[quote="Ocean8"]Tamyra is making four cookies and has exactly four chocolate chips.  If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie?  Express your answer as a common fraction.
[hide]
Answer: 51/64[/hide][/quote]

Very nice problem. :)

[hide]There are $ 4^{4}$ ways total.

First, it doesn't matter where you put the first chip. There are $ 4$ ways no matter what.

Let's consider cases.

CASE 1: Put 2nd chip on the same cookie that you put the 1st chip on.

There's one way to do this, and the 3rd and 4th chips have 3 ways each, so 9 total ways.

CASE 2: Put 2nd chip on a different cookie

There are 3 ways to do this.

CASE 2-1. Put 3rd chip on a cookie with one chip already
There are 2 ways for the 3rd chip and 3 ways for the 4th chip.
6 ways.
CASE 2-2. Put 3rd chip on a different cookie.
There are 2 ways for the 3rd chip and 4 ways for the 4th chip.
8 ways.

For case 2, there are $ 3\cdot (6+8) = 42$ ways.

Adding up, we get $ 4\cdot 51$ ways, and since there are $ 4^{4}$ total, we get the desired probability of:
$ \frac{4\cdot 51}{4^{4}}= \boxed{\frac{51}{64}}$[/hide]

I wonder if there is a better way to do it...


Decide whether these two numers :
[b]8k+3 [/b]   and   [b]13k+5[/b]
can have common divisor greater than 2.
k is a natural number

Is that homework?
Just see that $8(13k+5)-13(8k+3) = 1$... ;)

Pierre.


help me solving this prob.

Show that $ tan\frac{\pi}{10}$ is a root of the equation $ 5x^2\minus{}10x^2\plus{}1\equal{}0$.

$ tan\frac{\pi}{10}$ is $ tan 18$ but i don't know how to get the value of $ tan 18...$

I think you typed the equation wrong. tan 18 isn't a root.

Anyway $ \tan\left(\frac{\pi}{10}\right)=\sqrt{\frac{3-\sqrt{5}}{5+\sqrt{5}}.}$

(see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202083[/url])

No, he means $ 18^{\circ}$, his $ \text{\LaTeX}$ is just bad.

For future reference, express $ \tan{18^{\circ}}$ as
[code]\tan{18^{\circ}}[/code]

Letters in math mode are never roman, so try to avoid that.

I don't think this is supposed to be in Basics...

I saw that in some (very hard) puzzle book...

[hide="Small hint"] Rewrite as $ 5 \sin^4 \frac{\pi}{10} \minus{} 10 \sin^2 \frac{\pi}{10} \cos^2 \frac{\pi}{10} \plus{} \cos^4 \frac{\pi}{10} \equal{} 0$ and start using trig identities.[/hide]

yeah Kunny's way was the first way that i thought of as well

OOPS THERE WAS A TYPO

its not $ 5x^2\minus{}10x^2\plus{}1$.... it was $ 5x^4\minus{}10x^2\plus{}1$...

thats why i was wrong...

ahhh so $ \tan{18^{\circ}}$ is $ \sqrt\dfrac{5-2\sqrt5}{5}$....

let $ y$ be $ x^2$, and the eqaution becomes $ 5y^2-10y+1$.

so $ y$ is $ 1\pm\dfrac{2\sqrt5}{5}$.

SO $ X$ IS $ \sqrt{1\pm\dfrac{2\sqrt5}{5}}$ AND $ \sqrt\dfrac{5-2\sqrt5}{5}$ IS A SOLUTION OF THE EQUATION!!!

YAY I SOLVED IT!!!

Thanks kunny and other ppls!!


[quote="236factorial"]What is $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{999\cdot1000}$?

[hide="small hint"]look for a pattern[/hide][/quote]

[hide]
1x2=2
2x3=6
3x4=12
4x5=20

+4, +6, +8, +10...

i see the pattern of increase.. but i dont see one in addition.. :/
[/hide]

Add the first two together, then the first three, then the first four. See if you can find one now  :P

[quote="G-UNIT"][quote="236factorial"]What is $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{999\cdot1000}$?

[hide="small hint"]look for a pattern[/hide][/quote]

  Try telescoping, its much easier.[/quote]

What's telescoping, like this?

$(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{999}-\frac{1}{1000})$?


geometry
geometric transformation
reflection

[hide="Answer"]36[/hide]
If that's right:

[hide="Hint"]Reflect![/hide]

[quote="mna851"]There are two buildings one with a height of $12$ more than the other.  A light beam is sent from the top of the larger building reflected off the ground and onto the top of the smaller building.  The lightbeam travelled a total of $91$ units.  The straight line distance between the top of one building to the top of the other is $37$ units.  Find the height of the smaller building.[/quote]
[hide]
Yes, reflection is good.
Also, when you do that, you can get two equations.

Of course those aren't random, they are needed to solve the problem. :wink: [/hide]

[quote="mathgeniuse^ln(x)"]Of course those aren't random, they are needed to solve the problem. :wink: [/quote]
 :roll: I was referring to the Pythagorean Theorem.

Anyway, you should get something like this:

[hide]
$(x+(x+12))^{2}+35^{2}=91^{2}\implies x=36$. [/hide]

Yep, that is what you get.

Sorry, for upsetting you in any way.

[quote="mathgeniuse^ln(x)"]Sorry, for upsetting you in any way.[/quote]
 :harhar:


[quote="fedja"]Here are a few guidelines as to what this subforum is for and how to post here. It is my current view on the things rather than an "official description". We'll have to figure out the more detailed rules in the process :).

There is no strict borderline between a "research topic" and an "olympiad topic", so definitely there will be a borderline cases that can be posted both here and in Open Questions sections. I will just outline a few criteria that a "research topic" is assumed to meet but they are by no means a dogma. So, here is what you want to look for if you desire to start a new project in this forum:

1) Your project should be an "open question" to the best of your knowledge, meaning that there is no full solution available
2) It is better if it is series of questions related to one theme rather than just a single question. If at all possible, there should be reasonably simple questions in that series that allow one some entrance place to start working on your project.
3) You should have some general idea of how you would like to approach the problem. It may amount to having done some partial case, having found some technique that "almost works, but", etc.

If you want just to join some existing project, please read everything that has already been posted in it before making your post. Use your common sense when deciding whether to start a new thread or to add to one of the existing threads. Keep in mind that while nobody really expects you to come up with the full solution, you are still expected to add something to the existing knowledge when making your post. Basically, just consider your post a "mini-article" on the subject: something that you want to share with others and make them to think of. Unlike a real research paper, you are not obliged to have a complete result in your mini-article, but, like it, you have to convince the others that what you are saying makes sense even if it is not a full solution. So, if you are just offering an idea, show an example of the situation when it works or clearly explain which particular difficulty in the problem it may help to overcome, etc.

There are absolutely no restrictions on the "difficulty", "importance", "publicability", and other similar things for the projects. Also there are no restrictions on who can post in each particular project. Just use your common sense and pick up something matching your current strength.

Now a couple bureaucratic formalities that may help to keep this forum in (relative) order. When starting a new project, assign a 6 letter code to it and place this six letter code in the beginning of the title followed by a space. Do the same (but with an existing code) when starting a new thread in one of the existing projects. This will allow everybody to use the search function with this code and the forum name to immediately see all threads related to one project. Also, please, try to come up with meaningful titles for your posts ("just one more idea" and "an example" are [b]not[/b] good titles)[/quote]


ratio
geometry
probability
3D geometry
perimeter
rectangle
trapezoid

1. A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form [a*pi + b*sqrt(c)]/[d*pi - e*sqrt(f)], where a, b, c, d, e, and f are positive integers, a and e are relatively prime, and neither c nor f is divisible by the square of any prime. Find the remainder when the product a*b*c*d*e*f is divided by 1000.

2. A jar has 10 red candies and 10 blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is m/n, where m and n are relatively prime positive integers, find m + n.

3. A solid rectangular block is formed by gluing together N congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly 231 of the 1-cm cubes cannot be seen. Find the smallest possible value of N.

4. How many positive integers less than 10,000 have at most two different digits?

5. In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100 workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional 100 workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the 800 workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

6. Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 3:2:1, what is the least possible total for the number of bananas?

7. ABCD is a rectangular sheet of paper that has been folded so that corner B is matched with point B' on edge AD. The crease is EF, where E is on AB and F is on CD. The dimensions AE = 8, BE = 17, and CF = 3 are given. The perimeter of rectangle ABCD is m/n, where m and n are relatively prime positive integers. Find m + n.

8 How many positive integer divisors of 2004^2004 are divisible by exactly 2004 positive integers?

9. A sequence of positive integers with a_1 = 1 and a_9 + a_10 = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n >= 1, the terms a_(2n-1), a_(2n), a_(2n+1) are in geometric progression, and the terms a_(2n), a_(2n+1), and a_(2n+2) are in arithmetic progression. Let a_n be the greatest term in this sequence that is less than 1000. Find n + a_n.

10. Let S be the set of integers between 1 and 2^40 whose binary expansions have exactly two 1's. If a number is chosen at random from S, the probability that it is divisible by 9 is p/q, where p and q are relatively prime positive integers. Find p + q.

11. A right circular cone has a base with radius 600 and height 200sqrt(7). A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is 375sqrt(2). Find the least distance that the fly could have crawled.

12. Let ABCD be an isosceles trapezoid, whose dimensions are AB = 6, BC = 5 = DA, and CD = 4. Draw circles of radius 3 centered at A and B, and circles of radius 2 centered at C and D. A circle contained within the trapezoid is tangent to all four of these circles. Its radius is [-k + m*sqrt(n)]/p, where k, m, n, and p are positive integers, n is not divisible by the square of any prime, and k and p are relatively prime. Find k + m + n + p.

13. Let ABCDE be a convex pentagon with AB || CE, BC || AD, AC || DE, < ABC = 120 degrees, AB = 3, BC = 5, and DE = 15. Given that the ratio between the area of triangle ABC and the area of triangle EBD is m/n, where m and n are relatively prime positive integers, find m + n.

14. Consider a string of n 7's, 7 7 7 7 . . . 7 7, into which + signs are inserted to produce an arithmetic expression. For example, 7 + 77 + 777 + 7 + 7 = 875 could be obtained from eight 7's in this way. For how many values of n is it possible to insert + signs so that the resulting expression has value 7000?

15. A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?[/code]

[quote="Fountainhead"]Hey Random, your profile says you live in Missouri. I'm also a Missourian and I'm kinda curious- what school do you go to?[/quote]

Please use [url=http://www.artofproblemsolving.com/Forum/privmsg.php?mode=post]PM[/url] for this.

If you like PDFs, the problems are [url=http://goodlymaths.0catch.com/Contests/AIME/AIME_archive.htm]here[/url]


IMO
functional equation
IMO 2002
algebra
IMO Shortlist

Find all functions $f$ from the reals to the reals such that \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz)  \] for all real $x,y,z,t$.

Set $x=y=z=t=0$ then we will get $f(0)(2f(0)-1)=0$.
So we will have two cases 1) $f(0)=o$ and 2)$f(0)=\frac{1}{2}$
1) $f(0)=0$
Set $z=t=0$ then we will have $f(x)f(y)=f(xy)$ $(*)$
we know that $(*)$ is Cauchy's equation and its solutions are:
(a) $f(x)=|x|^c$ (where $c$ is constant an we will look for it),
(b) $f(x)=sgnx\times x$   (c) $f(x)=0$
in (a) put instead of each function which is given we will yield:
$|xy|^c+|xt|^c+|zy|^c+|zt|^c=|xy-zt|^c+|xt+yz|^c$
now denote $|xy|=a, |xt|=b, |zy|=d$ and $|zt|=f$
$a^c+b^c+d^c+f^c=(a-f)^c+(d+b)^c$, so we have new function as example $f(a)=a^c$.
Set $b=d=f=a$ then we will have $4a^c=(2a)^c$
take $c$ root from both sides, and have $4^{\frac{1}{c}} a =2a$
cancel $a$ an have $4^{\frac{1}{c}} =2$ from here $c=2$ only solution, so $f(x)=x^2$
(b) cannot be solution because it will give $f(x)=|x|$
(c) is obviously solution
Now we will look in case 2) $f(0)=\frac{1}{2}$
Set $z=x$ and $y=t=0$ then $2f(x)=1\rightarrow f(x)=\frac{1}{2}$
So the solutions are:
$f(x)=x^2$,  $f(x)=0$, and $f(x)=\frac{1}{2}$, for all $x$ element of real numbers.

cauchy equation is valid only for rationals.
we need an extra condition in this case (we can show that f is monotone) ;)

can you show how to prove that $f(x)$ is monotone. I think that this is only dificulty of this problem.

$x=z,y=t=0 \Rightarrow 2f(x)\cdot 2f(0)=2f(0)$ for all $x\in\mathbb{R}$ $\Rightarrow f(0)=0$ or $f\equiv{1\over 2}$.
Assume $f(0)=0$.

$z=t=0 \Rightarrow f(xy)=f(x)f(y)$ for all $x,y\in\mathbb{R}$ (*)
$x=y=0 \Rightarrow f(-zt)=f(z)f(t)=f(zt)$ for all $z,t\in\mathbb{R}$
So $f(-x)=f(x)$ for all $x\in\mathbb{R}$ (**)
and $f(x)=f(\sqrt{x})^{2}\geq 0$ for all $x\geq 0$ (***)
and $f(x)=f(x\cdot 1)=f(x)f(1)$ for all $x\in\mathbb{R}$ $\Rightarrow f(1)=1$ or $f\equiv 0$.
Assume $f(1)=1$.

$x=y=z=t=1\Rightarrow f(2)=4$
Then (*) gives $f(2^{r})=4^{r}=(2^{r})^{2}$ for all $r\in\mathbb{Q}$. (****)

And now
$x=t,y=z \Rightarrow (f(x)+f(y))^{2}=f(x^{2}+y^{2})=f(\sqrt{x^{2}+y^{2}})^{2}\\ \hspace*{0.9in}\Rightarrow f(y)\leq f(x)+f(y)=f(\sqrt{x^{2}+y^{2}})$
for all $x,y\geq 0$ using (***).
So $f(x)=f(\sqrt{(\sqrt{x^{2}-y^{2}})^{2}+y^{2}})\geq f(y)$ for all $x\geq y\geq 0$.
This and (****) give $f(x)=x^{2}$ for all $x>0$
and (**) and $f(0)=0$ give
$f(x)=x^{2}$ for all $x\in\mathbb{R}$.

I'll skip $ f(0) \equal{} 0$.
Set $ x \equal{} y, z \equal{} t$, we have $ (f(x) \plus{} f(z))^2 \equal{} f(x^2 \minus{} z^2) \plus{} f(2xz)$.
Set $ x \equal{} t, z \equal{} y$, we have $ (f(x) \plus{} f(z))^2 \equal{} f(x^2 \plus{} z^2)$.

Therefore $ f(x^2 \plus{} z^2) \minus{} f(x^2 \minus{} z^2) \equal{} f(2xz)$.
\[ \lim_{z\rightarrow 0}\dfrac{f(x^2 \plus{} z^2) \minus{} f(x^2 \minus{} z^2)}{2z^2} \equal{} \lim_{z\rightarrow 0}\dfrac{f(2xz)}{2z^2}
\]
Let $ h(x) \equal{} \frac {f(x)}{x^2}$, then $ f'(x^2) \equal{} \lim_{z\rightarrow 0}h(2xz)\cdot 2x^2$. So $ \lim_{r\rightarrow 0}h(r) \equal{} \dfrac{f'(x^2)}{2x^2}$. This means $ \dfrac{f'(x^2)}{2x^2}$ is constant for all $ x$, therefore $ f(x)$ is constant or $ f(x) \equal{} kx^2$, and we can easily check that $ k \equal{} 1$.

[quote="MellowMelon"]Where did you show the function was differentiable? :([/quote]

I just found out the same problem. But it can be fixed easily:

$ \lim_{z\rightarrow 0}\frac{f(x^{2}\plus{}z^{2})\minus{}f(x^{2}\minus{}z^{2})}{2z^{2}}\equal{}\lim_{z\rightarrow 0}h(2xz)\cdot 2x^{2}$

Since the limit exist (we are fixing $ x$ and let $ z$ go to $ 0$), the funciton must be differentiable. Is that right?

That requires continuity at $ 0$, which you also have the burden of showing. (trying to use derivatives is almost certainly going to fail, by the way)

my solution is wrongly for increasing

$ x\equal{}t,y\equal{}z$   $ \implies$

 $ [f(a) \plus{} f(c)]^{2} \equal{} f(a^{2} \plus{} c^{2})$

$ \implies$  $ x > 0$  $ \implies$  $ f(x)\geq 0$

$ \implies$  $ f(a^{2} \plus{} c^{2})\geq f(a^{2})$  $ \implies$  $ f$ is increasing for $ x > 0$

the rest is easy

By setting $x=0=z$ we see that for all $y,t$ we must have $2f(0)(f(y)+f(t))=2f(0)$. Therefore $f=1/2$ or $f(0)=0$. (Note that $f=1/2$ works, so WLOG $f(0)=0$).

By setting $z=t=0$ we see that $f(x)f(y)=f(xy)$ for any $xy$. In particular we get that $f(p) \ge 0$ for any positive $p$, by setting $x=y=\sqrt{p}$.

By setting $x=\frac{zt}{y}$ for any $z,t,y$ and by using multiplicity we get that

$f(\frac{zt^2}{y}+yz)=(f(x)+f(z))(f(y)+f(t))=f(xy)+f(xt)+f(zy)+f(zt)=2f(zt)+f(\frac{zt^2}{y})+f(zy)$

and by multiplting by $f(y)/f(z)$ we get that

$f(t^2+y^2)=(f(t)+f(y))^2$

by using multiplicity. Let $g=\sqrt{f}$ be defined on $\mathbb{R}^{+}$. Then for all $y,t$ positive, we get that

$g(t^2+y^2)=g(t)^2+g(y)^2=g(t^2)+g(y^2)$

and so $g$ is additive. Since it's defined on the positives we get that $g(x)=kx$ for a constant $x$, and also $k$ must be positive (or $0$). So we have that $f(x)=Cx^2$ for positive $x$. Since $f$ is multiplicative we  get that $C=1$ or $C=0$, since $C \ge 0$. If $C=0$ then because of multiplicity, $f=0$. Otherwise, $f(x)=x^2$ for positive $x$.

Finally, in the original equation we have, by having $x,y,t$ very big positives and $z$ a small negative number such that $xt+yz > 0$, that

$(x^2+f(z))(y^2+t^2) = (xy-zt)^2 + (xt+yz)^2=(xy)^2+(xt)^2+z^2(y^2+t^2)$

and so $f(z)(y^2+t^2)=z^2(t^2+y^2)$ and so $f(z)=z^2$.

So we have $f=1/2$; $f=0$ or $f(x)=x^2$. Done.

The solutions are $f(x) = x^2$, $f(x) = 0$, and $f(x) = \frac{1}{2}$.

Set all variables to $0$ and we get $f(0) = 0$ or $f(0) = \frac{1}{2}$. If we have that $f(0) = \frac{1}{2}$, letting $x = z = 0$ and $t = y$ gives us $2f(y) = 1$, so we see that $\boxed{f(y) = \frac{1}{2}}$ for all $y \in \mathbb{R}$.

So now assume that $f(0) = 0$. Plugging in $y = t = 1, z = 0$, we have
$$2f(1)f(x) = 2f(x).$$
If $\boxed{f(x) = 0}$ for all real $x$, we have another solution. Otherwise, assume there exists an $x$ such that $f(x) \neq 0$, so we get $f(1) = 1$.

Now letting $y = t = z = 1$, we get
$$2f(x) + 2 = f(x+1) + f(x-1).$$
This is a recurrence relation, and from induction, we can easily show that $f(n) = n^2$ for all $n \in \mathbb{Z}$.

Now we prove multiplicity: set $z = t = 0$ and we get
$$f(x)\cdot f(y) = f(xy).$$
Setting $x = q$ and $y = \frac{p}{q}$ for integers $p$ and $q$ gives us
$$f\left( \frac{p}{q} \right) = \frac{f(p)}{f(q)} = \frac{p^2}{q^2},$$
so $f(x) = x^2$ for all $x \in \mathbb{Q}$.

To show that $f$ is even: let $x = z$ and $y = 0$ in the original FE, which gives us
$$2f(z)\cdot f(t) = 2f(zt) = f(-zt) + f(zt) \qquad \Rightarrow \qquad f(zt) = f(-zt)$$

Now we will show that $f$ is strictly increasing on the interval $[0, \infty)$. Let $x = y$ and $z = -t$. Then:
\[f(x^2 + t^2)=(f(x) + f(t))^2 = f(x^2) + 2f(xt) + f(t)^2 \ge f(x^2)\]
if $x \ge 0$ and $t > 0$. [We have $f(xt) \ge 0$, because $f(xt) = f(\sqrt{xt})f(\sqrt{xt}) \ge 0$]. So now we see that $f$ is strictly increasing.

Because the rationals are dense within the reals, $\boxed{f(x) = x^2}$ for all $x \ge 0$, and because $f$ is even, we're done. So in conclusion, our answers are: $f(x) = x^2$, $f(x) = 0$, and $f(x) = \frac{1}{2}$.

[i]Footnote:[/i] In case the density argument wasn't entirely clear, suppose for some positive irrational number $x$, we have $f(x) = x^2 + \varepsilon$ for some $\varepsilon > 0$ (the proof for $\varepsilon < 0$ is analogous). Then, take a rational number $x < r < \sqrt{x^2 + \varepsilon}$, which  is possible since between any two real numbers there is a rational number. Since we had shown $f(r) = r^2$, and we had shown $f$ is monotone, we must have
$$f(r) \ge f(x) \qquad \Rightarrow \qquad r^2 \ge x^2 + \varepsilon$$
which is a contradiction since $r < \sqrt{x^2 + \varepsilon}$.

Is Cauchy's equation's solution for monotone functions citable?

Let $P(x,y,z,t)$ be the given assertion. $P(0,0,0,0)$ implies that $f(0)=0$ or $f(0)=\dfrac 12$. If $f(0)=\dfrac 12$, then $P(0,0,0,t)$ gives us the solution $f(t)=\dfrac 12$.

Now, we can assume that $f(0)=0$. $P(x,y,0,0)$ gives that $f(x)f(y)=f(xy)$, and $P(0,0,z,t)$ gives that $f(z)f(t)=f(-zt)$, so $f$ is multiplicative and even. Let $Q$ be the assertion that $f(x)f(y)=f(xy)$. Then, $Q(1,1)$ gives that $f(1)=0$ or $f(1)=1$. If $f(1)=0$, then $Q(1,y)$ gives the solution $f(y)=0$. Otherwise, assume $f(1)=1$. Then $Q\left(x,\dfrac 1x\right)$ gives that $f(x)\neq 0$ for all $x\neq 0$.

Now, $P(x,1,1,1)$ gives that $2=f(x+1)-2f(x)+2f(x-1)$, so using the base cases $f(0)=0$ and $f(1)=1$, we can get that $f(n)=n^2$ for all integers $n$. Then, $Q\left(b, \dfrac ab\right)$ gives that $f(q)=q^2$ for all rationals $q$.

Now, $Q(x,x)$ gives that $f(x^2)=f(x)^2>0$ for all $x\neq 0$. Furthermore, for any positive reals $x<w$, $P\left(x,\sqrt{w^2-x^2},\sqrt{w^2-x^2},x\right)$ gives that $\left[f(x)+f\left(\sqrt{w^2-x^2}\right)\right]^2=f(w)^2$, which implies that $f$ is strictly increasing over the positive reals.

But since $f(q)=q^2$ over the positive rationals and $f$ is increasing over the positive reals, we can use a Cauchy-esque argument to show that $f(x)=x^2$ for all positive reals, and hence over all reals. Hence, the solutions are $f(x)=0, f(x)=\dfrac 12, f(x)=x^2$.

What

[hide = Solution]
The solutions are $f(x)\equiv 0,\ f(x)=\tfrac12,\ f(x)= x^2$, which can be easily verified.

As always, let $P(x,z,y,t)$ denote the assertion $(f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz)$.
$P(0,0,0,0)\implies 4f(0)^2=2f(0)\implies f(0)\in \{0,\tfrac12\}$.

Case 1: $f(0)=\tfrac12$:
$P(0,0,y,y)\implies 2f(y)=\tfrac12+\tfrac12\implies f(x)\equiv\tfrac12$, which is a solution.

Case 2: $f(0)=0$:
$P(1,0,1,0)\implies f(1)^2=f(1)\implies f(1)\in{0,1}$.
If $f(1)=0$,
$P(1,1,y,0)\implies 0=2f(y)\implies f(x)\equiv 0$, which is a solution.

So $f(0)=0$ and $f(1)=1$:
$P(x,0,y,0)\implies f(x)f(y)=f(xy)\implies f(x^2)=f(x)^2$.
This implies $f(x)\geq 0$ for $x\geq 0$.

Claim: $f$ is monotone increasing for $x\geq 0$.
[hide = Proof]
$P(x,z,z,x)\implies (f(x)+f(z))^2=f(x^2+z^2)$. Let $x,z\geq 0$. Then
\[f(x^2+z^2)=(f(x)+f(z))^2=f(x)^2+2f(x)f(z)+f(z)^2\geq f(x)^2= f(x^2).\]
Taking $x\rightarrow \sqrt{x}$ and $z\rightarrow \sqrt{z}$, we get $f(x+z)\geq f(x)$ for $x,z\geq 0$, which verifies the claim.
[/hide]

Let $g(x)=\sqrt{f(x)}$ for $x\geq 0$. Because of Claim 1, $g$ is also monotone increasing for $x\geq 0$. Note that
\[g(x)^2=f(x)=\sqrt{f(x)^2}=\sqrt{f(x^2)}=g(x^2).\]
Also, from $(f(x)+f(z))^2=f(x^2+z^2)$ we get
\[g(x^2)+g(z^2)=f(x)+f(z)=\sqrt{f(x^2+z^2)}=g(x^2+z^2).\]
Again taking $x\rightarrow \sqrt{x}$ and $z\rightarrow \sqrt{z}$, we get that $g(x)+g(z)=g(x+z)$, which is Cauchy's FE.
Since $g$ is monotone for $x\geq 0,\ $ $g$ must be linear (Cauchy's + monotone = linear).

Take $g(x)=ax+b$. Then $f(x)=ax^2+2abx+b^2$ for $x\geq 0$.

Since $f(0)=0$, then $b=0$ and $f(x)=ax^2$ for $x\geq 0$. Also, $f(1)=1=a\cdot 1\implies a=1$. Therefore $f(x)=x^2$ for $x\geq 0$. We now extend this to all reals:

$P(0,x,0,1)\implies f(x)=f(-x)$.

Take $x\geq 0$. Then $f(-x)=f(x)=x^2$. This means that $f(x)\equiv x^2$ is also a solution.

After checking the three solutions work, we are done. :coolspeak:

[/hide]

No monotonicity (don't know if I should be happy about this because I didn't notice that substitution at all)

The answer is $f(x)=x^2,0,\tfrac{1}{2}$ which clearly work. Denote the assertion as $P(x,y,z,t)$.
From $P(0,0,0,0)$ we have $4f(0)^2=2f(0) \implies f(0)=0, \tfrac{1}{2}$. If $f(0)=\tfrac{1}{2}$, then from $P(x,0,0,0)$ we obtain
$$f(x)+\tfrac{1}{2}=1 \implies f(x)=\tfrac{1}{2}$$
for all $x$. Henceforth suppose $f(0)=0$.
From $P(x,y,0,0)$ we get $f(x)f(y)=f(xy)$. From $P(0,0,x,y)$ we also get $f(x)f(y)=f(-xy)$, hence $f$ is multiplicative and even. In particular $f(-x^2)=f(x^2)=f(x)^2\geq 0$ for all $x$.
From $f(1)=f(1)^2$ we get $f(1)=0,1$. If $f(1)=0$, then from $P(1,x,1,0)$ we get $2f(x)=0 \implies f(x)=0$ for all $x$.
Now suppose $f(1)=1$, so from $P(1,1,1,1)$ we have $f(0)+f(2)=4 \implies f(2)=4$. Next, $P(x,x,y,y)$ gives
$$(f(x)+f(y))^2=f(x^2-y^2)+f(2xy) \implies f(x)^2+2f(x)f(y)+f(y)^2=f(x^2-y^2)=4f(x)f(y) \implies (f(x)-f(y))^2=f(x^2-y^2).$$
Now let $g(x)=\sqrt{f(x)}$ from $\mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$; note that $g$ is also multiplicative and even. For $x\geq y\geq 0$ the above equation becomes
$$g(x)^2+g(y)^2=g(x^2-y^2) \implies g(x^2)=g(x^2-y^2)+g(y^2).$$
Substituting $a=x^2-y^2$ and $b=y^2$, this is just Cauchy. As $g$ is bounded below due to its codomain, $g(x)=cx$ for some $c\geq 0$. Substituting $x=1$ gives $c=1$, so
$$g(x)=x \implies f(x)=x^2~\forall x\geq 0 \implies f(x)=x^2~\forall x \in \mathbb{R}$$
by the evenness of $f$, as desired. $\blacksquare$

EDIT: I think I am losing it. I did this problem about 8 months ago and I had a solution typed up above.

Set $x = z = y = t = 0$ to get $f(0) = 0$ or $f(0) = \frac{1}{2}$
[i]Case 1:[/i] When $f(0) = \frac{1}{2}$
Set $x=z=0$ to get \[f(y) + f(t) = 1\] and after setting $y=t$ we see that \[f(y) = \frac{1}{2} \; \forall y \in \mathbb{R}\]
[i]Case 2:[/i] When $f(0) = 0$
Set $y=t=1$ and $z=0$ to get \[2f(1)f(x) = 2f(x)\] If $f(x) = 0 \; \forall x \in \mathbb{R}$ then we will have a seperate solution. Else, assume that that there exists an $x$ such that $f(x) \ne 0$, hence we get that $f(1) = 1$ Now set $y=t=z=1$ to get \[2f(x) + 2 = f(x+1) + f(x-1)\] this is a recurrence relation and an easy induction shows that $f(n) = n^2 \; \forall n \in \mathbb{Z}$
[i]Claim:[/i] The function is multiplicative.
[i]Proof:[/i] Set $z=t=0$ to get $f(xy) = f(x) \cdot f(y)$
$\blacksquare$
[i]Claim:[/i] $f$ is even.
[i]Proof:[/i] Set $x = z$ and $y = 0$ to get \[2f(z)\cdot f(t) = 2f(zt) = f(-zt) + f(zt)\] which implies that $f(zt) = f(-zt)$
[i]Claim:[/i] $f$ is increasing on the interval $[0, \infty)$
[i]Proof:[/i] Set $x=y$ and $z = -t$ to get \[f(x^2 + t^2)=(f(x) + f(t))^2 = f(x^2) + 2f(xt) + f(t)^2 > f(x^2)\]
if $x \ge 0$ and $t > 0$ we will have \[f(xt) \ge 0\] this is because \[f(xt) = f(\sqrt{xt})f(\sqrt{xt}) \ge 0\] Thus we see that $f$ is strictly increasing. As rationals are dense within the reals we have $f(x) = x^2$ for all $x \ge 0$ and $f$ is even. Thus our answers are $f(x) = x^2 , f(x) = 0 , f(x) = \frac{1}{2}$

Let $P(x,y,z,t)$ denote the assertion  \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz)  \]

$P(0,0,0,0)$ gives us
$4f(0)^2=2f(0)$ Hence, $f(0)=0$ or $f(0)=\frac{1}{2}$

$P(0,y,0,t)$
$2f(0)(f(y)+f(t))=2f(0)$

If $f(0)$ is not equal to $0$,
$f(y)+f(t)=1$ for all $y,t$, which means
$f(x)=\frac{1}{2}x$ for all $x$

So, suppose that $f(0)=0$

$P(z,t,x,y)$:
\[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(zt-xy)+f(xt+yz)  \]

which means that $f(zt-xy)=f(xy-zt)$ for all $(x,y,z,t)$, hence we can conclude that $f$ is even, therefore for the remainder of our proof, we shall only consider non negative $x$ since $f(x)=f(-x)$ for all $x$

$P(x,y,0,0)$ :
$$f(x)f(y)=f(xy)$$

$P(x,1,0,0)$:
$f(x)f(1)=f(x)$
unless $f(x)=0$ for all $x$, it means that $f(1)=1$

Now, we shall proceed by induction to show that $f(x)=x^2$ for all integer $x$

$P(x,1,1,1)$:
$(f(x)+1)(2)=f(x-1)+f(x+1)$
$f(x+1)=2f(x)+2-f(x-1)$
Since $f(0)=0$, $f(1)=1$, suppose for $f(x)=x^2$ for two consectuive integers $x$
$f(x+1)=2x^2+2-(x-1)^2=(x+1)^2$
So, by induction , $f(x)=x^2$ for all integer $x$

However, note that this is easily extended into all rational $x$
$P(x,\frac{1}{x},0,0):$
$f(x)f(\frac{1}{x})=f(1)$
$f(\frac{1}{x})=\frac{1}{x^2}$

$P(\frac{1}{a},\frac{a}{x},0,0)$: where $a$ is some integer constant
$f(\frac{1}{a})f(\frac{a}{x})=f(\frac{1}{x})$
$f(\frac{a}{x})=\frac{a^2}{x^2}$

Therefore, $f(x)=x^2$ too for all rational.

Now, we shall show that $f$ is a non decreasing function

$P(x,x,x,x):$
$4f(x)^2=f(2x^2)$
$f(x) \geq 0$

$P(x,y,y,x)$:
$(f(x)+f(y))^2=f(x^2+y^2)$
$f(x^2)+2f(xy)+f(y^2)=f(x^2+y^2)$
which means that $f(x^2+y^2) \geq f(x^2)$

Hence, $f$ is a non-decreasing function

FTSOC, suppose that $f(x)$ is not equal to $x^2$ for some real values

Case 1: $f(x)<x^2$

Then, choose a rational value $y$ such that $y<x$
$$ f(x)<y^2<x^2 $$

Note $f(x)<f(y)$ implies that $x<y$ but we have that $y<x$ so contradiction

Case 2: $f(x)>x^2$

Choose a rational value $y$ such that $y>x$
$$f(x)>y^2>x^2$$
Note that $f(x)>f(y)$ which implies that $x>y$ but $y>x$ so we have a contradiction.

Hence, we can conclude that the only solutions are

$$ f(x)=0, f(x)=\frac{1}{2} or f(x)=x^2 $$

Let $P(x,y,z,t)$ be the assertion of the statement.

$P(0,y,0,y) \implies 4f(0)f(y)=2f(0).$

Case 1: $f(0)\neq 0.$
It's obvious that $\boxed{f(x) \equiv 1/2},$ which is a working solution.

Case 2: $f(0)=0.$
$P(x,y,0,0) \implies$ $$ f(xy)=f(x)f(y) ~~~~~~[\alpha]$$
If $f(y)=0,$ for any $y=0$, then $\boxed{f\equiv 0},$ which works as another solution.

Assume $f(y)=0$, when $y\neq 0$. It follows from $[\alpha]$ that $f(x)=f(\sqrt x)^2 \geq 0~~ \forall x \geq 0.$
$\text{So }P(x,y,y,x) \implies f(x^2+y^2)=(f(x)+f(y))^2 \geq f(x^2)~~\forall x,y \geq 0$.

Using $[\alpha]$ it's easy to see $f(1)=1$. So now $P(x,y,z,y) \implies$ $$2(f(x)+f(z))=f(x-z)+f(x+z)~~\forall x,z ~~~~~[\beta]$$
It also follows that $f(z)=f(-z).$

Further induction in $[\beta]$ leads to $f(nx)=n^2f(x)~~\forall n\in \mathbb{Z}$. But this can be extended to $\mathbb{Q}.$ Thus $f(q)=f(-q)=q^2 ~~\forall q \in \mathbb{Q}.$

Notice that $f$ is increasing for $x>0$. So $\boxed{f(x)=x^2} ~~ \forall x,$ which works. And we're done!

We claim that the only solutions are $f \equiv 0, \frac{1}{2}$ and $f(x)=x^2$ and they indeed work. Let $P(x, y, z, t)$ be the assertion. Now $P(0, 0, 0, 0)$ tells us that $f(0)$ is either $\frac{1}{2}$ or $0$. If $f(0)=\frac{1}{2}$, $P(y, 0, 0, 0) \rightarrow f(y) \equiv \frac{1}{2}$. So now let $f(0)=0$, now $P(x, y, 0, 0) \rightarrow f(xy)=f(x)f(y)$ so $f$ is multiplicative.
[b][color=#f00]Claim:[/color][/b] $f$ is non-decreasing on $\mathbb{R}^+$
[b][color=#00f]Proof:[/color][/b] First, notice that $f(x^2)=f(x)^2 \geq 0$. Now $$P(x, y, y, x) \rightarrow f(x^2)+2f(xy)+f(y^2)=f(x^2+y^2)$$ So $f(x^2) < f(x^2+y^2)$. $\blacksquare$

Now let $g(x)=\ln(f(a^x))$, notice that $g(x+y)=g(x)+g(y)$ and  $g(x)=cx$ and this means $f(x)=x^c$, on plugging in the original equation, we see that only $f(x)=x^2$ works. $\blacksquare$

Le thonk

Let $P(x,z,y,t)$ denote the assertion. $P(0,0,y,y)$ gives $4f(0)f(y)=2f(0)$ which shows either $f(y)=\frac{1}{2}$ for all $y$ or $f(0)=0.$ Note that $f(y)=\frac{1}{2}$ is a valid function. Now, suppose $f(0)=0,$ then $P(x,x,y,y)$ gives $4f(x)f(y)=f(2xy).$ Letting $x=y=\frac{1}{2}$ gives $4f(\frac{1}{2})^2=f(\frac{1}{2})$ so $f(\frac{1}{2})=0$ or $\frac{1}{4}.$
\\$~$\\
If $f(\frac{1}{2})=0$ then $P(0,\frac{1}{2},0,t)$ gives $0=f(-\frac{1}{2}t).$ Note that $-\frac{1}{2}t$ can be anything so $f(x)=0$ for all $x$ which is also a valid function, so suppose $f(\frac{1}{2})=\frac{1}{4}.$
\\$~$\\
Now, $P(x,x,y,-y)$ gives $2f(x)(f(y)+f(-y))=f(2xy)=4f(x)f(y).$ We can easily choose $f(x)\neq 0$ so $f(y)+f(-y)=2f(y),$ which implies that $f(y)=f(-y).$ Thus, we solve $f$ over only the positive numbers. We have $P(x,z,x,z)$ giving us $(f(x)+f(z))^2=f(x^2-z^2)+f(2xz)$ and $P(x,z,z,x)$ giving us $(f(x)+f(z))^2=f(x^2+z^2).$ Note that $(x^2-z^2)^2+(2xz)^2=(x^2+z^2)^2$ so $f(\sqrt{a})+f(\sqrt{b})=f(\sqrt{a+b}).$
\\$~$\\
Let $g:(0,\infty)\to \mathbb{R}$ such that $g(x)=f(\sqrt{x})$ which implies that $g(a)+g(b)=g(a+b).$ Thus, $g(2a)=2g(a)$ and $g(4a)=4g(a).$ Also, $4f(x)f(y)=f(2xy)$ so $4g(x^2)g(y^2)=g(4x^2y^2)=4g(x^2y^2)$ which implies $g(a)g(b)=g(ab).$ Thus, we have $g(x)=kx$ for some $k$ and so $f(x)=kx^2.$ Thus, plugging $x=\frac{1}{2}$ gives $f(x)=x^2$ which is indeed a valid function. Thus, our functions are $f(x)=0,\frac{1}{2},x^2.$

Similar problem from 2015 Korea [url]https://artofproblemsolving.com/community/c6h1158064p5501261[/url]

Solved with [b][url=https://artofproblemsolving.com/community/user/797730]Pranav1056[/url][/b] and [b][url=https://artofproblemsolving.com/community/user/849747]proxima1681[/url][/b].

[i]Solution:[/i] We will show that $f \equiv 0, \frac{1}{2}$ and $f(x) = x^2$ for all $x \in \mathbb{R}$ are the only possible solutions. It is easy to check all of these works. Denote $P(x,z,y,t)$ as the assertion to the functional equation.

Throughout the solution, assume that $f$ is a non-constant function since the constant ones are fairly trivial. We will prove $f(x) = x^2$ for all $x$.

From $P(0,0,0,0)$ we get $f(0) \in \left\{0,\frac{1}{2}\right\}$. If $f(0) = \frac{1}{2}$, then $P(x,0,0,0)$ gives $f \equiv \frac{1}{2}$ which is a contradiction since we have assumed $f$ as non-constant. This gives $f(0) = 0$.

[b][color=red]Claim:[/b][/color] $f$ is multiplicative and strictly increasing over $\mathbb{R}^+$.

[i]Proof:[/i] $P(x,0,y,0)$ gives
\[f(x)f(y) = f(xy)\]
as desired. Also note that this gives $f(x) = 0 \iff x = 0$ for non-constant $f$. For proving increasing nature over $\mathbb{R}^+$, we consider the substitution $P(x,y,y,x)$ with $x,y>0$.
        \[f(x^2 + y^2) = f(x^2) + f(y^2) + 2f(xy) > f(x^2)\]
which proves the claim. $\square$
----------
By a well-known result of Cauchy, we get to know that $f(x) = x^n$ for all $x \in \mathbb{R}^+$. $P(1,1,1,1)$ would yield $n = 2$. Since we already have $f(0) = 0^2$, we need to show $f(x) = x^2$ for negative $x$ as well.

This part is easy to see. Take $P(x,z,1,1)$ with $z$ some arbitrary negative $z$ and sufficiently large positive $x$ such that $x-z>0$. This would give $f(z) = z^2$ for all negative $z$. $\blacksquare$

Denote the assertion with $P(x, z, y, t)$.
Note that the constant solutions of $f(x) = \frac{1}{2}$ and $f(x) = 0$ work.
By $P(x, z, 0, 0)$ it follows that \[ (f(x) + f(z))2f(0) = 2f(0) \] so either $f(x)$ is constant for all $x$ or $f(0) = 0$. Assume the latter.

[b][color=pink]Claim:[/color][/b] $f(x) \ge 0$ for all $x$ and $f$ is even.
[i]Proof.[/i] By $P(x, x, x, x)$ it follows that \[ 4f(x)^2 = f(0) + f(2x^2) \] so $f(x) \ge 0$ for $x \ge 0$ and $f(-x) = \pm f(x)$.
Then, if $f(x) = -f(-x)$ for $x \ne 0$ then by $P(x, -x, t, t)$ it follows that \[ f(2xt) = 0 \] and thus $f(x)$ is uniformly $0$. $\blacksquare$

[b][color=pink]Claim:[/color][/b] If there is a nonzero $x$ such $f(x) = 0$ then $f(x)$ is uniformly $0$. As such, we can take the codomain to be ${\mathbb R}^+$ over the domain ${\mathbb R}^+$.
[i]Proof.[/i] Substitute $P(x, x, y, t)$ to get \[ f(xy - xt) + f(xy + xt) = 0 \] which implies that $f(x(y-t)) = f(x(y+t)) = 0$. $\blacksquare$
Assume that $0$ is the only root. Now, by $P(x, z, y, t)$ and $P(z, x, y, t)$ it follows that \[ f(yz - xt) + f(xy + zt) = f(yz + xt) + f(xy - zt) \]
If we substitute it follows that for $ad = bc$ it follows that \[ f(a - d) + f(b + c) = f(a + d) + f(b - c) \]
Over the positive reals, this reduces to [url=https://aops.com/community/p2729512]ELMO 2011/4[/url] which has the solution set of $f(x) = ax^2 + b$ for $a$ and $b$ both $\ge 0$ where equality does not hold for both.
Since $f$ is even, this holds for all $x \ne 0$. Then, it follows that since \[ f(x - x) + f(x^2 + 1) = f(x + x) + f(x^2 - 1), \] $b = 0$ and thus $f(x) = ax^2$ for $a > 0$ holds for all $x$.
It can be seen that by substituting in the original equation that only one value of $a$ can possible work.
Furthermore, \[ (x^2 + z^2)(y^2 + t^2) = (xy - zt)^2 + (xt + yz)^2 \] holds so $a = 1$ and thus $f(x) = x^2$.

I claim that the only solutions are $f(x) = x^2, f(x) = 0, f(x) = \frac 12$, and it is obvious that all of these work.

Let $P(x,z,y,t)$ denote the assertion, then $P(x,z,0,0)$ gives $2f(0)(f(x) + f(z)) = 2f(0)$, so either $f(x) + f(z) = 1$ for all $x$, or $f(0) = 0$. The former case gives $f(x) = \frac 12$ for all $x$. In the latter case, consider $P(x,0,0,t)$, which gives $f(x)f(t) = f(xt)$. This then immediately forces $f(x) = 0$ or $f(x)$ nonzero for $x$ nonzero. We assume the latter case. Then $P(1,0,1,0)$ combined with the condition before gives $f(1) = 1$. Then plugging in $P(1,1,1,1)$ gives $f(2) = 2$. Then we use induction to prove $f(n) = n^2$ for all positive integers. We have already proved the base cases. Now assume the claim is true for all $k \le n$. Notice $P(n,1,1,1)$ gives $2(n^2 + 1) = (n-1)^2 + f(n+1)$, so we have $f(n + 1) = (n+1)^2$. Then by multiplicity, it is easy to get $f(\frac 1n) = \frac{1}{n^2}$, and the claim for all positive rationals is easy. Then notice that $P(x,x,x,x)$ gives $4f(x)^2 = f(2x^2)$, so $f(x) = f(-x)$, so this claim is in fact true for all rationals.

Now we extend this to the reals. Firstly, notice that $f$ is always nonnegative. We first show that $f(x) \ge x^2$ for all $x$. Consider $P(x,r,r,x)$ for real $x$ and rational $r$, with $x>r$. This then gives $(f(x)+r^2)^2 = f(x^2 + r^2)$. We can write this as $f(x^2) = (f(\sqrt{x^2 - r^2}) + r^2)^2$. Now assume $f(x) < x^2$ for some real $x$. We can then find some rational $r^4$ between $f(x),x^2$, such that $r^4 > f(x) = (f(\sqrt{x - r^2}) + r^2)^2 > r^4$, contradiction. A flipped argument can show the other side, so $f(x) = x^2$.

Let $P(x,y,z,t):=(f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz)$

$P(0,0,0,0)$ yields $4f(0)^2=2f(0)\Longrightarrow 2f(0)(2f(0)-1)=0$ thus $f(0)=0$ or $f(0)=\frac{1}{2}$

[b][color=#F6B26B]Case 1:[/color][/b] $f(0)=0$
$P(x,y,0,0)$ yields $f(x)f(y)=f(xy)$ therefore $f$ is a multiplicative function, and $f(x^2)=f(x)^2$ for $x\neq0$, however we can extend this for all $x$ as $f(0)=0$
$P(x,x,x,x)$ yields $4f(x)^2=f(2x^2)$ thus $f(x)\ge0\text{ for }x\ge0$
$P(y,0,x,y)$ yields $f(y)^2+f(xy)=f(-xy)+f(y^2)\Longrightarrow f(xy)=f(-xy)$ therefore $f$ is even.
Furthermore let $x,y\ge0$, thus $P(\sqrt{x},\sqrt{y},\sqrt{y},\sqrt{x})$ yields $f(x+y)=\left(f(x)+f(y)\right)^2\ge f(\sqrt{x})^2=f(x)$ thus $f(x+y)\ge f(x)$ for $x,y\ge0$ therefore $f$ is monotonically increasing over positive reals. Thus $f(x)=x^c$, and by plugging into our original $fE$ we obtain that $c=2$ and $f(x)=x^2, \forall x\in\mathbb{R}$

[b][color=#c00]Case 2:[/color][/b] $f(0)=\frac{1}{2}$
$P(0,x,0,x)$ yields $2f(x)=1\Longrightarrow f(x)=\frac{1}{2}, \forall x\in \mathbb{R}$

[b][color=#6AA84F]Case 3:[/color][/b] $f(x)=0$
Assume that $f$ is constant and not equal to $\frac{1}{2}$, when we plug in $f(x)=c$ into our original $fE$ we obtain $4c^2=2c\Longrightarrow 2c(2c-1)=0$, this forces $c=0$
Therefore $f(x)=0, \forall x\in\mathbb{R}$

So, to sum up $\boxed{f(x)=x^2, f(x)=\frac{1}{2}\text{ and }f(x)=0, \forall x\in\mathbb{R}}$ $\blacksquare$.

"good easy problem. solved in my head in a few minutes."

    $x=z=0\implies 2f(0)(f(y)+f(t))=2f(0)\implies\{f\equiv\frac12,f(0)=0\}$. In the latter, we have f(-xy)=f(x)f(y)=f(xy) by setting two variables equal to 0 (ill omit which ones). With $x=t,y=z\implies(f(x)+f(y))^2=f(x^2+y^2)$ so f is bounded over $x\ge0$, whence multiplicity with bounded with f(-xy)=f(xy) implies $f(x)=x^2$. Otherwise, for any weird edge cases check that the other constant sol is f(x)=0.

The answers are

   \[f\equiv 0,f\equiv \frac{1}{2}  \text{ and } f(x)=x^2 \text{ for all}x\in \mathbb{R}\]

It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. \\

$P(0,0,0,0)$ gives us that
\[4f(0)^2 = f(0)\]
Thus, $f(0)=0$ or $f(0)=\frac{1}{2}$. We then break into two cases.\\

\textbf{Case 1 : }$f(0)=0$. Note that now, $f(x,y,0,0)$ gives,
\[f(x)f(y)=f(xy)\]
for all $x,y \in \mathbb{R}$. Thus, $f(1)=0$ or $f(1)=1$. If $f(1)=0$, we know that the above equation gives $f\equiv 0$ and if $f(1)=1$ then $f(x)=x^k$ for all $x \in \mathbb{R}$ for some constant $k \in \mathbb{R}$. Consider $P(a,a,a,a)$ for some $a>1 \in \mathbb{R}$. This gives,
\[4f(a)^2=f(2a^2)\]
plugging in $f(x)=x^k$ gives us that $4a^{2k}=2^ka^{2k}$ from which it is quite clear that we must have $k=2$. Thus, the only solutions from this case are $f\equiv 0$ and $f(x)=x^2$ for all $x\in \mathbb{R}$.\\

\textbf{Case 2 : } $f(0) = \frac{1}{2}$. Plugging in $y=z=t$ gives us that,
\[f(x)+\frac{1}{2} = \frac{1}{2} +\frac{1}{2} \]
for all $x\in \mathbb{R}$. Thus, $f\equiv \frac{1}{2}$ in this case.\\

This implies that indeed all solutions to the given functional equation are of the claimed forms and we are done.

The solutions are $f\equiv 0$, $f\equiv \dfrac{1}{2}$, $f\equiv x^2$.

$P(0,0,0,0)\implies 4f(0)^2 = 2f(0) \implies f(0) \in \left\{0,\frac{1}{2}\right\}$.

$\textbf{Case 1:}$ If $f(0) = 0$. Now if $f\not\equiv 0$, let $f(p) \neq 0$.

$P(0,y,z,0) \implies f(z) f(y) = f(zy)$.

Put $y=z=1$ to get $f(1) \in \left\{0,1\right\}$.
-----
$\textbf{Case 1.1:}$ If $f(1) = 1$.

\begin{align*}
    P(0,y,p,t)&\implies f(p)(f(y) + f(t)) = f(-pt) + f(yp)\\
    &\implies f(p)(f(y) + f(t)) = f(p) (f(y) + f(-t))\\
    &\implies f(t) = f(-t)
.\end{align*}

Now, $P(1,1,z,1) \implies (1+f(z))(2) = f(1-z) + f(1+z)$.

Now by induction, $f(n) = n^2$ and by using multiplicativity, $f(q) = q^2$ for all $q\in \mathbb Q$.

Also, $P(x,x,z,z) \implies f(2xz) = (f(x) + f(z))^2 \implies f \ge 0$.

Then for any $t\neq 0$,
\begin{align*}
    P\left(x,y,\frac{xy}{24},t\right)\implies f(xy)+f(xt) + f\left(\frac{xy}{2t} \cdot y\right) + f\left(\frac{xy}{2}\right) = f\left(\frac{xy}{2}\right) + f\left(xt + \frac{xy^2}{2t}\right)\\
    \implies f(2t^2 + y^2) = f(2t^2) + f(2yt) \ge f(2t^2) \\
    \implies f(x + \varepsilon) \ge f(x) \;\forall\; x,\varepsilon \in\mathbb R^+
.\end{align*}

Now using multiplicativity and using some $\log$ stuff, we can convert it into the multiplicativity into Cauchy and then use the monotone fact to show that $f=x^k$ and $f(2) = 4 \implies k = 2 \implies f\equiv x^2$.
-----
$\textbf{Case 1.2:}$ Otherwise if $f(1) = 0$.

Then $f(z) = f(z) \cdot f(1) = 0 \implies f\equiv 0$.
-----
$\textbf{Case 2:}$ Finally, if otherwise at the beginning we had $f(0) = \dfrac{1}{2}$, then $P(x,0,x,0)\implies 2f(x) = 1 \implies f(x) = \dfrac{1}{2} \implies f\equiv \dfrac{1}{2}$. :yoda:


What is the number of cubic feet in the volume of a cube that has a surface area of 96 square feet?

96/6=16 Thus area of side is 16 and sidelength is 4. 4^3=64.

Divide by $ 6$ because there are $ 6$ faces on a cube, obviously all with the same area.


[quote="Treething"]nope, read it carefully[/quote]

how did you edit it without it saying so?

[hide]$\frac{1500x}{5500+x}=666$

$x\approx 4400$[/hide]

[quote="Treething"]In a certain computer game, the fraction of damage blocked is determined by the function $\frac{x}{x+5500}$, where x is the armor of your character. In order to receive 666 damage from an attack originally doing 1500 damage, how much armor is required? Round your answer to the nearest hundred.[/quote]

[hide] $\frac{x}{x+5500}\cdot1500=1500-666\Longrightarrow\frac{x}{x+5500}=\frac{834}{1500}=\frac{139}{250}\Longrightarrow 250x=139x+139\cdot5500\Longrightarrow 111x=139\cdot5500\Longrightarrow x=\frac{139\cdot5500}{111}\approx\boxed{6900}$ [/hide]

[quote="Treething"]heh i should put up more problems requiring you to read carefully, then take all the credit when a lot of people at nationals next year thank aops for helping them so much

lotr is the only person who got it correct... splash reread it, bubka reread it.[/quote]

Huh?  :huh:

[quote="SplashD"]how did you edit it without it saying so?[/quote]If you edit it before someone else posts, it won't say anything.

Treething said "round your answer to the nearest hundred."

His name is still regular, not bolded and green.

[quote="Treething"]In a certain computer game, the fraction of damage blocked is determined by the function $\frac{x}{x+5500}$, where x is the armor of your character. In order to receive 666 damage from an attack originally doing 1500 damage, how much armor is required? Round your answer to the nearest hundred.[/quote]

6900 after the rounding :rotfl:

[quote="solafidefarms"]MATHCOUNTS
Moderators: kyyuanmathcount, LadyKn1ght, nebula42, rcv, solafidefarms

No, I don't believe he is a mod, certainly not of MC. No more discussion of Daesun's mod status in this thread, or I'll split the discussion of his power to GFF, and nobody wants to lose a post.[/quote]

I'm pretty sure you don't lose a post if a post if yours is moved to G&FF as you don't lose a post when you make a post and delete it.

By the way, all you have to read is "Round your answer to the nearest hundred" to basically know the answer.

[quote="bubka"]Ok I rounded it to the nearest hundredth.
Same thing!  :P  :rotfl:[/quote]
hundred, not hundredth  :wink:

[hide]$\frac{x}{x+5500}=\frac{1500-666}{1500}=\frac{139}{250}\\ 250x=139x+764500\\ x=\frac{764500}{111}\approx\boxed{6900}$[/hide]


After the action took in the Leaning Tower of Treething, another action broke out between the Shubertestein and Turtikobackzksi, the very evil cousins of the history. In order to beat Turtikobackzksi, Shubertestein planned to build a machine that can see the future. This machine, however, is not like any other machines of the past. This has 3 choices of color (blue, black, red), 5 choices of size (X-Large, Large, Medium, Small, X-Small), 4 choices of skill (invisibility, reducible-size, laser-beams, supercomputer-brainchip), and 7 choices of name (Aiko, Buro, Corui, Dyuil, Eylp, Fyop, Gylp). His plan is to build BLUE X-LARGE LASER-BEAMS FYOP. How many different possible kinds of machines are possible if that each one has to be only different by two choices from BLUE X-LARGE LASER-BEAMS FYOP. (BLUE X-LARGE REDUCIBLE SIZE, CORUI is one such arrangements)

No, it's not too hard. This problem or something similar may come up on the Nationals target rounds or sooner. Probably nationals level though. The problem is however a little complicated. It's nice that we're trying to make word problems that are interesting, but let's not go overboard. I'll clarify for anybody.

Bag A has 3 coins
Bag B has 5 coins
Bag C has 4 coins
Bag D has 7 coins

In each bag, there is one and only one winning coin. A boy selects one coin from each bag. If all 19 coins are unique, how many combinations are there where the boy selects two and only two winning coins?

Hint:[hide]Sometimes, doing it the obvious way is easier than looking for fancy approaches[/hide]

I wouldn't think so, it's just combinations  .

Color - three choices (two different)

Size - five choices (four different)

Skill - four choices (three different)

Name - seven choices (six different)

I'm probably wrong, as I usually am, but...

[hide]

You can either choose...

Color and Size - 8 choices

Color and Skill - 6 choices

Color and Name - 12 choices

Size and Skill - 12 choices

Size and Name - 24 choices

Skill and Name - 18 choices

.. to be different.

Add 'em all up and you get 80 choices.

[/hide]

EDIT: I realized I made a mistake.

Nope.

That isn't right. This question was derived from AHSME actually. :) So it would be HARD but I think most members here can solve this.

Grr  .

Was my previous answer of

[hide]131[/hide] correct?

Wait. I redid the problem and your first answer, 80, was right. :)

[hide]80=30+20+30=18+24+12+12+6+8.
There can only be two possible correct answers.  Therefore, we know that are 6=3+2+1 groups we need to try to keep the same(color-size, color-skill, color-name; size-skill, size-name; skill-name)  Since we only want two the same, from the ones we want to change around, we need to subtract one from each of the numbers we want to multiply, because not subtracting would cause three or four of the same.[/hide]

[color=darkred]Yeah, when I got my first answer, that subtracting one tripped me up. I redid the problem and got 80, though.[/color]


How do you solve the first question?
If $ \frac{4}{2001} < \frac{a}{a\plus{}b} < \frac{5}{2001}$, compute the number of integral values $ \frac{a}{b}$ can take on.

[quote="avshockeyw"]How do you solve the first question?
If $ \frac {4}{2001} < \frac {a}{a \plus{} b} < \frac {5}{2001}$, compute the number of integral values $ \frac {a}{b}$ can take on.[/quote]

Is the question supposed to ask for the number of integral values $ \frac{b}{a}$ can take on?

[hide]
$ \frac{4}{2001} < \frac{a}{a \plus{} b} < \frac{5}{2001}$

$ \frac{2001}{5} < \frac{a\plus{}b}{a}  \equal{} 1\plus{}\frac{b}{a}< \frac{2001}{4}$

$ \frac{1996}{5} \equal{} 399.2 < \frac{b}{a} < 499.25 \equal{} \frac{1997}{4}$

$ 0 < \frac{4}{1997} < \frac{a}{b} < \frac{5}{1996} < 1$

Therefore, $ \frac{a}{b}$ can take on $ \boxed{0}$ integer values, and $ \frac{b}{a}$ can take on $ \boxed{100}$ integer values.
[/hide]


Solve the system
$\begin{cases}x^3+y^3=9z^3\\x^2y+xy^2=6z^3\end{cases}$

$(x^3+y^3)+3(x^2y+xy^2)=27z^3$ $\implies x+y=27z$

so $xy(x+y)=6z^3$ $\implies xy=\frac{2}{9}z^2$ or $z=0$

When $z=0$ we have $x=-y$ so the solution is $(x,y,z)=(t,-t,0)$

If $z\not =0$ , then $x^3+y^3=9z^3$ $\implies (x+y)((x+y)^2-3xy)=9z^3$ $\implies 9(2185)z^3=9z^3$

which yields also $z=0$

So the only solution is $(x,y,z)=(t,-t,0)$ for $t\in\mathbb{R}$

** opps , i make a mistake when factoring in first line . haha . Well Farenhajt has done it anyway :P

Multiplying the second equation by $3$ and adding to the first, we get

$(x+y)^3=27z^3$, or $x+y=3z$

Factorizing the second equation we get

$xy(x+y)=6z^3\implies 3zxy=6z^3\implies xy=2z^2$

Therefore, $x,y$ are the solutions of the equation

$t^2-3zt+2z^2=0$, or $\{x,y\}\in\{2z,z\}$

All solutions are given by $\{(z,2z,z)|z\in\mathbb{R}\}\cup\{(2z,z,z)|z\in\mathbb{R}\}$

You both have errors.  Shyong, you took the cube root of 27 and got 27.  Farenheit, you divided by z without noting the possibility $z = 0$.

That's true  :( . Hence additional set of solutions is $\{(x,-x,0)|x\in\mathbb{R}\}$

EDIT: Shoyng and I gave a system of solutions :P


function
real analysis
integration
complex analysis
Functional Analysis
complex analysis unsolved

Let $f$ be a devellopable in power serie function with convergenge radius $R$.Supppose that there exists $r$,$0<r<R$ such that the function $f$ gets real values in every point of the circle of radius $r$.
Show that $f$ is constant.

$\mathcal D(z,t)$ will denote the open disk of radius $t$ around $z\in\mathbb C$, and $\mathcal C(z,t)$ will be the border of $\mathcal D(z,t)$ (i.e. the circle of radius $t$ around $z$). $I$ is the compact interval $f(\mathcal C(0,r))$. Finally, let $\gamma(x)=re^{2\pi ix},\ \forall x\in[0,1]$ (i.e. a positively-oriented closed path describing $\mathcal C(0,r)$).

Suppose $f$ is not constant. Then $f(\mathcal D(0,r))$ is an open subset of the complex plane, so there must be a $z_0\in\mathcal D(0,r)$ such that $f(z_0)\not\in I$. On the one hand, the number of zeros of $f(z)-f(z_0)$ inside $\mathcal D(0,r)$ is $\ge 1$, but, on the other hand, it's the index of the closed path $f\circ\gamma$ with respect to $f(z_0)$, and this index is clearly $0$ (since the path can be contracted to a point inside $I$, without passing through $f(z_0)$). We have our contradiction.

Edit:

Actually, here's a stronger assertion:

Consider the unit circle $\mathbb T$, and on it consider the normalized Lebesgue measure $\mu$ (i.e. $\mu(\mathbb T)=1$). The functions $e_n(t)=e^{nit},\ n\in\mathbb Z$ form an orthonormal basis of the Hilbert space of $2\pi$-periodic complex functions defined on $\mathbb R^1$ which are square-integrable on an interval of length $2\pi$ with the scalar product $(f,g)=\frac 1{2\pi}\int_{-\pi}^{\pi}f(t)\overline {g(t)}\ \text dt$, where $\mu$ is the usual Lebesgue measure on the real line, and let $H^2$ be the closed subspace of this Hilbert space spanned by $(e_n)_{n\ge 0}$. The assertion is that a function from $H^2$ which is real (this means "real almost everywhere", since the "functions" aren't really functions here, but classes of equivalence of functions modulo sets of measure zero), is actually constant.

This is stronger than the initial problem because in that problem we can assume that $r=1$, and the restriction of a function $f$ holomorphic in some disk centered in the origin of radius $R>1$ to the unit circle can be regarded in a natural way as an element of $H^2$: we identify this restriction with $\tilde f\in H^2$ defined by $\tilde f(t)=f(e^{it}),\ \forall t\in\mathbb R$.

As for the proof, $H^2$ consists of functions $f$ which have the property that $\frac 1{2\pi}\int_{-\pi}^{\pi}f(t)e^{nit}\text dt=0,\ \forall n\ge 1\ (*)$. Now, if such a function $f$ is real almost everywhere, the conjugate of $(*)$ reads $\frac 1{2\pi}\int_{-\pi}^{\pi}f(t)e^{-nit}\text dt=0,\ \forall n\ge 1$. Together with $(*)$, this implies that $f$ is in the orthogonal complement of the space spanned by $\{e_n\}_{n\in\mathbb Z^*}$, i.e. it’s a multiple of $e_0$, or, in other words, a constant (again, this means "constant almost everywhere").


induction
number theory unsolved
number theory

prove that if u>v>1 then u^x=1(mod v^x) has finite solution for x?

If v is prime.Let be n the smallest integer such that u^n=1[v^n] and m an integer such that u^m=1[v^m] and m>=n let m=qn+r be the euclidian division of m by n.
u^m=u^r[v^n] so r=0.
Note v(u^m) be the greater integer l such that v^l|u^m-1
Let v(u^n)=k we will prove that v(u^(j*n*v^r))=k(r+1) be induction:(v don't divise j)
it is right for r=0 because u^(j*n)-1=(u^n-1)(u^((j-1)n)+...+1)
u^((j-1)n)+...+1=j[v^k] so v don't divise u^((j-1)n)+...+1 and v(u^(j*n)-1)=k suppose that it is true forr:
u^(j*n*v^(r+1))-1=(u^(j*n*v^r)-1)(u^(j*n*v^r*(v-1))+...+1)
and u^(j*n*v^r*(v-1))+...+1=v[v^(k*(r+1)] so v(u^(j*n*v^r*(v-1))+...+1)=v and v(u^(j*n*v^(r+1))-1)=k*(r+2). So it is true for all r

So if m is such that u^m=1[v^m] then v(u^m)>=m but lim((j*n*v^r)/(k(r+1)),r,infinity)=+oo so only a finite umber of such m exist.

If v is not a prime let p be a prime divisor of v so u^m=1[v^m] implies u^m=1[p^m] so we make the same way to prove the result.

Soryy for my very very poor english

I did not understand anything past your first few lines, raf.  :?

If I'm nor wrong this follows from mathlinks contest problem 3, round 1.

Indeed the problem is similar to the mathlinks 3rd Ed round 1 pb 3 but it deals only a particular residue and asks for a finite number of solutions.

Yes, it is much easier (though not very much) and can be deduced from that one.

I don' t know you want speak but I think it is a problem isn't easy :?  :(


$\overline{AB}$ is a line segment in the plane, $O$ is on one side of the line $\overline{AB}$ and $A'$ and $B'$ are on the other side of the line $\overline{AB}$, such that, $\angle OAA'=\angle OBB'=90$ and that $BB'=OB$ and $AA'=OA$. Show that the midpoint of $\overline{A'B'}$ never changes when $O$ changes position.

Its wrong.

Take A = (0,0) B = (2,0)

Try O = (1,-2).  Then A', B' = (0,1), (2,1) which has midpoint (1,1).

Try O = (3,-1).  Then A', B' = (1,3), (3,1), which has midpoint (2,2).

However, its easy to show when the midpoints are the same depending on O using vectors/complex # and considering cases.

Ahh, very good point, I forgot to consider when the $x$ coordinate of point $O$ is greater than or less than that of $A$ or $B$. So how about prove the previous fact for when the $x$ coordinate of $O$ is between that of $A$ and $B$. Assuming that line $\overline{AB}$ is parallel to the $x$ axis, of course.
It can be done purely geometrically, without vectors or coordinate, etc.


invariant
geometry
geometric transformation
reflection
geometry proposed

let (X) be a circle, P a point outside the circle, Q on the circle. The pencil of lines through P intersect (X) in points B and C, H is the orthocentre of QBC. Prove as H varies, (i) there exists a point T s.t. HQ^2 + HT^2 is invariant of H (ii) such a point is unique.

(i).

As shown in part (i) of [url=http://www.mathlinks.ro/viewtopic.php?t=6336]the thread "some hard problems"[/url], the locus of the point H is a circle. Let S be the center of this circle; then, the length HS is a constant. Now, if T is the reflection of the point Q in the point S, then the point S is the midpoint of the segment QT. Therefore, the segment HS is a median in triangle QHT; hence, after the formula for the length of a median in a triangle, we have

$4\cdot HS^{2}=2\cdot HQ^{2}+2\cdot HT^{2}-QT^{2}$.

Now, since both distances HS and QT are fixed, the sum $HQ^{2}+HT^{2}$ must also be fixed. Hence, we have found one point T such that the sum $HQ^{2}+HT^{2}$ is fixed.

(ii).

Now we will show that this point T is the only point with this property. In fact, assume that we have another point T' such that the sum $HQ^{2}+HT^{\prime }\;^{2}$ is fixed. We want to show that T' = T.

Well, consider the midpoint S' of the segment QT', and apply the formula for the length of a median again:

$4\cdot HS^{\prime }\;^{2}=2\cdot HQ^{2}+2\cdot HT^{\prime }\;^{2}-QT^{\prime }\;^{2}$.

Now, this entails that HS' must be an invariant, i. e. the point H must lie on a fixed circle with center S'. But this implies S' = S, and thus T' = T. Proof complete.

  Darij

we can also prove (i) more easily :D (without any calculations) using the fact that the reflection of the orthocentre in each of the sides lies on the circumcircle. Consider the point Q' diametrically opposite Q. Let the reflection of H in BC be H'; the line through H' parellel to BC obviously passes through Q'. Let T be the point on q'P extended s.t. Q'P=PT. Now obviously the lines through H parellel to BC, perpendicular QH pass through T. Thus QH^2 + HT^2 = QT^2, invariant of H.
we can also prove (ii) like this: suppose 2 such points, T and T' existed s.t. HQ^2 + HT^2 is invariant of H. Then it follows the locus of H is 2 lines perpendicular to TT', which contradicts the fact that the locus of H is a circle. your proof of (ii) was very nice though.

vinoth


Name: AoPSWiki Problem Uploader 1.0
URL: http://mathideas.org/public/scripts/AoPSWiki
Description: Constructs the code to place on all pages related to a national olympiad problem page, given parameters like name of olympiad, # of problems, what the problems are, etc.

It has been tested a little but there could be bugs still. I originally wanted to automatically edit the wiki and include the files, but I would need certain information from the admins to do that. There are a few things that could be better, like if the divisions are grades, then it starts from grade 1. :D

If anyone would like to improve the source code, they can PM me.

I hope this little script is helpful to people other than me. :roll:

VERY NICE!  I'm sure that will be helpful for a lot of people.

Oh, do you think that there is any way that you could also make it so that there is a "previous problem"/"next problem" link on each of the pages like [url=http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_2]here[/url]. ?  For the first problem, you'd only want "Next Problem" and for the last you'd only want "Previous Problem", of course.

Great job.  Thanks.

[quote="joml88"]Oh, do you think that there is any way that you could also make it so that there is a "previous problem"/"next problem" link on each of the pages like [url=http://www.artofproblemsolving.com/Wiki/index.php/1983_AIME_Problems/Problem_2]here[/url]. ?  For the first problem, you'd only want "Next Problem" and for the last you'd only want "Previous Problem", of course.[/quote]

Good idea. The suggestion has been implemented. :lol:

awupu2 (AopsWiki Unofficial Problem Uploader 2.0)

it has some cool new features, like automatically linking to other pages in the wiki. :D

Nice!!

However, it does generate code even if you don't put anything in...


I use xemacs for editing. It's pretty nice, but not as nice as winEDT.

Thing is, winEDT is shareware and expires, while xemacs is free (I think. I got a copy w/ another program I know emacs is free, pretty much the same thing).

What does everyone else use?

[url=http://www.texniccenter.org/front_content.php?idcat=26&lang=4&client=3]TeXnicCenter[/url] is free (GPL), easy to use and looks good

TeXnic is wonderful, I haven't had any problems with it at all.

For TeXnicCentre, is there a quick way to clean all the output files (e.g. .aux, etc)? For WinEdt, there's a neat "recycling bin" button to do that.

Click "Clean Project" in the Build menu.

but "Clean Project" is always disabled (in gray)

I think TeXnicCenter is one of those programs where you must have an actual "project" (.tcp file) to do things like that.


[quote="236factorial"]how do you prove the midpoint theorem, the part that says a line connecting two midpoints of two sides of a triangle is parallel to the base?[/quote]
Let the mid-points of $AB$ and $AC$ be $M$ and $N$. Then it is easy to show that $\triangle{BMC}=\triangle{BNC}$. Thus $MN || BC$.

Yeah, or you can use analytic geometry, like in the AoPS books.

I was thinking about drawing a line going through the top vertex of the triangle, apply parallel lines to the base and the drawn line, then using the parallel postulate.

use the midpoints to get similiar triangles by SAS, then use congruent angles and if corresponding angles congruent, then lines parallel


function
quadratics
geometry
3D geometry
algebra
polynomial

I'm rather curious about the reciprocals of equations.
Say we have the general formula for a quadratic:

$Ax^2+Bx+C=0$

How would we find the sum of the squares of the reciprocals?
How would we find the sum of the cubes of the reciprocals?
How would we do the same for a cubic? or a quartic?

If anyone could answer my questions, I'd greatly appreciate it.

for the quadratic $Ax^2+ Bx + C= 0$, with roots $a,b$
the quadratic $Cx^2 + Bx + A = 0$ has roots $\displaystyle\frac {1}{a}, \frac {1}{b}$
so in general reverseing the coefficients of a polynomial will give you a new polynomial with reciprocal roots

so to find the square, cube, and quadratic, you could use newtons sums

newtons sums is like... $s_1 = \displaystyle\frac {1}{a}+ \frac {1}{b}$, $s_2 = \displaystyle(\frac {1}{a})^2 +(\frac {1}{b})^2$, $s_3 = ... $

$(C)(s_1) + A = 0$
$(C)(s_2) + B(s_1) + 2A = 0$
$(C)(s_3) + B(s_2) + A(s_1) = 0$
$(C)(s_4) + B(s_3) + A(s_2) = 0$

and i think those should be right

hope that helps

I do not know Newton's sum, so I chosed another way.
$a^2+b^2=(a+b)^2-2ab$
$a^3+b^3=(a+b)(a^2-ab+b^2)$

Ahh yes, I just learned Newton's sums. Thanks for the help. I'll probably have another few questions in a while but that seems to be all for now. I'm supposing that Newton's sums will work for all polynomials.

If you want, say the sum of the squares of the roots then we can say $x_i^2 = r_i$ which is equivalent to $x_i = \sqrt{r_i}$. Since $f(x) = 0$ is the equation for the roots, $f(x_i) = 0$ is true since we defined $x_i$ to be the roots. From that We have $0 = f(x_i) = f(\sqrt{r_i})$. Thus, we can apply viete's relations on our new polynomial in $r$, though we might have to do a little squaring with it at first. But, we wanted $\displaystyle\sum x_i^2 = S$ but we have, by our equality $S = \displaystyle\sum x_i^2 = \displaystyle\sum r_i$.
We can generalize to say if we want the sum of $g(x)$ for $x$ the solutions of $f(x) = 0$ then say $g(x_i) = r_i \Rightarrow x_i = g^{-1}(r_i)$. So, this tells us $0 = f(x_i) = f(g^{-1}(r_i))$ and we can use viete's relations on the polynomial in $r$.
A quick example. Find the sum of the reciprocals of 2 less than the roots of $f(x) = x^2 - 7x + 12 = (x-3)(x-4)$. Without the method, the roots are $x = 3,4$ and we get $S = \frac{1}{3-2} + \frac{1}{4-2} = \frac{3}{2}$. By our method, we have $g(x) = \frac{1}{x-2} \Rightarrow g^{-1}(x) = \frac{1}{x} + 2$. So, we have $\frac{1}{r_i} + 2 = x_i$ Plugging in $0 = f(\frac{1}{r_i} + 2) = (\frac{1}{r_i} + 2)^2 - 7(\frac{1}{r_i} + 2) + 12$

$= (\frac{1}{r^2} + \frac{4}{r} + 4) - 7(\frac{1}{r} + 2) + 12$

$\Rightarrow 0r^2 = 0 = 1 + 4r + 4r^2 - 7r - 14r^2 + 12r^2 = 2r^2 - 3r + 1$

Now, by Viete's relations, the sum of the solutions of $r$ is $-\frac{-3}{2} = \frac{3}{2} = S$.

P.S. I realize the example didn't need the method.

I have another question

Given the polynomial: $Ax^3+Bx^2+Cx+D=0$ and its roots are [b][i]a[/i][/b], [b][i]b[/i][/b], and [b][i]c[/i][/b] find $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}$

Also find $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}$

Say$\frac{1}{a+b} = \frac{1}{(a+b+c)-c} = \frac{1}{-\frac{B}{A}-c}$ So, we want $\displaystyle\sum_{cyc} f(a)$ where $f(a) = \frac{1}{-\frac{B}{A}-a}$. Apply the same to the second but instead we have $\frac{1}{ab} = \frac{c}{abc} = \frac{c}{\frac{-C}{A}}$.


An object is propelled upward with an initial velocity of 20 meters per second.  The distance s (in meters) of the object from the ground after t seconds is s = -4.9t^2 + 20t.

(a) When will the object be 15 meters above the ground?

(b) When will the object hit the ground?

(c) Will the object reach the height of 100 meters?

see the other post that you posted:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127578[/url]

a) Sub in $s=15$

$15=-4.9t^{2}+20t$
$4.9t^{2}-20t+15=0$
$t=\frac{20\pm\sqrt{400-294}}{9.8}$
$t=\frac{20\pm\sqrt{106}}{9.8}$

b) Sub in $s=0$

$0=-4.9t^{2}+20t$
$-4.9t\left(t-\frac{200}{49}\right)=0$
$t=0,\frac{200}{49}$

However $t=0$ shows tells us that the ball has been thrown up at that point so $t=\boxed{\frac{200}{49}}$

c) sub in $s=100$ (or see other post)

$100=-4.9t^{2}+20t$
$4.9t^{2}-20t+100=0$
discriminant$\implies(400-1960)<0 \therefore\boxed{NO}$ It is not possible for the ball to reach a height of 100 m.


calculus
derivative
geometry solved
geometry

A geom problem:
Given P in the plane a and Q outside the plane a find R in the plane a s.t. the value (PQ+PR)/QR is maximized.

Let H be the plane projection of Q. Let us show that optimal R is situated on line PH (P<>H).
Otherwise consider circle with center P and radius PR. For any point R' of this circle sum PQ+PR'=PQ+PR, and it is clear QR' is minimal if R' is situated on line PH ==> if QR' is minimal then R<>R' and
(PQ+PR')/QR'>(PQ+PR)/QR.

Let PQ=q, \pi/2 \geq angle QPH=angle QPR=a, R is situated on PH, PR=x \geq 0.
(PQ+PR)2/QR2=1+2qx.cos a/(x2+q2-2qx.cos a) ==> derivative is
(q2-x2)/QR2  ==>  x_opt=q .

Other case is angle QPH=\pi-angle QPR=a ==> we obtain R_opt' with x_opt'=q ==> QR_opt'>QR_opt and PQ+PR_opt=PQ+PR_opt'=2q ==> true optimal point is R with x_opt=x.

``````````Q
`````````/|
````````/ |
`._____/__|__.
`Ropt' P  H  Ropt

Thx, that's more or less what I did. I think this was proposed for some IMO, or sth like that. I had it written on some old notebook of mine.


Se considera sirul $(x_{n})$ definit prin $a,x_{1}>0$ si $x_{n+1}=x_{n}+\frac{1}{x_{n}^a}$, oricare ar fi n>0.
Sa se arate $\mathop {\lim }\limits_{n \to \infty }x_{n}= \infty$,si calculati $\mathop {\lim }\limits_{x \to \infty } \frac{1}{n}x_{n}^{1+a} $

Ai nevoie de ajutor sau o propui spre rezolvare? Limita aia e $1+a$.

[quote="silviu"]Se considera sirul $(x_{n})$ definit prin $a,x_{1}>0$ si $x_{n+1}=x_{n}+\frac{1}{x_{n}^a}$[/quote]

Consider the continuous analogue, $y'(t) = \frac{1}{y(t)^a}$.
This gives an idea of the asymptotic behavior.
Now consider $z = x^{a+1}$ and show that $z_{n+1} = z_n + (a+1) + o(1)$.

[quote="silviu"]Grobber as dori sa-mi explici cum ai ajuns la acest rezultat (este corect :) )[/quote]

Scuze, silviu, nu m-am mai uitat la topic-ul asta si nu am vazut intrebarea.

Am incercat sa folosesc Cesaro-Stolz, si deci trebuia sa calculez limita $x_{n+1}^{a+1}-x_n^{a+1}$. Am scris $x_{n+1}$ in functie de $x_n$ (cu relatia de recurenta), si am obtinut o expresie doar in $x_n$. Dupa aceea am calculat limita functiei respective la infinit (pt ca $x_n\to\infty$). Cam asta ar fi.


Find the maximum possible value of $ 8\cdot27^{\log_6x}\plus{}27\cdot8^{\log_6x}\minus{}x^3$ as $ x$ varies over the positive real numbers.

Let $ y = \log_6 x$. When $ x\in (0,\infty)$ we have $ y\in ( - \infty,\infty)$. We have
\begin{align*}f(x) & = 8\cdot 27^y + 27\cdot 8^y - 216^y = 8\cdot 27^y + 8^y\cdot 27 - 8^y\cdot 27^y \\
& = - (27^y - 27)(8^y - 8) + 216.\end{align*}
Note that when $ 27^y < 27$ it follows that $ y < 1$ so $ 8^y < 8$ as well. Therefore $ - (27^y - 27)(8^y - 8)\leq 0$ with equality if and only if $ y = 1$, or when $ x = 6$, and therefore $ \max f = \boxed{216}$.


[quote="dgs01"]If I take AMC 12A at my high school and barely qualify for AIME with it and also take AMC 10B at another place and got a very high score, and I take AIME at my high school, then will the AMC 10 score be used for USAMo index?

Sorry if this sounds stupid but I was worried if the system that calculates the index scores views my AMC scores as done by 2 separate ppl since I am taking them at 2 different places.[/quote]

i am pretty sure that they will use your AMC 10B score.

Ya your highest score will be used to calculate your index.

However, if you took it at two locations, they might not be able to match up your scores accordingly (they will probably then use your AMC score for the place you took your AIME).

However, dont worry, you can then contact the AMC (if you do happen to surpass the index) and tell them about another location that you took an AMC.

I know this because it happened to one of my friends last year.  He took the 10A and got a 138 at the same school (the middle school) where he took the AIME (where he got a 6).  However, he also took the 10B at my school (the high school) and he got a 144 there.  He should have qualified because 144+6 was above the <10th grade index, but his name didnt initially show up.  So he contacted the AMC offices and after he gave them the CEEB code for the high school, they were able to match up his scores and he ended up making it.


Show that $ \sqrt[n]{n}<1\plus{}\frac{1}{\sqrt{n}}$ where $ n$ is positive integer $ \geq 2$

let x>0,
(1+x)^n = 1 + n + ...
hence
(1+x)^n > n
 taking nth root you get what you wanted

[quote="tohach"]let x>0,
(1+x)^n = 1 + n + ...
[/quote]

I think there is an error in here

[hide="Solution"]

If $ n \geq 3$, we can expand as a binomial and ignore all but the first few terms to get

$ (1\plus{}\frac{1}{\sqrt{n}})^n \geq 1 \plus{} \sqrt{n} \plus{} \frac{n\minus{}1}{2} \plus{} \frac{(n\minus{}1)(n\minus{}2)}{6\sqrt{n}}$.

We're supposed to show that this is more than $ n$. The last term behaves like $ n^{3/2}/6$ which is obviously more than $ n$ when $ n$ is large enough, so we're on the right track - we just need to check the equality holds starting $ n \geq 2$.

So note that

$ 1 \plus{} \sqrt{n} \plus{} \frac{n\minus{}1}{2} \plus{} \frac{(n\minus{}1)(n\minus{}2)}{6\sqrt{n}} > n$

is equivalent to

$ \sqrt{n} \plus{} \frac{(n\minus{}1)(n\minus{}2)}{6\sqrt{n}} > \frac{n\minus{}1}{2}$

or

$ 6n \plus{} (n\minus{}1)(n\minus{}2) > 3\sqrt{n}(n\minus{}1)$

or

$ (n\plus{}1)(n\plus{}2) > 3\sqrt{n}(n\minus{}1)$.

In fact, the stronger inequality

$ (n\plus{}1)(n\plus{}2) > 3\sqrt{n}(n\plus{}1)$

holds for $ n \geq 5$ since it is just $ n\plus{}2> 3\sqrt{n}$. For $ n \equal{} 2, 3, 4$ the original inequality is easily checked as well.

[/hide]

[b]randomgraph[/b] has got a solution by using Newton binomail factor. Yesterday, my friend give me a hint that the above inequality can sovle by applying Cacuhy's inequality. What do you think about?

I think the problem can solve by apllying Cauchy' ine where the problem has edited as follow: $ \sqrt[n]{n}<1\plus{}\frac{2}{\sqrt{n}}$
In fact:
$ \sqrt[n]{n}\equal{}\sqrt[n]{1.1...1.\sqrt{n}.\sqrt{n}}<\frac{n\minus{}1\plus{}2\sqrt{n}}{n}<\frac{n\plus{}2\sqrt{n}}{n}\equal{}1\plus{}\frac{2}{\sqrt{n}}$


Find the real numbers a and b:

$ 2^{\frac 1a}\plus{}16^a\plus{}2^{\frac 1b}\plus{}16^b\equal{}8$ .

[hide="Solution?"]There are no real solutions. We must have $ 2^{1/a} \plus{} 2^{4a} \le 4$, $ 2^{4a} \le 4$ when $ a \le \frac{1}{2}$, whereas we have $ 2^{1/a} \le 4$ when $ a \ge \frac{1}{2}$. $ a \equal{} \frac{1}{2}$ yields a sum of $ 8$, an impossibility. Therefore, no real solutions exist.[/hide]

darkdieuguerre, your answer is correct but your solution is lacking. You seem to have forgotten to consider the negative numbers.

[hide="Correct proof:"]Assume that at least one of $ a,b$, say $ a$, is positive. Then we note that by AM-GM, $ 2^{\frac {1}{a}} \plus{} 2^{4a} \ge 2^{2a \plus{} 1 \plus{} \frac {1}{2a}} \ge 2^3 \equal{} 8.$ Since all terms of the sum are nonnegative, we have a contradiction.

However, if both $ a$ and $ b$ are negative, each term of the sum is less than $ 1$, meaning the total is less than $ 4$ and will never reach $ 8$. Therefore, there are no real solutions.

We are done.[/hide]


There are $ n$ men in a company.It is known that each of them knows less than $ 5$ men among the others.And if two men don't know each other,then there exist a man,who knows both of them.Find the maximum value of $ n$.

The maximum is 17.

Any man knows at most four other people, each of whom know at most three other people, so there are at most 17 people who (i) are the given man, (ii) know the given man or (iii) are known by someone who knows the given man. Clearly the conditions of the question require that these 17 must include all of the people. On the other hand, it is not hard to construct an incidence graph among the four groups of three people who are two degrees away from the original man which will ensure that the separation condition holds.

[quote="Diarmuid"]The maximum is 17.[/quote]
Your answer is incorrect :)
[quote="Diarmuid"] On the other hand, it is not hard to construct an incidence graph among the four groups of three people who are two degrees away from the original man which will ensure that the separation condition holds.[/quote]
Try to do it,and you will understand that this problem is not so simple. :wink:


geometry
geometric transformation
rotation

Hello!

I have what I believe is a rather simple math question.  I just want to make sure I'm getting the right answer.

Let's say I have a checkerboard as below, which is 5 x 5 places

X X X X X
X X X X X
X X X X X
X X X X X
X X X X X

For each of the places, the place can be a value between 1 and 10

For Example:

X 9 X X X
X X X 8 X
X X X 4 X
7 X X X X
X 1 X X X

X X X X X
X 2 X X X
X X X 3 X
X X X X X
X X X X X

9 9 9 9 9
9 9 9 9 9
9 9 9 9 9
9 9 9 9 9
9 9 9 9 9

How many different combinations can I have?  Could you please explain how to do this?

Thanks!

Well, for each square there are 10 choices (X, or 1-9), and there are 25 squares.  So in total there are $10^{25}$ combinations, a very large number.  Your description of the problem is a bit ambigous.  Do you mean all integers between 1 and 10, with 10 included?  If so then there $11^{25}$ possibilities.  Or maybe you count two arrangements that can be changed into each other w/ a rotation to be identical.

there are 10 choises to choose from

25 spaces

so it is 10*10*10.... 25 times=10^25


How do you have half a chicken, or lay half an egg?

[hide="Disreguarding that"]This means that 1 chicken can lay 1 egg in $\frac32$ days.  Thus, 3 chickens will lay 3 eggs in $\frac32$ days, so in 2 days, they can lay $3\times\frac{2}{\frac32}=4$ eggs.[/hide]

[hide]The constant of proportionality is

1.5*1.5/1.5=1.5

So we just need to figure of the value of a, such that

2*3/a=1.5, this is when a=4

$4$[/hide]

[quote="Andrew619"]If a chicken and a half can lay an egg and a half in a day and a half, how many eggs can three chickens lay in two days?[/quote]

The trick could be that half a chicken can't lay an egg so doesn't affect the question (saw this type of thing in a logic puzzle book once), but the half of an egg kinda messes things up.

You said:
$1.5 C$ can lay $1.5 E$ in $1.5 D$.
So,
$3 C$ can lay $3 E$ in $1.5 D$
Also,
$3 C$ can lay $4 E$ in $2 D$

So our answer is $4$ eggs.

Is that even possible to have half a chicken? :huh:\

If its possible then the solution would be...
C=chickens

1.5C takes 1.5 days to lay 1.5 eggs
so
3C takes 1.5 days to lay 3 eggs
so
3C in 2/3 of the time can lay 2 eggs
meaning they can lay 4 eggs in 2 days

I don't think cutting edge technology has gotten to that yet :D , but perhaps the problem was stated that way to confuse the solver. Or, maybe a half of a chicken is just a mutant chicken that is twice as small as normal :lol: . Same with the egg. Or a one and a half chicken could be a chicken that is $3/2$ times larger than a normal chicken... :lol:

But since a chicken and a half laying an egg and a half is equivalent to a chicken laying an egg (it takes the same time), that idiosyncrasy doesn't really affect the problem.

[hide]There are double the amount of chickens and 4/3 the amount of days.

$\frac{3}{2}\cdot 2 \cdot \frac{4}{3}=\boxed{4}$[/hide]


Mafia
FTW
AMC
AIME
geometry
geometric transformation
Beat the Mod

Hi, and welcome to the game of [b]Beat the Mod![/b]
In this game, you will be pitted against each other, and of course, me! How am I in this game, you ask? Well, in this game, there will be several rounds in which each participant will be competing for their lives! Whoever gets through all the rounds will be the winner. This, of course, allows for multiple winners. With some skill, good allies, and a bit of luck, you could be in the winners’ circle! But in the event that no one survives through all rounds, I will be crowned winner! :D
And no, don’t think you can form an unilateral alliance; for most rounds, I will set a minimum amount of people that [i]must[/i] be eliminated. If this minimum is not met, I will automatically be crowned winner! (Don't worry, this probably won't happen.)

And now, to set some rules (It will help immensely if you read these):
1. Communication, both inside and outside the thread, is allowed UNLESS I specifically say it is not.
2. Making friends is encouraged, and possibly necessary to get far in this game. However, I would appreciate it if you told me what groups you are in. That way, I can use it for a final post. In fact, I really would appreciate it. Please? Just a link to a QT...
3. Just because I am the mod and have powers over this game doesn’t mean I will be actively trying to win this game. Sure, I might [i]lie[/i] to you, but I will reveal any twists when the time is right.  Simply know that I won’t be rigging luck-based games... I swear!
4. When you join this game, you take a secret oath of silence. Don’t discuss this with anyone out of the game. This applies after you are eliminated.
5. Have a conscience... if you think you’re going to do something that I wouldn’t allow. Don’t. Don’t flame. And please, please don’t spam... I want this game to be able to stay out of the Games Factory.
6. This game will use many ideas from other games. Don’t complain, or I might eliminate you.
7. If the game slows down, I will set deadlines, replace, and/or modkill. Keep the game alive!
(8. Just because I write something in parentheses doesn’t make it any less of a rule. Look out.)
9. Please consider that please might mean a little more than please...

Signups are now closed!

Players:
1. pacman2812
2. chenhsi
3. dragon96
4. cf249
5. james4l
6. pythag011
7. ProtestanT
8. perfect628
9. kimmystar94
10. jjx1
11. 1=2
12. BOGTRO
13. AIMEman
14. crazychinesemaniac
15. zapi2007
16. (^_^)
17. budi713
18. mewto55555
19. mathelete9999
20. alexhhmun

ummm...

This reminds me of the Beat the Mod game on MafiaScum...

/in

Hmm... I dunno if I've seen that. Well, this will probably be different.
(By the way, my posts and sign-in posts may just affect the game.... be careful....)
(Yes that was a hint.)

Oh, and this game will be a series of rounds... like I said. People will go away every round... (I want more people so I can add more rounds!)

Sounds nice and confusing.   :ninja:
I'm in.

I'm in. Message is too small pleas shoot me.

I hate not being able to edit...
When the game starts, I'll ask Klebian to edit the first post...
Signups:
1. pacman2812
2. chenhsi
3. dragon96
4. cf249
5. james4l
6. pythag011
7. ProtestanT
8. perfect628
9. kimmystar94
10. jjx1
11. 1=2
12. BOGTRO
13. AIMEman
14. crazychinesemaniac
15. zapi2007
16. (^_^)
17. budi713
18. mewto55555

Maybe you should end sign-ups now as it seems no one else is joining.

Btw, this game is between people who are in the game, not including me. I just post first image.

umm, how do you post pictures on forum??

[img]http://judykitsune.files.wordpress.com/2008/06/14677scissors.jpg[/img]

CUT THA RED WIRE!

[img]http://www.geoscanners.com/images/rock_solid_solutions_small.jpg[/img]

Rock beats scissors :P

Bump.

You know, if you post a picture with the word "gullible" in it, you win a special prize and the round ends with no deaths!

Of course, I MIGHT be lying...


Prove:

$ \sin^2{(A\plus{}B)} \plus{}\sin^2{(B\plus{}C)} \plus{}\sin^2{(A\plus{}C)} \ge \sin{2A}\sin{2B} \plus{}\sin{2A}\sin{2C} \plus{}\sin{2B}\sin{2C}$

This may have been posted before but I discovered it just now.

[hide]
By AM-GM, $ \sin^2 (A\plus{}B) \equal{} (\sin A \cos B \plus{} \sin B \cos A)^2$ $ \ge 4\sin A \cos A \sin B \cos B \equal{} \sin 2A \sin 2B$, with equality iff $ \sin A \cos B \equal{} \sin B \cos A \Longrightarrow \sin (A\minus{}B) \equal{} 0$.
[/hide]


[hide]$4x+45-x^{2}=33\\x^{2}-4x-12=0$, so $x=6,-2$ we only want a positive number, so $x=6$ and the perimeter is $2\times11+2\times3=28$[/hide]

[quote]A rectangle has sides of $x+5$ and $9-x$. Its area is 33. What is its perimeter?[/quote]

[hide="Very slick solution"]Avoiding all quadratic calculations, the perimeter is $2(x+5+9-x)=2(14)=\boxed{28}$.[/hide]

[quote]A rectangle has sides of $x+5$ and $9-x$. Its area is 33. What is its perimeter?[/quote]

[hide]
9x-x^2+45-5x=33
x^2-4x-12=0
(x-6)(x+2)=0
x=6, -2
only 6 will work IN THIS case

11+3+11+3=28

[/hide]


a)Let $ P(x)\in\mathbb{Z}[x]$. Prove that there exists an infinite number of integers $ t$ such that $ P(t),P(t+1),\cdot\cdot\cdot, P(t+m)$ are all composite for any given positive integer $ m$.
b)Let $ P(x)=x^{2}+1$. Find a value of $ t$ satisfying part a) for $ m=5$.

See my post in http://www.mathlinks.ro/Forum/viewtopic.php?t=158042 ...


Let $ABCD$ be a parallelogram with $\angle ABC=72^{\circ}$, $E$ on $AB$ and $F$ on $CD$ such that $\frac{AE}{EB}=3$ and $\frac{CF}{FD}=4$. $CE$ intersects $BF$ at $X$. Find the ratio of the area of $\triangle BXC$ to the area of $ABCD$.

Note: this is pretty easy, but see if you can find the most elegant solution, it should only take a line or two if you see it!!

[quote="Ahduhduh"]Is it 2/21?[/quote]

i did not find a solution, i just made up the numbers...giving just an answer is MEANINGLESS, the process is all that matters...to get across my point let $\frac{AE}{EB}=m$ and $\frac{CF}{FD}=n$, find the ratio in terms of $m$ and $n$

[hide="Here is a easy generalization."]
We have no further use for $A=72^{\circ}\ !$ [u]Indication:[/u] $[BXC]^2=[BXE]\cdot [CXF]\ .$

[b]Generally,[/b] $\boxed {\frac{EA}{EB}=m\ ,\ \frac{FC}{FD}=n\Longrightarrow [BXC]=\frac{n}{2(mn+2n+1)}\cdot [ABCD]}$
[b]Remark[/b] $1.\blacktriangleright\frac 12\cdot [ABCD]=[ABC]=\left(\frac{BA}{BE}+\frac{CD}{CF}\right)\cdot [BXC]\ .$
[b]Remark[/b] $2.\blacktriangleright m=3\ ,\ n=4\Longrightarrow [BXC]=\frac{2}{21}\cdot [ABCD]\ .$[/hide]

firstly connect AC, since AE:EB=3:1, so area of EBC: AEC=3:1. since AC divide ABCD into 2 equal parts, so area of EBC=1/4 area of ABC=1/8 area of ABCD.
∵FC:DF=4:1, FC=4/5 CD=4/5 AB and EB=1/4 AB
since AB‖CD, so EBX∽FCX, then we have BX:FX=5:16
so ratio of the area of BXC and BCF= 5:21,
connect BD, similarly area of BCF=2/5 ABCD, so that 6BXC=5/21 ABCD, BXC=2/21 ABCD
the answer is 2/21


inequalities
logarithms
calculus
derivative
inequalities proposed

Prove that if $x\geq0$ then
\[\sqrt{\frac2{1+x}}\leq 1-\frac14\ln x.\]

[quote="pvthuan"]Prove that if $x\geq0$ then
\[\sqrt{\frac2{1+x}}\leq 1-\frac14\ln x. \]
[/quote]
If $x=0$ it's wrong! :D

purpose : $x > 0$ then  $x \to  \infty$ we have :
$1-1/4lnx \to -\infty$ , other hand we have : $\sqrt{\frac{2}{1+x}}> 0$ forall $x > 0$

sorry I made a mistake when I calculated the derivative of the function.

Anyway, I am looking for inequalities with $\ln$. Can you help?

[quote="pvthuan"]

Anyway, I am looking for inequalities with $\ln$. Can you help?[/quote]
For you, pvthuan. My very old inequality.
Let $x^{y}=y^{x},$ $x\neq y$ and $x>1,$ $y>1.$ Prove that $x^{y}>e^{e}.$


Μπορεί κανείς να αποδείξει [u]αυστηρά[/u] οτι το κενό σύνολο είναι υποσύνολο κάθε συνόλου :?:

Έστω $ A \neq \oslash$ τυχαίο σύνολο τότε είναι
$ A \cup \oslash \equal{} A \neq \oslash \Leftrightarrow \oslash \subset A$

Wow, παίζει να είναι και από τις πιο βαθειές και δύσκολες ερωτήσεις που έχω δει στο forum...

Anyway guys, συγγνώμη για την παρέμβαση, αλλά μιας και μιλάμε για θεωρία συνόλων, είναι απαραίτητο να είστε αρκετά πιο προσεκτικοί στις διατυπώσεις σας (τόσο ως προς την ερώτηση του Dimitris T όσο και ως προς την απάντηση του maniopas).

Συγκεκριμένα, maniopas, η απάντηση σου αποτελείται από μία και μόνο ισοδυναμία η οποία είναι τόσο διαισθητικά προφανής όσο και η προς απόδειξη πρόταση. Δε λέω, "εκ του σχήματος (ίσως και να) φαίνεται" μια χαρά, αλλα πώς δικαιολογείται η ισοδυναμία αυτή? Στη θεωρία συνόλων (δυστυχώς ή ευτυχώς), οι αποδείξεις που είναι διαισθητικά προφανείς πρέπει να αποφεύγονται διά ροπάλου και αυτός είναι ο λόγος που προκύπτουν παράδοξα όπως αυτό του Russell με το σύνολο όλων των συνόλων κλπ, κλπ :( Αν μιλάμε για [i]αυστηρές[/i] αποδείξεις, θα πρέπει να κάνεις πολλή δουλειά...

Τώρα, ως προς την ερώτηση αυτή καθαυτή, Dimitris T, τι εννοείς με το κενό σύνολο? Πώς το ορίζεις? Τι σύνολο αξιωμάτων χρησιμοποιείς? Όλα αυτά είναι ερωτήματα που πρέπει να απαντηθούν με πολύ ακριβή τρόπο, πολύ πριν πάμε να απαντήσουμε μία τέτοια ερώτηση...

Στην "standard" θεωρία συνόλων (ας πούμε σκέτη ZF (Zermelo-Frankel)) η ύπαρξη του κενού συνόλου εξασφαλίζεται από το αξίωμα (μου φαίνεται είναι το δεύτερο? αλλά δε θυμάμαι και καλά):
\[ \exists \varnothing: \forall y, \neg(y\in \varnothing)
\]
(το $ \neg$ σημαίνει άρνηση της πρότασης που ακολουθεί, δηλ. ότι είναι ψευδής; σε πιο απλή γλώσσα: $ \exists\varnothing: \forall y, y\notin\varnothing$.)

Δηλαδή, το κενό σύνολο "ορίζεται" ως το σύνολο εκείνο που είναι τέτοιο ώστε κανένα σύνολο να μην ανήκει σε αυτό. Ήδη όμως, σε αυτό το σημείο "κλέβουμε" επειδή, για να οριστεί μονοσήμαντα το κενό σύνολο, πρέπει να δείξουμε ότι είναι και [i]μοναδικό[/i] ως προς την παραπάνω ιδιότητα. Επειδή δε νομίζω να ενδιαφέρεται κανείς για τόση λεπτομέρεια, ας δεχτούμε ότι αυτό προκύπτει από άλλα αξιώματα της ZF και, συγκεκριμένα, από το θεμελιώδες αξίωμα της επεκτασιμότητας (μου φαίνεται είναι το πρώτο? αλλά δεν παίρνω και όρκο) το οποίο πρακτικά λέει ότι ένα σύνολο καθορίζεται πλήρως από τα στοιχεία του: $ x \equal{} y\Leftrightarrow(z\in x\Leftrightarrow z\in y)$.

Όπως καταλαβαίνεις, το μπέρδεμα είναι πολύ μεγάλο και για να έχεις μία πραγματικά αυστηρή απόδειξη (δηλ. σε αξιωματικό επίπεδο), θα πρέπει να γραφούν πολλά πράγματα (heh, κοτζάμ υπαρξιακή κρίση πέρασαν τα μαθηματικά για 60 χρόνια εξαιτίας τέτοιων ερωτήσεων, δεν μπορώ ούτε καν να ξύσω την επιφάνεια σε ένα forum post).

Πάντως, η βασική ιδέα της απόδειξης είναι η εξής ([b]όχι, δεν είναι αυστηρά τα παρακάτω[/b]... αλλά δεν έχω και πολύ χρόνο ;) ):
1.  Πρέπει να δείξουμε $ \forall x$ ότι $ \forall z\in\varnothing\implies z\in x$ (*). Σε απλούστερη γλώσσα, [color=darkred]πρέπει κάθε στοιχείο του $ \varnothing$ να είναι και στοιχείο του συνόλου $ x$[/color].
2.  Θα δείξουμε την πρόταση προς απόδειξη με εις άτοπο απαγωγή (sorry, δεν αντέχω να γράψω για τη θεμελίωση της απαγωγής σε άτοπο αν και κανονικά θα έπρεπε :( αν σε πειράζει πολύ, να σου πω μόνο ότι δε χρειάζεται πραγματικά, απλώς διευκολύνει το γράψιμο...). Anyway, η αντίθετη πρόταση για δεδομένο $ x$ είναι ότι $ \exists z\in\varnothing: z\notin x$. Δηλαδή, σε πιο απλά "ελληνικά", [color=darkred]έστω ότι υπάρχει στοιχείο $ z$ του κενού συνόλου ώστε το $ z$ να μην ανήκει στο σύνολο $ x$[/color].
3.  Ωστόσο, από το αξίωμα του κενού συνόλου έχουμε ότι $ \forall z: z\notin\varnothing$, άρα η αντίθεση της (*) είναι τετριμμένα ψευδής. Δηλαδή, [color=darkred]από τη στιγμή που το $ \varnothing$ δεν έχει στοιχεία, δε θα έχει και στοιχεία που να ικανοποιούν μία συγκεκριμένη ιδιότητα, πχ την αντίθεση της (*)[/color].
4.  Συνεπώς, $ \varnothing\subseteq x$, QED.

Αυτά ως αρχή ενός ψήγματος μίας ανακριβούς απάντησης... μία πραγματικά αυστηρή απόδειξη θέλει πολλή δουλειά για να γραφτεί :( Τι να πώ, έτσι είναι η θεωρία συνόλων: γράφεις 5 σελίδες για να αποδείξεις κάτι που είναι διαισθητικά προφανές και στο τέλος είσαι πιο γεμάτος από ποτέ με αμφιβολίες για το αν και κατά πόσο αυτά που γράφεις είναι σωστά και έχουν νόημα... και συνήθως καταλήγεις να αμφισβητείς το αν και κατά πόσο η αρχική σου ερώτηση έχει η ίδια νόημα ;)

Hey, that's math :) Cheers,

Durandal 1707

Ρε συ Μανιέ!!! Ποιά [color=red]ισότοπο[/color] απαγωγή??????!!!!!!!!!!!!
Τί είναι ρε???? Ραδιενεργό ισότοπο???????? :rotfl:  :rotfl:
Εις άτοπο απαγωγή είναι!!!!!! :lol:
Τεσπα, τέτοια typos σπάνε τη μονοτονια :wink:

Νομίζω (χωρίς να 'μαι και 100% σίγουρος), ο [b]κ. Μοσχοβάκης[/b] στο βιβλίο του "[b][i]Notes on Set Theory - Springer Verlag[/i][/b]", το εξηγεί. Αν το βρω , θα το βάλω κάποια στιγμή, ΌΤΑΝ βρω χρόνο.

Βασικά, είναι αρκετά specialized αυτό που λες και όπως είπε και ο Durandal, χρειάζονται και οι κατάλληλες διευκρινίσεις γιατί αλλιώς η διατύπωση είναι ανακριβής. Τα πάντα είναι πολύ λεπτά και αυστηρώς καθορισμένα :)

Και ναι, that's math :D:D

Στέλιος

[quote="maniopas"]Συγνώμη για την ανακρίβεια :blush: , αλλά έχω υπόψη την εξής τεκμητίωση:

Επειδή θεωρώ ως κενό ότι ορίζεται το σύνολο το οποίο δεν έχει κανένα στοιχείο, αν γράψουμε
Α ένωση κενό έχουμε όλα τα στοιχεία του Α, δηλαδή το Α. (Προϋποθέτω δηλαδή την ουδετερότητα του κενού στην ένωση.)

Επίσης χρησιμοποιώ ότι αν δύο σύνολα Α, Β έχουν ένωση ίση με Α, τότε το Β θα είναι υποσύνολό του Α, αφού αν είχε και στοιχεία εκτός από αυτά του Α η ένωση δε θα ήταν Α (θα περιείχε κι άλλα στοιχεία). (Βασικά το θεωρώ σαν θεωρία που προϋποθέτει την απόδειξη της ισότοπο απαγωγής, που όπως είπες δε χρειάζεται.)

Πείτε μου αν κάπου έχω χρησιμοποιήσει διαισθητισμό, γιατί στη θεωρία συνόλων (και γενικά σε όλες τις θεωρίς) είναι πολύ δύσκλολο να διαχωριστεί ο διαισθητισμός από το αυστηρό στα πρώτα στάδια θεμελίωσής τους. :o

Την παραπάνω παράσταση τη θεώρησα για έναν ορισμό του κενού συνόλου που γίνεται στο σχολείο.[/quote]

Maniopas, αρχίζοντας ανάποδα, να πω ότι από τη στιγμή που μπήκε στο παιχνίδι η λέξη [b]αυστηρά[/b] για μία ερώτηση που αφορά θεωρία συνόλων... δε νομίζω ότι αρκεί ο "ορισμός" του σχολείου.  :D  Εδώ δεν ορίζουμε τους φυσικούς αριθμούς στο σχολείο, με το κενό σύνολο θα ασχοληθούμε?  :P
(και για να μην παρεξηγηθώ στο παραπάνω: φυσικά ΕΥΤΥΧΩΣ που δεν ορίζουμε κάτι τέτοιο στο σχολείο, έτσι? Προφανώς δεν προτείνω κάτι τέτοιο ;) )

Tώρα, μιας και το έθιξες, να σου πω μερικές εμφανείς "τρύπες" στο συλλογισμό σου:

0. Μην χρησιμοποιείς λέξεις παρά μόνο ως επεξηγήσεις των συμβόλων. Πχ, λέγοντας ότι "ορίζω το κενό σύνολο ως το σύνολο που δεν έχει κανένα στοιχείο", δε λες απολύτως τίποτα στη θεωρία συνόλων, sorry... :( Καταλαβαίνω πόσο ενοχλητικός ακούγομαι, αλλά έτσι έχουν τα πράγματα :(

1. Σε ό,τι αφορά την ένωση: από πού συνάγεις ότι αν $ A, B$ σύνολα, μπορείς να κατασκευάσεις (δηλ. υπάρχει) πάντα το $ A\cup B$? Πώς το ορίζεις αυτό το σύνολο? (Στην standard ZF χρειάζεται [i]ξεχωριστό[/i] αξίωμα ύπαρξης για την ένωση δύο συνόλων... μάλιστα, χρειάζονται δύο αξιώματα αν βάλεις και το axiom of pairing στη συζήτηση...)

2. Έστω ότι έχεις ορίσει την ένωση (που χρειάζεται όπως είπαμε δικό της αξίωμα). Πρέπει να δείξεις ότι $ \forall x: x\cup\varnothing \equal{} x$. Πώς το αποδεικνύεις αυτό? Από τη στιγμή που μιλάμε για απόδειξη πρώτων εννοιών, θέλει γράψιμο, δεν αρκεί να "φαίνεται" κάτι :(

3. Η τρίτη σου πρόταση χρήζει οπωσδήποτε αναλυτικής απόδειξης. Δεν αντιλέγω ότι αυτή είναι μία σωστή ιδέα προς μια κατεύθυνση που θα οδηγήσει σε λύση, αλλά χρειάζεται πολλή δουλειά για να γραφεί αξιωματικά αυτό που λες.

Προφανώς, η ιδέα σου ήταν κατανοητή από την αρχή ;) Ωστόσο, για την αξιωματική διατύπωση αυτών που λες, χρειάζεται η εισαγωγή δύο ή τριών παραπάνω αξιωμάτων στη θεωρία συνόλων με την οποία ξεκινάς απ'ό,τι αν αποφύγεις την ένωση... γενικά, όσο λιγότερες καινούριες έννοιες χρησιμοποιείς, τόσο μεγαλύτερη πιθανότητα έχεις να αποφύγεις ανακρίβειες ;)

Cheers,

Durandal 1707

[quote="DIMITRIS T"]Durandal ευχαριστώ για τις ωραίες παρατηρήσεις σου!
Για αρχή θέλω να ρωτήσω εκεί που γράφεις τον ορισμό του κενού συνόλου  $ \exists\varnothing: \forall y, y\notin\varnothing$ το y τί είναι και που ανήκει?Ποιος ο ορισμός του καθολικού ποσοδείκτη?

Υ.Γ.:Μπορείς να με παραπεμπψεις σε βιβλιογραφία για συνολοθεωρία,θεμελίωση  συνολοθεωρίας και θεμελίωση απαγωγής ?[/quote]

Hmmm, από βιβλία υπάρχει ένα ευκολοδιάβαστο (πραγματικά δυσεύρετο χαρακτηριστικό στη θεωρία συνόλων ;) ) βιβλιαράκι που λέγεται Joy of Sets, αλλά δε θυμάμαι συγγραφέα. Φυσικά, υπάρχει και το Naive Set Theory του Halmos (τρομερό βιβλίο) αλλά εκεί είναι naive και όχι axiomatic ο φορμαλισμός (ωστόσο είναι επίσης πολύ ωραίο βιβλίο). Κάτσε να κοιτάξω αύριο και θα σου πώ... Γενικό pointer: απόφυγε βιβλία του Bourbaki (ιδίως Weil) σαν την πανούκλα. Είναι μεν το στοιχείο τους και η λογική δομή είναι τέλεια... αλλά δεν διαβάζονται :(

Tώρα, για την ερώτησή σου... ήλπιζα να περάσει κάτω απ'το χαλί ο "ορισμός" μου, αλλά μ'έπιασες. Αξιωματικά, θα έπρεπε να είχα γράψει: $ \exists\varnothing\forall x\neg(x\in \varnothing)$, αλλά είπα να λερώσω το συμβολισμό μου για να τον κάνω πιο ευνόητο.

Aυτό που θα πρέπει να θυμάσαι, είναι ότι στη θεωρία συνόλων τα [i]πάντα[/i] είναι σύνολα. Πχ, οι αριθμοί φτιάχνονται από τα ordinals ($ 0\equal{}\{\varnothing\}$, φτιάχνεις τον successor και προχωράς) κλπ... Βασικά, στο πιο θεμελιώδες επίπεδο θεωρείς ως πρωταρχικές έννοιες αυτήν του συνόλου και της σχέσης μέλους $ \in$. Οπότε, ένα σύνολο να ανήκει σε ένα άλλο σύνολο (όχι να είναι υποσύνολο!) είναι το πρώτο πράγμα που μπορείς να γράψεις, και αυτό σημαίνει η γραφή πχ $ x\in\varnothing$. Bγάζει λίγο καλύτερο νόημα τώρα? Οι quantifiers τώρα είναι θέμα λογικής...

Βασικά, sorry... μου φαίνεται κακώς γράφω και όσα γράφω, το καλύτερο θα είναι να πας να διαβάσεις κάπου για όλα αυτά από την αρχή (θα είναι και πιο καλά γραμμένα!), αλλιώς θα γράψουμε ολόκληρο βιβλίο στο forum ;) Αν σ'ενδιαφέρει, θα δω μήπως ξεθάψω κανένα άλλο reference, ok?

Cheers,

Durandal 1707

Για όσους ενδιαφέρονται, μετά από λίγο googling, φαίνεται ότι το [url=http://www.apronus.com/provenmath/]provenmath[/url] είναι μία αρκετά καλή online πηγή πληροφοριών για πράγματα όπως τα παραπάνω (αν και έχει αρκετά άβολο συμβολισμό). Αν το ήξερα, θα το είχα κάνει link νωρίτερα ;)

Cheers,

Durandal 1707


If $ a>0,b>0,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}\frac{5}{3}$, So prove this $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}<\frac{1}{abc}$

I think the problem is wrong. If $ a\equal{}b\equal{}c$, we get $ ab\plus{}bc\plus{}ca\equal{}a^2\plus{}b^2\plus{}c^2\equal{}\frac53$, so $ \frac1a\plus{}\frac1b\plus{}\frac1c\equal{}\frac5{3abc}>\frac1{abc}$.

[quote="Johan Gunardi"]I think the problem is wrong. If $ a \equal{} b \equal{} c$, we get $ ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac53$, so $ \frac1a \plus{} \frac1b \plus{} \frac1c \equal{} \frac5{3abc} > \frac1{abc}$.[/quote]

No man!!! the problem is written right.

We have to prove
$ ab\plus{}bc\plus{}ca<1$ using cauchy's we get $ ab\plus{}bc\plus{}ca<\frac{5}{3}$ and not 1


geometry
circumcircle
geometric transformation
reflection
trigonometry
geometry unsolved

Draw the circle $ (O_1)$ of diameter $ G_eX_a$, the circle $ (O_2)$ of diameter $ G_eX_b$. They meet again at $ M$. Let point $ P$ on the opposite ray of the ray $ G_eM$, such that $ G_eP\equal{}\frac{1}{2}G_eM$. Denote $ O_3$ be the midpoint of $ O_1O_2$. The perpendicular line to $ PO_3$ at $ P$ meets the circles $ (O_1),(O_2)$ at points $ R,Q$ respectively. Then, $ X_aR$ and $ X_bQ$ meet at a point, call $ X_c$. Draw the circumcircle $ (w)$ of triangle $ X_aX_bX_c$. The tangents of $ (w)$ at $ X_a,X_b,X_c$ form the sides of triangle $ ABC$.  :)

See you again in the December.  :wink:
Best regards, mr.danh.

[quote="jrrbc"]solved but there's a mistake with point B, is not from the perpendicular of (w) at Xc, is a perpendicular of (w) at Ge  :rotfl:[/quote]
Sorry, I understood the mrdanh's solution. But I dont' understand jrrbc's idea. Ge doesn't lie on (w).  :huh:

[quote="jrrbc"]perp (w) and Xa
perp (w) and Xb
perp Ge(w-center) at Ge is a mistake of this problem would be perp (w) and Xc[/quote]
Check your English  :D
"Tangents" and "perpendiculars" are different to each other.
(w) is a circle, and a circle has tangents, not "perpendiculars".  :rotfl:

talking about geometry problem, not about english geometry problem. Is not a semantic problem :rotfl:

Reflect $ X_b$ in $ X_a$ into $ X_b',$ draw a circle $ (O_a)$ through $ X_b',$ tangent to $ GeX_a$ at $ X_a.$
Reflect $ X_a$ in $ X_b$ into $ X_a',$ draw a circle $ (O_b)$ through $ X_a',$ tangent to $ GeX_b$ at $ X_b.$
Circles $ (O_a), (O_b)$ intersect at most at 2 points $ X_c, X_c',$ giving at most 2 solutions.
Tangents to circumcircle $ (I)$ of $ \triangle X_aX_bX_c$ at its vertices pairwise intersect at $ A, B, C.$

[hide="Proof"]By sine theorem for $ \triangle X_aX_b'X_c,$
$ \frac{d(Ge, X_aX_b)}{|X_aX_b|} \equal{} \frac{|GeX_a| \sin \widehat{GeX_aX_b}}{|X_aX_b|}  \equal{} \frac{|GeX_a| \sin \widehat{X_aX_cX_b'}}{|X_aX_b'|} \equal{}$
$ \equal{} \frac{|GeX_a| \sin \widehat{X_cX_b'X_a}}{|X_cX_a|} \equal{} \frac{|GeX_a| \sin \widehat{X_cX_aGe}}{|X_cX_a|} \equal{} \frac{d(Ge, X_cX_a)}{|X_cX_a|}.$
Similarly,  $ \frac{d(Ge, X_aX_b)}{|X_aX_b|} \equal{} \frac{d(Ge, X_bX_c)}{|X_bX_c|}$ $ \Longrightarrow$ $ Ge$ is symmedian point of the intouch triangle $ \triangle X_aX_bX_c.$ [/hide]