Show that reducing the dimensions of a cuboid we can't get another cuboid with half the volume and half the surface.
Problem
Source: Cono Sur Olympiad, Uruguay 1989, Problem#6
Tags: probability, inequalities
madatmath
14.05.2006 08:56
It's rather obvious. If the original volume was $s^3$ then, the new reqd. volume would be $s^3/2$ which means the side would be $s/ (\sqrt[3]{2})$ and the surface would be $6*s^2/(\sqrt[3]{2})^2$ which is most ceratinly not half of $6s^2$
t0rajir0u
14.05.2006 08:58
This was the sixth problem?
madatmath
14.05.2006 09:44
What is the probability M4RIO meant $cuboid$ instead? I would put it as $P \longrightarrow 1$
t0rajir0u
14.05.2006 09:51
If the dimensions have to be reduced? It's still easy.
Let $a, b, c$ be the original dimensions, $a', b', c'$ the new. $a \ge a', b \ge b', c \ge c'$.
$abc = 2a'b'c'$
$ab + bc = ca = 2a'b' + 2b'c' + 2c'a'$
Divide the second equation by the first.
$\sum_{cyc} \frac{1}{a} = \sum_{cyc} \frac{1}{a'}$
But because we can only reduce the dimensions,
$\sum_{cyc} \frac{1}{a} \le \sum_{cyc} \frac{1}{a'}$
Hence $a = a'$, etc., contradiction.
Now, altering the dimensions arbitrarily... does it still hold?
madatmath
14.05.2006 10:05
hey, i don't get your solution. what is wrong with: $\displaystyle\sum_{cyc} \frac{1}{a} \leq \sum_{cyc} \frac{1}{a^'}$ ?
t0rajir0u
14.05.2006 10:07
madatmath wrote: hey, i don't get your solution. what is wrong with: $\displaystyle\sum_{cyc} \frac{1}{a} \leq \sum_{cyc} \frac{1}{a^'}$ ? We've shown that they have to be equal, and according to the inequality, the equality case can only occur if $a = a', b = b', c = c'$. But then the original two conditions cannot be satisfied!
M4RI0
14.05.2006 18:23
madatmath wrote: What is the probability M4RIO meant $cuboid$ instead? I would put it as $P \longrightarrow 1$ Sorry, by the mistake.