floor function
number theory proposed
number theory

A Nice but small problem :

Let $\alpha$ be a positive irrational number. Consider the following sequance $[n\alpha], n \in \mathbb{N}$.
Prove that the prime factors of the elements of this sequance are not limited.

It doesn't really matter that $\alpha$ is irrational, but the rational case is trivially true, so that's probably why the condition that $\alpha$ be irrational was added.

Since the values $\lfloor(n+1)\alpha\rfloor-\lfloor n\alpha\rfloor$ are bounded from above, a standard argument shows that $\sum\frac 1{\lfloor n\alpha\rfloor}=\infty$ (the sum is taken over $n$ large enough that $\lfloor n\alpha\rfloor>0$), but for any finite set $S=\{p_1,\ldots,p_k\}$ of primes, the sum of the reciprocals of those positive integers divisible only by the primes in $S$ is $\prod_{i=1}^k\frac{p_i}{p_i-1}<\infty$.

Well another solution :

We will show that for every prime number like $P$ ,there exists a natural number like $n$ such that $P|[n\alpha]$ .

Its suffice to find a number with the from of $PK$ such that ,$[PK\alpha]=P[K\alpha]$ with is equal to

$\{PK\alpha\}<\frac{1}{P}$  ,but we know that $\{n\alpha\}$ is dense in $[0,1]$,hence there exist such a $K$ and problem is solved.


function
trigonometry
real analysis
real analysis unsolved

For every constant $ a\in \mathbb {R}$ solve differential equation
\[ (1\minus{}x^2)y''\minus{}xy'\plus{}a^2 y\equal{}0.\]
[hide]It has solutions in elementar functions  :wink: [/hide]

hello, is this the Gegenbauer differential equation?
Sonnhard.

I don't know :blush:. You talking about [url]http://en.wikipedia.org/wiki/Leopold_Gegenbauer[/url]?

It has solutions in elementary form? Depends on what you mean by elementary. This is not quite Legendre's equation, but the solution has numerous similarities.

If we look for series about $ x\equal{}0,$ we get solutions of the form

$ y\equal{}\sum_{n\equal{}0}^{\infty}c_nx^n,$ where $ c_{n\plus{}2}\equal{}\frac{n^2\minus{}a^2}{(n\plus{}2)(n\plus{}1)}\,c_n.$

This gives a two-dimensional solution space. Setting $ c_0\equal{}1,c_1\equal{}0$ gives an even solution and setting $ c_0\equal{}0,c_1\equal{}1$ gives an odd solution; these form a natural basis.

If $ a$ is an integer, one of these two solutions is a polynomial. In all other cases, the series has radius of convergence $ 1$ and the functions are presumably singular at $ x\equal{}\pm 1.$

Yes, this even and odd solutions, that you wrote, we can write in close form.

$ y_1(x)\equal{}\cos (a \arccos x)$
$ y_2(x)\equal{}\sin (a \arccos x)$


trigonometry
rotation
geometry
rectangle
geometric transformation
reflection

Hello again,

I am trying to understand the following with the unit circle:

$ \cos (90 \plus{} x) \equal{} \minus{} \sin x$

and

$ \cos (180 \minus{} x) \neq \sin x$.

Could anyone explain the identity and the false statement using the unit circle?

Thank you very much.

If you have a point (cosx, sinx) on the unit circle, notice that if you draw the perpendiculars from the point to BOTH axis, you have a rectangle. If you rotate the whole thing 90 degrees, the angle in question is now 90+/-x, and the rectangle is flipped sideways, meaning cosine and sine are switched.

Thanks for your reply.

However, when I flip that rectangle at $ \cos (90 \plus{} x)$, I still don't see why it is equal to $ \minus{} \sin x$ and and why it is NOT equal to $ \sin x$.

When you rotate an angle by 90 degrees, the quadrant it's in switches (from $ I$ to $ II$, for example, or from $ IV$ to $ I$) -- that means that either the adjacent side or the opposite side switches sign from $ \plus{}$ to $ \minus{}$ or vice versa.  That accounts for the negative sign in $ \minus{} sin(x)$.  Also, say you're rotating (by 90 degrees) a triangle with a leg that intersects the x-axis at some point.  That leg will now intersect the y-axis, not the x-axis; by definition, however, the adjacent side is the side that intersects the x-axis.  Thus, when you rotated it by 90 degrees, you swapped which side intersects which axis, thus swapping the adjacent side for the opposite side.  Therefore, you must also swap the cosine for the sine.

As for the second one, you know that $ cos( \minus{} x)$ = $ cos(x)$ because a negative angle is just the reflection across the x-axis of the same angle with a positive sign.  Thus, even though the sign of its height will change, the sign of its length won't, so the value of the cosine of the angle won't be affected.  So you can rewrite $ cos(180 \minus{} x)$ as $ cos(x \minus{} 180)$.  By rotating something through two quadrants (180 degrees), both the length and the height will switch signs.  So $ cos(180 \minus{} x)$ $ \equal{}$ $ cos(x \minus{} 180)$ $ \equal{}$ $ \minus{}cos(x)$ $ \neq$ $ sin(x)$.


Let  $ ABC$be an triangle. Prove or Disprove that
$ \cot^2\frac{A}{2}\cot^2\frac{B}{2} \plus{} \cot^2\frac{B}{2}\cot^2\frac{C}{2} \plus{} \cot^2\frac{C}{2}\cot^2\frac{A}{2}\ge \ 27$

____________
Ligouras

Let $r_a,r_b,r_c$ be the exradii and $s$ be the semiperimeter of $\triangle ABC$.

By AM-GM, $\left(\frac{\sum{r_a r_b}}{3}\right)^2\cdot\left(\frac{\sum{r_a ^2}}{3}\right)\ge \left(\prod{r_a}\right)^{\frac{4}{3}}\cdot\left(\prod{r_a}\right)^{\frac{2}{3}}= \left(\prod{r_a}\right)^2$

$\Longleftrightarrow \frac{\left(\sum{r_a r_b}\right)^2\cdot\sum{r_a ^2}}{\left(\prod{r_a}\right)^2}\ge 27$

Note the identities: $\begin{cases}\cot\frac{A}{2}=\frac{s}{r_a}\\\\ \sum{r_a r_b}=s^2\end{cases}$

$\implies \sum{\cot^2 \frac{A}{2}\cot^2\frac{B}{2}}=s^4\cdot\sum{\frac{1}{r_a ^2 r_b ^2}}=\frac{s^4\cdot\sum{r_a ^2}}{\left(\prod{r_a}\right)^2}=\frac{\left(\sum{r_a r_b}\right)^2\cdot\sum{r_a ^2}}{\left(\prod{r_a}\right)^2}$.

Thus, $\boxed{\sum{\cot^2 \frac{A}{2}\cot^2\frac{B}{2}}\ge 27}$.


function
inequalities
real analysis
real analysis unsolved

I'm having trouble showing $f_{n}(x) = \frac{x}{1+nx^{2}}$ is uniformly convergent or not on $[0,1]$.

I have that $|\frac{x}{1+nx^{2}}| < |\frac{x}{nx^{2}}| = |\frac{1}{nx}|$. But then I can't get rid of the x.

Help?

You also have that $\frac x{1+nx^{2}}<x$

The two estimate, $x$ and $\frac1{nx}$ are the same size when
$x=\frac1{\sqrt{n}}.$ At that point, $x=\frac1{nx}=\frac1{\sqrt{n}}.$

So we conclude that $0\le\frac{x}{1+nx^{2}}\le\frac1{\sqrt{n}}$ for all $x\in [0,1].$

Since $\frac1{\sqrt{n}}\to0$ as $n\to\infty,$ we conclude that this sequence of functions does converge uniformly to zero.

--

Another observation: the sequence decreases. That is, $f_{n+1}(x)\le f_{n}(x)$ for all $x$ in the domain.  It can be shown that a monotone sequence of continuous functions on a compact set that converges pointwise to a continuous limit mus converge uniformly.

[quote="Kent Merryfield"]So we conclude that $0\le\frac{x}{1+nx^{2}}\le\frac1{\sqrt{n}}$ for all $x\in [0,1].$[/quote]
It follows from AM-GM inequality that $\frac{x}{1+nx^{2}}\leq \frac{1}{2\sqrt n}$.

another question
$x_{0}>0$

$x_{n+1}=\frac{x_{n}}{1+n^{2}x_{n}}$

Find an equivalent of $x_{n}$

[quote="Kent Merryfield"]Another observation: the sequence decreases. That is, $f_{n+1}(x)\le f_{n}(x)$ for all $x$ in the domain.  It can be shown that a monotone sequence of continuous functions on a compact set that converges pointwise to a continuous limit mus converge uniformly.[/quote]
This is what we call "Dini's Theorem".

[quote="liyi"]It follows from AM-GM inequality that $\frac{x}{1+nx^{2}}\leq \frac{1}{2\sqrt n}$.[/quote]
Of course. I was never claiming that $\frac1{\sqrt{n}}$ was the best possible estimate; just that it was a sufficiently good estimate that made use of bythecliff's previous work.

[quote="quantum leap"][quote="liyi"]
It follows from AM-GM inequality that $\frac{x}{1+nx^{2}}\leq \frac{1}{2\sqrt n}$.[/quote]
How?[/quote]
For $x > 0, \frac{1+nx^{2}}{2}\ge \sqrt{1 \cdot nx^{2}}= \sqrt{n}x$

$\Rightarrow \frac{x}{1+nx^{2}}\le \frac1{2\sqrt{n}}.$


limit
trigonometry
logarithms
function
geometric series
real analysis
real analysis unsolved

find limit
$ \lim_{n \to \infty } n sin (2 \pi \e n!)$

Isn't $ \sin (2\pi n!)\equal{}0$ for all $ n$  :?:

[quote="Extremal"]yes, it's zero.
I think Hong Quy mean next problem : find:
$ \lim n \sin (2\pi e n!)$ :)[/quote]
Then because $ \sin 2\pi e n! \equal{} \sin 2\pi \left(\frac {1}{n \plus{} 1} \plus{} \frac {1}{(n \plus{} 1)(n \plus{} 2)} \plus{} ...\right)$$ \sim 2\pi \left(\frac {1}{n \plus{} 1} \plus{} \frac {1}{(n \plus{} 1)(n \plus{} 2)} \plus{} ...\right)$
We have the limit equal to $ \lim \frac {n}{n \plus{} 1} \equal{} 1$ (the remaining terms obviously goes to zero, by comparing with geometric series for example)

$ \lim_{  n  \to \infty }  \frac{ \sqrt[n]{n!}}{n}$

Let $ \textbf L_n \equal{} n\cdot \sin (2\pi e n!)$.
Now we know that: $ e\cdot n! \equal{} \sum_{k \equal{} 0}^{\infty}{\frac {n!}{k!}} \equal{} \sum_{k \equal{} 0}^{n}{\frac {n!}{k!}} \plus{} \frac {1}{n \plus{} 1} \plus{} \sum_{k \equal{} n \plus{} 2}^{\infty}{\frac {n!}{k!}}$ and
$ e\cdot n! \equal{} \textbf N \plus{} \frac {1}{n \plus{} 1} \plus{} \textbf O(n^{ \minus{} 2})$ for some integer $ \textbf N$.

$ 2\pi e\cdot n! \equal{} 2\pi \cdot\textbf N \plus{} \frac {2\pi}{n \plus{} 1} \plus{} \textbf O(n^{ \minus{} 2})$.

And because function $ \sin (x)$ has period $ 2\pi$, we can write: $ \textbf L_n \equal{} n\cdot \left(\frac {2\pi}{n \plus{} 1} \plus{} \textbf O(n^{ \minus{} 2})\right) \equal{} 2\pi\cdot\frac {n}{n \plus{} 1} \plus{} \textbf O(n^{ \minus{} 1})$.

Finally: $ \textbf L_{} \equal{} \lim_{n\to \infty}{\textbf L_n} \equal{} 2\pi$

[quote="Extremal"][quote="Hong Quy"]$ \lim_{ n \to \infty } \frac {\sqrt [n]{n!}}{n}$[/quote]
try to consider :
$ \ln n!$[/quote]

This limit is $ \frac{1}{e}$ because $ n!\equal{}\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\cdot \left(1\plus{}\textbf O(n^{\minus{}1})\right)$.

[quote="Yuriy Solovyov"][quote="Extremal"][quote="Hong Quy"]$ \lim_{ n \to \infty } \frac {\sqrt [n]{n!}}{n}$[/quote]
try to consider :
$ \ln n!$[/quote]

This limit is $ \frac {1}{e}$ because $ n! \equal{} \sqrt {2\pi n}\left(\frac {n}{e}\right)^{n}\cdot \left(1 \plus{} \textbf O(n^{ \minus{} 1})\right)$.[/quote]
\[ \lim_n\sqrt[n]{\frac{n!}{n^n}}\]
Or easier:
Lemma:
if $ \lim_n\frac{|a_{n\plus{}1}|}{|a_n|}\equal{}g$, then $ \lim_n\sqrt[n]{|a_n|}\equal{}g$.
Hence becouse $ \lim_{n}\frac{(n\plus{}1)!}{(n\plus{}1)^{n\plus{}1}}\cdot\frac{n^n}{n!}\equal{}\lim_n\left(\frac{n}{n\plus{}1}\right)^n\equal{}\lim_n\frac{1}{\left(1\plus{}\frac{1}{n}\right)^n}\equal{}\frac{1}{e}$, then our limit equals $ \lim_n\frac{\sqrt[n]{n!}}{n}\equal{}\frac{1}{e}$


calculus
integration
function
trigonometry
calculus computations

I remember my friend telling me about a certain order of operations type thing with integration by parts (what type of function to use for u and which for dv), unfortunately, I haven't used it since he told me (last summer) so I can't remember it.

It's the acronym LIATE. It stands for:
Logarithmic
Inverse trig
Algebraic
Trigonometric
Exponential

It gives an order of preference for what part of an integrand to make "u" when doing integration by parts.

It works most of the time, but there are some examples where it doesn't work and where a different choice for u does.

Awesome! Thanks a lot. I knew exponential would be first but the rest I was kind of gut-calling. Btw, how often does the LIATE system not work? Do you think an integral would come up on my final where it does not hold true?


trigonometry
USAMTS
complex numbers
number theory
\/closed

[quote="Mr.Ocax"]I think I read or heard or something somewhere that there would be another Intermediate Trig/ Complex numbers course in the summer.. Is this true??

Mr.Ocax[/quote]

Sorry I can't speak for the admin, but it seems like this class is running once every 3-4 months. So most likely there will be one during the summer. It's reasonable to assume that there would be more classes during the summer than the regular school year.

We have not yet set our summer schedule.  However, if a particular class seems in demand, we will almost certain hold it.  Other students who will want to take it during the summer might find it a good idea to post in this thread that they plan to take it so that we will have a good idea of what demand is.

[quote="mo"]I would be very interested in an Intermediate Trig/Complex Numbers course.  :)[/quote]

Since we are currently running that course, it's not likely we'll run it again before the Fall.

[quote="mo"]That's too bad. Are you going to be offering intermediate classes other than the single intermediate class listed currently for the summer?[/quote]

If we do, it will most likely be an Intermediate Number Theory course which would start later in the summer than other courses and continue into the school year.  There seems to be a very large amount of demand for that course, so those who take it during either of the first two times we run it will probably benefit from an extremely active class and message board.

I would definately be interested in such a course! I look forward to seeing it offered later this year! :)

Since the class enrollment page says that the next time that this class will be offered is Summer or Fall, 2005, is there any way I can obtain a copy of the pre and post tests for the inter trig/complex# class before the next time it is to be offered?

There's a possiblity that they've been copyrighted by AoPS, but I'm not sure.

We plan to eventually post them on a permanent page.  We haven't gotten to that only due to higher priorities (general school/site duties and all of the recent competition and USAMTS Math Jams).


calculus
integration
logarithms
calculus computations

Evaluate $ \int_1^e \frac{\sqrt[3]{x}}{x(\sqrt{x}\plus{}\sqrt[3]{x})}\ dx$.

[quote="kunny"]Evaluate $ \int_1^e \frac {\sqrt [3]{x}}{x(\sqrt {x} + \sqrt [3]{x})}\ dx$.[/quote]

let$ \sqrt[6]{x}=t,$then $ \sqrt[3]{x}=t^2,\sqrt[2]{x}=t^3$,when $ x$ goes from 1 to $ e$,$ t$ goes from 1 to $ \sqrt[6]{e}$,the given integral can be translated into

$ 6\int_{1}^{\sqrt[6]{e}}\frac{1}{t(t+1)}dt=6\int_{1}^{\sqrt[6]{e}}(\frac{1}{t}-\frac{1}{t+1})dt=6(\ln t-\ln(t+1)|_{1}^{\sqrt[6]{e}}
=1-6\ln(1+\sqrt[6]{e})+6\ln 2$


calculus
integration
analytic geometry
geometry
geometric transformation
reflection
combinatorics proposed

Given a set $ M$ of points $ (x,y)$ with integral coordinates satisfying $ x^2 + y^2\leq 10^{10}$. Two players play a game. One of them marks a point on his first move. After this, on each move the moving player marks a point, which is not yet marked and joins it with the previous marked point. Players are not allowed to mark a point symmetrical to the one just chosen. So, they draw a broken line. The requirement is that lengths of edges of this broken line must strictly increase. The player, which can not make a move, loses. Who have a winning strategy?

I assumed the problem reads: Given [b]the[/b] set $ M$ of points $ (x,y)$ with integral coordinates satisfying:

$ x^{2}\plus{}y^{2}\leq 10^{10}$.

In that case the first player (A) wins.
Proof:
First he places a point on the origin. Then, for every point his opponent (B) places, he places the point that is the reflection about the origin of B's point. Each time, B must increase the distance from the origin. When the distance becomes $ 10^{5}$, he will no longer be able to do so.

[quote="Umut Varolgunes"][quote="Chgk"]Oops, I just read the original Russian statement of the problem and it seems that we are not allowed to mark a point symmetrical to the one just chosen.[/quote]

Is this really true?[/quote]

Yes. I have modified the problem statement according to the new correction. I hope it is ok now.
Thanks to [b]Chgk[/b].

The first player always win.

The key idea is to strengthen the game into one where the set of points are invariant under rotation of $90^{\circ}$ and also contain the origin. Clearly the original set of points satisfies the conditions. Now we induct on $|S|$ where $S$ is the set of points. Consider all the longest line segments containing asymmetric points in $S$ and let them be $A_1B_1,A_2B_2,..,A_nB_n$ where $A_i,B_i$ need not be distinct.

First, clearly the origin is not one of the endpoints of those line segments as else say if $OA = d$ is one of them, then rotate $A$ by $90^{\circ}$ to $B$ and then we get $OB = d\sqrt{2} > d$ - a contradiction.

Now let $S' = S$ \ $ \{A_iB_i , i = 1,2,...,n \}$. Note that $S'$ is still invariant under rotation of $90^{\circ}$ since the origin isn't one of the points. By induction, the first player will win if the game is played within $S'$ and if the second player chooses any point outside $S'$, then the first player simply chooses the corresponding $A_i$ or $B_i$ and the second player is unable to make any move.


Paul plans to rotate the five tires on his new four-wheel truck
so that each tire will have the same number of miles when
the odometer reads 20,000 miles. How many miles will be on
each of the tires?

The total distance traveled was 20,000 miles, and there were 5 wheels, so each wheel would get 4000 miles. But since there are 4 wheels on a car, our answer is 4000 * 4, or 16,000 miles.

Sorry, I don't understand why we multiplied by 4. Can you please explain?

we multiplied by 4 because when switching the wheels, there has to be four tires on the car at any time
that means any tire is on the car for 4/5 of the trip (four on the car, one spare)
since the total milage is 20000, we multiply 20000 by 4/5 to get our answer of 16000

This is like that other problem we're still discussing.

Oh, and I think the "mysterious" formula works only if $ spots > things$.


Functional Analysis
advanced fields
advanced fields unsolved

Prove that a Banach space which is generated by a compact set is finite-dimensional.

Let $ K$ be a compact generating set of our banach space $ X$. Replacing $ K$ by $ \{x | |x|\leq 1\}\cdot K$ we can assume wlog that $ K$ is balanced.

Now let $ A_1: \equal{}K, A_2: \equal{}K\plus{}K, A_3: \equal{}K\plus{}K\plus{}K,\ldots$. Because $ K$ (and hence all $ A_n$) is balanced, $ span(K)\equal{}\bigcup_{n\geq 1} A_n$.
All $ A_n$ are compact because $ K$ is. In particular $ X$ is a union of closed sets. The Baire theorem tells us, that there is an $ A_k$ with non-empty interior. Hence $ A_k$ is a compact neighbourhood of some point. All points of $ X$ are topological the same, so $ X$ is locally compact. A locally compact vectorspace is finite dimensional.


geometry
rectangle
combinatorics proposed
combinatorics

Given an $ n\times n$ chessboard.
a) Find the number of rectangles on the chessboard.
b) Assume there exists an $ r\times r$ square (label $ B$) with $ r<n$ which is located on the upper left corner of the board. Define "inner border" of $ A$ as the border of $ A$ which is not the border of the chessboard. How many rectangles in $ B$ that touch exactly one inner border of $ B$?

(a) The number of rectangles is $ \binom{n\plus{}1}{2} ^2$.
(b) From the question, it is not clear if the rectangle is allowed to touch one of the borders of the chessboard in addition to it touching the inner border of $ B$. The number of rectangles that touch exactly one of the inner borders is $ \binom{r\plus{}1}{2} ^2 \minus{} \binom{r}{1} ^2$(assuming the rectangle is allowed to touch the border of the chessboard). If the rectangle is not allowed to touch the border of the chessboard, then the answer is $ \binom{r}{2} ^2 \minus{} \binom{r\minus{}1}{1} ^2$.


Can i ask a favour of you to help me in those questions below which have been baffling me long?thx

Prove each statements.
1.If A⊆B, then A∩C ⊆B∩C
2.A⊆B if and only if B'⊆A'
3.Express (AUB)-(A∩B) in terms of unions and intersections that involve A, A', B, and B'.

Hint: whenever you want to prove $ A\subseteq B$, let $ x\in A$ and prove that $ x\in B$. whenever you want to prove $ X\equal{}Y$, where $ X$ and $ Y$ are sets, prove both $ X\subseteq Y$ and $ Y\subseteq X$.

for 3, get the answer using Venn diagrams (this part is easy), then prove it using the ideas above.

Thx very much for ur hints and it does help me alot. However, sometimes even i know the concept but still i cannot be able to solve certain questions. For instance:

Let Z denote the set of all integer,and let
                          A = { x | x = 3p-2 for some p∈Z}
                          B = { x | x = 3q+1 for some q∈Z}.
Prove that A=B.

I'm getting stuck :( Can u please demonstrate how to deal with it???

this is really very elementary...
$ 3p \minus{} 2 \equal{} 3p \minus{} 3 \plus{} 1 \equal{} 3(p\minus{}1)\plus{}1 \equal{} 3q \plus{} 1$
where q = p-1


AMC
AIME
geometry
parallelogram
analytic geometry
graphing lines
slope

Answers:
[hide]1. 118
2. 185
3. 169
4. too lazy; an easy way?
5. 025
6. 489
7. 345
8. ---[/hide]

Brief solutions:
[hide]1. In order to cut the parallelogram in half, the line must pass through its center, which is (19,99).  Can you guess what year this is from?  So the slope is 99/19 => 118.

2. There are eight similar triangles not in the area common to both squares.  Each is right and has a hypotenuse of 43/99.  Let one of these triangle's legs be $a$ and $b$.  Then $a+b = \frac{56}{99}$ (since a+b+43/99 = 1) and $a^2 + b^2 = \left(\frac{43}{99}\right)^2$.  So $2ab = (a+b)^2 - (a^2+b^2) = \left(\frac{56}{99}\right)^2 - \left(\frac{43}{99}\right)^2$ = 13/99.  Since the area of the octagon is just a square minus four of these triangles, the answer is $1 - 4\left(\frac12 ab\right) = 1 - 2ab = \frac{86}{99}$.

3. We find that there are $\binom{8}{3}$ good orders (I did this with a bunch of counting and the hockey stick theorem, but there's a better way probably) out of a total of 6^4 possible orders, so 7/162 => 169.

5. When restricted to $x,y,z \ge 0$, the plane is an equilateral triangle with vertices at (1,0,0), (0,1,0), and (0,0,1).  Map to a 2D diagram.  As in the below diagram, I named the vertices of my equilateral triangle $X,Y,$ and $Z$, and the closer a point is to, say, $X$, the greater its $X$ value.  If a point is in $XBE$, $x \ge \frac12$, if a point is in $YAD$, $y \ge \frac13$, and if a point is in $ZFC$, $z \ge \frac56$.  Thus we see that the shaded triangles represent points supporting $(\frac12, \frac13, \frac16)$.  The shaded area is $\frac{7}{18}$ by a simple calculation (probably use parallelograms.)

[img]http://www.artofproblemsolving.com/Forum/files/aime17-5.jpg[/img]

6. There are $\binom{10}{2} = 45$ segments and $\binom{10}{3} = 120$ triangles.  Thus there are 120*42 (because we used three segments in forming the triangle) combinations of a triangle and a segment.  There are $\binom{45}{4}$ ways to choose 4 segments.  Divide the first by the second, get 16/473 => 489.  This was on another activity.

7. Let the incircle be tangent to $AC$ at $Q$ and $BC$ at $R$.  From power of a point, $AQ = AP = 23$ and $BR = BP = 27$.  Let $CQ = CR = x$.  The perimeter of the triangle is then 2(50+x) and the area is $21(50+x)$ by the formula area = inradius*perimeter/2.

But the area is also, by Heron's formula,
\[[ABC] = \sqrt{(50+x)(23)(27)(x)}\]
so
\[21(50+x) = \sqrt{(50+x)(23)(27)(x)}\]
\[21\sqrt{50+x} = \sqrt{621x}\]
\[441(50+x) = 621x\]
\[441\cdot 50 = 180x\]
\[x = \frac{245}{2}\]

So the perimeter is $2(50+\frac{245}{2}) = 345$.[/hide]

yo if you're not too busy with school starting back up, could you explain how you got the 8C3 thing for #3?

#4
the original quadrilateral is bounded by y = -x + 1200, y = -x + 2400, y = 3x and y = (1/3)x, so its image is bounded by y^2 = -x^2 + 1200, y = -x + 2400, y = (sqrt3)x and y = (1/sqrt3)x.. because if you square the coordinates of its image, they should fit the original equations
so you're finding the area of a slice of the part of the larger circle not contained in the smaller circle
the angle between the lines is 30 degrees
pi(R^2 - r^2)/12 = pi(2400-1200)/12 = 100pi, so the answer is 314.
unless something's wrong.


I'm confused - - Can you explain it in a much more easier way that a 13 year old would understand xD

See, I had to do 6 Histograms of the features of the people in my class that were numbered from 1-24; Which would be the total number of observations? (24) On The X Axis; it was the length of the feature/etc in cm, and there were 4 bars in the x axis, so 4 categories?... Uhh..So how does the formula work with that?

Did you intend to reply to my post in your first topic?  You want the "Post Reply" button, not the "New Topic" button.

Oh Oops! *-*; Sorry I dont know how to work this


calculus
derivative
inequalities proposed
inequalities

If $x,y,z$ be positive real numbers such that $x^{-2}+y^{-2}+z^{-2}\leq 3$, then
\[6xyz+x+y+z\geq 3(yz+zx+xy).\]

It can be solved by taking derivatives and find the minimum value but it's really just brute force. Hoping for a nice method :)


[quote="Totally Zealous"]How can you turn microsoft word files into pdf files?
Is there a program included on Microsoft Windows XP or can you download a safe program online?

Usually I attach my files here but this time I can't.[/quote]
Here are two free programs that can do that for you:
[url=http://www.cutepdf.com/Products/CutePDF/Writer.asp]CutePDF[/url]
(click on "Free Download")
and
[url=http://www.go2pdf.com/product.html]Go2PDF[/url]
(click on "Download Now").
You only need one of them though; they do basically the same thing.

You have also Adobe® Acrobat® 7.0 Professional, but it is not free.

http://www.adobe.com/products/acrobatpro/

I use a program called PDF995.

Advantages: No advertising is made on your PDF AT ALL.

Disadvantages: 1 "Popup ad" in program every time you convert a document.

However, I find PDF995 the best option, and I have all the above installed, and more.

Ah - one advantage for Mac users. Just choose "Print" and the standard printer-driver dialog box includes "Save as .pdf" as an option. And the "Preview" application reads and displays .pdf files.

[quote="joml88"]Hm, well there's also [url=http://www.primopdf.com/]PrimoPDF[/url].  No ads or anything (as far as I know).

Eh, I wouldn't go out and buy a mac just because their word processor program comes with a built in print to pdf function.  It takes less than 2 minutes to download and install a free program to do the same thing on windows.[/quote]
I don't think that's what Kent Merryfield meant  :P ; he was just pointing out that if you [i]already[/i] have a Mac it comes with a PDF converter.  :wink:

[quote="joml88"]Hm, well there's also [url=http://www.primopdf.com/]PrimoPDF[/url].  No ads or anything (as far as I know).

Eh, I wouldn't go out and buy a mac just because their word processor program comes with a built in print to pdf function.  It takes less than 2 minutes to download and install a free program to do the same thing on windows.[/quote]

I also have PrimoPDF.  However, sometimes it lags.  I will hit print, and the window comes up like 3 minutes later.  It only happens when I have a lot of windows open, but it's still annoying.

[quote="joml88"][quote="nebula42"][quote="joml88"]Hm, well there's also [url=http://www.primopdf.com/]PrimoPDF[/url].  No ads or anything (as far as I know).

Eh, I wouldn't go out and buy a mac just because their word processor program comes with a built in print to pdf function.  It takes less than 2 minutes to download and install a free program to do the same thing on windows.[/quote]
I don't think that's what Kent Merryfield meant  :P ; he was just pointing out that if you already have a Mac it comes with a PDF converter.  :wink:[/quote]

Huh?  Dr. Merryfield said "Ah - one advantage for Mac users."  And I just responded that I didn't think it was all that big of an advantage.[/quote]
Oh, okay, I see what you mean now. :oops:

And, by the way, why do all .PDF converters have to convert files by going through print?

I think CutePDF is best.

I disagree with Go2PDF because unless you're using TexnicCenter to do this, you'll get this message on the bottom of the every page:

To remove this from your paper, purchase the real one from Go2PDF.com.

Or something like that in blue letters.


6 point in a plane are organized such a right line who has 2 of this point should have a another point (aumoins).
Prove that this 6 point are in the same right line of the plane. (Hope  you   understand)

1) proofs don't belong here
2) um, right line?

Please either work on your grammar or go to your country's fourm and use your native language.

Right line, infinite line in the plane, $ (d_1) ,(d_2)......$

Maybe you can understand what I mean if i say "2 right lines cut each other" is the same to write   $ (d_1)\cap({d_2})$
right line or line , infinite line . What Euclide's postulate refere to........

You could just say line you know.
Because a line is infinite anyways.

The question says (i believe)

There are 6 points in a plane. If a line connects any two points it touches a third.
Show that they are all collinear.

And back to the old argument, no proofs in classroom math.

Frankeal formulation is beter than mine but that is what I mean . You can not do any exclusion and post the solution of the problem.   If you don't where can I post this?

High school basics, intermediate math...

I don't have time to find the real proof, but try this...

On a piece of paper, draw 2 points. Connect them with a line. Draw a third point on the line. Now, see if you can place more points so that the rules hold. Obviously, it can't be done, so maybe you'll come up with the proof as you do it.

assume these points are not collinear
call these six points a,b,c,d,e,f

assume that b lies on the line formed by a and c.

b makes a line with d.

If a,b,c,d are not collinear, there must be another point on the line bd different from a and c. Call this point e.

Now the line fb contains a third point, implies that f,a,b,c or f,b,d,e are collinear.

Now fe must have a third points, implies e,f,a,b,c are collinear.
But since bde are collinear, e,f,a,b,c are collinear.

If a,b,c,d are collinear then lines ea, and eb must contain a third point.
If a,b,c,d,e are not collinear then
if this point is not a,b,c,d then it is f. but a and b are distinct thus the lines are distinct and f cannot lie on both.
Thus a,b,c,d,e,f are collinear.

If a,b,c,d,e are collinear then af must contain a third point, but the only other points are collinear to b,c,d,e. Any choice would make a,b,c,d,e,f collinear.

I am in 7th grade and I am doing geometry.So what is wrong with putting that in the middle school section?

It's not like all geometry is in 7th grade.
Proofs don't belong in 7th grade geo.

Yeah, it's like regular math class, where you learn how to solve something like $ x\plus{}8\equal{}20$

[quote="zolojetto"]this isnt "middle shcool" section--I am in middle school and take high school math(lets say im in calculus which my friend--also in junior high is) so why couldnt i/(S)he put calculus in this forum???.......lets see.....well.....this is a forum created for like algebra and basic arithmatic--not for middle shcoolers, classroom math to me reminds me of sitting in an elementary school classroom learning addition and basic algebra![/quote]

first off, middle school classroom math is what you would expect in an ordinary math class.  remember, if you are taking high school math, then even if you're in middle school, it doesn't belong in this forum.  Secondly, instead of arguing about this here, just read the forum guidelines.


Let $ m, n \in Z$ and $ m>n\ge 1$. Prove that: $ \boxed{\displaystyle \left( 1\plus{}\frac{1}{m}\right)^m>\displaystyle \left( 1\plus{}\frac{1}{n}\right)^n}$

This proof is quite useful in analysis:

Lets prove that $ a_{n} > a_{n\minus{}1}$, where $ a_{n} \equal{} \left( 1 \plus{} \frac {1}{n}\right) ^n$, for $ n\geq 2$.

Using Bernoulli's inequality for $ n\geq 2$, we obtain:

$ \left( 1 \plus{} \frac {1}{n^2}\right) ^n > 1 \minus{} n\cdot\frac {1}{n^2} \equal{} 1 \minus{} \frac {1}{n}$

so

$ a_n \equal{} \left( 1 \plus{} \frac {1}{n}\right) ^n > \left( 1 \minus{} \frac {1}{n}\right) ^{1 \minus{} n} \equal{} \left( \frac {n}{n \minus{} 1}\right) ^{n \minus{} 1} \equal{} \left( 1 \plus{} \frac {1}{n \minus{} 1}\right) ^{n \minus{} 1} \equal{} a_{n \minus{} 1}$ qed.


limit
logarithms
calculus
integration
function
calculus computations

Compute: $\displaystyle \lim_{n\to \infty} \sum_{k=2}^{n} \frac{1}{k \ln k}$

Use the graph of $y=\frac{1}{x\ln x}$ and Rieman's Integral to compare the area.

is it 0? I don't know Rieman and I guess I am not allowed to use it! :D

How can it be 0? The first term is positive....

$f(x)=\frac1{x\ln x}$ is a decreasing positive function so

$\int_2^\infty\frac1{x\ln x}dx=\ln\ln x|_{x=2}^{x=+\infty}=\textrm{diverges}$

by integral test....

Here's an argument. Note that this is precisely the condensation test. I'm assuming that the proposer is not familiar with that test, so I am essentially proving it in this case.

$\frac1{2\ln2}+\frac1{3\ln3}+\frac1{4\ln4}+\frac1{5\ln5}+\cdots$

$=\frac1{2\ln 2}+\left(\frac1{3\ln3}+\frac1{4\ln4}\right) +\left(\frac1{5\ln5}+ \cdots\frac1{8\ln8}\right)+\left(\frac1{9\ln9}+\cdots\frac1{16\ln16}\right)+\cdots$

$\ge\frac1{2\ln 2}+\left(\frac1{4\ln4}+\frac1{4\ln4}\right) +\left(\frac1{8\ln8}+ \cdots\frac1{8\ln8}\right)+\left(\frac1{16\ln16}+\cdots\frac1{16\ln16}\right)+ \cdots$

$\ge\frac1{2\ln2}+\frac2{4\ln4}+\frac4{8\ln8}+\frac8{16\ln16}+\cdots$

$=\frac1{2\ln2}+\frac2{4\cdot2\ln2}+\frac4{8\cdot3\ln2}+\frac8{16\cdot 4\ln2} +\cdots$

$=\frac1{2\ln2}\left(1+\frac12+\frac13+\frac14+\cdots\right)$

That's a constant multiple of the harmonic series, and the harmonic series is known to diverge.

If you don't already know that the harmonic series diverges, repeat the same idea for it (this is Oresme's proof - nearly 800 years old - that the harmonic series diverges.)

Wow, this is marvelous and beautiful! This method is similar to prove that harmonic series diverges!  :)

We can summarize the method as a test, known as the condensation test:

Suppose that the terms $a_n$ are positive, decreasing, and tend to zero. Then

$\sum_n^{\infty}a_n$ converges if and only if $\sum_k^{\infty}2^ka_{2^k}$ converges.

Since the first few terms are just a finite sum, you can weaken "positive, decreasing" to "eventually positive, eventually decreasing." But you can do that sort of thing with any test.

The reason that it's a two-way test (if and only if) is that we could estimate the power-of-2 blocks of terms either downward (as I did above) or upward..

What other test has "decreasing" as one of the hypotheses? The integral test. It's not that surprising that the integral test and the condensation test are, for the most part, effective and useful in the same cases.

[quote="10000th User"]Wow, this is marvelous and beautiful! This method is similar to prove that harmonic series diverges![/quote]

Absolutely. The best arguments in mathematics take on a life of their own and can be used beyond the setting in which they were originally proposed. Oresme's argument (originally given during the 1300's in the context of the harmonic series) is one of those best arguments.


1.Find the general formula  of the sequence u_n defined by :
$ u_0 = a;u_1 = b$; $ u_{n + 2} = 2.(2n + 3)^2.u_{n + 1} - 4(n + 1)^2(2n + 1)(2n + 3)u_n \for all n > 0$
2.Find a natural n such that there is a permutation $ {x_1,..x_n}$ of  $ {1,2,..,n}$ such that $ k|x_1 + ... + x_k$
for all k from 1 to n

[quote="nkht-tk14"]1.Find the general formula  of the sequence u_n defined by :
$ u_0 = a;u_1 = b$; $ u_{n + 2} = 2.(2n + 3)^2.u_{n + 1} - 4(n + 1)^2(2n + 1)(2n + 3)u_n \for all n > 0$
[/quote]

Since the recurrence is only valid $ \forall n > 0$, I consider that the sequence begins with $ n = 1$ and that we have $ u_1 = a$ and $ u_2 = b$ (instead of $ u_0 = a$, useless, and $ u_1 = b$)/

Let then $ u_n = n(2n - 1)!v_n$. $ \forall n\ge 1$. The equation becomes :

$ (n + 2)(2n + 3)!v_{n + 2} =$ $ 2(n + 1)(2n + 3)^2(2n + 1)!v_{n + 1} - 4n(n + 1)^2(2n + 1)(2n + 3)(2n - 1)!v_n$

$ \iff$ $ (n + 2)v_{n + 2} =$ $ (2n + 3)v_{n + 1} - (n + 1)v_n$

$ \iff$ $ (n + 2)(v_{n + 2} - v_{n + 1}) =$ $ (n + 1)(v_{n + 1} - v_n)$

And so $ n(v_n - v_{n - 1}) = 2(v_2 - v_1)$ $ \forall n > 1$

So $ v_n = 2(v_2 - v_1)\sum_{k = 2}^n\frac 1n + v_1$ $ \forall n > 1$

This may also be written $ v_n = 2(v_2 - v_1)\sum_{k = 1}^n\frac 1n + 3v_1 - 2v_2$ $ \forall n\ge1$

So $ \boxed{u_n = n(2n - 1)!\left(3u_1 - \frac {u_2}6 + (\frac {u_2}{6} - 2u_1)\sum_{k = 1}^n\frac 1n \right)}$ $ \forall n\ge 1$

[quote="nkht-tk14"]2.Find a natural n such that there is a permutation $ {x_1,..x_n}$ of  $ {1,2,..,n}$ such that $ k|x_1 + ... + x_k$
for all k from 1 to n[/quote]
$ n=1$

[quote="nkht-tk14"]
2.Find a natural n such that there is a permutation $ {x_1,..x_n}$ of  $ {1,2,..,n}$ such that $ k|x_1 \plus{} ... \plus{} x_k$
for all k from 1 to n[/quote]

At $ k \equal{} n$ we have $ n | \frac {n(n \plus{} 1)}{2}$ so $ \frac {n \plus{} 1}{2}\in\mathbb{N}^*$, meanwhile $ n$ is odd.

At $ k \equal{} n \minus{} 1$, we have $ (n \minus{} 1) | \frac {n(n \plus{} 1)}{2} \minus{} x_n$, thus $ (n \minus{} 1) | n(n \plus{} 1) \minus{} 2x_n \equal{} n(n \minus{} 1) \plus{} 2(n \minus{} x_n)$, meanwhile $ (n \minus{} 1) | 2(n \minus{} x_n)$. Thus $ x_n \equal{} 1$ or $ x_n \equal{} \frac {n \plus{} 1}{2}$. But if $ x_n \equal{} 1$ then $ x_1 \plus{} x_2 \plus{} ... \plus{} x_{n \minus{} 1} \equal{} \frac {n(n \plus{} 1)}{2} \minus{} 1 \equal{} \frac {n^2 \plus{} n \minus{} 2}{2} \equal{} \frac {(n \minus{} 1)(n \plus{} 2)}{2}$, which is not a multiple of $ n \minus{} 1$, since $ \frac {n \plus{} 2}{2} \not \in \mathbb{N}^*$, contradiction. Thus $ x_n \equal{} \frac {n \plus{} 1}{2}$.

At $ k \equal{} n \minus{} 2$, we have $ (n \minus{} 2) | \frac {n^2 \minus{} 1}{2} \minus{} x_{n \minus{} 1}$. Hence $ (n \minus{} 2) | (n^2 \minus{} 4) \minus{} (2x_{n \minus{} 1} \minus{} 3)$ or $ 2x_{n \minus{} 1} \minus{} 3 \vdots (n \minus{} 2)$. But $ 2x_{n \minus{} 4} \leq 2n < 3(n \minus{} 2)$ (if $ n > 6$) so either $ 2x_{n \minus{} 1} \minus{} 3 \equal{} 2(n \minus{} 2)$ or $ 2x_{n \minus{} 1} \equal{} n \minus{} 2$. But since $ 2x_{n \minus{} 1} \minus{} 3$ is odd, it must not be $ 2(n \minus{} 2)$ and since $ x_{n} \equal{} \frac {n \plus{} 1}{2}$ so $ x_{n \minus{} 1} \neq \frac {n \plus{} 1}{2}$, meanwhile $ 2x_{n \minus{} 1} \minus{} 3 \neq n \minus{} 2$. The contradiction is followed. Thus $ n\leq 6$.

A few more tests (or argument) yields the result.

[quote="pco"]Good job, Mashimaru, but useless : the OP request was for "a value", not "all values" (then $ n\in\{1,3\}$). In a contest, if you answer beyond the problem's request, you lose time.[/quote]

Thank you Mr.[b]pco[/b] :blush: . I do not think in any Olympiad contest is there a question to find [i]an[/i] $ n$. There should be a typo of the problem poster.

I will finish the solution :P .

Since $ n\leq 6$ and $ n$ odd, we just need to consider $ n \equal{} 1,3,5$. But $ n \equal{} 1$ is a trivially solution and when $ n \equal{} 3$, we have $ \pi : \equal{} (3,1,2)$ is a permutation of $ \{1,2,3\}$ satisfies the condition. For $ n \equal{} 5$, the permutation must be $ \pi \equal{} (x_1,x_2,x_3,x_4,3)$. But this means $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 12$ and $ x_1 \plus{} x_2 \plus{} x_3 \vdots 3$ so $ x_4 \vdots 3$ and hence $ x_4\equal{}3$, contradiction. Therefore, $ n\in\{1,3\}$.


calculus
integration
algebra
polynomial
number theory proposed
number theory

Find (x+y) where x and y are the non zero solutions to the pair of simultaneous equations:

$ y^{2}x \equal{} 15x^{2}\plus{}17xy\plus{}15y^{2}$

$ x^{2}y\equal{} 20x^{2}\plus{}3y^{2}$

I am lost on how to do this...

This is algebra, not number theory. :maybe:

Sorry I thought this was a Diophantine equation which was classed under Number Theory >.<

Are you maybe asking for [i]integral[/i] non-zero solutions?

[quote="Gib Z"]Find (x+y) where x and y are the [color=red]integral[/color] solutions to the pair of simultaneous equations: $ y^{2}x \equal{} 15x^{2}\plus{}17xy\plus{}15y^{2}$ and $ x^{2}y \equal{} 20x^{2}\plus{}3y^{2}$[/quote]
From the second equation, $ x \equal{} 0$ iff $ y \equal{} 0$. So admit $ xy\ne 0$. Then dividing both equations by $ xy$ and setting $ q \equal{} x/y$ yield $ y \equal{} 15q\plus{}17\plus{}15/q$ and $ x \equal{} 20q\plus{}3/q$. So $ q \equal{} (20q^{2}\plus{}3)/(15q^{2}\plus{}17q\plus{}15)$, i.e. $ 15q^{3}\minus{}3q^{2}\plus{}15q\minus{}3 \equal{} 0$, and $ q$ is a rational root of the integral polynomial $ P(z) \equal{} 15z^{3}\minus{}3z^{2}\plus{}15z\minus{}3 \equal{} (15z\minus{}3)(z^{2}\plus{}1)$. Thus $ q \equal{} 1/5$ and $ y \equal{} 5x$ and then the second equation gives $ 5x^{3}\equal{} 20x^{2}\plus{}75x^{2}$, i.e. $ x \equal{} 19$ and $ y \equal{} 95$.

The answer was meant to be a positive integer between 1 and 999 :( Perhaps x and y needn't be integers, as long as x+y is. Ill post in algebra section.

Please, note I edited my last post. :wink:


There are two objects $A$ and $B$. They slide onto the wheeled device with initial velocity $2v_0$ and $v_0$, the coefficient of friction between the objects and the wheeled device is $\mu$ and the ground is frictionless. $m_A=m_B=m_{wd}=m$.
(1) If $A$ and $B$ don't collide, find the least possible value of the lengh of the wheeled device.
(2) From the beginning till the objects are travelling with the device with the same velocity, how long have $B$ travelled? How much time has it taken??

I know my English is poor and, if i made mistakes, please correct me!

i'm not quite sure on part B with what you're saying...but i think i get part A

the forces on each object are the same, mg down, normal force up which is just mg in the opposite direction, and the frictional force opposite each of their motions which is the normal times $\mu$.

$F = -\mu mg$
$a = -\mu g$

using our kinematics equation $v^2 = v_0^2 + 2ax$

$0 = 4v_0^2 - 2\mu gx_A$
$0 = v_0^2 - 2\mu gx_B$
$2\mu gx_A = 4v_0^2$
$2\mu gx_B = v_0^2$
$x_A = (2v_0^2)/(\mu g)$
$x_B = (v_0^2)/(2\mu g)$
$x = x_A + x_B = (2v_0^2)/(\mu g) + (v_0^2)/(2\mu g) = (5v_0^2)/(2\mu g)$

ffdbzathf, notice that the wheeled device can move.

The key to a simple solution is noticing that we can find the energy lost to friction without actually calculating the stopping point of either block.

Compare the momentum at the beginning of the experiment to the momentum when the blocks have ceased sliding on the wheeled device.
$P_i = m(2v) - mv = mv$
$P_f = 3m(v_f)$
By conservation of momentum $v_f = v_i/3$

Now make the comparison of energy.
$E_i = (1/2)mv^2 + (1/2)m(2v)^2 = (5/2)mv^2$
$E_f = (1/2)(3m)(v/3)^2 + mg\mu (d_a + d_b) = (1/6)mv^2 + mg\mu L$

Equating the expressions for energy
$(5/2)mv^2=(1/6)mv^2+mg\mu L$
$L = (14v^2)/(6g \mu)$

Correct answer for part A is: $\displaystyle\frac{7v_0^2}{3\mu g}$.

What frame are we using to view box B in part B?


trigonometry
real analysis
real analysis unsolved

Find all critical points of the following system and determine whether they are asymptotically stable:

$ x' \equal{} \sin{(y/100)}$
$ y' \equal{} \sin{(x/100)}$

Critical points are the solutions of
\[ \left( {\begin{array}{*{20}c}
   {\sin \frac{y}
{{100}}}  \\
   {\sin \frac{x}
{{100}}}  \\

 \end{array} } \right) \equal{} \left( {\begin{array}{*{20}c}
   0  \\
   0  \\

 \end{array} } \right).
\]
Solving it we easily obtain $ x \equal{} 100m\pi$, $ y \equal{} 100n\pi$, $ m,n \in \mathbb{Z}$.

At these points, its linearization is
\[ \left( {\begin{array}{*{20}c}
   0 & {\frac{1}
{{100}}\cos \left( {n\pi } \right)}  \\
   {\frac{1}
{{100}}\cos \left( {m\pi } \right)} & 0  \\

 \end{array} } \right)\left( {\begin{array}{*{20}c}
   x  \\
   y  \\

 \end{array} } \right) \equal{} \left( {\begin{array}{*{20}c}
   0 & {\frac{{\left( { \minus{} 1} \right)^n }}
{{100}}}  \\
   {\frac{{\left( { \minus{} 1} \right)^m }}
{{100}}} & 0  \\

 \end{array} } \right)\left( {\begin{array}{*{20}c}
   x  \\
   y  \\

 \end{array} } \right).
\]

By solving
\[ \det \left( {\begin{array}{*{20}c}
   { \minus{} \lambda } & {\frac{{\left( { \minus{} 1} \right)^n }}
{{100}}}  \\
   {\frac{{\left( { \minus{} 1} \right)^m }}
{{100}}} & { \minus{} \lambda }  \\

 \end{array} } \right) \equal{} \lambda ^2  \minus{} \frac{{\left( { \minus{} 1} \right)^{m \plus{} n} }}
{{10^4 }} \equal{} 0
\]
we get
\[ \lambda  \equal{}  \pm \frac{{\sqrt {\left( { \minus{} 1} \right)^{m \plus{} n} } }}
{{100}} \equal{} \left\{ \begin{gathered}
   \pm \frac{1}
{{100}},m \plus{} n \geqslant 0 \hfill \\
   \pm \frac{1}
{{100}}i,m \plus{} n < 0 \hfill \\
\end{gathered}  \right.
\]
this means that critical points are saddle provided $ m\plus{}n \geq 0$ and to study the case $ m\plus{} n < 0$, you must construct Lyapunov functions.

Of course, you mean saddle points for $ ( \minus{} 1)^{m \plus{} n} > 0,$ meaning $ m \plus{} n$ even, and the case requiring further investigation is $ m \plus{} n$ odd.

I would assume that the critical points with $ m \plus{} n$ odd are stable but not asymptotically stable. It would suffice to investigate one such point, as all are similar.


AMC
AIME
geometry
3D geometry
geometric transformation
rotation
linear algebra

Most students that take the AIME are unable to solve most of the problems on the AIME.  The really motivated AIME takers go to this web site and read solutions to the ones they missed.  But unless you really work through a problem that you missed, simply reading a solution to it either here or from the official solutions pamphlet may not be enough.  Sometimes students read a solution and think they get it, but would still be unable to reproduce that solution on their own.

So, here's what I suggest:  Take a few problems that you missed on the AIME 1 or 2 and write a similar problem.  Then see if you can solve the similar problem.  If a few people like this idea and take up the challenge, we can compile a collection of problems called AIME 2007 Follow Up Problem Set (or some other more catchy title).  The idea is that if a student read and understood every AIME problem from the two tests then he/she should be able to get all of the Follow-Up problems.  Some of the problems might be just re-writes with different numbers while others might be more creative extensions of the ideas found on the original contests.

So, any takers for AIME 1 follow-ups?  Post your problems below:

How many perfect cubes below 10^13 are a multiple of 1620?

this is okay and all, but remember the only thing that this results in is an understanding of the specific problems on this test, it only partially results in an ability to solve new problems

but it is a still a nice idea...i usually do these things on my own though...

But making sure you understand the solutions to problems is important to learning. If you do lots of problems but never figure out how to do the ones you can't do, it doesn't help. While you may never come across a problem of a very similar format again, there may still be a key idea in the solution that can apply to problems of all sorts of variety.

Uhh sure, and this might be able to force people to tighten up some neglected area of mathematics.

Like the ugly 12, I'm like "Aw if only I remember how to construct the rotation matrix so I can do this more brainlessly"

Although I think the "trick" to that problem is finding an acceptable clean solution instead of trying plain brute-forcing (I personally found 3 solutions before I pick the easy one...)

A few people responded and recognized the value of this exercise.  It's probably most useful to write problems for the ones you missed.  I made oversights on #10 and 11, so let's see some solutions to these:

10. Let S represent the set {1,2,3,4,5,6,7}.  Two subsets are chosen at random from S.  Compute the probability that they are disjoint.

11.  A tire with a diameter of 15" is rolling at a velocity of 1" per second towards a tire with diameter 45", which is rolling towards the smaller tire at a rate of 2" per second.  When the two tires reach other, they continue moving with the same rotational velocity and the larger tire rolls over the smaller one without slipping.  Compute the distance between the centers of the tires 1 minute after they first make contact.


Hi all. I am new to the site, just stumbled upon it, looking for LaTeX help. I am new to LaTeX but have already found many useful things here. On to the question that I haven't found an answer to.
I am trying to make an entire paragraph indented, actually several of them. The format is supposed to be bold headers that are left aligned normally with the entire paragraph after being indented much like a blockquote in HTML or "Increase Indent" on a word processor.
A few other items are that for some reason the first line after a section is treated differently than the following lines (I can't force it to indent at all) and I will eventually try to remove the numbering from the section headings, but I haven't truly looked into these things yet, so I won't officially ask for help on them.
In case it is necessary (since I have no experience with other LaTeX front-ends I am using Kile.

I just asked this question earlier today.  Here is the answer I got:
You could try using \leftskip, for example \setlength\leftskip{3em} which you change back to \setlength\leftskip{0em} after the paragraph.

But often it is best to use a package which takes into account all the possible consequences. In this case use the hanging package (available in MiKTeX) where you write for example \hangpara{3em}{0} at the start of the hanging paragraph.

[quote="Mitlik"]for some reason the first line after a section is treated differently than the following lines (I can't force it to indent at all) [/quote]The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=indentfirst]indentfirst package[/url] solves this problem, which is by design, though not everyone likes it.

[quote="Mitlik"]I will eventually try to remove the numbering from the section headings[/quote]Use \section* rather than \section

Awesome thanks guys, I shall try it as soon as I get Kile back up and working, I am currently trying to switch from tetex to texlive for my back-end.
And I shall see if the hangpara package is available, but MiKTeX fails to install using Wine. :(

[quote="Proctonator"]I just asked this question earlier today.[/quote]
Sorry about the redundant question but I tried searching here and elsewhere using different wording but didn't find anything.

MiKTeX's package manager can be used in Linux directly without Wine. See http://www.miktex.org/unx/

Also the hanging package, like other LaTeX packages, can be used on any operating system so doesn't depend on MiKTeX. You can download it from http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=hanging, though the package manager above makes it easier to install.

Thanks again guys, especially Steve M. with the info. I am so newb right now it hurts. Turns out texlive (tug is their main distributor me thinks) has the hanging package. I may have only been interested in MiKTeX to try out WinEdt not sure how it would do for a wine install front-end with a non-wine install back-end, probably not possible, nor do I care. Kile is up a running fine on texlive. No need to monkey around with 40 USD non open source software.

An image generated by Kile and exactly what I wanted.
[img]http://i122.photobucket.com/albums/o264/Mitlik/example.png[/img]
Fear not, soon there will be real text in there to do my lab reports. Your efforts won't get wasted on gibberish.

Thanks,
    Mitlik


I dont know why cardozo's score is not posted there (twice in a row!)

http://www.mathleague.com/reports/2004_05/NY5.HTM

who can I ask about this?
I kinda care about this because I have 30/30 so far hehe

nobody, it's a dictatorship!!!

(Actually Mr. Conrad).

comments@mathleague.com.

Send him my love ;)


Let a, b, c, d be non-zero real numbers. Show that

(ab+2bc+cd)/(a^2+b^2+c^2+d^2)<=[(sqrt2)+1]/2


Use the Law of Sines to deduce the addition formula:

$\sin(B+C) = \sin(B) \cos(C)+\sin(C) \cos(B)$

[hide="And another hint"]Draw the altitude from $A$.[/hide]

[hide="Solution"]Draw the altitude from $A$ - let its foot be $D$. Then $BD=c\cos B, CD=b\cos C$, hence $a=c\cos B+b\cos C$. Now plug in the formulas obtained by Sine Law: $c=2R\sin C, b=2R\sin B, a=2R\sin A$, cancel the common factor $2R$ and use t0rajir0u's hint.[/hide]


geometry
circumcircle
geometric transformation
reflection
Gauss
homothety
vector

Let M be a point on the circumcircle of ABC. and D and E be the feet of the perpendeculars from M to BC and AC. Let P and Q be the midpoints of AD and BE. prove that PQ is perpendecular to DE.

Well, The official solution is that suppose N is midpoint of AB.
Two triangles MDE and PQN are similiar and their sides are pependicular.
But I solved it in a beutiful way.
Suppose M' is reflection of M in line BC.
Because of well-known theorem,HM' is pararell to DE(H orthocenter)
We can easily prove that HM' is radial axis of two circles with diameters AD and BE.
And problem has been solved ;)  ;)  ;)

What should I say? It's a problem on triangle geometry, hence I must answer ;D , but the most "elementary" solution I have needs three lemmas, so let me just hope you know these lemmas and won't have problems with the lots of preknowledge used.

My first lemma is rather harmless:

[b]Lemma 1, the Simson line theorem.[/b] If M is a point on the circumcircle of triangle ABC, then the feet D, E, F of the perpendiculars from M to the sides BC, CA, AB lie on one line, the so-called SIMSON LINE s of the point M with respect to triangle ABC.

Hence, the line DE is nothing but our Simson line s. We must show that this Simson line s is perpendicular to the line PQ, where P and Q are the midpoints of the segments AD and BE.

Note that the line s is not only the Simson line of the point M with respect to triangle ABC, but it is also the Simson line of M with respect to the triangles AEF, BFD, CDE. This is because the point M lies on the circumcircles of these triangles (in fact, from < AEM = 90 and < AFM = 90, we see that the points A, E, F, M are concyclic, i. e. the point M lies on the circumcircle of triangle AEF, and similarly it lies on the circumcircles of triangles BFD and CDE), and the feet of the perpendiculars from the point M to the sides of these triangles trivially lie on s (to be honest, each of these triangles just has a sideline coinciding with s).

Lemma 1 helped us to identify the line DE as the Simson line s. Another lemma will give us some clues about PQ:

[b]Lemma 2, the Steiner and Gauss lines theorem.[/b] Four lines form four triangles. The orthocenters of these four triangles lie on one line, the so-called STEINER LINE of the complete quadrilateral formed by our four lines. Also, the midpoints of the diagonals of this complete quadrilateral lie on another line, the so-called GAUSS LINE of the complete quadrilateral. The Gauss line and the Steiner line are perpendicular to each other.

In our case, if we consider the complete quadrilateral formed by the lines BC, CA, AB and the Simson line s, then the diagonals of this quadrilateral are the segments AD, BE, CF. The midpoints of the first two diagonals are P and Q. Hence, the line PQ is the Gauss line of our complete quadrilateral.

Therefore, we can rewrite the assertion we must prove as follows: The Simson line s is perpendicular to the Gauss line of the complete quadrilateral formed by the lines BC, CA, AB and s. Now, from Lemma 2, this Gauss line is perpendicular to the Steiner line of this quadrilateral. Hence, it remains to prove that the Simson line s is parallel to the Steiner line of the complete quadrilateral formed by the lines BC, CA, AB and s.

Well, by definition, this Steiner line passes through the orthocenters of triangles ABC, AEF, BFD, CDE. Now a last (well-known?) lemma will help:

[b]Lemma 3.[/b] The Simson line s of the point M with respect to triangle ABC bisects the segment MH, where H is the orthocenter of triangle ABC.

Thus, the image of the Simson line s in the homothety with center M and factor 2 is a line s' passing through the orthocenter H of triangle ABC. Now, since s is also the Simson line of the point M with respect to the triangles AEF, BFD, CDE, this image line s' must also pass through the orthocenters of the triangles AEF, BFD, CDE. Altogether, the line s' contains the orthocenters of the four triangles ABC, AEF, BFD, CDE. Consequently, the line s' is the Steiner line of the complete quadrilateral formed by the lines BC, CA, AB and s. Now, since the line s' is the image of the line s in the homothety with center M and factor 2, the line s' is parallel to s. So the Steiner line of our complete quadrilateral is indeed parallel to the Simson line s. Proof complete.

  Darij

The official solution is shorter and uses nothing except similiarity of 2 triangles. :P  :P

Indeed, but anyone with the triangle geometry know-how immediately sees my solution, while the idea of the proposed one is not so immediate...

  Darij

I use complex number to solve it on unit circle ,really IQ free solution :D  :D

well i knew steiner line, but Gauss Line? whats that, could you explain what it is a little bit.

Thanks

Thank you very much Amir.
but to Pascual2004 Suppose four lines $d_1,d_2,d_3,d_4$
($d_i,d_j$ intersection of $d_i$ and $d_j$)
Then midpoint of these segments are collinear
$(d_1,d_2)(d_3,d_4)$
$(d_1,d_3)(d_2,d_4)$
$(d_1,d_4)(d_3,d_2)$

[quote="Omid Hatami"]Suppose four lines $d_1,d_2,d_3,d_4$
($d_i,d_j$ intersection of $d_i$ and $d_j$)
Then midpoint of these segments are collinear
$(d_1,d_2)(d_3,d_4)$
$(d_1,d_3)(d_2,d_4)$
$(d_1,d_4)(d_3,d_2)$[/quote]

Yes, and the line through these midpoints is called the [b]Gauss line[/b] of the complete quadrilateral formed by the four lines $d_1$, $d_2$, $d_3$, $d_4$. It is also called "Newton line". See

  http://www.cut-the-knot.org/ctk/CompleteQuadrilateral.shtml

and

  http://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml

however, on these two pages, the Gauss line is not named.

  Darij

I don't think this one is complex. Using vector and Simson line should help.
 Call K the intersection of DE and AB.
 We have $\overrightarrow{PQ} = \overrightarrow{AB}+\overrightarrow{ED}$
  Thus $\overrightarrow{PQ}. \overrightarrow{DE} =(\overrightarrow{AB} + \overrightarrow{ED}).\overrightarrow{DE} = \overrightarrow{AB} .
\overrightarrow{DE} - DE^2 = 0$
  which is equivalant to  $  DE= AB \cos(AKD)$ (obvious).
 Anyway, Darij 's performance reveals that he knows a lot in geometry. :lol:

[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/NewBitmapImage-20.jpg[/img]

Denote by $ N$ the midpoint of $ AB$.

We have :

$ \frac {NQ}{NP} \equal{} \frac {AE}{BD} \equal{} \frac {ME}{MD}$

And $ (NQ,NP)\equiv (CA,CB)\equiv (ME,MD) (mod \pi)$

So two triangles $ NQP$ and $ MED$ are similar.

But $ NQ\perp ME,NP\perp MD$ .

Therefore  $ PQ\perp ED$.
:)

Dear Mathlinkers,
for a proof that the steiner line is perpendicular to the Gauss line (of a quadrilateral) see
http://perso.orange.fr/jl.ayme vol. 4.
Sincerely
Jean-Louis


Can anyone answer the question below:

Define the cumulative degree of a polynomial to be the sum of degrees of its monomials.

Thats a definition, whats the question? :)
Or do you want to know what that means?

That is not a question.  If you had said "Define the cumulative degree of a polynomial" then that would have been a question.

I think your problem is more of an English one than a math one.

[quote="mat12"]the question says define "the cumulative degree of polynomials to be sum of degrees of its monomials"

the definition.

can you define it?[/quote]Er.. I think you already did define it. You defined the cumulative degree of a polynomial to be the sum of the degrees of its monomials. I have absolutely no idea what your question is here.

Understood My English is bad. Thank you all for the help..I should study English Grammar first before mathematics


Show that if X1,....,Xn, Y1,....,Yn are elements of the real numbers then
sum(Xi*Yi)<=(sum|Xi|^p)^(1/p)*(sum|Yi|^q)^(1/q)

It is called Holder inequality if I remember correct.
Check http://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality


Prove the following:
If $f(x)$ is a polynomial with integer coefficients, then for all integers $a$, $f(a+f(a))$ is a multiple of $f(a)$.

It seems not to be true:

Take $f(x)=x+C$. Then $f(a)=a+C, a+f(a)=2a+C, f(a+f(a))=f(2a+C)=2a+2C$ and that need not be divisible by $a$ for all integer $a$.

No, but it is a multiple of $f(a)$ which is almost certainly what the question was intended to ask.

[hide]

Let $f(x) = \sum c_{n}x^{n}$. It's easy to show that $(a-b)|(f(a)-f(b))$ because $a^{n}-b^{n}= (a-b)(a^{n-1}+a^{n-2}b+\cdots+b^{n-1})$ for any positive integer $n$, which means $(a-b)$ divides $\sum c_{n}(a^{n}-b^{n}) = f(a)-f(b)$.

Consequently, $((a+f(a))-a)|(f(a+f(a))-f(a)) \Rightarrow f(a)|f(a+f(a))$ as desired.[/hide]

This problem is very easy
Infact
$f(a+f(a))-f(a) \vdots (a+f(a)-a)=f(a)$
So $f(a+f(a)) \vdots f(a)$


trigonometry
geometry
angle bisector
geometry unsolved

Let $ABC$ be a triangle and let $AD$ and $BE$ be the angle bisector of $\angle{A}$ and $\angle{B}$. Prove that if $\angle{BED}=30$, then $ABC$ has got an angle of 60 or 120.

[hide="Here is a - ugly - trigonometrical proof of the proposed problem."]We observe that $m(\widehat {ADB})=C+\frac A2$, $m(\widehat {AEB})=C+\frac B2$, $m(\widehat {ADE})=60^{\circ}-\frac C2\ .$ The well-known relation (in the quadrilateral $ABDE$)

$\sin \widehat {BAD}\cdot\sin \widehat {DBE}\cdot\sin \widehat {EDA}\cdot\sin \widehat {AEB}=$

$\sin \widehat {DAE}\cdot\sin \widehat {EBA}\cdot\sin \widehat {ADB}\cdot\sin \widehat {BED}\Longleftrightarrow$

$\sin\frac A2\cdot\sin\frac B2\cdot\sin \left(60^{\circ}-\frac C2\right)\cdot\sin \left(C+\frac B2\right)=$
$\sin \frac A2\cdot\sin\frac B2\cdot\sin \left(C+\frac A2\right)\cdot\sin 30^{\circ}\Longleftrightarrow$
$2\cos \left(30^{\circ}+\frac C2\right)\cdot\sin \left(C+\frac B2\right)=\sin \left(C+\frac A2\right)\Longleftrightarrow$
$\sin \left( 60^{\circ}+\frac A2-C\right)+\cos \left(30^{\circ}+\frac A2\right)=\sin \left(C+\frac A2\right)\Longleftrightarrow$
$\cos \left(30^{\circ}+\frac A2\right)=\sin \left(C+\frac A2\right)-\sin \left(60^{\circ}+\frac A2-C\right)\Longleftrightarrow$
$\cos\left(30^{\circ}+\frac A2\right)=2\sin \left(C-30^{\circ}\right)\cdot\cos\left(30^{\circ}+\frac A2\right)\Longleftrightarrow$
$\cos\left(\frac A2+30^{\circ}\right)=0\ \vee\ \sin \left(C-30^{\circ}\right)=\frac 12\Longleftrightarrow A=120^{\circ}\ \ \vee\ \ C=60^{\circ}\ .$[/hide]

It's very easy. Let $K$ on $AB$: $BD=BK$. Then the triangle $KED$ is equilateral. Then \[ {\sin AKD\over AD}={\sin KAD\over KD}={\sin EAD\over ED}={\sin AED\over AD} \] and $\sin AKD=\sin AED$. If $\angle AKD=\angle AED$, then $\angle ACB=180^\circ-\angle BAC-\angle ABC=1/2(\angle AKE+\angle BKD)=60^\circ$. If $\angle AKD=180^\circ-\angle AED$, then the quadrilateral $AKDE$ is circled and $\angle BAC=2\angle DAE=2\angle DKE=120^\circ$.

A solution with no trig uses the following lemma:
Let $X$ be a point on the bisector of angle $\angle A$ and let $Y,Z$ be points on the angle's sides such that $XY=XZ$. Then either $\angle XYA=\angle XZA$ or $\angle XYA+\angle XZA=180^{\circ}$ (see figure).
Considering, as in the previous solution, the point $K$ on $AB$ such that $BK=BD$, it follows that $EKD$ is equilateral, hence $DE=DK$. Then either $\angle DEA=\angle DKA$ or $\angle DEA=\angle DKB$. In the first case we obtain $\angle C=60^{\circ}$ while in the second case it results $\angle A=120^{\circ}$.


Construct a sequence $A_1,A_2,\ldots$ of subsets of $\mathbb{N}$ with the following properties:
(i) For all $i \neq j$, $A_i \not\subseteq A_j$;
(ii) If $S \subseteq \mathbb{N}$ has a non-empty intersecion with each $A_i$, then there exists an element $x \in S$ such that $S':=S \setminus \{x\}$ also has a non-empty intersection with each set $A_i$.

[b]Alternative formulation. [/b] Construct an infinite antichain $\mathcal{A}$ of $2^{\mathbb{N}}$ such that there does not exist a minimal subset of $\mathbb{N}$ having a non-empty intersection with every member of $\mathcal{A}$.

--Vesselin

In figure1 a schema of our sets is shown and we define them as follows: for any nonnegetive $k$ and $0 \leq x_{1}<x_{2}<...<x_{2^{k}} \leq 2^{k+1}-1$we define $A(x_{1},x_{2},...,x_{2^{k}})=\{2^{k+1}r+x_{s}|0 \leq r , 1\leq s \leq 2^{k}\}$ and we call set $\{2^{k+1}r+x_{s}|0 \leq r  \}$ the $s$th part of $A(x_{1},x_{2},...,x_{2^{k}})$ and they correspond to gray parts shown in figure.let $H$ be the collection of all these $A$s.
We prove $H$is the desired antichain:

suppose to contrary $S$ is a set described in the problem and for every $x \in S$ the set $S \backslash \{ x \}$ does not intersect every member of $H$.(*)
suppose $x,y\in S,x \not =y$ .From (*) there exist set $B=A(x_{1},x_{2},...,x_{2^{k}})$ such that $B\cap S =\{ x \}$ and set $C=A(y_{1},y_{2},...,y_{2^{l}})$ such that $C\cap S =\{ y \}$.let $k \leq l$ and $U$ be the part of $C$ that contain $y$.Because $k \leq l$ every part of $C$ is either a subset of $B$ or disjoint from it(figure is helpful here).So $U \cap B = \Phi$ because $y \in U$ and $y\not \in B$.

let $m$ be an arbitrary number such that $l+1\leq m$.We construct a $D=A(z_{1},...,z_{2^{m}})$ as follows:include $z_{i}$s which their corresponding parts are subsets of $B$ and do not contain $x$. they are $2^{m}-1$ parts.add an arbitrary part which is subset of $U$ and do not contain $y$(this is an extra part because $U \cap B = \Phi$).Clearly $D \subset B\cup C$ and $x,y \not \in D$ and so $D \cap S=\Phi$ and It's a contradiction .So (*) is false and we are done.


While searching for math information one day, I browsed across a link to "[url=http://www.phys.uu.nl/~thooft/theorist.html]How to Become a Good Theoretical Physicist[/url]," a phenomenal online resource by a recent Nobel Prize winner. Of course, you have to learn a lot of math along the way, but this site will guide you in what math to learn. Great stuff.

i noticed that you two have similar usernames...
is that just a coincidence or, the word tokenadult have some interesting meanings?

My son is a member of AoPS, but I logged on to find information about MathPath and MathCamp, and took a look at some of the postings -- very interesting!!

Since I am an adult, I think it is good to identify myself as such, and have followed the lead of the original tokenadult by choosing a name as such. I was going to name myself tokenadult2 but then that implies some relationship with the original tokenadult, and since there is none, I just chose tokenadult6151. Although some of you may have discovered some meaning in the number '6151', there is none -- it was chosen at random.


geometry
circumcircle
symmetry
trigonometry
geometry solved

[quote="shobber"]If $P$ is the orthocentre of $\triangle{ABC}$, then the radius of the three triangles are equal.

I don't know if there is more $P$ on the plane...[/quote]
I think I proved that the orthocentre is the only solution.  My way is just angle chasing:

We have angle PBA=1/2 PFA = 1/2 PEA=PCA. By symmetry, we get PAB=PCB and PAC=PBC.
It follows that B, C, E`, F` are on a same circle, C, A, F`,D` are on a same circle , and A, B, D`, E` are on a same circle. Rest follows easily.

_________________________________________________
EDITED: I found a counter-example. Let ABCP be a square.
EDITED AGAIN: 'found', instead of 'fount' :blush:

We will name $\angle BPC, \angle CPA, \angle APB$ as
$\alpha, \beta, \gamma$, respectively. The condition that
the three circumcircles have the same radius is
\[ \sin \alpha = \lambda \sin \hat A,\quad \sin \beta = \lambda \sin \hat B,\quad \sin \gamma = \lambda \sin \hat C  \]
We will proof that $\lambda = 1$. In that case, all three
circles will have radius $R$, equal to the circumradius of
the circle $\triangle ABC$, but there are exacly four circles
of radius $R$ passing by two of the vertices of $\triangle ABC$,
its circumcircle, and the other three intersecting in its
orthocenter, so the solution of the problem is that $P$ can be
any point of the circumcircle or the orthocenter.

FIRST CASE: $P$ is an inner point of $\triangle ABC$
\[ \alpha + \beta + \gamma = 360^\circ \Rightarrow \sin\gamma = - \sin\alpha\cos\beta - \cos\alpha\sin\beta \]
\[ \hat A + \hat B + \hat C = 180^\circ \Rightarrow \sin\hat C = \sin\hat A\cos\hat B + \cos\hat A \sin\hat B \]
and a quick manipulation gives the equations
\[ \sin\hat A(\cos\hat B+\cos\beta)+\sin\hat B(\cos\hat A+\cos\alpha)=0 \]
\[ \sin\hat B(\cos\hat C+\cos\gamma)+\sin\hat C(\cos\hat B+\cos\beta)=0 \]
\[ \sin\hat C(\cos\hat A+\cos\alpha)+\sin\hat A(\cos\hat C+\cos\gamma)=0 \]
Suppose that $\lambda < 1$, then $\vert\cos\alpha\vert>\vert\cos\hat A\vert$,
$\vert\cos\beta\vert>\vert\cos\hat B\vert$,
$\vert\cos\gamma\vert>\vert\cos\hat C\vert$. As
$\alpha+\beta+\gamma=360^\circ$ two of the angles, wlog. $\beta$ and
$\gamma$ have $\cos\beta<0$, $\cos\gamma<0$, but then
\[ \sin\hat B(\cos\hat C+\cos\gamma)+\sin\hat C(\cos\hat B+\cos\beta)<0 \]
contradicting the second of the three equations above.
Suppose now that $\lambda > 1$, then
$\vert\cos\alpha\vert<\vert\cos\hat A\vert$,
$\vert\cos\beta\vert<\vert\cos\hat B\vert$,
$\vert\cos\gamma\vert<\vert\cos\hat C\vert$. Wlog. we can suppose
$\hat B, \hat C<90^\circ$, then
\[ \sin\hat B(\cos\hat C+\cos\gamma)+\sin\hat C(\cos\hat B+\cos\beta)>0 \]
contradicting the second of the three equations above.

SECOND CASE: $P$ is between the rays $AB$ and $AC$, but out of the
triangle.
In that case we can suppose wlog. that $\hat A, \hat B<90^\circ$, and
we arrive at the three equations
\[ \sin\hat A(\cos\hat B-\cos\beta)+\sin\hat B(\cos\hat A+\cos\alpha)=0 \]
\[ \sin\hat B(\cos\hat C-\cos\gamma)+\sin\hat C(\cos\hat B-\cos\beta)=0 \]
\[ \sin\hat C(\cos\hat A+\cos\alpha)+\sin\hat A(\cos\hat C-\cos\gamma)=0 \]
The first equation rules out the cases $\lambda>1$ and $\lambda<1, \cos\beta>0, \cos\alpha<0$.
The second equation rules out the case $\lambda<1, \cos\beta>0, \cos\alpha>0$ (that implies $cos\gamma>0$).
The third equation rules out the case $\lambda<1, \cos\beta<0$ (that
implies $\cos\alpha<0, \cos\gamma>0$.

THIRD CASE: $P$ is in the angle opposite to $\hat B$.
I that case the three equations are
\[ \sin\hat A(\cos\hat C-\cos\gamma)+\sin\hat C(\cos\hat A-\cos\alpha)=0 \]
\[ \sin\hat B(\cos\hat C-\cos\gamma)+\sin\hat C(\cos\hat B+\cos\beta)=0 \]
\[ \sin\hat A(\cos\hat B+\cos\beta)+\sin\hat B(\cos\hat A-\cos\alpha)=0 \]
The case $\lambda>1$ is ruled out, obviously, by one of the three
equations above.
The case $\lambda<1, \cos\alpha>0, \cos\gamma>0$ is ruled out by
the first of the equations above.
The case $\lambda<1, \cos\alpha<0$ is ruled out by the second
equation.
The case $\lambda<1, \cos\gamma<0$ is ruled out by the third equation.

QED

Well, let's not make the problem more non-trivial than it is:

[quote="Rushil"]Let $ABC$ be a triangle in a plane $\pi$. Find the set of all points $P$ (distinct from $A,B,C$ ) in the plane $\pi$ such that the circumcircles of triangles $ABP$, $BCP$, $CAP$ have the same radii.[/quote]

In fact, this set is the union of the circumcircle of triangle ABC and the set consisting of one single point, namely the orthocenter of triangle ABC (of course, without the vertices A, B, C). In order to prove this, we have to show two assertions:

[b]Assertion 1.[/b] If ABC is a triangle, and P is either a point on the circumcircle of triangle ABC, or the orthocenter of triangle ABC, then the circumcircles of triangles BCP, CAP and ABP have the same radius.

[b]Assertion 2.[/b] If ABC is a triangle, and P is a point in its plane such that the circumcircles of triangles BCP, CAP and ABP have the same radius, then the point P either lies on the circumcircle of triangle ABC, or coincides with the orthocenter of triangle ABC.

[i]Proof of Assertion 1.[/i] - If P is a point on the circumcircle of triangle ABC, then the circumcircles of triangles BCP, CAP and ABP all coincide with the circumcircle of triangle ABC and thus have the same radius.

- If P is the orthocenter of triangle ABC, then it is known (see http://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ) that the circumcircles of triangles BCP, CAP and ABP have the same radius.

Thus, Assertion 1 is proven.

[i]Proof of Assertion 2.[/i] If the point P lies on the circumcircle of triangle ABC, everything is clear. Thus, assume that the point P does not lie on the circumcircle of triangle ABC. Then, the circumcircles of triangles BCP, CAP and ABP are pairwisely distinct (in fact, if any two of these circumcircles would coincide, then the points A, B, C and P would lie on one circle, i. e. the point P would lie on the circumcircle of triangle ABC). Also, these three circumcircles are congruent (since they have the same radius). Thus, we can restate the assertion that we have to prove as follows:

Three congruent circles meet at one point P; let A, B, C be their pairwise points of intersection different from P. Then, prove that the point P is the orthocenter of triangle ABC.

But this was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=2033 . Hence, Assertion 2 is proven.

And the problem is solved.

  darij

The points are H, and any point on circumcircle.


conics
parabola
geometry
perpendicular bisector

I'm reviewing SAT IIc stuff, and the book doesn't have answer explanations. >_<

If P and Q are different points in a plane, the set of all points in this plane that are closer to P than to Q is:
a. interior of square
b. wedge shaped region of the plane
c. region of plane bounded by parabola
d. interior of circle
e. the region of the plane on one side of a line

In how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?

Can there be a least element in any set of positive rational numbers?

1) square i think. the points that are equally distance are along a line that contains a point A such that A is equidistant from P and Q and the line is perpendicular to PQ.

[u]Question 1[/u]:
By simply plotting points, I'm thinking it would be E, but I'm not sure.

[u]Question 2[/u]:
We have 10 people, and we want to choose 3 of them.  (Once we choose 3, the other 7 will be automatically determined).  So we do the operation "10 choose 3," or $\binom{10}{3}$.

$\binom{10}{3}=\frac{10\cdot9\cdot8}{3\cdot2\cdot1}=120$

#1 $\text{Primary Solution: }$

$\cdot$                                       $\cdot$
P  Q

We want the set of points closer to P than Q.  So anything to the left of P is closer to P than Q.  So the set must be infinite...

Draw a vertical line between P and Q, halfway in between.  Anything on the right is closer to Q, anything on the left is closer to P.  So that's why I think it's E.

#1 $\text{Alternate Solution: }$

We can use equality of generality here.  Since point P is not described differently than point Q, either point could be labeled P or Q.  In that case, we cannot do the problem, unless the answer would be the same either way.  E is the only answer that is the same both ways.

The set of points which are equidistant from two points (such as P and Q) is the [b]perpendicular bisector [/b]of the line segment connecting the two points (PQ). anything on the 'P' side of this line will be closer to P than to Q, and vice versa.

Can there be a least element in any set of positive rational numbers?

I would say yes, assuming I understood the question correctly.  In any set of positive rational numbers, one of them will be the smallest in the set.  But that sounds like a trivial interpretation...

[quote="Farenhajt"]No, sets of rational numbers need not have least elements. Consider the set of all rational $x$ such that $0<x<1$. If we assume there's some $x_{\text{min}}$ as a minimal member, we can always take $\frac{x_{\text{min}}}{2}$ which is also the member of the set, but is smaller than $x_{\text{min}}$.[/quote]

Well, that's assuming an infinte set is allowed...

"Any set" implies the most general case, which includes infinite sets by default.


limit
real analysis
real analysis unsolved

\[ \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k \equal{} 1}^n {\frac{{\left( { \minus{} k} \right)^{k \minus{} 1} }}
{{k!}}}\]

thanks

Edit: Misread the problem...  :blush:

Assuming I may use power series, I would simply write it as:

$ \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k \equal{} 1}^n {\frac {{\left( { \minus{} 1} \right)^{k \minus{} 1} }} {{(k \minus{} 1)!}}} \equal{} \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k \equal{} 0}^n {\frac {{\left( { \minus{} 1} \right)^{k} }} {{k!}}}$

Now, I'd just take power series expansion for $ e^x$ and plug in $ \minus{} 1$, hence the limit is $ e^{ \minus{} 1}$.

that is not correct    :|  notice

$ \sum\limits_{k \equal{} 1}^\infty {\dfrac{{\left( { \minus{} k} \right)^{k \minus{} 1} }} {{k!}}} \; \equal{} \; \minus{} \sum\limits_{k \equal{} 1}^\infty \left[{\dfrac{{\left( { \minus{} 1} \right)^k}} {k}}\;\cdot\; \dfrac{k^{k \minus{} 1}}{ (k \minus{} 1)! } \right]$

hsiljak, yes, the original limit will diverge. what i meant to say was to compute   the limit

$ \sum\limits_{k \equal{} 1}^\infty {\dfrac{{\left( { \minus{} k} \right)^{k \minus{} 1} }} {{(2k)!}}}$

instead, which will converge.

i have seen this particular limit appear   in several places before and am   assuming that is probably what the poster meant to ask.

sorry that i didn't mention that in my post initially.

:blush:  :blush: How didn't I see that? I'm too sleepy, obviously. Now I see it obviously cannot converge (nth member of the sum tends to infinity, not to zero)! How do you plan to compute it, misan?


Let $ n\in \mathbb{N^*}$,$ z_n \in \mathbb{C}$ , $ z_n\equal{}(1\plus{}i)(2\plus{}i)...(n\plus{}i)$ .Prove that the sequence $ Im(z_n)$ takes infinetly many positive values and infinetly many negative values.

I belive that this problem was given in Romanian TST many years ago.

It suffices to show that the series $ \sum_{k\equal{}1}^{\infty} \arctan \frac{1}{k}$ (the sequence of arguments) diverges, which is trivial by comparison to the harmonic series.


OK, no idea if this is easy, medium, or hard, but I'll put it here so more people can see it. I haven't solved it yet.

Everyone (?) has heard of the problem where you have heapsa water, a 3L jug, a 5L jug, and you want to make x L somehow.

Heres a variant.

You feel like having a cup of coffee. Unfortunately, all that you have is a three-cup and a five-cup jug, an unlimited water supply, and a pack of instant coffee. At any time, you can dump the entire contents of the coffee pack into one of the jugs (it will dissolve instantly and uniformly). The pack contains enough coffee so that if it is dumped into a single cup of water, it will make one cup of coffee at 100% strength. But you happen to like your coffee weak, really weak. Find a way that you can measure out exactly one cup of coffee at 1% strength.

I don't suppose "dump only 1% of the instant coffee into the cup" counts?  ;)

Who drinks instant coffee, anyhow?

Nice question.

Heres how id do it

[hide]First, dump one instant coffee into the 5 cup.  That is now 20% strength.  Make it so there is 1 cup in the 5 cup one.  Fill the three cup one with water and dump it into the 5 cup one making the coffee at 5 percent strength.  Then make it again so the five cup one has 1 cup.  Add four cups of water so it is at 1 percent.  Then make it again so there is 1 cup of coffee in the 5 cup one.  I hope you know how to make it so the 5 cup has one cup.[/hide] [/hide]

to make a 1 in 5 you would need to introduce fresh water...

Or you could do 5 cup pitcher into 3 cup pitcher, save 2 cups in the 5 cup pitcher, throw out the 3 cup, pour the 2 cups in the 5 cup pitcher into the empty 3 cup, fill up the 5 cup, pour 1 cup into the 3 cup, have 4 cups in the 5 cup pitcher, throw out the 3 cup, pour 3 cups from the 5 cup pitcher (which has 4) into the empty 3 cup, and voila 1 cup!

I just solved it myself, this is such a nice problem.

Scrambled and Ragingg, sorry, you are both wrong.

Scrambled, you can't just say "make the 5 cups worth into 1 cup", as ml2 said, you would need to introduce fresh water for that. But you've got the gist of how the problem works, now you just need to find the correct solution.

Ragingg, you misread the question. You don't just want one cup, you want one cup of 1% strength.

[quote="Ian"]But that still changes the strength of coffee mix in the cup doesnt it?  I dont think there is a way to start with five cups in the larger pitcher and end up with one cup [i]without[/i] mixing the five cups of water with any fresh water.[/quote]
There isn't , but you might do something in another way..

No no... scrambled said he wanted 1 cup, and ml2 said you have to introduce water... I showed them that you could make 1 cup.

Hmm, if I could figure out how to 1/2 it, I would be all set ...

Why doesnt my way work?  (yes i already knew how to make 1 cup)

no, ragging.. you intoduced fresh water...
scrambled..  show us how you make that final 1 cup... without spilling all 1% water away..

Okay -- think about the process of making 1L of water.  In order to do it, you have to have the 5 L, and then at later points you add additional water to the process and dump out the water you started with.  At these times when you add it, you will unfortunately be diluting your original amount by a factor of 2/5 or 1/3 or 2/3 or some other amount that ruins what you wanted to do.

I can get to 2% now -- I'm not sure how to get down to 1%.  Just one comment, though -- I was considering this, in theory:  If we fill a 3L jug and dump in the coffee, and then pour the (all-water) contents of the 5L jug into the already full 3L jug, what we should get is 3L of a mixture that is 1/8 coffee, right?  Now, I think you can use this step, in theory, to actually do it.  On the other hand, it's pretty definitely cheating.

OK, wow. I obsessed over this problem for almost a day now..

[hide]Fill 5L, pour into 3L. Now you have 2 in 5L. Empty 3L, pour 5L -> 3L. Fill 5L -> 3L. Now you have 4 water in 5L. Add coffee. You have 4L of 25% coffee in 5L.

Empty 3L. Pour from 5L -> 3L. You have 1L of 25% coffee in 5L. Empty 3L. Pour 5L -> 3L. You have 1L of 25% in 3L. Fill 3L to top. Now you have 3L of 25/3% in 3L. Pour 3L -> 5L.

Fill 3L with water. Pour 3L -> 5L. You have 5L of 5% in 5L, and 2L water in 3L. <b>Don't empty.</b> Now pour 5L -> 3L. You should have 4L of 5% in 5L, and 3L of 5/3% in 3L. Empty the 5L, pour 3L -> 5L.

You have 3L of 5/3% in 5L. Fill this to the top. You now have 5L of 1% in 5L.[/hide]

Now I would be very upset if I didn't multiply correctly somewhere, but I think this method works.

Edt: So, anyway, about this method working.. never mind.

[quote="Ian"]But now you face the same problem as scrambled -- how do you change 5L into 1L without changing the concentration?[/quote]

=(

Haha. Once more I have failed to read the question.

I'm kinda confused i think :? :

[quote][spoiler]You have 3L of 5/3% in 5L. Fill this to the top. You now have 5L of 1% in 5L. [/spoiler][/quote]

Dont you still need to change that from 5L of 1% into 1L of 1%?


algebra
function
domain
Ring Theory
superior algebra
superior algebra solved

Consider the ring $F[[X]], +, \cdot$, where $F$ is a field. (Here $F[[X]] = \left\{\sum_{i = 0}^{\infty}f_iX^i | f_i \in F\right\}$.)

(a) Show that $X$ is the only element of $F[[X]]$ which is irreducible.
(b) Is $F[[X]], +, \cdot$ a UFD (unique factorization domain)? A principal ideal domain?

Define for a power series $f=(f_i)_{i\in \mathbb{N}_0}$ the degree $\deg(f)$ as the lowest $i$ with $f_i \neq 0$. Then $\deg(f) \geq 0$ and $\deg(fg)=\deg(f)+\deg(g)$ by simple use of the definitions.

It's easy to construct (just use the definition of multiplication) an inverse for all power series with $f_0 \neq 0 \iff deg(f)=0$, so all these are invertible and only those by using that we need ${0=\deg(1)=\deg(f)+\deg(f^{-1})} \geq \deg(f) \geq 0$ which is only possible for $\deg(f)=0$.

Write $f \neq 0$ as $x^{\deg(f)} f'$ with $\deg(f')=0$. This shows that $x$ gives the only irreducible class and that we have a UFD. To prove that it is PID, take an element with lowest degree in an ideal giving that all others are multiples of it.

Well, consider first that every element with a $f_{0}$ nonzero, is a unit
the opposite holds too, if it is divisible by x, it has no inverse
but now every element can be written as a power of $X$ multiplied with a unit

now if it is a multiple of a bigger power of X, than it is certainly reducible, because $X$, a nonunit and nonassociate of the element divides it

so now the only one that remains is $X$ multiplied with a unit, or simply a formal power series with $f_{0}$ zero but $f_{1}$ not
explicitly : every divisor either is divisible by $X$ or even has $f_{0}$ nonzero
in the first case it is an associate of element, in the latter case it is a unit, and thus associated with the identity element of the ring

I think your statement is a bit incorrect, you forget the associates of $X$ also being irreducible?

b) A ring is UFD if two properties are satisfied :
every element is the product of irreducibles
every irreducible element is prime (where p being a prime means : if p divides $ab$ it divides $a$ or $b$)

the first is definitely the case here, as every element is the product of a power of $X$ and a unit
the latter is also true, if $ab$ is divisible by an irreducible element, it is the same as being divisible by $x$ anbd everything follows

[quote="fredbel6"]I think your statement is a bit incorrect, you forget the associates of $X$ also being irreducible?[/quote]

Obviously you are right :)


function
combinatorics unsolved
combinatorics

[quote="Vinxenz"]How many (not eqivalent) boolean function can you do with 3 variable?[/quote]

Such a function is completely defined by the ordered set of its 8 values $f(0,0,0)$, $f0,0,1)$, $f(0,1,0)$, ..., $f(1,1,0)$ and $f(1,1,1)$

So we have $2^{8}=256$ such functions


for which positive integer k does the equation $x^{2k-1}+y^{2k}=z^{2k+1}$ have a solution in positive integer x,y,z.

All of them. Look here :
http://www.kalva.demon.co.uk/short/soln/sh976.html

Pierre.


inequalities
geometry
inequalities unsolved

This can be written $ x_1^n+...+x_n^n-n x_1...x_n\geq\frac{(x_1+...+x_n)^n-n^n x_1...x_n}{(n-1)^{n-1}}$ and I remember to have posted a solution in the open question area as answer to a question of Pcalin. It is true that it remains to prove a monster one-variable inequality, but I remember to have managed to prove that one some long time ago. Try to look on it, maiybe there is something good there.

I remember now your posted solution to this inequality, Harazi.
But what you say about this inequality in the conditon $x_1+x_2+...+x_n=n$:

$(n-1)(x_1^2+x_2^2+...+x_n^2)+nx_1x_2...x_n \ge n^2$ ?

But if we take all numbers equal to $1/n$ then it is false. I imagine you wanted the sum equal to $n$ in this case, isn't it?

it looks like this:
http://www.mathlinks.ro/Forum/topic-5285.html

No, this one is much stronger. I remember to have proved it for $n=4$ or in any case something very close to it. I will try to solve this one, it is really nice.


To get a 'divides' sign, i know that you use the vertical bar. But how do you get the "not divide" sign (i.e. vertical bar with slash through it)

I am using amsmath and amssymb packages.

Use \nmid as in $ a\nmid b$
The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=mathabx]mathabx[/url] package provides another version which I think looks nicer using the command \notdivides


analytic geometry
algebra
polynomial
vector

What point is the centroid of the triangle with vertices (-2,6), (8,0), and (6,12)?

Centroid = (average x coordinate, average y coordinate)
So in this case, the x coordinate is (-2 + 8 + 6)/3 = 4
The y coordinate is (6 + 0 + 12)/3 = 6

So the answer is (4, 6)

How would you prove that the centroid is at (average x, average y)?

To prove it:

Suppose a triangle has vertices $ (x_1,y_1)$, $ (x_2,y_2)$ and $ (x_3,y_3)$.

The midpoints of the lines are $ \left(\frac{x_1\plus{}x_2}{2},\frac{y_1\plus{}y_2}{2}\right)$, $ \left(\frac{x_2\plus{}x_3}{2},\frac{y_2\plus{}y_3}{2}\right)$, $ \left(\frac{x_3\plus{}x_1}{2},\frac{y_3\plus{}y_1}{2}\right)$.

Then find two medians as linear polynomials and intersect them I guess.

Or translate so one vertex is at origin and one side is on x-axis to make easier?

You can do that too. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1458816#1458816]Here[/url]'s a vector proof...


Let $ T: H\rightarrow H$ be a bounded linear operator on a Hilber space $ H$. Show that

$ 1.$ there exists a unique bounded linear operator $ T^*$ on $ H$ such that $ <Tf, g>\equal{}<f,T^*g>$ for all $ f,g\in H$

$ 2.$ $ \|T\|\equal{}\|T^*\|.$

If we can find an $ h\in\mathcal{H}$ such that $ \langle h, g\rangle\equal{}\langle f, Tg\rangle$, then we are basically there.

Fix, $ f\in\mathcal{H}$. Let $ \phi_f(g)\equal{}\langle f, Tg\rangle$, then $ \phi_f$ is a bounded linear functional on $ \mathcal{H}$ such that $ \|\phi_f\|\leq \|T\|\|f\|$. By Riesz representation, we have a unique $ h\in\mathcal{H}$ such that $ \phi_f(g)\equal{}\langle h, g\rangle$.

Now just define $ T^*f\equal{}h$. I think the other properties shouldn't be too hard to check now.


The trees in a peach orchard were arranged in a square array
with $ N$ columns and $ N$ rows. To enlarge the orchard by one
row and one column to make a larger square, the owner bought
and planted 29 more peach trees. How many trees are in the
orchard after the enlargement?

Setting up a system of equations, we have $ (N \plus{} 1)(N \plus{} 1) \equal{} N^2 \plus{} 29$. Simplifying this, we have $ 2N \plus{} 1 \equal{} 29$, Thus $ 2N \equal{} 28$, and $ N\equal{}14$. Now adding one to N, and squaring, we have $ 225$.

Edit: Opps, failed....

I think you made a mistake.  When you said that $ 2N \equal{} 28$, you were still correct, but you somehow got that $ N \equal{} 18$.  $ N$ should be $ \frac {28}2 \equal{} 14$.  Therefore, our final answer should be $ (14 \plus{} 1)^2 \equal{} 15^2 \equal{} \boxed{225}$.  Another way you could have solved this was by using differences of squares.  We have $ (N \plus{} 1)^2 \minus{} N^2 \equal{} (2N \plus{} 1)(1) \equal{} 2N \plus{} 1 \equal{} 29$, which gets us back to the same equation, but maybe a bit faster.

Here is an easier way of doing this problem. The best method is actually solving it out and not using a trick or formula, but one that works for this problem is [size=150](x+1)/2[/size], x being the number of whatever added to the square (it has to be a square for this to work). In this problem, the number of added peaches are [size=150]29[/size]. You add [size=150]1[/size] and you get [size=150]30[/size]. Then divide that by [size=150]2[/size] to get the new number of rows and columns which in this case is [size=150]15[/size]. Then you do [size=150]15^2[/size] to get [size=150]225[/size]. It also works if you try it with a smaller figure such as a [size=150]2x2[/size] square. If the added number of something is [size=150]5[/size], that means the new square will be a [size=150]3x3[/size]. [size=150](5+1)/2=3[/size]. To find the area, just square it. Understanding why it works is also important. Why you add one is because when you add [size=150]5[/size] to the column and row of a [size=150]5x5[/size] square, one of the corners will only be counted once. If you add [size=150]5[/size], then the most upper left point will be filled (The [size=150]5[/size] is an example). If you add [size=150]5[/size] to the row, that uppercase point will also be one of these [size=150]5[/size] points, but it will already be in use. That is what causes the [size=150]x+1[/size] part. You divide the [size=150]x+1[/size]by [size=150]2[/size] because that will tell you how many are in the new column and row of this square. Half of that added value will be on the top while the other half will be on the side. Also, notice how [size=150]29[/size] is [size=150]14+15[/size]. The old side length added to the new side length. Is this a coincidence?


CUELGUEN AKI LSO RESULTADOS

IBERO 2006.

Saludos,

Resultados de PERU:

Daniel Soncco: Oro (perfecto)
Franco P.:        Plata.
Jossy A.:          Plata
. Veles:            Bronce

Buen trabajo chicos

Saludos,

PD: Alguien mas tiene resultados de sus equipos?
colgarlos ..

Ochoa y Cifuentes: Plata con 30puntos

Paez y Samper: Bronce con 27 y 22 puntos respectivamente

[quote="tipe"]Solo para completar los resultados de Perú:
Daniel Chen, oro, 42
Jossy Alva, plata, 32
Franco Vargas, plata, 32
Luis Velez, bronce,20

Jossy y Franco fueron las platas más altas (a uno del oro, snif!)

$Tipe$[/quote]

Realmente un buen trabajo de los chicos de Peru!!

mmm ...
 alguien sabe donde encontrar los resultados por paises??
cual fue el rankig este año?

Gracias ;

bueno no se exactamente los resultados  pero  fue algo asi:

  pablo  soberon : 37  puntos oro
  joshua  hernandez : 36 puntos oro
  fernando campos : 30 puntos  plata
  isaac buenrostro : 28  puntos plata

   mexico  primer  lugar!!  esta es la primera vez. mexico  sigue haciendo  historia este año  primero en la imo   y  luego en la  ibero!!!!

[color=blue]El equipo paraguayo también hizo historia. Por vez primera, cada uno de sus representantes obtuvo medallas (todas de bronce). Además, ha quedado como octavo país, y ha ganado la Copa Puerto Ricopor una amplísima diferencia.[/color]

[color=green]Todos haciendo historia, El equipo Salvadoreño tambien, es la  primera vez q somos los mejores de centroamerica quedamos en el lugar 11

Alfonso Alvarenga  20  bronce
Gabriel Chicas        19 bronce
Otto Tang               16 mencion
Tatiana Vasquez      14 mencion

Este ha sido nuestro mejor resultado en iberos[/color]

Chile:

Héctor Pasten ___ 22 ___ Bronce
Giancarlo Lucchini ___ 18 ___ Mención
Jaime Soza ___ 11 ___ Mención
Simón Legassa ___ 15 ___ saludos...

los resultados de la página oficial están [url=http://www.icm.espol.edu.ec/oim/resultados/index.htm]aquí[/url]


In Ninetown, every house has a three-digit address, where each digit (including the first digit) is between 0 and 9 inclusive.  If the sum of the digits of the address must always be 9, how many residents can Ninetown support before they run out of addresses?

I had an odd approach that worked well:

[hide]Using the divisibility rule for 9, I knew that there were 111 multiples of 9, all of whose digit sum is a multiple of 9.  I had to eliminate multiples of 9 whose digit sum was 27 (999) and 18 (harder).  I quickly realized there were 10 numbers from 900-1000 (909, 918, ... 990) whose digit sum is 18, there are 9 in the 800s, 8 in the 700s, etc... until only 1 between 0 and 100 (99) for a total of 10(11)/2 = 55 multiples of 9 less than 1000 whose digit sum is 18.  This leaves 111-55-1 or 55 numbers whose digit sum is 9.[/hide]

I think that it is meant to be solved the following way:

If the number is $ abc$, then $ a\plus{}b\plus{}c\equal{}9$. So we are looking for the number of nonnegative integer solutions to $ a\plus{}b\plus{}c\equal{}9$.

This is what is called a [b]distribution[/b]. It is covered in both  Intermediate C&P and Vol. 2. If you don't have these books, then here is a relatively short explanation of how to solve this problem:

[hide]
Represent the ordered pair $ (a,b,c)$ using dots and |s. For example, the ordered pair (4, 2, 3) would be represented as follows.

$ \cdot\cdot\cdot\cdot|\cdot\cdot|\cdot\cdot\cdot$.

Notice how there are 11 symbols (if $ a\plus{}b\plus{}c\equal{}9$) and 2 of them are |s. So the answer is $ \binom{11}{2}$ or $ 55$.[/hide]

so there are 10 ways the address can start,
0,1,2,3,4,5,6,7,8, or 9.

lets start with 0.  if the second number is also 0, then the only way for that to work is to have the third number be 9
if the second # is 1, then it would be followed by 8, and so on.  You can do this with all of the other numbers too.

so: 0,0,9; 0,1,8; 0,2,7; 0,3,6; 0,4,5; 0,5,4; 0,6,3; 0,7,2; 0,8,1; and 0,9,0.  so far we have 10 ways that the address can start with 0

but for one, you can only do 9 because there is no way for there to be a 1 and a 9 in the same address, because that would add up to 10, and 10 is not equal to 9.

so for 0 there are 10 ways, for 1 there are 9, for 2 there are 8... and for 9 there is only 1.

10+9+8+7+6+5+4+3+2+1=
10+(1+9)+(2+8)+(3+7)+(4+6)+5=55!

 :pilot:  :starwars:   :showoff: :juggle:


calculus
integration
logarithms
limit
calculus computations

Find $ \int_0^{\infty} (1\minus{}\sqrt{1\minus{}e^{\minus{}2x}})\ dx$.

Let $ k\geq 1$ be a natural number and let $ I(k) = \int\limits_{0}^{\infty}(1 - \sqrt {1 - e^{ - kx}})dx$.

Using the substitutions $ e^{ - x} = y$,  $ y^{k} = z$, we get that

${ I(k) = \frac {1}{k}\int\limits_{0}^{1}\frac {1 - \sqrt {1 - z}}{z}dz} = \frac {2 - 2\ln 2}{k}$.

I hope I made no mistakes. You are right, I edited it.

That should be:

$ I(k) \equal{} \frac{2 \minus{} 2 \ln 2}{k}$

But otherwise, that's the same result I got.

That's the correct answers, guys.  :)

Here is my solution for Japanese highschool students.

[quote="didilica"]Let $ k\geq 1$ be a natural number and let $ I(k) \equal{} \int\limits_{0}^{\infty}(1 \minus{} \sqrt {1 \minus{} e^{ \minus{} kx}})dx$.[/quote]

Let $ t \equal{} \sqrt {1 \minus{} e^{ \minus{} kx}}\Longrightarrow 1 \minus{} e^{ \minus{} kx} \equal{} t^2$ $ \therefore ke^{ \minus{} kx}dx \equal{} 2tdt$.

$ I(k) \equal{} \lim_{a\to\infty} \int_0^a (1 \minus{} \sqrt {1 \minus{} e^{ \minus{} kx}})\ dx$

$ \equal{} \lim_{a\to\infty} \int_0^{1 \minus{} e^{ \minus{} ka}} (1 \minus{} t)\cdot \frac {2}{k}\cdot \frac {t}{1 \minus{} t^2}\ dt$

$ \equal{} \frac {2}{k}\lim_{a\to\infty} \int_0^{1 \minus{} e^{ \minus{} ka}} \frac {t}{1 \plus{} t}\ dt$

$ \equal{} \frac {2}{k}\lim_{a\to\infty} \int_0^{1 \minus{} e^{ \minus{} ka}} \left(1 \minus{} \frac {1}{1 \plus{} t}\right)\ dt$

$ \equal{} \frac {2}{k}\lim_{a\to\infty} \left[t \minus{} \ln (1 \plus{} t)\right]_0^{1 \minus{} e^{ \minus{} ka}}$

$ \equal{} \frac {2}{k}(1 \minus{} e^{ \minus{} ka} \minus{} \ln (2 \minus{} e^{ \minus{} ka}))$

$ \equal{} \boxed{\frac {2}{k}(1 \minus{} \ln 2)}$.

Thus the original problem's answer is $ I(2) \equal{} \boxed{1 \minus{} \ln 2}$.


linear algebra
matrix
vector
linear algebra solved

Suppose A is an (nxn) matrix and that there is no nonzero vector x in R^n such that Ax=x. Show that (A-I) is nonsingular.

Suppose $A-I_n$ is singular then there exists $\alpha=1$ an eigenvalue of $A$. If $\alpha$ is an eigenvalue of $A$ then there exists a vector $v$ such that $Av=\alpha v$ which means that $Av=v$ contradiction!

Thanks for your help!

However, is there a way to show this without using eigenvalues? I ask this because we have not actually learned about eigenvalues in class, so I imagine that my professor is thinking along some other lines. How about if we start with the argument, if x is nonzero then Ax does not equal x.

Maybe this would help:
If $Ax\neq x$ then  $x(A-I_n)\neq 0$. For a matrix $B$ which has $\det(B)=0$ exists a matrix $X\neq O_n$ such that $BX=BA=O_n$(prove this using Hamliton Cayley). So my guess is that there should also exist a vector $x\neq O_{n,1}$ such that $xB=O_{n,1}$. Supposing $A-I_n$ is not invertible then take $B=A-I_n$ and from the preceding facts you get a contradiction. Anyway it`s just a guess of how it could be proved without eigenvalues.

with linear systems... if $A-I_n$ is singular then the system $(A-I_n)x=0$ would have a non trivial solution. Contradiction.