probability
calculus
integration
geometry
3D geometry
tetrahedron
algebra

If 3 real numbers within (0,2) are randomly picked what is the probability that
a. the sum of the numbers is less than 1
b. the product of the numbers is lesser than 1

a is elementary geometric probability, of course, but I don't see how to solve b without a triple integral, which, obviously, isn't HSB level.

[quote="mathemonster"][hide=" a) HINT"]a) plot the region $ 2 > (x,y) > 0$  and then find where it intersects with x + y = 1. The result follows from there[/hide][/quote]

You mean $ 2 > x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} 1$, forming a tetrahedron and a cube. :wink:

Temperal: Maybe that's the reason why there are so few numbers. I think the best way to approach this problem is with a tree diagram.
a) [hide]There is one possibilitiy, and that is  $ P(0 ,0,0)$, hence
$ 1/27$[/hide]

b) [hide]Here it's actually much easier to count the ones who are not less than zero, and that is
$ P(2,2,2), P(2,2,1), P(2,1,1), P(1,1,1)$
Then we have $ 1 \minus{} (1/27) \minus{} (1/27)3 \minus{} (1/27)3 \minus{} 1/27 \equal{} 19/27$[/hide]

Hope it's not me beeing stupid  :maybe:

[quote="befuddlers"]If 3 [b]real numbers[/b] within (0,2) are randomly picked what is the probability that
a. the sum of the numbers is less than 1
b. the product of the numbers is lesser than 1[/quote]

Not integers.

its like a little advanced version of 2d geometry.

in 2d geometry problem like this, if you select a value of x, then the value of y gets selected automatically.

by same method, if you select a value of x and y, then the value of z gets selected automatically.

[hide="Tree Diagram"]
A tree diagram just lists out possibilities. For example, if you were to have $ 2$ toppings of ice cream (say V and C) to choose from, $ 3$ times, your tree might look like

      [geogebra]60d7edcf00500ab4b5cb93edaf37f6f6c6c6ef0a[/geogebra]
[/hide]
[hide="Tetrahedron in a Cube"]
These are the graphs of $ 0 < x,y,z < 2$ (the cube) and $ x \plus{} y \plus{} z \equal{} 1$ (the tetrahedron) in the 3-D space.

If you have trouble visualizing it, try $ 0 < x,y < 2$ and $ x \plus{} y \equal{} 1$ on the 2-D plane first, and then see if you can find the 3-D analog.
[/hide]

[quote="befuddlers"][quote="Camlon"]Temperal: Maybe that's the reason why there are so few numbers. I think the best way to approach this problem is with a tree diagram. [/quote]
Can you explain what a tree diagram is.
@Math 154: Yes the numbers are real numbers not integers.
(@Fantasy Lover and Math 154:) Can you explain how you got a tetrahedron in a cube.

Edit: Guys could you please check your solutions as the options were 1/48, 1/24, 2/27, 1/8, 1/7[/quote]

I have to inform you that tree diagram wasn't appropriate in this case, because it was a continous interval. I thought they were integers. Anyway, a tree diagram is more like a diagram you make in your head to understand probability much easier. It's a very good thing to learn for starters. For instance if you have a coin, then you will have a starting point and it will branch into two, because there is two options, then to four, 8, 16, etc.. A tree diagram with three options will look like this http://math.youngzones.org/tree_diagram1.gif

You must have seen this before. Then you can use this quite easialy for any type of probability question, with or without replacement. For instance if you have a bag of 40 balls, 15 red and 25 blue. Then the probability to get
10 red and 5 blue is $ ((15!/5!)(25!/20!)/40!25!) (15C5), 15C5$, because there is $ 15C5$ ways you can organize 10 red and 5 blue and each of the probabilities are identical. Now, there are other ways for this question, but this tree diagram works in every simple case if there is discrete countable amount of choices.

[hide="b"]
Instead of finding the probability, we can approximate it and find a discrete analog.

In other words, we can blow it up to something like $ 0<x,y,z<100$ and $ xyz<50$, and then the approximation will be pretty accurate.
[/hide]

The point that Temperal was trying to make was the following: in order to get an exact solution, you really DO need to use a triple integral. I guess I'll set it up, for those of you who are interested in this solution.

So, as has been said before, the sample space is obviously the cube in the first quadrant with $ 0 < x,y,z < 2$
Now, the condition that the product of the three numbers = 1 can be represented as
$ z \equal{} \frac {1}{xy}$, which will be our surface.

So our triple integral for the volume formed by the surface $ z \equal{} \frac {1}{xy}$ and the planes $ y \equal{} 0, x \equal{} 0$ is the following:
$ \int_0^2 \int_0^\frac {1}{x} \int_0^{\frac {1}{xy}} 1 dz dy dx$

Now, this works out to $ \int_0^2 \minus{} \frac {\ln(x)}{x} dx$ which is divergent if done like that, but because the domain tells us we're not actually including 0, we're free to take the limit of the integral as the lower bound approaches 0.
$ \lim_{k \to 0^{ \plus{} }} \int_k^2 \minus{} \frac {\ln(x)}{x} dx$ =

Crap. Which still diverges... Did I do something wrong?

Anyway's here's the plot of the region in question:

[img]http://img181.imageshack.us/img181/4339/xyzlessthan1zd2.png[/img]

I have figured out the error in the previous post. It had to do with the bounds on the integrals I was using.

Let's look at it again.
In the entire [0,2] x [0,2] x [0,2] region of the surface we're looking at, we can break it down into two subsections:
The part from [0.25, 2] x [0.25, 2] x [0.25, 2] contains the "curvy" part of the surface, and the region below it is entirely rectangular in form.

We can treat this volume of this surface then as a sum of an integral and a bunch of rectangular solids.

The integral becomes

$ \int_{0.25}^{2} \int_{0.25}^{\frac{1}{x}} \int_{0.25}^{\frac{1}{xy}} 1 dz dy dx$,

which is easily workable to

$ \frac{1}{64} (7 \minus{} 48 \ln 2 \plus{} 480 (\ln 2)^2)$

We know that the volume of the weird looking base + the volume of the rest (7/4)^3 must be equal to the total volume, so

$ v_{\text{base}} \plus{} \left( \frac{7}{4} \right )^3 \equal{} 8$

$ v_{\text{base}} \equal{} \frac{169}{64}$

Now, the total volume of the region in question is

$ \frac{169}{64} \plus{} \frac{1}{64} (7 \minus{} 48 \ln 2 \plus{} 480 (\ln 2)^2)$

$ \equal{}\frac{(11 \plus{} 30 \ln 2^2 \minus{} \ln 8)}{4} \equal{} 5.83354$

So the probability that three numbers randomly chosen from the interval $ (0,2)$ have product less than 1 is $ \frac{\frac{(11 \plus{} 30 \ln 2^2 \minus{} \ln 8)}{4}}{8} \equal{} \frac{5.83354}{8} \equal{} 0.729$

Grah. This doesn't appear to be right either, by a quick java program
[code]public class ProductTest
{

	public static void main(String[] args)
	{
		final int numTries = 1000000;
		int numSuccesses = 0;
		for(int i = 0; i < numTries; i++)
		{
			double x = 2*Math.random();
			double y = 2*Math.random();
			double z = 2*Math.random();
			if(x*y*z < 1)
			{
				numSuccesses++;
			}
		}
		double probability = (double)numSuccesses / numTries;
		System.out.println(numSuccesses);
		System.out.println(numTries);
		System.out.println(probability);

	}
[/code]

The probability I'm getting with this code seems to be about 0.655~
What's wrong with the other argument? :(

Probably just the bounds.

If you invert it so that you have $ (2 \minus{} x)(2 \minus{} y)(2 \minus{} z) \equal{} 1$, then the integral:

$ \int_0^{7/4} \int_0 ^{2 \minus{} \frac {1}{2(x \minus{} 2)} } \int_0 ^{2 \minus{} \frac {1}{(x \minus{} 2)(y \minus{} 2)}} dz dy dx$

correctly gives the volume of the complement region.

Am I mistaken, or is the volume of the tetrahedron in (a) actually $ \frac{1}{6}$, giving $ \frac{1}{48}$...?


Αν $a,b,c$ θετικοί πραγματικοί αριθμοί να δείξετε ότι:

$\ abc \leq \frac{(a+b+c)^{3}-(a^{3}+b^{3}+c^{3})}{24}\leq \frac{(a+b+c)^{3}}{27 }$

Ελπίζω να μην είναι πολύ γνωστή  :roll:

Αριστερή:

Πράξεις... $\Leftrightarrow 6abc \leq a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b$ που ισχύει από AM-GM.

Δεξιά:

$\Leftrightarrow 9(a^{3}+b^{3}+c^{3}) \geq (a+b+c)^{3}$ που ισχύει αφού (εφ. 2 φορές Chebyshev):

$9(a^{3}+b^{3}+c^{3}) \geq 3(a+b+c)(a^{2}+b^{2}+c^{3}) \geq (a+b+c)^{3}$

ωραια ανισοτητα  :) .την ελυσα με Caushy-Schwarz  σαν τη λυση του cmad


algebra
polynomial
algorithm
advanced fields
advanced fields open

Hello,

I know this is not the exact subforum in the page to put this question, but I didn't know exactly where it should belong.

Some days ago a professor from a course in engineering told in class that he found an article about some guys in India that were able to prove that factoring is a P problem, that is, a problem that can be solved in polynomial time. He told that talking about RSA. He said that the problem is that the algorithm has to high degree and too high and that it becomes faster, or more efficient than the usual algorithms for numbers with more than 2^100 digits.

I asked after class and he didn't want to give me any other reference. BUT I DON'T BELIEVE IT!, I mean, I think that factoring is an NP hard problem, and therefore, if someone proved that can be solved in polynomial time, well, I am not from theorical computer sciences, but I belielve that this is equivalent to prove P = NP, isn't it?

I think that may be someone found a deterministic algorithm to test primality in polynomial time (all the algorithms that I have heard about that are polynomial time are probabilistic) and that wouldn't be that amazing for me, but that someone found an PT algorithm to factor, I don't think so!

Does anyone knows anything about it?

Best regards,

Thank you very much Koo. As I said, it is a deterministic algorithm to prime testing, not for factoring!

That is for prime testing only.  One  notable thing is the method is  easily understandable (original pre-print was  less than 10 pages and  required only simple tools).

There have been a few discussions  about this in this forum.

Factoring is a different matter though.

At least now we can prime factor all primes in polynomial time. :P

Not only that, for small primes, (eg $2^n$) we can find all  factors of $p^n$ . :)

Factoring is not known (or believed, I think) to be NP-hard.  It may be intermediate between P and NP-complete.

[quote]It is known that factoring is not NP complete.. so Factoring in P will not imply P = NP[/quote]
It would be astonishing if that were known, as it is probably more difficult than proving that NP differs from P.
It is certainly a plausible conjecture.

[quote="Ravi B"]Right.  It is not known whether factoring is NP-complete.  If factoring were not NP-complete, then $\textrm{P} \ne \textrm{NP}$.  Because the conclusion is not known, the hypothesis can't be known either.

It is believed that factoring is not NP-complete.  If factoring were NP-complete, then $\textrm{NP} = \textrm{co-NP}$, which would be a surprising result.[/quote]

you are right.. I actually meant that assuming $coNP \neq NP$ which is strongly believed, factoring cannot be NP complete

(by the way, is there any proof that if NP=coNP then P=NP?)

[quote="Peter VDD"]Didn't they invent a method to factor numbers  [ O(log(n)) ] by quantum computers? Not sure, it could also have been prime testing...[/quote]
They did.. it is Shor's algorithm. However a practical quantum computer does not exist yet and may not possibly exist for the next 10-15 years. (I may be wrong but it may not be possible to make a realistic quantum computer at all)

A Quantum Architecture isn't a traditional architecture. We should take that into account since a Turing Machine may not be able to model a Quantum Computer (at least without some serious overhead). So in fact, Factoring in polynomial time with a Quantum Machine may not necessarily put Factoring in P. Right?

There are some formal models of quantum computation, and what is efficiently computable in such models includes factoring (of course) but probably not all of NP.

[quote]Right. It is not known whether factoring is NP-complete. If factoring were not NP-complete, then \textrm{P} \ne \textrm{NP}.[/quote]

So is it the case that:
If someone creates a polynomial time factoring algorithm, thereby showing it is not NP-complete,
then they show P != NP?

If so, how does one prove that implication?

Your premise is mistaken.  Even if Factoring has a polynomial-time algorithm, it still could be NP-complete.  (After all, P could equal NP.)

Thanks for the quick reply.  Clears up one of my many misunderstandings.


superior algebra
superior algebra theorems

This is just a very easy question, however, I am having a hard time figuring out how to multiply permutations expressed in cyclic notation. Our professor is so fast so I can't catch up with him.

For example: In a group with 5 elements namely, ( 1 2 3 4 5 ),
if A = (1 3) and (B = 2 4 5) and C = (1 3 5), what is the value of AB, BC and AC.

Note: We know that the elements that do not appear in the parenthesis are fixed elements. That is, it is mapped to itsef.

Another question, permutations that are disjoint commute, how about those which are not? for example, A and C are not since they have the common element 3?

Thanks, I hope you can help me on this. Im confused.  :blush:

There's no particularly efficient way to do it- you just have to get grubby. For example, here's $BC$:
$1\to 3\to 3$
$2\to 2\to 4$
$3\to 5\to 2$
$4\to 4\to 5$
$5\to 1\to 1$
In cycle notation, that's $(1\,3\,2\,4\,5)$.

There are a few exceptions- products of disjoint cycles are nice.


Hi there. I'm having a problem getting TeXnicCenter to talk to Adobe Reader. I'm hoping someone here can help me. The problem is somewhere between building and viewing but I can't figure out exactly what.

I'm getting three different errors depending on what I do. Typically, as I work, I keep Adobe Reader open so I can easily reference what I'm working on and also periodically build-and-view to make sure everything is turning out right. But now, if I have the PDF open when I try to build, TeXnicCenter says "I can't write on file "<filename.pdf>" Please type another file name for output. No new PDF is created.

If I close Adobe Reader and then build-and-view, the build runs, then gives me an error box saying:
[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")] Cannot execute the command.
At this point, the new PDF has been created. I can open it separately through Adobe, but it will not open in Adobe through TeXnicCenter.

Based on advice I found through Google, I took the following steps and got a bit further but ran into a third error. I went to Build --> Define Output Profiles --> LaTeX => PDF Profile --> Viewer. The Commands under "View project's output" and "Forward search" were [DocOpen("%bm.pdf")][FileOpen("%bm.pdf")] with "DDE command" selected. Changing the "Forward search" fields to "Command line argument" and "%pm", then build-and-viewing gets as far as opening Adobe Reader, but it passes a .tex file to Adobe instead of a .pdf. Adobe tells me it cannot open <filename.tex> because it's not a supported file type. However, once again, a new PDF is created which I can open separately through Adobe. But the .tex file is being sent in place of that PDF.

Substituting "%bm" for "%pm" above changes the error message-- Adobe says it cannot find the file. But again, the new PDF was created, it's just not being sent to Adobe right.

Any help would be great. Thanks in advance for reading all of this and for helping!

PS Before anyone asks, I have double-checked that the "path to executable" is correct and is actually pointing to the Adobe Reader executable.

Could you show us a screen shot of the settings in Build --> Define Output Profiles --> LaTeX => PDF Profile --> Viewer? Then we can check that everything is OK. Which version of Adobe Reader are you using (in your googling you might have seen a bug in version 8.0 but earlier and later version should be fine)?

While you're doing this check that you get the same problem if your tex file is in a different directory such as C:\test.

Are you using XP or Vista? If it's Vista what happens if you run TexnicCenter as administrator (right-click on the icon and choose Run as administrator)?

Thanks for the quick reply!

I've attached two screenshots, the first is the settings after I changed it, the second is the way it was before I changed it.

I'm using "Adobe Reader 7.0, Version 7.1.0," TeXnicCenter 1 Beta 7.01 and Windows XP with SP3.

I tried saving the .tex file to a different directory and build-and-viewing from there and got the same results.

That first screenshot is exactly what I have and works well, so something else is happening. Need to check where the problem lies:

1) With TexnicCenter. Do LaTeX => DVI and see if the viewer works there, similarly for LaTeX => PS if you have GhostScript installed. If they work, it is a pdf problem and, if not, a TexnicCenter problem

2) Try another pdf viewer. There are plenty of free ones about. Ones that get good reviews are [url=http://www.foxitsoftware.com/pdf/reader/]Foxit[/url] and [url=http://blog.kowalczyk.info/software/sumatrapdf/index.html]Sumatra[/url] (this has [url=http://www.latex-community.org/forum/viewtopic.php?f=5&t=1616]forward and inverse search[/url]). If I remember correctly, neither need installing so easy to try and dispose of. Obviously you would have to change "Path of excutable" in TexnicCenter.

Make sure you do not have 2 versions of Acrobat Reader on your computer simultaneously. Sometimes such behavior can be caused by having an old version associated with the pdf extension and the new one in the TeXnic center profile definition or vice versa.

Thanks. I've double-checked, and although I do have two versions of Adobe reader on my computer, both TeXnic and Windows were pointing at the same version.

Build-and-viewing from TeXnic under LaTeX => DVI and LaTeX => PS both worked fine. I downloaded SumatraPDF and LaTeX => PDF worked just great with Sumatra, so it appears the problem was in Adobe Reader.

Just to cover my bases, I downloaded the latest versions of TeXnic (1.0 Stable Release) and Adobe (9.1) that I could find. The update for TeXnic didn't change anything, but the update for Adobe did. No idea just what the problem was, but it appears to have been fixed in Adobe Reader 9.1.

(That said, the problem only appeared with Adobe 7.0 after quite awhile. I'm not sure what triggered it then, but I guess it's possible the update just reset whatever it was so it's waiting to trigger the problem again. Either way, it works with Sumatra, so all's well.)

Thanks for your help, and for the pointers on the other PDF readers!

[quote="reddleman"]Thanks. I've double-checked, and although I do have two versions of Adobe reader on my computer, both TeXnic and Windows were pointing at the same version.[/quote]
I still suspect the cause was somehow related to this. The ways the programs call each other in Windows is quite fancy and besides two obvious places to check that I mentioned there are some hidden things concerning environmental variables, path priorities, error handling rules, etc. (most of which I don't really understand myself). Installing a new version usually resets all those and everything works fine but sometimes you can get such a glitch. Anyway, next time you get something like this, try to uninstall the version of Adobe Reader you no longer use completely and see if it helps  :).


What happens to Magnesium Sulfate if you leave it overnight?
(Does it react with oxygen)? :huh:

MgSO4 will not react with oxygen (even under drastic conditions) because sulphur in MgSO4 is in its maximum oxidation state(+6).
rather the salt may absorb moisture due to its deliquescent nature and form a hydrate.


just a quick question: what is cos(a)sin(a) equal to?

(I do not know how long this question may be so ignore the "quick question")...

$ \cos a \sin a \equal{} \frac{\sin(2a)}{2}$

hello, see also here:
http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html
Sonnhard.

[quote="Bachukas"]$ sin(a)cos(a)$ equal to a ratio between area and diagonal of rectagular.[/quote]

Let me clarify on that... in rectangle $ ABCD$, $ sin(a)cos(a)$, where $ a$ is the measure of $ \angle ABD$, is the ratio of the area of the rectangle to the square of the diagonal.


Hi everyone, I'm terrible in math. I'm sitting here and wondering about it and something came onto my mind.

let's say you start with $x_{1}$
and then make up an equation like in description, (or anything similar)
every consecutive x is like in a geometrical series, previous solution defines consecutive x, so for example if x = 50, then x_2 = 50 + Sqrt[50], x_3 = x_2 + Sqrt[x_2] and so on and on and on.

$x+\sqrt{x}+\sqrt{x+\sqrt{x}}+\sqrt{x+\sqrt{x}+\sqrt{x+\sqrt{x}}}+... = y$
example:

$x_{1}= 50$
$x_{2}= 50+\sqrt{50}$
$x_{3}= x_{2}+\sqrt{x_{2}}$
$x_{4}= x_{4}+\sqrt{x_{4}}$
$.....$

now, how do you solve for y?, and how would you write this under a summation sign?, can you write something like that?

Thanks for any reply,

p.s my main equation seems not to work, let me trace an error.

//me again, can't quite express myself. Rather than how to solve for y, is there any quick formula out there for solving such things? Or is it necessery to design one?

Generally, when you make up arbitrary problems using elementary functions, they will not have nice solutions.

But let me be clear:  You want to study the recurrence

$x_{n+1}= x_{n}+\sqrt{ x_{n}}$


calculus
integration
trigonometry
limit
complex analysis
complex analysis unsolved

First of all sorry for the double posting!

Show that

$ \int_{0}^{\infty} \frac{dx}{x^{\alpha}(x + 1)}$ $ = \frac {\pi}{\sin(\pi \alpha)}$   where $ 0 < \alpha < 1$

I tried to use something like a keyhole contourn... Thanks!

Yes, you should use a keyhole contour, which we'll call $ \Gamma$.  I'll leave it to you to make the integrals along the circular arcs vanish.  Let $ \gamma_{1}$ be the upper portion of the keyhole and $ \gamma_{2}$ the lower part.  Then
$ I\equal{}\int_{\gamma_{1}}f(z)\,dz$ and $ \int_{\gamma_{2}}f(z)\,dz\equal{}\minus{}e^{\minus{}2\pi i\alpha}\int_{\gamma_{1}}f(z)\,dz\equal{}\minus{}e^{\minus{}2\pi i\alpha}I$.
The singularities occur at $ 0$ and $ \minus{}1$, but only $ \minus{}1$ is within the contour.  So
$ \text{Res}(f,\minus{}1)\equal{}\lim_{z\rightarrow \minus{}1}\frac{1}{z^{\alpha}}\equal{}(\minus{}1)^{\minus{}\alpha}\equal{}e^{\minus{}\pi i\alpha}$ (why?).
Hence
$ \oint_{\Gamma}f(z)\,dz\equal{}I(1\minus{}e^{\minus{}2\pi i\alpha})\equal{}2\pi ie^{\minus{}\pi i\alpha}$.  With some algebra, we get
$ I\equal{}\frac{2\pi ie^{\minus{}\pi i\alpha}}{1\minus{}e^{\minus{}2\pi i\alpha}}\equal{}2\pi i\frac{e^{\pi i\alpha}}{e^{2\pi i\alpha}\minus{}1}\equal{}\pi\left(\frac{2ie^{\pi i\alpha}}{e^{2\pi i\alpha}\minus{}1}\right)\equal{}\boxed{\frac{\pi}{\sin(\pi\alpha)}}$


inequalities function integration induction

Prove that for all positive integers $n$ and $p$,

${ \sum_{k=1}^{n-1}k^{p}< \frac{n^{p+1}}{p+1}< \sum_{k=1}^{n}k^{p}}$

[hide="Main Idea."]Try to prove that
\[k^{p}\leq \frac{(k+1)^{p+1}-k^{p+1}}{p+1}\leq (k+1)^{p}. \]
You can do it with the Mean Value Theorem, but probably you can kill it in an elementary fashion too.[/hide]

[hide="Solution"]The function $f(x)= x^{p}$ is strictly increasing, so t is easy to see that
\[\sum_{k=1}^{n-1}f(k)<\int_{1}^{n}f(x)dx= \frac{n^{p+1}-1}{p+1}<\frac{n^{p+1}}{p+1}= \int_{0}^{n}f(x)dx< \sum_{k=1}^{n}f(k).\][/hide]

eeek...lol this identity was used in the early parts of the apostol calc book to prove some of those ideas...i believe there is an induction solution

[hide]
it's just putting appropriate (able to be summed easily) bounds on the terms. In particular, we will be done if we can prove
\[k^{p}<\frac{(k+1)^{p+1}-k^{p+1}}{p+1}<(k+1)^{p}\]
because adding from $k=1$ to $k=n-1$, it gives
\[\frac{1}{p+1}+\sum_{k=1}^{n-1}k^{p}<\frac{n^{p+1}}{p+1}<\frac{1}{p+1}+\sum_{k=2}^{n}k^{p}\]
However, the inequality
\[k^{p}<\frac{(k+1)^{p+1}-k^{p+1}}{p+1}<(k+1)^{p}\]
follows immediately from difference of $(p+1)$st powers:
\[(k+1)^{p+1}-k^{p+1}=\sum_{j=0}^{p}(k+1)^{p}k^{j-p}\]
where the sum has $p+1$ terms, all at least $k^{p}$ (equality only when $j=0$) and at most $(k+1)^{p}$ (equality only when $j=p$).
[/hide]


$x,y,z,w$ are real numbers and $xy=4$ and $z^2+4w^2=4$ prove this
   inequality:
   $(x-z)^2+(y-w)^2 \geq 1.6$

is there any one who can slove this problem? :?:


man, it is simple,

United States of America Gangstar Olympiad.  ;)

yeah, dude, i remember that thing!

i won that thing two years ago, it was awesome, yo'll goin do it?  i'll tell yo'll some tips some time if yo'll wanna win.

Where do you register?  Can international students register too? :D

as winner of the 2004 Waterville Open, i get a free entry right?

it's sad that he bothered to make 3 different usernames for this thread.

I disupte your win. My round hasn't been played yet, but I would have destroyed you.

[quote="texas137"][quote="southside"]can anyone tell me what is USAGO?[/quote]

Is it possible that your mean USACO? That's the USA Computing Olympiad. It isn't strictly a US thing. They have free training materials and internet programming contests open to anyone in the world. http://www.usaco.org[/quote]

Dude, his name is southside.  What do you think he means?   :lol:

And if international students can enter, I'm sure they'll be whomped on by the Americans.

what is up, Bo Hong, Steve,
man, who is tetrahedr0n and Magnara from River Forest?

fred, you can take it up with the Waterville Open Director.  but i'm pretty sure all decisions are final.

besides, I heard you got knocked out in the first round by wice.

yo, what is up, fred
is that yo, tetrahedron?
man, why did yo let Alex Wise knock yo out, man?  i thought yo were cool.
but anyway, if yo want to register for USAGO, just let me know, i will put yo in, man.

by the way, fred, if yo really want to do USAGO,(which i am sure will be good for your college appl.) you got to take the AGC and AIGE first, me and drahorn both got perfect on it.  Steve was the Waterville open winner, so he will receive free entrance to USAGO.

------------- USAGO winner and IGO director: JP

Letian, what was your USAGO score?
My AIGE and AGC scores were good, I had a USAGO index of 234, but I only got 1 on the USAGO.   :?

puhahahaha how come I only saw this now!!!

LT youre awesome.
I went to IGO as a member of korean national team once.


Is there some way to send official AMC scores to colleges? Is this done automatically? Do I have to tell AMC which schools I'm applying to? Thanks.

You don't have to send over scores officially, like you do with SAT's or w/e. That being said, if you say that you got something really high, and it catches someone's eye, they can find out if it's true or not. Some things you can do:

1. Record your score in your résumé/CV, if you like it under some section along the lines of "Honors and Awards"

2. Some schools will have blanks on the application where you can fill in your score.
- Some schools, like MIT, explicitly ask for AMC/AIME scores
- Some schools have a section that allows you to enter in any additional test scores you want to include. It is usually appropriate to include AMC scores in this kind of situation.

3. If you're getting a letter of rec from a math teacher, and they talk about your success in math competitions, they can throw it in there

Good luck with your college apps

which college are you planning on going to?


You used to be able to press the up arrow to get to past entered messages, which was useful for when I made mistakes putting in answers, now you can't do that.

Yea, I used that feature a lot in classes, it would be nice to have it back. Or maybe at least a button that when clicked shows a list of previous inputs (if not all of it, maybe the last 5 or 10?).

Hi,

We had forwarded this request to the creators of the software several weeks ago. I'm sure they have put it on their wish list, but there is no guarantee of it being implemented.


For P2 in R2:
Is it just me or is $a_n$ not used in the problem?
Also, by saying that $a_1, a_2 ...$ are consecutive, does that mean that they are consecutive in this order. i.e. if it's increasing, $a_k = a_{k-1}+1$ and if decreasing, $a_k=a_{k-1}-1$

They are in increasing order (ie $a_1<a_2<\ldots < a_n$). I'm sorry for forgetting to mention this!  :blush:  :oops:
Of course $a_n$ is not used (otherwise the problem would be ... trivial :D). I have corrected the statement.


Let $ a, b, c$ be positive real numbers such that $ a \plus{} b \plus{} c \leqslant 1$. Find maximum value of following expression
\[ a^2 (1 \minus{} a)^2 \plus{} b^2 (1 \minus{} b)^2 \plus{} c^2 (1 \minus{} c)^2 \minus{} a^2 (b \plus{} c)^2 \minus{} b^2 (a \plus{} c)^2 \minus{} c^2 (a \plus{} b)^2.\]

[quote="alexilic"]Let $ a, b, c$ be positive real numbers such that $ a \plus{} b \plus{} c \leqslant 1$. Find maximum value of following expression
\[ a^2 (1 \minus{} a)^2 \plus{} b^2 (1 \minus{} b)^2 \plus{} c^2 (1 \minus{} c)^2 \minus{} a^2 (b \plus{} c)^2 \minus{} b^2 (a \plus{} c)^2 \minus{} c^2 (a \plus{} b)^2.
\]
[/quote]
Before I post my solution I want to check the answer :)
Is it [hide]
$ \frac {51 \sqrt{17} \minus{} 107}{4096}$
[/hide] ?

You are wrong.
The maximum is achieved for triple $ \left(\frac13,\frac13,0\right)$.


function
inequalities
pigeonhole principle
floor function
Additive Number Theory

Show that every integer greater than $1$ can be written as a sum of two square-free integers.

"][b]Lemma 1.[/b]$n\geq 6$Let $p_{1}<\dots<p_{k}$ be all primes not greater than $n$
then $\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}<\frac{n-1}{2n}$
[b]proof[/b]
:
\[\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}=\frac{1}{4}+\frac{1}{9}+\sum_{i=3}^{k}\frac{1}{p_{i}^{2}}< \frac{1}{4}+\frac{1}{9}+\sum_{l=5}^{n}\frac{1}{l^{2}}=\frac{1}{4}+\frac{1}{9}+\frac12\sum_{l=5}^{n}{(\frac{1}{l-1}-\frac{1}{l+1})}=\frac14+\frac19+\frac18-\frac{1}{n+1}<\frac{1}{2}-\frac{1}{2n}=\frac{n-1}{2n}\]
so there at most $\sum_{p\in P}{ \frac{n-1}{p^{2}}}<\frac{n-1}{2}$integer less than $n$ not being square-free integer.so there must exist $k$ such that $k$ and $n-k$ both square-free.(by a famous thoerem with the first letter "P",I forgot its name :oops: )

Probably the stupidest proof possible: We can check that it holds for $n \le 10^4$. Now, let $S$ be the set of all squarefree integers, except for the primes larger than $10^4$. Then by the fact that the Schnirelmann density of the set of squarefree integers equals $\dfrac{53}{88}$ $^{[1]}$ and some decent estimate on the prime counting function$^{[2]}$, we have that the Schnirelmann density of $S$ must be larger than $\dfrac{1}{2}$. By Mann's Theorem$^{[3]}$ we now have that every positive integer can be written as sum of at most $2$ elements of $S$. In particular, every prime number can be written as sum of $2$ elements of $S$, and every integer that is not squarefree can be written as sum of $2$ elements of $S$. All there is now left, is proving the theorem for composite squarefree numbers; $n = pq = (p_1 + p_2)q = p_1q + p_2q$, where $p$ is the smallest prime dividing $n$ and $p_1, p_2$ are squarefree integers.

$^{[1]}$ http://www.jstor.org/pss/2034736
$^{[2]}$ http://en.wikipedia.org/wiki/Prime-counting_function#Inequalities
$^{[3]}$ http://mathworld.wolfram.com/MannsTheorem.html


[quote="kstan013"]At what time between 6:00 and 7:00 will the two hands of a clock be pointed in opposite directions of each other (form a line). Round your answer to the nearest second.[/quote]
If it's exclusive, never.
Inclusive, at 6:00

I think...

[quote="kstan013"]At what time between 6:00 and 7:00 will the two hands of a clock be pointed in opposite directions of each other (form a line). Round your answer to the nearest second.[/quote]

which two hands? if its the minute and hour, there are no time (for exclusive), or 1 time (for inclusive): 6:00
if its the minute and the second, there are multiple times.
if its the hour and the second, there are also multiple times.

Okay that was too easy. How about between 2:00 and 3:00?

[hide]The amount of degrees the minute hand travels is 6x (where x is the number of minutes passed) and the hour hand travels a total of 60+x/2. Now if they form a straight angle,  you find that the difference in the total number of degrees they travled must equal 180 degrees
6x-(60+x/2)=180

11x/2-60=180

11x/2=240
x=480/11

So the answer is 2 hours  43 min and 420/11 seconds [/hide]


limit
logarithms
calculus
calculus computations

These are all probably very simple to you, but I am having difficulties.

Evaluate:

a)
\[ \lim_{x\to0} \frac{x}{e^x - \sqrt{2 - e^{2x}}} \]
b)
\[ \lim_{x\to\infty} \left(\frac{\ln(1 + x))}{x}\right)^x  \]
c)
\[ \lim_{x\to\infty} \left(\frac{x+2}{x}\right)^x  \]

Thank you! I am really lost... I can't evaluate a single limit similar to b) and c) for some reason.

This is for the first one, and for the others they are well known, just circle around the forum and you'll find them

1) let $t=e^{x}$ then

$\underset{x\rightarrow 0}{\lim }\frac{x}{e^{x}-\sqrt{2-e^{x}}}=\underset{t\rightarrow 1}{\lim }\frac{\ln t}{t-\sqrt{2-t}}=\underset{t\rightarrow 1}{\lim }\frac{\ln t}{t-1}\frac{t-1}{t-\sqrt{2-t}}$

$\underset{t\rightarrow 1}{\lim }\frac{\ln t}{t-1}=1$ and

$\underset{t\rightarrow 1}{\lim }\frac{t-1}{t-\sqrt{2-t}}=\underset{u\rightarrow 0}{\lim }\frac{u}{u+1-\sqrt{1-u}}$
$=\underset{u\rightarrow 0}{\lim }\frac{1}{1+\frac{1-\sqrt{1-u}}{u}}=\underset{u\rightarrow 0}{\lim }\frac{1}{1+\frac{1}{1+\sqrt{1-u}}}=\frac{2}{3}$

so $\underset{x\rightarrow 0}{\lim }\frac{x}{e^{x}-\sqrt{2-e^{x}}}=\frac{2}{3}$

I'll do (b), or what I think you might have meant by (b). For the original problem, as $x\to\infty$, we have that $\frac{\ln(1+x)}x\to0.$ That means we have a "$(0^+)^{+\infty}$" case, which is not an indeterminate case - it is unambigously zero. I'm going to guess that you really meant the problem to be the limit as $x\to0.$ It turns out that that's not indeterminate either.

Let $L=\lim_{x\to0}\left(\frac{\ln(1+x)}x\right)^x$

Then $\ln L = \lim_{x\to0}x\ln\left(\frac{\ln(1+x)}x\right)$

But as $x\to0,\,\frac{\ln(1+x)}x\to1$, so its logarithm goes to zero, and multiplying by $x$ only makes it go to zero faster. $\ln L=0$ so $L=1.$


geometry
circumcircle
trigonometry
inequalities
geometry proposed

$ABC$ is a triangle.$O$ circumsentre of $ABC$.$AO,BO$ and $CO$ meet the circumcicle of $ABC$ at $A',B'$ and $C'$recpectively.$S_O$ is area of interseqtion of the interior triangle $ABC$ and the interior of triangle $A'B'C'$
      Prove  That    $S_O\leq  (2/3)S_{ABC}$ ;)
See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=388279]www.mathlinks.ro/Forum/viewtopic.php?p=388279[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65594]www.mathlinks.ro/Forum/viewtopic.php?t=65594[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58188]www.mathlinks.ro/Forum/viewtopic.php?t=58188[/url]

Let $w=C(O,R)$ be the circumcircle of the triangle $ABC$ and
the point $P\in (ABC)$ (the interior of the triangle $ABC$). [b]I note:[/b]
$\blacksquare$ the area $[XYZ]$ of the triangle $XYZ$;
$\blacksquare\ \{A,A'\}\subset AO\cap w,\ \{B,B'\}\subset BB'\cap w,\ \{C,C'\}\subset CO\cap w;$
$\blacksquare\ A_1\in AC\cap B'C',\ A_2\in AB\cap B'C',\ B_1\in BA\cap C'A'$,
$B_2\in BC\cap C'A',\ C_1\in CB\cap A'B',\ C_2\in CA\cap A'B';$
$\blacksquare\ x_1=m(\widehat {PAC}),\ x_2=m(\widehat {PAB}),\ y_1=m(\widehat {PBA})$,
$y_2=m(\widehat {PBC}),\ z_1=m(\widehat {PCB}),\ z_2=m(\widehat {PCA})$.
[b]Then[/b] $[A_1A_2B_1B_2C_1C_2]\ \ge (\le )\ \frac 23 S\Longleftrightarrow\sum [AA_1A_2]\ \le (\ge )\ \frac 13 S\Longleftrightarrow$
$\sum \frac{\sin^2 y_1\sin^2 z_2\sin A}{\sin (y_1+z_1)\sin (y_2+z_2)}\ \le (\ge )\ \frac 13 \sin A\sin B\sin C.$

[u]Two particular cases:[/u]
$1^{\circ}.\ \ P: =H\rightarrow \sum \cos^4A\sin^2A\le\frac 13\sin^2A\sin^2B\sin^2C$ $\Longleftrightarrow \sum a^2\cos^4A\le\frac 13\left(\frac SR\right)^2;$.
$2^{\circ}.\ \ P: =O\rightarrow \sum a^2\cos^2B\cos^2C\le \frac 13\left(\frac SR\right)^2$ $\Longleftrightarrow\sum (a^2+b^2-c^2)(a^2+c^2-b^2)\le \frac 13\left(4S\right)^4.$

[b]Tiks, solved you these last two "horrible" inequalities ???[/b]

Hi Virgil,for $1^0$we have$\sum \cos^4A\sin^2A\geq \frac 13\sin^2A\sin^2B\sin^2C$, because $S_H\leq 2/3S$,and I  proved it :) .
For $2^0$I can say that $S_O\leq 2/3S$<=>$(ctgA*ctgB)^2+(ctgB*ctgC)^2+(ctgA*ctgC)^2\geq 1/3$but we know that
$(ctgA*ctgB)+(ctgB*ctgC)+(ctgA*ctgC)=1$and$a^2+b^2+c^2\geq 1/3(a+b+c)^2$so$S_O\leq 2/3S$ :lol:


I have just downloaded and am learning LaTex.  The WiKi instructions include the following:
[quote]First, make sure that the bar near the top of the TeXnicCenter window reads "LaTeX => PDF"; if it doesn't, change it so that it does.[/quote]
How do you "change it so that it does?"  Until I manage to do this, I can't view any of my output, so am stuck for now.

Run TexnicCenter's configuration wizard. Go to Build, Define Output Profiles then press the Wizard button at the bottom left.

I've got the wizard started and get to the screen which says "Enter the full path of the directory, where the executables of your Tex-distribution are located."

It rejects every directory that I try and I can't get past this screen.

What is the "executable" of my "Tex-distribution," anyway?  Is this the source code for LaTex, or is it the user-file that I create from LaTex?

See [[LaTeX:Downloads#TeXnicCenter]] number 7 which tells you the directory to use.

OK - I seem to be back on track now.  Thank you very, very much for your help!


Let $ x,y,z\equal{}const$ and $ a,b,c$ are variables. Find the condition of $ x,y,z$ such that the following system equation have at least one root $ \neq 0$ (it means that $ (a,b,c)\neq (0,0,0)$) :
\[ xa^2\plus{}yab\plus{}zac\equal{}0\]
\[ xb^2\plus{}ybc\plus{}zba\equal{}0\]
\[ xc^2\plus{}yca\plus{}zcb\equal{}0\]

If $ a\not \equal{}0\not \equal{}b,c\not \equal{}0$, then $ x^3\plus{}y^3\plus{}z^3\equal{}3xyz$.
If $ a\equal{}0,b\not \equal{}0\not \equal{}c$, then $ x^2\equal{}yz$.
If $ a\equal{}b\equal{}0,c\not \equal{}0$, then $ x\equal{}0$.
Therefore condition is $ xyz(x^2\minus{}yz)(y^2\minus{}xz)(z^2\minus{}xy)(x^3\plus{}y^3\plus{}z^3\minus{}3xyz)\equal{}0$.


if $ |z^3\plus{}\frac{1}{z^3}|\leq 2$ $ \implies$ $ |z\plus{}\frac{1}{z}|\leq 2$ ?, $ z_1,z_2 \in C$(complex numbers)

Doesn't seem to be true.  You can't just "take away" factors like that.

JRav: you make an awful lot of posts just like this one that add almost literally nothing to the discussion of the problems posted.  Why?  Either this problem is true, in which case you should post part or all of a proof, or it's false, in which case you should post some $ z$ that is a counter-example.  Your vague intuitition that the result can't possibly be true is just worthless spacefilling.  Please stop doing this.

[hide]Let
$ |z \plus{} \frac {1}{z}| \equal{} t \\
now |z^3 \plus{} \frac {1}{z^3}| \equal{} |z \plus{} \frac {1}{z}\parallel{}z^2 \plus{} \frac {1}{z^2} \minus{} 1| \\
\equal{} t(t^2 \minus{} 3)$
 now clearly we can see if $ t > 2 ,\ t(t^2 \minus{} 3) > 2$ hence proved[/hide]


The three sides of a triangle have lengths 4,6, and x. For what values of x is this triangle acute?

A. $ 4 < x < 14$
B. $ 2\sqrt5 < x < 2\sqrt13$
C. $ 2\sqrt6 < x < 2\sqrt10$
D. $ 2\sqrt6 < x < 2\sqrt13$
E. Not Here

[hide]In order for the triangle to be acute, we must have $ 4^2\plus{}6^2>x^2$ and $ 4^2\plus{}x^2>6^2$. Solving, we find that $ 2\sqrt{5}<x<2\sqrt{13}$, which is B.[/hide]


Hi dears. I am a new Iranian member .I have 7 good geometry problems
for you.(I have solved them) .

G1. Let C be the circumcircle of ABC and another circle C' with center
O which passes through B,C and cuts AB , AC in E,F. Let EF cut BC at Q
and AQ intersect C again in P. Prove that <OPA=90.

thanks ... and as a good member you will help us solve other problems in the forum, currently listed as unsolved,k ? :D

nice having you guys around!

First of all, I made some other notations: E and F are the points where C intersects AB and AC respectively, Q is the intersection of EF and BC, M is the intersection of BF and CE.

We know that QM perpendicular to AO (just check the things I posted for problems 2 and 4; the idea is that QM is the polar of A). Consider the inversion of pole A and power AE*AB=AF*AC. This inversion turns the circumcircle of ABC in the line EF. Because Q is on line EF and P is the intersection of AQ and the circle ABC, it means that P is the image of Q by this inversion. Let O' be the image of O by the inversion. It's not hard to prove that the image of O by the inversion we considered must be on the polar of A (just remember that if A is on the outside of the circle its polar is the line XY where AX and AY are tangent to the circle, and the inversion we considered has pole A and power AX^2; if A is inside the circle C' we do something similar). Since the polar of A is QM, it means that O' is on QM. Because O' is also on AO and AO perpendicular to QM we have angle QO'O=90 degrees. We also have AB*AE=AO*AO'=AP*AQ (we established that P is the image of Q), so quadrilateral PQOO' is cyclic. Because we have angle QO'O=90 degrees we must also have angle QPO=90 degrees, but angle QPO=angle OPA because Q, P, A are collinear, so angle OPA=90 degrees.

Hope I made it clear enough.


ummm...howd i give up in 5.155 sec in a 5 sec game? lol?

oh and game # 244649 i think.

That happens; I believed it was mentioned before. You can also Give Up before the 1.4 second limit.

That, my friend, is the power of giving up.

...No, I am not kidding.


inequalities
combinatorics proposed
combinatorics

If $ A_1$, $ A_2$, ..., $ A_n$ are $ n$ nonempty subsets of the set $ \left\{1,2,...,n \minus{} 1\right\}$, then prove that

$ \sum_{1\leq i < j\leq n}\frac {\left|A_i\cap A_j\right|}{\left|A_i\right|\cdot\left|A_j\right|}\geq 1$.

Nice problem.
If we multiply both sides by $ \prod|A_i|$, then on the RHS we obtain
the number of $ n$-tuples $ (a_1,\dots, a_n)$ with $ a_i\in A_i$ for each $ i$.
For each such $ n$-tuple there are indexes $ i,j$ such that $ a_i\equal{}a_j$ and
such $ n$-tuple is counted at least in the sumand of the LHS that corresponds
to indexes $ i$ and $ j$ (it is $ |A_1|\dots|A_n|\cdot |A_i\cap A_j|/(|A_i|\cdot |A_j|))$.
So the conclusion follows.
I hope this is correct.

Yes, this is the proposed solution.

Here is the original formulation of the problem:

[color=blue][b]E2309.[/b] Prove the following: Let $ A_1$, $ A_2$, ..., $ A_n$ be finite sets. If

$ \sum_{1\leq i<j\leq n}\frac{\left|A_i\cap A_j\right|}{\left|A_i\right|\cdot\left|A_j\right|}<1$,

then the sets $ A_1$, $ A_2$, ..., $ A_n$ have a system of distinct representatives (i. e., there are $ a_1$, $ a_2$, ..., $ a_n$ such that $ a_i\in A_i$ and $ a_i\neq a_j$ for $ i\neq j$).[/color]

This is equivalent to the above problem by Hall's matching theorem.

  darij


Let be $n\geq 4$ an integer. Show that when a convex $n$-gon is divided into triangles (not only triangulated!) there are everytime two triangles that share one side.

[quote]Let be n\geq 4 an integer. Show that when a convex n-gon is divided into triangles (not only triangulated!) there are everytime two triangles that share one side.[/quote]

I don't really get the "two triangles sharing one side" part, clarification please?  :?

OK, let me try:

Let $n\geq 4$ be an integer. Show that, if a convex n-gon is subdivided into triangles, then there must be two triangles sharing one [i]complete[/i] side.

Hereby, sharing a [i]complete[/i] side means that two vertices of one triangle coincide with two vertices of the other one. This is something different from just bordering on each other: Two triangles can border on each other also if a side of one of them is [i]part[/i] of a side of the other one.

[b]PS.[/b] I don't want to discourage anyone, but the problem was the hardest one in the Matboj. Both teams had wrong solutions.

[b]PPS.[/b] The Matboj I am talking about was a math battle at the 6th German IMO training seminar at the International University Bremen (this is where the IMO 2009 is supposed to take place). Both the 6th seminar and the matboj are actually premiere (I think until last year the training always consisted of 5 seminars only). By the way, for many of us, including me, it was the first math battle in their life - here in Germany, this kind of competition is organized rather seldomly.

  Darij


geometry
3D geometry
number theory solved
number theory

Are there such pairwise distinct natural numbers $ m, n, p, q$ satisfying $ m \plus{} n \equal{} p \plus{} q$ and $ \sqrt{m} \plus{} \sqrt[3]{n} \equal{} \sqrt{p} \plus{} \sqrt[3]{q} > 2004$ ?

Nice and easy:
  Take the numbers $ m=(2x^2+2x)^2, n=1, p=(1+2x^2)^2, q=8x^3$ with x a very large integer. Then the conditions are satisfied.

Let $ n\equal{}b^3$, $ q\equal{}a^3$, then $ m\equal{}k\minus{}b^3$, $ n\equal{}k\minus{}a^3$ and we have to find a solution to
$ \sqrt{k\minus{}b^3}\plus{}b\equal{}\sqrt{k\minus{}a^3}\plus{}a$
Let $ k\equal{}b^3\plus{}y^2$, then $ m\equal{}y^2$, we have to solve
$ y\plus{}b\equal{}\sqrt{b^3\plus{}y^2\minus{}a^3}\plus{}a$
$ (y\plus{}b\minus{}a)^2\equal{}b^3\plus{}y^2\minus{}a^3$
$ (b\minus{}a)(b\minus{}a\plus{}2y)\equal{}(b\minus{}a)(b^2\plus{}ab\plus{}a^2)$
$ 2y\equal{}a^2\plus{}b^2\plus{}ab\plus{}a\minus{}b$
Let $ a\equal{}2r$, $ b\equal{}2$. Then $ y\equal{}2r^2\plus{}3r\plus{}1$
So $ n\equal{}8$, $ q\equal{}8r^3$, $ m\equal{}4r^4\plus{}12r^3\plus{}13r^2\plus{}6r\plus{}1\equal{}(2r^2\plus{}3r\plus{}1)^2$, $ p\equal{}4r^4\plus{}4r^3\plus{}13r^2\plus{}6r\plus{}9\equal{}(2r^2\plus{}r\plus{}3)^2$
And we can make $ r$ arbitrarily large.


anyone can solve this?

following inequalities are also true (i think).
$a,b,c\ge0, a+b+c=3$ ->

$2(a^{2}b+b^{2}c+c^{2}a)+3abc \ge 13-(5-ab-bc-ca)^{2}$

$a,b,c\ge0, a+b+c=9$ ->

$227+(ab+bc+ca-23)^{2}\ge 2(a^{2}b+b^{2}c+c^{2}a)+3abc$


Find all functions $ f(x)$ such that $ [f(x)]^2f\left(\dfrac{1\minus{}x}{1\plus{}x}\right)\equal{}64x$ for $ x\not\equal{} 0,1,\minus{}1$.

[quote="tonypr"]Find all functions $ f(x)$ such that $ [f(x)]^2f\left(\dfrac{1 \minus{} x}{1 \plus{} x}\right) \equal{} 64x...........(1)$ for $ x\not \equal{} 0,1, \minus{} 1$.[/quote]

Put $ x\equal{}\dfrac{1 \minus{} x}{1 \plus{} x}$ we get
$ [f\left(\dfrac{1 \minus{} x}{1 \plus{} x}\right)]^2f(x)\equal{}64\dfrac{1 \minus{} x}{1 \plus{} x}............(2)$

$ \sqrt[3]{(1) \times (2)}$, we have $ f(x)f\left(\dfrac{1 \minus{} x}{1 \plus{} x}\right)\equal{}16 \sqrt[3]{\dfrac{x(1 \minus{} x)}{1 \plus{} x}}..........(3)$

$ \frac{(1)}{(3)}$, we have $ f(x)\equal{}4 \sqrt[3]{\dfrac{x^2(1\plus{}x)}{1\minus{}x}}$ for all $ x \neq 0,1,\minus{}1$ is the answer.

Notive that the function $ g(x) = \frac {1 - x}{1 + x}$ is a function of order 2; this means that $ g^2(x) = x$

Knowing this, if you plug in $ \frac {1 - x}{1 + x}$ as $ x$, you get

$ [f(\frac {1 - x}{1 + x})]^2 f(x) = 64(\frac {1 - x}{x + 1})$   (*)

Now if you take the first equation of that the problem gives us and square it, we will be able to eliminate that ugly $ f({\frac {1 - x}{1 + x})}$.

$ [f(x)]^4[f(\frac {1 - x}{1 + x})]^2 = (64x)^2$

If we divide this by (*), we get

$ [f(x)]^3 = \frac{(64x)^2}{64(\frac {1 - x}{x + 1})}$

$ [f(x)]^3 = \frac{64x^2}{\frac {1 - x}{x + 1}}$

$ [f(x)]^3 = \frac{64x^2(x+1)}{(1-x)}$

$ f(x) = \sqrt[3]{\frac{64x^2(x+1)}{(1-x)}}$

$ \boxed{f(x) = 4\sqrt[3]{\frac{x^2(x+1)}{(1-x)}}}$


Let x,y,z>0 such that xyz=x+y+z+2. Prove that
a) xy+yz+zx>=2(x+y+z);
b) sqrt(x)+sqrt(y)+sqrt(z)<=3/2*sqrt(xyz).

I have nice proff for the first one. For the second one, I have sln, but not so nice, so I will looking for the better one.

For the first part: We always can choose two number, say x, and y which either both >=2, either both <=2. So
    (x-2)(y-2) >= 0 => xy + 4 >= 2(x+y)
We will show that
     (x+y)z >= 4 + 2z (*)
But is trivial because z = (x+y+2)/(xy-1) and (*) is equivalent to
(x+y)^2 >= 4xy.

Namdung

I must tell you that your solution is absolutely beautiful. Congratulations.

    My solutions are based (as said ) on calassic inequalities. We have an expression xyz=x+y+z+2. Thus, it is very probable that we could establish a relation of the form sum 1/(x+a)=b, for a and b independent of x,y,z. A quick check shows that in fact a=1 and b=1. Thus, we have sum 1/(x+1)=1. Take a=1/(x+1), b=1/(y+1) and c=1/(z+1). Then x=(1-a)/a=(b+c)/a and so on.
   Now, the first part becomes sum (b+c)(c+a)/ab>=2 sum (b+c)/a and after some computations reduces to Schur.
   The second part becomes sum sqrt( ab/(a+c)(b+c))<=3/2 and is very well known.


The square root of a positive integer in any base with 2n digits always has n digits.

If it's true/false, can someone give a proof or a site.

Thanks

The numbers have to be perfect square?
[hide="Solution for base 10"]The number is obviously in the interval[100...(2n-2 0s),100...(2n 0s)[.Thus it's square root must be in the interval [100..(n-1 0s),100..(n 0s)[(by x<y<=>x^2<y^2).Any number in this interval will have n digits.
may be this can also be applied for other base numbers?? :?: [/hide]

[hide="Inequality"]In base $ b$, we have $ b^{2n\minus{}1}\le k^2<b^{2n}$ or, since $ \sqrt{x}$ is increasing, $ b^{n\minus{}1}<b^{n\minus{}\frac12}\le k<b^n$.[/hide]


analytic geometry
trigonometry
rotation
geometry
geometric transformation
geometry unsolved

The point $ P$ has coordinates $ (x, y)$ with respect to the origin $ O$. By writing $ x\equal{}r\cos\theta$ and $ y\equal{}r\sin\theta$, or otherwise, show that, if the line $ OP$ is rotated by $ 60^{\circ}$ clockwise about $ O$, the new $ y\minus{}$coordinate of $ P$ is $ \frac{1}{2}(y\minus{}\sqrt{3}x)$. What is the new $ y\minus{}$coordinate in the case of a counter-clockwise rotation by $ 60^{\circ}$?

An equilateral triangle $ OBC$ has vertices at $ O, (1, 0)$ and $ (\frac{1}{2}, \frac{\sqrt{3}}{2})$, respectively. The point $ P$ has coordinates $ (x, y)$. The perpendicular distance from $ P$ to the line through $ C$ and $ O$ is $ h_{1}$; the perpendicular distance from $ P$ to the line through $ O$ and $ B$ is $ h_{2}$; and the perpendicular distance from $ P$ to the line through $ B$ and $ C$ is $ h_{3}$.

Show that $ h_{1} \equal{} \frac{1}{2}\mid y\minus{}\sqrt{3}x\mid$ and find expressions for $ h_{2}$ and $ h_{3}$.

Show that $ h_{1}\plus{}h_{2}\plus{}h_{3} \equal{} \frac{\sqrt{3}}{2}$ if and only if $ P$ lies on or in the triangle $ OBC$.


algebra
polynomial
abstract algebra
group theory
superior algebra
superior algebra unsolved

Find some p(x) in Q[X] whose Galois group is Z_6

What are the steps to finding such a polynomial

Thank you

The general way to find a polynomial if a group is given is open.
But all finite abelian groups can be realised, in fact they are subgroups of cyclotomic Galois groups.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=139296 for ho any finite cyclic/abelian group can be archieved.


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]IMO 2007 HKTST 1[/url]
Problem 2

Let $A$, $B$ and $C$ be real numbers such that
(i) $\sin A \cos B+|\cos A \sin B|=\sin A |\cos A|+|\sin B|\cos B$,
(ii) $\tan C$ and $\cot C$ are defined.
Find the minimum value of $(\tan C-\sin A)^{2}+(\cot C-\cos B)^{2}$.

Condition $(i)$ is equivalent to:
$(\cos B-|\cos A|)(\sin A-|\sin B|)=0$, that is $\cos B=|\cos A|$ or $\sin A=|\sin B|$.
Anyway, $\sin^{2}A+\cos^{2}B=1$, so the point $P: (\sin A, \cos B)$ is placed on the unit circle $C$ centered in the origin $O$.

We want to minimize the square of the distance from $P$, a point on the circle $C$, to the point $Q: (\tan C, \cot C)$,  defined by condition $(ii)$:
$\min (d_{PQ})^{2}=(\min d_{PQ})^{2}=(d_{CQ})^{2}=(d_{OQ}-1)^{2}=(\sqrt{\tan^{2}C+\cot^{2}C}-1)^{2}$.

Since $\min \left(t+\frac{1}{t}\right)=2$ for $t>0$, the desired minimum value is $(\sqrt{2}-1)^{2}=3-2\sqrt{2}$.


Let p is a prime number. Prove $ p^2\equal{}30.a\plus{}1$ or $ p^2\equal{}30.b\plus{}19$

[quote="Arrange your tan"]
"Let $ p$ be an odd prime number that is greater than or equal to 7.
Prove $ p^2 \equal{} 30a \plus{} 1$ or $ p^2 \equal{} 30b \plus{} 19$,
for (appropriate) integers $ a$ and $ b$.
[/quote]

Is the following the correct use of the contrapositive?

If $ p^2$ is not of the form $ 30a \plus{} 1$ nor $ 30b \plus{} 19$, then
$ p$ is a prime number not greater than or equal to 7.

Or:

If $ p^2$ is among the forms
$ 30a, 30a \plus{} 2, 30a \plus{} 3,...,30a \plus{} 18, 30a \plus{} 20, 30a \plus{} 21,...,30a \plus{} 29$,
then $ p$ is a prime number less than or equal to $ 5$.

If $ p$ is $ 2, 3$, or $ 5$, then $ p^2$ is of the form $ 30a \plus{} 4, 30a \plus{} 9$, or
$ 30a \plus{} 25$, respectively, where $ a \equal{} 0$.

[quote="sondhtn"]Let p is a prime number. Prove $ p^2 \equal{} 30.a \plus{} 1$ or $ p^2 \equal{} 30.b \plus{} 19$[/quote]
Sorry. Where p >5.Can  You help me??

Take quadratic conqruences of 1, 7, 11, 13, 17, 19, 23, 29 mod 30. If they are only 1 or 19 then you are done.


superior algebra
superior algebra unsolved

let $ H < G$ such that if $ Ha\not \equal{} Hb \Longrightarrow aH\not \equal{} bH$ then prove for all $ g\in G$ we have $ gH \equal{} Hg$

conditions
=>
a*b^(-1) belongs H iff b^(-1)*a belongs H

 g^(-1)*g*h=h belongs H,so g*h*g^(-1) belongs H  #

\[ ab^{\minus{}1}\not\in
H\Longleftrightarrow b^{\minus{}1}a\not\in H\\
 ab^{\minus{}1}\in H\Longleftrightarrow b^{\minus{}1}a\in H\\
 \forall g\in G,h\in
H,g^{\minus{}1}(gh)\equal{}h\in H\Rightarrow ghg^{\minus{}1}\in H
\]
$ gH\equal{}Hg$
end


Let $x,y,z>0$ such that $xyz=1$ . Prove or disprove the inequality $2\sqrt{3(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x})}+x+y+z\geq 9$

Sorry, but It's wrong. Try with $x=1,y=0.8,z=1.25$


AMC
USA(J)MO
USAMO
geometry
3D geometry
pyramid
AIME

Hello!!! I kind of stink at USAMO, and IMO, so could somebody pm me, or post some questions that are easy USAMO and easy IMO questions. It would also be helpful if somebody could give me some [b]really easy[/b] questions that require proofs, because I stink at proofs also.

Prove that $ \sqrt{2}$ is irrational.  That's really easy.

Prove that a natural number is either even or odd.  Pretty easy.

Prove that every number has a unique prime factorization up to order.  Not too hard.

Not sure if these are the types of "easy questions" you are looking for but it's a start.  If you want USAMO and IMO questions, there's a Contest link in the forum which should provide you with an ample supply.

[quote="Wickedestjr"][quote="JRav"]Prove that $ \sqrt {2}$ is irrational.  That's really easy.

Prove that a natural number is either even or odd.  Pretty easy.

Prove that every number has a unique prime factorization up to order.  Not too hard.

Not sure if these are the types of "easy questions" you are looking for but it's a start.  If you want USAMO and IMO questions, there's a Contest link in the forum which should provide you with an ample supply.[/quote]

The square root of 2 is irrational because it is a terminating decimal. (I guess)

Every natural number is either even or odd, because natural numbers are basically integers above zero, and all integers are either divisible by 2 or are not divisible by 2.

Every number has a unique prime factorization because the number is just the product of all the prime factors, and if two numbers have the same prime factorization, then they have the same prime factors, which means they have the same product, which would mean that they would be the same.

Are those good solutions?[/quote]

Other than the fact that $ \sqrt 2$ is NOT terminating, those are pretty good explanations to a middle school student.

... HOWEVER... the USAMO and IMO require rigorous proofs, along the lines of this:

Assume for the sake of contradiction that $ \sqrt 2$ was rational. Then $ \sqrt 2$ can be expressed as $ \frac ab$, where $ a$ and $ b$ are relatively prime integers.

So we have
\begin{align*}\frac ab & = \sqrt 2 \\
\frac {a^2}{b^2} & = 2 \\
a^2 & = 2b^2\end{align*}
Now, since the RHS is even, that means that the LHS is even as well, and since the LHS is also a square, that implies that $ a$ is even as well. So let $ a = 2c$, so we have $ (2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$. Using the same argument as above, that means that $ b$ must be even as well.

But wait... we had previously required that $ a$ and $ b$ be relatively prime... so contradiction!

Therefore, $ \sqrt 2$ is irrational.

[quote]Every natural number is either even or odd, because natural numbers are basically integers above zero, and all integers are either divisible by 2 or are not divisible by 2.[/quote]

That is not a proof.  Maybe that helps you understand why it's true, but it doesn't prove that it's true.

[quote]Every number has a unique prime factorization because the number is just the product of all the prime factors, and if two numbers have the same prime factorization, then they have the same prime factors, which means they have the same product, which would mean that they would be the same.[/quote]

How do you know that?  Show that it's true.  By this, I mean suppose that two different numbers have the same prime factorization.  Then you can eventually show that there is a contradiction.

What Yongi781 provided is a rigorous proof.

[color=green]Pyramid quoting hurts my eyes.[/color]

Oh yeah, I remember seeing that proof somewhere. Wasn't that in one of the AoPS books, like Volume 1?

Does anyone have any USAMO questions or IMO questions that are really easy and don't require proofs?

[quote="BanishedTraitor"][quote="Wickedestjr"] Does anyone have any USAMO questions or IMO questions that are really easy and don't require proofs?
[/quote]

They do not exist. The whole point of IMO/USAMO Problems is for the students to provide rigorous proofs.[/quote]

You mean there isn't a single problem in the USAMO that is not a proof?  :wow:

duh do some research. And seeing your level I doubt you could even do aime problems let alone USAMO

I think there are some, but they're not straightforward calculation type problems.  I doubt you could solve any of them.

People don't do problems just like that. You need to prepare. So you should work on the lower levels if the USAMO scares you. Right now, the USAMO doesn't scare me, I could probably do 1 or two problems on each USAMO given enough time, but I feel more comfortable with AIME problems. Set yourself at a pace.

Well I did solve two really hard AIME geometry problems. Well, can somebody post some easy AIME questions then?

As someone mentioned, there is a link that says "Contests" at the top of this forum, with almost all of the AIME and USAMO problems ever. Do your own searching.

Actually, I've been carrying around a problem that is very easy, but that I think provides a good introduction to what it means to "prove" a statement, so now would be a good time to state it:

Between classes, students at a school try to get to their next class as fast as possible.  Assume that the corridors of the school are arranged in a rectangular grid and that every student takes a route that minimizes the total distance between his previous and next class.  Prove that no pair of students ever encounters each other twice.

Hmm, does the fact that (x-y)(x+y) = x^2 - y^2 have anything to do with this problem?

OH!!!!

Iff x^2 = y^2(mod ?), then that means x^2 - y^2 = 0 (mod ?)

Iff x = y (mod ?), then that means x-y = 0 (mod ?)

Iff (x-y)(x+y) = X^2 - Y^2, then that means that 0(x + y) = 0, and we know that this is true because 0=0.

Ha ha!!! If I solved this one correctly, then this is my very first proof.  :lol:

EDIT: NOW IS IT RIGHT?

"iff" means "if and only if". In your approach though, you only showed the "if" direction, which is trivial. (Instead, you want to start from $ x^2 \equiv y^2 \pmod{p}$, and find as an end result that $ x \equiv \pm y \pmod{p}$). See [[Euclid's Lemma]].

Wickedestjr had a start on the if direction, the only thing he was missing was the case $ x \equal{} \minus{} y$. (He only showed it for $ x \equal{} y$.)

[b]"Only If"[/b]: [i]If $ x^2 \equiv y^2 \pmod p$, then $ x \equiv \pm y \pmod p$.[/i]
The equation $ x^2 \equiv y^2 \pmod p$ is equivalent to $ x^2 \minus{} y^2 \equiv 0 \pmod p$, which factors into $ (x \plus{} y)(x \minus{} y) \equiv 0 \pmod p$.

This implies that $ p \mid (x \plus{} y)(x \minus{} y)$. By Euclid's Lemma, since $ p$ is prime, this implies that $ p \mid (x \plus{} y)$ or $ p \mid (x \minus{} y)$.

Case $ p \mid (x \plus{} y)$: This means that $ x \equiv \minus{} y \pmod p$.
Case $ p \mid (x \minus{} y)$: This means that $ x \equiv y \pmod p$.

It follows from the above results that if $ x^2 \equiv y^2 \pmod p$, then $ x \equiv \pm y \pmod p$.

[b]"If"[/b]: [i]If $ x \equiv \pm y \pmod p$, then $ x^2 \equiv y^2 \pmod p$.[/i]
This is trivial. $ x \equiv \pm y \pmod p \implies x^2 \equiv (\pm y)^2 \pmod p \implies x^2 \equiv y^2 \pmod p$, as desired.

Thus, combining the above results, we get that $ x^2 \equiv y^2 \pmod p$ if an only if $ x \equiv \pm y \pmod p$, for integers $ x, y$ and prime $ p$.

[b]Note[/b]: $ a \mid b$ means $ a$ divides $ b$, or in other words, $ b$ is divisible by $ a$. For example, $ 2 \mid 10$ because 10 is divisible by 2.

[hide="Clarification of the ONLY IF direction"]
So we have $ (x\plus{}y)(x\minus{}y) \equiv 0 \pmod p$. This basically means that the product $ (x\plus{}y)(x\minus{}y)$ is divisible by $ p$. Now, we were given that $ p$ is prime.

If you haven't already, go check up Euclid's Lemma by the link in the above post.

Okay? You did it? Okay. So by that lemma, we can say that one of the following is true.

1. $ (x\plus{}y)$ is divisible by $ p$.
2. $ (x\minus{}y)$ is divisible by $ p$.

Or,

1. $ x \equiv \minus{}y \pmod p$
2. $ x \equiv y \pmod p$.

Combining it into a single statement, $ x \equiv \pm y \pmod p$.
[/hide]

[quote="JRav"]Prove that $ \sqrt {2}$ is irrational.  That's really easy.

Prove that a natural number is either even or odd.  Pretty easy.

Prove that every number has a unique prime factorization up to order.  Not too hard.

Not sure if these are the types of "easy questions" you are looking for but it's a start.  If you want USAMO and IMO questions, there's a Contest link in the forum which should provide you with an ample supply.[/quote] A natural number is an integer more than 0. When deciding whether a natural number is a multiple of 2 or not is just a simple yes or no question. The answer has to be  yes or no. So all natural numbers are either even or odd.

Also, you can't really say a natural number is a positive integer, since techically, natural numbers are the starting point.  So you'd have to define integers based on natural numbers, not the other way around.

But Yongyi's point is more important.  You need to prove what would happen if a natural number were both even and odd.

One way to define "positive integer" is "natural number," so that's not a particularly valid criticism.  The second criticism depends on what definition of "odd" mathelete has in mind:  if it's just "not even," there's nothing wrong with his explanation (although it does become vacuous).  If the definition of odd is "leaves a remainder of $ 1$ upon division by $ 2$," the more pressing problem is what happens when a natural number is [i]neither[/i] even nor odd.

[quote="Yongyi781"][quote="mathelete"]A natural number is an integer more than 0. When deciding whether a natural number is a multiple of 2 or not is just a simple yes or no question. The answer has to be  yes or no. So all natural numbers are either even or odd.[/quote]

Are you new to this forum? This explanation would be perfectly fine to a math student, but what we need is a [i]rigorous[/i] explanation. (As this is an Olympiad forum).[/quote]
You cannot get more rigorous. We are just restating the definition of a natural number.

Again, that's not the issue.  The problem, as has been repeatedly stated, is to prove the following:  every integer is either of the form $ 2k\plus{}1$ or $ 2k$ for some integer $ k$.

[quote="mathelete"]You cannot get more rigorous. We are just restating the definition of a natural number.[/quote]

This is as rigorous as I could get it.

Consider some $ n\in\mathbb{N}$.
$ n\equal{}2q\plus{}r$ by the division algorithm.
If $ n$ is even, $ r\equal{}0$.
If $ n$ is odd, $ r\equal{}1$.
For $ n$ to be neither even nor odd, $ r>1$ or $ r<0$.
The division algorithm states that $ r\geq{0}$, so $ n$ is neither even nor odd only if $ r>1$.
Proceeding with mathematical induction, establish a base case: $ 1/2\equal{}0q\plus{}1$. $ r\equal{}1$ here, so 1 is odd.
Now assume that some natural number $ k$ is either even or odd: $ k\equal{}2q\plus{}r$, $ r\equal{}1$ or $ r\equal{}0$.
Add 1 to both sides: $ k\plus{}1\equal{}2q\plus{}r\plus{}1$.
Case 1: $ r\equal{}1$, then $ k\equal{}2q\plus{}2$, which factors into $ k\equal{}2(q\plus{}1)$, so that $ r\equal{}0$.
Case 2: $ r\equal{}0$, then $ k\equal{}2q\plus{}1$, so that $ r\equal{}1$.
It follows that it is impossible to have $ r\geq{2}$ because 1 being odd implies that $ r\equal{}1$ when 1 is divided by 2, and 1+1 implies that $ r\equal{}0$ when 2 is divided by 2, and so on and so forth.
Thus any natural number must be even or odd, Q.E.D.

You can't get more rigorous than induction  :lol:

[quote="leoxnlin"]I propose that this thread be locked. It has become the most popular thread in the Pre-Olympiad forum, and yet does not contain Pre-Olympiad material. The purpose of the thread seems just to be to train a single user for mathematical Olympiads, but this is not even a very effective way to learn and train oneself.[/quote]

Yes it is!!! People who read this thread learn from what they read.

There is very little meaningful discussion going on, and most of the thread is very spammy. If you'd like to continue this sort of discussion, it should be in HSB.

Wickedestjr, your best bet is to hang out in HSB answering basic proof questions.

Or just buy AoPS 1.


i received a number sequence in my psychology class and I was told to find why the numbers were in this sequence

8,5,4,1,7,6,3,2,0

Presumably, if it's for a psychology class, the point was [i]not[/i] to go ask other people what the answer is.  Think about it more.  It'll probably come to you eventually.  If not, what's the worst that could happen?

No i have thought about it I am confused i need help on another way to look at it? i am stuck on the same way of doing it . i did not mean tio come off as if i was looking for someone to answer it for me.

My suggestion: your psychology professor wouldn't have given you the question if it were just mathematical.  Unfortunately, it's really hard to give suggestions for this problem without giving it all away -- it's a "either you've seen it before or you haven't" type of thing.

ok this is my problem I keep going to figure it out with something to do with the average of the numbers. the sequence confuses me because of 1,7, and 2,0 how are they related?

The answer has nothing to do with the numerical properties of the numbers.

I thought that this was a web site to recieve help??  Is there anyone out there that can help?  Because I have not see it before and I am not the best at math!!!

Ok this may sound stupid but what are the numerical properties?

Numerical just means mathematical.  If you're not the best at math, that's okay!  The answer has nothing to do with math.

please, I don't have to answer this question I want to know just because it is bothering me!

what I don't understand is what other way is there to look at it what is the means of the numbers then?

so what you are saying is it has nothing to do with adding or subtracting, multiplying or dividing?

Yes, it has nothing to do with any of those arithmetic operations. There's lots of other ways to look at numbers.  You can look at their shapes, their pronunciations, their spellings, etc.

When I first saw the question, I thought for a minute and couldn't figure it out.  Joel's hint to look outside math turned out to be a big help, and that's how I figured it out.

does it have something to do with prime numbers or odd and even?

thanks for all your help!!!!  you did exactly what I asked to make me look at in a new light I was stuck with the math thing!


Right now I am hovering around 11-12 correct on AIMEs that I time. The problem is usually not because I cannot do the remaining problems (except #15 which happens to be unusually elusive on every contest within a reasonable amount of time), it is because I run out of stamina, and I start to fidget, work slower and make careless mistakes. Does anyone have suggestions for building stamina (eating a lot of carbohydrate-rich foods, exercising more, etc)?

[quote="fairyofwind"]Does anyone have suggestions for building stamina (eating a lot of carbohydrate-rich foods, exercising more, etc)?[/quote]

I would definitely put exercising more high on the list of stamina-building tricks.

Getting a good night's sleep the night before is what my mom always recommended any time a test was coming up. That worked for me.

I'd be curious to hear from other participants here what their suggestions are.

Practice doing tests for that period of time?
I don't know if it helped, but as training for the IMO (4.5 hour exams) we did many (like 6 to 8) 4.5 hour mock tests, not only to get more practice doing problems but to get practice taking that amount of time.


floor function
number theory proposed
number theory

Let $ p>2$ be a prime and let $ a,b,c,d$ be integers not divisible by $ p$,such that:
$ \{ \frac{ra}{p} \} \plus{} \{ \frac{rb}{p} \} \plus{} \{ \frac{rc}{p} \} \plus{} \{ \frac{rd}{p} \} \equal{}2$
for any integer r not divisible by $ p$
Prove that at least two of the numbers $ a\plus{}b,a\plus{}c,a\plus{}d,b\plus{}c,b\plus{}d,c\plus{}d$ are divisible by $ p$

i am able to prove it if $ a,b,c,d,r$ are all greater than $ p$ but i am not sure.
i will post what i did if u tell me whether the condition is satisfied.

Never mind my previous post i dont think it matters
[hide="try1"]
assume $ a,b,c,d,r$ are greater than p.
Let $ r \equal{} pq \plus{} x$,
$ a \equal{} pl \plus{} a_1$,
and similarly define $ b,c,d$
substitute them in the given equation,
we get ,
$ \sum \{\frac {ra}{p}\} \equal{} \sum \{\frac {(pq \plus{} x)(pl \plus{} a_1)}{p}\} \equal{} \sum \{\frac {xa_1}{p}\} \equal{} \sum \frac {xa_1}{p} \minus{} \lfloor\frac {xa_1}{p}\rfloor \equal{} 2$
we have finally the equation,
$ x(a_1 \plus{} b_1 \plus{} c_1 \plus{} d_1) \equal{} p(\lfloor\frac {xa_1}{p}\rfloor \plus{} 2)$
but by definition,$ x$ is less than p so p does not divide x.
therefore $ p|a_1 \plus{} b_1 \plus{} c_1 \plus{} d_1$
or $ a_1 \plus{} b_1 \plus{} c_1 \plus{} d_1 \equiv\ 0 (mod p)$ so $ a_1 \plus{} b_1 \equiv\ \minus{} (c \plus{} d) (mod p)$ or , $ a_1 \plus{} b_1 \equal{} c_1 \plus{} d_1$ (since they are all less than p)
therfore we get that,
$ p|2(c_1 \plus{} d_1)$ but p is greater than 2 so $ p|c_1 \plus{} d_1$ and also, $ p|a_1 \plus{} b_1$,

for all of them to be odd the same method is used ,
Now for the other cases we will get a contradiction that $ p|a_1,b_1,c_1,d_1$ if 3 are negative and if 2 are negative we get$ p| \minus{} 2(c_1 \plus{} d_1)$
so thus $ p$ divides atleast two of the given set.
[hide="doubt!"] if we do like this dont we get that it divides all of them??[/hide][/hide]

...
$ \left\{ {\frac{{a}}{p}} \right\} \plus{} \left\{ {\frac{{b}}{p}} \right\} \plus{} \left\{ {\frac{{c}}{p}} \right\} \plus{} \left\{ {\frac{{d}}{p}} \right\} \equal{} 2 \Rightarrow a \plus{} b \plus{} c \plus{} d \vdots p$
Thus,
$ \left[ {\frac{{ra \plus{} rb}}{p}} \right] \plus{} \left[ {\frac{{rc \plus{} rd}}{p}} \right] \minus{} \left[ {\frac{{ra}}{p}} \right] \minus{} \left[ {\frac{{rb}}{p}} \right] \minus{} \left[ {\frac{{rc}}{p}} \right] \minus{} \left[ {\frac{{rd}}{p}} \right] \equal{} 1.$
wlog $ \left[ {\frac{{ra \plus{} rb}}{p}} \right] \minus{} \left[ {\frac{{ra}}{p}} \right] \minus{} \left[ {\frac{{rb}}{p}} \right] \equal{} 1 \Rightarrow \left\{ {\frac{{ra}}{p}} \right\} \plus{} \left\{ {\frac{{rb}}{p}} \right\} \equal{} 1 \plus{} \left\{ {\frac{{ra \plus{} rb}}{p}} \right\}$
Because $ {ra}$ is a complete set of residue classes modulo p, that gives $ p|a\plus{}b.$ hence $ p|c\plus{}d.$

I know I am being totally stupid ,but can geniusbliss explain why $ a_1 \plus{} b_1 \equal{} c_1 \plus{} d_1$
That needn't be true always

[quote]geniusbliss wrote:
is what i did correct ?
i dont understand the solution of tronghieu at all! wat is that symbol with 3 dots??
[/quote]
it means that $ p|a\plus{}b\plus{}c\plus{}d$


calculus
integration
logarithms
calculus computations

Are there any other ways of doing this?

Gibt es noch andere Wege dieses Problem zu loesen?

To Moderators, could you edit the part of the context in post #2 as follows?

[quote="kunny"]$ \int_0^1 \frac {1}{\sqrt {16 \plus{} x^2}}\ dx \equal{} \frac {1}{3}\int_0^1 \frac {1}{\sqrt {x^2 \plus{} \frac {16}{9}}}\ dx$[/quote]

$ \Longrightarrow \int_0^1 \frac {1}{\sqrt {16 \plus{} 9x^2}}\ dx \equal{} \frac {1}{3}\int_0^1 \frac {1}{\sqrt {x^2 \plus{} \frac {16}{9}}}\ dx$

Japanese high school students don't study the formula $ \boxed{\int \frac {1}{\sqrt {x^2 \plus{} a^2}}\ dx \equal{} \ln |x \plus{} \sqrt {x^2 \plus{} a^2}| \plus{} Const.}$, so I show another approach later.

[quote="AndrewTom"]$ \int_{0}^{1} \frac {1}{\sqrt {16 + 9x^{2}}}\, dx\, + \int_{0}^{2} \frac {1}{\sqrt {9 + 4x^{2}}}\, dx = \ln a$.

Find the exact value of $ a$.[/quote]

$ y = \sqrt {x^2 + m^2}\ (m > 0)\Longrightarrow y^2 - x^2 = m^2$

$ \Longrightarrow \left\{\begin{array}{ll} y + x = me^t & \quad \\
y - x = me^{ - t} & \quad \end{array} \right.\Longleftrightarrow \left\{\begin{array}{ll} x = \frac {m}{2}(e^t - e^{ - t}) & \quad \\
y = \frac {m}{2}(e^t + e^{ - t}) & \quad \end{array} \right.$

Let $ \boxed{I_ < m,\ p > = \int_0^p \frac {1}{\sqrt {x^2 + m^2}}\ dx\ (p > 0)}$

$ = \int_0^{t_{m,\ p}} \frac {2}{m(e^t + e^{ - t})}\cdot \frac {m}{2}(e^t + e^{ - t})\ dt = t_{m,\ p}: = \ln \left\{\frac {p}{m} + \sqrt {\left(\frac {p}{m}\right)^2 + 1}\right\}$

$ \therefore \int_{0}^{1} \frac {1}{\sqrt {16 + 9x^{2}}}\ dx + \int_{0}^{2} \frac {1}{\sqrt {9 + 4x^{2}}}\ dx = \frac {1}{3}I_{ < \frac {4}{3},\ 1 > } + \frac {1}{2}I__{ < \frac {3}{2},\ 2 > }$

$ = \frac {1}{3}\ln 2 + \frac {1}{2}\ln 3 = \ln 2^{\frac {1}{3}}\cdot 3^{\frac {1}{2}}$, yielding $ a = \sqrt [3]{2}\sqrt {3}.$

[quote="AndrewTom"]Arigatoo gozaimasu, kunny-san.[/quote]

Du bist willkommen. :lol:


limit
function
algebra solved
algebra
Recurrence

Consider sequence ${x_n}$ satisfying:  $ x_1=1; x_{n+1}=x_{n}+\frac {2003}{\sqrt[3]{x_n}}+\frac {2004}{\sqrt[4]{x_{n}}}$.

Find all $m \in \mathbb{R}$ such that $ \exists \lim_{n \to \infty} \frac {x_{n}^m}{n} \not= 0$.

Is this so hard? I mean, the sequence is clearly increasing and unbounded (otherwise, it  would be convergent and this is not the case; just take the limit in the reccurence relation) so it tends to infinity. Then, just compute the limit of $\frac{(x_n)^{\frac{4}{3}}}{n} $ using Cesaro-Stolz theorem. The answer I think is $ m=1.25 $. It is enough to compute the limit when x tends to infinity of the function $ f(x)=(x+\frac{2003}{\sqrt[3]{x}}+\frac{2004}{\sqrt[4]{x}})^{\frac{5}{4}}-x^{\frac{5}{4}} $ which can be easily done by using the fact that $ \frac{x^t-1}{x-1} $ tends to t when x tends to 1. I hope Im not wrong.


[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that: $ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le 1$[/quote]

Put $ a \equal{} x^3,b \equal{} y^3,c \equal{} z^3$,we have $ xyz \equal{} 1$
[b]proof 1.[/b]
Note that $ x^3 \plus{} y^3 \plus{} 1 \ge xy(x \plus{} y) \plus{} 1 \equal{} xy(x \plus{} y \plus{} z)$

so
$ \sum \frac {1}{a \plus{} b \plus{} 1} \equal{} \sum \frac {1}{x^3 \plus{} y^3 \plus{} 1} \le \sum \frac {1}{xy(x \plus{} y \plus{} z)} \equal{} 1$

[b]proof 2[/b].By Cauchy-Schwazt,we have
$ a \plus{} b \plus{} 1 \equal{} \frac {(x^3 \plus{} y^3 \plus{} 1)(\frac {1}{x} \plus{} \frac {1}{y} \plus{} z^2)}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} z^2}$
$ \ge \frac {(x \plus{} y \plus{} z)^2}{yz \plus{} zy \plus{} z^2}$
so
$ \sum \frac {1}{a \plus{} b \plus{} 1} \le \sum \frac {yz \plus{} zy \plus{} z^2}{(x \plus{} y \plus{} z)^2} \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} yz \plus{} zx)}{(x \plus{} y \plus{} z)^2} \equal{} 1$

 :)

[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that: $ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le 1$ (*)[/quote]

Old problem

Let : $ a \equal{} x^3$ ; $ b \equal{} y^3$ and $ c \equal{} z^3$ $ \Longrightarrow$ $ x$,$ y$,$ z$ $ \ge 0$ and $ xyz \equal{} 1$

$ \bullet$ Because $ x^3 \plus{} y^3 \ge xy(x \plus{} y$) $ \Longrightarrow \frac {1}{x^3 \plus{} y^3 \plus{} 1} \le \frac {1}{xy(x \plus{} y) \plus{} 1} \equal{} \frac {1}{xy(x \plus{} y) \plus{} xyz} \equal{} \frac {z}{x \plus{} y \plus{} z}$

$ LHP_{(*)}\le \frac {x \plus{} y \plus{} z}{x \plus{} y \plus{} z} \equal{} 1$

hope this is correct ;

By AM >= GM

$ \frac {a \plus{} b \plus{} 1}{3} \ge (ab)^{1/3}$

 $ \frac {a \plus{} b \plus{} 1}{3} \ge c^{ \minus{} 1/3}$

similarly ,

$ \frac {c \plus{} b \plus{} 1}{3} \ge (bc)^{1/3}$

 $ \frac {c \plus{} b \plus{} 1}{3} \ge a^{ \minus{} 1/3}$

$ \frac {a \plus{} c \plus{} 1}{3} \ge (ac)^{1/3}$

 $ \frac {a \plus{} c \plus{} 1}{3} \ge b^{ \minus{} 1/3}$

therefore ,

$ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le \frac {a^{1/3} \plus{} b^{1/3} \plus{} c^{1/3}}{3} \le (abc)^{1/9} \le 1$ .

hence proved .

[quote="vijaymenon"]$ \boxed{\frac {a^{1/3} \plus{} b^{1/3} \plus{} c^{1/3}}{3} \le (abc)^{1/9}}$ [/quote]

Reversed

[quote="Null"] and since $ a \plus{} b \plus{} c\le ab \plus{} bc \plus{} ac$ [/quote]

I think you're wrong.

[quote="Null"]Can you find a counterexample? I'm pretty sure I've seen that inequality before.[/quote]

Ok, for you $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

Ah, true, and I found another one, even having $ abc\equal{}1.$

[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that: $ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le 1$[/quote]
I also have a stronger following results
Let $ a,b,c$ be positive real numbers such that $ abc \ge 1$. Prove that
$ \frac {1}{a \plus{} b \plus{} 1} \plus{} \frac {1}{b \plus{} c \plus{} 1} \plus{} \frac {1}{c \plus{} a \plus{} 1} \le 1$


A triangle has vertices at coordinates (2,2), (5,6) and (6,2).
What is the number of units in the length of the longest side
of the triangle?


modular arithmetic
algebra
binomial theorem
number theory
number theory unsolved

help me :
find n min such that $ 3^{n}\equiv {1} ( mod 2^{2009})$
thanks.

i believe it's been posted before.

by the fact that, if $ p$ is prime and $ p\not| a, b$, then $ a\equiv b \pmod{p^k} \Leftrightarrow a^p\equiv b^p\pmod{p^{k \plus{} 1}}$, and the fact that $ 3^2\equiv 1 \pmod{2^3}$ but $ 3^2\not\equiv 1 \pmod{2^4}$, we get $ 3^{2^{2007}}\equiv 1 \pmod{2^{2009}}$ but $ 3^{2^{2006}}\not\equiv 1 \pmod{2^{2009}}$, so the answer is $ 2^{2007}$.

sorry but i can't see why if  $ p$ is prime and $ p\not| a, b,$ then $ a\equiv b \pmod{p^k} \Leftrightarrow a^p\equiv b^p\pmod{p^{k \plus{} 1}}$

I should have also said that $ a^2\equiv b^2 \pmod{2^{k\plus{}1}} \Rightarrow a\equiv b \pmod{2^k}$ is only true if $ a\equiv b \pmod{2^2}$: that's why, in this problem, we stop at $ 3^2\equiv 1 \pmod{2^3}$. In all other cases, the claim is true, and here's a proof.

To show $ a^p\equiv b^p \pmod{p^{k\plus{}1}} \Rightarrow a\equiv b \pmod{p^k}$: Let $ b \equal{} a\plus{}rp^m$, where $ p\not|r$. First, note that $ m\geq 1$: By Fermat's Little Theorem, $ a\equiv a^p\equiv b^p \equiv b \pmod{p}$.

Then by the Binomial Theorem, $ p^{k\plus{}1}|a^p\minus{}b^p\equal{}a^p\minus{}(a\plus{}rp^m)^p\equal{}\minus{}a^{p\minus{}1}rp^{m\plus{}1}\plus{}\cdots$, where, by the fact that $ m\geq 1$, all terms after the first are divisible by a power of $ p$ greater than $ p^{m\plus{}1}$ -- unless $ m\equal{}1, p\equal{}2$. Thus, since $ p\not| a$, the highest power of $ p$ dividing the RHS of the last equation is $ p^{m\plus{}1}$. It follows that $ m\plus{}1\geq k\plus{}1$, i.e. $ m\geq k$.

To show $ a\equiv b \pmod{p^k}\Rightarrow a^p\equiv b^p\pmod{p^{k\plus{}1}}$: Let $ b\equal{}a\plus{}rp^k$. Then by the Binomial Theorem, $ a^p\minus{}b^p\equal{}a^p\minus{}(a\plus{}rp^k)^p\equal{}a^p\minus{}(a^p\plus{}a^{p\minus{}1}rp^{k\plus{}1}\plus{}\cdots)\equal{}a^{p\minus{}1}rp^{k\plus{}1}\plus{}\cdots$, where every term is divisible by $ p^{k\plus{}1}$.

Or use the fact that,
\[ \forall l \ge 1,\,\,\,{v_2}({3^{{2^l}}} \minus{} 1) \equal{} l \plus{} 2\]
The result follows.

yes, which can be proven the same way as my above posts.

or, actually even more simply, we can prove it by writing $ 3^{2^l}\minus{}1\equal{}(3^{2^{l\minus{}1}}\plus{}1)(3^{2^{l\minus{}2}}\plus{}1)\cdots (3^2\plus{}1)(3\plus{}1)(3\minus{}1)$


Given that function f is represented as f(x) = x/10 is x is divisible by 10, and f(x) = x+1 if x is not divisible by 10. Let $a_0 = 1993$ and $a_{n+1} = f(a_{n})$. What is the smallest n such that $a_n = 1$?

This was on math chat today. I thought this problem was quite fun.

Except it's TOO EASY> :P

[quote="isaacchao"]im not even sure if i can get it right
how do you define easy?
[hide]
37?
[/hide][/quote]
I would define 'easy' as once you consider that you already know how to solve a problem in a straightforward sense, in my opinion. Don't take me wrong if you don't agree with me.
[hide]
37? REALLY?
[/hide]

I consider "easy" when the question includes the basic concept that most people on getting started would have.

This question, for instance, is just a series problem.

haha, knew it, counted extra sets of numbers


For solely math purposes, which one is better and why?


If 8 bugs are situated on a unit cube, and they all suddenly start to chase each other on the [i]shortest path possible[/i] (they can't go through the cube) will they ever meet?  Where will they end up?  How far will they have traveled?

Be sure to post your proof.

One method is to lay out the cube in a net, and then solve. But it's too complicated... :(


function calculus derivative limit integration trigonometry inequalities

Let $f$ be a continuous function with continuous first derivative on $(a,b)$. and let

$\lim_{x\to a+}f(x) =+\infty$ and
$\lim_{x\to b-}f(x) =-\infty$
and $f'(x)+f^{2}(x) \geq 1$, for x on the interval (0,1).
Prove that $b-a \geq \pi$ and give an example where $b-a = \pi$

[quote="Pi-J"]This is the second problem of IMC 1994.[/quote] Thanks Pi-J for pointing this out. There is an error in the statement of the problem, something that I suspected all along: it should read $f'(x)+f^{2}(x) \ge-1.$

Here is my solution.

$f'(x)+f^{2}(x) \ge-1 \Rightarrow 1+f^{2}(x) \ge-f'(x)$

$\Rightarrow 1 \ge-\frac{f'(x)}{1+f^{2}(x)}\Rightarrow \int_{a}^{b}\,dx \ge-\int_{a}^{b}\frac{f'(x)}{1+f^{2}(x)}\,dx$

$\Rightarrow b-a \ge-\tan^{-1}(f(x))\Big|_{a}^{b}=-(-\frac{\pi}{2}-\frac{\pi}{2})$

$\Rightarrow b-a \ge \pi.$

A classic example for which $b-a = \pi$ is $f(x) = \tan x .$ Here , $a =-\frac{\pi}{2}$ and $b = \frac{\pi}{2}.$ In fact, I first guessed this example, and then worked my way towards the solution. As soon as I saw that we had nice integrable functions along with the fact that we get $\tan^{-1}(f(x))$ somewhere in our solution, I was convinced it would work!

[quote="Brambila"]well the inequality is valid just in the interval (0,1) . isnt there any problem about it? or was this condition wrong?[/quote]

it was already stated in the problem that x is in the interval $(0,1)$

[quote="FieryHydra"][quote="Brambila"]well the inequality is valid just in the interval (0,1) . isnt there any problem about it? or was this condition wrong?[/quote]Thanks for pointing this out.  I should have been more careful. The original question does say that the inequality holds for all $x \in (a,b)$ and not $(0,1).$

Violetcraze, please make those corrections.[/quote]

oh, okay, i get it. sorry, i didn't notice any of those errors on the problem set. thanks  :)


function
inequalities
algebra
functional equation
algebra proposed

Find all functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x \minus{} f(y)) \equal{} 4f(x) \minus{} f(y) \minus{} 4x\] for all $ x,y \in \mathbb{R}.$

$ f(x \minus{} f(y)) \equal{} 4f(x) \minus{} f(y) \minus{}4x \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}(1)$

$ x \equal{} f(0)$ in $ (1) \Rightarrow f(f(0) \minus{} f(y)) \equal{} 4f(f(0)) \minus{} f(y) \minus{} f(0) \equal{} c \minus{} f(y)$ where $ c \equal{} 4f(f(0)) \minus{} f(0)$.

$ x \equal{} 0$ in $ (1) \Rightarrow f(\minus{} f(y)) \equal{} 4f(0) \minus{} f(y) \equal{} d \minus{} f(y)$ where $ d \equal{} 4f(0)$.

Setting $ x \equal{} f(0) \minus{} f(u), y \equal{} 0$ in $ (1)$, we get

$ f(\minus{}f(u)) \equal{} 4f(f(0) \minus{} f(u)) \minus{} f(0) \minus{}4(f(0) \minus{} f(u))$

$ \Rightarrow d \minus{} f(u) \equal{} 4(c \minus{} f(u)) \minus{}f(0) \minus{}4f(0) \plus{} 4f(u)$

$ \Rightarrow f(u) \equal{} d \plus{} 5f(0) \minus{}4c \Rightarrow f$ is a constant function.  But $ (1)$ does not have any constant function as a solution.
Hence, the given functional equation does not have any solution.

[quote="nicetry007"]$ f(x \minus{} f(y)) \equal{} 4f(x) \minus{} f(y) \minus{} 4x \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (1)$

$ x \equal{} f(0)$ in $ (1) \Rightarrow f(f(0) \minus{} f(y)) \equal{} 4f(f(0)) \minus{} f(y) \minus{} f(0) \equal{} c \minus{} f(y)$ where $ c \equal{} 4f(f(0)) \minus{} f(0)$.

$ x \equal{} 0$ in $ (1) \Rightarrow f( \minus{} f(y)) \equal{} 4f(0) \minus{} f(y) \equal{} d \minus{} f(y)$ where $ d \equal{} 4f(0)$.

Setting $ x \equal{} f(0) \minus{} f(u), y \equal{} 0$ in $ (1)$, we get

$ f( \minus{} f(u)) \equal{} 4f(f(0) \minus{} f(u)) \minus{} f(0) \minus{} 4(f(0) \minus{} f(u))$

$ \Rightarrow d \minus{} f(u) \equal{} 4(c \minus{} f(u)) \minus{} f(0) \minus{} 4f(0) \plus{} 4f(u)$

$ \Rightarrow f(u) \equal{} d \plus{} 5f(0) \minus{} 4c \Rightarrow f$ is a constant function.  But $ (1)$ does not have any constant function as a solution.
Hence, the given functional equation does not have any solution.[/quote]

Nice !
:)

$ \forall x,y \in \mathbb{R}$ we have $ 4x \equal{} 4f(x)\minus{}f(y)\minus{}f(x\minus{}f(y))$

So the $ f(x)$ is surjective. Hence $ \exists u \in \mathbb{R}$ such that $ f(u)\equal{}0$.

$ y\equal{}u$ $ \implies f(x)\equal{}4f(x)\minus{}4x \implies f(x)\equal{}\frac{4x}3$

$ f(x)\equal{}\frac{4x}3$ is not a solution of original equation

[b]Hence there are no solution.[/b]

[i](I am not sure that we can assume that $ f(x)$ is surjective....what do you think about it?)[/i]

[quote="prester"]$ \forall x,y \in \mathbb{R}$ we have $ 4x \equal{} 4f(x) \minus{} f(y) \minus{} f(x \minus{} f(y))$

So the $ f(x)$ is surjective. ...

[i](I am not sure that we can assume that $ f(x)$ is surjective....what do you think about it?)[/i][/quote]

$ 4x \equal{} 4f(x) \minus{} f(y) \minus{} f(x \minus{} f(y))$ does not imply that $ f(x)$ is surjective  :)

Let $ a,c$ be arbitrary real numbers and say $ f(a)\equal{}b$ and $ f(c)\equal{}d$. Take y=a to get:
$ f(x\minus{}f(a))\equal{}4f(x)\minus{}f(a)\minus{}4x$, so $ f(x\minus{}b)\equal{}4f(x)\minus{}b\minus{}4x$
Take y=c to get:
$ f(x\minus{}f(c))\equal{}4f(x)\minus{}f(c)\minus{}4x$ so $ f(x\minus{}d)\equal{}4f(x)\minus{}d\minus{}4x$
Take x as x-b and y=c to get:
$ f(x\minus{}b\minus{}f(c))\equal{}4f(x\minus{}b)\minus{}f(c)\minus{}4(x\minus{}b)$ so $ f(x\minus{}b\minus{}d)\equal{}4(4f(x)\minus{}b\minus{}4x)\minus{}d\minus{}4(x\minus{}b)\equal{}16f(x)\minus{}20x\minus{}d$
Take x as x-d and y=a to get:
$ f(x\minus{}d\minus{}f(a))\equal{}4f(x\minus{}d)\minus{}f(a)\minus{}4(x\minus{}d)$ so $ f(x\minus{}d\minus{}b)\equal{}4(4f(x)\minus{}d\minus{}4x)\minus{}b\minus{}4(x\minus{}d)\equal{}16f(x)\minus{}20x\minus{}b$
Thus
$ 16f(x)\minus{}20x\minus{}d\equal{}f(x\minus{}b\minus{}d)\equal{}f(x\minus{}d\minus{}b)\equal{}16f(x)\minus{}20x\minus{}b$ so $ b\equal{}d$.
Therefore $ f$ is a constant function, ie $ f(x)\equal{}b$ for every real x. Thus $ b\equal{}4b\minus{}b\minus{}4x$, so $ 4x\equal{}2b$ and $ x\equal{}b/2$ for every real x; contradiction. Thus no such function exists.

[quote="pco"][quote="ozgurkircak"]Find all $ f: R\rightarrow R$ such that:
\[ f(x \minus{} f(y)) \equal{} 4f(x) \minus{} f(y) \minus{} 4x\]
for all $ x,y \in \mathbb{R}.$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \minus{} f(y)) \equal{} 4f(x) \minus{} f(y) \minus{} 4x$
Let $ I \equal{} f(\mathbb R)$

$ P(f(x),x)$ $ \implies$ $ f(0) \equal{} 4f(f(x)) \minus{} 5f(x)$ and so $ 4f(f((x)) \equal{} 5f(x) \plus{} f(0)$
$ P(f(x),y)$ $ \implies$ $ f(f(x) \minus{} f(y)) \equal{} 4f(f(x)) \minus{} f(y) \minus{} 4f(x)$ and so, using previous line : $ f(f(x) \minus{} f(y)) \equal{} f(x) \minus{} f(y) \plus{} f(0)$
$ \implies$ $ f(x) \minus{} f(y) \plus{} f(0)\in I$ $ \forall x,y$
$ \implies$ $ f(x) \minus{} (f(y) \minus{} f(z) \plus{} f(0)) \plus{} f(0) \equal{} f(x) \minus{} f(y) \plus{} f(z)\in I$ $ \forall x,y,z$
$ \implies$ $ (f(x) \minus{} f(y) \plus{} f(x)) \minus{} (4f(x) \minus{} f(y) \minus{} 4x) \plus{} (f(x) \minus{} f(y) \plus{} f(x)) \equal{} 4x \minus{} f(y)\in I$ $ \forall x,y$
$ \implies$ $ 0\in I$ (just set $ x \equal{} \frac {f(y)}4$ in previous line).

Using then $ y$ such that $ f(y) \equal{} 0$ in $ P(x,y)$, we get $ f(x) \equal{} 4f(x) \minus{} 4x$ and so $ f(x) \equal{} \frac {4x}3$ $ \forall x$

Then, plugging this back in original equation, we get that this necessary condition does not fit the equation.

Hence $ \boxed{\text{no solution for this functional equation}}$[/quote]
I do not understand, why we can for $f(y) = 0$
Thanks you pco!

[quote="namdan"]
I do not understand, why we can for $f(y) = 0$
Thanks you pco![/quote]
Because, just line above, I proved that $0\in I$

[quote="namdan"]Why we can not infer $4x-f(y)$ immediately?[/quote]
If you want some help, make some effort.

What is the meaning of this question ?
What is the meaning of "infer a quantity" ? I understand "infer a proposition / equality / inequality" from another ?

Make some full sentences and I'll try to answer.

$g(x)=x-f(x)$ implies $g\left( x-y+g(y) \right) = 4g(x)+x$
Then, $y=x$ implies $g(g(x))=4g(x)+x$, hence $g$ is injective.
Again, $g(g(0))=4g(0)+0=g(0-x+g(x))$. So injectivity of $g$ implies $g(x)=x+g(0)$, which is not a solution indeed.
So, there is no solution.

[quote="namdan"]I want to ask you: Why we prove $ 4x-f(y) \in I \forall x,y $?
Thanhks you![/quote]
Just above the line where I wrote that $4x-f(y)\in I$, I proved that $f(x)-f(y)+f(z)\in I$ $\forall x,y,z$

And just as the beginning of the line where I wrote that $4x-f(y)\in I$, I wrote :
$4x-f(y)=$ $(f(x)-f(y)+f(x))-(4f(x)-f(y)-4x)+(f(x)-f(y)+f(x))$

And since $f(x)-f(y)+f(x)\in I$ and $4f(x)-f(y)-4x=f(x-f(y))\in I$, we get the required result.

Btw, bappa1971's solution is quite shorter and nicer than mine :)

[quote="namdan"]$I=f(\mathbb{R})$: Valued function
We have $f(x-f(y))\in I$ so $f(f(x)-f(y))\in I$?[/quote]

$f($anything$)\in I$ by definition of $I$

I think namdan try to ask you for your idea to solve this problem,pco.

[quote="namdan"]I don't know here:
$f(x-f(y))\in I$ so $f(f(x)-f(y))\in I$?
It's true for all cases?
Thanks all.[/quote]
just replace $x$ by  $f(x)$

Pco explained for you $f(anything)\in I$. I can't understand you


[quote][color=darkred]$ 1.\blacktriangleright$ Let  $ ABC$ be a triangle for which exists $ M\in (BC)$ so that $ \odot\begin{array}{c}
\nearrow\\\
\searrow\end{array}$ $ \begin{array}{c}
MB\equal{}3\cdot MC\\\\
m(\widehat {AMB})\equal{}\frac B2\end{array}$ . Prove that $ a\equal{}2(b\minus{}c)$ .

$ 2.\blacktriangleright$ Let  $ \triangle ABC$  for which exists $ M\in (BC)$ so that $ m(\widehat {AMB})\equal{}\frac B2$ . Prove that $ a\cdot MC\plus{}c\cdot (a\plus{}c)\equal{}b^2$ .[/color][/quote]

[quote=Virgil Nicula][quote][color=darkred]$ 1.\blacktriangleright$ Let  $ ABC$ be a triangle for which exists $ M\in (BC)$ so that $ \odot\begin{array}{c}
\nearrow\\\
\searrow\end{array}$ $ \begin{array}{c}
MB\equal{}3\cdot MC\\\\
m(\widehat {AMB})\equal{}\frac B2\end{array}$ . Prove that $ a\equal{}2(b\minus{}c)$ .[/color][/quote][/quote]

If we prove $DA$ is symmedian of $\triangle BDE$, we are done ($E\in AC|AE=AB$)!

Merry Christmas


A positive multiple of 45 less than 1000 is randomly selected.
What is the probability that it is a two-digit integer? Express
your answer as a common fraction.

So, you know that $ 90\equal{}45*2$(obvious)  so you know that $ 990\equal{}90*11\equal{}45*2*11\equal{}45*22$
So you know that there are $ 22$ less than 1000, $ 2$ 2-digit, so answer is $ \frac{2}{22}\equal{}\frac{1}{11}$


conics
parabola
geometry
3D geometry
calculus
calculus computations

Given sequence of points $ A_n\ (a_n,\ a_n^2)\ (n \equal{} 1,\ 2,\ \cdots)$ and $ B_n\ (b_n,\ 0)\ (n \equal{} 0,\ 1,\ 2,\ \cdots)$. Suppose that $ A_n$ are on the parabola $ y \equal{} x^2$ and approach to the origin from right hand; $ B_n$ are on the $ x$ axis and approach in such way as well. If $ \triangle{A_nB_nB_{n \minus{} 1}}$ form an equilateral triangle for every $ n \equal{} 1,\ 2,\ \cdots$ and $ a_1\equal{}1$, then
Find
\[ \sum_{n \equal{} 1}^{\infty} a_n^2\ \text{and}\ \sum_{n \equal{} 1}^{\infty} a_n^3.\]

$ (a_n\minus{}b_n)\sqrt 3\equal{}a_n^2\equal{}(b_{n\minus{}1}\minus{}a_n)\sqrt 3$

$ b_{n\minus{}1}\minus{}b_n\equal{}\frac{2a_n^2}{\sqrt 3}$

so $ \sum_{n\equal{}1}^N a_n^2\equal{} \frac{\sqrt 3}{2}\sum_{n\equal{}1}^N (b_{n\minus{}1}\minus{}b_n)\equal{}\frac{\sqrt 3}{2}(b_0\minus{}b_{N\minus{}1}) \to \frac{\sqrt 3}{2}b_0\equal{}\frac{1\plus{}\sqrt 3}{2}$

i dont see the cube part though =/ something to do with the area?


induction
analytic geometry
strong induction
Pascal\u0027s Triangle

In one of my textbooks it describes a few strategies for proofs by induction when working with 2 variables (call them m and n), i.e. double induction.
The first is the following, which seems to make perfect sense to me:

STRATEGY 1:
a) prove P(1,1) - base case
b) assume P(m,1) - first inductive hypothesis
c) deduce p(m+1,1) -hence you know the statement is true for all (m,1)
d) assume p(m,n) - for all values of m
e) deduce p(m,n+1) - for all values of m, hence the statement is true for all (m,n)

The next one, however, I am having a little difficulty with

STRATEGY 2:
a) prove P(1,n) for all values of n    (doesn't this require an induction on n?)
b) prove P(m,1) for all values of m (similarly an induction on m?)
c) for m,n > 1 assume P(m-1,n) and P(m,n-1)  (i would have thought these were requisite stages in the previous inductions?)
d) deduce P(m,n)

To prove something for all natural numbers $n$ does not mean you need to use an induction, but it does mean you should consider using induction. Also, when you proof does not use induction you were probably using some fact that does need induction.

However, back to the point. When they say prove $P(1,n)$ for all $n$ they do not care how you prove it. You may feel free to use any method you want including induction on $n$. All that matters is that it can be proven.

Here's a problem which succumbs to the second method of double-induction but not to the first:

A [i]good lattice path[/i] is a finite sequence of steps, each step of the form $(a, b) \to (a+1, b)$ or $(a, b) \to (a, b+1)$.  Show that the number of good lattice paths which begin at $(0, 0)$ and end at $(m, n)$ is $\binom{m+n}{m}$.  (That's a binomial coefficient; you may have instead seen the notation $(m+n)Cm$ or $C_{m}^{m+n}$ or $C_{m+n}^{m}$.)

Two comments:
1. In this case, the base cases are with 0s, not with 1s.
2. This problem also has nice non-inductive solution(s).

Hrm, double induction is interesting.
[hide="Proof"]Base cases:
For the case for $(0,n)$, we only have one path: the path that goes vertically up $n$ units. $1=\dbinom{0+n}{0}$, and the hypothesis holds.
Similarly, for the case $(m,0)$, we have: $1=\dbinom{m+0}{m}$.
Assume that the hypothesis holds for $(m-1, n)$ as well as $(m, n-1)$. Since all "good lattice paths" to $(m,n)$ must pass through either one of these points, but not both, the number of ways to get to $(m,n)$ is $\dbinom{m+n-1}{m-1}+\dbinom{m+n-1}{m}$, which, by Pascal's identity, is equal to $\dbinom{m+n}{m}$. [i]QED.[/i][/hide]

Exactly (although it would be nice if you hid your solution, for others who might want to try).

In essence, induction works when you have a bunch of statements in a line: each follows from the one to its left.  (In a strong induction, each statement follows from the conjunction of all the statements to its left.)  In a double induction, you have a lattice (like, say, the first quadrant).  The first kind of double induction outlined in the original post works if one can handle the $x$-axis, and every other statement depends only on those statements below it.  However, it should be clear that in general, this might not be enough: you might need to rely on things both below and to the left.

By the way, the simplest statement of the theorem davidyko proved is, "If you turn Pascal's triangle on its side, it's still Pascal's triangle."   :lol:

Thanks. I think I get it now, although I'm not really sure what lattices are?

In this case, I mean a square grid, like a chessboard or the like.  Or the set of lattice points (points with integer coordinates) in the first quadrant.


limit
real analysis
real analysis unsolved

Let $ f: [0,1]\rightarrow [0,\infty)$ be a continuous function. Find the limit:

$ \lim_{n\rightarrow \infty}\sqrt[n]{\left(f\left(\frac{1}{n}\right)\right)^{n}+\left(f\left(\frac{2}{n}\right)\right)^{n}+\cdots+\left(f\left(\frac{n}{n}\right)\right)^{n}}=?$.

I think the answer can't be unique. If $ f(x)=0$ then the limit is equal to $ 0$ and if $ f(x)=1$ then the limit is $ 1$.

Yes, of course the answer depends on $ f.$ The question is exactly what that dependence is. My first guess is that the limit is the maximum value of $ f$ on $ [0,1],$ although I don't have a proof for that yet. The continuity of $ f$ will be important; without that the problem would be nonsense.

Kent,

You are, as always, right.

The limit is $ \max f(x)$.

Here is a slightly sketchy proof:  Let $ x_{0}\in [0, 1]$ s.t. $ f(x_{0})$ is maximal.  Choose $ \epsilon > 0$, and pick $ \delta > 0$ s.t. $ |x-x_{0}| < \delta$ implies $ f(x) > f(x_{0})-\epsilon$.  Choose $ N$ such that $ (2\delta)^\frac{1}{N}> 1-\epsilon$.  Then for $ n > N$ we have that our expression is (approximately) at least $ \sqrt[n]{2\delta (f(x_{0})-\epsilon)^{n}}> (1-\epsilon)(f(x_{0})-\epsilon)$.  Then since we can choose $ \epsilon$ as small as we like, we have that the limit is at least $ f(x_{0})$.  But also the given expression is less than $ \sqrt[n]{nf(x_{0})}\to f(x_{0})$, so it's at most $ f(x_{0})$, and we're done.