. Otherwise, it means that $f(M) = A$ and that $g(M) = B \in U(M)$. Thus $f(M) = A$ and $M \in U(B)$ which is not allowed from (1).

It follows that the result holds for $n+1$, which ends the induction step and the proof.

Pierre.


For n=3 it is true and comes from Schur. For n=4 it is false: a_1=1/2, a_2=1/2, a_3=1 and a_4=4. And I strongly believe that it is false for any n>=4. But I still cannot find an counterexample.

I think that we can choose a such that 1/2, 1/2, 1, 4, a,a,...,a is a counterexample.

Thank you,Harazi.
Note :when n=3,this inq appeared in IMO 2000

[quote="khoa"]Let n>=2 and a_1,...,a_n >0,satisfy a_1*a_2...*a_n=1.We have
      sum(a_1+1/a_1) \leq n+(a_2/a_1)+(a_3/a_2)+...+(a_1/a_n)
   I don't know it true or false.Can you help me[/quote]
I have a question:
this problem is not right however if we know a1 can we know the minimum value of (a_2/a_1)+(a_3/a_2)+...+(a_1/a_n)


[quote="inom"]Solve the equation

$cos^{2}(3x)+\frac{1}{4}cos^{2}(x)=cos3xcos^{4}(x)$[/quote]
Try using the command $\$$\cos x $\$$ for $\cos x$.  The problem is to solve \[\cos^{2}3x+\frac{1}{4}\cos^{2}x=\cos 3x\cos^{4}x.\]

Trigonometry -- ha!

[hide="Big hint"]Use AM-GM[/hide]

[hide] Maybe I'm missing something but ... AM-GM: $\cos^{2}3x+\frac{1}{4}\cos^{2}x\ge\cos3x\cos x$. How do we proceed?

$\cos3x\cos x\le\cos3x\cos^{4}x$??[/hide]

[hide]Under what conditions does equality hold?[/hide]

This is my guess:
$n\frac{\pi}{2}$, where n is any INTEGER.

I think that all you have to do is let:
$\cos(x)=\cos^{4}(x)$

Well,[hide]by AM-GM we have that \[\cos(3x)\cos^{4}(x) = \cos^{2}(3x)+\frac{1}{4}\cos^{2}(x) \geq |\cos(3x)\parallel \cos(x)|.\] RHS is positive, so LHS must also be positive, i.e. $\cos(3x)\geq 0$. Therefore, $|\cos(3x)|=\cos 3x$.
Clearly, $\cos(3x) = 0$ is a solution to the inequality, but then, we should have that $0 = 4cos(3x)\cos^{4}(x) = 4\cos^{2}(3x)+\cos^{2}(x) = \cos^{2}(x)$, so $\cos(x) = 0 = \cos(3x)$. But if $\cos(x) = 0$, then $\cos(3x) = \cos(x)(\cos^{2}(x)-3\sin^{2}(x)) = 0$. Therefore, $\cos(x) = 0$ yields a solution to the given inequality.
Now suppose $\cos(3x)\neq 0$, so $\cos(x)\neq 0$. We should have that $\cos(3x)\cos^{4}(x) \geq |\cos(3x)\parallel \cos(x)| = \cos(3x)|\cos x|$, i.e. $\cos^{4}(x) \geq |\cos x|$. As $\cos(x)\neq 0$, we should have that $%Error. "sign" is a bad command.
(\cos(x))\cdot\cos^{3}(x) \geq 1$, which is only possible if $\cos(x) = \pm 1$.
Therefore, $\cos(x) \in\{-1,0,1\}$. It's easy to conclude now.[/hide]

[quote="vishalarul"]I think that all you have to do is let:
$\cos(x)=\cos^{4}(x)$[/quote]
You got the correct answer, but $x=\pi$ doesn't satisfy this. :maybe:


Got a quick problem. I've been trouble shooting my miktex/texniccenter set-up and I think I have it but in setting up the filepaths so that there's no spaces in the address TeXnic Center now can't find the actual LaTeX program. How do I fix?

In TeXnicCenter, Build, Define OUtput Profiles, Wizard (in bottom left corner)


In a class of 25 students, 18 like pizza and 12 like spaghetti. What is the greatest possible number of students who dislike both pizza and spaghetti?

18 like pizza, and 12 like spaghetti. So 18+12=30 students like something. We don't know how many like both and how many don't like any. We want to find the greatest amount of kids who don't like either, and that would be if 12 kids like both, so the answer is 30-12=18.

Err, the answer is 7, not 18. How do you solve this problem?

This is a worst-case scenario problem.  We want to find the maximum number of students who dislike both pizza and spaghetti.

Since 18 students like pizza, there must be 7 students who dislike pizza.  However, suppose that the 12 students who like spaghetti are part of the set of 18 students who like pizza.  This means that there are no additional students who like either pizza or spaghetti.  In this scenario, which we are looking for, there are $\boxed{7}$ students who dislike pizza.


ABC is a triange .the bisector of the angle ACB,intersect AB in point D.point Kis on AB,inwich if AK and BK intersect

BC and AC in points E and F,the angles ADC and FEC will be equal.

prove that:

                angle DEF=90

ohhhh,,,sorry :(  :blush: .ure right shobber.K is on CD.


[hide="a start but its pretty obvious"]
the corner travels a semicurcular path

i've been working on a solution
but i screwed up
back to the drawing board...[/hide]

[quote="miyomiyo"]Compute the distanced traveled by the upper-left hand corner of a 9 in. wide by 12 in. tall sheet of paper when the upper-right hand corner is mapped onto the bottom-left hand corner.

Hide solutions and answers.  Show all justification.[/quote]

do you mean that you move the UR corner in a straight line towards BL corner?

When the upper right hand side is folded to the bottom left hand side, ...
a) how far did the upper left hand corner travel?
b) how far is it from its original position?

[quote="LordoftheMorons"][hide]First find the length of the diagonal. The distance formula yields:
$\sqrt{9^2 + 12^2} = 15$
The distance is a $180$ degree arc of a circle with diameter $15$. We thus find the distance to be $\boxed{\frac{15\pi}{2}}$[/hide][/quote]
Wrong corner. :D


Can someone tell me how to make matrices in Latex

\begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{pmatrix}
gives $\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$

If you want this in a document then ensure that you have \usepackage{amsmath} in the preamble.

Variations:

pmatrix: $\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$

bmatrix: $\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$

(I tend to like bmatrix.)

vmatrix: $\begin{vmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$

just plain matrix: $\begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}$

And what would happen if I did this:

$\begin{pmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}&5\\6&7\end{pmatrix}$

[quote="Kent Merryfield"]And what would happen if I did this:

$\begin{pmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}&5\\6&7\end{pmatrix}$[/quote]
Or $\begin{pmatrix}\left[\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right]&5\\6&7\end{pmatrix}$

It's a shame that there is no command like \smallmatrix that includes the outer symbals, because tacking brackets on afterwards looks rather ugly.

$\begin{bmatrix}\begin{bmatrix}\left[\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right]&5\\6&7\end{bmatrix}&8\\9&0\end{bmatrix}$

Just how far can matrices stack?

Does it not look the same? For example with normal size matrices
bmatrix: $\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$
matrix plus [ and ]: $\left[\begin{matrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}\right]$
Of course if you were to use smallmatrix a lot you could define a new environment smallbmatrix with
[code]\newenvironment{smallbmatrix}{\left[\begin{smallmatrix}}{\end{smallmatrix}\right]}[/code]which saves typing the brackets.

Apparently it is the same with regular-sized matrices, but, in a smallmatrix environment, the brackets are to the sides of the numbers, whereas, in the normal matrix environment, they extend over and under the numbers.

I suppose that's because smallmatrix is meant to be used in inline mode ie in the middle of a line of text so the brackets shouldn't stray too far up or down. You could try \big. Using \left[: $\left[\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right]$ $\bigl[\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\bigr]$ :using \big

Or, more in the style of the larger matrices: $\bigl[\!\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\!\bigr]$ (using negative spaces) $\bigl(\!\begin{smallmatrix}5&6\\7&8\end{smallmatrix}\!\bigr)$

Is there any way to make the brackets slightly bigger?

You can make the automated sizing brackets \left & \right bigger by pretending that the matrix is larger using \mathstrut $\left[\!\begin{smallmatrix}\mathstrut 1&2\\3&4\mathstrut \end{smallmatrix}\!\right]$


algebra polynomial inequalities AMC AIME trigonometry USA(J)MO

I'm going to start a new contest. Except it's a very special contest , because it require me to do no work. And this makes it infinitely better than any other contest I could start.

Here's how it goes:
I propose a hard algebra problem (systems of equations, polynomials, etc, just NO INEQUALITIES, and NO NUMBER THEORY)

Someone solves it and then posts another problem. This person gets a point for solving the problem. Points must be kept track of by the contestant.

Criteria for problem posting: NO INEQUALITIES, and NO NUMBER THEORY, anything equal to above a mid-difficultly AIME problem.

If your problem isn't solved in a reasonable amount of time (12 hours from posting), then another person may post an easier problem and no points are given.

So, let's get this thing rolling:

[b][size=150]Problem 1 (Easy)[/size][/b]
Clara compunted the product of the first n positive integers. Val computed the product of the first m even positive integers. Prove these products can never be equal, assuming m>=2.

The problems asks us to prove that $n! \neq 2^{m}m!, m \geq 2$. It's clear that equality is impossible if $n \leq m$, so we'll assum $n \geq m$. Divide by $m!$ to get $(n)(n-1)...(n-m+1)=2^{m}$. If one of the numbers $n,n-1...n-m+1$ is not a power of 2, the product won't be, so they all must be powers of 2. The only consecutive powers of 2 are 0 and 1, so either $n=m+1$ or $n=2=m+2$. Since $m \geq 2$, the latter is impossible. Thus, $n=m+1$ and $n=2^{m}$. Once again, since $m \geq 2$, this has no solutions.

Problem: Let a,b,c be distinct real numbers. Prove that $\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}\neq 0$.

[quote="PenguinIntegral"][b][size=150]Problem 1 (Easy)[/size][/b]
Clara compunted the product of the first n positive integers. Val computed the product of the first m even positive integers. Prove these products can never be equal, assuming m>=2.[/quote]
[hide] $n!=2^{m}\cdot m!$
Obviously, $n>m\ge2$.
$\frac{n!}{m!}=n\cdot n-1\cdot ... \cdot m+1=2^{m}$
If $n$ is a power of 2, $n-1$ cannot be, since $n\ge3$.
[hide="Proof"] $n=2^{a}$, $n-1=2^{b}$, $b>a$
$2^{a}-1=2^{b}$
$2^{a-b}-\frac{1}{2^{b}}=1$
LHS isn't an integer unless $b=0\Longrightarrow n=2$ [/hide]
Thus, $n=2^{m}=m+1\Longrightarrow m=1$. Contradiction. [/hide]

EDIT: Too late. :(

[quote="Phelpedo"]Problem: Let a,b,c be distinct real numbers. Prove that $\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}\neq 0$.[/quote]

[hide="One-liner"] $( \sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a})( \text{Junk}) = 3 \sqrt[3]{(a-b)(b-c)(c-a)}\neq 0$ [/hide]

[b][size=167]Problem 3[/size][/b]

Given an algebraic number $c$, let $M(x)$ be the monic polynomial with [b]rational[/b] coefficients of smallest degree such that $M(c) = 0$.

Show that if any polynomial with [b]rational[/b] coefficients satisfies $P(c) = 0$ then $M | P$.

[quote="Phelpedo"]This probably isn't rigorous
[hide] By some well-known theorem whose name escapes me at the moment, if a polynomial with real coefficients has a complex root, its conjugate is also a root. Therefore, we can say $M(x)=(x-c)(x-(what'sthecodeforconjugate)c)$. Furthermore, all polynomials with real coefficients with $c$ as a root have these two factors.[/hide][/quote]
Post another quickly before I get bored and decide to post one myself...

Whoops, I made the problem way too easy.  Edited.

[quote="hydro"][quote="t0rajir0u"]
Given an algebraic number $c$, let $M(x)$ be the monic polynomial with [b]rational[/b] coefficients of smallest degree such that $M(c) = 0$.
[/quote]

Just a question, but isn't that polynomial $M(x)=x-c$?[/quote]

No, because all the coefficients must be [b]rational[/b], but $c$ isn't necissarily rational.

I think it's been 12 hours. T0r, post the solution, and someone post a new problem.

[hide="Solution"] All polynomials mentioned here have rational coefficients.

Let $M(x)$ be our minimal polynomial as I have defined it.  Suppose we have $P(x)$ such that $P(c) = 0$.  Polynomial division:

$P(x) = M(x) Q(x)+R(x)$

Now, $deg(R) < deg(M)$ or $R(x) = 0$.  Moreover,

$P(c) = M(c) Q(c)+R(c) = R(c) = 0$

So $R(x)$ is a polynomial with rational coefficients with degree less than $M$, but such that $c$ is a root.  This means $R(x) = 0$, so $M | P$. [/hide]

Another problem anyone? Preferabley with some nice, chunky systems of equations?

[quote="Phelpedo"]A pretty nice one:

Solve in complex numbers: \[x^{3}-3x=y \\ \\ y^{3}-3y=z \\ \\ z^{3}-3z=x\] [/quote]
Good One!!!

Just remember it's problem 4...

[quote="pkerichang"]Uhh stole this problem from PSS:

[b][size=150]Problem 7[/size][/b]
P(x) is a polynomial, with $a$,$b$, $c$ being three reals.  Prove that the system:

$P(a) = b$
$P(b) = c$
$P(c) = a$

have no solutions.[/quote]

As stated, clearly false.  Let $a = b = c$ and of course there exist many polynomials such that $P(a) = a$ has solutions :P

[quote="Virgil Nicula"][quote="Phelpedo"]Problem: Let $a\ ,\ b\ ,\c$ be distinct real numbers. Prove that $\sqrt[3]{a-b}+\sqrt[3]{b-c}+\sqrt[3]{c-a}\neq 0$.[/quote]
From the well-known identity (over $C$) $x^{3}+y^{3}+z^{3}=3xyz+(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$ results that for any three real numbers $x\ ,\ y\ ,\ z$ exists the following equivalence : $\boxed{\ x^{3}+y^{3}+z^{3}=3xyz\Longleftrightarrow x=y=z\ \ \vee\ \ x+y+z=0\ }\ .$ For $x: =\sqrt [3]{a-b}$, $y: =\sqrt [3]{b-c}$, $z: =\sqrt [3]{c-a}$ a.s.o.[/quote]

Already gave this exact proof  :P  :P

[quote="pkerichang"][hide]
Following from Phelpedo's hint:

$t^{54}-t^{28}-t^{26}+1 = 0$
$(t^{28}-1)(t-1) = 0$
So $t=1$ and the 28th roots of unity of 1.  If $t=1$, then $x = 2$, $y = 2$, and $z = 2$ is a solution.
Otherwise, note that $t$ and $\frac{1}{t}$ are conjugates, so the imaginary part nicely cancels and we're left with a trig function stuff.
[/hide]

Uhh stole this problem from PSS:

[b][size=150]Problem 7[/size][/b]
P(x) is a polynomial, with $a$,$b$, $c$ being three reals.  Prove that the system:

$P(a) = b$
$P(b) = c$
$P(c) = a$

have no solutions.[/quote]
Yeah, you have the wording wrong. Does someone care to post the real problem, I need to go to school now...

[quote="Phelpedo"][quote="K81o7"][quote="Elemennop"]Honestly I'm too lazy to do Phelpedo's problem, but since no one posted a solution i'm just gonna point out it's obvious to substitute $x=2\cos A$ and so on[/quote]

Remember it's in $\mathbb{C}$.  I think that substitution would complicate things...[/quote]

A clever substitution is the way to go, and $2\cos A$ is a good one, but I don't think it's immediately obvious that the roots are between -2 and 2 (depending on your perspective of course.)[/quote]
Can you explain your answer?

Also, can someone either post another problem or get the correct citation from engel?

I can't find it in Engel: but it may be the old USAMO question:

$P(x)$ is a polynomial with integral coefficents. Prove that there are no solutions to the equations $P(a)=b$, $P(b)=c$, $P(c)=a$ with $a,b,c$  [b]distinct integers[/b]

(edit: I have now found it in Engel (example 6 in the polynomial chapter- how did I miss it?))

[hide="Solution"] By a well-known lemma we have

$(b-a) | (c-b)$
$(c-a) | (b-a)$
$(c-b) | (c-a)$

Let $u = b-a, v = c-b$.  Then $u+v = c-a$ and we have the system

$u | v$
$u+v | u$
$v | (u+v) \implies v | u$

So we have $u = \pm v$.

[b]Case:[/b]  $u = v$.  Then $2u | u \implies u = 0$.  Non-distinct.

[b]Case:[/b]  $u =-v$.  Then $0 | u \implies u = 0$.  Non-distinct.

Hence no non-distinct solutions exist.  QED. [/hide]

[quote="Phelpedo"]You know a lot of well-known lemmas, t0rajir0u  :lol:  What's the name of this one?[/quote]

That's a good question; we should name this lemma.

[quote="Phelpedo"]You know a lot of well-known lemmas, t0rajir0u  :lol:  What's the name of this one?[/quote]

it's the basic $a-b|P(a)-P(b)$ for integer coefficient polynomials

New problem:

Show that if $e^{i\theta}$ satisfies the equation $z^{n}+a_{n-1}z^{n-1}...+a_{1}z+a_{0}=0$, where $a_{i}$ are real, then $a_{n-1}\sin{\theta}+a_{n-2}\sin{2\theta}+...+a_{1}\sin{(n-1)\theta}+a_{0}\sin{n\theta}=0$.

I'd still like to know how that trig substitution helps...

[quote="K81o7"]Well, if all coefficients are real, then if $e^{i\theta}$ is a root, then so is its conjugate, $e^{-i\theta}$.  Plugging that into the equation gives us \[a_{n}e^{-ni\theta}+a_{n-1}e^{-[n-1]i\theta}+...+a_{0}=0\] Multiplying both sides by $e^{ni\theta}$ gives us our result.[/quote]

I think you left out a minor step- you have to take the imaginary part of the product.    $e^{ni\theta}(a_{n}e^{-ni\theta}+a_{n-1}e^{-[n-1]i\theta}+...+a_{0})=a_{n}+a_{n-1}e^{i\theta}...$


[color=blue][size=150]Problema:[/size][/color]
Sean $a,b$ enteros positivos tal que $b^{n}+n$ es un multiplo de $a^{n}+n$
para todo entero positivo $n$.

Probar que $a=b$.

Saludos,

[color=blue]¿Tendría sentido intentar resolverlo por cálculo de límites?[/color]

[quote="carlosbr"]Sean $ a,b$ enteros positivos tal que $ b^{n} \plus{} n$ es un multiplo de $ a^{n} \plus{} n$
para todo entero positivo $ n$.

Probar que $ a \equal{} b$.[/quote]

[hide="Solución"][u]Lema:[/u] Para todo primo $ p\not | a$ se puede encontrar $ N_p$ tal que $ p|a^{N_p}\plus{}N_p$ y $ (N_p,p\minus{}1)\equal{}1$
[u]Demostración:[/u] Sea $ k$ un natural con $ (k,p\minus{}1)\equal{}1$. El sistema de congruencias:
$ N \equiv k (\bmod (p\minus{}1))$
$ N \equiv \minus{}a^k (\bmod p)$
Tiene solución por el teorema chino. Sea $ N_p$ una de estas soluciones, entonces:
$ (N_p,p\minus{}1)\equal{}(k,p\minus{}1)\equal{}1$
$ a^{N_p}\plus{}N_p \equiv a^k\minus{}a^k \equiv 0$

Sea $ q>a\minus{}b$ un numero primo. Sabemos que $ q|a^{N_q}\plus{}N_q|b^{N_q}\plus{}N_q$ es decir:
$ a^{N_q}\plus{}N_q \equiv b^{N_q}\plus{}N_q (\bmod q)$
$ a^{N_q}\equiv b^{N_q} (\bmod q)$
$ a\equiv b (\bmod q)$ (Porque $ (N_q,q\minus{}1)\equal{}1$)
Entonces $ q|a\minus{}b$ y $ q>a\minus{}b$
$ a\equal{}b$
Q.E.D.[/hide]

[b]Otro problema:[/b]
Sea $ p$ un primo. Demuestre que existe un entero $ q$ tal que para todo natural $ n$, $ n^p\minus{}p$ no es divisible por $ q$.


inequalities geometry circumcircle trigonometry rotation

Let $ P$ be a point on the plane of an equilateral triangle $ ABC$. Show that $ PA,PB,PC$ are the side lengths of a triangle.

[quote="nayel"]
[color=darkblue][u][b]Pappus theorem.[/b][/u][/color] [color=darkblue] Let $ P$ be a point on the plane (not on the circumcircle) of an equilateral triangle $ ABC$. Show that $ PA,PB,PC$ are the side lengths of a triangle.[/color][/quote]

[color=darkred][b]Proof.[/b] [/color]

[color=darkblue][i]Interior.[/i]
Consruct on the side $ AB$ triangle $ P'AB$, such that $ \triangle{P'AB}\equiv \triangle{PCB}$.  Then $ \angle{P'BA} \plus{} \angle{ABP} \equal{} \angle{PBC} \plus{} \angle{ABP} \equal{} 60^\circ$, so $ \triangle{P'PB}$ is equilateral. Then in $ \triangle{P'AP}$ we have $ \left\| \begin{array}{cc} AP' \equal{} PC \\
PP' \equal{} PB \end{array} \right\|$. So it is required triangle with sides $ PA,PB,PC$.

[i]Exterior.[/i]
Call $ AB \equal{} a$. Apply to the quadrilateral $ PACB$ Ptolemey's inequality and obtain $ PA \cdot a \plus{} PB \cdot a > PC \cdot a \Leftrightarrow PA \plus{} PB > PC$.
[/color]

[color=brown][b][u]Remark.[/u] [/b]If $ P \in \widehat{AB}\in \omega_{ABC}$ then $ PA \plus{} PB \equal{} PC$. [/color]

Ptolemy's inequality, nice!
I think I can modify that to the interior too.
Let PB,PC be the shortest of PA,PB,PC.
Start out constructing the triangle you gave.
P'C is greater than PC.
But by Ptolemy, as you showed, P'C, PA, and PB make a triangle.
Hence, PC,PA, and PB must too.

That is a great solution. Yes I missed that $ P$ must not lie on the circumcircle of $ ABC$. Here's my solution:

Let $ D,E,F$ be the feet of the perpendiculars from $ P$ on to $ BC,CA,AB$ respectively. Then the quadrilaterals $ AEPF,BFPD$ and $ CDPE$ are cyclic. Therefore $ DE\equal{}PC\sin C\equal{}\frac{a\cdot PC}{2R}$, and $ EF\equal{}\frac{a\cdot PA}{2R}, FD\equal{}\frac{a\cdot PB}{2R}$ similarly, where $ R$ is the circumradius of $ ABC$ and $ AB\equal{}BC\equal{}CA\equal{}a$. Therefore from triangle $ DEF$ we get $ DE\plus{}EF>DF\implies \frac{a(PC\plus{}PA)}{2R}>\frac{a\cdot PB}{2R}$ which implies $ PC\plus{}PA>PB$. Similarly $ PA\plus{}PB>PC$ and $ PB\plus{}PC>PA$. Therefore $ PA,PB,PC$ are the side lengths of a triangle.

Remark: This also shows that the triangle with sides $ PA,PB,PC$ is similar to the pedal triangle $ DEF$ of the point $ P$.

For $P$ inside or outside the triangle:

[hide] Rotate triangle $\triangle APB$ $60^\circ$ about $B$ so that $A$ goes to $C$ and $P$ to $P'$; $\triangle BPP'$ is equilateral and $\triangle CPP'$ is the required triangle: $PP'=BP$ (from equilateral $\triangle BPP'$), $P'C=PA$ (from rotating triangle $\triangle APB$) and third side being $PC$, done. [/hide]

Best regards,
sunken rock


function
induction
algebra unsolved
algebra

[img]http://www.gnux.be/latex/data/694a852decedd37f506ddd840b082ce4.png[/img]
Thanks  :rotfl:

It can be read $f(x)=f(x-3)*f(x+3)$. If $f(x)=0$ by induction $f(x+3k)=0 \ \forall k\in Z$.
If $f(x)\not =0$ we get $f(x+3k)\not =0 \ \forall k\in Z$ and $f(x)=f(x-3)f(x+3)=f(x-3)f(x)f(x+6)\to f(x-3)f(x+6)=1\to f(x)=f(x+18)$.
Therefore 18 is period.


AMC
USA(J)MO
USAMO
inequalities
induction
geometry
analytic geometry

I voted for my lower end. I know I solved #1, 2, 4, 6, but my proofs for #2 and 6 may or may not be watertight enough to deserve a 6-7, so I voted for 15-21, but I'm hoping that I'll get a much higher score than a 15. I'm also almost certain that I got 1 point on #5 and possibly one on #3, so we'll see.

This is actually my first year taking USAMO, and I'm a freshman, so I'm very happy.

8-14 maybe higher if the partials work my way.

How hard was #6?  I didn't even both with it, really.

i solved the easiest part of #6, proving it for an equilateral triangle....

lol me too...and then i wrote some kiss-up crap that wasn't much of a proof at all...I was just like "The further the angles are from each other the bigger it gets"...with like no explanation why...i had a good start on it but then it all went bad. You know how when you're making something up it sounds like you're saying the same stuff over and over again like someone from the Department of Redundancy Department?

hehe i said something was ALMOST isoceles and therefore angles B and C should be equal as A approached 0.

don't know if something can be "almost isoceles" but whatever :rotfl:

either 14-16 or 20+ depending on how #2 comes out.

noneoftheabove: forget about partials.  the graders almost never give partial credit unless you're quite a ways on the right track.  and if you're that far, you usually can finish.

how about a solid proof with a hole or two or somethings are trivialized?

meh...like 4*7 +1{maybe} which would be 29...

If I'm REALLY lucky, maybe a 14 or 15? But I'd be satisfied with any score over 5. Was fun though.

i typed these up yesterday...i was too lazy to type up 4,6 today...

1.
Define s_k=a_1+a_2+…a_k. If we have that s_k=j*k where 0=<j=<k, then we can take s_{k+1}=j, and we are done.
We always know that s_k is a multiple of k, so we know that there is some integer j there, but we also need that inequality for j to be done...i.e. s_k=<k^2
However, we also know that s_k=a_1+a_2+...+a_k=<n+1+2+...+k-1=n+(k-1)(k)/2
So we note that for sufficiently large k, we have that n+(k-1)(k)/2=<k^2, hence s_k=<k^2,
Then we are done, since we can take that suitable j and keep adding it.

2.
Lemma: among 3 tangent circles with radius greater than or equal to 5, one can always fit a circle with radius greater than 1/sqrt 2 between those 3 circles.
Proof: descartes' circle theorem states that if a is the curvature of a circle [a=1/radius, positive for externally tangent, negative for internally tangent], then we have that

(a+b+c+d)^2=2(a^2+b^2+c^2+d^2)...solving for a, we get

a=b+c+d+2*sqrt(bc+cd+db)   [take positive root, negative root corresponds to externally tangent circle]

now clearly, we have b+c+d=<3/5, and bc+cd+db=<3/25
summing/square root/multiply appropriately shows that a=<3/5+2*root(3)/5. Incidently, 3/5+2*root(3)/5<root 2, so a<root2, radius>1/root 2, as desired
---

For sake of contradiction, assume that we have a satisfactory placement of circles. Consider 3 circles, p,q,r where there are no circles in between. By appolonius' problem, there exists a circle, t, tangent to p,q,r externally that is between those 3 circles. Clearly, if we move p,q,r together, t must decrease in radius. Hence it is sufficient to consider 3 tangent circles. By lemma 1, there is always a circle of radius greater than 1/root 2 that lies between p,q,r. However, any circle with radius >1/root 2 must contain a lattice point. [Consider placing a circle between a unit square]. That is a contradiction. Hence no tiling exists.

You see, that's not my argument at all. I didn't prove it misses a lattice points, I proved something else entirely.

[quote="Hamster1800"][quote="zeb"]WTF? Since when does SHRINKING THE RADII not introduce a gap?

If you were right my solution would actually work, but what you just said makes no sense to me.[/quote]

I'm only focusing on a specific gap. Obviously we need to reconfigure the other circles to maintain optimal, but the specific gap is all we need to consider to show that it misses a lattice point.[/quote]
Except how to prove that (without using curvature, that I didn't know)?
This was a monumental handwave that I think almost everyone who thinks they solved the problem made, and I think its quite far from rigorous, seeing as I spent the last hour and a half trying to fix my proof (Maybe I'm just dumb). Analytic geometry bash probably works (after fixing the point of intersection and finding that each line in the gap of the smaller circles is shorter than the gap in the larger circles), but I timed out on that =(

EDIT: My brother tells me inversion works, but for the geo noobs like me, anyone care to post a solution thats intelligible for the non-advanced like me?

[quote="Scrambled"]Except how to prove that (without using curvature, that I didn't know)?
This was a monumental handwave that I think almost everyone who thinks they solved the problem made, and I think its quite far from rigorous, seeing as I spent the last hour and a half trying to fix my proof (Maybe I'm just dumb). Analytic geometry bash probably works (after fixing the point of intersection and finding that each line in the gap of the smaller circles is shorter than the gap in the larger circles), but I timed out on that =(

EDIT: My brother tells me inversion works, but for the geo noobs like me, anyone care to post a solution thats intelligible for the non-advanced like me?[/quote]

Same thing here.  I thought simply stating that it was sufficient to show that the tangent circle to three pairwise tangent circles of radii at least 5 has radius less than 1/sqrt(2) would not be rigorous enough, and so I spent much of my time at the end trying to improve on that, to no avail, so I stated what I could...

[quote="Hamster1800"]Do you really think that the USAMO would require someone to know about curvature? The result of shrinking a circle is very intuitive, and I think rigorizing it is very easy. If you shrink a circle, keeping it tangent to the other two circles, then the points of tangency are moved in. However, it's clear that these are now outside of the circle, and the point between them on the other side of the circle is on the interior of the old circle, so the arc of the new circle is completely outside the old circle. Unfortunately, I didn't see this during the test.[/quote]
Hence the existance of other proof. Analytics and locusese look like they work out nicely. And something "intuitive" doesnt really work out well. For example, a lot of IMO number 3's are actually quite intuitive, but the proof is the monster. Also, everyone who took USAMO here at my area spent the rest of the time not spent doing problem 1 making problem 2 airtight; everyone in our local chapter got the 1/sqrt 2 part. Its just no one got any further (and we have a fair share of blue and black MOPpers in our region).
And if the official proof has anything about it, then we all get owned. If you claim its easy to rigorize, then do so and post it here please. It would be appreciated.

[quote="Scrambled"][quote="Hamster1800"][quote="me from other thread"]
I believe that this will rigorously show that using tangent circles of radius 5 will be optimal.

Consider three circles in the plane. If they are not all tangent, then we may move one of the circles such that they are. It is quite clear that the gap from before the move contains the gap from after the move. This can be made rigorous by noting that the point of tangency is moved out of the original circle, and so the arc that limits the new gap lies completely outside of the old circle. Therefore, we only wish to have gaps between 3 tangent circles.

Now consider one of these such gaps. Suppose you decrease the radius of one of the circles, while keeping it tangent to the other two. Then, the points of tangency both lie outside of the circle, but there is a point on the other side of the circle inside the old circle, unless they never intersect. In both cases, the new circle's arc that limits the gap lies completely outside of the old circle, and thus the gap only gets smaller, and no new gap is introduced.
[/quote][/quote]
Actually, you assume the circles (the one that gets "bigger" and the original) don't intersect. But while that seems intuitively correct, the radius does get larger and thus you don't know if the circles intersect or not straight-off. Demonstrate please?[/quote]

I don't quite follow your scenario. Can you try to explain what doesn't work again?

[quote="Altheman"]i guess the problem is that i assume that you are able to push the circles together...like consider circles radius 5 centered at the verticies of a unit square...

so we need a better way to argue that circles spaced apart, or 4 tangent circles [not mutually tangent, but tangent like a ring], etc, have a larger circle between them

i think the best way to argue this is we want to find positioning of circles such if f(r) is the largest radius between those circles, then f(r) is minimized[/quote]

Simple! All you have to do is note that if you take away one of the four circles, so only three remain, then move one of them such that they form the triangle, you have not introduced any new gap, and therefore the gap between four circles contains the gap between three circles.

[quote="Scrambled"]Your claim is that if you fix two circles, say O1 and O2, and you make the other, O3, larger into O4. You claim that the arc of O3 in between the points where it is tangent to O1 and O2 never intersects with that of the arc of O4 between the points where O4 is tangent to O1 and O2.
But, how do you know? it sure looks like it, but since when was that a proof?

And taking away one of the four circles doesnt necessarily work. If I had three nearly collinear circles and a large circle that is tangent to all 3 of the nearly collinear ones, then the area not covered is monumentally changed (although this one barrier isnt particularly hard to overcome)[/quote]

Okay, here's the trick in the first part. When you enlarge O3 into O4, the points of tangency of O3 with O1 and O2 lie outside O4. However, we have 2 cases. Either there exists a point on O3 in O4, but this point is on the wrong side to intersect the arc limiting the gap, and since two circles can only intersect at at most 2 points, the arc cannot intersect with the circle. The other case is that they don't intersect, and since the points of tangency are outside of O4, O3 is not contained in O4, therefore the arc never intersects the circle.

For the second point, disregard one of the three nearly collinear ones and only the large one and two of the small ones. You have the gap between three tangent circles.

Two circles can only intersect at two points >.<
Dang I gotta remember these things.

I thought that while the radius could vary, all the circles had to have the same radius ie it could 6.7, but the it would be that for all of them.


$ AP, AQ, AR, AS$ are chords of a given circle with the property that

$ \angle PAQ \equal{} \angle QAR \equal{} \angle RAS$.

Prove that

$ AR(AP\plus{}AR) \equal{} AQ(AQ\plus{}AS)$.

Let $ B,C$ be the point on $ AR,AQ$ respectively such that $ QC \equal{} AS,RB \equal{} AP$. The problem is equivalent to prove that $ QRBC$ cyclic.It's easy to see $ \triangle QRC \equal{} \triangle ARS$ => $ \angle QCR \equal{} \angle RAS$.Similarly We'll get the result  :)
 Sorry for my bad english  :)


That's awesome! So, yif man12, has your "maybe" become a "yes" yet?

Well, I still haven't sent in my tryout problems yet, but I've done the first two in physics class over the last week and hoping to finish the rest this weekend.

I might go as well. Currently waiting for a reply from them.

Excellent! I wonder how much AoPS will dominate at PROMYS...

does anyone know when is the registration deadline? Currently I am not sure whether I will be going(50%)...I need some more time

[quote]PROMYS 2005 is from July 3 - August 13. Student, counselor and teacher applications are now available. The student application deadline has been extended to June 15.[/quote]

I have a very cool friend who is going to be a counselor there, so if you like, say hi to Megan for me :)

Sure will, JBL.

So...who's going to take advantage of that extension? Beta?

I just wish to revive this topic...it just seems to me that PROMYS is not a very popular program here. Please prove me wrong, but I am getting an impression that PROMYS is almost an AAST inclusive camp (at least 8 people from AAST are going for first year, returning students, and counselors)...    :huh:

[quote="corinao"]what's AAST?  :blush:[/quote]

Academy for Advancement in Science and Technology or something like that... it's a magnet (right?) school in NJ.

yes, that is correct.

I am curious at the fact that there are few AoPSers eager to apply (or reply the message?) for the program. If this presumption is not entirely true, I then wonder if there are other people here in the forum that are going to attend the program.

Yeah, I was wondering about the same thing, but I was afraid to ask.  :blush:

I have noticed, however, that other camps have faculty starting threads to promote their camp. Surely that's not the issue...right?

Please tell me that the PROMYS AoPSers are only shy...

[quote="mel"]
Please tell me that the PROMYS AoPSers are only shy...[/quote]

Well at least last year when I went there wasn't an over-abundance of AoPSers... actually you'd be surprised at how many people are not AoPSers (there are people outside of AoPS :o). Though it would be nice to meet AoPSers, you'll still meet a bunch of great (and maybe some not so great) people when you go.

[quote="mel"]Haha...yes...I know there are people outside of AoPS.  :oops:

I guess maybe the few of us going will just have to convert them. :P

So, paladin8, is there any chance that you're returning?[/quote]

Sorry, I will be unable to... too many ongoing things for the summer. I had originally planned on returning but really my schedule won't allow it... :(

[quote="j.chung"]I'm going to PROMYS this year!

I just received my acceptance letter today. I'm so happy :)

I hope to see a lot of you there[/quote]

Yay! So...does that bring the AoPS count to...[size=59]2[/size]?

YAY! So that brings our AoPS attendence number to 3?!

hahah i was trying to figure out what "aopser" was... i understand now :P

I just recently joined this forum :P

However, I have been part of College Confidential for the longest time!

I hope to see you guys around :D

Welcome to the forum, j.chung! You'll find that it's quite cozy around here, and it's a pretty hard place to leave. So...hope you stick around!

[quote="smbelcas"][quote="mel"]Haha...yes...I know there are people outside of AoPS.  :oops:

I guess maybe the few of us going will just have to convert them. :P
[/quote]

You should definitely try to convert them!   My suggestion for this: when you arrive, ask your counselor(s) if there's some way you could either make a program-wide announcement about AoPS or give a 10-min demo of the site or something.  It might be that you can schedule a computer-projector room during some free time.  (This summer, at the program with which I'm affiliated we'll be passing out AoPS brochures.  By my count, ~1/6 of our students will be AoPS folk.  PROMYS is a larger program, so it's not surprising that the fraction seems to be lower.  Or it could be that PROMYS attendees for this year just happen not to frequent this portion of the AoPS forum.)[/quote]

[quote="qvc122"]mel and other AoPSers should let people in the camp know of the AoPS site. Lot of classes here in AoPS may be interesting to other PROMYS attendees through the year round.[/quote]

Sure! I definitely don't mind doing something like that. If any of the admins happen to be passing by this thread, is there anything that you'd like me to say/do?

Also, seeing that I'm going to have to keep up with my AoPS classes, I definitely don't think there will be any shortage of AoPS presence in my summer.  :P

Awesome! That brings our AoPSer attendance to 4.

I'm sure many people who are attending PROMYS have accounts on AoPS, just that not many of them post often.


let f(n) defined for positive integers n have the properties:
(i) f(f(n))=4n+9 for each positive integer n
(ii) f($2^k$)=$2^{k+1}$+3 for each non-negative integer k

determine f(1987)...

Theres a few solutions to that function, you cant assume its that particular one. You have to try get the answer from the initial conditions...


proof that : every increasing multiplicative function from the positive integers to the reals is of the form $ f(n)=n^{a}$ for some real $ a>1$.

See here http://www.mathlinks.ro/Forum/viewtopic.php?p=346551#346551 and i must move this topic .  :wink:


Tryouts for the Chicago ARML Teams will be [b]Thursday, Feburary 23, 2006[/b], at Evanston Township High School in Evanston, IL from 4pm until 10pm (with a dinner break).
Tryouts are open to any student who has not yet completed high school and lives in the northern 2/3 or so of Illinois.
For more information, see http://chicagoarml.org/

Alright!  Who all from AoPS is tryin out?  and were you on it before?

I was on B or C last year, never really knew which color was which, and I don't remember which color it was  :P

[quote="Altheman"]i might, but i have to see what the time commitment is[/quote]

A brief description of the time commitment is on the web site, [url=http://chicagoarml.org/faq-teams.php#a5]in the FAQ[/url].

Essentially, after tryouts, there are four evening practices (two Wednesdays, two Thursdays), and then the trip to Iowa from Thursday night to Saturday night, June 1-3.

I'm trying out, and of course, really looking forward to it!

JB

grr..i will have to miss my aops class again  :mad:

Whoa didn't realize it was so soon.  Anyways, see you guys all there!


calculus
integration
trigonometry
derivative
function
algebra
rational function

calculate:
I=$ \int_{\frac { \minus{} 1}2}^0 \frac {dx}{1 \plus{} \sqrt { \minus{} x^2 \minus{} x}}$

$ \left(\underbrace{\int_{ \minus{} \frac {1}{2}}^0 \frac {1}{\sqrt { \minus{} x^2 \minus{} x} \plus{} 1} \, dx}_{x \equal{} \frac {y \minus{} 1}{2}} \equal{} \int_0^1 \frac {1}{\sqrt {1 \minus{} y^2} \plus{} 2} \, dy \equal{} \underbrace{\int_0^1 \frac {2}{y^2 \plus{} 3} \, dy}_{y \equal{} \sqrt {3} \tan (\theta )} \minus{} \int_0^1 \frac {\sqrt {1 \minus{} y^2}}{y^2 \plus{} 3} \, dy \equal{} \int_0^{\frac {\pi }{6}} \frac {2}{\sqrt {3}} \, d\theta \minus{} \ldots\right)$
Couldn't finish, trading a integral for another and another.

[quote="Arrigo-Sacchi"]I think we can substitute $ y \equal{} \sin t$ in
$ \int_{0}^{1}\frac {1}{\sqrt {1 \minus{} y^{2}} \plus{} 2}\, dy$ and obtain $ \int_0^1 (1 \minus{} {2\over 2 \plus{} \cos t})dt.$ Then the usual substitution
$ t \equal{} 2\arctan z$......[/quote]

I think we should have $ \int_0^{\frac{\pi}{2}} \left(1\minus{} {2 \over 2 \plus{} \cos t}\right)dt.$

Perform the substitution

$ x \equal{} \minus{}\frac{1}{1\plus{}u^2}, \quad dx \equal{} \frac{2u}{(1\plus{}u^2)^2} \, du$.

This gives

$ \minus{}x^2\minus{}x \equal{} \frac{\minus{}1\plus{}(1\plus{}u^2)}{(1\plus{}u^2)^2} \equal{} \left(\frac{u}{1\plus{}u^2}\right)^{\!\!2}$,

and hence

$ \int_{x\equal{}\minus{}1/2}^0 \frac{dx}{1\plus{}\sqrt{\minus{}x^2\minus{}x}} \equal{} \int_{u\equal{}1}^\infty \frac{2u/(1\plus{}u^2)^2}{1\plus{}u/(1\plus{}u^2)} \, du$

$ \equal{} \int_{u\equal{}1}^\infty \frac{2u \, du}{(1\plus{}u^2)(1\plus{}u\plus{}u^2)} \equal{} 2 \int_{u\equal{}1}^\infty \left(\frac{1}{1\plus{}u^2} \minus{} \frac{1}{1\plus{}u\plus{}u^2}\right) \, du$

$ \equal{} 2\left[ \tan^{\minus{}1} u \right]_{u\equal{}1}^\infty \minus{} 2\int_{u\equal{}1}^\infty \frac{du}{(\sqrt{3}/2)^2 \plus{} (u\plus{}1/2)^2}$

$ \equal{} 2\left( \frac{\pi}{2} \minus{} \frac{\pi}{4} \minus{} \frac{2}{\sqrt{3}} \left[\tan^{\minus{}1} \frac{1\plus{}2u}{\sqrt{3}} \right]_{u\equal{}1}^\infty \right)$

$ \equal{} 2 \left( \frac{\pi}{4} \minus{} \frac{2}{\sqrt{3}} \left(\frac{\pi}{2} \minus{} \frac{\pi}{3}\right) \right) \equal{} \frac{9\minus{}4\sqrt{3}}{18}\pi$.

[quote="replay"][quote="atomicwedgie"]Perform the substitution

$ x \equal{} \minus{} \frac {1}{1 \plus{} u^2}, \quad dx \equal{} \frac {2u}{(1 \plus{} u^2)^2} \, du$.[/quote]

How can you find it? :maybe:[/quote]
Answer: $ dx \equal{} d \left(\frac{ \minus{}1}{1\plus{}u^2}\right) \equal{} f'(u) du \equal{} \frac{2u}{(1\plus{}u^2)^2}\, du$

[quote="Hong Quy"][quote="replay"]How can you find it? :maybe:[/quote]
Answer: $ dx \equal{} d \left(\frac { \minus{} 1}{1 \plus{} u^2}\right) \equal{} f'(u) du \equal{} \frac {2u}{(1 \plus{} u^2)^2}\, du$[/quote]

I don't think replay was asking for how to compute the derivative of the substitution; rather, I think replay was asking what motivated this particular choice of substitution, as it is not obvious.

I will explain how I arrived at this choice.  It is actually a combination of three distinct substitutions.

1.  We know we must complete the square underneath the square root.  This means $ \minus{} x^2 \minus{} x \equal{} \minus{} (x \plus{} 1/2)^2 \plus{} (1/2)^2$, so we know a substitution of the form $ t \equal{} x \plus{} 1/2$ is the first step.  We don't need to compute the entire effect of the substitution on the integrand.

2.  We know that the square root expression requires a trigonometric substitution, something of the form $ t \equal{} (1/2) \cos \theta$.  This clears out the square root but in exchange expresses the integrand in sines/cosines.

3.  In order to convert the integrand into a rational function, we need to use the familiar substitution $ u \equal{} \tan \frac {\theta}{2}$.

Now we combine all three substitutions together.  $ x \plus{} \frac {1}{2} \equal{} \frac {1}{2} \cos \theta$, so $ x \equal{} \minus{} \frac {1 \minus{} \cos\theta}{2} \equal{} \minus{} \cos^2 \frac {\theta}{2} \equal{} \minus{} \cos^2 (\tan^{ \minus{} 1} u) \equal{} \minus{} \frac {1}{1 \plus{} u^2}$.  We know this substitution will immediately transform the integrand into a rational function of $ u$.  That it happens to be easily decomposed and integrated as a difference of inverse tangents is a bonus.

Thank atomicwedgie so much! You really know what I asked for :lol:


Triangle ABC has AB=3, BC=4 and AC=5. AB is extended beyond B to point R so that AR=8, BC is extended beyond C to point S so that BS=8 and AC is extended beyond A to point T so that CT=8. What is the area of triangle RST?

Area of $\triangle RST$= $\triangle ABC$ + $\triangle BSR$ +$\triangle STC$ + $\triangle ART$.

Let $\angle CAB = \theta$.

1. Area of $\triangle ABC$= $\frac{4 \times 3}{2}= 6$

2. Area of $\triangle BSR$= $\frac{5 \times 8}{2}= 20$

3. Area of $\triangle STC$= $\frac{1}{2}\times 4 \times 8 \sin(\frac{\pi}{2}+\theta)$= $16 \cos \theta$= $\frac{48}{5}$

4. Area of $\triangle ART$= $\frac{1}{2}\times 8 \times 3 \sin(\pi-\theta)$= $\frac{48}{5}$

Therefore by adding together the four values we obtain that the area of $\triangle RST$= $\frac{226}{5}$.

Doesn't look like a very clean answer. Did I make a mistake? :maybe:

[quote="pkothari13"]Nice solution! I wonder though, since this was written by Sam Baethge, if there is a purely geometric solution, with no trig involved.[/quote]

the stuff he did was not really trig, the same results can be done with using the base to area ratios of triangles


All of the ordered quadruples of positive integers that satisfy:
\[ a\plus{}b\plus{}c\plus{}d\equal{}15\]
are written on a sheet of paper.  Laura selects one at random.  What is the probability that $ a>1$ in the randomly selected quadruple?

[hide="Solution"]

[hide="Calculating Total Possibilities"]Our goal is to find the total number of solutions to $ a\plus{}b\plus{}c\plus{}d\equal{}15$. If we extend this to a more word-problem scenario, this is similar to dealing out 15 cards amongst 4 friends. We can depict this with stars and dividers: *|***|***|********. Everything left of divider 1 is for Person A, everything between dividers 1 and 2 is for Person B, everything between dividers 2 and 3 is for Person C, and lastly everything right of divider 3 is for Person D.

So, the problem essentially boils down to arranging 3 dividers amongst a total of (15 stars + 3 dividers) 18 objects. This is nothing but:
$ \binom{18}{3}$
[/hide]

[hide="Calculating for when a>1"]We have a similar scenario, except this time we're forced to give at least 1 star to A. That means we have 14 stars left to deal out normally as before. Aha! This problem is the same except with arranging 3 dividers amongst a total of (14 stars+3 dividers) 17 objects.
$ \binom{17}{3}$
[/hide]

[hide="Probability"]
$ \dfrac{\binom{17}{3}}{\binom{18}{3}}\equal{}\dfrac{17*16*15}{18*17*16}\equal{}\boxed{\dfrac{5}{6}}$
[/hide][/hide]

Unless I'm mistaken, that works for giving them 0+, not 1+. You have to subtract 4 from 15 first.

[quote="The Archer"]
[color=#0000FF][u]Calculating Total Possibilities[/u][/color]
Our goal is to find the total number of solutions to $ a+b+c+d=15$. If we extend this to a more word-problem scenario, this is similar to dealing out 15 cards amongst 4 friends. We can depict this with stars and dividers: *|***|***|********. Everything left of divider 1 is for Person A, everything between dividers 1 and 2 is for Person B, everything between dividers 2 and 3 is for Person C, and lastly everything right of divider 3 is for Person D.

So, the problem essentially boils down to arranging 3 dividers amongst a total of (15 stars + 3 dividers) 18 objects. This is nothing but:
$ \binom{18}{3}$
[/quote]
Remember that the variables are positive.  Therefore, there is a slot to choose from between each of the $15$ cards and $3$ dividers which makes $\binom{14}{3}=364$ possible solutions to $a+b+c+d+e=15$.
[quote="The Archer"]

[color=#0000FF][u]Calculating for when a>1[/u][/color]
We have a similar scenario, except this time we're forced to give at least 1 star to A. That means we have 14 stars left to deal out normally as before. Aha! This problem is the same except with arranging 3 dividers amongst a total of (14 stars+3 dividers) 17 objects.
$ \binom{17}{3}$[/quote]
Again, please note that the ordered quadruple contains all positive integers.  So instead of counting how many ways can $a>1$, let's count the number of ways $a=1$.  This is just choosing $2$ slots from the $13$ slots which is $\binom{13}{2}=78$.  Subtracting from the total number of possibilities, we have the number of ways $a>1$ is $364-78=286$.

Therefore, the probability that $a>1$ in the randomly selected quadruple is:
\[\frac{286}{364}=\boxed{\frac{11}{14}}\]


What is Balls and Urns? How do you use it and in which situations would it be applicable? I reallly dont get the AOPS wiki article about it. I need someone to explain it to me easily. ALso, would this question require the use of balls and urns, and if so, how?: If you roll 5 die, how many ways can you roll a 16?

Well, balls and urns is really a quick way to count the ways that $ N$ distinguishable objects can be put into $ X$ distinguishable containers. For example, how many ways can you put $ 10$ distinguishable balls into $ 2$ distinguishable containers? The total number of ways would be $ 2^{10}$ as there are 2 ways for each of the $ 10$ objects. In your dice problem, i don't think that would require the use of balls and urns.

There are 4 types of balls and urns.

$ \text{Distinguishable balls and distinguishable urns.}$

[b]Distinguishable balls and indistinguishable urns.[/b]

[u]Indistinguishable balls and distinguishable urns.[/u]

[i]Indistinguishable balls and indistinguishable urns.[/i]

Check the chapter about prob in AoPS volume 1, it covers the balls and urns principle.


Hey guys, my teacher gave me a list of 45 trig identities to prove, and I did get each and everyone of them, except of course the 3 below. If anyone could help me out, I would really appreciate it!

1. Cot (x+y)= cot x cot y - 1
                   ____________
                    cot x + cot y

2. Cos (A + B) Cos (A - B)= cos^2A - sin^2B

3. Cos 3x= cos x (1 + 2sin x) (1 - 2sin x)

Thanks in advance guys!

Hints:

[hide="1"]
Express all the $ \cot$ terms as $ \frac\cos\sin$, simplify, and use some identities you probably have proven already.
[/hide]
[hide="2"]
Expand the left side, then use the identity $ \cos^2\alpha \equal{} 1 \minus{} \sin^2\alpha$ multiple times.
[/hide]
[hide="3"]
Note that $ 3 \equal{} 1 \plus{} 2$. (You could also prove this by expanding $ (\cos x \plus{} i\sin x)^3$ and equating real parts to the expansion with DeMoivre's Theorem, but you might not have learned that yet.)
[/hide]

These are all derived from sine and cosine angle addition formulas. Try expanding things out using the identities:

\[ \cos(x\pm y)\equal{}\cos x\cos y\mp \sin x\sin y\]
\[ \sin(x\pm y)\equal{}\sin x \cos y\pm \cos x \sin y\]

Also, recall that $ \cot x\equal{}\cos x/\sin x$

I still cant find any of them, I give up.  :(

Can someone help me out?

Alright I found one..

Cos3x= cos x (1 + 2sinx) (1 - 2sinx)
cos (x + 2x)= cos x (1-4sin^2x)
cosx cos2x - sinx sin2x= cos x (1 - 4sin^2x)
cosx (cos^2x - sin^2x) - sinx (2sinxcosx)= cos x (1-4sin^2x)
cos^3x - sin^2xcosx - 2sin^2xcosx= cos x (1 - 4sin^2x)
cos^3x - 3sin^2xcosx= cos x (1-4sin^2x)
cosx (cos^2x - 3sin^2x)= cos x (1-4sin^2x)
cosx (1-sin^2x - 3sin^2x)= cos x (1-4sin^2x)
cosx ( 1-4sin^2x)= cos (1-4sin^2x)

The others are too hard though... :(

Here's a start on #2:

Apply $ \cos(a\pm b)\equal{}\cos a\cos b\mp\sin a\sin b$ to the LHS to get:

$ (\cos A\cos B\minus{}\sin A\sin B)(\cos A\cos B\plus{}\sin A\sin B)\equal{}\cos^2 A\minus{}\sin^2 B$.

Try expanding even more. I managed to prove it without touching the RHS, so...
Also, as math154 mentioned before, applying $ cos^2 a\equal{}1\minus{}\sin^2 a$ is a good idea.

For #1, the best idea would be to expand all the $ \cot a$ terms into $ \frac{\cos a}{\sin a}$.

[hide="number three"]
Write as
$ \cos(3x) \equal{} \cos(x)(1\minus{}4\sin^2(x)) \equal{} 4\cos^3(x) \minus{} 3\cos(x)$
And we know this is true  :)

...but in case you don't know that Identity then we just need to expand $ \cos(2x\plus{}x)$ with the double angle formula and it falls out.
[/hide]


I looked on their website and found out it was a web browser. Do you guys recommend it? Will it mess up anything?
-jorian

could you go a little more "in-depth"
thank you
-jorian

Yes, I recommend it as it's a better browser with more features than IE 6. I don't know about IE 7.

What do you consider "messing up things"

I have sometimes found links that do not work with Firefox, or symbols that only show up correctly on IE, and for this reason I would not recommend firefox. IE7 certainly has more features than IE6 (e.g. tabbed browsing), though perhaps not as much as Firefox.

It's normally a problem of those sites, not of FireFox, if it doesn't work.
The only sites I know that don't work best with FireFox are stupid online games whose programmers aren't able to write good code  :roll:

In fact, there are websites that don't work under IE...

I think that, if you don't know the differences in features and performance of the two browsers, than it doesn't matter which one you have. I use firefox, and I've never had any problems with it, but if you don't have any problems with IE, then there's no reason to change.

People who care about computers generally, in my experience, use firefox, but if you don't want obsessively excellent performance from your PC, then any of IE's deficiencies won't bother you. Ditto for firefox's deficiencies if that's what you choose.

If you're really worried, get both.

[quote="akkaren"]...but if you don't have any problems with IE, then there's no reason to change.[/quote]

Wrong. I've heard this argument so many times about so many different things, but it doesn't make sense. Just because you don't have problems with something doesn't mean if there's something better, you should just keep using whatever you're using.

And yes, I recommend Firefox, and no it won't mess up anything. In fact, it can only make things better. As a web developer, I realize how much better Firefox is because I always have to test my sites in IE to make sure they render correctly, and most of the time IE doesn't understand my CSS because it doesn't fully support the new CSS standards. It's really quite annoying. :(

go FIREFOX!

I got it a long long time ago when it wasn't popular yet

I saw it in USATODAY advertising about safe computers and saw about it coming out.

So yeah.. and it works sometimes, but like when you want to listen to things, you have to go to IE  :(

[quote="now a ranger"]but like when you want to listen to things, you have to go to IE  :([/quote]

Hmm... I don't even have IE but I have no problem listening to things... :maybe:

[quote="chess64"][quote="akkaren"]...but if you don't have any problems with IE, then there's no reason to change.[/quote]

Wrong. I've heard this argument so many times about so many different things, but it doesn't make sense. Just because you don't have problems with something doesn't mean if there's something better, you should just keep using whatever you're using.
[/quote]

Just cuz I hate people calling me wrong...

If you don't have problems with something and aren't attentive enough to tell the difference between it and an improvement (which describes me since switching to firefox), then the two different things are functionally equivalent. To me, firefox is NOT better than IE, because my perceptions are not precise enough to tell the difference. If for some reason I hit the IE icon, I'll use that browser instead.

Most likely, changing the browser isn't a signifcant chore to someone too inattentive to distinguish IE from Firefox, so I can't argue that changing would be going out of the way or sacrificing something for nothing. So, there is absolutely nothing in favor of my position. However, to me, there is nothing in favor of your position--to someone like you, there is, because you apparently have a strong drive for improvement or approaching perfection.

So, now I've admitted you're right but not that I'm wrong, which is apparently contradictory since your point was that I'm wrong... but your argument holds under a different set of premises, whereas both sides have no meaning under my premises. How, morally and pragmatically speaking, can I prefer a lazy, non-perfectionist approach to things over a way of thinking that doubtless increases efficiency? Quite simply because giving weight to the little things, the gradations that don't matter until they add up into a personal philosophy of efficiency and productivity, can and does backfire on me frequently, and I need to find a different force to impel me through life--I'll pick and choose a few things to care about, and the other stuff may drag me down a little, all told, but not as much as the psychological effects of trying to pay attention to everything.

xP Maybe one day I'll write this explanation/argument coherently. If it bothers anyone, PM me and I'll start being more careful about how I post--I'm not yet sure about the level of structure and logical flow that's acceptable here.

Just get Firefox!!!  :thumbup: . The reason is that it has no spyware, is more secure, supports css, and much more. It can also edit the source with an additional extension and be able to easily just edit the source code of a webpage and view it live. GET FIREFOX!!!

[img]http://www.graphicsguru.com/300x250FF2b.gif[/img]
I don't have IE, as I am on SuSE 10.0.  :)

will the way u go to webpages change...anything major?
-jorian

When I got Firefox it was back when IE was extremely insecure. However, after having it for a while I realize that Firefox is more powerful than IE because of its plugin support. Barebones, firefox may not be better than IE, but its plugins (such as Adblock, editCSS, StumbleUpon, NoScript, MeasureIt, etc.) make it superior.   Adblock dynamically removes ads from websites by using filters. editCSS allows you to dynamically edit CSS on webpages (useful for developers like me, and for anyone else who wants to know what AoPS looks like in different colors). StumbleUpon gives you random webpages that match your interests that you select (for when you're bored). I also have a Google/Wikipedia search box in the corner of my screen.

Firefox does take a bit more time to start up than IE, but it's worth it. Also, since FF has gained much market share, most sites now make themselves compatible with FF as well as IE. In fact, there are now some sites that run well on FF and have problems on IE.

As the interface goes, it is about the same as IE. At first it is a bit jarring because FF only has 5 buttons as opposed to IE's who-knows-how-many (back, forward, refresh, stop, home), but you realize it's better this way.

For the couple of websites that still only work with IE, I also have a plugin that allows me to view the page from IE with a single click (two clicks, actually)

Firefox has more security than IE. You can't get keyloggers and other unwanted spyware just by clicking on a link. It has a feature which is an add on called all-in-one-movement so that you can browse pages using only the mouse , without clicking the back button, and other cool stuff. I recommend it

Even without the add-ons and extensions and themes, I still like Firefox better than IE. I have never gotten a pop-up ad on Firefox. Ever. Firefox almost never crashes or freezes.

I know of a few websites that do not work with Firefox, but I rarely frequent them, so it's not a hassle to switch to IE for those isolated cases.

[quote="Bictor717"]
I know of a few websites that do not work with Firefox, but I rarely frequent them, so it's not a hassle to switch to IE for those isolated cases.[/quote]

That's what the IE Tab extension for firefox is for ;)

Of course, you could all use Opera and not waste space with silly extensions that Opera mostly have built-in....  :lol:
Firefox2(with some extensions)=30mb
Firefox2(without)=25mb
Opera 9=10mb  :wink:
And plus Opera has mouse gestures.  :)

dude opera's handling of javascript [b]sucks[/b]

Firefox is known for stopping 99% of all popups that occur on Internet Explorer. Plus, there is less clutter on your toolbar. I do recommend Firefox.

I think firefox is 20 times better than internet explorer. Cooler, faster, and better looking!

[quote="Ignite168"][quote="anirudh"]The reason is that it has no spyware[/quote]

That's a lie, as FireFox becomes more and more popular, the amount of reported outbreaks of adware and spyware being specifically for Firefox increases and it's definitely not as "safe" as it was considered when it just came out.[/quote]
However, Mozilla has a reputation of fixing holes much faster than Microsoft. [url=http://www.informationweek.com/internet/showArticle.jhtml?articleID=193005378&subSection=Browsers]This one incident seems to suggest that Firefox fixes bugs 9 times faster than Internet Explorer[/url].

[quote="jmadsen"][quote="anirudh"]It can also edit the source with an additional extension and be able to easily just edit the source code of a webpage and view it live.[/quote]
That sounds useful. What's the name of that extension?[/quote]
It's called the web developer extension.  It also provides you with a nice and easy button to steal blog css if you're into that kind of stuff. [hide="_:mad:_"]*cough* anirudh *cough*[/hide]

:lol: That's in mine too, but mine isn't formatted for easy reading.

It's also interesting how his css incorporates the same random numbers I put in mine. :roll:

[quote="chess64"]Dude i included the following random css into my blog

[code]
#anirudh {
   blog-background: ugly;
   blog-color: ugly;
   blog-decoration: ugly;
}
[/code]

And in a suspiciously short amount of time I get a PM from anirudh about it...

What does that tell you? :roll:[/quote]

that's because I programmed a CSS browser for myself, and it manipulates css, and just popped out  :P

Deleting linebreaks is fun.

Let's get back on topic maybe.

[quote="kheg_k"]How is the SBC Yahoo! Browser?

That is what I sometimes use since it came with DSL.[/quote]
Most of the branded browsers are just fancy interfaces for Internet Explorer.

there was an article in either fortune or business 2.0, i dont remember which one, about this kid who helped improve mozilla firefox


function
arithmetic sequence
number theory solved
number theory
BritishMathematicalOlympiad

Let $f$ be a bijection of the set of all natural numbers on to itself. Prove that there exists positive integers $a < a+d < a+ 2d$ such that $f(a) < f(a+d) <f(a+2d)$

I think this has been posted before but I don't manage to find the link.

So, here is an outline of my solution:

Take $a$ such that $f(a) = 1$ and take $d$ such that $f(a + d) = \min\{f(a + 1),f(a + 2),\dots\}$, and we're done.

there might not exist such a that f(a)=1
natural numbers is an infinite set
con sider f(x) = x+1
it is bijective :?:  :!:

we call a function $ f: A\to B$ bijective iff for all $ y$ in $ B$ there exists one and only one $ x$ in $ A$ such that $ f(x) \equal{} y$.
so there exists exactly one natural number $ a$ such that $ f(a) \equal{} 1$.
and for your information $ f(x) \equal{} x \plus{} 1$ is not bijective on $ \mathbb{N}$.

Doesn't this generalize to us being able to find an infinite (or at least arbitrarily long... infinite might not be true) arithmetic progression mapping to a strictly increasing sequence under f?

[quote="Rushil"]Let $ f$ be a bijection of the set of all natural numbers on to itself. Prove that there exists positive integers $ a < a \plus{} d < a \plus{} 2d$ such that $ f(a) < f(a \plus{} d) < f(a \plus{} 2d)$[/quote]
Let $ x$ be such that $ f(x) \equal{} 1$. Then let $ m \in \mathbb{N}$ be the smallest number such that $ f(1),f(2),f(3),...,f(x\minus{}1) \neq m$. And let $ d>0$ be such that $ f(x\plus{}d) \equal{} m$. Then $ f(x\plus{}2d) > f(x\plus{}d) > f(x)$.

Mathias i think urs is similar to arne's solution.

[quote="Aravind Srinivas L"]How is $ f(x) \equal{} x \plus{} 1$ for Natural numbers not bijective?[/quote]
There is no solution to $ f(x) \equal{} 1$.

exist a $$f(a)=1$$
$$f(a+k)>1.f(a+2k)<f(a+k)$$
$$f(a+k*2^t)<f(a+k)$$
contradiction injective

This is a problem from round 2 of the [url=https://artofproblemsolving.com/community/c6t164f6h107928_permutation]British Math Olympiad 2003[/url]


Hey guys

We just saw the results posted:

AUS1 7  7  0  7  1  0  22
AUS2 7  1  1  7  2  0  18
AUS3 7  1  0  1  0  0   9
AUS4 7  1  0  7  0  0  15
AUS5 7  7  0  7  1  0  22
AUS6 7  7  1  3  4  0  22

total 42 24 2  32 8 0  108

we will find out the medal cutoffs either tonight or tomorrow morning

Congrats Charles! :) (and anyone else who sees this)

I think Valentin's excel document shows how many of each medal/award were given out in each country. You could establish cutoffs for those.. So that's 3 silvers, 2 bronzes, 1 HM?

hey guys good work.
three silvers, two bronze, HM  is heaps good

[quote="shobber"]Congratulations to the Australian Team!  :D[/quote]

Congratulations too! Good work!

my international team was the seventh prize with 3 gold, 3 silver. I'm so happy


Tetris knowledge:

T blocks (shown): 4 configurations
L blocks: 8 configurations
S blocks: 8 configurations
Square Blocks: 1 configuration
Line Blocks: 2 configurations

For a total of 23 blocks.

[quote="james4l"]
T blocks (shown): 4 configurations
L blocks: 8 configurations
S blocks: 8 configurations
[/quote]

Could you please explain in more detail? Also, what are S blocks?

I think he means

 [geogebra]98547f449fa51d85f65f20023f9602a8f89bd351[/geogebra]

Tetris FTW.

There are $ 19$ configurations, shown below:

 [geogebra]6c5c270e92949bb3ea24ae4a8f891e800ec56e68[/geogebra]

(gah don't worry, this won't take [i]too[/i] long)

The number of stamps needed is $ 4\times19\equal{}\boxed{76}$


This is used for showing that $\mathbb{N}^2$ is the same cardinality as $\mathbb{N}$ (although I don't know that that is the context here).  I would just call it counting down the diagonals, or something -- I don't know that it has an "official" name.

In the theory of computation, we used to call it "dovetailing".

I don't really know what it is applicable to, but it makes interesting designs for certain sequences


Prove that if the equation $ x^4\plus{}ax^3\plus{}2x^2\plus{}bx\plus{}1$ has at least one root, then $ a^2\plus{}b^2$ $ \ge 8$

What do you mean by having at least one root. It is a quartic so it must have at least one root, either complex or real. Do you mean one real root?

We will prove the contrapositive, that if $ a^2 \plus{} b^2 < 8$, then the equation has no real roots.

To do this, we will show that $ x^4 \plus{} ax^3 \plus{} 2x^2 \plus{} bx \plus{} 1 > 0$ when $ a^2 \plus{} b^2 < 8$.

$ x^4 \plus{} ax^3 \plus{} 2x^2 \plus{} bx \plus{} 1$
$ \equal{} (x^2 \plus{} 1)^2 \plus{} x(ax^2 \plus{} b)$
$ \ge(x^2 \plus{} 1)^2 \minus{} |x|(ax^2 \plus{} b)$
$ \equal{} (x^2 \plus{} 1)^2 \minus{} |x|\sqrt {(a^2 \plus{} b^2)(x^4 \plus{} 1)}$
$ > (x^2 \plus{} 1)^2 \minus{} |x|\sqrt {8(x^4 \plus{} 1)}$
$ \equal{} (x^4 \plus{} 1) \plus{} 2x^2 \minus{} 2\sqrt {2x^2(x^4 \plus{} 1)}$
$ \equal{} (\sqrt {x^4 \plus{} 1} \minus{} \sqrt {2}|x|)^2$
$ \ge 0$

Q.E.D.

Cheers,

Rofler

well, proving the contraopositive doesn't really prove the real thing, does it?

It does. If the statement is true, then the contrapositive is always true.

[quote="Rofler"]$ \equal{} (x^2 \plus{} 1)^2 \plus{} x(ax^2 \plus{} b)$
$ \ge(x^2 \plus{} 1)^2 \minus{} |x|(ax^2 \plus{} b)$
[/quote] When $ x>0, ax^2\plus{}b<0$ this isn't true.


Today I use Mathematica 5 to calculate these values . Maybe the answers I got are correct , but I cannot prove it .

$\sum_{n=0}^{\infty}\left[\frac{42n+5}{2^{12n+4}}{2n\choose n}^{3}\right]=\frac{1}{\pi}$

$\sum_{n=1}^{\infty}\frac{(n!)^{2}}{n^{2}(2n)!}=\frac{\pi^{2}}{18}$


If 6 moles of chlorine gas are placed in a 3 liter flask at 1250K.  At this temperature, the chlorine begins to dissociate into chlorine atoms.  What is the value for K if 50% of the chlorine molecules dissociate when equilibrium has been achieved?

A Reaction has a very large equilibrium constant of 4*10^12.  Which statement is not true about this reaction?
A the reaction is very fast.
B the reaction is essentially complete
C the reaction is spontaneous
D the equilibrium constant will change if the temperature is changed

Appropriate use of a visible-light spectrophotometer include which of the following?
A Determining he concentration of a solution of Cu(NO3)2
B Measuring the conductivity of a solution KMnO4
C determining which ions are present in a solution that may contain Na+, Mg2+ and Al3+

T or F
reaction order can only be a whole number
Reaction order can only be determined by experiment

[hide]for the first one I think K=2/3 I'm not too confident though
second one i think is A
third one i think is C
and both are true for the seperate post[/hide]

Reaction order does not have to be a whole number. But it's true that it is experimentally determined.

I've never seen a reaction with a noninteger order, that might be because I'm not that advanced.  Are all the other answers I posted correct though?

[quote="cdymdcool"]PbS +O2 => PbO + SO2  heat = -414kJ the equilibrium can be shifted to the products by?
decreasing the pressure, and keep the temperature constatn
heating the reaction, and keep the pressure constant
decreasing the volume and keep the temperature constant[/quote]

[hide]decreasing the volume and keep the temperature constant[/hide]


trigonometry
analytic geometry
graphing lines
slope
geometry
rotation
circumcircle

Supposed that $C$ be a unit circle with centered at origin and $0<a<1,b>1$.
Let $P$ be the intersection point such that the line passing through $A(a,0)$ and $N(0,1)$ has an intersection point with $C$,different from $N$,
and $Q$ be that one passing through $B(b,0)$ and $N$ has an intersection point with $C$,different from $N$.

If $AQ\parallel PB,\angle ANB=\frac{\pi}{4}$,then find the value of $a$.

The equations of the lines $AN, BN$ passing through the points $A = (a, 0),\ N = (0, 1)$ and $B = (b, 0),\ N = (0, 1)$ are:

$AN:\ \ \frac x a + y = 1,\ \ y = 1 - \frac x a = 1 + x \tan \alpha$

$BN:\ \ \frac x b + y = 1,\ \ y = 1 - \frac x b = 1 + x \tan \beta$

Since the lines $AN, BN$ for the angle of $\beta - \alpha = \frac \pi 4$,

$\tan \beta = \tan{\left( \frac \pi 4 + \alpha \right) = \frac{1 + \tan \alpha}{1 - \tan \alpha}}$

$-\frac 1 b =  \frac{1 - \frac 1 a}{1 + \frac 1 a},\ \ \ b = \frac{1 + a}{1 - a}$

The unit circle has an equation $x^2 + y^2 = 1$. Coordinates of the intersections $P, Q$ of the lines $AN, BN$ with the unit circle $(O)$ are obtained by solving the circle equation paired with the equation of the corresponding line. For the point $P$, substitute the $y = 1 - \frac x a$ into the equation of the unit circle:

$x_P^2 + \left(1 - \frac{x_P}{a} \right) = 1,\ \ \ x_P \left( 1 + \frac{1}{a^2}\right) - \frac{2}{a} = 0$

$x_P = \frac{2a}{a^2 + 1},\ \ \ y_P = 1 - \frac{x_P}{a} = \frac{a^2 - 1}{a^2 + 1}$

and similarly

$x_Q = \frac{2b}{b^2 + 1},\ \ \ y_Q = \frac{b^2 - 1}{b^2 + 1}$

The lines $AQ, BP$ have equations:

$y = \frac{y_Q}{x_Q - a}(x - a) = \frac{b^2 - 1}{2b - a(b^2 + 1)}(x - a)$

$y = \frac{y_P}{x_P - b}(x - b) = \frac{a^2 - 1}{2a - b(a^2 + 1)}(x - b)$

Since these 2 lines are parallel, their slopes are equal:

$\frac{a^2 - 1}{2a - b(a^2 + 1)} = \frac{b^2 - 1}{2b - a(b^2 + 1)}$

$2a^2b - a^3b^2 - a^3 - 2b + ab^2 + a = 2b^2a - b^3a^2 - b^3 - 2a + ba^2 + b$

$b^3 - a^3 + a^2b^2(b - a) - ab(b - a) - 3(b - a) = 0,\ \ \ a \neq b$

$b^2 + ab + a^2 + a^2b^2 - ab = 3$

$b^2 + a^2 + a^2b^2 = 3$

which is the 2nd equation for $a, b$. Substituting $b = \frac{1 + a}{1 - a}$ into this equation:

$\frac{(1 + a)^2}{(1 - a)^2} + a^2 + a^2 \frac{(1 + a)^2}{(1 - a)^2} = 3$

$1 + 2a + a^2 + a^2(1 + 2a + a^2 + 1 - 2a + a^2) = 3(1 - 2a + a^2)$

$2a^4 + 3a^2 + 2a + 1 = 3 - 6a + 3a^2$

$a^4 + 4a - 1 = 0$

The only rational roots this quartic equation could have are $\pm 1$ and obviously, neither is a root. Hence, it is necessary to solve numerically: $a \doteq -1.663$ (not acceptable) or $a \doteq 0.249$ (OK, but not quite 0.25).

Thank you for your reply, yetty. I am impressed by your solution and good skil of calculation.

Your solution is complete. In particular the equation which you obtained  $a^4+4a-1=0$ is right,
regrettably I can't explain how to solve this quartic equation.The answer to this problem is as follows.

The answer is $a=\frac{\sqrt{4\sqrt{2}-2}-\sqrt{2}}{2}=\sqrt{\sqrt{2}-\sqrt{2\sqrt{2}-1}}$.

The point is $AN\cdot BN=2$ which is obtained by $AQ\parallel PB$,needles to say, which is obtained from the equation obtained by you $a^2+b^2+a^2b^2=3$ moreover remarking to $\triangle ABN$ .

kunny

[quote="kunny"]...In particular the equation which you obtained $a^2 - 4a + 1$ is right, regrettably I can't explain how to solve this quartic equation.The answer to this problem is as follows.

The answer is $a=\frac{\sqrt{4\sqrt{2}-2}-\sqrt{2}}{2}=\sqrt{\sqrt{2}-\sqrt{2\sqrt{2}-1}}$.

The point is $AN\cdot BN=2$ ...[/quote]

This is a constructible number and the quartic equation can be factored by following the construction.

$AN^2 = OA^2 + ON^2 = a^2 + 1$, [color=white].[/color] $BN^2 = OB^2 + ON^2 = b^2 + 1$

$AN^2 \cdot BN^2 = (a^2 + 1)(b^2 + 1) = a^2 + b^2 + a^2b^2 + 1 = 3 + 1 = 4$

$AN \cdot BN = 2$

Since $AQ \parallel PB$, the triangles $\triangle AQN \sim \triangle PBN$ are similar and

$\frac{AN}{QN} = \frac{PN}{BN}$, [color=white].[/color] $PN \cdot QN = AN \cdot BN = 2$

Hence, the area of the triangle $\triangle PQN$ is

$|\triangle PQN| = \frac 1 2 PN \cdot QN \sin{\frac \pi 4} = \frac{\sqrt 2}{2}$

Since $\measuredangle PNQ \equiv \measuredangle ANB = \frac \pi 4$, the corresponding central angle $\measuredangle PON = \frac \pi 2$ is right and consequently, the base $PQ$ of the triangle $\triangle PQN$ is equal to $PQ = \sqrt{OP^2 + OQ^2} = \sqrt 2$. Since the base and area of this triangle are known, its altitude $NR$ from the vertex $N$ to the base $PQ$ is obtained:

$NR = \frac{2 |\triangle PQN|}{PQ} = \frac{2 \cdot \frac{\sqrt 2}{2}}{\sqrt 2} = 1$

Hence, it is easy to construct a congruent triangle $\triangle P'Q'N' \cong \triangle PQN$ inscribed in the unit circle $(O)$ and then rotate it around the origin (the circumcenter) $O$ to make the points $N', N$ coincide. Let $P', Q'$ be the intersections of the coordinate axes x, y with the unit circle $(O)$, such that $P' = (0, -1)$ is diametrally opposite to $N = (0, 1)$ and $Q' = (1, 0)$. Obviously, $P'Q' = \sqrt 2$. Construct a line parallel to $P'Q'$ at the distance equal to the radius $ON$ of the unit circle $(O)$ (i.e., to 1), intersecting the unit circle $(O)$ at points $N', M'$ above and below the x-axis. Rotate the triangle $\triangle P'Q'N'$ by the angle $\measuredangle N'ON$ around the origin $O$ into the triangle $\triangle PQN$. The line $PN, QN$ intersect the x-axis at the desired points $A, B$. The rotation is not even absolutely necessary, because the lines $P'N', PN$ are symmetrical with respect to the x-axis and as a result, the line $P'N'$ already intersects the x-axis at the point $A$.

The line $P'Q'$ has an equation

$P'Q':\ \ \ y = x - 1$

Let $K$ be the point on a normal to the line $P'Q'$ erected at the point $Q'$, such that $Q'K = ON = 1$. From this and from using the right angle triangle $\triangle P'Q'K$ with the legs $P'Q' = \sqrt 2$, $Q'K = 1$, we get 2 equations for the coordinates of the point $K = (x_K, y_K)$

$Q'K^2 = (x_K - 1)^2 + y_K^2 = 1$

$P'K^2 = P'Q'^2 + Q'K^2$

$x_K^2 + (y_K + 1)^2 = 2 + 1 = 3$

Solving this set of 2 equations for $x_K, y_K$,

$x_K^2 - 2x_K + y_K^2 = 0$

$x_K^2 + y_K^2 + 2y_K = 2$

$x_K + y_K = 1$

$2x_K^2 - 4x_K  + 1 = 0$

$2y_K^2 - 1 = 0$

$x_K = 1 \pm \frac{\sqrt 2}{2}$, [color=white].[/color] $y_K = \pm \frac{\sqrt 2}{2}$

Since the point $K$ obviously is in the 1st quadrant inside of the unit circle, $0 < x_K < 1,\ 0 < y_K < 1$, we have to take the roots

$x_K = 1 - \frac{\sqrt 2}{2}$, [color=white].[/color] $y_K = \frac{\sqrt 2}{2}$

The line through the point $K$ parallel to the line $P'Q'$ has an equation

$y - y_K = x - x_K$

$y = x - 1 + \sqrt 2$

This line intersects the unit circle at the point $N'$. The coordinates of this point are obtained by solving the equations of the line and of the unit circle $(O)$:

$y_{N'} = x_{N'} - 1 + \sqrt 2$

$x_{N'}^2 + y_{N'}^2 = 1$

$x_{N'}^2 + x_{N'}(\sqrt 2 - 1) - (\sqrt 2 - 1) = 0$

$y_{N'}^2 - y_{N'}(\sqrt 2 - 1) - (\sqrt 2 - 1) = 0$

$x_{N'} = \frac{1 - \sqrt 2 \pm \sqrt{2 \sqrt 2 - 1}}{2}$, [color=white].[/color] $y_{N'} = \frac{\sqrt 2 - 1 \pm \sqrt{2 \sqrt 2 - 1}}{2}$

Since $x_{N'} > 0, y_{N'} > 0$, we have to take the roots with the plus signs. Using $P' = (0, -1)$, the equation of the line $P'N'$ is

$y + 1 = \frac{y_{N'} + 1}{x_{N'}} x = \frac{1 + \sqrt 2 + \sqrt{2 \sqrt 2 - 1}}{1 - \sqrt 2 + \sqrt{2 \sqrt 2 - 1}} x$

and the x-coordinate $a$ of its intersection $A$ with the x-axis is obtained for $y = 0$:

$a = \frac{1 - \sqrt 2 + \sqrt{2 \sqrt 2 - 1}}{1 + \sqrt 2 + \sqrt{2 \sqrt 2 - 1}} = \frac{\sqrt 2}{2} \left(\sqrt{2 \sqrt 2 - 1} - 1 \right)  = \sqrt{\sqrt 2 - \frac 1 2} - \frac{\sqrt 2}{2} \doteq 0.24904$

Squaring $a$ and comparing the expressions for $a^2$ and $a$, we get

$a^2 = \sqrt 2 \left( 1 - \sqrt{\sqrt 2 - \frac 1 2} \right) = \sqrt 2 \left(1 - a - \frac{\sqrt 2}{2} \right)$

$a^2  + a \sqrt 2 + 1 - \sqrt 2 = 0$

Thus we obtained one quadratic factor of the quartic equation $a^4 + 4a - 1 = 0$. The other quadratic factor is obtained by long division:

$a^4 + 4a - 1 = \left(a^2  + a \sqrt 2 + 1 - \sqrt 2 \right) \left(a^2  - a \sqrt 2 + 1 + \sqrt 2 \right) = 0$

Playing smart after the fact, the quartic equation really can be written as a difference of 2 squares:

$a^4 + 4a - 1 = a^4 + 2a^2 + 1 - 2(a^2 - 2a + 1) = (a^2 + 1)^2 - 2(a - 1)^2 = 0$

[b]Thank you for your taking much time for explaining to this problem,yetti.[/b]

Alternative solution:

In using $AN\cdot BN=2$,  $\triangle ABN=\frac{1}{2} AB\cdot ON=\frac{1}{2}AN\cdot BN\sin \frac{\pi}{4}$

$\Longleftrightarrow \frac{1}{2}(b-a)=\frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{2}}$

$\Longleftrightarrow b-a=\sqrt{2}\ \cdots\ [1]$, then for $\triangle ANB$, applying to cosine rule,we have

$AB^2=AN^2+BN^2-2AN\cdot BN\cos \frac{\pi}{4}$

$\Longleftrightarrow (b-a)^2=(a^2+1)+(b^2+1)-2\cdot 2\cdot \frac{1}{\sqrt{2}}$

$\Longleftrightarrow ab=\sqrt{2}-1\ \cdots\ [2]$, from $[1],[2]$, we obtain $(a+b)^2=(a-b)^2+4ab=(\sqrt{2})^2+4(\sqrt{2}\1)=4\sqrt{2}-2$.

Since $0<a<1,0<b<1$, yielding $a+b=\sqrt{4\sqrt{2}-2}\ \cdots\ [3]$.

Therefore solving the system of equation $[1]$ and $[3]$, the desired answer is  $a=\frac{\sqrt{4\sqrt{2}-2}-\sqrt{2}}{2}$.

kunny


[b]Compute the following sum:
$ \frac{1}{2\cdot 4}\plus{}\frac{1}{4\cdot6}\plus{}\frac{1}{6\cdot 8}\plus{}...\plus{}\frac{1}{18\cdot 20}$[/b]
i tried to use partial fraction decomposition but that failed miserably. how would u do this?

$ \frac{1}{n(n\plus{}2)}\equal{}\frac{A}{n}\plus{}\frac{B}{n\plus{}2}$
$ 1\equal{}A(n\plus{}2)\plus{}Bn$
$ A\equal{}1/2, B\equal{}\minus{}1/2$

The easiest thing to do is to multiply by $ 4$, so that we're computing the sum

$ \frac{1}{1 \cdot 2} \plus{} \frac{1}{2 \cdot 3} \plus{} \frac{1}{3 \cdot 4} \plus{} ... \plus{} \frac{1}{9 \cdot 10}$.

[hide="Now..."] We want to use the decomposition $ \frac{1}{k(k\plus{}1)} \equal{} \frac{1}{k} \minus{} \frac{1}{k\plus{}1}$.  Can you do it from here?  :) [/hide]

[quote="t0rajir0u"]The easiest thing to do is to multiply by $ 4$, so that we're computing the sum

$ \frac {1}{1 \cdot 2} \plus{} \frac {1}{2 \cdot 3} \plus{} \frac {1}{3 \cdot 4} \plus{} ... \plus{} \frac {1}{9 \cdot 10}$.

[hide="Now..."] We want to use the decomposition $ \frac {1}{k(k \plus{} 1)} \equal{} \frac {1}{k} \minus{} \frac {1}{k \plus{} 1}$.  Can you do it from here?  :) [/hide][/quote]
of course. thanks for the great hint. :-)
thanks for the help as well mountaindew


I'm kinda weak at inequalities, so here goes...

Find all values of x such that 3x² – 11x + 6 > 0

This was a question at a local math contest and I got it, but I did guess and check. I'd appreciate it if you could show me how to solve it without having to guess and check.

If you have time, please help me solve this one as well. Only my team mate got it and it's a shame since I was just watching him do all the work lol.

Find values of x and y that will make 2x– 1/3 – 3(y+1)/5 = 3/2 and 2x– y/2 – y –1/3 = 4 1/12 true.

The inequality can be factored into $ (3x\minus{}2)(x\minus{}3)>0$ so there are two values that $ x$ cannot be, $ \frac{2}{3}$ and $ 3$. By the fact that the LHS has to be greater than the RHS, the solution is $ \boxed{x<\frac{2}{3},x>3}$

Basically what hunter34 said; we need both factors to be of the same sign, or else the LHS would be less than 0 (when you multiply by -1, the inequality sign changes). For the second one, it's just about not making stupid errors; all you have to do is subtract the 1st equation from the second (or vice versa) and solve for y, and then plug the value of y you obtained into one of the equations to find x.

to make stuff clearer, you should translate the second question into LaTeX
the code
[code]$\frac{x}{y}$[/code]
makes the fraction $ \frac {x}{y}$

anyway, here goes...
multiply both sides of the first equation by 30 to zap *:D* all the denominators
$ 60x \minus{} 10 \minus{} 6(3(y \plus{} 1)) \equal{} 45$
simplify...
$ 60x \minus{} 10 \minus{} 6(3y \plus{} 3) \equal{} 45$
$ 60x \minus{} 10 \minus{} 18y \minus{} 18 \equal{} 45$
$ 60x \minus{} 18y \minus{} 28 \equal{} 45$
$ 60x \minus{} 18y \equal{} 73$
tada! our first equation

multiply both sides of the second eqation by 12...
$ 24x \minus{} 6y \minus{} 12y \minus{} 4 \equal{} 49$
simplify...
$ 24x \minus{} 18y \equal{} 53$
our second equation!

using elimination, we subtract the second equation from the first one to get...
$ 36x \equal{} 20$
$ x \equal{} \frac {20}{36}$
$ x \equal{} \frac {5}{9}$
and plugging this into the first equation, we find y to be...
$ 60 \cdot \frac {5}{9} \minus{} 18y \equal{} 73$
$ \frac {100}{3} \minus{} 18y \equal{} 73$
$ \frac {100}{3} \minus{} 18y \equal{} \frac {219}{3}$
$ \minus{} 18y \equal{} \frac {119}{3}$
$ y \equal{} \frac {119}{ \minus{} 54}$
$ y \equal{} \minus{} \frac {119}{54}$

$ x \equal{} \frac {5}{9}$
$ y \equal{} \minus{} \frac {119}{54}$


The numbers 2 and 3 are consecutive primes, as are 19 and 23, because there are no prime numbers between them. Given that $ x$ and $ y$ are consecutive primes less than 100, what is the greatest possible difference between $ x$ and $ y$?

We find that these $ 2$ numbers are $ 89$ and $ 97$.  $ 97\minus{}89\equal{}8$


hey guys try to solve this

Find a closed form of the sum:

1.1! + 2.2! + 3.3! + ......... n.n! in terms of n and constants.

plz post the solns as soon as u find them.........

plz reply

 :)  :wink:

Nice solution keshav

But instead of adding and subtracting 1 i guess this can be written as a telescoping series.

Observe that $ 1! \equal{} (2\minus{}1)*1!$ = $ 2!\minus{}1!$

Similarly nth term is $ (n\plus{}1)! \minus{}n!$

We observe that in this telescoping series all the terms get cancelled except the first and the last term.

So the answer is $ (n\plus{}1)! \minus{} 1!$

which is $ (n\plus{}1)! \minus{} 1$

hey guys nice soln both of u befuddlers and keshav

now plz post some nice problems under this very thread........

Prove the following identity:

Summation of i from 1 to n ( i.nCi ) = 1.nC1 + 2.nC2 +..... + n.nCn =
n. 2^(n-1)

where nCi ,means n choose i or the the no. of ways of choosing i items
from a set of n
= n! / [i! * (n-i)!]

 :)  :wink:  :lol:

FOr the first sum:  Befuddlers wasnt this asked in the AMTi 2nd level Junior exam when you were in 10th da?
Jaswinder where did u find this sum? They have copied even this kind of simple sum.....

2nd sum: Just use combinatorial arguments for selecting some given no. of objects of a total constant and no. of ways of choosing a special one among those selcted.

plz don't post incomplete ideas

i want full solutions

 :wink:

Here is a nice proof.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1328831#1328831[/url].

Thanks,
Befuddlers

Well this is not something like tough but this concept is very useful and torajirou told me this concept was used in IMO 2008

Find $ nC_0  \plus{}nC_3  \plus{} nC_6...................$

Hey befuddlers it is nothing special.
If you want to get rid of the Nc1, Nc2 etc substitute $ \omega^3 \equal{} 1$. And substitute it in the Binomial theorem.
And Jaswinder I am not patient enough to write complete solutions in the Latex Code as I feel it is very time consuming for me. I am ot that fast too like Akashnil who is so used to typing in latex nowadays.
Thats why I didnt post my tig solution too in the INMOTC problems thread . It is such an inelegant solution and will take atleast 35-45 mins for me ven to tyoe it without the latex. So sorry. You people may better consider that I didnt solve that problem at all. Had it been in INMO, I would have got 0 credit for my solution.


linear algebra
matrix
induction
algebra unsolved
linear algebra unsolved

Assume that $X\in\mathcal M_n(\mathbb R)$. Prove that $\mbox{Tr}(A)=0$ iff there exist $P\in\mathcal M_n(\mathbb R)$ which all coeeficents oon the main diagnal of $PAP^{-1}$ are zero.

This has to be the most posted problem on the Linear Algebra forum, the last time was posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=86676]here[/url], by you, in one of its forms. There, spix put a pdf (read it!), which I put in some other topic, also.

Furthermore, at the moment, there are three threads named "Zero trace" on the first page of Linear Algebra - Unsolved Problems, all about the same thing.

As it is stated, it's not exactly correct. Just think of a field $\mathbb F$ of characteristic $p$, and take $n=p$ and $X=I_p$.

Maybe the field is supposed to have characteristic zero? Or maybe $X$ should not be a scalar matrix? Either way, some hypotheses are missing.

Well, induction on $n$ could give a solution.


calculus
integration
logarithms
calculus computations

How to evaluate such integrals?

1)  Integral of {log(a+bcosx)dx}
2)  Integral of {log(a+bsinx)dx}

[quote="Abhishek"]How to evaluate such integrals?

1)  $\int \log (a+b\cos x) dx$
2)  $\int \log (a+b\sin x) dx$[/quote]

Would someone answer the man?

Thank you for fulfilling my request, puuhikki.


In how many ways can we pick three different numbers out of the group \[ 1,2,3,\ldots,100\] such that the largest number is larger than the product of the two smaller ones?  (The order in which we pick the numbers does not matter.)

$ (1,a,b)$--for any $ 2\leq a\leq99$, there are $ 100\minus{}a$ possible values for $ b$, for a total of $ 1\plus{}2\plus{}\cdots\plus{}98\equal{}49\cdot99$ choices.
$ (2,a,b)$--for any $ 3\leq a\leq49$, there are $ 100\minus{}2a$ possible values for $ b$, for a total of $ 2(1\plus{}2\plus{}\cdots\plus{}47)\equal{}47\cdot48$ choices.
$ (3,a,b)$--for any $ 4\leq a\leq33$, there are $ 100\minus{}3a$ possible values for $ b$, for a total of $ 1\plus{}4\plus{}7\plus{}\cdots\plus{}88\equal{}30\plus{}(0\plus{}3\plus{}6\plus{}\cdots\plus{}87)\equal{}30\plus{}3(1\plus{}2\plus{}\cdots\plus{}29)\equal{}30\plus{}3\cdot29\cdot15$ choices.
$ (4,a,b)$--for any $ 5\leq a\leq24$, there are $ 100\minus{}4a$ possible values for $ b$, for a total of $ 4(1\plus{}2\plus{}\cdots\plus{}20)\equal{}2\cdot20\cdot21$ choices.
$ (5,a,b)$--for any $ 6\leq a\leq19$, there are $ 100\minus{}5a$ possible values for $ b$, for a total of $ 5(1\plus{}2\plus{}\cdots\plus{}14)\equal{}5\cdot7\cdot15$ choices.
$ (6,a,b)$--for any $ 7\leq a\leq16$, there are $ 100\minus{}6a$ possible values for $ b$, for a total of $ 2(2\plus{}5\plus{}\cdots\plus{}29)\equal{}2(2\cdot10\plus{}(0\plus{}3\plus{}\cdots\plus{}27))\equal{}2(20\plus{}3\cdot9\cdot5)$ choices.
$ (7,a,b)$--for any $ 8\leq a\leq14$, there are $ 100\minus{}7a$ possible values for $ b$, for a total of $ 2\plus{}9\plus{}\cdots\plus{}44\equal{}2\cdot7\plus{}(0\plus{}7\plus{}\cdots\plus{}42)\equal{}14\plus{}7\cdot3\cdot7$ choices.
$ (8,a,b)$--for any $ 9\leq a\leq12$, there are $ 100\minus{}8a$ possible values for $ b$, for a total of $ 4(1\plus{}3\plus{}\cdots\plus{}7)\equal{}4(4^2)$ choices.
$ (9,a,b)$--for any $ 10\leq a\leq11$, there are $ 100\minus{}9a$ possible values for $ b$, for a total of $ 1\plus{}10\equal{}11$ choices.

Finally, adding, we get $ \boxed{10353}$.

This is the most brutal casework problem (maybe because it was so boring) I have ever seen.

Extremely confusing solution

don't get a single thing

This problem should not be given at this stage of lesson!!!!!!!

[quote=yao136]Why can't we use complementary counting?  For, (x,y,z), z should be larger than the product of x and y according to the problem.  Well, what if we find all the values of z not conforming to this rule?  Therefore the product of x and y has to be 100.  There are only four pairs of two distinct numbers that meet this requirement: (1,100), (2,50), (4,25),and (5,20).  Each pair of x and y, means that 98 values of z do not follow the rule.  If we have four pairs of 98 possibilities each, there are 4*98 possible combinations that do not  follow the rule.  The total number of ways to choose three distinct numbers is C(100,3).  Therefore C(100,3) - 4*98 should be the number of possible combinations.  Right?[/quote]

This is not correct. Consider $(x,y,z)=(1,4,3)$.

Is there a quicker way to do this than Running through all the possibilities?

[quote=asdf334]you've gotta be joking
that was so messed up
not even level 25 worthy but i'll take the free rating[/quote]

me whose rating went up 0.4 after doing this:

[url=https://www.wolframalpha.com/input?i=98*99%2F2+%2B+47*48+%2B+%283*30*31%2F2-2*30%29+%2B+2*20*21+%2B+5*7*15+%2B+%283*10*11-2*10%29+%2B+%287*7*4-5*7%29+%2B+%284*4*5-4*4%29+%2B+10]Y'all, I did it by hand the first time and using a calculator the second, making it the first Alcumus problem I've done with a calculator since reset, and guess what happened.[/url]


Let $ D,E,F$ be tangency point of incircle of triangle $ ABC$ with sides $ BC,AC,AB$. $ DE$ and $ DF$ intersect the line from $ A$ parallel to $ BC$ at $ K$ and $ L$. Prove that the Euler line of triangle $ DKL$ passes through Feuerbach point of triangle $ ABC$.

See Theorem 1 from [url=http://forumgeom.fau.edu/FG2006volume6/FG200621index.html]Bogdan Suceava and Paul Yiu, The Feuerbach point and Euler lines, [i]Forum Geom.[/i], 6 (2006) 191-197[/url].


I have already an answer to this problem, but i just want to know if there are other ways to solve it.

Find the remainder when $ 15!$ is divided by $ 17$.

Yuriy, you are thinking too hard!
$ 15!\equiv a\pmod{17}$
$ 16!\equiv 16a \equiv \minus{} 1\equiv 16\pmod{17}$ (Wilson theorem)
$ a\equiv 1\pmod{17}$

Why $ 16a\equiv \minus{}1\mod 17$? :blush:
Small Fermat`s theorem: $ a^{16}\equiv 1\mod 17$

sorry  :blush: i have fixed it. it is wilson's

Yes, now i see :w00t: : $ (p \minus{} 1)!\equiv \minus{} 1\mod p$ for prime $ p$. Thanks  :wink:


A mile is 5280 feet. How many miles per hour is equal to 528 feet per minute?

Use conversion factors!  You'll need $ \frac {1 \text{mile}}{5280 feet} = 1$  and $ \frac {60 \text{minute}s}{1 \text{hour}} = 1$

Multiply $ \frac {528 \text{feet}}{\text{minute}}$ by these two 1's, and "cancel units":

$ \frac{528 \text{feet}}{\text{minute}}\cdot\frac {60 \text{minute}}{1 \text{hour}}\frac {1 \text{mile}}{5280 \text{feet}} = \frac {528\cdot 60}{5280}\frac {\text{mile}}{\text{hour}} = \boxed{6\frac {\text{mile}}{\text{hour}}}$

[color=red][size=75]Please do not use plain TEXT in $ \text{\LaTeX}$ tags. Use the \text{} instead. -Isabella2296[/size][/color]

Or...
528 feet per minute is the same as 5280 feet per 10 minutes is the same as 1 mile per 10 minutes is the same as 6 miles per hour.


Let $ f: \mathbb{R}\to\mathbb{R}$. prove that if $ f(xy\plus{}y\plus{}x)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$, then $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.

[quote="Jure the frEEEk"]Let $ f: \mathbb{R}\to\mathbb{R}$. prove that if $ f(xy + y + x) = f(xy) + f(x) + f(y)$ than $ f(x + y) = f(x) + f(y)$[/quote]

A not_so_simple solution :

Let $ P(x,y)$ be the property $ f(xy + y + x) = f(xy) + f(x) + f(y)$.

$ P(0,0)$ implies $ \boxed{f(0)=0}$

$ P(x,-x)$ implies $ \boxed{f(-x)=-f(x)}$ $ \forall x$

Let $ x\neq -1$. Then :
$ P(x,\frac{x}{x+1})$ gives $ f(2x) = f\left(\frac{x^2}{x+1}\right) + f(x) + f\left(\frac{x}{x+1}\right)$
$ P(x,-\frac{x}{x+1})$ gives $ f(0) = -f\left(\frac{x^2}{x+1}\right) + f(x) - f\left(\frac{x}{x+1}\right)$
Adding  these two equalities, we have $ f(2x)=2f(x)$ $ \forall x\neq -1$. So, for $ x=1$, we have $ f(2)=2f(1)$ and so $ -f(2)=-2f(1)$ and so $ f(-2)=2f(-1)$ and so
$ \boxed{f(2x)=2f(x)}$ $ \forall x$

$ P(-x,-y)$ gives then $ f(xy - y - x) = f(xy) - f(x) - f(y)$. Adding with $ P(x,y)$, we get $ f(xy+x+y)+f(xy-x-y)=2f(xy)$ and, since $ 2f(xy)=f(2xy)$ :
$ \boxed{f(xy+x+y)=f(2xy)+f(x+y-xy)}$ $ \forall x,y$

Let then $ u\leq 0$ and $ v\in \mathbb{R}$. It is always possible to find $ x$ and $ y$ such that $ xy=\frac{u}{2}$ and $ x+y=v+\frac{u}{2}$
Hence, since ${ f(xy+x+y)=f(2xy)+f(x+y-xy)}$, we have $ f(u+v)=f(u)+f(v)$.
Now, if $ u>0$, we have $ -u<0$ and also $ f(-u-v)=f(-u)+f(-v)$ and so $ f(u+v)=f(u)+f(v)$

And so $ \boxed{f(u+v)=f(u)+f(v)}$ $ \forall u,v$

thanks very much. i had problems with proving it for irationals

I've found a nice solution..

Plugging $ y \equal{} 1$ to the equation we get
$ f(2x \plus{} 1) \equal{} 2f(x) \plus{} c$, where $ c \equal{} f(1)$. Call this property $ (*)$.

Now substituting $ y$ with $ 2y \plus{} 1$ in the initial equation we get
$ f(2xy \plus{} 2x \plus{} 2y \plus{} 1) \equal{} f(2xy \plus{} x) \plus{} f(x) \plus{} f(2y \plus{} 1)$, and now using $ (*)$ after some simplification we get
$ 2f(xy) \plus{} f(x) \equal{} f(2xy \plus{} x)$, which obviously means $ f(2a \plus{} b) \equal{} 2f(a) \plus{} f(b), \forall a,b$. Call this property $ (**)$.

Now substituting $ x$ with $ 2x \plus{} 1$ and $ y$ with $ 2y \plus{} 1$ in the initial equation, we get
$ f(4xy \plus{} 4x \plus{} 4y \plus{} 3) \equal{} f(4xy \plus{} 2x \plus{} 2y \plus{} 1) \plus{} f(2x \plus{} 1) \plus{} f(2y \plus{} 1)$. Now using $ (*)$ and $ (**)$ this equality leads to $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$, and that is what we wanted...


LaTeX
analytic geometry
Pythagorean Theorem
geometry

I made up this question.  It's pretty easy  ;)

If an ant starts in the center of a circle with a radius of 100 mm and goes 30 mm up and 25 mm to the left, how far away is it from the center of the circle?  Round to the nearest hundredth of a mm.

You probably want to use a calculator.

[hide="Here is the answer"]
30 squared + 25 squared=1525
root 1525 is 39.05124838...
39.05 mm[/hide]

[hide]
It's a right triangle, so
$\sqrt{25^2+30^2}=\sqrt{1525}=39.05$ mm
You don't even need to include the circle into the question.
Exact answer is $5\sqrt{61}$ mm.  :D [/hide]

[quote="daermon"]I got that too but how did u get [hide]5 square root 61?[/hide]

by the way, how do you do the square root symbol :huh:[/quote]

You can simplify the radical $\sqrt{1525}=\sqrt{25}\sqrt{61}=5\sqrt{61}$

And the square root is \sqrt{number}, just put $\$$ signs around it :D and there's a forum on it called LaTeX under Site&General Information. You can also go to LaTeX in the sidebar on the left. I found "Guide" most helpful.

[quote="daermon"]oh [color=blue]now[/color] I get it  :oops:

the square root thing still doesn't work :mad:
 \sqrt{61} see?[/quote]

You have to put dollar signs around it. :D

[hide]pythagorean theroem.  It forms a right triangle so just use the pythagorean thereom to get:
39.05[/hide]

use pythogerean theorem. 30 squared plus 25 squared= 1525. take the square root of that and simply radicals to get 5 square root of 61
 :ninja:  :spider:

[hide="my solution"]Denote the answer D.
$D=\sqrt{30^2+25^2}=\sqrt{900+625}=\sqrt{1525}$
Simplify....
$D=5\sqrt{61}$[/hide]

[hide]$30^2+25^2=1525$
$sqrt(1525)$= about 39.05mm[/hide]

How do you make a square root sign with latex?

[quote="Arvind_sn"]I used a different way, which some think is pretty strange--I use a view from the coordinate plane.

[hide]I really did not need the circle. I just imagined the coordinate plane, so basically the ant moved from the origin $(0,0)$ to $(-30, 25)$. Using the distance formula, I got 39.05.[/hide][/quote]
Well, the distance formula is derived from the Pythagorean Theorem, so it's basically the same.

And it's \sqrt{number}.  ;)

Uh, I think you forgot to put the second bracket for $25^{2}+30^{2}$.
It should look like this.
$\sqrt{25^{2}+30^{2}}=\sqrt{1525}=5\sqrt{61}$
[code]\sqrt{25^{2}+30^{2}}=\sqrt{1525}=5\sqrt{61}[/code]

Here's a helpful hint:
If you see someone do something $\mathfrak{cool}$ with Latex, then quote it, see what the code is, and change what they wrote inside to what you want to write inside.
If you see someone put $\sqrt{3}$ and you want to do $\sqrt{6}$, then quote them, see that the code is [b]\sqrt{[color=blue]3[/color]}[/b], and change it to [b]\sqrt{[color=darkred]6[/color]}[/b]

[quote="mimar"]The answer is sqrt(900 + 225)

= 5sqrt(61) :)[/quote]
I recommend you use $\LaTeX$.

Just read the latex guide and you'll be all set. :wink:

[quote="daermon"]I made up this question.  It's pretty easy  ;)

If an ant starts in the center of a circle with a radius of 100 mm and goes 30 mm up and 25 mm to the left, how far away is it from the center of the circle?  Round to the nearest hundredth of a mm.

You probably want to use a calculator.

[hide]25^2+30^2=x^2
solve x[/hide]