calculus
derivative
induction
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ such that
 \[ f(x)=f\left(\frac{x}{2}\right)+\frac{x}{2}f'(x)\]
for all $x \in \mathbb R$.

So we have to solve $ f(x) \equal{} f(\frac x2) \plus{} \frac x2f'(x)$ (and $ f(x)$ is continuous and it's derivative exists all over $ \mathbb R$)

Let $ f(x)$ any solution, let $ u\neq 0$ and let $ g(x) \equal{} f(x) \plus{} (f(0) \minus{} f(u))\frac xu \minus{} f(0)$. Obviously, $ g(x)$ is a solution and $ g(u) \equal{} g(0) \equal{} 0$
Let then $ M \equal{} \max_{x\in[0,u]}g(x)$ (I write $ [0,u]$ even if $ u < 0$) and any $ x_0\in[0,u]$ such that $ g(x_0) \equal{} M$ ($ M$ and $ x_0$ exist since $ g(x)$ is continuous)
If $ M\neq 0$, $ x_0\neq 0$ and $ f'(x_0) \equal{} 0$. We have then $ f(x_0) \equal{} f(\frac {x_0}{2}) \equal{} M$ and so $ f'(\frac {x_0}{2}) \equal{} 0$.
An immediate induction give us $ f(\frac {x_0}{2^n}) \equal{} M$ and so $ M \equal{} 0$ (since $ g(0) \equal{} 0$ and $ g(x)$ is continuous at $ 0$).

Same, Let then $ m \equal{} \min_{x\in[0,u]}g(x)$ and any $ x_1\in[0,u]$ such that $ g(x_1) \equal{} m$. The same method implies $ m \equal{} 0$

So $ g(x) \equal{} 0$ $ \forall x\in[0,u]$. and so $ f(x) \equal{} \frac {f(u) \minus{} f(0)}{u}x \plus{} f(0)$ $ \forall x\in[0,u]$

So  $ \frac {f(u) \minus{} f(0)}{u} \equal{} \frac {f(v) \minus{} f(0)}{v}\equal{}a$ $ \forall u,v$ with same sign.

So $ f(x) \equal{} ax \plus{} b$ $ \forall x\geq 0$ and $ f(x) \equal{} cx \plus{} d$ $ \forall x\leq 0$. But continuity at $ 0$ implies $ b \equal{} d$ and existence of $ f'(0)$ implies $ a \equal{} c$

And so the only solutions are $ f(x) \equal{} ax \plus{} b$ $ \forall x\in\mathbb R$ (and we easily verify that these solutions fit).


Find all three continuous functions $f,g,h: R\to R$ such that $: f(xy)=g(x).g(y)-h(x+y)+1, \forall x,y\in R$

It seems like solutions are infinite... for example, $f(x) = g(x) = x, h(x) = 1$, and we can also have $f(x) = g(x) = h(x) = 1$.

thanks,please,you get the solution for this problem.I think this problem is very difficult.


Given 3 possitive real $a,b,c$ satisfying $abc+a+c=b$. Find the maximum value of
$P=\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$

If ac = 1, then a = -c and 1/(a^2 + 1) - 2/(b^2 + 1) needs to be maximized, choose a, b tending to 0, infinity to get maximum of 1.

From now on, ac isn't 1.

We have $b = \frac{a+c}{1-ac}$ given.

Take a look at the negative term: $\frac{2}{b^2 + 1} = \frac{2(1-ac)^2}{(a+c)^2 + (1-ac)^2} = \frac{2(1-ac)^2}{(a^2 + 1)(c^2 + 1)}$.

So we can get a common denominator fairly easily.  Let's do that.

Now:

$P(a^2 + 1)(c^2 + 1) = 2(c^2 + 1) + 3(a^2 + 1) - 2(1-ac)^2 = 3 + 3a^2 + 2c^2 + 4ac -2(ac)^2$.

i dont see a nice way so im just going to give up, but ya you should be able to work from there since this is only 2 vars

Hmmm I don't think it's easy from there Singular.
Here is my oponion : Let $b'=-b$ so we have $a+b'+c=ab'c$ thus we can let $a=tgA,b'=tgB,c=tgC$ with $A,B,C$ are 3 angle of a triangle and $\angle B>\pi/2$
Now we have to find max of :
$\frac{2}{tg^2A+1}-\frac{2}{tg^2B+1}+\frac{3}{tg^2C+1}$
But I can't sovle it yet. What do you think?

[quote="Stun"]Hmmm I don't think it's easy from there Singular.
Here is my oponion : Let $b'=-b$ so we have $a+b'+c=ab'c$ thus we can let $a=cotgA,b'=cotgB,c=cotgC$ with $A,B,C$ are 3 angle of a triangle.
Now we have to find max of :
$\frac{2}{cotg^2A+1}-\frac{2}{cotg^2B+1}+\frac{3}{cotg^2C+1}$
But I can't sovle it yet. What do you think?[/quote]
I think it should be $a=tanA,b'=tanB,c=tanC$

I'm sorry it's a stupid mistake . I did corect it  :D .
And now do you get any idea? (Huh!!!! I'm really bad at this section :( )

Oh really!!! Could we see some proof for this claim :lol:

hello, by inserting $c=\frac{b-a}{1+ab}$ in $P$ we get the following expression:
$\frac{2-2a^2+6ab+2b^2+3a^2b^2}{\left(1+a^2\right)\left(1+b^2\right)}$.
Now we will show, that
$\frac{2-2a^2+6ab+2b^2+3a^2b^2}{\left(1+a^2\right)\left(1+b^2\right)} \le \frac{10}{3}$. But this inequality is equivalent with
$0 \le 4\left(2a-b\right)^2+\left(ab-1\right)^2$.
For $a=\frac{1}{\sqrt{2}}$ and $b=\sqrt{2}$ we get equality.
This completes the proof.

source:VietnamMO and it's true for a,b,c are real


probability
analytic geometry
geometry
similar triangles

I'm having trouble with these four problems. I know the answers but I want detailed solutions on each of these. Thanks.

21. A circle with a radius of 2 units has its center at (0, 0). A circle with a radius of 7 units has its center at (15, 0). A line tangent to both circles intersects the x-axis at (x, 0) to the right of the origin. What is the value of x? Express your answer as a common fraction.
[hide]Answer: 10/3[/hide]

26. Kevin will start with the integers 1, 2, 3, and 4 each used exactly once and written in a row in any order. Then he will find the sum of the adjacent pairs of integers in each row to make a new row, until one integer is left. For example, if he starts with 3, 2, 1, 4, and then takes sums to get 5, 3, 5, followed by 8, 8, he ends with the final sum 16. Including all of Kevin's possible starting arrangements of the integers 1, 2, 3, and 4, how many possible final sums are there?
[hide]Answer: 5[/hide]

29. Given that x^2-5x+8=1, what is the possible value of x^4-10x^3+25x^2-9?
[hide]Answer: 40[/hide]

30. An o-Pod MP3 player stores and plays entire songs. Celeste has 10 songs stored on her o-Pod. The time length of each song is different. When the songs are ordered by length, the shortest song is only 30 seconds long and each subsequent song is 30 seconds longer than the previous song. Her favorite song is 3 minutes, 30 seconds long. The o-Pod will play all the songs in random order before repeating any song. What is the probability that she hears the first 4 minutes, 30 seconds of music-there are no pauses between songs-without hearing every second of her favorite song? Express your answer as a common fraction.
[hide]Answer: 79/90[/hide]

Huge hints:

[hide="21"]Draw the circles and line, and use similar triangles.[/hide]

[hide="26"]Write a, b, c, and d in a row, and if you add it up, you get that the top number is $ a\plus{}3b\plus{}3c\plus{}d\equal{}a\plus{}d\plus{}3(b\plus{}c)$, so you just have to find the number of possible values of $ a\plus{}d$, since the value of $ b\plus{}c$ will be forced from that.[/hide]

There is already a topic on 29.

[hide="30"]Find the probability that she [b]will[/b] be able to hear [b]all[/b] of her favorite song (it's really not that complicated) and subtract it from 1. It might help to make a list all of the song lengths in seconds.[/hide]

I don't get using similar triangles for 21.

I tried to, as well as finding the internal tangent's length as 12, but I still couldn't get it.

[hide="30"]I spent like 30 seconds on #30 in the middle of the test and thought it was too easy. But I was right so yay.

We only need to worry about the first minute of music that plays.

There is a $ \frac{1}{10}$ chance that the first minute is the song one minute long.  Then the next song played must be one of the 8 other choices that is not her favorite song.

$ \frac{1}{10}*\frac{8}{9}\equal{}\frac{8}{90}$

Another possibility is that the first song is 30 seconds long, $ \frac{1}{10}$ of a chance.

Then the next song can be any other song except her favorite song, $ \frac{8}{9}$, giving us the same probability, $ \frac{8}{90}$.

Now the final probability is if the first song is any song except her favorite song and the two other songs, $ \frac{7}{10}$.

Add these up to get $ \boxed{\frac{79}{90}}$[/hide]

[hide="21"]Let the center of the circle with radius 2 be $ C_2$ and the center of the other circle be $ C_7$
Let the point tangent to the circle radius 2 be $ t_2$ and similarly denote it for the other circle.
Now let the point with coordinate $ (x,0)$ be $ A$.

Then $ \bigtriangleup C_2t_2A\sim \bigtriangleup C_7t_7A$ because by vertical angles, $ \angle C_2At_2\equal{}\angle C_7At_7$ and the tangent line is perpendicular to the radius at those points.

Thus $ \frac{C_2t_2}{C_7t_7}\equal{}\frac{C_2A}{C_7A}$
$ \frac{2}{7}\equal{}\frac{x}{15\minus{}x}$
$ 7x\equal{}30\minus{}2x$
$ x\equal{}\boxed{\frac{10}{3}}$[/hide]
[hide="26"]
Consider the arrangement $ a,b,c,d$
Then the next row will be $ a\plus{}b,b\plus{}c,c\plus{}d$
Then the next row will be $ a\plus{}2b\plus{}c,b\plus{}2c\plus{}d$
And the last row will be   $ a\plus{}3b\plus{}3c\plus{}d$

Thus, we only need to worry about the middle two numbers.

There are 4 numbers, and thus $ \binom{4}{2}\equal{}6$ ways to arrange them.

But $ 1\plus{}4\equal{}2\plus{}3$, which is the overcounted because $ a\plus{}d$ as well as $ b\plus{}c$ is the same.

Thus there are $ \boxed{5}$ sums.[/hide]

[hide="a much easier way for 26"]

It can only be even numbers, after a little effort, you find 16 is lowest possible and 24 is highest, that is 5 even numbers between 16 and 24 inclusive.[/hide]

Ok. I get the first three, but I still don't get #30. Any faster ways?

For thirty, I counted the ways she DOES hear every second of her favorite song and then subtracted from 1.

First off, since her favorite song is 3:30 long, the song must start within the first minute for here to hear all of it. So you only have to consider the first two songs that the iPod chooses.

Case 1: iPod chooses her favorite song first and then any other song : $ \frac{1}{10}\cdot{1}\equal{}\frac{1}{10}$

Case 2: iPod chooses the 30 sec or 1 minute song and then her favorite song: $ \frac{2}{10}\cdot\frac{1}{9}\equal{}\frac{1}{45}$

$ \frac{1}{45}\plus{}\frac{1}{10}\equal{}\frac{11}{90}$

$ 1\minus{}\frac{11}{90}\equal{}\boxed{\frac{79}{90}}$

[quote="tinytim"]For thirty, I counted the ways she DOES hear every second of her favorite song and then subtracted from 1.

First off, since her favorite song is 3:30 long, the song must start within the first minute for here to hear all of it. So you only have to consider the first two songs that the iPod chooses.

Case 1: iPod chooses her favorite song first and then any other song : $ \frac {1}{10}\cdot{1} \equal{} \frac {1}{10}$

Case 2: iPod chooses the 30 sec or 1 minute song and then her favorite song: $ \frac {2}{10}\cdot\frac {1}{9} \equal{} \frac {1}{45}$

$ \frac {1}{45} \plus{} \frac {1}{10} \equal{} \frac {11}{90}$

$ 1 \minus{} \frac {11}{90} \equal{} \boxed{\frac {79}{90}}$[/quote]

It's [b]o-pod[/b], not ipod. :P

[quote="314.math"]Sorry, but I still don't get #30.  Could someone please explain it in greater detail?  Thanks![/quote]
I'll try expanding upon a previous explanation.
[hide]
If the very first song is Celeste's favorite, she gets to hear it all.  Since there are 10 songs, the chance that her song plays first is $ \frac{1}{10}.$

Otherwise, in order for Celete's favorite song to play before the end of 4 minutes 30 seconds, the first song must be 1 minute or less.  There are two such choices -- the 30 second song and the 1 minute song.  If one of these is first, then Celeste's song must be second.  The combined probability is the product of these two events:  $ \frac{2}{10}\times\frac{1}{9}\equal{}\frac{1}{45}.$

Even if the two shortest songs played first, and Celeste's favorite played third, she won't hear every second of her favorite song.  So the previous two ways are the only ways in which Celeste will hear every second of her favorite song.  Adding the above... $ \frac{1}{10}\plus{}\frac{1}{45}\equal{} \frac{9}{90}\plus{}\frac{2}{90}\equal{}  \frac{11}{90}.$

But since we want the chance she [b]won't[/b] get to hear every second, we subtract from 1:
$ 1\minus{}\frac{11}{90}\equal{}\boxed{\frac{79}{90}}.$[/hide]

[quote=Max0815]#26:
If Kevin's original four numbers are $a$, $b$, $c$, and $d$, then here are the numbers that he writes down:
\[
\begin{array}{ccccccc}
a & & b & & c & & d \\
& a + b & & b + c & & c + d & \\
& & a + 2b + c & & b + 2c + d & & \\
& & & a + 3b + 3c + d & & &
\end{array}
\]

Since Kevin always starts with the numbers 1, 2, 3, and 4 (in some order), we have that $a + b + c + d = 10$. Then Kevin's final number is
\[a + 3b + 3c + d = 2(b + c) + (a + b + c + d) = 2(b + c) + 10.\]
The possible values of $b + c$ are 3, 4, 5, 6, and 7, so there are $\boxed{5}$ possible final sums.[/quote]

lol gotta love that 9.5 year bump.... :)

Lol that is hilarious.... I bet he/she would look back at this forum and be like "gee that bump lol xD troll"

I love trolling.......I did that on several fourums..... xD
they probably were like...... ( -_-) bro....no

There seems to be solutions for everything but 29, so:
[hide=Solution for 29]
Since x^2-5x+8=1, then x^2-5x=-7. If we square both sides of the equation and simplify, we have x^4-10x^3+25x^2=49, so x^4-10x^3+25x^2-9=49-9=40.
[hide]

for problems like 29, they are all engineered a special way, otherwise it would be too tedious. As my father always said, "Find the grand truth." Translation: all competition problems are engineered specially, so translate the problem into a easier problem.


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Mod delete.


function algebra polynomial calculus Germany

Note first that the polynomial $ z^3 \plus{} Az \plus{} B \equal{} 0$ has at least one real solution for any real $ A,B$, and this solution is unique when $ A\geq0$ (the derivative of the polynomial never changes sign), so the function is surjective; indeed, for any real $ r$ take $ y > 0$ and $ z \equal{} \minus{} r^3 \minus{} yr$, then $ x^3 \plus{} xf(y) \plus{} f(z) \equal{} 0$ has at least one root whose image satisfies $ f(x)^3 \plus{} yf(x) \plus{} z \equal{} 0$, with unique solution $ f(x) \equal{} r$.

Take now any nonzero $ x$, since $ f$ is surjective we may select $ y$ such that $ f(y) \equal{} \minus{} x^2 \minus{} \frac {f(x)}{x}$, or $ f(x)^3 \plus{} yf(x) \plus{} x \equal{} 0$, and $ f(x)$ may not be zero.  Since $ f$ is surjective, then $ f(0) \equal{} 0$.

Take now $ y \equal{} f(y) \equal{} 0$ and $ x \equal{} \minus{} \sqrt [3]{f(z)}$.  Then, $ z \equal{} \minus{} f(x)^3$ is uniquely determined by $ f(z)$, and $ f$ is injective.  Moreover, taking $ z \equal{} f(z) \equal{} 0$ and $ f(y) \equal{} \minus{} r^2 < 0$ any negative real, then $ \pm r$ are roots of $ x^3 \plus{} f(y)x \plus{} f(z) \equal{} 0$, and $ \pm f(r)$ are the roots of $ f(x)^2 \equal{} \minus{} y$; this proves two things: first, if $ f(y) < 0$, then $ y < 0$, and second, $ |f(r)| \equal{} |f( \minus{} r)|$ for any real $ r$, and since $ f$ is injective, then $ f(r) \equal{} \minus{} f( \minus{} r) > 0$ for any positive real $ r$.

For any reals $ u,v$, take $ f(y) \equal{} \minus{} u^2 \minus{} uv \minus{} v^2 \equal{} uv \minus{} (u \plus{} v)^2$ and $ f(z) \equal{} uv(u \plus{} v)$.  Clearly $ x \equal{} u,v, \minus{} u \minus{} v$ are the roots of $ x^3 \plus{} f(y)x \plus{} f(z) \equal{} 0$, or $ f(u),f(v),f( \minus{} u \minus{} v) \equal{} \minus{} f(u \plus{} v)$ are the roots of $ f(x)^3 \plus{} yf(x) \plus{} z \equal{} 0$, and by Cardano, $ f(u \plus{} v) \equal{} f(u) \plus{} f(v)$, as well as $ y \equal{} f(u)f(v) \minus{} f(u)f(u \plus{} v) \minus{} f(v)f(u \plus{} v) \equal{} \minus{} f(u)^2 \minus{} f(u)f(v) \minus{} f(v)^2 \equal{} f(u)f(v) \minus{} f(u \plus{} v)^2$ and $ z \equal{} f(u)f(v)f(u \plus{} v)$.

Take now $ f(y) \equal{} \minus{} 1$, $ z \equal{} 0$, $ x \equal{} 1$.  Clearly, $ y \equal{} \minus{} f(1)^2$, or $ 1 \equal{} f(f(1)^2)$.  Take also $ f(z) \equal{} \minus{} 1$, $ y \equal{} 0$, $ x \equal{} 1$.  Clearly $ z \equal{} \minus{} f(1)^3$, or $ 1 \equal{} f(f(1)^3)$.  Since $ f$ is injective and $ f(0) \equal{} 0$, then $ f(1)^3 \equal{} f(1)^2$ or $ f(1) \equal{} 1$.

Take now in the previous result $ v \equal{} 1$, then $ v \equal{} \minus{} 1$, resulting in $ f(y) \equal{} \minus{} u^2 \minus{} u \minus{} 1$ when $ y \equal{} \minus{} f(u)^2 \minus{} f(u) \minus{} 1$ and $ f(y') \equal{} \minus{} u^2 \plus{} u \minus{} 1$ when $ y' \equal{} \minus{} f(u)^2 \plus{} f(u) \minus{} 1$.  Then $ f(2u)\equal{}f(y'\minus{}y)\equal{}f(y')\minus{}f(y)\equal{}2u$ for all real $ u$, done!

As the equation $ x^{3} \plus{} ax \plus{} b \equal{} 0$ always has at least one real solution (the solution is unique if $ a$ is nonnegative), we easily conclude the surjectivity of $ f$. (For any real $ m$, set $ y \equal{} 1$, $ z \equal{} \minus{} m(m^{2} \plus{} 1)$ and $ x$ such that $ x^{3} \plus{} xf(y) \plus{} f(z) \equal{} 0$. Than from the given condition we have $ f(x) \equal{} m$). If we take $ t$ such that $ f(t) \equal{} 0$ and set $ (x,y,z) \equal{} (0,a,t)$ in condition, we get that $ af(0)$ is fixed, which means $ f(0) \equal{} 0$ and from that we get $ t \equal{} 0$. It means that for any real $ x$, different from $ 0$, $ f(x)$ is also different from $ 0$. Assume there are some different $ a,b$ such that $ f(a) \equal{} f(b)$. Than, plugging any $ z$, different from $ 0$ and such $ x$, that $ x^{3} \plus{} xf(a) \plus{} f(z) \equal{} 0 \equal{} x^{3} \plus{} xf(b) \plus{} f(z) \equal{} 0$, the condition gives $ af(x) \equal{} bf(x)$, but $ (a \minus{} b)$ can't be $ 0$. So $ f(x) \equal{} 0$, but this imply $ x \equal{} 0$ and $ z \equal{} 0$, which contradicts the assumption. Hence, $ f$ is injective. As $ f$ is surjective, we can easily prove that  the equalities $ f(x)^{3} \plus{} yf(x) \plus{} z \equal{} 0$ and $ x^{3} \plus{} xf(y) \plus{} f(z) \equal{} 0$ are equivalent. That means that $ f( \minus{} f(x)^{3} \minus{} yf(x)) \equal{} \minus{} x^{3} \minus{} xf(y)$. (1)
  When $ x \equal{} 1$, we have: $ f(y) \plus{} f(z) \equal{} \minus{} 1$ iff $ y \plus{} z \plus{} 1 \equal{} 0$. So, $ f( \minus{} y \minus{} 1) \equal{} \minus{} f(y) \minus{} 1$.(2)  If $ y$ is such that $ f(y) \equal{} \minus{} 1$, than plugging in the last condition $ z \equal{} 0$, we get $ y \equal{} \minus{} 1$.(3)   Setting $ x \equal{} \minus{} 1$ in (1), we get: $ f(y \plus{} 1) \equal{} f(y) \plus{} 1$. (4). It means that for all integers $ m$, $ f(x \plus{} m) \equal{} m \plus{} f(x)$. (*)  From (2) and (4) we conclude that $ f( \minus{} x) \equal{} \minus{} f(x)$, for all $ x$ and as $ f(0) \equal{} 0$, from $ (4)$ we get $ f(m) \equal{} m$, for all integer values of $ m$. Substituting $ x$ by $ ( \minus{} x)$ in (1), we get:  $ f(f(x)^{3} \plus{} yf(x)) \equal{} x^{3} \plus{} xf(y)$ (5). If $ x$ is integer in (5), we have
$ x^{3} \plus{} xf(y) \equal{} f(x^{3} \plus{} xy) \equal{} x^{3} \plus{} f(xy)$. Hence, for all integers $ m$: $ f(mx) \equal{} mf(x)$. (6).  If $ x \equal{} \frac {n}{m}$, where $ n,m$ are integers ($ m > 0$), (6) yields $ f(\frac {n}{m}) \equal{} \frac {n}{m}$, which means that $ f(q) \equal{} q$ for all rational $ q$.
  When $ y \equal{} \minus{} f(x)^{2}$, (5) leads to $ f( \minus{} f(x)^{2}) \equal{} \minus{} x^{2}$, but $ \minus{} f(x)^{2}$ can get any negative value. It means that for all $ x < 0$, we have $ f(x) < 0$, but as $ f$ is odd, we have $ f(x) > 0$ when $ x > 0$. From (6) and (*), we get:  $ f(x \plus{} \frac {m}{n}) \equal{} \frac {f(nx \plus{} m)}{n} \equal{} \frac {nf(x) \plus{} m}{n} \equal{} f(x) \plus{} \frac {m}{n}$. It means that for all rational $ q$, we have $ f(x \plus{} q) \equal{} f(x) \plus{} q$. (**).
   Assume there exists such $ x$ that $ f(x) < x$. It's clear that there is such rational number $ q$ that $ f(x) < q < x$. In this case, $ f(x \minus{} q) > 0$ and   $ q > f(x) \equal{} f(q \plus{} (x \minus{} q)) \equal{} q \plus{} f(x \minus{} q) > q$, which is impossible. We will get the similar contradiction if we assume that $ f(x) > x$ holds for some real number $ x$.
   Finally, $ f(x)$ can't be neither bigger, nor smaller than $ x$, for any $ x$. So, $ f(x) \equal{} x$ holds for any $ x$. It's easy to verify that the function $ f(x) \equal{} x$ indeed satisfies the given conditions.  Lasha Lakirbaia :)

[i]Yes , i have checked Lasha's solution and it's really a correct solution

   I don't have time , can anybody check Daniel's solution ?[/i] :lol:

[quote=Germany TST 2009/3]Find all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that $ \forall x,y,z \in \mathbb{R}$ we have: If
\[ x^3 \plus{} f(y) \cdot x \plus{} f(z) \equal{} 0,\]
then
\[ f(x)^3 \plus{} y \cdot f(x) \plus{} z \equal{} 0.\][/quote]

Nice problem :)

[hide=solution] Answer: The only function which works is the identity.

It is clear to check that it works. I prove that no other function does, so for the sake of argument, let $f$ be a function which satisfies the given condition. Henceforth, we will refer to a triple $(x,y,z)$ satisfying $x^3+f(y)x+f(z)=0$ as [i]valid.[/i] We will use subtly, the fact that any cubic real polynomial has at least one real root.

[b]Claim 1:[/b] There exists a real number $x_0$ such that $f(x_0)=0$.

(Proof) Let $x_0$ be a real solution of $x^3+xf(0)+f(0)=0$. Evidently, $(x_0,0,0)$ is valid so $f(x_0)^3=0$ so our claim holds. $\square$

[b]Claim 2:[/b] $f(0)=0$.

(Proof) Note that $(0, y, x_0)$ is valid where $x_0$ satisfies $f(x_0)=0$ so $f(0)^3+yf(0)+x_0=0$ for all reals $y$ which yields $f(0)=0$. $\square$

[b]Claim 3:[/b] $f$ is injective.

(Proof) Let $f(a)=f(b)$ and take $x, y$ as real numbers for which $(x, y, a)$ is valid. Consequently, $(x,y,b)$ is also valid, so $$f(x)^3+yf(x)+a=f(x)^3+yf(x)+b=0 \Longrightarrow a=b$$ so injectivity follows. $\square$

Now that we have proved the easy claims, it's time to make bolder ones.

[b]Claim 4:[/b] $f$ is surjective.

(Proof) Fix an arbitrary $t \in \mathbb{R}$. Consider the polynomial $$P(X)=(X-t)\cdot \left(X^2+tX+(1+t^2)\right).$$ Set $z_0=-t(1+t^2)$ and $y_0=1$. Take $x_0$ to be a real number for which $(x_0, y_0, z_0)$ is valid and observe that $$P(x_0)=f(x_0)^3+y_0f(x_0)+z_0=0.$$ Since $P$ has exactly one real root $t$ we conclude $f(x_0)=t,$ establishing the claim. $\square$

[b]Claim 5:[/b] For all $y \ge 0$ we have $f(y) \ge 0$ as well.

(Proof) For the sake of contradiction, let $y_0>0$ be a real number for which $f(y_0)<0$. Let $x_0$ be a solution to $x^2=-f(y)$ and see that $(x_0, y_0, 0)$ is a valid triple so $f(x_0)^3+yf(x_0)=0$ which fails to occur since $x_0 \ne 0 \Longrightarrow f(x_0) \ne 0$ by injectivity and $f(x_0)^2+y_0>0$. Our initial assertion fails so we get a contradiction! $\square$

[b]Claim 6:[/b] $f$ is additive.

(Proof) Fix any three distinct real numbers $a, b, c$ such that $a+b+c=0$. Note that $f$ is surjective so we can choose real numbers $y_0, z_0$ such that $a, b, c$ are the roots of the polynomial $f(X)=X^3+Xf(y)+f(z)=0$. Evidently, $f(a), f(b), f(c)$ are distinct real numbers (since $a, b, c$ are distinct), which happen to be the roots of the equation $X^3+yX+z=0$. It follows that $f(a)+f(b)+f(c)=0$ so for all real numbers $a, b$ we get $$f(-(a+b))=-(f(a)+f(b)).$$ Put $b=0$ to get $f(-x)=-f(x)$ for all $x$, yielding $f(a+b)=f(a)+f(b)$ for all distinct reals $a, b$. For $a=b$ simply consider $$f(2a)=f(a+m)+f(a-m)=(f(a)+f(m))+(f(a)-f(m))=2f(a)$$ where $m$ is a sufficiently large number. Thus, $f$ is completely additive. $\square$

[b]Claim 7:[/b] $f$ is the identity function.

(Proof) By [b]Claim 5[/b] and [b]Claim 6[/b] it is clear that $f$ is additive and bounded on an interval, so $f$ is linear. Since $f(0)=0$ we get $f(x)=cx$ for all reals $x$. Notice that $c \ne 0$ so we get from a valid triple $(1,y,z)$ that $z=\left(\frac{-1}{c}-y\right)=-c^3-cy$. This clearly fails to hold for $y \rightarrow \infty$ unless $c=1$ in which case we are done! $\square$

The End.

[/hide]


logarithms
real analysis
real analysis unsolved

1) Find a sequence $ \{a_k\}_{k\ge1}$ such that $ a_k\ne 0,1$ and $ \sum_{k \equal{} 1}{1\over a_k}$ converges but $ \sum_{k \equal{} 1}{1\over a_k \minus{} 1}$ diverges

2) Find a conditionally convergent sequence $ \sum a_k$ such that $ \sum \vert a_k\vert^c \cdot sgn(a_k)$ diverges
for any $ c > 0,$ $ c\ne1.$

For Problem 1, I think the following sequence works: $ \sqrt{2}$, $ \minus{}\sqrt{2}$, $ \sqrt{3}$, $ \minus{}\sqrt{3}$, $ \sqrt{4}$, $ \minus{}\sqrt{4}$, and so on.  The first sum is an alternating sequence with decreasing terms that tend to zero, so by somebody's theorem, the series converges.  (Or we can just see that its partial sums hover around zero.)  For the second sum, we can group the terms into alternating pairs to see that the series diverges.

Another sequence that works for Problem 1 is $ a_n \equal{} (\minus{}1)^n n^{1/2}$.  The proof is similar.

For the second problem:

$ (a_k)\equal{}\frac2{\ln 2},\frac{\minus{}1}{\ln 2},\frac{\minus{}1}{\ln 2},\frac2{\ln 3},\frac{\minus{}1}{\ln 3},\frac{\minus{}1}{\ln 3},\frac2{\ln 4},\dots.$


I have a quick question somebody can hopefully clarify for me.

If there is a contest that runs every hour for 40 days. I enter 30 entries during one hour of a each day for 40 days what are my odds of winning during the course of 40 days?

This is a math question. Moving to math forum.

[quote="peeta"]what kind of contest?[/quote]

To win an xbox 360.

This guy believes that if I enter 30 a day and on average there are 2000 entrie so 30/2000 are mine that after 40 days my chances of winning are 40%

You have posted the same spamming message in different forums:

 :spam:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55546]http://www.mathlinks.ro/Forum/viewtopic.php?t=55546[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55547]http://www.mathlinks.ro/Forum/viewtopic.php?t=55547[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55549]http://www.mathlinks.ro/Forum/viewtopic.php?t=55549[/url]

I urgently ask you to stop. Thank you.

[quote="10000th User"]You have posted the same spamming message in different forums:

 :spam:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55546]http://www.mathlinks.ro/Forum/viewtopic.php?t=55546[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55547]http://www.mathlinks.ro/Forum/viewtopic.php?t=55547[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55549]http://www.mathlinks.ro/Forum/viewtopic.php?t=55547[/url]

I urgently ask you to stop. Thank you.[/quote]

Alright but answer the question please.

[quote="Bictor717"]If a billion other people do the exact same thing, your chances would be one in a billion (and one). If you're the only person trying, your chances would be one in one.
Obviously, more information is needed. The best you could probably give are many assumptions that may not be based on fact, so it might not even be possible to calculate it. Would the odds be stated somewhere in the contest rules, like the fine print, or something like that?[/quote]

I'll do my best to explain this fully.

Ok there is a contest and they are giving away 1 xbox every hour for 40 days. In total there are like 900+ xbox's to give out. On average there are 3000 entries for every contest.

You put in 30 entries outta 3000 once a day for 40 days what are you odds of winning overall.


Any typo? There is no solution of course.


let x,y,z be positive real numbers.
find the minimum value of (x+y+z) \sum  \sqrt (x 2 +xy+y 2 ) -1

[quote="wpolly"]Let $ x,y,z$ be nonnegative real numbers such that $ yz \plus{} zx \plus{} xy > 0.$

Find the minimum value of $ \sum{\frac {x \plus{} y \plus{} z}{\sqrt {y^2 \plus{} yz \plus{} z^2 }}}.$[/quote]$ \sum{\frac {1}{\sqrt {y^2 \plus{} z^2 \plus{} yz}}}\geq\sum{\frac {y^2 \plus{} z^2 \plus{} 2\left(\sqrt {3} \minus{} 1\right)yz}{(y \plus{} z)\left(y^2 \plus{} z^2 \plus{} yz\right)}}\geq\cfrac{4 \plus{} \cfrac{2}{\sqrt {3}}}{x \plus{} y \plus{} z},$

with equality if $ x \equal{} y \minus{} z \equal{} 0.$


probability
algebra
polynomial
geometry
rectangle
arithmetic sequence

Hello Everyone,

Since marathons are such a great way to prepare for competitions and exams, I have decided to start one!

The following are the rules:
1) If you are deriving the problems from previous competitions, then state the competition name, year, and question number.
2) Provide a [i]full[/i]solution to your answers (extremely trivial steps may be skipped but it is good to include them). It is easier for people to understand your steps like that.
3) Read the guidelines for posting problems in this forum and follow them.
4) Post a problem when you submit a solution. This will keep the marathon moving.
5) You may use calculators only on problems labeled as "brute force".
6) You must simplify all fractions except those labeled as "brute force".
These are all basically the guidelines you find in other marathons so there should be no problem in following them.

Problem:
[b]Brute Force[/b]
What is the value of $ x$ in the equation $ \frac{97x}{83}\equal{}\frac{727}{863}$?

I'm just wondering here, I may not know, but to me problems like the one you just posted will not serve any purpose whatsoever except:
A. Calculator training
B. SIMPLE linear equations

Yes I do believe we should refrain from posting these simple problems, as they don't belong in this forum. But anyway... here's the answer and some new problems to work on.

97x / 83 = 727/863

97x times 863 = 727 times 83

83711x = 60341

x = 60341/83711

My next problems

An easy one:

Let a, b, c, d, and e be five consecutive terms in an arithmetic sequence and suppose that a + b+ c + d + e = 30. Which of the five terms can we find?

A harder question:

Integers a, b, c, and d not necessarily distinct, are chosen independently and at random from 0 to 2007 inclusive. What is the probability that ad-bc is even?

Easy:
You can let $ x \equal{} c$. And rewrite the equation given the fact that they are in an arithmetic sequence:

$ x \minus{} 2d \plus{} x \minus{} d \plus{} x \plus{} x \plus{} d \plus{} x \plus{} 2d \equal{} 30 \\
5x \equal{} 30\\
x \equal{} 5$

So $ \fbox{c}$ can be found.

[hide="Harder question(Solution)"]
First,
odd * odd = odd
even * odd = even
even * even = even

odd - odd = even
odd - even = odd
even - odd = odd
even - even = even

O = Odd
E = Even
Possible even out comes:
O*E - O*E (number of outcomes O*E*O*E = $ 1004^4$)
O*E - E*O (number of outcomes ... = $ 1004^4$)
E*O - O*E (number of outcomes ... = $ 1004^4$)
E*O - E*O (number of outcomes ... = $ 1004^4$)
O*E - E*E (number of outcomes ... = $ 1004^4$)
E*O - E*E (number of outcomes ... = $ 1004^4$)
E*E - O*E (number of outcomes ... = $ 1004^4$)
E*E - E*O (number of outcomes ... = $ 1004^4$)
E*E - E*E (number of outcomes ... = $ 1004^4$)
O*O - O*O (number of outcomes ... = $ 1004^4$)

Total odd numbers from 0-2007: 1004
Total even numbers from 0-2007: 1004
Total outcomes: $ 2008^4$
Probability: $ \dfrac{10*1004^4}{2^4*1004^4}\equal{}\dfrac{10}{2^4}\equal{}\dfrac{5}{8}$
Probability I am wrong: $ 60$%
[/hide]

Question(Easy):
The Real Numbers: $ a\not\equal{}0$ and $ b\not\equal{}0$ have a specification such that $ ab\equal{}a\minus{}b$.

What is the value of $ \dfrac{a}{b}\plus{}\dfrac{b}{a}\minus{}ab$?

Question(hard):
Simplify the following Polinomial
$ P(n)\equal{}(2\plus{}1)(2^2\plus{}1)(2^4\plus{}1)(2^8\plus{}1)\cdots(2^{2^n}\plus{}1)$

[hide]

For the easier problem:

(a^2 + b^2)/a-b - (a-b)

= (a^2 + b^2 - (a-b)^2) / (a-b)

Then you expand and combine terms on the numerator to get:

2ab/(a-b) which is 2ab/ab = 2 [/hide]

Yes it is.
[hide="Hint to harder question"]
$ (2 \minus{} 1)P(n)$
[/hide]

Another question:
What digit does the number $ 2^{2008} \plus{} 2^{2009}$ end in?

[hide]

This post addresses the problem posted earlier.

tonypr: your answer is correct, but your method is somewhat too complicated

You can have ad and bc both odd or both even

If you have them both odd, then both a, d, b, and c must all be odd. The probability of this happening is (1/4) ^4 = 1/16.

Then if you want ad and bc to be even, you just have to subtract the probability that both ad and bc are odd. This is equal to 1 - (1/4) which is 3/4. Squaring 3/4 (we're considering both ad and bc) yields 9/16.

9/16 + 1/16 = 10/16 = 5/8. [/hide]

$ P(n)\equal{}(2\minus{}1)P(n)\equal{}2^{2^{n\plus{}1}}\minus{}1$ after doing $ a^2\minus{}b^2\equal{}(a\plus{}b)(a\minus{}b)$

Harder question:
If we divide a square with 8 lines, 4 parallel to 2 sides, and other 4 to other 2 sides, then if there are 4 squares among 25 little rectangles, there's a 5th square among them. Prove.


For a positive integer $n$ let $d(n)$ be the sum of its digits.when written in demical form.Prove that for each positive integer $k$ there exists a positive integer $m$ such that the equation    $x+d(x)=m$ has exactly $k$ positive integer solutions.

This is a Romanian ?Year Unknown? Problem 6.
It was unsolved on the forum (if I remember right) so if you have a solution it will be a joy for us to know it !


What is the radius of the largest circle that can be contained in a cube of side length 1?

[hide="hint"]
This circle would be inscribed in a regular hexagon whose sides lie on the faces of the cube
[/hide]
I'm actually not sure that i can prove that that case is the maximum, but I'm pretty sure.
Assuming that this condition is ture...
[hide]
This hexagon would be formed by taking the midpoints of 6 edges of the cube such that each face would have exactly 2 adjacent dges among the 6 chosed edges and connection them in order to make a regular hexagon.
this would lead to an edge length of $ \frac{\sqrt2}{2}$, and since the radius of a circle inscribed in a hexagon is the side length of teh hexagon, this is also our answer.
(this fact is found by drawing the longest diagonals in a hexagon and forming equilateral triangles.
[/hide]

I think you'll have to multiply the side length of the hexagon by $ \frac {\sqrt{3}}{2}$ to get the radius of the circle inscribed in the hexagon, so the correct answer would be $ \frac{\sqrt{6}}{4}$. I also strongly feel that this is the correct answer, but I don't know how to prove it.. How do we know that by tilting the axis of the circle elsewhere we can enlarge the circle a bit while keeping it within the cube?

I agree The answer should be $ r \equal{} \frac {\sqrt {6}}{4}$, here's an idea for a proof...

[hide="solution"]
Firstly, we must show that the center of the circle must also be the center of the cube(which i'll call the centroid for brievity). Suppose the center of the circle was not the centroid, then the circle would be closer to one face of the cube than the other (and also closer than the centroid is to that face). Therefore we would be able to shift the center of the circle onto the centroid and it would still be contained within the cube. Hence wlog we may assume that the center of the circle is also the center of the cube

Suppose there is a sphere inscribed inside our unit cube. On this sphere there is an infinite number of circles - all with radius $ \frac {1}{2}$
Now suppose the sphere increases, but its center and the cube remains constant.
as the sphere increases it protrudes from each face of the cube; creating a circles on each face

Eventually we wont be able to draw a complete circle arround the sphere such that it never leaves the interior of the cube, and we must find this point.

Now, if we have a circle running arround the sphere then we can also extend and trace that circle onto the cube making a straight line. So let's fold the cube out
 [geogebra]aa5e4c448e18adc38ec839fb15691c0bc8e7a551[/geogebra]

The radius of our circle will be maximised when the circles on the face of each cube are maximised (and we can still draw a straight line between them)
Therefore from here it is clear that the maximum size of the circles on the face of each cube occurs in the situation above - where they are tangent to line runing though the midpoints of each edge.
This happens to be the same situation described by [b]steven meow[/b]
And the maximum radius is $ \frac {\sqrt {6}}{4}$
[/hide]


It looked kinda scary at first when I saw the factorial symbol but got better when I saw the subtraction.  :)

[hide="answer"]With some log rules, we have $\log_{10} \frac{225!}{223!}=1 + \log_{10} n!$

Next, $\log_{10} 5040 = \log_{10} n!$
$n!=5040$ and $n=7$
[/hide]

Solution:
[hide]
$\\\log 225!-\log 223!=1+\log (n!)\\\log 50400=1+\log (n!)\\n!=10^{\log 50400-1}=5040\\n=7\\\mathbb{QED}$
[/hide]

EDIT: 4everwise was faster. :(

Masoud Zargar


Another proof:
  We have  \sum 1/x 2  \geq 4/ \sqrt xyzt \geq 4. But 4(xy+yz+zt+tx) \leq (x+y+z+t) 2  (well known)=16. So, xy+yz+zt+tx \leq 4 \leq  \sum 1/x 2


geometry
parallelogram
geometry proposed

Points $ K,M$ are taken on diagonal $ AC$ at points $ P,T$ on diagonal $ BD$ of convex quadrilateral $ ABCD$ such that $ AK=MC=\frac{1}{4}AC$ and $ BP=TD=\frac{1}{4}BD$. Prove that the line passing through the midpoints of $ AD,BC$ bisects the segments $ PM$ and $ KT$.

[quote="cancer"][color=darkred]Points $ K,\,M$ are taken on diagonal $ AC$ at points $ P,\,T$ on diagonal $ BD$ of convex quadrilateral $ ABCD$ such that $ AK=MC=\frac{1}{4}AC$ and $ BP=TD=\frac{1}{4}BD.$
Prove that the line passing through the midpoints of $ AD,\,BC$ bisects the segments $ PM$ and $ KT$[/color][/quote]
[color=indigo][b]Proof.[/b] Denote $ G,\,H,\,I,\,J,\,S$ be the midpoints of $ AD,\,BC,\,AC,\,BD,\,IJ,$ respectively.
We have $ GIHJ$ is a parallelogram, so $ S\in GH\quad (*)$

We also have: $ MHPS$ is a parallelogram, so $ SH$ bisects the segment $ PM$
Similarly, $ SG$ bisects the segment $ TK$
Hence the line passing through the midpoints of $ AD,\,BC$ bisects the segments $ PM$ and $ KT$[/color]


calculus
integration
geometry
3D geometry
sphere
analytic geometry
calculus computations

In Euclidean space, take the point $ N(0,\ 0,\ 1)$ on the sphere $ S$ with radius 1 centered in the origin. For moving points $ P,\ Q$ on $ S$ such that $ NP \equal{} NQ$ and $ \angle{PNQ} \equal{} \theta \left(0 < \theta < \frac {\pi}{2}\right),$ consider the solid figure $ T$ in which the line segment $ PQ$ can be passed.

(1) Show that $ z$ coordinates of $ P,\ Q$ are equal.

(2) When $ P$ is on the palne $ z \equal{} h,$ express the length of $ PQ$ in terms of $ \theta$ and $ h$.

(3) Draw the outline of the cross section by cutting $ T$ by the plane $ z \equal{} h$, then express the area in terms of $ \theta$ and $ h$.

(4) Pay attention to the range for which $ h$ can be valued, express the volume $ V$ of $ T$ in terms of $ \theta$, then  find the maximum $ V$ when let $ \theta$ vary.

[quote="kunny"]consider the solid figure $ T$ by forming the line segment $ PQ$ passes.[/quote]

I don't understand this statement.

[quote="kunny"](4) Pay attention to the range for which $ h$ can be valued, express the volume $ V$ of $ T$ in terms of $ \theta$, then  find the maximum $ V$ when let $ \theta$ move.[/quote]

I also don't understand this statement.

Sorry for my poor English.I have just edited. How about this time?


inequalities
graph theory
combinatorics proposed
combinatorics

Given a graph with $n\ge 2$ vertices s.t. the degree of every vertex is $>\frac n2$, prove that any two vertices are the ends of some hamiltonian chain.

It comes straightforward from the Ore's Theorem [url=http://mathworld.wolfram.com/OresTheorem.html]mathworld.wolfram.com/OresTheorem.html[/url].
An easy, indirect proof can be found in many discrete math books.

Could you still post a complete proof though? I don't think it's [b]that[/b] obvious how it follows from Ore's Theorem :). And I didn't even apply Ore's Theorem "per se".

Ok, soon I'll place a proof of this theorem, note that the graph you're talking about fulfills that theorem's conditions.

Since every vertex has degree more than $n/2$ then the graph is connected.

Consider the maximum path $P=v_0,...,v_k$. Since $P$ is maximm all the neighbors of $v_0$ and of $v_k$ lie in $P$. SO at least $n/2$ of the vertices $v_0,...,v_{k-1}$ are adjacent to $v_k$ and at least $n/2$  of these same vertices $v_i$ are such that $v_0 v_{i+1}$ is an edge. By pigeon hole principle,  there exists a vertex say $v_i$ that has both properties, so we have that $v_0 v_{i+1}$  and $v_i v_k$ are both edges. So now it is easy to see that the cycle $C=v_0 v_{i+1}Pv_kv_iPv_0$ is a hamilton cycle in the graph.  Otherwise, since the graph is conncted, $C$ would have a neighbor in $G_C$ but this neighbor plus a spannig path of $C$ makes a path longer than $P$ contradicting its maximality. So the graph must conain a hamilton cycle.

People, did you understand what the problem I posted asks? :)

It does not ask for a proof that the graph is hamiltonian. For this, it suffices (it follows immediately from Ore's Theorem, cited above) to have all the degrees $\ge\frac n2$. The hypothesis here is stronger: all the degrees are $>\frac n2$, and we have to prove something else: given [b]any[/b] two vertices $x_1,x_n$ of the graph, there is a hamiltonian chain (not cycle, i.e. $x_1$ need not be connected to $x_n$) $x_1,x_2,\ldots,x_n$ beginning with $x_1$ and ending with $x_n$.

Oops!  :blush:
I'll think on it.
Grobber, is a halmilton chain the same thing as a spannig path right?

here is an approach(i still haven't finished it):
the question is equivalent to showing that in a graph where $\delta(G)>n/2$, every edge is contained in a Hamiltonian cycle(if $u,v$ are not adjacent, then consider the graph $G\cup\{e\}$ where $e=\{u,v\}$.then this new graph would have a hamiltonian cycle containing the edge $e$ and hence the original graph has a hamiltonian path between $u$ and $v$).
 for such a pair of vertices $u,v$ consider all the cycles of $G$ that contain the edge $\{u,v\}$. this isn't empty since the degree condition implies that $N(u)\cap N(v)\neq \emptyset$.Let $C$ be a maximum such cycle. Now every vertex $x\in V(G)\setminus V(C)$ can have atmost $\frac{|C|+1}{2}$ neighbors in $C$(or else there are 2 'consecutive' vertices,other than $u,v$ of $C$ adjacent to $x$ and this admits a larger cycle containing $\{u,v\}$) and so every vertex $x\in V(G)\setminus V(C)$ has $\geq (n/2-c/2)=(n-c)/2$ neighbors off $C$ and so by Ore's thm $G\setminus C$ is hamiltonian.
now we seek to enlarge the cycle $C$ to get a contradiction. if 2 consecutive vertices of $C$(other than $u,v$) have neighbors off $C$ then such an enlargement of $C$ comes immediately(since $G\setminus C$ is Hamiltonian).

after a few futile attempts along that direction i proposed(which seemed promising) i was more and more convinced that there must be another way out...!
  Let us do the following: For a given pair of vertices $u\neq v$ we form a new graph $G_{u,v}$ with vertex set $V(G)\cup\{x_{uv}\}$ and the edges being the edges of $G$ along with a couple of new edges-joining $u,v$ to the new vertex $x_{uv}$.now this new graph on $n+1$ vertices is Hamiltonian iff the original graph $G$ has a hamiltonian path between $u$ and $v$.
 To see that this new graph $G_{u,v}$ is hamiltonian we take its Hamiltonian closure,i.e. for every pair of nonadjacent vertices $x,y$ such that $d(x)+d(y)\geq (n+1)$ we add the edge $\{x,y\}$ and keep doing this till we no longer can add newer edges.
now a graph is hamiltonian iff its closure is.denoting by $\overline{G_{u,v}}$ the closure, we note that in the closure, since for every $x,y\in V(G)$,nonadjacent, we have $d(x)+d(y)>n\Rightarrow d(x)+d(y)\geq n+1$, so $\overline{G_{u,v}}$ restricted to $V(G)$ gives the complete graph $K_n$, and hence $G_{u,v}$ is hamiltonian and that finishes the proof.
note that all this proof needs is that for every pair of nonadjacent vertices $x\neq y$ of $G$, we have $d(x)+d(y)>n$.(so the given condition is stronger).

sorry, I've mistaken the definition of a chain :)

Here is my proposal for a solution:

Let's use induction. For $n=3$ it's obvious.
Now let's have a graph with $n (n>3)$ vertices. Suppose that there exists a pair of vertices $x,y$ which are not the ends of a a hamiltonian chain. The number of edges equals $k < n $. Let's name the group of vertices belonging to the chain by $X$, let $Y$ be the group of vertices which not belong to $Y$.
Let's note that the degree of every vertice in $Y$ is bigger than $\frac{n-k}{2}$ cause if there is a vertice with a smaller degree in $Y$ then it is connected with at least $\frac{k}{2}$ vertices from the chain. That means that     there exist two vertices from the chain conected with that vertice from $Y$ and wich are consecutive edges in the chain. That means that we can expand our chain with that element from $Y$ - contradiction coming from maximality of $k$. Apllying the inductive assumption subset $Y$ satisfies the theorem of the problem (inductive condition) -.
Now let's solve two cases (which solve the problem):
(1)If $ k=|X| \leq \frac{n}{2}$ then it's easy cause picking two consecutive edges $a,b$ from the chain -
each of them is connected with an edge with some elements in $Y$ so we can expand our chain replacing $a,b$ with
$a,k_1,...,k_l,b$ were $k_1,...,k_l$ is the fragment of the hamilton cycle in $Y$ - so we have
a contradiction because of the maximality of $k$.
(2)If $ k > \frac{n}{2}$ then each vertice from $Y$ is conected with more then $\frac{n}{2}-(n-k)+1$
 vertices in the chain in $X$. Because $Y$ satisfies the inductive condition - in $Y$ for every two vertices
there exists a hamiltonian chain with ends in those two vertices. So if there exist two vertices from $Y$
$a,b$ when $a$ is connected with $c$ from the chain and $b$ is connected with $d$ from the chain ($c,d$ are from $X$ and in addition there are less then $n-k$ vertices in the chain which lie between $a$ and $b$ - then we can expand our chain adding to $X$ the chain from $Y$ - so we have a contradiction with the maximality of $k$.
We will show that such $c,d$ in chain $X$ must exist. There are more than $(n-k)(\frac{n}{2}-(n-k))$ vertices conecting $X$ with $Y$. The maximal value of vertices connecting $X$ with $Y$ for which there is no pair of vertices in $X$ which have less than $n-k$ vertices between them in the chain is $\frac{(n-k)k}{n-k+1}$ (every vertice from $X$ connected with some vertice from $Y$ is connected with all from $Y$) but(!!) $(n-k)(\frac{n}{2}-(n-k)+1) \geq \frac{(n-k)k}{n-k+1}$. In order to show that let's put $k=|X|=a(t+1)+r$, $|Y|=t$ for $t,a$ positive integer and $0 \leq r \leq t$. The inequality is equivalent to $\frac{a(t+1)+r+t+1}{2}-a(t+1)-r-t+a(t+1)+r+1 \geq \frac{a(t+1)+r}{t+1}\"=\"a+1$-if $r>0$,$a$ if $r=0$. What is equivalent to $\frac{a(t+1)+r+t+1}{2}+1 \geq a+1+t$ or $a+t$, but $(a+1)(t+1) \geq 2a+2t$ - this inequality holds for all $a,t$ positive integers. That means that we have have a contradiction (it's against the maximality of $k$) - we can indeed expand our chain $X$ adding the chain $Y$ - of lenght $t$.

suppose that it isn't true, so there exist $a,b$ such that there doesn't exist hamiltonian chain
if it is not true, we can add more edges untill every next adding of edge will give us hamiltonian chain beween $a$ and $b$, thus let us add such number of edges, we have now
$a \rightarrow x_1 \rightarrow x_2 \rightarrow ... \rightarrow x_{n-3} \rightarrow b$
but $x_n$ is connected with more than $\frac{n}{2}$ vertices so it must be connected with one of pairs
$\{a,x_1\},\{x_1,x_2\},..,\{x_{n-3},b\}$. contradiction, becouse there shouldn't be any chain between $a,b$


calculus
integration
real analysis
real analysis unsolved

Compare the integrals $\int_{0}^{1}x^{x}dx$ and $\int_{0}^{1}\int_{0}^{1}(xy)^{xy}dxdy.$
(V. Radchenko)


[quote="Michael"]Hi,

$n\in{\Bbb N}\setminus\{0\}$, $x\in{\Bbb R}$. Prove that :
\[\lfloor x\rfloor^{2}+\left\lfloor x+{1\over n}\right\rfloor^{2}+\cdots+\left\lfloor x+{{n-1}\over{n}}\right\rfloor^{2}=\lfloor nx\rfloor^{2}\]
if and only if $\lfloor nx\rfloor=\lfloor x\rfloor$.[/quote]
Let $\lfloor x\rfloor =a\in Z$ and $\frac{k}{n}\le\{x\}<\frac{k+1}{n},0\le k<n$. Then left expression is $(n-k)a^{2}+k(a+1)^{2}=na^{2}+k(2a+1)$
and right expression is $(na+k)^{2}$.
Therefore $n(n-1)a^{2}+k(n-2)a+k^{2}-k=0$. Let n>1, then this equation had only solution $a=k=0$, because $D=k^{2}(n-2)^{2}<4n(n-1)k(k-1)$ for k>1, k=0 give a=0, k=1 had non integer solution.
$a=0=k$ mean  $\lfloor x\rfloor =\lfloor nx\rfloor$.


hello, $ 4^3\plus{}6^2\equal{}64\plus{}36\equal{}100\equal{}10^2$, so we get $ 10^2\equal{}10^x \Leftrightarrow x\equal{}2$.
Sonnhard.


for those who know more about the history of mathcounts than i do, has a score of 46 ever been achieved by an individual at states? at national? if not what are the highest scores ever for state and national?

also, what was the highest team score ever achieved, state and national?

im pretty sure there were people with perfects mathcounts aint the hardest math competition

but, mathcounts is only US, and not like china or something.

and if bobby shen didn't get a perfect, then....yeah.

Well, what about in 1984? Of course people got perfects.

Yeah, even at Nationals. Mr. Crawford did that.

Hard to tell. Some years have easier/harder tests than others. The skill of the participants also varies from year to year.

@math154: Can't be entirely sure. The nats participants in 1984 were probably not as smart as recent years' participants, partly because people now know what to expect (from being able to study past tests) and because MATHCOUNTS has "created" smarter participants. Parent and schools now teach their kids problem-solving math from early ages, to prepare them for contests like MATHCOUNTS. Thus, we can't be sure.

EDIT: People did get perfect scores.

Um no, word got out wrong. DWu got a 42. Texas 3-way tie was definitely top score in the nation. But still, DWu is very beast.

[quote="mathking123"][quote="tinytim"]DWu got a perfect at states this year.[/quote]seriously? i thought the highest was that 3 way tie in texas....wow

dwu is beast, looking forward to next year mathcounts to (probably) seeing him and bobby in final round of cd :D that will be exiting and interesting[/quote]
Actually, it might not be that, since if they're in the same half of the bracket, you'll probably see them sooner.

good point, but still it will be interesting.

on that topic, does anyone know where i can find a video of the 2008 countdown round?

i think mathcounts took it down(?)

I think at a national level, people would get it perfect[/i]

If you are a test-writer, you ideally want your test to be such that nobody gets a perfect score and nobody gets a score of 0. Think about it. If 3 people had perfect scores at nats, what would you do? How do you decide who gets the 1st place trophy/award? Also, a test in which 25% of people score 0 would be bad as well (not good for comparing skills accurately). The best tests are not the ones on which a perfect score CAN be achieved on or ones that many people fail, but the ones that can accurately seperate the smartest people from the others.

This is why the MATHCOUNTS tests are so good. It is as hard to get a score of 0 as it is to get a score of 46 (provided you try your hardest and use all your skills and knowledge). Tests like these are rare and very good for competitions.

They would use a tiebreaker round, which considers time. Also, I believe the highest score in California was also a 44.


Outline reasonable syntheses of 6-acetyl-1,2,3,4-tetrahydropyridine, the principal compound responsible for the aroma of bread, and of 2-acetyl-1-pyrroline, the principal component of cooked rice.

for second one- pyrrole + houben housche or vilsmeier haack reaction to get 2 acetyl pyrrole foll. by birch reduction.
not sure abt the reduction step.

Yes, that reduction procedure doesn't also look reliable to me.

for the first one - maybe u cud use an aza diels alder reaction.

ne of these two reaction ofc with the acyl group attached to the substrate.


Let $a_1,a_2,...,a_n$ be distinct integers with $n\geq1$. Show that the polynomial
$f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)-1$
cannot be decomposed as a product of two integral polynomials

[quote="Altheman"]$f(x) = (x-1)(x+1)-1$
$f(x) = x^{2}$
$f(x) = (x)(x)$
but if you mean that $a_{n} \ge 1$
that would be different[/quote]um $(x-1)(x+1)-1=x^2-1-1=x^2-2...$

anyone want to try (HINT: the statement is true)

anybody want to try? Come on its not too hard

Well lets suppose that it can:

then $f(x)=p(x)q(x)$ whicn have integral coeficients, but if we evaluate $x$ like $a_1$, $a_2$, ...,$a_n$ we have:

$f(a_i)=-1=p(a_i)q(a_i)$  but both are integrers so one has to be $1$ and the other $-1$ let´s say that the grade of $p(x)$ is less of equal than the grade of $q(x)$.
And let´s see that we have n values of $p(x)$ some 1´s and some (-1)´s but there are half or more of one....let´s say there are more 1´s (the other case is equal).

Then the number of 1´s is $\ge$ than n/2 and that is $\ge$ than the grade of $p(x)$ but if the both of them arent  $n/2$ then $p(x)-1$ would have more roots than its grade so it would be constant wich is impossible...then $p(x)$ is of grade n/2 and it has n/2 q´s and -1´s being evaluated in the roots of $f(x)+1$...

but then $p(x)+1$ divides $f(x)+1$ because they have the same roots and also $p(x)-1$  but they have different roots so $(p(x)-1)(p(x)+1) divides f(x)+1$ but they have he same grade so $p(x)^2-1=f(x)+1$ $(1)$
and also is the same for $q(x)$ because they have the same grade so $q(x)^2-1=f(x)+1$ and then $p(x)=q(x)$ but then p(x)^2=f(x)$ that´s absurd becuase of $(1)$

sorry about the last thing

but then $p(x)^2=f(x)$ whcih is absurd because of the $(1)$

JoshuaMex, instead of making a new post to correct something you did wrong in your previous one, you can edit your previous one.  Simply click "edit" at the bottom of the post you want to edit.

I like the first part of the solution, but theres an easier way:

call $g(x)=p(x)+q(x)$, and since we established that $p(a_i)=\pm1$ and $q(a_i)=\mp1$, then $g(a_i)=0$.  This means that either:

since there are at least $n$ zeros, $\deg[g(x)]\geq{n}$, which means that at least one of $p(x)$ or $q(x)$ is of degree at least $n$ also, which already contradicts.

or that $g(x)=0$. In this case, $p(x)=-q(x)$, so $f(x)=-q(x)^2$, which can't be true becuase of the  leading coefficient of $f(x)$ is positive


I'm afraid this might take a while....there is no integral for this function!
This is the "Bell Curve", which cannot be integrated (right?).

RicOh

I would say definitely not HSB... but you write that as a maclaurin series and integrate no?

They don't even allow calculus problems on the Intermediate board.  They definitely won't allow them on HSB.

[b]Please post calculus problems in the Calculus Computations and Tutorials forum.[/b]

[url]http://www.artofproblemsolving.com/Forum/index.php?f=296[/url]


function
real analysis
real analysis unsolved

Let $ f: [a,b] \rightarrow \mathbb{R}$ be monotone.
Prove $ f$ has a countabe set of discontinuities.

Here's an idea.

Let's say $ f$ is non-decreasing.  Let $ S$ be the set of points where $ f$ is discontinuous.  For each $ x$ in $ S$, associate with $ x$ a rational number in between $ f(x\minus{})$ and $ f(x\plus{})$.  Prove that this correspondence is $ 1\minus{}1$, using the fact that $ f$ is non-decreasing.  $ S$ has now been put into a $ 1\minus{}1$ correspondence with a subset of the rational numbers, so $ S$ is countable.

ok nice. i have one solution that sounds like it starts off the same as this and will post it soon.

Well here it goes
[hide]
     Assume WOLOG $ f$ to be increasing and set $ f(a \minus{} ) \equal{} f(a)$ and $ f(b \plus{} ) \equal{} f(b)$. Then $ \forall x \in [a,b]$ $ f(x \minus{} )$ and $ f(x \plus{} )$ exist and $ f(x \minus{} ) \leq f(x) \leq f(x \plus{} )$ with a discontinuity at $ x$ meaning $ f(x \minus{} ) < f(x \plus{} )$.
     Now consider the function $ s(x) \equal{} f(x \plus{} ) \minus{} f(x \minus{} )$ which represents the "jump". We say $ f$ is discontinuous at $ x$ iff $ s(x) > 0$. Now we want to look at the set $ s_n \equal{} \{x \in [a,b] : s(x) \geq \frac {1}{n}\}$ for $ n \in \mathbb{N}$.  Say we have $ x_1,x_2, \ldots,x_k \in s_n$ with $ a \leq x_1 < x_2 \cdots < x_k \leq b$.
     Select points $ y_0 \equal{} a \leq x_1 < y_1 < x_2 < y_2 \cdots < x_k < y_k \equal{} b$. Clearly, we have $ f(y_k) \minus{} f(y_{k \minus{} 1}) \geq f(x_k \plus{} ) \minus{} f(x_k \minus{} ) \equal{} s(x_k) \geq \frac {1}{n}$.
Now we add them all up,
\[ \left[f(y_1) \minus{} f(y_0)\right] \plus{} \left[f(y_2) \minus{} f(y_1) \right] \plus{} \cdots \plus{} \left[f(y_k) \minus{} f(y_{k \minus{} 1})\right] \equal{} f(y_k) \minus{} f(y_0) \equal{} f(b) \minus{} f(a) \geq \frac {k}{n}
\]
This implies that $ s_n$ has at most $ n \left[f(b) \minus{} f(a)\right]$ elements, that is $ s_n$ is finite.
     Now let $ \mathfrak{D} \equal{} \{ x \in [a,b] : f$ discontinuous at $ x \}$. If $ x \in \mathfrak{D}$ then $ s(x) > 0$ hence, $ \exists n \in \mathbb{N}$ such that $ s(x) \geq \frac {1}{n}$ therefore $ x_n \in s_n$. If $ x \in s_n$ for some $ n$ then $ s(x) \geq \frac {1}{n} > 0$ hence $ x \in \mathfrak{D}$. Thus $ \mathfrak{D} \equal{} \displaystyle \bigcup_{n \equal{} 1}^{\infty} s_n$. And since $ s_n$ is finite $ \mathfrak{D}$ is countable.
[/hide]

I would get rid of that first thing you do where you say $ f(a\minus{})\equal{}f(a)$ etc. Notationally that gets confusing because you then say $ f(x\minus{})\leq f(x)\leq f(x\plus{})$ which is true, but doesn't really cohere with that first part.

Next. I can't really tell what you did to get $ s_n$ to be finite. It seems that when you "select points" you assume that there are only finitely many because you write them down up to $ k$. I could be misunderstanding what you did there, though.  :huh:

First off, it may be strange that I define $ f(a\minus{}) \equal{} f(a)$ and so on, but I do so to make sure the function is well behaved on the interval $ [a,b]$ and it doesn't contradict the statement $ f(x\minus{}) \leq f(x) \leq f(x\plus{})$ I believe  :blush: .

Next, I know that $ s_n$ must be finite because were dealing with a finite interval and the sum of the jumps cannot exceed $ f(b) \minus{} f(a)$ and it must be finite. i think? :maybe:

I'm having trouble coming up with an counterexample for $ s_2$, say. Let me think about that some more. Maybe it is true. I have a feeling it isn't, though.

Sorry about the other thing. I had forgotten that the interval was labeled $ [a,b]$, I was thinking for some reason that you had just taken two arbitrary letters (I was doing it on $ [0,1]$).  :blush:

Wow. I'm being really thick-headed for some reason. That fact that I've been trying to dispute is obviously true. Suppose there are infinitely many, then by the same technique used later you have that $ f(b)\minus{}f(a)\geq\infty$ a contradiction. Thus each $ s_n$ is finite.

so does this mean you agree with what i said or is there still an error in my work?

I like the idea. I haven't seen this done this way before. I think you are trying too hard at parts, though. You are trying to bound the number of elements in $ s_n$ to get it to be finite. This is all you need. So, I would take the sentence $ s_n\equal{}\{x\in [a,b]: s(x)\geq\frac{1}{n}\}$ for $ n\in\mathbb{N}$ and alter it to: Let $ s_n\equal{}\{x\in [a,b]: s(x)\geq\frac{1}{n}\}$.

Now, I'd erase all of the next section to the definition of $ \mathfrak{D}$ and replace it with:

Claim: $ s_n$ is finite for each $ n\in\mathbb{N}$. Suppose not for some $ s_m$. Then just looking at the "gaps" in $ s_m$ we have $ f(b)\minus{}f(a)\geq |s_m|\cdot\frac{1}{m}\equal{}\infty$. (Where $ |s_m|$ is the order of that set). This contradicts our assumption that both $ f(a)$ and $ f(b)$ are finite, so each $ s_n$ is finite.

Your proof that if $ x\in\mathfrak{D}$ then $ x\in s_n$ for some $ n$ is fine. This gives you that $ \mathfrak{D}\subset\bigcup s_n$, or $ |\mathfrak{D}|\leq \big|\bigcup s_n\big|$, so $ \mathfrak{D}$ is at most countable.

A weird thing about this is that it seems like you count things too many times. I don't think this leads to error, but it bothers me. A simple redefining of $ s_n$ should do the trick. Here is what I mean: Suppose there is an $ x_0$ such that $ s(x_0)\equal{}1$, then $ x_0\in s_1$ and $ x_0\in s_2$ and $ x_0\in s_3$ etc.

This question has been asked recently, and kenn4000 gives a proof along the same lines here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=210782[/url]

A monotonic function only has discontinuity points of the first kind so you can use [url=http://www.mathlinks.ro/viewtopic.php?t=31694]this problem[/url].  :)

Thanks Hilbert for your insight. Sorry ive been away and havent responded but i definitely see where you are coming from. i have very minimal experience writing formal proofs and is something ive been working on just recently with rudin.
thanks again.


inequalities
geometry
incenter
circumcircle
inradius
rotation
geometry proposed

Let $ I$ be the incenter of a bicentric quadrilateral $ ABCD$ with diagonals $ AC$ and $ BD.$ Show that, if  $R, r$ denote its circumradius and inradius, then we have
\[2R^2\ge IA \cdot IC \plus{} IB \cdot ID\ge4r^2\]

$ \boxed{ IA*IC \plus{} IB*ID \equal{} 4Rr^2(\frac{1}{AC}\plus{}\frac{1}{BD})}$

1. $ AC \le 2R$ and $ BD \le 2R$ => $ IA*IC \plus{} IB*ID \ge 4r^2$

2. Let $ AB\equal{} a$ , $ BC\equal{} b$ , $ CD\equal{} c$ ,  $ DA\equal{} d$ , $ AC\equal{} d_1$ , $ BD\equal{} d_2$

$ \boxed{4Rrs\equal{}\sqrt{d_1d_2}\sqrt{(ab\plus{}cd)(bc\plus{}ad)}}$ ( $ s$ - semiperimeter )

$ \frac{4Rrs}{\sqrt{d_1d_2}} \le \frac{(a \plus{} c)(b\plus{}d)}{2}\equal{}\frac{s^2}{2}$

$ IA*IC \plus{} IB*ID \equal{} 4Rr^2\frac{d_1\plus{}d_2}{d_1d_2} \le \frac{s^2}{4} \le2R^2$ :)

Let $ (I,r),(O,R)$ be the incircle and circumcircle of $ABCD.$ $(I)$ touches its side-segments $AB,BC,CD,DA$ at $X,Y,Z,W.$ Simple angle chase gives $ \angle XAI \equal{} \angle YIC$ and also  $ \angle BYI \equal{} \angle DIZ.$ Rotate  $\triangle  AXI$ about $ I$ by angle $\angle XIY$ counterclokwise, thus $ X \mapsto Y$ and we get a right triangle with legs $\overline{IA}$ and $\overline{IC}.$ Similarly, rotating $ \triangle IBY$ about $I$ by angle $\angle  YIZ$ we obtain a right triangle with legs $\overline{ IB}$ and $\overline{ ID}.$ Thereby, area of $ABCD$ can be expresed as twice the sum of the areas of the two said right triangles. In other words, $ [ABCB] \equal{} IA \cdot IC \plus{} IB \cdot ID.$ Now, using the fact that in a fixed circle, the inscribed/circumscribed quadrilateral with maximum/minimum area is the square, it follows that

$2R^2\ge [ABCD] \ge 4r^2 \Longrightarrow  2R^2 \ge IA \cdot IC \plus{} IB \cdot ID \ge 4r^2.$


Hallo,

I've a problem wich I'm not able to solve it.

Find all integers for wich

(x 2 +y 2 )/(xy+1)

is an integer.

$ pgcd (xy, xy+1) = 1$ implies that $\frac{x^2+y^2}{xy+1}$ is integer ssi $\frac{x^3y+y^3x}{xy+1}$ is a integer. But,

$\frac{x^3y+y^3x}{xy+1} = x^2 + y^2 - \frac{x+y}{xy+1}$

That implies that :

$\frac{x+y}{xy+1}$ is a integer

But $(x-1)(y-1) = xy-x-y+1$

=> $x+y=<xy+1$ ssi x, y>=1

=> There is a unique solution $(x=1, y=1)$

It is known that the expression is integer iff its a perfect square, provable by infinite decent.
But the general case is crazily hard..

Bomb

My short solution for it:
[tex]x^2-kxy+y^2-k=0[/tex]
[tex](x_0,y_0)[/tex] is a root...
1/[tex]x=y->x=y=1[/tex] and k=1
2/WLOG ,assume [tex]x_0+y_0[/tex] is min  and [tex]x_0>y_0.[/tex]
Then [tex]x_1=ky_0-x_0[/tex] is another root
Easily prove [tex]x_1>0[/tex]
[tex]-> x_1>x_0>y_0>0[/tex]
[tex]->y_0^2<x_0.x_1=y_0^2-k[/tex]
[tex]->k<0->[/tex](not satisfy)

I'm Sorry. I made a mistake. The correct expression must be

(x²+y²)/(xy-1)

my conjecture is that (x²+y²)/(xy-1) is always equal to 5.
f.e. (x,y)=(67,14)


trigonometry
geometry
inequalities
inequalities unsolved

Prove that  :  In triangle  with $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$ then$ \sqrt {3}\max\lbrace h_a,h_b,h_c\rbrace\ge p$.

What about a triangle with angles $ 30, 30, 120$? Then (assuming the short side is 1) $ p \equal{} 1 \plus{} \frac {\sqrt {3}}{2}$ but the maximal altitude has height only $ \frac{\sqrt {3}}{2}$. (and $ \frac {3}{2} < 1 \plus{} \frac {\sqrt {3}}{2}$.)

Does the triangle have to be acute?

[quote="Hong Quy"]Prove that  :  In triangle  with $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$ then$ \sqrt {3}\max\lbrace h_a,h_b,h_c\rbrace\ge p$.[/quote]
 i forgot .
 You're  good  , not_trig
 The problem's  true in acute triangle.
i wait to see  a nice solution.

[hide]

Let $ k$ be the area of $ \triangle ABC$. Assume WLOG that $ a$, and hence $ \alpha$, is minimal. ($ a,b,c$ are sides and $ \alpha,\beta,\gamma$ are angles.)

From $ 2k = ah_a$, we see that $ h_a$ is maximal. Then, as $ k = rp$, the desired inequality is equivalent to
\[ 2\sqrt {3}r \ge a
\]
Assume for contradiction that $ 2\sqrt {3}r < a$. Then we will show that $ a$ is not minimal.

Fix $ a$ and $ r$. Then, the maximum of $ \min(b,c)$ occurs when $ b = c$. (I have yet to prove this; I will edit my post when I do :( )

So, it suffices to show it for $ b = c$. But then, $ \angle IBC = \angle ICB < 30$, so $ \angle ABC = \angle ACB < 60$; hence $ AB = BC < BC$, contradiction.
[/hide]

EDIT: I have proved it, but with a slightly different method.

[hide]
We want to show that $ 2\sqrt{3}r \ge a$. Considering triangles $ DIC, DIB$, where $ D$ is the point of tangency of the incircle with $ BC$, we see that $ a=r(\cot \frac{\beta}{2} + \cot \frac{\gamma}{2})$. Hence it suffices to show that

$ \cot \frac{\beta}{2} + \cot \frac{\gamma}{2} \le 2\sqrt{3}$.

Since $ \cot$ is convex, the maximum of $ \cot \frac{\beta}{2} + \cot \frac{\gamma}{2}$ for fixed $ \beta+\gamma$ occurs when $ \beta$ and $ \gamma$ are as far apart as possible. We know that $ \alpha+\beta+\gamma=180$, and the triangle is acute. (We may assume it is just non-obtuse, as $ cot$ is convex and monotonic on $ [0,90]$. Hence, the maximum is either when $ \alpha = \beta, \gamma=180-2\alpha$, or when $ \beta = 90-\alpha, \gamma=90$.

The former case is maximal when $ \alpha \ge 45$. As $ \alpha \le \beta, \gamma$, $ \alpha \le 60$. So, for $ 45 \le \alpha \le 60$, we want to show that

\[ \cot 90-\alpha  + \cot \frac{\alpha}{2} \le 2\sqrt{3}\]

Equality occurs at $ 60$, and when $ \alpha = 45$, the LHS is $ 2+\sqrt{2} < 2\sqrt{3} \Leftrightarrow 32 < 36$, as $ \cot \frac{45}{2} = \sqrt{\frac{1+\cos{45}}{1-\cos{45}} } = 1+\sqrt{2}$.. As the LHS is concave in $ \alpha$, the maximum occurs at the extrema - hence, the inequality is true here.

When $ \alpha \le 45$, the maximum is when $ \beta = 90-\alpha$, $ \gamma=90$. But then we just need to show that $ \cot 45-\frac{\alpha}{2} + \cot 45 \le 2\sqrt{3}$. As $ \cot$ is decreasing, it suffices to show that $ \cot \frac{45}{2} + \cot 45 = 2+\sqrt{2} < 2\sqrt{3}$. Hence we are done.
[/hide]

I have solved it, in the post above. (I edited it)


let $x,y>0 ,x+y=1$ , prove

$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}} \geq \sqrt2$

$f(x)=\frac{1}{\sqrt{x}}$ is convex in $(0,1)$ so applying Jensen's inequality
we have: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}} \geq \frac{1}{\sqrt{2xy}}$ but
$\frac{1}{\sqrt{2xy}} \geq \frac{2}{\sqrt{2}(x+y)}=\sqrt{2}$, q.e.d.

yes ,as you have said it will be easily killed by the power of jensen

i was trying to find the solution by using the means ,any idea about it?

Sure.

Substitute $x=a^2$, etc.  The LHS is transformed to

$\frac {a^3+b^3}{ab}$

By Power-Mean, $a^3+b^3\geq \frac {\sqrt {2}}{2}(a^2+b^2)^{\frac 32}$

Now $\frac {a^2+b^2}{ab}\geq 2$, so we get

$\frac {a^3+b^3}{ab}\geq \frac {\sqrt {2}}{2}\cdot \frac {a^2+b^2}{ab}\cdot \sqrt {a^2+b^2}\geq \frac {\sqrt {2}\cdot 2}{2}=\sqrt {2}$

thanks i really enjoyed it
there is no need of using strong tool of jensen
its like killing an ant with a cannon(jensen) , i by myself prefare to kill it by hand(means)

having not much time on the olympiad ...I'm not sure whether you would be even aware of killing an ant with a cannon :)


If a,b,c are positive integers such that $ 0<a^{2}\plus{}b^{2}\minus{}abc\leq c$, show that $ a^{2}\plus{}b^{2}\minus{}abc$ is a perfect square.


induction
function
search
algebra
binomial theorem

Could somebody show how to derive or prove, without using induction, the following identity:

\[ \binom{n}{1}+\binom{n}{3}+\cdots+\binom{n}{n-1}=2^{n-1}\]

is this assuming n is even? i.e., is n-1 odd?

[hide="Solution 1"]
We know,
$ {n\choose 0}+{n\choose 1}+...+{n\choose n}=2^{n}$
We also know that (use generating functions),
$ {n\choose 0}-{n\choose 1}+...\pm{n\choose n}= 0$
So let,
$ x={n\choose 0}+{n\choose 2}+...$
$ y={n\choose 1}+{n\choose 3}+...$
And we have,
$ x+y=2^{n}$
$ x-y=0$
Solve those systems.
[/hide]

[hide="Solution 2"]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160946]Here[/url]
[/hide]

Generating functions?  How about the binomial theorem?
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1765673554&t=130066
Related problems
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1765673554&t=155187
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1765673554&t=154452


geometry
circumcircle
trigonometry
cyclic quadrilateral
geometry proposed

Suppose $ABCD$ is a quadrilateral in which: $\frac{AB}{BC}=\frac{AD}{DC}$

$E,F$ are two points on $AB,AD$, in which line$EF$, is antiparalell with $BD$.

If $AC$,bisects $EF$, prove that, the quadrilateral is cyclic.

For easiness, we can prove the follows alternative problem.

     A triangle ABD is given and let C be, the intersection point of it’s circumcircle, from the symmedian through the vertex A. In cyclic
quadrilateral ABCD prove that:  (AB)/(BC) = (AD)/(DC)  (1).

     But this is clear because of, from the point C on the symmedian of ABD through A, we have that (B'C)/(D'C) = (AB)/(AD),  (2), where B', D',  are the orthogonal projections of C on the sidelines  AB, AD,   respectively.

    From the similar right triangles BB'C, DD'C, we have also that (B'C)/(DC) = (BC)/(DC),   (3).

    From (2), (3), =>  (1) and the proof is completed.

PS. It is well known that if E, F, are two points on AB, AD respectively, such that the line EF to be antiparallel to BD, with respect to the lines AB, AD, then the line AC bisects the segment EF.

    Kostas Vittas.

$AC$ is the symmedian of $\triangle BAD$ so $\frac{\sin{DAC}}{\sin{BAC}}=\frac{DC}{BC}=\frac{DA}{AB}\rightarrow \sin{ADC}=\sin{ABC}$ so either ABCD is cyclic either ABC=ADC in this last case the quadrilateral is a kite  :|

sorry dears, i dont have much experience to design hard problems :D


Does the AMC or the AIME help someone qualify for the ARML???

That depends 100% on how your local ARML organization chooses its team-members.

Check the [url=http://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team]wiki page.[/url] The information there is not universal, but there is something about selection for quite a few of the teams.


[quote]Copyright © 2007 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others[/quote]
so i can't post this on the internet, share with friends, etc.? i think this has been asked before, but i just want to double-check.

Basically, just don't post it publicly on the web or give it to anyone outside your family.
 :P

does that mean i cant take problems from the class and use them anywhere else? say...at school?
am i allowed to change the numbers around to make a similar problem? please let me know.

Well...it really depends on which class. For example, if it's from the AIME class, you can obviously share those, since they're contest problems. If it's a problem directly from an AOPS book, I think you can change the wording and the numbers around...In general, just don't post things publicly on the internet.

Can you post questions you have on problems in AOPS texts?  Or can you post problems from mathcounts books in the mathcounts marathons in the mathcounts forum?  Can you modify a problem and then post it?

If you have a question about a problem in one of the AoPS books, you can feel free to post it on the AoPS forum to ask about it.

[quote="rrusczyk"]You definitely [b]cannot[/b] post it on the internet or make copies of the transcripts or problem sets to give to others.  You certainly can discuss problems and ideas from the class with friends.[/quote]
thanks i really appreciate it


Can anyone post solution  official USAMO 2009?

The MAA/AMC office will not post the official solutions to the 2009 USAMO until after the grading is completed, on May 11, 2009.

Steve Dunbar
MAA Director, American Mathematics Competitions


I don't think this is easy and don't sure it is right,but I need it to do a difficult problem:
Let  \sqrt a, \sqrt b, \sqrt c are three sides of a triangle.Prove that:
 a 2 b+b 2 c+c 2 a-ab 2 -bc 2 -ca 2  \leq 3/4abc.

Dear Anh Cuong

Your ineq is wrong. Let's take  a = 36, b = 9, c = 49  we will find that

a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2 - (3/4)abc = 2844.

Let's know your original problem.

What a pity!!!  :(  My original problem is from 'From an IMO problem' in 'Advanced Section -> Inequalities -> Unsolved Problems'.But i want to find a greater condition for that one and it lead me up to this inequality.


Let $ x,$ $ y,$ $ z,$ $ \alpha,$ $ \beta$ and $ \gamma$ are real numbers such that $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi.$ Prove that
\[ x^2 \plus{} y^2 \plus{} z^2\geq2xy\cos\gamma \plus{} 2xz\cos\beta \plus{} 2yz\cos\alpha
\]
There is a short proof. :wink:

$ \sum_{cyc}( x^{2} \minus{} 2yz \cos \alpha) \equal{} (z \minus{} (x \cos \beta \plus{} y \cos \alpha))^{2} \plus{} (x \sin \beta \minus{} y \sin \alpha)^{2}$

Beautiful, pardesi!
What about the following problem:
Let $ x,$ $ y$, $ z,$  $ t$ $ \alpha,$ $ \beta,$ $ \gamma$ and $ \delta$ are real numbers such that $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \equal{} \pi.$ Prove that
\[ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2\geq\sqrt2xy\cos\alpha \plus{} \sqrt2yz\cos\beta \plus{} \sqrt2zt\cos\gamma \plus{} \sqrt2tx\cos\delta
\]
I haven't short proof for this one.


logarithms
inequalities
inequalities unsolved

Taking logarithms on both sides, the inequality is equivalent to

$ (x^2 \plus{} 2yz) log x \plus{} (y^2 \plus{} 2zx) log y \plus{} (z^2 \plus{} 2xy) log z \geq (xy\plus{}yz\plus{}zx) (log x \plus{} log y \plus{} log z)$
Sending all terms to the lefthand side, we wish to prove that

$ (x^2 \minus{} xy \minus{} xz \plus{} yz) log x \plus{} (y^2 \minus{} yz \minus{} yx \plus{} xz) log y \plus{} (z^2 \minus{} zx \minus{} zy \plus{} xy) log z \geq 0$

$ (x\minus{}y)(x\minus{}z) log x \plus{} (y\minus{}x)(y\minus{}z) log y \plus{} (z\minus{}x)(z\minus{}y) log z \geq 0$

Since this inequality is symmetric in the three variables $ x,y,z$, assume that $ x \geq y \geq z$

Due to the assumption $ x,y,z > 1$, $ log x , log y , log z > 0$.

Now, the inequality holds because
$ (x\minus{}y)(x\minus{}z) log x \plus{} (y\minus{}x)(y\minus{}z) log y \plus{} (z\minus{}x)(z\minus{}y) log z$
$ \equal{}(x\minus{}y)((x\minus{}y)\plus{}(y\minus{}z)) log x \plus{} (y\minus{}x)(y\minus{}z) log y \plus{} (z\minus{}x)(z\minus{}y) log z$
$ \equal{}(x\minus{}y)^2 log x \plus{} (x\minus{}y)(y\minus{}z)(logx \minus{} logy) \plus{} (z\minus{}x)(z\minus{}y) log z \geq 0$.


you have a polymerase that is making copies of a single strand of DNA in a closed system (say
the strand of DNA is double stranded and is 300 base pairs long, though I
don't think the total length matters for our problem).  So, the original
copy of the double stranded template becomes 2 double stranded molecules
during the first cycle of PCR, then the 2 double stranded molecules
produced during the first cycle of PCR become 4 double stranded molecules
during the second cycle of PCR, and so on.  The polymerase makes copies of
the DNA by inserting the correct complementary base across from each base
in the template strand until it reaches the end of the 300bp template
strand, but the polymerase also makes mistakes- 1/10,000 times it inserts
a base, it will insert the wrong base.  Now, say we are interested in a
particular base- say position 150 out of 300 is a "T" that we are
particularly interested in because it leads to cancer.  How frequently
will the majority product at the end of 40 cycles of PCR be "wrong" (i.e.
have a base other than T at that position).  Realize that the A that pairs
with that T is equally important to consider (the A mutating to some other
base would also make the strand "wrong" at that position).

okay i really dont understand exactly whats going on, but im thinking the tools you need for this are

markov chains and the concept of coupling ?


$ S \equal{} 1 \plus{} \frac{5}{11} \plus{} \frac{12}{11^2} \plus{} \frac{22}{11^3} \plus{} \frac{35}{11^4} \plus{}  \cdots$

Find the value of $ S$ numerically in the form of $ x \plus{} \frac{y}{z}$


allora le avete fatte le gare stamani?
io penso di aver fatto una strage dopo un po avevo già il cervello fuso..

e cmq si potrebbero anche sbrigare a pubblicare le sol no? :?

Quanto pensate di aver fatto all'incirca?
In effetti l'archimede di quest'anno era più difficile del solito...

ehm diciamo male... 50 pt veramente schifoso come punteggio dato che mi ero allenata su quelli precedenti ed erano andati molto meglio....vabe cmq era la prima volta ;)


1. Find the sum of the squares of the real roots of:

x^20 - x^18 - x^16 - ... - x^2 - 2

2. Calculate (without using a calculator) the largest integer less than:

:sqrt:2500 - :sqrt:2501 + :sqrt:2502 - ... - :sqrt:2999 + :sqrt:3000

1. We substitute y=x^2 to get y^10-y^9-y^8-...-2=0, or y^10-1=y^9+y^8+...+y+1=(y^10-1)/(y-1), which can only be equal when y^10-1=0 or y-1=1. In the first case, when y is real, y= :pm: 1, and in the second case, y=2. But y=-1 doesn't work because then x=:pm: i, which isn't real. Thus we have x=:pm: 1 and x=:pm: :rt2:, so the sum of the squares of the real roots is 6.

Not quiiite right for the first one. Take a look at it again.

Ah yes, sorry. That doesn't work when y=1 because then we have 0=10, which is false. Thus the only solution are :pm: :rt2:, so the sum of the squares of the roots is 4.

right. :)

You could also refer to Descartes' Rule of Signs to tell that there are only two real roots, one positive and one negative.

Just looking back at some old problems, saw nobody tried the second part here..
:sqrt:2500 is 50. If we put the rest of the terms as (:sqrt:2502 - :sqrt:2501)+(:sqrt:2504 - :sqrt:2503)...
Well :sqrt:(a+1)-:sqrt:a = 1/(:sqrt:(a+1) + :sqrt:a).. (just after multiplying by its conjugate)

So we get 50 + 1/(:sqrt:2502 + :sqrt:2501) + ... + 1/(:sqrt:3000 + :sqrt:2999).

Each of those 1/ terms is certainly less than 1/(50+50) = 1/100, so the sum is less than 50 + 250/100 (since there are 250 terms there) which is 52.5.
Each of them is also a lot bigger than 1/(60+60) = 1/120, so the sum is greater than 50 + 250/120 which is bigger than 52..

So the answer is 52.


probability
function
probability and stats

I'm not sure how to go about this - any help would be appreciated:

Suppose we choose a random point in the interval (-2,1) and denote the distance to 0 by X.  Show that X is a continuous random variable.

That's a little weird.
Show that any interval in the range of possible values has a positive probability? It's not hard to calculate this probability.

The thing is, the phrase "choose a random point in the interval $ ( \minus{} 2,1)$" is a little loose and imprecise, but the most likely interpretation is to make it a well-known type of random variable: let $ U$ be uniformly distributed on $ ( \minus{} 2,1).$ That's a continuous distribution; its density is $ \frac13$ on that interval and zero off of it; the cumulative distribution function is a piecewise linear function whose graph includes the points $ ( \minus{} 1,0)$ and $ (2,1).$

Now you want to let $ X \equal{} |U|.$

[hide="Oops. Not sure what I was thinking."]Theorem: Suppose $ U$ is a random variable with an absolutely continuous distribution. (You probably don't know why I said "absolutely continuous," but it's the same as saying that it has a density). Suppose $ g$ is a Lipschitz function defined on the range of $ U.$ Let $ X \equal{} g(U).$ Then $ X$ has an absolutely continuous distribution.
[/hide]
---

In this particular case, we can be quite concrete. We can just go ahead and calculate the cdf and the density of $ X.$

What is $ P(X\le x)?$

If $ x\le0,$ this is zero.

If $ 0 < x\le1,\ P(X\le x) \equal{} P( \minus{} x\le U\le x) \equal{} \frac {2x}3.$

If $ 1 < x\le 2,\ P(X\le x) \equal{} P( \minus{} 1 < U\le x) \equal{} \frac {x \plus{} 1}3.$

If $ x > 2,\ P(X\le x) \equal{} 1.$

So that's the cdf, and it is continuous. Differentiate to find that the density is

$ f_X(x) \equal{} \begin{cases}\frac23, & 0 < x < 1 \\
\frac13, & 1 < x < 2 \\
0, & \text{otherwise}\end{cases}.$

[quote="Kent Merryfield"]
Theorem: Suppose $ U$ is a random variable with an absolutely continuous distribution. (You probably don't know why I said "absolutely continuous," but it's the same as saying that it has a density). Suppose $ g$ is a Lipschitz function defined on the range of $ U.$ Let $ X \equal{} g(U).$ Then $ X$ has an absolutely continuous distribution.
[/quote]
Erm... How about $ g\equal{}\text{const}$ :? :o

But this leads to an interesting question (possibly more interesting than the original one): what conditions on a map $ f\colon \mathbb R^n\to\mathbb R^n$ imply that it pushes any absolutely continuous measure to another such?


geometry
vector
complex number geometry
geometry proposed

Hi,
I'm pretty new to complex-number geometry.this is an easy problem on my book. I read the solution but have some troubles understanding it :blush:  Ok, here's how it goes
A regular n-gon $A_{1}...A_{n}$ is inscribed in a circle with center O and radius R.  X is any point with $d=|OX|$. Then $\sum^{n}_{i=1}|A_{i}X|^{2}=n(R^{2}+d^{2})$.

Solution:We have $A_{1}+...+A_{n}=O$, $|A_{i}X|^{2}=A_{i}^{2}+X^{2}-2A_{i}\cdot X=R^{2}+d^{2}-2A_{i}\cdot X$, and $|A_{1}X|^{2}+..+|A_{n}X|^{2}=n(R^{2}+d^{2}).$

It's this part that confuses me $|A_{i}X|^{2}=A_{i}^{2}+X^{2}-2A_{i}\cdot X=R^{2}+d^{2}-2A_{i}\cdot X$ what operation rule is used here? $A_{i}^{2}=R^{2}$ why? $|A_{i}X|^{2}=(A_{i}X)^{2}$ how? :(

Can anyone help me? Thanks :)

[quote="N.T.TUAN"][quote="andy_tok"] $A_{i}^{2}=R^{2}$ [/quote]
Because modul of vector $A_{i}$ equal $R$.[/quote]

Thanks. but does $A_{i}^{2}=|A_{i}|^{2}$? for example $A_{i}=1+i$,  $A_{i}^{2}=2i$ but $|A_{i}|^{2}=2$

[quote="andy_tok"]but does $A_{i}^{2}=|A_{i}|^{2}$? [/quote]
Yes, but for vectors!

[quote="andy_tok"]Thanks. ok, the solution is using vectors. Because the chapter before the questions is about complex-number geometry, I though the $A_{i}$ and $X$ are points on a complex plane :blush:[/quote]

Please read carefuly, before send question to everyone. I thinked a lot of :P

Sorry by English of mine :D


inequalities
inequalities unsolved
inequalities proposed
inequalities open
convex-concave inequalities

[quote="radio"][quote="libra_gold"]

$\sum\frac{(x-y)^{2}}{xy}\geq 3\sum\frac{(x-y)^{2}}{(x+z)(y+z)}$

[/quote]

I don't think this line is correct  :wink:[/quote]
I think, it's correct:
$\sum\frac{(x-y)^{2}}{xy}\geq 3\sum\frac{(x-y)^{2}}{(x+z)(y+z)}\Leftrightarrow\sum_{cyc}(x-y)^{2}S_{z}\geq0,$ where
$S_{x}=\frac{1}{yz}-\frac{3}{(x+y)(x+z)},$ $S_{y}=\frac{1}{xz}-\frac{3}{(x+y)(y+z)},$ $S_{z}=\frac{1}{xy}-\frac{3}{(y+z)(x+z)}.$
Let's assume $2\geq x\geq y\geq z\geq1.$ Hence, $S_{x}\geq0$ and $S_{y}\geq0\Leftrightarrow y^{2}+y(x+z)-2xz\geq0.$
But $y^{2}+y(x+z)-2xz\geq z^{2}+z(x+z)-2xz=2z^{2}-xz\geq2z-xz=z(2-x)\geq0.$
Thus, $S_{y}\geq0.$ Hence, $\sum_{cyc}(x-y)^{2}S_{z}\geq(x-z)^{2}S_{y}+(x-y)^{2}S_{z}\geq(x-y)^{2}(S_{y}+S_{z}).$
Remain to prove that $S_{y}+S_{z}\geq0.$ But $S_{y}+S_{z}\geq0\Leftrightarrow$
$\Leftrightarrow\frac{1}{xz}-\frac{3}{(x+y)(y+z)}+\frac{1}{xy}-\frac{3}{(y+z)(x+z)}\geq0\Leftrightarrow$
$\Leftrightarrow\frac{y+z}{xyz}-\frac{3(2x+y+z)}{(x+y)(x+z)(y+z)}\geq0\Leftrightarrow$
$\Leftrightarrow(y+z)^{2}x^{2}+(y+z)^{3}x+(y+z)^{2}yz-6yzx^{2}-3(y+z)yzx\geq0\Leftrightarrow$
$\Leftrightarrow(y-z)^{2}x^{2}+(y-z)^{2}(y+z)x+(y+z)^{2}yz+(y+z)yzx-2yzx^{2}\geq0.$
But $(y+z)^{2}+(y+z)x-2x^{2}\geq4+2x-2x^{2}=2(1+x)(2-x)\geq0.$
Id est, our inequality is proved. :)

this was already discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=triangle&t=84210]here[/url]

[quote="radio"][quote="libra_gold"]

$\sum\frac{(x-y)^{2}}{xy}\geq 3\sum\frac{(x-y)^{2}}{(x+z)(y+z)}$

[/quote]

I don't think this line is correct  :wink:[/quote]

This one is equivalent to the original one. Do you think so?

[quote=radio]Prove that for all real numbers $x,y,z \in [1,2]$ the following inequality always holds:
\[(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}). \]
When does the equality occur?[/quote]
[url=http://www.artofproblemsolving.com/community/c6h84210p487332]Vietnam Team Selection Tests 2006, Day 2, Problem 1[/url]
[b]Reverse inequality[/b]
Prove that for all real numbers $x,y,z \in [1,2]$ the following inequality always holds:
\[(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\leq 10. \][b][url=http://www.artofproblemsolving.com/community/c6h478312p2678288]Inference[/url][/b]
Prove that for all real numbers $x,y,z \in [1,2]$ the following inequality always holds:
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\leq\frac{5}{3}$$
[url=http://www.artofproblemsolving.com/community/c6h623533p3731246]here,[/url][url=http://www.artofproblemsolving.com/community/c6h422978p2391981]here,[/url][url=http://www.artofproblemsolving.com/community/c6h538481p3096039]here,[/url][url=http://www.artofproblemsolving.com/community/c6h344149p1841164]here,[/url][url=http://www.artofproblemsolving.com/community/c6h398201p2215590]here,[/url][url=http://www.artofproblemsolving.com/community/c4h1061866p4682502]here[/url]

If $ x,y $ and  $z$ are sides of triangle.  Prove that$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}).$$[url=http://www.artofproblemsolving.com/community/c275h116646p662466]Romania[/url]

[quote=sqing]Let $a,b,c>0.$  Prove that
$$(a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right) \geq  9\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}},$$
[hide=*]$$\sqrt{\frac{a(a+c)}{b(b+c)}}+\sqrt{\frac{b(b+a)}{c(c+a)}}+\sqrt{\frac{c(c+b)}{a(a+b)}}+\frac{8abc}{(a+b)(b+c)(c+a)} \geq  4,$$
$$\frac{a^3+b^3+c^3}{abc}+36\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\geq57,$$
$$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 2\sqrt{(ab+bc+ca)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}.$$
[size=50](Rahim Shahbazov, CYCLIC-INEQUALITY-680  676  675 673)[/size][/hide][/quote]
Let $a,b,c$ be positive real numbers . Prove that$$(a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right) \geq 5+\frac{4(a^2+b^2+c^2)}{ab+bc+ca}.$$
[b]Proof:[/b]
$$\sum_{cyc}\frac {bc(b+c)}{a}=\sum_{cyc}\left(\frac {c}{b}+\frac {b}{c}\right)a^2\geq \sum_{cyc}a^2\iff \sum_{cyc}\frac {(b+c)}{a} \sum_{cyc}ab\geq 4 \sum_{cyc}a^2+2 \sum_{cyc}ab\iff$$
$$(a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right) \geq 5+\frac{4(a^2+b^2+c^2)}{ab+bc+ca}$$
Let $a,b,c$ be positive real numbers . Prove that$$(a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right) \geq 3+6\left(\frac{a+b+c}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}\right)^2.$$
[url=https://artofproblemsolving.com/community/c6h234212p1295275]here[/url]


$ n$ different  books $ (n\geq 3)$ are put at random in a shelf. Among these books there is a book $ A$ and a book $ B$. Determine the probability of which there are exactly $ r$ books between these two books.

[hide="My result"]

$ \frac{2}{n(n\minus{}1)}$[/hide]

I dont think you result can be correct because depending on the size of $ r$ there will be a differnet probability.
Consider: $ r \equal{} n \minus{} 2$ then the books must be on either end of the shelf. But if $ r \equal{} 0$ then there are many places ($ n \minus{} 1$ places to be precise) where this can occur.

So I believe the answer is; $ P_n(r) \equal{} \dfrac{2(n\minus{}(r\plus{}1))}{n(n\plus{}1)}$

[quote="ocha"]So I believe the answer is; $ P_n(r) \equal{} \dfrac{2(n \minus{} (r \plus{} 1))}{n(n \plus{} 1)}$[/quote]

[b]ocha[/b] i think you are right, i saw my mistake.

But i got

[list]$ P_n(r) \equal{} \dfrac{2(n \minus{} (r \plus{} 1))}{n(n \minus{} 1)}$[/list]

There are $ {n\minus{}2\choose r}$ ways to choose $ r$ books.

Hence, consider the block with $ r\plus{}2$ books $ (A,\,B$ and $ r$ books). There are $ [n\minus{}(r\plus{}2)\plus{}1]!\equal{}(n\minus{}r\minus{}1)!$ permutations considering $ n$ books with the block mentioned. Besides, we have $ 2!\equal{} 2$ permutations of the books $ A$ and $ B$, and $ r!$ permutations of the $ r$ books.

The probability desired is given by

[list]$ \frac{{n\minus{}2\choose r}\cdot(n\minus{}r\minus{}1)!\cdot 2\cdot r! }{n!}$[/list]

Remark:
[list]$ {n\choose p}\equal{}\frac{n!}{(n\minus{}p)!p!}$[/list]

Sorry for my poor english.

Faster is...

There are $ (n\minus{}2)!$ ways to order all of the books except $ A$ and $ B$. There are $ n\minus{}r\minus{}1$ ways to choose the two spots to insert $ A$ and $ B$ so that there are $ r$ books between the two. There are two ways to place $ A$ and $ B$ into these spots ($ A$ first or $ B$ first). Thus the number of ways to place the $ n$ books with $ r$ between $ A$ and $ B$ is just
$ 2(n\minus{}r\minus{}1)(n\minus{}2)!$.
and the desired probability is
$ \dfrac{2(n\minus{}r\minus{}1)(n\minus{}2)!}{n!} \equal{} \dfrac{2(n\minus{}r\minus{}1)}{n(n\minus{}1)}$.


Continue discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=6150


Welcome to Middle Earth! This mafia game has 12 players, which is relatively few so we'll hopefully have a quick game. Let me state a few rules for this game. Most will be pretty familiar: [list][*]As in any mafia game, your team wins when all other teams are eliminated (or when your team has a winning strategy).
[*]It is [b]strictly forbidden[/b] to modify posts once someone has replied to them!
[*]Everyone is on the block, at any moment. So you can simply vote aye on someone and/or retract your vote later. Voting nay is pointless, but allowed (e.g. to stress your opinion).
[*]To prevent stalling games, we will have minority lynching. That means: next to the normal lynching, if a day lasts for more than 336 hours (14 days), the day will end. At that point, the person having most ayes is lynched, if that amount is stricly greater than 1. If two or more persons tie for the highest, the person who got his last aye first is lynched. If both are made in the same post, I will take them in the order written.
[*]Just so you know what to claim, Citizens are called Ents in this game: [quote]You are an Ent. You belong to the good team.
At night, you may target one player. If Saruman is targetted by at least half of the Ents alive, he dies.[/quote]And of course you may claim anything else as well.
[*]Sometimes lynches may fail. This can happen on any team, given the right circumstances. In this case day will simply end without lynch. Same for nightkills.
[*]Day lynches will be revealed if succesful, nightkills will not.
[*]If there are any questions, please ask them to me in PM. By asking them here, you spread clues on your identity.[/list]

It is day 1. Players remaining: [b]12[/b][list][*]142857[*]abacadaea[*]ajai[*]Ars[*]BoesFX[*]CheeseIsGood[*]hunter34[*]miyomiyo[*]mz94[*]pianoforte[*]pieterminate[*]Temperal[/list]

Role PM's are sent. Don't be surprised if you don't understand it completely, [b]mystery[/b] will be part of the fun in this game. :)

You may start discussing.

yo so I believe that during the day we should agree on a player for the ents to target, possibly 2
if we do that it is more likely that we will hit saruman at night.

 the 2 people would work if there is an even number and both have the same amount we could check both

now this is all I have to say at the moment

Confirming that I'm here.  I'm excited for this game! :lol:

About the editing rule, it has been seen in prior games (take Game 21, for example) that when bubka edited his roleclaim post even before anybody responded, it caused an uproar in the course of the game.  This gives an unfair advantage to those who have enough time on their hands to be constantly on this thread.  Thoughts?  Ideas?

[quote="miyomiyo"]About the editing rule, it has been seen in prior games (take Game 21, for example) that when bubka edited his roleclaim post even before anybody responded, it caused an uproar in the course of the game.  This gives an unfair advantage to those who have enough time on their hands to be constantly on this thread.  Thoughts?  Ideas?[/quote]I highly encourage not to edit at all, but the software doesn't keep track of editing before anyone replied, so I will not punish it since I have no proof. Do at own risk of making yourself suspicious.

[quote="Ars"]yo so I believe that during the day we should agree on a player for the ents to target, possibly 2
if we do that it is more likely that we will hit saruman at night.[/quote]  I would say one target a night (whoever is most suspicious).  Two would be too difficult because we don't know who the Ents are.  For example, if we think two people are the mafia, we can't evenly divvy up the unknown Ents.  Also, don't call me suspicious for saying this, but in this game maybe a few Ents dying is sometimes a good thing.  (That way it's easier to get half the Ents to do something.)

[quote="miyomiyo"][quote="Temperal"]Heh, I wonder who Sauron is. I wish we knew all the roles...[/quote]We're assuming there is a Sauron (which is rather obvious, I know).  Just to get everybody talking and to rise some suspicions, I suggest everybody explain what they think this third power is and what that third power can do.  I just don't want this to turn into another endless Day 1.[/quote]

Maybe im just being stupid, but where is this "third power" mentioned?...

Uh, if you're stupid, then I'm stupid too. Unless it's something in miyomiyo's role PM...?

Lets just target one person each night, randomly. But what is a Sauron? For all we know, it could be a good role.

Sauron was the big evil Eye guy from Lotr, so Id assume he's evil

Lol, I never read/watched LOTR. And I noticed we've been spelling Saruman as Sauron.

So far, I think miyomiyo is the most suspicious.

Sauron is a great evil EYE thing
Sarumon is a powerful, but corrupt, wizard
They're different characters

And I agree, miyomiyo is the most suspicious. But let's give him a chance to defend himself...

No, I was just following up on Temperal's guess that there was a Sauron.  I wanted to get people talking about what they think this "third power" (probably Sauron) is.  I do NOT know that there exists a third power, but I'm just guessing (like temperal was).  Don't leap to conclusions and assume I'm guilty.

I did something stupid.  the night before hunter was almost lynched i had put on the the ring. i lied about the powers of the ring
1. the ring also protects from lynch
2. the ring can only be used once
so i obviously couldnt reveal 2....and 1. i forgot about it so i didnt really need to roleclaim.

that was silly of me........but the answer was 2/4.  not sure about last night.

BoesFX was the Witch-King of Angmar. ;) Ars was the real Gandalf.

I will post the other roles and the triggered effects if you like, when I have more time.

[quote="pianoforte"]Well of course I was lying. Honestly, I'm surprised that Temperal was sarumon. I would be surprised if Ars was not gandalf.

Hunter, how many more steps toward Mt. Doom were there?

Darnit, I probably woulda won too.[/quote]

With this night included, 1 step.

I guess I have to pay attention to the other mafia game.

no peter I think you got me confused with someone I was a foolish dwarf that wasn't in the game at all and wanted to kill everyone
BoesFX was Gandalf and was my other personality and mafia partner
I made 6 nightkills a night of people not in the game
and couldn't be killed
I also investigated everyone.
When killed I would comeback the next day

Ok, I am the Witch King of Angmar....
I can kill anyone except the ringbearer, I don't know who my allies are :mad:

I don't kill in the first day just for making confusion :P

Nice kill, pianoforte :wink:

[quote="pianoforte"]eh, i actually targeted you that night (2nd night), though i didn't even notice until i went through my sent pms

anyways, after hunters death i would be able to kill anyone at night, if no other effects. and there would be no way to kill me.[/quote]

Double Strike :roll:

pie probably shoulda just let hunter die that day

Actually hunter won't have died, since Pie used the ring and hunter was immortal for the day. It was purely a tactical mistake to reveal themselves.

[quote="BoesFX"]Nice kill, pianoforte :wink:[/quote]

Uh thanks?

[quote="pianoforte's signature"]I'm Gandalf guys  :( [/quote]You can take that off now.

Peter, I actually would like to see all the roles.

yes, that was the mistake i was speaking of....hunter was already immune...and after that i couldnt use the ring any more so it was silly.  I DID suspect that someone would go for me that night actually, which was i used the ring :)  I think i may have left some hints to some rules that may have provoked some people..and ars also said that i was innocentish so yeah..


It's cute, but in my opinion, not as funny as some of their other stuff.

McCain doesn't look realistic; Palin does. :P

Pretty funny.  The guy they got to do McCain sounded pretty good, but the voice was a bit high.  Palin was too fat.

I agree, and the first part is good, but the second part kinda...  isn't.