(Frankly, I had no idea where to put this thread, so feel free to move it on to appropriate place if needed.)

I'm currently preparing for National Olympiad (Croatia), and there is a month or so time left. When it comes to algebra or other branches, I'm pretty good and still improving.

But when it comes to geometry, I'm fully blocked. I just can't achive the way of thinking that is required in order to solve geometrical problems.

I somehow get lost in galore of information that's provided in geo based problems.

Is there any suggestion on how to solve geometrical problems? Should I start off from the very basics or engage in hard problems? Should I solve or just read the solutions? Are there any other mathlinks members with the very same "weakness".

I would like to hear some response, so thanks in advance.

Regards.

I would appreciate any information too as I have the Canadian Math Olympiad in 5 days.

Thank you.

Reviewing AoPS2 and Geometry Revisited may help for more advanced things like Simpson lines and Ceva. Challenging Problems in Geometry is good for practice.

In general, find a contest that is "your level", (like do not do IMO problems if you are AIME level), find a compilation of problems, mark out all the geometry, and do them in order. Solving problems is the best way to get better, but the problem is finding good problems and lots of them.

I too suck at geometry.

I used to think I was good at geometry, but then I hit the analytic geo chapter of AoPS 2, and discovered that I had never seen truly difficult geometry problems.

geometry doesn't like me too. i have tried to learn using complex numbers in geometry, but didn't manage :oops:

Oh god, I get killed by geometry.

The problem I have is that whenever I see a geometry problem, I have no idea where to start. There are just too many different possibilities.


Heard on some commercial that scent is strongest sense tied to memory.

So what scent do u associate wit math? For some reason, I tend to associate the scent of hotel soaps with math... That and tacos.

[quote="mcalderbank"]yes some deodorant commercial: (Axe or Old Spice)?

Fresh, crisp, math paper.  :D hot of the printer[/quote]

Old Spice

What scent do I associate with math...bad question...*nods to self*

I will not answer this because the answer is rather embarrassing...

Flatulence after beans for me. It's a great scent to carry( or release) on long road trips with the whole family! :D

Warm carpet.

For three years, I've had math right after lunch at school... Guess that stuck.

I must be strange, but memories associate with taste for me...for example, I went on a Butterfinger(tm) (If you spell that any other way, Nestle WILL HUNT YOU DOWN AND [i]KILL YOU![/i]) spree while reading a book, I forget the title now, but it was by John Christopher,   I believe, (this was a few years ago) and now Butterfingers associate with that book for me.

sillypuddy because i got high off the raytheon brain they handed out at state

Am I the only one here who doesn't associate a smell with math?  When I think of math, I think of numbers, not smells.

Tastee-Freez Hamburgers for Math Team Practice!


For each real number $x,$ let $\lfloor x\rfloor$ denote the greatest integer that does not exceed $x.$  For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_2 n\rfloor$ is a positive even integer.

[hide="Answer"]The conditions are satisfied for $4\le x<7$, $16\le x<32$, $64\le x<128$, and $256\le x<512$. Therefore, the total is $4+16+64+256=\boxed{340}$.[/hide]

Firstly, let's begin the problem by listing the powers of 2 under 1000. We have:

$2,4,8,16,32,64,128,256,512$

Let's make a table if these powers of 2's indices are odd or even

\begin{tabular}{|c|c|c|} \hline
Powers of 2 under 1000 & even or odd exponent \\ \hline
2 & odd \\ \hline
4 & even \\ \hline
8 & odd \\ \hline
16 & even \\ \hline
32 & odd \\ \hline
64 & even \\ \hline
128 & odd \\ \hline
256 & even \\ \hline
512 & odd \\ \hline
\end{tabular}
Wait! The question says that$\lfloor \log_{2} n \rfloor$ must have an even exponent when 2 is raised to a power which is less than 1000, which means we have to create inequalities for $n$.

$n$ must satisfy one of these inequalities.

$4 \leq n < 8$

$16 \leq n < 32$

$64 \leq n < 128$

$256 \leq n < 512$

After calculating, we get that there are exactly 340 values of $n$ which satisfy one of the inequalities (I said this because $n$ will never satisfy 2 inequalities). So $\boxed{340}$ is our answer

[quote=yofro]Yup, just a casual 15 year bump.[/quote]

nothin to see here folks, just pass along

[quote=yofro]Yup, just a casual 15 year bump.[/quote]

yes and just a casual post farm

but was there really a need to bump this? This is gonna be locked anyways tho.
Also, what were you doing at 4:38AM CST? If you are in Asia that is understandable but now it is 3M CST :P


Triangle ABC has vertices A(0,0), B(0,3), and C(5,0). A point, P, inside the triangle is $ \sqrt{10}$ units from point A and $ \sqrt{13}$ units from point B. How many units is P from point C? Express your answer in simplest radical form?

[hide]i graphed it to get a sense of the boundaries for the x-coordinate and y-coordinate. assuming they're lattice points there were only a few reasonable points to try. the coordinate of point P is (3,1) therefore the distance from (3,1) to (5,0) is $ \sqrt{5}$ units[/hide]

[hide="probably not the best solution..."]

First you draw the triangle on a diagram.  By looking at these radicals, 10=1^2+3^2 and 13=2^2+3^2.  If we try P= (3,1) we get that it works for the distance for both A and B.  Since side AC=5 units, and AP's change in x is 3, the distance of P from C would be square root (1^2+2^2)= $ \sqrt{5}$

[/hide]

[quote="infinity4ever"][hide="probably not the best solution..."]

First you draw the triangle on a diagram.  By looking at these radicals, 10=1^2+3^2 and 13=2^2+3^2.  If we try P= (3,1) we get that it works for the distance for both A and B.  Since side AC=5 units, and AP's change in x is 3, the distance of P from C would be square root (1^2+2^2)= $ \sqrt {5}$

[/hide][/quote]

that's what i did too. the only question i was asking myself that may cause this to be the wrong answer is whether (3,1) is actuallyi n the triangle. im too lazy to check though.  :P

[hide]well from the information we can create two circles with centers at $ (0,0)$ and $ (0, 3)$ with radius $ \sqrt {10}$ and $ \sqrt {13}$ respectively.

So we get to equations

$ x^{2} \plus{} y^{2} \equal{} 10$
$ x^{2} \plus{} (y \minus{} 3)^{2} \equal{} 13$

Plugging in some values quickly yields $ (3, 1)$ as an intersect

Now applying the distance formula, to $ (3, 1)$ and $ (5, 0)$ gives

$ d \equal{} \sqrt {(0 \minus{} 1)^{2} \plus{} (5 \minus{} 3)^{2}}$ which tells us that the point is $ \sqrt {5}$ away from $ c$.

$ (\minus{}3, 1)$ also satisfies, but is not in the triangle.[/hide]

[quote="SnipedYou"][hide]well from the information we can create two circles with centers at $ (0,0)$ and $ (0, 3)$ with radius $ \sqrt {10}$ and $ \sqrt {13}$ respectively.

So we get to equations

$ x^{2} \plus{} y^{2} \equal{} 10$
$ x^{2} \plus{} (y \minus{} 3)^{2} \equal{} 13$

Plugging in some values quickly yields $ (3, 1)$ as an intersect

Now applying the distance formula, to $ (3, 1)$ and $ (5, 0)$ gives

$ d \equal{} \sqrt {(0 \minus{} 1)^{2} \plus{} (5 \minus{} 3)^{2}}$ which tells us that the point is $ \sqrt {5}$ away from $ c$.

$ ( \minus{} 3, 1)$ also satisfies, but is not in the triangle.[/hide][/quote]

what does  x and y represent?

Those are just the equations of a circle.

Generally an equation for a circle is $ (x \minus{} a)^{2} \plus{} (y \minus{} b)^{2} \equal{} r^{2}$ where $ (a, b)$ is the center of the circle.

In my first equation the center is at $ (0, 0)$ so instead of writing $ (x \minus{} 0)^{2} \plus{} (y \minus{} 0)^{2}$ I just wrote $ x^{2} \plus{} y^{2}$.

Notice that the point is $ \sqrt{10}$ away from the vertex too. So $ \sqrt{10}$ is the radius and all points $ \sqrt{10}$ units from $ (0, 0)$ lay on the circle.

$ x^{2} \plus{} y^{2} \equal{} 10$

Sorry if that was a bit confusing.

[hide]
$ x^{2} \plus{} y^{2} \equal{} 10$
$ x^{2} \plus{} (3 \minus{} y)^{2} \equal{} 13 \implies x^{2} \plus{} y^{2} \minus{} 6y \plus{} 9 \equal{} 13 \implies 10 \plus{} 9 \minus{} 6y \equal{} 13 \implies (3,1) \text{or} (\minus{}3, 1)$. Now plug the coordinates of point P and C back into the distance formula and you get $ \sqrt {(5 \minus{} 3)^{2} \plus{} 1} \implies \boxed{\sqrt {5}}$.[/hide]

[quote="pascal_1623"]Triangle ABC has vertices A(0,0), B(0,3), and C(5,0). A point, P, inside the triangle is $ \sqrt {10}$ units from point A and $ \sqrt {13}$ units from point B. How many units is P from point C? Express your answer in simplest radical form?[/quote]

[hide]
the square pairs...(haha)

3,1
3,2

-
- -
- - - *-*    those are the integer points within the triangle
- - -   - - -

so we find that pretty easily, since it's the only way to get 3,2 from point a (if you get what i mean), and then we just say rt5 away from c

[/hide]

Let the coordinates of point $P$ be $(a,b)$. We have $a^2+b^2=10$ since $AP = \sqrt{10}$, and $a^2+(b-3)^2=13$ since $AB = \sqrt{13}$. Expanding $(b-3)^2$ gives us \[a^2 +b^2 - 6b + 9 = 13.\] Since $a^2 + b^2 = 10$, we have $10-6b+9=13$, so $b=1$. From $a^2+b^2=10$, we have $a^2=9$, so $a=\pm 3$. If $a$ is $-3$, the point is not inside the triangle, so $a=3$. So the point is $(3,1)$ and the distance from $C$ is $$\sqrt{(3-5)^2+1^2}=\boxed{\sqrt{5}}$$


[quote="FMako"]Suppose you had a problem like,
$\cos{x}+\cos{2x}+\cos{3x}+\cos{4x} = 3$
How would you solve?
How would you solve a similar problem with $\sin{x}$?
How would you solve a similar problem with $\tan{x}$?
$\cot{x},\sec{x},\csc{x}$?
What if the coefficients inside the trig function were not linear?
Let's say $x,2x,4x,...2^kx$ or even $x,4x,9x,16x,...k^2x$?[/quote]

For the first three, I think you could rewrite in cis or $e^{i\theta}$ form and sum. Or just do it manually. :)
If the coefficients were not linear, it will depend on the situation. Sometimes, it will be impossible to solve it easily.

[quote="FMako"]Suppose you had a problem like,
$\cos{x}+\cos{2x}+\cos{3x}+\cos{4x} = 3$
How would you solve?
How would you solve a similar problem with $\sin{x}$?
How would you solve a similar problem with $\tan{x}$?
$\cot{x},\sec{x},\csc{x}$?
What if the coefficients inside the trig function were not linear?
Let's say $x,2x,4x,...2^kx$ or even $x,4x,9x,16x,...k^2x$?[/quote]

[hide]
$2\cos{5x/2}\cos{3x/2} + 2\cos{5x/2}\cos{x/2} =3$
$2\cos{5x/2}(\cos{3x/2} + \cos{x/2}) =3$
$2\cos{5x/2}(2cos{x}\cos{x/2}) = 3$

and so on...i need to go to sleep, but should be  more reasonable from here.[/hide]


[quote="agripa"]\sqrt {x + 2 - x^2 } < \sqrt {x^2 - 3x + 2} + x. please help me[/quote]

I think you mean : solve inequality $ \sqrt {x \plus{} 2 \minus{} x^2 } < \sqrt {x^2 \minus{} 3x \plus{} 2} \plus{} x$ for real $ x$.

If so :
$ 1)$ you must have $ x \plus{} 2 \minus{} x^2\geq 0$ and $ x^2 \minus{} 3x \plus{} 2\geq 0$, which gives $ x\in[\minus{}1,1]$

$ 2)$ you must have $ \sqrt {x^2 \minus{} 3x \plus{} 2} >\minus{}x$  which indeed is true $ \forall x\in[\minus{}1,1]$ (at least).

$ 3)$ then, since both sides of inequality are nonnegative, you can raise to power 2 :

$ (\sqrt {x \plus{} 2 \minus{} x^2 })^2 < (\sqrt {x^2 \minus{} 3x \plus{} 2} \plus{} x)^2$

$ 2x\sqrt {x^2 \minus{} 3x \plus{} 2}>4x\minus{}3x^2$
We can then consider $ x\neq 0$ since inequality is strict and we have two cases :

$ 3.1)$ $ x\in[\minus{}1,0)$, so $ x<0$ and inequality becomes $ 2\sqrt {x^2 \minus{} 3x \plus{} 2}<4\minus{}3x$
Both sides are $ \geq 0$ so we can raise to power 2 with same inequality : $ 4x^2 \minus{} 12x \plus{} 8<16\minus{}24x\plus{}9x^2$
So $ x\in[\minus{}1,0)$ and $ 5x^2\minus{}12x\plus{}8>0$ which is always true when $ x\in[\minus{}1,0)$

$ 3.2)$ $ x\in(0,\plus{}1]$, so $ x>0$ and inequality becomes $ 2\sqrt {x^2 \minus{} 3x \plus{} 2}>4\minus{}3x$
Both sides are $ \geq 0$ so we can raise to power 2 with same inequality : $ 4x^2 \minus{} 12x \plus{} 8>16\minus{}24x\plus{}9x^2$
So $ x\in(0,\plus{}1]$ and $ 5x^2\minus{}12x\plus{}8<0$ which is always wrong when $ x\in[\minus{}1,0)$

And so the solution is $ x\in[\minus{}1,0)$


inequalities
calculus
algorithm
inequalities solved

For a,b,c real positive such that a^2+b^2+c^2=1 prove that

   a/(1+bc) + b/(1+ac) + c/(1+ab) \leq  \sqrt 2

Are you sure it's like this? Do you have any idea when the equality is fulfilled?

I was thinking: maybe it's  \sum a^2=3/2 so that we get equality when a=b=c=1/2^1/2. It doesn't have to be like this, of course.

Equality occurs for a = 0, b=c= \sqrt (2)/2.
I know a is supposed to be positive, but using continuity, it shows that the inequality, if true, is best possible.

This is a very difficult inequality. It has been proved only by calculus.
Many mathematicians are trying to find an NCB( No Calculus Based )
solution. I don't know if it will be a vain attempt. Do tou think it will be possible to find an NCB solution ? It would be interesting to find an algorithm by computer to decide if an equality is NCB solvable.
Is it a reasonable or silly idea?

Hello,

I use "cutting" technics of physician to solve this problem (see the similar inequality in this forum)

Asumme that a  \leq b  \leq c, then

LHS  \leq (a+b+c)/(1+ab) (1)

We now try to prove that
             a+b+c  \leq  \sqrt2 (1+ab)  (2)

By squaring both sides (2) is equivalent to
            a 2 + b 2 + c  2 + 2ab+2bc+2ca  \leq 2(1+2ab+a 2b 2)
<=> (a+b-c) 2 + 2a 2 b 2  \geq 0 that holds.

From (1) and (2) we get the result. The euality holds when a = 0, b = c = 1/ \sqrt 2.

Jichen... Have you got use computer? :D
By UCT we can find the following inequality easily:
\[\frac{a}{1+bc}\le \frac{\sqrt{2}a}{a+b+c}\]


A charity sells $140$ benefit tickets for a total of $\$$ $2001$. Some tickets sell for full price ( a whole dollar amount), and the rest for half price. How much money is raised by the full-price tickets?

Here's a [hide="hint"]It's pretty straightforward; just set up a system of equations and solve.[/hide]

I don't think you can just set up equations...you don't get the required information...you could always set up equations that would solve for integers, but I'm clueless on those.

Yeah, you are right
I tried setting up the system of equations but we are missing information. Any suggestions?

nice problem

at first I didn't think there was enough info, but playing around with it allowed me to eliminate a lot of options and keep moving forward...

hint-[hide]
Try factoring. Keep in mind the amount of tickets is an integer, as is the amount of the full price ticket. After prime factoring use constraints within the equation to make eliminations.

This is assuming I did it correctly, lol.
[/hide]

[hide]I got 34 full price tickets at 23, and 106 at 23/2, for a profit on full tickets of $782$

I'm a bit tired now, and probably wrong, but I may post how I did it tomorrow. I don't think it's necessary though.[/hide]

Could you describe what your variables mean? I'm having a hard time following what you did, and I don't want to mislead you.

oh ok.

$x=$ full price of ticket
$y=$ amount of full price tickets sold

Factored after expanding. (not two seperate equations).

I know that x and y have to be integers. i finally solved it by just plugging the equation into a calculator and going to table and finding values with two integers (under 140).

[hide]

alright, for convenience multiplying your equation by 2 yields

$x(y + 140) = 4002$

now, is there anything special about $x$ and $y$? $x$, the price, must be a whole dollar amount, aka integer. $y$, the amount of tickets sold, must be an integer as well. This should simplify things. Looking back at $x(y + 140) = 4002$, we notice 2 factors are being multiplied to 4002. The first factor is an integer because x is. The second is an integer because y is, only that it will be an integer 140 more than y.

All you have to do now is prime factor 4002. There should only be one factor, $q$, such $q>140$ and $q - 140 < 140$

happy solving
[/hide]

thanks pocketsand!  :D

I get it now. One of the factors has to be both greater than 140 (because $y$ is a positive integer) and less than 140 when subtracted by 140. This incorporates all of the different limits that are given.


Let be $ x,y,z\in \mathbb{R}_\plus{}$ such that : $ \frac{1}{1\plus{}x}\plus{}\frac{1}{1\plus{}y}\plus{}\frac{1}{1\plus{}z}\equal{}2$ . Prove that :

$ 8xyz\le 1$

[quote="alex2008"]Let be $ x,y,z\in \mathbb{R}_ \plus{}$ such that : $ \frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z} \equal{} 2$ . Prove that :

$ 8xyz\le 1$[/quote]
We have: $ \frac{1}{1\plus{}x} \equal{}\frac{y}{1\plus{}y}\plus{}\frac{z}{1\plus{}z} \ge 2 \sqrt{\frac{yz}{(y\plus{}1)(z\plus{}1)}}$
Similar,we have Q.E.D

very nice quykhtn-qa1! this is my solution,
let put, $ \frac{1}{1+x}=a$ and $ \frac{1}{1+y}=b$ and $ \frac{1}{1+z}=c$, we have $ a+b+c=2$
therefore, $ x=\frac{1-a}{a}$ and ....
then, $ xyz=\frac{(1-a)(1-b)(1-c)}{abc}=\frac{1-\sum a +\sum ab -abc}{abc}=\frac{%Error. "sumab" is a bad command.
-abc-1}{abc}$
the inequality becomes, $ \frac{%Error. "sumab" is a bad command.
-abc-1}{abc}\leq \frac{1}{8}$
$ \Leftrightarrow 8\sum ab -8abc-8\leq abc$
$ \Leftrightarrow 8\sum ab \leq 9abc+ 8$ wich is Shur,
this one is more easy, and more nice, if $ x,y,z \in \mathbb{R^+}$ and $ \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$,
prove that, $ xyz \geq 8$

[quote="peine"]very nice quykhtn-qa1! this is my solution,
let put, $ \frac {1}{1 + x} = a$ and $ \frac {1}{1 + y} = b$ and $ \frac {1}{1 + z} = c$, we have $ a + b + c = 2$
therefore, $ x = \frac {1 - a}{a}$ and ....
then, $ xyz = \frac {(1 - a)(1 - b)(1 - c)}{abc} = \frac {1 - \sum a + \sum ab - abc}{abc} = \frac {%Error. "sumab" is a bad command.
- abc - 1}{abc}$
the inequality becomes, $ \frac {%Error. "sumab" is a bad command.
- abc - 1}{abc}\leq \frac {1}{8}$
$ \Leftrightarrow 8\sum ab - 8abc - 8\leq abc$
$ \Leftrightarrow 8\sum ab \leq 9abc + 8$ wich is Shur,
this one is more easy, and more nice, if $ x,y,z \in \mathbb{R^ + }$ and $ \frac {1}{x + 1} + \frac {1}{y + 1} + \frac {1}{z + 1} = 1$,
prove that, $ xyz \geq 8$[/quote]
We have: $ \frac{x}{x+1} =\frac{1}{y+1}+\frac{1}{z+1} \ge 2 \sqrt{\frac{yz}{(y+1)(z+1)}}$
Then,we have Q.E.D

nice, now, we have $ x_1,x_2,...,x_n\in \mathbb{R^\plus{}}$ and,$ \sum_{i\equal{}1}^n \frac{1}{x_i\plus{}1}\equal{}1$, prove that,
$ x_1x_2...x_n \geq (n\minus{}1)^n$

[quote="peine"]nice, now, we have $ x_1,x_2,...,x_n\in \mathbb{R^ \plus{} }$ and,$ \sum_{i \equal{} 1}^n \frac {1}{x_i \plus{} 1} \equal{} 1$, prove that,
$ x_1x_2...x_n \geq (n \minus{} 1)^n$[/quote]
We have: $ \frac {x_1}{x_1 \plus{} 1} \equal{} \frac {1}{x_2 \plus{} 1} \plus{} ... \plus{} \frac {1}{x_n \plus{} 1} \ge (n \minus{} 1) \sqrt[n\minus{}1] {\frac {1}{(x_2 \plus{} 1)....(x_n \plus{} 1)}}$
Similae,we have Q.E.D :wink:

[quote="quykhtn-qa1"][quote="peine"]nice, now, we have $ x_1,x_2,...,x_n\in \mathbb{R^ \plus{} }$ and,$ \sum_{i \equal{} 1}^n \frac {1}{x_i \plus{} 1} \equal{} 1$, prove that,
$ x_1x_2...x_n \geq (n \minus{} 1)^n$[/quote]
We have: $ \frac {x_1}{x_1 \plus{} 1} \equal{} \frac {1}{x_2 \plus{} 1} \plus{} ... \plus{} \frac {1}{x_n \plus{} 1} \ge (n \minus{} 1) \sqrt {\frac {1}{(x_2 \plus{} 1)....(x_n \plus{} 1)}}$
Similae,we have Q.E.D :wink:[/quote]
yes very nice but it's $ \frac {x_1}{x_1 \plus{} 1} \equal{} \frac {1}{x_2 \plus{} 1} \plus{} ... \plus{} \frac {1}{x_n \plus{} 1} \ge (n \minus{} 1) \sqrt[n\minus{}1] {\frac {1}{(x_2 \plus{} 1)....(x_n \plus{} 1)}}$ not $ \frac {x_1}{x_1 \plus{} 1} \equal{} \frac {1}{x_2 \plus{} 1} \plus{} ... \plus{} \frac {1}{x_n \plus{} 1} \ge (n \minus{} 1) \sqrt {\frac {1}{(x_2 \plus{} 1)....(x_n \plus{} 1)}}$


Is it just me, or is [url=http://mathcircle.berkeley.edu/mpgeo.pdf]this article[/url] about mass points really confusing?

If I knew the theory behind why mass points work, I could probably understand it, but right now, I'm lost. Help is appreciated.

Thanks.  :)

Here's some basic pictures detailing everything I know about mass points, which is enough to apply it to any of the problems in that PDF.

[url=http://img.photobucket.com/albums/v734/Elemennop/massp1.jpg]1st Mass Points Info[/url]
[url=http://img.photobucket.com/albums/v734/Elemennop/massp2.jpg]2nd Mass Points Info[/url]

[url=http://staff.imsa.edu/math/journal/volume4/articles/TriangleCenters.pdf]This might help some more[/url] but it doesn't tell you why it works. But, its kind of obvious why it works so I think you should be fine.


Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be 100 times larger than the magnitude of A - B, what must be the angle between them?

I am having trouble figuring out the correct way to approach this. I have tried an analytical approach by letting [b]A[/b] = Ai + Aj etc... and writing expressions in terms of other expressions using the given information in the problem. I also am aware of the formula for the angle between two vectors in terms of the magnitudes and the dot products but my professor has not lectured about that so i am assuming he does not want us to use that method.

The analytic approach is much harder than the geometric approach.  Consider:

Draw vector $\vec{A}$ starting from the origin.  Draw vectors $\vec{B},-\vec{B}$ starting from the end of $\vec{A}$.  This will give you two new resultant vectors $\vec{C}= \vec{A}+\vec{B}, \vec{D}= \vec{A}-\vec{B}$.

$\triangle OCD$ is a [b]right triangle[/b] (this is the important result - can you convince yourself that this must be true?), where $O$ is the origin.  We want $\frac{OD}{OC}= \tan \angle OCD = 100$, so $\angle OCD = \arctan 100$.

Next, try to convince yourself geometrically that $\angle OCD$ is half the angle between $\vec{A}$ and $\vec{B}$, so our desired angle is $\boxed{ 2 \arctan 100 }$.

ah, ok. Thanks alot! However, i believe that OD/OC = 1/100, or OC/OD = 100.

Right, sorry, I'd forgotten which letter I'd assigned to which vector.

I am having a problem with this exact problem.

I am completely new to vectors and only had one day to figure out this exact problem, but I still can not figure it out... even with your hints.

What is the exact angle in °'s and how did you find this out? Is there anything you could do to help me here?

It would also help if there was a way you could help me visual the process.

Thanks.

Nevermind, I got it. 2arctan(1/100)... i missed the last comment :]


A number is randomly selected from the one-digit prime numbers. What is the probability that its square has a units digit of 9? Express your answer as a common fraction.

The one-digit prime numbers are 2,3,5, and 7. The square of 3 and 7 have a units digit of 9, so the answer is 1/2.


geometry
3D geometry
sphere
prism
function
trigonometry

A small sphere of radius r and mass m is held in a cylindrical  laser parallel beam of radius R>r (the sphere is levitating) .Fiind the power of the laser if f reprezents the energy absorbed by the sphere from the total energy that interacts with the sphere.

I think this was an IPHO problem not so long ago. But I may be wrong.
Anyway Radac slow down a bit with posting the problems we cant keep up with you  :D  :D

I think that it was a prism at the IPhO, but maybe there was a sphere as well...

Btw, [b]Thaakisfox[/b], are you going to the IPhO? What does the selection look like in Hungary?

No. I would have to every week to some extra classes, and then there are about 3 qualifieing competitions before it, and half of my summer would go waste.
So instead I just study physics at home. I'm not so interested in the competions. I just like pure physics  :D
and you? are you going?

Well, I hope so, our national competition will be held in two weeks, and after that we have two more selection tests and some preparations for the IPhO. But the alternative is going to school  :D And, besides the great feeling of going to the IPhO, our government has started awarding olympiad medalists with enormous sums of money, so it really pays off...

But anyway, I'll open up a new topic regarding the IPhO, and we could continue discussing it there.

I got W=sqr(R/r)*m*g*c.
Are you all in school or something? :)

Here is what I got for the power:
$P=\frac{R^{2}}{r^{2}}\frac{mgc}{2-f}$

Unfortunately i don't have the answer.
Did you reach dF=[4(1-f)Wsinx (cos^4x) dx]/c for the force exerted by the reflected photons(the vertical component)?
where W is the power on the sphere and x is the angle that must be integrated from 0 to pi/2

I got the same answer as [b]Thaakisfox[/b]. I just used the fact that the pressure on the sphere is \[p = \frac{I}{c}(2-f),\] where $I = P/R^{2}\pi$ is the intensity of the laser...

By considering the elastic collisions between the photons and the sphere, but you don't have to derive it :)

well i started by finding the force that a photon exerts by colliding elastically at a certain angle-x then i found the F component on the vertical and then i introduced the concentration of photons so that the number of electrons that fall on the sphere and make a angle x with vector(r) is dS*N/sqr(r)*pi where N is the number of reflected photons and dS=ds*cos x where ds is the surface of the sphere where the same number of photons fall then i obtained the dF as a function of dx.isn't it right?

could one of you please post a sollution?

Well, I've already posted the most important part of my solution, but here's how it would end:

Ok, you now the pressure exerted on the ball. The force is $F = p r^{2}\pi$ (the pressure times the area of the cross-section of the ball). Since $mg = F$, you can now easily retrieve our solution.

the part with the deduction of the pressure is troubleing me....

Well, see here: if you analyze each photon, you will conclude that it will supply a pressure of $\text dI/c \cdot (2-f)$; then the corresponding force is $\text dF = p\text dS \cos\phi$; you can regard this as the pressure $p$ acting on a projection of the elementary surface $\text dS$ of the sphere on the plane normal to the ray of photons. Summing them all up, you'll get $I/c (2-f) r^{2}\pi$ on one side, since each photon contributes the same pressure, and the sum of all the projections of elementary surfaces equals to the cross-section of the sphere.


let $ A$ be a subset of $ \{1,2,\ldots,2n\}$ such that $ |A|\equal{}n\plus{}1$. Show that there exists $ a,b\in A$ such that $ a$ and $ b$ are coprime.

Partition the set into $ n$ subsets : $ \{1,2\},\{3,4\}....\{2n\minus{}1,2n\}$
Then in each subset $ A$ having $ |A|\equal{}n\plus{}1$ will contain $ 2$ elements from at least $ 1$ of the $ n$ partitions. These two elments will differ by $ 1$ and hence they are coprime.


vector
linear algebra
matrix
analytic geometry

1) Let V be a finite dimensional space over the field F. If L is a linear functional on the dual space V* of V, then there is a unique vector a such that L(f) = f(a).

2) Let V be a finite dimensional space over the field F. Each basis for V* is the dual of some basis for V.

I am trying to interpret these statements using matrices in the following way.

Let $ B' = \{ f_{1},...,f_{n}\}$be a basis for V* and let $ C = \{ b_{1},...,b_{n}\}$be any basis for V. Let S be the n x n invertible matrix with ith row as the matrix of $ f_{i}$relative to the ordered basis C and {1}. Let the kth column of $ S^{-1}$be the coordinate matrix of some vector $ a_{k}$ in the ordered basis C ; $ B = \{ a_{1},...,a_{n}\}$is a basis for V and B' is the dual of B.

However my book uses statement 1 to prove statement 2. Although what I've done here seems fine, I can't see how to demonstrate 1. I've tried writing $ SS^{-1}= I$ and taking the transpose of both sides but I can't seem to relate it to S. I have proved the second statement using S but I would like to know how to prove the first similarly.

I would handle this a little differently.

First, we show that for any vector space $ V$ there is a canonical linear map from $ V$ to $ V^{**}$ given by $ (x^{**})(f)=f(x)$. This map is injective because for any given nonzero $ u$, we can find some linear functional which is not zero at $ u$.
Second, we show that $ V^{*}$ has the same dimension as $ V$ if $ V$ is finite dimensional; this forces the map defined above to be an isomorphism.

As for the second part, let's be constructive:
Choose a basis $ f_{1},\dots,f_{n}$ of $ V^{*}$. For each $ i$, let $ U_{i}$ be the null space of $ f_{i}$. Each $ U_{i}$ has dimension $ n-1$, so the intersection of any $ n-1$ of them is nontrivial. In addition, the intersection of all $ n$ $ U_{i}$ is $ \{0\}$; since the $ f_{i}$ are a basis of $ V^{*}$, every $ x\in V$ has $ f_{i}(x)$ nonzero for some $ i$. Therefore, we can choose $ x_{i}\in \bigcap_{j\neq i}U_{j}\setminus \{0\}$, and $ f_{i}(x_{i})\neq 0$ for each $ i$. Let $ v_{i}=\frac{x_{i}}{f(x_{i})}$. The $ f_{i}$ are a dual basis to the $ v_{i}$.


vector
linear algebra
matrix
superior algebra
superior algebra unsolved

Let $ A$ be a finite dimensional algebra over a field $ k$. Let $ I$  be a two sided ideal generated by nilpotent elements.
Prve that $ I$ is nilpotent.

I suppose that you mean generated as a $ k$-vector space by nilpotent elements.
First suppose $ A$ is semisimple.
Then by Artin-Wedderburn theorem it follows that $ A\simeq M_{n_1}(D_1)\times\dots\times M_{n_r}(D_r)$ where $ D_i$ are $ k$-algebras which are division rings.
So any two sided ideal in $ A$ must be of the form: $ M_{n_1}(D^\prime_1)\times\ldots\times M_{n_r}(D^\prime_r)$ where, $ D^\prime_i$ means either $ D_i$ or $ \{0\}$. (example: $ M_{n_1}(D_1)\times \{0\}\times M_{n_3}(D_3)$ ).
Therefore the only ideal which can be generated (as a $ k$-vector space) by nilpotent elements is $ (0)$ (the zero ideal) which is nilpotent.

If $ A$ is not semisimple, then $ A/J(A)$ is semisimple and looking at $ I$ inside $ A/J(A)$ we see that it must be $ 0$. This means that $ I\subset J(A)$.
Since $ A$ si artinian (because it is finite dimensional over $ k$) we have that $ J(A)$ is nilpotent, hence $ I$ is nilpotent.

[quote="Kuba"]So any two sided ideal in $ A$ must be of the form: $ M_{n_1}(D^\prime_1)\times\ldots\times M_{n_r}(D^\prime_r)$ where, $ D^\prime_i$ means either $ D_i$ or $ \{0\}$. (example: $ M_{n_1}(D_1)\times \{0\}\times M_{n_3}(D_3)$ ).
Therefore the only ideal which can be generated (as a $ k$-vector space) by nilpotent elements is $ (0)$ (the zero ideal) which is nilpotent.
[/quote]
Why  cannot a nontrivial two sided ideal be generated by nilpotent elements?

In $ M_n(k)$ where $ k$ is a field any two-sided ideal is $ (0)$ or the whole ring.
So you want to see that $ M_n(k)$ cannot be generated (as a $ k$-vector space) by nilpotent matrices.
This can be seen by looking at the trace map: if any matrix were a linear combination of nilpotent matrices then it would have trace 0. But it is obvious that there are matrices with non-zero trace.

That's how the solution goes but first you have to prove that $ A$ is of the form $ M_{n_1}(k)\times\dots\times M_{n_s}(k)$. First you need to show that the problem reduces to the case when $ k$ is algebraically closed by considering extension of scalars. And then that the only finite dimensional division algebra over an algebraically closed field is the field itself, so at last $ D_i \equal{}k$.
And then as you said :roll:


Let k be a posotive integer number. Suppose $\large x_1 \geq  x_2 \geq  . .. \geq x_n$ be positive number roots of equation:
$\large x^2_{1}+x^2_{2}+. . .+x^2_{n}=kx_{1}x_{2}. . .x_{n}$
satisfying $\large x_{1}+x_{2}+. . .+x_{n}$ be maximum.
Prove that:
$\large x_{3}x_{4}. . .x_{n} \leq  \frac{2(n-1)}{k}$.

[quote="toumaf"]How do you prove \[ x_1^2 \leq (x_2^2+ \cdots + x_n^2)  \]?[/quote]
Because $x_1+x_2+...+x_n$maximum.This is a property of Markov'equation


Find the ordered triple of [i]positive[/i] integers $ (a,b,c)$ that satisfy $ (\sqrt{5}\plus{}\sqrt{2}\minus{}\sqrt{3})(3\sqrt{a}\plus{}\sqrt{b}\minus{}2\sqrt{c})\equal{}12$. (NYSML)[/i]

Solution

[hide] $ (2,30,3)$[/hide]

Explanation
[hide]Divide $ 12$ by $ (\sqrt{5} \plus{} \sqrt{2} \minus{} \sqrt{3})$.  Now simplify the fraction by multiplying by the conjugate $ (\sqrt{5} \plus{} \sqrt{2} \plus{} \sqrt{3})$, but don't expand yet.  Reducing yields $ \dfrac{6(\sqrt{5} \plus{} \sqrt{2} \plus{} \sqrt{3})}{2 \plus{} \sqrt{10}}$.  Multiply by the conjugate again to obtain $ \dfrac{(\sqrt{5} \plus{} \sqrt{2} \plus{} \sqrt{3})(2 \minus{} \sqrt{10})}{ \minus{} 1}$.
Expanding and simplifying yields the term $ 3\sqrt{2} \plus{} \sqrt{30} \minus{} 2\sqrt{3}$.[/hide]


For how many people here is today (Wednesday the 13th) a school day?  Finals do not count.

LAST DAY TOMORROW
(like spanish finals and 100 minutes of slacking off in pe)
W00T

4 days of classes, then canobie lake park on wednesday and walk around and do nothing on thursday

woo hoo schools been out for a week.  i just finished my math final at a college nearby today.

i havent really had much free time so far because the first day i had to go to my graduation and then my sister had her 12th grade graduation and then i had to go to a physics thing and study for the final so yea

like two weeks, three if not include finals
Edit: i mean we got out two weeks ago

We still have two more days (school ends on 6/15), but we really don't do much. (No finals...)

School might end in 2008 or something, when the Mathcounts season is over.

You guys are so lucky.
Oh well, these last days of school aren't bad, three out of five of my classes are just watching movies, and they're not all bad.
We have a field trip to some Kimball's Farm place... I envy the 8th graders, they get to go hiking on Mt. Wachusett.
Oh, and we get to watch the end-of-year relay races.

[quote="alkjash"]We have a field trip to some Kimball's Farm place... [/quote]

Isn't that some minigolf place? I seem to remember having gone there when I was like 7...

anyway 6/12 was my last day. my math teacher gave a REALLY long speech about me when presenting the math awardy thing at graduation. it was at least 3 times as long as all the other ones. i just missed the hghest gpa award-1 A- vs. an A...

Technically our last day of school is on the 18th but for me, 8th grade graduation is on the 15th, so the last "work" day of school was kind of today...(not really, I just got some friends to play Texas Hold 'Em in class).

Still got some finals... Last one is next Thursday.

Tomorrow is last day and graduation (from MS). :D

Out since May 16th.
And no, I'm not a senior  :)

13 WEEK SUMMER VACATION

[quote="The Zuton Force"]13 WEEK SUMMER VACATION[/quote]*Cough* WHAT??!?!?!?!?  I get 7 weeks!!! :mad:

ok, I think Dad's being nice to me, and letting me have one month break(a long time, one month), ending in August since I have a 3 week( :( ) class at a College.


analytic geometry
conics
hyperbola
geometry proposed
geometry

It is given a circle centered at $ O$ and a chord $ AB$ on it. Construct a chord $ CD$ perpendicular to $ AB$ such that $ E\in CD \cap AB$ and $ EC=EO$

[img]10118[/img]

Let G be foot of a normal from O to AB and denote $ a \equal{} OG.$ Let P be any point on the line AB and Q a point on a normal to AB at P, such that $ OP \equal{} PQ.$ Put the coordinate origin at the circle center O, x-axis parallel to AB. The x-coordinate of Q is $ x$ and the y-coordinate of Q is $ a \plus{} \sqrt {a^2 \plus{} x^2}.$ Then $ (y \minus{} a)^2 \minus{} x^2 \equal{} a^2$ and locus of Q is a rectangular hyperbola $ \mathcal H$ centered at $ G \equal{} (0, a)$ with semimajor axis $ a.$ The circle $ \mathcal C$ centered at the origin O and with radius $ r$ has equation $ x^2 \plus{} y^2 \equal{} r^2.$ To find intersections $ \mathcal C \cap \mathcal H,$ solve for the 2 equations for $ y.$

$ (y \minus{} a)^2 \minus{} (r^2 \minus{} y^2) \equal{} a^2,\ \ 2y^2 \minus{} 2ay \minus{} r^2 \equal{} 0,\ \ y \equal{} \frac {a \pm \sqrt {a^2 \plus{} 2r^2}}{2}$

Construct square OKLM with sides $ r \equal{} OK \equal{} KL \equal{} ...,$ transfer its diagonal OL to ON on the x-axis, $ ON \equal{} OL \equal{} r \sqrt {2},$ $ GL \equal{} \sqrt {a^2 \plus{} 2r^2}.$ Circle with center G and radius GL meets the y-axis OG at U, U', such that $ OU,\ OU' \equal{} a \pm \sqrt {a^2 \plus{} 2r^2}.$ Take midpoints V, V' of OU, OU' and draw parallels to the x-axis AB through V, V', one intersecting the circle $ \mathcal C$ at C, the other at C'. Drop normals from C, C' to AB, their feet labeled E, E'. Then $ OE \equal{} CE,\ OE' \equal{} CE'.$

Here is my solution:

[quote="yetti"]Let G be foot of a normal from O to AB and denote $ a \equal{} OG$[/quote]

$ EC \cdot ED \equal{} R^2 \minus{} EO^2$

Let $ x\equal{}EO\equal{}EC\equal{}>$

$ x \cdot (x\plus{}2a) \equal{} R^2 \minus{} x^2\equal{}>$

$ 2x^2\plus{}2ax \minus{} R^2 \equal{} 0 \equal{}>$

$ x \equal{} \frac{\minus{}a\plus{}\sqrt{a^2\plus{}2R^2}}{2}$, constructible

We have a solution only iff $ x>a \Leftrightarrow \frac{\minus{}a\plus{}\sqrt{a^2\plus{}2R^2}}{2}>a \Leftrightarrow \boxed{R>2a}$


prove for each $n$.

there exist a prime number $P$,inwhich :

$n<P<n!$

It's a direct consequence of Bertrand's Postulate (http://mathworld.wolfram.com/BertrandsPostulate.html ) but in this case there is a shorter proof :

$(n!, n! - 1) = 1$ which means that a prime divisor of $n!-1$ must be $> n$ and conclusion follows.

[quote="ZetaX"]And for all others its rather trivial...
(I think it should be put down to Intermediate or Getting Started)[/quote]

 i my self decide ,where to put the problems... :)


modular arithmetic
number theory unsolved
number theory

If $ m$ is a product of distinct primes, show that $ x\equiv b^u\pmod{m}$ is always a solution to $ x^k\equiv b\pmod{m}$, even if $ \gcd(b,m)>1$

[quote="ZetaX"]Some description of $ k,u$ would be usefull ;)
Probably $ uk \equiv 1 \mod \varphi(m)$.[/quote]

hmm... but it no longer becomes true that $ b^{\phi(m)}\equiv 1\pmod{m}$ if $ gcd(b,m)>1$. So I won't be able to effectively usee the fact that $ uk\equiv1\pmod{\phi(m)}$

By my post, I wanted to _ask_ you what they are supposed to fulfill. Obviously, it's not true for arbitrary $ u,k$.

For the case I gave, just check $ \mod$ all prime factors, this is sufficient.


stie cineva cine a propus problemele la juniori :?:

Nu stiu. Felicitari Radu pentru problema de geometrie.

Asta e ce stiu eu:

Ziua 2, [color=darkblue]Problema 3[/color] http://www.mathlinks.ro/viewtopic.php?p=820961#820961

Bozgan Francisc.

Ziua 3.

[color=darkblue]Problema 1 [/color] http://www.mathlinks.ro/Forum/viewtopic.php?p=840301#840301

Mircea Lascu.

[color=darkblue]Problema 2[/color] http://www.mathlinks.ro/Forum/viewtopic.php?p=839754#839754

Flavian Georgescu.

[color=darkblue]Problema 3 [/color] http://www.mathlinks.ro/Forum/viewtopic.php?p=840302#840302

Ilie Livia.

[color=darkblue]Problema 4[/color] http://www.mathlinks.ro/Forum/viewtopic.php?p=839813#839813

Ciupan Andrei.

Ziua 4- Maky a avut 2 probleme, si problema 4 http://www.mathlinks.ro/Forum/viewtopic.php?p=840302#840302 a fost a lui dzeta. (aici s-ar putea sa ma insel :oops: )

Ziua 5, [color=darkblue]Problema 1[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=152598

Cosmin Pohoata

Ziua 6, [color=darkblue]Problema 4 [/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=152839

Turea Lucian (a.k.a maky)

Cam asta e tot ce stiu eu. :)

P.S Cum vi s-au parut problemele anul asta la juniori?

ziua 1, [color=blue]Problema 1[/color]  http://www.mathlinks.ro/viewtopic.php?p=818552#818552   - Adrian Zahariuc  :)

Multumesc la toti... si felicitari andrei - a fost misto problema facuta de tine!

Merci :)   ..oricum, au fost frumoase problemele de anul asta la juniori.

Am doua rectificari de facut:

1. [quote] nush... anyway felicitari radu ai fost singuru care ai facut geometria aia care cik era o lu persu[/quote]
Geometria din ultima zi a fost a lui Dinu Serbanescu: [url]http://www.mathlinks.ro/viewtopic.php?t=152838[/url]

2. [quote]
Ziua 4- problema 3 propusa de Munteanu Sinziana si Georgescu Flavian[/quote]
Nu stiu cine a propus problema, dar nu a fost propusa de cei doi.

Ca niste adaugari, PB2/ziua 5 - Viorel Vajaitu; PB3/ziua 5 - Mihai Baluna; PB4/ziua 5 - Calin Popescu


factorial
function
calculus
calculus computations

$\textit{ Mathematica}$ $5.2$ Says that:

${\prod_{\epsilon =1}^{\frac{n}{2}}\frac{1}{n-(2 (\epsilon-1)+1)}=\frac{i^{-n}(n+1)^{-\frac{n}{2}}}{\Gamma \left(\frac{n}{2}+1\right)}}$

But the right answer calculated by me is:

(Only valid for Even $n$)

${\prod_{\epsilon =1}^{\frac{n}{2}}\frac{1}{n-(2 (\epsilon-1)+1)}=\frac{2^{-\frac{n}{2}}\sqrt{\pi }}{\Gamma \left(\frac{n+1}{2}\right)}}$

(Only valid for Odd $n$)

${\prod_{\epsilon =1}^{\frac{n-1}{2}}\frac{1}{n-(2 (\epsilon-1)+1)}=\frac{2^{\frac{1}{2}-\frac{n}{2}}}{\Gamma \left(\frac{n+1}{2}\right)}}$

Why mathematica made this simple error and dont compute the double factorial?!

If we set before n:= 7, it calculate is true, but if we just use n -> 7 the result is different.

I will never more thrust mathematica output

Seems to me to be a pretty clear bug in Mathematica.  After some investigating, I observed the following behavior.

In[1] := Product[2x+1, {x,0,n-1}]

Out[1] : = 2^n Gamma[1/2 + n]/Sqrt[Pi]

This evaluation is correct.  However, a simple shift in the index gives the strange result

In[2] := Product[2(x-1)+1, {x,1,n}]

Out[2] := n!

This is clearly wrong.  But then if we simplify beforehand, we find

In[3] := Product[2x-1, {x,1,n}]

Out[3] := 2^n Gamma[1/2 + n]/Sqrt[Pi]

which is again correct.  Whether this is a known bug, I do not know.  My advice is to always check with numerical computations whenever possible, and to submit a bug report to Wolfram Research.

This is a strange bug.

I we make ExpandAll Before calculations the result changes!!

i hope that Mathematica 6.0 is better.

But test numerically could be impossible sometimes.

One possible solution is try to solve the problem in another way.
Don't usig the Product function, but RSolve function, and solve a recursive sequence analagous to the Product Function.

hmm... c ya


Let $ x,y,z$ be 3 non-negative reals,such that $ x\plus{}y\plus{}z\equal{}1$
Find the minimum value of
$ S\equal{}2(\sqrt{x\plus{}y}\plus{}\sqrt{y\plus{}z}\plus{}\sqrt{z\plus{}x})\minus{}(\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})$.

The answer is $ 2\minus{}\sqrt{3}$,right?

[quote="kunny"]The answer is $ 2 \minus{} \sqrt {3}$,right?[/quote]
If so, then $ 2\minus{}\sqrt2.$ :wink:


Hey, I have just completed my sophomore year at a USNews Top 25 college, and I've recently become interested in pursuing a career as a quant. Looking through the forums, I realize that such firms seem to recruit Physics and science majors. My problem: I've never taken a science course so far in college. I am a math and economics double major (CS minor), and I was wondering if I will seem unactractive to quantitative trading firms. I do have a decent amount of math ability, however. I qualified for the USAMO in my senior year of high school, and placed first in the Putnam among students in my college this year.

But does not taking sciences ruin any chance of pursuing this career?
Also: how might a land an intership over the summer next year at a big firm?

Thanks!

Not having a science background doesn't preclude you from a quant career, but you should make sure you have very good programming credentials.


any four of five circles have a common point. prove that :all five circles have a common point.

$C_1 \cap C_2 \cap C_3 \cap C_4 = A$ (1)

$C_2 \cap C_3 \cap C_4 \cap C_5 = B$ (2)

So it means that $C_3 \cap C_4 = \left\{A,B\right\}$, adn these are indeed the only points, for two circles have at most two intersections.

$C_1 \cap C_3 \cap C_4 \cap C_5 = P$. (3)
$P \in C_3$ and $P \in C_4$, so we make two cases.

Case 1: $P=B$
This means that $B \in C_2,C_3,C_4$ and $C_5$ (we know this from (2) )
But from (3) follows that $B \in C_1$
The five circles have point $B$ in common.

Case 2: $P=A$
This means that $A \in C_1,C_2,C_3$ and $C_4$ (we know this from (1) )
But from (3) follows that $A \in C_5$
The five circles have point $A$ in common.

So QED


Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i,-2+i$ and $-1-2i$. The fourth number is

A. $2+i$
B. $2-i$
C. $1-2i$
D. $-1+2i$
E. $-2-i$

[hide="Answer"]From $B$ to $A$, we go up 1 and right 3, so from $C$ to $D$, we do the same and get $2-i\Rightarrow \boxed{B}$.[/hide]

[hide]Complex plane is just similar to $(x,y)$ plane. So we will have three points as in the $(x,y)$ plane as $(1,2), (-2,1), (-1,-2)$ and the fourth point we can easily find out which is (2,-1)

$\boxed{B}$[/hide]

[hide="Third Alternate Solution"]
Using vector representation of complex numbers we see that if $z_1,  z_2,  z_3$ and $ z_4$ are 4 complex numbers forming any parralelogram in order then $z_1-z_2=z_4-z_3$ as the operations on both sides of the equality yield the same complex in length and inclination pointing from $z_2$ to $z_1$ and $z_3$ to $z_4$.

Solving we get the answer as $2-i$.

Here we use the fact that a vector and therefore a complex can shifted in the x-y plane without any change.

[/hide]

[quote=Math1331Math][hide=Alternate Solution]Consider a complex number $z$. Then the product $zi$ will rotate the complex number $90$ degrees. We now can just multiply the numbers by $i$ to find a complex number that is not one of the three vertices given. $(-2i-1)(i)=2-i$ and the answer is $\boxed{B}$[/hide][/quote]

This is a great solution, but requires one caveat.

[hide="Solution"]You have to ensure that the square is centered around the origin.  If you had $(0, 1), (1, 1), (1, 0)$ as three of your vertices, multiplying one vertex in complex form, such as $1 + i$, by $i$ to rotate it $90$ degrees would never get you the fourth vertex.

So, to fix this, note that if we label $1+2i, -2 + i, -1 - 2i$ as $A, B, C$, we quickly note that $AB^2 = BC^2 = 3^2 + 1^2 = 10$.   Any three vertices of a square form an isosceles right triangle, so these two must be the legs, and $AC$ must be the diagonal.  The midpoint of the diagonal of a square is the center, and this midpoint is $\frac{(1 + 2i) + (-1 - 2i)}{2} = 0 + 0i$, so the square is indeed centered about the origin.

From there, multiplying $(1 + 2i)$ by $i, i^2, i^3$ will give the other three vertices; you just have to find the one that doesn't match the other two given vertices.  Alternately, since we know $AC$ is a diagonal, we know that $BD$ is the other diagonal, and multiplying $B$ by $i^2 = -1$ will give the $180$ degree rotation.

There's another important question:  What if the square hadn't been centered about the origin?  Can we use this method?  Absolutely.  From any three vertices of a square, find the diagonal, which should be the side that's not equal to the other two.  Find the midpoint of the diagonal, which is the center of the square.  Then, take all three vertices and translate them so that the center of the square goes on the origin.  Find the fourth vertex by multiplying by $i$, rotating the points $90$ degrees.  Finally, untranslate this fourth vertex back to the original position.

This is actually an important problem solving strategy:  If you have a problem you don't know how to solve, change it into one that you can solve, and then change it back to answer the original question.[/hide]


How do you create graphical design on your computer? Do you need some special kind of software? Or is it already available on most computers? :o Sorry, but I'm a newbie, even though I really like the arts, and it's one of the first times I've encountered graphic art fascination. All of you guys here must be experts at this craftmanship, and help me, the apprentice! :)

You need some sort of graphics software. Some common ones are listed below:

- GIMP: GNU Image Manipulation Problem. 100% open source and the best one available. Some examples of things created with it:

[img]http://i618.photobucket.com/albums/tt263/Stigmatised_/SOTW/summer2h.jpg[/img]

[img]http://i28.tinypic.com/15cfw3t.png[/img]

[img]http://i31.tinypic.com/25zkow8.png[/img]

- Photoshop: Expensive. Essentially it's GIMP with a few features that make it a tad easier to use in some cases, with a large price tag. (I'm not the best person to ask about Photoshop because I'm biased.)

- Cinema 4D: You can create 3D stuff with it, but it's definitely not for beginners.

Some fora you should check out:

gimper.net

graphicforce.org

justgfx.com

gimptalk.com

It also depends on what you're using. Windows comes with MS Paint.

beneton movie gif:  good for animations, user friendly
MS paint is one of my favorites. sure, it has only a few features, but it really helps for rookie image editors. I made my avatar using it because it was on a sprite sheet. as for gimp, I think you shouldnt use it as a beginner. It is not at all user friendly. for like user ferocious. :-P

If you don't want to pay for software, then get [url=http://www.gimp.org/]GIMP[/url].  It is really difficult to learn how to use, but once you learn it, you won't forget it.  If you don't want to learn how to use it, you could use [url=http://www.getpaint.net/]Paint.NET[/url].

If you want the best program for Graphic Design, you should undoubtedly use [url=http://www.adobe.com/products/photoshopextended.html]Photoshop[/url].  At a $500 dollar price, this is the best program to design graphics with.

[quote="isabella2296"]You need some sort of graphics software. Some common ones are listed below:

- GIMP: GNU Image Manipulation Problem. 100% open source and the best one available. Some examples of things created with it:

[/quote]Quick note, this may seem like an opinion, but this is quite simply not true.

Photoshop is among the best, and  Gimp is no where near the best. Gimp is good if you want to change the colors of an image, but if you're talking about heavy-duty image manipulation, then you're better off with Photoshop since it has all of the proper features that produce better results.

And do not buy Photoshop Elements. I understand that it's cheap, but it's essentially Gimp. Photoshop Elements isn't even the same as the other versions of photoshop.

I paid for it... At a student price of course. The thing about Photoshop is that it has loads more features than gimp or other open source/paid applications. However, to the untrained user (like me) not too many features seem to be added. All in all, if you know how to use it, Photoshop is definitely worth teh moola.

Wow... 500$ is a lot to pay for a graphical program. It must be
[b][size=200][color=#FF0000]REALLY[/color][/size][/b]
awesome.
Can someone give examples of GIMP-created images?

[quote="PythonNut"]But who wants to pay that much for photoshop? It may be powerful, but who wants to pay that much for it? (my computer came with elements and I am happy with it for now.)[/quote]
Elements is generally for drawing(that's why it comes with most tablets) and not actually photoshopping. Photoshop allows you to manipulate images, while elements lacks the basic features needed for effective image editing.

Gimp is kind of wonky with its tools, like the curves tool, it's not too great at producing curves, and it's hard to use.

So which is better? GIMP or elements. Because it came with my computer so elements is already on it.. And my computer is not a tablet. (And I know how to use elements completely already)

Gimp is more versatile, while Elements is more polished off. It really depends on what suits you, but I would go with elements, just because it's better for drawing/effects.

Not to say that Gimp isn't good at that stuff either. I just find Gimp really irritating. The switch from Photoshop 7 to Gimp had me stuck. It was a bit difficult to transition.

You actually can do most PhotoShop stuff on Gimp.

It's through scripts.

Adobe constantly makes new scripts for PhotoShop, as they make scripts as their job. After simple tools, add scripts.

Gimp, however, has scripts that are user made. However, if you can make scripts for Gimp, go for it!

I don't know if it's called a script, but in Gimp, go to Filters or Dialogs (forgot which one) and play around with it.

I use the Linux version, BTW.

Pssh $500 is nothing compared to some programs out there. I think it costs like $10,000 to get the Wii SDK  :o

Anyway, I would suggest making do with GIMP and investing a good graphics tablet, like a Wacom Intuos. IMO, its nearly impossible to anything graphics related using only a mouse.

[quote="Octavian"]If you don't need anything better than MS Paint then you might as well download paint.net for free. It is like MS paint but better. For example, it can handle transparency.[/quote]
But it handles anti-aliasing very poorly. A lot of the other tools are poorly made. For instance, the curve tool is absolutely terrible and the chances of you actually getting it to look right are slim to nil.

I don't think buying a tablet is a solution to anything though. I've trained myself and I can produce good results with the mouse. An essential is the curve/bezier/pen tool that's included in many applications, not including ps elements which allows you to make accurate, straight curves.

Well, you can make do with the pen tool, but a tablet makes things significantly faster. Also, I know of no other way to efficiently blend colors without a tablet.

iv never done graphic design before which one should i get?

Okay so the best one is probably Photoshop... which is REALLY expensive. But there are some cheaper Photoshops, like Photoshop Elements. And Gimp is a free program, it's okay... so if you are just starting out I recommend Gimp or Pixlr.com (the advanced option). Pixlr.com is a website and it's free and easy to use. :) (By the way I use pixlr and photoshop cs2.)

Since most of y'alls are probably in high school or college or so, purchasing Photoshop might not be too bad. If you apply for the student discount, you can probably get it for a somewhat reasonable price. You can also get it packaged with some other great Adobe graphic design tools like Illustrator (for vector design) and InDesign (for what I think is a print media deal).

It looks like someone revived an 8-month old thread. Requesting mod lock


geometry
incenter
circumcircle
geometry unsolved

[quote="77ant"]For a triangle ABC, D is the incenter of triangle ABC. Point E is taken on side AC, G and F are the centers of circles that touch BE, AC and the circumcircle of triangle ABC. Prove that G, D, and F are collinear.
 :help: please~[/quote]
It is a nice problem. It was mentioned by an American mathematican, but i don't his name.
It was solved by using Lyness theorem. My solution is rather long, so tomorrow I will post it if nobody gives ones.

This is the famous theorem of Victor Thébault

Thanks so much,everyone.
I will find out "The Bault (?)". :ewpu:

The name of the mathematician is "Thebault"

See [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003) 225--229[/url].


We list all permutations of [n] (the set containing all naturals between 1 and n inclusive) in lexicographic (dictionary) order, starting with 123...n. To get from one permutation to the next, we can only swap numbers. So 1234 -> 1243 requires one swap, and they are in fact adjacent in lexicographic order.

So if we let S[n] represent the number of swaps required to list all permutatiotns of [n] in lexicographic order, find

lim[n->+inf] (S[n]/n!)

For example, for n=3

123
132 - 1 swap
213 - 2 swaps
231 - 1 swap
312 - 2 swaps
321 - 1 swap

So S[3] = 7, and s[3]/3! = 7/6.

I assume you want the fewest number of swaps.  For example, to go from 1432 to 2134, we can swap 1 and 2, and then 1 and 4.

We can get a recurrence relation for S(n).  For n :ge: 2, we have S(n) = nS(n-1) + (n-1) floor((n+1)/2).  Also S(1) = 0.  Using that recurrence formula, we can show that the limit of S(n)/n! is (e + 1/e)/2.

By the way, if you didn't care about lexical order, there is an algorithm to generate all the permutations using just one swap from one permutation to the next.  In fact, all swaps can be made adjacent swaps.

Yeah thats the same answer I saw before... and I will work on the non lexical algorithm too. Looks interesting.


In a 30-year period, what is the maximum possible number of
leap years?


Hey, i was just interested in India's IMO selection test: how well do the IMO participants do generally in the RMO and NMO to be selected (roughly, what scores do they get?)

Hello,
  You can just read the thread INMO 2006 Results.
Sathej


1. Each of the following sets of points could be the intersection of a line and a plane EXCEPT
A) the empty set  B) a single point  C) a line  D)a plane  E) the intersection of two lines.
I chose A) empty set because when a line intersects a plane, there should not be an empty set, but the answer was 4), why can't a plane be the intersection of a line and a plane?
2. Given a point P, a line L containing a point Q, and a plane E containing a point M. If P, L, and E do not intersect, then which of the following is false?
1) There is exactly one line perpendicular to L at Q. 2) If segment PM is perpendicular to plane E, then PM<PX, where X is any other point of E.   These are two choices out of five. I chose 2) to be false because X can be any point on E, then why PX must be longer than PM?  1) seems to me is a true statement, but it turns out 1) is false. Why?

Thanks.

1. We define "intersection" of two objects as something that belongs to both objects. So, the intersection of two non-parallel lines in a plane is a point, because that point belongs to the two lines.

With that definition, the intersection of a plane and a line can't be a plane because a whole plane can't belong to a line.

2. Well in fact 1) is true when we are working on a plane but in this case we are in 3D space so there are infinite perpendicular lines to $ L$ at $ Q$. And 2) is true, to see it think of Pythagoras' Theorem.


For a give-away at the mall, the color of the entry forms changes every hour on the hour, and once a color has been used, it is not used again that day. To limit the number of entries, a person may not fill out more than one form of any color. Using the table below, what is the maximum number of total entries Sarah and her three friends could have submitted during the day?

\begin{tabular}{|c|c|c|}
\hline
\textbf{Who}&\textbf{Arrival Time at}&\textbf{Departure Time}\\
&\textbf{Mall}&\textbf{from Mall}\\\hline
Sarah&8:50 a.m.&4:02 p.m.\\\hline
Davie&9:03 a.m.&3:15 p.m.\\\hline
Zeb&10:52 a.m.&4:10 p.m.\\\hline
Alishe&12:03 p.m.&4:57 p.m.\\\hline
\end{tabular}

[quote="GameBot"]For a give-away at the mall, the color of the entry forms changes every hour on the hour, and once a color has been used, it is not used again that day. To limit the number of entries, a person may not fill out more than one form of any color. Using the table below, what is the maximum number of total entries Sarah and her three friends could have submitted during the day?

$ \begin{tabular}{|c|c|c|} \hline \textbf{Who} & \textbf{Arrival Time at} & \textbf{Departure Time} \\
& \textbf{Mall} & \textbf{from Mall} \\
\hline Sarah & 8: 50 a.m. & 4: 02 p.m. \\
\hline Davie & 9: 03 a.m. & 3: 15 p.m. \\
\hline Zeb & 10: 52 a.m. & 4: 10 p.m. \\
\hline Alishe & 12: 03 p.m. & 4: 57 p.m. \\
\hline \end{tabular}$[/quote]

If I interpreted the question correctly, it's just (convert to military) $ (16 \minus{} 8 \plus{} 1) \plus{} (15 \minus{} 9 \plus{} 1) \plus{} (16 \minus{} 10 \plus{} 1) \plus{} (16 \minus{} 12 \plus{} 1) \equal{} 9 \plus{} 7 \plus{} 7 \plus{} 5 \equal{} \boxed{28}$ (this works since none of these start on the hour).

Edit: Sorry, I missed something stupid. Izzy is correct. :blush:


Oznaczmy przez $F_n$ n-ty wyraz ciagu fibonnaciego.
Pokaz ze
\[ \sum_{n = 2}^{\infty} \frac{1}{F_{n - 1} \cdot F_{n + 1}} = 1  \]


What is the unique value numerical value of each latter in the addition problem shown? (See attached Drawing)

i always did terribly on these in middle school, but i believe [hide="hint"]looking at the leftmost column is helpful[/hide]

You know a+1=e...or do you?  I'm confused now.

Try to look at the rightmost part of the given. What can you say about the digits?

Here's a HUGE hint:

[hide]The rightmost column consists of how many different letters?[/hide]

hmmm... I dont know what that implies but there's also another pattern (if you haven't recognized it yet...) each time another letter is added starting from the 100,000,000ths place... ie: A, then A + L, then A + L + R then A + L + R + E...
but you should probably listen to Farenhajt  :roll:

Try to concentrate on the rightmost part of the given.  How many rows are there? What can you observe regarding the letters that represent digits? If you don't have a clue, you may want to plug-in some values. Sorry, Im going to give hints, but I'm not just gonna post the answer. We are here to solve problems anyway... :P

I'm like Altheman, I've always stinked at these.  Is there some particular strategy someone can show us to do it?

And I noticed that same thing Apprentice, except that is goes from 7 rows to 9...

OK, I give up. Here's the beginning of the solution, and I hope it will be enough for someone to finish it.

[hide]In the rightmost column we have TEN different letters, therefore ALL the digits from zero to nine. Their sum is $0+1+\dots+9=45$, which means that $S=5$ and the carryover is $4$.

In the next column to the left we have all the letters BUT $Y$. Therefore their sum is $45-Y$, but we must take the carryover into account. Hence $45-Y+4$ ends in $5$ (the value of $S$), which means that $Y=4$, giving the sum $45$ and the new carryover $4$.

The next column contains all the letters BUT $Y, S, T$, hence the sum with the carryover is $(45-Y-S-T)+4$ and it must end in $4$ (the value of $Y$). Putting the values for $S$ and $Y$, we get $49-4-5-T=\overline{x4}\iff 40-T=\overline{x4}$. This means that $T=6$ and the sum of that column is $36$, which gives the new carryover $3$. (Note: notation $\overline{x4}$ denotes the two-digit number with the units digit $4$ and the tenths digit $x$.)

You're free to go on :)[/hide]

Ta da :D
[hide]
y=4
s=5
t=6
x=7
p=0
n=1
e=3
r=8
l=9
a=2

*sigh* too many variables.... overload....ahhhh :P
  [/hide]


calculus
integration
trigonometry
calculus computations

Evaluate $ \int \frac{x\minus{}1}{\left( \sqrt{x}\plus{}x\plus{}1\right) \sqrt{x(x\plus{}1)}}dx$

[b]I[/b] = $ \int \frac {x \minus{} 1}{\left( \sqrt {x} \plus{} x \plus{} 1\right) \sqrt {x(x \plus{} 1)}}dx$

$ x \equal{} t^2 \ \Rightarrow \ dx \equal{} 2t.dt$

$ I \equal{} 2 \int \frac{t^2\minus{}1}{(t^2\plus{}t\plus{}1) \ \sqrt{t^2\plus{}1}} \ dt$

dividing the numerator and denominator by $ t^2$ gives

$ I \equal{} 2 \int \frac{1\minus{}\frac{1}{t^2}}{(t\plus{}1\plus{}\frac{1}{t}) \ \sqrt{t\plus{}\frac{1}{t}}} \ dt$

$ t\plus{}\frac{1}{t} \equal{} u^2 \ \Rightarrow \ (1\minus{}\frac{1}{t^2}) \ dt \equal{} 2u \ du$

$ I \equal{} 4 \int \frac{u}{(u^2\plus{}1)u} \ du \\ \\ I \equal{} 4 \int \frac{1}{u^2\plus{}1} du \ \equal{} \ 4 \tan ^{\minus{}1}(u) \plus{} C$

$ \equal{} 4 \tan ^{\minus{}1} (\sqrt{t\plus{}\frac{1}{t}}) \plus{} C \equal{} 4 \tan ^{\minus{}1} (\sqrt{\sqrt{x}\plus{}\frac{1}{\sqrt{x}}}) \plus{} C$

hello, i think your solution is wrong, because $ \sqrt{x(x\plus{}1)}\equal{}\sqrt{x^2\plus{}x}\equal{}\sqrt{t^4\plus{}t^2}$
Sonnhard.

[quote="Nora.91"]

dividing the numerator and denominator by $ t^2$ gives

$ I \equal{} 2 \int \frac {1 \minus{} \frac {1}{t^2}}{(t \plus{} 1 \plus{} \frac {1}{t}) \ \sqrt {t \plus{} \frac {1}{t}}} \ dt$

[/quote]this step is wrong. :wink:

hello, no, this was not the error sorry,but when we dividing numerator and denominator by $ t^2$ i get
$ \frac{1\minus{}\frac{1}{t^2}}{(1\plus{}\frac{1}{t}\plus{}\frac{1}{t^2})\sqrt{t^2\plus{}1}}$
Sonnhard.

Though an error occurred but Nora's answer is correct :P

@ indochina

How is that answer correct. When we differentiate NORA 's answer i don get the integral .
i get that as

 $ \frac {x \minus{} 1}{\sqrt {\sqrt {x}}\left(\sqrt {x} \plus{} x \plus{} 1\right)\sqrt {x(x \plus{} 1)}}$

Oh no... I have made two consecutive errors in my most recent postings...

It should be $ \int \frac{x\minus{}1}{\sqrt{x^{3/2}\left( x\plus{}1\right) }\left( \sqrt{x}\plus{}x\plus{}1\right) }dx$

Apologies :blush:

Ya now the integral is easy  :P

Jus put $ \sqrt {x} \plus{} \frac {1}{\sqrt {x}} \equal{} t$

Thanks Indochina  :)

Yes, vijaymenon, it is.

Using the substitution you suggest, I get

$ \frac{(x\minus{}1)\,dx}{2x\sqrt{x}} \equal{} dt$ and

$ I \equal{} \int \frac{\frac{(x\minus{}1)\,dx}{2x\sqrt{x}}}{\frac{1}{2x\sqrt{x}}\sqrt{x^\frac{3}{2}(x\plus{}1)}(\sqrt{x} \plus{}x\plus{}1)}$

$ \equal{}\int \frac{2\,dt}{\sqrt{\frac{x^\frac{3}{2}(x\plus{}1)}{x^{2}}}(\frac{\sqrt{x} \plus{} x \plus{} 1}{\sqrt{x}})}$

$ \equal{}\int \frac{2\,dt}{\sqrt{\sqrt{x}(1\plus{}\frac{1}{x})}(1\plus{}\sqrt{x} \plus{}\frac{1}{\sqrt{x}})}$

$ \equal{}\int\frac{2dt}{\sqrt{t}(1\plus{}t)}$

Now let $ u\equal{}\sqrt{t}$, so $ 2du\equal{} \frac{dt}{\sqrt{t}}$

Therefore the integral becomes

$ \int \frac{4du}{1\plus{}u^{2}}$

$ \equal{}4\arctan u$

$ \equal{}4\arctan\sqrt{t}$

$ \equal{}4\arctan \sqrt{\sqrt{x} \plus{} \frac{1}{\sqrt{x}}}$

Differentiating this, I get

$ \frac{4}{1\plus{}(\sqrt{x}\plus{}\frac{1}{\sqrt{x}})}\times\frac{\frac{1}{\sqrt{x}}\minus{}\frac{1}{x\sqrt{x}}}{4\sqrt{\sqrt{x}\plus{}\frac{1}{\sqrt{x}}}}$

$ \equal{} \frac{\frac{x\minus{}1}{x\sqrt{x}}}{\frac{1}{\sqrt{x}}(\sqrt{x}\plus{}x\plus{}1)\sqrt{\sqrt{x}\plus{} \frac{1}{\sqrt{x}}}}$

$ \equal{}\frac{x\minus{}1}{(\sqrt{x}\plus{}x\plus{}1)\sqrt{x^{2}(\sqrt{x}\plus{} \frac{1}{\sqrt{x}})}}$

$ \equal{}\frac{x\minus{}1}{(\sqrt{x}\plus{}x\plus{}1)\sqrt{x^\frac{5}{2} \plus{} x^\frac{3}{2}}}$

$ \equal{}\frac{x\minus{}1}{(\sqrt{x}\plus{}x\plus{}1)\sqrt{x^\frac{3}{2}(x\plus{}1)}}$

[quote="Nora.91"]sorry about my mistake  :oops:[/quote]

It's ok, Nora. This is an interesting occurrence. My wrongly posted problem + your calculation error gave the right result!! :P

Who says two wrongs don't make a right?


just a quick yes or no answer, can we use microsoft word? I'm assuming we can, but on the website, it seems to say either handwrite or latex.

I know it'll be in pdf, etc.

Also, if we do use word, we have to have the right format and everything, right? (like the format for the handwritten solutions)

For format, you'll probably want each solution to start on a new page. And make the whole thing readable. Otherwise, it probably shouldn't matter.

And as you said, it has to be in .pdf.

[quote="Differ"]I've been using Word for the past year with latex, and let me tell you that the formatting is absolutely horrible. I pity the graders who must attempt to decipher my poorly rendered documents.

On the other hand, I haven't gotten marked off for it.

If anyone can recommend a free latex creator/renderer/whatever, I would be grateful.[/quote]

[url=http://www.lyx.org/]Lyx[/url]?
I've never tried it out myself, but I have heard things about it.
Using LaTeX by itself is not altogether too hard. The USAMTS files take care of almost all of the formatting; the only the left to you is to import graphics into your LaTeX as needed.

http://www.lyx.org/

It has some kinks, but it's the best click'n'play editor for TeX on the net. (It can make problems with template files, though, if they're too advanced)

http://www.texniccenter.org/

Almost no kinks at all, perhaps not as user friendly as Lyx, but most certainly got more quality in it.


search
projective geometry
geometry solved
geometry

[hide]use Pascal's theorem http://en.wikipedia.org/wiki/Pascal_theorem[/hide]

[Moderator edit: The problem also is part of (2) from
http://www.mathlinks.ro/Forum/viewtopic.php?t=77663 and a part (the concurrence of the lines AC, PQ, RS) of http://www.mathlinks.ro/Forum/viewtopic.php?t=104155 .]

Another interesting ( well known ) collinearity about this configuration, is as follows:

The intersection points of the opposite sidelines of the quadrilaterals $ABCD,$ $PQRS,$ are collinear.

I don't know if this problem has already been posted in the Forum. There is a reference about it (without proof), in the problem ''Maclaurin theorem'', presented by [b][size=109]''hucht'',[/size][/b] last August.

http://www.mathlinks.ro/Forum/viewtopic.php?t=75425

Kostas Vittas.

Sorry dear dondigo. I didn't saw that this is a problem in unsolved area.

I search my archive and I will post here later, the proofs I have in mind.

Kostas Vittas.

This problem is nice. I will use the [b]Pole-Polar theorem[/b] to kill it.
Let $ X$ be the intersection of $ AD$ and $ BC$, $ Y$ be the intersection of $ AC$, $ BD$. It is well- known that $ XY$ is the polar of $ E$ wrt (O). Now, it is easy to see that $ AD,BC$, respectively, is the polar of $ S,Q$. Obviously, $ AD,BD,XY$ concurent, then, $ S,Q,E$ is collinear.
Our proof is completed. :D


search
induction
function
geometry
3D geometry
algebra
polynomial

find all f N--->N
for all x;y
f(x²+y²)=f²(x)+f²(y)

[quote="erdos"]Well you said "For all $x,y$" but for all $x,y$ belonging to which set ??  :?[/quote]
I believe it is $x,y \in N$.
BTW, is $0 \in N$?

I only get:
[hide]$f(x)=x$[/hide]
is that correct?? i am sure this is one of the answers but am i missing some really important things??

If the set is the set of positive integers, the problem has infinitely many solutions.
indeed the problem would be very simple and to give you the ideas i give a construction for one solution:
take f(1)=a1. Now knowing f(1) we can calculate f(1),f(2),f(5),f(29),...
the smallest number not belonging to this sequence is 3.
define f(3)=a2. So Now we know f(x) for all x in A1={1,2,3,5,10,13,18,26,...}
define B1={1,4,9,25,100,169,...}
Now we search for the smallest number x not in A1 and such that (B1-x^2) :cap: A1= 0
here it is 4. then we define f(4)=a4. we form the corresponding A2 and B2 and ...

-Ali

sorry, Forget it  :blush:

There is one step in my solution that i need to work on it. That step was missed above.

ali, i don't understand you solution and i don't see where is wrong......i have little knowledge about sequence and set, perhaps you can introduce something??

And i shall post my solution:
By induction, we have: $f(nx^2)=nf^2(x)$
Let $x=1$, we obtain $f(n)=nf^2(1)$. Then let $n=1$, we have: $f(1)=1$.
So $f(x)=x$.

Hope it's correct.

[quote]
shobber wrote:
By induction, we have:  f(nx^2)=nf(x)^2

[/quote]

I don't see how you get this. indeed f(2x^2)= 2f(x)^2 which is correct.
but how do you prove f(3x^2)=3f(x)^2???
in fact 3x^2 can't be written as sum of two squares and so nothing can be said about
its f-value.
Anyway I will expand my solution soon.
-A

oh,yes, that's my HUGE mistake :blush:  :P ...
But at least, $f(x)=x$ is one of the answers... :lol:

This really seems to be a tough problem.
After a couple of hours, the only result that i got is this: (which may be the key)

if f(1)=b then
32b^6 - 16b^4 -2b^2 +2 is a perfect square.
and also 32b^6 +16b^4 +2b^2 -1 is a perfect square.
I couldn't show that this implies that b=1, but if someone can do that
then I'm pretty sure i can finish the problem.
-Ali

Take f(0)=a and f(1)=b
Then: f(1)=f(1)2+a2  --->b2-b +a2=0 wich has a solution only if a=0 (Delta=1-4a2) then For all x : f(x2)=f(x)2 (*)
So b2=b
Case 1: b=0 then All x : f(x2+1)= f(x)2 .Try, using (*) to show that means  f(x)=0 for all x in N
Case 2: b=1 then All x : f(x2+1)= f(x)2+1. Using (*) Show that means f(x)=x for all x in N

Just Ideas! Am i Wrong?

x=0 is not allowed. The set of positive integers

Anyway I further modified my proof to show that f(x)=bx for all x
where for every x,  b is a number such that

  32b^6 - 16b^4 -2b^2 +2 is a perfect square.
and also 32b^6 +16b^4 +2b^2 -1 is a perfect square.

I'm pretty sure that the only choice of b is 1 in which case problem is
solved.
-Ali

OK! x=0 is not allowed as ALI said.

[u]Another idea:[/u]

Put S1= {x in N such that f(x)=x}
Suppose S1 is different of N. S=N\S1.
If s is in S then for all x in S1, s2+x 2 is in S.

Analyse:
a-S1 Finite : elementary contradition.
b-S1 Infinite, show that it is not possible to find a bijection from S1 to S
U'll find a contradition. N is Countable!!

what if S1 is empy?
say there is no x with f(x)=x
i don't see how this gives you a contradiction.
please be more detailed in your ideas.

The idea is :By induction ,we prove that : with an integer n >10 ,we can write n^2 as  a^2+b^2-c^2 with a,b,c<n.It follow (f(n))^2=(f(a))^2+(f(b))^2-(f(c))^2
                  so by induction we have   f(n)^2 =a^2*f(1)^2+b^2*f(1)^2-c^2*f(1)^2=n^2*f(1)^2->f(n)=nf(1)->f(n)=n

but $ naluvhunter $
How did you prove that:
\[f(x^2-y^2)=f(x)^2-f(y)^2\] :?

Source: Concours Général des lycées 1994.

Some hints:
If $f(1)>0$:

1.Calculate $f(i)$ for $1\leq i \leq 12$
2. Try to find a cool equality with two sums of two squares so that you could use induction !

When you finish this, find all functions

$f: \mathbb{Z}  \rightarrow \mathbb{Z}$ such that $\forall (x,y,z) \in \mathbb{Z}^3, f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f (z)^3$

The prove uses the same idea as the previous problem  :)

That right! Your problem is VietNam second round 2005!

I have miss that all $a_i$ < $k$
And i have just find this:
\[(2k+1)^3=(2k-1)^3+(k+4)^3+(4-k)^3+(-5)^3+(-1)^3\]
And now the proof is complete :D

It was also used in the french correspondance program in 2003-2004 !

The problem with the cubes already appeared many times here.
I think it first appeared in the A.M.M., proposed by Titu Andreescu, but I do not remember the year. But clearly before 2004.

Pierre.

whats the primitif solutions to
$x^2$-$y^2$=$2x-1$

oh!!
sorry
whats the primitif solutions to
$x^2$-$y^2$=2$x$-1+$z^2$


geometry
rectangle
algorithm
analytic geometry

http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=3548

Any hints on how to solve this ??

I'm not sure, but what if you started by including the tree with the least area to include another tree, then take the next one with least average area weighted by area required to include the first tree, etc.? Would that be a good approach?

[Sorry, I'm more clueless than you are, and I'd like an expert to enlighten me.]

Not completely sure if this runs in time, but it should be reasonably successful (I hope?). First store an array tree[x][y] of the number of trees in the rectangle from $ (1, 1)$ to $ (x, y)$. This can be calculated by simple DP in $ O(nm)$ time (where $ n$ and $ m$ are the side lengths of the garden). Then we can calculate the number of trees in the rectangle bounded by $ (a, b)$ and $ (c, d)$ with $ c > a$ and $ d > b$ by tree[c][d]-tree[a][d]-tree[c][b]+tree[a][b]. This is a classic technique, something like PIE (it's easy to see by drawing it out).

Now I had an idea from here, but I realize that the area less than or equal to bound makes it too slow. I'll keep thinking. I suspect that there is an $ O(F^{2})$ algorithm, but I can't quite figure it out.

I got it accepted.

The statement F=1000 is misleading in this problem.
After doing the preprocessing i tried all possible rectangles which can be done in
O(F^3). Total complexity is O(F^3 + n^2) where n is maximum x coordinate (in this case it is 1000).

Do update me if there is a better solution.


If $ (a,b)$ is the only solution to the equation

$ x\plus{}\frac{9}{x}\plus{}y \plus{}\frac{25}{y}\equal{}4$

find $ 4a \plus{} 7b$

thanks in advance.

Yet another question that both of you've posted :P  (you and athunder).
2nd one in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243689]this thread.[/url]


linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

What is the group generated by the matrices $ \bmatrix{0 & -1 \\ 1 & 0}$ and $ \bmatrix{1 & -1 \\ 1 0}$?

Just a note, my group theory is very limited, so I probably wouldn't be able to understand many solutions. I just saw this problem show up in one of my approaches on the PUMaC Power Test.

Thanks.

[quote="Zhero"]What is the group generated by the matrices $ \left[ \begin{array}{cc} 0 & \minus{} 1 \\
1 & 0 \end{array} \right]$ and $ \left[ \begin{array}{cc} 1 & \minus{} 1 \\
1 & 0 \end{array} \right]$? [/quote]
Don't take my word for it, but I'm reasonably sure these matrices generate all of $ SL_2(\mathbb{Z})$, i.e. every integer matrix with determinant $ 1$.  Either way, some related results are discussed [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].

Edit:  If the matrices you gave are $ A$ and $ B$, then $ A^3 B$ and $ BA^3$ generate $ SL_2(\mathbb{Z})$ by "take 2" of the post I linked to.


Express $ \frac{0.\overline{666}}{1.\overline{333}}$ as a common fraction.

This is just $ \frac{\frac{2}{3}}{\frac{4}{3}}\equal{}\frac{2}{4}\equal{}\boxed{\frac{1}{2}}$.


A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is:
$ \textbf{(A)}\ \frac{1}{4} \text{ the area of the original square}\\
\textbf{(B)}\ \frac{1}{2} \text{ the area of the original square}\\
\textbf{(C)}\ \frac{1}{2} \text{ the area of the circular piece}\\
\textbf{(D)}\ \frac{1}{4} \text{ the area of the circular piece}\\
\textbf{(E)}\ \text{none of these}$

[hide][geogebra]047aa7c024d138819f9fc74737acb8f301623e94[/geogebra]
See geogebra.
$ \boxed{\text{B}}$
Or you can see that the length of the large square is the same as the length of diagonal of the small square, then use the rectangle area formula for the large square and the rhombus area formula for the small square to get 1/2.[/hide]


[i]           This is my frist proposed prolem
   I think it's not easy but very beautiful[/i] :blush:

 [i] Let a,b,c,d be positive real numbers such that  [/i]$ a\plus{}b\plus{}c\plus{}d \equal{} 3$
 [i]Find the minimum value of  this expression[/i]
[i] A[/i] = $ \sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{b\plus{}c\plus{}d} \plus{}\sqrt{c\plus{}d\plus{}a} \plus{}\sqrt{d\plus{}a\plus{}b}$

        [i] Love An forever[/i]

[quote="nkht-tk14"]
 [i] Let a,b,c,d be positive real numbers such that  [/i]$ a \plus{} b \plus{} c \plus{} d \equal{} 3$
 [i]Find the minimum value of  this expression[/i]
[i] A[/i] = $ \sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {b \plus{} c \plus{} d} \plus{} \sqrt {c \plus{} d \plus{} a} \plus{} \sqrt {d \plus{} a \plus{} b}$
   [/quote]
Thank you, nkht-tk14 for your gift! :)
$ \left(\sum_{cyc}\sqrt {a \plus{} b \plus{} c}\right)^2 \equal{} \sum_{cyc}(a \plus{} b \plus{} c) \plus{} \frac {1}{2}\sum_{sym}\sqrt {(a \plus{} b \plus{} c)(a \plus{} b \plus{} d)}\geq$
$ \geq3(a \plus{} b \plus{} c \plus{} d) \plus{} \frac {1}{2}\sum_{sym}(a \plus{} b \plus{} \sqrt {cd}) \equal{} 27 \plus{} \frac {1}{2}\sum_{sym}\sqrt {ab}\geq27.$
Thus, $ A\geq3\sqrt3.$ But for $ a \equal{} 3$ and $ b \equal{} c \equal{} d \equal{} 0$ the equality holds.
Id est, $ \inf A \equal{} 3\sqrt3$ and the minimum of $ A$ is not exists.

note that $ \sqrt{3\minus{}x}\plus{}\sqrt{3\minus{}y}\geq \sqrt{3\minus{}(x\plus{}y)}\plus{}\sqrt{3}$, for all positive reals $ x,y$ with $ x\plus{}y\leq 3$. then, we have that

$ \sum_{cyc} \sqrt{a\plus{}b\plus{}c}\equal{}\sum_{cyc} \sqrt{3\minus{}a}\geq 3\sqrt 3\plus{}\sqrt{3\minus{}(a\plus{}b\plus{}c\plus{}d)}\equal{}3\sqrt 3$

and we're done.

Sorry for my mistake
  It should be " nonegative real numbers "
     Sorry :blush:
 I have just corrected


Find right-angled triangles whose sides are in arithmetic progression.

Here's a proof that this is the only ratio of sides there can be.

\[ x^2\plus{}(x\plus{}y)^2\equal{}(x\plus{}2y)^2\]\[ 2x^2\plus{}2xy\plus{}y^2\equal{}x^2\plus{}4xy\plus{}4y^2\]\[ 4y^2\equal{}x^2\minus{}2xy\plus{}y^2\]\[ (2y)^2\equal{}(x\minus{}y)^2\]\[ x\equal{}3y\]

[quote="varun.tinkle"]my solution is suprisingly different

let the sides be a-d ,a,a+d    the are in ap since

so solving it

(a+d)^2=a^2+(a-d)^2
solving it we get
a=2d
so  we know the siders  as  d ,2d,3d[/quote]

$ (a\plus{}d)^{2}\equal{}a^{2}\plus{}(a\minus{}d)^{2}$
$ a^{2}\plus{}2ad\plus{}d^{2}\equal{}a^{2}\plus{}a^{2}\minus{}2ad\plus{}d^{2}$
$ a^{2}\minus{}4ad\equal{}0$
$ a(a\minus{}4d)\equal{}0$
$ a\equal{}4d$

So, the sides are $ 3d,4d,5d$

Primitive triples are of the form $ r^2\plus{}s^2$, $ 2rs$, and $ r^2\minus{}s^2$.
If they're in arithmetic progression, $ 2rs\minus{}r^2\minus{}s^2\equal{}r^2\minus{}s^2\minus{}2rs$ (because the first finite difference of an arithmetic sequence is constant).

Then $ 4rs\equal{}2r^2\implies 2s\equal{}r$, which makes the triple $ 5s^2$, $ 4s^2$, and $ 3s^2$ by substituting in. The primitive triple, therefore, would be $ (3,4,5)$, and all others are a multiple of this.


geometric transformation
vector
ceiling function
analytic geometry

Let $n$ be an integer greater than 2, and $P_1, P_2, \cdots , P_n$ distinct points in the plane. Let $\mathcal S$ denote the union of all segments $P_1P_2, P_2P_3, \dots , P_{n-1}P_{n}$. Determine if it is always possible to find points $A$ and $B$ in $\mathcal S$ such that $P_1P_n \parallel AB$ (segment $AB$ can lie on line $P_1P_n$) and $P_1P_n = kAB$, where (1) $k = 2.5$; (2) $k = 3$.

The case $k=2.5$ is trivial. There is counterexample!

Let $k=3$.
Without loss of generality let $P_1 \equiv (-3,0)$ and $P_n \equiv (3,0)$.

Let $M$ be $S$ under translation by the vector $-\frac{1}{3} . \vec{P_1P_n}$.
Let $N$ be $S$ under translation by the vector $\frac{1}{3} . \vec{P_1P_n}$.

Suppose that we can't find $A$ and $B$ such that $k=3$ and $AB||P_1P_n$.
This means that $M\cap S=\emptyset$ and $N\cap S=\emptyset$

Now we have three broken lines $M,S$ and $N$.
Let's continue $S$ in both directions so that we divide the plane into two parts $\mu$ and $\nu$, and M is in the part $\mu$ and N is in the part $\nu$.

Lets look at the $Ox$ axis. Each point of it is either in $\mu$ or $\nu$ or in the border $S$. The point $-3$ is on the border, so $-1$ is in $\nu$. $3$ is on the border, so $1$ is in $\mu$. So there exist $-1\leq  B_1 \leq  1$ so that $B_1$ is on the border. So $-3\leq B_1 -2\leq -1$ and $M_1=B_1 -2$ is in $\mu$. So there exist $M_1\leq  B_2 \leq  N$ with $B_2$ on the border. Then $N_2= B_2 +2$ is between $B_1$ and $1$. There exist $B_3$ between $N_2$ and $1$ on the border and we build infinite sequence $B_{2k+1}$ of points on the border and between every two there is point not on the border.

That means that the broken line  S intersects the $Ox$ axis infinite number of times- \textbf{CONTRADICTION}

[quote="imortal"]The case $ k \equal{} 2.5$ is trivial. There is counterexample!

[/quote]

How?

If $k\ge 0$, this is always possible iff $k$ is a nonnegative integer. If $k\le1$, this is all obvious; from now on, suppose that $k>1$.

First suppose $k\notin\mathbb{Z}$, and let $m=\lceil{k}\rceil\ge2$ so that $m-1<k<m$. For $1\le i\le m-1$, we take
\begin{align*}
&P_{8i-7}=(m(i-1),0), P_{8i-6}=(m(i-1),m-i), P_{8i-5}=((m-1)i,m-i), P_{8i-4}=((m-1)i,0), \\
&P_{8i-3}=((m-1)i,0), P_{8i-2}=((m-1)i,-i), P_{8i-1}=(mi,-i), P_{8i}=(mi,0),
\end{align*}where the fact that $m(i-1)<(m-1)i<mi$ ensures that $S$ is a simple curve (although the problem does not stipulate this, it's convenient for the construction). It's easy to see that $P_1=(0,0)$, $P_n=(m(m-1),0)$, and the resulting figure $S$ is symmetric about $(m(m-1)/2,0)$.

Hence if we go by contradiction (i.e. assume we can find $A,B$ such that $AB\in(m-1,m)$), we can WLOG work with $A,B$ on or above the $x$-axis. Note that the region of $S$ above the $x$-axis is a sequence of $m-1$ disjoint squares (bottom sides missing) with decreasing height (from left to right), so we can assume that $A$ lies on the $i^{th}$ square and $B$ lies on the $j^{th}$, where $1\le i<j\le k-1$. Since $A,B$ have the same $y$-coordinate, the $x$-coordinate of $A$ is either $m(i-1)$ or $(m-1)i$, and the $x$-coordinate of $B$ lies in the closed interval $[m(j-1),(m-1)j]$. If $A_x=m(i-1)$, then $AB\ge m(j-1)-m(i-1)\ge m$, contradiction, so
\[A_x=(m-1)i\implies AB\in[m(j-1)-(m-1)i,(m-1)(j-i)].\]But if $j=i+1$, then $AB\le m-1$, and if $j\ge i+2$, then $m(j-1)-(m-1)i\ge m(i+1)-(m-1)i=m+i\ge m$, contradiction.
------------------------------------
It remains to show that when $k\ge2$ is an integer, there exist $A,B\in S$ such that $P_1P_n=kAB$. Suppose otherwise; WLOG set $P_1=(0,0)$ and $P_n=(k,0)$. For any curve $C$ (possibly a point) and real $r$, let $C^{(r)}=C+(r,0)$. By assumption, $C^{(r)}$ and $C^{(r+1)}$ are disjoint for all $r$. The key idea is to define "left/right" properly: take indices $M$ and $m$ such that the $y$-coordinates of $P_M$ and $P_m$ are maximal and minimal, respectively, and $|M-m|$ is as small as possible; WLOG $m<M$, and $J$ is the region bounded by the two horizontal lines through $P_M,P_m$ (inclusive); note that $y=0$ lies in $J$. Let $T$ denote the broken line $P_mP_{m+1}\ldots P_M$, the exterior $E$ be the set of points $X\ne P_M\in J$ such that there exists a continuous path from $P_M$ to $X$ intersecting $T$ only at the beginning (at $P_M$), and the interior $I=J\setminus E$. Define the left $L$ of $T$ as the set of points $X\in J$ such that there exists a continuous path from $(-\infty,0)$ to $X$ that stays in $J$ and never meets $T$; define the right $R$ analogously (note that the points in $I$ are in neither $L$ nor $R$). Clearly $E=L\cup R$.

If $S^{(1)}$ intersects both $I$ and $E$, then $S^{(1)}$ clearly hits $T$, contradiction, and if $S^{(1)}$ lies entirely in $I$, then the uppermost $y$-coordinate in $S^{(1)}$ is strictly less than that of $I$ (since $S^{(1)}$ does not intersect $T$), contradicting the fact that $P_M^{(1)}\in S^{(1)}$. Hence $S^{(1)}$ lies entirely in either $L$ or $R$ (since $S^{(1)}$ never leaves $J$ and never hits $T$). But it's not hard to show that $P_M^{(1)}\in R$, so $S^{(1)}$ lies entirely in $R$. In particular, $I^{(1)}\subset R$, so $R^{(1)}\subset R$ by the continuity of $I^{(1)}$ and $R$ (in other words, a path staying in $J$ from $(\infty,0)$ to some $X\in R^{(1)}$ cannot meet $I$ without first hitting $I^{(1)}$, since $I$ and $I^{(1)}$ are disjoint). Hence $S^{(2)}\subset R^{(1)}\subset R$; by a simple induction, $S^{(i)}$ lies entirely in $R$ for $i\ge1$, and by symmetry, $S^{(-i)}$ lies entirely in $L$ for $i\ge1$. Thus $P_1^{(k-1)}\in R$ (recall that $k\ge2$) and $P_n^{(-1)}\in L$, contradicting the fact that $P_1^{(k-1)}=P_n^{(-1)}$ and $L\cap R=\emptyset$.


Let $ a,b,c\in[0;1]$.Prove the following inequality:

$ \frac{1}{5\minus{}ab}\plus{}\frac{1}{5\minus{}bc}\plus{}\frac{1}{5\minus{}ac}\ge \frac{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{4}$

I proposed it in my national forum,but i don't receive any solutions.

From $ \sqrt a \plus{}\sqrt b \le 1\plus{}\sqrt {ab}$, we get $ \frac 1{5\minus{}ab} \ge \frac {\sqrt a\plus{}\sqrt b}{8}$. Done. :)

[quote="Maxim Bogdan"]Let $ a,b,c\in[0;1]$.Prove the following inequality:

$ \frac {1}{5 \minus{} ab} \plus{} \frac {1}{5 \minus{} bc} \plus{} \frac {1}{5 \minus{} ac}\ge \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{4}$

I proposed it in my national forum,but i don't receive any solutions.[/quote]
The general one
\[ \sum^{n}_{i \equal{} 1} \frac {1}{2k \plus{} 1 \minus{} x_i x_{i \plus{} 1}} \ge \frac {\sum^{n}_{i \equal{} 1} \sqrt [k]{x_i}}{2k}
\]
holds for all $ x_i \in [0,1]$ and $ k \ge \frac {1}{2}$.
:)
And there is a nice solution for it. :)

I'm very curious to see a proof for that inequality.Can you post your solution,please? :)

Applying the same way posted by Vasc the inequality could also be proved.

We need to show

$ \frac {1}{2k \plus{} 1 \minus{} xy}\ge \frac {\sqrt [k]{x} \plus{} \sqrt [k]{y}}{4k}$

Applying $ \sqrt [k]{x} \plus{} \sqrt [k]{y}\le 1 \plus{} \sqrt [k]{xy}$ it suffices to show

$ (1 \plus{} t)(2k \plus{} 1 \minus{} t^k)\le 4k$

where $ t \equal{} \sqrt [k]{xy}\ge 0$. The inequality above is equivalent to

$ t^{k \plus{} 1} \plus{} t^k \plus{} (2k \minus{} 1)\ge (2k \plus{} 1)t$

which is valid (particularly, for integer $ k$ just applying AM-GM).

I wonder whether or not Can_hang2007 got another proof different from Vasc's  :)

[quote="shalex"]Applying the same way posted by Vasc the inequality could also be proved.

We need to show

$ \frac {1}{2k \plus{} 1 \minus{} xy}\ge \frac {\sqrt [k]{x} \plus{} \sqrt [k]{y}}{4k}$

Applying $ \sqrt [k]{x} \plus{} \sqrt [k]{y}\le 1 \plus{} \sqrt [k]{xy}$ it suffices to show

$ (1 \plus{} t)(2k \plus{} 1 \minus{} t^k)\le 4k$

where $ t \equal{} \sqrt [k]{xy}\ge 0$. The inequality above is equivalent to

$ t^{k \plus{} 1} \plus{} t^k \plus{} (2k \minus{} 1)\ge (2k \plus{} 1)t$

which is valid applying AM-GM.

I wonder whether or not Can_hang2007 got another proof different from Vasc's  :)[/quote]
Yeah, my proof is different :)
I used the following inequality for all $ t \in [0,1]$
\[ \frac{2k}{2k\plus{}1\minus{}t^k} \ge \frac{1}{2}(t\plus{}1)\]
:)

[quote="can_hang2007"]
Yeah, my proof is different :)
I used the following inequality for all $ t \in [0,1]$
\[ \frac {2k}{2k \plus{} 1 \minus{} t^k} \ge \frac {1}{2}(t \plus{} 1)
\]
:)[/quote]

But yours is just the same as

$ (1\plus{}t)(2k\plus{}1\minus{}t^k)\le 4k$

in my post  :roll:

[quote="shalex"][quote="can_hang2007"]
Yeah, my proof is different :)
I used the following inequality for all $ t \in [0,1]$
\[ \frac {2k}{2k \plus{} 1 \minus{} t^k} \ge \frac {1}{2}(t \plus{} 1)
\]
:)[/quote]

But yours is just the same as

$ (1 \plus{} t)(2k \plus{} 1 \minus{} t^k)\le 4k$

in my post  :roll:[/quote]
Oh, I see, sorry.
But I found it indepently ;)


is the following statement true ?

for every integer $ n>0$ there is a prime between $ 10^n$ and $ 10^n\plus{}10^{n\minus{}1}\plus{}1$

A generalization of Bertrand's postulate states that

[color=blue]For any $ c > 0$ there exists positive integer $ N_c$ such that for any integer $ n\geq N_c$
there exists at least a prime between $ n$ and $ (1 \plus{} c)n$.[/color]

The problem is that that $ N_c$ may be quite large ...

For your case, since $ 10^n \plus{} 10^{n \minus{} 1} \plus{} 1 > (1 \plus{} 1/10)10^n$, we can take $ c \equal{} 1/10$.

Also see [url]http://en.wikipedia.org/wiki/Bertrand%27s_postulate[/url]

EDIT. In reply to the post following: NO (just rough asymptotic value).

thanks

is there any method to effectively find this number $ N_c$ for fixed $ c$ ?


Basically, I am working on a computer assignment that consists of 4 tough questions. Thus, I really need to have backup of my program. Here is what I do:

the filename "inprogress" is where I save my current progress. Whether I am going to take a break, or just want to save, I will save my current progress there.

the filename "q1" is where I save when I am working on q1. So when I am taking a break after working on q1, I will save my current progress in "inprogress" and "q1." Sometimes a question might be too long for me to just save it in "q1," in that case, I will save it in "q1v1", or "q1v2", etc.

the filename "BU" is where I save when I am done with the program for the day. e.g. after a whole day of working on the assignment which I end at question 2, I will save my current program in "inprogress," "q2", and "BU."

I am not sure if this scheme is good or not. Some of you might have a better idea of doing it (such as using version #). So I would like to hear your ideas :roll:

Use SVN? Host it on an FTP server somewhere?

thx for mentioning SVN. Once I enter SVN in the wikipedia, and it brings out a link to "revision control." I think this might need to what I'm trying to look for :)


Find the natural number $ n$ for which $ 5(\log_3 n)^2\minus{}\log_3 n^{12}\plus{}9$ is minimized.

[quote="kunny"]Find the natural number $ n$ for which $ 5(\log_3 n)^2 \minus{} \log_3 n^{12} \plus{} 9$ is minimized.[/quote]

First rewrite the formula to $ 5x^2\minus{}12x\plus{}9$ with $ x \equal{} \log_3 n.$ It is well-known that $ ax^2\plus{}bx\plus{}c$ has an extremum at $ x\equal{}\minus{}\frac{b}{2a}$ which gives $ n\equal{}3^\frac{6}{5}\approx 3,7.$

This number isn't natural though, but as $ 5x^2\minus{}12x\plus{}9$ is convex its minimum must be at 3 or 4. For $ n\equal{}3$ the function has the value 2 and for $ n\equal{}4$ it has the value $ 1,81\ldots$ so the minimum occurs at $ n\equal{}4.$

The point of the problem is to determine which is bigger, $ \log_3 4$ or $ \frac{6}{5}$ without using calculator. :D

$ \log_3{4} > \frac{6}{5} \Leftrightarrow 4^5 > 3^6 \Leftrightarrow 1024 > 729$


calculus
integration
function
abstract algebra
limit
real analysis
real analysis solved

For a function $f(x)$ on $[a, \infty )$, we define:

$S(x) = \int_{a}^{x} f(s)ds$

and

$\sigma(x) = \frac{\int_{a}^{x} S(t)dt}{(x-a)}$

What happens to $\sigma(x)$ as $x \rightarrow \infty$ ?

I'm just trying to find out if there are any special properties of this process (like if this Cesaro Integral can consider a larger class of integrals). It appears that the only interesting results are if $\int f(x)dx$ does not exists. For example:

$\int_{a}^{\infty} sin(x)dx = DNE$

Cesaro Integral:

$S(x) = \cos(a) - \cos(x)$

$\sigma(x) = \frac{\int_{a}^{x} S(t)dt}{(x-a)} = \cos(a) - \frac{\sin(x) - \sin(a)}{(x-a)}$

As $x \rightarrow \infty$, $\sigma(x) = cos(a)$

Any thoughts? Anyone have a better example?

Of course this will be very much analogous to the same things for sums. Let me pose some problems for you.

The first thing you want to do is establish the [b]Abelian Theorem[/b]:

If $\lim_{x\to\infty}S(x)=L$, then $\lim_{x\to\infty}\sigma(x)=L.$

Once you have that, you'd like to see if you can prove one or more [b]Tauberian Theorems[/b]:

If $\lim_{x\to\infty}\sigma(x)=L$ and (condition to be named later), then $\lim_{x\to\infty}S(x)=L.$

Things to try for the condition to be named later:

1. $f(x)=o\left(\frac1x\right)$ as $x\to\infty.$

2. $f(x)=O\left(\frac1x\right)$ as $x\to\infty.$

Assume that integration by parts will be the crucial technique.

Your example is fine for demonstrating that this method can sometimes give a definite number for a divergent integral. The equivalent example for Cesaro summability of series would be the sum of $(-1)^n.$


geometry
inradius
area of a triangle
Heron\u0027s formula

Find sides of triangle whose inradius is 1!  :roll:

Are there no other requirements?

I believe a $ 3\minus{}4\minus{}5$ triangle works quite well.

From the formula, $ A \equal{} rs$, where r is the inradius and s is the semiperimeter.

Actually, this was just a cool fact to know, so I learned it.  If you wanted a right triangle and know the inradius, then $ \frac {1}{2}ab \equal{} \frac {a \plus{} b \plus{} \sqrt {a^2 \plus{} b^2}}{2}\rightarrow \minus{} \sqrt {a^2 \plus{} b^2} \equal{} a \plus{} b \minus{} ab\rightarrow 0 \equal{} ab(ab \plus{} 2 \minus{} 2a \minus{} 2b)\rightarrow ab \minus{} 2a \minus{} 2b \equal{} \minus{} 2\rightarrow (a \minus{} 2)(b \minus{} 2) \equal{} 2$.WLOG, and assuming the answers are integers, then $ a \equal{} 4, b \equal{} 3$.

And I don't know why the post has a bunch of space.

Another way to show that a 3-4-5 triangle works:

Let $ r$ be the inradius.  Then $ 3 \equal{} r \plus{} x$, $ 4 \equal{} r \plus{} y$, $ 5 \equal{} x \plus{} y$ for some $ x$ and $ y$.  The solution is $ r \equal{} 1$, $ x \equal{} 2$ and $ y \equal{} 3$

Of course, there are an infinite number of triangles whihc have an inradius of $ 1$.  The restriction is that the area of the triangle is half the perimeter.  Thus $ s\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ from heron's formula, where $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$ and $ a,b,c$ are the side lengths.

Alternately, we can construct any such triangle by drawing a unit circle, adding three tangents to the circle and using the points of intersection of these tangents as the vertices; the only restriction is that such a triangle must inclose the original circle.

A = absenC/2, right! How do you know that C = 90º?

Gabrielz, the triangle is not necessarily right - it is just one of the solutions.

Another example: consider the equilateral triangle with inradius $ 1$.  Then the area can be written in two ways: $ A\equal{} 3s/2 \equal{} s^2 \sqrt{3}/4$ where $ s$ is the side length.  Then we have $ \sqrt{3}s^2 \minus{} 6s \equal{} 0$ so $ (\sqrt{3}s\minus{}6)s\equal{}0 \implies s\equal{}6/\sqrt{3} \equal{} 2\sqrt{3}$


And thus people can edit other's posts, make them say other than what they want, change rules, etc... I don't think that's a good idea at all.

[quote="cbk"][quote="AMC37.5"]I propose that the AOPS Wiki forum be deleted and archived. Wiki-related discussion should go on Wiki itself, where it is immediately accessible to editors.[/quote]

What are you saying?  :o  :o[/quote]

i think he is saying that AoPS Wiki forum is useless and he doesnt want it anymore. The discussion should just go on wiki or something and accessed  by only editors??

Bad translation lol

[quote="AMC37.5"]I propose that the AOPS Wiki forum be deleted and archived. Wiki-related discussion should go on Wiki itself, where it is immediately accessible to editors.[/quote]

*see what solafidefarms said*

editors can just come here and post here. If the admins made this forum, it was for a reason.

Forums are for discussions. Wikis have a different purpose. All discussions should take place on the Forum rather than on the Wiki.