From 11 positive integer scores on a 10-point quiz, the mean is 8, the median is 8, and the mode is 7. Find the maximum number of perfect scores possible on this test.

That can't be correct because the mode must be $ 7$, not $ 10$.

If there is 11 scores, then the total of those scores must be 88. We want as many 10s as possible, so you can try a few combinations. There has to be at least 2 7s, but as noted before, that didn't work. So we try 3 7s, but that would result in only 3 7s and a whole lot more 10s (use the 8s and 9s to get to a units digit of 8.) Next we try 4 7s, but that makes alot of 10s, more than 7s!. Finally, we try the maximum, 5 7s. That makes 35, and you add a few 8s to get closer to a units digit of 8. So that makes 43, 51, 59, aha! We can add a nine to finish it up! 68, add 10s, 78, 88 would make the total of the possible scores. Now we need to know if the numbers add up to 11. Theres 5 sevens, 3 8s, 1 9, and 2 10s. 5+3+1+2 is... 11! Thus the correct answer is 2.

Total no of integers is 11.Mean is 8 , so sum of all 11 integers is 88. Median is 8,so there are 5 integers greater than 8, and 5 integers smaller than 5. So,frequency of  score of 9 or 10 may be 2 or 3 each(as total no is 5). Now Mod is 7, so no of frequency of mod number 7 Will be greater than 3, it may be 4 or 5. If frequency of Mod no. 7 is 5 then sum of all 5 integers greater 8 will be 88-5*7-8=45, it is only possible with 5 nos of 9 but as Mod is 7 so, 5 nos of score 9 is not possible. If frequency of modal no. 7 is 4 then with one number between 0 to 6 the sum of the numbers greater than 8 may be 88-4*7-0-8=52 or  88-4*7-1-8=51or 88-4*7-2-8=50 or 88-4*7-3-8=49or 88-4*7-4-8=48,or 88-4*7-5-8=47 or  88-4*7-6-8=46, so out of set of numbers {52,51,50,49,48,47,46} it is only possible make by total 5 nos either 9 or 10 is 3*10 + 2*9. Hence maximum possible perfect score , that is "10" is 3.

[hide=Solution]Note we want the maximum perfect scores, but the mode is $7$, so we can have at most $4$ from the last two constraints. Now we consider the first constraint. So far, our sum is $7*5+8=43$. We have total sum as $88$, so $45$ as the sum of last $5$. If we have $4$ $10$'s, we have a $5$ left, which should be less than the median, which cannot happen. If we have $3$ tens, we get $15$ as the sum of the other two. Notice this is actually possible because there are three tens now, so we can have $4$ $7$'s, and making the sum for the other two greater.[/hide]

[quote=math154]Since the median is $ 8$, there can be at most $ 5$ perfect scores. There are also at least $ 2$ sevens, so we have $ 8\cdot11\minus{}8\minus{}5\cdot10\minus{}2\cdot7\equal{}16$ more to distribute to $ 3$ of the lowest $ 5$ scores. We can have $ 7\plus{}6\plus{}3\equal{}16$, for example, so $ \boxed5$ perfect scores is achievable.[/quote]

hide the text when presenting a solution!!!


Find $ \displaystyle\sum_{n \equal{} 1}^{\infty} \frac {n^6}{2^n}$.

Here is an ugly solution:
Consider $ f(x) \equal{} \frac {1}{1 \minus{} x} \equal{} \sum_{n \equal{} 0}^{\infty} x^n$, which is valid whenever $ |x| < 1$.
Differentiate f and multiply the result by x and then differentiate both sides and multiply both sides of the result by x. Do this until you get n^6 inside the sum. Now set $ x \equal{} \frac {1}{2}$ and you should get the answer.

After messy differentiations, I got:
$ \sum_{n \equal{} 1}^{\infty} n^6 x^n \equal{} \minus{} \frac {x^6 \plus{} 57x^5 \plus{} 302x^4 \plus{} 302x^3 \plus{} 57x^2 \plus{} x}{(x \minus{} 1)^7} \implies \sum_{n \equal{} 1}^{\infty}\frac {n^{6}}{2^{n}} \equal{} 9366$

Define $ F(k) \equal{} \sum_{n \equal{} 1}^{\infty}\frac{n^{k}}{2^{n}}$

Observe that, using an index shift:

$ F(k) \equal{} \sum_{n \equal{} 0}^{\infty}\frac{(n\plus{}1)^{k}}{2^{n\plus{}1}} \equal{} \frac{1}{2} \plus{} \frac{1}{2} \sum_{n \equal{} 1}^{\infty}\frac{(n\plus{}1)^{k}}{2^{n}}$

$ \equal{} \frac{1}{2} \plus{} \frac{1}{2} \sum_{n \equal{} 1}^{\infty} \sum_{j\equal{}0}^k \binom{k}{j} \frac{n^j}{2^{n}} \equal{} \frac{1}{2} \plus{} \frac{1}{2}  \sum_{j\equal{}0}^k \binom{k}{j} \sum_{n \equal{} 1}^{\infty}  \frac{n^j}{2^{n}}$

$ \equal{} \frac{1}{2} \plus{} \frac{1}{2}  \sum_{j\equal{}0}^k  \binom{k}{j} F(j)$

$ \equal{} \frac{1}{2} \plus{} \frac{1}{2} F(k) \plus{} \frac{1}{2}  \sum_{j\equal{}0}^{k\minus{}1}  \binom{k}{j} F(j)$

Now, solving for $ F(k)$, we get:

$ F(k) \equal{} 1 \plus{} \sum_{j\equal{}0}^{k\minus{}1}  \binom{k}{j} F(j)$

Now, if you know $ F(1)$, you can find $ F(2)$, then etc.

If you wanted to do things that way, it would be easier to define

$ G(x) \equal{} \sum_{k \ge 0} \frac {F(k)}{k!} x^k \equal{} \sum_{k \ge 0} \sum_{n \ge 0} \frac {(nx)^k}{2^n k!}$

and exchange the order of summation, getting

$ G(x) \equal{} \sum_{n \ge 0} \frac {1}{2^n} e^{nx} \equal{} \frac {2}{2 \minus{} e^x} \equal{} \frac{2}{1 \minus{} (e^x \minus{} 1)}$

which means $ F(k)$ is exactly twice the rather curious sequence [url=http://www.research.att.com/~njas/sequences/A000670]here[/url].  I have yet to figure out a direct explanation for this.

Interesting. I just meant to offer a non-calculus approach.

So wait -- how do you derive F from your simplified form for G?

Multiplying by $ 2 \minus{} e^x$ gives your recursion.  Anyway, differentiating formal power series doesn't have to be "calculus."  It has a perfectly straightforward combinatorial meaning, or you can just view it as an abstract linear operator satisfying certain identities.


algebra
polynomial
calculus
function
derivative
calculus computations

x^4 - 2x + 3 = 0 : how many reel zeros has the given equation.

$ f(x)\equal{}x^4 \minus{} 2x \plus{} 3 \equal{} 0$
$ f'(x)\equal{}4x^3\minus{}2$

$ f'(x)\equal{}0 \Rightarrow 4x^3\equal{}2 \Rightarrow x\equal{}\frac {1}{\sqrt[3]{2}}$
Therefore $ \frac {1}{\sqrt[3]{2}}$ is minimum of the function.
Hence there are no real root.

Calculus is not necessary.

$ x^4 \minus{} 2x \plus{} 3$

$ \equal{} x^4 \plus{} ( \minus{} x^2 \plus{} x^2) \minus{} 2x \plus{} \left( \frac {1}{4} \plus{} 1 \plus{} \frac {7}{4} \right)$

$ \equal{} \left( x^4 \minus{} x^2 \plus{} \frac {1}{4} \right) \plus{} \left( x^2 \minus{} 2x \plus{} 1 \right) \plus{} \frac {7}{4}$

$ \equal{} \left( x^2 \minus{} \frac {1}{2} \right)^2 \plus{} (x \minus{} 1)^2 \plus{} \frac {7}{4}$

Since squares are nonnegative, the function value cannot be less than 7/4, and thus cannot be 0.

good job. that solution is better than using derivative
thank you :lol:

$ f(x)\equal{}x^4\minus{}2x\plus{}3$
Let $ x\in (0,1)$. It follows that $ 3\minus{}2x>0$ for this entire interval, as $ 2x<2$.

$ 2x\equal{}x^4\Rightarrow x_i\equal{}\sqrt[3]{2}$
$ f(1)\equal{}2, f(x_i)\equal{}3$

Since $ 4x^3\minus{}2>0$ for all $ x\in (1,x_i)$, our function is strictly increasing on the interval $ (1,\infty)$. The only other possible location of a zero is for some $ x\le 0$; thus we may observe that $ f(\minus{}x)\equal{}x^4\plus{}2x\plus{}3$ is strictly positive for positive $ x$ implying $ x\le 0\Leftrightarrow f(x)\ge 3$.

$ f(x)$ has no zeros on the interval $ (\minus{}\infty, 0]$ or on $ (0,\infty)$ and therefore no zeros in the set of all reals.


hmm... sorry, this is just to say hi to the old people I know :-)
I'm David, from Australia, 19, IMO 3-timer ('00, '01 - bronze, '02 - silver). After becoming too old (joke!) I was preparing to go to the US, but I ended up staying in Australia to study medicine. So yeah, I'm now a second year medical student... with plenty more years to come... I also do a bit of maths in my spare time, split around 50-50 between teaching (high school tutoring + olympiad camps) and doing courses... of course, at the slow rate I'm doing them I'll never finish a maths degree,but it's fun anyway! I just finished analysis I, which makes around 1/3 a year of 3rd year maths done...  I also play piano (chopin + rachmaninoff) and pipe organ in my spare time...

Apart from the introduction, there's always the apologies + hi's... hi to: Valentin (of course!), who I think I actually took on at basketball in '01... fred, charlie, zhihong, people from the 3 years who made it so brilliant.. whoever else I missed, sorry :-)
I'll keep on checking from time to time, if you want to contact me it's turtle_2468AThotmail.com... just in case :-)

David

Wow. Another '02! Saw you fellas playing Mao at the club in Glasgow. Wasn't really very social at the IMO though, except with the Indonesians.
(Uh oh, is that sound I hear the sound of Dept of homeland security moving me to the 'do not allow study in the US' list?)

Miss those days... Wonder how many '02 people are here. *Sigh* Should have interacted more with the other teams besides the ASEAN ones.


For numbers $x,y,z\geq2$ prove that:

$\boxed{\left(y^{3}+x\right)\left(z^{3}+y\right)\left(x^{3}+z\right)\geq125xyz}$

i worked 30 minutes on this one, pretty easy and beautiful :lol:

[quote="sinajackson"]For numbers $x,y,z\geq2$ prove that:

$\boxed{\left(y^{3}+x\right)\left(z^{3}+y\right)\left(x^{3}+z\right)\geq125xyz}$
[/quote]
$x^{3}\geq4x$ :D

Are you sure that'll solve the problem?!!! :wink:
Explain more about your solution...

Since we have the condition $x,y,z \geq 2$, we also have $x^{3}\geq 4x$, like arqady said before (this is clearly true since we can divide by $x$ which is positive and $x^{2}\geq 4$ holds.

Now, because of this we have
$(y^{3}+x)(z^{3}+y)(x^{3}+z) \geq (4y+x)(4z+y)(4x+z)$

It's enough to show that
$(4y+x)(4z+y)(4x+z)\geq 125xyz$

By weighted AM-GM we have

$\frac{4y+x}{5}\cdot \frac{4z+y}{5}\cdot \frac{4x+z}{5}\geq \sqrt[5]{y^{4}x}\sqrt[5]{z^{4}y}\sqrt[5]{x^{4}z}= xyz$

Multiplying with 125 we get the desired inequality!

ok now i get it, nice solution :) ... my solution is a little bit similar to yours:

$y^{3}+x=\frac{y^{3}}{4}+\frac{y^{3}}{4}+\frac{y^{3}}{4}+\frac{y^{3}}{4}+x\geq5\sqrt[5]{\frac{y^{12}x}{2^{8}}}$
and also
$z^{3}+y\geq5\sqrt[5]{\frac{z^{12}y}{2^{8}}}$
$x^{3}+z\geq5\sqrt[5]{\frac{x^{12}z}{2^{8}}}$

By multiplying these ineQualities we have:

$\left(y^{3}+x\right)\left(z^{3}+y\right)\left(x^{3}+z\right)\geq125\sqrt[5]{\frac{x^{5}y^{5}z^{5}\cdot x^{8}y^{8}z^{8}}{2^{24}}}$

but as we know $x,y,z\geq2$ so $x^{8}y^{8}z^{8}\geq2^{8}\cdot2^{8}\cdot2^{8}=2^{24}$

$\Longrightarrow \left(y^{3}+x\right)\left(z^{3}+y\right)\left(x^{3}+z\right)\geq125xyz$


If $x^2+3x+5=9$, and x is positive, what is $x^2+3$?

[hide="My answer....."]I think this is wrong but I factored it out to $x(x+3)=4$ so got x as 1 and $x^2+3$ as 4....[/hide]


geometry
geometric transformation
homothety
geometry unsolved

On a circle $ O$, there are five points $ A, B, C, D$, and $ E$ such that $ AC \equal{} CE$. Segment $ BD$ intersects segments $ AC,CE$ at points $ P,Q$, respectively. Circle $ O_1$ is tangent to segments $ AP$ and $ BP$ and arc $ AB$ (that does not contain point $ C$). Circle $ O_2$ is tangent to segments $ DQ$ and $ EQ$ and arc $ DE$ (that does not contain point $ C$). Let circles $ O_1$ and $ O_2$ be internally tangent to circle $ O$ at points $ R$ and $ S$, respectively. If $ X$ is the intersection point of lines $ RP$ and $ SQ$, prove that line $ XC$ bisects the angle $ \angle ACE$.

Call K the center of the circle O.Call s the tangent to the arch AC of O (that contain D) parallel to AC, t the tangent to the arch CE of O (that contain B) parallel to CE, r the tangent to the arch BD of O (that contain C) parallel to BD. Call $ C' : s \cap t$, $ P': r \cap s$ and $ Q': r \cap t$. By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=201342]this[/url] P',P,R are collinear and Q',Q,S are collinear. But CPQ and C'P'Q' are homothetic so X is the center of homothety, so C,X,K,C' are collinear. Now $ \angle ACX\equal{}\angle CC'P' \equal{} \angle CC'Q' \equal{} \angle ECX$


function
real analysis
real analysis unsolved

the application of thsi has more to do with  physics than with maths :maybe:
[b]The theorem:[/b]by an corollary to the uniqueness theorem if for a  given volume we know the potential at it's surface and also the charge density in the volume then we can uniquely specify the $V(x,y,z)$ that is the potential of each point assuming it's value to be zero at $\infty$
now [b]is this true[/b]
we have two similar volumes with simlar surfaces similar charge distribution within the volume but they differ in the potential on their surfaces by an amount which can be made as small as possible but not exactly $0$ then does this imply that the potential functions through out the volume differ by a very small amount.

It sounds like the difference of potentials you're thinking of is a harmonic function- its Laplacian is zero.

Harmonic functions have many nice properties. One is that they have no local maxima or minima; the maximum and minimum over any compact set occurs on the boundary. This means that your boundary size estimate applies uniformly over the whole region.

yes i was somehat excluding certain points thinking of harmonic functions...and yes thanks :)


Asymptote conics parabola geometry hyperbola function articles

I know I sound like  a noob but how do you graph parabolas and hyperbolas.
Also how do you draw out the x and y axis and make tick marks.
I could the the last one by raw calculating and labeling but there must be a better way. :(
As for the first one, there really isn't a way to brute force calculate it.

You do not sound like a noob, not at all :)

I still have to learn about various Asymptote packages myself to figure out for what there are standard routines and for what there aren't. Right now I can recommend that you take a look at Ph.Ivaldi's geometry_dev package. It allows you to handle the conic sections quite easily. For instance, the following code draws the part of the hyperbola with foci (1,2) and (0,-1) and eccentricity 0.7 inside the box with corners (-4,-4) and (4,4):
[code]
import geometry_dev;
size(250);
hyperbola h;
h.init((1,2),(0,-1),0.7);
h.bmin=(-4,-4); h.bmax=(4,4);
path v=h;
draw(v);
[/code]
[asy]import geometry_dev;
size(250);
hyperbola h;
h.init((1,2),(0,-1),0.7);
h.bmin=(-4,-4); h.bmax=(4,4);
path v=h;
draw(v);[/asy]
I'm not sure whether he finished writing the English documentation for geometry_dev (see one of the announcements in this subforum for the links) but the package is fairly self-explanatory and if you have any trouble with some particular task, you can always ask here how to do it (but, please, be specific: just saying "draw a hyperbola" is not enough; you have many ways to specify the hyperbola and it is an unbounded curve, so it would be a good idea also to tell how large part of it you want). As to the ticks, just read the documentation about the graph package in the standard Asymptote help file: it is all there and I do not see any reason to copy and paste the corresponding paragraphs to this post :).

Okay three questions.
1) How do you download the geometry_dev package? All I found was the following:
[url]http://piprim.tuxfamily.org/asymptote/travaux/modules/geometry_dev.asy[/url]
That just has the copyright stuff.

2) Let's say I have the geometry package installed.
How would you graph equations like $ x^2 \plus{} 2x \plus{} 1$ or $ x^2 \plus{} 5x \plus{} 6$
Is there a general form to this?

3)If all you were drawing were circles, parabolae, hyperboale, asymptotes and that level related algebra, which would you recomend? cse5 or geometry_dev?

1) Sometimes it is a good idea to use the scroll bar before deciding what is and what is not in the file you are viewing. :P

Anyway, the last post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197974]this thread[/url] has the links to the versions of geometry_dev and graph_pi used on the site.

2) If you want just to draw a graph of an explicit function $ y\equal{}f(x)$, the standard graph package is already good enough. For instance, to draw the graph of $ y\equal{}x^2\minus{}2x$ between $ x\equal{}0$ and $ x\equal{}2$, you can write
[code]
import graph;
size(200);
xaxis(-1,3);
yaxis(-2,1);
real f(real x)
{
return x^2-2*x;
}
draw(graph(f,0,2));
[/code]
[asy]import graph;
size(200);
xaxis(-1,3);
yaxis(-2,1);
real f(real x)
{
return x^2-2*x;
}
draw(graph(f,0,2));[/asy]
(For some reason, this code doesn't work on the site right now but it works perfectly well on my home computer)

3) geometry_dev is designed to handle conic sections and other fancy stuff but if all you need is graphing explicit functions, then graph or graph_pi are more than enough :)

So is this graph_pi a seperate download?
If it is, where can I get it?

Also I know this sounds a little much but if you could make a short little guide on basic graphing for me that would be nice. (Parbolae, Hyperbolae, Asymptotes, etc...)

Yes, graph_pi is separate from geometry_dev.

...the last post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197974]this thread[/url] has the links to the versions of geometry_dev and [b]graph_pi[/b] ...

As to a short guide, let's do it this way: you tell what exactly you want to draw and I or somebody else will show you how to do it. If you do not understand the code, we'll go over it line by line. Next time when you need another picture, you'll try to do as much of it as possible yourself and tell what you have difficulties with, and so on. This way you'll quickly learn the commands that fit your particular needs. To give you a comprehensive course of Asymptote graphing would be a bit too much, indeed (plus I can easily spend my time trying to describe the things that you'll not really need and miss the ones you are looking for entirely). By now, you should see how to graph any explicit function on any given interval (see my example above; thanks to Valentin, it works now). So, what is the next thing you want to learn? (please, be as specific as possible in your requests). :)

[asy]import graph;
size(200);
xaxis(-2,2);
yaxis(-2,2);
real f(real x)
{
return x^5+x^3;
}
draw(graph(f,-1,1));[/asy]

Well it works.
However when I try to compile it on TeXnic center, (I use alt a) it doesn't work.
Here is the full code I used:
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
\begin{figure}[h]
 \begin{asy}
import graph;
size(200);
xaxis(-2,2);
yaxis(-2,2);
real f(real x)
{
return x^5+x^3;
}
draw(graph(f,-1,1));
\end{asy}
\end{figure}
\end{document}[/code]

Well, it works for me, so there is no error in the $ \text{\LaTeX}$ or Asymptote code. Has to do something with how you configured the Asymptote tool in TeXnic center. Is there any picture that works for you as a part of the $ \text{\LaTeX}$ document? If no, we'll have to go over configuration settings; if yes, show the file that works and we'll try to see what is different. Also, don't forget that you shoud build the file, then run the Asymptote tool, and then build the file again to get the picture.

OK. So here are a few questions:
1) Do you get any error messages for the code that doesn't work?
2) Do you have the file that works and the file that doesn't in the same directory? (it should work from every directory, of course, but let us try to make the files as similar as possible; it may make sense even to try to give the file that doesn't work the same name as the file that does; you'll have to rename the latter one first, of course).
3) Are you completely sure that the file that works indeed does work now? (Try to rename it and do everything again.)
4) Is the .asy file created after you build the file that does not work?
5) If so, can you compile it with Asymptote (not from TeXnic center but just directly with the Asymptote program)?
6) Is the pdf file with the picture there?
This should give you an idea of which stage in the process goes astray. As to the files themselves, try to change the first into the second line by line (so, replace include by import as the first step, change the size() command as the second, add the xaxis() command as the third and so on) and see where it stops working. Also try to add some text around your picture (again, it should be irrelevant, but it doesn't hurt to explore this opportunity).

By the way, there is no need to do manual computation of the coordinates in Asymptote (but it is a separate story; let's try to figure out what's wrong with your TeXnic center first :)).

1) I knew this was the error but don't know how to fix it.
When I build first on TeXnic center, I just get the warning that it couldn't read Grpah_1 (Graph was my file name) but this is what is supposed to happen. When I press alt-a, some words flash up (which I know is an error but I can't read it) and asy closes on me, not making the file work.  :(

2) Both files are in the same directory.

3) Actually the one that works right now was one I built seperately, by copying and pasting.

4) Yes but it says that there is no variable matching x axis and y axis (I do not have any other downloads) and says to press any key then it exits out.

5) Well on just asy, it only draws the x and y axis, not the graph. It says something about abounded y scale, unexpencted end of input, and function could not return a value.  :(

6) When I try to make the pic, an asy error flashes and no pdf is created. When I build after that, it displays a blank page.

OK, the first thing to do is to rename the file to anything but graph, math, etc. (something neutral like ttt.tex). The second thing is to delete the file graph.asy created by $ \text{\LaTeX}$ from the directory where you keep your TeX file. The last thing is to compile your newly named file and enjoy the picture. In general [b]never ever[/b] name your $ \text{\LaTeX}$ files that have Asymptote code in them by the names of standard Asymptote packages!

Thanks Fedja! You're a lifesaver!
Anyway what were you saying about not having to calculate roots?
Also how do you make the tick marks? I went to the tick mark thread but I don't know what loops are...
I would look it up under help in asy but I don't know where mine went... (I pressed help but it said there was no file to open :( )

[quote="Dojo"]Thanks Fedja! You're a lifesaver!
Anyway what were you saying about not having to calculate roots?
[/quote]
Asymptote has a lot of built-in mathematical functions (and many other things), so you never need to use your calculator and punch in long decimals. If you can describe your picture in words, you can program it.
[quote]
Also how do you make the tick marks? I went to the tick mark thread but I don't know what loops are...
[/quote]
Something like that is enough for most purposes. If you want more, read the documentation or tell exactly what you are looking for.
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-10,10,Ticks(f, 2.1, 0.7));[/asy]
[quote]
I would look it up under help in asy but I don't know where mine went... (I pressed help but it said there was no file to open :( )[/quote]
You can find it online: [url]http://asymptote.sourceforge.net/doc/[/url]

So can you write in (something like) sqrt{2} in the values or something like that?

Sure. [b]sqrt(2)[/b] to be exact. For some fancier functions you need to add [b]import math;[/b] to your code but sin, cos, sqrt and many others do not require even that.

The final image!  :D
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return x^3;
}
draw(graph(f,-2,2));[/asy]
Phew it works!
Thanks fedja

Okay new question.
How would you graph $ y=\frac{1}{x}$?
I get errors when I use 1/x and frac(1)(x) on the return line.
[code]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return 1/x;
}
draw(graph(f,-2,2));[/code]
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return 1/x;
}
draw(graph(f,-2,2));[/asy]
Like that.
It doesnt work on either my home comp or the forum. (Which is kinda obvious)

I'll answer your question with a question: how would you draw it on paper (or, more precisely, what will you do with the singularity at $ 0$?). If you do not guess in 20 minutes, I'll show you the answer :).

I think the answer is: I would draw the graph as two same curves oppisite from each other across the x and y axis. This is because 0 is undefined in that function.

Exactly. So, here you go:
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return 1/x;
}
draw(graph(f,-2,-0.2));
draw(graph(f,0.2,2));[/asy]
Of course, you can specify any other intervals as well; just make sure that there are no singularities inside any of the intervals.

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return 1/x;
}
draw(graph(f,-8,-0.125));
draw(graph(f,8,0.125));[/asy]

Aha!!!  :showoff:  :showoff:  :showoff:

what about graphing $ \sqrt{x-5}$? I just get:
[asy]import graph;
Label f;
f.p=fontsize(6);
xaxis(-5,7,Ticks(f, 6, 0.5));
yaxis(-3,11,Ticks(f, 7, 0.5));
real f(real x)
{
return sqrt(x-5);
}
draw(graph(f,-2,4),red);[/asy]

wait a minute, figured it out:
[asy]import graph;
Label f;
f.p=fontsize(6);
xaxis(-3,11,Ticks(f, 7, 0.5));
yaxis(-3,3,Ticks(f, 6, 0.5));
real f(real x)
{
return sqrt(x-5);
}
draw(graph(f,5,11),red);[/asy]

I'm a COMPLETE noob at asymptote

anyways I need to know how to label the equation of a parabola at its vertex (or some other way to identify it)

Sorry for reviving  :oops:

Weeeellllllllllllllll you could just label?
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-4,4,Ticks(f, 8, 0.5));
yaxis(-4,4,Ticks(f, 8, 0.5));
real f(real x)
{
return x^2;
}
draw(graph(f,-2,2),red);[/asy]
Just find a point or something:
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-4,4,Ticks(f, 8, 0.5));
yaxis(-4,4,Ticks(f, 8, 0.5));
real f(real x)
{
return x^2;
}
draw(graph(f,-2,2),red);
label("$x^2$",(1.5,2.25),SE);[/asy]

[asy]import graph;
    size(300);
    Label f;
    f.p=fontsize(6);
    xaxis(-4,4,Ticks(f, 8, 0.5));
    yaxis(-4,4,Ticks(f, 8, 0.5));
    real f(real x)
    {
    return x^2;
    }
    draw(graph(f,-2,2),red);
    label("$x^2$",(1.5,2.25),SE);
dot((0,0),blue);[/asy]

I think he meant this.

thanks guys, I was also wondering how to graph several parabalas on the same graph?

This might be self explanatory but for some reason I can't get it to work.

ok: i got it working but is there any ways to make the ticks mark on this graph a little more "visually aligned"?

[asy]import graph;
size(250);
Label f;
Label g;
f.p=fontsize(5);
xaxis(-6,20,Ticks(f, 10, 1));
yaxis(-5,15,Ticks(f, 10, 1));
real f(real x)
{
return (x-2)^2+5;
}
real g(real x)
{
return (x-15)^2 +5;
}
real h(real x)
{
return -(x-15)^2 +5;
}
draw(graph(f,-1,5));
draw(graph(g,12,18));
draw(graph(h,12,18));
label("$(x-2)^2+5$",(5,5),S);
label("$\pm(x-15)^2+5$",(16,5),SE);
dot((2,5),blue);
dot((15,5),blue);[/asy]

[color=#9a00ff]How in the world did you figure that out?[/color] :noo:


Happy Thanksgiving!

First, Math Zoom Academy would like to thank all participants!  You all did great on a very challenging round!

The following students qualified for the Grand Challenge Round (Names are arranged in alphabetical order by last name):

Tuan Le
George Xing
Kevin Yin

CONGRATULATIONS!! YOU WERE THE TOP THREE SCORERS!!!!

The following students qualified for the Team Challenge (Names are arranged in alphabetical order by last name):

Di Ai
Lisa Beckmann
Aritro Biswas
Bryan Brzycki
Jacob Calistio
Jackson Chang
Ruiqi Chen
Aaron Cheng
Richard Cho
David Ding
Brian Edwards
Rachel Garfinkel
Marco Gentili
Kevin Hayakawa
Sarah Herrmann
Edward Huang
Alex Jow
Bryan Kam
Fengyu Ke
Edward Kim
Hyun Jong Kim
Jasper Lee
Patrick Lee
Michelle Li
Yun Qing Lu
Vinson Luo
Steven Ly
Henry Maltby
Patrick Manalo
Christopher Miro
Eric Murray
Igor Raush
Hyun Jae Seo
Roger Sheu
Kyle Sitton
Daniel Suryakusuma
Thomas Swayze
Jeff Van
Matthew Voss
Tony Wang
Raku Watari
Moshe Willner
Quinn Wong
Richmond Wong
Raymond Wu
Simon Wu
Patrick Xiao
Dai Yang
Leon Yao
Stephen Yoo
Guanyang Yu
Leon Zhang
Leon Zhou

CONGRATULATIONS!!

QUALIFYING STUDENTS: PLEASE INFORM US ON WHETHER OR NOT YOU CAN ATTEND THE EVENT BY, DECEMBER 2, 2009.
THOSE STUDENTS THAT DO NOT RSVP, WILL NOT BE ASSIGNED A TEAM FOR THE COMPETITION.
ALSO, WE ASK ALL QUALIFYING STUDENTS TO BE AT THE EVENT ½ HOUR EARLY.  ORIENTATION WILL BEGIN PROMPTLY AT 12:30!

Information about the SCMO:

Date: Sunday, December 6th, 2009
Time: 1:00 PM – 5:00 PM
Location: Sage Hill High School
    20402 Newport Coast Dr.
Newport Beach, CA

Brief agenda of the SCMO event on December 6th:

The event will begin with a Math lecture, entitled "How Computers Use Mathematics to Solve Real World Problems”, given by Dr. Hongkai Zhao, Math Professor in UCI.   This lecture relates the use of Mathematics in real-world problems.  Dr. Zhao will provide answers to interesting questions, such as, “From the standpoint of mathematics, is there a limit to how fast and efficient computers will become?”

After the lecture, the event will go on to the second chapter, the SCMO Team Challenge.  While monitoring the scoreboard of the game, we'll be talking about the various mathematics competitions, sample problems for those competitions, and the benefits of participating in such competitions. Some rare to find photos and video clips of some of the competitions will be shown as well.

Next, on the agenda, is the Grand Challenge Round, where the top three scorers from the qualifying test present the solutions to given problems in front of a panel of judges and the audience.

The finale is the Award Ceremony.

Please contact Math Zoom for any additional information.

[quote="bengal"]Hey Mr. Wang, I emailed you but you did not reply. Anyways, I will be able to make it to SCMOO. Thanks a lot!
-from Aritro in the math zoom classes over weekends[/quote]
I just came back from a trip.  I received your email, so don't worry.


Prove that the function $f: \mathbb{R}\to\mathbb{R}$ is periodic if:
                               $f(x+\frac{13}{42})+f(x)=f(x+\frac16)+f(x+\frac17)$

Without having tried it: is the last term correct, or should it be $f(x + \frac17)$?

Yes, you're wright. Sorry for my carelessness, edited. :oops:

[quote="JBL"]Okay, but now it is not true: the function $f(x) = x$ satisfies the given equation but is not periodic.[/quote]

hmm... my friend gave it to me, it's from ISL, there maybe a typo here, I'll check the question again...  :(

Right problem:

[quote]Prove that the function $f: \mathbb{R}\to\mathbb{R}$, which satifies $|f(x)|\le 1$, is periodic if:
                               \[ f(x+\frac{13}{42})+f(x)=f(x+\frac16)+f(x+\frac17) \][/quote]

Sorry again.


Find the pair of positive integer solution $ (x,\ y,\ z)$ such that $ yz\plus{}zx\plus{}xy\equal{}xyz$.

[hide]
Dividing both sides by $ xyz$ gives
\[ \frac1x \plus{} \frac1y \plus{} \frac1z \equal{} 1\]
And now there are $ (3,3,3)$ first, and if not, one of them should be $ 2$, and hence $ \frac1x \plus{} \frac1y \equal{} 1/2$ and this gives $ x\equal{}y\equal{}4$ or $ x\equal{}3$ and $ y\equal{}6$, since one of them should be at least 4. Therefore $ (3,3,3)$, six arrangements of $ (2,3,6)$, and three arrangements of $ (2,4,4)$.
[/hide]

[hide]$ yz\plus{}zx\plus{}xy\equal{}xyz$
Now by $ AM \geq GM$
$ \frac{yz\plus{}zx\plus{}xy}{3} \geq \sqrt[3]{yz.zx.xy}$
$ \frac{xyz}{3} \geq \sqrt[3]{x^2y^2z^2}$
$ xyz \geq 27$
equality when $ x\equal{}y\equal{}z\equal{}3$

again $ yz\plus{}zx\plus{}xy\equal{}xyz$
clearly we have three cases a) All odd b) two odd c) one odd d) no odd

if two are odd, LHS is odd but RHS is even, not possible! [/hide]

[quote="kunny"]Find the pair of positive integer solution $ (x,\ y,\ z)$ such that $ yz \plus{} zx \plus{} xy \equal{} xyz$.[/quote]

[hide]WLOG let $ x\le y\le z$. Verify that $ x \equal{} 1\implies y \plus{} z \equal{} 0$, impossible. Thus $ x > 1$.

Note that $ xy \plus{} yz \plus{} zx \equal{} xyz$ implies $ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \equal{} x \plus{} y \plus{} z \minus{} 1 \equal{} (x \minus{} 1) \plus{} (y \minus{} 1) \plus{} (z \minus{} 1) \plus{} 2$. Hence we get $ abc \equal{} a \plus{} b \plus{} c \plus{} 2$, where $ a \equal{} x \minus{} 1,b \equal{} y \minus{} 1,c \equal{} z \minus{} 1$. If $ a\ge 3$, the rate of increase of the left side is more than that of the right side. Hence $ a\le 2$. $ a \equal{} 1$ implies $ bc \equal{} b \plus{} c \plus{} 3\implies (b \minus{} 1)(c \minus{} 1) \equal{} 4$. Hence $ (b,c) \equal{} (5,2),(3,3)$. $ a \equal{} 2$ yields $ 2bc \equal{} b \plus{} c \plus{} 4\implies \frac {b \plus{} c}{bc \minus{} 2} \equal{} 2$. But $ (b \minus{} 1)(c \minus{} 1)\ge 0\implies b \plus{} c\le bc \plus{} 1$. Therefore $ 2\le\frac {bc \plus{} 1}{bc \minus{} 2}\implies bc\le 5$. This is possible only if $ b \equal{} c \equal{} 2$. Thus all possible triples are $ (a,b,c) \equal{} (1,2,5),(1,3,3),(2,2,2)$ which imply $ (x,y,z) \equal{} (2,3,6),(2,4,4),(3,3,3)$[/hide]

EDIT: [hide="Easier Solution"]Note that $ x\equal{}\min(x,y,z)\implies x\le 3$. $ x \equal{} 2$ yields $ 2(y \plus{} z) \equal{} yz\implies (y \minus{} 2)(z \minus{} 2) \equal{} 4$. Thus $ (y,z) \equal{} (4,4),(3,6)$. Now $ x \equal{} 3$ implies $ 3(y \plus{} z) \equal{} 2yz$. But $ (y \minus{} 3)(z \minus{} 3)\ge 0\implies yz \plus{} 9\ge 3(y \plus{} z) \equal{} 2yz$. Thus $ 9\ge yz$ which is possible only if $ y \equal{} z \equal{} 3$. Thus $ (x,y,z) \equal{} (2,4,4),(2,3,6),(3,3,3)$[/hide]


projective geometry
geometry
parallelogram
geometry solved

in quadrilateral $CDEF$ $CF\cap DE=A$ $CD \cap EF=B$ $P,N,Q,M$are at $CD,DE,EF,FA$ such that $PQ$pass through $A$ and $MN$ pass through $B$ .$MN\cap PQ=R$ prove that $DQ,FN,CR$ are concurrent.

After we apply the reciprocal of Desargues for the triangles $CDF,RQN$, we see that the statement is equivalent to $PM,DF,NQ$ being concurrent.

Apply a projective transformation which sends $AB$ to the infinity line. Now $CDEF$ is a parallelogram, $MN,PQ$ are parallel to $CD,DE$ respectively, and the concurrence of $PM,DF,NQ$ follows immediately from an application of Menelaus in $CDF,EDF$ with transversal lines $MP$ and $NQ$ respectively.

Mathx, your problem would be slightly better readable if you put commas between different formulas...

[color=blue][b]Problem.[/b] In a quadrilateral CDEF, let $CF\cap DE=A$ and $CD\cap EF=B$. Furthermore, let P, N, Q, M be points on the lines CD, DE, EF, FC such that the line PQ passes through the point A and the line MN passes through the point B. Let $MN\cap PQ=R$. Prove that the lines DQ, FN and CR are concurrent.[/color]

[i]Solution.[/i] (Actually, a simplification of Grobber's one.)

Applying the Pappos theorem to the collinear point triples (B; M; N) and (A; Q; P), we see that the points $BQ\cap MA$, $MP\cap NQ$ and $NA\cap BP$ are collinear. Since $BQ\cap MA=F$ and $NA\cap BP=D$, it follows that the points F, $MP\cap NQ$ and D are collinear. In other words, the point $MP\cap NQ$ lies on the line FD. Equivalently, the lines MP, NQ and FD concur. In other words, the point $FD\cap NQ$ lies on the line MP. Equivalently, the points $FD\cap NQ$, P and M are collinear. Since $P=DC\cap QR$ and $M=CF\cap RN$, we thus see that the points $FD\cap NQ$, $DC\cap QR$ and $CF\cap RN$ are collinear. Thus, by the Desargues theorem, the triangles CFD and RNQ are perspective; so the lines CR, FN and DQ are concurrent. And the problem is solved.

  darij


AMC
AMC 12
AMC 10
ARML
AIME
induction
AMC 10 A

Is getting a 100 on amc12 harder than getting a 140+ on Amc10?

(in other words if ur in <=10 grade who got a 100~ on amc12, and you just found out you could have gotten a 140+ on amc10, would u feel bad about it ?)

getting a 140 on the AMC10 is much harder.

But the AMC 12 is so much more fun.  I mean, isn't that what it's really all about?

By the way, if by "just found out" you mean "took it at home in your spare time," that's not an accurate assessment of how you would've done under test conditions.  For example, on practice ARML individuals I can get about $7$ or $8$, but last year in Las Vegas I got $5$ (I still blame the heat, but you see my point).

> Is getting a 100 on amc12 harder than getting a 140+ on Amc10?

A meaningful, quantitative way to answer this question would be to look at past statistics on the AMC website and compute the percentage of students in your grade who have reached each respective level.  Use the yearly contest score distribution by grade at
http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml

Steve Dunbar
Director, American Mathematics Competitions

Yeah, I agree with t0rajir0u; I could've taken the 10 this year, and would've probably ended up with around 140 (138 last year). But the AMC 12 is just a much more interesting test, and I learned a lot more from it even though I only got a 112.5.

Getting a 140+ on AMC 10 is much harder, infact I think getting a 140+ on AMC 10 is harder than getting 120+, let along 100+, on AMC 12.  ANd since Im not very fast at calculation and at noticing tricks, a 140+ is hard considereing the time limit.

Since Im only a 9th grader, I took both to get a feel for the AMC 12s..and really, the first like 18ish on the AMC 10 and 12 dont have that much difference in difficulty, but the last couple the AMC 12 is way harder...but also the AMC 12 has more "book"(like amc 10a 23, 24) type of questions, whereas the 10 has more word logic questions....

I think getting a 140 on AMC 10 is harder than getting a 100 on AMC 12, but easier than getting 120+ on AMC 12.

[quote="Twiz"]I don't know why, but I generally do about the same on each test. Last year I made a 107.5 of the 10A and a 108 on the 12B, and this year, I made a 126 on the 10A (but guessed 2 for 4) and a 114 on the 12B (but guessed 0 for 6). I know that guessing is bad, but I do it because it adds an element of mystery to my score. :)

I know that I need to stay away from gambling when I'm older. I would quickly become an addict.[/quote]
Hah, I feel like a wimp if I leave problems blank.

I don't.  I have bad experiences of guessing and getting them wrong and getting like 1 point off of first or tying first with somebody that I should have beat.  I call it being safe.  Sure, the average score is the same, but let's say there are two people with the same score up past #24.  One guesses on #25, one doesn't.  The chance that the guesser will win is 1/5.  Which isn't good.

Yeah, but on the 10A, I beat 2nd place at my school by 24 points anyways.

Practice is always different from the competitions. No matter how calm you feel, you'll always do worse. At least for me. xD

Who cares of you are the school winner?(you is not an answer)

The AMC 12 has much more interesting stuff. You get more acknowledgement if you get something like a 100.5. You still made the AIME, but on a harder test.

P.S. I can't get a 140+ on the AMC 10.

I personally think that getting 100+ on the 12 is harder, but that's probably just because the 12 is harder in general.
The only thing that I can think of that goes against that is that I keep making stupid mistakes (i.e. I got a perfect on the 10A last year when I was in 7th grade but I only got a 144 this year.)
But, then, I got in the 120's for the 12B when I took it this year, and it  seemed a lot harder.

I dont think the questions on the amc's are that hard, but its just that Im not really fast at math and the AMC's are all about speed- so my amc 12 score (126-if i didnt bubble in something wrong) and amc 10 score-132 are very close because I really dont think the difficulty is that much different(aside from last 5 or so)-I jsut have a hard time doing the problems in in average of 3 mins..

In my opinion, a 140 on the amc10 is about the same as a 110 on the amc12. I've gotten a few perfect scores on practice amc 10's (probably mid 140's on avg), but I probably only average upper 120's on 12's. I think in general it depends on [b]how[/b] you do math. If you like to memorize as many formulas/tricks as possible you can get away with that and own the amc10. But in order to own the 12 you have to regularly work your brain to think a little more creatively.

it doesn't even matter what you get as long as you make it into AIME.

So which test do you believe is better? If I am going to be a 10th grader next year, and our school only lets us take the A-version of the test, which should I take?

I got a 112.5 this year on the AMC 10-A (stupid mistake on one of the problems, or I would have made the cut), and I guess I'm OK right now at precalc, not that good.

Any recommendations?

Thanks.

Unless you improve a lot at math before the test next year, I'd say take the 10A again; a 112.5 on the 10A may or may not give you a qualifying score on the 12A.
I took the 10A last year and got 138, and only managed a 112.5 on the 12A this year. (Well, that's also partly because I didn't study at all, but you see my point.)

i'd say that getting a 140 on amc 10 is about as hard as getting 120 on amc 12.
but i may be biased because while taking the amc, i use the calculator frequently to write programs to solve problems quickly so that i can move on the more difficult problems (#20+)  :ninja:
it also depends on how well you can do simpler problems relative to how well you can do the more complex high-level algebra/geometry problems
(on the aime, instead of running a recursive or looping program, i use induction instead, which works the same except it takes much longer)

Though I can't really talk from personal experience (137 on AMC10A 2006, then 138/141 on AMC12AB 2007), I would say getting 140+ on AMC10 is much harder.  That means that in practices, you have to be pretty comfortable with getting 150's most of the time.  However, AMC12, the first 18-22 problems are not much harder than those of the AMC10.

well, i got a 120 on amc 10 last year and got a 114.5 on amc 12.

but last year i made a lot of silly mistakes.

this year, i got a 144 on amc10 but would've gotten a 120-126 on amc12.

so i'd say getting 100 on amc 12 isn't that hard.

obviously 140 for AMC 10A is harder, recalling that only a very small percent of people get 140+ on the AMC 10(estimating 0.05%). While approximately 5% of people get 100+ on the AMC 12(cutoff for AIME)

But... the AMC 10 can only be taken by 10 grade and under, while the AMC 12 can be taken by older kids. The percentages cannot be compared.

Yes, but still. The AMC 10 is roughly proportionally easier compared to the AMC 12.

[quote="junggi"]i see many people saying it's harder to get a 140 on the amc10....well duh last time i checked it's impossible to get a 140  :P[/quote]
i'm assuming they meant 140+  :P
also, keep in mind that this is the site where everybody is shooting for a 16 on the AIME, so you never know...  :rotfl:

[quote="123s"]davidyko, I was not saying that you can't compare the difficulty of scoring various scores on the two tests. I was saying that you can't use the percentages that Allan Z quoted (unless you are asking whether is it harder for a 10 grader and under to get 140+ or a 12 grader and under to get a 120+). For any one person, these statistics don't work, because although a higher percentage of people score above 120+ on the 12 than 140+ on the 10, there are also better people taking the 12.[/quote]

I see; I must've misread and read "scores" instead of "percentages" Sorry.  :)

Thanks for your comments. I think I'll just take some practice AMC tests and compare my results on both.


[quote="citistarz"][hide="solution"]If you extend DC to K(5 units away from D) you will see that $\triangle{EAK}$ and $\triangle{DBE}$ are similar by AA and have a $5: 4$ ratio. So $AE=5\times\frac{AB}{9}$ and you know that $AB=3\sqrt5$ and so $AE=\frac{5(3\sqrt{5})}{9}=\frac{5\sqrt{5}}{3}$. So the answer is $\boxed{B}$.[/hide][/quote]

Where is the point K? You say it is on the extension line of DC 5 units from D, but shouldn't K be such that AK is parallel to DB? Could you explain your solution?

[hide="solution2"]
$\text{Mark the point 2 units above D as F.}$
$\text{ Thus }DF\text{ = 2. We see that the slope of }$$DE\text{ is 1,}$
$\text{and the slope of }AE\text{ is -1/2. Mark the horizontal distance from A to E as x.}$
$\text{Thus x + (1/2)x = 2 so x = 4/3. We see that (10/3)/(2) = (AE)/(}\sqrt{5}\text{)}$
$\text{So}\ \frac{5\sqrt5}{3}\text{ = AE, (B)}$
[/hide]

Hello this will be an 11 year bump :)

So what I did was set $A=(0,3)$, and marked the rest of point as if it were a graph, found the intersection of $AB$ and $CD$, and then distance formulated it home.

My questions:
Is what I did valid, why or why not?
Can anyone provide a synthetic solution (or a diagram of the first one)?

I synthetic Geo from mathcounts minis :)
 Best,
m

For this problem, things come out especially nicely, if you draw in the vertical lines from $A$ and $B$ and continue $DC$ to intersect those lines. It makes some nice similar triangles.

[quote=catcurio]The diagram show $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.

[asy]
path seg1, seg2;
seg1=(6,0)--(0,3);
seg2=(2,0)--(4,2);
dot((0,0));
dot((1,0));
fill(circle((2,0),0.1),black);
dot((3,0));
dot((4,0));
dot((5,0));
fill(circle((6,0),0.1),black);
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((4,1));
dot((5,1));
dot((6,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));
fill(circle((4,2),0.1),black);
dot((5,2));
dot((6,2));
fill(circle((0,3),0.1),black);
dot((1,3));
dot((2,3));
dot((3,3));
dot((4,3));
dot((5,3));
dot((6,3));
draw(seg1);
draw(seg2);
pair [] x=intersectionpoints(seg1,seg2);
fill(circle(x[0],0.1),black);
label("$A$",(0,3),NW);
label("$B$",(6,0),SE);
label("$C$",(4,2),NE);
label("$D$",(2,0),S);
label("$E$",x[0],N);[/asy]

$\text{(A)}\ \frac{4\sqrt5}{3}\qquad\text{(B)}\ \frac{5\sqrt5}{3}\qquad\text{(C)}\ \frac{12\sqrt5}{7}\qquad\text{(D)}\ 2\sqrt5 \qquad\text{(E)}\ \frac{5\sqrt{65}}{9}$[/quote]

Typo, should be "shows" firstly.

Secondly, time for the solution.

We will use similar triangles. Extend EC to point F (the point $\sqrt{2}$ away from C, that is up to the right) so that AEF and BED are similar with ratio $5:4$. We now have $AE+\frac{4}{5}AE=\sqrt{3^2+6^2}=3\sqrt{5}$. So, $\frac{9AE}{5}=3\sqrt{5}$. Solving this gives $AE=\frac{5\sqrt{5}}{3}$, or $\boxed{\text{(B)}}$. $\blacksquare$

We coordinate bash :)

B is at (6,0) and A is at (0,3). D is at (2,0) and C is at (4,2). Slope of AB is $-\frac12$. Equation is $y=-\frac12x+3$. Slope of CD is $1$. Equation is $y=x-2$. $-\frac12x+3=x-2\rightarrow \frac32x=5\rightarrow x=\frac{10}{3}\rightarrow y=\frac43$. Distance formula gives $\boxed{\textbf{(D)}\frac{5\sqrt5}{3}}$. Not very nice but ok.

This is [url]https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_16[/url] right?

I simply just drew lines to find a familiar area. Not sure if it is a good solution tho.

I was so dumb when I first mocked this that I tried to use a ruler :rotfl:

Lmao that's hilarious
Altho the diagram [i]is[/i] to scale...

[quote=CoolJupiter][s]Why are we posting on an 14 year old thread[/s][/quote]

Why did I just realize that?

The thread is old, but it was "only" a 3 year bump from the last post, which was in 2017.

I honestly don't even know why bumping is bad, as long as it is contributing to the discussion

[quote=mop][quote=OlympusHero][hide=Solution]We will form a similar triangle by extending $EC$. This gives $\frac{AE}{EB}=\frac{5}{4}$, and $AE+EB=3\sqrt5$. Let $AE=5x, EB=4x$. Then $x=\frac{\sqrt5}{3}$, and $AE=\frac{5\sqrt5}{3} \implies \boxed{\text{(B)}}$.[/hide][/quote]

You already posted a solution...[/quote]
The exact same one also :spam:

Oh sorry! I was speedrunning the AMC 10 and did not look. :blush:

[hide=Solution]We begin by assigning the lattice point in the bottom left corner the origin, or $(0,0)$. This means that we know the following points:
$$A(0,3)\qquad B(6,0)\qquad C(4,2)\qquad D(2,0)$$
We will now proceed by coordinate bashing. We figure out that line $AB$ has equation $y=-\frac{2}{3}x+3$ and line $DC$ has equation $y=x-2$. Since $E$ is the intersection of $AB$ and $CD$, we proceed by finding $E$:
$$-\frac{1}{2}x+3=x-2$$
$$\frac{3}{2}x=5$$
$$x=\frac{10}{3}$$
By plugging our value for $x$ into any of the equations of the lines, we will get that $E$ has a y-coordinate of $\frac{4}{3}$. Therefore, $E$ has coordinates $(\frac{10}{3},\frac{4}{3})$. We now use the distance formula to find the distance (denoted as $d$) of segment $AE$:
$$d=\sqrt{\frac{10}{3}^2+(\frac{4}{3}-3)^2}=\frac{\sqrt{100+25}}{3}=\frac{5\sqrt{5}}{3}$$
Therefore, the correct answer is $\boxed{\text{(B)}\ \frac{5\sqrt5}{3}}$[/hide]

[hide="Coordinate Sol"]The equation of line $AB$ is $y=-\frac{x}{2}+3$ and the equation of line $CD$ is $y=x-2$. Their intersection point is $E$, then the coordinates of point $E$ are found as follows, $-\frac{x}{2}+3=x-2\implies x=\frac{10}{3}\implies y = \frac{4}{3}$. Then, we simply use Pythagoras to find $AE$.
$$AE^2=(\frac{5}{3})^2+(\frac{10}{3})^2\implies AE^2=\frac{125}{9}\implies AE=\frac{5\sqrt{5}}{3}$$ which is answer choice $B)$[/hide]


inequalities
calculus
derivative
inequalities proposed

This inequality holds for $ n$ numbers if I am not misremember, since I got it from the article ehku sent to me along time ago. I will check my computer and attach his solution soon. I remember that his solution is to use directly derivative, very nice.

As I promise, I will post the solution by ehku (Le Huu Dien Khue). Best thank to him for collaborating with me  :lol:

The language is Vietnamese but it is enough to understand :)


[quote="Ramanujan47"]Hi,

One day, a member (Nguyen Thuc Vu Hoang) wrote a PDF file about a technic in order to demonstrate inequalities in 3 variables. He wrote something like "The case t = 0 of Schur ineq <=> $ pq \minus{} 9r \geq 0$, with p = a+b+c, q = ab+ac+bc and r = abc"

But when I expand both, I don't find the same result ... What's wrong ?[/quote]
$ pq\equal{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\equal{}(a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}abc\geq 8abc\plus{}abc\equal{}9abc\equal{}9r$ from AM-GM. :lol:


Being a new member and an untrained problem solver, can anyone please describe me the following inequalities?? :oops:

1. Schur's inequality
2. Muirhead
3. Jensen's inequality.
4. Minkowski's inequality

I won't fret even if quite a bit of CALCULUS/COORDINATE GEOMETRY is required to prove these.  :maybe:

[A few spelling mistakes are possible but please, I need help on these subjects.]

May I recommend reading Topics in Inequalities by Hojoo Lee?

http://compactorange.googlepages.com/tin2006new.pdf

or Ivan Matić's

http://www.imomath.com/tekstkut/ineq_im.pdf

It will most certainly answer all of your questions concerning these inequalities. (The quality of these texts is in the fact that they're written for olympiad contestants, without 'obsolete' facts and with examples.)

Many many thanks for your suggestions! The materials might be able to clear my doubts on these subjects! :)


This forum seemed like it needed some life, so...
How'd everyone do on the AIME I?  (or II, if this thread last that long  :rotfl: )

Oh, hey, look, an old AIME thread!  So how did Iowa people do on the AIME I?

Decided to take the II so I could have more time to procrastinate this year.

:O
Really?
Hmmm
$%Error. "dontquote" is a bad command.
$

This forum really is dead, right?

Even Saphira's amazing signature has no impact on the amount of people active on the forum.


articles
inequalities
function
analytic geometry

\documentclass{article}
\begin{document}
Minimize Z=X+3Y

subject to

2x + y\leq 10

5X + 2y \geq 20

-X + 2y \geq 0

x\geq0,  y\geq0
\end{document}

I got (5,0) for the answer. Is that right

[quote="dvcalculus"]\documentclass{article}

Minimize $ Z\equal{}X\plus{}3Y$

subject to

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

$ 5X \plus{} 2y \geq 20$

$ \minus{}X \plus{} 2y \geq 0$

$ x\geq0,  y\geq0$

I got (5,0) for the answer. Is that right[/quote]

There i just latexified your post

PS: you don't need to add the doucumentclass thing when you post on AoPs

Sorry to double post but i am pretty sure its correct.....

hello, i have got
$ x\equal{}\frac{10}{3},y\equal{}\frac{5}{3}$.
Sonnhard.

The pairwise intersections of the boundary lines of those inequalities are:

$ (0,10)$
$ (4,2)$
$ \left( \frac{10}{3}, \; \frac{5}{3} \right)$

Verify that the triangle in the Cartesian plane (edge-inclusive) with these vertices is equivalent to the solution set of the system of inequalities.

Then, we know that if the function x+3y achieves a minimum value in that region, that minimum value can be achieved at a vertex.

We try out the three vertices to find that the minimum happens when $ (x,y) \equal{} \left( \frac{10}{3}, \; \frac{5}{3} \right) \implies z(x,y) \equal{} \frac{25}{3}$


Two brothers share a flock of x sheep. They take the sheep to the market and sell each sheep for dollars x. At the end of the day they put the money from the sales on the table to divide it equally. All money is in dollars 10 bills, except for less than ten excess 1 doller bills. One at a time they take out 10 dollar bills. The brother who draws first also draws last. The second brother complains about getting one less 10 dollar bill so the first brother offers him all the 1 dollar bills. The second brother still received a total less than the first brother so he asks the first brother to write him a check to balance the things out. How much was the check?

sorry to say but your answer is wrong.Can you give a solution?

K81o7 just made a slight mistake at the end.

Let's assume $x$ is a single digit number, equal to $a$. The brothers make $a^2$ dollars. In order for the first brother to get 1 more 10 dollar bill than his brother, there must be an odd number of tens. Thus, after all the 10s have been taken, it is the second brother's turn to draw. Since the first brother draws last, there must be an even number of 1s left. This tells us that $a$ must be even. So, we look at possible values for a:

$0^2 = 0$
$2^2 = 4$
$4^2 = 16$
$6^2 = 36$
$8^2 = 64$

Of these 5 possible values, the tens digit is only odd when $a = 4, 6$. We see that the second digit is always 6 for these two cases. Thus, when dividing the money, the first brother initially gets $10(y + 1) + 3$ and the second brother gets $10y + 3$, for some $y$. After giving all the ones, the first brother has $10(y + 1)$ and the second has $10y + 6$. Thus, the diffence is 4 dollars and the older brother writes a two dollar check.

Now, what about the cases where $x \ge 10$? We can write $x = (10b + 1)$, so $x^2 = 100b^2 + 20ba + a^2$. We can see that the first two terms are divisible by 20 (implying both brothers get an even number of 10s in that subset). Thus, we are reduced to $a^2$ which is the first case. Similarly, we can break any value of $x$ down into the $a^2$ case.

So, the answer is 2 dollars.

[quote="riddler"][quote="chesspro"] The brothers make $a^2$ dollars[/quote]
Why? it is not given in the question? :?  :([/quote]

The problem states that there are $x$ sheep, and each is sold for $x$ dollars. This is a total of $x \times x = x^2$ dollars. Since we defined $a$ such that $a = x$, the fact that the brothers made $x^2$ dollars is the same as saying that they made $a^2$ dollars.

But is the check necessarily taken from the first brother's money?  :D
Okay yeah, i made a pretty stupid mistake, in order to balance out the 4$ difference, it has to be 2$...

I'm still trying to get the question,but thanx for the solution chesspro :D


inequality system
Inequality
Sequence
Linear Recurrences
IMO Shortlist

Let $ \{a_k\}^{\infty}_1$ be a sequence of non-negative real numbers such that:
\[ a_k \minus{} 2  a_{k \plus{} 1} \plus{} a_{k \plus{} 2} \geq 0
\]
and $ \sum^k_{j \equal{} 1} a_j \leq 1$ for all $ k \equal{} 1,2, \ldots$. Prove that:
\[ 0 \leq a_{k} \minus{} a_{k \plus{} 1} < \frac {2}{k^2}
\]
for all $ k \equal{} 1,2, \ldots$.

First, we will prove that $0 \leq a_{k}-a_{k+1}$ for all $k\geq 1$.
If not, let $ a_{k}-a_{k+1}=d<0$ for some $k\geq 1$. As we are given $ a_{k}-a_{k+1}\geq a_{k+1}-a_{k+2}$, it follows $a_{k+i-1}-a_{k+i}\le d<0$ for all $i>0$. Moreover, by summing them all, $a_{k}-a_{k+i}\le id$. For large enough $i$, we must have $a_{k+i}<0$, which contradicts the fact the sequence is made of non-negative real numbers. Therefore $0 \leq a_{k}-a_{k+1}$ for all $k\geq 1$.
Second, we will prove that $a_{k}-a_{k+1} <\frac{2}{k^{2}}$.
We have $1\ge \sum_{i=1}^{k}a_i\ge \sum_{i=1}^{k}a_i-ka_{k+1}= \sum_{i=1}^{k} i(a_{i}-a_{i+1}) \ge \frac{k(k+1)}{2}(a_{k}-a_{k+1})$.
Therefore $a_{k}-a_{k+1}\le \frac{2}{k(k+1)} <\frac{2}{k^{2}}$.

The argumentation in the first part is wrong. From $a_{k}-a_{k+i}\le id < 0$, even for large enough $i$, it does not follow $a_{k+i}<0$. A simple example is for $a_n=n$ for all $n\geq 1$, when $d=-1<0$ for all $n$.

In order to obtain that result, we need also use the second condition, namely $1\ge \sum_{i=1}^{k}a_i$ for all $k$. As above, if $a_{k}-a_{k+1} = d < 0$ for some $k$, it follows $a_{k}-a_{k+i}\le id < 0$ for all $i$; but then $a_{k+i} \geq a_k - id \geq i|d|$, and so $\lim_{n\to \infty} a_n = +\infty$, absurd.


Find all $ x \in R$ such that: $ a^x\plus{}b^x\equal{}a\plus{}b$ and $ ab\equal{}1$

Please post real exponential problems in college playground.
I think all of this kind problems, could be solve only by real analysis.

hello, with $ b\equal{}\frac{1}{a}$ we get $ a^x\plus{}\left(\frac{1}{a}\right)^x\equal{}a\plus{}\frac{1}{a}$, setting $ a^x\equal{}t$ we have $ t^2\minus{}\left(a\plus{}\frac{1}{a}\right)t\plus{}1\equal{}0$, this quadratic equation gives $ x\equal{}1$ or $ x\equal{}\frac{\ln(\frac{1}{a})}{\ln(a)}$.
Sonnhard.

Thank you madadpoor and Dr Sonnhard Graubner.
P/S: I'm just a 9th grader and my teacher gave it to me so I can't post it in college playground, madadpoor :D


algebra
polynomial
superior algebra
superior algebra unsolved

Show that for every prime $p$ and integer $n$, there is an irreducible polynomial of degree $n$ in $\mathbb{Z}_p[x]$ and use that to show there is a field of size $p^n$.

Please post such problems in Superior Algebra!
This problem was posted there already!

I don't know who moved this subject from Combinatorics but I think he went up so fast.
I know that it has an algebraic solution but it has a combinatorial solution too.
So please just move it back to the Combinatorics.
Still if you believe there isn't any combinatorial proof I can post it here.

You can look http://www.mathlinks.ro/Forum/viewtopic.php?highlight=irreducible&t=68075#

@Nima Ahmadi Pour can you tell us your proof? :)


On a 5-question multiple choice quiz, each question has 5 choices. For each correct answer, the student earns 10 points and for each incorrect asnwer the student loses 2.5 points.  Thus, the highest score possible is 50 and the lowest is -12.5.  If Jorge is able to eliminate one choice from each question and then randomly guesses from the remaining 4 choices, what is his expected score?

According to probability, he would get one right and three wrong from the first four, summing to a score of 2.5.
For the last question the odds are against him getting it right, ending up with a score of 0.

actully his expected score is $\frac{25}{8}$ or $3\frac{1}{8}$
P.S. if you don't know what an expected score is try using the wiki

For each question, the expected value is 1/4 * (1 * 10 - 3 * 2.5) = 1/4 * (2.5) = 5/8.

Since there are 5 questions, the expected score is 25/8.


PEN: Problems in Elementary Number Theory (by Hojoo Lee)

http://my.netian.com/~ideahitme/pen.html

Nice source of olympiad number theory problems (no solutions though :( , but that's what this forum is for I guess :D )

Hmm...Hojoo Lee...where have I seen that name?

Oh, that's right.  He's done stuff like [url=http://myhome.personaldb.net/ideahitme/syllabusct.html]this[/url].


calculus
integration
function
real analysis
real analysis solved

Let $f,\,g$ be two non-negative real $\mathrm{L}^1$-functions on a finite interval $I$. If

\[\int_I f(x)\, dx = \int_I g(x)\, dx = 1\]
show that for every $n \in \mathbf{Z^+}$ there exists a subinterval $I'$ of $I$ such that

\[\int_{I'} f(x)\, dx = \int_{I'} g(x)\, dx = \frac{1}{n}.\]

Edit: The integrals' values should have been 1 in the premise, which I forgot.

[quote="Mindspa"]
\[\int_{I'} f(x)\, dx = \int_{I'} g(x)\, dx = \frac{1}{n}.\][/quote]

Are you sure that $\frac{1}n$ stands alone?

Is it $\frac{1}n \cdot \int_I f(x)dx$? or is it: prove that for every large enough n and $\int_I f(x) dx>0$?

I am incredibly sorry. I left out the premise that the value of the integrals over $I$ is 1!

1) Suppose mes{t|f(t)=0}=0. Then we can define function T(x) by condition
\[\int_x^{T(x)}f(t)dt=1/n,\]
where $x\in[0,X],\ X<1,\ T[X]=1$. Due to absolute continuity of Lebesgue's integral and to fact that f(t)>0 we have T(x) is continuous function.
Now consider G(x)=$\int_x^{T(x)}g(t)dt$. It is clear G(x) is continuous. It implies that if G(x)<>1/n for all $x\in[0,X]$ then we necessarily obtain G(x)>1/n or G(x)<1/n for all x. It follows that
\[\int_0^1g(t)dt=\sum_{k=1}^n\int_{x_k}^{x_{k+1}}g(t)dt\neq 1,\]
where $x_1=0,\ x_{i+1}=T(x_i),\ x_{n+1}=1$.

2) Now suppose mes{t|f(t)=0}>0. Consider $f_a(x)=f(x)+ax$ and $g_a(x)=g(x)+ax$, a>0. According to 1) we can find interval $I_a$ s.t. $\int_{I_a}f(t)dt=\int_{I_a}g(t)dt=\frac{1+a/2}{n}$. Let $a\to 0$. Then we choose converging sequence of intervals (i.e. sequences of left and right ends are converges): $I_{a_k}\to I'$. Due to absolute continuity of Lebesgue's integral we obtain $I'$ is desired interval.


factorial
number theory proposed
number theory

For all $ n\in\mathbb{Z}_{+}$, let $ s(n)$ denote the rightmost nonzero digit of $ n$. Prove that $ s((5n)!)=s(2^{n}n!)$.


Let's see:

AoPS volume 1 & 2 1st, 2nd, and 7th edition.
AoPS intro to Geo, Counting & Probability, NT 1st edition.
AoPS intermediate C&P, Algebra 1st edition
Beginning Algebra
Intermediate Algebra
PreCalculus
Calculus
1,2,3, infinity

That's the ones I can think off the top of my head.

My dad's a tutor, and he teaches at our house, so a lot

idk i have lots so i put 5-7... must go count lol jk

more than 16 ...

i basicly learnt all the math i know from books

[quote]
I'm not very good at making polls, am I?[/quote]

maybe you should create a poll about that :lol:

[quote="Wichking"]

[quote]
I'm not very good at making polls, am I?[/quote]

maybe you should create a poll about that :lol:[/quote]

 :rotfl:

[quote="Urc"][quote="leoxnlin"]I'm not very good at making polls, am I?[/quote]

No, just the wrong forum.
Try asking it on an english forum.[/quote]

Then everyone would put 0 or 1. It'd be a bad poll too. *sigh*

Maybe I [i]should[/i] make a poll about how bad I am at making polls...


When a class ends, how long does the message board for that class stay before it gets deleted?

At the point the Message Boards remain indefinitely because after the class ends, the students are added to the archives.  This is a new feature that aggregates students from last semester on into a permanent forum in which they can hunt down other solutions to challenging problems and have indefinite access to the message board problems.

Wait so are there multiple forums for the same class?  There was a Introductory Number Theory class held in the Fall and now there is one being taught.  Will there be 2 separate forums?  Or will both classes use the same forum?  Sorry I get confused easily..

There is a forum for each ongoing class.  Before the next class begins we shift members of that forum into being members of the archive forum.  We also transfer all posts from that class to the archives.  Then we populate the class forum with a new class and the cycle begins anew.

[quote="sran"]How about transcripts?[/quote]

We still leave transcripts up for a month after class.  It is up to students to either be responsible in saving transcripts or to find other students in their class who will email them to them.


how do we prove that
sum of reciprocals of  first n natural numbers  where n>1 is not an integer?

this is of course been discussed in this forum and so let me direct you to the correct topic by clicking
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37336]HERE HERE HERE HERE HERE[/url]


induction
number theory solved
number theory

1) Prove, using induction, that every positive integer greater than one may be written as a product of prime numbers.

2) Bertrand's postulate states that for every number x > 1, there is a prime number between x and 2x. Use this fact to show that every positive integer can be written as a sum of distinct primes.

I don't get the induction in the first one? :?
Is it a prime? Yes: done., No: split off factor that causes compositeness. Loop cycle.

The second one is false. 4 cannot be written as the sum of distinct primes.

The second is well-known to hold for $n \geq 7$.

Pierre.


If $ a>0$ and $ b<0$, which is larger, $ (2a\minus{}2b)$ or $ (2a\minus{} |2b|)$?

Since $ b<0$, $ 2a\minus{}2b$ is lager than $ 2a\minus{}|2b|$.

$ b < 0 \implies b < |b| \implies \minus{}2b > \minus{} |2b| \implies (2a \minus{} 2b) > (2a \minus{} |2b|)$

Hence, $ \boxed{(2a \minus{} 2b)}$ is larger.


[quote="pbornsztein"]...and it would be cooler if you didn't post it directly into the solved problems section... ;)

Pierre.[/quote]
thanks but what do you mean here is unsolved problems. ;)

Sure. Because I moved it here...

Pierre.

[quote="jensen"]we have $100$ coins such that the weight of coins are from $1$ to $100$ and none of them have equal weight.we have a scale such that it shows the weight of coins on it.(you can put one coin,two coin,...,one hundred coin on it).with at least how many scaling we can find the coin that it weight is $1$? :D[/quote]i think this problem is cool and it seems like a game so i want you to think about it.(i promise that you will enjoy it) :D

[quote="cm"]is it 99? I know my answer is stupid though...[/quote]
no your answer is not stupid,but i think the answer is very smaller thatn your answer(may be the answer is less than or equal to $10$.).i don't know the answer,so if you find the answer,tell me. :D

[quote="Mathx"]Just one measuring[/quote]
how?

yes,i really don't how should i measure such that gave us exactly the true answer,i mean may be we get some answers but there is smaller one?what should we do?


The number $2300000_4$ is a number written in base 4.

For example, $123_4 = 1\times 4^2 +  2\times 4 + 3 = 27$

Write the number that is 1 less than $2300000_4$ in base 4.

Consider, first, an arbitrary number such as 2300 in base 10. Clearly, if one were to subtract 1 from this number, the resulting number would be 2299. What has been done is each zero that comes before a non-zero integer is reduced to 9, which is the largest single digit value in base 10, and the first integer encountered is reduced by 1.

Applying this same idea to base 4, we know 3 is the largest single digit value, so $2300000_4-1=2233333_4$.

since it is in base 4, 0-3 are the numbers used to represent any quantity.
One less than 2300000 in base 4 is the question.  since 3 is the greatest number digit in the base , 3+1=0 so  2300000b4-1b4=2233333
[b]
2233333 base4
[/b]


integration
real analysis
real analysis solved

Show that an=\int{0->n}sinx*sinx2dx has a finite limit as n->00.


I'm trying to write the shorthand notation for a periodic continued fraction, which involves a bar going above the repeated section - much like the notation for complex conjugate. How do I do this in Latex?

[quote="stevem"]\overline as in $ 0.1\overline{23}$[/quote]

Thank you very much!


geometry
3D geometry
quadratics
algebra
quadratic formula

There is only one set of four consecutive positive integers such that the sum of the cubes of the three of them is equal to the cube of the fourth number. What is the smallest of these four integers?

[quote="The QuattoMaster 6000"][hide] Well, since you will get a cubic if you try to assign variables to the numbers (which I would prefer to avoid), I did guess and check: Try {1,2,3,4}: That doesn't work.  Try {2,3,4,5}: That doesn't work.  Try {3,4,5,6}: This works.  Therefore, the smallest number of the set is $3$. [/hide][/quote]
That way would probably take a long time if the number was high.

By the way, how do you factor $x^{3}-6x-9=0$?

Cubic formula(not worht it), or rational root theorem.

Actually, the cubic formula is much simpler when there's no quadratic term, that's how Descartes? derived it, he got rid of the $x^{2}$ term first.

Anyway, we are told that they are positive integers, so RRT tells us that 1, 3, 9 are the only possibities (we rule out the negatives since it says positive)

Then we can find that 3 is a root and apply the factor theorem:
$x^{3}-6x-9$
$(x-3)(x^{2}+3x+3)$
Then apply the quadratic formula from there, to find that it's no longer factorable over the reals.

I may memorize the cubic and quartic formulas again...it's tough to keep them memorized unless you practice it a ton...


When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a wholenumber quotient with a reminader of 1. How many distinct values are possible for this three digit dividend?

[hide]I got $51$
so obviously the number dividing the dividend thingy can't be even, 1, or 5. So the options are 3,7, and 9. For 3, when the hundreds digit is 1, there are three options for the second digit, if hundreds is 2, then 4, if hundreds is 3, then 3, if hundreds is 4, 3, so the pattern is 3+4+3+3+4+3+3+4+3 (kinda guessed on that.) For 7, since the only number divisible by 7 with units digit 2 is 42, and you keep adding 70, there are 13 of those with 3 digits and units digit 2. For 9, there is 172, 262, 352, etc. until 802, which is 8 numbers. 30+13+8=51[/hide]

So... didn't you overcount the ones with 9, as every multiple of 9 is a multiple of 3?

[hide]Oh. Right. Hehe
Is it just 43? [/hide]

All we have to do is find the number of three-digit integers that have a units digit of 2 with a remainder of 1 when divided by 3 or 7.

We have 30 three digit integers that have a units digit of two and have a remainder of 1 when divided by three namely the numbers in the set of ${112, 142, 172, ..., 982}$

We have 12 three digit integers that have a units digit of two and have a remainder of 1 when divided by seven namely the numbers in the set of ${162, 232, 302, ..., 932}$

We also have 4 three digit integers that have a units digit of two and have a remainder of 1 when divided by three or seven namely the numbers in the set of ${232, 442, 652, 862}$

So the number of three digit integers is $30+12-4$=$38$


function
logarithms
trigonometry
inequalities
limit
complex numbers
real analysis

HI! I need to check the following series for convergence $ \sum_{n \equal{} 2}^{\infty}\frac {( \minus{} i)^n}{\log(n)}$, where the log-function is taken to base greater than 1.

We have:
$ \sum_{n \equal{} 2}^{\infty}\frac {( \minus{} i)^n}{\log(n)} \equal{} \sum_{k \equal{} 0}^{\infty}(\frac { \minus{} 1}{\log(4k \plus{} 2)} \plus{} \frac {i}{\log(4k \plus{} 3)} \plus{} \frac {1}{\log(4k \plus{} 4)} \plus{} \frac { \minus{} i}{\log(4k \plus{} 5)}) \equal{} \\
\sum_{k \equal{} 0}^{\infty}(\frac {\log(4k \plus{} 2) \minus{} \log(4k \plus{} 4)}{\log(4k \plus{} 2)\log(4k \plus{} 4)} \plus{} i\frac {\log(4k \plus{} 5) \minus{} \log(4k \plus{} 3)}{\log(4k \plus{} 3)\log(4k \plus{} 5)}) \equal{} \\
\sum_{k \equal{} 0}^{\infty}(\frac {\log\frac {4k \plus{} 2}{4k \plus{} 4}}{\log(4k \plus{} 2)\log(4k \plus{} 4)} \minus{} i\frac {\log\frac {4k \plus{} 3}{4k \plus{} 5}}{\log(4k \plus{} 3)\log(4k \plus{} 5)})$

Now, since $ \frac {\log(n)}{\sqrt {n}} < \varepsilon$ for large $ n$, then:

$ \frac {\log\frac {4k \plus{} 2}{4k \plus{} 4}}{\log(4k \plus{} 2)\log(4k \plus{} 4)}\ge\frac {\log\frac {1}{2}}{(\log(4k \plus{} 4))^2} > \frac {\log\frac {1}{2}}{(4k \plus{} 4){\varepsilon}^2}$ and
$ \frac {\log\frac {4k \plus{} 3}{4k \plus{} 5}}{\log(4k \plus{} 3)\log(4k \plus{} 5)}\ge\frac {\log\frac {3}{5}}{(\log(4k \plus{} 5))^2} > \frac {\log\frac {3}{5}}{(4k \plus{} 5){\varepsilon}^2}$

So the divergence of the initial series follows from the divergence of the harmonic series.

Second method:
Since $ ( \minus{} i)^n \equal{} \cos\frac {3\pi{}n}{2} \plus{} i\sin\frac {3\pi{}n}{2}$, we have:
$ \sum_{n \equal{} 2}^{\infty}\frac {( \minus{} i)^n}{\log(n)} \equal{} \\
\sum_{n \equal{} 2}^{\infty}\frac {\cos\frac {3\pi{}n}{2}}{\log(n)} \plus{} i\sum_{n \equal{} 2}^{\infty}\frac {\sin\frac {3\pi{}n}{2}}{\log(n)} \equal{} \\
\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^{k \plus{} 1}}{\log(2k \plus{} 2)} \plus{} i\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^{k \plus{} 2}}{\log(2k \plus{} 3)}$

But what is it? It strikes that $ \frac {1}{\log(m)}$ falls to 0. It turns out, that we can apply the Leibniz criterion to get two convergent alternating series. Where is the error?

It is true that if $ |z| \equal{} 1,z\ne 1$ and $ b_n$ is sequence that is eventually positive, decreasing, and tends to zero, then $ \sum b_nz^n$ converges.

This statement includes the alternating series test as a special case, and it includes this series directly with $ z \equal{} \minus{} i.$ All you need to know about $ \frac1{\log n}$ is that it decreases and tends to zero.

The proof is by (Abel's) summation by parts. Sum the $ z^n$ and difference the $ b_n.$

[quote="JBL"]Looks to me like the error is in replacing $ \log \frac {4k \plus{} 2}{4k \plus{} 4}$ with $ \log \frac {1}{2}$ -- the former expression tends to 0 as $ k \to \infty$!

(By the way, if you draw a picture, it becomes immediately clear that this is converging, in the same way that the alternating series test is immediately clear if you draw a picture, i.e., it's not actually obvious how to prove it but it allows you to convince yourself that it really happens.  The partial sums of this series are trapped in a shrinking rectangle.)[/quote]

This is what I can't grasp at the moment: $ \log \frac {4k \plus{} 2}{4k \plus{} 4}$ is negative and tends to 0 from below as long as $ \frac {4k \plus{} 2}{4k \plus{} 4}$ tends to 1 from below. But the first element in the sequence is $ \log\frac{1}{2}\equal{}\minus{}\log(2)$, which is the lower bound for $ \log \frac {4k \plus{} 2}{4k \plus{} 4}$, so I don't see why the inequality should not hold.

On the other hand I see the convergence with the picture at hand, which you described above.

 [quote]The proof is by (Abel's) summation by parts. Sum the  and difference the [/quote]

Hm, could you make it explicit, please.

The real part of the series converges by the alternating series test, as does the imaginary part. Therefore the series converges.

OK, more explicit. Assume $ b_n$ is positive and decreasing for $ n\ge n_0$ and that $ \lim b\equal{}0.$ And assume that $ |z|\le 1$ but $ z\ne 1.$

$ \sum_{n\equal{}n_0}^{N}b_nz^n\equal{}b_N\frac{1\minus{}z^{N\plus{}1}}{1\minus{}z}\plus{}\sum_{n\equal{}n_0}^{N\minus{}1}(b_n\minus{}b_{n\plus{}1})\frac{1\minus{}z^{n\plus{}1}}{1\minus{}z}.$

Now, take the limit of that as $ N\to\infty.$ Note that since $ \left|\frac{1\minus{}z^{N\plus{}1}}{1\minus{}z}\right|\le\frac2{|1\minus{}z|}$ and $ b_N\to 0,$ the "boundary term" tends to zero. That gives us

$ \sum_{n\equal{}n_0}^{\infty}b_nz^n\equal{}\sum_{n\equal{}n_0}^{\infty}(b_n\minus{}b_{n\plus{}1})\frac{1\minus{}z^{n\plus{}1}}{1\minus{}z},$ provided the latter sum converges.

We claim that the right hand sum converges absolutely. We estimate the absolute values, and use that $ |b_n\minus{}b_{n\plus{}1}|\equal{}(b_n\minus{}b_{n\plus{}1}).$ That's the one place we use the hypothesis that $ b_n$ is decreasing. (We don't need strictly decreasing - I should have said "nonincreasing".)

\[ \left|\sum_{n\equal{}n_0}^{\infty}(b_n\minus{}b_{n\plus{}1})\frac{1\minus{}z^{n\plus{}1}}{1\minus{}z}\right|\le\frac{2}{|1\minus{}z|}\sum_{n\equal{}n_0}^{\infty}(b_n\minus{}b_{n\plus{}1})\le\frac{2b_{n_0}}{|1\minus{}z|}.\]

So we have absolute convergence by comparison to a convergent telescoping series. (Of course, we haven't said that the original series converges absolutely, and indeed this whole theorem would be pointless if it did.)

We even have an error estimate - a statement about how close you get if you add up so-and-so many terms. The error is no bigger than twice the next term divided by $ |1\minus{}z|.$ Note that if you put $ z\equal{}\minus{}1$ here you get exactly the alternating series test, including the error estimate. The error estimate also tells us that this convergence is uniform on intervals bounded away from $ 1$ but not uniform on all of $ \{|z|\equal{}1,z\ne 1\}.$

[quote="JBL"]Ah, you're right -- the inequality is valid.  It's just the conclusion that you've drawn from it that's invalid ;)  You've shown that your real series is termwise larger than a divergent series -- what theorem are you trying to apply to conclude divergence?  What are the precise conditions of that theorem, and are they satisfied here?  (Hint: the word "negative" is really important ;))

Also: summation by parts:
http://en.wikipedia.org/wiki/Summation_by_parts
http://mathworld.wolfram.com/SummationbyParts.html[/quote]

Alright, I completely forgot the fact, that the comparison test can only be applied to series with nonnegative terms!

Thank you, JBL!

Also thanks to you Kent for writing down the complete proof. But I'll study it tomorrow.

Cheers.


D is a point on BC in triangle ABC such that BD=2DC. E is the midpoint of AD. when line BE is extended, it meets point F on line AC. thus find the ratio of the area of triangle BED to the area of triangle AEF

[hide="Solution"]
$\frac{AE}{ED}\cdot\frac{DB}{BC}\cdot\frac{CF}{FA}= 1 \iff \frac{CF}{FA}=\frac{3}{2}$
by Menelaus' theorem
$\triangle BED = \frac{ED}{AD}\triangle BAD = \frac{ED}{AD}\left (\frac{BD}{BC}\triangle ABC \right ) = \frac{1}{3}\triangle ABC$
$\triangle AEF = \frac{AF}{AC}\triangle AEC = \frac{AF}{AC}\left ( \frac{AE}{AD}\triangle ADC \right ) = \frac{AF}{AC}\left ( \frac{AE}{AD}\left ( \frac{DC}{BC}\triangle ABC \right ) \right ) = \frac{1}{15}\triangle ABC$
$\therefore \triangle BED : \triangle AEF = \frac{1}{3}\triangle ABC : \frac{1}{15}\triangle ABC = 5 : 1$
[/hide]

can someone explain Menelaus' theorem to me? how do you determine which sides to take?

$\{\begin{array}{c}\frac{\boxed{\cdot}\ A}{\boxed{\cdot}\ B}\cdot\frac{\boxed{\cdot}\ B}{\boxed{\cdot}\ C}\cdot\frac{\boxed{\cdot}\ C}{\boxed{\cdot}\ A}=1\\\\ P\in AB\ ,\ M\in BC\ ,\ N\in CA\end{array}$ $\implies$ $\frac{PA}{PB}\cdot \frac{MB}{MC}\cdot\frac{NC}{NA}=1$.

okay.. thks Virgil Nicula(:
i think i understand more now
how about using other methods to solve this prob? not using the theorem

yay that is so how i use menelaus' theorem also

you can also use mass points

[use google search]

note that the methods are really the same

[quote="pigley"][color=darkred]$D$ is a point on $BC$ in triangle $ABC$ such that $BD=2\cdot DC$ and $E$ is the midpoint of $AD$.
When line $BE$ is extended, it meets point $F$ on line $AC$. Find the ratio of the area of $\triangle BED$ to the area of $\triangle AEF$.[/color][/quote]
[color=darkblue][b][u]Proof[/u][/b] (without the Menelaus' theorem). Define $[XYZ]$- the area of $\triangle XYZ$.

Denote $[AEF]=x$, $[BED]=y$ and the point $G\in BE$ for which $DG\parallel AC$.

Observe that $[ABE]=[ADC]=y$, $[DGE]=x$, $[BDG]=[BEFC]=y-x$. Therefore,

$\frac{[BGD]}{[BCF]}=\left(\frac{BD}{BC}\right)^{2}$ $\implies$ $\frac{y-x}{2y-x}=\frac{4}{9}$ $\implies$ $y=5x$, i.e. $[BED]=5\cdot [AEF]$.[/color]

Let $(XYZ)$ be the area of triangle $XYZ$
Note that $(BEC)= \frac{3}{2}(ABE)$
So that $(ECF)= \frac{3}{2}(AEF)$
But $(AEC)= \frac{1}{6}(ABC)$
Thus $(AEF)= \frac{1}{15}(ABC)$
It is obvious that $(BED)= \frac{1}{3}(ABC)$
And we get the desired result.


let n: positive integer. Prove that $2<(n+\frac{1}{n})^n<3$         :rotfl:

I think what was meant is that $2<(1+\frac{1}{n})^n<3$, which is true for all $n>1$, or $2\le(1+\frac{1}{n})^n<3$ for all $n$.

By the way, $(1+\frac{1}{n})^n$ approaches $e$ as $n$ approaches $\infty$, which is how we can tell that it's less than $3$ for all $n$.

[quote="JesusFreak197"]By the way, $(1+\frac{1}{n})^n$ approaches $e$ as $n$ approaches $\infty$, which is how we can tell that it's less than $3$ for all $n$.[/quote]

That fact is not enough to prove that $2\le (1+\frac{1}{n})^n<3$ for $n\in\mathbb{N}$. You must also show that $n=1$ satisfies $2\le (1+\frac{1}{n})^n<3$.

OK, let me try:

If $n=1$, $(1+\frac{1}{n})^n=2$. $(1+\frac{1}{n})^n$ approaches $e$ as $n$ approaches $\infty$. Q.E.D.

[quote="pontios"][quote="mathcat"]$2<(n+\frac{1}{n})^n<3$[/quote]

$(2+\frac{1}{2})^2 = (\frac{5}{2})^2 =  \frac{25}{4} = 6.25>3$[/quote]

what a fool i am! i got tricked by it  :ewpu:
again, i did Search, and this is already discussed in here several times

[quote="NoSoupForYou"]That fact is not enough to prove that $2\le (1+\frac{1}{n})^n<3$ for $n\in\mathbb{N}$.[/quote]

Umm... if I had been trying to completely prove the whole thing, I would have done so. I considered the lower bound easy enough to test.

$(1+\frac{1}{n})^n=1+1+\frac{n-1}{2n}+\frac{(n-1)(n-2)}{6n^2}+..$
Btw, the original inequality is not true because n=1->value is 2.  It has to be $\ge 2$.  But anyways....
That entire expression...let it be $S$.  Then
$S \le 2+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}+...<2+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=3$
QED.

Dear JesusFreak.
Thanks for correctting this problem.
I use Cauchy to prove sequence:
$u_{n}=(1+\frac{1}{n})^{n}$ is a increase sequence\\
So, $u_{1}\leq u_n \leq u_\infty $
Now,too easy and tasty

Proof of convergence for the number $e$

\[a_n=\left( 1+\frac{1}{n} \right)^n
=1+ \frac{1}{n} \cdot {n \choose 1}+\frac{1}{n^2} \cdot {n \choose
2} +...+ \frac{1}{n^k} \cdot {n \choose k}\]
\[=\sum_{k=0}^n {n \choose k} \cdot \frac{1}{n^k}
=\sum_{k=0}^n \frac{n!}{(n-k)! \cdot k! \cdot n^k}\]
\[=1+1+\frac{1}{2!} \cdot \frac{n(n-1)}{n^2}+\frac{1}{3!} \cdot
\frac{n(n-1)(n-2)}{n^3}+...+\frac{1}{n!} \cdot \frac{n!}{n^n}\]

Pulling out a $1$ and factoring we obtain
\[a_n=1+\sum_{k=1}^n
\frac{1}{k!} \cdot \left(1-\frac{1}{n}\right) \cdot
\left(1-\frac{2}{n}\right) \cdot ... \cdot
\left(1-\frac{k}{n}\right)\]

Therefore \[a_{n+1}=1+\sum_{k=1}^{n+1} \frac{1}{k!} \cdot
\left(1-\frac{1}{n+1}\right) \cdot \left(1-\frac{2}{n+1}\right)
\cdot ... \cdot \left(1-\frac{k}{n+1}\right)\]

Since each term in the Binomial expansion of $a_n$ is less than the term in the expansion of $a_{n+1}$ we conclude that $a_{n+1}>a_n$ for all $n$. Also, we can derive the following:

\[a_n=1+\sum_{k=1}^n
\frac{1}{k!} \cdot \left(1-\frac{1}{n}\right) \cdot
\left(1-\frac{2}{n}\right) \cdot ... \cdot
\left(1-\frac{k}{n}\right)<1+\sum_{k=1}^n \frac{1}{k!}\]

This is true since we replaced each term in the form of $ \left(1-\frac{k}{n} \right)$ with $1$. The RHS is convergent with a sum less than three, therefore $a_n<3$ and it is a monotone increasing sequence. Also, since $a_1=2$ then $a_n>2$. Thus we can conclude that $2<a_n<3$ and by definition this number is called $e$.

\[(*) \quad S=\sum_{k=0}^{\infty}
\frac{1}{k!}=1+1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{(n+1)!}+...\]
\[S<1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}+...=3\]

Be sure to mention that $\left(1+\frac{1}{n}\right)^n$ is strictly increasing for a certain interval. It is not enough to mention that the expression satisfies the inequality for $n=1$ if it isn't actually the lower bound.
Similarly, failing to mention this fact nullifies any point in stating that as $n\rightarrow\infty$, the expression approaches $e$. Horizontal asymptotes can be crossed.

DPopov, your first $\frac{1}{n^k}\binom{n}{k}$ should actually be $\frac{1}{n^n}\binom{n}{n}$. Also, I'm not sure, but you might be off by a term in your factorization: $\displaystyle a_n=1+\sum_{k=1}^n\frac{1}{k!} \cdot \left(1-\frac{1}{n}\right) \cdot\left(1-\frac{2}{n}\right) \cdot ... \cdot\left(1-\frac{k}{n}\right)$.

Also, I would like to see your Cauchy proof, mathcat, if you wouldn't mind. :)

That's fine!!! I'm very glad to help you!!!
We prove $u_n \leq u_{n+1}$
$\Leftrightarrow (1+\frac{1}{n})^n\leq(1+\frac{1}{n+1})^{n+1}$
$\Leftrightarrow n+2 \geq (n+1)\sqrt[n+1]{(1+\frac{1}{n})^n}$
Use Cauchy with (n+1) terms:
$n+2=(1+\frac{1}{n})+...+(1+\frac{1}{n})+1\geq (n+1)\sqrt[n+1]{(1+\frac{1}{n})^n.1}$

[quote="JesusFreak197"]What exactly is Cauchy?[/quote]
USE THE SEARCH FEATURE


real analysis
function
real analysis theorems

"Fubini's theorem was stated for Borel functions; [color=red][i]naturally[/i][/color], we want to use these theorems when the given function $ f$ is [color=red][i]only [/i][/color]known to be Lebesgue measurable."

-why would we "naturally" want to use these theorems when $ f$ is only known to be lebesgue measurable? And why "only" Lebesgue measurable?

I don't understand the question. Of course it's desirable to be able to use the theorem with weaker conditions than borel measurable?

thanks for your answer. i see now that we would want a weaker condition. :0)


[hide]Solution for those with poor math skills:

$(7x+y+3z)^{3}-(x-2y+z)^{2}-13$[/hide]


Let {Xn, n>= 0} be a Markov chain with M states. Suppose that i,j are in the state space E. Prove that there is an m <= M such that Pij^(m) > 0.

How do I prove this?

" Pij^(m) > 0 " is the probability of transitioning from state i to state j in exactly m steps.

The key observation is that if you can reach state $ j$ from state $ i$ at all, then you can reach it in at most $ M$ steps. Indeed, let $ E(m)$ be the set of states reachable from $ i$ in not more than $ m$ steps. Note that if $ E(m) \equal{} E(m \plus{} 1)$, then all $ E(k)$ with $ k\ge m$ coincide. On the other hand, we can have a strict inclusion $ E(m)\subset E(m \plus{} 1)$ only $ M \minus{} 1$ times (otherwise the last set $ E(m)$ will have more states than $ M$).

[quote="fedja"] Indeed, let $ E(m)$ be the set of states reachable from $ i$ in not more than $ M$ steps..[/quote]

I think there is something incomplete :maybe:

Just a stupid misprint :blush:. I fixed it.

JimmyT
Your Markov chain is irreducible isn't it?!


k be a square-free,k>1,l be an odd interger.if (m,n) is a root of this interger equation:
     x^2-k(k.l^2-1)y^2=1prove n is even

:roll: Your problem is nice but it isn't difficult.


Find all primes $ p$ such as $ 2^p\plus{}p^2$ is a prime as well.

Right, right, right... Was just about to edit.

[quote="coquitao"]- If $ p$ is a prime number, the expression $ 2^{p} \plus{} p^{2}$ is always a composite number divisible by $ p$.

The claim is certainly true for $ p \equal{} 2$. If $ p > 2$, the well-know FLT allows to conclude that $ 2^{p} \equiv p \mod p$, whence $ 2^{p} \plus{} p^{2} \equiv p \plus{} p^{2} \equiv 0 \mod p$. Hence, $ p | 2^{p} \plus{} p^{2}$ and the veracity of our claim follows.[/quote]That's an erroneous application of Fermat's Little Theorem. If you think about it... $ 2^p$ is divisible by $ p$ when $ p>2$?  :huuh:

The expression is not a prime number if $ p \equal{} 2$ and it is equal to $ 17$ when $ p \equal{} 3$. We claim that $ p \equal{} 3$ is the only prime number for which [b]hasan[/b]'s formula yields a prime number.

If $ p$ is  prime number greater than $ 3$, then $ p$ is of the form $ 6k \plus{} 1$ or $ 6k \plus{} 5$. In any of the previous cases, we have that $ 2^{p} \equiv 2 \mod 3$. Since $ p^{2}$ is certainly equal to $ 1$ modulo $ 3$, we conclude that $ 2^{p} \plus{} p^{2}$ is divisible by $ 3$ whenever that $ p$ is a prime number greater than $ 3$, and the result follows.


The limit of the sum of an infinite number of terms in a geometric progression is $ \frac {a}{1 \minus{} r}$ where $ a$ denotes the first term and $ \minus{} 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is:

$ \textbf{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textbf{(E)}\ \text{none of these}$

[hide="Solution"]
The original infinite geometric series is $ a \plus{} ar \plus{} ar^2 \plus{} ar^3 \plus{} \ldots$ and the geometric series of the sum of squares is $ a^2 \plus{} a^2r^2 \plus{} a^2r^4 \plus{} a^2r^6 \plus{} \ldots$. So the limit of the sum of their squares is

$ \boxed{(C)\frac {a^2}{1 \minus{} r^2}}$
[/hide]


Let real numbers $ a,b\in (0,1)$; $ a \geq b$,Prove that: $ a^{a}\plus{}b^{b}\geq a^{b}\plus{}b^{a}$

See here: http://www.mathlinks.ro/Forum/viewtopic.php?t=118722

 :)   :maybe:


Prove that there are infinitive many integers k such that $\lfloor\sqrt{2}\;*n\rfloor = 2^{k}$ where n is a positive integer.

Hm... I'm sure I've done this task before but don't seem to remember the solution for it... Anyone have any clue?

This is actually a shortlist problem from the IMO of 1985. You can find a solution (it is problem 19) here:

[url]http://www.kalva.demon.co.uk/short/sh85.html[/url]

Oh!

Thank you very much...

Heh... I knew I had seen it somewhere before... ;)


Let $ x_1,\ x_2,\ x_3,\ x_4,\ x_5$ be real numbers such that $ \left\{
\begin{array}{ll}
x_1x_2\plus{}x_1x_3\plus{}x_1x_4\plus{}x_1x_5\equal{}\minus{}1 &\quad  \\
x_2x_1\plus{}x_2x_3\plus{}x_2x_4\plus{}x_2x_5\equal{}\minus{}1 &\quad  \\
x_3x_1\plus{}x_3x_2\plus{}x_3x_4\plus{}x_3x_5\equal{}\minus{}1 &\quad  \\
x_4x_1\plus{}x_4x_2\plus{}x_4x_3\plus{}x_4x_5\equal{}\minus{}1 &\quad  \\
x_5x_1\plus{}x_5x_2\plus{}x_5x_3\plus{}x_5x_4\equal{}\minus{}1 &\quad
\end{array}
\right.$

Find the possible value of $ x_1$.

[hide]
For brevity let $ \sum_{i\equal{}1}^5 x_i \equal{} k$

So from what we are told; $ k\minus{}x_i \equal{} \frac{\minus{}1}{x_i}$

$ \Rightarrow x_i^2 \minus{} kx_i \minus{}1 \equal{} 0 \Leftrightarrow x_i \equal{} \frac{k \pm \sqrt{k^2\plus{}4}}{2}$

$ \Rightarrow k \equal{} 5\left(\frac{k \pm \sqrt{k^2\plus{}4}}{2}\right) \leftrightarrow k\equal{}\pm \frac{1}{\sqrt{2}}$

$ \Rightarrow \frac{\minus{}1}{x_1} \equal{} \frac{1}{\sqrt{2}} \minus{} x_1$

$ \Rightarrow x_1 \equal{} \frac{1\pm 3}{2} \leftrightarrow x_1 \equal{} \minus{}1,2$[/hide]

[quote="ocha"]$ \frac { - 1}{x_1} = \frac {1}{\sqrt {2}} - x_1$

$ \Rightarrow x_1 = \frac {1\pm 3}{2} \leftrightarrow x_1 = - 1,2$[/quote]

$ \frac { - 1}{x_1} = \pm \frac {1}{\sqrt {2}} - x_1$

$ \iff %Error. "sqer" is a bad command.
{2}x_1^2 \pm x_1 - \sqrt{2} = (\sqrt{2}x_1 \mp 1)(x_1 \pm \sqrt{2}) = \leftrightarrow x_1 = \pm \frac{1}{\sqrt{2}}, \ \pm \sqrt{2}$


One mole of nitrogen ($\gamma=\frac 75)$ and one mole of helium ($\gamma=\frac 53)$ are mixed and contained in a volume of $15\times 10^{-3}m^3$ at $250K$ ,The mixture is then compressed adiabatically to a volume of $5.0\times 10^{-3}m^3$. Calculate , for the mixture ,

(i)the effective value of $\gamma$
(ii)the final temperature
(iii)the ratio of the initial to the final pressure

** I dont know how to calculate the effective value for $\gamma$ .So if someone can help me on the first one , then the second and third can be done .Thanks . :)

[quote="azereyvazov"]Adiabatic compression constant $\gamma$ can be found such a way
$\gamma=\frac{c_p}{c_v}$
$(\nu_1+\nu_2)c_p=\nu_1 c_{p1}+\nu_2 c_{p2}$
$(\nu_1+\nu_2)c_v=\nu_1 c_{v1}+\nu_2 c_{v2}$
$\gamma=\frac{\nu_1 c_{p1}+\nu_2 c_{p2}}{\nu_1 c_{v1}+\nu_2 c_{v2}}=\frac{c_{p1}+c_{p2}}{c_{v1}+c_{v2}}$
From the values of $c_p$ and $c_v$'s we find that
$\gamma=\frac{3}{2}$
You can find other parts yourself.[/quote]

Honestly , I cant get what you mean .  :huh:
What is $\nu_1,\nu_2$ ? and what is $c_{v1}$,$c_{v2}$ those thing ??

$\nu_1,\nu_2$ are mole numbers of gases,
$c_v, c_p$ are the specific heat capacity of the gases in
constantvolume and constant pressure respectively.

ok thanks . But I never saw the formula

$(\nu_1+\nu_2)c_p=\nu_1c_{p1}+\nu_2c_{p2}$ before  :(

And my teacher just tell me that the way to get the effective $\gamma$ is just the mean of the two $\gamma$ ..... though they are very near  :D ($1.5333$ and $1.5$)


floor function
number theory
least common multiple

How many positive integers less than $2008$ are not divisible by $6$ or $10$?

[hide] by inclusion-exclusion we get

2008-(334 + 200)+33 = 1507 [/hide]

hold on, you got the 33 form 2007/60 rounddown right? I don't think that's right

you have to be a bit more careful...

it is $2008-([\text{numbers divisible 6}]+[\text{numbers divisible 10}]-[\text{numbers divisible 6 and 10}])$

this you have to be careful with the last term, it is:

$\left\lfloor \frac{2008}{\text{lcm}(6,10)} \right\rfloor$

the rest follows

[quote="jli"]first of all, it is 2007-blah blah blah

second, it is 60 not lcm(6,10)[/quote]

Why would it be 60? It would be 30 because we overcounted all the multiples of 30.

Yes, I believe that according to PIE, you must add back in the numbers you subtract twice, which are the multiples of 6 and 10, so use lcm(6,10).

[hide]$\lfloor\frac{2007}{6}\rfloor + \lfloor\frac{2007}{10}\rfloor - \lfloor\frac{2007}{30}\rfloor = 468 \implies 2007 - 468 = \boxed{1539}$
It's subtracted from $2007$ because we want the number [b]less than[/b] $2008$[/hide]

[hide]I used PIE, and I got 1539

$1539$

LCM(6,10)=30[/hide]

Do you learn at school to solve this kind of problems??

No, unfortunately my school doesn't have that many advanced math students, so contest-like math problems like this aren't taught. You have to learn on your own from a book like The Art of Problem Solving or maybe on a math team (if your school has a good one).

So, do I have do buy this book? I am from Azores (google it!), it is difficult for me to buy it. Isn't there a pdf document with tutorials explaining this kind of problems?

I would really try to buy the book. It would help you more than most things.
You could take classes from this site. Also, you could look at local exams from across the u.s. There are lots of accesible (on the internet) exams that you could practice with. Just go to the list of exams on this site (not the olympiad list). If you practice enough, you'll start discovering things on your own.


Hi, first of all i dont know where to post this so sorry

secondly, i have two problems:

suppose you have the choice of investing 1000 in just one of the two schemes. each scheme will pay you amount listed at the end of every year for next 5 years

year              1        2        3        4         5
scheme1      225    215     250      225      205
scheme2      220    225     250      250      210

assume bank rate at 4.33% compounded yearly. which scheme you would choose???

really i dont know what to do with this! should i choose scheme2 since after 5 years the scheme2 will pay 210 ???

the second problem is:
i give you 100 today OR 110 after 10 months with bank rate 10% p.m
which offer you should take?

here i cant use the c= p(1+r/m)^mt so what to do ?

thats all thanks :)

thnx mr Peter but the link does not work :(


inequalities
LaTeX
inequalities proposed

Indeed easy.

Expand, it comes out to
\[
\sum_\mathrm{cyc}\left(\frac{a^2}{b^2} + \frac{b}{a}\right) \geq 3 + \sum_\mathrm{cyc} \frac{a}{b}
\]

Now $\frac{a^2}{b^2} + \frac{a^2}{b^2}  + \frac{b}{a} \geq 3\frac{a}{b}$ and $\frac{a^2}{b^2} + \frac{b}{a}  + \frac{b}{a} \geq 3$ by AM-GM. Sum its three cyclic forms yields the desired.

[quote="billzhao"]Now $\frac{a^2}{b^2} + \frac{a^2}{b^2}  + \frac{b}{a} \geq 3\frac{a}{b}$ and $\frac{a^2}{b^2} + \frac{b}{a}  + \frac{b}{a} \geq 3$ by AM-GM. Sum its three cyclic forms yields the desired.[/quote]

Alternatively, I noted that $\frac{a}{b}, \frac{a}{b}, 1, \frac{b}{a}$ are in order of size, so by rearrangement:

$\frac{a}{b} * \frac{a}{b} + 1 * \frac{b}{a} \geq \frac{a}{b}*1 + \frac{a}{b} * \frac{b}{a}$

PS: What is latex for multiplying?

here's an other way
we have $2\frac{a}{b}+\frac{b}{c}=\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\frac{a}{\root{3}\of{abc}}$
the same $2\frac{b}{c}+\frac{c}{a}\geq 3\frac{b}{\root{3}\of{abc}}$
and $2\frac{c}{a}+\frac{a}{b}\geq 3\frac{c}{\root{3}\of{abc}}$
by suming the three we get $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{\left( a+b+c\right) }{\root{3}\of{abc}}$ (1)
on the other hand : $\frac{a}{b}+2\frac{b}{c}=\frac{a}{b}+\frac{b}{c}+\frac{b}{c}\geq 3\frac{\root{3}\of{abc}}{c}$
the same : $\frac{b}{c}+2\frac{c}{a}\geq 3\frac{\root{3}\of{abc}}{a}$
and $\frac{c}{a}+2\frac{a}{b}\geq 3\frac{\root{3}\of{abc}}{b}$
by suming the three we get $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \root{3}\of{abc}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $ (2)
finally by multiplying (1) and (2) we get the inequality desired.

I like this solution, because it doesn't include so many calculations... :
since it is homogeneous, let $abc=1$. By Cauchy we have that \[(\frac ab + \frac bc + \frac ca)(a+b+c) = (\frac ab + \frac bc + \frac ca)(\frac 1{ab}+\frac 1{bc}+\frac 1{ca})\geq (\frac 1a+\frac 1b+\frac 1c)^2\]
\[(\frac ab + \frac bc + \frac ca)(\frac 1a+\frac 1b+\frac 1c) = (\frac ab + \frac bc + \frac ca)(ab+bc+ca)\geq (a+b+c)^2\]
Multiply both ineq and the conclusion follows  :lol:


calculus
integration
trigonometry
geometry
geometric transformation
reflection
real analysis

For what is worth here are some calculations:

$ \int_{0}^{\pi/2}\sqrt[4]{1-\sin x}dx=\int_{0}^{\pi/2}\left(\left(\cos(x/2)-\sin(x/2)\right)^{2}\right)^{1/4}dx$

$ =\int_{0}^{\pi/2}\sqrt{\cos(x/2)-\sin(x/2)}dx$.

Using the substitution $ x=2t$ we get that

$ =2\int_{0}^{\pi/4}\sqrt{\cos(t)-\sin(t)}dt=2\int_{0}^{\pi/4}\sqrt{\cos(\pi/4-t)-\sin(\pi/4-t)}dt$

$ =2^{5/4}\int_{0}^{\pi/4}\sqrt{\sin t}dt$.

Perhaps the last integral can be expressed in terms of eliptic functions.

[quote="didilica"]Perhaps the last integral can be expressed in terms of elliptic functions.[/quote]
Mathematica confirms this. However, it seems that
\[ \int_{0}^\frac{\pi}2 \sqrt{\sin x}\, dx = \sqrt{\frac2{\pi}}\cdot \Gamma^{2}\left( \frac34 \right) . \]
Can anyone prove this? :maybe:

[quote="didilica"]
$ =2^{5/4}\int_{0}^{\pi/4}\sqrt{\sin t}dt$.

Perhaps the last integral can be expressed in terms of eliptic functions.[/quote]

Sure. By Maple, $ \int_{0}^{\pi/4}\sqrt{\sin x}\,dx=-\sqrt{2}\,{\text{EllipticF}}\left( \sqrt{2}\sqrt{{\frac{\sqrt{2}}{2+\sqrt{2}}}},1/2\,\sqrt{2}\right)+\sqrt{2}\,{\text{EllipticPi}}\left( \sqrt{2}\sqrt{{\frac{\sqrt{2}}{2+\sqrt{2}}}},1/2,1/2\,\sqrt{2}\right)$

You are right perfect-radio.

$ I=\int_{0}^{\pi/2}\sqrt{\sin(x)}dx=\sqrt{\frac{2}{\pi}}\Gamma^{2}\left(\frac{3}{4}\right)$.

Here is the proof of your result:

Since  $ \beta(p,q)=2\int_{0}^{\pi/2}\sin^{2p-1}\theta \cos^{2q-1}\theta d\theta$,  $ Re p>0$, $ Req>0$ we get that

$ I=\frac{1}{2}\beta\left(\frac{3}{4},\frac{1}{2}\right)=2\sqrt{\pi}\frac{\Gamma\left(3/4\right)}{\Gamma\left(5/4\right)}$.

Now $ \Gamma(5/4)=\frac{1}{4}\Gamma(1/4)$ which combined with the reflection formula

$ \Gamma(1/4)\Gamma(3/4)=\frac{\pi}{\sin \pi/4}$ gives the desired result.