In a simple graph on $ n$ points, the orders of the points are $ 0 < d_1 \leq d_2 \leq ... \leq d_n$. Prove that it is possible to select at least $ \sum_{i\equal{}1}^n \frac {2}{d_i \plus{} 1}$ points, such that the subgraph formed by these points contains no loop.

This is untrue for example if d_i = 0 for each i. Is there no missing hypothesis?

Pierre.

What are you talking about when you say 'loop'. Do you mean cycle?

Are you sure that the probabilistic method does not work? The sum looks like an expectation.

Pierre.

I don't manage to come up to sol,because of ...,the original format mention loop and I realize cycle like you ,any way u can find it on komal

I think we should use in some way the idea that if there are given d1,d2,....dn, integers > 0 and Sum(di)=2n-2, then there exists a tree with these degrees. So maybe we choose somehow some vertices, then add some fake edges :D

Here it is:

[url]http://www.komal.hu/info/bemutatkozas.e.shtml[/url]

There are [b]lots[/b] of beautiful problems. You'll have to browse a bit to find the problems. Go to Contest (top side of the page)$\to$Problems (left side of the page).

But it is possible that after deletion some $d_i=0$.

Don't mind my previous post please. I completely realize andrew's idea.

I think I'm missing something.. How do we use it? Which are the points we select?

[quote="sam-n"]any help ,I can not come up to sol with probablitistic method ,plz devote your idea to this :D[/quote]
  I think the probabilistic method still works here:
 We choose a subgraph S from the initial graph G so that $ \Delta(S)<2$. Of course S contain no cycle and the expression of the individual expectation of each vertex A in G is $ E(A)\equal{} {n\choose {d_i\plus{}1}}\cdot 2\cdot d_i!\cdot (n\minus{}d_i\minus{}1)!/n!$ which is precisely $ \frac{2}{d_i\plus{}1}$
   Maybe I'm making a big mistake but maybe not !!  :maybe:

As usual it stands for the maximum order of vertice of S.

I suspected it.
But I doubt you can compute $ E(A)$ in this way, since it means if $ d_i\equal{}1$ then $ E(A)\equal{}1$, so $ A$ is contained in any subgraph $ S$ satisfying $ \Delta(S)\leq 1$. Right?

please correct me if i'm wrong, but I find it quite trivial since it is known that we can find at least$ \sum \frac {1}{d_i \plus{} 1}$ vertices that are not adjacent to each other. Now for each such vertex add to this subset one of his neighbour. This subgraph contains at least $ \sum \frac {2}{d_i \plus{} 1}$ vertices and cannot have any loop.

Since the proof of the first result is probabilistic, we can merge those two ideas and get strictly probabilistic proof:

Consider random permutation of vertices. For each vertex $ v$ let $ X_v$ be random variable equal $ 1$ if $ v$ has at most $ 1$ of his neighbours at lower positions in this permutation, and $ 0$ otherwise.
Now consider number of all such vertices - $ X$. We have $ EX \equal{} \sum E X_v$. Obviously $ EX_v \equal{} \frac {2}{d_v \plus{} 1}$. So we can find permutation with at least $ EX \equal{} \sum \frac {2}{d_i \plus{} 1}$ vertices that have at most $ 1$ neighbour before. Those vertices cannot create a loop, because one of vertices would have $ 2$ neighbours before him.

That's exactly what I said in my previous post. I believe that's fine !

[quote="Zorro"]please correct me if i'm wrong, but I find it quite trivial since it is known that we can find at least$ \sum \frac {1}{d_i \plus{} 1}$ vertices that are not adjacent to each other. Now for each such vertex add to this subset one of his neighbour. This subgraph contains at least $ \sum \frac {2}{d_i \plus{} 1}$ vertices and cannot have any loop.[/quote]

Sorry, but I can't confirm this argument - how do you prevent these neighbours from coinciding?

Anyway I have now posted the same probabilistic solution that you gave further below in your post in http://www.mathlinks.ro/Forum/viewtopic.php?t=240139 . Andrew's proof is great.

  darij

yes you're right, such simple argument doesn't work


analytic geometry
calculus
integration
topology
absolute value
advanced fields
advanced fields unsolved

Let $ X$ be an affine variety over $ \mathbb{C}$ ,and let $ R\equal{}\mathbb{C}[x_1,\cdots,x_n]/I$ be its coordinate ring.Show that if $ p(x)/q(x)$ is an element of the quotient field of $ R$ that is integral over $ R$,then for each point $ x \in X$ there is a neighborhood $ U$ of $ x$(in the usual topology) of $ x$ and a real number $ B$ such that the absolute value $ |p(x)/q(x)|$ is bounded by $ B$ at all the point of $ U$.
($ U$ is an open set $ X$)

For example,consider $ X\equal{}\{x^3\equal{}y^2|(x,y)\in \mathbb{C}^2 \}$
then $ R\equal{}\mathbb{C}[x,y]/(x^3\minus{}y^2)$ and $ y/x$ is integral over $ R$ because $ (y/x)^2\minus{}x\equal{}0$
and $ y/x$ such the conclusion.

Did you locally consider the set of solutions of the integral relation?

Note that the quotient needs to be inspected such that  is can be evaluated at every point of $ X$. In your example how is the quotient evaluated at $ (0,0)$?


pigeonhole principle
combinatorics unsolved
combinatorics

In an $ m\times n$ rectangular board, numbers $ 1,2,...,mn$ are written, each once. Numbers $ i$ and $ i\plus{}1$ are always adjacent, meaning the squares containing them share a side ($ i\equal{}1,2,...,mn\minus{}1$). Prove that there's a number $ t$ such that $ t$ and $ t\plus{}3$ are adjacent.

Basically, we have a board in which we could move a piece one square at a time from $ 1$ to $ 2$ to $ 3$ etc. all the way to $ mn$ (basically, we have a path).  proving $ t$ is next to $ t \plus{} 3$ is the same as $ t$, $ t \plus{} 1$, $ t \plus{} 2$ and $ t \plus{} 3$ being in a $ 2$ by $ 2$ square.  Perhaps pigeonhole now?  Or maybe induction... both on $ m$ and on $ n$.  The base case is just a two by two, for which the claim is clearly true.  Now if we add a row or column to an $ m$ by $ n$, how do our possible paths change?


limit
inequalities
absolute value
real analysis
real analysis unsolved

Here is a problem:

Suppose $\sum_{n=1}^{\infty}a_{n}$ converges; show that

$\lim_{n,\infty}\frac{1}{n}\sum_{k=1}^{n}ka_{k}= 0$

Here is my solution:
Given $\epsilon > 0$ we may take $n_{0}$ such that for every $n,m>n_{0}$
$| \sum_{k=m+1}^{n}a_{k}| < \frac{\epsilon}{2}$ by cauchy criterion.
Then
$| \frac{1}{n}\sum_{k=1}^{n}ka_{k}| = \frac{1}{n}| \sum_{k=1}^{n_{0}}ka_{k}+\sum_{k=n_{0}+1}^{n}ka_{k}| \leq \frac{1}{n}| \sum_{k=1}^{n_{0}}ka_{k}|+\frac{1}{n}|\sum_{k=n_{0}+1}^{n}ka_{k}| < \frac{n_{0}^{2}M}{n}+|\sum_{k=n_{0}+1}^{n}a_{k}| = \epsilon$
if $n > \max \{n_{0},n_{1}\}$;
where $M = \max \{| a_{k}|\}$ for $1\leq k \leq n_{0}$ and for all $n>n_{1}$
$\frac{n_{0}^{2}M}{n}< \frac{\epsilon}{2}$

...is there any problem with this solution?  Do you have a nicer one?  I think there might be a different approach that uses the fact that a cesaro sum of $a_{n}$ has the same limit as $a_{n}$ because in this same exercise that was shown.  (PS how do i make nice absolute value signs)

What can we get from summation by parts?

Let $A_{n}=\sum_{k=1}^{n}a_{k}$ and $A=\lim_{n\to\infty}A_{n}.$

$\frac1n\sum_{k=1}^{n}ka_{k}=\frac1n(nA_{n})-\frac1n\sum_{k=1}^{n-1}A_{k}(k+1-k)= A_{n}-\frac1n\sum_{k=1}^{n-1}A_{k}.$

Now take the limit as $n\to \infty.$ The right hand side tends to $A-A=0.$

[quote="arp"] (PS how do i make nice absolute value signs)[/quote]
Use \left| xxx \right| instead of | xxx |.

Example: $\left|\sum_{k=1}^{n}x_{k}\right|\leq\sum_{k=0}^{n}|x_{k}|$

@arp: I think that at some point you use the inequality $\left| \sum_{k = n_{0}+1}^{n}k a_{k}\right| < \left| \sum_{k = n_{0}+1}^{n}n a_{k}\right|$ (correct me if I'm wrong). But I don't think it's always true :maybe:

hmmm....

you're right, i don't know what happened there...is there any nice way to repair it


i know there are multiple programs for mac, but what is the best?

[mac users only please]

I put [url=http://www.uoregon.edu/%7Ekoch/texshop/texshop.html]TexShop[/url] on our Mac and am pretty happy with it.

I haven't tried anything else, but TeXShop works well for me- I followed the AOPS recommendation. It's free and reliable.


Part 1
The Gibb's Hill LIghthouse, Southampton, Bermuda, in operation since 1846, stands 117 feet high on a hill that is 245 feet high, so its beam of light is 362 feet above sea level.  A brochure states that the light itself can be seen on the horizon about 26 miles distant.

a) Verify the correctness of the information in Part 1 above.

Part 2
The brochure further states that ships 40 miles away can see the light of the lighthouse and planes flying at 10, 000 feet can see the light of the lighthouse 120 miles away.

b) Verify the accuracy of these statements in Part 2 above.

c) What assumption did the brochure make about the height of the ship?

I found that there are mistakes on that brochure
Or I made some mistake. Read and conclude

[i][color=olive](See the sketch below)[/color][/i]
Let $O,R$ be the center and the radius of the earth (respectively).
$B:$ the top of the lighthouse
$BC:$ a tangent segment from $B$.

We'll calculate the length of the arc $AC$, where $A$ is the projection of $B$ onto the earth's surface
(it would be more easy to find the length of the segment $BC$ but if we want an accurate distance on the surface of earth - the sea - we have to find the lengh of the arc)

Let $\theta = \angle BOC$

From the right triangle $OBC$ we have $\cos \theta = \frac{OC}{OB}= \frac{R}{R+AB}$.

Applying the values $R=3955.77187 \text{ miles}$ and $AB = 0.068560606 \text{ miles}$

we get $\theta=0.005887536$ in radians

and finally the length of the arc $AC$ is $R\cdot\theta \approx 23.29 \text{ miles}$

So 26 miles is a [color=red][u][b]lie[/b][/u][/color].
It should be $AB \approx 451 \text{ ft}$ but it is not

[size=134][color=blue]Part 2[/color][/size]

At the part 2 there is another mistake.

The airoplane acts like a new lighthouse, we can apply the same formula and find the maximum distance where the airoplane can be shown.

If $B'$ is the airoplane and $A'$ its projection , then we find that the length of the arc $A'C$ is $122.3846741 \text{ miles}$. So the distance of 120 miles would be true if the lighthouse had a [color=red][b]zero height[/b][/color]!
From the scetch you can see that the points $B,B'$ can see each other
So the total distance is $145.6744229 \text{ miles}$

Now assume that a ship is on the position $A'$ and the observer is on $B'$

solving for $AB$ in the above formulas we get $A'B' = \frac{R}{cos\theta}-R = R(\csc\theta-1)$

For $CA'=40 \text{ miles}\Rightarrow \theta = \frac{40}{R}$ we get $A'B' = 1067.852269 \text{ ft}\approx 325.5 m$

For $AA'=40 \text{ miles}\Rightarrow \theta = \frac{17.66258632}{R}$ we get $A'B' = 208.2019974 \text{ ft}\approx 63.5 m$

So the truth is that a ship with height 208 ft = 63.5 m can see the lighthouse in a maximum of 40 miles

But I think they assume that the height of the ship is 1067.85 ft, because they forgot that the lighthouse has a non-zero height

There are two tables below. The first calculates the distance $AC$ if the height $AB$ is given. The second does the inverse

What an excellent reply!!  Tell me, how did you make such a beautiful geometric drawing?

How can I do that?  Do I need to buy a special program for the computer?

I use the program [b] Euklides Dynamic Geometry[/b], but the link
[url]http://www.euklides.hu/[/url] doesn't work anymore.

By the way, I made the shape in that program and then I colored it in mspaint

Pontios,

Is there another geometry link that allows people to make shapes for geometry and trigonometry questions?

LINKERS:
http://www.chartwellyorke.com/
http://www.mathsnet.net/software.html
http://www.mathsnet.net/dynamic/index.html
http://www.google.com/Top/Science/Math/Geometry/Software/

OTHER:
Geometry:
[url=http://www.euclidraw.com/]EucliDraw[/url]
[url=http://www.geogebra.org/download/install.htm]Geogebra[/url]
[url=http://math.exeter.edu/rparris/wingeom.html]WinGeom[/url]
[url=http://www.dynageo.com/eng/index.html]DynaGeo[/url]

Graph drawing:
[url=http://www.download.com/ZGrapher/3000-2053_4-10490185.html]ZGrapher[/url]
[url=http://www.mathgv.com/]MathGV[/url]

I will try those sites.

I think students can benefit more from math question if replies are posted with diagrams.

Agree?

Of course  :wink:

Especially with geometry a scetch is necessary for the reader to understand the solution (or at least it helps to make the solution more clear).

By the way, if the picture is colored, the reading is also funny.  :)
It's the first time that I tried to color a scetch, and I saw that it is much more beautiful than the un-colored


calculus
integration
inequalities
function
LaTeX
real analysis
real analysis solved

Hello I've got this problem to solve.
Let be  $f:[0,1]\to\mathbb{R}$ such that
\[
f(x)=\begin{cases} x & \text{if }\quad x\in\mathbb{Q}\\
x+1 & \text{if }\quad x\notin\mathbb{Q}\end{cases}
\]
If $\Delta$ is any partition of $[0,1]$ prove that
\[
S(f,\Delta)\geq\frac{3}{2}\qquad\text{and}\qquad s(f,\Delta)\leq\frac{1}{2}
\]
where $S$ is a superior sum and $s$ an inferior sum of $f$ for $\Delta$.
Sebas.

What is the supremum of this function on an interval? What is the infimum?
Answer those questions, and you'll have what you need to calculate upper and lower sums. All you really need to show is that $\sup_{a<x<b} f(x)>1+\frac{a+b}2$ and $\inf_{a<x<b} f(x)<\frac{a+b}2.$

What do you think about this?

Let $\Delta=\{0=x_0,\ldots,1=x_n\}$ be a partition of [0,1]. We know that
\[x_{k-1}<\frac{x_{k-1}+x_k}{2}<x_k\]
so
$
\begin{aligned}
\sup_{t\in [x_{k-1},x_k]}f(t)&=x_k+1>\frac{x_k+x_{k-1}}{2}+1\\
\text{and }
\inf_{t\in [x_{k-1},x_k]}f(t)&=x_{k-1}<\frac{x_k+x_{k-1}}{2}
\end{aligned}
$
then
$
\begin{aligned}
S(f,\Delta)=\sum_{k=1}^n (x_k+1)(x_k-x_{k-1})&>
\sum_{k=1}^n (x_k-x_{k-1})+\frac{1}{2}\;\sum_{k=1}^n (x_k^2-x_{k-1}^2)\\
&=x_n-x_0+\frac{1}{2}(x_n^2-x_0^2)=\frac{3}{2}
\end{aligned}
$
Similarly $s(f,\Delta)<\frac{1}{2}$. Hence $f$ is not Riemann integrable...
Thanx!!!!!!!

[b] Moderator edit: Edited for $\LaTeX$ [/b]

I think we can prove $S(f,\Delta)\geq \int_0^1  1+x dx$
and  $s(f,\Delta)\leq \int_0^1  x dx$

in fact, both inequalities follow from the density of both $\mathbb{Q}$ and $\mathbb{R} - \mathbb{Q}$.
and the problem is same as showing that $I^* = 1$ and $I_* = 0$ for dirichlet function on the interval $[0;1]$, where  $I^* = 1$ and $I_* = 0$ are superior sum and inferior sum for "simple" functions of dirichlet functions... hope you understand what i mean...


Ok ive got a test coming up in a couple of weeks and im stuck on the math section.Heres some of the type problems im stuck on.If anybody could just help me understand these and walk me through them i would greatly appreciate it!

1.How many miles has flight X75635 flown after 35 min. at a speed of 250MPH?

2.If a plane was descending at 60 feet per second from a altitude of 13,500 feet to 9,000 feet at 250MPH how many miles would it had traveled?

3.A plane just left 15,000 feet descending and will reach 9,000 feet in 10 min.What is the decent rate per min?

4.A plane is taking off and gaing altitude at 50 feet per second at 300MPH.What is the Altitude of the plane after traveling 100 miles?

Thats all ive got for now.If you could plz help me i would appreciate it.Thanks in advance!

#1: THE FLIGHT NUMBER DOESN'T MATTER, so ignore that. We know that the airplane flies for 35 min, which is $ \frac {35}{60} \equal{} \frac {7}{12}$ hours. Since in $ \frac {12}{12}$ hours the plane flies 250 miles, in 35 minutes it flies $ \frac {7}{12}* 250\approx\boxed{145.833}$

#2: We see that the total descent is $ 13500 \minus{} 9000 \equal{} 4500$. We know that it descends 60 feet per second. To find the total flight time, we divide 4500 by 60 and get 75 seconds. 75 seconds is $ \frac {75}{3600} \equal{} \frac {1}{48}$ of an hour. Multiply this by 250 and get approxamitely $ \boxed{5.20833}$.

#3: We find the total descent to be 15000-9000=6000. If it descends 6000 ft. per 10 minutes, dividing by ten, we get that it descends $ \boxed{600 ft}$ per second.

#4: We find the total flight time $ \frac {1}{3}hour \equal{} 20mins \equal{} 1200 secs$ because it travels 100 miles at a rate of 300 MPH. The altitude is just the rate times the time (rate * time=distance). That is $ 1200*50 \equal{} \boxed{60000ft}$.

PS: Are these problems on worksheets or something, because they don't come out to be very even numbers, unless I am making a mistake.

Be sure to check your units (secs, mins, hours).
Also, if you didn't understand anything, ask your question and I will explain.

Okay,thanks so far.

In the first question were did you get the 7 from?

I have to do all this without a calculator or any scratch paper so its all mental math and plus im timed as well. so im still a little slow.lol I only get a minute per question.

In the second question do i multiply 250 by 75? When i do that i get 18750.

To be honest with you,im actually studying for a Air Traffic Control test,and this is just off ones of the practice test.

Thats my main issue is being able to do all this in my head within a minute.

Thanks again for your help i really appreciate it.

I edited some of them so it should look right now.

I think i got those.How about this one

1.At 800Z flight L37182 at 200MPH is 40 miles from flying over Airport DLQ.What isthe ETA for the airport?

Thanks again.

What's an ETA?  :maybe:

EDIT: Duh. Never mind.

But that question doesn't give us enough info. How can we find the Estimated Time Arrival if we don't know which path it is going to take?

And what is 800Z?

EDIT: If that means (I don't know terminology for flight) 800 ft. altitude, then I am assuming it is like a Pythagorean problem.

800Z is the time.

It dosent give a path,it just says 40 miles away,i guess they figure in the air that can be in any direction i reckon.

This stuff is so confuseing sometimes.lol

OK. Check the book. Okay? Cause you said the ETA for the AIRPORT not the airplane. Now that's weird. An airport can travel at 200 miles per hour above itself. And the other thing. What its asking for doesn't make sense. You need to have to measurements. You can't have time and length because those don't go together. You need Time and 2 lengths at least.

I meant ETA for the plane not the Air Port my bad.lol

Thats all it gives,im starting to think that the disk i have is messed up or somthing.

I think the altitude doesn't matter and that the plane can change altitude with a constant forward speed of 200 MPH. (of course, you should know more about that than me) Anyway, this is how the problem would look like that:

If the plane needs to travel 40 miles and has a speed of 200 MPH, then the total travel time is $ \frac{40}{200}\equal{}\frac{1}{5}hours\equal{}12mins$. So the arrival time is $ \boxed{8: 12}$

Air traffic Control Test...

That's interesting. I'm guessing you must have passed college?


I need to brush up on problems 20-25 on the AMC.
Will taking the AIME class help with those problems?

Early AIME problems are often roughly the same difficulty level of late AMC problems.  If you're comfortable with 1-20 on the AMC, you should be spending a lot of your time at the AIME level, anyway, so the AIME class would probably be a better fit.


Find all integers solutions for
$ xy\plus{}yz\plus{}zx\minus{}xyz\equal{}2$

$ i)x\geq y\geq z \geq 0$  $ \implies$   $ \frac {1}{z} \plus{} \frac {1}{x} \plus{} \frac {1}{y} \equal{} 1 \plus{} \frac {2}{xyz}$
$ \frac {1}{z} \plus{} \frac {1}{x} \plus{} \frac {1}{y}\leq\frac {3}{z}$  $ \implies$  $ z\leq 3$
$ z \equal{} 3$  $ \implies$  $ xy \plus{} 3y \plus{} 3x \minus{} 3xy \equal{} 2$ $ \implies$  $ 3x \plus{} 3y \equal{} 2xy \plus{} 2$  $ \implies$
$ 6x \plus{} 6y \equal{} 4xy \plus{} 4$  $ \implies$  $ 2x(3 \minus{} 2y) \minus{} 3.(3 \minus{} 2y) \equal{} \minus{} 5$ $ \implies$  $ (2x \minus{} 3)(2y \minus{} 3) \equal{} 5$  $ \implies$
$ 2y \minus{} 3 \equal{} 1,2x \minus{} 3 \equal{} 5$  $ \implies$  $ z > y$  contradiction!
$ z \equal{} 2$  $ \implies$  $ xy \plus{} 2y \plus{} 2x \minus{} 2xy \equal{} 2$  $ \implies$  $ 2x \plus{} 2y \equal{} xy \plus{} 2$
$ x(2 \minus{} y) \minus{} 2(2 \minus{} y) \equal{} \minus{} 2$  $ \implies$  $ (x \minus{} 2)(y \minus{} 2) \equal{} 2$  $ \implies$  $ x \equal{} 4,y \equal{} 3$ ,$ (4,3,2)$ is a solution
$ z \equal{} 1$  $ \implies$    $ xy \plus{} x \plus{} y \minus{} xy \equal{} 2$  $ \implies$  $ x \plus{} y \equal{} 2$  $ (x\geq y\geq z)$  $ (1,1,1)$ is a only solution.
$ z \equal{} 0$  $ \implies$    $ xy \equal{} 2$  $ \implies$  $ (2,1,0)$ is a solution
$ ii)x\geq y \geq 0$ ,$ z < 0$  $ \implies$  $ z \equal{} \minus{} r$.
$ xy \minus{} yr \minus{} rx \plus{} xyr \equal{} 2$  $ \implies$  $ \frac {1}{r} \plus{} 1 \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {2}{xyr}$
$ \implies$  $ x,y,r > 2$  $ \implies$  $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {2}{xyr} < 1$  contradiction!
$ y \equal{} 2$  $ \implies$   $ 2x \minus{} 2r \minus{} rx \plus{} 2rx \equal{} 2$  $ \implies$  $ 2x \minus{} 2r \plus{} rx \equal{} 2$  $ \implies$
$ x(2 \minus{} r) \minus{} 2(2 \minus{} r) \equal{} \minus{} 2$  $ \implies$  $ (x \minus{} 2)(r \minus{} 2) \equal{} 2$  $ \implies$  $ x \equal{} 4,r \equal{} 3$ and $ x \equal{} 3,r \equal{} 4$
$ y \equal{} 1$  $ \implies$    $ x \minus{} r \minus{} rx \plus{} rx \equal{} 2$  $ \implies$  $ x \minus{} r \equal{} 2$  $ x \equal{} r \plus{} 2$  $ \implies$  $ (r \plus{} 2,1,r)$
similarly $ r \equal{} 2,r \equal{} 1$
similarly, $ x\geq 0$  $ y\leq z < 0$ $ \implies$ $ y \equal{} \minus{} p,z \equal{} \minus{} r$ $ \implies$  $ \minus{} px \plus{} pr \minus{} rx \minus{} prx \equal{} 2$  but $ \minus{} px \plus{} pr \minus{} rx \minus{} prx < 0$ for $ x\geq 1$  $ \implies$  $ x\equal{}0$  $ pr\equal{}2$  $ \implies$   $ (2,1,0)$
and $ z\leq y \leq x < 0$  $ \implies$  $ z \equal{} \minus{} r,y \equal{} \minus{} p,x \equal{} \minus{} q$  $ \implies$  $ pr \plus{} rq \plus{} qp \plus{} pqr \equal{} 2$
$ p,q,r > 0$  $ \implies$  $ pr \plus{} qr \plus{} qp \plus{} pqr\geq 4$contradiction!
$ r \equal{} 0$  $ \implies$  $ (2,1,0)$ is a solution  :wink:

[quote="lambruscokid"]Find all integers solutions for
$ xy \plus{} yz \plus{} zx \minus{} xyz \equal{} 2$[/quote]
See here:

http://www.mathlinks.ro/viewtopic.php?t=265297


I know there's also another thread on AIME scores, but it doesn't have a poll.  What do you think you got?  Vote, and maybe we can guestimate what the cutoff might be for this year...

I got a 3.... if you ask me "do you feel bad?" you know the answer buddy..

could have gotten 15... would have gotten 15... should have gotten 15...  :(

I got a 4, should have gotten a 6 without stupid mistakes.  :mad:

"I got a 4, should have gotten a 6 without stupid mistakes."

Exactly the same with me. I actually answered 8, with 6 of them answered with the completely correct method (10 was a partial guess, and on 4 I used an incomplete method). However, the following were my silly mistakes

2. Perhaps the second easiest question on the test (after #3) but I misread the part about Cy walking beside the walkway instead of on the walkway :wallbash_red:
9. This was probably the most disappointing stupid mistake. I had spent a good amount of time on this problem and had come to the verge of solving it correctly, up to the equation
coefficient*radius = hypotenuse
However, in the last equation, I suddenly revived a mistake I had made on an old equation and set the hypotenuse as 30, instead of 34.  :mad: I thus got 200/19 instead of 680/57.

I ended up correctly answering 1, 3, 7, and 11.

"Solved" 1-7, 9, 11, but only got 1,3, and 7 correct to to extremely careless mistakes...

Answered 1-10.
Only got 2,3,7,8,9 = 5.
It's all those counting questions...
At least I know what to study for next year.

grrr...

i got 1,2, and 11

i somehow misdid number 3.

stupid (sorta) on number 5.

stupid on number 4.

got 1,2,3,8

Missed 4,5,6,11,13

4:  Didn't spend enough time thinking about it
5: missed it by one
6: I must have left out a case
11:  Too many computations messed me up (255)
13:  I still don't know how this isn't 20

I got NONE of the geometry yesterday. I'm sort of sad...

(unless you count the planets question as geometry, which I didn't)

Also, you should have put a 16 option there :P

I got questions 1, 3, 4, 6, 7, 8, 11 (score: 7)

Misread #2. Who walks beside the walkway? Besides, the guy isn't actually at the midpoint of the segment, since they aren't collinear (I'm just pissed at my misreading)

5: I was too lazy to brute force. I knew it cycled in 9s and I tried 32 and 34 worked, so I put 214 <_<.

9: My answer made no sense, I figured (correctly) that the centers were on the angle bisectors, but then I got 1/127 or something which doesn't work.

10: Tried case analysis, must've miscomputed something

12-14 were guesses (012)

15: i tried something, but my answer didn't really seem to work anyway since I got $\frac{3}{2}\pm \frac{1}{2}\sqrt{97}$, so yeah.

[quote="b-flat"][quote="bpms"]I got an 8.
Two 15s?!?!?!?[/quote]

No, 3.[/quote]
It was 2 earlier, unless I'm losing it.

Lies, obviously. This AIME was rather hard not to make stupid mistakes on. And considering every year there are less than 5 perfect papers...  :P

yeah I always lie on polls

sometimes higher, sometimes lower

I voted "6" this year

we should've put a 16 option to filter out some of the liars :D

I got a 7, made stupid mistakes on number 5 and 6....
number 5, I forgot to count the cases between 995 and 1000, so I put 536, and number 6, I forgot to count the extra case with the 13, so I put  140 instead of 169. DARN! So my index is 196...not good enough.

That's higher than mine, I had 120/7. Evidently there was a 150/7, so unless the index cutoff is more than 220, you'll get in (but do you have to take the AMC 10 to qualify through the floor?).

I'm in grade 12, so I don't get the benefit of the floor value  :(
Btw, what do you guys think the index is going to be this year? I'm thinking between 205-210

[quote="minsoens"]why would anyone lie? just as a joke?[/quote]

Well suppose you didn't take the AIME and stumbled onto this poll (intentionally or unintentionally) and simply feel the urge to vote. The most popular choice for these people is probably 15.

Well, compared to everyone else, I'm apparently a complete failure. I don't even want to share my score, but I got 4 and 5 wrong. Enough said

9 people got a 15, eh?

Ha.  As if.  :P

I personally got a 7.  Hopefully that's good enough...

[i]very[/i] nice use of reverse psychology on the poll, op.

[quote="Ignite168"]Why would the most common scores, 0-5. be lumped together into only 2 categories? It makes no sense at all.[/quote]
This person wanted to see how many people got each distinct potential floor, 6, 7, 8, 9.
Also, on AoPS, those aren't the most common scores.


Your mama is so stupid she used the pythagorean thereom for an obtuse triangle!!!  :rotfl:
It's not that funny actually.

Your Momma's so fat, when she went to the store,[hide] the people over there told her that she was fat.[/hide]

Your mama's house is so small, she has to eat a large pizza outside!

Yo' mama sounds so [hide]similar to "Yo Yo Ma" that someone thought she was a famous cellist! :D[/hide]

your mom is so smart she's smater than my pet dog! :D

yo mama is so stupid.....she used yo mama jokes on her brother

Yo mama is so stupid when she read the back of AoPS shirt, she thought that there were two different parts, A. and B., for #10.

[hide="best your mama joke ever"]your mama is the last digit of pi[/hide]

[quote="Chigr"]mcalderbank, it should be on her son, not brother[/quote]

No, brother works as well, because if she says yo momma jokes to her brother, she is insulting her own mother as well.


[quote="Tare"]Just wondering. If you could also write when and how much you skipped, that'll be great.[/quote]

My skip was across the board. The school district I lived in at the time was very much against grade skips. Until a few years before when I had my grade skip, the district had an "accelerated" program in junior high that compressed three years into two, equivalent to one year of grade-skipping. I knew an alumna of that policy in my college days, an older girl from my same school district. It seems more civilized, if one must be in school (my children are not, as they are homeschooled) to keep a whole group of bright kids together than to have each one skip (or not) piecemeal at differing ages.

Of course the math standards in my alma mater school district were pathetic, hence the tagline in my signature . . .

Well, I've skipped one grade in everything and then two more in math... it kind of varies between schools, though. Our school offers AP Calculus BC in 12th grade (which is what I'm taking), but I know other schools where juniors in the honors program take BC...

Edit.. oh yeah, when I skipped.. skipped all of third grade, which doesn't count for much... skipped seventh-grade and tenth-grade math

What grade are you in right now? Don't want to subtract right now...

[quote="Chinaboy"]i guess 3??? since Alg. II is taught normally in 11th and I'm in 8th... but I don't really know...[/quote]
11th!? By that time I'd be well into Calculus...I'd say 1 or 2

well... 11th grade Alg. II is for most of the people at our school... the normal ones.

average is doing algebra I at our school in 9th grade. and trig in senior

Wow...

Oh, I'm not sure about the other classes. I'm only interested in Honors curriculum :)  :twisted:
Guess I skipped 4 grades then :) </bragging>

i took algebra 2 in 8th grade also, but i was only considered 2 yrs advanced of our usual curriculum. i also skipped a year of chemistry.

[quote="bookworm271828"]I'm doing Algebra II now, but am going to do trig through CTY this spring, so that puts me 2 or 3 grades ahead.[/quote]

You're going to hate it. The first four chapters (half of the course) have nothing to do with trig, and the same for the eighth and final one. Also, the first chapter is excruciatingly long.  :(

yee, from geometry to calculus...so like 3.

Calculus as a 9th grader. You can decide how many grades ahead that is...

3 or 4 I guess...wow, you guys are all geniuses compared to others...

[quote="Mindspa"]Skipping grades does not make you a genious.[/quote]

That's certainly true. Really, the mistake in the school system in the first place is assuming that most learners learn all subjects at the same pace as most other learners of approximately the same age. My very first online FAQ,

http://learninfreedom.org/age_grading_bad.html

now more than a decade old, is about how ridiculous that assumption is.

Skipping a grade doesn't prove a person smart, because staying at grade level doesn't prove that a person is making adequate progress. What I noticed when I moved from state to state in the United States (which I did twice at school age) and when I lived in a country (Taiwan) outside the United States (once from 1982 through 1985, once from 1998 through 2001) is that there ISN'T any set standard for when to study different subjects. Is algebra a fifth-grade subject, an eighth-grade subject, or a tenth-grade subject? How about geometry? How about calculus?

The TIMSS studies

http://nces.ed.gov/timss/

http://timss.bc.edu/

provide interesting information about just how different the curriculum expectations are in different countries. That's what is cool about this online forum: we have participants here from all over the world, the better to get a reality check on how much math is enough to know by various ages. For the people like me who grew up as monolingual native speakers of English, this forum is also a good reality check on use of English as a second language by people from a variety of countries. Americans who can use a second language to discuss mathematics are quite rare.

Exactly... because you could've skipped kindergarten or 1st grade.

I'm taking Precalc, and I'm in 9th grade, but that's pretty irrelevant considering I didn't make nats.

[quote="Tare"]Just wondering. If you could also write when and how much you skipped, that'll be great.[/quote]

When I took EPGY,

http://epgy.stanford.edu/

I went through the fifth grade, sixth grade, prealgebra, and beginning algebra courses in one calendar year. When I was starting I was nine, and when I got done I would just have been finishing the fifth grade. So if algebra 1 was a normal eighth-grade course, I skipped three grades.

This year I am repeating algebra 1 and doing algebra 2 through UMTYMP

http://www.itcep.umn.edu/umtymp/

and I'm taking Honors Geometry through Johns Hopkins. (I took the Johns Hopkins prealgebra course--which I don't recommend--back when I was taking EPGY prealgebra.) So that means by the end of this year I'll be skipping four grades in math.


geometry MATHCOUNTS function calculus AMC USA(J)MO USAMO

How many years ahead are you in school math?  I am four years ahead.  Might go to five ahead next year.

you should prob specify what "regular school math standards" are cause differnt places are different. I am assuming you mean algebra in 8th grade though... im only one year ahead but next year i will be one year ahead/2-4 years ahead if that makes sense...

In our school Pre-algebra is standard for 8th graders, but I'm taking Precalculus, so I'm 4 years ahead.

Prealgebra is standard for seventh graders here,. but I'm taking Geometry, so two. My old school didn't teach me half the things I needed to know in geometry, so I had to repeat it. So two.

Same as NoSoup, but I started pre-calc about 1.5 years ago, and then stopped in favor of AoPS, so I guess I was about 5 years ahead back then.

what do you mean? i'm supposed to think that in the US you can take which lessons you want?am i right? if it is so, it isn't the same here in italy. we have to take the lessons that all the guy of your age take, you can't choose which ine you want to take and which one not. but i study math my self at home and i'm 2 years ahead. :)

[quote="solafidefarms"]At least for me, it's called homeschooling. Don't know about NoSoup[/quote]
yes homeschooling now i remember it my english teacher said us about that...unfortunately you can't do it in Italy :? so if the lessons you take are too easy you have to take them, and if you want to study more you'll have to do it self that is to say: every year you'll get more bored because you are ahead, i've done so since i was 9

I'm taking Geometry in 8th grade.

Here in Nebraska, that's one year ahead. If I had arrived early in the 6th grade, I would be taking Elements... some math course. Ask ElementsofMathKid or Icantpeeloranges about it.

I am also enrolled in geometry, which is two years ahead because the "on-level" course is a repeat of Pre-Algebra called BAG.  Regardless of this, I think that most of our grade is in Algebra I.  All this will change with some weird reform they're introducing that will mess everything up.  However, I have learned Algebra II on my own this year, and am currently studying trigonometry.  By the end of the summer, I hope to have learned some multivariable calculus.  My parents are math professors, so I have access to a rich library of undergraduate and graduate math books.

Wow, another nat mc poll! :)

To answer the question, next year, taking Alg II in ninth grade, I will be one year ahead, and I was two years head a couple of years ago taking Alg I in sixth, and Geometry in seventh stretch out to cover part of eighth.

But about the topic, I just want everybody to realize that how far one is ahead in math does not prove anything about one's math abilities or even the solidity of one's math education. Oftentimes it is better for a math education to be built horizontally (spending more time at each level, expanding on these levels) than vertically (trying to reach as high as possible without building such a strong foundation). Still, taking pre-Calc in the seventh or eighth grade, is pretty impressive, I'm not going to deny it! Just mull over that for a moment or two.

[quote="solafidefarms"]At least for me, it's called homeschooling. Don't know about NoSoup[/quote]

No, I go to a public school. They just see that I'm advanced and let me take classes at the local high school.

Im in geometry seventh grade. For my school thats 3 years ahead

At my school:

8-Algebra I
9-Geometry
10-Algebra II
11-Pre-Calculus
12-AP Calculus

[quote="JesusFreak197"]Hmm... Statistics is also normally a college class, but apparently that doesn't require Calculus... I know it can be done before Calc. Well, wherever you go to school is just a little abnormal, I guess. :P[/quote]

Hey, I go to school in Lexington. (Looks over at AMC forum) We're the school with the seventh most USAMO qualifiers :D

Right now in my 7th grade we are still learning how to make [i]bar graphs[/i]!
And we just finished similar triangles, and I cannot tell you how many people are clueless about what's going on. This is supposed to be an enriched class of students!

[quote="chess64"][quote="superman"]it doesent have to be ap calculus at our school, you can take regular calculus[/quote]

AP calculus means college-level calc in high school. the only way you can take "regular" calculus is in college.  ;)[/quote]

maybe u r right...but i don't think so...

They offer both at our school, and I think I'm going to take AP calc next year

For my school:
7th: 3 different levels (Math 7, Pre Alg, and Alg)
8th: same (Pre Alg, Alg, and Geometry)
9th: same (Alg, Geometry, and Alg II)
10th: same (Geometry, Alg II, and Pre Calc)
11th: same (Alg II, Pre Calc, and Calculus)
12th: same (Pre calc, Calculus, and Stats{or Concurrent Enrollment})

So I'm in Algebra II in 8th grade, and since I go to a junior high (7-9 grades), I don't have to go to the high school yet--I go to the 9th grade advanced math class.  I feel really wierd in it, though--I already skipped a grade...I'm glad that I went to the school I went to, because I believe that it is one of three schools in our school district that has Algebra II in the junior high.  I go to my school by special permit, meaning that I'm not going to the school I'm supposed to go to.  This school is 15 miles away from my house, while the one I'm supposed to is within walking distance...the school that I go to is a public school, so the only thing I like about it is that it does teach Algebra II.  I think that it is also better because the high school that it feeds into has an International Baccleareatte(probably spelled wrong) program.
I would take Stats in 11th, but I think instead I'll do concurrent enrollment at the University of Utah(Advanced Calculus classes).  Sincce the majority of students in 8th grade are in Pre Algebra, I consider myself three years ahead in math.

I was considering going to a school next year that has the IB program (the program you mentioned), and it's apparently very good from an academic standpoint. I ended up choosing something different, though.

Also, I know what you mean about 15 miles away... it's a little farther than what I'm used to, but since 2nd grade I've been going to schools about 10 miles from my house. :huh:

[quote]umm.. i suppose im 0 years ahead actually, in enriched class.. 7th we finish pre-algebra first half of the year (start in 6th) and we finish i think half of Algebra I by year end.. i think we'll start geometry sometime in 8th.[/quote]

I think we're looking at how far ahead of a normal schedule we are, not how far ahead of the special program we're in; if that was the case, I would only be two years ahead, rather than four. ;) Also, Geometry in 8th is (generally) two years ahead.

the farthest you can go up in math is three years in my school  :(  thats what i am in

I'm doing Independent Study Precalculus from [url=http://thinkwell.com]Thinkwell[/url] along with normal school. (Which I'm also ahead in. I'm in Geometry, while the average 8th grader is in Pre-algebra, smarter ones are in Algebra I, and GT (gifted and talented) is in Algebra II. Geometry is meant to be a GT 9 class.) Next year, I'm taking AP stats as an elective, along with AP calc (Hopefully they'll let me skip pre-calc.)

[quote="addicted2sharpiz"][quote="superman"]why is that?

couldn't you just take a class w/ older stuedents?[/quote]

yeah i do take a class with older people. But you just can't go up any further....[/quote]

 :huh: You should be able to! If your parents complain enough, they'll probably do something about it. BTW, how [i]will[/i] you go up further if you haven't completed the math level you're in right now? That's why I did distance education from JHU and then from Thinkwell.

Then I told the school that I took a course, so I wanted to skip the class. So they gave me a final exam (which I got 100% on, btw  :showoff:) so I got an itinerant teacher since I was in 8th grade and they didn't want me to go to the high school (darn!).


[hide="answer"]
Rachel, Jane, Kim, Shawn
because J is older than K and K is older than S, and R is older than J[/hide]

[hide]Rachel, Jane, Kim, Shawn  :idea: [/hide]"Shawn is younger than Jane" was completely unnescasary ;)

[quote="math92"]If Jane is older than Kim, Kim is older than Shawn. Shawn is younger than Jane and Rachel is older than Jane
List the people from oldest to youngest.[/quote]
[hide]
J>K
K>S
S<J
R>J
R>J>K>S

Rachel, Jane, Kim, Shawn
[/hide]


$ \leftrightarrow (a \plus{} b \plus{} c \minus{} 1)( \sum \frac {1}{b \plus{} c} ) \ge 3$
By Am-Gm , we can prove :
$ LHS \ge (a \plus{} b \plus{} c \minus{} 1)\frac {9}{2(a \plus{} b \plus{} c)}$.
So we need to prove this ineq :
$ 3(a \plus{} b \plus{} c \minus{} 1) \ge 2(a \plus{} b \plus{} c)$
<=>$ a \plus{} b \plus{} c \ge 3$ (True) .
I'm a spamer  :ninja:


Let $ a,b \in N$ and $ a.b\equal{}1991^{1995}$. Prove that $ a\plus{}b$ indivisible$ 1995$.

[hide]$ ab\equal{}1991^{1995}\equal{}11^{1995}181^{1995}$, so $ a\equal{}11^c181^d$ and $ b\equal{}11^{1995\minus{}c}181^{1995\minus{}d}$ where $ 0\le c,d \le 1995$ and $ c,d\in N$.

Note that $ 1995\equal{}3\cdot 5\cdot 7\cdot 19$.

$ \frac{a\plus{}b}{1995} \equal{} \frac{11^c181^d \plus{} 11^{1995\minus{}c}181^{1995\minus{}d}}{3\cdot 5\cdot 7\cdot 19}$.

This is obviously irreducible since the numerator and denominator have no common primes.[/hide]

i think your last statement needs some proving


I was wondering if I am correct in saying that to apply to Seimens-Westinghouse, you have to be in 12th grade at the time you apply? If so, does that mean, if a student is in the 11th grade and is in RSI (which i am not), can he start the project at RSI and enter that project into the Westinghouse competion

So you have to be in 12th grade to apply?

To submit a project as an individual, you must be a senior.

You can submit a project anytime else as long as you are with a team, I think. I'm not sure if the team requires a senior on it, though.


What is the value of $ (895 \minus{} 54)^2 \minus{} (54 \minus{} 895)^2$?

We know that 54-895 is the negative value of 895-54, the square of a negative number is the same as the square of its positive value.  Therefore, $ (895\minus{}54)^2$ = $ (54\minus{}895)^2$, so $ (895\minus{}54)^2\minus{}(54\minus{}895)^2\equal{}0$.


algebra
function
domain
inequalities
topology
calculus
calculus computations

given that an open set B is star-shaped if there exist a point P in B s.t.
the line segment between P and any point
tP + (1-t)Z    belongs in B , for every Z in B and t in [0,1]

1) Prove that D = { z=x+iy | x>0 & [x>y+1 or x>1-y] } is star shaped
2) Prove that D = {z: |z|>1} is not star shaped

for 1) ...I said choose P to be any point on  Re(z) and P>1
but then...I don't know how to show the line segment between P and other z is in the set D

and for 2)...I know for any point P=x+iy, P'=-x-iy......the line segment PP' can't be within D...but how to show it in a proof?

thanks

You just need to pick another point and use the parametric form for a line.  Let me do a related example:

Say we have the domain D = {z = x + iy  |  y > x or y < - x)}.  (This is similar to your first question.)  I claim this is a star-shaped domain, using the point $z_0 = -1$.  Let $z = x + iy$ be any point in D.  The segment connecting $z$ to $z_0$ is the set of all points of the form $tz + (1-t)z_0$ where $0 \leq t \leq 1$.  For any such point, we can re-arrange this expression by substituting in for $z$, $z_0$ to make it look like
$tx + ity - (1 - t) = (tx + t - 1) + (ty)i$.  Now, we know $y > x$ or $y < - x$.  In the first case, $ty \geq tx \geq tx + t -1$, but the first inequality is equality if and only if $t = 0$ while the second inequality is equality if and only if $t = 1$, so we have $ty > tx + t -1$ and the point is in our domain.  In the second case, $y < -x$ so $ty \leq -tx \leq -tx -t + 1$, but again we cannot have equality all the way through so that $ty < -(tx + t - 1)$ and the point is again in our domain.  So each segment is entirely in the domain and we are done by definition.

Your second problem is particularly easy -- you know a point on that segment that is not in D, so you just have to say so.  (You probably want to phrase it as a proof by contradiction: Assume D is a star-shaped domain.  Then there exists a point $z = x + iy$ such that ...)


is it possible to restart all your stats is you are unsatisfied?

I don't think so...there will be quite a few cases resulting where people reset to get a higher Alcumus score and "memmed" all the questions.

Restarting would negatively impact the AI so there is no way to do this.

So if you miss a question, can your alcumus score ever be 99.99?

I don't think your alcumus score can ever be that high.

Wait, so can your Alcumus score only go down, since you miss problems? And do those problems count as missed even when you answer them correctly the second time you solve them?

when you go through them the second time, any problems you got right the second time will be marked as right
as you get more problems right, your Alcumus score can go up

So then once everyone has finished and answered all the problems correctly, won't everyone have the same Alcumus score?

No, since the Alcumus score is adjusted over time. One person may have had a score of 70 when they ran out of problems and started repeating while another will have a score of 60. If both these people then answered all repeated questions correctly, the person who originally had 70 as an alcumus score would still be higher.

The problem is many people are too accustomed to percentage scores, like you would have in a school setting. This is not what the Alcumus score is, and it is purely an estimation. Similar to when you search on some sites and you get "94% relevant" and such.


how can we prove that time taken for a body thrown horizontally from a height h

is equal to time taken for a body freely falling from the same height

[color=blue]Consider the motion in the vertical direction[/color]


MIT college Support probability articles HCSSiM geometry

who do you think will win? please vote truthfully

I think that Bush will win. Bush had a 5 point lead to Kerry prior to the debate; now he has a 2 point loss. However, Bush still does have a brain, and he probably learned something in that debate with Kerry, so I wager that Bush will do better in the debate and recover his gain.

About 60~70% of those saying they are voting for Kerry say they are really just voting against Bush rather than for Kerry, as opposed to less than 50% that will vote for Bush, and Bush is still up by a couple of points.

And if you look at the electoral vote polls (those done in each state) Bush has over 300 electoral votes. If the election were held today he'd win.

Well...If you saw Frank Wilczek of MIT   the other day,  when  he found out that he won the nobel prize in physics .. he was wearing Kerry/Edwards Button ...So I guess at least in  Cambridge (If not  the Mass. state) bush has no chance :)  (BTW  when asked, he said  something to the effect that Bush/Chenny are not good for scientific research ...)

[quote="Nukular"]And if you look at the electoral vote polls (those done in each state) Bush has over 300 electoral votes. If the election were held today he'd win.[/quote]

Oh really? Where do you get your information? [url=http://www.electoral-vote.com/]This site[/url] doesn't agree with you.

Several sources, but just looking quickly on the web I can find a site

[url]http://www.electionprojection.com/elections2004.html[/url]

whose current poll is based on The Polling Report and RealClearPolitics that has Bush up right now.

I guess it depends which poll you look at, but its definitely close and changes almost daily at this point.

I think what's weird about the Electoral college is that a candaidate can only win in 11 states, but still win the election.

[quote="h_s_potter2002"]I think what's weird about the ELectoral college is that a candaidate can only wim in 11 states, but still win the election.[/quote]

But that is unlikely to happen (or at least be nonrepresentative) assuming a rather even dispersal and that a candidate doesn't win by very large landslides in the states in which he does win.

I don't think it is likely for a candidate to win the electoral college and lose by a lot in the popular vote anyway. (and I emphasize <i>a lot</i>)

this isn't for who you will vote for, its for who you think will win (to the 2 who voted Nader...do you [i]really[/i] think he will win?) :lol:

Not that I support Bush but Kerry is not the type u want for President!

Saw a bumper sticker"[quote]Bush/Cheney 04 - This time elect us [/quote] :)

[quote="Nukular"]Well thats true Bush is definitely not winning Mass, but then again has any republican won mass in recent history?[/quote]

Actually, Reagan won massachusetts in 1984, only 20 years ago.  That was also the year where Reagan got 97.6% of the electoral vote, but only 58.8% of the popular vote.

[quote]Quote:
I think that evreyone is overly critical about George Bush.

I believe that we should all be as critical as possible of the people who make the decisions that influence our life deeply.
[/quote]  So why do people focus so much onGeorge Bush?  Kerry has just as many, if not more faults than Bush.

[quote]Quote:
Is there anything wrong with protecting our country and making the world a better place? We went into Iraq and got rid of Saddam. Isn't that a major accomplishment? he was killing thousands of his own people, poisoning them, torturing them.

There is nothing wrong with protecting our country. Unfortunately, when there is no evidence that another country is in the position to attack our country, then defending ourselves from them does not make sense. While I am glad that Saddam is no longer in power, the current situation in Iraq is not far better, and millions of Iraqis have died in the transition from the former regime to the current. So the deaths of Iraqis caused by Saddam are comparable to the deaths of Iraqis caused by the invasion of American forces.

Also, a real threat to our country is North Korea. There is significant evidence (accepted by the US, the UN, and the international community) that shows they have recently developed nuclear weapons. They are also testing missiles to increase their attack range. It seems hypocritical to run a preemptive war against a country that cannot harm ours while not going to war with a country that can.
[/quote]
Terrorists attacked our country.  We are not going to just sit back and say"Oh shoot.  Oh well, they're not going to do it again" and do nothing about it.  We had to let them know that we are going to protect our country. If we hadn't, there could have been more attacks and more lost lives.  Innocent lives.  The Bush administration felt that Saddam was a threat, so they did something about it.  Better safe than sorry.  Many of the Iraqi deaths were Iraqis that are loyal to Saddam.  Are you saying that we shouldve just left Saddam alone and let him be?  We are also trying to negotiate with North Korea.

[quote]Quote:
Kerry just went to Vietnam, got a scratch, went to the doctor to get a band aid and received a purple heart, came back and made up 'Heroic War Stories' and gets some medals.

Do you believe the proliferations of a harsh anti-democratic, anti-Kerry group? Also, if he was just "scratched" how did he win 3 Purple Hearts? Also, do you choose to disregard the reports of those who were close to Kerry in the war and testify of his bravery? Even if you do not, Kerry is a person who has been to war and experienced it first hand - whether harmed or not. It isn't your place to trivialize the war experience of a combat veteran or to mock his "heroic war stories."

[/quote]
He must have got scratched thrice.  Those people made them up because they were close to him.  He easily could have paid them.  Or he couldve written stories for them too.  And no one says that you have to fight in a war, or even serve in the military, to become president.

[quote]Quote:
If he were elected, all the terrorists would laugh.

Why? Lets look at what terrorists think of the Bush administration. It has promised to catch Bin Laden for the last three years, yet has not been able to accomplish this, world's best military and information or not. We also note that we have invaded a country like Afghanistan to rid it of terrorists, and yet, those with the real power there are war lords who terrorists can ally themselves to. After Iraq's invasion, any terrorist can now enter the country (which has for the most part uncontrolled borders) and fight as resistance/rebel. Essentially, we have provided an entire nation for terrorists to train in. We have also supplied targets for them - American soldiers. We also have provided them with sympathy in the Arab world - sympathy for the innocent Arabs who have died in the invasion of Iraq. To conclude, if I were a terrorist, I would keep in mind the ineffectiveness of the antiterrorist measures of the Bush administration be more scared of a possibly different Kerry administration.
[/quote]
Bush has gotten rid of many of the terrorists leaders.  He has also gotten rid of Saddam, who has been helping them.  Any terrorist could have enterd Iraq before the war, Saddam wasn't going to stop them.  Now we have guards on the borders.  Even though there is some chaos now, the situation will get better.  Its better than having  entire cities of Kurds beeing killed off with mustard gas.  Why would people give them sympathy?  Many Less people hae been killed by US soldiers than by Saddam and terrorists.  Just because someone hates you, does not mean you are doing the wrong thing.  THe French did not go to war with us becasue they had a deal with Saddam about their oil supplies.  As for the Kerry administration, different does not mean better.  I bet the first thing terrorists would have done if Kerry was elected would be another attack to test Kerry.  They already know not to mess with Bush.  We can not have a president that flip flops on issues, one that sends troops into war, then say we shouldn't support them.

[quote]Quote:
. Plus, look at this picture. HOw can we elect Kerry? http://www.ebaumsworld.com/bushkerryfootball.html[/url]

Is this really something that should influence ANYONE's political beliefs? Do you depend on this to determine who represents you for the next four years?
[/quote]

Alright, alright.  THat was a joke.  I guess Kerry supporters cant take jokes.  I dno't no why because expecting Kerry to win the election was a joke.  Haha, just kidding, that was a joke too.

[quote]So why do people focus so much onGeorge Bush? Kerry has just as many, if not more faults than Bush. [/quote]
who sez?

[quote][quote]Quote:
Kerry just went to Vietnam, got a scratch, went to the doctor to get a band aid and received a purple heart, came back and made up 'Heroic War Stories' and gets some medals.  [/quote]

Do you believe the proliferations of a harsh anti-democratic, anti-Kerry group? Also, if he was just "scratched" how did he win 3 Purple Hearts? Also, do you choose to disregard the reports of those who were close to Kerry in the war and testify of his bravery? Even if you do not, Kerry is a person who has been to war and experienced it first hand - whether harmed or not. It isn't your place to trivialize the war experience of a combat veteran or to mock his "heroic war stories." [/quote]
that is the most ignorant post i have ever seen.
where the hell do u get this kind of crock?
the army doesnt give out purple hearts like that.

[quote="SirErnest"][quote]who sez?[/quote]
Who says he doesn't?[/quote]
think about it
kerry supporters are attacking this
bush supporters arent attacking this.
why do u think so?

[quote="Scrambled"][quote][quote]Quote:
Kerry just went to Vietnam, got a scratch, went to the doctor to get a band aid and received a purple heart, came back and made up 'Heroic War Stories' and gets some medals.  [/quote]

Do you believe the proliferations of a harsh anti-democratic, anti-Kerry group? Also, if he was just "scratched" how did he win 3 Purple Hearts? Also, do you choose to disregard the reports of those who were close to Kerry in the war and testify of his bravery? Even if you do not, Kerry is a person who has been to war and experienced it first hand - whether harmed or not. It isn't your place to trivialize the war experience of a combat veteran or to mock his "heroic war stories." [/quote]
that is the most ignorant post i have ever seen.
where the - do u get this kind of crock?
the army doesnt give out purple hearts like that.[/quote]

The US Army gives out Purple Hearts like that.  Many of the people that have fought with Kerry have said that.

A little off topic, but has anyone seen the SNL episode where the Swiftboat Veterans are criticizing Kerry? That part was hilarious. A couple quotes from that.

[quote]When we were going down the river, John suddenly turned to me and started talking about the pollution of burning oil and global warming. He then took off our boat's engine, tossed it into the river, and made us complete the mission on windpower. Of course, then he flip-flopped, and made us fish the engine out of the river and attach it on the boat again."[/quote]

[quote]Once, we were ambushed by two Vietcong. John ran after them, and then came back half an hour later. When we asked him what had happened, he told us that he didn't kill the two men. They had only wanted to be married, so he played priest to the gay marriage ceremony.[/quote]

It was funnier if you had seen it.

[quote]Alright, alright. THat was a joke. I guess Kerry supporters cant take jokes. I dno't no why because expecting Kerry to win the election was a joke. Haha, just kidding, that was a joke too.[/quote]
Okay, that was stupid. You don't need to attack Kerryites in general because you posted the comic in a context that suggested you weren't just joking.

[quote]The US Army gives out Purple Hearts like that. Many of the people that have fought with Kerry have said that.[/quote]
And many have stood by him. Why are they not as trustworthy?

[quote]When we were going down the river, John suddenly turned to me and started talking about the pollution of burning oil and global warming. He then took off our boat's engine, tossed it into the river, and made us complete the mission on windpower. Of course, then he flip-flopped, and made us fish the engine out of the river and attach it on the boat again."[/quote]
LOL that's funny...what's SNL?

I'm glad you posted a reply to my earlier post SirErnest.

[quote]So why do people focus so much onGeorge Bush? Kerry has just as many, if not more faults than Bush. [/quote]

Someone challenged your statement here, and you answered with "Who says he doesn't?" Note that you made this original statement, backed with no evidence, except your vague response. I don't know who has more faults. The reason people focus on George Bush is that he has been president for the past 4 years and some of the most radical decisions this country has made in its existance have been made by his administration. Other reasons for focusing on George Bush: the country has the worst financial record since Herbert Hoover (in a little period of time known as the Great Depression.) There are all excellent reasons for focusing on George Bush's mistakes. Is Kerry faultless? No. Is it possible that Kerry will become a worse president? Possible, but by my analysis, unlikely.

[quote]Terrorists attacked our country. We are not going to just sit back and say"Oh shoot. Oh well, they're not going to do it again" and do nothing about it. We had to let them know that we are going to protect our country. If we hadn't, there could have been more attacks and more lost lives. Innocent lives. The Bush administration felt that Saddam was a threat, so they did something about it. Better safe than sorry. Many of the Iraqi deaths were Iraqis that are loyal to Saddam. Are you saying that we shouldve just left Saddam alone and let him be? We are also trying to negotiate with North Korea.
[/quote]

I agree that terrorists attacked our country. I agree we are not going to just sit back and say "Oh shoot. Oh well, they're not going to do it again" and do nothing about it. I agree that we had to let them know that we are going to protect our country. I agree if we had'nt there could have been more attacks and more lost innocent lives. Completely agree with you, in every respect so far. So does John Kerry. In order to stop the terrorists, we logically attacked the country that we had evidence was aiding terrorists. Where? Afghanistan. And soon after we succeeded in destroying the government there, we quickly moved on to another war, letting the country go to ruin. So, now the quesion is, why Saddam? Well, to quote George Bush: "He tried to kill my daddy." Great, I'm glad our national agenda includes personal vendettas for family revenge. There is and was no evidence whatsoever that Saddam was involved in 9/11 or that he aided terrorists. On the other hand, we have enormous amounts of evidence that Iran has been aiding terrorists. So, why not Iran? Why Iraq? Better safe then sorry? Well, I am deeply sorry for the more than a thousand lives lost in the armed forces, and the innocent civilian lives lost by hostages who are brutally beheaded everyday, and by the millions of innocent Iraqis. Easy for you to say that all the Iraqis lost lives were of Saddam loyalists. Arguably, every one in Iraq had to be a Saddam loyalist openly, to avoid execution. On the other hand, most of them were not. Your argument sounds like a sour-grapes argument - "Eh, too bad they died, but they were bad anyway." Not a responsible way to run an offensive. By the way, much like George Bush, you have bypassed an enormous argument about North Korea. You simply say that we are in peace talks. Why don't we attack though? I'm not going to elaborate, since I already have, and you have not answered it.

[quote]He must have got scratched thrice. Those people made them up because they were close to him. He easily could have paid them. Or he couldve written stories for them too. And no one says that you have to fight in a war, or even serve in the military, to become president.
[/quote]

Okay, if you are going to simply argue reality and make up stories, defending them with - "Oh, he paid them." Okay here's what I believe: George Bush paid the 9/11 terrorists to attack, since his presidency would fail otherwise, without a nation-uniting attack and war. And he paid everybody who knew not to tell. And I'm going to pay 1000 solid democrats to agree with me and make commercials on TV about it. Yeah, there we go.

By the way, your statements about "getting scratched thrice" would offend any veteran, even those who were not injured. War is not something simple to cope with. And while I agree you don't have to be in the military to become president, I don't quite see why the candidate who did not have the courage to go serve in a way is more convincing to me about going to fight for my country. It's like someone who's never risked their life telling you to do something that involves risking your life.

[quote]Bush has gotten rid of many of the terrorists leaders. He has also gotten rid of Saddam, who has been helping them. Any terrorist could have enterd Iraq before the war, Saddam wasn't going to stop them. Now we have guards on the borders. Even though there is some chaos now, the situation will get better. Its better than having entire cities of Kurds beeing killed off with mustard gas. Why would people give them sympathy? Many Less people hae been killed by US soldiers than by Saddam and terrorists. Just because someone hates you, does not mean you are doing the wrong thing. THe French did not go to war with us becasue they had a deal with Saddam about their oil supplies. As for the Kerry administration, different does not mean better. I bet the first thing terrorists would have done if Kerry was elected would be another attack to test Kerry. They already know not to mess with Bush. We can not have a president that flip flops on issues, one that sends troops into war, then say we shouldn't support them.
[/quote]

Bush promised to get rid of the BIGGEST terrorist leader. Well, what happened to that? How's that going to happen with all of our efforts being concentrated in Iraq? Furthermore, no, not any terrorist could enter Iraq before the war. There is no evidence of that, nor of Saddam's aiding terrorists. On the other hand, we have evidence of terrorists entering  Iraq just about every day now. Conclusion: terrorism is worse now than before. I'm not saying that popularity determines what is right and what is wrong. However, I believe, and have explained why Bush is wrong on these issues. I am unaware of whatever oil treaty France had with pre-war Iraq (altought I am curious; perhaps you could give us a link?) Okay, since when do terrorists know not to mess with Bush? Do the hundreds of terrorist acts in the last 6 months mean anything to you? No? Terrorists have no fear of Bush - they have been living their dreams during his administration. What do terrorists want, except attention, a world-recognized cause, monetary backing, and a place to assemble an offensive? Bush has given all this to them. He is literally a terrorist's best
friend. Scary, no?

By the way, Kerry simply said that, given the current information (note that as a senator, he has limited information as opposed to Bush) that a war is an option. However, Kerry's plan for war in Iraq is far better than Bush's, which has empirically been proven - terrible. What do you mean by "not supporting our troops?" Kerry has a plan to succesfully finish the war and bring our troops home. That's the kind of support I would want if I were a soldier.

[quote]Alright, alright. THat was a joke. I guess Kerry supporters cant take jokes. I dno't no why because expecting Kerry to win the election was a joke. Haha, just kidding, that was a joke too.[/quote]

I'm glad jokes are such a big part of your political arguments. (Sorry, that was a joke.)

[quote="SirErnest"]Don't tell me you don't believe them. Who Are you going to believe? Kerry?  He hasn't done much to deny it.[/quote]
[size=150][b]YES HE HAS!!!!![/b][/size]
OMFG
if you believe in them
your a total loon (no offense)
the news repeatedly denounced those and bush conceded they were total bull
and go to [url]http://swiftvets.eriposte.com/[/url]
and
[url]http://swiftvets.eriposte.com/bushcampaign.html[/url]

[quote="tetrahedr0n"]
By the way, Kerry simply said that, given the current information (note that as a senator, he has limited information as opposed to Bush) that a war is an option. [/quote]

Oh, so now Bush's information reliability is an EXCUSE for Kerry now?  Instead of being Bush's handicap?

[quote="tetrahedr0n"]Kerry has a plan to succesfully finish the war and bring our troops home. [/quote]

And that would be...?

Mind you, we're looking for a plan that isn't a fairy tale.  It's easy to concoct a story to tell the people will do everything they hope will happen--the terrorists will be defeated, the war will be over, and everyone would come home.  Plans don't always go the way they were conceived.  You think Bush was actively looking for potential terrorists, pointing at them, and going, "I Want YOU--For the Jihadist Army"?  Kerry has not explained anything recognizable as realistically feasible.

Robinhood3000, please look back to 1968 and tell me how Nixon got elected.


Λοιπόν, καλησπέρα σε όλους. Χθες με πήρε ο Γιώργος τηλ και μου λέει:<<Αντί να βάζουμε διάσπαρτες ασκήσεις δε φτιάχνουμε τέσσερα τόπικ Άλγεβρα , Γεωμετρία , Συνδιαστική, Θεωρία Αριθμών που όμως δε θα είναι όπως τα παλιά τόπικ αλλά θα είναι συγκροτημένα .Δηλαδή :Σε κάθε τόπικ από αυτά τα τέσσερα μετά από συνεννόηση μεταξύ όλων των μελών του <<Ελληνικά>> θα αναλαμβάνει κάποιος κάθε φορά να βάζει ένα βασικό θεώρημα, την απόδειξή του αν είναι στοιχειώδης και στη συνέχεια τρεις-πέντε εφαρμογές του θεωρήματος(με αύξουσα δυσκολία και κατά κύριο λόγο τα θέματα να προέρχονται από Ολυμπιάδες) και τέλος  να βάζει κάποιες στοιχειώδεις αλλά και πιο hard ασκήσεις στα προτεινόμενα >> .Μόλις το άκουσα μου άρεσε πάρα πολύ ως ιδέα .Συνοψίζοντας θα ήθελα να σας ρωτήσω για τους κλάδους στους οποίους θα θέλατε να γράψετε κάτι ώστε άτομα στον ίδιο κλάδο να συννενοούνται μέσω ΡΜ για να μην μπαίνουν τα ίδια πράγματα. Ελπίζω να σας αρέσει ως ιδέα και να την προχωρήσουμε μεθοδικά ώστε να κερδίσουμε όλοι .

ΒΤW, προτείνω να βγάλουμε κάτι σαν moderators σε κάθε είδος μετά από ψηφοφορία ή όποιο τρόπο άλλο θέλετε(αριθμό post,ηλικία ιδιότητα , επάγγελμα κοκ).
Περιμένω απόψεις. Ελπίζω να μην ναυαγίσει αυτή η ιδέα.
Θα τα πούμε

ΓΕΙΑ ΣΟΥ ΣΙΛΟΥΑΝΕ ΕΙΜΑΙ Ο ΑΜΒΡΟΣΙΟΣ ΑΝ ΜΕ ΘΥΜΑΣΑΙ ΤΙ ΚΑΝΕΙΣ?

Πως δε σε θυμάμαι ρε. Αλλά ότι θέλεις για μένα να τα στέλνεις σε Private Message, οκ?

Τι θα συμβεί αν βάλουμε μια άσκηση η οποία θα παραμείνει άλυτη για καιρό;

Ή θα σταματήσει το τόπικ εκεί ή κάποιος θα βάλει καινούρια άσκηση και θα ξεχαστεί η προηγούμενη.

Ας κάνουμε αυτό που λέτε αλλά παράλληλα να βάζουμε και μεμονωμένες ασκήσεις.

Το καλύτερο θα ήταν να ζητούσαμε από τον admin να μας βάλει sub-forums, όπως είναι χωρισμένο το φόρουμ της Ρουμανίας ή της Ισπανίας

Ναι αυτό θα ήταν το καλύτερο. Όταν το είχαμε ζητήσει αρχικά ο mr Vornicu δεν μπορούσε να το πραγματοποιήσει καθώς είμασταν λίγα μέλη έλληνες. Όμως τώρα γίναμε αρκετοί. Κάποιος που τον γνωρίζει ας του το ζητήσει παρακαλώ.Νομίζω ο nickolas τον γνωρίζει

ΜΙΑ ΧΑΡΑ ΣΙΛΟΥΑΝΕ ΟΚ ΓΙΑ ΑΥΤΟ ΠΟΥ ΖΗΤΑΣ ΘΑ ΠΡΟΣΠΑΘΗΣΩ ΝΑ ΛΥΣΩ ΤΗΝ ΠΟΛΥΩΝΥΜΙΚΗ ΑΣΚΗΣΗ ΒΥΕ

ΦΙΛΟΙ ΜΟΥ ΓΙ ΑΥΤΟ ΠΟΥ ΛΕΤΕ ΘΑ ΠΡΕΠΕΙ ΝΑ ΣΥΜΦΩΝΗΣΟΥΜΕ ΟΛΟΙ ΣΤΗΝ ΚΑΤΩΤΕΡΗ ΥΛΗ ΤΙ ΕΝΟΟΩ ? ΠΧ ΣΥΝΔΥΑΣΤΙΚΗ ΚΙ ΕΓΩ ΚΙ ΠΟΛΛΟΙ ΑΛΛΟΙ ΦΑΝΤΑΖΟΜΑΙ ΔΕΝ ΞΕΡΟΥΝ ΑΠΟΛΥΤΩΣ ΤΙΠΟΤΑ ΕΚΤΟΣ ΑΠΟ 1,2 ΞΕΡΟΥΣ ΤΥΠΟΥΣ ΕΔΩ ΘΑ ΠΡΕΠΕΙ ΝΑ ΓΙΝΕΙ ΔΙΔΑΣΚΑΛΙΑ ΣΕ ΕΙΣΑΓΩΓΙΚΑ ΠΟ ΤΗΝ ΑΡΧΗ ΑΠΟ ΚΑΠΟΙΟΝ ΑΨΟΓΟ ΑΥΤΟ ΠΟΥ ΛΕΩ ΕΙΝΑΙ Η ΒΑΣΗ ΓΙΑ ΤΟ ΕΓΧΕΙΡΗΜΑ ΑΡΚΕΙ ΝΑ ΤΟ ΣΚΕΦΤΕΙΤΕ :thumbup:
ΧΑΙΡΕΤΑΙ


Write a simple formula in which you use the variable x only once, such that when any positive integer is substituted for x, the formula is sure to be a prime number. :)

Thanks man, that problem was driving me crazy and now I can finally impress my algebra teacher

[quote="fedja"]This is an old joke. $ 2 \plus{} 0\cdot x$, $ 1 \plus{} 1^x$, and many other similar things are most trivial examples. You can do a non-constant formula: $ 4 \plus{} ( \minus{} 1)^x$ works fine. There are even some crazy formulae like $ [\alpha\cdot 10^{10^{10^x}}]$ with some real $ \alpha$ constructed in a special way that give you infinitely many primes (the problem is that $ \alpha$ is defined using the primes that are guaranteed to exist but cannot be found explicitely).[/quote]

all these work well;
more examples are:

2+1^x works (similar to one already given)

([[[[2^2]^2]^2]^2]....^2])-(x^0)

[[[[[(2*x-1)^x-1]^x-1]!]^1-x]^2]^2]^2]^2]...^2]-1  :huh:  (i derived this wild one myself); someone please convert this to Latex

and etc. Unfortunately, this is not a "simple formula  :oops:

Wow... How do you guys do these? My brain feels like it is gonna expload... :blush:

im only in 8th grade.............
so the derivation is not too difficult (only 'bout 2.5 pages long)
once i learn latex (this is pretty much all i can do  :oops: : $ x$) then i will post the proof
think it through :idea:
 :P

yeah any formula where x doesnt make much of a difference works

I derived a very wild one...
$ 2\times \frac{\frac{d}{dx}\left(\displaystyle\sum_{i\equal{}0}^{\infty}\frac{x^i}{i!}\right)}{(\log_{\pi  i}{(\minus{}1)})^x}$

oh this is sorta funny. (enh kinda) i have some formulas.
1. any prime number + 0*x
2. (any prime number -1) + 1^x
3. any prime number * 1^x

You can derive pretty simple formulae, like $ 2 \cdot 1^x$, or something of the sort.  You could also do something slightly more complicated but still inexplicably simple.  Here's a more complicated simple formula:
\[ \log_{10} 100\sin([2x \minus{} 1] \pi )\]
You can easily derive a formula to get a prime number in this manner.

[quote="Lil whopper"]wow thats strange..
[/quote][/quote]

I think you clicked submit too many times.


Click "View more Emoticons" (which links to http://www.artofproblemsolving.com/wforum/posting.php?mode=smilies) on the left when you write a post.

Or type :sqrt:.  :rt2:, :rt3:, and :rt5: are built-in symbols for the square roots of 2, 3, and 5 respectively.

Note: they will normally appear as the appropriate symbol.  I just turned that feature off for this post.


trigonometry
geometry
angle bisector
exterior angle

In $\triangle ABC$, $AC=6$, $BC=8$ and $\angle BAC=2\angle ABC$. Find the length of $AB$.

Probably trig (double angle formula etc.) is a standard approach, but there is another way that you don't need any knowledge of trig. Can you figure it out?
(That's why I posted this problem here, not Intermediate.)

Draw the angle bisector of $<BAC$ and let it meet $BC$ at $D$. Call the length of $AB$ the variable $x$.
Then $CD=\frac{48}{6+x}$ and $BD=\frac{8x}{6+x}$ by the angle bisector theorem. Since $ADB$ is isoceles, $AD=\frac{8x}{6+x}$.
Applying Stewart's theorem, $x*\left(\frac{48}{6+x}\right)*x+6*\left(\frac{8x}{6+x}\right)*6=\left(\frac{48}{6+x}\right)\left(\frac{8x}{6+x}\right)*8+\left(\frac{8x}{6+x}\right)\left(\frac{8x}{6+x}\right)*8$
Multiply by $(6+x)$ and divide by $8x$ since $x\neq0$.
$6x+36=\frac{48(8)+64x}{6+x}$
$3x^2+4x-84=0$
$(x+6)(3x-14)=0$
$x=\frac{14}{3}$ since $x>0$.

[quote="riddler"]Yes! so My answer is right, it is just the same as yours. But I used similarity. and there was no need for $\text{Stewart's Theorem}$.
It is possible without this hard theorem.[/quote]Can you post your solution? :)

[quote="riddler"]Yes! so My answer is right, it is just the same as yours. But I used similarity. and there was no need for $\text{Stewart's Theorem}$.
It is possible without this hard theorem.[/quote]
riddler is correct

Draw the angle bisector of $\angle BAC$ and let it meet $BC$ at $D$. $ABD$ is isosceles and $\angle BAD=\angle ABD=\alpha$. Exterior angle is sum of remote interior angles and get $\angle ADC=\angle BAD+\angle ABD=2\alpha$ and $\angle DAC=\alpha$ because $BD$ is angle bisector. $\triangle DAC\sim\triangle ABC$.
\[\frac{DC}{AC}=\frac{AC}{BC}=\frac{AD}{AB}\]
$DC=\frac{AC^2}{BC}$

$BD=BC-DC=\frac{BC^2-AC^2}{BC}$

$AD=BD$ because $\triangle ABD$ is isosceles

$AB=\frac{AC\times AD}{DC}=\frac{AC\times\frac{BC^2-AC^2}{BC}}{\frac{AC^2}{BC}}=\frac{BC^2-AC^2}{AC}=\frac{14}{3}$

[color=blue]Theorem: In a triangle ABC, if <C=2<B, then: $c^2=b^2+ab$      [/color]

Proof. Construct the angle bisector of <C, and let the intersection of bisector and AB be D.
Hence <ACD=<BCD=<B. Since <A=<A, so:
\[\triangle{ACD} \sim \triangle{ABC}\]
Thus: $\frac{AC}{CD}=\frac{AB}{BC}$.

$\frac{b}{CD}=\frac{c}{a}$

<BCD=<B gives $CD=BD$. Use the Angle Bisector Theorem:
\[\frac{b}{a}=\frac{c-BD}{BD}\]
\[BD=\frac{ac}{a+b}\]

Plug it in we get:
\[c^2=b^2+ab\]

I prefer to call it the $\text{Double Angle Theorem}$.

[quote="shobber"][quote="riddler"]no need for Double angle theorem aswell, you see shobber that this is a Getting Started problem and only basic stuff should be the solution.[/quote]
Well, this theorem is also the solution to this problem,and it is basic......
By the way, is it really a well-known theorem? I mean, I created the name Double angle theorem......[/quote]
I dont think it is that well known, I have never heard of it.

[quote="riddler"]slightly different from my method but still the same answer. :D[/quote]
Show your method please

[quote="gunes"]another[/quote]
well it is the same as Gunes I proved these to triangles similiar, so then found $AB$. Simple.


Let $ a,b,c > 0$ and $ k\ge 0$ such that $ a \plus{} b \plus{} c \plus{} k \equal{} (k \plus{} 3)abc$. Show that:
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge 3\]

[quote="keyree10"][quote="Abdek"]From $ \text{AM \minus{} GM Inequality}$ we have $ a \plus{} b \plus{} c \plus{} k \ge (k \plus{} 3)\times \sqrt [k \plus{} 3]{abc} \implies abc \ge 1$
[/quote]

That assumes that $ k$ is an integer doesn't it ?[/quote]

It's Weighted AM-GM inequality It's more general  :P

[quote="Abdek"]From $ \text{AM \minus{} GM Inequality}$ we have $ a \plus{} b \plus{} c \plus{} k \ge (k \plus{} 3)\times \sqrt [k \plus{} 3]{abc} \implies abc \ge 1$

Thus :

$ \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \ge3\left( \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\right) \equal{} 3\left[k \plus{} 3 \minus{} \frac {k}{abc}\right] \ge 9$

And we are done ![/quote]

Ok, it's easier than i thought. The following one is also true:

Let $ a,b,c>0$ and $ k\ge 1$ such that $ a\plus{}b\plus{}c\plus{}k\equal{}(k\plus{}3)abc$. Prove that:

\[ \frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\le 3\]

[quote="Pain rinnegan"]

Let $ a,b,c > 0$ and $ k\ge 1$ such that $ a \plus{} b \plus{} c \plus{} k \equal{} (k \plus{} 3)abc$. Prove that:
\[ \frac {1}{\sqrt {ab}} \plus{} \frac {1}{\sqrt {bc}} \plus{} \frac {1}{\sqrt {ca}}\le 3\]
[/quote]

Any idea for this one?

I have an idea , let S=1\rac(ab)+1\rac(ac)+1\rac(bc) . we have a+b+c+=3abc + k(abc-1) , with abc >=1 . this equality prove that if the values of k increases , the value of a , b and c  increases to  , which make the value of S decreases , then S be maximum when abc=1 , or a+b+c=3 .  and since S =< S_{max}   we need to prove that S_{max} remains smaller than 3  (when a+b+c=3  and abc=1 ) we have S_{max} = root(a) + root(b) +root(c)  and by C-S   : (S_{max})²=< 3(a+b+c) = 9 , thus  S_{max}=< 3   as desired . Thk's .

[quote="Pain rinnegan"] The following one is also true:

Let $ a,b,c>0$ and $ k\ge 1$ such that $ a\plus{}b\plus{}c\plus{}k\equal{}(k\plus{}3)abc$. Prove that:

\[ \frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\le 3\][/quote]

Suppose that  $k=7.$  then we have \[a+b+c+7=10abc\ \ \ \Longrightarrow \ \ \ \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{7}{abc}=10\]

[b]But what if [/b] \[c\to \infty , \ \ \ \frac{1}{ab}\to 10 \ \ \ ?\]

For k=7, this counterexample works:

\[a=b=\frac{10}{31}, \ \ \ c=\frac{7347}{39}\]


[hide="Possible answer"]From the descriptions of 1. and 2., its obvious that A is thionyl chloride or phosphorus oxychloride. But from descriptions 6., 7., and 8. A seems to be thionyl chloride.[/hide]

[quote="valeriummaximum"]1.When SO2 and PCl5 react together,a sub 'A' is formed.
2.'A' is fuming liquid which dissolves in water to give a pungent smelling liquid which turns blue   litmus red
3.5.95 g of 'A' was dissolved in water and the solution is made upto i litre.
4.dil HNO3 and excess of AgNO3 soln was added to 100 ml of this soln and the ppt was collected,dried and weighed.
5.the wt of the precipitate was 1.435 gm.
6.Dil HCl foll by BaCl2 soln was added to the second 100 ml portion of the solution.
7.No ppt is formed.
8.But on addition of H2O2 soln to the mixture,a white ppt is formed.
9.After collecting and drying,the ppt weighed 1.165 gm.

Deduce the emperical formula of 'A'.[/quote]
is it SOCl2???only a guess!!!! dei Ranga post the answer no???and how to go about the steps??? :lol:

Jagadamba singh??? nice name :rotfl:  for mechanisms , there is a book by jerry march. havent seen much of it tho, so dunno how good it is.

maybe srinivas has made an allibi fer his own name  :P   :rotfl:
And btw srinivas see the pm i sent you.  :D

[quote="srinitrs"][quote="valeriummaximum"]see your personal message...........................
SRINIVAS[/quote]
you didnt send me even one i have a doubt speaking about SOCl2 in SNi mechanism  when the rearrangement is taking place anyway a carbocation is going to take place then wont you get racemisation why does he say no inverion in configuration???for the mechanism i referred Jagadamba singh???is there any good book to refer for mechanism of organic reactions??? :D[/quote]
yes it ian an SNi mechanism and since it is an Sni there is no inversion(twice inversion)

HEY madness ok ok i cant fool you obviously da there is no book such as jagadamba singh for your kind information we people in chennai here chumma for fun think of names and keep speaking about it to kindle the curiosity of others just like what prasad did in the indian forum ie physice olympiaci??? :rotfl:  :rotfl:  :rotfl:

[quote="VARUNARASU"][quote="srinitrs"]
HEY madness ok ok i cant fool you obviously da there is no book such as jagadamba singh for your kind information we people in chennai here chumma for fun think of names and keep speaking about it to kindle the curiosity of others just like what prasad did in the indian forum ie physice olympiaci???
[/quote]
unfortunately for you, your bluff wont work, coz there is a book by jagadamba singh. my friend possesses a copy and i saw it rite now. its really not worth readding tho.  :ninja:[/quote]
Oh is that so i didnt know about it long time back madness told me jokingly i thought he was bluffing!!!!so what book do you prefer for studying organic???

for jee, i read arihant. and bcoz i wanna learn organic chem, morisson, solomons etc.


This has been driving me nuts for like 20 minutes. It doesn't seem like enough information is given. I got all the others, I just can't figure out #35 and 36.
[img]http://i48.tinypic.com/230avt.jpg[/img]

[hide="35"]First, note that since $ YU$ and $ YT$ are angle bisectors, we know that $ \angle UYZ \equal{} \angle UYW$

We also know that $ \angle WYX \plus{} \angle ZYW \equal{} 180$. So, since $ \angle ZYW \equal{} 2 \angle UYZ$, we have that

$ \angle WYX \plus{} 2 \angle UYZ \equal{} 180$

$ 2(12b \plus{} 7) \plus{} 2(9b \minus{} 1) \equal{} 180$

$ 21b \plus{} 6 \equal{} 90$

$ 21b \equal{} 84$

$ b \equal{} 4$

Therefore, $ \angle UYW \equal{} \angle UYZ \equal{} 9(4) \minus{} 1 \equal{} \boxed{35^{\circ}}$.[/hide]

[hide="36"]$ \angle ZYW \equal{} 90 \equal{} 2 \angle ZYU \equal{} 2(13a \minus{} 7)$

$ 90 \equal{} 26a \minus{} 14$

$ 104 \equal{} 26a$

$ \boxed{a \equal{} 4}$[/hide]

Hint for #35:

Note that since XYZ is a straight line, it has 180 degrees. Furthermore, note that $ \angle WYX \plus{} \angle UYW \plus{} \angle ZYU \equal{} \angle XYZ$.

[hide="Hints"]35: They have to add to $ 180^\circ$. Also note that $ \angle UYW\equal{}\angle ZYU$, so you can find $ b$.

36: If $ \angle ZYW\equal{}90^\circ$, what must $ \angle ZYU$ necessarily be...?[/hide]

Thanks guys.
I thought about the 180 degrees thing, but I wasn't sure if I could use that. I didn't know I could throw in degree numbers with all those other numbers because I didn't know if they'd be compatible. I don't know why I thought that since it was using the "angle measure" symbol. For some reason, in my head, I separated the numbers they were giving me from actual degrees numbers.

Thank you very much.


If a,b,c are roots of $x^3-rx+20=0$, r E IR, find $a^3+b^3+c^3$

Since we have $a^3=ra-20,b^3=rb-20,c^3=rc-20$, we get

$a^3+b^3+c^3=r(a+b+c)-60=\boxed{-60}\ (\because from \ vieta's\ formula, a+b+c=0)$

In this case, it says the sum of the roots, $a+b+c$ is $0$ since the coefficient of $x^2$ is zero.

[quote="white_horse_king88"]What does vieta's say?[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=40153 ;)

From the equation given, we know that it can be written as $(x-a)(x-b)(x-c)=0$, and thus we have
$a+b+c=0$, $abc=-20$, $ab+bc+ac=r$.  Thus
$a^3+b^3+c^3=(a+b+c)^3-6abc-3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2=0-3(a+b+c)(ab+bc+ac)+3abc=0-0-60=-60$
Btw, very nice solution, kunny!

Thanks, K81o7.  :)  You can also use the following equalities.

\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\]

\[(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)\]

kunny

[quote="Rushil"]Obviously, the solution using vieta's formulae is the better one!!! :rotfl:  :rotfl:[/quote]
Well technically, they both use it.  It's a common formula which is easily derived.  The solutions are in essence the same except kunny's is a lot quicker :D so yes it is better.

I didn't know this was called Vieta's formula, but i used
if the roots are a, b and c
$(x-a)(x-b)(x-c)=x^3-(a+b+c)x+(ab+bc+ac)x-abc$


topology
advanced fields
advanced fields unsolved

In my topology course, we defined the degree of a continuous map $ f: S^1\rightarrow S^1$ where $ S^1\equal{}\{z\in \mathbb{C}: |z|\equal{}1\}$.
If $ q: \mathbb{R}\rightarrow S^1; t \rightarrow e^{2i\pi t}$,  there is a continuous mapping (unique up to an integer) $ g: [0,1]\rightarrow \mathbb{R}$ such that \[ f\circ q_{|[0,1]}\equal{}q \circ g\]
The degree of $ f$ is then \[ \deg(f)\equal{}g(1)\minus{}g(0).\]
Can someone prove that if $ f,g : S^1 \rightarrow S^1$ are continuous then \[ \deg{(f \circ g)}\equal{} \deg{(f)}\cdot \deg{(g)} .\]
I wonder if this is true without any further hypthesis like differentiability. Thank you

In topology, differentiability has no meaning  :wink:

The theorem is true in general:
Note that the degree you define is exactly the winding number (by definition). As two loops with same winding number are homotopic, you may homotop $ g$ into $ w: t \mapsto e^{2\pi i t \cdot \mathrm{deg}(g)}$. Now $ f \circ w$ is nothing else than doing $ f$ exactly $ \mathrm{deg}(g)$ times, the result follows.

That's a quick and very useful answer ! I didn't know that two loops are homotopic iff they have same winding number. I also didn't know this terminology. Thank you for your clarifying reply.


Asymptote
LaTeX
geometry
3D geometry
tetrahedron

I have produced 3D images in Asymptote, which show up for me
correctly as pdf files.  But, when I include them in Latex files,
they don't produce an error, but there's just a blank space
where they should appear in the final product.  Non-3D images
have no such difficulty.  Any ideas what could cause this?

Give a minimal example, please (TeX code including 3D asy code). Also, tell which version of Asymptote you are using. I'll try to run it on my PC and see if I can reproduce the problem.

Here's what I ran in Asymptote:

access settings;
if(!settings.multipleView)
 settings.batchView=false;
settings.tex="pdflatex";

eval(quote{
defaultfilename='rectangularprism';

size(150); import three; import graph; defaultpen(linewidth(0.8));
currentprojection = orthographic(2,3,2.5);
picture Tetrahedron(){
  picture a;
  draw(a,(0,2,0)--(0,0,0)--(3,0,0), dashed);
  draw(a,(3,0,0)--(3,2,0)--(0,2,0));
  draw(a,(3,0,1)--(3,2,1)--(0,2,1));
  draw(a,(0,2,1)--(0,0,1)--(3,0,1));
  draw(a,(3,0,0)--(3,0,1));
  draw(a,(3,2,0)--(3,2,1));
  draw(a,(0,2,0)--(0,2,1));
  draw(a,(0,0,0)--(0,0,1),dashed);
  return a;
}
add(currentpicture,Tetrahedron());

});

It does produce for me a good pdf file (attached).

Yes, it produces a good pdf file for me too :). So, where does the trouble begin?

The trouble comes with including the image in a Latex file.
It doesn't produce an error or warning, but it just doesn't
show up.  There's just a blank white space where it should be.

I'm using the \includegraphic command.  And, all the 2D images
do show up fine--just not 3D ones.

Any ideas?

OK, make sure that

a) you have the latest version of asymptote
b) you have updated asymptote.sty and asycolors.sty in your latex directory to the newest versions
c) you are using Acrobat Reader 8.0 or higher.

I remember now that I had a similar problem when they just switched to the new rendering of 3D images and the reason it didn't work for me was (b) (it is easy to forget and, alas, it doesn't happen automatically).

Let me know if it helps.

Thanks very much for your help.  (c) checks out--I'm using Acrobat 8.0.
As for the asymptote and asymptote.sty, I updated to 1.65 (from 1.57,
I think).  But, that didn't solve the problem :(
Just to clarify, I'm not trying to run the asymptote code inside a latex
file, just to include the pdf file in the latex file.  I'm doing this in Texnics
Center, in case that might matter.

Any other ideas?

Ashley

I was looking at the documentation for the [b]three[/b] package:
[url]http://asymptote.sourceforge.net/doc/three.html[/url]

If you scroll down a little less than halfway, there's a section that starts "There are four choices for viewing 3D Asymptote output:...".

I think you want option #4 on that list (render the image as two-dimensional PDF).  So you want to try to set the option "render=0" in your asymptote compilation.  (You should be able to do this by adding "-render 0" to your command line arguments.)  Basically, I think that you need to strip out the fancy 3D-rendering stuff in order to get a version of the diagram that will import correctly into a 2D document.

The other option is to switch to an older version of asymptote.  I have version 1.36 (too lazy to update) and your file works fine for me. :)

(Of course, I'm just guessing.)

Here you are cheating me: I'll never believe that
[code]
access settings;
if(!settings.multipleView)
settings.batchView=false;
settings.tex="pdflatex";

eval(quote{
defaultfilename='rectangularprism';

});
[/code]
is something you wrote yourself. That is the standard preamble produced by LaTeX when extracting the asy code.

Now, the point is that it is a preamble that was produced by [b]old[/b] asymptote.sty. With the new one, you should have
[code]
access settings;
if(!settings.multipleView)
settings.batchView=false;
settings.tex="pdflatex";

// Beginning of Asymptote Figure 1
eval(quote{
defaultfilename='rectangularprism';
if(settings.render < 0) settings.render=4;
settings.inlineimage=true;
settings.embed=true;
settings.outformat='';
settings.toolbar=false;

 size(150); import three; import graph; defaultpen(linewidth(0.8));
currentprojection = orthographic(2,3,2.5);
picture Tetrahedron(){
picture a;
draw(a,(0,2,0)--(0,0,0)--(3,0,0), dashed);
draw(a,(3,0,0)--(3,2,0)--(0,2,0));
draw(a,(3,0,1)--(3,2,1)--(0,2,1));
draw(a,(0,2,1)--(0,0,1)--(3,0,1));
draw(a,(3,0,0)--(3,0,1));
draw(a,(3,2,0)--(3,2,1));
draw(a,(0,2,0)--(0,2,1));
draw(a,(0,0,0)--(0,0,1),dashed);
return a;
}
add(currentpicture,Tetrahedron());

});
// End of Asymptote Figure 1
[/code]
The difference may seem not that great but the pdf file produced from the first code (the one you posted) does not include correctly and the pdf file produced from the second one does.

So, I suggest that you make a new test TeX-file with asy code in it, execute the build command, look at the automatically created .asy file, check that its preamble looks like the second code, not like the first one, run Asymptote on that file (from TC or outside of it, does not matter), execute the build command again, and view the output (or just execute the asy file I posted instead of what you had and see if it includes OK).

brilliant.  absolutely brilliant.
Actually, I just got the preamble from a file someone shared with me.
I substituted the one you included, and that did the trick.
Thank you very, very much.

Ashley

[quote="ashleyahlin"]brilliant.  absolutely brilliant.
[/quote]

I suspect fedja hears this a great deal
[quote="ashleyahlin"]
Thank you very, very much.
[/quote]

That too.


Let M be a midpoint of the side AC of an acute-angle triangle ABC; $\omega$ is a circle with diameter BM. Let D and E be points of intersection $\omega$ with AB and BC respectively. Let P is a point of intersection of tangents to $\omega$ through D and E. Prove that PB=PC.

We had this problem discussed in an equivalent form at http://www.mathlinks.ro/Forum/viewtopic.php?t=16779 , with four solutions.

  Darij


consider the following scenario: let's say the AMC 12B has more qualifiers for AIME than AMC 12A.  However, only like 20-50 people got more than 130 on the 12B.  Ignoring any differences there might be between AIME and alternate AIME, is this possible to produce two different USAMO cut off based on the AMC 12 test??

Or even, let's say AIME and alternate AIME has significant difficulty gap too, will there be 4 different USAMO cut off??

Because I (actually the So. Cal. geniuses and me) thought it is a lot harder to get high scores on the 12B, but the first 11-14 questions are easier to answer.


[hide]By AM-GM on $n,n-1,\ldots,2,1$,

$\frac{n(n+1)}{2n}\ge\sqrt[n]{n!}$
$(\frac{n(n+1)}{2n})^{n}\ge n!$
$(\frac{n+1}{2})^{n}\ge n!$

By the equality case of AM-GM, equality occurs if and only if all our terms are equal. Clearly, this is not the case when $n>1$, hence

$n!<(\frac{n+1}{2})^{n}$, as desired.
[/hide]

[quote="drunner2007"]Prove that for all positive integers greater $n>1$,

${ n!< \left( \frac{n+1}{2}\right)^{n}}$[/quote]
[hide="induction"]
Base Case: n=2.
2<2.25
Inductive step.
$k!<(\frac{k+1}{2})^{k}$
$k!(k+1)<(\frac{k+1}{2})^{k}(k+1)$
Now if we divide the RHS by 2, the desired inequality is proven, and clearly 2*RHS>RHS, since k is positive.

Thus our induction is complete[/hide]

I don't see it diophantient
[hide]

wouldn't the inductive step be
assume:
$k! < (\frac{k+1}{2})^{k}$

prove
$k!(k+1) < (\frac{k+2}{2})^{k+1}$

you appear to be proving it for $(\frac{k+1}{2})^{k+1}$
[/hide]


Can anyone explain the term like base of ___ on the problem because I don't get those questions by myself.

Please use example problem as one in Problem Solving Marathon if you don't mind.

Thank you.

Take comp sci and base problems will make a lot of sense.

Basically, numbers that people are used to dealing with are in base 10. This means that, for example, the number:

34601

Is equivalent to (3 * 10^4) + (4 * 10^3) + (6 * 10^2) + (0 * 10^1) + (1 * 10^0) because it is a base 10 number system.

Written in base 9, that number means  (3 * 9^4) + (4 * 9^3) + (6 * 9^2) + (0 * 9^1) + (1 * 9^0). In general, written in base b, it is (3 * b^4) + (4 * b^3) + (6 * b^2) + (0 * b^1) + (1 * b^0). Note the following:

Base b must be a positive integer.
Base b must be GREATER THAN (not greater than or equal to) the largest digit. This means that 6543216 makes  no sense.

So basically, take this simple question.

Show that there exists no such b such that 5610 = 156b.

In base 10, 156b = b^2 + 5b + 6.

b^2 + 5b + 6 = 56
b^2 + 5b - 50 = 0
(b+10)(b-5) = 0
b = 5, -10

b must be positive, so b=5. However, it's 156b, so b :ge: 7 is the domain for b. QED.


function
limit
calculus
calculus computations

Let $f:\mathbb{R}^+_0 \rightarrow \mathbb{R}^+_0$ be a strictly decreasing function.

(a)   Be $a_n$ a sequence of strictly positive reals so that $\forall k \in \mathbb{N}_0:k\cdot f(a_k)\geq (k+1)\cdot f(a_{k+1})$
Prove that $a_n$ is ascending, that $\displaystyle\lim_{k\rightarrow +\infty} f(a_k)$ = 0and that $\displaystyle\lim_{k\rightarrow +\infty} a_k =+\infty$

(b)   Prove that there exist such a sequence ($a_n$) in $\mathbb{R}^+_0$ if you know $\displaystyle\lim_{x\rightarrow +\infty} f(x)=0$.

$f(a_k) \geq \frac{k+1}{k} f(a_{k+1}) = (1+\frac{1}{k})f(a_{k+1}) > f(a_{k+1})$. Since $f$ is strictly decreasing, we have $a_k < a_{k+1}$.

$f(a_k) \geq \frac{1}{k} \cdot (k+1)f(a_{k+1}) \geq \frac{1}{k} \cdot (k+2)f(a_{k+2}) \geq \cdots \geq \frac{1}{k} \cdot (k+m)f(a_{k+m})$
we have $f(a_{k+m}) < \frac{k}{k+m} f(a_k)$. Let $m\to\infty$, we get $f(a_{k+m})\to 0$ which is that $\lim_{n\to\infty} f(a_n) \to 0$...

if $\{a_n\}$ is upper-bounded, then $\{f(a_n)\}$ is bounded too. suppose $a_n<M$ then $f(a_n) > 1/M$, it contradicts with $f(a_n) \to 0$. thus $a_n$ is not upper-bounded and since it is increasing, we know that $a_n\to \infty$

Did I miss something or is the problem wrong?

(b)
suppose have $f(a_k)$, we are now going to find $f(a_{k+1})$.
it is required that $f(a_{k+1}) \leq \frac{k+1}{k} f(a_k)$, in fact, we can find $f(a_{k+1}) < f(a_k)/2$.

since $f(x) \to 0$, thus for $\epsilon > 0$ there exists $X>0$ such that $|f(x)|<epsilon, \forall x>X$. Take $\epsilon = f(a_k)/2$ here and find a $a_{k+1} > X$.

corrected the problem. I had a misprinted paper apparently :D


I wanted to know what is everyones personality here.

How many of you have a stutter problem? And how many are too shy to speak in person?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=88374]This thread[/url] is a much more interesting poll.

Your questions appear focused on the introvert/extrovert axis, and introversion is certainly very common here.


(1+3+5+7+9...)/(2+4+6+8+...)

Call the top A

Call the bottom B

B = 2A + 4(A+B)

-3B = 6A

A/B = -1/2

WHY?!?!  This is killing me.

Someone please point out my error?  Or explain it.

How would you define $\infty-\infty$? :wink:

[quote="Hamster1800"][quote="minsoens"]you would seem to think that if $A=1+3+5+\cdots$, $B=2+4+6+\cdots$, then $A+B=1+2+3+4+5+6+\cdots$ and $B = 2+4+6+\cdots = 2(1+2+3+\cdots) = 2(A+B)$. But it really depends on how you define $A$ and $B$. those definitions are ambiguous; if both sequences go on forever, the fraction approaches $1$ and that is [hide="because"]$A = n^{2}$, $B=2\left(\frac{n(n+1)}{2}\right)=n(n+1)$[/hide][/quote]

That depends on how you're defining it. I could make a definition such that the limit was $0$ if I wanted (let B be the complete sum and A be $n^{2}$, the limit is 0, or more rigorously ($A_{n}$ and $B_{n}$ denote the sum of the first n terms.)

$\lim_{n\to\infty}(\lim_{m\to\infty}\frac{A_{n}}{B_{m}})= 0$[/quote]oh hmm..  :blush: ok so there is no one particular answer for this one, unless it is more specifically defined.. (?)

Note that by defining B = 2(A+B), you still get A/B = -1/2.

Sorry about my weird formatting.  I was working off a generalization.

Also, I think the reason is because both sums are divergent.  Thus the answer depends on the approach, since it is equal to infinity/infinity.

Alright you guys, thanks.

Well, the fact that it is equal to 0 or 1, as you guys "showed", is less puzzling than that it is equal to -1/2, lol.

So i guess there is no answer to this.


we have 5 A and 5 B... How many triad B groups can you get ?  ( For ex:   [size=150][u]BB[/u]AAA[u]B[/u]AA[u]BB[/u], AAA[u]BBB[/u]A[u]B[/u]A[u]B[/u], A[u]BB[/u]AA[u]B[/u]A[u]BB[/u]A[/size], etc... )

the number of ways to divide 5 B's into 3 ordered groups is 6. at least one $ A$ is required between groups 1 and 2 and one between groups 2 and 3, so we only need to count the number of ways that the remaining 3 $ A$'s can be distributed among 4 ordered bins (the bins being the 4 spaces around the 3 $ B$ groups). this is just $ \binom{6}{3}$. so the answer is $ 6\cdot \binom{6}{3} \equal{} 120$