AoPS - Problem

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Problem

Source: Cono Sur Olympiad, Uruguay 1989, Problem#6

Tags: probability, inequalities

M4RI0

14.05.2006 08:47

Show that reducing the dimensions of a cuboid we can't get another cuboid with half the volume and half the surface.

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.