function
algebra
functional equation
algebra unsolved

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

It is easy(ish) to see that $f$ is either identically zero or bijective. In either case we must have $f(0)=0$.

In the latter case, suppose $f(z)=1$, $x=0$ and we get $f(y)=zf(y)$ for all $y$, so $f(1)=1$.

Letting $x=0,y=1$, we get $f(f(z))=z$ for all $z$.

Letting $x=0,z=f(w)$, we get $f(yf(f(w)))=f(yw)=f(y)f(w)$ for all $y,w$, so $f$ is multiplicative.

Now for $x\ne0$, $y=z=0$, we get $f(x^2)=\left(f(x)\right)^2=xf(x)$.  Now since $f$ is bijective and $f(0)=0$, $f(x)\ne0$, and we get $f(x)=x$.

I believe my solution is similar(probably there's only one) . Only a few different steps(effectively the same!!!

The INMO problems you've posted are quite nice.

I have a different solution:

As Diarmuid said, it can easily be seen that if $f$ is not identically zero, it is surjective (by varying $z$). Bijectivity is non-essential, except at $x=0$, which we will now prove. Suppose $f(z)=0$ for some $z \neq 0$. Then we have $f(x^2)=xf(x)+zf(y)$, a contradiction since $f(y)$ takes every real value.

Substituting $-x$ for $x$ gives that $xf(x)=-xf(-x)$, and therefore that $f(x) = -f(-x)$.

Now fix $x=y \neq 0$ and put $z = -x$. We get that $f(x^2+xf(-x))=0$ and therefore $x^2=xf(x)$ (by bijectivity at 0) and $x=f(x)$.

Can you explain how do you get that $f$ is surjective, because I couldn't get it.
Also I didn't understand why $f$ is bijective.
Thanks in advance.

Anyway, I understood the solution  :blush:

I have another (but strange  :rotfl: ) solution to this problem...
Given is that \[ f(x^2 \plus{}yf(z)) \equal{} xf(x) \plus{}zf(y) \ \ \ \cdots (*)\]
Taking $ x\equal{}y\equal{}z\equal{}0$,
$ f(0) \equal{} 0$
Now taking $ y\equal{}z\equal{}0$
$ f(x^2) \equal{} xf(x)  \ \ \ \ \cdots \boxed{1}$

Putting $ \minus{}x$ instead of $ x$ in $ \boxed{1}$, we have
$ f(x^2) \equal{} \minus{}xf(\minus{}x)$

Therefore, $ xf(x) \equal{} \minus{}xf(\minus{}x) \iff f(\minus{}x) \equal{} \minus{}f(x)  \ iff \ x \not \equal{} 0 \ \ \ \cdots \boxed{2}$
For the case $ 0$, $ \boxed{2}$ is also true, so it is true for all real $ x$

Taking $ y\equal{}z\equal{}k$ in $ (*)$, we have,
$ f(x^2 \plus{}kf(k)) \equal{} xf(x) \plus{} kf(k)$
$ \Rightarrow f(x^2 \plus{} f(k^2)) \equal{} f(x^2) \plus{} f(k^2) \ (From \ \boxed{2})$
Let $ x^2 \equal{} a, k^2 \equal{} b$ with $ a,b \ge 0$
$ f(a\plus{}f(b)) \equal{} f(a) \plus{} f(b) \ \ \ \cdots \boxed{3}$

Now, taking $ y\equal{}1, z \ge 0$ in $ (*)$
$ f(x^2 \plus{} f(z)) \equal{} xf(x) \plus{}zf(1)$
Therefore, by $ \boxed{3}$, we have
$ f(x^2) \plus{} f(z) \equal{} xf(x) \plus{} zf(1)$
$ \Rightarrow xf(x) \plus{}f(z) \equal{} xf(x) \plus{} zf(1)$ By $ \boxed{1}$

Let $ f(1) \equal{} c$ for some constant $ c$
$ \Rightarrow f(z) \equal{} cz \ \forall z \ge 0$
From $ \boxed{2}$,
$ f(\minus{}z) \equal{} \minus{}f(z) \equal{} c \cdot (\minus{}z)$ so $ f(z) \equal{} cz$ is true for all real $ z$.

Substituting this information in $ (*)$, we get
$ cx^2 \plus{} c^2yz \equal{} cx^2 \plus{}cyz$
$ \Rightarrow c^2 \equal{} c$
$ \Rightarrow c \equal{} 0,1$

$ \boxed{f(x) \equal{}0\ or\  x  \forall  \ x \in \mathbb{R}}$

\[f(x^2+yf(z))=xf(x)+zf(y)                       (1)\]

Taking y=z=0 in (1) we get
         \[f(x^2)=xf(x)                 (2)\]
Taking x=0 in (2) we get f(0)=0
Taking x=0 and y=z=x in (1) we get
       \[f(xf(x))=xf(x)                 (3)\]
Taking y=z=x in (1) we get,
      \[f(x^2+xf(x))=2xf(x)       (4)\]
Now using the fact that \[f(x^2)=xf(x)\] we get
     \[f(x^2+xf(x))=f(x^2+f(x^2))=xf(x)+x^2f(1)     (5)\]
Comparing (4) and (5) we get,
    \[2xf(x)=xf(x)+cx^2\] where $c=f(1)$
or,\[f(x)=cx\] considering x not equal to 0.
Substituting this back in (3) we get
    \[c^2x^2=cx^2\]
or,\[c^2=c\]
From this we get that c=0 or 1 and consequently
\[f(x)=0\]or
\[f(x)=x\] for all real x

Somebody, [b]please check my solution[/b]:
$f(x^2+yf(z))=xf(x)+zf(y)$
Let $P(x,y,z)$ be the assertion for the functional equation.
$P(0,0,0)\implies f(0)=0$
$P(x,0,0)\implies f(x^2)=xf(x)$
$\implies f(x^2+yf(z))=x^2+zf(y) \cdots \star$
$P(0,y,z)\implies f(yf(z))=zf(y)$
$\implies f(x^2+yf(z))=f(x^2)+f(yf(z))$
Which is the Cauchy's functional equation and has solutions $f(x)=cx,f(x)=0$
from here we can prove $c=1$ as Sayantan(previous poster) did and this gives us solutions $f(x)=x \text{ or }f(x)=0\forall x\in \mathbb{R}$
$%Error. "blackbox" is a bad command.
$
From the equation marked with a star we may also proceed as follows:
We can prove $f(x)$ is injective except the solution $f(x)=0\forall x\in \mathbb{R}$
Now $P(x,y,-\frac{x^2}{f(y)})\implies f(x^2+yf(-\frac{x^2}{f(y)}))=0$
As the function is injective for the non-zero solutions, also $f(0)=0$,
We have,
$x^2+yf(-\frac{x^2}{f(y)})=0\implies f(-\frac{x^2}{f(y)})=-\frac{x^2}{y}$
Hence the solution $f(x)=x\forall x\in \mathbb{R} \text{ or } f(x)=0\forall x\in \mathbb{R}$

[b]Yet another solution[/b]:

Let $P$ be the given assertion. Then we get:
$P(0, 0, 0)\implies f(0)=0$
$P(x, 0, 0)\implies f(x^2)=xf(x)$
$P(0, x, x)\implies f(xf(x))=xf(x)$
$P(0, 1, x)\implies f(f(x))=xf(1)$

Then from the statements above, we have:
$x^2f(1)=f(f(x^2))=f(xf(x))=xf(x)\implies f(x)=xf(1)$.
$P(1, 1, 1)\implies f(1)+f(1)^2=f(1+f(1))=2f(1)\implies f(1)^2=f(1)$

Therefore, $f(1)=0$ or $1$, so:
$f(x)=0$ or $f(x)=x$.

Seriously? This was InMO #6?

[quote="Ashutoshmaths"]
$P(x,0,0)\implies f(x^2)=xf(x)$
$\implies f(x^2+yf(z))=x^2+zf(y) \cdots \star$
[/quote]
I think I found a mistake
It probably should imply $ f(x^2+yf(z))=f(x^2)+zf(y)$

I have got a solution apparently different

Please confirm if it is correct !

$f(x^2+yf(z))=xf(x)+zf(y)$
Let $P(x,y,z)$ be the assertion of the functional equation

$P(x,0,0) \implies f(x^2)=xf(x)$

$f(x)=0$ clearly satisfies the equation.
Now let $f(x)$ be an identically non zero function.

$P(0.y.z) \implies f(yf(z))=zf(y)$

Then $f(f(yf(z)))=f(zf(y))=yf(z)$
For each $x$ we can chose $y,z$ appropriately (since it is a non zero function) such that,
$yf(z)=x$

Thus $f(f(x))=x$ that is $f(x)$ is involutive

Also $xf(x)=f(xf(x))$ (Since $ f(yf(z))=zf(y)$ as above)

Combining these results
$xf(x)=f(xf(x))=f(f(x^2))=x^2$

That is $f(x)=x$ for all $x$

Thus the only solutions are
$f(x)=0$ and $f(x)=x$

It's easy and nice function.Here my solution:
Let $ P(x,y,z) $ be the assertion of the function.
1) $ P(0,0,0) \implies f(0)=0 $.
2) $ P(x,0,0) \implies f(x^2)=xf(x) (*) $ and $ P(0,1,z) \implies f(f(z))=f(1)z $.It's easy to see that $ f $ is zero function or bijective fuction.In least identity we take $ z=1 \implies f(f(1))=f(1) $ so by bijectivite we get $ f(1)=1 $ and so $ f(f(z))=z (**) $.
3) From $ (*)+(**) $ we have :
$ f(f(x)^2)=f(x)f(f(x))=f(x)x=f(x^2) \implies f(x)=x,-x $. But we see that $ f(x)=-x $ isn't solution.
Answers:
1) $ f(x)=0 $
2) $ f(x)=x $

We claim that $f(x)=0$ $\forall x \in R$ or $f(x)=x$ $\forall x \in R$ are the only two solutions. It is easy to check that these 2 work. We now show that these are the only 2 solutions.
$[1]$ Set $x=y=z=0$ to get $f(0)=0$.
$[2]$ Set $y=0$ to get $f(x^{2})=xf(x)$ $\forall x \in R$. Thus the equation now takes the form $f(x^{2}+yf(z))=f(x^{2})+zf(y)$.
$[3]$ Set $x=0$ to get $f(yf(z))=zf(y)$ $\forall y,z \in R$.
$[4]$ Set $x=-x$ in $[2]$ to get $f(-x)=-f(x)$.
$[5]$ Set $x=\sqrt{y(1-f(z))}$ in $[2]$ to get (according to the sign of $1-f(z)$, we choose the sign of $y$ to ensure that $x \in R$, and hence either $y \in R^{+}$, or $y \in R^{-}$. We do not consider $y=0$.)
$$f(y)=f \left( \left (\sqrt{y(1-f(z))} \right)^{2} +yf(z) \right)=f(y(1-f(z)))+zf(y) \Leftrightarrow (1-z)f(y)=f \left( y(1-f(z)) \right)$$
In this equation, set $z=1$ to get $f \left( y(1-f(1)) \right)=0$ $\forall y \in R^{+}$ or $R^{-}$, depending on whether $1 \geq f(1)$ or $1 < f(1)$. Since this is true for all $R^{+}$ (or $R^{-}$), either $1-f(1)=0$ or $f(x)$ is a constant on one of these 2 domains. In the second case, we can use $[4]$ and hence easily get $f(x)=0$ $\forall x \in R$.
So now assume $f(1)=1$. This is our result $[6]$.
$[7]$ Now, substituting $y=1$ in $[3]$ and using $[6]$ we get $f(f(y))=y$ $\forall y \in R$.
Hence $f(f(x^{2}))=x^{2}$ $\forall x \in R$. But setting $y=z$ in $[3]$ (and using $[2]$) gives $f(f(y^{2}))=f(y^{2})$ $\forall y \in R$. Thus, $f(a)=a$ $\forall a \in R^{+}$. Then using $[4]$ gives $f(x)=x$ $\forall x \in R$. (We can check that this is true for the case $x=0$ as well).

Amusing one for showing for the positive reals.

Trivial to note $f(0)=0$, then $P(x,y,0)$ gives $f(x^2)=xf(x)$ so that $P(f(x),y,y)$ gibes $f(f(x)^2+f(y)^2)=f(x)^2 + f(y)^2$ and varying $x,y$ we get the result for positive reals

Allzero works, so assume henceforth that $\exists u$, $f(u)\neq 0$
$P(1,u,z)\implies f(1+uf(z))=f(1)+zf(u)$
$P(0,0,0)\implies f(0)=0 \implies u\neq 0  \implies$ bijective
$P(x,0,0)\implies f(x^2)=xf(x)$ and $P(0,x,x)\implies f(xf(x))=xf(x)$
Hence, $f(x^2)=f(xf(x))$ and since bijective, $x^2=xf(x) \implies f(x)=x, \forall x\neq0$
So, $f(x)=x, \forall x\in\mathbb{R}$.
Hence two solutions:
$$f(x)=0, \forall x\in\mathbb{R}$$ and $$f(x)=x \forall x\in\mathbb{R}$$

clearly, if $f$ is constant then it's zero so suppose that it's not constant
let $P(x,y)$ be the assertion of the equation.
$P(0,y,z)\implies f(yf(z))=zf(y)$ so $f(xf(x))=xf(x)$
subbose that $f(a)=f(b)$ for some $a,b$ and compare$P(x,y,a)$ with $P(x,y,b)$
$\implies f$ is bejective
now , $P(x,0,0)\implies f(x^2)=xf(x)=f(xf(x))$
so $\boxed{f(x)=x \quad \text{or}\quad 0 \quad \forall x\in\mathbb{R}}$

[b][color=#f00]Answer: [/color][/b]$f(x) = x \quad \text{or} \quad f(x) = 0$

Proof:
Let $P(x,y,z)$ denote the given assertion.
$P(0,0,0) \to f(0) = 0$
$P(x,0,0) \to \color{red}{f(x^2) = xf(x)}$
$P(x,x,x) \to f(x^2 + xf(x)) = xf(x) + xf(x)$
$\implies \color{red}{f(x^2 + f(x^2)) = 2xf(x)}$
$P(x,1,x^2) \to f(x^2 + f(x^2)) = xf(x) + x^2f(1)$
$\implies 2xf(x) = xf(x) + x^2f(1)$
$\implies xf(x) = x^2f(1)$
$\implies f(x) = xf(1)$ (Holds even when $x=0$)
If $f(1)=c \implies f(x) = cx$
Plugging this back into our original equation.
$c(x^2 + cyz) = cx^2 + cyz$
$\implies c^2 = c$
$\implies c=0,1$
After checking we see that both values of $c$ work.
Hence $f(x) = x \quad \text{or} \quad f(x) = 0$.

$P(x,0,0)\Rightarrow f(x^2)=xf(x)\Rightarrow f$ is odd.
$P(x,y,1)\Rightarrow f(x^2+y)=f(x^2)+f(y)$ so since $f$ is odd, it is additive.
This is now USAMO 2002 P4.

Alternatively:
$P(0,x,y)\Rightarrow Q(x,y):f(xf(y))=yf(x)$
$Q(1,x)\Rightarrow f(f(x))=xf(1)$
If $f(1)=0$ then $Q(x,f(x))\Rightarrow f(x)^2=0$, so $\boxed{f(x)=0}$, which works.
Otherwise, $f$ is bijective.
$Q(1,1)\Rightarrow f(f(1))=f(1)\Rightarrow f(1)=1$
$Q(1,x)\Rightarrow f(f(x))=x$
$Q(x,f(y))\Rightarrow f(xy)=f(x)f(y)$ and since $f$ is additive and multiplicative, $f(x)=cx,c\in\{-1,1\}$. Checking, we see that $\boxed{f(x)=x}$.

Oh, I really overcomplicated it :O But yeyy, I solved an INMO fe!

Let $P(x,y)$ denote the assertion.
Then $$P(x,0,0)\implies f(x^2)=xf(x)\implies f(x)=-f(-x)$$

And $$f(x^2+y^2)=f(x^2)+f(y^2)\implies f(a+b)=f(a)+f(b), a,b \in \mathbb{R}^{+}$$

Also $f(0)=0.$ Now we show that $f$ is injective at $0.$ So let $f(t)=0, t\ne 1.$ So  $$P(0,1,t)\implies 0=f(f(t))=tf(1)\implies f(1)=0\text{ or } f=0$$
Assuming $f$ is non constant, we get $f(1)=0.$ Then $$P(0,a,1)\implies 0=f(a)\forall a.$$

Not possible, hence $f$ is injective at $0.$
Moreover, $$P(0,1,x)\implies f(f(x))=xf(1)$$

Note that $$f(f(x))=xf(1)\implies f \text { is bijective }$$

Actually, we don't need the following claim, but I found it too good to not add.

[b]Claim:[/b] $f(1)=1$
[b]Proof: [/b]$$P(0,y,1)\implies f(yf(1))=f(y)\implies yf(1)=y\implies f(1)=1$$

[b]Alternate proof:[/b] Using subjectivity, $$\exists k, f(k)=1\implies P(0,k,k)\implies 1=f(kf(k))=k\implies k=1$$

Now, fix $x.$ We carry $y$ and choose $z$ such that  $$f(x^2+yf(z)=xf(x)+zf(y)=0$$
Then $$x^2+yf(z)=0, xf(x)+zf(y)=0$$

Take $y=x\implies z=-x$ in $xf(x)+zf(y)=0.$ So, we get $$x^2+xf(-x)=0\implies f(-x)=-x\implies f(x)=x$$

Hence the solutions are $\boxed{ f=0,f(x)=x}$

Denote the assertion by $P(x,y,z).$ Clearly $f(0)=0.$ Then $P(0,1,x)$ gives $f(f(x))=cx$ so $f$ is bijective otherwise if $c=0$ then $P(0,x,f(x))$ gives $f(x)\equiv 0$ this works. Now comparing $P(x,x,x)$ and $P(x,0,x)$ we get $f(x)\equiv x,$ which works.

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

$P(0,x,y): f(xf(y))=yf(x)$.

$P(0,1,x): f(f(x))=xf(1)$.

If $f(1)=0$, then $f(f(x))=0$, $P(0,x,f(y))$ gives $f(0)=f(x)f(y)$. Setting $y=1$ gives $f(0)=f(x)f(1)=0$. So $f(x)f(y)=0\implies f(x)^2=0\implies \boxed{f\equiv 0}$.

Henceforth assume $f(1)\ne 0$. Then $f$ is bijective.

Note that $f(f(1))=f(1)$, so $f(f(f(1)))=f(f(1))=f(1)$.

$P(0,1,f(1)): f(1) = f(1)^2\implies f(1)=1$ since $f(1)\ne 0$.

Thus, $f(f(x))=x$.

$P(0,x,f(y)): f(xy)=f(x)f(y)$, so $f$ is multiplicative.

$P(x,y,0): f(x^2)=xf(x)$, so $f$ is odd.

$P(x,y,1): f(x^2+y)=f(x^2)+f(y)$. Since $f$ is odd, $f$ is additive.

Now, $f$ is additive and multiplicative, so $f(x)=cx$ for some constant $c\ne 0$, checking gives $\boxed{f(x)=x}$ or $\boxed{f(x)=-x}$, which work.

[b][color=#00f]Claim: [/color][/b]
$\boxed{f(x)=cx, \forall x\in \mathbb{R}}$ where $c=0$ or $c=1$

[b][color=#00f]Proof:[/color][/b]
Let $P(x,y,z)$ denote the assertion: $f(x^2+yf(z))=xf(x)+zf(y)$
$P(0,0,0)$ yields $f(0)=0$
$P(x,0,0)$ yields $f(x^2)=xf(x)$
Furthermore $P(x,x,x)\implies f(x^2+f(x^2))=2xf(x)$ (1)
$P(x,1,x^2)$ yields $f(x^2+f(x^2))=xf(x)+x^2$
However notice that from (1) we have:
$2xf(x)=xf(x)+x^2f(1)\Longrightarrow f(x)=xf(1)$
Thus, since $f(1)$ is a constant, we have that $f(x)=cx$
$P(1,1,1)$ yields $f(c+1)=2c\Longrightarrow c^2=c$
Thus $c$ can only be equal to $1$ or $0$.QED
And we are done! :)

Let $P(x,y,z)$ be the assertion $f(x^2+yf(z))=xf(x)+zf(y)$
$P(0,0,0)$ gives $f(0)=0$
$P(x,y,0)$ gives $f(x^2)=xf(x)$
$P(x,y,c_1)$ and $P(x,y,c_2)$ gives $f$ is injective or $f(x)=0$
$P(0,x,x)$ gives $f(xf(x))=xf(x)$ so $f(f(x^2))=f(xf(x))=xf(x)=f(x^2)  f(x^2)=x^2$ and from $f : \mathbb{R} \longrightarrow \mathbb{R}$ solutions are $f(x)=x$ and $f(x)=0$

Very similar to IMO 1992/2. We use pretty much the same idea here too!

[i]Solution.[/i] The answers are $f \equiv 0$ and $f \equiv \mathrm{id}$. All of them work. Denote the assertion to the functional equation by $P(x,y,z)$. We will now proceed towards to the other direction.

Its clear that $f \equiv 0$ is the only constant. $P(0,0,0)$ reveals $f(0) = 0$. If $f$ attains $0$ at any non-zero constant $\alpha$, then $P(0,y,\alpha)$ would yield $f \equiv 0$. Henceforth, assume that $f$ is non-zero everywhere except $0$.

$P(x,0,0)$ gives $f(x^2) = xf(x)$. $P(0,y,1)$ gives us $f(yf(1)) = f(y) \implies f(1) = 1$ and finally $P(0,1,z)$ gives us that $f$ is bijective and an involution. Here's the nice part, $P(x,y,1)$ gives
    \[f(x^2 + y) = xf(x) + f(y) = f(x^2) + f(y) \implies f(a+b) = f(a) + f(b)\]
for all real $a$ and $b \ge 0$. This condition is well-known over $\mathbb{R}$ and can be extended to full a Cauchy. $P(0,y,f(y))$ would give $f(y^2) = f(y)^2$. So, $f > 0$ for positive values of domain which due to Cauchy means that $f$ is strictly increasing. Since $f$ is strictly increasing and $f$ is additive, $f(x) = ax$ for some real constant $a$. Just checking it once, we're done. $\blacksquare$

Claim 1: $f(0)=0$
Proof: trivial
Claim 2: $f(x^2)=xf(x)$
Proof: Putting $y=z=0 \implies f(x^2)=xf(x)$
Claim 3: $f$ is an involution
Proof: Putting $x=0 \implies f(yf(z))=zf(y) \implies f(f(yf(z))=f(zf(y))=yf(z)$
            But $yf(z)$ is surjective in $\mathbb{R}$
            Therefore $f(f(x))=x \forall x \in \mathbb{R}$
Claim 3: $f$ is odd
Proof: Putting $x=f(z), y=-f(z)$ in original equation
            We get $f(f(z)^2-f(z)^2)=f(0)=0=f(z)f(f(z))+zf(-f(z))=f(z)z+zf(-f(z)) \implies f(-f(z))=-f(z)$ now since $f$ is surjective therefore $f$ is also odd.
Claim 4: f(x)=x
Proof: taking $y=z$
We get $f(x^2+y^2)=x^2+y^2 \implies f(x)=x$ for all positive real x. But since $f$ is odd therfore we can extend this to the negative numbers as well and hence we are done.

How was this INMO P$6$ wth. Just tried out random stuffs and got my solution.
$P(0,0,0) \implies f(0) = 0$.
$P(x,0,0) \implies f(x^2) = xf(x)$.
Comparing $P(x,0,0)$ with $P(-x,0,0)$ gives us $f(x) = -f(-x)$, so $f$ is odd.
Now, consider $a,b \in \mathbb{R}$ where $f(a) = f(b)$.
Then, $xf(x) + af(y) = xf(x) + bf(y)$
Clearly $f(x) = 0$ is a solution, then assume $f(x) \neq 0$.
This gives out $a = b$, and thus $f$ is injective.
$P(0,y,1) \implies f(yf(1)) = f(y)$.
$\implies yf(1) = y$.
$\implies f(1) = 1$.
Now, $P(0,1,z) \implies f(f(z)) = z$.
So, $f$ is involutive.
$P(0,x,x) \implies f(xf(x)) = f(x^2) = xf(x)$ [From $xf(x) = f(x^2)$]
Or, $f(x) = x$.
Overall, we get $2$ solutions, $f(x) = 0$ and $f(x) = x$, both seem to work.

Really interesting problem.

We claim the only solutions are $f(x) \equiv 0 \forall x$ and $f(x) = x \forall x$. Clearly, they satisfy the given equation. We now prove these are the only solutions.
Since $f(x) \equiv 0 \forall x$ satisfies, the equation, assume $f$ is not identically zero, and let $P(x, y, z)$ denote the assertion $f(x^2 + yf(z)) = xf(x) + zf(y)$.
$P(0, 0, 0): f(0) = 0$
$P(0, y, z): f(yf(z)) = zf(y) \implies f(f(yf(z))) = yf(z)$.
Note that we can choose z such that $f(z) \ne 0$, and therefore $yf(z)$ can represent any real, say $a$.
Therefore we have $f(f(a)) = a$, i.e. $f$ is an involution and therefore bijective.
Now $P(0, y, 1): f(yf(1)) = f(y) \implies yf(1) = y \implies f(1) = 1$.
Now, $P(x, y, 1): f(x^2 + y) = xf(x) + f(y)$ (Denote this by (1))
Now note that $f$ is odd, as $P(x, 0, 0): f(x^2) = xf(x) = (-x)f(-x)$.
Therefore (1) simplifies as $f(x^2 + y) = f(x^2) + f(y)$.
Now since this holds for all real $x$ and $y$, it holds for all positive reals $x, y$, so we assume $x, y>0$, and let $b = x^2$.
Then we have $f(b+y) = f(b) + f(y)$, and since we are assuming f to be bounded (in the positive reals) , this satisfies Cauchy's Functional Equation, i.e. we get $f(x) = cx \forall$ positive real x.
Also, $f(-x) = -f(x) = -cx = c(-x)$ for positive $x$, so $f(x) = cx$ for all nonzero x, and as $f(0) = 0$, we get $f(x) = cx \forall x$.
Substituting into the original equation, we get $cx^2 + c^2 yz = cx^2 + cyz \implies c^2 = c,$ and as $c \ne 0 $we have $c = 1$, i.e. $f(x) = x$.
Therefore, $f(x) \equiv 0$ and $f(x) = x$ are the only solutions.

Bruh

First notice that $f(x)=0$ obviously works, now assume that $f(x)$ is nonzero for atleast one input value, say $c \neq 0$
$P(0,0,0) \implies f(0) = 0$.
Now $P(0,c,z) \implies f(cf(z))=zf(c)$ which implies that $f$ is bijective. Finally $$P(x,1-x,x) \implies f(x^2+xf(1-x))=f(x) \implies x^2+xf(1-x)=x \implies f(1-x)=1-x \implies f(x)=x \ \ \forall \ x \in R-0 $$ and since we know that $f(0)=0$ we can conclude that $f(x)=0$ and $f(x)=x$ are the only solutions $\blacksquare$

Also, INMO doesn't order its problems difficulty-wise, so this being p6 (in 2005) is not a big deal  :)


Let $ a,b,c>0$.Prove that
$ \sqrt{\frac{a^3\plus{}abc\plus{}b^3}{c^3\plus{}abc}}\plus{}\sqrt{\frac{b^3\plus{}abc\plus{}c^3}{a^3\plus{}abc}}\plus{}\sqrt{\frac{c^3\plus{}abc\plus{}a^3}{b^3\plus{}abc}}\geq\ 3\sqrt{\frac{3}{2}}$

[quote="PHONGVAN"]Let $ a,b,c > 0$.Prove that
$ \sqrt {\frac {a^3 \plus{} abc \plus{} b^3}{c^3 \plus{} abc}} \plus{} \sqrt {\frac {b^3 \plus{} abc \plus{} c^3}{a^3 \plus{} abc}} \plus{} \sqrt {\frac {c^3 \plus{} abc \plus{} a^3}{b^3 \plus{} abc}}\geq\ 3\sqrt {\frac {3}{2}}$[/quote]
After AM-GM we need to prove that
$ \prod_{cyc}\frac {a^3 \plus{} abc \plus{} b^3}{c^3 \plus{} abc}\geq\frac{27}{8},$ which is equivalent to
$ \sum_{cyc}(8a^6b^3 \plus{} 8a^6c^3 \plus{} 8a^7bc \minus{} 11a^5b^2c^2 \minus{} 3a^4b^4c \minus{} 10a^3b^3c^3)\geq0,$ which is obvious.

Can't we assume $ abc\equal{}1$ since it's homogeneous.

[quote="PHONGVAN"]Let $ a,b,c > 0$.Prove that
$ \sqrt {\frac {a^3 \plus{} abc \plus{} b^3}{c^3 \plus{} abc}} \plus{} \sqrt {\frac {b^3 \plus{} abc \plus{} c^3}{a^3 \plus{} abc}} \plus{} \sqrt {\frac {c^3 \plus{} abc \plus{} a^3}{b^3 \plus{} abc}}\geq\ 3\sqrt {\frac {3}{2}}$[/quote]
It's similar to
   $ \sqrt {\frac {a^2 \plus{} ab \plus{} b^2}{c^2 \plus{} ab}} \plus{} \sqrt {\frac {b^2 \plus{} bc \plus{} c^2}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {c^2 \plus{} ca \plus{} a^2}{b^2 \plus{} ca}}\geq\ 3\sqrt {\frac {3}{2}}$

[quote="PHONGVAN"]Let $ a,b,c > 0$.Prove that
$ \sqrt {\frac {a^3 \plus{} abc \plus{} b^3}{c^3 \plus{} abc}} \plus{} \sqrt {\frac {b^3 \plus{} abc \plus{} c^3}{a^3 \plus{} abc}} \plus{} \sqrt {\frac {c^3 \plus{} abc \plus{} a^3}{b^3 \plus{} abc}}\geq\ 3\sqrt {\frac {3}{2}}$[/quote]
Since $ a^3\plus{}abc\plus{}b^3 \ge ab(a\plus{}b\plus{}c)$, it suffices to prove that
\[ \sum \sqrt{\frac{ab}{c(c^2\plus{}ab)}} \ge 3\sqrt{\frac{3}{2(a\plus{}b\plus{}c)}}\]
By Holder Inequality, we have that
\[ LHS^2 \cdot \sum a^2b^2c(c^2\plus{}ab) \ge (ab\plus{}bc\plus{}ca)^3\]
It suffices to prove that
\[ (ab\plus{}bc\plus{}ca)^3(a\plus{}b\plus{}c) \ge \frac{27}{2} abc(a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}abc(a\plus{}b\plus{}c))\]
\[ \Leftrightarrow 2(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^3 \ge 27abc(a\plus{}b\plus{}c)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}abc(a\plus{}b\plus{}c))\]
By AM-GM Inequality, we have
\[ RHS \le \frac{27}{8}[ 2abc(a\plus{}b\plus{}c)\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}abc(a\plus{}b\plus{}c)]^2 \equal{}\frac{27}{8}(a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}3abc(a\plus{}b\plus{}c))^2 \le \frac{27}{8} \cdot \frac{16}{9} (ab\plus{}bc\plus{}ca)^4 \equal{}6(ab\plus{}bc\plus{}ca)^4 \le 2(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^3\]
We have done. :)


If $a,b,c,d$ are non-negative and $a+b+c+d=1$, prove that $|ab-cd| \leq \frac{1}{4}$

[hide]If a, b, c, and d are non-negative, $cd \ge 0 $. Assume that the statement is false, then $ab \ge \frac{1}{4}$. Let $c = d = 0$, since we want to maximize $a$ and $b$. By AM-GM, $\frac{a+b}{2} \ge \sqrt{ab} \Rightarrow ab \le \frac{1}{4}$. We have to consider $ab < cd$ which is just the case that $cd - ab$ is positive. And we apply the same argument. Clearly, $ab = cd$ does not work. Thus, we have our contradiction and the statement is true. [/hide]

yeah,
like white_horse_king88 pointed out
$0\le ab,cd \le \frac14$
therefore will the diffrence be at max $\frac14 - 0$
this is maybe easier to see ; )

peeta


(a) Let $a_1,a_2,...$ be a sequenceof reals with $a_1=1$ and $a_{n+1}>\frac32 a_n$ for all $n$. Prove that $\lim_{n\rightarrow\infty}\frac{a_n}{\left(\frac32\right)^{n-1}}$ exists. (finite or infinite)

(b) Prove that for all $\alpha>1$ there is a sequence $a_1,a_2,...$ with the same properties such that $\lim_{n\rightarrow\infty}\frac{a_n}{\left(\frac32\right)^{n-1}}=\alpha$

I doubt there's any number 1 problem you can't do. ;)

Well, I suck at Analysis, so I'm sure you can find one I can't do ;)

[quote=Adamsawyer][quote=Arne]I can do this one :D

Take $x_n = \frac{a_n}{\left(\frac{3}{2}\right)^{n - 1}}$. Then $x_1 = 1$ and $x_n < x_{n + 1}$ so the sequence $\{x_n\}$ is strictly increasing. Thus, it either converges or tends to infinity.

For the second part, since $\alpha > 1$, we can choose a monotonically increasing sequence $\{x_n\}$ such that $\lim_{n \rightarrow + \infty} x_n = \alpha$. Choosing $a_n = \left(\frac{3}{2}\right)^{n - 1}x_n$, we are done.[/quote]
why we can choose such sequence. can someone explain me[/quote]

Because that's the sequence we want to know the limit of. The reason for writing $x_n$ instead of $\frac{a_n}{\left (\frac{3}{2}\right )^{n-1}}$ of just simplifying the notation.


Does anybody have Aops Book 1, and Book 2, and the new Counting and Probablility book?

How are all three?

Because I am planning to get one, two, or all three of them so I can work on them over the summer...

Thanks!

I would get AoPS 1 to start out. Really, you need problems more than you need knowledge. A lot of techniques and strategies come only through practice. If you can breeze through MathCounts and first 15 AMC 10 problems, but aren't at the AIME level, I would reccomend doing lots of AMC's and reading the solutions carefully.

Tell me more about your math ability.


Does anyone know where to find the first post of AoPS?

It was most likely an admin test post that was later deleted.

I think that you would find somewhere where says "Search", and find in ages, I think :|

How about http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2

The very first post is a bot one that's automatically posted so it probably doesn't count and its like

Welcome to phpBB.

etc etc

AoPS/MathLinks used to be two separate forums, with I think MathLinks starting before AoPS.  So if you just look at the dates, you'll probably only find the first posts from MathLinks.


inequalities
geometry
triangle inequality

This problem is from aops vol 1, I have been playing around with it for over an hour, with only minor progress, I was wondering if someone could give me a hint?

proves that if h_a , h_b, h_c are alltitudes of a triangle, then 1/h_a < 1/h_b + 1/h_c .

I know it reminds me of a triangle inequality, I tried re-aranging that with am-gm, to no sucess, I also manages to find that h_a < ( a + b + c)/2 ( with a, b, c triangle side lengths, simmilarly for h_b, h_c), but can't seem to get that anywhere. I also tried to make three right triangles that each had one leg congruent to one hieght and a hypotenoose congruent to the coresponding side.

thanks

It's actually not always true...here's a counterexample...
Suppose you have a right triangle such that one leg is much longer than the other, say, more than twice as long.  Those two legs are two altitudes, while the altitude to the hypotenuse, shorter than either of the two legs, is the third.  Then that altitude plus the shorter leg is less than the longer leg.
It seems that you can have this for acute and obtuse triangles as well: i.e, if one altitude is disproportionately larger than the other two, in that its angle is much less than the other two.

woops, bad typo should be 1/h_a < 1/h_b + 1/h_c . i'll edit it. sorry.

ok. that's much better.

[hide="hint"]it involves areas[/hide]

it seems easier to work with if you look at it as [hide="hint 1"]$h_ah_b < h_ah_c+h_bh_c$[/hide]

If that doesn't help so much,
[hide="hint 2"] Try to relate each side of the inequality to the area of the triangle.[/hide]

[quote="jasamper88"]We have that $(ABC) = \frac{h_a \cdot c}{2} =\frac{h_b \cdot b}{2} = \frac{h_c \cdot c}{2}$

It is the same if we prove:

$\frac{(ABC)}{h_a} < \frac{(ABC)} {h_b} + \frac{(ABC)}{h_c}$
And it is the same as:
$a<b+c$  so we are done[/quote]

That's one way.

K81o7 , I got the same answer the same way as jasamper88 did, whats your way?

I never thought of doing it that way.  I was thinking if the altitudes are $AD$, $BE$ and $CF$, then \[ h_ah_b = (AD)(BE) \le (AD)(BC)=2{\triangle{ABC} }\] and then, \[ h_c(h_a+h_b) = (CF)(AD+BE) \ge CF(AB) = 2\triangle{ABC} \] for non-obtuse triangles, and for obtuse triangles, apparently, it doesn't always work...
Well i'm trying something else for obtuse triangles at this point...that case might cover acuts and right triangles as well.


Find all function $ f: {R^\plus{}} \to\ {R^\plus{}}$ such that
$ f(x\plus{}f(y))\equal{}f(x\plus{}y)\plus{}f(y)$    $ \forall x,y \in R^\plus{}$

$ f(x)\equal{}2x$. It is not hard to prove for continiosly function.

[quote="Rust"]$ f(x) \equal{} 2x$. It is not hard to prove for continiosly function.[/quote]

Can you show your solution??

this is only I know
$ f(x\plus{}nf(y)) \equal{} f(x\plus{}ny)\plus{}nf(y)$  $ \forall x,y \in R^\plus{}$ and $ n \in N$

$ f(nf(y)) \equal{} f(ny)\plus{}nf(y)$  $ \forall x,y \in R^\plus{}$ and $ n \in N$ $ n>1$

Let $ f(x)\equal{}g(x)\plus{}x$, then your condition is $ g(z\plus{}g(y))\equal{}g(z)\plus{}y, z\equal{}x\plus{}y.$
Therefore $ g(y)\equal{}0\to y\equal{}0$. Let $ x_n\equal{}z\plus{}ng(y)$, then $ g(x_n)\equal{}g(x_{n\minus{}1})\plus{}y\to g(x\plus{}ng(y))\equal{}g(x)\plus{}ny.$
Let $ x_1\plus{}ng(y_1)\equal{}x_2\plus{}mg(y_2)$, then $ g(x_1)\plus{}ny_1\equal{}g(x_2)\plus{}my_2.$
It give $ g(x_2\plus{}mg(y_2)\minus{}ng(y_1))\equal{}g(x_2)\plus{}my_2\minus{}ny_1$, by continiosly we get $ g(x)\equiv x$ or $ f(x)\equal{}2x$.

[quote="Rust"]Let $ f(x) \equal{} g(x) \plus{} x$, then your condition is $ g(z \plus{} g(y)) \equal{} g(z) \plus{} y, z \equal{} x \plus{} y.$[/quote]
We have $ g(x \plus{} g(y)) \equal{} g(x) \plus{} y$ for all $ x > y > 0$.
If $ g(y) \equal{} g(z)$ for some $ y,z > 0$, then for any $ x > y,z$
we have $ g(x) \plus{} y \equal{} g(x \plus{} g(y)) \equal{} g(x \plus{} g(z)) \equal{} g(x) \plus{} z\Rightarrow y \equal{} z$.
So $ g$ is injective.
Then for all $ y,z > 0$ and $ x > y \plus{} z$ we have
$ g(x \plus{} g(y \plus{} z)) \equal{} g(x) \plus{} y \plus{} z \equal{} g(x \plus{} g(y)) \plus{} z \equal{} g(x \plus{} g(y) \plus{} g(z))$,
(here $ x\plus{}g(y)\equal{}x\minus{}y\plus{}f(y)>x\minus{}y>z$)
which gives $ g(y \plus{} z) \equal{} g(y) \plus{} g(z)$ for all $ y,z > 0$.
And now $ f(x \plus{} y) \equal{} g(x \plus{} y) \plus{} x \plus{} y \equal{} g(x) \plus{} g(y) \plus{} x \plus{} y \equal{} f(x) \plus{} f(y) > f(x)$ for all $ x,y > 0$.
So $ f$ is a strictly increasing solution of the Cauchy equation,
and therefore $ f(x) \equal{} \alpha x$ for all $ x > 0$, with $ \alpha : \equal{} f(1) > 0$.
This leaves to show $ \alpha \equal{} 2$, which is trivial.

Good solution.
Shorter:
Let $ y\equal{}z\plus{}f(t)$, then $ f(y)\equal{}f(z\plus{}t)\plus{}f(t)$, therefore $ f(z\plus{}t)\plus{}f(t)\plus{}f(x\plus{}z\plus{}f(t))\equal{}f(x\plus{}f(y))\equal{}f(x\plus{}f(t)\plus{}f(z\plus{}t))$ or
$ f(z\plus{}t)\plus{}f(t)\plus{}f(x\plus{}z\plus{}t)\plus{}f(t)\equal{}f(x\plus{}f(z\plus{}t)\plus{}f(t))\equal{}f(x\plus{}t\plus{}f(z\plus{}t))\plus{}f(t)\equal{}f(t)\plus{}f(x\plus{}z\plus{}2t)\plus{}f(z\plus{}t).$
Let $ u\equal{}x\plus{}z\plus{}t\ge t$, then it equavalent to $ f(u\plus{}t)\equal{}f(u)\plus{}f(t)$ (therefore it is true for any u,t). Therefore $ f$ is nonnegative solution $ Gauch's$ equation and must be $ f(x)\equal{}ax$. then $ a\equal{}2$.


Let $ f\equal{}\frac{|x|\minus{}2}{x\minus{}2}$. Determine the range of f.

Here's what I did:

If $ x \ge 0$ then $ y\equal{}\frac{x\minus{}2}{x\minus{}2}\equal{}1$ when x is not 2 and undefined if x=2

If $ x<0$ then $ y\equal{}\frac{\minus{}x\minus{}2}{x\minus{}2}$
$ x\equal{}\frac{\minus{}2\plus{}2y}{y\plus{}1}$

In the expression for x in terms of y, y+1 is in the denominator so the only restriction is that y can't be -1

So I said that the range is all reals except -1. And I don't understand why that's wrong. Any explanation would be appreciated!

[quote="redcomet46"]...If $ x < 0$ then $ y \equal{} \frac { \minus{} x \minus{} 2}{x \minus{} 2}$
$ x \equal{} \frac { \minus{} 2 \plus{} 2y}{y \plus{} 1}$

In the expression for x in terms of y, y+1 is in the denominator [b]so the only restriction is that y can't be -1[/b][/quote]

This is not true. Because of the restriction on $ x$, $ x < 0$, there are some other implicit restrictions on $ y$.

Another way to solve this problem is to graph it (as a piecewise curve), and look at the possible y-values on the curve.

Thanks Gauss202. So should I say that

$ \frac{2(y\minus{}1)}{y\plus{}1}<0$

Using a sign chart, it is only less than 0 when -1<y<1

So then combining -1<y<1 and y=1 except at x=2, the range would be $ \minus{}1<y \le 1$ ?


I median of 5 consecutive integers is 4. What is the mean of the five integers?

Since they are consecutive the mean is lso $ 4$

Also   the set of integers is 2 3 4 5 6  so from tht we see the  mean is $ 4$


the highest order doesn't match, it must be wrong isn't it

Dear Soarer,
I think you make a mistake with your mind.
For example with the case $n=3$,the problem becomes to prove:
         $(a+b^{2})(b+a^{2})(c+a^{2})(c^{2}+a)(b+c^{2})(c+a^{2}) \geq abc(a+1)(b+1)(c+1)(a+bc)(b+ca)(c+ab)$    (the degree of both side are $12$)
And this problem can be done with holder's inequality.
For the general case you can see easily that both side have degree $4n$
So i don't think it's wrong.Do you agree with me? :)  :)

sorry, i misread the left hand side  :blush:


algebra
polynomial
calculus
integration
modular arithmetic
algebra unsolved

Among the prime divisors of the elements of the set $\{a_1,\ldots,a_n\}$, let $p_j,\ j\in\overline{1,k}$ be those for which there exist $i,t$ s.t. $p_j^t|a_i$, but $p_j^t$ does not divide $P(x)$ for all integral values of $x$.

For each $i$, there must be some $j(i)\in\overline{1,k}$ and a $t_i$ s.t. $p_{j(i)}^{t_i}|a_i$, but there is some $x$ for which $p_{j(i)}^{t_i}\not|P(x)$. Now, for each $j\in\overline{1,k}$ for which there is some $i$ s.t. $j(i)=j$, consider only an index $i_0$ for which $t_{i_0}$ is minimal among the $t_i$ with $j(i)=j$. This means that, in general, in what follows we won't be working with all the $i$'s from $1$ to $n$, but with such a restricted system of distinct $i$'s. However, for simplicity (otherwise I'd have to use even nastier index notations), we may assume that the $j(i)$ are distinct as $i$ runs through $1,\ldots,n$.

Assuming that $p_{j(i)}^{t_i}\not|P(x_i)$, we can find $x\equiv x_i\pmod{p_{j(i)}^{t_i}},\ \forall i\in\overline{1,n}$. We find that, for all $i$'s, we have $p_{j(i)}^{t_i}\not|P(x)$, but $p_{j(i)}^{t_i}|a_i,\ \forall i$, so $\forall i,\ a_i\not|P(x)$.

[quote="N.T.TUAN"]Let be given [tex]P(x) \in Z[x][/tex] and [tex]a_1,a_2,...,a_n[/tex] are positive integers numbers.We have that [tex]\forall k\in Z[/tex] there is [tex] a_i[/tex] such that [tex]a_i[/tex] divide [tex]P(k)[/tex].Prove that there is [tex]a_i[/tex] such that [tex]a_i[/tex] divide [tex]P(k) [/tex] [tex]\forall k\in Z[/tex][/quote]
Initially, the problem was stated for pairwise coprime numbers, and, indeed, it was very simple.
I don't know how this statement appeared. My supposition that someone forgot to write "pairwise coprime" :P But we have a nice problem as the result. :) First time I see it 10 years ago, but I have known the solution only since last summer (actually, I had some doubts it is true :blush: ).

:)

Last night, when I pressed the "post reply" button, I started writing something like "Tuan, I think this is wrong; people might have spent lots of time on it with no result" blahblahblah.. Then, I realized that it was me who was wrong :).


given a triangle ABC with AB=BC . in interior triangle choosen two points M and N such that ^MAB=^NCA and ^NAC=^MCB .  prove that points M , N and B are collinear

[quote="Alisher"]given a triangle $ ABC$ with$ AB\equal{}BC$. in interior triangle choosen two points $ M$ and$ N$such that $ \angle MAB\equal{}\angle NCA$ and $ \angle NAC\equal{}\angle MCB$ .  prove that points $ M , N$and$ B$ are collinear[/quote]

very easy:

Let $ AM$ and $ AN$  intersect $ BC$ at $ A_1$ and $ A_2$ respectively. And $ CM$ and $ CN$ intersect $ AB$ at $ C_1$ , $ C_2$ respectively. By using Cheva's theorem we sufficient to show that
$ \frac {AC_1}{BC_1}\frac {BA_1}{CA_1} \equal{} \frac {AC_2}{BC_2}\frac {BA_2}{CA_2}$
It's equivalent with
$ \frac {sin(\angle ACC_1)}{sin(\angle BCC_1)}\frac {sin(\angle BAA_1)}{sin(\angle CAA_1)} \equal{} \frac {sin(\angle ACC_2)}{sin(\angle BCC_2)}\frac {sin(\angle BAA_2)}{sin(\angle CAA_2)}$ it's true.


vector
linear algebra
linear algebra unsolved

can you please give me an example of projection
$ P: \mathbb{C}^2\to\mathbb{C}^2$ over C-vector Space

such that

$ \text{Im}(P)\equal{}\langle(3,5),(9,4),(5,1),(7,0)\rangle$

I know The only such map is the identity map on C^2 but how do i solve the above.

thanks for any help

Is $ P$ linear ?
Your question is unclear. $ [3,5]^T,[9,4]^T$ span $ \mathbb{C}^2$ (as a $ \mathbb{C}$ vector space).


search
number theory unsolved
number theory

a) Find all positive integers $n$ such that $3^{n}-1$ is divisible by $2^{n}$.

b) Find all positive integers $n$ such that $9^{n}-1$ is divisible by $7^{n}$.

c) Find all $n$ such that $x^{n}+2^{n}+1$ is a divisor of $x^{n+1}+2^{n+1}+1$

For C) it has been discussed before , just search it .


trigonometry
algebra
polynomial
modular arithmetic

This is a nice problem but to solve it i need a hint.

Find the "rest" (i don't know how to say :D) of division  of $7\cdot\sin^{31}x+8\cdot\sin^{13} x- 5\cdot\sin^5x\cdot\cos^4x - 10\cdot\sin^7x+5\cdot\sin^5x - 2\end)$ by $\sin^4x-(1+\sin x) \cdot (\cos^2x-2)$ without doing the division.

By substituting $\cos x$  I obtained an ecuation in $\sin x$ and ($\sin x=t$) all resumes at an polynomial problem. From this point im stucked... :?

Sorry for my imperfect english.  :blush:  Hope you will understand the problem  :P

Let $\sin x=t$,we have $7\cdot\sin^{31}x+8\cdot\sin^{13} x- 5\cdot\sin^5x\cdot\cos^4x - 10\cdot\sin^7x+5\cdot\sin^5x - 2\end)=7t^{31}+8t^{13}-5t^9-2$,on the other hand, $\sin^4x-(1+\sin x) \cdot (\cos^2x-2)=t^4+t^3+t^2+t+1$,then we are to notice $t^5\equiv 1\pmod {t^4+t^3+t^2+t+1}$

:huh: ... I must read something about $a\equiv b\pmod {m}$. This is something new for me  :oops:

O.K. Never mind.  :) I rewrite as follows.
$t^5-1$ is devisible by $t^4+t^3+t^2+t+1$.


trigonometry
inequalities proposed
inequalities

Is there something wrong? At a glance,
\[\sin \frac{A}{2}\geq \sin\frac{A}{2}\sin \frac{B}{2},\]
since $|\sin x|\leq 1$

[quote="Redriver"]Given triangle $ABC$. Prove that : $sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}\ge sin\frac{A}{2}sin\frac{B}{2}+sin\frac{B}{2}sin\frac{C}{2}+sin\frac{A}{2}sin\frac{C}{2}$[/quote]

Maybe he meant
\[\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq \frac{2}{\sqrt{3}}\times \left(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2}\right). \]
It is true.

Yes, I'm sorry for my mistakes .... The problems is from VMO !

Moreover, for all $0\leq k\leq 1$ we have
\[\sin^{k}\frac{A}{2}+\sin^{k}\frac{B}{2}+\sin^{k}\frac{C}{2}\geq \left[\frac{2}{\sqrt{3}}\right]^{k}\times \left(\sin^{k}\frac{A}{2}\sin^{k}\frac{B}{2}+\sin^{k}\frac{B}{2}\sin^{k}\frac{C}{2}+\sin^{k}\frac{A}{2}\sin^{k}\frac{C}{2}\right). \]


No Calculators (not that you need one).

1. In a square $ABCD$, $E$ is the midpoint of $CD$, $F$ is the midpoint of $AD$. Connect $BE, CF$, they intersect at $P$. Angle $CBP$ is 27 degrees. Then what is the measure of Angle $BAP$?

2. In Trapezoid ABCD, AD is parallel to BC. AC and BD intersect at M, and AB=AC, AB is perpendicular to AC, BC=BD. Then what is the measure of Angle AMB?

3. In Triangle $ABC$, $E, F$ are points on $AB, AC$ respectively, and $BE=CF$. $M, N$ are the midpoints of $BF, CE$. A line segment through $M, N$ meets $AB$ at $P$ and $AC$ at $Q$. Angle $A$ is 70 degrees. Then what is the measure of the angle APQ?

4. IN a trapezoid ABCD, AB is parallel to DC, CE is the bisector of Angle BCD, CE is perpendicular to AD, $DE=2AE$. CE splits the area of ABCD into 2 unequal parts with the areas $S_{1}$ (above CE) and $S_{2}$ (below CE). Then what is value of $\frac{S_{1}}{S_{2}}$?

5. In a convex pentagon $A_{1}A_{2}A_{3}A_{4}A_{5}$, $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ are the midpoints of $A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, A_{4}A_{5}, A_{5}A_{1}$. $M$ is the midpoint of $B_{1}B_{3}$, N is the midpoint of $B_{2}B_{4}$. If $A_{5}A_{1}=a$, find the length of $MN$.

tere is a fault in Q1 angle CBP must be 30 right?

Nope. In Triangle BCE, $BC=2CE$. Therefore angle $EBC=arctan(\frac{1}{2})$ which is around 27.

Now if BE instead of BC was 2CE then it would be 30, but that is not the case.

[hide="1"] 54 degrees because $AP = AB$. Dunno how to prove it... :dry: [/hide]

How did you come to that conclusion then?

Or is it one of those conjectures? :)

[hide="1"]Clearly $ABPF$ is cyclic ($\Delta BCE \equiv \Delta CDF$, from which $\measuredangle PBA = 180^{\circ}-\measuredangle PFA$). Hence $\measuredangle BAP = \measuredangle BFP = 180^{\circ}-2\measuredangle CBE = 54^{\circ}$.[/hide]

question 1 is wrong as it is stated...27 is not a constructable angle (furthermore, $\tan^{-1}(1/2)$ is not equal to 27 degrees)

Then use $arctan(\frac{1}{2})$ which is somewhere around 26.565 if you feel that 27 degrees is unbearable to work with.

Or you can skip to number 2, or 3, or 4, or 5.

[hide="2"]75 degrees because $CD=CM$ Dunno how to prove it...   :dry: [/hide]


calculus
derivative
trigonometry
function
real analysis
calculus computations

This one is harder than we would expect at a first glance!

$g: \mathbb{R}^2 \to \mathbb{R}$ is defined by

$g\left( {x,y} \right) = \left\{ \begin{gathered} y^2 \,\,\,\,if\,\,\,x = 0 \hfill \\ x^3 \cdot \sin \left( {x^{ - 1} } \right) + y^2 \,\,\,\,\,\,\,otherwise \hfill \\ \end{gathered} \right.$

Is   $\frac{{\partial g}}{{\partial x}}$  continuous at the origin ?

This is the same problem as asking whether the function $h(x)=x^3\sin(x^{-1}),$ with $h(0)=0,$ has a continuous derivative as zero.

In other words, asking this a partial derivative question isn't adding anything to the problem beyond what is there in the ordinary derivative version, and $y^2,$ being a perfectly smooth function, isn't doing anything to alter the nature of the problem.

By the way, the answer is yes, $h'(x)$ is continuous at zero. But $h''(0)$ doesn't exist.

Good point, Mr. Merryfield. Looking this way, the problem is much easier, for sure.

Thanks for the reply,
fskilnik.


In the table shown, the formula relating $ x$ and $ y$ is:
\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \\
\hline y & 3 & 7 & 13 & 21 & 31 \\
\hline \end{tabular}
\]$ \textbf{(A)}\ y \equal{} 4x \minus{} 1 \qquad\textbf{(B)}\ y \equal{} x^3 \minus{} x^2 \plus{} x \plus{} 2 \qquad\textbf{(C)}\ y \equal{} x^2 \plus{} x \plus{} 1$
$ \textbf{(D)}\ y \equal{} (x^2 \plus{} x \plus{} 1)(x \minus{} 1) \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
By guess and check, the answer is $ \boxed{\textbf{(C)}}$.
[/hide]


If $ a,b,c,d \in \mathbb{R}$ such that :

$ a\plus{}b\sqrt{2}\plus{}c\sqrt{3}\plus{}d\sqrt{4}\ge \sqrt{10(a^2\plus{}b^2\plus{}c^2\plus{}d^2)}$ ,

show that $ a^2\plus{}d^2\equal{}b^2\plus{}c^2$

By Cauchy-Schwarz, $ (1\plus{}2\plus{}3\plus{}4)(a^2\plus{}b^2\plus{}c^2\plus{}d^2) \ge (|a|\plus{}|b|\sqrt{2}\plus{}|c|\sqrt{3}\plus{}|d|\sqrt{4})^2$
Equality holds iff $ \frac{a}{1}\equal{}\frac{b}{\sqrt{2}}\equal{}\frac{c}{\sqrt{3}}\equal{}\frac{d}{\sqrt{4}}$ and the result directly follows.


if x,y,z are +ve integers ,then (xy+1)(yz+1)(zx+1)is a perfect square if and only if (xy+1), (yz+1),(zx+1) are perfect squares


Consider a circle $ S$ and four points $ A,B,C,D$ found on $ S$ s.t. $ ABCD$ is an issosceles trapezoid, with $ AD\equal{}BC$. Suppose that the lines $ AD$ and $ BC$ intersect at the point $ P$. Let $ EF$ be the polar of $ P$ wrt the circle $ S$. Prove that $ AC$, $ BD$ and $ EF$ concur.

the problem is a special case of the following lemma
IF ABCD is concyclic, AC,BD,EF are concurrent

Please give a proof to your lemma, plane geometry.
Thanks.

I redefine the point
BK/KF=PBsin∠BPQ/PFsin∠FPQ
FL/LQ=EF sin∠FEB/EQ sin∠QEB
QM/MB=QC sin∠QCF/BC sin∠BCF
BP/PF=AP/AF   EF/BC=AF/AC  QC/EQ=AC/AQ
∠BPQ=∠QEB  ∠FPQ=∠QCF ∠FEB=∠BCF
BK/KF* FL/LQ* QM/MB=1
DONE!

Any trivial solutions for the particular case $ ABCD$- issosceles trapezoid? ($ AD\equal{}BC$)


The two digit numeral $ AB_x$, where $ x$ is the base of numeration and with $ A \not\equal{} 0$ represents a prime integer if $ x$ is ten. $ AB_x$ also represents a prime integer if $ x$ is either twelve or twenty-one. Find the value of $ A \plus{} B$.

I don't think this should be in basic topics. Would someone please move this?

I'm quite sure that this is the appropriate forum.

[hide="What I have so far"]
Let $ p,q,r$ be primes, not necessarily distinct. From the given information, we have the following:

$ 10A \plus{} B \equal{} p\\
12A \plus{} B \equal{} q\\
21A \plus{} B \equal{} r$

Obviously, $ r > q > p$ from above. Eliminating and solving for $ A$, we have:

$ A \equal{} p \minus{} r \plus{}q$

Substituting that back in and solving for $ B$ in all three equations, we can set up a system of three equations of $ p,q,r$ but I highly doubt that was necessary since this is only a 20 minute team competition with 20 problems.

Does anyone have any better ideas?
[/hide]

[hide="My solution"]
From the problem, we know that the following three must all be prime:
$ 10A \plus{} B$
$ 12A \plus{} B$
$ 21A \plus{} B$

From the first expression, we know that $ B \not\equiv 0\mod{2}$, since $ 10 \equiv 0\mod{2}$, and if $ B \equiv 0\mod{2}$, the whole expression would be congruent to $ 0\mod{2}$ and thus be divisible by $ 2$. By similar logic, we get $ B \not\equiv 0\mod{5}$ from the first expression, $ B \not\equiv 0\mod{3}$ from the second, and $ B \not\equiv 0 \mod{7}$ from the third. The above congruences tell us that $ B$ cannot be $ 0, 2, 3, 4, 5, 6, 7, 8, 9$, so $ B \equal{} 1$.

Now let's look at $ A$. We know that $ 21A \plus{} 1 \equiv 1\mod{2} \Rightarrow 21A \equiv 0\mod{2} \Rightarrow A \equiv 0\mod{2}$. This gives us $ 2, 4, 6, 8$ as possible values for $ A$. We can immediately rule out $ 2$ and $ 8$, since $ 21$ and $ 81$ are both divisible by $ 3$. Trying $ 4$ gives us $ 4(12) \plus{} 1 \equal{} 49$, which is not prime, so $ A \equal{} 6$. Checking, we find that our three numbers are $ 61, 73,$ and $ 127$, which are all prime.

$ A \plus{} B \equal{} 7$
[/hide]


geometry
3D geometry
real analysis
real analysis unsolved

Hi,

I have tried to figure out what the intersection is between a four dimensional cube and a hyperplane but in vain. Can you help me out?

I'm not too sure, but wouldn't it just be a parallelpaloid (probably spelled that wrong).  My only basis for this guess is that the itersection of a plane and a cube is a parallelogram.

and what do you mean by hyperplane here? is it smth 3-dimensional?

Here's the definition from mathworld: http://mathworld.wolfram.com/Hyperplane.html


What do you guys think about the AMC 8?
There was one perfect in our state, yet I still don't know my score!

http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2009-8a/2009statestats/AL.HTML

a seventh grader from AL got perfect score. So you must be the one

congratulations!!!


(Pretty easy)

The digits 1 through 9 are assigned uniquely to the variables $a$ through $i$. $\frac{ab}c - \frac{de}f + \frac{gh}i = x$, where $x$ is a positive integer that is as large as possible. What is $x$?

Given a table of the possible values of the denominators and their subsequent numerators, you find the lowest set of numbers that yield 1 are $\frac{2 \times 3}{6}$ followed by $\frac{2 \times 4}{8}$. Since you want higher numbers for the other ones because they're positive, use the 2, 3, 6 combination. The numbers left are 1, 4, 5, 7, 8, 9, and because the lowest denominators are 1 and 4, and because 8 has to be above the 4, by testing each case the highest products comes out to be $\frac{8 \times 5}{4}-\frac{2 \times 3}{6}+\frac{9 \times 7}{1}=72$

I'm not sure if this is right. It was done in a somewhat crude way. Any elegant solutions?


Show $\lim_{n\to\infty}\frac{p_{n}}{n\ln n}=1$ , where $p_{n}$ is the $n$th prime .

I think that the following site is useful.
Please try to refer to references of the site.
[url=http://mathworld.wolfram.com/RossersTheorem.html]Rosser's Theorem[/url]


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

Each point of a circle is painted either Balck or white.

(a):Prove that there exists an isosceles triangle iscribed in this circle such that all this vertices are painted the same color.

(b):Does there exist an equilateral triangle inscribed in this circle such that all its vertices are painted the same color.

Consider the vertices of a regular $ 9$-gon inscribed in the circle, and label them $ 1,\cdots,9$ in the clockwise sense. Any two-colouring of the vertices induces a two-colouring of the integers $ 1,\cdots,9$. Since Van der Waerden's number $ V(2,3)=9$, there exists a monochromatic arithmetical progression of length $ 3$, say $ a,b,c$. This leads to a monochromatic isoscele triangle.

Note that you can use the same argument on any rotation of the above $ ç$-gon with sufficientely small angle. Thus, there exists a uncountable number of monochromatic isoscele triangles.

b) No. Colour a semi-circle in black and the other white.
There is no equilateral triangle.

Pierre.


The mean of a set of five different positive integers is 15.
The median is 18.  What is the maximum possible value of the largest of these
five integers?

since its asking for maximum value, other numbers should be the smallest.

the set is _, _, 18, _, _

so smallest = 1, 2, 18, 19, _

15*5-1-2-18-19=35

answer : 35


hi,
 how can a cartoon face be created using trignometry graphs. Using at least 3 different types of trig functions to add detail.

Ummm...this should probably be in a different forum.  Not only because of its difficulty, but because its not the right kind of question to put in this forum.

I'll move it to Other Problem Solving Topics.


Mafia function MATHCOUNTS ratio calculus

Match up two characters, real or fictional, to a fight! Choose who would win in a fight (giving explanations if necessary) and then choose a new challenger for the winner!

Base rules:

1) Chuck Norris is not allowed to challenge anyone.

Okay. ROUND 1:

Nurse Joy (from Pokemon) vs. Monica Lewinsky

MONICA WOULD SEDUCE NURSE JOY SO SHE WINS.

Nurse Joy vs. Chancey

The [i]winner[/i] is supposed to continue, though I'm not really sure what you meant by the unclear pronoun "she".

Chansey has Egg bomb, but Nurse Joy has tranquilizer darts so she wins.

Nurse Joy vs. House (from the T.V. show House)

Nurse Joy would seduce the House so she wins.

Nurse Joy v. randomdragoon

randomdragoon would be to random for nurse joy to do anything to but would most likely from randomness knock himslef out
Nurse Joy vs Tony Soprano

Nurse Joy would seduce Tony Soprano and win

Nurse Joy vs. Goliath

Goliath would not be able to fight because David killed him. Nurse Joy wins.

Nurse Joy v.s. $FRANZ$ $JOSEPH$

Nurse Joy would win because no one has heard of Franz Joseph.

Nurse Joy vs. G-d

Pokemon is a godless society, so Nurse Joy can't even begin to comprehend G-d, much less hit him.

G-d vs. Buddha

G-d would win because he's G-d.

G-d vs. Moses

new rule:

g-d can't compete...

Moses vs. Darth Sideous

No you can't make up rules

So far it's
Nurse Joy: 7
G-d: 2

G-d vs. Moses

G-d of course because he is G-d
G-d vs. Ignite

well, this goes on forever, so g-d is in forever.

On the other side of the universe...

Al-quieda vs. G.W.B.

OK. G.W.B.? Al-quieda's done well so far...

Al-Queida vs. Mr. T

Yahoo wins because wth is that other person?

Yahoo vs Yahoo Mail

Yahoo contains Yahoo mail, so they cancel out.  So Yahoo wins.

Yahoo vs. Ignite168

big al because the gangsters get afraid of big als bigness.

Big Al v.s. Fat albert

big al because fat albert is slow

big al vs calculus

big al wins because he's the master of gangsters

big al vs. Your mom

Your Mom wins because she had crazy Your Mom jokes that makes Big Al laugh to death

Your Mom V.S. [size=150][b][u][i]YOUR GRANDMA[/i][/u][/b][/size]

your mom wins for reasons unknown

your mom vs. swiss cheese

Your mom because she eats it.

Your mom versus Deep Blue.

WTF is with people on AoPS?

Deep Purple > Deep Blue

Anyway, so your mom was like playing chess against Deep Blue and complained about it being extremely long and hard and Deep Blue was all like "THATS WHAT SHE SAID" so Deep Blue wins.

Deep Blue vs. Deep Throat

Deep Throat is older.

Deep Throat vs. Throat Mints

Hey mods why didn't you take down this thread? :P

samrocksnature vs ObjectZ

Nature obviously wins so samrocksnature proceeds to the next round!


With $ a,b,c>0$. Prove that
  $ \sum \sqrt{ \frac{a^3\plus{}abc}{b\plus{}c}} \geq a\plus{}b\plus{}c$

$ \sqrt {\frac {1}{a}.\frac {b \plus{} c}{a^2 \plus{} bc}}\leq\frac {\frac {1}{a} \plus{} \frac {b \plus{} c}{a^2 \plus{} bc}}{2} \equal{} \frac {(a \plus{} b)(a \plus{} c)}{2(a^3 \plus{} abc)}$
$ \Rightarrow\sum\sqrt {\frac {a^3 \plus{} abc}{b \plus{} c}}\geq 2\frac {\sum(a^3 \plus{} abc)(b \plus{} c)}{\prod(a \plus{} b)}$
We have $ 2\sum(a^3 \plus{} abc)(b \plus{} c)\geq\sum a^3(b \plus{} c) \plus{} 4abc\sum a \plus{} 2\sum a^2b^2 \equal{} \sum a\prod(a \plus{} b)$,OK :lol:


Find all positive integer $ n$ such that there exist a permutation $ (x_{i})$ of $ {1,2,...,n }$ such that
  $ \sqrt{x_{1}+\sqrt{x_{2}+\sqrt...+\sqrt{x_{n-1}+\sqrt x_{n}}}}$ is a rational number.

If the right verion is:
$ \sqrt{x_{1}+\sqrt{x_{2}\dots\sqrt{x_{n-1}+\sqrt{x_{n}}}}}$.It is from BMO 2007 problem 3.

Only now I realized that instead of $ _n$ there should be $ x_n$ and not $ \sqrt{x_n}$,as I thought one year ago...
That doesn't make the problem much harder than it was at BMO,but a little bit different,though idea is essentially the same.

Find all $ n$,for those there exist a permutation $ \sigma=\{\sigma_1,\dots,\sigma_n\}$ of $ \{1,2,\dots,n\}$,such that
\[ \sqrt{\sigma_1+\sqrt{\sigma_2+\dots+\sqrt{\sigma_{n-1}+\sigma_{n}}
}}\]
is a rational number

Here's where the BMO problem was posted (it was actually posted with $ \sqrt{\sigma_{n\minus{}1}\plus{}\sqrt{\sigma_n}}$ at the end)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827183#827183


solve in $ \mathbb{R}$:
$ \sqrt [3]{x \plus{} 8} \plus{} \sqrt [3]{x \minus{} 8} \equal{} \sqrt [3]{64 \minus{} x^{2}}$

hello, because of the range of definition we have $ x\geq 8$ and $ 64\minus{}x^2 \geq 0$. From here we get $ x\equal{}8$. But $ x\equal{}8$ is not the solution of our equation.
Sonnhard.

hello, $ \mathbb{R}$ is the set of real numbers.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, because of the range of definition we have $ x\geq 8$ and $ 64 \minus{} x^2 \geq 0$. From here we get $ x \equal{} 8$. But $ x \equal{} 8$ is not the solution of our equation.
Sonnhard.[/quote]

Why do we have this? Can't you take the cube root of a negative number?

[hide="start... probably wrong"]note that $ \sqrt [3]{x} \plus{} \sqrt [3]{y} \equal{} \sqrt [3]{x \plus{} y \plus{} 3\sqrt [3]{xy}(\sqrt [3]x \plus{} \sqrt [3]y)}$

then $ \sqrt [3]{2x \plus{} 3\sqrt [3]{(x \plus{} 8)(x \minus{} 8)}(\sqrt [3]{x \plus{} 8} \plus{} \sqrt [3]{x \minus{} 8})} \equal{} \sqrt [3]{64 \minus{} x^2}$

i am too lazy to go from there[/hide]

hello, i have used that if $ x$ is a real number it must be $ x^{\frac{1}{3}}\equal{}e^{\frac{1}{3}\cdot\ln(x)}$.
Sonnhard.

Let $ a\equal{}\sqrt [3]{x \plus{} 8},b\equal{}\sqrt [3]{x \minus{} 8}$ with $ a^3\minus{}b^3\equal{}16$,then the eqt is equiv to

$ a\plus{}b\equal{}\minus{}ab$.Thus we should solve the system

$ a^3\minus{}b^3\equal{}16$,
$ a\plus{}b\equal{}\minus{}ab$.

but I can't go ahead from here,,,is this really easy??

hello, from you second equation you get $ b\equal{}\minus{}\frac{a}{1\plus{}a}$, by substituting this in your first equation we have
$ a^6\plus{}3a^5\plus{}3a^4\minus{}14a^3\minus{}48a^2\minus{}48a\minus{}16\equal{}0$.
Sonnhard.

$ a^3 \minus{} b^3 \equal{} 16$
$ a \plus{} b \equal{} \minus{}ab$

Okay, so we can factor our first equation as:

$ (a \minus{} b)(a^2 \plus{} ab \plus{} b^2) \equal{} 16$

Hmm...

Square the second equation.

$ a^2 \plus{} 2ab \plus{} b^2 \equal{} \minus{}a^2b^2$

$ a^2 \plus{} b^2 \equal{} \minus{}a^2b^2 \minus{} 2ab$

Go back to the first equation.

$ (a \minus{} b)(\minus{}a^2b^2 \minus{} ab) \equal{} 16$

$ \minus{}ab \equal{} a \plus{} b$, so...

$ (a \minus{} b)(\minus{}a^2b^2 \plus{} a \plus{} b) \equal{} 16$

Ahh, I don't know what to do now. Am I on the right track or was this all completely useless? :maybe:

Heh, I was gonna say $ 0$, and I thought it worked, until I realized that $ \sqrt[3]{\minus{}8} \equal{} \minus{}2$, not $ 2$...

[quote="Bovy11"]If the equation is $ \sqrt [3]{8 \plus{} x} \plus{} \sqrt [3]{8 \minus{} x} \equal{} \sqrt [3]{64 \minus{} x^2}$,

let $ a \equal{} \sqrt [3]{8 \plus{} x},b \equal{} \sqrt [3]{8 \minus{} x}$, we would have the system

$ a^3 \plus{} b^3 \equal{} 16,a \plus{} b \equal{} ab$ and a unique solution $ a \equal{} b \equal{} 2$ and finally get $ x \equal{} 0$.

Is this the true problem?[/quote]

yes that is the problem...

Wow, ernie, how'd you get cube root of -8?  The second term simplifies to $ \sqrt [3] {8 \minus{} 0}$...

EDIT: checked, and I think bovy is right.

roben i don't think this was so easy....

[quote="Bovy11"]If the equation is $ \sqrt [3]{8 \plus{} x} \plus{} \sqrt [3]{8 \minus{} x} \equal{} \sqrt [3]{64 \minus{} x^2}$,

let $ a \equal{} \sqrt [3]{8 \plus{} x},b \equal{} \sqrt [3]{8 \minus{} x}$, we would have the system

$ a^3 \plus{} b^3 \equal{} 16,a \plus{} b \equal{} ab$ and a unique solution $ a \equal{} b \equal{} 2$ and finally get $ x \equal{} 0$.

Is this the true problem?[/quote]

$ \sqrt [3]{x \plus{} 8} \plus{} \sqrt [3]{x \minus{} 8} \equal{} \sqrt [3]{64 \minus{} x^{2}}$
unfortunately, \sqrt [3]{x - 8}, not \sqrt [3]{8 - x}.
a+b=-ab

AIME15, I don't know how you checked this.
$ \sqrt [3]{10} \plus{} \sqrt [3]{6} \not\equal{} \sqrt [3]{60}$

roben, care to explain?  This doesn't seem to be such a trivial problem.

Oh whoops, i wrote the equation out wrong...

[quote="roben123"][quote="Poincare"]from all this evidence, the problem is probably wrong.... :([/quote]
S={16/V5}[/quote]

Is that supposed to mean $ 16\sqrt{5}$?


Find all positive integers $n=p_1p_2 \cdots p_k$ which divide $(p_1+1)(p_2+1)\cdots (p_k+1)$ where $p_1 p_2 \cdots p_k$ is the factorization of $n$ into prime factors (not necessarily all distinct).

surprising, it's not hard!  :)
Assume that $p_k$ is a maximal  of $\{p_1,\ldots,p_k\}$. So we have $p_k\mid p_j+1$ for some $j$. If $p_k>3$ then $p_j>2$ and then $p_j+1$ must be a composite! So $p_k<p_j+1$ so $p_k\leq p_j$, which implies $p_k=p_j$, contract to the fact $p_k\mid p_j+1$. So $p_k=3$.
Assume we have $r\geq0$ numbers $p_j$ which equal to $2$ and $s=k-r\geq0$ ones equal to $3$. So we have $2^r3^{s}\mid 3^r\cdot 4^s$. So we have $r\leq 2s$ and $s\leq r$.
Answer $n=2^r3^s$ where $0\leq s$ $\leq r\leq 2s$.


polynomial
superior algebra
superior algebra unsolved

Hint: it's irreducible over $\mathbb Q$ and has prime degree.

Applying Eisensteins criterion with $p=3$; $\beta=x^{5}-9x^{3}+15x+6$ is irreducible over $\mathbb{Q}$. Then $\deg(\beta,\mathbb{Q})=5$.
Also $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ has a basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ which implies its degree is $4$. But $5$ does not divide $4$. Hence $\beta$ is irreducible.


[quote="sri340"]Suppose that w = 2^1/2,  x = 3^1/3,   y = 6^1/6, and z = 8^1/8. From among these numbers, list the biggest first and the second biggest second

(1) w, x      (2) x,w   (3) y, z     (4) x, z[/quote]
[hide]$ w\equal{}2^{\frac{1}{2}}$, $ x\equal{}3^{\frac{1}{3}}$, $ y\equal{}6^{\frac{1}{6}}$ and $ z\equal{}8^{\frac{1}{8}}$
Comparing $ \sqrt{2}$ and $ \sqrt[3]{3}$, raising both sides to the 6th power results in $ 8$ and $ 9$, so we know that $ x>w$. Similarly, for $ y$ and $ z$, raising both sides to the 24th power results in $ 1296$ and $ 512$ so $ y>z$. Comparing $ x$ and $ y$, raise both sides to the 6th power to see that $ x\equal{}9$ and $ y\equal{}6$ so $ x>y$. Also you can do this method to find that $ w>z$ and $ w>y$ to have the inequality $ x>w>y>z$ so the two largest values are $ x$ and $ w$. (2)[/hide]


calculus
integration
derivative
number theory unsolved
number theory

find all intregal solutions to the equation $ x^2\plus{}y^2\equal{}(x\minus{}y)^3$

Let $ a \equal{} x \minus{} y, b \equal{} x \plus{} y$; then we want to solve $ a^2 \plus{} b^2 \equal{} 2a^3 \implies b^2 \equal{} a^2(2 \minus{} a)$, so $ 2 \minus{} a$ is an integer which must be the square of a rational, hence must itself be a square, and $ a | b$, so write $ 2 \minus{} a \equal{} n^2$; then $ a \equal{} 2 \minus{} n^2$ and $ b \equal{} na \equal{} 2n \minus{} n^3$.

A general method for finding out whether a cubic Diophantine has infinitely many integer solutions (which can occur if and only if it has genus zero by [url=http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points]Siegel's theorem[/url]) is to figure out if it has any singularities.  The relevant partial derivatives are $ 2x \minus{} 3(x \minus{} y)^2$ and $ 2y \minus{} 3(x \minus{} y)^2$, and so there is a singularity at the origin.

This equation is true for $ x\equal{}y\equal{}0$
If at least one of the numbers $ x,y$ are nonzero we have:
$ (x\minus{}y)^2\equal{}(x\minus{}y)^3\minus{}2xy$
Which is
$ (x\minus{}y)^2(x\minus{}y\minus{}1)\equal{}2xy$
Let us asume that:
$ (x,y)\equal{}m$.We denote by $ x_1\equal{}\frac{x}{m},and,y_1\equal{}\frac{y}{m}$
So $ (x_1,y_1)\equal{}1$
If we put this into the equation we obtain:
$ m^2(x_1\minus{}y_1)(mx_1\minus{}my_1\minus{}1)\equal{}2xym^2$
So $ mx_1\plus{}my_1\minus{}1\equal{} \frac{2x_1y_1}{(x_1\minus{}y_1)^2}$
So $ x_1\minus{}y_1\equal{}1$.Beucause if $ x_1\minus{}y_1>1$ Then there will exist a prime $ p$ so that $ p|x_1\minus{}y_1$..But then $ p|x_1$ and $ p|y_1$ (contradiction),or $ p^2|2$ contradiction......
So $ x_1\equal{}y_1\plus{}1$
So by the last equation $ mx_1\plus{}my_1\minus{}1\equal{} \frac{2x_1y_1}{(x_1\minus{}y_1)^2}$ we have that $ m(y_1\plus{}1)\plus{}my_1\minus{}1\equal{}2(y_1\plus{}1)y_1 \leftrightarrow m\equal{}1\plus{}2(1\plus{}y_1)y_1$
So $ x\equal{}mx_1\equal{}[2y_1(1\plus{}y_1)\plus{}1](1\plus{}y_1)$
And $ y\equal{}my_1\equal{}[2y_1(1\plus{}y_1)\plus{}1]y_1$


algorithm
absolute value
algebra unsolved
algebra

[quote="harazi"]Someone gave me some time ago an absolutely fantastic problem, but far too difficult for my poor head. Yet, I think it will be a challenge for everyone on this forum, since it seems to be really terribly difficult:
   Suppose we have $x_1,...,x_n$ some real numbers of absolute value at most $1$ and having their sum $0$. Prove that we can arange them in a row such that  no matter how we choose some consecutive numbers on this row, their sum is smaller than or equal to $2-\frac{1}{n}$ Does anyone know a solution to this problem?[/quote]

I suppose you rather meant "the absolute value of their sum is smaller than or equal to $2-\frac{1}{n}$", because that's surely a more complicated one (as the bound is for two sides, not one).

Because without the absolute value assumption we can pick (suppose all $x_i$ all non-zero) from $x_1,...,x_n$ the subset $S_1$ with positive $x_i$ and subset $S_2$ with negative $x_i$. Then we construct our row using an algorithm :
First we take the biggest $x_m$ from $S_1$ - it's our first number in the row:
1) if the biggest $x_j$ from subset $S_1-\{x_m\}$ satisfies
$x_m+x_j>2-\frac{1}{n}$ then the second number in the row will be the smallest $x_i$ from $S_2$.
If $x_j$ satisfies $x_m+x_i+x_j \leq 2-\frac{1}{n}$ then it's our third number in the row,
if not we take the the smallest number from the set $S_2-\{x_i\}$ and put in the row and start 1) again.
2)if the biggest $x_j$ from subset $S_1-\{x_m\}$ satisfies
$x_m+x_j \leq 2-\frac{1}{n}$ then we start the algorithm again starting from subset $S_1-\{x_m,x_j\}$

Now I don't have time to proove the corectness but it's actually a good construction of the row.

Here is the solution from the book "Problems and Solutions From Around the World 1999-2000"

Is there a link from whom the book could be downloaded ? :roll:

Thank you very much, Enescu. Fantastic problem, fantastic solution.


A airplane takes off ata ground velocity of 375 km/h on a heading of 375km/h,climbing at a angle of 15 degrees. A 20 km/h wind is blowing fromm south.determine the magnitude of the resultant air and ground velocities.Also how would you draw this situation.

****sorry so the plane is flying ona heading of N45 Degrees W

[quote]A airplane takes off ata ground velocity of 375 km/h on a heading of 375km/h,climbing at a angle of 15 degrees. A 20 km/h wind is blowing fromm south.determine the magnitude of the resultant air and ground velocities.Also how would you draw this situation.[/quote]

First we need to know which direction the plane is flying in ^_^


I am stuck yet again.... :oops:

Find the geometric image of the complex number z
$ |z \minus{} 2| \minus{} |z \plus{} 2| < 2$

[hide="My Attempt"]Well, you have a difference of distances, which defines a hyperbola. I think that if you draw the hyperbola defined by $ |z \minus{} 2| \minus{} |z \plus{} 2| \equal{} 2$, or $ \frac {x^2}{1} \minus{} \frac {y^2}{\sqrt {3}} \equal{} 1$, your region will be all the points to the left of the right branch of this hyperbola, not including the branch itself.

EDIT: Sorry, I flipped the half-planes, as the two posters below noted.  :oops: [/hide]

ThinkFlow's equation should be [hide="this"]$ \frac{x^2}{1^2}\minus{}\frac{y^2}{(\sqrt3)^2}\equal{}1$[/hide]

The region indicated is wrong. Try $ z\equal{}0$. That should indicate immediately the correct open half-plane.

ThinkFlow, you overlooked the fact that this is true for all $ z$ with non-negative real part.


Let  $ ABC$ be an triangle. Prove o Disprove that
$ \sqrt{\sin(A)} \plus{} \sqrt{\sin(B)} \plus{} \sqrt{\sin(C)}\le\frac{3}{4}\sqrt[4]{\frac{3}{4}}[\frac{\sin(2A) \plus{} \sin(2B) \plus{} \sin(2C)}{\sin(A)\sin(B)\sin(C)}]$ .

______________
Nice  :)

$ \sin{2A} \plus{}  \sin{2B} \plus{} \sin{2C} \equal{} 4 \sin{A} \sin{B} \sin{C}$ so $ RHS \equal{} \frac{3 \sqrt[4]{3}}{\sqrt{2}}$

now for AM-QM $ LHS \le 3 \sqrt{\frac{\sin A \plus{} \sin B \plus{} \sin C}{3}}$ but we know also $ \sin{A} \plus{} \sin{B} \plus{} \sin{C} \le \frac{3\sqrt{3}}{2}$ ando so we are done.


geometry
trapezoid
circumcircle
algebra
polynomial
function
number theory

I'll post the problems of the Greek selection exam for IMO 2001:

1) Let P and Q be points on a given circle k. Let M be the midpoint of PQ, and suppose A and C are points on k for which AC passes through M. Let ABCD be a trapezoid with k as its circumcircle such that AB // CD // PQ. Prove that AD and BC intersect in a point X which is independent of the position of A on k.

2) Find all integers n for which p(x) = x5 - nx - n - 2 can be written as the product of two non-constant polynomials with integer coefficients.

3) Prove that there exists a positive integer N such that the decimale representation of 2000N starts with the digits 200120012001.

4) An equilateral triangle ABC of side 1 is given and S is the set of all its interior points (including the boundary points). For any M, let a(M), b(M) and c(M) denote the distances from M to BC, CA and AB respectively. Define a function by
f(M) = a(M)(b(M) - c(M)) + b(M)(c(M) - a(M)) + c(M)(a(M) - b(M)).
- For which points M in S is f(M) :ge: 0?
- Determine the extremal values of f(M) for all possible positions of M.

Just an idea about 3:

It's obviously equivalent to proving that thing for 2^n. The following holds: given any positive integer, there exists a power of 2 which begins with the digits of this number (in the same order). This result was obtained by Jordan (I think), and it's not that hard to prove. You guys think about it. I don't want to spoil it for you. I'd post the solution, but I have no time right now.

Oh, we did the equivalent question in math this year.  It's true for any number that isn't a power of 10.  Fun stuff.  </false sense of superiority>

What was the name of that density theorem, used to solve this?
I know the theorem but not the name ... maybe Kronecker? :?

The way I have it in my notes is as a corollary to Dirichlet's Theorem.

Anyway, if :alpha: is irrational then {n:alpha:} is dense on [0,1], where n ranges over all integers and {} means fractional part.

Problems 2 and 4 are nice ... no attempt?

For 4, we know a + b + c = :rt3:.  I'm sure that's important.

I've met that result that Arne posted (the density of {n :alpha: } in [0,1] ) both as Kronecker's thm and as Jacobi's thm. I don't know which one of them found it first, but it might have been something almost simultaneous :)

That theorem is most often known as Kronecker's theorem - but it's actually a corollary of the more general Kronecker Approximation Theorem (see the book [u]Geometric and Analytic Number Theory[/u] by Hlawka, et. al. for a statement and proof). But according to [u]A Course in Number Theory[/u] by H.E. Rose, Kronecker's theorem was first considered (and proven?) by Chebyshev.

This whole thing is even very confusing though, because throughout his entire life Kronecker didn't "believe" in irrational numbers. For example, when Lindemann proved that :pi: was transcendental Kronecker said that it was a beautiful proof but that it meant nothing because transcendental numbers didn't exist!!!

A good account of Kronecker's life and beliefs on this matter can be found at the MacTutor Biographies of Mathematicians website - http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html

So I'd be very interested to know more about the history of this problem. How Kronecker got involved involved, and what role he actually had in it.

By the way, an interesting exercise from Rose's book based on this theorem states:
[i]Let S be a square in the plane and suppose the sides of S act as a mirror to a ray of light which is emitted from a point P inside the square with angle :alpha: to one of the sides, and which is reflected repeatedly by the mirrors. If the path passes through a corner assume that it is reflected directly back along the same path. Show that the path either passes arbitrarily close to every point in the square or is periodic, in which case tan :alpha: is rational[/i]

This exercise is also incidentally related to the 1992 NYSML Power Question.

Here's something stronger than this theorem, and, usually, much more useful (but I don't know a proof for this one):

Let  k be irrtional and let a < b be numbers in [0,1]. Let a_n = {n*k} and let S_n(a,b) be the number of terms of the set {a_1, a_2, .., a_n} which are in (a,b). Then S_n(a,b)/n -> b-a when n->infinity.

This is called Weyl's theorem, I think.

I don't understand your post.  What are the ai?

Hmmm.. My post was indeed somewhat incoherent. I have edited it. Hope it makes sense now.

Well, here's a proof for 1 (it's better than nothing... :)):

Let T be the intersection point between XB and PQ.
XC/XB=CD/AB (*) because triangles XDC and XAB are similar. MT/AB=TC/CB (1) and MT/CD=TB/CB (2). We divide (1) by (2) and we get CD/AB=TC/TB (**).

From (*) and (**) we get XC/XB=TC/TB.

X is obviously on the perpendicular bisector of PQ. Let's assume the tangents from X to the circle touch it at points K and L, K near P and L near Q. KL || PQ. KL intersects XB at T'. Since KL is the polar of X with respect to the circle, it means that XC/XB=T'C/T'B=TC/TB, which means hat T=T' because T and T' are on the segment [BC]. From here we get (K,L)=(P,Q), which means that X is the intersection of the tangents to the circle drawn at P and Q.

Partial soln for 4:

f(M)=(b(M)-a(M))(c(M)-b(M))(a(M)-c(M))(a(M)+b(M)+c(M)). JBL said a(M)+b(M)+c(M)= :rt3: , but it's actually  :rt3: /2 (the length of a height of the triangle), so f(M)= (:rt3: /2)*(b(M)-a(M))(c(M)-b(M))(a(M)-c(M)).

This gives us a quick solution for (a): Let D, E, F be the midpts of BC, CA, AB, respectively and let O be the center of the triangle. The locus of M s.t. f(M)>=0 is the reunion of all the interior and border points of the triangles OAF, OBD, OCE.

I'll think more about (b), but I have the feeling the soln isn't far...
:)

First of all, it's obvious that the minimum value is -k and the maximum value is k for some positive k...


I don't understand chess terminology such as Qxf7 What does that mean!!! And then it starts adding plus signs in there!!! Please help

Read a chess book or search "chess notation" on google.

Qxf7 means that the queen captures the peice on f7. A '+' indicates check.

[quote="chess64"]"chess notation" on google.[/quote]
ahhh...i see now...thanks chess


geometry
power of a point
similar triangles

Point $P$ lies at the intersection of chords $AB$ and $CD$ inside circle $O$. Let $E$ denote the intersection of lines $AD$ and $BC$. If $AP=3$, $PC=6$, $DE=5$, and $BE=3$, then find $AD$.

Yeah. It was the first real geometry problem I made up.    :D

Are you sure about the measures?, because it seems to be negative  :?

$\frac{EB}{ED} = \frac{EA}{EC}$ $\Rightarrow$ $\frac{EA}{EC} = \frac{3}{5}$

$\frac{AP}{PC} = \frac{AD}{BC}$ $\Rightarrow$ $\frac{AD}{BC} = \frac{1}{2}$

$AD = ED - EA = 5 - EA$ and $BC = 3 - EC$

$AD = 5 - \frac{3}{5}EC = 5 - \frac{3}{5}(3 - BC) = 5 - \frac{3}{5}(3 - 2AD)$ $\Longrightarrow$ $AD = -16$.  :?

check $AD=ED-EA$. It should be $AD=EA-ED=EA-5$. Same for the other side: $BC=EB-3$. That explains  ;)

I don't get it, how could the lines intersect twice? diagram would help  :)  tnx

[quote="drunner2007"]well you have $AB$ and $CD$ intersecting inside circle $O$. Then a lines $AD$ and $BC$ intersect outside the circle. Notice that the original chords don't lie on the two lines that intersect.[/quote]
I'm confused now :o. Could you post a diagram?

I've tried posting pictures before, but it never takes paint or geometer sketchpad sketches. This time I'm using a pdf.

It doesn't take .bmp, but you can use paint to save in other formats (.jpg, .gif) that it does take.  You can get GSP diagrams by copying them into paint and then saving in one of those formats.

[quote="lotrgreengrapes7926"][quote="Chocoboom"]i just got 16 =/ is that right?

$\bigtriangleup ADP \sim \bigtriangleup CBP$
$2AD=BC, AD=x, BC=2x$
using power of a point
$(DE)(AE)=(EB)(EC)$
$5(5+x)=3(3+2x)$
$25+5x=9+6x$
$\fbox{x=16}$[/quote]
Wouldn't it be $5(5-x)=3(3-2x)$?
But that gives the answer $-16$.[/quote]
I got -16 too.

look at the diagram again you guys. your thinking AE=5, but its DE=5 (unless you're applying power of a pt. wrong)
Chocoboom has it right

ok, I think it is $\frac{16}{11}$ instead... first we see that from similar triangles $BC = 2AD$, setting $AD = x$ and using power of the point we have $5(5-x) = 3(3+2x)$ and solving gets $x = \frac{16}{11}$. The problem is the ambiguity of the picture: i get $B$ to be between $C$ and $E$ and $D$ between $A$ and $E$, but you can draw otherwise...

[quote="Chocoboom"]i just got 16 =/ is that right?

$\bigtriangleup ADP \sim \bigtriangleup CBP$
$2AD=BC, AD=x, BC=2x$
using power of a point
$(DE)(AE)=(EB)(EC)$
$5(5+x)=3(3+2x)$
$25+5x=9+6x$
$\fbox{x=16}$[/quote]
Sorry, but I'm still confused. Could you please explain Chocoboom? This is a trick question :ninja:
Isn't 5+x = DE+AD, and 3+2x=BE+BC?
P.S. I haven't done geometry in a while :oops:

ok, this is exactly why i am posting these problems. I want to catch mistakes or issues before I give them to anyone else. I guess the problem description is ambiguous (there are different ways to set it up). On an actual exam the picture should clear up some ambiguities.

Refering to the diagram (the one that I've posted), the power of a point thm. tells us that $(AE)(DE)=(CE)(BE)$. If we let $AD=x$ (meaning $BC=2x$ from similar triangles) then we can substitute:
$(5+x)(5)=(3+2x)(3)$. $16$ is the answer.

How should I reword the problem description to rule out anyone drawing the wrong diagram?

I would suggest giving them the picture and saying underneath "Find AD"  :D
Your diagram was different than mine, no wonder I got a negative number. I switched A & D, and B & C. :P
Very nice porblem :)

:yup: You gotta say that $D$ is on $AE$ and $B$ is on $CE$ i guess.


For $a,b,c$ positive real numbers such that $abc=7$ prove that

$  \frac{1}{\root3\of {a(1+b)}} + \frac{1}{\root3\of{b(1+c)}} + \frac{1}{\root3\of{c(1+a)}} > \frac{3}{2} $

[quote="manlio"]For $a,b,c$ positive real numbers such that $abc=7$ prove that

$\frac{1}{\root3\of {a(1+b)}} + \frac{1}{\root3\of{b(1+c)}} + \frac{1}{\root3\of{c(1+a)}} > \frac{3}{2}$[/quote]

[color=darkred]Happy Christmas day!![/color]

 :)

where is manlio?
does anybody know?

for the problem:
can we prove:
$\sum \frac 1{2+a+b} > \frac 12$


integration
search
trigonometry
calculus
calculus computations

$a_{n}=\sqrt{2 + \sqrt[3]{3+ ... + \sqrt[n-1]{n-1+ \sqrt[n]{n}}}}$ and
$b_{n}=\sqrt{2 + \sqrt[3]{3+ ... + \sqrt[n-1]{n-1+ \sqrt[n]{n+2}}}}$
1) show that $a_{n}<b_{n}$
2) show that $a_{4}>1,9$
3) show that $2^{n+1}>n+3$ , n from N, n>=2
4) find the monotony of $a_{n},b_{n}$
5) show that $a_{n},b_{n}$ are convergent
6) show that $a_{n},b_{n}$ have the same limit, and that this limit is from the interval (1,9;2)
7) Find $\int a_{x} dx$
8) Find $\int (cos(x))^n dx$

Most of the questions concerning your $a_n$ and $b_n$ were posted recently ... try a search of the forum.

Yes, I no. There isn't a clear solution there. [url=http://www.mathlinks.ro/Forum/topic-90328.html]www.mathlinks.ro/Forum/topic-90328.html[/url]

I've solved 1,2,3,4,5 , but i don't know how to solve 6,7 and 8.

(6) is done in the thread you mentioned (if something still seems unclear, please, tell what exactly your difficulty is)

(7) is very strange as stated: your $a_x$ make sense only for integer $x$, so what are we supposed to integrate?

For (8) it may be a good idea to use Euler's formula $\cos x=\frac 12(e^{ix}+e^{-ix})$. By the way, this problem has nothing to do with (1)-(6), so it should be a separate topic.

[quote="nemesis"]I've solved 1,2,3,4,5[/quote]

Then, please, post your solutions


trigonometry
calculus
calculus computations

How do I go about solving this differential equation?
y''=a-by where a and b are constants.

Rewrite this as $y''+by=a.$

That reveals its identity. It is a second order linear constant coefficient differential equation. It is nonhomogeneous, but the nonhomogeneity is about as simple as possible.

You didn't mention whether $b$ is positive or negative. We can handle it either way, but for the sake of the argument, I will assume $b>0.$

The homogeneous equation $y''+by=0$ has as its solutions $y=c_1\cos(\sqrt{b}\,t)+c_2\sin(\sqrt{b\,}t).$

We find a particular solution by the method of undetermined coefficients. Assume that $y=k$ (a constant) is a solution. Then $0+bk=a,$ so $k=\frac ab.$

The general solution: $y(t)=\frac ab + c_1\cos(\sqrt{b}\,t)+c_2\sin(\sqrt{b}\,t).$

W are to classify the three cases of $1^\circ\ a<-\frac{b^2}{4},\ 2^\circ\ a=-\frac{b^2}{4}$ and $3^\circ\ a>-\frac{b^2}{4}$


How many ways can you arrange "decambir" if there must be a vowel on both ends and d cannot be next to b?

[hide="solution"]First, let's place the vowels. There are 3 vowels in decambir. So there are 3 choices for the first letter, and 2 choices for the second.

Now let's see how many ways that d CAN be next to b:

3*2*10*4*3*2*1=1440

Now let's find all possibilities:

3*2*720=4320

We subtract:

4320
1440
-----
2880[/hide]

[quote="Necrowarrio0"]huh? Is this Maths or ENglish?[/quote]
What country are you from? Anyways, this is a combinatorics problem, read the question again :)

[quote="DarkKnight"][quote="13375P34K43V312"]Extension: What bout the word December?[/quote]

Do the sames rule apply concering the vowels and the D and the B?[/quote]

Sure.

[hide="extension"]Would it just be 2880/3!, so 480 arrangements.[/hide]


The problem asked for the sum of the numbers. :wink:

So the answer is 12.

isnt there some algebraic way to do it? i thought thered be one algebraic way.

gah i hate brute force.

you can try 1/x+1/y+1/z=yz+xz+xy/xyz=47/60 and simplify that....and not many 3 integers multiplied together equals 60....

In most problems like this, you do need some amount of brute force.

You can write equations for 3 variables, but you'll get 2 equations, plus the fact that the numbers are whole numbers. It yields one solution, but the problem is that the last statement cannot be used algebraically in the solution. (At least, I don't know how it can...)

[quote="FantasyLover"]brute force ftw

$ \frac {47}{60} \equal{} \frac {20}{60} \plus{} \frac {15}{20} \plus{} \frac {12}{60} \equal{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{5}$

answer : 3, 4, 5[/quote]

I believe you meant $ \frac {15}{60}$ but yeah your answer only needs to be summed and thats it

[quote="Urc"]The sum of the reciprocals of three numbers is equal to 47/60. What is the sum of the numbers?[/quote]

$ \frac1x \plus{} \frac1y \plus{} \frac1z \equal{} \frac {47}{60}$. WLOG, assume $ \frac1x\geq\frac1y\geq\frac1z$.
Note that the average of $ \frac1x,\frac1y,\frac1z$ must be equal to $ \frac13 < \frac {47}{180} < \frac14$, so at least one term must be greater or equal to this. Since $ x,y,z\in\mathbb{Z}$, we have a mandatory $ x\leq3$. We also know that since $ \frac {47}{60} < 1$, $ x\geq2$. So we have $ x \equal{} \{2,3\}$
We can proceed with cases, using the same general method.
$ x \equal{} 2$: Simplifying, we have $ \frac1y \plus{} \frac1z \equal{} \frac {17}{60}$. Setting bounds, we get $ 4\leq{y}\leq7$. We find solutions of $ (x,y,z) \equal{} \{(2,4,30),(2,5,12)\}$
$ x \equal{} 3$: $ \frac1y \plus{} \frac1z \equal{} \frac9{20}$. We can set bounds on $ \frac1y$ as we did above, getting $ y \equal{} \{3,4\}$. This is simple plug and chug and check now, and we get $ (x,y,z) \equal{} (3,4,5)$.
Thus we have 3 solutions: $ \boxed{12,19,36}$.
Remember this technique...

EDIT: Owned brute force much?

what do you mean we have 3 solutions (12,19,36)??

EDIT: oh oops sry :D

??? :?: The question is what is the sum of the numbers; there are 3 possible sums...

The individual sets of 3 numbers are
$ (3,4,5)$
$ (2,4,30)$
$ (2,5,12)$

So there is not only one solution...try these solutions if you want.

Note to people reading this: Solution 2 posts above.


A cone is a right cone that can hold 2 quarts and has a lateral height of 10 inches. Rob cuts off the tip of the cone and fills it up with one quart. What is the lateral height of this new cone?

A. $ 5^3\sqrt4$

B. $ 1\frac{1}{4}$

C. $ 5\sqrt10$

D. $ 5$

E. $ 8\frac{3}{4}$

[hide]@xxxxx: I'm assuming you mean $ 5\sqrt[3]{4}$ instead of $ 5^3 \sqrt{4}$.

Since the ratios of the volumes of two similar objects is the cube of the ratio of the lengths of two similar objects, we have:
\[ \left(\frac {x}{10}\right)^3 \equal{} \frac {1}{2} \\
\implies \frac {x^3}{10^3} \equal{} \frac {1}{2} \\
\implies x \equal{} 5 \sqrt [3]{4} \implies \boxed{A}
\]
[/hide]


Prove that if $ p \equiv 1 (mod 3)$ $ p$ can be expressed as $ x^2 \plus{} xy \plus{} y^2$ where $ x$ and $ y$ are integers

This is a classical result. See e.g.
http://arxiv.org/PS_cache/math/pdf/0408/0408107v1.pdf

How about this one:
If a prime $ p$ is 5 or 11 mod 24, then $ p\equal{}2x^2\plus{}3y^2$ for some integers $ x$ and $ y$. :D

Notice that : if p can be expressed as $ 3a^2 \plus{} b^2$ then p also can be expressed as $ x^2 \plus{} xy \plus{} y^2$
I doubt whether this reuslt below is true :-?
Given an integer number n and a prime number $ p$ such that : $ ( \frac { \minus{} n}{p}) \equal{} 1$ (where () is the Legendere symbol))then exist two integers number a,b such that : $ p \equal{} a \plus{} nb^2$
any idea ;)

nkht, yes, the problem of $ a^2\plus{}nb^2$ gets complicated for larger values of $ n$.

$ a^2\plus{}5b^2$ represents a prime $ p$ if and only if $ (\minus{}1|p)\equal{}(5|p)\equal{}1$, i.e. $ p$ is 1 or 9 mod 20.
$ a^2\plus{}5b^2$ represents $ 2p$ if and only if $ (\minus{}1|p)\equal{}(5|p)\equal{}\minus{}1$, i.e. $ p$ is 3 or 7 mod 20.

There is another "genus" of forums in the same "order" as $ a^2\plus{}5b^2$. The other genus contains (for example) the form $ 2a^2\plus{}2ab\plus{}3b^2$.

$ 2a^2\plus{}2ab\plus{}3b^2$ represents $ p$ if any only if $ p$ is 3 or 7 mod 20.
$ 2a^2\plus{}2ab\plus{}3b^2$ represents $ 2p$ if any only if $ p$ is 1 or 9 mod 20.


1. In trapezoid ABCD, AB is parallel to CD and the area of triangle ABC divided by the area of triangle ACD is 1/3. If E and F are midpoints of BC and AD respectively, find the numerical value of the area of region ABEF divided by the area of region EFDC.

[hide]since the altitudes of triangles ABC and ACD are the same,  the ratio of side AB to CD is 1:3. The ratio AB:EF:CD is 1:2:3 ... therefore the ratio of the areas is 3:5[/hide]

[hide]Let AB = x and the altitude perpendicular to AB and DC be segment MN.  Therefore A(ABC) = x * MN * 1/2 and A(ACD) = CD * MN * 1/2

Since 3A(ABC) = A(ACD)m

3x * MN * 1/2 = CD * MN * 1/2

So CD = 3x.

Since EF is the median of trap ABCD, its length is the average of x and 3x, that is, 2x.

Finally, the area of trap ABEF = 1/2 * (x + 2x) * MN/2=3x*MN/4

and area trap CDFE =1/2 * (2x + 3x) * MN/2 = 5x*MN/4

The ratio of areas is 3 to 5, or 3:5[/hide]


Let $ n+2$ positive intergers $ 1\leq a_{1}<a_{2}<....<a_{n+2}\leq 3n$. Prove that there exist $ 1\leq i<j<k \leq n+2$
such that  $ a_{i}\mid(a_{j}+a_{k})$

[hide="Is it helpful?"]The case of $ n=1$ is trivial
Suppose that is true for every $ k<n$ then let us prove for $ n$
if for some $ k$,  $ a_{k+2}\le 3k$ then we are done
if not then
$ 3<a_{3}$
$ \cdots$
$ 3n-3<a_{n+1}$
and $ a_{n+1}<a_{n+2}\le3n$[/hide]


where i find Japanese MO of 2004,2005 in english?

The Japanese version of JMO problems from 2000 to 2005 can be found here:

http://www.mmjp.or.jp/jmo/challenge/index.html

Can someone translate it into English?


Sorry if this is spam, but I was wondering how all of you are going to pay for some of these summer programs. This year I decided that I realy wanted to go to a summer institute, but all of the programs I've seen are really expensive. Any advice on how to pay for them?

Most of them offer financial aid of some sort, look at their websites for more information.  If you're a junior then there is RSI which is free but pretty hard to get in.

Definitely apply!  Many of the programs offer a lot of aid.  I'm from Mathcamp and I know that we have substantial funds available to help you come, and I'm sure it's true of many other programs as well.


Find the function that is finally graphed after the following transformations are applied to the graph of y = sqrt{x}.

(a) reflected about the x-axis

(b) shifted right 3 units

(c) shift down 2 units

[hide="(a)"]

$\text{f(x)}=-\sqrt{x}$

[/hide]

[hide="(b)"]

$\text{f(x)}= \sqrt{x-3}$

[/hide]

[hide="(c)"]

$\text{f(x)}= \sqrt{x}-2$

[/hide]

explanation for these problems are at

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129278[/url]

don't feel like repeating my self and i have to eat dinner =p

I thank you for your reply and the link provided.


[quote="Silverfalcon"]$(-1)^{5^2} + 1^{2^5} =$[/quote]

the exponent of -1 is $5^2=25$ which is odd, so the first term is -1.For the 2nd: (real) powers of 1 are always 1

So -1+1=0, $\boxed{C}$

[hide="generalization"]all powers of 5 are odd so -1 has a power of somthing odd and 1 will always be 1 no matter how many 1s there are so it is -1 + 1 = 0  C[/hide]

[hide]
Since -1 is raised to an odd power it will stay -1, and 1 to any positive power is always 1 so 1+-1=0.  THe answer is C.

[/hide]


Why is every third Fibonacci number even?

What is the sum of all the even Fibonacci numbers up to any term?

a1=1,a2=1    a3=a1+a2=2,a4=a3+a2 and so on...
Because a1  and a2 are odd then the sum of 2 odd numbers is even.(the numbers a(3k-2) and a(3k-1) )
After that you get odd number+ even number=odd number
After that you get even number+odd number= odd number
After that you get odd number+ odd number= even number
  so p(k) has a parity  imply p(k+3) has the same parity
Bye induction you get that every number a(3k+1)  and a(3k+2) is odd and every number a(3k) is even.
About the 2nd question i dont know how to do it  :P  ,its pretty hard to find the general form (the one with [x]) the sum would be even harder.

[hide="Hint for (II)"]Define:
$ \blacktriangleright \; \; S_n \equal{} F_3 \plus{} F_6 \plus{} \ldots \plus{} F_{3n}$
$ \blacktriangleright \; \; T_n \equal{} F_1 \plus{} F_2 \plus{} \ldots \plus{} F_{3n}$.
What is $ T_n \minus{} S_n$?
[hide="Further Hint"]Recall that $ F_{3k} \equal{} F_{3k \minus{} 1} \plus{} F_{3k \minus{} 2}$ for each $ k$[/hide][/hide]

[hide="(a)"]The first three Fibonacci numbers are 1, 1, and 2: O, O, E. The next one will be the sum of the last two, O and E, which is O. The next one will be the sum of the last two, E and O, which is O. The next one will be the sum of the last two, O and O, which is E. The next will be O+E=O, the next E+O=O, the next O+O=E, and so forth:
1-O
1-O
2-E
3-O
5-O
8-E
13-O
21-O
34-E
And so forth. Because an odd number plus an even number equals an odd number and two odd numbers make an even number, every third Fibonacci number will be even.[/hide]
No time to post answer for B now, but I'll edit my comment to do so later. :)


[quote="Zhero"]Suppose $ a$, $ b$, and $ c$ are positive integers satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} ac \plus{} bc)$. Show that $ a$, $ b$, and $ c$ are perfect squares.

Korea 2006, I believe.[/quote]

It IS Korea 2006, but you forgot $ gcd(a, b, c)\equal{}1$.

I'll post the solution anyway.

$ a^2\minus{}2(b\plus{}c)a\plus{}(b\minus{}c)^2\equal{}0$. But, $ \frac{D}{4}$ must be square.

So, $ 4bc\equal{}k^2$.

So, $ bc$ is a square.

Now, we have to prove that $ gcd(b, c)\equal{}1$.

Let $ p|gcd(b, c)$ and $ p$ is prime.

Since $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)$, $ p|a$.

So, it isn't $ gcd(a, b, c)\equal{}1$.(X)

So, we can know that $ gcd(b, c)\equal{}1$! (QED)

[quote="Zhero"]Suppose $ a$, $ b$, and $ c$ are positive integers satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} ac \plus{} bc)$. Show that $ a$, $ b$, and $ c$ are perfect squares.

Korea 2006, I believe.[/quote]
You must add condition $ \gcd(a,b,c) \equal{} 1$ . Here is my solution :
The first we will show that $ a,b,c$ are pairwise relative prime . If not , we can wlog that $ \gcd(a,b) > 1$ , which means that exist a prime p satisfy $ p|\gcd(a,b)$ ,by the condition we have $ p|c^2$ and because it is a prime number so $ p|c$ . This gives contradiction . Therefore $ a,b,c$ are pairwise relative prime .
By the condition $ 2|a^2 \plus{} b^2 \plus{} c^2$ ,it is not hard to show that among three numbers ,there is exactly one even number . We can assume that $ a,b$ are odd and $ c$ is even . From the condition we have :
\[ a(2(b \plus{} c) \minus{} a) \equal{} (b \minus{} c)^2
\]
Let $ d \equal{} \gcd(a.2(b \plus{} c) \minus{} a)$ then we have $ d|b \minus{} c$ , because a is odd and $ \gcd(b \plus{} c,b \minus{} c) \equal{} 1$ so $ d \equal{} 1$ . Hence $ a$ must be a perfect square . Similarly , we can prove that $ b$ is a perfect square .
Let's rewrite the condition :
\[ c(2(a \plus{} b) \minus{} c) \equal{} (a \minus{} b)^2
\]
Because c is an even number so we can write $ c \equal{} 2m$ and $ m(a \plus{} b \minus{} m) \equal{} (\frac {a \minus{} b}{2})^2$
From above result ,a and b are odd perfect squares so $ \frac {a \minus{} b}{2}$ must be an even number , hence $ m$ is an even number . Notice that  $ \gcd(\frac{a \plus{} b}{2},\frac{a \minus{} b}{2}) \equal{} 1$ then it is easy to prove that $ m \equal{} 2n^2$ ,this follows that $ c$ is a perfect square too .
That ends the proof .

[quote="Zhero"]Suppose $ a$, $ b$, and $ c$ are positive integers satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} ac \plus{} bc)$. Show that $ a$, $ b$, and $ c$ are perfect squares.

Korea 2006, I believe.[/quote]
$ \iff (a\minus{}b\minus{}c)^2 \equal{} 4bc, (b\minus{}a\minus{}c)^2 \equal{} 4ac, (c\minus{}a\minus{}b)^2 \equal{} 4ab$... and we are done ;)

[quote="Mathias_DK"][quote="Zhero"]Suppose $ a$, $ b$, and $ c$ are positive integers satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} ac \plus{} bc)$. Show that $ a$, $ b$, and $ c$ are perfect squares.

Korea 2006, I believe.[/quote]
$ \iff (a \minus{} b \minus{} c)^2 \equal{} 4bc, (b \minus{} a \minus{} c)^2 \equal{} 4ac, (c \minus{} a \minus{} b)^2 \equal{} 4ab$... and we are done ;)[/quote]

You are right...a better solution than mine.


On five consecutive days, the low temperatures in Sioux City, SD
were 2, -9, 9, 7 and -4, measured in degrees Celsius. How many
degrees Celsius was the mean of these low temperatures?

add the numbers like u would positive mean

2 + (-9) + 9 + 7 + (-4) = 5

then divide by 5

ther4- the mean is 1


geometry
ratio
similar triangles
geometry unsolved

Intersect a given triangle ABC by a straight line DE parallel to BC so that the area of triangle BDE is of a given magnitude k^2. What relationship between k^2 and the area of the triangle ABC guarantees the solvability of the problem and how many solutions has the problem? :?:

Let the point D lie on AB, AD:AB=l.
Then the ratio of area of BDE and ABC is
l(1-l) (using similar triangles), so $l(1-l)=\frac{k^{2}}{S}=\left(\frac{k}{\sqrt S}\right)^{2}$,
$\sqrt Sl.\sqrt S(1-l)=k^{2}$. From there we can construct lenght kl using Euclidean theorem:
We draw a halfcircle of  diameter $\sqrt S$ and a line parallel to the diameter in distance $k$.
-If it doesn't intersect the halfcircle,  the problem has no soln (thats iff $ k>\frac{\sqrt S}{2}$).
-If it meets the halfcircle, the problem has one soln - D is a midpoint of AB (thats iff $ k=\frac{\sqrt S}{2}$).
-If it intersects the circle in X,Y, perpendicular projections of X,Y to the diameter divide the dimeter in the ratios in which D should divide AB, so construction of D is easy.(thats iff $ k<\frac{\sqrt S}{2}$).

EDIT: i've fixed the errors in the ratios.

(a) $0\leqslant k\leqslant{1\over 2}$ is a necessary condition, (b) the problem has two solutions for $k\neq{1\over 2}$ and a unique solution for $k={1\over 2}$

To Kondr...
I get...  $l(\sqrt S) .(1-l)(\sqrt S) = k^{2}$.
Hence    $S \leq 4k^{2}$.

I,m not sure...


geometry
trigonometry
geometric transformation
homothety
geometry solved

let DEFGH be a pentagon wich has equal sides.
the pentagon is in the area of triangle ABC in wich point H be on line AB. and point E be on line AC and the segment GF be on line BC. let name the mid point of GF , A'.
we will find B' and C' in  that way.
(the pentagons may change but ABC is fixed)
prove that AA' and BB' and CC' are cocurrent .

Could you make your wording a bit more clear? You said DEFGH may change but [i]how[/i] does it change?

i think ashegh means this:

Let $ABC$ be a triangle and
$A(x_1x_2x_3x_4x_5)$ be a regular pentagon such that
$x_2 \in AB,x_3x_4 \in BC,x_5 \in AC $
Let $A`=\text{mid point of } x_3x_4$.
We will define $B(y_1y_2y_3y_4y_5)$ in the same way and $B`=\text{mid point of}   y_3y_4$.
and $C(z_1z_2z_3z_4z_5)$ in the same way and $C`=\text{mid point of}  z_3z_4$

Prove that lines $AA`,BB`,CC`$ are concurrent.

PS.Ashegh,darij has already told you,try to write your problems in an understanable manner,And i think you can ,if you want ,Schoolmate(these guys are putting their time to help you (this is your homework,should you post it?)You should not make fun of them ,Its not nice :|  :|   :|

mecrazywong, i mean to find points B' and C', we should draw two other pentagons.
for example we want to find point B'.
we should consider a pentagon D'E'F'G'H' wich has equal sides in wich H' is on line BC. and E' is on line BA.
and the segment G'F' is on line AC.

and i said that the triangle ABC is fixed.

OK, this makes sense now :)
Let x be the sidelength of the pentagon.
Let M be the midpoint of GF. $BM=BG+\frac{GF}{2}=\sin(B+72^\circ)\frac{HG}{\sin B}+\frac{GF}{2}=x(\frac{\sin(B+72^\circ)}{\sin B}+\frac{1}{2})$. Now it is easy by Ceva.

[quote="lomos_lupin"]Let $ABC$ be a triangle and
$A(x_1x_2x_3x_4x_5)$ be a regular pentagon such that
$x_2 \in AB,x_3x_4 \in BC,x_5 \in AC $
Let $A`=\text{mid point of } x_3x_4$.
We will define $B(y_1y_2y_3y_4y_5)$ in the same way and $B`=\text{mid point of}   y_3y_4$.
and $C(z_1z_2z_3z_4z_5)$ in the same way and $C`=\text{mid point of}  z_3z_4$

Prove that lines $AA`,BB`,CC`$ are concurrent.[/quote]

This can be proven in an almost philosophical way:

Let $P_1P_2P_3P_4P_5$ be an arbitrary regular pentagon. Let M be the midpoint of its side $P_2P_3$. Then, denote $\phi=\measuredangle MP_1P_4$. We actually won't need to know what the value of $\phi$ really is, but we only have to know that $\phi$ is a constant, i. e. it is the same angle for every regular pentagon $P_1P_2P_3P_4P_5$ (what is obvious since all regular pentagons are similar).

Since $A\left(x_1x_2x_3x_4x_5\right)$ is a regular pentagon, $x_2x_5\parallel x_3x_4$. In other words, $x_2x_5\parallel BC$. Also, we know that $x_2\in AB$ and $x_5\in AC$. Hence, there exists a homothety with center A mapping the points $x_2$ and $x_5$ to the points B and C. Thus, this homothety will map the regular pentagon $A\left(x_1x_2x_3x_4x_5\right)$ to a regular pentagon $\left(x^{\prime}_1Bx^{\prime}_3x^{\prime}_4C\right)$, and thus it will map the midpoint A' of the segment $x_3x_4$ to the midpoint of the segment $x^{\prime}_3x^{\prime}_4$. Denoting the midpoint of the segment $x^{\prime}_3x^{\prime}_4$ by $A_1$, we thus see that our homothety maps the point A' to the point $A_1$. Since the center of the homothety is A, this entails that the points A, A' and $A_1$ are collinear, i. e. the line AA' coincides with the line $AA_1$. Similarly, the lines BB' and CC' coincide with the lines $BB_1$ and $CC_1$, respectively, where the points $B_1$ and $C_1$ are defined similarly to the point $A_1$. Hence, instead of proving that the lines AA', BB', CC' are concurrent, it will be enough to prove that the lines $AA_1$, $BB_1$, $CC_1$ are concurrent (since they are the same lines).

Since $\left(x^{\prime}_1Bx^{\prime}_3x^{\prime}_4C\right)$ is a regular pentagon, and $A_1$ is the midpoint of its side $x^{\prime}_3x^{\prime}_4$, we have $\measuredangle A_1BC=\phi$. Similarly, $\measuredangle A_1CB=\measuredangle B_1CA=\measuredangle B_1AC=\measuredangle C_1AB=\measuredangle C_1BA=\phi$. Hence, the triangles $BA_1C$, $CB_1A$, $AC_1B$ are three directly similar isosceles triangles (in fact, they are all isosceles with the same base angle $\phi$) erected on the sides BC, CA, AB of triangle ABC (with bases BC, CA, AB, respectively). Hence, by the Kiepert theorem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 post #2), the lines $AA_1$, $BB_1$, $CC_1$ are concurrent. And we are done.

Sorry for the length - the argument was pretty simple, but rather hard to put into words.

By the way, Ashegh, a "pentagon with equal sides" is not the same as a "regular pentagon" (but it is clear that you meant a regular pentagon although, since the problem would have been wrong else).

  darij