function
algebra
functional equation
IMO Shortlist

It is from the short-list 2005 (i think A2) and it's offisial solution is very beautiful

You can also see this problem similar it http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=f%28x%29f%28y%29%3Df%28x%2Byf%28x%29%29&t=14110[/list]

I think the only solution is the constant function $f\equiv 2$.

Suppose we can find $x$ such that $f(x)<1$. Then $y=\frac x{1-f(x)}>0$, and $x+yf(x)=y$. Plug these values of $x,y$ into the equation to get $f(x)f(y)=2f(y)\Rightarrow f(x)=2$, contradicting $f(x)<1$. We thus find $f(x)\ge 1,\ \forall x>0\ (*)$. Now assume we can find $y$ such that $f(y)<2$. Choose $x_0>0$ arbitrarily, and put $x_{n+1}=x_n+yf(x_n),\ \forall n\ge 0$. We have $f(x_n)=\left(\frac{f(y)}2\right)^nf(x)\to 0$ as $n\to\infty$, but this contradicts $(*)$ applied to $x_n$. This means that $f(x)\ge 2,\ \forall x>0\ (**)$. $(**)$ applied to $y$ instead of $x$, together with the initial equation, show that $f(x)\le f(x+yf(x)),\ \forall x,y>0$, i.e. $f$ is non-decreasing.

If $x_1<x_2$ and $f(x_1)=f(x_2)$ (i.e. if $f$ is not one-to-one), then it follows from the equation that $\forall t>0,\ f(x_1+t)=f(x_2+t)$, and together with the fact that $f$ is non-decreasing, this implies that $f$ is eventually constant. This constant can only be $2$, so $f\equiv 2$.

Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, we get $x+yf(x)=y+xf(y),\ \forall x,y>0$. This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$, but this contradicts $(**)$.

[quote="#20002 pwns"][hide="hmmm..."]I hope 0 is in the domain... (or else I don't have the solution)...
Take $y=0$, we have $f(x)=0$ or $f(0)=2$. So we have a partial solution, now we take other option of $f(0)=2$. Now $f(x)=f(2x)$, which makes the only solutions f(x)=0, f(x)=2. I guess if 0 isn't in the domain, we only have $f(x)=2$[/hide][/quote]
Nope, $0$ is not in the domain ($\mathbb R^+=\{x\in\mathbb R|x>0\}$)

[quote="abdurashidjon"][quote="grobber"]
Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, we get $x+yf(x)=y+xf(y),\ \forall x,y>0$. This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$, but this contradicts $(**)$.[/quote]
 Can  you explain this part[/quote]

@abdurashidjon
I think he replaced $x$ by $y$ and $y$ by $x$ and got $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, and from property one to one he got
$x+yf(x)=y+xf(y)$ $\Longrightarrow \frac{f(x)-1}{x}=\frac{f(y)-1}{y}=constant=a$
So $f(x)=ax+1$,contradiction. ;)

Let $P(x,y)$ be the assertion that $f(x)f(y) = 2f(x+yf(x))$. $P(x+zf(x), y)$ yields $f(x+zf(x))f(y) = 2f(x+zf(x) + yf(x+zf(x))$. $f(x+zf(x)) = \frac{f(x)f(z)}{2}$, so $f(x)f(y)f(z) = 4f(x+zf(x) + \frac{yf(x)f(z)}{2})$. Similarly, $P(x, z+yf(z))$ yields $f(x)f(y)f(z) = 4f(x+(z+yf(z))f(x)) = 4f(x + zf(x) + yf(x)f(z))$. Hence, $f(x+zf(x) + \frac{yf(x)f(z)}{2}) = f(x+zf(x) + yf(x)f(z))$ for all positive $x,y,z$.

Let $a$ and $b$ be any positive reals such that $2a > b > a$. Let $x = a - \frac{b}{2}$, $z = \frac{2a-b - x}{f(x)}$, and $y = \frac{2(b-a)}{f(x)f(z)}$. Then $x + zf(x) + \frac{yf(x)f(z)}{2} = 2a-b + (b-a) = a$, and $x + zf(x) + yf(x)f(z) = 2a-b + 2(b-a) = b$, so $f(a) = f(b)$. It follows that $f$ is constant over all intervals of the form $(c, 2c)$.

For all integers $k$, let $S_k = ((3/2)^k, 2 (3/2)^k)$. From the above, $f$ is constant over each $S_k$. Since $2(3/2)^k > (3/2)^{k+1}$, $S_{k+1} \cap S_k \neq \emptyset$, so the constant value of $f$ over any $S_k$ is equal to that of $S_{k+1}$. Since $\bigcup_{k=-\infty}^{\infty} S_k = \mathbb{R}^+$, $f$ must be constant over $\mathbb{R}^+$. It follows that $f(x) = 2$ for all $x$.

Here is a solution I came up with a long time ago:
Note that from the given, we have:
\[f\left(x\right)=\frac2{f\left(y\right)}f\left(x+yf\left(x\right)\right)\]
Thus we have:
\begin{eqnarray*}2f\left(x\right)f\left(y+2z\right)&=&f\left(x\right)f\left(y\right)f\left(\frac{2z}{f\left(y\right)}\right)\\
&=&2f\left(x+yf\left(x\right)\right)f\left(\frac{2z}{f\left(y\right)}\right)\\
&=&4f\left(x+yf\left(x\right)+\frac{2z}{f\left(y\right)}f\left(x+yf\left(x\right)\right)\right)\\
&=&4f\left(x+yf\left(x\right)+zf\left(x\right)\right)\\
&=&2f\left(x\right)f\left(y+z\right)
\end{eqnarray*}
Thus dividing both sides by $f\left(x\right)$ gives $f\left(y+z\right)=f\left(y+2z\right)$ for all positive reals $y,z$. Now repeatedly replacing $y$ with $y+z$ gives $f\left(y+z\right)=f\left(y+nz\right)$ for all reals $y,z$ and positive integers $n$. Now suppose $a,b$ are arbitrary positive real numbers with $a>b$. Let $n$ be an integer greater than $a/b$, and let
\[y=\frac{nb-a}{n-1}, z=\frac{a-b}{n-1}\]
Thus $y+nz=a$ and $y+z=b$, so we have $f\left(a\right)=f\left(b\right)$. However, since $a,b$ were arbitrary, $f\left(x\right)\equiv c$ for some $c$, so plugging this back into the given equation gives $f\left(x\right)\equiv 2$ as the only solution. $\blacksquare$

[hide="Not particularly nice"]

Obviously if $f$ is constant, $f(x)=2$.

Otherwise, use Zhero's notation again.

[b]Lemma[/b]: $f(x) \geq 1$ for all $x$.

[b]Proof[/b]: Assume there is some $a$ with $f(a)<1$. Then, $b=\frac{a}{1-f(a)} \in \mathbb{R}^{+} $ and thus the domain of $f$.

$P(a,b)$ gives $f(a)f(b)=2f(a+bf(a))=2f(\frac{a-af(a)+af(a)}{1-f(a)})=2f(b)$. As $f(b)>0$, we have $f(a)=2$, contradiction.

Thus, $f(x) \geq 1$.

Also, if there is some $c$ such that $f(c)=1$, then $[f(c)]^2=2f(c+cf(c)) \implies 1=2f(2c) \implies f(2c)=\frac{1}{2}$, but $f(2c) \geq 1$ by our lemma, contradiction. Thus, $f>1$.

We now have two cases.

[b]Case 1: $f$ is one-to-one, so $f(m)=f(n) \Longleftrightarrow m=n$[/b]

$P(m,n): f(m)f(n)=2f(m+nf(m))$
$P(n,m): f(n)f(m)=2f(n+mf(n))$

Thus, $2f(m+nf(m))=2f(n+mf(n)) \implies m+nf(m)=n+mf(n) \implies m(1-f(n))=n(1-f(m)) $
$\implies \frac{m}{f(m)-1}=\frac{n}{f(n)-1}$ for any $m,n$, so $\frac{x}{f(x)-1}$ equals some constant $k$.

$x=kf(x)-k \implies f(x)=1+\frac{x}{k}$.

However, $P(k,k): 4=[f(k)]^2=2f(k+kf(k))=2f(k+2k)=2f(3k)=8$, contradiction. Thus, no one-to-one functions satisfy the conditions.

[b]Case 2: $f$ is not one-to-one, so there is some pair $m,n$ such that $n<m$ and $f(m)=f(n)$.[/b]

$P(m,x): f(m)f(x)=2f(m+xf(m))$
$P(n,x): f(n)f(x)=2f(n+xf(n))$

As $f(m)=f(n)$, $f(m+xf(m))=f(n+xf(n))=f(n+xf(m))$. Letting $x=\frac{y}{f(m)}$, we have $f(n+y)=f(m+y)$. Thus, $f$ is periodic. Let it have period $p \ge 0$ (if $p=0$, $f$ is constant)

$P(x,y+p): f(x)f(y+p)=2f(x+(y+p)f(x)) \implies f(x)f(y)=2f(x+yf(x)+pf(x)$
$P(x,y): f(x)f(y)=2f(x+yf(x))$.

Thus, $f(x+yf(x))=f(x+yf(x)+pf(x)) \implies p|pf(x)$, so $f(x)$ is a positive integer for all $x$.

$P(x,p): f(x)f(p)=2f(x+pf(x))=2f(x) \implies f(p)=2$

$P(p,\frac{p}{2}): f(p)f(\frac{p}{2})=2f(p+\frac{p}{2}f(p))$
$\implies 2f(\frac{p}{2})=2f(p) \implies f(\frac{p}{2})=f(p) \implies p|\frac{p}{2}$, which is impossible. Thus, no $f$ exist for this case either.

Our only solution is then $\boxed{f(x)=2}$[/hide]

[quote="Mewto55555"]Thus, $f$ is periodic. Let it have period $p \ge 0$ (if $p=0$, $f$ is constant)
[...]
$\implies 2f(\frac{p}{2})=2f(p) \implies f(\frac{p}{2})=f(p) \implies p|\frac{p}{2}$, which is impossible.[/quote]
This is only a contradiction if there exists a minimal period $p$, which is not necessarily true. (You don't rule out the case in which $f$ takes one value over the rationals and another over the irrationals, in which case $p$ could be any rational number.)

Whoops, but that's easily fixed if we replace that last $\frac{p}{2}$ with an $x$, meaning $p$ divides all real $x$, impossible.

I don't see how you're getting that $p$ divides all real $x$. Your solution doesn't address the case I mentioned in my last post.

[b]Edit:[/b] Actually, I missed an even more glaring mistake:
[quote="Mewto55555"]Thus, $f(x+yf(x))=f(x+yf(x)+pf(x)) \implies p|pf(x)$, so $f(x)$ is a positive integer for all $x$.[/quote]You're going to need some more algebraic manipulation with the original equation to get more information.

Quite cute. Let $P(x,y) $ denote the given equation. We claim $f\equiv 2$ is the only solution (which trivially works).

If $f$ is injective, changing the positions of $x\leftrightarrow y$ gives $x+yf(x)=y+xf(y)$ and $y=1$ gives linearity. Checking, we see there aren't any solutions.

Assume $f(a)=f(b)$ with $a>b$. $P(b, \frac{a-b}{f(a)})$ gives $f(c)=2$ for $c=\frac{a-b}{f(a)}$. \[P(c,y)\Rightarrow f(y)=f(2y+c) \textrm{ and comparing } P(y,\frac{x}{f(y)}) \textrm{ with }P(2y+c,\frac{x}{f(y)})\Longrightarrow f(x+y)=f(x+2y+c)\] or $f(z)=f(z+y+c)$ for $z>y$ and thus, $f\equiv \textrm{constant}$ in $(z+c,2z+c)$. Varying $z$ we get $f$ to be constant for all sufficiently large inputs. Setting $x$ large in $P(x,y)$ we finish.

Beautiful!

The only solution is $f\equiv 2$, which obviously works. Now, let $P(x,y)$ denote the assertion.

Note that if for any $x$ we have $f(x)<1$ then
\[\displaystyle{P\left(x,\frac{x}{1-f(x)}\right)}\implies f(x)=2,\]
a contradiction. The key is to now consider the infimum $m\ge 1$ of $f$. Clearly if $m<2$ then if we take $f(x)\to m$ and $P(x,x)$ we obtain a contradiction. Therefore $m\ge 2$. However this now implies
\[f(y)\ge 2,\, f(x)\le f(x+yf(x))\]
so that $f$ is nondecreasing. The final step is to now take $x\to 0$ and $P(x,x)$: we get
\[f(x)^2=2f(x+xf(x))\implies m^2=2m\implies m=2\]
so then $P(x,y)$ and $x\to 0$ gives that $f(y)=f(2y)$. Then $f$ is eventually constant at $2$ and we're done. $\blacksquare$

Cleanest finish to a "set of injectivity-breaking pairs" fe from yours truly

The only solution is $f \equiv 2$, which clearly works. We now prove that it is the only one.

First off, if $f$ is injective, then by swapping $x$ and $y$ we obtain $x+yf(x)=y+xf(y) \implies \frac{f(x)-1}{x}=\frac{f(y)-1}{y}$ for all $x, y \in \mathbb{R}^+$, we have $f(x)=cx+1$ for some constant $c$. However, if we plug this in we find that this function does not work.

Therefore, $f$ is not injective, so there exist $b>a$ such that $f(a)=f(b)=r>0$. Let $S$ be the set of all $d>0$ such that for all $x>a$ we have $f(x)=f(x+d)$. Then we have
$$2f(a+yr)=rf(y)=2f(b+yr),$$
so we have $f(b+x)=f(a+x)$ for all $x>0$, hence $b-a \in S$. In fact, by plugging in $x=b-a$ into the previous equality, we get $f(2b-a)=f(a) \implies f(2b-a+y)=f(a+y)$ for all $y$, so $2(b-a) \in S$, and continuing in this way we may easily prove by induction that $k(b-a) \in S$ for all $k \in \mathbb{Z}^+$.

Now consider an arbitrary $x$ which is massive compared to $b$ and $a$. I claim that $f(x) \in \{1,2\}$. Indeed if $f(x)>1$ we may select some massive $k$ such that the solution to
$$x+y(f(x)-1)=k(b-a)$$
is greater than $a$, which is definitely possible. In this case we have $f(y)=f(x+yf(x)) \implies f(x)=2$. Likewise if $f(x)<1$, since $x$ is sufficiently large, the solution in $y$ to
$$x+y(f(x)-1)=b-a$$
will be greater than $a$, since it's at least $x-(b-a)$ as $f(x)>0$, so we have $f(x)=2$. The final case of $f(x)=1$ is trivial.

We are almost done. Since if $x$ is large, $x+yf(x)$ is larger, by plugging in large $x$ and letting $y$ be any positive real, we obtain $f(y) \in \{1,2,4\}$ in general. On the other hand, if there exists $r$ with $f(r)=1$, we obtain $2f(r+rf(r))=1$ by plugging in $x=y=r$, which is absurd. Likewise if $f(r)=4$ we obtain $2f(r+rf(r))=16$. Thus $f \equiv 2$. $\blacksquare$

Really cool! First, the $f \equiv 2$ makes me really uncomfortable, so I'll change it by $g \equiv \frac{f}2$, the new equation becomes $g(x)\cdot g(y)=g(x+2yg(x))$

Claim 1: $g(x) \ge 1; \forall x \in \mathbb{R^+}$

Indeed, assume otherwise that there exists some positive real $t$, such that $c=g(t)<1$. Then $cg(x)=g(k(x))$, where $k(x)=t+2xg(t)$. Now, a quick induction, implies for all positive integers $i$, and all positive reals $x,y$: $c^ig(x)g(y)=g(k^i(x)+2c^iyg(x))$. But then there is some large enough $i$ such that $g(x)<\frac{1}{2c^i}$. And then take, $y=\frac{k^i(x)}{1-2g(x)c^i}$, we get that $\frac{1}{c^i}=g(x)<\frac{1}{2c^i}$. Clearly impossible.

Claim 2: $g$ is non-decreasing
Let $a>b$. Take $x=b, y=\frac{a-b}{2g(b)}$, and use claim $1$.

Claim 3: if there exists some t such that $g(t)=1$, then $g \equiv 1$
Of course, taking $y=t$ implies $g(x)=g(x+2tg(x))$ for all positive $x$. This implies , by Claim 2, that $g$ is constant on the interval $[x;x+2tg(x)]$, and since $g(x)\ge 1$, we have that $g$ is constant on the interval $[x;x+2t]$, which quickly implies that it is constant on every interval of the form $[a;+\infty[$, and thus on $\mathbb{R^+}$.

The Finale:
Now, we can suppose that $g(x)>1$ for all positive $x$. Then, the proof of claim $2$, gives us that $g$ is increasing on $\mathbb{R^+}$.

And $\forall x \in \mathbb{R^+} g(x)=\frac{g(x+2g(x))}{g(1)}>\frac{g(x+2)}{g(1)}>\frac{g(2)}{g(1)}=M>1$,

Now, $g(x+2yg(x))>M^2$, which implies $g(x)>M^2$, for all positive $x$. Actually, $g(x)>M^k$ for all positive integers $k$, which is impossible since $M>1$.

Denote the assertion with $P(x, y)$.
Note that if $f(x) < 1$ then by $P\left(x, \frac{x}{1 - f(x)}\right)$, $f(x) = 2$. Thus, either $f(x) \ge 1$.
Suppose that $f(x) \ge c$ holds. Then, by $P(x, x)$ it follows that $f(x) \ge \sqrt{2c}$. Repeating this gives that $f(x) \ge 2$.
As such, the condition implies that $f(x) \le f(x + yf(x))$ so $f$ is nondecreasing.
Note that by symmetry \[ f(x + yf(x)) = f(y + xf(y)) \]

[b][color=pink]Claim:[/color][/b] $f$ is either injective or constant.
[i]Proof.[/i] Suppose there exist $a, b$ such $f(a) = f(b)$ but $a > b$.
Then, by $P(a, x)$ and $P(b, x)$ it follows that \[ f(a + yf(a)) = f(b + yf(b)) \]
As such, $f(x) = f(a - b + x)$ holds for $x > b$ so $f$ is eventually periodic with period $a - b$. This with the nondecreasing condition gives that
Thus, $f$ must be eventually constant, which implies $f$ is totally constant through $P(x, y)$ for sufficiently large $y$. $\blacksquare$
The only constant solution is $f(x) = 2$.
Else, there can't exist a $u$ such $f(u) = 2$ or else $P(x, u)$ would imply that $x = x + uf(x)$, contradiction.
Injectivity gives that $x + yf(x) = y + xf(y)$ or \[ x(1 - f(x)) = y(1 - f(y)) \] so $x(1 - f(x))$ is fixed as $x$ varies.
Thus, $f(x) = 1 - \frac{C}{x}$ for some positive constant $C$. However, then $f(C) = 0$. As such, no other such function $f$ exists.

[quote=YaoAOPS]
Injectivity gives that $x + yf(x) = y + xf(y)$ or \[ x(1 - f(x)) = y(1 - f(y)) \][/quote]
I don't think this is quite right. But if you follow through correctly, you will get that injectivity $\implies f(x) = ax + b$, which does not work for any $a, b$

The solution is $\boxed{f(x)=2}$
Let $P(x,y)$ be the assertion $f(x)f(y)=2f(x+yf(x))$
$1:$ If there exist $f(k)<1$ $P(k,\frac k{1-f(k)})$ gives $f(k)=2$ contradiction so for all $x\in\mathbb R^+$ $f(x) \geq 1$
$2:$ Define $t$ as $f(t)$ is minimum. $P(t,t)$ gives $f(t) \geq 2$ so so for all $x\in\mathbb R^+$ $f(x) \geq 2$
$3:$ From step $2$ and $f(x)f(y)=2f(x+yf(x))$ $f(x) \leq (x+yf(x))$ so the function is non decreasing.
$4:$ From symmetry $f(x+yf(x))=f(y+xf(y))$
$i:$ If $x+yf(x)=y+xf(y)$ $\Longrightarrow \frac{f(x)-1}{x}=\frac{f(y)-1}{y}=k$ so $f(x)=yk+1$ and contradiction.
$ii:$ If the expressions are not equal, function is periodic and non decreasing so constant at $2$

[hide=Solution]
The answer is $f\equiv \boxed{2}$ only, which clearly works. Now, we will show that this is the only solution.

Let $P(x,y)$ denote the assertion. First, we will show that $f(x) \geq 1$ for all $x\in \mathbb{R}^{+}$. To show this, assume for the sake of contradiction that there existed an $x\in \mathbb{R}^{+}$ with $f(x) < 1$. Then, $$P\left(x, \frac{x}{1-f(x)}\right) \implies f(x)f\left(\frac{x}{1-f(x)}\right) = 2f\left(\frac{x}{1-f(x)}\right),$$ so since $f > 0$, it follows that $$f(x) = 2,$$ contradiction. Thus, $f(x)\geq 1$ for all $x\in \mathbb{R}^{+}$. Then, note that $$P(x,x)\implies f(x)^2 = 2f(x+xf(x)),$$ so since $f(x+xf(x)) \geq 1$, $$f(x)^2 \geq 2,$$ so since $f > 0$, $$f(x) \geq 2^{1/2}$$ for all $x\in \mathbb{R}^{+}$. Then, $f(x+f(x)) \geq 2^{1/2}$ for all $x\in \mathbb{R}^{+},$ so $$f(x)^2 = 2f(x+xf(x)) \geq 2\cdot 2^{1/2} = 2^{3/2},$$ so $f(x) \geq 2^{3/4}$ for all $x\in \mathbb{R}^{+}$. Applying this logic again, we get that $$f(x)^2 = 2f(x+f(x)) \geq 2\cdot 2^{3/4} = 2^{7/4},$$ so $f(x) \geq 2^{7/8}$ for all $x\in \mathbb{R}^{+}$, and repeating this yields that $f(x) \geq 2^{1 - \frac{1}{2^n}}$ for all $x\in \mathbb{R}^{+}$ and for any $n\geq 1$. Therefore, we have that $$f(x) \geq \lim_{n\rightarrow \infty}2^{1-\frac{1}{2^n}} = 2$$ for all $x\in \mathbb{R}^{+}$.

Now, we will show that $\inf_{x>0}f = 2$. Assume the contrary. Since $f$ is bounded below, $\inf_{x>0}f$ exists. Note that since $f(x) \geq 2$ for all $x > 0$, it follows that $\inf_{x>0}f \geq 2$, so if we had $\inf_{x>0}f \neq 2$, then we must have $\inf_{x>0}f > 2$. Let $\inf_{x>0}f = d > 2$. Then, by the definition of infinum, we have that for any $\epsilon > 0$, there exists an $x_0 > 0$ (depending on the choice of $\epsilon$). for which $d \leq f(x_0) < d + \epsilon$. Thus, taking some $0 < r < x_0$, we have that $$P\left(r, \frac{x_0-r}{f(r)}\right)\implies f(r)f\left(\frac{x_0-r}{f(r)}\right) = 2f(x_0) < 2(d+\epsilon).$$ Since $\inf_{x>0}f = d$, it follows that $f\geq d$ on $\mathbb{R}^{+}$, so $f(r)$ and $f\left(\frac{x_0-r}{f(r)}\right)$ are both at least $d$. Therefore, we have that $$2(d+\epsilon) > f(r)f\left(\frac{x_0-r}{f(r)}\right) \geq d^2.$$ Thus, $2(d+\epsilon) \geq d^2$ for any choice of $\epsilon > 0$, so it follows that $2d\geq d^2$. Thus, since $d > 0$, it follows that $2\geq d$, contradicting the fact that $d > 2$. Thus, we must have $\inf_{x>0}f = 2$, as claimed.

Now, I claim that $f$ is increasing. To show this, take some $0 < r \leq s$. If $r = s$, then clearly $f(r) \leq f(s)$, so assume that $r < s$. Then, note that $$P\left(r, \frac{s-r}{f(r)}\right) \implies f(r)f\left(\frac{s-r}{f(r)}\right) = 2f(s).$$ Since $f\geq 2$ on $\mathbb{R}^{+}$, $f\left(\frac{s-r}{f(r)}\right) \geq 2$, so $$2f(r) \leq f(r)f\left(\frac{s-r}{f(r)}\right) = 2f(s),$$ implying that $f(r) \leq f(s)$, so $f$ is increasing.

Therefore, we have that for any $x, y\in \mathbb{R}^{+}$, $f(x)\geq 2$, so $x+yf(x) \geq x+2y$, so $f(x+yf(x)) \geq f(x+2y)$. Thus, we have that $$P(x,y)\implies f(x)f(y) = 2f(x+yf(x)) \geq 2f(x+2y)$$ and $$P(y,x)\implies f(x)f(y) = 2f(y+xf(y)) \geq 2f(2x+y).$$ Now, note that $\max(x+2y, 2x+y) \geq \frac{(x+2y) + (2x+y)}{2} = \frac{3}{2}(x+y)$, so since $f$ is increasing, we have that $$f(x)f(y) \geq \max(2f(x+2y),  2f(2x+y)) = 2\max(f(x+2y), f(2x+y)) = 2f(\max(x+2y, 2x+y)) \geq 2f\left(\frac{3}{2}(x+y)\right).$$ Thus, we have that $$f(x)f(y) \geq 2f\left(\frac{3}{2}(x+y)\right).$$ Thus, since $f(x+y) \geq f(x)$, $$f(x+y)f(y) \geq 2f\left(\frac{3}{2}(x+y)\right)$$ as well.

Now, fix some $r > 0$. Then, for any $y < r$, plugging in $x = r-y$ and $y$ gives $$f(r)f(y) \geq 2f\left(\frac{3}{2}r\right).$$ Since $\inf_{x>0} f = 2$, we have that for any $\epsilon > 0$, there exists a $y_0 > 0$ for which $2 \leq f(y_0) < 2+\epsilon$. Then, taking $y < \min(y_0, r)$, we have that since $f$ is increasing, $f(y) \leq f(y_0) < 2+\epsilon,$ so $$f(r)(2+\epsilon) > f(r)f(y) \geq 2f\left(\frac{3}{2}r\right),$$ and this holds for any choice of $\epsilon > 0$, so it follows that in fact $$2f(r) \geq 2f\left(\frac{3}{2}r \right),$$ so $$f(r) \geq f\left(\frac{3}{2}r\right).$$ However, since $f$ is strictly increasing, $f\left(\frac{3}{2}r\right) \geq f(r).$ Therefore, we have that $f(r) = f\left(\frac{3}{2}r\right)$ for all $r > 0$.

Now, assume for the sake of contradiction that there existed a $t > 0$ with $f(t) \neq 2$, then $f(t) > 2$ as we have $f(t) \geq 2$. Then, $$f(t) = f\left(\frac{3}{2}t\right) = f\left(\frac{9}{4}t\right) = \cdots,$$ so there exist arbitrarily large $s$ with $f(s) = f(t)$. Take such an $s$ with $s > t+tf(t)$. Then, $f(t) = f(s) \geq f(t+tf(t)) \geq f(t),$ since $f$ is stictly increasing, so $f(t+tf(t)) = f(t)$. Thus, $$P(t,t)\implies f(t)^2 = 2f(t+tf(t)) = 2f(t),$$ so since $f(t) > 0$, it follows that $f(t) = 2$, a contradiction. Therefore, it follows that $f\equiv 2$, as desired. $\blacksquare$.

[quote=Nima Ahmadi Pour]We denote by $\mathbb{R}^\plus{}$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ \plus{} \rightarrow\mathbb R^ \plus{}$ which have the property:
\[f(x)f(y)\equal{}2f(x\plus{}yf(x))\]
for all positive real numbers $x$ and $y$.

[i]Proposed by Nikolai Nikolov, Bulgaria[/i][/quote]

Our preliminary checks show no immediate solutions except for the constant f(x)=2. As such, we claim that this is the only solution. First note the very easy and obvious to spot symmetry with $$f(x+yf(x))=f(y+xf(y)).$$ [hide=thought process and solution]There isn't much suggested here since it's hard to read off any types that this function can be, other than trying to remove the f by caseworking on injectivity. Indeed, if f(a)=f(b) with a>b then P(a,y) and P(b,y) yield $$f(a+yf(a))=f(b+yf(b))\implies f(a+yc_1)=f(b+yc_1),$$ implying that f is periodic.

We want to show f is constant, so some bounding might help. Indeed, we find that $f(x)\ge d\implies f(x)\ge \sqrt{2d}$ from P(x,x). After infiinite iterations if d is less than 2 it strictly increases infinitely until it reaches 2, so f(x) is always greater or equal to 2. Now substituting back we get $f(x)\le f(x+yf(x)),$ and since yf(x) is nonnegative, f is nondecreasing. Then to repeat f must be constant, which yields the function $\boxed{f(x)=2.}$

Finally, if f is injective then $$x+yf(x)=y+xf(y)\iff x(1-f(x))=y(1-f(y))\implies x(1-f(x))=c_2\iff f(x)=1-c_2/x\implies f(c_2)=0,$$ contradiction! So f(x)=2 is our only solution. $\blacksquare$[/hide]

@above
We also have that f is periodic, hence it either must decrease again after increasing (contradiction) or it will be constant from then on. It's easily checked that c^2=2c where f(x)=c

Solution w/o proving non-decreasing:

Let $P(x,y)$ denote the given assertion.

Comparing $P(x,y)$ and $P(y,x)$ gives that $f(x+yf(x))=f(y+xf(y))$.

If $f$ is injective, then $x+yf(x)=y+f(x)$. Subbing $y=1$ gives that $f$ is linear, and it is easy to check that no solutions work.

So $f$ is non-injective, and there exist $a<b$ with $f(a)=f(b)=k$
Comparing $P(a,y)$ and $P(b,y)$ gives $f(a+yk)=f(b+yk) \implies f(y)=f(y+b-a)$ for $y>a$. Let $b-a=T$.

Let $n$ be a positive integer. For $x>a$,
$P\left(x,\frac{nT}{f(x)}\right)\implies f\left(\frac{nT}{f(x)}\right)=2$
So there exist sufficiently large $\alpha$ with $f(\alpha)=2$. Taking $x=\alpha$ gives
$f\left(\frac{nT}2\right)=2$ for any $n$.

$P\left(\frac{nT}2,y\right) \implies f(y)=f\left(2y+\frac{nT}2\right)$
Letting $n>4$, $f(y)=f\left(2y+\frac{nT}2\right)=f\left(2y+\frac{(n-4)T}2+2T\right)=f(y+T)$, so $f$ is periodic for all $y$.
Then setting $n=2$ gives that $f(y)=f(2y+T)=f(2y)$.

For any $x$, let $m$ be a positive integer such that $2^m-f(x)>0$. Then
$P\left(x,\frac x{2^m-f(x)}\right) \implies f(x)\equiv 2$, which works.

Solved with [b]john0512[/b].

The answer is $f\equiv 2$ which clearly works.

[b]Claim:[/b] $f(x)\ge 1$.
[b]Proof:[/b] If not then write $y=\frac{x}{1-f(x)}$ giving $f(x)=2$, contradiction.

[b]Claim:[/b] $f(x)\ge 2$.
[b]Proof:[/b] If not then take some $f(y)<2$ and notice that $f(x)\cdot \frac{f(y)}{2}=f(x+yf(x))$. Fix $y$ and continue replacing $x\to x+yf(x)$ to derive some infinite geometric series with ratio $\frac{f(y)}{2}<1$, a contradiction to the first claim.

[b]Claim:[/b] $f$ is nondecreasing.
[b]Proof:[/b] Consider $a<b$. Choose $x,y$ with $x=a$ and $x+yf(x)=b$. Then $f(b)\ge f(a)$.

[b]Claim:[/b] If there exists $f(x_0)=2$, then $f\equiv 2$.
[b]Proof:[/b] Write $x_{i+1}=x_i+x_if(x_i)$. Verify that $f(x_i)=2$ always. Hence $x_{i+1}=3x_i$ which is unbounded. So there are arbitrarily large values which output $2$. Hence $f\equiv 2$ by nondecreasing.

[b]Claim:[/b] $f>2$ cannot hold everywhere.
[b]Proof:[/b] Verify that $f(x+yf(x))=f(y+xf(y))$ by swapping $x$ and $y$. [b]Suppose we have found[/b] values $x,y$ where $x+yf(x)<y+xf(y)$. The interval $[x+yf(x),y+xf(y)]$ is constant. Now fix $x$ and increase $y\to y'$ where $x+y'f(x)=y+xf(y)$.

The key is that the interval $[x+y'f(x),y'+xf(y')]$ is constant, hence $[x+yf(x),y'+xf(y')]$ is constant. If we continue applying the increment $y\to y'$ infinitely many times, then I claim that the upper bound of the interval is unbounded.

The proof is quite simple: assume the current interval has size $s$. Then we increase $y$ by exactly $\frac{s}{f(x)}$, and the size of the interval increases by at least $\frac{s}{f(x)}$. Hence the new interval has size at least
\[s\cdot \left(1+\frac{1}{f(x)}\right)\]
which is unbounded. Then all large values are constant, hence take such $x,y$ and now $f(x)=f(y)=2$, a contradiction.

It remains to show that $x+yf(x)=y+xf(y)$ does not hold everywhere. This is left as an exercise to the reader (happy new year, it's 12:10 AM and an hour past bedtime, my parents are yelling, will update if I can)

Hi I'm back. If this holds always then
\[xf(y)-x=yf(x)-y\implies x(f(y)-1)=y(f(x)-1)\]
hence $\frac{x}{f(x)-1}=c$ so then $f(x)=\frac{x}{c}+1$. Taking $x\to 0$ contradicts $f(x)\ge 2$. Yay!

The solution is $f\equiv 2$.

Now, if for some $x_0$, we have $f(x_0) < 1$, then $P\left(x_0,\dfrac{x_0}{1-f(x_0)}\right) \implies f(x_0) f\left(\dfrac{x_0}{1-f(x_0)}\right) = 2f\left(\dfrac{x_0 - x_0f(x_0) + x_0f(x_0)}{1-f(x_0)}\right) \implies f(x_0) = 2$, contradiction.

Therefore $f \ge 1$. Now $P(x,x)$ gives that $f(x)^2 = 2f(x+xf(x)) \ge 2\cdot 1 \implies f(x) \ge \sqrt{2 \cdot 1}$.

Now define $a_{n+1} = \sqrt{2a_n}$ with $a_1 = 1$.

So, $a_n \le 2 \implies a_{n+1} = \sqrt{2a_n} \le \sqrt{2\cdot 2} = 2$.

Also, $a_{n+1} = \sqrt{2a_n} \ge \sqrt{a_n \cdot a_n} = a_n$.

Therefore by Monotone Convergence Theorem, we get that $a_n$ converges and that the limit is $=2$.

Therefore we can repeat the step we did to go from $f \ge 1$ to $ f\ge \sqrt{2\cdot 1}$ repeatedly to get the RHS arbitrary close to $2$ using which we conclude $f \ge 2$.

Now $P\left(x,\dfrac{\varepsilon}{f(x)}\right) \implies 2f(x+\varepsilon) = f(x)f\left(\dfrac{\varepsilon}{f(x)}\right) \ge 2f(x) \implies f(x+\varepsilon) \ge f(x)$.

So $f$ is non-decreasing. Thus we can conclude that $\lim_{x\rightarrow 0^+} f(x)$ exists say to $a$. So $a \ge 2$.

Firstly note that $\displaystyle\lim_{x \rightarrow 0^+} (x+xf(x)) = (0 + 0 \cdot a) = 0$. Now note that, \[ \displaystyle\lim_{x \rightarrow 0^+} f(x)^2 = \displaystyle\lim_{x \rightarrow 0^+} 2f(x+xf(x)) \implies a^2 = 2a \implies a = 2. \]

Now fix $x_0$ arbitrarily.

$f(x_0)f(y) = 2f(x_0 + yf(x_0)) \implies \displaystyle\lim_{y \rightarrow 0^+} f(x_0)f(y) = 2f(x_0)$.

Therefore, $f$ is continuous at $x_0$ by choosing $\dfrac{y}{f(x_0)} \rightarrow 0$.

Now fix $y$,
\begin{align*}
    &\displaystyle\lim_{x \rightarrow 0^+} f(x)f(y) = 2 \displaystyle\lim_{x \rightarrow 0^+} f(x + yf(x))\\
    \implies &2f(y) = 2f(2y) \implies f(y) = f\left(\frac{y}{2}\right) = f\left(\frac{y}{4}\right) = \cdots = f\left(\frac{y}{2^n}\right)
.\end{align*}

Now taking $n \rightarrow \infty$, we get that $f$ is constant using the continuity. :yoda:

The answer is $f(x)=2$ for all $x\in \mathbb{R}_{>0}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We first prove the following.

[color=#f00][b]Claim : [/b]$f(x)\geq 2$ for all $x\in \mathbb{R}_{>0}$.[/color]
[i]Proof : [/i]First, assume there exists $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha) <1$. Then, plugging in $x=\alpha$ and $y= \frac{\alpha}{1-f(\alpha)}$ gives
    \begin{align*}
        f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\alpha + \frac{\alpha}{1-f(\alpha)}\cdot f(\alpha)\right) \\
        f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\frac{\alpha}{1-f(\alpha)}\right)
    \end{align*}
    from which it follows that $f(\alpha)=2$ which is a very clear contradiction. Thus, it follows that $f(x) \geq 1$ for all $x\in \mathbb{R}_{>0}$. Now, note that we have
    \[f(x)^2 = 2f(x+xf(x))\geq 2\]
    from which it follows that $f(x) \geq \sqrt{2}$ for all $x\in \mathbb{R}_{>0}$. Similarly, a straightforward induction gives,
    \[f(x) \geq 2^{\sum{\frac{1}{2^i}}}=2\]
    for all $x\in \mathbb{R}_{>0}$,  which proves the claim.

Now, we show that there exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$. First note that swapping $x$ and $y$ gives us that
\[f(x+yf(x))=f(y+xf(y))\]
for all positive real numbers $x$ and $y$. Then, say $x+yf(x)=y+xf(y)$ for all $x,y \in\mathbb{R}_{>0}$. Then, plugging in $y=1$ gives $f(x)=cx+1$ for some constant $c$ for all positive reals $x$ which is clearly false for any $c\in \mathbb{R}_{>0}$ upon substitution. This means, there exists $x_0,y_0 \in \mathbb{R}_{>0}$ such that $x_0+y_0f(x_0) \neq y_0+x_0f(y_0)$. WLOG, assume that $x_0+y_0f(x_0)>y_0+x_0f(y_0)$. Then, let $b= y_0+x_0f(y_0)$ and $a=\frac{x_0+y_0f(x_0) - (y_0+x_0f(y_0))}{f(y_0+x_0f(y_0))}>0$.
Then, plugging in $x=b$ and $y=a$ yields,
\[f(b)f(a)=2(f(b+af(b)))=2f(x_0+y_0f(x_0)) = 2f(b)\]
from which it follows that $f(a)=2$. Thus, there indeed exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$.
We now have our final key claim.

[color=#f00] [b]Claim :[/b]  $f$ is non-decreasing.[/color]
 [i] Proof : [/i]Simply note that
    \[2f(x+yf(x))=f(x)f(y)\geq 2f(x)\]
    since $f\geq 2$. Thus, $f(x) \leq f(x+yf(x))$ for all $x,y \in\mathbb{R}_{>0}$. Thus, it follows that $f(x)\leq f(x+\epsilon)$ for any $x, \epsilon \in \mathbb{R}_{>0}$ (by varying $y$) which implies that $f$ is non-decreasing as claimed.

 Now, consider $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha)=2$. Plugging in $x=y=\alpha$ gives
 \[4=f(\alpha)^2=2f(\alpha+\alpha f(\alpha))=2f(3\alpha)\]
 which implies that $f(3\alpha)=2$. Thus, it follows by a straightforward induction that $f(3^ra)=2$ for all $r\in \mathbb{N}$, from which it follows that there exists arbitrarily large values of $x\in\mathbb{R}_{>0}$ such that $f(x)=2$. Putting this together with the non-decreasing nature of $f$ it follows that $f$ is constant 2, as desired.


limit
integration
logarithms
function
calculus
trigonometry
real analysis

Ascertain $L\equiv\lim_{n\to \infty}\ n\cdot\int^1_0 \frac{x^n}{\sqrt{1+x^2}}\ln \left(x+\sqrt{1+x^2}\right)\ dx.$[hide="Answer"]$L=\frac{\sqrt 2}{2}\ln (1+\sqrt 2).$[/hide]

Suppose $f$ is a continuous, summable function on $(0, 1]$ with $f(1)$ finite so that $\lim_{n \to \infty} n \int_0^1 x^n f(x)\,dx - f(1) = \lim_{n \to \infty} n \int_0^1 x^n f(x)\,dx - \lim_{n \to \infty} \frac{n}{n+1} f(1) = \lim_{n \to \infty} \int_0^1 x^n \left(f(x) - f(1)\right)\,dx$. Fix any $\varepsilon > 0$ and consider the $\delta$ guaranteed by continuity such that $y > 1 - \delta$ implies $|f(1) - f(y)| < \varepsilon$. Then $|\lim_{n \to \infty} n \int_0^1 x^n (f(x) - f(1))\,dx|$ $\le$ $|\lim_{n \to \infty} n \int_0^{1 - \delta} x^n (f(x) - f(1))\,dx|$ $+$ $|\lim_{n \to \infty} n \int_{1 - \delta}^1 x^n (f(x) - f(1))\,dx|$ $\le$ $\lim_{n \to \infty} n (1 - \delta)^n \int_0^{1-\delta} |f(x) - f(1)|\,dx$ $+$ $\lim_{n \to \infty} n\int_{1-\delta}^1 x^n \varepsilon\,dx$. Note that the integrand of the first term is less than $|f(x)| + |f(1)|$; the integral of this is finite due to the assumption of summability and finiteness of $f(1)$ and so the limit of this finite quantity times something that goes to $0$ is $0$, and thus the limit is bounded above by $\lim_{n \to \infty} \frac{n}{n+1} \varepsilon (1 - (1 - \delta)^{n+1}) = \varepsilon$. Since this holds for any $\varepsilon > 0$, we find that the limit is $0$ and thus that $\lim_{n \to \infty} \int_0^1 x^nf(x)\,dx = f(1)$ for any such function $f$ satisfying the conditions outlined at the beginning. Let us check these conditions for the particular $f(x) = \frac{\ln\left(x + \sqrt{1+x^2}\right)}{\sqrt{1+x^2}}$. Certainly $f(1) = \frac{\ln \left(1 + \sqrt{2}\right)}{\sqrt{2}}$ is finite and $f$ is continuous, so it remains to check that $f$ is summable, i.e. that $\int_0^1 f_{+}, \int_0^1 f_{-} < \infty$ (the positive and negative parts of $f$ both have finite integrals). In fact, note that our particular $f$ is always nonnegative, so it suffices merely to check that $\int_0^1 f(x)\,dx < \infty$, and this is now just a nice integral to evaluate (well we don't actually need to evaluate it; we just need to check it's finite). Unfortunately, I am trying to evaluate it and am in the process of realizing how bad I am at doing random integrals...I'll keep working on this last step.

Aha! Incredibly, the antiderivate is simply $\frac{\left(\ln \left(x + \sqrt{1+x^2}\right)\right)^2}{2} + C$, and this is just fine for our purposes.

To evaluate
\[ \lim_{n\to\infty} \left|n\int^{1-\delta}_0 x^n(f(x)-f(1))dx\right|  \]
you can just use the dominated convergence theorem, since $f(x)-f(1)$ is continuous (hence bounded), and $nx^n\to 0$ for all $x\in [0,1-\delta]$ as $n\to\infty$, so that if $f_n(x)=nx^n(f(x)-f(1))$ for $n\in\mathbb{N}$, then $f_n(x)\leq M$ for all $n,x$ and some $M$, and $f_n(x)\to 0$ as $n\to\infty$

I'm curious if we can use one of the convergence theorems to prove the rest of it, but I haven't had any luck so far.

This is essentially a case for Laplace's method for asymptotic evaluation of integrals. How about this one:
$\lim_{n \to \infty} \sqrt{n}\int_0^\pi \left( \sin x\right)^n e^{x^2} dx$.

[quote="hpe"]This is essentially a case for Laplace's method for asymptotic evaluation of integrals. [/quote]
What is Laplace's method?

[quote=liyi]What is Laplace's method?[/quote]
A method for evaluating asymptotically (in the limit of large $x$) integrals like $\int_a^b e^{xf(t)}g(t) dt$.


geometry
calculus
email
Harvard
college
MIT
linear algebra

I will be a senior this fall, and I would like to begin a research project.  I don't have any opportunity to use a lab in my area, so it would be mostly on my own.  I need to find a topic that doesn't require extensive lab work and supervision (a lot of people have mentors).  Does anyone have any suggestions, or know where I could start looking?

Hmmm...there are no colleges or universities nearby?  Those are a good resource, you can try to find a professor who can be your mentor.  Hmm...but I guess a math topic wouldn't really require much lab stuff.  But I wouldn't really know, I'm just a freshman with no experience in research.

There are colleges / universities in my area, but I've already checked with them.  Like you said, it would be interesting to do something with mathematics, but I'm not that experienced (compared to many of you here).  I have completed math through Multivariable Calculus and Linear Algebra, and will either be taking Diff. Eq. or doing an independent study this fall.  Are there any math research topics that might suit me at this point?

Thanks again,

Alex

i have never done math research myself, but there should be plenty of options with that much math background.  i've seen projects that use at most single variable calculus.  why don't you look for something in number theory?

That's a good point.  I just need a place to begin looking.  It seems that everything that pops into my head is either too difficult, or has already been done.  I have gone to planetmath.org to look through some of their unproven things, but they are too advanced.  I figure anything on the internet that is easy enough for a high school student to work on has already been done by a professional, but I may be wrong (hopefully).  Does anyone know a website maybe?

Thanks again, I appreciate your help.

your AoPS profile says you live in Massachusetts.  i don't know how far you are from boston, but there are plenty of colleges there that you could e-mail professors at.  even if it's an hour away, you can get most of the work done and basically just have the professor to go to in order to fill out any necessary forms or guide you on your project in any way.

I live near Springfield, roughly 1.5 hours from Boston.  If I were to do that, would I need a topic already picked out or could I just send an email to a few professors and ask what they would suggest?  I just didn't think that they would do something like that without being paid (by a program or whatnot).  I could definitely get down to Boston on the weekends though if I found a professor.  Which colleges would you suggest trying?

I say go for it.  You'd be surprised how many professors wouldn't mind getting into something like that with a high school student.  Of course when you e-mail them they will not all respond to you, and you might very well get negative responses from some of them, but you are bound to find someone who will be willing to help you out.  Professors are typically in love with their field of work and they will be glad to enlighten a high school student who is interested in their work.  You don't have to have a project planned previously.  You can simply e-mail some professors saying who you are, how devoted you are, that you love math, etc., and ask if they would be willing to work with you by e-mail and once in a while face to face on a high school research project along their lines of interest.  I recommend going on the math department websites of schools such as Harvard, MIT, and Boston University, and taking a look at some professors research interests.  Their e-mail addresses should all be listed there.  Hope this works for you!

That sounds great.  I am doing it right now.  Thank you very much for all of your help!

Alex


errr i was looking at #13 on the 1984 AIME and i saw someones solution and i just wanted to know if someone could show me why

$\cot^{-1}3 = \tan^{-1}\displaystyle\frac {1}{3}$

or errr if someone could just show me the general proof involving $x$

Just look at it geometrically...

Or, if you want it in all algebra:

Let $\theta=\cot^{-1}3\Rightarrow \cot \theta = 3$.  Since $\tan x =\frac 1{\cot x}$, $\tan \theta =\frac 13\Rightarrow \tan^{-1}\frac 13 =\theta.$


The number m=999...9 consists of 999 nines.  The sum of the digits of number m^2 is equal to:

[hide="hint"]

$(9...9)^{2}$ with n 9's is $9...980...01$ for $n-1$ 9's and $n-1$ 0's. Unfortunately, I have no clue how to prove it.

EDIT- actually i just realized t0rajir0u's hint can be used as a proof

[/hide]

You guys, I know the answer, all I want is the answer from you.


ok,this problem is the current Mathnfriend Contest,MOD or ADMIN please lock this topic until the contest ends. :)

Done. Let me know when it's over so that it can be unlocked.


algebra function calculus minimization polynomial

Let $F = \max_{1 \leq x \leq 3} |x^3 - ax^2 - bx - c|$.  When $a$, $b$, $c$ run over all the real numbers, find the smallest possible value of $F$.

maybe it helps to put $y = x-2$, so that $|y| \leq 1$.

Okay, I am back home from school.. problem doesn't look that hard, which means I am doing something wrong.. someone please check my soln?

EDIT:  The last part was bad, but I think its fixed now.
--------

Use above trick, substituting y  = x+ 2.

Now the coefficents of y are independent, so we can consider the equivalent problem of minimizing G, the absolute value of a general monic cubic $y^3 + ay^2 + by + c$ on the domain $|y| \leq 1$, over all reals a,b,c.

Write it in the form:  $ay^2 + c  + (y^3 + by)$.

It is easy to check that $(a,b,c) = (0,\frac{-3}{4},0)$ gives us a maximum of $1/4$, because $y^3 - (3/4)y = 1/4 \Leftrightarrow (y-1)(y+ 1/2)^2$.

Suppose [for a contradiction] that (a,b,c) is a tuple that gives a lower maximum.

By differentiation, it is clear that the extremal values of $y^3 + by$ occur at $y = \pm \sqrt{-b/3}$ (if b<0) and $y = \pm 1$, (in the domain $|y| \leq 1$).

Now, $ay^2 + c \leq 0$ whereever maximums of $y^3 + by$ occur, since if it was $> 0$, we would have the maximum larger.  Also, $ay^2 + c \geq 0$ whereever minimums of $y^3 + by$ occur.  By symmetry, both in the places where extremal values of $y^3 + by$ occur, as well as the graph of $ay^2 + c$, we have to conclude $ay^2 + c = 0$ at the maximums of $y^3 + by$ (**).

--

If b < -3/4, then y = 1 gives us a maximum > 1/4, and y = -1 gives us a minimum < 1/4.  Since $ay^2 + c$ is an even function, if it is nonzero, then for some value of y in $\{-1, 1\}$, the signs of $ay^2 + c$ and $y^3 + by$ must be the same, and the magnitude of the entire cubic exceeds 1/4, if $ay^2 + c$ doesn't have roots at $y = \pm 1$.  Now suppose $ay^2 + c$ has roots at $y = \pm 1$.  Clearly, $y = \sqrt{-b/3}$ is a local maximum of $y^3 + by$, and by the supposition b < -3/4, we have infact that it is a real number in [0, 1/2].  Now consider $\sqrt{-b/3}^3 + -ab/3 + b\sqrt{-b/3} + c$.  By "$ay^2 + c$ has roots at $y = \pm 1$", we have $-a = c$; by considering the cubic at $y=0$, we have $|a| < 1/4$.  So we have $\sqrt{-b/3}(4b/3) + c(b/3 + 1) < (1/2)(-1) + c(3/4) = 3c/4 - 1/2$.  Now $|3c/4| < 3/16$, whereby $3c/4 - 1/2 < -5/16$.  So the general cubic has a magnitude above $1/4$ at that point, and it is inadmissible.

Similarily, consider b > -3/4 at the point $y = \sqrt{-b/3}$.   Similarily, we must have $ay^2 + c$ with roots at $y = \pm \sqrt{-b/3}$.  This implies ab = 3c.  Consider $y = 1$ in the general cubic.  We have $1 + a + b + c > 1/4 + a + c$.  Now $3c = ab > -3a/4$, whereby $c > -a/4$.  So $1/4 + a + c > 1/4 + 3a/4$.  Obviously, due to the symmetry (oddness) of $y^3 + by$, it doesn't matter whether a was positive or negative.  So the magnitude of the entire cubic is clearly $> 1/4$, and so it is inadmissible.

So $b = -3/4$.  By (**), $ay^2 + c$ is zero at 4 places, thus it must be the zero function.  So we have $(a,b,c) = (0, -3/4, 0)$, with a maximum of 1/4.  This is the same maximum as in the original problem, so we are done.

I think I solved a much more general problem of Alexandru Lupas posted on the forum and which asked for the minimal value of a monic polynomial on a compact interval $[a,b]$. Anyway, it is a direct consequence of Chebyshev's theorem for polynomials on $[-1,1]$.

[b]Chebyshev's theorem[/b] Let $f \in \mathbb{R}[x]$ be a monic polynomial of degree $n$. Then

$$\max_{x\in [-1,1]}|f(x)| \ge \frac{1}{2^{n-1}}$$

[b]Original problem:[/b] Denote $y=x-2$. $(y\in [-1,1]) \implies F=\max_{y \in [-1,1]} |y^3 +Ax^2 +Bx+C|$
 By [i]Chebyshev's theorem[/i],  $$F\ge \frac{1}{2^{3-1}}=\frac{1}{4}$$
[b]Answer: [/b]$\frac{1}{4}$

Equality holds $f(x)=\frac{1}{4}T_{3}(x)$.  $T_{3}(x)$ is [i]Chebyshev polynomials[/i] $(T_{3}(x)=4x^3-3x) (\because cos3x=4cos^3x-3cosx)$


superior algebra
superior algebra unsolved

How many permutations of $ S_{7}$ are product of three disjoint transpositions? $ (**)(**)(**)$
What about two disjoint transpositions? $ (**)(**)$.

[hide="My answers"]The answer to both question is $ 105$
First : $ \binom{7}{2}\binom{5}{2}\binom{3}{2}\frac{1}{6}$

Second : $ \binom{7}{2}\binom{5}{2}\frac{1}{2}$
 [/hide]

My question is. Assuming that I've computed it correctly the sets of permutation of these types have the same number of elements. It's against my intuition. I believe the first case is larger than the second one.
Could you give me some kind of bijection between them??

You've computed correctly.  Using a bijective proof to show that $ 105 \equal{} 105$ seems a little silly to me, but here we go.  Three times the first is the number of set partitions of a seven element set into a singleton and three pairs, where one of the pairs is specially marked.  Three times the second one is the number of set partitions of a seven element set into two pairs and three singletons, where two of the singletons are specially marked.  From here, the bijection is obvious: compare $ (**)(**)\textcolor{green}{(**)}*$ and $ (**)(**)\textcolor{green}{**}*$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a subgroup of $S_{5}$ containing a 5-cycle and a 2-cycle. Show that $G=S_{5}$.

I did not try to do it, but I was thinking. If you can show that $G$ properly contains $A_{5}$ then since $(S_{5}: A_{5})=2$ it cannot be that $A_{5}<G<S_{5}$ thus $G=A_{5}$


The number A = 200420052006......2040 is formed by putting the comsecutive integers from 2004 to 2040 together. What is the remainder when A is divided by 9??


calculus
integration
function
geometry
3D geometry
sphere
calculus computations

suppose that f is continuous function . show that

INT INT(surface) f(ax+by+cz)dS=2pi INT(from -1 to 1) f( u*sqrt(a^2+b^2+c^2))du

where s is the sphere x^2+y^2+z^2=1(a>0,b>0,c>0)

perhaps you would do better with :

 Show that :
 $\displaystyle\int \oint f( ax+ by+ cz)\ ds = 2 \pi \int^1_{-1} f( u* \sqrt{a^2 + b^2 + c^2 })\ du$ :rotfl:


geometry
circumcircle
incenter
geometric transformation
homothety
symmetry
cyclic quadrilateral

Let the incircle of triangle $ ABC$ touch sides $ BC,\, CA$ and $ AB$ at $ D,\, E$ and $ F,$ respectively. Let $ \omega,\,\omega_{1},\,\omega_{2}$ and $ \omega_{3}$ denote the circumcircles of triangle $ ABC,\, AEF,\, BDF$ and $ CDE$ respectively.

Let $ \omega$ and $ \omega_{1}$ intersect at $ A$ and $ P,\,\omega$ and $ \omega_{2}$ intersect at $ B$ and $ Q,\,\omega$ and $ \omega_{3}$ intersect at $ C$ and $ R.$

$ a.$ Prove that $ \omega_{1},\,\omega_{2}$ and $ \omega_{3}$ intersect in a common point.

$ b.$ Show that $ PD,\, QE$ and $ RF$ are concurrent.

From cyclic quadrilateral $ APFE,$ we have that $ \angle EPF = \angle A$ $ ,(1)$

Froma cyclic quadrilateral $ APBC,$ we have that $ \angle BPC = \angle A$ $ ,(2)$ and $ \angle ABP = \angle ACP$ $ ,(3)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \angle EPC = \angle FPB$ $ ,(4)$

From $ (3),$ $ (4)$ we conclude the similarity of the triangles $ \bigtriangleup EPC,$ $ \bigtriangleup FPB$ and so, we have $ \frac{PB}{PC}= \frac{BF}{CE}= \frac{BD}{CD}$ $ ,(5)$

From $ (5)$ we conclude that the segment line $ PD,$ is the angle bisector of $ \angle BPC$ and so, it passes through the point $ A',$ as the intersection point of $ (\omega)$ from $ AI,$ where $ I$ is the incenter of $ \bigtriangleup ABC.$

Similarly the segment line $ QE,$ passes through the point $ B'\equiv (\omega)\cap BI.$

$ \bullet$ We denote the points $ M\equiv OI\cap PA',$ $ M'\equiv OI\cap QB'$ and we will prove that $ M'\equiv M.$

Because of $ ID\parallel OA'$ $ \Longrightarrow$ $ \frac{MI}{MO}= \frac{ID}{OA'}$ $ ,(6)$ and similarly $ \frac{M'I}{M'O}= \frac{IE}{OB'}$ $ ,(6)$

From $ (6),$ $ (7)$ $ \Longrightarrow$ $ \frac{MI}{MO}= \frac{M'I}{M'O}$ $ ,(8)$ $ ($ because of $ ID = IE$ and $ OA' = OB'$ $ ).$

From $ (8)$ $ \Longrightarrow$ $ M'\equiv M$ $ ,(9)$

From $ (9)$ we conclude that the intersection point of the segment lines $ PA'\equiv PD,$ $ QB'\equiv QE,$ lies the constant segment line $ OI.$

By the same way we can prove that the intersection point of the segment lines $ PD,$ $ RF,$ lies also on $ OI.$

Hence, the segment lines $ PD,$ $ QE,$ $ RF,$ are concurrent at one point and the proof is completed.

Kostas Vittas.

PS. I don’t understand well the part [b][size=100](a)[/size][/b] of the problem. If by ''comment point'', you mean ''common point'', then it is clear that the concurrency point of $ (\omega_{1}),$ $ (\omega_{2}),$ $ (\omega_{3}),$ is the incenter $ I$ of $ \bigtriangleup ABC.$

Sorry, I will post here later a figure, because now I am very tired.

[b]Problem statement posted by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=40604]tobeno_1:[/url]

$ \odot I$ and $ \odot O$ are the incircle and circumcircle of $ \triangle ABC$ respectively. $ D,E,F$ are the tangent points.

$ \odot O_1$ is the circumcirlce of $ \triangle AEF$,$ \odot O_2$ is the circumcirlce of $ \triangle BDF$ ,$ \odot O_3$ is the circumcirlce of $ \triangle CDE$.

$ \odot O_1 \cap \odot O \equal{} P(P\neq A)$, similarly define $ Q,R$.

Prove: $ PD,QE,RF$ are concurrent. (see image below)

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=26129]QuattoMaster 6000:[/url]

We consider the circle, $ T_A$ that is tangent to the incircle at $ D$ and tangent to $ (O)$ at a point, say $ P'$.  Notice that the homothety that maps $ T_A$ to $ (O)$ maps $ D$ to the midpoint of arc $ BC$, say $ K$.  Thus, $ P'DK$ is a line.  We also have that $ \angle PBF \equal{} \angle PCE$ and $ \angle PFB \equal{} 180 \minus{} \angle PFA \equal{} 180 \minus{} \angle PEA \equal{} \angle PEC$, so $ \triangle PBF\sim \triangle PCE$.  This gives us that $ \frac {PB}{PC} \equal{} \frac {FB}{EC} \equal{} \frac {BD}{DC}$, so $ PD$ bisects $ \angle BPC$, which implies that $ PD$ passes through the midpoint of arc $ BC$.  Hence, $ PDK$ is a line, which means that $ P \equal{} P'$.  Now, $ P$ is the center of homothety between $ (O)$ and $ T_A$ and $ D$ is the center of homothety between $ (I)$ and $ T_A$.  This implies that the center of homothety between $ (O)$ and $ (I)$ lies on $ PD$ by Monge's Theorem.  Similarly, this center of homothety lies on $ FR$ and $ EQ$, which implies that $ PD$, $ FR$, and $ EQ$ concur, as desired.

[b]Approach by[/b] [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=18812]pohoatza:[/url]

Consider the inversion $ \Psi$ with respect to the incircle of triangle $ ABC$. The vertices $ A$, $ B$, $ C$ go into the midpoints $ A'$, $ B'$, $ C'$ of the segments $ EF$, $ FD$, $ DE$, respectively, and the circumcircle of $ ABC$ goes into the circumcircle of $ A'B'C'$. Since the circumcircle of $ AEF$ passes through the pole of inversion $ I$, its image is the line $ EF$, and therefore the image $ P'$ of the second intersection $ P$ of the circles $ AEF$ and $ ABC$ is the intersection of $ EF$ and the nine-point circle $ (A'B'C')$ of $ DEF$ (the one different from $ A'$). This means that $ P'$ is the foot of the perpendicular from $ D$ to $ EF$. Similarly, we get that the images $ Q'$, $ R'$ of $ Q$, $ R$ under $ \Psi$ are the feet of the perpendiculars from $ E$, $ F$ to $ FD$, $ DE$, respectively, and thus $ P'Q'R'$ is the orthic triangle of $ DEF$. Now, since the orthocenter of $ DEF$ has the same power with respect to the circles $ (DIP')$, $ (EIQ')$ and $ (FIR')$, it is easy to see that they are coaxal, and thus they have another common point, different then $ I$. This means that the lines $ DP$, $ EQ$, $ FR$ are concurrent.

Excellent,excellent....This was the solution I looked for...Thanks a lot,nick.I think even before approaching you thought of radical center...But how can you say that $PEDQ$ is cyclic?

Consider the spiral symmetry that takes $E \to F$ and $B \to C$
It is well known that $P$ is the centre of this spiral symmetry.
Thus we have $\triangle PEB \sim \triangle PFC$
$\implies \frac{PB}{PC}=\frac{BE}{CF}=\frac{BD}{CD}$
$\implies PD$ is the angle bisector of $\angle BPC$

Let $X$ be the midpoint of arc $BC$ of circumcircle of $\triangle ABC$

We have $PDX$ collinear.

Define $Y,Z$ similarly.

Thus we are left to prove $XD,YE,ZF$ are collinear which is easy :)

[quote="nickthegreek"]Let me give another,quicker approach to this beautiful problem...

For the first part, it is easy to observe that all of the circles must pass through the incenter ( to see this just drop the perpendiculars from $D,E,F$ to the sides of $BC,CA,AB$ respectively). In fact, it is not even necessary to use the incenter. We could just use the well-known [b]Miquel point of the triangle[/b] with respect to the points $D,E,F$.

Now, as far as the second part is concerned, it is easy to prove (using nothing more than angle-chasing) that the three quadrilaterals $PEDQ,QFER,PFDR$
are cyclic. Hence , the three lines $PD,QE,RF$ are the radical axes of the circles taken two at a time. Finally, the three lines concur at the radical center of the three circles!

Best,
Nick[/quote]
How is $PEDQ$ cyclic ? :maybe:

[quote=61plus]U can show that $DE$ is parallel to the line joining midpoint of arcs $BC,AC$, then from $P,D,X$ collinear etc. we can deduce that those $4$ points are concyclic.[/quote]

I don't understand you proof @61plus , caan you explain more ?

[quote=utkarshgupta]
Thus we are left to prove $XD,YE,ZF$ are collinear which is easy :)[/quote]

This is tricky. We use the Cevian Nest Theorem. Observe that $AX$, $YE$, and $ZF$ are concurrent at the incenter $I$, and that $AD$, $BE$, and $CF$ are concurrent at the Gergonne point. Hence, by the Cevian Nest theorem, we are done.

Revival, as the answer I'm looking for, wasn't answered very well in here.
How do we show that $PEDQ$ is a cyclic quadrilateral?

[quote=Kezer]How do we show that $PEDQ$ is a cyclic quadrilateral?[/quote]
\begin{align*}
	\measuredangle QPE &= \measuredangle QPA + \measuredangle APE \\
	&= \measuredangle QBA + \measuredangle AIE \\
	&= \measuredangle QBA + \measuredangle ABI + \measuredangle IDE \\
	&= \measuredangle QBI + \measuredangle IDE \\
	&= \measuredangle QDI + \measuredangle IDE \\
	&= \measuredangle QDE.
\end{align*}
All angles directed. With this, of course, the problem falls to radical axis.

A shorter but more advanced way to show the quadrilateral is cyclic is to note that $PQDE$ becomes the circle with diameter $DE$ after inverting around the incircle. (The circumcircle of $ABC$ maps to the nine-point circle of $DEF$.)

(A little addendum: for anyone that's reading this because of [url=http://www.mit.edu/~evanchen/geombook.html]my book[/url], the above angle chase is actually a lot harder to find than it looks, or at least it took me half an hour to work out again just now. I wish now that I had included another hint or two afterwards, but too late now.)

We notice that $PD$ passes through the midpoint af arc $BC$ not containing $A$ since $\angle API=90$. (This follows since $P,D$ are in verses under inversion about $M$ with radius $MB=MC=MI$) Now, this concurrency point has to be centre of positive Homothety sending the in touch triangle $DEF$ to the triangle formed by the midpoint of arcs $MKL$. Thus, we are done.

[hide=Solution]
a) Note that $I$, the incenter of $\triangle{ABC}$, lies on all three of these circles.

b) Let $I$ be the incenter of $\triangle{ABC}$ and note that $I$ is on all of $\omega_1$, $\omega_2$, and $\omega_3$.

Perform inversion about the incircle of $\triangle{ABC}$. The problem rewords to the following: Let $DEF$ be a triangle and $PQR$ the orthic triangle of $\triangle{DEF}$. Let $I$ be the circumcenter of $\triangle{DEF}$. Prove that the circumcircles of $\triangle{DIP}$, $\triangle{EIQ}$, and $\triangle{FIR}$ are coaxial.

I will rename points so it is easier to deal with: Let $ABC$ be a triangle and $DEF$ the orthic triangle of $\triangle{ABC}$. Let $O$ be the circumcenter of $\triangle{ABC}$. Prove that the circumcircles of $\triangle{AOD}$, $\triangle{BOE}$, and $\triangle{COF}$ are coaxial.

Let $H$ be the orthocenter of $\triangle{ABC}$. Then since $BFEC$ is cyclic, we have that \[HB\cdot HE=HC\cdot HF.\] Similarly, \[HA\cdot HD=HB\cdot HE,\] so \[HA\cdot HD=HB\cdot HE=HC\cdot HF.\] This implies that $H$ has equal power with respect to each of the circumcircles of $\triangle{AOD}$, $\triangle{BOE}$, and $\triangle{COF}$, so they are coaxial since $O$ also has equal power. In particular, their radical axis is the Euler line of $\triangle{ABC}$.

Thus, we have proved the modified problem statement which we have shown to be equivalent to the actual problem statement.
[/hide]

Let me post an unenlightening solution since trumpeter already took the time to bump this  :-D

[hide="solution"]
a) By definition of miquel point, they concur at the incenter $I$.
b) It's well-known that $P, D$ and the minor arc midpoint of $BC$ concur. But this means $PD$ passes through the exsimilicenter of $(I)$ and $\omega$, which is the desired concurrency point.
[/hide]

Very nice!
[hide=Solution]It is obvious that the common point is $I$ the incenter of $\triangle ABC$
Now what do we do when we have a thousand circles passing through $I$?
We invert about $I$  :D

The inversion about the incircle maps $$\omega \iff \text{Nine point circle of DEF}$$ and $\omega_1$ is mapped to $EF$ and similarly the other circles map to their respective sides.
With this we get that the points $P,Q,R$ get swapped with the feet of $D,E,F$ altitudes in $\triangle DEF$ (Lets call them $P'Q'R'$ respectively)
Now we see that $$PD \iff \odot(P'ID)$$ $$QE \iff \odot(Q'IE)$$ $$RE \iff \odot(R'IF)$$ Now our goal is to prove that these three circles are concurrent at some point other than $I$
Notice that $I$ is the circumcenter of $\triangle DEF$
$$\text{This is EGMO Problem 2.36! (My proof is the same as evan's proof in the book. Geogebra helped in noticing the point H)}$$  :D and that finishes the proof!  :D $\blacksquare$[/hide]

By convention, we denote $I$ as the incentre of triangle $ABC$.
Denote $A'$, $B'$ and $C'$ as the intersection of rays $AI$, $BI$, and $CI$, respectively with $\omega$.
Since $AI$, $BI$, and $CI$ are the angle bisectors of triangle $ABC$, $A'$, $B'$ and $C'$ are the midpoints of arcs $BC$ (not containing $A$), $CA$ (not containing $B$), and $AB$ (not containing $C$), respectively.

Lemma 1: $PD$, $QE$ and $RF$ passes through points $A'$, $B'$, and $C'$ respectively.
Proof: By angle chasing, we have $\angle BPC = \angle BAC = \angle FAE = \angle FPE$ and hence,$$\angle FPB = \angle BPC - \angle FPC = \angle FPE - \angle FPC = \angle EPC$$Moreover, $\angle PBF = \angle PCE$ as $ACBP$ is cyclic, and hence $\triangle FPB \sim \triangle EPC$.
This shows that $\frac{PB}{PC} = \frac{BF}{CE} = \frac{BD}{CD}$, and hence by angle bisector theorem, $PD$ bisects $\angle BPC$.
Therefore, $PD$ would intersect $\omega$ at the midpoint of arc $BC$ (not containing $A$), i.e. $A'$.
Similarly, $QE$ and $RF$ would intersect $\omega$ at $B'$ and $C'$ respectively, which completes the proof.

Lemma 2: $EF \parallel B'C'$, $FD \parallel C'A'$ and $DE \parallel A'B'$.
It is well known that $I$ is the orthocentre of triangle $A'B'C'$.
Therefore, we have $B'C' \perp AA'$. Since it is obvious that $EF \perp AA'$, we can conclude that $EF \parallel B'C'$.
Similarly, $FD \parallel C'A'$ and $DE \parallel A'B'$.

Finally, we show that $EFRQ$, $FDPR$ and $DEQP$ are cyclic.
Let $B'Q \cap C'R = T$.
As $EF \parallel B'C'$, it can be deduced that
$$ TQ \cdot TB' = TR \cdot TC' \iff \frac{TQ}{TR} = \frac{TC'}{TB'} = \frac{TF}{TE} \iff TF \cdot TR = TE \cdot TQ$$Therefore $EFRQ$ is cyclic and similarly, $FDPR$ and $DEQP$ are cyclic.
Note that the three circles $(EFRQ)$, $(FDPR)$ and $(DEQP)$ have radical axes $PD$, $QE$, and $RF$, which are clearly not parallel, which suggests that $PD$, $QE$, and $RF$ must be concurrent (at $T$).

For first part just add incenter and that's the common point and can be easily proved by angle chasing.

For second part if we prove PFDR and PEDQ and QFER are cyclic then PD,QE,RF will be radical axis and then they are concurrent.
I will just prove one of these because the rest can be proved the same way.

∠PED + PQD = ∠A/2 + ∠C/2 + ∠B + ∠PQF + ∠PAB so we just need to prove ∠PQF + ∠PAB = ∠A/2 + ∠C/2
∠PQF + ∠PAB = 180 - ∠PQB + PQF = ∠FDB = 90 - ∠B/2 = ∠A/2 + ∠C/2.
we're Done.

Pitiful problem.

a) is trivial; the common point is the incenter.

for b), it's well known that line $PD$ passes through the arc midpoint $M$ of minor arc $BC$, and since the positive homothety sending the incircle to the circumcircle also sends $D$ to $M$, $PD$ passes through the exsimilicenter between the incircle and the circumcircle, and since that point is symmetrically defined we are done.

We are just going to bary bash this entire problem (yes, even part a).

Set $\triangle ABC$ as the reference triangle with $A=(1,0,0),B=(0,1,0), C=(0,0,1).$ We also have $I=(a:b:c),D=(0:s-c:s-b),E=(s-c:0:s-a),F=(s-b:s-a:0)$.

Part (a): We claim that the common point is $I$. Recall that the circle formula is $$0=-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z).$$ Plugging in $A$ gives $u=0$, plugging in $E$ gives $w=b(s-c)$, and plugging in $F$ gives $v=c(s-b)$. Hence, $$\omega_1=-a^2yz-b^2xz-c^2xy +(c(s-b)y+b(s-c)z)(x+y+z).$$ Finally, note that $$\text{Pow}_{\omega_1} (I)=-a^2bc-b^2ca-c^2ab +(c(s-b)b+b(s-c)c)(a+b+c)=(a+b+c)[-abc+bc(s-b+s-c)]=0.$$ Similar work shows that $I$ is on $\omega_2$ and $\omega_3$.

Part (b): First, we will prove the following lemma.

[b]Lemma:[/b] Let $M_a$ be the midpoint of $\widehat{BC}$ not containing $A$. Then $P,D,M_a$ are colinear.

[i]Proof.[/i] Clearly, $$\omega=a^2yz+b^2xz+c^2xy$$ so adding $\omega$ and $\omega_1$ gives $$(c(s-b)y+b(s-c)z)=0.$$ Suppose $y=b(s-c)$ and $z=-c(s-b)$. Then putting this into the equation for $\omega$ gives $x = \frac{a^2(s-b)(s-c)}{(b-c)(s-a)}$. Thus, $$P=(a^2(s-b)(s-c):(b^2-bc)(s-a)(s-c):(c^2-bc)(s-a)(s-b)).$$

Let the $A$-excenter have coordinates $I_a=(-a:b:c)$. It is well-known that $M_a$ is the midpoint of $II_a$ so $E=\frac{I+I_a}2=\frac12\cdot\frac{1}{(a+b+c)(-a+b+c)}[(-a^2+ab+ac,-ab+b^2+bc,-ac+bc+c^2)+(-a^2-ab-ac,ab+b^2+bc,ac+bc+c^2)].$ Thus,$$M_a=(-a^2:b^2+bc:c^2+bc).$$

Now observe \begin{align*}\begin{vmatrix}0&s-c&s-b\\-a^2&b^2+bc&c^2+bc\\a^2(s-b)(s-c)&(b^2-bc)(s-a)(s-c)&(c^2-bc)(s-a)(s-b)\end{vmatrix} &=(s-c)[(c^2+bc)a^2(s-b)(s-c)-(-a^2)(c^2-bc)(s-a)(s-b)]+(s-b)[-a^2(b^2-bc)(s-a)(s-c)-(b^2+bc)(a^2)(s-b)(s-c)]\\ &=a^2(s-b)(s-c)[c(b+c)(s-c)+(s-a)(c-b)(c+b)-b(b+c)(s-b)]\\&=a^2(s-b)(s-c)(b+c)[c(s-c)+(c-b)(s-a)-b(s-b)]\\&=a^2(s-b)(s-c)(b+c)[c(b)-b(c)]\\&=0\end{align*} so $P,D,M_a$ are colinear. $\square$

Of course, permuting variables shows that this colinearity holds wrt to the other vertices of $\triangle ABC$. Hence, we just need to show that $DM_a,EM_b,FM_c$ concur.

The equation of line $DM_a$ is given by $$\begin{vmatrix}x&y&z\\0&s-c&s-b\\-a^2&b^2+bc&c^2+bc\end{vmatrix}=0.$$ Factoring a $2$ from the second row and expanding, the equation becomes $$x[(b+c)(b-c)(-a+b+c)]+y[-a^2(a-b+c)]+z[a^2(a+b-c)]=0$$. We permute the variables to get the equations of $EM_b$ and $FM_c$.

For these three lines to concur, we must have $$\begin{vmatrix}(b+c)(b-c)(-a+b+c)&-a^2(a-b+c)&a^2(a+b-c)\\ b^2(-a+b+c)&(c+a)(c-a)(a-b+c)&-b^2(a+b-c)\\ -c^2(-a+b+c)&c^2(a-b+c)&(a+b)(a-b)(a+b-c)\end{vmatrix}=0.$$ But notice \begin{align*}\begin{vmatrix}(b+c)(b-c)(-a+b+c)&-a^2(a-b+c)&a^2(a+b-c)\\ b^2(-a+b+c)&(c+a)(c-a)(a-b+c)&-b^2(a+b-c)\\ -c^2(-a+b+c)&c^2(a-b+c)&(a+b)(a-b)(a+b-c)\end{vmatrix}&=(-a+b+c)(a-b+c)(a+b-c)\begin{vmatrix}(b+c)(b-c)&-a^2&a^2\\ b^2&(c+a)(c-a)&-b^2\\ -c^2&c^2&(a+b)(a-b)\end{vmatrix}\\&= [(-a+b+c)(a-b+c)(a+b-c)][(b^2-c^2)(a^2c^2-b^2c^2-a^4+a^2b^2+b^2c^2)-a^2(b^2c^2-a^2b^2+b^2)+a^2(b^2c^2+c^4-a^2c^2)]\\&= [a^2(-a+b+c)(a-b+c)(a+b-c)][(b^2-c^2)(c^2-a^2+b^2)-(b^2c^2-a^2b^2+b^2)+(b^2c^2+c^4-a^2c^2)]\\&= [a^2(-a+b+c)(a-b+c)(a+b-c)][b^2c^2-a^2b^2+b^4-c^4+a^2c^2-b^2c^2+a^2b^2-b^4+c^4-a^2c^2]\\&=0\end{align*} and we are finally done!

a) Trivially the incircle.
b) This common point is the exsimilicenter by Miquel's incircle configuration.

Invert about the incircle.

[quote=Canadian MO 2007 Problem 5, Inverted]Let $\triangle ABC$ be a triangle with circumcenter $O$ and orthic triangle $\triangle DEF$. Show that the circumcircles of $\triangle ADO, \triangle BEO, \triangle CFO$ are coaxial.[/quote]

Note that the orthocenter of $\triangle ABC$ has equal power to all three circles, which finishes. $\blacksquare$

In todays world of config geo it is a shame to use angle chasing to prove that $PQDE$ and other variations cyclic.

Dude $P,Q$ are the $A$ and $B \text{ Sharkydevil point}$.If $M_A,M_B$ are the midpoints of minor arcs $\widehat{BC}'\widehat{CA}$ respectively, it is well known that, $\overline{P-D-M_A}$ and $\overline{Q-E-M_B}$.
Now as it is well known that $M_AM_B \perp CI ,DE \perp CI$ we have $M_AM_B \parallel DE$.Since $PQM_AM_B$ is cyclic, by converse of Reim's Theorem $PQDE$ is also cyclic , thats all $\blacksquare$

[b]Solution (a):[/b] This part of the problem is trivialized by Miquel's theorem.

[b]Solution (b):[/b] We claim that $PDEQ$ is cyclic, effectively proving that both $EQRF$ and $RFPD$ are cyclic. Thus by the existence of the radical center, we prove the desired concurrency as was necessary in the problem.

[b]Claim:[/b] $PDEQ$ is cyclic.

[b]Proof:[/b] We use directed angles modulo $\pi$:
\begin{align*}
\measuredangle EDQ &= \measuredangle EDF + \measuredangle FDQ \\
&= \measuredangle EPA + \measuredangle FBQ \\
&= \measuredangle EPA + \measuredangle APQ \\
&= \measuredangle EPQ.\phantom{A}_{\square}
\end{align*}

Therefore by symmetry we also must have that $EQRF$ and $RFPD$ are cyclic, so that their circumcircles have a radical center determined by the concurrency of the lines $PD, QE, RF$, as desired. $\phantom{A}_\blacksquare$

(a) All three circles pass through $I$.

(b) All three lines pass through the exsimilicenter of $\omega$ and the incircle.

Disappointing; I thought I'd have to guess the concurrency point ($H$?) or make some other clever observation, but it turns out you can just read off the concurrent lines as radical axes and prove that the resulting quadrilaterals are cyclic.

(a) Clearly, they concur at $I$.

(b) It suffices to prove that $PDQE$ and cyclic variations are cyclic, whence $PD,QE,RF$ concur at the radical center. But
\[\measuredangle PQD=\measuredangle PQI+\measuredangle IQD=\measuredangle PQI+\measuredangle IBD\]
and
\[\measuredangle PED=\measuredangle PEI+\measuredangle IED=\measuredangle PAI+\measuredangle ICD=\measuredangle PAB+\measuredangle FAI+\measuredangle ICD,\]
which, after rearranging, reduces the condition to
\[\measuredangle PAB=\measuredangle PQI+\measuredangle IBD+\measuredangle ICE+\measuredangle IAF=\measuredangle PQI+\tfrac{\pi}{2}.\]
The above is true because $\measuredangle PQI+\tfrac{\pi}{2}=\measuredangle PQI+\measuredangle IQB=\measuredangle PQB$. $\square$

Here's my solution for the second part (the first one being obvious).

Let $M,N,P$ be the midpoints of the minor arcs $\stackrel{\frown}{BC}$,$\stackrel{\frown}{AC}$,$\stackrel{\frown}{AB}$. We want to show that $(MD),(NE)$ and $(FP)$ concur at one point. This is true since the triangles $\triangle DEF$ and $\triangle MNP$ are directly similar and the sides of the triangles are parallel to each other, hence $(MD),(NE)$ and $(FP)$ concur at the center of homothety that sends $\triangle DEF$ to $\triangle MNP$.

$$\mathbb{Q.E.D.}$$

A twist to this classic,
Sketch: Extend $AP,BQ,CR$ to form a big triangle, by trig ceva and cyclic quads it is easy to see $GQ,JR,HP$ are concurrent, and obviously $JF,GE,HD$ concur at $I$, by cevian nest we're done.

The desired concurrency is just the incenter $I$ in part (a).

For part (b), note that we can classify $P$, $Q$, and $R$ as the $A$-, $B$-, and $C$-Sharkydevil points, respectively. Hence $PD$, $QE$, and $RF$ each pass through the exsimilicenter of the incircle and circumcircle. $\blacksquare$

For part (a), the circles just concur at the incenter. For part (b), the points $P, Q, R$ are the incenter Miquel points for each of the vertices. It follows that $\overline{PD}$, $\overline{QE}$, $\overline{RF}$ concur at the exsimilicenter of the incircle and circumcircle.


Sea $a_n$ una sucesion de enteros postivos tal que $(a_m, a_n)=a_{(m,n)}$ para todo $m,n\in N$. Demuestre que existe una sucesion $b_n$ de enteros positivos tal que \[a_n=\prod_{d|n} b_d\]

Este es realmente un problema de Leonardo, ademas sirve mucho a la hora de atacar problemas de series como Fibonaccis etcetera que cumplen esa propiedad del amximo comun divisor. Como lo hizo?

Pascual2005, su mensaje fue editado. Intente moderar su lenguaje.

tratare, no se si lo logre! ;)  ah y tambien un consejo: Ya que es moderador, cuando edite, deje una nota de lo que edito, y no cambie la estrtura del mensaje! inicialemnte comente que este era un problema muy bueno, usted lo cambio a "este si es un problema de leonardo", el problema NO ES de leonardo, es de Iran 2001, Leonardo solo lo compartio con nosotros. Si lo que usted queria decir es que este es el tipo de problemas de Leonardo pues le digo que a pesar de que conozco muy bien a Leonardo y somos bastante amigos desde hace ams de 2 años, aun no podria identificar sus problemas :D  :D  :D  Ojala este mensaje no este muy editable (o censurable) :D

Buehj, el problema me paecio bien interesante, pero mi solucion es poco escribible. Para resolverlo hice varios casos hayando varios b_n. Ademas es facil ver que los b_n, dados por esa formula, enteros o no, estas determinados de manera unica. Luego conjeture una formula general para los b_n, y luego con induccion demostre que servia. Por ultimo con manipulacion demostre que los b_n´s son enteros. Es bastante larga, y la notacion no es nada bonita.

Leo!, si este problema es asi, digamos tome los enteros y vea lo que va a hacer, y pues uno intuitivamente sabe como hacerlo...no? pues lo que yo hice fue definir los $b_n$ en los primos, en $n=1$, luego en potencias de primos, y luego ver como los usaba en compuestos y eso....finalmente todo cuadraba, pero nada agradable para escribir! jaja! bueno, suerte mis hermanos, nos vemos!

Define $b_1=a_1$ and $b_n=\frac{a_n}{\prod_{d\mid n,d<n}b_d}$ for $n>1$.
Then we see that everything is Ok, but we don't know that $b_n$ are integer numbers.
We will show it by induction.

Suppose that $b_i$ is an integer for all $i<n$ and $b_n$ is not an integer, then $b_n=\frac{u}{pv}$, $p$ is a prime number.
A simple observation: if $i<m$, $j<m$, $i\nmid j$ and $j\nmid i$ then $(b_i,b_j)=1$.
The rest is clear. We know $a_n$ is an integer, so $b_n\prod_{d\mid n,d<n}$ is an integer. It means that there are $b_{i_l}$ ($i_l\mid n$) s.t. $p\mid b_{i_l}$. Suppose wlog $i_1<i_2<...<i_k$. From our observation we conclude that $i_1\mid i_2\mid ...\mid i_k$.
Consider $a_{i_k}$. We see that $(a_n,a_{i_k})=a_{i_k}$, so $\frac{a_n}{a_{i_k}}$ is an integer. But it is impossible, since $\frac{a_n}{a_{i_k}}=b_n\cdot \prod_{d\mid n,d\nmid i_k} b_{d}$ and $\prod_{d\mid n,d\nmid i_k} b_{d}$ is not divisible by $p$. Contradiction.

[Edit: I have made a stupid typo here. Corrected]

Just something....

[quote]A simple observation: if $i<m$ and $j<m$, $i\nmid j$ and $j\nmid i$ then (b_i, b_j)=1[/quote]

Why is that so?, thanks in advance

Consider $a_i$ and $a_j$. We know that $(a_i,a_j)=a_{(i,j)}=\prod_{d\mid (i,j)}b_d$, and if $i\nmid j$ and $j\nmid i$ then $b_i$ and $b_j$ are not in this product, so $a_i=b_i\cdot x\cdot \prod_{d\mid (i,j)}b_d$ and $a_j=b_j\cdot y\cdot \prod_{d\mid (i,j)}b_d$, therefore $(b_i,b_j)>1$ implies $(a_i,a_j)>a_{(i,j)}$.


How does one find sin/cos/tan of 15 degrees by hand?

Half-angle formulas...
I get $\sin {15}=\frac{\sqrt{2-\sqrt{3}}}{2}$

$tan(\frac x 2) = \frac {1-cos x} {sin x}$   [hide="or"] $tan(\frac x 2) = \frac  {sin x}{1+cos x}$ [/hide]once you know tan, you can find sin (or  cos)  of any
[hide=" angle "]$sin x = \frac {tan x} {\sqrt { 1+ tan^2 x} } $ etc..[/hide]

And you  do know tan (or sin or cos) of 30 degrees .. right?

Ah, I see, thanks.  I was stuck on an AIME problem (#15) because I couldn't find $\sin 15^{\circ}$.

[quote="Ravi B"]The only problem with using the half-angle formulas is that we might get a radical inside a radical.  To avoid that, we can use the angle-subtraction formulas (45 - 30).

[Replying to chess64.][/quote]

Nice!

I like to solve it like Ravi B said: $\sin{15^{o}}=\sin{(45^{o}-30^{o})}$.

The results for sin 15 from the 2 methods are indeed equivalent.

[quote="Ravi B"]The only problem with using the half-angle formulas is that we might get a radical inside a radical [/quote]

that's not a problem if you know how to work with radical inside a radical.

take the numerator part of chess64's answer, $\sqrt{2-\sqrt{3}}$. You can see that we could turn this into a form of $\sqrt{(a+b)^2}$ which then turns to $a+b$. Then do the matching game.

$a^2+2ab+b^2 = 2-\sqrt{3}$

$a^2+b^2 =2, 2ab=-\sqrt{3}$
$ab=-\frac{\sqrt{3}}{2}$

Then use substitution to find $a=\frac{\sqrt{6}}{2}$, $b=-\frac{\sqrt{2}}{2}$

then the result is ${\frac{\frac{\sqrt{6}-\sqrt{2}}{2}}{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Maybe that took a long time, and the sum/difference formula works faster, but i just have a too much love for radicals :D

[quote="K81o7"]Without really using any trig, you can do it using an angle bisector.  Just take a 30-60-90 triangle, and bisect the angle.  You then know that the two segments formed have lengths (if the shorter leg has length 1) $\frac{2}{2+\sqrt{3}}$ and $\frac{\sqrt{3}}{2+\sqrt{3}}$.  From there, you know tan15 degrees is $\frac{1}{2+\sqrt{3}}$, and you can find sin and cos.[/quote]
Editted by Shobber.

[quote="Ravi B"]The only problem with using the half-angle formulas is that we might get a radical inside a radical.  To avoid that, we can use the angle-subtraction formulas (45 - 30).

[Replying to chess64.][/quote]
But to get 7.5, then we need half angle

[quote="peeta"](btw: nice method :) )[/quote]

 :first: weeee~~~

[quote="chess64"][quote="peeta"]heres what xxreddevilzxx said without variables (btw: nice method :) ):

$\sqrt{2-\sqrt 3}=\sqrt{\frac32 - \sqrt3 +\frac12}=\sqrt{\left(\frac{\sqrt6}2 - \frac{\sqrt2}2\right)^2}=\frac{\sqrt6-\sqrt2}2$[/quote]

awesome signature :D how did you get from step #2 to step #3? :?[/quote]$a^2-2ab+b^2=(a-b)^2$

Let $a^2=\frac{3}{2}$, and $b^2=\frac{1}{2}$.

as a problem solver, one should develop an eye for such things ;) :)

peeta

[quote="chess64"]Oh. Well i think it's much easier to just use the difference formulas. How are you supposed to see that decomposition/factorization?[/quote]You can use reddevilz's method.  There's a little section about this method in AOPS V.1


Pedro and Cecilia play the following game: Pedro chooses a positive integer number $a$ and Cecilia wins if she finds a positive integrer number $b$, prime with $a$, such that, in the factorization of $a^3+b^3$ will appear three different prime numbers. Prove that Cecilia can always win.

$a^3+b^3 = (a+b)(a^2-ab+b^2)$, so Cecilia just chooses $b$  such that $a+b$ has three different prime factors not dividing $a$.


maybe this is classic but its interestng:

How many monomials of total degree $d$ are in $\mathbb{Z}[x_{1},..,x_{r}]$ ??
(where the total degree of $x_{1}^{\alpha_{1}}\cdot\ldots\cdot x_{r}^{\alpha_{r}}$ is $\alpha_{1}+\ldots+\alpha_{r}$)


Prove that the  equation $ a^{2}\plus{}b^{2}\equal{}c^{2}\plus{}3$ has  infinite  solutions.  :)

[hide="Hidden by mod"]Let $ a$ be an even number and $ c \equal{} \frac {a^{2} \minus{} 2}{2}$ and $ b \equal{} \frac {a^{2} \minus{} 4}{2}$. Then:

$ c^{2} \minus{} b^{2} \equal{} (c \minus{} b)(c \plus{} b) \equal{} a^{2} \minus{} 3$[/hide]


I really dont know where to ask this...

Does anybody know how to create matices on the TI-89 titaniums? I know many of you have TI-89 titaniums here... and I was wondering if anyone knew. I need to use them for some problems on Markov Chains.

Thanks in advance!

I think I got it now. you use the data/matrix editor, correct?  :maybe:


advanced fields
advanced fields unsolved

It is not difficult ot show that the punctured disk $ D\minus{}0$ in the complex plane has a universal cover, conformally isomorphic to the upper half plane, in other words $ D\minus{}0$ is a hyperbolic Riemann surface. How it can be showed that $ D\minus{}0$ with the induced hyperbolic metric has no closed geodesic.

The covering map is $ \exp\colon\mathbb H\to D\setminus\{0\}$, where $ \mathbb H$ is the left halfplane. The geodesics on $ \mathbb H$ are horizontal half-lines (mapped onto radial segments) and left semicircles (mapped onto some weird curves). The weird curves have both endpoints on the unit circle, so in that sense they are not closed. However, they can self-cross forming something that can be called a closed geodesic: a geodesic arc $ \gamma\colon [0,1]\to D\setminus\{0\}$ such that $ \gamma(0)\equal{}\gamma(1)$. But from another viewpoint, such $ \gamma$ is not a closed geodesic because $ \gamma'(0)\ne\gamma'(1)$.


linear algebra
matrix
linear algebra unsolved

Let A be a symmetric matrix of order $n$. Show that there is a symmetric invertible matrix B such that $B - B^{-1} = A$.

I suppose this is a trivial problem for you guys :)

with diagonalisation it is straightforward.

Yes, I didn't know the spectral theorem when I asked about this problem. It is hard without the spectral theorem :P


IMC 2001 Longlist is in attached djvu file (rename pdf to djvu).

[quote="N.T.TUAN"]Sorry :oops: I can't open it, please help me.[/quote]

After downloading, extension of file should be changed to djvu
General information about format of djvu: http://en.wikipedia.org/wiki/DjVu , http://filext.com/file-extension/djvu
Viewer for djvu files: http://lizardtech.com/download/dl_options.php?page=viewers (DjVu Browser Plug-in)


number theory
least common multiple
calculus
integration
prime numbers
algebra
linear equation

Let  [tex]a_1,a_2,\ldots,a_{10}[/tex] be integers with [tex]1\leq a_i\leq25[/tex], for [tex]1\leq i\leq10[/tex]. Prove that there exist integers [tex]n_1,n_2,\ldots,n_{10}[/tex], not all zero, such that [tex]\displaystyle{\prod_{i=1}^{10}a_i^{n_i}=1.}[/tex]

I'm surprised that it doesn't go up to 28 instead of 25.

Nice problem. Where does it come from ?

A try :

[hide]If 1 is among the ai, or if two numbers are equal, it's obvious.

Otherwise, if 13,17,19, 23 are among the ai, choose ni=0 for them.

If there is a number which is the only one to have one particular prime factor, take ni=0 for it. There can't be 10 such numbers (only 9 prime numbers until 25).

Now we are left with >= 2 numbers with at most 2 prime factors (2*3*5 > 25), and with no numbers with a unique prime factor.

Then we can split the numbers in two groups such for any number in one group there is a number in the other sharing one of its prime factor (otherwise there would be a number with 3 prime factors).

Now it's just the matter of taking the lcm of every prime factors both sides.[/hide]

[quote="belenos"]
Then we can split the numbers in two groups such for any number in one group there is a number in the other sharing one of its prime factor (otherwise there would be a number with 3 prime factors).
[/quote]

If we have 6, 10, and 15 among these 10 numbers, how do we split into two groups?

Another try

[hide] For each prime p=2,3,5,7,11,13,17,19,23 (9 primes that are less than 25) we can write linear equation for n_i of form  b_1*n_1+...+b_10*n_10=0, where p^b_i|a_i. So we have 9 equations for 10 variables, and this system defines some non-trivial subspace of 10-dimesional space. This subspace clearly contains at least one point with integral components, that gives us the solution.  [/hide]

[quote="fuzzylogic"]  If we have 6, 10, and 15 among these 10 numbers, how do we split into two groups?[/quote]

Bump !


Let $ f: \mathbb(0; 2)\rightarrow\mathbb{R},   f(x)\equal{}(2x\minus{}1)[x^2]$
Investigate the function on continuity.

This function isn't continue at only $ 1$ and $ \sqrt{2},\sqrt{3}$ because, for example:
$ \lim_{x\rightarrow 1^\minus{}}f(x)\equal{}0\neq 1\equal{}\lim_{x\rightarrow 1^\plus{}}f(x)$


Μία πηγή στέλνει τρία γράμματα x,y,z με συχνότητα 30%, 20% και 50% αντίστοιχα. Ένας δέκτης λαμβάνει τα γράμματα με ένα σφάλμα, έτσι ώστε: Η πιθανότητα να πάρει κάποιος x, y, z, όταν στέλνεται x, είναι 0.7, 0.2, 0.1 αντίστοιχα, η πιθανότητα να πάρει κάποιος x, y, z, όταν στέλνεται y, είναι 0.3, 0.6, 0.1 αντίστοιχα και τέλος, η πιθανότητα να πάρει κάποιος x, y, z, όταν στέλνεται z, είναι 0.1, 0.5, 0.4 αντίστοιχα. Τα γράμματα εκπέμπονται ανεξάρτητα το ένα από το άλλο και παίρνονται από το δέκτη στη σειρά που στάλθηκαν. Ορίζοντας τα γεγονότα απαντήστε στα κάτωθι ερωτήματα:

(α) Ποια η πιθανότητα να πάρουμε το σήμα yyx όταν εκπέμπεται yyx;
(β) Ποια η πιθανότητα να πάρουμε το σήμα yyx; (Υπόδειξη: Χρησιμοποιείστε το Θεώρημα Ολικής
     Πιθανότητας).

$x$ endexomeno na steilei h phgh $x$
$y$ endexomeno na steilei h phgh $y$
$z$ endexomeno na steilei h phgh $z$
$X$ endexomeno na paroume $x$
$Y$ endexomeno na steilei h phgh $y$
$Z$ endexomeno na steilei h phgh $z$
$P(x)=0.3$ $P(y)=0.2$ $P(z)=0.5$
$P(X|x)=0.7$ $P(Y|x)=0.2$ $P(Z|x)=0.1$
$P(X|y)=0.3$ $P(Y|y)=0.6$ $P(Z|y)=0.1$
$P(X|z)=0.1$ $P(Y|y)=0.5$$P(Z|y)=0.4$

$P(Y)= P(Y|x)P(x)+P(Y|y)P(y)+P(Y|z)P(z)=\hdots$
$P(X)= P(X|x)P(x)+P(X|y)P(y)+P(X|z)P(z)=\hdots$

a) To y antistoixei sto y k to x sto x? Tote h apanthsh einai $P(Y|y)P(Y|y)P(X|x)$. Alliws an to y antistoixei sto x, to y sto y kai to x sto y h apanthsh einai $P(X|y)P(Y|y)P(Y|x)$.
b) $\hdots=P(Y)P(Y)P(X)$

Από ότι κατάλαβα και εγώ αντιστοιχεί το y στο y και το χ στο χ...Ευχαριστώ φίλε...Με ξεκόλλησες.... :lol:  :lo

Μπορείτε να παρουσιάσετε τις λύσεις του Λευκοπούλειου?


[quote="spider_boy"]Find all possible values of $ f\left(x,y,z\right) \equal{} \left\{\frac {xyz}{xy \plus{} yz \plus{} xz}\right\}$, where $ x$, $ y$, $ z$ are natural numbers.

Hereby, the notation $ \left\{u\right\}$ means the fractional part of a real $ u$.[/quote]

I guess the answer is evry rational number between 0.,1  (include0)

[quote="11235813"][quote="spider_boy"][quote="11235813"]I guess the answer is evry rational number between 0.,1  (include0)[/quote]
Absolutely true :D (but not $ 0$ ;) )[/quote]

INCLUDE 0           take f(3,3,3)=0[/quote]
Very very sorry when writing my mind was at another place! :lol:


Determine the number of square units in the area of the
trapezoid whose bases are $ (5 \minus{} x)$ units and $ (13 \plus{} x)$ units and
whose height is 10 units.

The formula for the area of a trapezoid is $ \frac{1}{2}(b_1\plus{}b_2)h$.

We know the two bases and the height, so we just plug it into the formula.

$ \frac{1}{2}([5\minus{}x]\plus{}[13\plus{}x])10$.

$ \frac{1}{2}(18)10$.

Therefore, we get $ \boxed{90}$ square units as the answer.


I believe your definition of "perfect" is wrong. Note that:

D(6) = {1, 2, 3, 6}
1 + 2 + 3 = 6

D(28) = {1, 2, 4, 7, 14, 28}
1 + 2 + 4 + 7 + 14 = 28

The sum of the proper divisors is the number itself.

EDIT: Oh, I get it now.  :oops:

[quote="bigman"]A natural number is perfect if it is the sum of its divisors smaller than itself; for example,
6 and 28 are perfect. Prove that if 2^n − 1 is prime, then 2^(n−1)*(2^n − 1) is perfect.[/quote]

Say we have a prime number in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We wish to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of a prime number $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \sigma(a) \equal{} \sigma(2^{k\minus{}1}) \cdot \sigma(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2a$

$ \mathbb{Q.E.D}$

[quote="ffao"]
The sum of the proper divisors is the number itself.[/quote]

That is what the OP meant when he said the "divisors smaller than itself."

[quote="isabella2296"][quote="bigman"]A natural number is perfect if it is the sum of its divisors smaller than itself; for example,
6 and 28 are perfect. Prove that if 2^n − 1 is prime, then 2^(n−1)*(2^n − 1) is perfect.[/quote]

Say we have a prime number in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We wish to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of a prime number $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \sigma(a) \equal{} \sigma(2^{k \minus{} 1}) \cdot \sigma(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2a$

$ \mathbb{Q.E.D}$[/quote]

As your interpretation , what is related to the concept of "perfect" in the original questions ?
 :D

A number is perfect if sum of all of his divisors (including 1 and itself) equals 2 times that number.
6,28,496 are perfect.

Isabella can you explain the very last line? why is sigma(2^(k-1))=2^k-1

She used the identity $ 1\plus{}2\plus{}4\plus{}8\plus{}...\plus{}2^{k\minus{}1}\equal{}2^k\minus{}1$ which is easily showed.

It's true for k=1
Assume that it's true for some k:
$ 1\plus{}2\plus{}...\plus{}2^{k\minus{}1}\equal{}2^k\minus{}1$ and then for k+1 it is:
$ 1\plus{}2\plus{}4\plus{}...\plus{}2^k\equal{}2^{k\plus{}1}\minus{}1$ which is true because $ 1\plus{}2\plus{}...\plus{}2^k\equal{}1\plus{}2\plus{}...\plus{}2^{k\minus{}1}\plus{}2^k\equal{}2^k\minus{}1\plus{}2^k\equal{}2*2^k\minus{}1\equal{}2^{k\plus{}1}\minus{}1$ using the above.
It's now proved using mathematical induction.


Let $P(x)$ be a polynomial of degree n.If  $P(k)=\frac{k}{k+1}$  for $k=0,1,...,n.$
Evaluate $P(m)$, where $m>n$.

How fast can you say Lagrange interpolation? :D It's just a first thought. I don't know if it works out nicely.

how abt considering the polynomial q(x)=(x+1)p(x)-x  .
alll its roots are know

Now I realize how easy was this problem.I just need more experience in algebra. :)


In isosceles triangle ABC, AB=AC and the altitude to BC is 20. If the perimeter of the triangle is 80, the triangle area is:

A.400
B.300
C.200
D.100
E.NOTA

[hide]Let the length of $ BC$ be $ a$. Then, the altitude bisects $ BC$ into two segments of length $ \frac{a}{2}$. Using the Pythagorean Theorem, each leg of the isosceles triangle is
\[ \sqrt{{\left(\frac{a}{2}\right)}^2\plus{}{20}^2}\]From Since the perimeter is $ 80$, we have
\[ 2\sqrt{{\left(\frac{a}{2}\right)}^2\plus{}{20}^2}\plus{}a\equal{}80\]
Rearranging, and simplifying, we have
\[ 2\sqrt{\frac{a^2\plus{}1600}{4}}\equal{}80\minus{}a \\
\Rightarrow a^2\plus{}1600\equal{}(80\minus{}a)^2 \\
\Rightarrow a^2\plus{}1600\equal{}6400\minus{}160a\plus{}a^2 \\
\Rightarrow 160a\equal{}4800 \\
\Rightarrow a\equal{}30\]Finally, the area of the triangle is $ \frac{20\cdot 30}{2}\equal{}300 \Rightarrow \boxed{B}$.[/hide]


Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
\[f(f(x)+f(y)+2xy)=yf(x)+xf(y)\]
for all $x,y \in \mathbb R$.

[quote="reason"]find all f such that:

$ f(f(x) \plus{} f(y) \plus{} 2xy) \equal{} yf(x) \plus{} xf(y)$

$ f: \mathbb{IR}\to\mathbb{IR}$

thanx[/quote]

Wow, I found this one difficult. I hope somebody will find a simpler solution than mine (and I'm very surprised that such a problem was given in some contest).

Let $ P(x,y)$ be the assertion $ f(f(x)\plus{}f(y)\plus{}2xy)\equal{}yf(x)\plus{}xf(y)$

1) Consider first that $ f(0)\neq 0$
$ P(x,0)$ $ \implies$ $ f(f(x)\plus{}f(0))\equal{}xf(0)$ $ \implies$ $ f(x)$ is bijective.
Then $ P(\frac 12,\frac 12)$ $ \implies$ $ f(2f(\frac 12)\plus{}\frac 12)\equal{}f(\frac 12)$ and, since $ f(x)$ is bijective, $ 2f(\frac 12)\plus{}\frac 12 \equal{}\frac 12$ and so $ f(\frac 12)\equal{}0$
$ P(0,0)$ $ \implies$ $ f(2f(0))\equal{}0$ and so, since $ f(x)$ is bijective and $ f(\frac 12)\equal{}0$, we get $ f(0)\equal{}\frac 14$

$ P(x,\frac 12)$ $ \implies$ new assertion $ Q(x)$ : $ f(f(x)\plus{}x)\equal{}\frac 12f(x)$
$ P(x,0)$ $ \implies$ new assertion $ R(x)$ : $ f(f(x)\plus{}\frac 14)\equal{}\frac x4$

Then $ Q(f(2x)\plus{}\frac 14)$ $ \implies$ $ f(f(f(2x)\plus{}\frac 14)\plus{}f(2x)\plus{}\frac 14)\equal{}\frac 12f(f(2x)\plus{}\frac 14)$ $ \implies$ $ f(\frac x2\plus{}f(2x)\plus{}\frac 14)\equal{}\frac x4$ $ \equal{}f(f(x)\plus{}\frac 14)$ and so $ f(2x)\equal{}f(x)\minus{}\frac x2$

But $ R(f(x)\plus{}\frac 14)$ $ \implies$ $ f(f(f(x)\plus{}\frac 14)\plus{}\frac 14)\equal{}\frac {f(x)}4\plus{}\frac 1{16}$ $ \implies$ $ f(\frac {x\plus{}1}4)\equal{}\frac {f(x)}4\plus{}\frac 1{16}$. Using then twice the fact that $ f(2u)\equal{}f(u)\minus{}\frac u2$, we get $ f(x\plus{}1)\equal{}\frac{f(x)}4\minus{}\frac{3x}8\minus{}\frac 5{16}$

So we got :
$ U(x)$ : $ f(2x)\equal{}f(x)\minus{}\frac x2$
$ V(x)$ : $ f(x\plus{}1)\equal{}\frac{f(x)}4\minus{}\frac{3x}8\minus{}\frac 5{16}$

From this we can compute $ f(2x\plus{}2)$ in two different ways :
$ V(x\plus{}1)$ $ \implies$ $ f(x\plus{}2)\equal{}\frac{f(x\plus{}1)}4\minus{}\frac{3(x\plus{}1)}8\minus{}\frac 5{16}$ $ \equal{}\frac{f(x)}{16}\minus{}\frac{3x}{32}\minus{}\frac 5{64}\minus{}\frac{3(x\plus{}1)}8\minus{}\frac 5{16}$ $ \equal{}\frac{f(x)}{16}\minus{}\frac{15x}{32}\minus{}\frac {49}{64}$
Using $ 2x$ instead of $ x$ in this lats equality, we get $ f(2x\plus{}2)\equal{}\frac{f(2x)}{16}\minus{}\frac{15x}{16}\minus{}\frac {49}{64}$ $ \equal{}\frac{f(x)}{16}\minus{}\frac {x}{32}\minus{}\frac{15x}{16}\minus{}\frac {49}{64}$ and so :

$ f(2x\plus{}2)\equal{}\frac{f(x)}{16}\minus{}\frac{31x}{32}\minus{}\frac {49}{64}$

But $ U(x\plus{}1)$ $ \implies$ $ f(2x\plus{}2)\equal{}f(x\plus{}1)\minus{}\frac {x\plus{}1}2$ and so $ f(2x\plus{}2)\equal{}\frac{f(x)}4\minus{}\frac{3x}8\minus{}\frac 5{16}\minus{}\frac {x\plus{}1}2$ and so $ f(2x\plus{}2)\equal{}\frac{f(x)}4\minus{}\frac{7x}8\minus{}\frac {13}{16}$

And so $ \frac{f(x)}{16}\minus{}\frac{31x}{32}\minus{}\frac {49}{64}$ $ \equal{}\frac{f(x)}4\minus{}\frac{7x}8\minus{}\frac {13}{16}$

And $ f(x)\equal{}\frac 14\minus{}\frac{x}{2}$

Plugging this mandatory condition in the original equation, we can easily see that this indeed is a solution.

2) Consider now that $ f(0)\equal{}0$
(1) : $ P(x,0)$ $ \implies$ $ f(f(x))\equal{}0$ $ \forall x$
(2) : $ P(\minus{}\frac 12,f(\minus{}\frac 12))$ $ \implies$ $ f(\minus{}\frac 12)\equal{}0$ (using 1)
(3) : $ P(x,\minus{}\frac 12)$ $ \implies$ $ f(f(x)\minus{}x)\equal{}\minus{}\frac{f(x)}{2}$ (using 2)
(4) : $ P(f(x),\minus{}\frac 12)$ $ \implies$ $ f(\minus{}f(x))\equal{}0$ $ \forall x$ (using 1 and 2)
(5) : $ P(\frac 12,\minus{}f(\frac 12))$ $ \implies$ $ f(\frac 12)\equal{}0$ (using 1)
(6) : $ P(x,\frac 12)$ $ \implies$ $ f(f(x)\plus{}x)\equal{}\frac{f(x)}2$ (using 5)
(7) : $ P(x,f(y))$ $ \implies$ $ f(f(x)\plus{}2xf(y))\equal{}f(x)f(y)$ (using 1) and so $ f(f(x)f(y))\equal{}0$ $ \forall x,y$ (using 1 again)

(8) : If $ f(x)\neq 0$, then $ P(\minus{}\frac{1}{2f(x)},f(x)f(\minus{}\frac{1}{2f(x)}))$ $ \implies$ $ f(\minus{}\frac{1}{2f(x)})\equal{}0$ (using 7). Then, since $ f(f(x)\minus{}x)\equal{}\minus{}\frac{f(x)}{2}\neq 0$ (using 3), and replacing $ x$ with $ f(x)\minus{}x$, we get $ f(\frac{1}{f(x)})\equal{}0$

(9) : If $ f(x)\neq 0$, $ P(y,\frac{1}{f(x)})$ $ \implies$ $ f(f(y)\plus{}\frac{2y}{f(x)})\equal{}\frac{f(y)}{f(x)}$ (using 8) and so $ f(\frac{f(y)}{f(x)})\equal{}0$ $ \forall x,y$ such that $ f(x)\neq 0$

$ P(x,x)$ $ \implies$ $ f(2f(x)\plus{}2x^2)\equal{}2xf(x)$ and so (using 6) $ f(z)\equal{}xf(x)$ with $ z\equal{}2f(x)\plus{}2x^2\plus{}f(2f(x)\plus{}2x^2)$

Using then 9, we get $ f(x)\neq 0$ $ \implies$ $ f(\frac{f(z)}{f(x)})\equal{}0$ with the above $ z$ such that $ f(z)\equal{}xf(x)$ and so $ f(x)\equal{}0$, so contradiction and so $ f(x)\equal{}0$ $ \forall x$

3) hence the solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}\frac 14\minus{}\frac{x}{2}$ $ \forall x$

good problem and amazing solution , Patrick  :lol:

[b]the mine :[/b] (simpler , i think ! )

[quote]
Let $ P(x,y)$ be the assertion $ f(f(x) \plus{} f(y) \plus{} 2xy) \equal{} yf(x) \plus{} xf(y)$

1) Consider first that $ f(0)\neq 0$
$ P(x,0)$ $ \implies$ $ f(f(x) \plus{} f(0)) \equal{} xf(0)$ $ \implies$ $ f(x)$ is bijective.
Then $ P(\frac 12,\frac 12)$ $ \implies$ $ f(2f(\frac 12) \plus{} \frac 12) \equal{} f(\frac 12)$ and, since $ f(x)$ is bijective, $ 2f(\frac 12) \plus{} \frac 12 \equal{} \frac 12$ and so $ f(\frac 12) \equal{} 0$
$ P(0,0)$ $ \implies$ $ f(2f(0)) \equal{} 0$ and so, since $ f(x)$ is bijective and $ f(\frac 12) \equal{} 0$, we get $ f(0) \equal{} \frac 14$

$ P(x,\frac 12)$ $ \implies$ new assertion $ Q(x)$ : $ f(f(x) \plus{} x) \equal{} \frac 12f(x)$
$ P(x,0)$ $ \implies$ new assertion $ R(x)$ : $ f(f(x) \plus{} \frac 14) \equal{} \frac x4$

Then $ Q(f(2x) \plus{} \frac 14)$ $ \implies$ $ f(f(f(2x) \plus{} \frac 14) \plus{} f(2x) \plus{} \frac 14) \equal{} \frac 12f(f(2x) \plus{} \frac 14)$ $ \implies$ $ f(\frac x2 \plus{} f(2x) \plus{} \frac 14) \equal{} \frac x4$ $ \equal{} f(f(x) \plus{} \frac 14)$ and so $ f(2x) \equal{} f(x) \minus{} \frac x2$
[/quote]

we have previously proved that : $ \boxed{f(2x)\equal{}f(x)\minus{}\frac{x}{2}}$ : $ M(x)$

but we have : $ R(2f(x))\rightarrow f(f(2f(x))\plus{}\frac 14)\equal{}\frac{f(x)}{2}$

but from $ Q(x)$ we have $ f(f(x)\plus{}x)\equal{}\frac{f(x)}{2}$

so : $ f(f(2f(x))\plus{}\frac 14)\equal{}f(f(x)\plus{}x)\implies f(2f(x))\plus{}\frac 14\equal{}f(x)\plus{}x$

and we have $ M(f(x))\implies f(2f(x))\equal{}f(f(x))\minus{}\frac{f(x)}{2}$

so : $ f(2f(x))\equal{}f(f(x))\minus{}\frac{f(x)}{2}\equal{}f(x)\plus{}x\minus{}\frac 14$

$ \iff \boxed{f(f(x))\minus{}\frac{3}{2}f(x)\minus{}x\plus{}\frac 14\equal{}0}$  $ (E)$

now $ (E)$ can easily be solved  and $ f(x)\equal{}\frac{\minus{}x}{2}\plus{}\frac{1}{4}$

2) for $ f(0)\equal{}0$

$ P(x,0)\implies f(f(x))\equal{}0$

$ P(\frac{\minus{}1}{2},f(\frac{\minus{}1}{2}))\implies f^2(\frac{\minus{}1}{2})\equal{}0 \implies f(\frac{\minus{}1}{2})\equal{}0$

$ P(x,\frac{\minus{}1}{2})\implies f(f(x)\minus{}x)\equal{}\minus{}\frac{f(x)}{2}\implies f(\minus{}\frac{f(x)}{2})\equal{}0$

$ P(1,\minus{}\frac{f(1)}{2})\implies f(f(1)\minus{}f(1))\equal{}\frac{\minus{}f^2(1)}{2}\implies f(1)\equal{}0$

$ P(x,1)\implies f(f(x)\plus{}2x)\equal{}f(x)$ $ (*)$

now put $ g(x)\equal{}f(x)\plus{}2x$

$ (*)\iff g(g(x))\minus{}3g(x)\plus{}2x\equal{}0$   $ (E')$

$ (E')$ is also easy to solve , $ g(x)\equal{}2x$ $ \implies f(x)\equal{}0$

:)

Mehdi

Hello Mehdi !

[quote="mehdi cherif"] $ \boxed{f(f(x)) \minus{} \frac {3}{2}f(x) \minus{} x \plus{} \frac 14 \equal{} 0}$  $ (E)$

now $ (E)$ can easily be solved  and $ f(x) \equal{} \frac { \minus{} x}{2} \plus{} \frac {1}{4}$[/quote]

Huh, could you give us the easy solution to this equation ?

[quote="mehdi cherif"] $ (*)\iff g(g(x)) \minus{} 3g(x) \plus{} 2x \equal{} 0$   $ (E')$

$ (E')$ is also easy to solve , $ g(x) \equal{} 2x$ $ \implies f(x) \equal{} 0$[/quote]

And also the easy solution to this one ?

Thanks

[quote="mehdi cherif"][quote="pco"]

Huh, could you give us the easy solution to this equation ?

[/quote]

not easy , but known  :)

we can use this result : http://www.mathlinks.ro/viewtopic.php?p=1222512#1222512

Mehdi

 :)[/quote]

No, you cant use this result, because the key point of the demonstration in the link you gave is that $ f(x)>0$. And we have no such constraint here.

[quote="mehdi cherif"][quote="pco"]

Huh, could you give us the easy solution to this equation ?

[/quote]

not easy , but known  :)

:)[/quote]

Not known by be ...  :(

So you mean that equation $ f(f(x))\minus{}bf(x)\minus{}cx\equal{}0$ always has as unique solutions $ f(x)\equal{}rx$ with $ r$ root of $ x^2\minus{}bx\minus{}c\equal{}0$  ?

If so, it's obviously wrong. Chosse for example $ b\equal{}0$ and $ c\equal{}1$ : $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$ are certainly not the unique solutions of $ f(f(x))\equal{}x$