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We have 3 hemispheres and 3 spheres each of diameter 6 units. Stack them in such a way that the height of the tower formed is maximum.

So, I know this question seems stupid because to form a tower of maximum height we just need to stack the hemispheres along the diameter and the spheres in whatever way we want to and by doing that we find that the maximum area of the tower is 36 units. But the thing is that this question appeared in a competition, so it cant be so simple...moreover i remember that the solution had a hint to use the cauchy-schwarz inequality. Maybe our answer is correct but maybe a rigorous proof needs to be provided using the cauchy-schwarz inequality

Tupac Shakur
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It doesn't seem so hard to formalize. I guess that the setting is formalizable as follows:

We are in $\mathbb{R}^3$ with coordinates $x,y,z$. We have an object $T$ that is a union of translation and rotations of 3 spheres and 3 hemispheres. We assume that $T$ is connected and $\inf\{z|z\text{ is the third coordinate of a point in $T$}\}=0$ ($T$ is placed on the ground). We measure the height of the object as $h:=\sup\{z|z\text{ is the third coordinate of a point in $T$}\}$. Both these quantities are actually $\max$ and $\min$ by compactness (let's assume the hemispheres are closed).

Proof by contradiction: suppose that $h>36$. Then there must be at least a center of a sphere (or of a hemisphere) that has $z$ coordinate $>33$, otherwise, you can easily prove that $h\leq 36$. Since T is connected, there must be a center of one of the remaining figures such that its z coordinate is $>27$, otherwise the highest ball would be disconnected from the tower... Iterating, you end up saying that the center of the sixth sphere (hemisphere) has $z$ coordinate greater than $3$. But by the 'ground hypothesis', there is another sphere or hemisphere that has center with $z$ coordinate at most $3$. Contradiction.

Lorenzo Pompili
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